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Re: remarks on significance arithmetic implementation [Was: Re: Numerical accuracy/precision - this is a bug or a feature?]

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  • Subject: [mg120340] Re: remarks on significance arithmetic implementation [Was: Re: Numerical accuracy/precision - this is a bug or a feature?]
  • From: Richard Fateman <fateman at eecs.berkeley.edu>
  • Date: Tue, 19 Jul 2011 06:58:03 -0400 (EDT)

On 7/18/2011 11:54 AM, Daniel Lichtblau wrote:
>
> [Caution: This may increase the entropy of the thread, and almost 
> certainly will increase the ennui.]
Yep
>
>
... snip ... a bunch of things I think we agree upon.
> To expand on "properly rounded floats" I will mention that 
> Mathematica's significance arithmetic, in certain settings, is as 
> conservative as interval arithmetic. Specifically this works out when 
> the operations used are addition, multiplication, integer powering, or 
> division with both operands approximate. Also required is that they 
> have more significant digits than can be represented by the error 
> recorder (that is to say, more than 16 or so digits when error is 
> recorded as a 53-bit-mantissa float). The gist is that precision is 
> altered downward in a way that will even account for second order 
> error (and the higher order error terms that exist in computing x/y 
> where x and y are both approximate bignums).
I am unclear on this zeroth, first, second  order error categorization.  
It seems to me that the critical problem, and one that is not addressed 
by Mathematica, in interval arithmetic, is addressing dependence of 
operands.  Consider the two expressions
f[x_]:= (x-1)*(x+1)  and g[x_]:=(x^2-1)

which are analytically the same function.

given the value v=Interval[{-1/2,1/2}]    f[v] returns an interval with 
width 2.  g[v] returns an interval with width 1/4.

The function g is obviously a lot better than f.
Same thing happens for v=0.1`2.  f[v] is -0.99`269..     but g[v]  is 
-0.99`3.69...  So according to Mathematica, g evaluates
to a full decimal digit higher accuracy.

A true devotee of interval arithmetic (or of significance arithmetic)  
would take the function body for f and rearrange it as g,
and also would require for comparisons such concepts as possibly-equal, 
certainly-greater-than  etc.  And would need
to resolve meanings for empty intervals and other such things.  Some of 
these may in fact be in Mathematica's interval
arithmetic suite.  They seem to be absent in the significance arithmetic 
world, though there is a transition from one to the
other possible.
Note that Interval[0.1`2] returns the fairly nonsensical   
Interval[{0.1,0.10}] but InputForm[%] shows
Interval[{0.098046875`1.9914337561691884,
   0.101953125`2.008400542026431}], maybe overkill.
Interval[{0.98,0.102}]  might be OK.


There are ways of reducing semi-automatically, the number of repeat 
occurrences of operands in such expressions, thereby improving the 
interval widths.  This (that is, computing "single use expressions") is 
the kind of thing that would, I think, be useful and for which computer 
algebra systems are especially appropriate.  That, in combination with 
some other tricks of interval "reliable" arithmetic would be helpful 
(and for which I have written some programs, not in Mathematica 
though).  So the real contribution to interval arithmetic (or 
significance arithmetic) of Mathematica would be if it were able to 
reformulate some subset of expressions so that they were automatically 
computed faster and more accurately.

(One could argue that boosting the software precision will get the 
answer accurately, especially if one ignores exponentially longer 
run-times.)


For some discussion in the context of a (different) programming language 
that is not much like Mathematica, see

http://developers.sun.com/sunstudio/products/archive/whitepapers/tech-interval-final.pdf

There are other, more expensive techniques. E.g.  if p(x) is a 
polynomial, what is p[Interval[{a,b}]] ?
Instead of just interval-alizing the + and * and ^,  find the extrema of 
p between a and b.
What is Sin[Interval[...]], etc.

>
> It would not be hard to extend this to the taking of reciprocals and 
> hence division of integer by float e.g. x --> 1/x. But as best I can 
> tell from looking at the code, Mathematica does not do this, so error 
> tracking in that case is purely first order.

Maybe I'm not following here, but computing an interval for 1/x does not 
seem difficult.
>
>
>> [...]
>> My point remains:
>>    WRI could have a much more direct notion of number, and equality, 
>> that
>> would make it easy to implement my choice of arithmetic or theirs or 
>> yours.
>> They didn't.  The default is not completely reliable. People write to
>> this newsgroup, periodically, saying Huh? what's going on? I found a 
>> bug!
>
> This is, I am afraid, nonsense. Nothing would make it "easy" to 
> implement significance arithmetic across the spectrum of Mathematica's 
> numerical engine. To do so efficiently, reliably, maintainably, etc. 
> is, well, a significant task. Jerry Keiper could do it. Mark Sofroniou 
> could do it. I do not think I could do it, at least not very well. 
> And, as the saying goes, "I have me doubts about thee".

OK, I did not mean it would be easy to reproduce what Jerry Keiper did,  
but after all, he did do it using ordinary arithmetic at its base.
What I did mean is that programs to do arithmetic "the machine way" 
would run at machine speed (= easy); the programs to do software 
bigfloats without significance tracking would run faster (=easier) than 
including significance tracking.   And that if one wished to patch-up 
such code with built-in tracking ,after the fact by re-stating the 
precision after each operation, that would be slower (= not as easy) 
than avoiding it in the first place.
So, easy= easy for the computer.

I do think that people write to the newsgroup reporting what they think 
of as bugs, but are actually the deliberate consequences of the design 
of arithmetic.
>
>
>> As for the experimental and computational notions of (im)precision, I
>> think there is an issue of using the same words with different meanings.
>> Similar but unfortunately different. Precision of a floating point
>> number F is simply the number of bits in the fraction.  If you impute
>> some uncertainty to F, you can store that in another piece of data D in
>> the computer. If you want to assert that F and D together represent a
>> distribution of a certain kind  you can also compute with that.
>>
>> RJF
>> [...]
>
> Related to the above (I think), you stated earlier in this thread that 
> the differences in terminology usage can be troublesome. I agree that 
> the meaning of Mathematica's Precision and, to a lesser extent, 
> Accuracy, might cause initial confusion to one unfamiliar with the 
> software but well versed in Numerical Analysis. I would expect such 
> people to catch on after being bitten once or twice, and either get 
> used to the distinctions, or use different software.
yes, but there are not too many well versed in numerical analysis.
>
> This strikes me as similar to the situation wherein Append and Prepend 
> are O(n) rather than O(1) complexity operations, since they rewrite 
> their List arguements in full. A CS person new to Mathematica might 
> well assume List is akin to the CS notion of a (singly or doubly) 
> linked list. After learning otherwise they'd probably not be too 
> flustered, other than perhaps to mutter that the naming of "List" was 
> problematic. Even that would perhaps abate once they get accustomed to 
> the very general Mathematica notion of a List. After all, a linked 
> list is a modestly arcane structure to those outside CS, and most 
> general software users are not from inside CS, and most CS people will 
> understand that.
Similar, yes.  Dissimilar in the sense that the List operations return 
expected results, just slower. But arithmetic returns (on occasion) 
unexpected results.

RJF


From mathgroup-adm at smc.vnet.net  Wed Jul 20 04:37:43 2011
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From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
Subject: [mg120357] Re: Find position of nonzero elements
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On Tue, 19 Jul 2011, Dominic Wörner wrote:

> Hi,
>
> How can I find the indices i and j of all nonzero elements of a matrix.
> I want that because there are only some nonzero elements in a huge
> matrix.
>
> Best regards,
> Dominic

Dominic,

how about

sa = SparseArray[IdentityMatrix[4]]
sa["Properties"]
sa["NonzeroPositions"]

Oliver

>
>


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From: Sebastian Hofer <sebhofer at gmail.com>
Subject: [mg120373] Re: Find position of nonzero elements
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Does ArrayRules work?



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From: carlos at colorado.edu
Subject: [mg120365] Total plot width
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How can I specify the total plot width in Mathematica?  I was looking for
something like AbsolutePlotWidth->250  (points implied) but havent found
that option in the documentation.  There is AspectRatio, but that is
not a dimension. Thanks.



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From: Elizabeth Latta <elizabeth at elizabethlatta.com>
Subject: [mg120374] Re: Using Plot3D; rotating the plot but, getting a different image after exporting.
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I`ve never had a problem. I rotate then right click and choose "Save Graphic As..."
-Elizabeth
-----
www.elizabethlatta.com




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From: Sseziwa Mukasa <mukasa at gmail.com>
Subject: [mg120377] Re: Find position of nonzero elements
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Position[matrix,Except[0]]

If it's machine precision you may need to do

Position[matrix,Except[0.]]

On Jul 19, 2011, at 6:53 AM, Dominic W=F6rner wrote:

> Hi,
>
> How can I find the indices i and j of all nonzero elements of a matrix.
> I want that because there are only some nonzero elements in a huge
> matrix.
>
> Best regards,
> Dominic
>




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From: "Tong Shiu-sing" <sstong at phy.cuhk.edu.hk>
Subject: [mg120356] Plotting a star map
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Dear All,

I have the coordinates (RA and Decl) of some stars and nebulae in a small 
part of the celestial sphere and want to plot them as a star map. What 
projection method should I use? Does Mathematica has a standard package for 
doing it (just like that for the geo-data)? How do specify the boundary of 
the plot.

Regards,
Dominic 




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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120378] Imported Data in Mathematica Player or CDF Documents
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Hello everyone,

I'm in the process of creating a set of Mathematica Player documents 
for my students, and for many of them, I need to import data from text 
files.  When I convert the notebooks to CDF, is the imported data 
cached?  Can I simply leave the Import function in the player document?  
Or do I have to copy and paste the data manually into a variable?

Any advice would be much appreciated,

Gregory



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From: Murray Eisenberg <murray at math.umass.edu>
Subject: [mg120358] Re: Find position of nonzero elements
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There must be a simpler way, but here's one way.

   mat = {{-2, -3, 0, -1}, {3, -2, 0, 2}, {0, -1, 1, 2}};
   mn = Dimensions[mat];
   ij = Table[{i, j}, {i, First@mn}, {j, Last@mn}];
   Pick[Flatten[indices, 1], Thread[0 != Flatten@mat]]
{{1,1}, {1,2}, {1,4}, {2,1}, {2,2}, {2,4}, {3,2}, {3,3}, {3,4}}

The difficulty is that Mathematica, in contrast to some programming
languages, really has no sense of higher-order arrays, just of
1-dimensional lists, so that it really has no such thing as a matrix,
just a list of lists.


On 7/19/11 6:53 AM, Dominic W=F6rner wrote:
> Hi,
>
> How can I find the indices i and j of all nonzero elements of a matrix.
> I want that because there are only some nonzero elements in a huge
> matrix.
>
> Best regards,
> Dominic
>

--
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305




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From: =?iso-8859-1?Q?Dominic_W=F6rner?= <dominic.woerner at mpi-hd.mpg.de>
Subject: [mg120376] Re: Find position of nonzero elements
References: <201107191053.GAA10147 at smc.vnet.net> <Pine.LNX.4.63.1107190532290.28777 at wopr.wolfram.com>
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Thanks Oliver and also Andy. This is very good to know!

Am 19.07.2011 um 12:33 schrieb Oliver Ruebenkoenig:

> On Tue, 19 Jul 2011, Dominic W=F6rner wrote:
>
>> Hi,
>>
>> How can I find the indices i and j of all nonzero elements of a matrix.
>> I want that because there are only some nonzero elements in a huge
>> matrix.
>>
>> Best regards,
>> Dominic
>
> Dominic,
>
> how about
>
> sa = SparseArray[IdentityMatrix[4]]
> sa["Properties"]
> sa["NonzeroPositions"]
>
> Oliver
>
>>
>>




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From: gonzorascal <djw22 at duke.edu>
Subject: [mg120362] Inverse of Interpolating Function?
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Hello All,

I am working with an Interpolating Function of an oscillating solution (time series) to a differential equation.  I am trying to find the period and pulse width of the oscillation.  To do this I would like to have an inverse of my function y[t] (so I would have t[y]).  I realize this function would be multivalued (that's what I want to find the period).  I am not having success using Mathematica's InverseFunction[] or Reduce[] commands.  Does anyone have any experience or suggestions with this sort of thing (either finding Inverse Interpolating Functions or another method for period and pulse width)?  Thank you.

-GR



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From: Bill Rowe <readnews at sbcglobal.net>
Subject: [mg120370] Re: Find position of nonzero elements
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On 7/19/11 at 6:53 AM, dominic.woerner at mpi-hd.mpg.de (Dominic W=C3=B6rner)
wrote:

>How can I find the indices i and j of all nonzero elements of a
>matrix. I want that because there are only some nonzero elements in
>a huge matrix.

I will assume your matrix is not currently a SparseArray. If so,
do the following

Most[ArrayRules@SparseArray@m] /. HoldPattern[a_ -> _] -> a

where m is your matrix. ArrayRules applied to a SparseArray
returns a list of the form

{i,j}->x

where i,j are the indices of a non-zero entry and x is the value
of that entry. The last rule in this list will be {_,_}->0 which
is default rule for elements with a zero value you aren't
interested in. Finally, the replacement rule simply extracts the
{i,j} portion of each rule for a non-zero entry.




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From: blamm64 <blamm64 at charter.net>
Subject: [mg120363] Re: Unexpected Behavior: SetDelayed versus Set
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Thanks Daniel L., I have implemented essentially as you suggested,
with a further check to Select only real and greater than zero
results, choose Max of what remains, and to Print a diagnostic if Max
returns -Infinity.

Thanks Bobby, what can I write except Exp[WOW]!?


Thanks Leonid, I am quite careful with the scoping in the notebook
from which this example, paired down, came.  In the notebook, as it
currently exists, from which this example was extracted, aLinQS and
bQuadQS are parameterized, I just opted to make the example less
cluttered.  I did read the links you provided and they were very
helpful.


Now, if I can just find that baseball cap with "DA" on it, I have to
wear it for a while ... .





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From: "Hobbs, Sylvia (DPH)" <sylvia.hobbs at state.ma.us>
Subject: [mg120375] Re: Using Plot3D; rotating the plot but, getting a different image after exporting.
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It's always the obvious. So there are two ways of rotating, using the rotate function command or just taking an unrotated graph and manually rotating it with the mouse rather than function. So it appears that the exported mouse rotated jpg saves as rotated, but the exported function rotated jpg exports unrotated.

Sylvia Hobbs
________________________________________
From: Elizabeth Latta [elizabeth at elizabethlatta.com]
Sent: Tuesday, July 19, 2011 9:10 AM
To: Hobbs, Sylvia (DPH); ss2011 at wolframscience.com; mathgroup at smc.vnet.net
Subject: [mg120375] Re: RE: Using Plot3D; rotating the plot but, getting a different image after exporting.

I`ve never had a problem. I rotate then right click and choose "Save Graphic As..."
-Elizabeth
-----
www.elizabethlatta.com



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From: Richard Fateman <fateman at eecs.berkeley.edu>
Subject: [mg120372] Re: remarks on significance arithmetic implementation [Was: Re: Numerical accuracy/precision - this is a bug or a feature?]
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On 7/19/2011 6:49 PM, DrMajorBob wrote:
> As before, I think much of this discussion is useless. But this 
> example DOES give me pause:
>
>> Consider the two expressions
>> f[x_]:= (x-1)*(x+1)  and g[x_]:=(x^2-1)
>>
>> which are analytically the same function.
>>
>> given the value v=Interval[{-1/2,1/2}]    f[v] returns an interval with
>> width 2.  g[v] returns an interval with width 1/4.
>
> I don't see why we'd want significance tracking to act this way.
>
> Bobby
>
I think that it is considered too expensive to do it this way.
If you use significance arithmetic most of the time, it can't be
so expensive.

Also, it is difficult to write a program to do it, in general.

See
http://www.cs.berkeley.edu/~fateman/papers/interval.pdf

and look for SUE  or Single Use Expressions.

RJF




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From: PA32R <tio540s1 at gmail.com>
Subject: [mg120371] Re: Replace, test question
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On Jul 19, 4:12 am, Yves Klett <yves.kl... at googlemail.com> wrote:
> Hi,
>
> you need to supply x as argument to PrimeQ... alternatively use PatternTest:
>
> {58, 61, 15, 66, 10, 2, 24, 81, 45, 84} /. x_ /; PrimeQ[x] -> P
>
> {58, P, 15, 66, 10, P, 24, 81, 45, 84}
>
> {58, 61, 15, 66, 10, 2, 24, 81, 45, 84} /. x_?PrimeQ -> P
>
> {58, P, 15, 66, 10, P, 24, 81, 45, 84}
>
> Regards,
> Yves
>
> Am 18.07.2011 12:15, schriebPA32R:
>
>
>
>
>
>
>
> > I'm very new to Mathematica and trying to work my way through it. I
> > don't understand why:
> > {58, 61, 15, 66, 10, 2, 24, 81, 45, 84} /. x_ /; PrimeQ -> P
> > returns:
> > {58, 61, 15, 66, 10, 2, 24, 81, 45, 84}
>
> > I would expect it to take the list, replace everything in the list
> > that matches any pattern and passes the PrimeQ test, i.e., is prime,
> > with P to yield:
> > {58, P, 15, 66, 10, P, 14, 81, 45, 84}
>
> > Can someone explain my misunderstanding?
>
> > Thanks.

Thanks everyone, I got it done. The assistance is much appreciated.



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From: "Hobbs, Sylvia (DPH)" <sylvia.hobbs at state.ma.us>
Subject: [mg120368] Re: Using Plot3D; rotating the plot but, getting a different image after exporting.
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Ditto that problem and it is not limited to 3D plots. I've had that problem with other rotated Mathematica graphics that I import as a jpg. I end up pasting the jpg into PowerPoint re-rotating it and saving it as the jpg I had hoped Mathematica would rotate. Any insight on Mathematica commands that fix this without bringing the graphic into other program would be great. If there isn't one, then maybe, just maybe this is a teeny weeny bug.



Sylvia Hobbs

________________________________________
From: Gilmar Rodriguez-pierluissi [peacenova at yahoo.com]
Sent: Tuesday, July 19, 2011 6:57 AM
To: mathgroup at smc.vnet.net
Subject: [mg120368] Using Plot3D; rotating the plot but, getting a different image after exporting.

Sometimes, when I build a plot using Plot3D (or similar 3D plotting program); rotate the plot on the Mathematica platform and then export it; the jpeg or png picture that I get after exporting the plot is different than the original image generated by Plot3D.  Is there any way to avoid this? Thank you for your help! - Gilmar Rodriguez Pierluissi




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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120364] Possible Bug in Dates Supplied with FinancialData Function
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Hello everyone,

Not sure whether I've misunderstood how FinancialData works, but it 
seems to me that the wrong date is attached to prices if they are 
downloaded using a monthly frequency.

I used the following start and end dates.

	seriesStartDate = {2009, 01, 01};
	seriesEndDate = {2011, 6, 30};

I download daily prices for Apple Inc. by running

	dailyPrices = FinancialData["AAPL", {seriesStartDate, seriesEndDate, =93Day=94}]

If I then download monthly prices by running

	monthlyPrices = FinancialData["AAPL", {seriesStartDate, seriesEndDate, =93Month=94}]

the observations are dated at the beginning of each month but correspond to the price on the last day of the month.  The price of Apple on 2009-01-30, for example, when downloaded as daily is $90.13 but when downloaded as monthly, that same price is date stamped 2009-01-02, that is, at the beginning of the month.  So, when downloading using a monthly frequency, the date stamps may mislead you into thinking that you are getting beginning-of-month prices when, in fact, you are getting end-of-month prices.

I almost always need end-of-month prices, so to get the correct dates, I download as daily and extract the last daily observation for each month

	GatherBy[dailyPrices, #[[1, {1, 2}]] &]
	Last /@%

You could do the same using Cases.

Regards,

Gregory




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From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
Subject: [mg120361] Re: Transforming an expression to publication form
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On 19 Jul 2011, at 11:57, carlos at colorado.edu wrote:

> Suppose I have the expression
>
> C11=A*Em*(4-3*omega^2+nu*(-8+3*omega^2)+(-4+8*nu)*Cos[kappa]);
>
> I want to transform this into
>
> C11pub=Em*A*(4*(1-2*nu)*(1-Cos[kappa])-3*(1-nu)*omega^2);
>
> which is the exactly the way it has to appear in a journal
> publication,
> once mapped to LaTeX.  Both C11 and C11pub have the
> same LeafCount (27), and C11 is invariant under Simplify
> and FullSimplify.
>
> How do I accomplish  C11 -> C11pub within Mathematica,
> without using any extra packages?  BTW this is part of one
> of 36 matrix entries, so transforming all by hand takes a while.
>

On the one hand: in general, it is not reasonable to expect Mathematica 
to do such things. It is not really intended for this purpose and while 
one can often succeed by using special tricks, there is no general 
approach and I don't think it is worth spending time on uncovering the 
tricks needed in an individual case.

On the other hand, in this particular case it is rather easy to see 
these "tricks".  Namely:

C11 = A*Em*(4 - 3*omega^2 +
    nu*(-8 + 3*omega^2) + (-4 + 8*nu)*Cos[kappa])

Collect[Collect[C11, omega, Factor], {Em, A}]

A Em (4 (2 nu-1) (cos(kappa)-1)+3 (nu-1) omega^2)

This is essentially your C11pub except for some signs in a few places.

Andrzej Kozlowski



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From: Daniel Lichtblau <danl at wolfram.com>
Subject: [mg120360] Re: numeric Groebner bases et al [Was Re: Numerical accuracy/precision - this is a bug or a feature?]
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On 07/19/2011 05:54 AM, Richard Fateman wrote:
> [...]
> I gather from previous correspondence (DanL), that what significance
> arithmetic is used for is a decision point as to whether some value is
> or is not possibly zero.
> The value is presumably the result of evaluating a (multivariate)
> polynomial whose coefficients are exact rationals [or maybe not] at a
> point where the coefficients are known as intervals [or maybe exact].
> Perhaps DanL can advise on this characterization.

It's not that, but simple coefficient arithmetic e.g. in running 
Buchberger's algorithm or one of the more recent alternatives. One has 
to know when some linear combination of coefficients cancels.


> [...]
> I don't think there is a need to produce an independent implementation
> of anything; the same abstract structure of algorithms should hold.

It does, in running Buchberger's algorithm. Not sure if that was 
specifically what you had in mind here but it seems relevant to mention.


> The idea that one must come up with a different proof (that is, by not
> emulating the other proof) is irrelevant,

Slightly tangential: Kondratyev et al proposed a variant to Groebner 
bases, which uses a variant of Buchberger's algorithm. In his thesis, 
Kondratyev did require a "different" set of proofs to show termination 
and correctness. Related to the usual proofs, but by no means identical 
and not exactly what i would view as emulation.


> but as it happens, the
> algorithm for evaluating a polynomial with guaranteed error does NOT
> emulate significance arithmetic by carrying along extra baggage at each
> operation. Crudely speaking, it evaluates the polynomial in
> floating-point, and then it evaluates the polynomial with the signs of
> all the coefficients changed to +.
>
> see slide 21 in
> http://www-pequan.lip6.fr/~graillat/papers/slides_MIMS.pdf
> which is something I just found with Google.
>
> RJF

Certainly there are various methods of gauging precision loss in numeric 
evaluation of polynomials, and some may be better than blind 
significance arithmetic. The slides look like something I should check 
more closely.


Daniel Lichtblau
Wolfram Research





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From: Ray Koopman <koopman at sfu.ca>
Subject: [mg120367] Re: MultinormalDistribution Question
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On Jul 17, 3:03 am, Steve <s... at epix.net> wrote:
> [...]
> What I really need to do is perform this analysis on test data for
> which I have only a few data points, hence the Student T distribution
> would be more appropriate than the Normal distribution. Secondly,
> values for the "independent" and "dependent" variables have no
> physical meaning below zero. So this implies that I need truncated
> distributions. I'm hoping that the solution Andrzej  provided can be
> generalized for these added complications.
> Here are my 9 {F,t} data points where "F" is considered "independent"
> and t considered "dependent".

> {{1.01041, 0.3152}, {10.455, 0.3386}, {17.9032, 0.2534}, {24.9581,
>    0.5412}, {26.4688, 0.3251}, {27.4651, 0.4428}, {30.1682,
>    0.3402}, {36.6174, 0.2106}, {45.6129, 0.2154}}

> Would someone be so kind as to plop this data into their notebook to
> confirm a solution or two for me ? My results are below which are
> based on truncating the Student T distribution, 8 degrees of freedom
> and a calculated rho of -0.2327.

> [...]

Another approach is to regress u = log[t] on f linearly. This
solves the problem of keeping the conditional distributions of
t non-negative, but makes the regression of t on f nonlinear.

For your data, switching to u increases the correlation, but
the conditional s.d. is still bigger than the marginal s.d.,
so there is still room for questioning the whole exercise.

FWTW, here are the numbers I got:

{mf, sf} = {Mean@f, StandardDeviation@f}

{24.5177, 13.3704}

{mu, su} = {Mean@u, StandardDeviation@u}

{-1.1482, .311145}

r = Correlation[f,u]

-.322776

b = r*su/sf  (* slope *)

-.00751142

a = mu - b*mf  (* intercept *)

-.964034

se = su*Sqrt[(1-r^2)(n-1)/(n-2)]  (* conditional s.d. *)

.314825

Table[{x, CDF[StudentTDistribution[n-2],
      (a+b*x-Log[.5])/se]}, {x,0,50,5}]
(* f, Pr[(t|f) > .5] *)

{{0, .209020},
{ 5, .179928},
{10, .154056},
{15, .131283},
{20, .111422},
{25, .0942435},
{30, .0794909},
{35, .0669001},
{40, .0562108},
{45, .0471758},
{50, .0395667}}



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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120366] Re: Generating Arbitrary Linear Combinations of Functions
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You say "this works well", but it actually can't, since f -- as you used  
it in defining "Data" -- is undefined.

(You didn't give us all the code.)

So I'll build the model, only.

Clear[f, a, z, c]
f[a_?NumberQ, x_?NumberQ, xi_?NumberQ, c_?NumberQ] :=
   c Exp[-a (x - xi)^2/2]/Sqrt@a;
f[a_, z_, c_][x_] := f[a, x, z, c]
n = 10;
aVec = Array[a, n];
zVec = Array[z, n];
cVec = Array[c, n];
model = Total@Through[(f @@@ Transpose@{aVec, zVec, cVec})@x]

f[a[1], x, z[1], c[1]] + f[a[2], x, z[2], c[2]] +
  f[a[3], x, z[3], c[3]] + f[a[4], x, z[4], c[4]] +
  f[a[5], x, z[5], c[5]] + f[a[6], x, z[6], c[6]] +
  f[a[7], x, z[7], c[7]] + f[a[8], x, z[8], c[8]] +
  f[a[9], x, z[9], c[9]] + f[a[10], x, z[10], c[10]]

Then, after f[oneArgument_] and "data" have ALSO been defined:

nlm = NonlinearModelFit[data, model, Flatten@{aVec, zVec, cVec}, x]

Bobby

On Tue, 19 Jul 2011 05:55:32 -0500, Thomas Markovich  
<thomasmarkovich at gmail.com> wrote:

> Hi All,
>
> I'm trying to curve fit an arbitrary linear combination of distributed
> gaussians to a function -- each with their own set of parameters.  
> Currently,
> if I want to use twenty functions, I do the following
>
> f[a_?NumberQ, x_?NumberQ, xi_?NumberQ, c_?NumberQ] := (
>   c E^(- a (x - xi)^2/2))/Sqrt[a/=F0];
> Data := Table[{n/50, N[f[n/50]]}, {n, -300, 300}];
>
> model = f[a1, x, x1, c1] + f[a2, x, x2, c2] + f[a3, x, x3, c3] +
>    f[a4, x, x4, c4] + f[a5, x, x5, c5] + f[a6, x, x6, c6] +
>    f[a7, x, x7, c7] + f[a8, x, x8, c8] + f[a9, x, x9, c9] +
>    f[a10, x, x10, c10] + f[a11, x, x11, c11] + f[a12, x, x12, c12] +
>    f[a13, x, x13, c13] + f[a14, x, x14, c14] + f[a15, x, x15, c15] +
>    f[a16, x, x16, c16] + f[a17, x, x17, c17] + f[a18, x, x18, c18] +
>    f[a19, x, x19, c19] + f[a20, x, x20, c20];
> nlm = NonlinearModelFit[Data,
>    model, {a1, x1, c1, a2, x2, c2, a3, x3, c3, a4, x4, c4, a5, x5, c5,  
> a6,
> x6,
>      c6, a7, x7, c7, a8, x8, c8, a9, x9, c9, a10, x10, c10, a11, x11,  
> c11,
>     a12, x12, c12, a13, x13, c13, a14, x14, c14, a15, x15, c15, a16, x16,
> c16,
>      a17, x17, c17, a18, x18, c18, a19, x19, c19, a20, x20, c20}, x];
>
>
>
> This works well but it becomes tedious to create these linear  
> combinations
> by hand. It would be wonderful to create a linear combination of  
> functions
> with a vector of coefficients for a, xi, and c. I am just unsure of how  
> to
> approach this and I was hoping that you guys could offer some insight  
> into
> this.
>
>
>
> Best,
>
>
>
> Thomas


-- 
DrMajorBob at yahoo.com



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Subject: [mg120359] Re: sorting a nested list of 4-tuples
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On 07/19/2011 05:55 AM, Luis Valero wrote:
> Dear Sirs,
>
> I want to sort a list of 4-tuples of real numbers, so that, the second 4-tuple has minimum distance to the first, the third, selected from the rest, ha minimum distance to the second, and so on.
>
> The distance is the euclidean distance, calculated with the first two elements of each 4-tuple.
>
> I have defined a function that work:
>
> orderedList[list_] := Module[{nearest},
> 	Flatten[ Nest[{
>     	    Append[ #[[1]] , nearest = Flatten @@ Nearest[ Map[ Rule[ #[[{1, 2}]], #]&, #[[2]] ], Last[ #[[1]] ] [[{1, 2}]], 1] ],
>      	    Delete[ #[[2]], Position[ #[[2]], nearest ] ]  }&, { {list[[1]] }, Delete[ list, 1 ] }, Length[ list ] - 2], 1] ];
>
>
> but I need to improve the temporal efficiency
>
> Thank you

The complexity will show up either in needing to obtain a growing number 
of nearest items (to account for the fact that many may already ahve 
appeared), or removing selected items from further consideration. There 
may be a fancy way around these but I'm not seeing it offhand. At least 
not if we want to couple with a fast search for nearest items, using 
Nearest[].

A way to alleviate this is to gather newly selected items every so often 
and remove them in one shot. This adds to the coding complexity 
(required some debugging to get it to work, for one thing). But it does 
seem faster.

I made no effort to tune for the removal interval (which need not be 
constant). There ar also some parts of the code that might be made 
slightly faster in other ways.

n = 3000;
data = RandomReal[{-100, 100}, {n, 4}];

chunksize = Ceiling[Log[n]]^2;
denom = 10000.;
moddata =
   Map[Take[#, 3] &, MapThread[Prepend, {data, Range[n]/denom}]];
nf = Nearest[moddata];
next = moddata[[1]];
taken = ConstantArray[False, n];
result = ConstantArray[{0., 0., 0., 0.}, n];
remove = ConstantArray[{0., 0., 0.}, chunksize];
result[[1]] = data[[1]];
remove[[1]] = moddata[[1]];
taken[[1]] = True;

Timing[Do[
   nextset = nf[next, 1 + Mod[j, chunksize, 1]];
   posns = Round[denom*Map[First, nextset]];
   k = 1;
   While[k < Length[nextset] && taken[[posns[[k]]]], k++;];
   taken[[posns[[k]]]] = True;
   result[[j]] = data[[posns[[k]]]];
   next = nextset[[k]];
   remove[[Mod[j, chunksize, 1]]] = next;
   If[Mod[j, chunksize] == 0,
    done = True;
    posns = Apply[Alternatives, remove];
    moddata = DeleteCases[moddata, posns];
    nf = Nearest[moddata];
    ];
   , {j, 2, n}]]

Out[639]= {0.57, Null}

By comparison:

In[640]:= Timing[result2 = orderedList[data];]
Out[640]= {14.71, Null}

In[641]:= result2 === result
Out[641]= True

Daniel Lichtblau
WOlfram Research





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Subject: [mg120379] Re: sorting a nested list of 4-tuples
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Your function does not appear to work correctly, sometimes the list is 
in the wrong order, but orderedList2:

orderedList2[list_] :=
 Module[{first = First[list], rest = Rest[list], nearest, result},

  result = {first};
  Do[nearest = Nearest[rest -> Automatic, Last[result]];
   result = Join[result, rest[[nearest]]];
   rest = Drop[rest, nearest], {Length[rest]}]; result]

 seems faster anyway.:

In[11]:= Module[{list = RandomReal[{0, 1}, {20, 4}], result1, 
result2},
 result1 = Timing[orderedList[list]];
 result2 = Timing[orderedList2[list]]; {result1[[2]] == 
result2[[2]],
  result1[[1]], result2[[1]], {list, result1[[2]], result2[[2]]}}]
Out[11]= {False, 0.002556, 0.0007, {{{0.0780386, 0.220901, 0.748457,
    0.548342}, {0.722099, 0.308747, 0.33661, 0.304516}, {0.623358, 0.761312,
    0.884318, 0.21337}, {0.14872, 0.284139, 0.244084, 0.582692}, {0.0829757,
    0.911987, 0.448848, 0.606659}, {0.312903, 0.703568, 0.747021,
    0.126831}, {0.847102, 0.437041, 0.0924343, 0.0333101}, {0.189339,
    0.419192, 0.498754, 0.846607}, {0.661185, 0.455712, 0.31286,
    0.387921}, {0.566059, 0.331815, 0.412978, 0.84236}, {0.714704, 0.335434,
    0.417758, 0.0569088}, {0.855757, 0.0118403, 0.310144,
    0.312894}, {0.278001, 0.874878, 0.289541, 0.199531}, {0.682503, 0.387263,
    0.931379, 0.662633}, {0.8045, 0.264645, 0.595204, 0.534895}, {0.645246,
    0.915467, 0.167849, 0.210847}, {0.789694, 0.987745, 0.0405679,
    0.0957831}, {0.811736, 0.77314, 0.73452, 0.0771508}, {0.784869, 0.582503,
    0.0986897, 0.152786}, {0.426251, 0.135863, 0.465908,
    0.228462}}, {{0.0780386, 0.220901, 0.748457, 0.548342}, {0.14872,
    0.284139, 0.244084, 0.582692}, {0.189339, 0.419192, 0.498754,
    0.846607}, {0.312903, 0.703568, 0.747021, 0.126831}, {0.278001, 0.874878,
    0.289541, 0.199531}, {0.0829757, 0.911987, 0.448848, 0.606659}, {0.623358,
     0.761312, 0.884318, 0.21337}, {0.645246, 0.915467, 0.167849,
    0.210847}, {0.789694, 0.987745, 0.0405679, 0.0957831}, {0.811736, 0.77314,
     0.73452, 0.0771508}, {0.784869, 0.582503, 0.0986897,
    0.152786}, {0.847102, 0.437041, 0.0924343, 0.0333101}, {0.714704,
    0.335434, 0.417758, 0.0569088}, {0.722099, 0.308747, 0.33661,
    0.304516}, {0.682503, 0.387263, 0.931379, 0.662633}, {0.661185, 0.455712,
    0.31286, 0.387921}, {0.566059, 0.331815, 0.412978, 0.84236}, {0.426251,
    0.135863, 0.465908, 0.228462}, {0.8045, 0.264645, 0.595204,
    0.534895}, {0.855757, 0.0118403, 0.310144, 0.312894}}, {{0.0780386,
    0.220901, 0.748457, 0.548342}, {0.189339, 0.419192, 0.498754,
    0.846607}, {0.14872, 0.284139, 0.244084, 0.582692}, {0.566059, 0.331815,
    0.412978, 0.84236}, {0.8045, 0.264645, 0.595204, 0.534895}, {0.722099,
    0.308747, 0.33661, 0.304516}, {0.661185, 0.455712, 0.31286,
    0.387921}, {0.784869, 0.582503, 0.0986897, 0.152786}, {0.847102, 0.437041,
     0.0924343, 0.0333101}, {0.714704, 0.335434, 0.417758,
    0.0569088}, {0.426251, 0.135863, 0.465908, 0.228462}, {0.855757,
    0.0118403, 0.310144, 0.312894}, {0.682503, 0.387263, 0.931379,
    0.662633}, {0.623358, 0.761312, 0.884318, 0.21337}, {0.811736, 0.77314,
    0.73452, 0.0771508}, {0.312903, 0.703568, 0.747021, 0.126831}, {0.278001,
    0.874878, 0.289541, 0.199531}, {0.645246, 0.915467, 0.167849,
    0.210847}, {0.789694, 0.987745, 0.0405679, 0.0957831}, {0.0829757,
    0.911987, 0.448848, 0.606659}}}}

From the above output it seems to me orderedList is incorrect:

In[12]:= Nearest[{{0.7220990188851284`, 0.30874655560999353`, 0.33661039115480085`,
   0.30451607189733565`}, {0.623358140247426`, 0.7613121754790066`,
   0.8843184964161348`, 0.21336965042896106`}, {0.14871956929750207`,
   0.2841388256243189`, 0.2440839157054575`,
   0.5826918077236027`}, {0.08297574175354927`, 0.9119868514832306`,
   0.44884815651517496`, 0.606658888854545`}, {0.3129032093692543`,
   0.7035682306379043`, 0.7470209852828422`,
   0.126831036636313`}, {0.8471019825138335`, 0.4370412344560497`,
   0.09243429503478429`, 0.03331005052172231`}, {0.18933936353633118`,
   0.4191916563275504`, 0.49875370996337387`,
   0.8466068811542358`}, {0.6611848899498884`, 0.455712283905654`,
   0.31286031890780386`, 0.3879208331484807`}, {0.566058771065481`,
   0.3318146887998126`, 0.41297814980274516`,
   0.8423595336937795`}, {0.7147038318708883`, 0.33543423510640613`,
   0.4177577945972242`, 0.056908774958363884`}, {0.8557567595586966`,
   0.011840287587700837`, 0.31014365477384986`,
   0.31289360223925855`}, {0.27800072513915963`, 0.8748776068932764`,
   0.28954067774830894`, 0.19953109481447573`}, {0.6825027823624006`,
   0.38726329607605536`, 0.9313788703031116`,
   0.6626327289353637`}, {0.8045002220254145`, 0.26464489073341646`,
   0.5952036781362782`, 0.5348945499518614`}, {0.6452463375631783`,
   0.9154672494486176`, 0.16784916226023205`,
   0.21084656084695386`}, {0.7896939168446158`, 0.9877454808787693`,
   0.04056790678073363`, 0.09578312387470667`}, {0.8117363160995603`,
   0.7731396349254072`, 0.7345204475002354`,
   0.0771508252866615`}, {0.784869093354617`, 0.5825033331936824`,
   0.09868967226141323`, 0.1527856209885663`}, {0.426250694413443`,
   0.1358634032399666`, 0.46590788139431916`,
   0.22846239863279472`}}, {0.07803858379747108`, 0.22090093831902768`,
  0.7484574646075761`, 0.5483416965515358`}]
Out[12]= {{0.189339, 0.419192, 0.498754, 0.846607}}

Regards,
	Ssezi

On Jul 19, 2011, at 6:55 AM, Luis Valero wrote:

> Dear Sirs,
>
> I want to sort a list of 4-tuples of real numbers, so that, the second 4-tuple has minimum distance to the first, the third, selected from the rest, ha minimum distance to the second, and so on.
>
> The distance is the euclidean distance, calculated with the first two elements of each 4-tuple.
>
> I have defined a function that work:
>
> orderedList[list_] := Module[{nearest},
> 	Flatten[ Nest[{
>   	    Append[ #[[1]] , nearest = Flatten @@ Nearest[ Map[ Rule[ #[[{1, 2}]], #] &, #[[2]] ], Last[ #[[1]] ] [[{1, 2}]], 1] ],
>    	    Delete[ #[[2]], Position[ #[[2]], nearest ] ]  } &, { {list[[1]] }, Delete[ list, 1 ] }, Length[ list ] - 2], 1] ];
>
>
> but I need to improve the temporal efficiency
>
> Thank you
>
>
>




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Date: Wed, 20 Jul 2011 06:33:07 -0400 (EDT)
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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120369] Re: remarks on significance arithmetic implementation [Was: Re: Numerical accuracy/precision - this is a bug or a feature?]
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As before, I think much of this discussion is useless. But this example  
DOES give me pause:

> Consider the two expressions
> f[x_]:= (x-1)*(x+1)  and g[x_]:=(x^2-1)
>
> which are analytically the same function.
>
> given the value v=Interval[{-1/2,1/2}]    f[v] returns an interval with
> width 2.  g[v] returns an interval with width 1/4.

I don't see why we'd want significance tracking to act this way.

Bobby

On Tue, 19 Jul 2011 05:58:03 -0500, Richard Fateman  
<fateman at eecs.berkeley.edu> wrote:

> On 7/18/2011 11:54 AM, Daniel Lichtblau wrote:
>>
>> [Caution: This may increase the entropy of the thread, and almost
>> certainly will increase the ennui.]
> Yep
>>
>>
> ... snip ... a bunch of things I think we agree upon.
>> To expand on "properly rounded floats" I will mention that
>> Mathematica's significance arithmetic, in certain settings, is as
>> conservative as interval arithmetic. Specifically this works out when
>> the operations used are addition, multiplication, integer powering, or
>> division with both operands approximate. Also required is that they
>> have more significant digits than can be represented by the error
>> recorder (that is to say, more than 16 or so digits when error is
>> recorded as a 53-bit-mantissa float). The gist is that precision is
>> altered downward in a way that will even account for second order
>> error (and the higher order error terms that exist in computing x/y
>> where x and y are both approximate bignums).
> I am unclear on this zeroth, first, second  order error categorization.
> It seems to me that the critical problem, and one that is not addressed
> by Mathematica, in interval arithmetic, is addressing dependence of
> operands.  Consider the two expressions
> f[x_]:= (x-1)*(x+1)  and g[x_]:=(x^2-1)
>
> which are analytically the same function.
>
> given the value v=Interval[{-1/2,1/2}]    f[v] returns an interval with
> width 2.  g[v] returns an interval with width 1/4.
>
> The function g is obviously a lot better than f.
> Same thing happens for v=0.1`2.  f[v] is -0.99`269..     but g[v]  is
> -0.99`3.69...  So according to Mathematica, g evaluates
> to a full decimal digit higher accuracy.
>
> A true devotee of interval arithmetic (or of significance arithmetic)
> would take the function body for f and rearrange it as g,
> and also would require for comparisons such concepts as possibly-equal,
> certainly-greater-than  etc.  And would need
> to resolve meanings for empty intervals and other such things.  Some of
> these may in fact be in Mathematica's interval
> arithmetic suite.  They seem to be absent in the significance arithmetic
> world, though there is a transition from one to the
> other possible.
> Note that Interval[0.1`2] returns the fairly nonsensical
> Interval[{0.1,0.10}] but InputForm[%] shows
> Interval[{0.098046875`1.9914337561691884,
>    0.101953125`2.008400542026431}], maybe overkill.
> Interval[{0.98,0.102}]  might be OK.
>
>
> There are ways of reducing semi-automatically, the number of repeat
> occurrences of operands in such expressions, thereby improving the
> interval widths.  This (that is, computing "single use expressions") is
> the kind of thing that would, I think, be useful and for which computer
> algebra systems are especially appropriate.  That, in combination with
> some other tricks of interval "reliable" arithmetic would be helpful
> (and for which I have written some programs, not in Mathematica
> though).  So the real contribution to interval arithmetic (or
> significance arithmetic) of Mathematica would be if it were able to
> reformulate some subset of expressions so that they were automatically
> computed faster and more accurately.
>
> (One could argue that boosting the software precision will get the
> answer accurately, especially if one ignores exponentially longer
> run-times.)
>
>
> For some discussion in the context of a (different) programming language
> that is not much like Mathematica, see
>
> http://developers.sun.com/sunstudio/products/archive/whitepapers/tech-interval-final.pdf
>
> There are other, more expensive techniques. E.g.  if p(x) is a
> polynomial, what is p[Interval[{a,b}]] ?
> Instead of just interval-alizing the + and * and ^,  find the extrema of
> p between a and b.
> What is Sin[Interval[...]], etc.
>
>>
>> It would not be hard to extend this to the taking of reciprocals and
>> hence division of integer by float e.g. x --> 1/x. But as best I can
>> tell from looking at the code, Mathematica does not do this, so error
>> tracking in that case is purely first order.
>
> Maybe I'm not following here, but computing an interval for 1/x does not
> seem difficult.
>>
>>
>>> [...]
>>> My point remains:
>>>    WRI could have a much more direct notion of number, and equality,
>>> that
>>> would make it easy to implement my choice of arithmetic or theirs or
>>> yours.
>>> They didn't.  The default is not completely reliable. People write to
>>> this newsgroup, periodically, saying Huh? what's going on? I found a
>>> bug!
>>
>> This is, I am afraid, nonsense. Nothing would make it "easy" to
>> implement significance arithmetic across the spectrum of Mathematica's
>> numerical engine. To do so efficiently, reliably, maintainably, etc.
>> is, well, a significant task. Jerry Keiper could do it. Mark Sofroniou
>> could do it. I do not think I could do it, at least not very well.
>> And, as the saying goes, "I have me doubts about thee".
>
> OK, I did not mean it would be easy to reproduce what Jerry Keiper did,
> but after all, he did do it using ordinary arithmetic at its base.
> What I did mean is that programs to do arithmetic "the machine way"
> would run at machine speed (= easy); the programs to do software
> bigfloats without significance tracking would run faster (=easier) than
> including significance tracking.   And that if one wished to patch-up
> such code with built-in tracking ,after the fact by re-stating the
> precision after each operation, that would be slower (= not as easy)
> than avoiding it in the first place.
> So, easy= easy for the computer.
>
> I do think that people write to the newsgroup reporting what they think
> of as bugs, but are actually the deliberate consequences of the design
> of arithmetic.
>>
>>
>>> As for the experimental and computational notions of (im)precision, I
>>> think there is an issue of using the same words with different  
>>> meanings.
>>> Similar but unfortunately different. Precision of a floating point
>>> number F is simply the number of bits in the fraction.  If you impute
>>> some uncertainty to F, you can store that in another piece of data D in
>>> the computer. If you want to assert that F and D together represent a
>>> distribution of a certain kind  you can also compute with that.
>>>
>>> RJF
>>> [...]
>>
>> Related to the above (I think), you stated earlier in this thread that
>> the differences in terminology usage can be troublesome. I agree that
>> the meaning of Mathematica's Precision and, to a lesser extent,
>> Accuracy, might cause initial confusion to one unfamiliar with the
>> software but well versed in Numerical Analysis. I would expect such
>> people to catch on after being bitten once or twice, and either get
>> used to the distinctions, or use different software.
> yes, but there are not too many well versed in numerical analysis.
>>
>> This strikes me as similar to the situation wherein Append and Prepend
>> are O(n) rather than O(1) complexity operations, since they rewrite
>> their List arguements in full. A CS person new to Mathematica might
>> well assume List is akin to the CS notion of a (singly or doubly)
>> linked list. After learning otherwise they'd probably not be too
>> flustered, other than perhaps to mutter that the naming of "List" was
>> problematic. Even that would perhaps abate once they get accustomed to
>> the very general Mathematica notion of a List. After all, a linked
>> list is a modestly arcane structure to those outside CS, and most
>> general software users are not from inside CS, and most CS people will
>> understand that.
> Similar, yes.  Dissimilar in the sense that the List operations return
> expected results, just slower. But arithmetic returns (on occasion)
> unexpected results.
>
> RJF
>
>


-- 
DrMajorBob at yahoo.com



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From: "Joseph O'Rourke" <orourke at scinix.smith.edu>
Subject: [mg120381] ExampleData[{"Geometry3D", "Torus"}]
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Is it possible to control the radii of a torus in Mathematica 8 using the ExampleData[] function?  I cannot see how to get anything but the default torus.  Thanks!



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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120395] Computable Document Format Replaces Player Notebooks
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Hi Everyone,

Probably a silly question but I just want to be sure: Has the computable document format that is built into Mathematica replaced player notebooks, so that I no longer have to upload my notebooks to Wolfram for conversion?

Regards,

Gregory





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From: Myeong Ae Kang <makang at purdue.edu>
Subject: [mg120383] How to set the format of every output to traditionalform in txt files.
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Hello.

I wonder how I can set the default data format to "traditionalform" to run Mathematica on the command-line.
I want to print all values in traditonalform, not including "traditonalform" in every "Print" command.

Again, on the command line. Not on the GUI.



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From: Daniel Lichtblau <danl at wolfram.com>
Subject: [mg120382] Re: sorting a nested list of 4-tuples
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On 07/19/2011 05:55 AM, Luis Valero wrote:
> Dear Sirs,
>
> I want to sort a list of 4-tuples of real numbers, so that, the second 4-tuple has minimum distance to the first, the third, selected from the rest, ha minimum distance to the second, and so on.
>
> The distance is the euclidean distance, calculated with the first two elements of each 4-tuple.
>
> I have defined a function that work:
>
> orderedList[list_] := Module[{nearest},
> 	Flatten[ Nest[{
>     	    Append[ #[[1]] , nearest = Flatten @@ Nearest[ Map[ Rule[ #[[{1, 2}]], #]&, #[[2]] ], Last[ #[[1]] ] [[{1, 2}]], 1] ],
>      	    Delete[ #[[2]], Position[ #[[2]], nearest ] ]  }&, { {list[[1]] }, Delete[ list, 1 ] }, Length[ list ] - 2], 1] ];
>
>
> but I need to improve the temporal efficiency
>
> Thank you

I wanted to improve on my last proposed method and also say a bit about 
the computational complexity.

First I'll note that a straightforward double iteration can be used. If 
coupled with Compile it should be reasonably fast, albeit still O(n^2) 
for a list of n elements. Also it will scale nicely with dimension in 
cases where we are not restricting the distance measure to the first two 
components of the tuples.

Here is code for this.

findClosestUniqueC = Compile[{{ll, _Real, 2}},
    Module[{n = Length[ll], res = ll, posn, max, min, diff, dist, last,
       tmp},
     max = (4.*Max[ll[[All, 1 ;; 2]]])^2;
     Do[
      last = res[[j - 1, 1 ;; 2]];
      posn = 0;
      min = max;
      Do[
       diff = res[[k, 1 ;; 2]] - last;
       dist = diff.diff;
       If[dist < min, min = dist; posn = k];
       , {k, j, n}];
      res[[{j, posn}]] = res[[{posn, j}]];
      , {j, 2, n - 1}];
     res
     ]
    ];

The method I used in an earlier response, and will refine below, was 
based on Nearest[]. First some explanatory comments.

It takes the first two components of each tuple and augments with a 
small "counter" value that allows us to determine which is the 
corresponding element in the original list. We start with a large 
"denominator", then prepend 1/denom to the first 2-tuple, 2/denom to the 
second 2-tuple, etc. The presence of this counter value can in fact lead 
to erroneous results by slightly offsetting distances based on the 
2-tuple values alone. I now use a heuristic based on list size that 
should make such a failure unlikely in the version below. But it is not 
terribly clever and certainly can fail. A viable method would have to be 
based on minimal separation, which could be underestimated by sorting 
the flattened list of all values, and taking the smallest nonzero gap.

We'll assume that the complexity of creating a one-argument Nearest[] 
function, given n values, is around O(n*log(n)). For modest dimension 
this seems to be about right, if I recall correctly. Once we have such a 
function we'll assume that the cost of looking up the k nearest values 
to a given value is O(k*log(n)). Again, this seems to be born out with 
experience as best I recall. Also I'll add that the implied 
multiplicative constant is much larger for lookup than for building the 
list.

At this points there are various ways to go about the lookup in order to 
get the next closest value to a given one, from amongst all elements not 
already chosen. A guaranteed way is, if j elements have been chosen, 
find the j+1 nearest elements to the last one (so at least one will not 
already have been selected).

If we build a Nearest function once, then we end up doing O(n) lookups, 
each finding O(n) closest elements. The overall complexity would be 
O(n^2*log(n)). (If we just call it anew every iteration, thats n alls of 
n*log(n) complexity, so no better.)

A heuristic to reduce that, not guaranteed, would be to only look up the 
7 or so closest, and try the full size needed only if all 7 have already 
been selected.

Another way, this one guaranteed to bring down the complexity, is to 
discard those elements already chosen from further consideration, at 
intervals that are an appropriate function of n. If we do this every 
sqrt(n) elements, and rebuild the Nearest function, then we never need 
find more than sqrt(n) closest elements from amongst those still under 
consideration. So overall lookup cost will be no worse than 
O(n*sqrt(n)*log(n)). Meanwhile the cost of removing sqrt(n) elements 
form a list of n elements is O(n), and we do this sqrt(n) times so that 
comes to O(n^(3/2)). Finally the cost of building these Nearest 
functions sqrt(n) times, with O(n) elements each time, is 
O(n*log(n)*sqrt(n)). Overall cost: O(n^(3/2)*log(n)).

In my earlier code I used a different "chunk" size, one that seemed like 
it might be sensible. It did work well for ranges I tried, but sqrt(n) 
seems to be the value with the guarantee. The use of a new Nearest 
function at intervals means I want to use an index computed modulo the 
interval length (the chunk size, in the code terminology).

I also use the heuristic of trying only a few closest elements at first, 
grabbing more only when needed. Some testing indicates this happens 
relatively infrequently.

So here is the updated code. It should be properly Module-ized, but I'm 
not spending more time to do that.

n = 3200;
data = RandomReal[{-100, 100}, {n, 4}];

chunksize = Ceiling[Sqrt[n]];
denom = 2^3*2^Ceiling[Log[2, n]];
moddata =
   Map[Take[#, 3] &, MapThread[Prepend, {data, Range[n]/denom}]];
nf = Nearest[moddata];
next = moddata[[1]];
taken = ConstantArray[False, n];
result = ConstantArray[{0., 0., 0., 0.}, n];
remove = ConstantArray[{0., 0., 0.}, chunksize];
result[[1]] = data[[1]];
remove[[1]] = moddata[[1]];
taken[[1]] = True;

Timing[Do[
   modj = Mod[j, chunksize, 1];
   nextset = nf[next, Min[7, 1 + modj]];
   posns = Round[denom*Map[First, nextset]];
   k = 1;
   While[k <= Length[nextset] && taken[[posns[[k]]]], k++;];
   If[k > Length[nextset],
    nextset = nf[next, 1 + modj];
    posns = Round[denom*Map[First, nextset]];
    k = 1;
    While[k < Length[nextset] && taken[[posns[[k]]]], k++;];
    ];
   taken[[posns[[k]]]] = True;
   result[[j]] = data[[posns[[k]]]];
   next = nextset[[k]];
   remove[[modj]] = next;
   If[modj == chunksize,
    done = True;
    posns = Apply[Alternatives, remove];
    moddata = DeleteCases[moddata, posns];
    nf = Nearest[moddata];
    ];
   , {j, 2, n}]]

Out[1294]= {0.49, Null}

Comparison:

In[1336]:= Timing[result2 = findClosestUniqueC[data];]
Out[1336]= {1.89, Null}

In[1337]:= result === result2
Out[1337]= True

Daniel Lichtblau
Wolfram Research




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After importing a huge SAS data set into Mathematica that contains 400 data elements, is there a global validation command where Mathematica could return a report of on the blanks and content frequencies in all of the data fields or do you have to run separate queries for every single data element?



Sylvia Hobbs



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From: Helen Read <readhpr at gmail.com>
Subject: [mg120392] Re: Total plot width
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On 7/20/2011 6:35 AM, carlos at colorado.edu wrote:
> How can I specify the total plot width in Mathematica?  I was looking for
> something like AbsolutePlotWidth->250  (points implied) but havent found
> that option in the documentation.  There is AspectRatio, but that is
> not a dimension. Thanks.
>

I suspect you will get many replies to this. The option you want is 
ImageSize


-- 
Helen Read
University of Vermont



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From: dimitris <dimmechan at yahoo.com>
Subject: [mg120397] unable to save notebook
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Hello to all.
When my student tries to save a document (e.g. hh its name) the
following message appears...

"Mathematica could not find enough disk space to save the file C:\Users
\=D7\[Rho]=DE\[Sigma]\[Tau]\[Omicron]\[FinalSigma]\Downloads\hh.nb. The
file was left unchanged. The disk is full, or if you are using a file
server you may have reached your disk space quota."

Any ideas on how to solve the problem?

Thank you very much in advance for your replies.



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From: "Gabriel T. Landi" <gtlandi at gmail.com>
Subject: [mg120380] Re: Total plot width
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You can use ImageSize-> {{left, right},{bottom,top}}. 
Check out ImageSize page for more details.

Cheers

Gabriel

On 20/07/2011, at 07:32, carlos at colorado.edu wrote:

> How can I specify the total plot width in Mathematica?  I was looking for
> something like AbsolutePlotWidth->250  (points implied) but havent found
> that option in the documentation.  There is AspectRatio, but that is
> not a dimension. Thanks



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From: Jay Bee <jiri.bocan at gmail.com>
Subject: [mg120391] Re: Total plot width
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On Jul 20, 11:35 am, car... at colorado.edu wrote:
> How can I specify the total plot width in Mathematica?  I was looking for
> something like AbsolutePlotWidth->250  (points implied) but havent found
> that option in the documentation.  There is AspectRatio, but that is
> not a dimension. Thanks.

Hard to say, what do you mean exactly; do you mean Plot[] or
Listplot[], or something of higher dimension? Resulting plot-width is
influenced by AspectRatio, PlotRange, ImageSize, ...



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Subject: [mg120390] Re: Total plot width
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The option:

PlotRange->{Width, Height}

should help. Try various cases specifying the Width say, 250 or
whatever you need, and the Height that may be either a fixed number or
Automatic.

Have fun. Alexei





How can I specify the total plot width in Mathematica? I was looking for

something like AbsolutePlotWidth->250 (points implied) but havent found

that option in the documentation. There is AspectRatio, but that is

not a dimension. Thanks.





Alexei BOULBITCH, Dr., habil.

IEE S.A.

ZAE Weiergewan,

11, rue Edmond Reuter,

L-5326 Contern, LUXEMBOURG



Office phone: +352-2454-2566

Office fax: +352-2454-3566

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From mathgroup-adm at smc.vnet.net  Thu Jul 21 04:08:37 2011
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Erroneous prices are randomly returned:

FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
   "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
   "PTTAX", "RGACX", "STRFX"}

{8.81, 72.4, 14., 10.64, 46.49, 12.48,
  5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
  5.03241*10^8, 11.04, 31.44, 32.47}

FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
   "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
   "PTTAX", "RGACX", "STRFX"}

{5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
  5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
  5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}

Bobby

-- 
DrMajorBob at yahoo.com




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From: dr DanW <dmaxwarren at gmail.com>
Subject: [mg120387] Warning: OS X Lion and compilation
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I have not dug thoroughly into this yet, but I wanted to get the warning out quickly.  Apparently, OS X Lion does not include the gcc compiler.  I upgraded to Lion this morning, then on a lark tried to compile a function in Mathematica with CompilationTarget -> "C".  I got an error that no compiler was found.

I am sure that it is only a matter of either obtaining a copy of gcc or Xcode (with its own c compiler, not gcc) and configuring Mathematica to use that compiler.  I have not tried either, since I am not using anything but opcode compilation.  If you are successful, please post instructions for others on this group.

Thanks,
Daniel



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From: Murray Eisenberg <murray at math.umass.edu>
Subject: [mg120394] Re: Total plot width
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The option you want is ImageSize.

On 7/20/11 6:32 AM, carlos at colorado.edu wrote:
> How can I specify the total plot width in Mathematica?  I was looking for
> something like AbsolutePlotWidth->250  (points implied) but havent found
> that option in the documentation.  There is AspectRatio, but that is
> not a dimension. Thanks.
>

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305



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From: Gary Wardall <gwardall at gmail.com>
Subject: [mg120388] Re: locator
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On Jul 13, 2:19 am, Heike Gramberg <heike.gramb... at gmail.com> wrote:
> I'm not sure what it is what you're asking for, but if you want to use a locator to trace a curve you can do something like
>
> Module[{p = {0, 1}, temp},
>  temp[t_] := Exp[- t];
>  Row[{
>    LocatorPane[Dynamic[p, (p[[2]] = temp[#[[1]]]; p[[1]] = #[[1]]) &],
>      Plot[temp[t], {t, 0, 5}, PlotRange -> {Automatic, {0, 1}},
>      ImageSize -> 400],
>     {{0, 0}, {5, 1}}],
>    Dynamic[Grid[{{"Time:", p[[1]]}, {"Temp:", p[[2]]}}]]}]]
>
> Heike.
>
> On 12 Jul 2011, at 11:59, raga wrote:
>
> > hi,
> > i created a front end in mathematica which provides the user the
> > independence of providing the input to the code  and my code plots the
> > graph for variation of temperature with time for newtons law of
> > cooling but can any body please help me out by telling how to use a
> > locator to locate the individual points which shows the respective
> > temperature on the graph

Would the "Drawing Tools" under "Graphics" be of help. I find it
useful when I want to locate the coordinates on Plots.

Gary Wardall



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Subject: [mg120399] Re: sorting a nested list of 4-tuples
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On Jul 19, 4:00 am, Luis Valero <luis.val... at mac.com> wrote:
> Dear Sirs,
>
> I want to sort a list of 4-tuples of real numbers, so that, the second 4-tuple has minimum distance to the first, the third, selected from the rest, ha minimum distance to the second, and so on.
>
> The distance is the euclidean distance, calculated with the first two elements of each 4-tuple.
>
> I have defined a function that work:
>
> orderedList[list_] := Module[{nearest},
>         Flatten[ Nest[{
>             Append[ #[[1]] , nearest = Flatten @@ Nearest[ Map[ Rule[ #[[{1, 2}]], #] &, #[[2]] ], Last[ #[[1]] ] [[{1, 2}]], 1] ],
>             Delete[ #[[2]], Position[ #[[2]], nearest ] ]  } &, { {list[[1]] }, Delete[ list, 1 ] }, Length[ list ] - 2], 1] ];
>
> but I need to improve the temporal efficiency
>
> Thank you

This does the job nicely without Nearest. For your real data,
change DistanceMatrix@xy to DistanceMatrix[list[[All,{1,2}]]].

n = 10
xy = Table[Random[Real],{n},{2}]
ddx = 2 Max[dd = DistanceMatrix@xy];
Do[dd[[i,i]] = ddx,{i,n}];
ddx = Table[ddx,{n}];
NestList[(dd[[#]] = ddx;
 Ordering[dd[[All,#]],1][[1]])&, 1, n-1]

10

{{0.923827,0.822223}, {0.369167,0.352365},
 {0.309981,0.792447}, {0.375829,0.953723},
 {0.680294,0.594137}, {0.411011,0.098692},
 {0.295584,0.0860474},{0.749339,0.927127},
 {0.263566,0.0431098},{0.0684959,0.230974}}

{1,8,5,2,6,7,9,10,3,4}



From mathgroup-adm at smc.vnet.net  Thu Jul 21 04:18:35 2011
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From: Gareth Edwards <gareth.edwards at cubicmotion.com>
Subject: [mg120393] newbie Map, Dynamic, etc
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Hi,

Still playing around with Mathematica as a newbie and wrestling with the
'different'  way one has to think about things as opposed to other
languages. I think (hopefully) that an informative example for me would be a
solution (or at least some hints) to the following:

Say I have an image sequence (mySequence_0000.jpg, etc.) I'd like to create
a viewer with a slider that scrubs through the images. That I can do.

What I'd also like is an array of points (just one point per image, and
initialized to {0,0} say ), such that on each image I can move a locator
around and thereby modify the corresponding stored point in the array. In
other words, store the location of a point (as manipulated by me) against
each image.

My gut tells me that this is trivial to the average Mathematica user- the
trouble is I remain well below average right now!

Any help appreciated.

Best,

Gareth

-- 
Dr. Gareth Edwards
Director
www.cubicmotion.com

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From: Ray Koopman <koopman at sfu.ca>
Subject: [mg120385] Re: sorting a nested list of 4-tuples
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On Jul 19, 4:00 am, Luis Valero <luis.val... at mac.com> wrote:
> Dear Sirs,
>
> I want to sort a list of 4-tuples of real numbers, so that, the second 4-tuple has minimum distance to the first, the third, selected from the rest, ha minimum distance to the second, and so on.
>
> The distance is the euclidean distance, calculated with the first two elements of each 4-tuple.
>
> I have defined a function that work:
>
> orderedList[list_] := Module[{nearest},
>         Flatten[ Nest[{
>             Append[ #[[1]] , nearest = Flatten @@ Nearest[ Map[ Rule[ #[[{1, 2}]], #] &, #[[2]] ], Last[ #[[1]] ] [[{1, 2}]], 1] ],
>             Delete[ #[[2]], Position[ #[[2]], nearest ] ]  } &, { {list[[1]] }, Delete[ list, 1 ] }, Length[ list ] - 2], 1] ];
>
> but I need to improve the temporal efficiency
>
> Thank you

This is faster than the code in my previous post (in which the
line 'Do[dd[[i,i]] = ddx,{i,n}];' should be deleted because it
was left over from an earlier version and is no longer needed):

n = Length[xy =
 {{0.9238265948180384,  0.8222225655013509},
  {0.36916681632329446, 0.352365112953319},
  {0.30998090503173714, 0.7924465475420762},
  {0.3758287909737646,  0.9537228544636929},
  {0.6802940452860295,  0.5941374424537628},
  {0.41101146869486416, 0.09869197753233201},
  {0.2955842108802197,  0.08604736979206176},
  {0.7493387235460567,  0.9271268053705936},
  {0.26356623115061834, 0.04310983468962257},
  {0.06849589850565166, 0.23097371847465945}}]

10

dd = DistanceMatrix@xy; q = Range@n;
NestList[(q = SparseArray[q,Automatic,#] /.
              SparseArray[_,_,_,x_] :> x[[3]];
  q[[Ordering[dd[[#,q]],1][[1]]]])&, 1, n-1]

{1,8,5,2,6,7,9,10,3,4}



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From: Gary Wardall <gwardall at gmail.com>
Subject: [mg120386] Re: Inverse of Interpolating Function?
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On Jul 20, 5:34=C2 am, gonzorascal <dj... at duke.edu> wrote:
> Hello All,
>
> I am working with an Interpolating Function of an oscillating solution (time series) to a differential equation.  I am trying to find the period and pulse width of the oscillation.  To do this I would like to have an inverse of my function y[t] (so I would have t[y]).  I realize this function would be multivalued (that's what I want to find the period). =C2 I am not having success using Mathematica's InverseFunction[] or Reduce[] commands.  Does anyone have any experience or suggestions with this sort of thing (either finding Inverse Interpolating Functions or another method for period and pulse width)?  Thank you.
>
> -GR

gonzorascal,

First make sure the function you wish to have an inverse function is a
function that has an inverse function. A quick visual test is the
Horizontal Test for Inverse Functions. If any horizontal line
intersects the graph twice or more then the inverse is NOT an inverse
function. Graph the function you wish to create an inverse function
for. If it fails the horizontal test you may have ti restrict  your
function like is done in Trigonometry. If it passes try something like
in my example.


points = {{0, -1}, {1, 1}, {2, 3}, {3, 4}, {4, 6}, {5, 10}};=E2=80=A8ifun=
Interpolation[points]
Does the function pass the Horizontal Test?

Plot[ifun[x], {x, 0, 5}]

It does. Define:

invifun[x_] := FindRoot[ifun[y] == x, {y, 0, 5}][[1]][[2]]

Note that:

ifun[invifun[4]]

is 4.

and

invifun[ifun[3]]

is 3.

and look at the sketch.

Plot[{invifun[x], ifun[x], x}, {x, 0, 5}, PlotRange -> {0, 5},
AspectRatio -> Automatic]

We have an inverse function.

I hope this will help.

Good Luck

Gary Wardall



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Date: Thu, 21 Jul 2011 05:48:05 -0400 (EDT)
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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120398] Re: sorting a nested list of 4-tuples
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Sseziwa,

The OP wanted distances to take into account only the 1st two elements of  
each vector, not all four, so your orderings usually won't match those of  
the original code.

A simple modification works well, however. I call it "treat" below, but  
it's still your code, mostly.

valero[list_] :=
   Module[{nearest},
    Flatten[Nest[{Append[#[[1]],
         nearest =
          Flatten @@
           Nearest[Map[Rule[#[[{1, 2}]], #] &, #[[2]]],
            Last[#[[1]]][[{1, 2}]], 1]],
        Delete[#[[2]], Position[#[[2]], nearest]]} &, {{list[[1]]},
       Delete[list, 1]}, Length[list] - 2], 1]];

sseziwa[list_] :=
  Module[{first = First[list], rest = Rest[list], nearest, result},
   result = {first};
   Do[nearest = Nearest[rest -> Automatic, Last[result]];
    result = Join[result, rest[[nearest]]];
    rest = Drop[rest, nearest], {Length[rest]}]; result]

treat[list_] :=
  Module[{first = First@list, rest = Rest@list, nearest, result},
   result = {first};
   Do[nearest =
     Nearest[rest[[All, ;; 2]] -> Automatic, result[[-1, ;; 2]], 1];
    result = Join[result, rest[[nearest]]];
    rest = Drop[rest, nearest], {Length[rest]}]; result]

Here is Daniel Lichtblau's code, but I may have broken it somehow. It  
sometimes agrees with Valero's code and sometimes does not.

daniel[data_List] :=
  Module[{n = Length@data, chunksize, denom = 1000., moddata, nf, next,
     taken, result, remove, nextset, posns, k},
   chunksize = Ceiling[Log[n]]^2;
   moddata =
    Map[Take[#, 3] &, MapThread[Prepend, {data, Range[n]/denom}]];
   nf = Nearest[moddata];
   next = moddata[[1]];
   taken = ConstantArray[False, n];
   result = ConstantArray[{0., 0., 0., 0.}, n];
   remove = ConstantArray[{0., 0., 0.}, chunksize];
   result[[1]] = data[[1]];
   remove[[1]] = moddata[[1]];
   taken[[1]] = True;
   Do[nextset = nf[next, 1 + Mod[j, chunksize, 1]];
    posns = Round[denom*Map[First, nextset]];
    k = 1;
    While[k < Length[nextset] && taken[[posns[[k]]]], k++;];
    taken[[posns[[k]]]] = True;
    result[[j]] = data[[posns[[k]]]];
    next = nextset[[k]];
    remove[[Mod[j, chunksize, 1]]] = next;
    If[Mod[j, chunksize] == 0, done = True;
     posns = Apply[Alternatives, remove];
     moddata = DeleteCases[moddata, posns];
     nf = Nearest[moddata];], {j, 2, n}];
   result
   ]

n = 10^2;
data = RandomReal[1, {n, 4}];
Timing[one = valero@data;]
Timing[two = sseziwa@data;]
Timing[three = treat@data;]
Timing[four = daniel@data;]
{one == two, one == three, one == four}

{0.025656, Null}

{0.004861, Null}

{0.005198, Null}

{0.007332, Null}

{False, True, False}  <----- Daniel's and Sseziwa's code disagreed.

n = 10^2;
data = RandomReal[{-100, 100}, {n, 4}];
Timing[one = valero@data;]
Timing[two = sseziwa@data;]
Timing[three = treat@data;]
Timing[four = daniel@data;]
{one == two, one == three, one == four}

{0.02548, Null}

{0.004597, Null}

{0.005269, Null}

{0.007438, Null}

{False, True, True}   <------ Daniel's code agreed.

n = 10^3;
data = RandomReal[{-100, 100}, {n, 4}];
Timing[one = valero@data;]
Timing[two = sseziwa@data;]
Timing[three = treat@data;]
Timing[four = daniel@data;]
{one == two, one == three, one == four}

{2.26575, Null}

{0.357105, Null}

{0.404373, Null}

{0.150693, Null}

{False, True, False}  <------ Daniel's code disagreed, again.

n = 10^4;
data = RandomReal[{-100, 100}, {n, 4}];
Timing[one = valero@data;]
Timing[two = sseziwa@data;]
Timing[three = treat@data;]
Timing[four = daniel@data;]
{one == two, one == three, one == four}

{254.635, Null}

{41.0605, Null}

{43.1114, Null}

{3.94189, Null}

{False, True, False}    <------ Again.

Daniel's code is faster, and it scales better. But there may be a  
discrepancy in the ordering. It usually agrees with "valero" and "treat"  
for small data sets, but not for larger ones.

Bobby

On Wed, 20 Jul 2011 05:34:56 -0500, Sseziwa Mukasa <mukasa at gmail.com>  
wrote:

> Your function does not appear to work correctly, sometimes the list is
> in the wrong order, but orderedList2:
>
> orderedList2[list_] :=
>  Module[{first = First[list], rest = Rest[list], nearest, result},
>
>   result = {first};
>   Do[nearest = Nearest[rest -> Automatic, Last[result]];
>    result = Join[result, rest[[nearest]]];
>    rest = Drop[rest, nearest], {Length[rest]}]; result]
>
>  seems faster anyway.:
>
> In[11]:= Module[{list = RandomReal[{0, 1}, {20, 4}], result1,
> result2},
>  result1 = Timing[orderedList[list]];
>  result2 = Timing[orderedList2[list]]; {result1[[2]] ==
> result2[[2]],
>   result1[[1]], result2[[1]], {list, result1[[2]], result2[[2]]}}]
> Out[11]= {False, 0.002556, 0.0007, {{{0.0780386, 0.220901, 0.748457,
>     0.548342}, {0.722099, 0.308747, 0.33661, 0.304516}, {0.623358,  
> 0.761312,
>     0.884318, 0.21337}, {0.14872, 0.284139, 0.244084, 0.582692},  
> {0.0829757,
>     0.911987, 0.448848, 0.606659}, {0.312903, 0.703568, 0.747021,
>     0.126831}, {0.847102, 0.437041, 0.0924343, 0.0333101}, {0.189339,
>     0.419192, 0.498754, 0.846607}, {0.661185, 0.455712, 0.31286,
>     0.387921}, {0.566059, 0.331815, 0.412978, 0.84236}, {0.714704,  
> 0.335434,
>     0.417758, 0.0569088}, {0.855757, 0.0118403, 0.310144,
>     0.312894}, {0.278001, 0.874878, 0.289541, 0.199531}, {0.682503,  
> 0.387263,
>     0.931379, 0.662633}, {0.8045, 0.264645, 0.595204, 0.534895},  
> {0.645246,
>     0.915467, 0.167849, 0.210847}, {0.789694, 0.987745, 0.0405679,
>     0.0957831}, {0.811736, 0.77314, 0.73452, 0.0771508}, {0.784869,  
> 0.582503,
>     0.0986897, 0.152786}, {0.426251, 0.135863, 0.465908,
>     0.228462}}, {{0.0780386, 0.220901, 0.748457, 0.548342}, {0.14872,
>     0.284139, 0.244084, 0.582692}, {0.189339, 0.419192, 0.498754,
>     0.846607}, {0.312903, 0.703568, 0.747021, 0.126831}, {0.278001,  
> 0.874878,
>     0.289541, 0.199531}, {0.0829757, 0.911987, 0.448848, 0.606659},  
> {0.623358,
>      0.761312, 0.884318, 0.21337}, {0.645246, 0.915467, 0.167849,
>     0.210847}, {0.789694, 0.987745, 0.0405679, 0.0957831}, {0.811736,  
> 0.77314,
>      0.73452, 0.0771508}, {0.784869, 0.582503, 0.0986897,
>     0.152786}, {0.847102, 0.437041, 0.0924343, 0.0333101}, {0.714704,
>     0.335434, 0.417758, 0.0569088}, {0.722099, 0.308747, 0.33661,
>     0.304516}, {0.682503, 0.387263, 0.931379, 0.662633}, {0.661185,  
> 0.455712,
>     0.31286, 0.387921}, {0.566059, 0.331815, 0.412978, 0.84236},  
> {0.426251,
>     0.135863, 0.465908, 0.228462}, {0.8045, 0.264645, 0.595204,
>     0.534895}, {0.855757, 0.0118403, 0.310144, 0.312894}}, {{0.0780386,
>     0.220901, 0.748457, 0.548342}, {0.189339, 0.419192, 0.498754,
>     0.846607}, {0.14872, 0.284139, 0.244084, 0.582692}, {0.566059,  
> 0.331815,
>     0.412978, 0.84236}, {0.8045, 0.264645, 0.595204, 0.534895},  
> {0.722099,
>     0.308747, 0.33661, 0.304516}, {0.661185, 0.455712, 0.31286,
>     0.387921}, {0.784869, 0.582503, 0.0986897, 0.152786}, {0.847102,  
> 0.437041,
>      0.0924343, 0.0333101}, {0.714704, 0.335434, 0.417758,
>     0.0569088}, {0.426251, 0.135863, 0.465908, 0.228462}, {0.855757,
>     0.0118403, 0.310144, 0.312894}, {0.682503, 0.387263, 0.931379,
>     0.662633}, {0.623358, 0.761312, 0.884318, 0.21337}, {0.811736,  
> 0.77314,
>     0.73452, 0.0771508}, {0.312903, 0.703568, 0.747021, 0.126831},  
> {0.278001,
>     0.874878, 0.289541, 0.199531}, {0.645246, 0.915467, 0.167849,
>     0.210847}, {0.789694, 0.987745, 0.0405679, 0.0957831}, {0.0829757,
>     0.911987, 0.448848, 0.606659}}}}
>
> From the above output it seems to me orderedList is incorrect:
>
> In[12]:= Nearest[{{0.7220990188851284`, 0.30874655560999353`,  
> 0.33661039115480085`,
>    0.30451607189733565`}, {0.623358140247426`, 0.7613121754790066`,
>    0.8843184964161348`, 0.21336965042896106`}, {0.14871956929750207`,
>    0.2841388256243189`, 0.2440839157054575`,
>    0.5826918077236027`}, {0.08297574175354927`, 0.9119868514832306`,
>    0.44884815651517496`, 0.606658888854545`}, {0.3129032093692543`,
>    0.7035682306379043`, 0.7470209852828422`,
>    0.126831036636313`}, {0.8471019825138335`, 0.4370412344560497`,
>    0.09243429503478429`, 0.03331005052172231`}, {0.18933936353633118`,
>    0.4191916563275504`, 0.49875370996337387`,
>    0.8466068811542358`}, {0.6611848899498884`, 0.455712283905654`,
>    0.31286031890780386`, 0.3879208331484807`}, {0.566058771065481`,
>    0.3318146887998126`, 0.41297814980274516`,
>    0.8423595336937795`}, {0.7147038318708883`, 0.33543423510640613`,
>    0.4177577945972242`, 0.056908774958363884`}, {0.8557567595586966`,
>    0.011840287587700837`, 0.31014365477384986`,
>    0.31289360223925855`}, {0.27800072513915963`, 0.8748776068932764`,
>    0.28954067774830894`, 0.19953109481447573`}, {0.6825027823624006`,
>    0.38726329607605536`, 0.9313788703031116`,
>    0.6626327289353637`}, {0.8045002220254145`, 0.26464489073341646`,
>    0.5952036781362782`, 0.5348945499518614`}, {0.6452463375631783`,
>    0.9154672494486176`, 0.16784916226023205`,
>    0.21084656084695386`}, {0.7896939168446158`, 0.9877454808787693`,
>    0.04056790678073363`, 0.09578312387470667`}, {0.8117363160995603`,
>    0.7731396349254072`, 0.7345204475002354`,
>    0.0771508252866615`}, {0.784869093354617`, 0.5825033331936824`,
>    0.09868967226141323`, 0.1527856209885663`}, {0.426250694413443`,
>    0.1358634032399666`, 0.46590788139431916`,
>    0.22846239863279472`}}, {0.07803858379747108`, 0.22090093831902768`,
>   0.7484574646075761`, 0.5483416965515358`}]
> Out[12]= {{0.189339, 0.419192, 0.498754, 0.846607}}
>
> Regards,
> 	Ssezi
>
> On Jul 19, 2011, at 6:55 AM, Luis Valero wrote:
>
>> Dear Sirs,
>>
>> I want to sort a list of 4-tuples of real numbers, so that, the second  
>> 4-tuple has minimum distance to the first, the third, selected from the  
>> rest, ha minimum distance to the second, and so on.
>>
>> The distance is the euclidean distance, calculated with the first two  
>> elements of each 4-tuple.
>>
>> I have defined a function that work:
>>
>> orderedList[list_] := Module[{nearest},
>> 	Flatten[ Nest[{
>>   	    Append[ #[[1]] , nearest = Flatten @@ Nearest[ Map[ Rule[ #[[{1,  
>> 2}]], #] &, #[[2]] ], Last[ #[[1]] ] [[{1, 2}]], 1] ],
>>    	    Delete[ #[[2]], Position[ #[[2]], nearest ] ]  } &, {  
>> {list[[1]] }, Delete[ list, 1 ] }, Length[ list ] - 2], 1] ];
>>
>>
>> but I need to improve the temporal efficiency
>>
>> Thank you
>>
>>
>>
>
>


-- 
DrMajorBob at yahoo.com



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From: Sseziwa Mukasa <mukasa at gmail.com>
Subject: [mg120396] Re: sorting a nested list of 4-tuples
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On Jul 20, 2011, at 6:33 AM, Luis Valero wrote:

> Thank you very much for your time
>
> The function Nearest does not use only the two first elements of each 4-tuple. This is the reason for the two different results

Sorry, I missed that part in your first post, but it's easy to modify my example to handle that case:

 orderedList2[list_] :=
 Module[{first = First[list], rest = Rest[list], nearest, result},
  result = {first};
  Do[nearest = Nearest[rest[[All, ;; 2]] -> Automatic, Last[result][[;; 2]]];
   result = Join[result, rest[[nearest]]];
   rest = Drop[rest, nearest], {Length[rest]}]; result]

In[5]:= Module[{list = RandomReal[{0, 1}, {1000, 4}], result1, result2},
 result1 = Timing[orderedList[list]];
 result2 = Timing[orderedList2[list]]; {result1[[2]] == result2[[2]],
  result1[[1]], result2[[1]]}]
Out[5]= {True, 4.07419, 0.644008}

So it still seems an order of magnitude faster.  Daniel Lichtblau's solution is faster still, but doesn't agree on the ordering:

orderedList3[data_] :=
 Module[{n = Length[data], chunksize, denom, moddata, nf, next, taken, result,
    remove, nextset, posns, k, done}, chunksize = Ceiling[Log[n]]^2;
  denom = 10000.;
  moddata = Map[Take[#, 3] &, MapThread[Prepend, {data, Range[n]/denom}]];
  nf = Nearest[moddata]; next = moddata[[1]]; taken = ConstantArray[False, n];
   result = ConstantArray[{0., 0., 0., 0.}, n];
  remove = ConstantArray[{0., 0., 0.}, chunksize]; result[[1]] = data[[1]];
  remove[[1]] = moddata[[1]]; taken[[1]] = True;
  Do[nextset = nf[next, 1 + Mod[j, chunksize, 1]];
   posns = Round[denom*Map[First, nextset]];
   k = 1;
   While[k < Length[nextset] && taken[[posns[[k]]]], k++;];
   taken[[posns[[k]]]] = True;
   result[[j]] = data[[posns[[k]]]];
   next = nextset[[k]];
   remove[[Mod[j, chunksize, 1]]] = next;
   If[Mod[j, chunksize] == 0, done = True;
    posns = Apply[Alternatives, remove];
    moddata = DeleteCases[moddata, posns];
    nf = Nearest[moddata];];, {j, 2, n}]; result]

In[9]:= Module[{list = RandomReal[{0, 1}, {1000, 4}], result1, result2, result3},
 result1 = Timing[orderedList[list]];
 result2 = Timing[orderedList2[list]];
 result3 = Timing[orderedList3[list]]; {result1[[2]] == result2[[2]],
  result1[[2]] === result3[[2]], result1[[1]], result2[[1]], result3[[1]]}]
Out[9]= {True, False, 4.24234, 0.664731, 0.20674}

>
> Regards
>
> Luis
>
>
> El 19/07/2011, a las 18:05, Sseziwa Mukasa escribi=F3:
>
>> Your function does not appear to work correctly, sometimes the list is in the wrong order, but orderedList2:
>>
>> orderedList2[list_] :=
>> Module[{first = First[list], rest = Rest[list], nearest, result},
>> result = {first};
>> Do[nearest = Nearest[rest -> Automatic, Last[result]];
>>  result = Join[result, rest[[nearest]]];
>>  rest = Drop[rest, nearest], {Length[rest]}]; result]
>>
>> seems faster anyway.:
>>
>> In[11]:= Module[{list = RandomReal[{0, 1}, {20, 4}], result1, result2},
>> result1 = Timing[orderedList[list]];
>> result2 = Timing[orderedList2[list]]; {result1[[2]] == result2[[2]],
>> result1[[1]], result2[[1]], {list, result1[[2]], result2[[2]]}}]
>> Out[11]= {False, 0.002556, 0.0007, {{{0.0780386, 0.220901, 0.748457,
>>   0.548342}, {0.722099, 0.308747, 0.33661, 0.304516}, {0.623358, 0.761312,
>>   0.884318, 0.21337}, {0.14872, 0.284139, 0.244084, 0.582692}, {0.0829757,
>>   0.911987, 0.448848, 0.606659}, {0.312903, 0.703568, 0.747021,
>>   0.126831}, {0.847102, 0.437041, 0.0924343, 0.0333101}, {0.189339,
>>   0.419192, 0.498754, 0.846607}, {0.661185, 0.455712, 0.31286,
>>   0.387921}, {0.566059, 0.331815, 0.412978, 0.84236}, {0.714704, 0.335434,
>>   0.417758, 0.0569088}, {0.855757, 0.0118403, 0.310144,
>>   0.312894}, {0.278001, 0.874878, 0.289541, 0.199531}, {0.682503, 0.387263,
>>   0.931379, 0.662633}, {0.8045, 0.264645, 0.595204, 0.534895}, {0.645246,
>>   0.915467, 0.167849, 0.210847}, {0.789694, 0.987745, 0.0405679,
>>   0.0957831}, {0.811736, 0.77314, 0.73452, 0.0771508}, {0.784869, 0.582503,
>>   0.0986897, 0.152786}, {0.426251, 0.135863, 0.465908,
>>   0.228462}}, {{0.0780386, 0.220901, 0.748457, 0.548342}, {0.14872,
>>   0.284139, 0.244084, 0.582692}, {0.189339, 0.419192, 0.498754,
>>   0.846607}, {0.312903, 0.703568, 0.747021, 0.126831}, {0.278001, 0.874878,
>>   0.289541, 0.199531}, {0.0829757, 0.911987, 0.448848, 0.606659}, {0.623358,
>>    0.761312, 0.884318, 0.21337}, {0.645246, 0.915467, 0.167849,
>>   0.210847}, {0.789694, 0.987745, 0.0405679, 0.0957831}, {0.811736, 0.77314,
>>    0.73452, 0.0771508}, {0.784869, 0.582503, 0.0986897,
>>   0.152786}, {0.847102, 0.437041, 0.0924343, 0.0333101}, {0.714704,
>>   0.335434, 0.417758, 0.0569088}, {0.722099, 0.308747, 0.33661,
>>   0.304516}, {0.682503, 0.387263, 0.931379, 0.662633}, {0.661185, 0.455712,
>>   0.31286, 0.387921}, {0.566059, 0.331815, 0.412978, 0.84236}, {0.426251,
>>   0.135863, 0.465908, 0.228462}, {0.8045, 0.264645, 0.595204,
>>   0.534895}, {0.855757, 0.0118403, 0.310144, 0.312894}}, {{0.0780386,
>>   0.220901, 0.748457, 0.548342}, {0.189339, 0.419192, 0.498754,
>>   0.846607}, {0.14872, 0.284139, 0.244084, 0.582692}, {0.566059, 0.331815,
>>   0.412978, 0.84236}, {0.8045, 0.264645, 0.595204, 0.534895}, {0.722099,
>>   0.308747, 0.33661, 0.304516}, {0.661185, 0.455712, 0.31286,
>>   0.387921}, {0.784869, 0.582503, 0.0986897, 0.152786}, {0.847102, 0.437041,
>>    0.0924343, 0.0333101}, {0.714704, 0.335434, 0.417758,
>>   0.0569088}, {0.426251, 0.135863, 0.465908, 0.228462}, {0.855757,
>>   0.0118403, 0.310144, 0.312894}, {0.682503, 0.387263, 0.931379,
>>   0.662633}, {0.623358, 0.761312, 0.884318, 0.21337}, {0.811736, 0.77314,
>>   0.73452, 0.0771508}, {0.312903, 0.703568, 0.747021, 0.126831}, {0.278001,
>>   0.874878, 0.289541, 0.199531}, {0.645246, 0.915467, 0.167849,
>>   0.210847}, {0.789694, 0.987745, 0.0405679, 0.0957831}, {0.0829757,
>>   0.911987, 0.448848, 0.606659}}}}
>>
>> From the above output it seems to me orderedList is incorrect:
>>
>> In[12]:= Nearest[{{0.7220990188851284`, 0.30874655560999353`, 0.33661039115480085`,
>>  0.30451607189733565`}, {0.623358140247426`, 0.7613121754790066`,
>>  0.8843184964161348`, 0.21336965042896106`}, {0.14871956929750207`,
>>  0.2841388256243189`, 0.2440839157054575`,
>>  0.5826918077236027`}, {0.08297574175354927`, 0.9119868514832306`,
>>  0.44884815651517496`, 0.606658888854545`}, {0.3129032093692543`,
>>  0.7035682306379043`, 0.7470209852828422`,
>>  0.126831036636313`}, {0.8471019825138335`, 0.4370412344560497`,
>>  0.09243429503478429`, 0.03331005052172231`}, {0.18933936353633118`,
>>  0.4191916563275504`, 0.49875370996337387`,
>>  0.8466068811542358`}, {0.6611848899498884`, 0.455712283905654`,
>>  0.31286031890780386`, 0.3879208331484807`}, {0.566058771065481`,
>>  0.3318146887998126`, 0.41297814980274516`,
>>  0.8423595336937795`}, {0.7147038318708883`, 0.33543423510640613`,
>>  0.4177577945972242`, 0.056908774958363884`}, {0.8557567595586966`,
>>  0.011840287587700837`, 0.31014365477384986`,
>>  0.31289360223925855`}, {0.27800072513915963`, 0.8748776068932764`,
>>  0.28954067774830894`, 0.19953109481447573`}, {0.6825027823624006`,
>>  0.38726329607605536`, 0.9313788703031116`,
>>  0.6626327289353637`}, {0.8045002220254145`, 0.26464489073341646`,
>>  0.5952036781362782`, 0.5348945499518614`}, {0.6452463375631783`,
>>  0.9154672494486176`, 0.16784916226023205`,
>>  0.21084656084695386`}, {0.7896939168446158`, 0.9877454808787693`,
>>  0.04056790678073363`, 0.09578312387470667`}, {0.8117363160995603`,
>>  0.7731396349254072`, 0.7345204475002354`,
>>  0.0771508252866615`}, {0.784869093354617`, 0.5825033331936824`,
>>  0.09868967226141323`, 0.1527856209885663`}, {0.426250694413443`,
>>  0.1358634032399666`, 0.46590788139431916`,
>>  0.22846239863279472`}}, {0.07803858379747108`, 0.22090093831902768`,
>> 0.7484574646075761`, 0.5483416965515358`}]
>> Out[12]= {{0.189339, 0.419192, 0.498754, 0.846607}}
>>
>> Regards,
>> 	Ssezi
>>
>> On Jul 19, 2011, at 6:55 AM, Luis Valero wrote:
>>
>>> Dear Sirs,
>>>
>>> I want to sort a list of 4-tuples of real numbers, so that, the second 4-tuple has minimum distance to the first, the third, selected from the rest, ha minimum distance to the second, and so on.
>>>
>>> The distance is the euclidean distance, calculated with the first two elements of each 4-tuple.
>>>
>>> I have defined a function that work:
>>>
>>> orderedList[list_] := Module[{nearest},
>>> 	Flatten[ Nest[{
>>> 	    Append[ #[[1]] , nearest = Flatten @@ Nearest[ Map[ Rule[ #[[{1, 2}]], #] &, #[[2]] ], Last[ #[[1]] ] [[{1, 2}]], 1] ],
>>>  	    Delete[ #[[2]], Position[ #[[2]], nearest ] ]  } &, { {list[[1]] }, Delete[ list, 1 ] }, Length[ list ] - 2], 1] ];
>>>
>>>
>>> but I need to improve the temporal efficiency
>>>
>>> Thank you
>>>
>>>
>>>
>>
>




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On Jul 20, 11:34 am, gonzorascal <dj... at duke.edu> wrote:
> Hello All,
>
> I am working with an Interpolating Function of an oscillating solution (time series) to a differential equation.  I am trying to find the period and pulse width of the oscillation.  To do this I would like to have an inverse of my function y[t] (so I would have t[y]).  I realize this function would be multivalued (that's what I want to find the period).  I am not having success using Mathematica's InverseFunction[] or Reduce[] commands.  Does anyone have any experience or suggestions with this sort of thing (either finding Inverse Interpolating Functions or another method for period and pulse width)?  Thank you.
>
> -GR

So, when you will find out the period, compute sufficient number of
points with the help of this interpolation function, e.g., as a
Table[], change coordinates, and interpolate again... Of course, if
you know exact formula, worth using it...



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From: John Fultz <jfultz at wolfram.com>
Subject: [mg120401] Re: Computable Document Format Replaces Player Notebooks
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On Thu, 21 Jul 2011 05:47:32 -0400 (EDT), Gregory Lypny wrote:
> Hi Everyone,
>
> Probably a silly question but I just want to be sure: Has the computable
> document format that is built into Mathematica replaced player notebooks,
> so that I no longer have to upload my notebooks to Wolfram for=
 conversion?
>
> Regards,
>
> Gregory

Correct.

Sincerely,

John Fultz
jfultz at wolfram.com
User Interface Group
Wolfram Research, Inc.



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On Thu, 21 Jul 2011, Joseph O'Rourke wrote:

> Is it possible to control the radii of a torus in Mathematica 8 using the ExampleData[] function?  I cannot see how to get anything but the default torus.  Thanks!
>
>

Hello Joseph,

would this be of use

Manipulate[
  ParametricPlot3D[{ Cos[t] (r2 + r1 Cos[u]), Sin[t] (r2 + r1 Cos[u]),
    r1 Sin[u]}, {t, 0, 2 Pi}, {u, 0, 2 Pi}, Mesh -> None], {{r1, 1}, 0,
    2},
  {{r2, 3}, 0, 5}]

Oliver



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From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
Subject: [mg120403] Re: unable to save notebook
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On Thu, 21 Jul 2011, dimitris wrote:

> Hello to all.
> When my student tries to save a document (e.g. hh its name) the
> following message appears...
>
> "Mathematica could not find enough disk space to save the file C:\Users
> \=D7\[Rho]=DE\[Sigma]\[Tau]\[Omicron]\[FinalSigma]\Downloads\hh.nb. The
> file was left unchanged. The disk is full, or if you are using a file
> server you may have reached your disk space quota."
>
> Any ideas on how to solve the problem?
>
> Thank you very much in advance for your replies.
>
>

Dimitris,

a wild guess: does saving to a path without any special characters work?

Oliver



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From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
Subject: [mg120404] Re: Inverse of Interpolating Function?
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On Thu, 21 Jul 2011, Gary Wardall wrote:

> On Jul 20, 5:34=C2 am, gonzorascal <dj... at duke.edu> wrote:
>> Hello All,
>>
>> I am working with an Interpolating Function of an oscillating solution (time series) to a differential equation.  I am trying to find the period and pulse width of the oscillation.  To do this I would like to have an inverse of my function y[t] (so I would have t[y]).  I realize this function would be multivalued (that's what I want to find the period). =C2 I am not having success using Mathematica's InverseFunction[] or Reduce[] commands.  Does anyone have any experience or suggestions with this sort of thing (either finding Inverse Interpolating Functions or another method for period and pulse width)?  Thank you.
>>
>> -GR
>
> gonzorascal,
>
> First make sure the function you wish to have an inverse function is a
> function that has an inverse function. A quick visual test is the
> Horizontal Test for Inverse Functions. If any horizontal line
> intersects the graph twice or more then the inverse is NOT an inverse
> function. Graph the function you wish to create an inverse function
> for. If it fails the horizontal test you may have ti restrict  your
> function like is done in Trigonometry. If it passes try something like
> in my example.
>
>
> points = {{0, -1}, {1, 1}, {2, 3}, {3, 4}, {4, 6}, {5, 10}};=E2=80=A8ifun=
> Interpolation[points]
> Does the function pass the Horizontal Test?
>
> Plot[ifun[x], {x, 0, 5}]
>
> It does. Define:
>
> invifun[x_] := FindRoot[ifun[y] == x, {y, 0, 5}][[1]][[2]]
>
> Note that:
>
> ifun[invifun[4]]
>
> is 4.
>
> and
>
> invifun[ifun[3]]
>
> is 3.
>
> and look at the sketch.
>
> Plot[{invifun[x], ifun[x], x}, {x, 0, 5}, PlotRange -> {0, 5},
> AspectRatio -> Automatic]
>
> We have an inverse function.
>
> I hope this will help.
>
> Good Luck
>
> Gary Wardall
>
>

You could also use NDSolve for that - that advantage would be that it will 
generate an interpolation function

init = x /. FindRoot[ifun[x] == 0, {x, 0, 1}]

inf = t /.
   First[NDSolve[{t'[a] == 1/(ifun'[t[a]]), t[0] == init},
     t, {a, 0, 5}]]

Plot[{inf[x], ifun[x], x}, {x, 0, 5}, PlotRange -> {0, 5},
  AspectRatio -> Automatic]

Oliver



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Subject: [mg120406] Re: ExampleData[{"Geometry3D", "Torus"}]
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Thanks, Oliver.  I know how to make a torus via ParametricPlot3D[], and by using the deprecated Torus[] function. I was wondering if ExampleData[] could be adjusted so that one would have the same control as with the old Torus[] function.  The reason I wanted this is that the torus produced by ExampleData[] looks much nicer than that produced by Torus[].  Or, at least, I cannot get the latter to look as smooth when rendered as the former.  But maybe I should just use ParametricPlot3D[] with Mehs->None as you suggest.  Thanks again for your attention.



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Subject: [mg120420] Re: Warning: OS X Lion and compilation
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I just installed Xcode (which is now free in the App Store) and was able to compile using C as the compilation target.

David



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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120411] How Can I Eliminate Line Wrapping When Using the Row Function
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Hello everyone,

Sometimes when I use Row for four or five items, the result wraps over two lines.  How can I force the result to be displayed on one line?  I couldn't find an option for this in Documentation Centre.

Gregory




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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120413] How Can I Include a Button in a Manipulate
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Hello everyone,

I have a Manipulate that generates a random integer or integers 
depending on the user's choice of parameters in a number of setter 
bars.  I'd like to include a button in the Manipulate's output pane 
that allows the user to generate a new result by generating a new random 
number while keeping all of the parameters in the setter bar the same.  
Can that be done?


Regards,

Gregory



From mathgroup-adm at smc.vnet.net  Thu Jul 21 19:18:34 2011
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From: dr DanW <dmaxwarren at gmail.com>
Subject: [mg120416] Re: Warning: OS X Lion and compilation
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Ok, I won't call it a total false alarm.  I had Xcode 4.0 installed prior to installing Lion, and the scenario I described in my previous posting occured.  At that time, I also tried typing "gcc" in the Terminal and it was not found.

I have now installed Xcode 4.2 (free from the App Store) and I am now able to compile functions in Mathematica.  All is well.

Daniel



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From: Gilmar Rodriguez-pierluissi <peacenova at yahoo.com>
Subject: [mg120415] Producing an image that only contains its interior and boundary but, no exterior.
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This is an old issue mentioned by the late Jens-Peer Kuska in:
http://forums.wolfram.com/mathgroup/archive/2006/Mar/msg00037.html
How can a user produce a graphic image that includes only its interior and boundary but,
without the exterior. This would facilitate embedding an image within another without
having to waste time having to eliminate the annoying edges. Thank you!
Gilmar Rodriguez-Pierluissi


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From: "DaleJenk" <dale.jenkins8 at deletegooglemail.com>
Subject: [mg120423] mandelbrot in version 7 & Ruskeepaa version 3
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I bought Mathematica 7 Home Edition last year. I'm trying to plot the 
Mandelbrot Set following the instructions in Ruskeepaa's Mathematica 
Navigator version 3. I'm getting an error message.

Here's what I am introducing:

mandelbrot = Compile[{{c,_Complex}},
Module[{z=0+0I, n=0}, While[Abs[z]<2 && n<50, z=z^2+c; n++]; n]];

DensityPlot[mandelbrot[x+Iy], {x,-2,1},{y,-1.5,1.5}, PlotPoints->200, 
Mesh->False, FrameTicks->{{-2,-1,0,1},{-1,0,1}}]; //Timing

I get the message:

CompiledFunction::cfsa:
Argument -1.99998 + Iy at position 1 should be a machine-size complex 
number.

Where am I going wrong?

Thanks.
 




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From: David Reiss <dbreiss at gmail.com>
Subject: [mg120410] Re: Imported Data in Mathematica Player or CDF Documents
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One approach is to place the data into a variable and then use the
SaveDefinitions->True option for Manipulate.  Then when that
Manipulate is placed in a CDF it will have the data embedded.

Best,
David

On Jul 20, 6:42 am, Gregory Lypny <gregory.ly... at videotron.ca> wrote:
> Hello everyone,
>
> I'm in the process of creating a set of Mathematica Player documents
> for my students, and for many of them, I need to import data from text
> files.  When I convert the notebooks to CDF, is the imported data
> cached?  Can I simply leave the Import function in the player document?
> Or do I have to copy and paste the data manually into a variable?
>
> Any advice would be much appreciated,
>
> Gregory




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From: George Woodrow III <georgevw3 at mac.com>
Subject: [mg120417] Re: Warning: OS X Lion and compilation
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The good news is that Xcode is now free.

george

On Jul 21, 2011, at 5:46 AM, dr DanW wrote:

> I have not dug thoroughly into this yet, but I wanted to get the warning out quickly.  Apparently, OS X Lion does not include the gcc compiler.  I upgraded to Lion this morning, then on a lark tried to compile a function in Mathematica with CompilationTarget -> "C".  I got an error that no compiler was found.
> 
> I am sure that it is only a matter of either obtaining a copy of gcc or Xcode (with its own c compiler, not gcc) and configuring Mathematica to use that compiler.  I have not tried either, since I am not using anything but opcode compilation.  If you are successful, please post instructions for others on this group.
> 
> Thanks,
> Daniel
> 




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From: paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Subject: [mg120414] TransformedDistribution, for a sum of M iid variables
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Using the TransformedDistribution function it is easy to show that the
sum of two normal variables is normal, e.g.,

Input:
TransformedDistribution[Z1+Z2,{Z1\
[Distributed]NormalDistribution[0,1],Z2\
[Distributed]NormalDistribution[0,1]}]

Output:
NormalDistribution[0, Sqrt[2]]


Likewise it wouldn't be hard to show that the sum of 3 normal
variables is normal, or 4.

But how do I show that the sum of M normal variables is normal, where
M is an arbitrary positive integer.

I'm thinking I should use the Sum function inside
TransformedDistribution, but I don't know how the notation would work.
I see that there's a UniformSumDistribution function, but that's
limited to uniform variables.

Later I will want to do similar calculations for nonnormal
distributions.

Thanks for any pointers.



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From: carlos at colorado.edu
Subject: [mg120424] Re: Total plot width
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On Jul 21, 3:52 am, Murray Eisenberg <mur... at math.umass.edu> wrote:
> The option you want is ImageSize.
>
> On 7/20/11 6:32 AM, car... at colorado.edu wrote:
>
> > How can I specify the total plot width in Mathematica?  I was looking for
> > something like AbsolutePlotWidth->250  (points implied) but havent found
> > that option in the documentation.  There is AspectRatio, but that is
> > not a dimension. Thanks.
>
> --
> Murray Eisenberg                     mur... at math.umass.edu
> Mathematics & Statistics Dept.
> Lederle Graduate Research Tower      phone 413 549-1020 (H)
> University of Massachusetts                413 545-2859 (W)
> 710 North Pleasant Street            fax   413 545-1801
> Amherst, MA 01003-9305

Thanks to all who replied.  ImageSize works fine, I had missed that
option. Sometimes one need to sync widths between multiple plots
to simplify downstream editing by Illustrator and placing on a single
LaTeX page with [p].



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From: David Reiss <dbreiss at gmail.com>
Subject: [mg120409] Re: Warning: OS X Lion and compilation
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I believe that XCode 4.1 for Lion is now free and I believe that the
compilers are installed with an installation of XCode...

http://developer.apple.com/xcode/

I haven't installed Lion yet since I use Eudora and can't find a
better email client ...  and Eudora is PowerPC...   :-(

--David

On Jul 21, 5:49 am, dr DanW <dmaxwar... at gmail.com> wrote:
> I have not dug thoroughly into this yet, but I wanted to get the warning out quickly.  Apparently, OS X Lion does not include the gcc compiler. I upgraded to Lion this morning, then on a lark tried to compile a function in Mathematica with CompilationTarget -> "C".  I got an error that no compiler was found.
>
> I am sure that it is only a matter of either obtaining a copy of gcc or Xcode (with its own c compiler, not gcc) and configuring Mathematica to use that compiler.  I have not tried either, since I am not using anything but opcode compilation.  If you are successful, please post instructions for others on this group.
>
> Thanks,
> Daniel




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From: Daniel Lichtblau <danl at wolfram.com>
Subject: [mg120407] Re: Inverse of Interpolating Function?
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On 07/20/2011 05:31 AM, gonzorascal wrote:
> Hello All,
>
> I am working with an Interpolating Function of an oscillating solution (time series) to a differential equation.  I am trying to find the period and pulse width of the oscillation.  To do this I would like to have an inverse of my function y[t] (so I would have t[y]).  I realize this function would be multivalued (that's what I want to find the period).  I am not having success using Mathematica's InverseFunction[] or Reduce[] commands.  Does anyone have any experience or suggestions with this sort of thing (either finding Inverse Interpolating Functions or another method for period and pulse width)?  Thank you.
>
> -GR
>

Here is an approach that might be suitable, at least if you can sample 
from a fairly large number of periods.

(1) Sample your function in equal intervals.

(2) Take the Fourier transform.

(3) Find the largest frequency, not counting the first (DC) component.

(4) ue that to estimate the period.

Here is an example. We'll use a reference function that is a simple 
sinusoid.

y[t] /. First[
    DSolve[{y''[t] == -2*y[t], y[0] == 0, y'[0] == 1}, y[t], t]]

Out[1500]= Sin[Sqrt[2] t]/Sqrt[2]

The period is Sqrt[2]*Pi, or around 4.44288

We will now obtain this as an InterpolatingFunction from NDSolve, add a 
DC component, sample at intervals, and add some random noise.

soln = y[t] /.
   First[NDSolve[{y''[t] == -2*y[t], y[0] == 0, y'[0] == 1},
     y[t], {t, 0, 150}]]

separation = .1;
vals = Table[soln + 3, {t, 0., 150., separation}];
vals = vals + RandomReal[{-.1, .1}, Length[vals]];

ft = Abs[Fourier[vals]];

mainfreq = Position[Rest[ft], Max[Rest[ft]]][[1, 1]]
Out[1563]= 34

Here is the period estimate.

In[1564]:= N[Length[vals]/mainfreq]*separation
Out[1564]= 4.41471

So this agrees with the period of the unperturbed signal to 2+ decimal 
places.

A related approach may be found at

Documentation Center > Fourier > Applications > Frequency Identification

Daniel Lichtblau
Wolfram Research



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Suppose plots is your list of imported images, then you could do something like

pts = Table[{0, 0}, {Length[plots]}]

Manipulate[LocatorPane[Dynamic[pts[[k]]], plots[[k]]],
 {{k, 1, "Index"}, Range[Length[plots]], ControlType -> Slider,
  Appearance -> "Labeled"}]

The k-th entry in pts will correspond to the coordinates of  the locator on the k-th
image. You can even see a live update of the coordinates  while you manipulate the
locators by doing something like

Dynamic[pts // TableForm]

Heike

On 21 Jul 2011, at 10:47, Gareth Edwards wrote:

> Hi,
>
> Still playing around with Mathematica as a newbie and wrestling with the
> 'different'  way one has to think about things as opposed to other
> languages. I think (hopefully) that an informative example for me would be a
> solution (or at least some hints) to the following:
>
> Say I have an image sequence (mySequence_0000.jpg, etc.) I'd like to create
> a viewer with a slider that scrubs through the images. That I can do.
>
> What I'd also like is an array of points (just one point per image, and
> initialized to {0,0} say ), such that on each image I can move a locator
> around and thereby modify the corresponding stored point in the array. In
> other words, store the location of a point (as manipulated by me) against
> each image.
>
> My gut tells me that this is trivial to the average Mathematica user- the
> trouble is I remain well below average right now!
>
> Any help appreciated.
>
> Best,
>
> Gareth
>
> --
> Dr. Gareth Edwards
> Director
> www.cubicmotion.com
>
> Mobile: +44 7813 534111
> Office: +44 1925 606430
> Fax: +44 161 884 0095
>
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>
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[ I am helping to put on this international conference and am wanting to spread the word. The conference may be of interest to Mathematica users since it is part of the work of ComputerBasedMath.org which was created by Conrad Wolfram and is initially being supported by Wolfram Research. Separately from the conference, members of this community may want to get involved by helping to develop courseware. ]

How do we fix math education? The importance of math to jobs, society, and thinking has exploded over the last few decades. Meanwhile, math education has gotten stuck or has even slipped backward. Why has this chasm opened up? It's all about computers: when they do the calculating, people can work on harder questions, try more concepts, and play with a multitude of new ideas.

computerbasedmath.org is a project founded by Conrad Wolfram and initially supported by Wolfram Research to build a completely new math curriculum with computer-based math at its heart -- alongside a campaign to refocus math education away from historical hand-calculating techniques and toward relevant and conceptually interesting topics.

computerbasedmath.org is hosting a two-day summit to address the worldwide math education crisis by answering the question, In an era of ubiquitous computing, how should we rebuild math education from the ground up, to keep pace with and drive progress in the real world?

If you are a leader in industry, technology, government or education with astake in math we urge you to request an invitation to the summit and to join us in addressing math (and associated STEM) education from all perspectives to kick-start a rational, computer-based new direction.

Please see the Computer-Based Math Education Summit site ( http://computerbasedmath.org/events/londonsummit2011/index.html) for more information. Contact the summit organizers at summit at computerbasedmath.org with questions.

Whether or not you can attend the inaugural summit we welcome your participation in the Computer-Based Math project. Click the Sign Up link at the Computer-Based Math home page (http://computerbasedmath.org) to keep informed. If you know of other leaders who should be involved in the project we would greatly appreciate knowing about them through the Sign Up link or via the summit at computerbasedmath.org email.



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Disappointed...

So much things I heard about CDF applications, and then, well, I have
to classify them as documents... Very powerful documents, and almost
touching the application world but... just interactive documents for
an OPEN market...

Somehow nothing new than what already existed. Not that it is bad (I
personally have already commented a lot of good on the subject).

The problem is that I'm still unable to see how could I use this in an
cooperation environment.

Let's suppose two different scenarios:
1. An application to be distributed to 100 colleagues of the same
company, that, well, to make it simple, just needs to import a gif
image, apply some sort of simple filter, and export it to a gif file.
Very simple application, but not feasible. There isn't even concrete
information on the subject (besides "please contact us").  Have
contacted several times on the subject, and always no solution... Has
this changed? Why so much mystery on the PROfessional use of this
technology (this confirms what I keep hearing from others: academic,
institutes, governments, etc, but not corporate...)? 100 special
licenses for such a simple application?

2. Let's suppose I developed a package that does some awesome stuff,
and really toke me some time to develop it. I use it to write a report
for a client, and I want to give the report as an interactive document
to the client (nothing could be better than CDF!). But the interaction
needs my package (there's no way of previously generating all possible
outputs). So, I can't send it to my client, since I risk having it
copied to my competitors. Each client to whom I send a report needs a
professional version? I mean, each person in the company of my client,
and their consultant partners that will evaluate my report...? Or is
this now the other way around: each report or report revision needs to
be converted at Wolfram for pro functionalists activation on the
standard player (I sign confidential agreements with my client...)?
After all the announcements I have no idea what's the strategy. Did I
missed some information on WR site?

I just resent some questions to WR on the "please contact us", to see
if something changed in the past couple of months (when I last
insisted). Waiting for the answer, but why all the mystery on the PRO
version? What can't WR figure out about a professional strategy?

Can someone from Wolfram clarify this here for everyone? Is there a
strategy for these two very common and simple examples of my world? Is
this only meant for publishing documents on an OPEN environment, or on
a direct revenue strategy (selling each document/application copy for
a not so small price?).

It's interesting that all the examples presented in the site are of
very simple applications, but how can these simple application be used
on a closed corporate environment, if the licensing of the
applications cost more than their internal development cost?

I'm completely in the dark...

I know that I already asked this before, without great success (just
one person answered), but, am I the only one here?

Regards,
P. Fonseca



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From: Arturo Amador <amador at tf.phys.ntnu.no>
Subject: [mg120405] Phase Diagrams and problem probably with NIntegrate (or FindRoot)
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Hi,

I am trying to make some phase diagrams on Mathematica. I have the 
following code:

(*Free parameters*)n=4
start=150.
(*Experimental quantities (MeV)*)
mpi=139.
msi=600.
fpi=93.
(*Renormalisation scale*)
scale=mpi Sqrt[n]
(*Dimensionless auxiliary function*)
Jint[x_,y_]:=y^2/Sqrt[x^2+y^2]/(Exp[Sqrt[x^2+y^2]]-1)
J[x_]:=NIntegrate[Jint[x,y],{y,0,Infinity}]
(*Sum-integral*)
qT[m_,t_]:=t^2  J[m/t]/(2 Pi^2)
qT[m_,0.]:=0.
p0[m_]:=m^2 Log[m^2/scale^2]/(16 Pi^2)
p0[0.]:=0.
p[m_,t_]:=p0[m]+qT[m,t]
(*Physical point*)
lp=n (msi^2-mpi^2)/fpi^2
fp=fpi^2 (msi^2-3 mpi^2)/(msi^2-mpi^2)/n
hp=mpi^2 fpi/Sqrt[n]
(*Chiral limit*)

lc=n msi^2/fpi^2
fc=fpi^2/n
(*Pion condensate (squared)*)
rhop[mui_,t_]:=fp-hp^2/mui^4+2 mui^2/lp-p0[mui]-qT[mui,t]
rhoc[mui_,t_]:=fc+2 mui^2/lc-p[mui,t]
(*Normalised plot of chiral condensate at mu_I0 MeV*)
rnp=Sqrt[rhop[200.,0.]]
rnc=Sqrt[rhoc[200.,0.]]
rhoplot:=Plot[{Sqrt[rhop[200.,t]]/rnp,Sqrt[rhoc[200.,t]]/rnc},{t,0.,300.},PlotRange->All,AxesOrigin->{0.,0.}]
(*Critical temperature and phase diagram*)
tcp[mui_]:=FindRoot[rhop[mui,t],{t,start}][[1]][[2]]
tcc[mui_]:=FindRoot[rhoc[mui,t],{t,start}][[1]][[2]]
phaseplot:=Plot[{tcp[mui],tcc[mui]},{mui,0.01,300.},PlotRange->All,AxesOrigin->{0.,0.}]
Grid[rhoplot,phaseplot]


It is supposed to give me a pair of plots but I am only getting one and 
some error messages for the second one (phaseplot). The error that 
Mathematica gives in the terminal says as follows:

NIntegrate::inumr:
                                              2
                                             y
    The integrand --------------------------------------------------------
                                           2    2
                         Sqrt[0.000260124/t  + y ]       0.000260124    2
                  (-1 + E                         ) Sqrt[----------- + y ]
                                                              2
                                                             t
      has evaluated to non-numerical values for all sampling points in the
      region with boundaries {{Infinity, 0.}}.

NIntegrate::inumr:
                                              2
                                             y
    The integrand --------------------------------------------------------
                                           2    2
                         Sqrt[0.000260124/t  + y ]       0.000260124    2
                  (-1 + E                         ) Sqrt[----------- + y ]
                                                              2
                                                             t
      has evaluated to non-numerical values for all sampling points in the
      region with boundaries {{Infinity, 0.}}.

NIntegrate::inumr:
                                              2
                                             y
    The integrand --------------------------------------------------------
                                           2    2
                         Sqrt[0.000260124/t  + y ]       0.000260124    2
                  (-1 + E                         ) Sqrt[----------- + y ]
                                                              2
                                                             t
      has evaluated to non-numerical values for all sampling points in the
      region with boundaries {{Infinity, 0.}}.

General::stop: Further output of NIntegrate::inumr
      will be suppressed during this calculation.

FindRoot::lstol:
    The line search decreased the step size to within tolerance specified 
by
      AccuracyGoal and PrecisionGoal but was unable to find a sufficient
      decrease in the merit function. You may need more than 
MachinePrecision
      digits of working precision to meet these tolerances.




The error message that Mathematica gives in the front end is this:


NIntegrate::inumr: The integrand y^2/((-1+E^Sqrt[0.000260124 
Power[<<2>>]+y^2]) Sqrt[0.000260124/t^2+y^2]) has evaluated to 
non-numerical values for all sampling points in the region with boundaries 
{{\[Infinity],0.}}. >>
NIntegrate::inumr: The integrand y^2/((-1+E^Sqrt[0.000260124 
Power[<<2>>]+y^2]) Sqrt[0.000260124/t^2+y^2]) has evaluated to 
non-numerical values for all sampling points in the region with boundaries 
{{\[Infinity],0.}}. >>
NIntegrate::inumr: The integrand y^2/((-1+E^Sqrt[0.000260124 
Power[<<2>>]+y^2]) Sqrt[0.000260124/t^2+y^2]) has evaluated to 
non-numerical values for all sampling points in the region with boundaries 
{{\[Infinity],0.}}. >>
General::stop: Further output of NIntegrate::inumr will be suppressed 
during this calculation. >>
NIntegrate::vars: Integration range specification Compile`$19 is not of 
the form {x, xmin, ..., xmax}. >>
NIntegrate::vars: Integration range specification Compile`$19 is not of 
the form {x, xmin, ..., xmax}. >>
NIntegrate::vars: Integration range specification Compile`$19 is not of 
the form {x, xmin, ..., xmax}. >>
General::stop: Further output of NIntegrate::vars will be suppressed 
during this calculation. >>
FindRoot::njnum: The Jacobian is not a matrix of numbers at {t} = {150.}. 
>>
FindRoot::njnum: The Jacobian is not a matrix of numbers at {t} = {150.}. 
>>






When running this in the terminal it seems to keep working for a while, 
but in the front end it just interrupts all further evaluation.

Any comment would be of great help.


Thanks,

--
Arturo



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From: Gary Wardall <gwardall at gmail.com>
Subject: [mg120421] Re: Inverse of Interpolating Function?
References: <201107210945.FAA06489 at smc.vnet.net>
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Yes, that is so. I kind of thought that's what GR should do,

Gary

On Thu, Jul 21, 2011 at 5:17 AM, Oliver Ruebenkoenig
<ruebenko at wolfram.com>wrote:

> On Thu, 21 Jul 2011, Gary Wardall wrote:
>
>  On Jul 20, 5:34 am, gonzorascal <dj... at duke.edu> wrote:
>>
>>> Hello All,
>>>
>>> I am working with an Interpolating Function of an oscillating solution
>>> (time series) to a differential equation.  I am trying to find the period
>>> and pulse width of the oscillation.  To do this I would like to have an
>>> inverse of my function y[t] (so I would have t[y]).  I realize this function
>>> would be multivalued (that's what I want to find the period).  I am not
>>> having success using Mathematica's InverseFunction[] or Reduce[] commands.
>>>  Does anyone have any experience or suggestions with this sort of thing
>>> (either finding Inverse Interpolating Functions or another method for period
>>> and pulse width)?  Thank you.
>>>
>>> -GR
>>>
>>
>> gonzorascal,
>>
>> First make sure the function you wish to have an inverse function is a
>> function that has an inverse function. A quick visual test is the
>> Horizontal Test for Inverse Functions. If any horizontal line
>> intersects the graph twice or more then the inverse is NOT an inverse
>> function. Graph the function you wish to create an inverse function
>> for. If it fails the horizontal test you may have ti restrict  your
>> function like is done in Trigonometry. If it passes try something like
>> in my example.
>>
>>
>> points = {{0, -1}, {1, 1}, {2, 3}, {3, 4}, {4, 6}, {5, 10}}; ifun=
>> Interpolation[points]
>> Does the function pass the Horizontal Test?
>>
>> Plot[ifun[x], {x, 0, 5}]
>>
>> It does. Define:
>>
>> invifun[x_] := FindRoot[ifun[y] == x, {y, 0, 5}][[1]][[2]]
>>
>> Note that:
>>
>> ifun[invifun[4]]
>>
>> is 4.
>>
>> and
>>
>> invifun[ifun[3]]
>>
>> is 3.
>>
>> and look at the sketch.
>>
>> Plot[{invifun[x], ifun[x], x}, {x, 0, 5}, PlotRange -> {0, 5},
>> AspectRatio -> Automatic]
>>
>> We have an inverse function.
>>
>> I hope this will help.
>>
>> Good Luck
>>
>> Gary Wardall
>>
>>
>>
> You could also use NDSolve for that - that advantage would be that it will
> generate an interpolation function
>
> init = x /. FindRoot[ifun[x] == 0, {x, 0, 1}]
>
> inf = t /.
>  First[NDSolve[{t'[a] == 1/(ifun'[t[a]]), t[0] == init},
>    t, {a, 0, 5}]]
>
> Plot[{inf[x], ifun[x], x}, {x, 0, 5}, PlotRange -> {0, 5},
>  AspectRatio -> Automatic]
>
> Oliver
>


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From: Gareth Edwards <gareth.edwards at cubicmotion.com>
Subject: [mg120419] Re: newbie Map, Dynamic, etc
References: <201107210947.FAA06542 at smc.vnet.net> <8A9117AD-B871-4686-9CCB-FF12EC28921A at gmail.com>
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Thanks Heike, that's extremely helpful to understand the way to build these
things!

Best,

Gareth

On 21 July 2011 14:58, Heike Gramberg <heike.gramberg at gmail.com> wrote:

> Suppose plots is your list of imported images, then you could do something
> like
>
> pts = Table[{0, 0}, {Length[plots]}]
>
> Manipulate[LocatorPane[Dynamic[pts[[k]]], plots[[k]]],
>  {{k, 1, "Index"}, Range[Length[plots]], ControlType -> Slider,
>  Appearance -> "Labeled"}]
>
> The k-th entry in pts will correspond to the coordinates of  the locator on
> the k-th
> image. You can even see a live update of the coordinates  while you
> manipulate the
> locators by doing something like
>
> Dynamic[pts // TableForm]
>
> Heike
>
> On 21 Jul 2011, at 10:47, Gareth Edwards wrote:
>
> > Hi,
> >
> > Still playing around with Mathematica as a newbie and wrestling with the
> > 'different'  way one has to think about things as opposed to other
> > languages. I think (hopefully) that an informative example for me would
> be a
> > solution (or at least some hints) to the following:
> >
> > Say I have an image sequence (mySequence_0000.jpg, etc.) I'd like to
> create
> > a viewer with a slider that scrubs through the images. That I can do.
> >
> > What I'd also like is an array of points (just one point per image, and
> > initialized to {0,0} say ), such that on each image I can move a locator
> > around and thereby modify the corresponding stored point in the array. In
> > other words, store the location of a point (as manipulated by me) against
> > each image.
> >
> > My gut tells me that this is trivial to the average Mathematica user- the
> > trouble is I remain well below average right now!
> >
> > Any help appreciated.
> >
> > Best,
> >
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> >
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From: "Hobbs, Sylvia (DPH)" <sylvia.hobbs at state.ma.us>
Subject: [mg120426] Re: mlink SASImport
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See: http://library.wolfram.com/infocenter/MathSource/7808/

Sylva Hobbs
________________________________________

From: Gary Owen [jloughlin at loughlinconsulting.com]
Sent: Thursday, July 21, 2011 7:36 PM
To: Hobbs, Sylvia (DPH)
Subject: [mg120426] Re: mlink SASImport

On Jul 21, 2:50 am, "Hobbs, Sylvia (DPH)" <sylvia.ho... at state.ma.us>
wrote:
> After importing a huge SAS data set into Mathematica that contains 400 data elements, is there a global validation command where Mathematica could return a report of on the blanks and content frequencies in all of the data fields or do you have to run separate queries for every single data element?
>
> Sylvia Hobbs

How did you import the SAS data?  Did you first export the SAS data to
excel or csv and then import that data into Mathematica?  Or were you
able to import the SAS data directly? If so, what Mathematica code did
you use?

I use SAS a lot and it would be great if I could import SAS data
dircectly into Mathematica.



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From: "Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com>
Subject: [mg120425] Re: mandelbrot in version 7 & Ruskeepaa version 3
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There is a space missing between I and y in the expression mandelbrot[x +  
Iy] (should be mandelbrot[x + I y]). Also, you should omit the semicolon  
after DensityPlot, otherwise the plot will not be shown.

Incidentally, tabulating the values and then plotting them with ArrayPlot  
is faster for this example than using DensityPlot. That is,

ArrayPlot@Table[
   mandelbrot[x + I y],
   {y, -1.5, 1.5, 0.005}, {x, -2.0, 1.0, 0.005}
  ]

On Fri, 22 Jul 2011 02:15:27 +0100, DaleJenk  
<dale.jenkins8 at deletegooglemail.com> wrote:

> I bought Mathematica 7 Home Edition last year. I'm trying to plot the
> Mandelbrot Set following the instructions in Ruskeepaa's Mathematica
> Navigator version 3. I'm getting an error message.
>
> Here's what I am introducing:
>
> mandelbrot = Compile[{{c,_Complex}},
> Module[{z=0+0I, n=0}, While[Abs[z]<2 && n<50, z=z^2+c; n++]; n]];
>
> DensityPlot[mandelbrot[x+Iy], {x,-2,1},{y,-1.5,1.5}, PlotPoints->200,
> Mesh->False, FrameTicks->{{-2,-1,0,1},{-1,0,1}}]; //Timing
>
> I get the message:
>
> CompiledFunction::cfsa:
> Argument -1.99998 + Iy at position 1 should be a machine-size complex
> number.
>
> Where am I going wrong?
>
> Thanks.
>



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From: Bill Rowe <readnews at sbcglobal.net>
Subject: [mg120427] Re: TransformedDistribution, for a sum of M iid variables
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On 7/21/11 at 9:07 PM, paulvonhippel at yahoo.com (paulvonhippel at
yahoo) wrote:

>Using the TransformedDistribution function it is easy to show that
>the sum of two normal variables is normal, e.g.,

<code snipped>

>Likewise it wouldn't be hard to show that the sum of 3 normal
>variables is normal, or 4.

>But how do I show that the sum of M normal variables is normal,
>where M is an arbitrary positive integer.

Rather than approach this problem using TransformedDistribution,
I would approach this using MomentGeneratingFunction. The moment
generating function for the sum of n things selected from a
given distribution is the same as the moment generating function
for the distribution raised to the nth power. That is for the
sum of n standard normal deviates the moment generating function is:

In[3]:= MomentGeneratingFunction[NormalDistribution[], t]^n

Out[3]= (E^(t^2/2))^n

Comparing this to

In[4]:= MomentGeneratingFunction[NormalDistribution[0, s], t]

Out[4]= E^((s^2*t^2)/2)

it is clear the distribution for the sum of n deviates drawn
from a standard normal distribution is a normal distribution
with mean = 0 variance = n.

I should point out the moment generating function doesn't exist
for all distributions. In those cases you can use the
characteristic function which always exists. And since the
characteristic function is essentially the Fourier transform of
the distribution function, you can use the inverse Fourier
transform to recover the distribution function from the
characteristic function.

Use of the characteristic function or moment generating function
is a very useful technique for answering these types of
questions. Each distribution has a unique characteristic
function (and moment generating function when it exists). Also,
you can easily compute the moments of the distribution directly
from the characteristic function or moment generating function
without actually finding the distribution function.




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Subject: [mg120428] Re: TransformedDistribution, for a sum of M iid variables
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Clear[distSN]

distSN[1] = NormalDistribution[0, 1];

Assume that

distSN[n_] = NormalDistribution[0, Sqrt[n]];

Proof by induction

distSN[n + 1] == 
 TransformedDistribution[
  x + y, {Distributed[x, distSN[n]], 
   Distributed[y, NormalDistribution[0, 1]]}]

True

More generally,

Clear[distN]

distN[1] = NormalDistribution[m, s];

Assume

distN[n_] = NormalDistribution[n*m, Sqrt[n*s^2]];

Proof by induction

distN[n + 1] == 
  TransformedDistribution[
   x + y, {Distributed[x, distN[n]], 
    Distributed[y, NormalDistribution[m, s]]}] // ExpandAll

True

More general forms cannot be as proved as readily; however, the extensions are obvious.

Clear[dist]

dist[n_Integer?Positive] =
  NormalDistribution[Sum[m[k], {k, n}], Sqrt[Sum[s[k]^2, {k, n}]]];

dist[m_List, s_List] :=
  
  NormalDistribution[Total[m], Norm[s]] /; Length[m] == Length[s];

For example,

dist[4]

NormalDistribution[m[1] + m[2] + m[3] + m[4], 
 Sqrt[s[1]^2 + s[2]^2 + s[3]^2 + s[4]^2]]

With[{n = RandomInteger[{10, 100}]},
 Simplify[
  dist[n] == dist[Table[m[k], {k, n}], Table[s[k], {k, n}]],
  Thread[Table[s[k], {k, n}] > 0]]]

True

Testing consistency with the simpler cases:

With[{n = RandomInteger[{10, 100}]},
 (dist[n] /. {m[_] -> m, s[_] -> s}) == distN[n]]

True

With[{n = RandomInteger[{10, 100}]},
 (dist[n] /. {m[_] -> 0, s[_] -> 1}) == distSN[n]]

True


Bob Hanlon



---- paulvonhippel at yahoo <paulvonhippel at yahoo.com> wrote: 

=============
Using the TransformedDistribution function it is easy to show that the
sum of two normal variables is normal, e.g.,

Input:
TransformedDistribution[Z1+Z2,{Z1\
[Distributed]NormalDistribution[0,1],Z2\
[Distributed]NormalDistribution[0,1]}]

Output:
NormalDistribution[0, Sqrt[2]]


Likewise it wouldn't be hard to show that the sum of 3 normal
variables is normal, or 4.

But how do I show that the sum of M normal variables is normal, where
M is an arbitrary positive integer.

I'm thinking I should use the Sum function inside
TransformedDistribution, but I don't know how the notation would work.
I see that there's a UniformSumDistribution function, but that's
limited to uniform variables.

Later I will want to do similar calculations for nonnormal
distributions.

Thanks for any pointers.


--

Bob Hanlon




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From: Paul von Hippel <paulvonhippel at yahoo.com>
Subject: [mg120429] Re: TransformedDistribution, for a sum of M iid variables
References: <20110721225621.9H9BM.1153349.imail@eastrmwml34>
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Thanks for the neat proof. Sorry I wasn't clear about my goals. I want to include a sum of iid variables in a TransfornedDistribution, and then calculate the Mean and Variance of the distribution.

The iid variables need not be normal, and the TrasformedDistribution will contain other variables besides the iid sum.



Sent from my iPhone.
Please excuse my brevity.

On Jul 21, 2011, at 9:56 PM, Bob Hanlon <hanlonr at cox.net> wrote:

> Clear[distSN]
>
> distSN[1] = NormalDistribution[0, 1];
>
> Assume that
>
> distSN[n_] = NormalDistribution[0, Sqrt[n]];
>
> Proof by induction
>
> distSN[n + 1] ==
> TransformedDistribution[
>  x + y, {Distributed[x, distSN[n]],
>   Distributed[y, NormalDistribution[0, 1]]}]
>
> True
>
> More generally,
>
> Clear[distN]
>
> distN[1] = NormalDistribution[m, s];
>
> Assume
>
> distN[n_] = NormalDistribution[n*m, Sqrt[n*s^2]];
>
> Proof by induction
>
> distN[n + 1] ==
>  TransformedDistribution[
>   x + y, {Distributed[x, distN[n]],
>    Distributed[y, NormalDistribution[m, s]]}] // ExpandAll
>
> True
>
> More general forms cannot be as proved as readily; however, the extensions=
 are obvious.
>
> Clear[dist]
>
> dist[n_Integer?Positive] =
>  NormalDistribution[Sum[m[k], {k, n}], Sqrt[Sum[s[k]^2, {k, n}]]];
>
> dist[m_List, s_List] :=
>
>  NormalDistribution[Total[m], Norm[s]] /; Length[m] == Length[s];
>
> For example,
>
> dist[4]
>
> NormalDistribution[m[1] + m[2] + m[3] + m[4],
> Sqrt[s[1]^2 + s[2]^2 + s[3]^2 + s[4]^2]]
>
> With[{n = RandomInteger[{10, 100}]},
> Simplify[
>  dist[n] == dist[Table[m[k], {k, n}], Table[s[k], {k, n}]],
>  Thread[Table[s[k], {k, n}] > 0]]]
>
> True
>
> Testing consistency with the simpler cases:
>
> With[{n = RandomInteger[{10, 100}]},
> (dist[n] /. {m[_] -> m, s[_] -> s}) == distN[n]]
>
> True
>
> With[{n = RandomInteger[{10, 100}]},
> (dist[n] /. {m[_] -> 0, s[_] -> 1}) == distSN[n]]
>
> True
>
>
> Bob Hanlon
>
>
>
> ---- paulvonhippel at yahoo <paulvonhippel at yahoo.com> wrote:
>
> =============
> Using the TransformedDistribution function it is easy to show that the
> sum of two normal variables is normal, e.g.,
>
> Input:
> TransformedDistribution[Z1+Z2,{Z1\
> [Distributed]NormalDistribution[0,1],Z2\
> [Distributed]NormalDistribution[0,1]}]
>
> Output:
> NormalDistribution[0, Sqrt[2]]
>
>
> Likewise it wouldn't be hard to show that the sum of 3 normal
> variables is normal, or 4.
>
> But how do I show that the sum of M normal variables is normal, where
> M is an arbitrary positive integer.
>
> I'm thinking I should use the Sum function inside
> TransformedDistribution, but I don't know how the notation would work.
> I see that there's a UniformSumDistribution function, but that's
> limited to uniform variables.
>
> Later I will want to do similar calculations for nonnormal
> distributions.
>
> Thanks for any pointers.
>
>
> --
>
> Bob Hanlon
>



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Subject: [mg120431] Re: How Can I Eliminate Line Wrapping When Using the Row Function
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On Thu, 21 Jul 2011 21:06:54 -0400 (EDT), Gregory Lypny wrote:
> Hello everyone,
>
> Sometimes when I use Row for four or five items, the result wraps over
> two lines.  How can I force the result to be displayed on one line?  I
> couldn't find an option for this in Documentation Centre.
>
> Gregory

This is the intended behavior of Row.  If you don't want word-wrapping,
then you probably want a single row Grid[].  So, instead of doing
Row[{items}], instead do Grid[{{items}}].

If Row is really doing everything else you want, then I suppose you
could also do...

ExpressionCell[Row[{items}], LineBreakWithin->False]

Sincerely,

John Fultz
jfultz at wolfram.com
User Interface Group
Wolfram Research, Inc.



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Hi, once more I'm with you 100%.
-Francesco



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You might like to know that the 2012 International Mathematica Symposium 
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Kind regards

William






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The option ImageSize in Row dictates the total size of the row, so you 
could try setting that to a large enough value.

Heike

On 22 Jul 2011, at 02:06, Gregory Lypny wrote:

> Hello everyone,
>
> Sometimes when I use Row for four or five items, the result wraps over 
two lines.  How can I force the result to be displayed on one line?  I 
couldn't find an option for this in Documentation Centre.
>
> Gregory
>
>




From mathgroup-adm at smc.vnet.net  Fri Jul 22 18:00:07 2011
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From: Gregory Lypny <gregory.lypny at jmsb.concordia.ca>
Subject: [mg120437] Preventing In-line Math Typesetting From Being Scaled Down in Text Cells
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Hi everyone,

When I include fractions as inline math typesetting, they are scaled down to fit the effective line height of the cell.  How can I prevent this or, I guess, make the line height automatically expand to accommodate the math?  If my regular text in the cell is 12-point times, I'd like all math variables that are not subscripts or superscripts to be 12-point as well.

Incidentally, other big typeset objects like matrices are not scaled down, or at least they down't appear to be.

Sincerely,

Gregory



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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120432] Re: How Can I Include a Button in a Manipulate
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Hi Heike,

Thank you.  I'll certainly experiment with this one.

Gregory



On Fri, Jul 22, 2011, at 4:51 AM, Heike Gramberg wrote:

> You could do something like
>
> Manipulate[Module[{p},
>  q; p = RandomInteger[max, {height, width}];
>  p // TableForm],
> {{width, 3}, Range[8], SetterBar},
> {{height, 3}, Range[8], SetterBar},
> {{max, 4}, Range[10], SetterBar},
> {{q, False}, ControlType -> None},
> Control[Button["Regenerate", q = Not[q]]]
> ]
>
> Heike
>
> On 22 Jul 2011, at 02:07, Gregory Lypny wrote:
>
>> Hello everyone,
>>
>> I have a Manipulate that generates a random integer or integers
>> depending on the user's choice of parameters in a number of setter
>> bars.  I'd like to include a button in the Manipulate's output pane
>> that allows the user to generate a new result by generating a new random
>> number while keeping all of the parameters in the setter bar the same. 
>> Can that be done?
>>
>>
>> Regards,
>>
>> Gregory
>>
>




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From: paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Subject: [mg120439] Re: TransformedDistribution, for a sum of M iid variables
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To clarify what I'm looking for, suppose I have the following
distribution:

TransformedDistribution[Z1+Sqrt[U/(n-1)]*Z2,{Z1\
[Distributed]NormalDistribution[0,1],Z2\
[Distributed]NormalDistribution[0,1],U\
[Distributed]ChiSquareDistribution[n-1]}]

I want to replace Z2 with the average of M independent variables that
are distributed like Z2.
Or I might want to replace U with the average of M independent
variables that are distributed like U.

Is there a convenient way to do this? Thanks!



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From: Heike Gramberg <heike.gramberg at gmail.com>
Subject: [mg120442] Re: How Can I Include a Button in a Manipulate
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You could do something like

Manipulate[Module[{p},
  q; p = RandomInteger[max, {height, width}];
  p // TableForm],
 {{width, 3}, Range[8], SetterBar},
 {{height, 3}, Range[8], SetterBar},
 {{max, 4}, Range[10], SetterBar},
 {{q, False}, ControlType -> None},
 Control[Button["Regenerate", q = Not[q]]]
 ]

Heike

On 22 Jul 2011, at 02:07, Gregory Lypny wrote:

> Hello everyone,
> 
> I have a Manipulate that generates a random integer or integers 
> depending on the user's choice of parameters in a number of setter 
> bars.  I'd like to include a button in the Manipulate's output pane 
> that allows the user to generate a new result by generating a new random 
> number while keeping all of the parameters in the setter bar the same.  
> Can that be done?
> 
> 
> Regards,
> 
> Gregory
> 




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From: Heike Gramberg <heike.gramberg at gmail.com>
Subject: [mg120443] Re: ExampleData[{"Geometry3D", "Torus"}]
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I'm not familiar with Torus[] but another option might be to use 
RevolutionPlot3D, i.e.

torus[r1_, r2_] :=
 RevolutionPlot3D[{r1 + r2 Cos[t], 0, r2 Sin[t]}, {t, 0, 2 Pi},
  Mesh -> False, PlotPoints -> 60, Boxed -> False, Axes -> False]

torus[1, 0.5]

Heike

On 22 Jul 2011, at 02:06, Joseph O'Rourke wrote:

> Thanks, Oliver.  I know how to make a torus via ParametricPlot3D[], 
and by using the deprecated Torus[] function. I was wondering if 
ExampleData[] could be adjusted so that one would have the same control 
as with the old Torus[] function.  The reason I wanted this is that the 
torus produced by ExampleData[] looks much nicer than that produced by 
Torus[].  Or, at least, I cannot get the latter to look as smooth when 
rendered as the former.  But maybe I should just use ParametricPlot3D[] 
with Mehs->None as you suggest.  Thanks again for your attention.
>




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From: Yves Klett <yves.klett at googlemail.com>
Subject: [mg120444] Re: Producing an image that only contains its interior and boundary but, no exterior.
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Not really sure what you are looking for, but perhaps this can help:

g = Graphics3D[Cuboid[], SphericalRegion -> True]

Show[g, Method -> {"ShrinkWrap" -> True}]

I presume this is similar to right-clicking on a graphic and choosing
"Trim Bounding Box"

The shrinkwrapping thing can sometimes affect the interactive
performance, i.e. I had cases where rotation via mouse becomes sluggish,
perhaps not surprisingly so.

Regards,
Yves

Am 22.07.2011 03:11, schrieb Gilmar Rodriguez-pierluissi:
> This is an old issue mentioned by the late Jens-Peer Kuska in:
> http://forums.wolfram.com/mathgroup/archive/2006/Mar/msg00037.html
> How can a user produce a graphic image that includes only its interior and boundary but,
> without the exterior. This would facilitate embedding an image within another without
> having to waste time having to eliminate the annoying edges. Thank you!
> Gilmar Rodriguez-Pierluissi



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From: "Kevin J. McCann" <kjm at KevinMcCann.com>
Subject: [mg120445] Re: mandelbrot in version 7 & Ruskeepaa version 3
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Two things. The error is due to your argument x+Iy in the call to 
mandelbrot. You need a space between the I and the y, i.e. x+I y. 
Second, you have a semicolon at the end of the DensityPlot call; this 
prevents the plot from being shown.

Kevin

On 7/21/2011 9:15 PM, DaleJenk wrote:
> I bought Mathematica 7 Home Edition last year. I'm trying to plot the
> Mandelbrot Set following the instructions in Ruskeepaa's Mathematica
> Navigator version 3. I'm getting an error message.
>
> Here's what I am introducing:
>
> mandelbrot = Compile[{{c,_Complex}},
> Module[{z=0+0I, n=0}, While[Abs[z]<2&&  n<50, z=z^2+c; n++]; n]];
>
> DensityPlot[mandelbrot[x+Iy], {x,-2,1},{y,-1.5,1.5}, PlotPoints->200,
> Mesh->False, FrameTicks->{{-2,-1,0,1},{-1,0,1}}]; //Timing
>
> I get the message:
>
> CompiledFunction::cfsa:
> Argument -1.99998 + Iy at position 1 should be a machine-size complex
> number.
>
> Where am I going wrong?
>
> Thanks.
>
>
>



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From: Curtis Osterhoudt <cfo at lanl.gov>
Subject: [mg120438] Re: mandelbrot in version 7 & Ruskeepaa version 3
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Looks as though the "Iy" in the code needs to be I (that's a capital aye) times why: I*y. If there wasn't a space in the original code, Mathematica interprets it as a new, two-character symbol, not the product of the imaginary basis times y. 

   Hope that helps!

 

On Thursday, July 21, 2011 19:09:03 DaleJenk wrote:
> I bought Mathematica 7 Home Edition last year. I'm trying to plot the 
> Mandelbrot Set following the instructions in Ruskeepaa's Mathematica 
> Navigator version 3. I'm getting an error message.
> 
> Here's what I am introducing:
> 
> mandelbrot = Compile[{{c,_Complex}},
> Module[{z=0+0I, n=0}, While[Abs[z]<2 && n<50, z=z^2+c; n++]; n]];
> 
> DensityPlot[mandelbrot[x+Iy], {x,-2,1},{y,-1.5,1.5}, PlotPoints->200, 
> Mesh->False, FrameTicks->{{-2,-1,0,1},{-1,0,1}}]; //Timing
> 
> I get the message:
> 
> CompiledFunction::cfsa:
> Argument -1.99998 + Iy at position 1 should be a machine-size complex 
> number.
> 
> Where am I going wrong?
> 
> Thanks.
>  
> 
> 
> 




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From: =?iso-8859-1?Q?S=F6nmezer at smc.vnet.net,
        _D=2E?= <d.sonmezer at student.tue.nl>
Subject: [mg120447] Code is not working...
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 Hello,

I am in trouble with below codes:

I run this code but Program is not finsh evertime It is running and running...
Could you possible help me.

dim = Dimensions[image]
r = 1;
k = 1;
n = 10;
cost = Function[{positionValues, neighborValues},
   First[neighborValues] == 1];

(* Image that only contains the value zero *)
imageNew = Table[0, {i, 1, dim[[1]]}, {j, 1, dim[[2]]}];

(* Walk through the rows of the image *)
While[r < dim[[1]],
  (* Walk through the columns of the image*)
  k = 1;
  While[k < dim[[2]],
   (* Find a white pixel*)
   If[imgfilter[[r, k]] == 1,
    (* Proceed RegionGrowing and give them a identical value by \
multiplying it with n *)
    region = RegionGrowing[{imgfilter}, {{r, k}}, 1, cost]*n;
    (* Add the identical cell to imageNew *)
    imageNew = imageNew + region;
    (* Change n, so each cell has a different value *)
    n = n + 2;
    (* Take the cell out of the original image,
    so he will stop when there is no white pixel anymore *)
    imageFilled = imgfilter - region;
    ];
   k++;
   ];
  r++;
  ];




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From: Vince Virgilio <blueschi at gmail.com>
Subject: [mg120435] Re: CDF limitations and unclear professional strategy
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No complaints to WRI. And I am very glad that I can now create nbp/cdfs locally, without going through WRI's website. (Aside: This CDF player release confuses me a bit. Was the CDF player available before this release only a reader, without interactivity?)

Nonetheless, dittos to P. Fonseca.

The whole data management scheme stymies me. Some recent threads touch on this. Notebooks just aren't convenient containers for large amounts of data.  Say, tens of MBs? Certainly the Notebook Save's become lethargic. And for much larger data volumes, I notice latencies in the Front End's refresh of the notebook. That might have been while scrolling, though not displaying large imagery. But that was a while ago. I've since stopped trying to keep so much data in notebooks, and am now trimming my problems/presentations to fit the format's obvious capabilities.

My nagging suspicion is there are certain inefficiencies and limitations as sociated with Notebooks' encoding in text instead of binary. In a way, isn't Mathematica still character-based because of this? (Even though it eventually translates to something else represented by those characters, a la a universal computer.)


Vince Virgilio



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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120433] Re: How Can I Include a Button in a Manipulate
References: <201107220107.VAA16785 at smc.vnet.net> <4E2910D8.9080302 at tue.nl>
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Hello Fred,

Good point, and that appears to have come up in some of the other 
responses.  I use Manipulate only because I am most familiar with it.  
Now it looks like it's time to work through the tutorials on some of the lower level functions!

Gregory


On Fri, Jul 22, 2011, at 1:55 AM, Fred Simons wrote:

> Hi Gregory,
>
> Not an answer, just a question (or suggestion): why do you use Manipulate? My experience is that for the things you seem to want to do you are much more flexible when you use the functions one level lower than Manipulate: DynamicModule for the interactivity and Grid for the layout. You can place the button in any field of the grid you like.
>
> Best wishes,
>
> Fred
>
> Op 22-7-2011 3:07, Gregory Lypny schreef:
>> Hello everyone,
>>
>> I have a Manipulate that generates a random integer or integers
>> depending on the user's choice of parameters in a number of setter
>> bars.  I'd like to include a button in the Manipulate's output pane
>> that allows the user to generate a new result by generating a new random
>> number while keeping all of the parameters in the setter bar the same.
>> Can that be done?
>>
>>
>> Regards,
>>
>> Gregory
>>
>>
>>
>> -----
>> Geen virus gevonden in dit bericht.
>> Gecontroleerd door AVG - www.avg.com
>> Versie: 10.0.1390 / Virusdatabase: 1518/3778 - datum van uitgifte: 07/21/11
>>
>>
>>
>




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From: "DaleJenk" <dale.jenkins8 at deletegooglemail.com>
Subject: [mg120446] Re: mandelbrot in version 7 & Ruskeepaa version 3
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Many thanks - to you, and to the people who replied to me privately.

"Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com> wrote in message 
news:j0b9l4$jjj$1 at smc.vnet.net...
> There is a space missing between I and y in the expression mandelbrot[x +
> Iy] (should be mandelbrot[x + I y]). Also, you should omit the semicolon
> after DensityPlot, otherwise the plot will not be shown.
>
> Incidentally, tabulating the values and then plotting them with ArrayPlot
> is faster for this example than using DensityPlot. That is,
>
> ArrayPlot@Table[
>   mandelbrot[x + I y],
>   {y, -1.5, 1.5, 0.005}, {x, -2.0, 1.0, 0.005}
>  ]
>
> On Fri, 22 Jul 2011 02:15:27 +0100, DaleJenk
> <dale.jenkins8 at deletegooglemail.com> wrote:
>
>> I bought Mathematica 7 Home Edition last year. I'm trying to plot the
>> Mandelbrot Set following the instructions in Ruskeepaa's Mathematica
>> Navigator version 3. I'm getting an error message.
>>
>> Here's what I am introducing:
>>
>> mandelbrot = Compile[{{c,_Complex}},
>> Module[{z=0+0I, n=0}, While[Abs[z]<2 && n<50, z=z^2+c; n++]; n]];
>>
>> DensityPlot[mandelbrot[x+Iy], {x,-2,1},{y,-1.5,1.5}, PlotPoints->200,
>> Mesh->False, FrameTicks->{{-2,-1,0,1},{-1,0,1}}]; //Timing
>>
>> I get the message:
>>
>> CompiledFunction::cfsa:
>> Argument -1.99998 + Iy at position 1 should be a machine-size complex
>> number.
>>
>> Where am I going wrong?
>>
>> Thanks.
>>
> 



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From: DANA DELOUIS <dana.del at gmail.com>
Subject: [mg120430] Re: FinancialData still broken
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...> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8...

Hi.  This function has lots of problems.
In addition to the problems you point out, here's a slightly different view.
Sometimes the Price, and Latest Price match.
Other times, the Price is corrupt, but the latest price is correct, and sometimes it's the other way around.
Sometimes the CIK codes or Exchange are missing, etc.
Other issues are bugs in Option pricing, etc.
I really hope they fix this soon, but I guess they never issue bug fixes.


stocks={"ACEIX","FCNTX","JACNX","JANFX","JANWX","JMSCX","JNGIX","JNGLX", "JNMCX=94,
        "JNOSX","JNSGX","JNSTX","PTTAX","RGACX","STRFX=94};

Transpose[
{
Map[FinancialData[#,"Price"]&,stocks],
Map[FinancialData[#,"LatestTrade"]&,stocks],
Map[FinancialData[#,"CIK"]&,stocks],
Map[FinancialData[#,"Company"]&,stocks]
}][[;;6]]

< Just the first 6 for posting
< Looks better in MatrixForm, or something similar...>

{5.0324*10^8,{Missing[NotAvailable],5.0324*10^8},0000080832,Van Kampen Equity Income Fund A}
{5.0311*10^8,{{2011,7,20,17,11,0.},72.17},0000024238,Fidelity Contra Fund}
{13.97,{{2011,7,20,17,50,0.},13.97},0000277751,Janus Contrarian Fund D Shares}
{5.0324*10^8,{{2001,12,22,20,42,0.},5.0324*10^8},0000277751,Janus Flexible Bond Fund D Shar}
{5.0324*10^8,{{2011,7,20,18,24,0.},46.74},0000277751,Janus Worldwide Fund D Shares}
{12.48,{{2011,7,20,17,51,0.},12.48},Missing[NotAvailable],Janus Smart Portfolio Conservat}

= = = = = = = = = =
Dana DeLouis
$Version
8.0 for Mac OS X x86 (64-bit) (November 6, 2010)


On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
> Erroneous prices are randomly returned:
>
> FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>    "PTTAX", "RGACX", "STRFX"}
>
> {8.81, 72.4, 14., 10.64, 46.49, 12.48,
>   5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
>   5.03241*10^8, 11.04, 31.44, 32.47}
>
> FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>    "PTTAX", "RGACX", "STRFX"}
>
> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
>   5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
>   5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}
>
> Bobby
>
> --
> DrMajor... at yahoo.com





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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120434] Re: How Can I Include a Button in a Manipulate
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Thank you, David.  I will check it out.  Where can I find the Presentations application?

Gregory



On Thu, Jul 21, 2011, at 8:59 PM, David Park wrote:

> Gregory,
> 
> I'm not certain how easy it is to do with Manipulate, but it is certainly
> possible to do it with a custom dynamic display. I find custom dynamic
> displays much easier to write because you can format them any way you wish
> without having to override defaults.
> 
> The Presentations application, in the PlaneGeometry section, has a tutorial
> that shows how to write custom dynamics, including one example with a button
> that performs an action on the displayed dynamic.
> 
> The trick is to think of primary dynamic variables (operated by controls)
> and secondary dynamic variables that depend on the primary variables. Write
> a routine, or routines, that calculate the secondary variables from the
> primary variables. Invoke these by using the second argument in the Dynamic
> statement. The random number could be an initially set primary variable. The
> button could then regenerate it and recalculate all the secondary variables.
> 
> 
> David Park
> djmpark at comcast.net
> http://home.comcast.net/~djmpark/  
> 
> 
> 
> 
> From: Gregory Lypny [mailto:gregory.lypny at videotron.ca] 
> 
> 
> Hello everyone,
> 
> I have a Manipulate that generates a random integer or integers 
> depending on the user's choice of parameters in a number of setter 
> bars.  I'd like to include a button in the Manipulate's output pane 
> that allows the user to generate a new result by generating a new random 
> number while keeping all of the parameters in the setter bar the same.  
> Can that be done?
> 
> 
> Regards,
> 
> Gregory
> 




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Subject: [mg120450] call Fortran subroutine from Mathematica // use NETLink and DLL
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This post explains how to call a Fortran subroutine as a DLL from Mathematica. I don't think the hints in this post exist in explicit form in the Wolfram documentation or in the MathArchive, at least so far as I could find.  

There is information in a post from Wolfram for how to use Mathlink via C wrappers to get to Fortran subroutines (version 3 of Mathematica? http://library.wolfram.com/infocenter/TechNotes/174/). *BUT* I think since the introduction of NETLink into Mathematica (version 6?), the matter is now much more straightforward, at least in a Windows environment. 

The purpose of this post is therefore to a future user who might want to follow what I have done.

>From Windows7, 64-bit, and using the gnu compiler (http://gcc.gnu.org/wiki/GFortranUsage), here is the trace of the CMD commands:

C:\Demo>dir
      22-Jul-11 21:16 336 SumAndDifference.f90

C:\Demo>more SumAndDifference.f90

subroutine sum_and_difference(i,j,resultArray)
      implicit none
      integer*4, intent (in) :: i 
      integer*4, intent (in) :: j 
      integer*4, dimension(2), intent (out) :: resultArray 
      resultArray(1) = i + j 
      resultArray(2) = i - j 
      return
end subroutine sum_and_difference

C:\Demo>gfortran -c SumAndDifference.f90

C:\Demo>gfortran -s -shared -mrtd -o SumAndDifference.dll SumAndDifference.
o

C:\Demo>dir
      22-Jul-11 21:21 12,800 SumAndDifference.dll
      22-Jul-11 21:16 336 SumAndDifference.f90
      22-Jul-11 21:20 429 SumAndDifference.o


Now follows the Mathematica part:

In[1]:= Needs["NETLink`"]

In[2]:= SetDirectory@NotebookDirectory[]
Out[2]= "C:\\MyFortran"

In[3]:= $pathToDLL = FileNameJoin[{Directory[], "SumAndDifference.dll"}]
Out[3]= "C:\\MyFortran\\SumAndDifference.dll"

[To understand below, read:  http://www.wolfram.com/learningcenter/tutorialcollection/NETLinkUserGuide/ ]
[Note addition of "_" to "sum_and_difference" that is done by compiler.]
In[4]:= SumAndDifference =  DefineDLLFunction["sum_and_difference_", $pathToDLL,  "void", {"Int32*", "Int32*", "Int32[]"}]

In[5]:= results = MakeNETObject[{0, 0}, "System.Int32[]"]

In[6]:= SumAndDifference[10, 20, results]

In[7]:= NETObjectToExpression@results
Out[7]= {30, -10}





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From: James Stein <mathgroup at stein.org>
Subject: [mg120449] Re: FinancialData still broken
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DANA DELOUIS wrote:
I really hope they fix this soon, but I guess they never issue bug fixes.

Are these bugs in Mathematica, or bugs in Yahoo?
FWIW, I do not see the bug reported by the OP (DrMajorBob).


On Fri, Jul 22, 2011 at 4:44 PM, DANA DELOUIS <dana.del at gmail.com> wrote:

> ...> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8...
>
> Hi.  This function has lots of problems.
> In addition to the problems you point out, here's a slightly different
> view.
> Sometimes the Price, and Latest Price match.
> Other times, the Price is corrupt, but the latest price is correct, and
> sometimes it's the other way around.
> Sometimes the CIK codes or Exchange are missing, etc.
> Other issues are bugs in Option pricing, etc.
> I really hope they fix this soon, but I guess they never issue bug fixes.
>
>
> stocks={"ACEIX","FCNTX","JACNX","JANFX","JANWX","JMSCX","JNGIX","JNGLX",
> "JNMCX=94,
>        "JNOSX","JNSGX","JNSTX","PTTAX","RGACX","STRFX=94};
>
> Transpose[
> {
> Map[FinancialData[#,"Price"]&,stocks],
> Map[FinancialData[#,"LatestTrade"]&,stocks],
> Map[FinancialData[#,"CIK"]&,stocks],
> Map[FinancialData[#,"Company"]&,stocks]
> }][[;;6]]
>
> < Just the first 6 for posting
> < Looks better in MatrixForm, or something similar...>
>
> {5.0324*10^8,{Missing[NotAvailable],5.0324*10^8},0000080832,Van Kampen
> Equity Income Fund A}
> {5.0311*10^8,{{2011,7,20,17,11,0.},72.17},0000024238,Fidelity Contra Fund}
> {13.97,{{2011,7,20,17,50,0.},13.97},0000277751,Janus Contrarian Fund D
> Shares}
> {5.0324*10^8,{{2001,12,22,20,42,0.},5.0324*10^8},0000277751,Janus Flexible
> Bond Fund D Shar}
> {5.0324*10^8,{{2011,7,20,18,24,0.},46.74},0000277751,Janus Worldwide Fund D
> Shares}
> {12.48,{{2011,7,20,17,51,0.},12.48},Missing[NotAvailable],Janus Smart
> Portfolio Conservat}
>
> = = = = = = = = = =
> Dana DeLouis
> $Version
> 8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
>
>
> On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
> > Erroneous prices are randomly returned:
> >
> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
> >    "PTTAX", "RGACX", "STRFX"}
> >
> > {8.81, 72.4, 14., 10.64, 46.49, 12.48,
> >   5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
> >   5.03241*10^8, 11.04, 31.44, 32.47}
> >
> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
> >    "PTTAX", "RGACX", "STRFX"}
> >
> > {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
> >   5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
> >   5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}
> >
> > Bobby
> >
> > --
> > DrMajor... at yahoo.com
>
>
>
>


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Subject: [mg120448] Re: FinancialData still broken
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I'm used to some of these problems, and I wrote code to try something else  
when Missing comes back.

This is the first time I've seen prices on the order of 5 x 10^8.

And so... I've coded an infinite loop to call over and over (on each  
stock) until I get a price less than 10^8. I may never get rid of that  
kludge, since it doesn't hurt anything unless it's TRULY infinite, or  
semi-infinite, if you will. So far, the error seems truly random, so it  
only takes an extra call or two to get past it.

I don't think there's a problem in Mathematica on our machines. It's  
probably a problem at the curated database end of things, and if so, there  
couldn't be a bug fix.

Bobby

On Fri, 22 Jul 2011 18:44:52 -0500, DANA DELOUIS <dana.del at gmail.com>  
wrote:

> ...> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8...
>
> Hi.  This function has lots of problems.
> In addition to the problems you point out, here's a slightly different  
> view.
> Sometimes the Price, and Latest Price match.
> Other times, the Price is corrupt, but the latest price is correct, and  
> sometimes it's the other way around.
> Sometimes the CIK codes or Exchange are missing, etc.
> Other issues are bugs in Option pricing, etc.
> I really hope they fix this soon, but I guess they never issue bug fixes.
>
>
> stocks={"ACEIX","FCNTX","JACNX","JANFX","JANWX","JMSCX","JNGIX","JNGLX",  
> "JNMCX=94,
>         "JNOSX","JNSGX","JNSTX","PTTAX","RGACX","STRFX=94};
>
> Transpose[
> {
> Map[FinancialData[#,"Price"]&,stocks],
> Map[FinancialData[#,"LatestTrade"]&,stocks],
> Map[FinancialData[#,"CIK"]&,stocks],
> Map[FinancialData[#,"Company"]&,stocks]
> }][[;;6]]
>
> < Just the first 6 for posting
> < Looks better in MatrixForm, or something similar...>
>
> {5.0324*10^8,{Missing[NotAvailable],5.0324*10^8},0000080832,Van Kampen  
> Equity Income Fund A}
> {5.0311*10^8,{{2011,7,20,17,11,0.},72.17},0000024238,Fidelity Contra  
> Fund}
> {13.97,{{2011,7,20,17,50,0.},13.97},0000277751,Janus Contrarian Fund D  
> Shares}
> {5.0324*10^8,{{2001,12,22,20,42,0.},5.0324*10^8},0000277751,Janus  
> Flexible Bond Fund D Shar}
> {5.0324*10^8,{{2011,7,20,18,24,0.},46.74},0000277751,Janus Worldwide  
> Fund D Shares}
> {12.48,{{2011,7,20,17,51,0.},12.48},Missing[NotAvailable],Janus Smart  
> Portfolio Conservat}
>
> = = = = = = = = = =
> Dana DeLouis
> $Version
> 8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
>
>
> On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
>> Erroneous prices are randomly returned:
>>
>> FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>>    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>>    "PTTAX", "RGACX", "STRFX"}
>>
>> {8.81, 72.4, 14., 10.64, 46.49, 12.48,
>>   5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
>>   5.03241*10^8, 11.04, 31.44, 32.47}
>>
>> FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>>    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>>    "PTTAX", "RGACX", "STRFX"}
>>
>> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
>>   5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
>>   5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}
>>
>> Bobby
>>
>> --
>> DrMajor... at yahoo.com
>
>
>


-- 
DrMajorBob at yahoo.com



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From: Markus <markus.seywerd at seywerd.com>
Subject: [mg120453] CDF Files Break when being mailed or uploaded
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Hi I have created a Computable Document File with a Barchart. When I
transfer the file to a machine using a flash drive or over the office
network the receiver can read the file with the CDF reader and
everything looks good. However if I email or ftp the file to another
location the barchart is broken. Does anyone have any ideas?



From mathgroup-adm at smc.vnet.net  Sat Jul 23 18:00:56 2011
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From: Heike Gramberg <heike.gramberg at gmail.com>
Subject: [mg120454] Re: Code is not working...
References: <201107222347.TAA28685 at smc.vnet.net>
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If you want any useful help you'll probably need to give a bit more 
information. For example, what are
image, RegionGrowing and imgfilter? And what is the function of 
imageFilled (it doesn't seem to do anything
at the moment)?

Heike.

On 23 Jul 2011, at 00:47, =?iso-8859-1?Q?S=F6nmezer at smc.vnet.net wrote:

>
> Hello,
>
> I am in trouble with below codes:
>
> I run this code but Program is not finsh evertime It is running and running...
> Could you possible help me.
>
> dim = Dimensions[image]
> r = 1;
> k = 1;
> n = 10;
> cost = Function[{positionValues, neighborValues},
>   First[neighborValues] == 1];
>
> (* Image that only contains the value zero *)
> imageNew = Table[0, {i, 1, dim[[1]]}, {j, 1, dim[[2]]}];
>
> (* Walk through the rows of the image *)
> While[r < dim[[1]],
>  (* Walk through the columns of the image*)
>  k = 1;
>  While[k < dim[[2]],
>   (* Find a white pixel*)
>   If[imgfilter[[r, k]] == 1,
>    (* Proceed RegionGrowing and give them a identical value by \
> multiplying it with n *)
>    region = RegionGrowing[{imgfilter}, {{r, k}}, 1, cost]*n;
>    (* Add the identical cell to imageNew *)
>    imageNew = imageNew + region;
>    (* Change n, so each cell has a different value *)
>    n = n + 2;
>    (* Take the cell out of the original image,
>    so he will stop when there is no white pixel anymore *)
>    imageFilled = imgfilter - region;
>    ];
>   k++;
>   ];
>  r++;
>  ];
>
>




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From: "Dr. Robert Kragler" <kragler at hs-weingarten.de>
Subject: [mg120452] Editing/modifying palettes
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My question ist : How can I merge resp. copy some useful buttons from one 
palette into another one?
I am aware that there exists a thread "How edit a saved palette?" [mg85234], 
but the recipes given do not 
really work with Mathematica V7. Any suggestions are appreciated.

Robert Kragler

Email : kragler at hs-weingarten.de
URL : http://portal.hs-weingarten.de/web/kragler





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From: "Harvey P. Dale" <hpd1 at nyu.edu>
Subject: [mg120459] From Reduce to List
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	What is the easiest and most efficient way to transform the
output from Reduce into a list of values?  For example, if Reduce
produces
(a==1001&&b==1001000)||(a==1002&&b==501000)||(a==1004=
&&b==251000) and I
want {{1001,1001000},{1002,501000},{1004,251000}}, what is the best way
to go from the former to the latter?

	Harvey

Harvey P. Dale
University Professor of Philanthropy and the Law
Director, National Center on Philanthropy and the Law
139 MacDougal Street
New York, N.Y. 10012-1076
Tel: 212-998-6161
Fax: 212-995-3149 



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From: "Ted Sariyski" <tsariysk at craft-tech.com>
Subject: [mg120458] how to ListPlot3D large data sets
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Hi, 
I guess I am pushing the limits of ListPlot3D. I try to ListPlot3d a data
set with many millions of records. I was not able to get an image from the
full dataset, it takes forever. If I use e.g. every fifth record, although
slow, I get an image. The machine running Mathematica is Windows 7 (64 bit),
has 24 GB RAM and there was no swapping. I wonder what is considered as a
reasonable data size for ListPlot3D and are there other tools in Mathematica
for visualization of large data sets? 
Thanks in advance,
--Ted 




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From: paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Subject: [mg120451] random variables
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I would like to define a symbol representing a random variable, and
then use that symbol in other expressions. It's not clear to me how to
do that. For example, when I type the following input --

 U = ChiSquareDistribution[n - 1]
 V = 2*U
 W = 3*U
 Mean[V]
 Mean[W]

-- I do not get the mean of the random variables V and W. I know I can
get the desired answer by typing:

V = TransformedDistribution[2*U,
  U \[Distributed] ChiSquareDistribution[n - 1]]
W = TransformedDistribution[3*U,
  U \[Distributed] ChiSquareDistribution[n - 1]]
 Mean[V]
 Mean[W]

But that requires me to redefine the distribution of U every time I
need it.

Is there a way to define U just once?



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From: Murray Eisenberg <murray at math.umass.edu>
Subject: [mg120455] Re: Producing an image that only contains its interior and boundary but, no exterior.
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How did you discover that option for Show?

I find no entry for "ShrinkWrap" in the Documentation Center, see no 
mention of that option value on the page for Show. On the Graphics3D ref 
page, the option Method is listed in the More Information section, but 
that Method option is not listed in the Options section of Examples.

And of course the ref page for Method is essentially useless in such a 
case. It's long been a documentation lacuna that there are no "inverse 
references" for such options as Method that apply to multiple functions 
-- from the general option such as Method back to all the functions for 
which it's an option, along with some information as to what settings 
are allowed for each such function (at least in cases where within the 
target function page itself the settings are not all listed).


On 7/22/11 7:47 PM, Yves Klett wrote:
> g = Graphics3D[Cuboid[], SphericalRegion ->  True]
>
> Show[g, Method ->  {"ShrinkWrap" ->  True}]

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305



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From: Bill Rowe <readnews at sbcglobal.net>
Subject: [mg120457] Re: TransformedDistribution, for a sum of M iid variables
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On 7/22/11 at 3:42 AM, paulvonhippel at yahoo.com (Paul von Hippel)
wrote:

>Thanks for the neat proof. Sorry I wasn't clear about my goals. I
>want to include a sum of iid variables in a TransfornedDistribution,
>and then calculate the Mean and Variance of the distribution.

>The iid variables need not be normal, and the TrasformedDistribution
>will contain other variables besides the iid sum.

While it is possible to do this using TransformedDistribution, I
think it is far easier to to what you want with the various
generating functions built into Mathematica. Use the exponential
distribution as an example. For the exponential distribution:

In[2]:= mg =
CumulantGeneratingFunction[ExponentialDistribution[a], t]

Out[2]= Log[a/(a - t)]

Cumulants sum. So, the cumulant generating function for the sum
n exponential deviates is simply n times the cumulant generating
function of the original exponential distribution. And the mean
is the first derivative of the cumulant generating function
evaluated at t = 0. So, the mean of the sum of n exponential
deviates is:

In[3]:= D[n mg, t] /. t -> 0

Out[3]= n/a

And the variance is given by the second derivative. So, the
desired variance is:

In[4]:= D[n mg, {t, 2}] /. t -> 0

Out[4]= n/a^2

And to check the validity of the above, note:

In[5]:= MomentGeneratingFunction[ExponentialDistribution[a], t]

Out[5]= a/(a - t)

In[8]:= MomentGeneratingFunction[GammaDistribution[n, 1/a], t]

Out[8]= (1 - t/a)^(-n)

That is, the sum of n exponential deviates is distributed as a
gamma distribution with parameters n, 1/a. So,

In[9]:= Mean[GammaDistribution[n, 1/a]]

Out[9]= n/a

In[10]:= Variance[GammaDistribution[n, 1/a]]

Out[10]= n/a^2

which verifies the earlier result using the cumulant generating
function. Note, this technique of using generating functions
does not require the sum to be of identically distributed
values. I could just as easily find the mean and variance of say
the sum of an exponential deviate and the sum of a normal
deviate using this technique.




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I usually don't see the "price > 10^8" bug, either. It may have been  
permanently gone by the time you looked for it. But it was there for a day  
or so.

I know the occasional spurious Missing coming back is a months-old issue  
at least, and Dan Delouis confirmed it still exists.

I've inoculated my code against both bugs, so I may not notice when they  
go away.

Bobby

On Sat, 23 Jul 2011 01:29:09 -0500, James Stein <mathgroup at stein.org>  
wrote:

>
> DANA DELOUIS wrote:
> I really hope they fix this soon, but I guess they never issue bug fixes.
>
> Are these bugs in Mathematica, or bugs in Yahoo?
> FWIW, I do not see the bug reported by the OP (DrMajorBob).
>
>
> On Fri, Jul 22, 2011 at 4:44 PM, DANA DELOUIS <dana.del at gmail.com> wrote:
>
>> ...> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8...
>>
>> Hi.  This function has lots of problems.
>> In addition to the problems you point out, here's a slightly different
>> view.
>> Sometimes the Price, and Latest Price match.
>> Other times, the Price is corrupt, but the latest price is correct, and
>> sometimes it's the other way around.
>> Sometimes the CIK codes or Exchange are missing, etc.
>> Other issues are bugs in Option pricing, etc.
>> I really hope they fix this soon, but I guess they never issue bug  
>> fixes.
>>
>>
>> stocks={"ACEIX","FCNTX","JACNX","JANFX","JANWX","JMSCX","JNGIX","JNGLX",
>> "JNMCX=94,
>>        "JNOSX","JNSGX","JNSTX","PTTAX","RGACX","STRFX=94};
>>
>> Transpose[
>> {
>> Map[FinancialData[#,"Price"]&,stocks],
>> Map[FinancialData[#,"LatestTrade"]&,stocks],
>> Map[FinancialData[#,"CIK"]&,stocks],
>> Map[FinancialData[#,"Company"]&,stocks]
>> }][[;;6]]
>>
>> < Just the first 6 for posting
>> < Looks better in MatrixForm, or something similar...>
>>
>> {5.0324*10^8,{Missing[NotAvailable],5.0324*10^8},0000080832,Van Kampen
>> Equity Income Fund A}
>> {5.0311*10^8,{{2011,7,20,17,11,0.},72.17},0000024238,Fidelity Contra  
>> Fund}
>> {13.97,{{2011,7,20,17,50,0.},13.97},0000277751,Janus Contrarian Fund D
>> Shares}
>> {5.0324*10^8,{{2001,12,22,20,42,0.},5.0324*10^8},0000277751,Janus  
>> Flexible
>> Bond Fund D Shar}
>> {5.0324*10^8,{{2011,7,20,18,24,0.},46.74},0000277751,Janus Worldwide  
>> Fund D
>> Shares}
>> {12.48,{{2011,7,20,17,51,0.},12.48},Missing[NotAvailable],Janus Smart
>> Portfolio Conservat}
>>
>> = = = = = = = = = =
>> Dana DeLouis
>> $Version
>> 8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
>>
>>
>> On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
>> > Erroneous prices are randomly returned:
>> >
>> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>> >    "PTTAX", "RGACX", "STRFX"}
>> >
>> > {8.81, 72.4, 14., 10.64, 46.49, 12.48,
>> >   5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
>> >   5.03241*10^8, 11.04, 31.44, 32.47}
>> >
>> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>> >    "PTTAX", "RGACX", "STRFX"}
>> >
>> > {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
>> >   5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
>> >   5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}
>> >
>> > Bobby
>> >
>> > --
>> > DrMajor... at yahoo.com
>>
>>
>>
>>


-- 
DrMajorBob at yahoo.com



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From: Simon <simonjtyler at gmail.com>
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Reduce[expr, vars]
vars /. ToRules[%]



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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120462] Re: From Reduce to List
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(a == 1001 && b == 1001000) || (a == 1002 &&
     b == 501000) || (a == 1004 && b == 251000) /. {Or -> List,
   And -> List, Equal[_, x_] :> x}

{{1001, 1001000}, {1002, 501000}, {1004, 251000}}

or

expr = (a == 1001 && b == 1001000) || (a == 1002 &&
      b == 501000) || (a == 1004 && b == 251000);
{a, b} /. List[expr // ToRules]

{{1001, 1001000}, {1002, 501000}, {1004, 251000}}

Bobby

On Sat, 23 Jul 2011 18:54:16 -0500, Harvey P. Dale <hpd1 at nyu.edu> wrote:

> 	What is the easiest and most efficient way to transform the
> output from Reduce into a list of values?  For example, if Reduce
> produces
> (a==1001&&b==1001000)||(a==1002&&b==501000)||(a==1004=
> &&b==251000) and I
> want {{1001,1001000},{1002,501000},{1004,251000}}, what is the best way
> to go from the former to the latter?
>
> 	Harvey
>
> Harvey P. Dale
> University Professor of Philanthropy and the Law
> Director, National Center on Philanthropy and the Law
> 139 MacDougal Street
> New York, N.Y. 10012-1076
> Tel: 212-998-6161
> Fax: 212-995-3149
>


-- 
DrMajorBob at yahoo.com



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Subject: [mg120465] TransformedDistribution -- odd problem
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I'm having a very strange problem with TransformedDistribution, where
I can calculate the mean of an F distribution but I cannot calculate
the mean of a constant multiplied by an F distribution. That is, if I
type

 Mean[TransformedDistribution[F, F \[Distributed]
FRatioDistribution[v, v]]]

Mathematica gives me an answer. But if I type

 Mean[TransformedDistribution[k*F ,  F \[Distributed]
FRatioDistribution[v, v]]]

Mathematica just echoes the input. I swear I got an answer for the
second expression earlier today. What am I doing wrong?



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I am trying to numerically solve systems of ODEs using Mathematica 7.

The desired accuracy that I am striving for depends on the number of solution functions so I want to structure my solution algorithm to allow different numbers of solution functions. Solving the problem as a system of ODEs in matrix form seems the logical choice:

     dX/dt = f[X[t]],

where f[X[t]] is a specified function, and X[t] is a vector of solution functions {x1[t],x2[t],...}. The initial conditions are specified by a vector X[0] = {x1[0],x2[0],...}.

In my problem, the number of solution functions will usually vary from 3 to 7, but could be any number.

One approach that doesn't work is:

     f := {f[[1]][t], f[[2]][t], f[[3]][t]}
     fInit := {0, 0, 2}
     mat := {(-3*f[[1]][t] - f[[2]][t]), 26.5*f[[1]][t] - f[[2]][t] - f[[1]][t]*f[[3]][t], f[[1]][t]*f[[2]][t] - f[[3]][t]}
     solution =  NDSolve[{Derivative[1][f][t] == mat, f[0] == fInit}, f, {t, 0, 17}]

I don't know if the problem is in the way f is specified or in my expression for NDSolve.

I'd appreciate any assistance.

Thanks and regards,

Glenn



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I am trying to numerically solve systems of ODEs using Mathematica 7.

The desired accuracy that I am striving for depends on the number of 
solution functions so I want to structure my solution algorithm to 
allow different numbers of solution functions. Solving the problem as 
a system of ODEs in matrix form seems the logical choice:

      dX/dt = f[X[t]],

where f[X[t]] is a specified function, and X[t] is a vector of 
solution functions {x1[t],x2[t],...}. The initial conditions are 
specified by a vector X[0] = {x1[0],x2[0],...}.

In my problem, the number of solution functions will usually vary from 
3 to 7, but could be any number.

One approach that doesn't work is:

      f := {f[[1]][t], f[[2]][t], f[[3]][t]}
      fInit := {0, 0, 2}
      mat := {(-3*f[[1]][t] - f[[2]][t]), 26.5*f[[1]][t] - f[[2]][t] - 
f[[1]][t]*f[[3]][t], f[[1]][t]*f[[2]][t] - f[[3]][t]}
      solution =  NDSolve[{Derivative[1][f][t] == mat, f[0] == fInit}, 
f, {t, 0, 17}]

I don't know if the problem is in the way f is specified or in my 
expression for NDSolve.

I'd appreciate any assistance.

Thanks and regards,

Glenn



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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120461] Re: random variables
Reply-To: drmajorbob at yahoo.com
References: <201107232352.TAA08648 at smc.vnet.net>
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U=ChiSquareDistribution[n-1];
twoU=TransformedDistribution[2*u,u\[Distributed]U];
Through[{Mean,Variance,PDF}@twoU]

{2 (-1+n),8 (-1+n),Function[\[FormalX],\[Piecewise]	(2^(1-n)  
\[FormalX]^(-1+1/2 (-1+n)) E^(-\[FormalX]/4))/Gamma[1/2  
(-1+n)]	\[FormalX]>0
0	True
,Listable]}

Bobby

On Sat, 23 Jul 2011 18:52:50 -0500, paulvonhippel at yahoo  
<paulvonhippel at yahoo.com> wrote:

> I would like to define a symbol representing a random variable, and
> then use that symbol in other expressions. It's not clear to me how to
> do that. For example, when I type the following input --
>
>  U = ChiSquareDistribution[n - 1]
>  V = 2*U
>  W = 3*U
>  Mean[V]
>  Mean[W]
>
> -- I do not get the mean of the random variables V and W. I know I can
> get the desired answer by typing:
>
> V = TransformedDistribution[2*U,
>   U \[Distributed] ChiSquareDistribution[n - 1]]
> W = TransformedDistribution[3*U,
>   U \[Distributed] ChiSquareDistribution[n - 1]]
>  Mean[V]
>  Mean[W]
>
> But that requires me to redefine the distribution of U every time I
> need it.
>
> Is there a way to define U just once?
>


-- 
DrMajorBob at yahoo.com



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From: Fred Simons <f.h.simons at tue.nl>
Subject: [mg120466] Re: From Reduce to List
References: <201107232354.TAA08719 at smc.vnet.net>
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In[8]:= 
expr=(a==1001&&b==1001000)||(a==1002&&b==501000)||(a==1004&&b==251000);
{a,b}/.{ToRules[expr]}

Out[9]= {{1001,1001000},{1002,501000},{1004,251000}}

If expr is indeed the result of a Reduce command, I think it is very 
likely that instead of Reduce you could have used Solve (it has quite a 
lot of new features) in order to obtain the rules immediately.

Kind regards,

Fred Simons
Eindhoven University of Technology

Op 24-7-2011 1:54, Harvey P. Dale schreef:
> 	What is the easiest and most efficient way to transform the
> output from Reduce into a list of values?  For example, if Reduce
> produces
> (a==1001&&b==1001000)||(a==1002&&b==501000)||(a==1004=
> &&b==251000) and I
> want {{1001,1001000},{1002,501000},{1004,251000}}, what is the best way
> to go from the former to the latter?
>
> 	Harvey
>
> Harvey P. Dale
> University Professor of Philanthropy and the Law
> Director, National Center on Philanthropy and the Law
> 139 MacDougal Street
> New York, N.Y. 10012-1076
> Tel: 212-998-6161
> Fax: 212-995-3149




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On Jul 24, 1:25 am, Markus <markus.seyw... at seywerd.com> wrote:
> Hi I have created a Computable Document File with a Barchart. When I
> transfer the file to a machine using a flash drive or over the office
> network the receiver can read the file with the CDF reader and
> everything looks good. However if I email or ftp the file to another
> location the barchart is broken. Does anyone have any ideas?

Hi, Markus. Have you tried to send it archived?
-jb-



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From: Heike Gramberg <heike.gramberg at gmail.com>
Subject: [mg120469] Re: From Reduce to List
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You would use ToRules:

{a, b} /. {ToRules[(a == 1001 && b == 1001000) || (a == 1002 && 
      b == 501000) || (a == 1004 && b == 251000)]}

out:

{{1001, 1001000}, {1002, 501000}, {1004, 251000}}

Heike


On 24 Jul 2011, at 00:54, Harvey P. Dale wrote:

> 	What is the easiest and most efficient way to transform the
> output from Reduce into a list of values?  For example, if Reduce
> produces
> (a==1001&&b==1001000)||(a==1002&&b==501000)||(a==1004=
> &&b==251000) and I
> want {{1001,1001000},{1002,501000},{1004,251000}}, what is the best way
> to go from the former to the latter?
> 
> 	Harvey
> 
> Harvey P. Dale
> University Professor of Philanthropy and the Law
> Director, National Center on Philanthropy and the Law
> 139 MacDougal Street
> New York, N.Y. 10012-1076
> Tel: 212-998-6161
> Fax: 212-995-3149 
> 




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From: Jay Bee <jiri.bocan at gmail.com>
Subject: [mg120467] Re: Code is not working...
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On Jul 24, 1:25 am, Heike Gramberg <heike.gramb... at gmail.com> wrote:
> If you want any useful help you'll probably need to give a bit more
> information. For example, what are
> image, RegionGrowing and imgfilter? And what is the function of
> imageFilled (it doesn't seem to do anything
> at the moment)?
>
> Heike.
>
> On 23 Jul 2011, at 00:47, =?iso-8859-1?Q?S=F6nme... at smc.vnet.net wrot=
e:
>
>
>
>
>
>
>
>
>
> > Hello,
>
> > I am in trouble with below codes:
>
> > I run this code but Program is not finsh evertime It is running and run=
ning...
> > Could you possible help me.
>
> > dim = Dimensions[image]
> > r = 1;
> > k = 1;
> > n = 10;
> > cost = Function[{positionValues, neighborValues},
> >   First[neighborValues] == 1];
>
> > (* Image that only contains the value zero *)
> > imageNew = Table[0, {i, 1, dim[[1]]}, {j, 1, dim[[2]]}];
>
> > (* Walk through the rows of the image *)
> > While[r < dim[[1]],
> >  (* Walk through the columns of the image*)
> >  k = 1;
> >  While[k < dim[[2]],
> >   (* Find a white pixel*)
> >   If[imgfilter[[r, k]] == 1,
> >    (* Proceed RegionGrowing and give them a identical value by \
> > multiplying it with n *)
> >    region = RegionGrowing[{imgfilter}, {{r, k}}, 1, cost]*n;
> >    (* Add the identical cell to imageNew *)
> >    imageNew = imageNew + region;
> >    (* Change n, so each cell has a different value *)
> >    n = n + 2;As
> >    (* Take the cell out of the original image,
> >    so he will stop when there is no white pixel anymore *)
> >    imageFilled = imgfilter - region;
> >    ];
> >   k++;
> >   ];
> >  r++;
> >  ];

As Heike asked, what are those functions you use? If the computation
lasts forever, there must be some bug... Check all braces - correct
number and position, if all the inputs and outputs have an expected
format, and so on...



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From: "Kevin J. McCann" <kjm at KevinMcCann.com>
Subject: [mg120470] Re: NDSolve in matrix form
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Glenn,

First of all, you got into a recurrence problem with you definition of 
f. Second, the form of the input to NDSolve isn't quite right, you need 
to use Thread to transform something like

{f1[t],f2[t],f3[t]}=={0,0,2} into {f1[t]==0,f2[t]==0,f3[t]==2}

Take the code below apart and you will see how this is used.

Finally, note that f3 does not enter the first two de's and that it is 
fairly simple to see that f1=f2=0.

Here is my take:

Clear[f,fInit,mat]
f[t_]={f1[t],f2[t],f3[t]}
fInit={0,0,2}
mat={-3 f1[t]-f2[t],26.5 f1[t]-f2[t]-f1[t] f2[t],f1[t] f2[t]-f3[t]}
eqns=Flatten[{Thread[(f^\[Prime])[t]==mat],Thread[f[0]==fInit]}]
solution=NDSolve[eqns,f[t],{t,0,17}][[1]]

Cheers,

Kevin

On 7/24/2011 3:22 AM, Glenn Carlson wrote:
> I am trying to numerically solve systems of ODEs using Mathematica 7.
>
> The desired accuracy that I am striving for depends on the number of solution functions so I want to structure my solution algorithm to allow different numbers of solution functions. Solving the problem as a system of ODEs in matrix form seems the logical choice:
>
>       dX/dt = f[X[t]],
>
> where f[X[t]] is a specified function, and X[t] is a vector of solution functions {x1[t],x2[t],...}. The initial conditions are specified by a vector X[0] = {x1[0],x2[0],...}.
>
> In my problem, the number of solution functions will usually vary from 3 to 7, but could be any number.
>
> One approach that doesn't work is:
>
>       f := {f[[1]][t], f[[2]][t], f[[3]][t]}
>       fInit := {0, 0, 2}
>       mat := {(-3*f[[1]][t] - f[[2]][t]), 26.5*f[[1]][t] - f[[2]][t] - f[[1]][t]*f[[3]][t], f[[1]][t]*f[[2]][t] - f[[3]][t]}
>       solution =  NDSolve[{Derivative[1][f][t] == mat, f[0] == fInit}, f, {t, 0, 17}]
>
> I don't know if the problem is in the way f is specified or in my expression for NDSolve.
>
> I'd appreciate any assistance.
>
> Thanks and regards,
>
> Glenn
>



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From: Glenn Carlson <g.crlsn at gmail.com>
Subject: [mg120473] Re: NDSolve in matrix form
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Thanks for your response, Kevin.

I figured out how to do this late last night.  I attribute differences between the solution in terms of f[1][t], f[2][t], and f[3][t] and the solution from the tutorial in terms of x[t], y[t], and z[t] to be due to internal numerics in solving the nonlinear system.

Applying either approach to the differential-algebraic system in the tutorial gives the same result.

Thanks again very much.

Glenn
------------

(* System of ordinary differential equations. 
http://reference.wolfram.com/mathematica/tutorial/NumericalSolutionOfDifferentialEquations.html *)

n := 3
fInit := {0, 2, 0}
mat[t_] = {-3 (f[1][t] - f[2][t]), -f[1][t] f[3][t] + 26.5 f[1][t] - f[2][t], f[1][t] f[2][t] - f[3][t]};

(* Create a list of the solution functions excl. the argument t. *)
fsol = Table[f[i], {i, 1, n}];

(* Create a list of ODEs and ICs for NDSolve. *)
sys = Join[Table[fsol[[i]]'[t] == mat[t][[i]], {i, 1, n}], Table[fsol[[i]][0] == fInit[[i]], {i, 1, n}]];

solution1 = NDSolve[sys, fsol, {t, 0, 200}, MaxSteps -> Infinity]

(* Create a list of solution functions incl. the argument t for ParametricPlot3D. *)
fsolt = Table[f[i][t], {i, 1, n}]
ParametricPlot3D[Evaluate[fsolt /. solution1], {t, 0, 200}, PlotPoints -> 10000, ColorFunction ->(ColorData["Rainbow"][#4] &)]

(* Plot solutions as functions of time. *)
Plot[Evaluate[f[1][t] /. solution1], {t, 0, 200}]
Plot[Evaluate[f[2][t] /. solution1], {t, 0, 200}]
Plot[Evaluate[f[3][t] /. solution1], {t, 0, 200}]
Plot[Evaluate[fsolt /. solution1], {t, 0, 200}]



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From: icegood <icegood1980 at gmail.com>
Subject: [mg120475] default font
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everytime with new norebook i change font size. How to change it in
default manner?



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From: rych <rychphd at gmail.com>
Subject: [mg120479] Simplify[ArcTan[Tan[a] + Tan[b]]]
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Can't Mathematica tell anything about ArcTan[Tan[a] + Tan[b]]? Maybe
there exists no useful identity for that? By useful I mean the
expression with functions of the "angles" a and b other than tangents.
Thanks
Igor



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From: Richard Fateman <fateman at cs.berkeley.edu>
Subject: [mg120482] And now for something completely different
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Can you start a paragraph with "And"?

http://answers.yahoo.com/question/index?qid090413160418AAyfRO2

says, 'No.'


Stephen Wolfram often does so, and I find it annoying.
example
http://www.wolframscience.com/nksonline/page-42?firstview=1

This construction seems to have been removed from the online documentation.


--RJF



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From: paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Subject: [mg120484] Re: TransformedDistribution -- odd problem
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I also have a more complicated example where Mathematica will give the
mean of A and the mean of B, but when I ask for the mean of A+B
Mathematica just echoes the input. (Here A and B are complicated
functions of several random variables.) This strikes me as really odd,
since of course Mean(A+B)=Mean(A)+Mean(B).

What could be going on here? Under what circumstances does Mathematica
echo the input when asked to provide a Mean? Is there a setting I need
to change? I suspect this is a software issue rather than a
mathematical issue.



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From: Don <donabc at comcast.net>
Subject: [mg120477] Grid Divider Style With Background Colors
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Mathematica  permits divider lines between rows and columns in a table
(grid)  such
as this example from the documentation:

Grid[Table[x, {4}, {11}], Dividers -> {{{{True, False}}, -1 -> True},
False}]

But, the dividers are single lines only.

Sometimes  double line dividers are desired
as shown in this table on the Web:

http://www.assetcorrelation.com/user/correlations/90

Notice that the background color for an item  is only within
the inside line of a double line divider.

Is there any way to produce (a) this  kind of divider (double line)
and (b) with background colors for specific items that don't cross
the inside line of the double line divider?

Thank you in advance.

Don



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From: Markus <markus.seywerd at googlemail.com>
Subject: [mg120476] Re: CDF Files Break when being mailed or uploaded
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On Jul 24, 12:39 pm, Jay Bee <jiri.bo... at gmail.com> wrote:
> On Jul 24, 1:25 am, Markus <markus.seyw... at seywerd.com> wrote:
>
> > Hi I have created a Computable Document File with a Barchart. When I
> > transfer the file to a machine using a flash drive or over the office
> > network the receiver can read the file with the CDF reader and
> > everything looks good. However if I email or ftp the file to another
> > location the barchart is broken. Does anyone have any ideas?
>
> Hi, Markus. Have you tried to send it archived?
> -jb-

Yes I have tried to send it as a zip file and the same thing happens.



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From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
Subject: [mg120487] Re: how to ListPlot3D large data sets
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On Sat, 23 Jul 2011, Ted Sariyski wrote:

> Hi,
> I guess I am pushing the limits of ListPlot3D. I try to ListPlot3d a data
> set with many millions of records. I was not able to get an image from the
> full dataset, it takes forever. If I use e.g. every fifth record, although
> slow, I get an image. The machine running Mathematica is Windows 7 (64 bit),
> has 24 GB RAM and there was no swapping. I wonder what is considered as a
> reasonable data size for ListPlot3D and are there other tools in Mathematica
> for visualization of large data sets?
> Thanks in advance,
> --Ted
>
>
>

Ted,

is your data packed?

Developer`PackedArrayQ[yourData]

Oliver



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From: Alan <alan.isaac at gmail.com>
Subject: [mg120481] Re: CDF Files Break when being mailed or uploaded
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Are you running on the same version of Mathematica at both ends? If it
is Mathematica on one end and the Mathematica Player on the other, it could be a player bug.

If the former, either the file is changed or it is different than
expected at the other platform, probably the latter.

You can tell if the file changes by using a check sum tool at both
ends.

The most likely cross-platform problem would involve handling of end-
of-line conventions.
http://www.rfc-editor.org/EOLstory.txt
This is pure speculation: I don't know how Mathematica handles end-of-lines
across platforms.
Hopefully it transparently handles any common end-of-lines (like many
popular editors).

fwiw,
Alan Isaac



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From: Joseph Gwinn <joegwinn at comcast.net>
Subject: [mg120486] Re: how to ListPlot3D large data sets
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In article <j0fou1$8pm$1 at smc.vnet.net>,
 "Ted Sariyski" <tsariysk at craft-tech.com> wrote:

> Hi, 
> I guess I am pushing the limits of ListPlot3D. I try to ListPlot3d a data
> set with many millions of records. I was not able to get an image from the
> full dataset, it takes forever. If I use e.g. every fifth record, although
> slow, I get an image. The machine running Mathematica is Windows 7 (64 bit),
> has 24 GB RAM and there was no swapping. I wonder what is considered as a
> reasonable data size for ListPlot3D and are there other tools in Mathematica
> for visualization of large data sets? 

How long did you wait for the full dataset picture to form?

I have plotted some very large datasets over the years, and although 
slow, it usually worked.  So long as there was enough memory.

But printing the image is quite another matter.


Joe Gwinn



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On 7/23/2011 8:25 PM, Markus wrote:
> Hi I have created a Computable Document File with a Barchart. When I
> transfer the file to a machine using a flash drive or over the office
> network the receiver can read the file with the CDF reader and
> everything looks good. However if I email or ftp the file to another
> location the barchart is broken. Does anyone have any ideas?

Your FTP client and email client (or web server, if you are using 
webmail) may be sending the CDF files as ASCII, which is often the 
default if the file type is not recognized, and may be throwing in extra 
line breaks that wreck the file.

Try setting your FTP client to send CDF files (or all files, for that 
matter) as Binary.

As for email, if it is webmail, ask your Sys Admin to set a Mime type 
for CDF so that it is recognized as an application.

If you are using some an e-mail client (Outlook or whatever), you may 
have to look around in the settings to see how it handles attachments.


-- 
Helen Read
University of Vermont



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Subject: [mg120480] Re: Preventing In-line Math Typesetting From Being Scaled Down in Text Cells
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On Fri, 22 Jul 2011 19:46:07 -0400 (EDT), Gregory Lypny wrote:
> Hi everyone,
>
> When I include fractions as inline math typesetting, they are scaled down
> to fit the effective line height of the cell.  How can I prevent this or,
> I guess, make the line height automatically expand to accommodate the
> math?  If my regular text in the cell is 12-point times, I'd like all
> math variables that are not subscripts or superscripts to be 12-point as
> well.
>
> Incidentally, other big typeset objects like matrices are not scaled
> down, or at least they down't appear to be.
>
> Sincerely,
>
> Gregory

If you look in Core.nb, you'll find a style called "InlineCell".  This style is automatically applied to all inline cells everywhere.  One of the options it has set is:

ScriptLevel->1

This is what's causing the behavior you're seeing.  You can override this with a custom stylesheet.  For example, in a given notebook, you can make a private override by doing Format->Edit Stylesheet..., and pasting and interpreting the following cell expression at the end of the resulting stylesheet notebook:


Cell[StyleData["InlineCell"], ScriptLevel->0]


Sincerely,

John Fultz
jfultz at wolfram.com
User Interface Group
Wolfram Research, Inc.



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From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
Subject: [mg120474] Re: From Reduce to List
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I think the only exceptions are non-generic cases like this one:

{Reduce[a x == 0, x] // ToRules}

{{a->0},{x->0}}

Solve[a x == 0, x]

{{x->0}}

Andrzej Kozlowski


On 24 Jul 2011, at 12:38, Fred Simons wrote:

> In[8]:= 
> expr=(a==1001&&b==1001000)||(a==1002&&b==501000)||(a==1004&&b==251000);
> {a,b}/.{ToRules[expr]}
> 
> Out[9]= {{1001,1001000},{1002,501000},{1004,251000}}
> 
> If expr is indeed the result of a Reduce command, I think it is very 
> likely that instead of Reduce you could have used Solve (it has quite a 
> lot of new features) in order to obtain the rules immediately.
> 
> Kind regards,
> 
> Fred Simons
> Eindhoven University of Technology
> 
> Op 24-7-2011 1:54, Harvey P. Dale schreef:
>> 	What is the easiest and most efficient way to transform the
>> output from Reduce into a list of values?  For example, if Reduce
>> produces
>> (a==1001&&b==1001000)||(a==1002&&b==501000)||(a==1004=
>> &&b==251000) and I
>> want {{1001,1001000},{1002,501000},{1004,251000}}, what is the best way
>> to go from the former to the latter?
>> 
>> 	Harvey
>> 
>> Harvey P. Dale
>> University Professor of Philanthropy and the Law
>> Director, National Center on Philanthropy and the Law
>> 139 MacDougal Street
>> New York, N.Y. 10012-1076
>> Tel: 212-998-6161
>> Fax: 212-995-3149
> 
> 




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From: Yves Klett <yves.klett at googlemail.com>
Subject: [mg120471] Re: Producing an image that only contains its interior and boundary but, no exterior.
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Murray,

not sure anymore if someone from WRI helped me out (John Fultz comes to
mind) at some point or if I found it somewhere else. And yes, I do not
think it is documented properly for whatever reason.

Googling for "mathematica shrinkwrap" turns up several hits around 2008.

Regards,
Yves

Am 24.07.2011 02:26, schrieb Murray Eisenberg:
> How did you discover that option for Show?
> 
> I find no entry for "ShrinkWrap" in the Documentation Center, see no 
> mention of that option value on the page for Show. On the Graphics3D ref 
> page, the option Method is listed in the More Information section, but 
> that Method option is not listed in the Options section of Examples.
> 
> And of course the ref page for Method is essentially useless in such a 
> case. It's long been a documentation lacuna that there are no "inverse 
> references" for such options as Method that apply to multiple functions 
> -- from the general option such as Method back to all the functions for 
> which it's an option, along with some information as to what settings 
> are allowed for each such function (at least in cases where within the 
> target function page itself the settings are not all listed).
> 
> 
> On 7/22/11 7:47 PM, Yves Klett wrote:
>> g = Graphics3D[Cuboid[], SphericalRegion ->  True]
>>
>> Show[g, Method ->  {"ShrinkWrap" ->  True}]
> 



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From: paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Subject: [mg120485] Re: TransformedDistribution -- odd problem
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A little more experimenting shows that the TransformedDistribution
function will also not provide the mean of k+F where k is a constant
and F has an F distribution -- i.e.,
 Mean[TransformedDistribution[k*F ,  F \[Distributed]
FRatioDistribution[v, v]]]

If I changce the distribution of F to NormalDistribution or
ChiSquareDistribution, I can get a mean for k*F or k+F. So the problem
only occurs when I define a simple function of an F variable using the
TransformedDistribution function.
This all strikes me as very strange, and I'd be curious to know if
others can reproduce my results. If you can't reproduce my results,
I'd be interested in theories about why my results differ from yours.
E.g., is there a setting I should change in the software?

I am using version 8.0.0.0 and looking to upgrade to 8.0.1, if that
makes a difference.

Many thanks for any pointers.

On Jul 24, 2:22 am, paulvonhippel at yahoo <paulvonhip... at yahoo.com>
wrote:
> I'm having a very strange problem with TransformedDistribution, where
> I can calculate the mean of an F distribution but I cannot calculate
> the mean of a constant multiplied by an F distribution. That is, if I
> type
>
>  Mean[TransformedDistribution[F, F \[Distributed]
> FRatioDistribution[v, v]]]
>
> Mathematica gives me an answer. But if I type
>
>  Mean[TransformedDistribution[k*F ,  F \[Distributed]
> FRatioDistribution[v, v]]]
>
> Mathematica just echoes the input. I swear I got an answer for the
> second expression earlier today. What am I doing wrong?




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From: Gabriel Landi <gtlandi at gmail.com>
Subject: [mg120488] Re: NDSolve in matrix form
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Hi Glenn.

I solve a similar system in the following way:

Say it is defined as

X'(t) = a X(t) + b

Then once i know a,b and my initial conditions ic, i do:

vars = Table[f[i][t],{i,n}];
varsD = D[#,t	] &/@vars;
eqns = Flatten@{Thread[varsD == a.vars + b],
    Thread[(vars /. t -> 0) == ic]};
sol = NDSolve[eqns,vars,{t,0,tf}]

Good luck,

Gabriel

On Jul 24, 2011, at 4:19 AM, Glenn Carlson wrote:

> I am trying to numerically solve systems of ODEs using Mathematica 7.
>
> The desired accuracy that I am striving for depends on the number of 
solution functions so I want to structure my solution algorithm to allow 
different numbers of solution functions. Solving the problem as a system 
of ODEs in matrix form seems the logical choice:
>
>     dX/dt = f[X[t]],
>
> where f[X[t]] is a specified function, and X[t] is a vector of 
solution functions {x1[t],x2[t],...}. The initial conditions are 
specified by a vector X[0] = {x1[0],x2[0],...}.
>
> In my problem, the number of solution functions will usually vary from 
3 to 7, but could be any number.
>
> One approach that doesn't work is:
>
>     f := {f[[1]][t], f[[2]][t], f[[3]][t]}
>     fInit := {0, 0, 2}
>     mat := {(-3*f[[1]][t] - f[[2]][t]), 26.5*f[[1]][t] - f[[2]][t] - 
f[[1]][t]*f[[3]][t], f[[1]][t]*f[[2]][t] - f[[3]][t]}
>     solution =  NDSolve[{Derivative[1][f][t] == mat, f[0] == 
fInit}, f, {t, 0, 17}]
>
> I don't know if the problem is in the way f is specified or in my 
expression for NDSolve.
>
> I'd appreciate any assistance.
>
> Thanks and regards,
>
> Glenn
>




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From: Robert Rosenbaum <robertr at math.uh.edu>
Subject: [mg120483] Re: TransformedDistribution, for a sum of M iid variables
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>> Thanks for the neat proof. Sorry I wasn't clear about my goals. I
>> want to include a sum of iid variables in a TransfornedDistribution,
>> and then calculate the Mean and Variance of the distribution.



I know I'm pointing out the obvious here, but the mean and variance of a 
sum of independent variables is the sum of the means and variances.  No 
need to use TransformedDistribution or CumulantGeneratingFunction or 
anything else. 

Perhaps you're trying to provide an illustration to a classroom or 
something along those lines and you want to plot the density of the sum, 
then find its mean and variance.  You can achieve this by convolving the 
individual densities.



On Jul 23, 2011, at 6:53 PM, Bill Rowe wrote:

> On 7/22/11 at 3:42 AM, paulvonhippel at yahoo.com (Paul von Hippel)
> wrote:
>
>> Thanks for the neat proof. Sorry I wasn't clear about my goals. I
>> want to include a sum of iid variables in a TransfornedDistribution,
>> and then calculate the Mean and Variance of the distribution.
>
>> The iid variables need not be normal, and the TrasformedDistribution
>> will contain other variables besides the iid sum.
>
> While it is possible to do this using TransformedDistribution, I
> think it is far easier to to what you want with the various
> generating functions built into Mathematica. Use the exponential
> distribution as an example. For the exponential distribution:
>
> In[2]:= mg =
> CumulantGeneratingFunction[ExponentialDistribution[a], t]
>
> Out[2]= Log[a/(a - t)]
>
> Cumulants sum. So, the cumulant generating function for the sum
> n exponential deviates is simply n times the cumulant generating
> function of the original exponential distribution. And the mean
> is the first derivative of the cumulant generating function
> evaluated at t = 0. So, the mean of the sum of n exponential
> deviates is:
>
> In[3]:= D[n mg, t] /. t -> 0
>
> Out[3]= n/a
>
> And the variance is given by the second derivative. So, the
> desired variance is:
>
> In[4]:= D[n mg, {t, 2}] /. t -> 0
>
> Out[4]= n/a^2
>
> And to check the validity of the above, note:
>
> In[5]:= MomentGeneratingFunction[ExponentialDistribution[a], t]
>
> Out[5]= a/(a - t)
>
> In[8]:= MomentGeneratingFunction[GammaDistribution[n, 1/a], t]
>
> Out[8]= (1 - t/a)^(-n)
>
> That is, the sum of n exponential deviates is distributed as a
> gamma distribution with parameters n, 1/a. So,
>
> In[9]:= Mean[GammaDistribution[n, 1/a]]
>
> Out[9]= n/a
>
> In[10]:= Variance[GammaDistribution[n, 1/a]]
>
> Out[10]= n/a^2
>
> which verifies the earlier result using the cumulant generating
> function. Note, this technique of using generating functions
> does not require the sum to be of identically distributed
> values. I could just as easily find the mean and variance of say
> the sum of an exponential deviate and the sum of a normal
> deviate using this technique.
>
>


Best,
Robert








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From: Paul von Hippel <paulvonhippel at yahoo.com>
Subject: [mg120478] Re: TransformedDistribution, for a sum of M iid variables
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References: <20110721225621.9H9BM.1153349.imail@eastrmwml34> <j0b9p4$jl1$1 at smc.vnet.net> <201107222346.TAA28623 at smc.vnet.net> <4E2D4909.813B.006A.0 at newcastle.edu.au>
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I meant question ii -- i.e., how do I code this in Mathematica. The problem I am trying to solve is complicated and it wouldn't be productive to lay out all its details. But a simpler version would be this: I wish to calculate the mean and variance of the distribution U+avgMF, where U is a chi-square variable and avgMF is the average of M iid F variables. I can solve this problem by hand, but I wish solve this and similar, more complicated problems using the TransformedDistribution function. Many thanks for any pointers.



________________________________
From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
To: mathgroup at smc.vnet.net; paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Sent: Sunday, July 24, 2011 7:44 PM
Subject: [mg120478] Re: TransformedDistribution, for a sum of M iid variables

Hi Paul

I for one am not clear on the exact question you are asking.

If by "Is there a convenient way to do this?" and you mean by "this" "... replace Z2 with the average of M independent variables", do you mean:

(i) how do I calculate the distribution of the average of M independent variables (a technical statistical question, the answer to which can be found an any statistics textbook), or

(ii) how do I code the whole calculation in Mathematica (a Mathematica coding/representation question).

Maybe, as with many questions posted on this group, more background details will help respondents to your question(s) give you better answers. For example, do you have a particular problem in mind (i.e., application), or are you just generally curious (always a great idea) about how to do such and similar things?

(I often think there's a monotonically increasing relation between usefulness of answer and quality of question, if I may put it like that).

Willing to help with either question ...

Cheers

Barrie

>>> On 23/07/2011 at 9:46 am, in message <201107222346.TAA28623 at smc.vnet.net>,
paulvonhippel at yahoo <paulvonhippel at yahoo.com> wrote:
> To clarify what I'm looking for, suppose I have the following
> distribution:
> 
> TransformedDistribution[Z1+Sqrt[U/(n-1)]*Z2,{Z1\
> [Distributed]NormalDistribution[0,1],Z2\
> [Distributed]NormalDistribution[0,1],U\
> [Distributed]ChiSquareDistribution[n-1]}]
> 
> I want to replace Z2 with the average of M independent variables that
> are distributed like Z2.
> Or I might want to replace U with the average of M independent
> variables that are distributed like U.
> 
> Is there a convenient way to do this? Thanks!


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Subject: [mg120493] Re: ExampleData[{"Geometry3D", "Torus"}]
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Thanks, Heike, for the suggestion to use RevolutionPlot3D[]. I didn't even know about that function! So your suggestion is very useful.



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From: Heike Gramberg <heike.gramberg at gmail.com>
Subject: [mg120500] Re: default font
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The default FontSize of a notebook is set by the global option FontSize. 
You can change it by running something like

SetOptions[$FrontEnd, FontSize -> 14]

Heike.

> everytime with new norebook i change font size. How to change it in
> default manner?
>




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From: Gary Wardall <gwardall at gmail.com>
Subject: [mg120495] Re: Simplify[ArcTan[Tan[a] + Tan[b]]]
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On Jul 25, 6:31 am, rych <rych... at gmail.com> wrote:
> Can't Mathematica tell anything about ArcTan[Tan[a] + Tan[b]]? Maybe
> there exists no useful identity for that? By useful I mean the
> expression with functions of the "angles" a and b other than tangents.
> Thanks
> Igor

Igor,

I'm not quite sure what you want Mathematica to do with it. It's not
like ArcTan and Tan are truly inverses.

ArcTan[Tan[W]]  my not be  W  while

Tan[ArcTan[W]] is  W.

Personally I think

ArcTan[Tan[a] + Tan[b]]

is simplified enough. Of course there are many alternate and equal
expressions to it that could be of some use. But, which one?

Gary



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From: JonSilverman <jonsilverman2006 at gmail.com>
Subject: [mg120496] Can't make Input[] work the way I want it to.
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When prompted for an input by Input[], I want to input a list of
numbers. But when i try to enter more than one number I get something
like this:

RowBox[{"45", ",", "65", ",", "56"}]


What is that?

If I added spaces then the space got automatically turned into a
multiplication sign!

How can I enter a list at the input prompt?



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From: Brett Champion <brettc at wolfram.com>
Subject: [mg120506] Re: And now for something completely different
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On Jul 25, 2011, at 6:29 AM, Richard Fateman wrote:

>
> Can you start a paragraph with "And"?
>
> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>
> says, 'No.'
>
>
> Stephen Wolfram often does so, and I find it annoying.
> example
> http://www.wolframscience.com/nksonline/page-42?firstview=1
>

http://www.wolframscience.com/nksonline/page-849c-text?firstview=1

Other links, which show up above Yahoo! Answers in a Google search:

http://wiki.answers.com/Q/Can_you_start_a_sentence_with_the_word_And
=
http://www.dailywritingtips.com/can-you-start-sentences-with-=93and=94-and=
-=93but=94/
http://www.gpuss.co.uk/english_usage/start_sentence_conjunction.htm

Brett=



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From: Jack L Goldberg 1 <jackgold at umich.edu>
Subject: [mg120494] Re: And now for something completely different
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As long as we are talking about poor written English, a particularly  
annoying habit of Wolfram (or his editors) is the use of the word  
"essentially" throughout his tome, "Mathematica". It appears so often  
that it both disturbing and unessential (so to speak) :-)

Jack


Quoting Richard Fateman <fateman at cs.berkeley.edu>:

>
> Can you start a paragraph with "And"?
>
> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>
> says, 'No.'
>
>
> Stephen Wolfram often does so, and I find it annoying.
> example
> http://www.wolframscience.com/nksonline/page-42?firstview=1
>
> This construction seems to have been removed from the online documentation.
>
>
> --RJF
>
>
>
>





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From: "Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com>
Subject: [mg120501] Re: Simplify[ArcTan[Tan[a] + Tan[b]]]
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On Mon, 25 Jul 2011 12:31:40 +0100, rych <rychphd at gmail.com> wrote:

> Can't Mathematica tell anything about ArcTan[Tan[a] + Tan[b]]? Maybe
> there exists no useful identity for that? By useful I mean the
> expression with functions of the "angles" a and b other than tangents.
> Thanks
> Igor

It can certainly tell you something:

In[1] :=
f = Function[{expr},
   100 Count[expr, _Tan, {0, Infinity}] +
   100 Count[expr, _ArcTan, {0, Infinity}] +
   LeafCount[expr]
  ];

In[2] :=
FullSimplify[
   ArcTan[Tan[a] + Tan[b]],
   ComplexityFunction -> f
  ]

Out[2] =
1/2 I (Log[1-I Sec[a] Sec[b] Sin[a+b]]-Log[1+I Sec[a] Sec[b] Sin[a+b]])

However, this result is arguably neither interesting nor useful.

Best,

O. R.



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From: "Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com>
Subject: [mg120490] Re: And now for something completely different
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On Mon, 25 Jul 2011 12:33:12 +0100, Richard Fateman  
<fateman at cs.berkeley.edu> wrote:

> Can you start a paragraph with "And"?
>
> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>
> says, 'No.'
>
> Stephen Wolfram often does so, and I find it annoying.
> example
> http://www.wolframscience.com/nksonline/page-42?firstview=1
>
> This construction seems to have been removed from the online  
> documentation.
>
> --RJF

A curious choice of topic for this mailing list, if I may say so.  
Nevertheless, I do agree with you that Stephen Wolfram seems to have a  
rather awkward writing style.

However, in my humble opinion, one should be wary of using Yahoo! Answers  
as a source, as it is populated almost entirely by trolls. Of course, I am  
sure you would not wish to be associated with such a thing. :-)

Best,

O. R.



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From: ccarter at mit.edu
Subject: [mg120502] Re: And now for something completely different
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And, how is this relevant to the group topic?

And, for what it is worth:

 	HAMLET
ACT III
SCENE I	A room in the castle.
 	[Enter KING CLAUDIUS, QUEEN GERTRUDE, POLONIUS,
 	OPHELIA, ROSENCRANTZ, and GUILDENSTERN]

KING CLAUDIUS	And can you, by no drift of circumstance,
 	Get from him why he puts on this confusion,
 	Grating so harshly all his days of quiet
 	With turbulent and dangerous lunacy?



On Mon, 25 Jul 2011, Richard Fateman wrote:

> Date: Mon, 25 Jul 2011 07:29:07 -0400 (EDT)
> From: Richard Fateman <fateman at cs.berkeley.edu>
> To: mathgroup at smc.vnet.net
> Subject: And now for something completely different
> 
>
> Can you start a paragraph with "And"?
>
> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>
> says, 'No.'
>
>
> Stephen Wolfram often does so, and I find it annoying.
> example
> http://www.wolframscience.com/nksonline/page-42?firstview=1
>
> This construction seems to have been removed from the online documentation.
>
>
> --RJF
>
>



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From: Heike Gramberg <heike.gramberg at gmail.com>
Subject: [mg120489] Re: Grid Divider Style With Background Colors
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You could do something like

With[{gap = 1, elemWidth=3, h = 8, w = 7},
 mat = RandomInteger[10, {h, w}];
 colours = Table[Blend[{Lighter[Blue], Pink}, j/w], {i, h}, {j, w}];
 Framed[Grid[
   MapThread[
    Framed[Pane[#1, Scaled[1]], ImageMargins -> gap,
      Background -> #2] &, {mat, colours}, 2], Spacings -> {0, 0},
   ItemSize -> elemWidth], FrameMargins -> gap]]

elemWidth is the width of the elements in the Grid and gap the distance between the lines.

Heike

On 25 Jul 2011, at 12:28, Don wrote:

> Mathematica  permits divider lines between rows and columns in a table
> (grid)  such
> as this example from the documentation:
>
> Grid[Table[x, {4}, {11}], Dividers -> {{{{True, False}}, -1 -> True},
> False}]
>
> But, the dividers are single lines only.
>
> Sometimes  double line dividers are desired
> as shown in this table on the Web:
>
> http://www.assetcorrelation.com/user/correlations/90
>
> Notice that the background color for an item  is only within
> the inside line of a double line divider.
>
> Is there any way to produce (a) this  kind of divider (double line)
> and (b) with background colors for specific items that don't cross
> the inside line of the double line divider?
>
> Thank you in advance.
>
> Don
>




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From: Murray Eisenberg <murray at math.umass.edu>
Subject: [mg120507] Re: And now for something completely different
Reply-To: murray at math.umass.edu
References: <201107251129.HAA25514 at smc.vnet.net>
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Disallowing conjunctions such as "and" and "but" to introduce sentences 
is one of those prissy prohibitions promulgated by grammar police, 
especially in school.

But it's quite common and acceptable today to violate that, at least in 
informal writing. And certainly in speaking. Yet the usage is long 
established in English: check the OED.

And indeed Stephen Wolfram often does begin sentences with "and". Which 
you find annoying. But you seem to find so much that Stephen Wolfram 
does annoying, so why should this be any different?

On 7/25/11 7:29 AM, Richard Fateman wrote:
> Can you start a paragraph with "And"?
>
> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>
> says, 'No.'
>
>
> Stephen Wolfram often does so, and I find it annoying.
> example
> http://www.wolframscience.com/nksonline/page-42?firstview=1
>
> This construction seems to have been removed from the online documentation.
>
>
> --RJF
>

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305



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From: Paul von Hippel <paulvonhippel at yahoo.com>
Subject: [mg120498] Re: TransformedDistribution -- odd problem
Reply-To: Paul von Hippel <paulvonhippel at yahoo.com>
References: <j0gh7s$bd7$1 at smc.vnet.net> <201107251129.HAA25532 at smc.vnet.net> <4E2EA10E.813B.006A.0 at newcastle.edu.au>
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That problem went away, thanks. I upgraded to 8.0.1, and I learned that Mathematica can have trouble evaluating symbols with subscripts.



________________________________
From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
To: mathgroup at smc.vnet.net; paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Sent: Monday, July 25, 2011 8:12 PM
Subject: [mg120498] Re: TransformedDistribution -- odd problem

Hi Paul

Could we see an example of "a more complicated example where Mathematica will give the mean of A and the mean of B, but when I ask for the mean of A+B Mathematica just echoes the input", please?

Barrie

>>> On 25/07/2011 at 9:29 pm, in message <201107251129.HAA25532 at smc.vnet.net>,
paulvonhippel at yahoo <paulvonhippel at yahoo.com> wrote:
> I also have a more complicated example where Mathematica will give the
> mean of A and the mean of B, but when I ask for the mean of A+B
> Mathematica just echoes the input. (Here A and B are complicated
> functions of several random variables.) This strikes me as really odd,
> since of course Mean(A+B)=Mean(A)+Mean(B).
> 
> What could be going on here? Under what circumstances does Mathematica
> echo the input when asked to provide a Mean? Is there a setting I need
> to change? I suspect this is a software issue rather than a
> mathematical issue.


From mathgroup-adm at smc.vnet.net  Tue Jul 26 05:34:52 2011
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From: Gregory Lypny <gregory.lypny at jmsb.concordia.ca>
Subject: [mg120504] Re: Preventing In-line Math Typesetting From Being Scaled Down in Text
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Thank you, John.  Excellent.

Gregory


On Mon, Jul 25, 2011, at 4:39 AM, jfultz at wolfram.com wrote:

> On Fri, 22 Jul 2011 19:46:07 -0400 (EDT), Gregory Lypny wrote:
> > Hi everyone,
> >
> > When I include fractions as inline math typesetting, they are scaled down
> > to fit the effective line height of the cell. How can I prevent this or,
> > I guess, make the line height automatically expand to accommodate the
> > math? If my regular text in the cell is 12-point times, I'd like all
> > math variables that are not subscripts or superscripts to be 12-point as
> > well.
> >
> > Incidentally, other big typeset objects like matrices are not scaled
> > down, or at least they down't appear to be.
> >
> > Sincerely,
> >
> > Gregory
> 
> If you look in Core.nb, you'll find a style called "InlineCell". This style is 
> automatically applied to all inline cells everywhere. One of the options it has 
> set is:
> 
> ScriptLevel->1
> 
> This is what's causing the behavior you're seeing. You can override this with a 
> custom stylesheet. For example, in a given notebook, you can make a private 
> override by doing Format->Edit Stylesheet..., and pasting and interpreting the 
> following cell expression at the end of the resulting stylesheet notebook:
> 
> 
> Cell[StyleData["InlineCell"], ScriptLevel->0]
> 
> 
> Sincerely,
> 
> John Fultz
> jfultz at wolfram.com
> User Interface Group
> Wolfram Research, Inc.
> 
> and pasting and interpreting the following cell expression at the end of the resulting stylesheet notebook: Cell[StyleData["InlineCell"], ScriptLevel->0] Sincerely, John Fultzjfultz at wolfram.com User Interface Group Wolfram Research, Inc.


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From: Glenn Carlson <g.crlsn at gmail.com>
Subject: [mg120491] Re: NDSolve in matrix form
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Here's a more complicated example.

The system is of the form A[i]'[t] + a[[i]] A[i][t] = Sum[G[t][[i]][[j]] A[j][t], {j,1,n}], where A is an array of solution functions with i = 1 to n. G[t][[]][[]] and a[[]] are specified arrays.

Note how I can work with different numbers of solution functions just by assigning a different value to n and defining the inputs G, a, AInit, and tf.

I'm still trying to make sense out why the single and double brackets have to be specified as they are, and why sometimes the index is after t and sometimes before t.

Thanks to all.

Glenn

-----
ClearAll;
n := 4
AInit := {1, 0, 0, 0};
G[t_] = {{0, t, 1, 0}, {0, 0, t, 1}, {1, 0, 0, t}, {t, 1, 0, 0}};
a = {5, 3, 2, 1};
tf := 5

(* Create a list of the solution functions excl. the argument t. *)
Asol = Table[A[i], {i, 1, n}];
(* Create a list of ODEs and ICs for NDSolve. *)
Table[Asol[[i]][t], {i, 1, n}];

eqns = Table[
   Asol[[i]]'[t] + a[[i]] Asol[[i]][t] == 
    Sum[G[t][[i]][[j]] Asol[[j]][t], {j, 1, n}], {i, 1, n}];
ic := Table[Asol[[i]][0] == AInit[[i]], {i, 1, n}];
sys = Join[eqns, ic];

Table[Asol[[i]]'[t] + a[[i]] Asol[[i]][t] == 
   Sum[G[t][[i]][[j]] Asol[[j]][t], {j, 1, n}], {i, 1, 
   n}] // MatrixForm
Table[Asol[[i]][0] == AInit[[i]], {i, 1, n}] // MatrixForm

solution1 = NDSolve[sys, Asol, {t, 0, tf}, MaxSteps -> Infinity]

(* Create a list of solution functions incl. the argument t for ParametricPlot3D. *)
(*
fsolt=Table[A[i][t],{i,1,n}]
ParametricPlot3D[Evaluate[fsolt/.solution1],{t,0,tf},PlotPoints->1000,ColorFunction->(ColorData["Rainbow"][#4]&)]
*)

(* Plot solutions as functions of time. *)
(*
Plot[Evaluate[A[1][t]/.solution1],{t,0,tf},PlotRange->{0,2}]
Plot[Evaluate[A[2][t]/.solution1],{t,0,tf},PlotRange->{0,2}]
Plot[Evaluate[A[3][t]/.solution1],{t,0,tf},PlotRange->{0,2}]
Plot[Evaluate[A[4][t]/.solution1],{t,0,tf},PlotRange->{0,2}]
*)

Asolt = Table[A[i][t], {i, 1, n}];
Asum[t_] := Sum[(A[i][t] /. solution1), {i, 1, n}]
Plot[{Evaluate[Asolt /. solution1], Asum[t]}, {t, 0, tf}, PlotRange -> Automatic]



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From: Jacopo Bertolotti <J.Bertolotti at utwente.nl>
Subject: [mg120503] Roots of a Jacobi polynomial
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Dear MathGroup,
Mathematica implements Jacobi polynomials as JacobiP[n,a,b,x] where n is 
the order of the polynomial. As it can be checked plotting it a Jacobi 
polynomial has n real roots in the interval [-1,1] and it goes rapidly 
to infinity outside this interval (at least when both a and b are >-1).
The problem arise when you try to find the roots of such a polynomial 
for a relatively high value of n. As an example the command 
NSolve[JacobiP[20, -0.5, -0.5, x] == 0, x] correctly returns

{{x -> -0.99702}, {x -> -0.972111}, {x -> -0.92418}, {x -> -0.852398}, 
{x -> -0.760555}, {x -> -0.649375}, {x -> -0.522529}, {x -> -0.382672}, 
{x -> -0.233449}, {x -> -0.0784582}, {x -> 0.0784591}, {x -> 0.233445}, 
{x -> 0.382684}, {x -> 0.522504}, {x -> 0.649423}, {x -> 0.760466}, {x 
-> 0.852539}, {x -> 0.924002}, {x -> 0.972267}, {x -> 0.996958}}

while NSolve[JacobiP[25, -0.5, -0.5, x] == 0, x] gives

{{x -> -1.01869}, {x -> -0.979859 - 0.0479527 I}, {x -> -0.979859 + 
0.0479527 I}, {x -> -0.870962 - 0.070991 I}, {x -> -0.870962 + 0.070991 
I}, {x -> -0.71378 - 0.0505783 I}, {x -> -0.71378 + 0.0505783 I}, {x -> 
-0.571283}, {x -> -0.486396}, {x -> -0.367829}, {x -> -0.248377}, {x -> 
-0.125513}, {x -> -0.0000434329}, {x -> 0.125442}, {x -> 0.2489}, {x -> 
0.365644}, {x -> 0.496977}, {x -> 0.555743}, {x -> 0.717741- 0.0573399 
I}, {x -> 0.717741+ 0.0573399 I}, {x -> 0.87423- 0.0652273 I}, {x -> 
0.87423+ 0.0652273 I}, {x -> 0.977876- 0.0422422 I}, {x -> 0.977876+ 
0.0422422 I}, {x -> 1.01494}}

i.e. both complex roots and roots outside the [-1,1] interval. 
Substituting any of these values back into the polynomial easily show 
that these values are not roots at all. Also notice that using the 
command Root (e.g. Sort@Table[Root[JacobiP[25, -0.5, -0.5, x], i], {i, 
1, 25}]) gives different but still wrong results.
On a related note: NIntegrate sometimes gives wrong results when 
integrating function of the form (1-x)^a (1+x)^b f[x] and I feel this 
might related to the Gauss-Jacobi quadrature failing to retrieve the 
correct roots of a Jacobi polynomial.

Do anyone have a solution for that?

Thank you

Jacopo



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From: Paul von Hippel <paulvonhippel at yahoo.com>
Subject: [mg120497] Re: TransformedDistribution -- odd problem
Reply-To: Paul von Hippel <paulvonhippel at yahoo.com>
References: <j0gh7s$bd7$1 at smc.vnet.net> <201107251129.HAA25540 at smc.vnet.net> <4E2EA04E.813B.006A.0 at newcastle.edu.au>
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Thanks -- that fixes it!

Bonus question: if we don't specify v2>2, why doesn't Mathematica return a two part solution:
 k v2 / (v2-2) v2>2

 Indeterminate True

That's what it does if we request the mean of F -- why doesn't it do the same if we request the mean of k*F.




________________________________
From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
To: mathgroup at smc.vnet.net; paulvonhippel at yahoo <paulvonhippel at yahoo.com>
Sent: Monday, July 25, 2011 8:09 PM
Subject: [mg120497] Re: TransformedDistribution -- odd problem

Hi Paul

There are conditions on the v1 and v2, the degrees of freedom of the F distribution:

Assuming[v2 > 2, 
Mean[TransformedDistribution[F , 
  F \[Distributed] FRatioDistribution[v1, v2]]]]

{Assuming[v2 > 2, 
 Mean[TransformedDistribution[k*F , 
  F \[Distributed] FRatioDistribution[v1, v2]]]], k*v2/(-2 + v2)}

{Assuming[v2 > 2, 
  Mean[TransformedDistribution[k + F , 
   F \[Distributed] FRatioDistribution[v1, v2]]]], 
 k + v2/(-2 + v2)} // FullSimplify

which shows precisely what you expect for k*F and k+F.

Cheers

Barrie

>>> On 25/07/2011 at 9:29 pm, in message <201107251129.HAA25540 at smc.vnet.net>,
paulvonhippel at yahoo <paulvonhippel at yahoo.com> wrote:
> A little more experimenting shows that the TransformedDistribution
> function will also not provide the mean of k+F where k is a constant
> and F has an F distribution -- i.e.,
>  Mean[TransformedDistribution[k*F ,  F \[Distributed]
> FRatioDistribution[v, v]]]
> 
> If I changce the distribution of F to NormalDistribution or
> ChiSquareDistribution, I can get a mean for k*F or k+F. So the problem
> only occurs when I define a simple function of an F variable using the
> TransformedDistribution function.
> This all strikes me as very strange, and I'd be curious to know if
> others can reproduce my results. If you can't reproduce my results,
> I'd be interested in theories about why my results differ from yours.
> E.g., is there a setting I should change in the software?
> 
> I am using version 8.0.0.0 and looking to upgrade to 8.0.1, if that
> makes a difference.
> 
> Many thanks for any pointers.
> 
> On Jul 24, 2:22 am, paulvonhippel at yahoo <paulvonhip... at yahoo.com>
> wrote:
>> I'm having a very strange problem with TransformedDistribution, where
>> I can calculate the mean of an F distribution but I cannot calculate
>> the mean of a constant multiplied by an F distribution. That is, if I
>> type
>>
>>  Mean[TransformedDistribution[F, F \[Distributed]
>> FRatioDistribution[v, v]]]
>>
>> Mathematica gives me an answer. But if I type
>>
>>  Mean[TransformedDistribution[k*F ,  F \[Distributed]
>> FRatioDistribution[v, v]]]
>>
>> Mathematica just echoes the input. I swear I got an answer for the
>> second expression earlier today. What am I doing wrong?


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From: Richard Fateman <fateman at eecs.berkeley.edu>
Subject: [mg120492] Re: And now for something completely different
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On 7/25/2011 7:36 AM, Brett Champion wrote:
> On Jul 25, 2011, at 6:29 AM, Richard Fateman wrote:
>
>> Can you start a paragraph with "And"?
>>
>> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>>
>> says, 'No.'
>>
>>
>> Stephen Wolfram often does so, and I find it annoying.
>> example
>> http://www.wolframscience.com/nksonline/page-42?firstview=1
>>
> http://www.wolframscience.com/nksonline/page-849c-text?firstview=1

Thanks for the pointer!  I now know that Wolfram knows that it will be 
annoying to some people.   I find it interesting that on this particular 
page, striking out each of the "And"s at the beginning of the sentences 
has no negative consequences, at least as I read it.

There is a reason, one that some may dismiss as pedantry, for removing 
the "And"s and fixing some of the
other oddities of Wolfram's prose.  That is,  formal English discourse 
has sentences with "subject" and "predicate". They
adhere to rules of syntax and can be formally diagrammed.  Humans and 
(especially of interest to some people)
computer programs can sometimes read and parse sentences, assign 
semantic interpretations, etc. Sentence fragments
and "extra" words are not helpful.   By the way, a sentence beginning 
with "But" which is not part of a subordinate clause
is also a no-no traditionally...

  Using the word "And" at the beginning of a sentence to indicate that 
this sentence is related to the one before it should not be required 
since adjacent sentences in the same paragraph should be related.  
Adjacent paragraphs in the same chapter should be related as well, so they
need not begin with "And".

The occasional relaxation of rules as some kind of device or flourish 
may be effective.  I find Wolfram's use of "And" to be annoying.
But how did he know he would annoy me?

RJF

>
> Other links, which show up above Yahoo! Answers in a Google search:
>
> http://wiki.answers.com/Q/Can_you_start_a_sentence_with_the_word_And
> http://www.dailywritingtips.com/can-you-start-sentences-with-?and?-and-?but?/
> http://www.gpuss.co.uk/english_usage/start_sentence_conjunction.htm
>
> Brett




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Subject: [mg120505] Re: And now for something completely different
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On 7/25/2011 6:12 AM, ccarter at mit.edu wrote:
> And, how is this relevant to the group topic?
>
> And, for what it is worth:
>
>     HAMLET
> ACT III
> SCENE I    A room in the castle.
>     [Enter KING CLAUDIUS, QUEEN GERTRUDE, POLONIUS,
>     OPHELIA, ROSENCRANTZ, and GUILDENSTERN]
>
> KING CLAUDIUS    And can you, by no drift of circumstance,
>     Get from him why he puts on this confusion,
>     Grating so harshly all his days of quiet
>     With turbulent and dangerous lunacy?

One can find arguments on both sides, e.g.
http://www.dailywritingtips.com/can-you-start-sentences-with-%E2%80%9Cand%E2%80%9D-and-%E2%80%9Cbut%E2%80%9D/

For example, I find the quote from Shakespeare to be non- annoying, 
perhaps because as
"spoken" English it is informal, it is not repetitive, and in verse. 
Presumably Shakespeare
had some motive based on cadence.  On the other hand, your first 
sentence beginning
with "And" is grating, though understandable as humor/snark..
The second one actually operates as a conjunction, sort of.  Email 
probably falls into the
category of "informal", so some syntax might be more relaxed.

I have noticed many of my colleagues begin all their initial sentences 
with "So, .."  perhaps as a
way of clearing their throats and seeing if their voices still work.  A 
reference claiming that this
originated in Microsoft, is, in my opinion, bogus:

http://boingboing.net/2010/06/17/origins-of-using-so.html

As for why this is relevant:  I was reading some documentation for 
Mathematica.
And it had a link to Wolfram's A New Kind of Science. :)
RJF

>
>
> On Mon, 25 Jul 2011, Richard Fateman wrote:
>
>> Date: Mon, 25 Jul 2011 07:29:07 -0400 (EDT)
>> From: Richard Fateman <fateman at cs.berkeley.edu>
>> To: mathgroup at smc.vnet.net
>> Subject: And now for something completely different
>>
>>
>> Can you start a paragraph with "And"?
>>
>> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>>
>> says, 'No.'
>>
>>
>> Stephen Wolfram often does so, and I find it annoying.
>> example
>> http://www.wolframscience.com/nksonline/page-42?firstview=1
>>
>> This construction seems to have been removed from the online 
>> documentation.
>>
>>
>> --RJF
>>
>>




From mathgroup-adm at smc.vnet.net  Tue Jul 26 05:42:59 2011
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From: DANA DELOUIS <dana01 at me.com>
Subject: [mg120499] Re: FinancialData still broken
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Hi.  Here's some added thoughts on the issue you bring up.
I added some stocks just in case the issue is related to mutual funds.

stocks = {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX", "JMSCX=94, "JNGIX=94,
       "JNGLX", "JNMCX=94, "JNOSX", "JNSGX", "JNSTX", "PTTAX", "RGACX=94,
       "STRFX=94, "AAPL", "IBM", "INTL", "VZ", "GE"};

I've noticed that if I ask for price about 100 times, then about 80% of the time I get a correct value,
and 20% an incorrect value (n*^8).
Here, a value of 8.91 was returned 77 times, or 77% (out of 100 tries)

f[n_]:=Table[FinancialData[stocks[[n]],"Price"],{100}]//Tally//Sort

f[1]
{{8.91,77},{5.032406*10^8,23}}

This took a little while, but I did it 100 times on each stock.

m=Table[f[j],{j,stocks//Length}];
m//TableForm

Here's a count of the successful returns:

good=m[[All,1,-1]]
{76,74,77,77,87,77,90,79,81,82,76,79,79,77,85,77,80,78,76,81}

Here's the Mean and SD.
As you can see, there's only about a 80% success rate, with a small Stand Deviation.

{Mean[good],StandardDeviation[good]}//N
{79.4, 4.031455}

With such a consistent failure rate, it sounds like a communication problem between servers

Out of curiosity, here are the different values that were high.
I don't see any pattern to these nonsense values.

m[[All,2,1]]//Union

5.031051*10^8
5.031396*10^8
5.031565*10^8
5.031669*10^8
5.031676*10^8
5.031677*10^8
5.031678*10^8
5.032022*10^8
5.032406*10^8
5.055450*10^8
5.055453*10^8
5.055455*10^8
5.055456*10^8
5.055457*10^8

Thanks for bringing this up.

= = = = = = = = = =
Dana DeLouis
$Version
8.0 for Mac OS X x86 (64-bit) (November 6, 2010)



On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
Erroneous prices are randomly returned:

FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
"JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
"PTTAX", "RGACX", "STRFX"}

{8.81, 72.4, 14., 10.64, 46.49, 12.48,
5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
5.03241*10^8, 11.04, 31.44, 32.47}

FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
"JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
"PTTAX", "RGACX", "STRFX"}

{5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}

Bobby

--
DrMajor... at yahoo.com





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Hi,

how do I delete a mistakenly added folder from the $Path variable?

Thanks,
Max



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From: Jay Bee <jiri.bocan at gmail.com>
Subject: [mg120508] Re: Simplify[ArcTan[Tan[a] + Tan[b]]]
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Since
tg(a) + tg(b) = sin(a+b)/[cos(a)*cos(b)] = tg(a+b)*[1-tg(a)*tg(b)] =
[tg(a)*tg(b)]*[cotg(a) + cotg(b)]
maybe Mathematica is right...



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From: Leo Ditolaghi <leoditolaghi at gmail.com>
Subject: [mg120512] Re: And now for something completely different
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And really, gentlemen.

;-)

Leo


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Subject: [mg120522] Re: work the way I want it to.
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Input it as a List, i.e., include the List brackets:  {1,2,3,4,5}


Bob Hanlon

---- JonSilverman <jonsilverman2006 at gmail.com> wrote: 

=============
When prompted for an input by Input[], I want to input a list of
numbers. But when i try to enter more than one number I get something
like this:

RowBox[{"45", ",", "65", ",", "56"}]


What is that?

If I added spaces then the space got automatically turned into a
multiplication sign!

How can I enter a list at the input prompt?





From mathgroup-adm at smc.vnet.net  Wed Jul 27 04:21:33 2011
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Hi,

I am often working in two kernels with Mathematica Programs coming from different directories also from different drives. After working some time, I very 
often don't remember the origin of a particular open program.

On opening a notebook the name of the program always appears on top of 
the notebook, but not the origin. In my latest installation of Mathematica (8.01 on Win 7, 64 bit) I am no longer able to find the origin of an open program.

Please tell, how to recover this feature.


Thanks   E.F.



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From: Simon <simonjtyler at gmail.com>
Subject: [mg120524] Re: Can't make Input[] work the way I want it to.
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Put brackets {} around the input.

1,2,3 is not a good Mathematica expression, so Input can only return the Box expression. You could do some simple syntax checking to find when an expression like "1,2,3" (x=RowBox[{"1", ",", "2", ",", "3"}]) has been entered and then fix it using, e.g., ToExpression@RowBox[{"{", x, "}"}]

Or you could prompt for the correct form using something like
x = Input["Input a list", {Placeholder[Subscript["e", 1]], Placeholder[Subscript["e", 2]], Placeholder[\[Ellipsis]]}]



From mathgroup-adm at smc.vnet.net  Wed Jul 27 04:25:44 2011
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From: roby <roby.nowak at gmail.com>
Subject: [mg120520] SystemDialogInput["RecordSound"] automatisation
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Hi all.

Is it possible to automate the call to
SystemDialogInput["RecordSound"] (Windows XP SP3) ?

Normaly when SystemDialogInput["RecordSound"] is evaluated a dialog
pops up and the user must interactivly select a recording device as
well as the record format. After the selcetion has been done the user
must click the recod button and finaly the stop button and the ok
button to obtain a recorded sample.

I would like to select the device and format by passing them as
paramters to SystemDialogInput["RecordSound"] and then start the
recording for a specifyed time interval.

Any hints ?

Regards Robert



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Subject: [mg120526] Why won't this sum evaluate?
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In[120]:= $Version
Out[120]= "7.0 for Mac OS X x86 (32-bit) (January 30, 2009)"

In[122]:= Sum[c^n/(1 + c^(2*n)), {n, 1, Infinity}]
Out[122]= -(1/4) + 1/4 EllipticTheta[3, 0, c]^2

In[123]:= Sum[c^n/(1 + c^(2*n)), {n, 0, Infinity}]
Out[123]= (won't simplify)

The only thing different in the two sums is that the second sum is from 0 to Infinity rather than 1 to Infinity. Clearly, the n=zero term is 1/2. 

I have tried various Regularizations and Methods, (not exhaustively) but none seem to work on either of the sums, much less the last.

A side problem - Is there a way to determine what Regularization and Method were used when none were specified?

Thanks



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From: Yves Klett <yves.klett at googlemail.com>
Subject: [mg120527] Re: And now for something completely different
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Funny enough that an author who is clearly concerned about too many
"help vampires" should post messages that are clearly off-topic.

Nevertheless, he do does so, and I find it annoying.

Yves

Am 25.07.2011 13:33, schrieb Richard Fateman:
> Can you start a paragraph with "And"?
> 
> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
> 
> says, 'No.'
> 
> 
> Stephen Wolfram often does so, and I find it annoying.
> example
> http://www.wolframscience.com/nksonline/page-42?firstview=1
> 
> This construction seems to have been removed from the online documentation.
> 
> 
> --RJF
> 



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From: "Christoph Lhotka" <christoph.lhotka at univie.ac.at>
Subject: [mg120531] Re: work the way I want it to.
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hello,

if you enter 45,65,56 into the input dialog it is converted to a rowbox,
if you enter it as 45 65 56 it is assumed to be the expression 45*65*56,
finally if you enter it as {45,65,56} it will be seen as a list of numbers.

think of typing it directly into your notebook...it is the usual notation

best,

christoph

On 26/07/2011 13:06, JonSilverman wrote:
> When prompted for an input by Input[], I want to input a list of
> numbers. But when i try to enter more than one number I get something
> like this:
>
> RowBox[{"45", ",", "65", ",", "56"}]
>
>
> What is that?
>
> If I added spaces then the space got automatically turned into a
> multiplication sign!
>
> How can I enter a list at the input prompt?





From mathgroup-adm at smc.vnet.net  Wed Jul 27 04:34:01 2011
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From: "Ted Ersek" <ersekt at md.metrocast.net>
Subject: [mg120511] Genomes Mathematica knows about and how to use them?
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I know someone who is a researcher in biochemistry and I wanted to show them
how Mathematica could be useful in their research.  I read somewhere that
besides the human genome,  Mathematica has the genome of a certain species
of mouse and fruit fly in its  curated data.  Then I figured Mathematica
might also have the genomes of the bacteria they research in its curated
data.

 

How does one find out what genomes are included in curated data that comes
with Mathematica?  Once you know a certain genome besides the human genome
is available, how do you access it?

 

If one is going to Import a genome from a source independent of Wolfram
Research, what are likely file formats one would Import?

 

Thanks,

   Ted Ersek


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From: Gilmar Rodriguez-pierluissi <peacenova at yahoo.com>
Subject: [mg120514] Fusing two 3-D surfaces
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How can I "fuse" the following two 3-D surfaces,
as a single 3-D surface? I need to join them where
the downward funnel of plt1 meets the upward funnel of plt2;
as a single 3-D surface.
 
plt1= Plot3D[-1/(x^2 + y^2 + 0.5), {x, -2, 2}, {y, -5, 5},
   PlotStyle -> Directive[Orange, Specularity[White, 40]],
   Boxed -> False, Axes -> False, Mesh -> False, ImageSize -> 500,
   ViewPoint -> {-22.197, 0.387, 11.0}]
 
plt2 = ParametricPlot3D[{E^(v/(2 Sqrt[3])) Sqrt[u]
     Cos[1/2 (v - Sqrt[3] Log[u])],
   E^(v/(2 Sqrt[3])) Sqrt[u] Sin[1/2 (v - Sqrt[3] Log[u])],
   E^(-(v/Sqrt[3]))/u}, {u, .1, 5}, {v, -3, 3},
  PlotStyle -> Directive[Orange, Specularity[White, 40]],
  Boxed -> False, Axes -> False, Mesh -> False ,
  ViewPoint -> {-22.197, 0.387, 11.0}]
 
Thank you for your help!
 
Gilmar Rodriguez-Pierluissi



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From: Simon <simonjtyler at gmail.com>
Subject: [mg120525] Re: And now for something completely different
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As other people have already mentioned; in informal or conversational writing, starting a sentence with "And" is fine. That said, the writing style of NKS does not annoy me, because I never got around to reading it!

This thread does remind me, however, of the infamous "Lord" Timothy Dexter.
To quote from: http://en.wikipedia.org/wiki/Timothy_Dexter
"""
At the age of 50 he wrote a book about himself - A Pickle for the Knowing Ones or Plain Truth in a Homespun Dress. He wrote about himself and complained about politicians, clergy and his wife. The book contained 8,847 words and 33,864 letters, but no punctuation, and capital letters were seemingly random. At first he handed his book out for free, but it became popular and was re-printed in eight editions. In the second edition Dexter added an extra page which consisted of 13 lines of punctuation marks. Dexter instructed readers to "peper and solt it as they plese".
""""

Simon



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From: JUN <noeckel at gmail.com>
Subject: [mg120510] Re: default font
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I have the same issue on a computer with a high-resolution display -
everything (not just text but also graphics) looks too small to me. So
what I did is to change the setting in

Preferences > Advanced > "Open Option Inspector" > Notebook Options >
Display Options > Magnification

I set the Magnification factor to 1.5 there. This factor affects the
display of a Notebook even when its individual Magnification setting
(as it appears on the status line) shows 100%; it's a factor that
multiplies the size of displayed graphics as well as font size, so the
relative proportions of the two don't change when you use the Notebook
on different computers (with different display resolutions).

Jens

On Jul 26, 4:11 am, Heike Gramberg <heike.gramb... at gmail.com> wrote:
> The default FontSize of a notebook is set by the global option FontSize.
> You can change it by running something like
>
> SetOptions[$FrontEnd, FontSize -> 14]
>
> Heike.
>
>
>
>
>
>
>
> > everytime with new norebook i change font size. How to change it in
> > default manner?




From mathgroup-adm at smc.vnet.net  Wed Jul 27 04:41:41 2011
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From: Eli Fenichel <Eli.Fenichel at asu.edu>
Subject: [mg120523] getting ride of 0.i
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I am trying to use FindMinimum to minimize the square of a function over many variables (i.e., parameters).  The function itself has the potential to generate imaginary numbers and involves numerical integration (the parameter values that minimize the function are real).  For clarification the function in evaluated at multiple points so it can be written as vector. By squaring the vector the numerical values of the elements are always real with no imaginary parts.  However, Mathematica often writes x + 0.i, where x is a some numerical value, for some of the elements.

Typically, this can be ignored, the Chop command can be used, or it simply does not cause problems.  However, I keep getting an error: [cid:image003.png at 01CC4BBC.E87A0F30]

NIntegrate::nlim: "t = Y[1.] is not a valid limit of integration."

Y is an array with the parameters to be minimized.

However, if evaluate the objective function to be minimized using replacement rules I get
x + 0.i.

Is there a way for me tell Mathematica to always treat 0.i as 0 and drop it?  It seems to be causing problems in the FindMinimum call.

Thanks,
Eli


From mathgroup-adm at smc.vnet.net  Wed Jul 27 04:43:36 2011
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From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
Subject: [mg120529] Re: And now for something completely different
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Or, perhaps this is even more appropriate ;-) :

http://kingjbible.com/genesis/1.htm


Andrzej Kozlowski




On 26 Jul 2011, at 13:07, ccarter at mit.edu wrote:

> And, how is this relevant to the group topic?
>
> And, for what it is worth:
>
> 	HAMLET
> ACT III
> SCENE I	A room in the castle.
> 	[Enter KING CLAUDIUS, QUEEN GERTRUDE, POLONIUS,
> 	OPHELIA, ROSENCRANTZ, and GUILDENSTERN]
>
> KING CLAUDIUS	And can you, by no drift of circumstance,
> 	Get from him why he puts on this confusion,
> 	Grating so harshly all his days of quiet
> 	With turbulent and dangerous lunacy?
>
>
>
> On Mon, 25 Jul 2011, Richard Fateman wrote:
>
>> Date: Mon, 25 Jul 2011 07:29:07 -0400 (EDT)
>> From: Richard Fateman <fateman at cs.berkeley.edu>
>> To: mathgroup at smc.vnet.net
>> Subject: And now for something completely different
>>
>>
>> Can you start a paragraph with "And"?
>>
>> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
>>
>> says, 'No.'
>>
>>
>> Stephen Wolfram often does so, and I find it annoying.
>> example
>> http://www.wolframscience.com/nksonline/page-42?firstview=1
>>
>> This construction seems to have been removed from the online =
documentation.
>>
>>
>> --RJF
>>
>>
>




From mathgroup-adm at smc.vnet.net  Wed Jul 27 04:45:56 2011
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From: Ted Sariyski <tsariysk at craft-tech.com>
Subject: [mg120528] Re: how to ListPlot3D large data sets
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Hi,

In one attempt MaxPlotPoints did not help, but I was not persistent 
because I have to resolve other issues before I return to ListPlot3D.  
One issue is that I cannot pack data. On the big data set I verify that 
it is numeric array and still  I cannot get it packed. Here is an example:

mydata={{0.00100,10.,0.},{0.00100,10.,0.00100},{0.00100,10.,0.00200},{0.00100,10.,0.00300},{0.00100,10.,0.00400}};

MatrixQ[mydata,NumericQ]
True

myPackedData=ToPackedArray[mydata,Real];

MatrixQ[myPackedData, NumericQ]
True

PackedArrayQ[myPackedData]
False

What I am doing wrong?
Thanks,
--Ted

On 7/25/2011 7:30 AM, Oliver Ruebenkoenig wrote:
> On Sat, 23 Jul 2011, Ted Sariyski wrote:
>
>> Hi,
>> I guess I am pushing the limits of ListPlot3D. I try to ListPlot3d a data
>> set with many millions of records. I was not able to get an image from the
>> full dataset, it takes forever. If I use e.g. every fifth record, although
>> slow, I get an image. The machine running Mathematica is Windows 7 (64 bit),
>> has 24 GB RAM and there was no swapping. I wonder what is considered as a
>> reasonable data size for ListPlot3D and are there other tools in Mathematica
>> for visualization of large data sets?
>> Thanks in advance,
>> --Ted
>>
>>
>>
> Ted,
>
> is your data packed?
>
> Developer`PackedArrayQ[yourData]
>
> Oliver
>




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On Jul 26, 9:05 pm, Heike Gramberg <heike.gramb... at gmail.com> wrote:
> You could do something like
>
> With[{gap = 1, elemWidth=3, h = 8, w = 7},
>  mat = RandomInteger[10, {h, w}];
>  colours = Table[Blend[{Lighter[Blue], Pink}, j/w], {i, h}, {j, w}];
>  Framed[Grid[
>    MapThread[
>     Framed[Pane[#1, Scaled[1]], ImageMargins -> gap,
>       Background -> #2] &, {mat, colours}, 2], Spacings -> {0, 0},
>    ItemSize -> elemWidth], FrameMargins -> gap]]
>
> elemWidth is the width of the elements in the Grid and gap the distance between the lines.
>
> Heike
>
> On 25 Jul 2011, at 12:28, Don wrote:
>
>
>
>
>
>
>
> > Mathematica  permits divider lines between rows and columns in a table
> > (grid)  such
> > as this example from the documentation:
>
> > Grid[Table[x, {4}, {11}], Dividers -> {{{{True, False}}, -1 -> True},
> > False}]
>
> > But, the dividers are single lines only.
>
> > Sometimes  double line dividers are desired
> > as shown in this table on the Web:
>
> >http://www.assetcorrelation.com/user/correlations/90
>
> > Notice that the background color for an item  is only within
> > the inside line of a double line divider.
>
> > Is there any way to produce (a) this  kind of divider (double line)
> > and (b) with background colors for specific items that don't cross
> > the inside line of the double line divider?
>
> > Thank you in advance.
>
> > Don

nicely done!



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From: Gregory Lypny <gregory.lypny at jmsb.concordia.ca>
Subject: [mg120515] Re: Preventing In-line Math Typesetting From Being Scaled Down in Text
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Hello Mr. Fultz,

Tried but it didn't work.  I executed the command in the stylesheet notebook corresponding to the notebook I'm working with.  Perhaps I've misunderstood.  Also, not sure where to find Core.nb.  Is it in the application 
package?


Gregory Lypny





> On Fri, 22 Jul 2011 19:46:07 -0400 (EDT), Gregory Lypny wrote:
> > Hi everyone,
> >
> > When I include fractions as inline math typesetting, they are scaled down
> > to fit the effective line height of the cell. How can I prevent this or,
> > I guess, make the line height automatically expand to accommodate the
> > math? If my regular text in the cell is 12-point times, I'd like all
> > math variables that are not subscripts or superscripts to be 12-point as
> > well.
> >
> > Incidentally, other big typeset objects like matrices are not scaled
> > down, or at least they down't appear to be.
> >
> > Sincerely,
> >
> > Gregory
>
> If you look in Core.nb, you'll find a style called "InlineCell". This style is
> automatically applied to all inline cells everywhere. One of the options it has
> set is:
>
> ScriptLevel->1
>
> This is what's causing the behavior you're seeing. You can override this with a
> custom stylesheet. For example, in a given notebook, you can make a private
> override by doing Format->Edit Stylesheet..., and pasting and interpreting the
> following cell expression at the end of the resulting stylesheet notebook:
>
>
> Cell[StyleData["InlineCell"], ScriptLevel->0]
>
>
> Sincerely,
>
> John Fultz
> jfultz at wolfram.com
> User Interface Group
> Wolfram Research, Inc.
>



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From: Bob Hanlon <hanlonr at cox.net>
Subject: [mg120521] Re: Roots of a Jacobi polynomial
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Don't use machine precision

eqn1 = JacobiP[25, -1/2, -1/2, x] == 0;

soln1 = NSolve[eqn1, x, WorkingPrecision -> 15];

Length[soln1]

25

And @@ (eqn1 /. soln1)

True

And @@ (-1 < # < 1 & /@ (x /. soln1))

True

eqn2 = JacobiP[25, -0.5`20, -0.5`20, x] == 0;

soln2 = NSolve[eqn2, x, WorkingPrecision -> 15];

soln1 == soln2

True


Bob Hanlon

---- Jacopo Bertolotti <J.Bertolotti at utwente.nl> wrote: 

=============
Dear MathGroup,
Mathematica implements Jacobi polynomials as JacobiP[n,a,b,x] where n is 
the order of the polynomial. As it can be checked plotting it a Jacobi 
polynomial has n real roots in the interval [-1,1] and it goes rapidly 
to infinity outside this interval (at least when both a and b are >-1).
The problem arise when you try to find the roots of such a polynomial 
for a relatively high value of n. As an example the command 
NSolve[JacobiP[20, -0.5, -0.5, x] == 0, x] correctly returns

{{x -> -0.99702}, {x -> -0.972111}, {x -> -0.92418}, {x -> -0.852398}, 
{x -> -0.760555}, {x -> -0.649375}, {x -> -0.522529}, {x -> -0.382672}, 
{x -> -0.233449}, {x -> -0.0784582}, {x -> 0.0784591}, {x -> 0.233445}, 
{x -> 0.382684}, {x -> 0.522504}, {x -> 0.649423}, {x -> 0.760466}, {x 
-> 0.852539}, {x -> 0.924002}, {x -> 0.972267}, {x -> 0.996958}}

while NSolve[JacobiP[25, -0.5, -0.5, x] == 0, x] gives

{{x -> -1.01869}, {x -> -0.979859 - 0.0479527 I}, {x -> -0.979859 + 
0.0479527 I}, {x -> -0.870962 - 0.070991 I}, {x -> -0.870962 + 0.070991 
I}, {x -> -0.71378 - 0.0505783 I}, {x -> -0.71378 + 0.0505783 I}, {x -> 
-0.571283}, {x -> -0.486396}, {x -> -0.367829}, {x -> -0.248377}, {x -> 
-0.125513}, {x -> -0.0000434329}, {x -> 0.125442}, {x -> 0.2489}, {x -> 
0.365644}, {x -> 0.496977}, {x -> 0.555743}, {x -> 0.717741- 0.0573399 
I}, {x -> 0.717741+ 0.0573399 I}, {x -> 0.87423- 0.0652273 I}, {x -> 
0.87423+ 0.0652273 I}, {x -> 0.977876- 0.0422422 I}, {x -> 0.977876+ 
0.0422422 I}, {x -> 1.01494}}

i.e. both complex roots and roots outside the [-1,1] interval. 
Substituting any of these values back into the polynomial easily show 
that these values are not roots at all. Also notice that using the 
command Root (e.g. Sort@Table[Root[JacobiP[25, -0.5, -0.5, x], i], {i, 
1, 25}]) gives different but still wrong results.
On a related note: NIntegrate sometimes gives wrong results when 
integrating function of the form (1-x)^a (1+x)^b f[x] and I feel this 
might related to the Gauss-Jacobi quadrature failing to retrieve the 
correct roots of a Jacobi polynomial.

Do anyone have a solution for that?

Thank you

Jacopo




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From: John Fultz <jfultz at wolfram.com>
Subject: [mg120516] Re: Preventing In-line Math Typesetting From Being Scaled Down in Text
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Looks like there's a minor bug in the updating mechanism.  Perhaps this is what you ran into.  If, after following my procedure, you save, close, and reopen the notebook, you'll see that the change has taken effect.

Core.nb is inside the Mathematica layout, but it's generally not a good idea to be changing things there.  If you're hoping to make the change globally for all notebooks on your system, let me know and I can walk you through that procedure.  However, doing so will not change the notebook when viewed on somebody else's system, where as the Format->Edit Stylesheet... change will persist with the notebook regardless of which system it's viewed on.

Sincerely,

John Fultz
jfultz at wolfram.com
User Interface Group
Wolfram Research, Inc.


On Tue, 26 Jul 2011 14:10:44 -0400, Gregory Lypny wrote:
> Hello Mr. Fultz,
>
> Tried but it didn't work.  I executed the command in the stylesheet
> notebook corresponding to the notebook I'm working with.  Perhaps I've
> misunderstood.  Also, not sure where to find Core.nb.  Is it in the
> application package?
>
>
> Gregory Lypny
>
>
>> On Fri, 22 Jul 2011 19:46:07 -0400 (EDT), Gregory Lypny wrote:
>>> Hi everyone,
>>>
>>> When I include fractions as inline math typesetting, they are scaled
>>> down
>>> to fit the effective line height of the cell. How can I prevent this
>>> or,
>>> I guess, make the line height automatically expand to accommodate the
>>> math? If my regular text in the cell is 12-point times, I'd like all
>>> math variables that are not subscripts or superscripts to be 12-point
>>> as
>>> well.
>>>
>>> Incidentally, other big typeset objects like matrices are not scaled
>>> down, or at least they down't appear to be.
>>>
>>> Sincerely,
>>>
>>> Gregory
>>>
>> If you look in Core.nb, you'll find a style called "InlineCell". This
>> style is
>> automatically applied to all inline cells everywhere. One of the
>> options it has
>> set is:
>>
>> ScriptLevel->1
>>
>> This is what's causing the behavior you're seeing. You can override
>> this with a
>> custom stylesheet. For example, in a given notebook, you can make a
>> private
>> override by doing Format->Edit Stylesheet..., and pasting and
>> interpreting the
>> following cell expression at the end of the resulting stylesheet
>> notebook:
>>
>>
>> Cell[StyleData["InlineCell"], ScriptLevel->0]
>>
>>
>> Sincerely,
>>
>> John Fultz
>> jfultz at wolfram.com
>> User Interface Group
>> Wolfram Research, Inc.





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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120533] Re: FinancialData still broken
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Interesting!! I had thought the failure rate was lower, but I had noticed  
the erroneous prices had several different values. I've considered  
computing tighter acceptable ranges for each stock, but so far it appears  
sufficient to reject anything > 10^8 and repeat the call until I get a  
usable answer.

I have no proof that smaller errors are not getting through unnoticed, of  
course.

Sigh...

Bobby

On Tue, 26 Jul 2011 06:07:02 -0500, DANA DELOUIS <dana01 at me.com> wrote:

>
> Hi.  Here's some added thoughts on the issue you bring up.
> I added some stocks just in case the issue is related to mutual funds.
>
> stocks = {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX", "JMSCX=94,  
> "JNGIX=94,
>        "JNGLX", "JNMCX=94, "JNOSX", "JNSGX", "JNSTX", "PTTAX", "RGACX=94,
>        "STRFX=94, "AAPL", "IBM", "INTL", "VZ", "GE"};
>
> I've noticed that if I ask for price about 100 times, then about 80% of  
> the time I get a correct value,
> and 20% an incorrect value (n*^8).
> Here, a value of 8.91 was returned 77 times, or 77% (out of 100 tries)
>
> f[n_]:=Table[FinancialData[stocks[[n]],"Price"],{100}]//Tally//Sort
>
> f[1]
> {{8.91,77},{5.032406*10^8,23}}
>
> This took a little while, but I did it 100 times on each stock.
>
> m=Table[f[j],{j,stocks//Length}];
> m//TableForm
>
> Here's a count of the successful returns:
>
> good=m[[All,1,-1]]
> {76,74,77,77,87,77,90,79,81,82,76,79,79,77,85,77,80,78,76,81}
>
> Here's the Mean and SD.
> As you can see, there's only about a 80% success rate, with a small  
> Stand Deviation.
>
> {Mean[good],StandardDeviation[good]}//N
> {79.4, 4.031455}
>
> With such a consistent failure rate, it sounds like a communication  
> problem between servers
>
> Out of curiosity, here are the different values that were high.
> I don't see any pattern to these nonsense values.
>
> m[[All,2,1]]//Union
>
> 5.031051*10^8
> 5.031396*10^8
> 5.031565*10^8
> 5.031669*10^8
> 5.031676*10^8
> 5.031677*10^8
> 5.031678*10^8
> 5.032022*10^8
> 5.032406*10^8
> 5.055450*10^8
> 5.055453*10^8
> 5.055455*10^8
> 5.055456*10^8
> 5.055457*10^8
>
> Thanks for bringing this up.
>
> = = = = = = = = = =
> Dana DeLouis
> $Version
> 8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
>
>
>
> On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
> Erroneous prices are randomly returned:
>
> FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
> "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
> "PTTAX", "RGACX", "STRFX"}
>
> {8.81, 72.4, 14., 10.64, 46.49, 12.48,
> 5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
> 5.03241*10^8, 11.04, 31.44, 32.47}
>
> FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
> "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
> "PTTAX", "RGACX", "STRFX"}
>
> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
> 5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
> 5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}
>
> Bobby
>
> --
> DrMajor... at yahoo.com
>
>
>


-- 
DrMajorBob at yahoo.com



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From: Armand Tamzarian <mike.honeychurch at gmail.com>
Subject: [mg120518] Re: FinancialData still broken
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On Jul 24, 10:26 am, DrMajorBob <btre... at austin.rr.com> wrote:
> I usually don't see the "price > 10^8" bug, either. It may have been  
> permanently gone by the time you looked for it. But it was there for a day  
> or so.
>
> I know the occasional spurious Missing coming back is a months-old issue 
> at least, and Dan Delouis confirmed it still exists.
>
> I've inoculated my code against both bugs, so I may not notice when they 
> go away.
>
> Bobby



Would it be easier to just get the data from Yahoo rather than have to
worry about fixes and innoculations?



>
> On Sat, 23 Jul 2011 01:29:09 -0500, James Stein <mathgr... at stein.org>  
> wrote:
>
>
>
>
>
>
>
>
>
>
>
> > DANA DELOUIS wrote:
> > I really hope they fix this soon, but I guess they never issue bug fixes.
>
> > Are these bugs in Mathematica, or bugs in Yahoo?
> > FWIW, I do not see the bug reported by the OP (DrMajorBob).
>
> > On Fri, Jul 22, 2011 at 4:44 PM, DANA DELOUIS <dana.... at gmail.com> wrote:
>
> >> ...> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8...
>
> >> Hi.  This function has lots of problems.
> >> In addition to the problems you point out, here's a slightly different
> >> view.
> >> Sometimes the Price, and Latest Price match.
> >> Other times, the Price is corrupt, but the latest price is correct, and
> >> sometimes it's the other way around.
> >> Sometimes the CIK codes or Exchange are missing, etc.
> >> Other issues are bugs in Option pricing, etc.
> >> I really hope they fix this soon, but I guess they never issue bug  
> >> fixes.
>
> >> stocks={"ACEIX","FCNTX","JACNX","JANFX","JANWX","JMSCX","JNGIX","JNGLX",
> >> "JNMCX=94,
> >>        "JNOSX","JNSGX","JNSTX","PTTAX","RGACX","STRFX=94};
>
> >> Transpose[
> >> {
> >> Map[FinancialData[#,"Price"]&,stocks],
> >> Map[FinancialData[#,"LatestTrade"]&,stocks],
> >> Map[FinancialData[#,"CIK"]&,stocks],
> >> Map[FinancialData[#,"Company"]&,stocks]
> >> }][[;;6]]
>
> >> < Just the first 6 for posting
> >> < Looks better in MatrixForm, or something similar...>
>
> >> {5.0324*10^8,{Missing[NotAvailable],5.0324*10^8},0000080832,Van Kampen
> >> Equity Income Fund A}
> >> {5.0311*10^8,{{2011,7,20,17,11,0.},72.17},0000024238,Fidelity Contra 
> >> Fund}
> >> {13.97,{{2011,7,20,17,50,0.},13.97},0000277751,Janus Contrarian Fund D
> >> Shares}
> >> {5.0324*10^8,{{2001,12,22,20,42,0.},5.0324*10^8},0000277751,Janus  
> >> Flexible
> >> Bond Fund D Shar}
> >> {5.0324*10^8,{{2011,7,20,18,24,0.},46.74},0000277751,Janus Worldwide 
> >> Fund D
> >> Shares}
> >> {12.48,{{2011,7,20,17,51,0.},12.48},Missing[NotAvailable],Janus Smart
> >> Portfolio Conservat}
>
> >> = = = = = = = = = =
> >> Dana DeLouis
> >> $Version
> >> 8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
>
> >> On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
> >> > Erroneous prices are randomly returned:
>
> >> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
> >> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
> >> >    "PTTAX", "RGACX", "STRFX"}
>
> >> > {8.81, 72.4, 14., 10.64, 46.49, 12.48,
> >> >   5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
> >> >   5.03241*10^8, 11.04, 31.44, 32.47}
>
> >> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
> >> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
> >> >    "PTTAX", "RGACX", "STRFX"}
>
> >> > {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
> >> >   5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
> >> >   5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}
>
> >> > Bobby
>
> >> > --
> >> > DrMajor... at yahoo.com
>
> --
> DrMajor... at yahoo.com




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From: Darren Glosemeyer <darreng at wolfram.com>
Subject: [mg120532] Re: TransformedDistribution -- odd problem
References: <j0gh7s$bd7$1 at smc.vnet.net> <201107251129.HAA25540 at smc.vnet.net> <4E2EA04E.813B.006A.0 at newcastle.edu.au> <201107261106.HAA09233 at smc.vnet.net>
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On 7/26/2011 6:06 AM, Paul von Hippel wrote:
> Thanks -- that fixes it!
>
> Bonus question: if we don't specify v2>2, why doesn't Mathematica return a two part solution:
>   k v2 / (v2-2) v2>2
>
>   Indeterminate True
>
> That's what it does if we request the mean of F -- why doesn't it do the same if we request the mean of k*F.
>
>
>
>
> ________________________________
> From: Barrie Stokes<Barrie.Stokes at newcastle.edu.au>
> To: mathgroup at smc.vnet.net; paulvonhippel at yahoo<paulvonhippel at yahoo.com>
> Sent: Monday, July 25, 2011 8:09 PM
> Subject: Re: TransformedDistribution -- odd problem
>
> Hi Paul
>
> There are conditions on the v1 and v2, the degrees of freedom of the F distribution:
>
> Assuming[v2>  2,
> Mean[TransformedDistribution[F ,
>    F \[Distributed] FRatioDistribution[v1, v2]]]]
>
> {Assuming[v2>  2,
>   Mean[TransformedDistribution[k*F ,
>    F \[Distributed] FRatioDistribution[v1, v2]]]], k*v2/(-2 + v2)}
>
> {Assuming[v2>  2,
>    Mean[TransformedDistribution[k + F ,
>     F \[Distributed] FRatioDistribution[v1, v2]]]],
>   k + v2/(-2 + v2)} // FullSimplify
>
> which shows precisely what you expect for k*F and k+F.
>
> Cheers
>
> Barrie
>
>>>> On 25/07/2011 at 9:29 pm, in message<201107251129.HAA25540 at smc.vnet.net>,
> paulvonhippel at yahoo<paulvonhippel at yahoo.com>  wrote:
>> A little more experimenting shows that the TransformedDistribution
>> function will also not provide the mean of k+F where k is a constant
>> and F has an F distribution -- i.e.,
>>    Mean[TransformedDistribution[k*F ,  F \[Distributed]
>> FRatioDistribution[v, v]]]
>>
>> If I changce the distribution of F to NormalDistribution or
>> ChiSquareDistribution, I can get a mean for k*F or k+F. So the problem
>> only occurs when I define a simple function of an F variable using the
>> TransformedDistribution function.
>> This all strikes me as very strange, and I'd be curious to know if
>> others can reproduce my results. If you can't reproduce my results,
>> I'd be interested in theories about why my results differ from yours.
>> E.g., is there a setting I should change in the software?
>>
>> I am using version 8.0.0.0 and looking to upgrade to 8.0.1, if that
>> makes a difference.
>>
>> Many thanks for any pointers.
>>
>> On Jul 24, 2:22 am, paulvonhippel at yahoo<paulvonhip... at yahoo.com>
>> wrote:
>>> I'm having a very strange problem with TransformedDistribution, where
>>> I can calculate the mean of an F distribution but I cannot calculate
>>> the mean of a constant multiplied by an F distribution. That is, if I
>>> type
>>>
>>>    Mean[TransformedDistribution[F, F \[Distributed]
>>> FRatioDistribution[v, v]]]
>>>
>>> Mathematica gives me an answer. But if I type
>>>
>>>    Mean[TransformedDistribution[k*F ,  F \[Distributed]
>>> FRatioDistribution[v, v]]]
>>>
>>> Mathematica just echoes the input. I swear I got an answer for the
>>> second expression earlier today. What am I doing wrong?

The reason is internal processing in the general moment/expectation code. The result for mean of FRatioDistribution is hard-coded, while TransformDistribution has to go through more general code. In simple cases like shifting or rescaling by a constant, correctly propagating indeterminacies and infinities is fairly straightforward, but for more complicated transforms this may not be as easy. This is an area we will look at for improvement in the future.

Darren Glosemeyer
Wolfram Research





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From: Daniel Lichtblau <danl at wolfram.com>
Subject: [mg120513] Re: Roots of a Jacobi polynomial
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On 07/26/2011 06:07 AM, Jacopo Bertolotti wrote:
> Dear MathGroup,
> Mathematica implements Jacobi polynomials as JacobiP[n,a,b,x] where n is
> the order of the polynomial. As it can be checked plotting it a Jacobi
> polynomial has n real roots in the interval [-1,1] and it goes rapidly
> to infinity outside this interval (at least when both a and b are>-1).
> The problem arise when you try to find the roots of such a polynomial
> for a relatively high value of n. As an example the command
> NSolve[JacobiP[20, -0.5, -0.5, x] == 0, x] correctly returns
>
> {{x ->  -0.99702}, {x ->  -0.972111}, {x ->  -0.92418}, {x ->  -0.852398},
> {x ->  -0.760555}, {x ->  -0.649375}, {x ->  -0.522529}, {x ->  -0.382672},
> {x ->  -0.233449}, {x ->  -0.0784582}, {x ->  0.0784591}, {x ->  0.233445},
> {x ->  0.382684}, {x ->  0.522504}, {x ->  0.649423}, {x ->  0.760466}, {x
> ->  0.852539}, {x ->  0.924002}, {x ->  0.972267}, {x ->  0.996958}}
>
> while NSolve[JacobiP[25, -0.5, -0.5, x] == 0, x] gives
>
> {{x ->  -1.01869}, {x ->  -0.979859 - 0.0479527 I}, {x ->  -0.979859 +
> 0.0479527 I}, {x ->  -0.870962 - 0.070991 I}, {x ->  -0.870962 + 0.070991
> I}, {x ->  -0.71378 - 0.0505783 I}, {x ->  -0.71378 + 0.0505783 I}, {x ->
> -0.571283}, {x ->  -0.486396}, {x ->  -0.367829}, {x ->  -0.248377}, {x ->
> -0.125513}, {x ->  -0.0000434329}, {x ->  0.125442}, {x ->  0.2489}, {x ->
> 0.365644}, {x ->  0.496977}, {x ->  0.555743}, {x ->  0.717741- 0.0573399
> I}, {x ->  0.717741+ 0.0573399 I}, {x ->  0.87423- 0.0652273 I}, {x ->
> 0.87423+ 0.0652273 I}, {x ->  0.977876- 0.0422422 I}, {x ->  0.977876+
> 0.0422422 I}, {x ->  1.01494}}
>
> i.e. both complex roots and roots outside the [-1,1] interval.
> Substituting any of these values back into the polynomial easily show
> that these values are not roots at all. Also notice that using the
> command Root (e.g. Sort@Table[Root[JacobiP[25, -0.5, -0.5, x], i], {i,
> 1, 25}]) gives different but still wrong results.
> On a related note: NIntegrate sometimes gives wrong results when
> integrating function of the form (1-x)^a (1+x)^b f[x] and I feel this
> might related to the Gauss-Jacobi quadrature failing to retrieve the
> correct roots of a Jacobi polynomial.
>
> Do anyone have a solution for that?
>
> Thank you
>
> Jacopo

This is not a well conditioned problem. To do better you'll need to
allow NSolve to use higher precision by not giving machine numbers in 
the input.

In[102]:= solns = Sort[NSolve[JacobiP[25, -1/2, -1/2, x] == 0, x]]

Out[102]= {{x -> -0.998027}, {x -> -0.982287}, {x -> -0.951057}, {x \
-> -0.904827}, {x -> -0.844328}, {x -> -0.770513}, {x -> -0.684547}, \
{x -> -0.587785}, {x -> -0.481754}, {x -> -0.368125}, {x -> \
-0.24869}, {x -> -0.125333}, {x -> 0.}, {x -> 0.125333}, {x ->
    0.24869}, {x -> 0.368125}, {x -> 0.481754}, {x -> 0.587785}, {x ->
    0.684547}, {x -> 0.770513}, {x -> 0.844328}, {x -> 0.904827}, {x ->
     0.951057}, {x -> 0.982287}, {x -> 0.998027}}

These might seem imperfect when you check residuals (see below). In fact 
they are not at all bad. We first show that by comparing to higher 
precision versions that do validate fairly well as giving smallish 
residuals.

In[104]:= solns30 =
  Sort[NSolve[JacobiP[25, -1/2, -1/2, x] == 0, x,
    WorkingPrecision -> 30]]

Out[104]= {{x -> -0.998026728428271561952336806863}, {x -> \
-0.982287250728688681085641742865}, {x -> \
-0.951056516295153572116439333379}, {x -> \
-0.904827052466019527713668647933}, {x -> \
-0.844327925502015078548558063967}, {x -> \
-0.770513242775789230803009636396}, {x -> \
-0.684547105928688673732283357621}, {x -> \
-0.587785252292473129168705954639}, {x -> \
-0.481753674101715274987191502872}, {x -> \
-0.368124552684677959156947147493}, {x -> \
-0.248689887164854788242283746006}, {x -> \
-0.125333233564304245373118759817}, {x -> 0}, {x ->
    0.125333233564304245373118759817}, {x ->
    0.248689887164854788242283746006}, {x ->
    0.368124552684677959156947147493}, {x ->
    0.481753674101715274987191502872}, {x ->
    0.587785252292473129168705954639}, {x ->
    0.684547105928688673732283357621}, {x ->
    0.770513242775789230803009636396}, {x ->
    0.844327925502015078548558063967}, {x ->
    0.904827052466019527713668647933}, {x ->
    0.951056516295153572116439333379}, {x ->
    0.982287250728688681085641742865}, {x ->
    0.998026728428271561952336806863}}

This shows the two results are quite close.

In[106]:= Max[Abs[(x /. solns) - (x /. solns30)]]
Out[106]= 4.47615*10^-10

This shows the high precision solutions give reasonable residuals.

In[108]:= Max[Abs[JacobiP[25, -1/2, -1/2, x] /. solns30]]
Out[108]= 0.*10^-12

The machine precision results will not give good residuals. There are 
two possible (and related) reasons. One is that the conditioning is such 
that we are bound to lose many digits in assessing residuals (as seems 
to happen above). The other is that the form of JacobiP polynomials 
might just be inducing cancellation error.

Certainly this second item is part of the problem:

In[109]:= Max[Abs[Expand[JacobiP[25, -1/2, -1/2, x]] /. solns30]]
Out[109]= 0.*10^-21

Here are the corresponding machine precision evaluations of residuals.

In[112]:= Max[Abs[JacobiP[25, -1/2, -1/2, x] /. solns]]
Out[112]= 17.

In[113]:= Max[Abs[Expand[JacobiP[25, -1/2, -1/2, x]] /. solns]]
Out[113]= 9.31323*10^-9

Daniel Lichtblau
Wolfram Research



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From: "McHale, Paul" <Paul.McHale at excelitas.com>
Subject: [mg120517] Re: CDF limitations and unclear professional strategy
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>> Disappointed...

>> So much things I heard about CDF applications,
>> and then, well, I have to classify them as
>> documents... Very powerful documents, and almost
>> touching the application world but... just
>> interactive documents for an OPEN market...

I agree.  CDF format is a PDF on steroids.  It is designed to PUBLISH on the cheap.  Admittedly, it is phenomenal for just a document reader. 

There are two things to do with Mathematica, publish papers or do work where work is defined as change to external data or task/report being based external data.  I think for Mathematica to get to the next level, sharing documents must be only one focus.  Engineers also use Mathematica to parse databases, read external files, process like a program and write data out.  Sharing of these efforts via a common library would be very beneficial.

Then again, they are just starting to approach the corporate/enterprise are na.  Who knows what they have in mind.

Paul


Paul McHale  |  Electrical Engineer, Energetics Systems  |  Excelit as Technologies Corp.

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-----Original Message-----
From: Fonseca [mailto:public at fonseca.info]
Sent: Thursday, July 21, 2011 9:07 PM
To: mathgroup at smc.vnet.net
Subject: [mg120517] CDF limitations and unclear professional strategy

Disappointed...

So much things I heard about CDF applications, and then, well, I have
to classify them as documents... Very powerful documents, and almost
touching the application world but... just interactive documents for
an OPEN market...

Somehow nothing new than what already existed. Not that it is bad (I
personally have already commented a lot of good on the subject).

The problem is that I'm still unable to see how could I use this in an
cooperation environment.

Let's suppose two different scenarios:
1. An application to be distributed to 100 colleagues of the same
company, that, well, to make it simple, just needs to import a gif
image, apply some sort of simple filter, and export it to a gif file.
Very simple application, but not feasible. There isn't even concrete
information on the subject (besides "please contact us").  Have
contacted several times on the subject, and always no solution... Has
this changed? Why so much mystery on the PROfessional use of this
technology (this confirms what I keep hearing from others: academic,
institutes, governments, etc, but not corporate...)? 100 special
licenses for such a simple application?

2. Let's suppose I developed a package that does some awesome stuff,
and really toke me some time to develop it. I use it to write a report
for a client, and I want to give the report as an interactive document
to the client (nothing could be better than CDF!). But the interaction
needs my package (there's no way of previously generating all possible
outputs). So, I can't send it to my client, since I risk having it
copied to my competitors. Each client to whom I send a report needs a
professional version? I mean, each person in the company of my client,
and their consultant partners that will evaluate my report...? Or is
this now the other way around: each report or report revision needs to
be converted at Wolfram for pro functionalists activation on the
standard player (I sign confidential agreements with my client...)?
After all the announcements I have no idea what's the strategy. Did I
missed some information on WR site?

I just resent some questions to WR on the "please contact us", to see
if something changed in the past couple of months (when I last
insisted). Waiting for the answer, but why all the mystery on the PRO
version? What can't WR figure out about a professional strategy?

Can someone from Wolfram clarify this here for everyone? Is there a
strategy for these two very common and simple examples of my world? Is
this only meant for publishing documents on an OPEN environment, or on
a direct revenue strategy (selling each document/application copy for
a not so small price?).

It's interesting that all the examples presented in the site are of
very simple applications, but how can these simple application be used
on a closed corporate environment, if the licensing of the
applications cost more than their internal development cost?

I'm completely in the dark...

I know that I already asked this before, without great success (just
one person answered), but, am I the only one here?

Regards,
P. Fonseca




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Subject: [mg120535] Re: File directory
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You could do something like

SetOptions[$FrontEnd, WindowTitle -> "FullFileName"]


Heike

On 27 Jul 2011, at 11:14, E. Frauendorfer wrote:

> Hi,
>
> I am often working in two kernels with Mathematica Programs coming 
from different directories also from different drives. After working 
some time, I very
> often don't remember the origin of a particular open program.
>
> On opening a notebook the name of the program always appears on top of

> the notebook, but not the origin. In my latest installation of 
Mathematica (8.01 on Win 7, 64 bit) I am no longer able to find the 
origin of an open program.
>
> Please tell, how to recover this feature.
>
>
> Thanks   E.F.
>




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From: Yves Klett <yves.klett at googlemail.com>
Subject: [mg120534] Re: File directory
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You can use NotebookDirectory[] to get the path to the current notebook.

In 8.01 the Save-As-Dialog behaves not as expected and opens in the path
of the last opened document as default instead of using the current
document´s path.

This results in widely scattered notebooks if you are not careful.

Regards,
Yves

Am 27.07.2011 12:22, schrieb E. Frauendorfer:
> Hi,
> 
> I am often working in two kernels with Mathematica Programs coming from different directories also from different drives. After working some time, I very 
> often don't remember the origin of a particular open program.
> 
> On opening a notebook the name of the program always appears on top of 
> the notebook, but not the origin. In my latest installation of Mathematica (8.01 on Win 7, 64 bit) I am no longer able to find the origin of an open program.
> 
> Please tell, how to recover this feature.
> 
> 
> Thanks   E.F.
> 



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From: Gregory Lypny <gregory.lypny at videotron.ca>
Subject: [mg120541] How Can I Make Plot Options Conditional?
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Hello everyone,

Is there a preferred way to make plot options conditional?  For Epilog 
points, for example, I've been using an IF statement within the Epilog 
list to determine whether the point size should be set to zero and 
therefore make the point invisible.

Epilog -> {If[x >= xMin, PointSize[.02], PointSize[0]], Orange, Point[ptX]},

Gregory=



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From: Zach Bjornson <zachb at wolfram.com>
Subject: [mg120540] Re: Genomes Mathematica knows about and how to use them?
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The GenomeData function in Mathematica only has the human genome. 
However, there are a number of other genomes available via the 
WolframAlpha interface present in Mathematica 8. At the end of this blog 
post, there are some examples -- rat, zebra fish, mouse, drosophila, ...

http://blog.wolframalpha.com/2010/03/10/did-you-know-that-wolframalpha-knows-your-dna/

Best,
Zach
W|A Computational Biology

On 7/27/2011 3:13 AM, Ted Ersek wrote:
> I know someone who is a researcher in biochemistry and I wanted to show them
> how Mathematica could be useful in their research.  I read somewhere that
> besides the human genome,  Mathematica has the genome of a certain species
> of mouse and fruit fly in its  curated data.  Then I figured Mathematica
> might also have the genomes of the bacteria they research in its curated
> data.
>
>
>
> How does one find out what genomes are included in curated data that comes
> with Mathematica?  Once you know a certain genome besides the human genome
> is available, how do you access it?
>
>
>
> If one is going to Import a genome from a source independent of Wolfram
> Research, what are likely file formats one would Import?
>
>
>
> Thanks,
>
>     Ted Ersek




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From: "Berthold Hamburger" <b-hamburger at artinso.com>
Subject: [mg120550] Timing progress bar
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Hi,

Is there any possibility to get visible feedback about the progress of a
calculation via the Timing function? Something like a progress bar or "%
completed".

Thanks

Berthold

-- 

Berthold Hamburger - Cellist/Spain

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Subject: [mg120546] Re: File directory
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On 7/27/2011 8:08 AM, Yves Klett wrote:
>
> In 8.01 the Save-As-Dialog behaves not as expected and opens in the path
> of the last opened document as default instead of using the current
> document's path.
>
> This results in widely scattered notebooks if you are not careful.

That's something I really wish they would fix. A simple Save-As requires 
me to remember what directory the file is in, and sometimes I have to 
browse all over the place for it. It's a huge nuisance.


-- 
Helen Read
University of Vermont



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From: "Christoph Lhotka" <christoph.lhotka at univie.ac.at>
Subject: [mg120536] Re: File directory
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thats easy, just use {$MachineName, NotebookDirectory[]} to get the name
of the computer and the absoulte file directory of the notebook in which
you type in the command.

best,

christoph

On 27/07/2011 12:14, E. Frauendorfer wrote:
> Hi,
>
> I am often working in two kernels with Mathematica Programs coming from
different directories also from different drives. After working some
time, I very
> often don't remember the origin of a particular open program.
>
> On opening a notebook the name of the program always appears on top of
> the notebook, but not the origin. In my latest installation of
Mathematica (8.01 on Win 7, 64 bit) I am no longer able to find the
origin of an open program.
>
> Please tell, how to recover this feature.
>
>
> Thanks   E.F.
>
>





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From: James Stein <mathgroup at stein.org>
Subject: [mg120547] Re: File directory
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On a Mac: command-click on the document name to see its complete hierarchy
of containing folders.  Surely Windows has a similar feature...

On Wed, Jul 27, 2011 at 3:14 AM, E. Frauendorfer <
he.frauendorfer at t-online.de> wrote:

> Hi,
>
> I am often working in two kernels with Mathematica Programs coming from
> different directories also from different drives. After working some time, I
> very
> often don't remember the origin of a particular open program.
>
> On opening a notebook the name of the program always appears on top of
> the notebook, but not the origin. In my latest installation of Mathematica
> (8.01 on Win 7, 64 bit) I am no longer able to find the origin of an open
> program.
>
> Please tell, how to recover this feature.
>
>
> Thanks   E.F.
>
>



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From: Richard Fateman <fateman at cs.berkeley.edu>
Subject: [mg120551] Re: And now for something completely different
References: <201107251129.HAA25514 at smc.vnet.net> <201107261107.HAA09286 at smc.vnet.net> <j0oosn$kte$1 at smc.vnet.net>
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On 7/27/2011 3:22 AM, Andrzej Kozlowski wrote:
> Or, perhaps this is even more appropriate ;-) :
>
> http://kingjbible.com/genesis/1.htm
>
>
> Andrzej Kozlowski
>

I suppose SW would like NKS to be compared with the King James 
translation of the Bible.  The KJV's version of Genesis tends
to be rather poetical, and hardly the touchstone for conveying
clarity to a modern reader. It does have verses beginning with
"And".  A good catch.

However, the initial "And"s are lacking
in more contemporary translations intended to express the clear
content of the Bible to a modern English reader: e.g.
  http://net.bible.org/#!bible/Genesis+1:1

I see one spot though..  "And there was light." in that translation.
For drama, don't you think?







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From: Alexey Popkov <lehin.p at gmail.com>
Subject: [mg120548] Why FullDefinition does not work in MathLink mode?
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Hello,

When working with MathKernel in interactive session without of the
FrontEnd, FullDefinition works as expected:

In[1]:= a=2;FullDefinition[a]

Out[1]= a = 2

But when trying to use it in MathLink mode one get ReturnPacket with
unevaluated FullDefinition:

In[1]:= kernel=LinkLaunch[First[$CommandLine] <> " -mathlink"];
LinkRead[kernel];
LinkWrite[kernel,Unevaluated[a=2;FullDefinition[a]]]
LinkRead[kernel]//FullForm

Out[4]//FullForm= ReturnPacket[FullDefinition[a]]

The built-in parallelization functions have the same problem:

In[2]:= ParallelEvaluate[a=2;FullDefinition[a]]//FullForm
Out[2]//FullForm= List[FullDefinition[a],FullDefinition[a]]

One workaround is to use ToString:

In[3]:= ParallelEvaluate[a=2;ToString[FullDefinition[a]]]
Out[3]= {a = 2,a = 2}

Why FullDefinition returns unevaluated in MathLink session although it
works as expected in interactive MathKernel session? Is it possible to
make it working without any glitches?

Any ideas?



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From: James Stein <mathgroup at stein.org>
Subject: [mg120549] Re: Genomes Mathematica knows about and how to use them?
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Genomic data grows daily; many different archives exist, addressing
different needs by people with keen interest. (I suspect different archives
also have different conventions for metadata.) I doubt Wolfram could or
should compete  against such specialists. If a few genomes are available,
it's likely for hobbyists.

On Wed, Jul 27, 2011 at 3:11 AM, Ted Ersek <ersekt at md.metrocast.net> wrote:

> I know someone who is a researcher in biochemistry and I wanted to show
> them
> how Mathematica could be useful in their research.  I read somewhere that
> besides the human genome,  Mathematica has the genome of a certain species
> of mouse and fruit fly in its  curated data.  Then I figured Mathematica
> might also have the genomes of the bacteria they research in its curated
> data.
>
>
>
> How does one find out what genomes are included in curated data that comes
> with Mathematica?  Once you know a certain genome besides the human genome
> is available, how do you access it?
>
>
>
> If one is going to Import a genome from a source independent of Wolfram
> Research, what are likely file formats one would Import?
>
>
>
> Thanks,
>
>   Ted Ersek
>




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From: Armand Tamzarian <mike.honeychurch at gmail.com>
Subject: [mg120542] Re: getting ride of 0.i
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On Jul 27, 8:19 pm, Eli Fenichel <Eli.Fenic... at asu.edu> wrote:
> I am trying to use FindMinimum to minimize the square of a function over 
many variables (i.e., parameters).  The function itself has the potential
 to generate imaginary numbers and involves numerical integration (the parameter values that minimize the function are real).  For clarification the function in evaluated at multiple points so it can be written as vector. By squaring the vector the numerical values of the elements are always real with no imaginary parts.  However, Mathematica often writes x + 0.i, where x is a some numerical value, for some of the elements.
>
> Typically, this can be ignored, the Chop command can be used, or it simply does not cause problems.  However, I keep getting an error: [cid:image003.... at 01CC4BBC.E87A0F30]
>
> NIntegrate::nlim: "t = Y[1.] is not a valid limit of integration."
>
> Y is an array with the parameters to be minimized.
>
> However, if evaluate the objective function to be minimized using replacement rules I get
> x + 0.i.
>
> Is there a way for me tell Mathematica to always treat 0.i as 0 and drop it?  It seems to be causing problems in the FindMinimum call.
>
> Thanks,
> Eli

Re[3 + 2 I]



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From: Tomas Garza <tgarza10 at msn.com>
Subject: [mg120552] Re: And now for something completely different
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And, could one quote a better example?
-Tomas


[This thread is now over - moderator]



> Date: Wed, 27 Jul 2011 06:14:29 -0400
> From: akoz at mimuw.edu.pl
> Subject: Re: And now for something completely different
> To: mathgroup at smc.vnet.net
>
> Or, perhaps this is even more appropriate ;-) :
>
> http://kingjbible.com/genesis/1.htm
>
>
> Andrzej Kozlowski
>
>
>
>
> On 26 Jul 2011, at 13:07, ccarter at mit.edu wrote:
>
> > And, how is this relevant to the group topic?
> >
> > And, for what it is worth:
> >
> > 	HAMLET
> > ACT III
> > SCENE I	A room in the castle.
> > 	[Enter KING CLAUDIUS, QUEEN GERTRUDE, POLONIUS,
> > 	OPHELIA, ROSENCRANTZ, and GUILDENSTERN]
> >
> > KING CLAUDIUS	And can you, by no drift of circumstance,
> > 	Get from him why he puts on this confusion,
> > 	Grating so harshly all his days of quiet
> > 	With turbulent and dangerous lunacy?
> >
> >
> >
> > On Mon, 25 Jul 2011, Richard Fateman wrote:
> >
> >> Date: Mon, 25 Jul 2011 07:29:07 -0400 (EDT)
> >> From: Richard Fateman <fateman at cs.berkeley.edu>
> >> To: mathgroup at smc.vnet.net
> >> Subject: And now for something completely different
> >>
> >>
> >> Can you start a paragraph with "And"?
> >>
> >> http://answers.yahoo.com/question/index?qid090413160418AAyfRO2
> >>
> >> says, 'No.'
> >>
> >>
> >> Stephen Wolfram often does so, and I find it annoying.
> >> example
> >> http://www.wolframscience.com/nksonline/page-42?firstview==1
> >>
> >> This construction seems to have been removed from the online ==
> documentation.
> >>
> >>
> >> --RJF
> >>
> >>
> >
>
>


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From: "Donald DuBois" <donabc at comcast.net>
Subject: [mg120545] Re: Grid Divider Style With Background Colors
Reply-To: "Donald DuBois" <donald at ieee.org>
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Heike,

That was a wonderful solution.  Many thanks.

Don


-----Original Message----- 
From: Heike Gramberg
Sent: Monday, July 25, 2011 5:19 AM
To: Don
Cc: mathgroup at smc.vnet.net
Subject: [mg120545] Re: Grid Divider Style With Background Colors

You could do something like

With[{gap = 1, elemWidth=3, h = 8, w = 7},
mat = RandomInteger[10, {h, w}];
colours = Table[Blend[{Lighter[Blue], Pink}, j/w], {i, h}, {j, w}];
Framed[Grid[
   MapThread[
    Framed[Pane[#1, Scaled[1]], ImageMargins -> gap,
      Background -> #2] &, {mat, colours}, 2], Spacings -> {0, 0},
   ItemSize -> elemWidth], FrameMargins -> gap]]

elemWidth is the width of the elements in the Grid and gap the distance 
between the lines.

Heike

On 25 Jul 2011, at 12:28, Don wrote:

> Mathematica  permits divider lines between rows and columns in a table
> (grid)  such
> as this example from the documentation:
>
> Grid[Table[x, {4}, {11}], Dividers -> {{{{True, False}}, -1 -> True},
> False}]
>
> But, the dividers are single lines only.
>
> Sometimes  double line dividers are desired
> as shown in this table on the Web:
>
> http://www.assetcorrelation.com/user/correlations/90
>
> Notice that the background color for an item  is only within
> the inside line of a double line divider.
>
> Is there any way to produce (a) this  kind of divider (double line)
> and (b) with background colors for specific items that don't cross
> the inside line of the double line divider?
>
> Thank you in advance.
>
> Don
> 



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From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
Subject: [mg120537] Re: how to ListPlot3D large data sets
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On Wed, 27 Jul 2011, Ted Sariyski wrote:

> Hi,
>
> In one attempt MaxPlotPoints did not help, but I was not persistent
> because I have to resolve other issues before I return to ListPlot3D.
> One issue is that I cannot pack data. On the big data set I verify that
> it is numeric array and still  I cannot get it packed. Here is an example:
>
> mydata={{0.00100,10.,0.},{0.00100,10.,0.00100},{0.00100,10.,0.00200},{0.00100,10.,0.00300},{0.00100,10.,0.00400}};
>
> MatrixQ[mydata,NumericQ]
> True
>
> myPackedData=ToPackedArray[mydata,Real];
>
> MatrixQ[myPackedData, NumericQ]
> True
>
> PackedArrayQ[myPackedData]
> False
>
> What I am doing wrong?
> Thanks,
> --Ted


Ted, possibly ToPackedArray are not on you context path

either try

<<Developer`

or

mydata = {{0.00100, 10., 0.}, {0.00100, 10., 0.00100}, {0.00100, 10.,
     0.00200}, {0.00100, 10., 0.00300}, {0.00100, 10., 0.00400}};
myPackedData = Developer`ToPackedArray[mydata, Real];
Developer`PackedArrayQ[myPackedData]

Oliver

>
> On 7/25/2011 7:30 AM, Oliver Ruebenkoenig wrote:
>> On Sat, 23 Jul 2011, Ted Sariyski wrote:
>>
>>> Hi,
>>> I guess I am pushing the limits of ListPlot3D. I try to ListPlot3d a data
>>> set with many millions of records. I was not able to get an image from the
>>> full dataset, it takes forever. If I use e.g. every fifth record, although
>>> slow, I get an image. The machine running Mathematica is Windows 7 (64 bit),
>>> has 24 GB RAM and there was no swapping. I wonder what is considered as a
>>> reasonable data size for ListPlot3D and are there other tools in Mathematica
>>> for visualization of large data sets?
>>> Thanks in advance,
>>> --Ted
>>>
>>>
>>>
>> Ted,
>>
>> is your data packed?
>>
>> Developer`PackedArrayQ[yourData]
>>
>> Oliver
>>
>
>
>



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From: "Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com>
Subject: [mg120538] Re: how to ListPlot3D large data sets
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It works if you either specify ToPackedArray/PackedArrayQ by context:

myPackedData = Developer`ToPackedArray[mydata,Real];

Developer`PackedArrayQ[myPackedData]
True

or add Developer` to $ContextPath:

$ContextPath = Prepend[$ContextPath, "Developer`"]

ToPackedArray[mydata,Real];

PackedArrayQ[myPackedData]
True

Best,

(another) O. R.

On Wed, 27 Jul 2011 11:21:44 +0100, Ted Sariyski <tsariysk at craft-tech.com>  
wrote:

> Hi,
>
> In one attempt MaxPlotPoints did not help, but I was not persistent
> because I have to resolve other issues before I return to ListPlot3D.
> One issue is that I cannot pack data. On the big data set I verify that
> it is numeric array and still  I cannot get it packed. Here is an  
> example:
>
> mydata={{0.00100,10.,0.},{0.00100,10.,0.00100},{0.00100,10.,0.00200},{0.00100,10.,0.00300},{0.00100,10.,0.00400}};
>
> MatrixQ[mydata,NumericQ]
> True
>
> myPackedData=ToPackedArray[mydata,Real];
>
> MatrixQ[myPackedData, NumericQ]
> True
>
> PackedArrayQ[myPackedData]
> False
>
> What I am doing wrong?
> Thanks,
> --Ted
>
> On 7/25/2011 7:30 AM, Oliver Ruebenkoenig wrote:
>> On Sat, 23 Jul 2011, Ted Sariyski wrote:
>>> Hi,
>>> I guess I am pushing the limits of ListPlot3D. I try to ListPlot3d a  
>>> data
>>> set with many millions of records. I was not able to get an image from  
>>> the
>>> full dataset, it takes forever. If I use e.g. every fifth record,  
>>> although
>>> slow, I get an image. The machine running Mathematica is Windows 7 (64  
>>> bit),
>>> has 24 GB RAM and there was no swapping. I wonder what is considered  
>>> as a
>>> reasonable data size for ListPlot3D and are there other tools in  
>>> Mathematica
>>> for visualization of large data sets?
>>> Thanks in advance,
>>> --Ted
>>>
>> Ted,
>>
>> is your data packed?
>>
>> Developer`PackedArrayQ[yourData]
>>
>> Oliver



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From: Emu <samuel.thomas.blake at gmail.com>
Subject: [mg120543] Re: Roots of a Jacobi polynomial
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On Jul 27, 8:18 pm, Bob Hanlon <hanl... at cox.net> wrote:
> Don't use machine precision
>
> eqn1 = JacobiP[25, -1/2, -1/2, x] == 0;
>
> soln1 = NSolve[eqn1, x, WorkingPrecision -> 15];
>
> Length[soln1]
>
> 25
>
> And @@ (eqn1 /. soln1)
>
> True
>
> And @@ (-1 < # < 1 & /@ (x /. soln1))
>
> True
>
> eqn2 = JacobiP[25, -0.5`20, -0.5`20, x] == 0;
>
> soln2 = NSolve[eqn2, x, WorkingPrecision -> 15];
>
> soln1 == soln2
>
> True
>
> Bob Hanlon
>
> ---- Jacopo Bertolotti <J.Bertolo... at utwente.nl> wrote:
>
> =============
> Dear MathGroup,
> Mathematica implements Jacobi polynomials as JacobiP[n,a,b,x] where n is
> the order of the polynomial. As it can be checked plotting it a Jacobi
> polynomial has n real roots in the interval [-1,1] and it goes rapidly
> to infinity outside this interval (at least when both a and b are >-1).
> The problem arise when you try to find the roots of such a polynomial
> for a relatively high value of n. As an example the command
> NSolve[JacobiP[20, -0.5, -0.5, x] == 0, x] correctly returns
>
> {{x -> -0.99702}, {x -> -0.972111}, {x -> -0.92418}, {x -> -0.852398},
> {x -> -0.760555}, {x -> -0.649375}, {x -> -0.522529}, {x -> -0.382672},
> {x -> -0.233449}, {x -> -0.0784582}, {x -> 0.0784591}, {x -> 0.233445},
> {x -> 0.382684}, {x -> 0.522504}, {x -> 0.649423}, {x -> 0.760466}, {x
> -> 0.852539}, {x -> 0.924002}, {x -> 0.972267}, {x -> 0.996958}}
>
> while NSolve[JacobiP[25, -0.5, -0.5, x] == 0, x] gives
>
> {{x -> -1.01869}, {x -> -0.979859 - 0.0479527 I}, {x -> -0.979859 +
> 0.0479527 I}, {x -> -0.870962 - 0.070991 I}, {x -> -0.870962 + 0.070991
> I}, {x -> -0.71378 - 0.0505783 I}, {x -> -0.71378 + 0.0505783 I}, {x ->
> -0.571283}, {x -> -0.486396}, {x -> -0.367829}, {x -> -0.248377}, {x ->
> -0.125513}, {x -> -0.0000434329}, {x -> 0.125442}, {x -> 0.2489}, {x ->
> 0.365644}, {x -> 0.496977}, {x -> 0.555743}, {x -> 0.717741- 0.0573399
> I}, {x -> 0.717741+ 0.0573399 I}, {x -> 0.87423- 0.0652273 I}, {x ->
> 0.87423+ 0.0652273 I}, {x -> 0.977876- 0.0422422 I}, {x -> 0.977876+
> 0.0422422 I}, {x -> 1.01494}}
>
> i.e. both complex roots and roots outside the [-1,1] interval.
> Substituting any of these values back into the polynomial easily show
> that these values are not roots at all. Also notice that using the
> command Root (e.g. Sort@Table[Root[JacobiP[25, -0.5, -0.5, x], i], {i,
> 1, 25}]) gives different but still wrong results.
> On a related note: NIntegrate sometimes gives wrong results when
> integrating function of the form (1-x)^a (1+x)^b f[x] and I feel this
> might related to the Gauss-Jacobi quadrature failing to retrieve the
> correct roots of a Jacobi polynomial.
>
> Do anyone have a solution for that?
>
> Thank you
>
> Jacopo


Here's one way to get them in Alpha.

http://www.wolframalpha.com/input/?i=numerical+solutions+to+JacobiP%5B25%2C+-1%2F2%2C+-1%2F2%2C+x%5D+%3D%3D+0

Sam



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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120539] Re: FinancialData still broken
Reply-To: drmajorbob at yahoo.com
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> Would it be easier to just get the data from Yahoo rather than have to
> worry about fixes and innoculations?

I don't know how to do that, but I suspect it's Yahoo sending erroneous  
prices, anyway.

Bobby

On Wed, 27 Jul 2011 05:12:30 -0500, Armand Tamzarian  
<mike.honeychurch at gmail.com> wrote:

> On Jul 24, 10:26 am, DrMajorBob <btre... at austin.rr.com> wrote:
>> I usually don't see the "price > 10^8" bug, either. It may have been
>> permanently gone by the time you looked for it. But it was there for a  
>> day
>> or so.
>>
>> I know the occasional spurious Missing coming back is a months-old issue
>> at least, and Dan Delouis confirmed it still exists.
>>
>> I've inoculated my code against both bugs, so I may not notice when they
>> go away.
>>
>> Bobby
>
>
>
> Would it be easier to just get the data from Yahoo rather than have to
> worry about fixes and innoculations?
>
>
>
>>
>> On Sat, 23 Jul 2011 01:29:09 -0500, James Stein <mathgr... at stein.org>
>> wrote:
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> > DANA DELOUIS wrote:
>> > I really hope they fix this soon, but I guess they never issue bug  
>> fixes.
>>
>> > Are these bugs in Mathematica, or bugs in Yahoo?
>> > FWIW, I do not see the bug reported by the OP (DrMajorBob).
>>
>> > On Fri, Jul 22, 2011 at 4:44 PM, DANA DELOUIS <dana.... at gmail.com>  
>> wrote:
>>
>> >> ...> {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8...
>>
>> >> Hi.  This function has lots of problems.
>> >> In addition to the problems you point out, here's a slightly  
>> different
>> >> view.
>> >> Sometimes the Price, and Latest Price match.
>> >> Other times, the Price is corrupt, but the latest price is correct,  
>> and
>> >> sometimes it's the other way around.
>> >> Sometimes the CIK codes or Exchange are missing, etc.
>> >> Other issues are bugs in Option pricing, etc.
>> >> I really hope they fix this soon, but I guess they never issue bug
>> >> fixes.
>>
>> >>  
>> stocks={"ACEIX","FCNTX","JACNX","JANFX","JANWX","JMSCX","JNGIX","JNGLX",
>> >> "JNMCX=94,
>> >>        "JNOSX","JNSGX","JNSTX","PTTAX","RGACX","STRFX=94};
>>
>> >> Transpose[
>> >> {
>> >> Map[FinancialData[#,"Price"]&,stocks],
>> >> Map[FinancialData[#,"LatestTrade"]&,stocks],
>> >> Map[FinancialData[#,"CIK"]&,stocks],
>> >> Map[FinancialData[#,"Company"]&,stocks]
>> >> }][[;;6]]
>>
>> >> < Just the first 6 for posting
>> >> < Looks better in MatrixForm, or something similar...>
>>
>> >> {5.0324*10^8,{Missing[NotAvailable],5.0324*10^8},0000080832,Van  
>> Kampen
>> >> Equity Income Fund A}
>> >> {5.0311*10^8,{{2011,7,20,17,11,0.},72.17},0000024238,Fidelity Contra
>> >> Fund}
>> >> {13.97,{{2011,7,20,17,50,0.},13.97},0000277751,Janus Contrarian Fund  
>> D
>> >> Shares}
>> >> {5.0324*10^8,{{2001,12,22,20,42,0.},5.0324*10^8},0000277751,Janus
>> >> Flexible
>> >> Bond Fund D Shar}
>> >> {5.0324*10^8,{{2011,7,20,18,24,0.},46.74},0000277751,Janus Worldwide
>> >> Fund D
>> >> Shares}
>> >> {12.48,{{2011,7,20,17,51,0.},12.48},Missing[NotAvailable],Janus Smart
>> >> Portfolio Conservat}
>>
>> >> = = = = = = = = = =
>> >> Dana DeLouis
>> >> $Version
>> >> 8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
>>
>> >> On Jul 21, 5:47 am, DrMajorBob <btre... at austin.rr.com> wrote:
>> >> > Erroneous prices are randomly returned:
>>
>> >> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>> >> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>> >> >    "PTTAX", "RGACX", "STRFX"}
>>
>> >> > {8.81, 72.4, 14., 10.64, 46.49, 12.48,
>> >> >   5.03202*10^8, 26.48, 23.72, 45.35, 12.48,
>> >> >   5.03241*10^8, 11.04, 31.44, 32.47}
>>
>> >> > FinancialData /@ {"ACEIX", "FCNTX", "JACNX", "JANFX", "JANWX",
>> >> >    "JMSCX", "JNGIX", "JNGLX", "JNMCX", "JNOSX", "JNSGX", "JNSTX",
>> >> >    "PTTAX", "RGACX", "STRFX"}
>>
>> >> > {5.03241*10^8, 72.4, 14., 10.64, 5.03241*10^8,
>> >> >   5.03168*10^8, 32.61, 26.48, 23.72, 5.03168*10^8,
>> >> >   5.03168*10^8, 3.1, 11.04, 5.03241*10^8, 32.47}
>>
>> >> > Bobby
>>
>> >> > --
>> >> > DrMajor... at yahoo.com
>>
>> --
>> DrMajor... at yahoo.com
>
>


-- 
DrMajorBob at yahoo.com



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Date: Thu, 28 Jul 2011 07:54:42 -0400 (EDT)
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From: Jacopo Bertolotti <J.Bertolotti at utwente.nl>
Subject: [mg120544] Re: Roots of a Jacobi polynomial
References: <201107261107.HAA09296 at smc.vnet.net> <4E2EFAAD.60403 at wolfram.com>
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On 07/26/2011 07:34 PM, Daniel Lichtblau wrote:
> On 07/26/2011 06:07 AM, Jacopo Bertolotti wrote:
>> Dear MathGroup,
>> Mathematica implements Jacobi polynomials as JacobiP[n,a,b,x] where n is
>> the order of the polynomial. As it can be checked plotting it a Jacobi
>> polynomial has n real roots in the interval [-1,1] and it goes rapidly
>> to infinity outside this interval (at least when both a and b are>-1).
>> The problem arise when you try to find the roots of such a polynomial
>> for a relatively high value of n. As an example the command
>> NSolve[JacobiP[20, -0.5, -0.5, x] == 0, x] correctly returns
>>
>> {{x ->  -0.99702}, {x ->  -0.972111}, {x ->  -0.92418}, {x ->  
>> -0.852398},
>> {x ->  -0.760555}, {x ->  -0.649375}, {x ->  -0.522529}, {x ->  
>> -0.382672},
>> {x ->  -0.233449}, {x ->  -0.0784582}, {x ->  0.0784591}, {x ->  
>> 0.233445},
>> {x ->  0.382684}, {x ->  0.522504}, {x ->  0.649423}, {x ->  
>> 0.760466}, {x
>> ->  0.852539}, {x ->  0.924002}, {x ->  0.972267}, {x ->  0.996958}}
>>
>> while NSolve[JacobiP[25, -0.5, -0.5, x] == 0, x] gives
>>
>> {{x ->  -1.01869}, {x ->  -0.979859 - 0.0479527 I}, {x ->  -0.979859 +
>> 0.0479527 I}, {x ->  -0.870962 - 0.070991 I}, {x ->  -0.870962 + 
>> 0.070991
>> I}, {x ->  -0.71378 - 0.0505783 I}, {x ->  -0.71378 + 0.0505783 I}, 
>> {x ->
>> -0.571283}, {x ->  -0.486396}, {x ->  -0.367829}, {x ->  -0.248377}, 
>> {x ->
>> -0.125513}, {x ->  -0.0000434329}, {x ->  0.125442}, {x ->  0.2489}, 
>> {x ->
>> 0.365644}, {x ->  0.496977}, {x ->  0.555743}, {x ->  0.717741- 
>> 0.0573399
>> I}, {x ->  0.717741+ 0.0573399 I}, {x ->  0.87423- 0.0652273 I}, {x ->
>> 0.87423+ 0.0652273 I}, {x ->  0.977876- 0.0422422 I}, {x ->  0.977876+
>> 0.0422422 I}, {x ->  1.01494}}
>>
>> i.e. both complex roots and roots outside the [-1,1] interval.
>> Substituting any of these values back into the polynomial easily show
>> that these values are not roots at all. Also notice that using the
>> command Root (e.g. Sort@Table[Root[JacobiP[25, -0.5, -0.5, x], i], {i,
>> 1, 25}]) gives different but still wrong results.
>> On a related note: NIntegrate sometimes gives wrong results when
>> integrating function of the form (1-x)^a (1+x)^b f[x] and I feel this
>> might related to the Gauss-Jacobi quadrature failing to retrieve the
>> correct roots of a Jacobi polynomial.
>>
>> Do anyone have a solution for that?
>>
>> Thank you
>>
>> Jacopo
>
> This is not a well conditioned problem. To do better you'll need to
> allow NSolve to use higher precision by not giving machine numbers in 
> the input.
>
> In[102]:= solns = Sort[NSolve[JacobiP[25, -1/2, -1/2, x] == 0, x]]
>
> Out[102]= {{x -> -0.998027}, {x -> -0.982287}, {x -> -0.951057}, {x \
> -> -0.904827}, {x -> -0.844328}, {x -> -0.770513}, {x -> -0.684547}, \
> {x -> -0.587785}, {x -> -0.481754}, {x -> -0.368125}, {x -> \
> -0.24869}, {x -> -0.125333}, {x -> 0.}, {x -> 0.125333}, {x ->
>    0.24869}, {x -> 0.368125}, {x -> 0.481754}, {x -> 0.587785}, {x ->
>    0.684547}, {x -> 0.770513}, {x -> 0.844328}, {x -> 0.904827}, {x ->
>     0.951057}, {x -> 0.982287}, {x -> 0.998027}}
>
> These might seem imperfect when you check residuals (see below). In 
> fact they are not at all bad. We first show that by comparing to 
> higher precision versions that do validate fairly well as giving 
> smallish residuals.
>
> In[104]:= solns30 =
>  Sort[NSolve[JacobiP[25, -1/2, -1/2, x] == 0, x,
>    WorkingPrecision -> 30]]
>
> Out[104]= {{x -> -0.998026728428271561952336806863}, {x -> \
> -0.982287250728688681085641742865}, {x -> \
> -0.951056516295153572116439333379}, {x -> \
> -0.904827052466019527713668647933}, {x -> \
> -0.844327925502015078548558063967}, {x -> \
> -0.770513242775789230803009636396}, {x -> \
> -0.684547105928688673732283357621}, {x -> \
> -0.587785252292473129168705954639}, {x -> \
> -0.481753674101715274987191502872}, {x -> \
> -0.368124552684677959156947147493}, {x -> \
> -0.248689887164854788242283746006}, {x -> \
> -0.125333233564304245373118759817}, {x -> 0}, {x ->
>    0.125333233564304245373118759817}, {x ->
>    0.248689887164854788242283746006}, {x ->
>    0.368124552684677959156947147493}, {x ->
>    0.481753674101715274987191502872}, {x ->
>    0.587785252292473129168705954639}, {x ->
>    0.684547105928688673732283357621}, {x ->
>    0.770513242775789230803009636396}, {x ->
>    0.844327925502015078548558063967}, {x ->
>    0.904827052466019527713668647933}, {x ->
>    0.951056516295153572116439333379}, {x ->
>    0.982287250728688681085641742865}, {x ->
>    0.998026728428271561952336806863}}
>
> This shows the two results are quite close.
>
> In[106]:= Max[Abs[(x /. solns) - (x /. solns30)]]
> Out[106]= 4.47615*10^-10
>
> This shows the high precision solutions give reasonable residuals.
>
> In[108]:= Max[Abs[JacobiP[25, -1/2, -1/2, x] /. solns30]]
> Out[108]= 0.*10^-12
>
> The machine precision results will not give good residuals. There are 
> two possible (and related) reasons. One is that the conditioning is 
> such that we are bound to lose many digits in assessing residuals (as 
> seems to happen above). The other is that the form of JacobiP 
> polynomials might just be inducing cancellation error.
>
> Certainly this second item is part of the problem:
>
> In[109]:= Max[Abs[Expand[JacobiP[25, -1/2, -1/2, x]] /. solns30]]
> Out[109]= 0.*10^-21
>
> Here are the corresponding machine precision evaluations of residuals.
>
> In[112]:= Max[Abs[JacobiP[25, -1/2, -1/2, x] /. solns]]
> Out[112]= 17.
>
> In[113]:= Max[Abs[Expand[JacobiP[25, -1/2, -1/2, x]] /. solns]]
> Out[113]= 9.31323*10^-9
>
> Daniel Lichtblau
> Wolfram Research
>
Thank you Daniel. And thank you also to everyone who wrote me 
suggestions (both here and in private). The combination of not using 
machine number in the input and increasing the WorkingPrecision seems to 
work fine. Apparently just using one of the two tricks do not yield any 
sensible result.

Jacopo



From mathgroup-adm at smc.vnet.net  Fri Jul 29 02:47:31 2011
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From: Sam McDermott <samwell187 at gmail.com>
Subject: [mg120556] hatched regions, shading, and fills
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Hi,

After looking over some old threads, it looks like there is no elegant
way to add hatch marked regions in the interior of a graphic. This is
regrettable, because it would really simplify things for my particular
task right now.

However, are there other easily accessed directives that might give a
similar result? For instance, filling in a region with polka dots?
Adding diagonal lines? Filling with a very simple texture?

Any help is much appreciated,
Thanks,
Sam



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From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
Subject: [mg120554] Re: Timing progress bar
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On Thu, 28 Jul 2011, Berthold Hamburger wrote:

> Hi,
>
> Is there any possibility to get visible feedback about the progress of a
> calculation via the Timing function? Something like a progress bar or "%
> completed".
>
> Thanks
>
> Berthold
>
>

Berthold,

here are two examples with NDSolve:

tEnd = 100;
ProgressIndicator[Dynamic[currentTime], {0, tEnd}]
currentTime = 0;
NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
   u[t, 0] == Sin[t], u[t, 5] == 0}, u, {t, 0, tEnd}, {x, 0, 5},
  EvaluationMonitor :> (currentTime = t;)]

showStatus[status_] :=
   LinkWrite[$ParentLink,
    SetNotebookStatusLine[FrontEnd`EvaluationNotebook[],
     ToString[status]]];
clearStatus[] := showStatus[""];
clearStatus[]

(* look in the lower left of the FrontEnd *)

NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
   u[t, 0] == Sin[t], u[t, 5] == 0}, u, {t, 0, 10}, {x, 0, 5},
  EvaluationMonitor :> showStatus["t = " <> ToString[CForm[t]]]]



Oliver



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From: Helen Read <readhpr at gmail.com>
Subject: [mg120565] Re: Timing progress bar
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On 7/28/2011 8:00 AM, Berthold Hamburger wrote:
> Hi,
>
> Is there any possibility to get visible feedback about the progress of a
> calculation via the Timing function? Something like a progress bar or "%
> completed".
>
> Thanks
>
> Berthold
>

Have you tried using ProgressIndicator?

-- 
Helen Read
University of Vermont



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From: "Christoph Lhotka" <christoph.lhotka at univie.ac.at>
Subject: [mg120566] Re: Timing progress bar
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hy,

i like to use it a lot:

Dynamic[ProgressIndicator[i]]

where i is a variables between 0 and 1. If you write somewhere in your
notebook the above expression and update i  in your peace of code
accordingly you will have a feeling about the status of the evaluation.

christoph

On 28/07/2011 13:55, Berthold Hamburger wrote:
> Hi,
>
> Is there any possibility to get visible feedback about the progress of a
> calculation via the Timing function? Something like a progress bar or "%
> completed".
>
> Thanks
>
> Berthold
>






From mathgroup-adm at smc.vnet.net  Fri Jul 29 02:52:18 2011
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From: barry brunson <mathisfun at mac.com>
Subject: [mg120562] CompressedData in Buttons
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Sorry if this is really obvious. 

I have some Buttons that display various data, e.g., area, population, density, etc. of U.S. states, and would like to update the data. I have only the buttons, not the Input code that produced them long ago. Using Show Expression on the button reveals that, instead of being in List form, much of the data appears as "CompressedData[ ... (gibberish) ...]. 

How can I recover the original, uncompressed, data so that I can revise it? I found nothing in the Documentation Center or Virtual Book, and none of the  74 hits on wolfram.com appear to shed any relevant light.

Thanks.



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From: "Chris Degnen" <degnen at cwgsy.net>
Subject: [mg120559] Re: FinancialData still broken
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DrMajorBob wrote:
> 
Armand Tamzarian wrote:  
>>
>> Would it be easier to just get the data from Yahoo rather than have to
>> worry about fixes and innoculations?
> 
> I don't know how to do that, but I suspect it's Yahoo sending erroneous  
> prices, anyway.

You can get quotes from Yahoo like this:

quotes = Import[
   "http://download.finance.yahoo.com/d/quote.csv?s=\
CLU11.NYM+^GSPC+^FTSE&f=abdlohg"];
(*www.mathematicacookbook.com/downloads/YahooDataDownload.htm*)
TableForm[
 Map[List, MapThread[StringJoin, {{"WTI - ", "S&P - ", "FTSE - "},
    StringReplace[quotes[[All, 4]], {"<b>" -> "", "</b>" -> ""}]}]],
 TableHeadings -> {None, {"Last Trade"}}]





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From: =?ISO-8859-1?Q?Iv=E1n_Pulido_Sanchez?= <ijpulido.s at gmail.com>
Subject: [mg120561] Mathematica 8 remote parallel kernels
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Hello,

When I try to configure remote kernels in Mathematica via
Evaluation>Parallel Kernel Configuration ... then I go to "Remote Kernels"
and add hosts, after that I try to Launch the remote kernels and only some
of them get launched (the number of them varies),and finally I get a msg
like the following.

KernelObject::rdead: Subkernel connected through remote[nodo2] appears dead.
>> LinkConnect::linkc: Unable to connect to LinkObject[36154 at 192.168.1.104,
49648 at 192.168.1.104,38,12]. >> General::stop: Further output of
LinkConnect::linkc will be suppressed during this calculation. >>

Any ideas how to get this working?

Take into account it sometimes does load some of the remote kernels but
never all of them. Thanks in advance I hope you can help me with this.




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The WeierstrassP function is doubly periodic. Do the numerical evaluation
routines for this function take this into account when evaluating for large
arguments, using a Mod function on the argument?  If not, would there be
something to be gained by the user writing a "front end" for
WeierstrassP[z,{g2,g3}] moving z to the principal period parallelogram?
Especially if g2 and g3 were fixed and z was scanned only along the
direction of one half-period.

 

I'm thinking of something like a high precision mod function followed by a
MachinePrecision evaluation of WeierstrassP.

 

My experience has been that calculating the half-periods from g2-g3 has a
fair loss of significance but if we only had to calculate them once we could
afford extra high precision.

 

David Park

djmpark at comcast.net

 <http://home.comcast.net/~djmpark> http://home.comcast.net/~djmpark/  


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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120564] FinancialData errors
Reply-To: drmajorbob at yahoo.com
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I've found a new (to me) species of error in FinancialData.

One example is the price of the following stock on Feb 10, 2011:

s = "JANFX";
FinancialData["JANFX", "Name"]
FinancialData["JANFX", {{2011, 2, 9}, {2011, 2, 11}}]

"Janus Flexible Bond Fund D Shar"

{{{2011, 2, 9}, 10.25}, {{2011, 2, 10}, 10.23}, {{2011, 2, 11},
   10.24}}

As you see, FinancialData says the price was 10.23.

But I BOUGHT the stock through Janus Funds on that day, and the price was  
10.36.

Moreover, if I go to Yahoo's interactive price chart for the stock at the  
following page, 10.36 is the price I find:

http://finance.yahoo.com/echarts?s=JANFX+Interactive#chart3:symbol=janfx;range=ytd;indicator=volume;charttype=line;crosshair=on;ohlcvalues=0;logscale=on;source=undefined

So... where are the errors really coming from?

This is not an astronomically large price or a Missing["Not Available"]  
price; it's within 2% of the correct price. But it's wrong.

These errors make it very difficult to trust FinancialData for any useful  
purpose.

Bobby

-- 
DrMajorBob at yahoo.com



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From: Leonid Shifrin <lshifr at gmail.com>
Subject: [mg120558] Re: Why FullDefinition does not work in MathLink mode?
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>From the documentation of Definition, in Possible Issues section:


Definition is an output form; it does not evaluate:

In[1]:= Definition[Plus]
Out[1]=
Attributes[Plus]={Flat,Listable,NumericFunction,OneIdentity,Orderless,Protected}

Default[Plus]:=0

In[2]:= FullForm[%]
Out[2]//FullForm= Definition[Plus]


Regards,
Leonid


On Thu, Jul 28, 2011 at 3:55 PM, Alexey Popkov <lehin.p at gmail.com> wrote:

> Hello,
>
> When working with MathKernel in interactive session without of the
> FrontEnd, FullDefinition works as expected:
>
> In[1]:= a=2;FullDefinition[a]
>
> Out[1]= a = 2
>
> But when trying to use it in MathLink mode one get ReturnPacket with
> unevaluated FullDefinition:
>
> In[1]:= kernel=LinkLaunch[First[$CommandLine] <> " -mathlink"];
> LinkRead[kernel];
> LinkWrite[kernel,Unevaluated[a=2;FullDefinition[a]]]
> LinkRead[kernel]//FullForm
>
> Out[4]//FullForm= ReturnPacket[FullDefinition[a]]
>
> The built-in parallelization functions have the same problem:
>
> In[2]:= ParallelEvaluate[a=2;FullDefinition[a]]//FullForm
> Out[2]//FullForm= List[FullDefinition[a],FullDefinition[a]]
>
> One workaround is to use ToString:
>
> In[3]:= ParallelEvaluate[a=2;ToString[FullDefinition[a]]]
> Out[3]= {a = 2,a = 2}
>
> Why FullDefinition returns unevaluated in MathLink session although it
> works as expected in interactive MathKernel session? Is it possible to
> make it working without any glitches?
>
> Any ideas?
>
>


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From: "Berthold Hamburger" <b-hamburger at artinso.com>
Subject: [mg120563] Re: Timing progress bar
References: <201107281155.HAA04128 at smc.vnet.net> <Pine.LNX.4.63.1107280631410.23644 at wopr.wolfram.com>
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Thanks Oliver!

I also found the progress.m package:
http://www.physics.ohio-state.edu/~jeremy/mathematica/progress/

Is this compatible with Mathematica 8?

Regards

Berthold
-- 
Berthold Hamburger - Cellist/Spain
http://www.artinso.com | http://www.astronomy.artinso.com |
http://www.artemis.artinso.com 
http://www.facebook.com/berthold.hamburger 

-----Original Message-----
From: Oliver Ruebenkoenig [mailto:ruebenko at wolfram.com] 
Sent: jueves, 28 de julio de 2011 13:34
To: Berthold Hamburger
Cc: mathgroup at smc.vnet.net
Subject: [mg120563] Re: Timing progress bar

On Thu, 28 Jul 2011, Berthold Hamburger wrote:

> Hi,
>
> Is there any possibility to get visible feedback about the progress of 
> a calculation via the Timing function? Something like a progress bar 
> or "% completed".
>
> Thanks
>
> Berthold
>
>

Berthold,

here are two examples with NDSolve:

tEnd = 100;
ProgressIndicator[Dynamic[currentTime], {0, tEnd}] currentTime = 0;
NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
   u[t, 0] == Sin[t], u[t, 5] == 0}, u, {t, 0, tEnd}, {x, 0, 5},
  EvaluationMonitor :> (currentTime = t;)]

showStatus[status_] :=
   LinkWrite[$ParentLink,
    SetNotebookStatusLine[FrontEnd`EvaluationNotebook[],
     ToString[status]]];
clearStatus[] := showStatus[""];
clearStatus[]

(* look in the lower left of the FrontEnd *)

NDSolve[{D[u[t, x], t] == D[u[t, x], x, x], u[0, x] == 0,
   u[t, 0] == Sin[t], u[t, 5] == 0}, u, {t, 0, 10}, {x, 0, 5},
  EvaluationMonitor :> showStatus["t = " <> ToString[CForm[t]]]]



Oliver




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From: Joseph Gwinn <joegwinn at comcast.net>
Subject: [mg120567] Re: getting ride of 0.i
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In article <j0rips$41r$1 at smc.vnet.net>,
 Armand Tamzarian <mike.honeychurch at gmail.com> wrote:

> On Jul 27, 8:19 pm, Eli Fenichel <Eli.Fenic... at asu.edu> wrote:
> > I am trying to use FindMinimum to minimize the square of a function over 
> many variables (i.e., parameters).  The function itself has the potential
>  to generate imaginary numbers and involves numerical integration (the 
>  parameter values that minimize the function are real).  For clarification 
>  the function in evaluated at multiple points so it can be written as vector. 
>  By squaring the vector the numerical values of the elements are always real 
>  with no imaginary parts.  However, Mathematica often writes x + 0.i, where x 
>  is a some numerical value, for some of the elements.
> >
> > Typically, this can be ignored, the Chop command can be used, or it simply 
> > does not cause problems.  However, I keep getting an error: 
> > [cid:image003.... at 01CC4BBC.E87A0F30]
> >
> > NIntegrate::nlim: "t = Y[1.] is not a valid limit of integration."
> >
> > Y is an array with the parameters to be minimized.
> >
> > However, if evaluate the objective function to be minimized using 
> > replacement rules I get
> > x + 0.i.
> >
> > Is there a way for me tell Mathematica to always treat 0.i as 0 and drop 
> > it?  It seems to be causing problems in the FindMinimum call.
> >
> > Thanks,
> > Eli
> 
> Re[3 + 2 I]

It sounds like the "0 I" is in fact a roundoff error times I, such as 
10^-18 I.  This problem is what Chop[] used for.

Chop[x+0 I] will yield x, if roundoff error is in fact the problem.  One 
can tell Chop[] what threshold to use.

Joe Gwinn



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From: Sebastian Hofer <sebhofer at gmail.com>
Subject: [mg120560] Indeterminate result (numerical problem)
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I have a large symbolic expression which is a sum of a large number of fractions, which is the result of an integral. I can get numerical values by NIntegrate, but eventually the expression should go into a minimization problem, which would be much faster starting from an analytic expression. My troubles start even earlier though. When I try to evaluate the expression by

N[expr/.parameters]

I get Indeterminate. This is true independently of the precision given to N(machine or arbitrary precision) or the chosen parameters, which I initially chose to be rational numbers from an interval of about 1/10 to 10. Simplifying the symbolic expression is not possible due to its sheer size. Simplifying the separate terms and evaluating them gives me a result, but only if I set $MinPrecision=$MaxPrecision as in

In[466]:= Block[{prec=MachinePrecision,$MaxPrecision,$MinPrecision},$MaxPrecision=$MinPrecision=prec;N[slist,prec]//Total]

Out[466]= -0.0000554539 + 0.000957437 I

where slist contains the simplified and evaluated terms. If I work with arbitrary precision instead I get

In[462]:= Block[{prec=16,$MaxPrecision,$MinPrecision},$MaxPrecision=$MinPrecision=prec;N[slist,prec]//Total]

Out[462]= -0.00001695456281093148 + 0.0009204297860912486 I

The latter result coincides with what I obtain from NIntegrate, which I guess is correct. (Is this a valid assumption?)
However, for different values of prec I get very strange results:
prec=17,18,21: same as prec=16
prec=19: -0.001922263621488465141 - 0.003379886739497078367 I
prec: Indeterminate
prec=22: -0.0006868461930781501884710 - 0.0006068980872473164707112 I
...

What is the best way to approach this problem? How do I know which results I can trust? Is there a better approach to start with? All tips are highly welcome!
Sebastian



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From: "Kristof Lebecki" <Kristof.Lebecki at uni-konstanz.de>
Subject: [mg120553] Re: NonlinearModelFit vector valued functions
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Hi everybody,

One year ago following post was placed in this forum (Date: Thu, 4 Feb 2010 06:26:24 -0500 (EST); From: janos). It describes shortly and exactly my problem. Maybe in the last year somebody has found a nice solution?
> NonlinearModelFit is able to fit to a scalar valued function,
> including complex valued functions.
> However, I would like to fit to a vector valued function.
> I can do this using NMinimize,
> but in that case I'll lose all the nice statistics NonlinearModelFit
> provides us.
> Any idea to circumvent this problem?

If you are interested I can describe my problem further, with pleasure ;)
Similarly, as in the case of Janos I use currently FindMinimum. I am happy with it, I use it together with ParallelSum. The only pain are mentioned already missing "nice statistics". 
One option would be to create statistics for FindMinimum. I have read that you can get parameter confidence intervals (mostly interesting for me) when you have Hessian matrix, invert it, etc. Anyhow you will also have to evaluate noise of your data. Shortly speaking: something terribly complex for me.
Other option would be to do some tricks with NonlinearModelFit. If I had 2D real vector function I could for instance transform it into 1D complex function - problem solved. But my function is 3D! I can still imagine other tricks, like mapping N three-dimensional data points into 3N scalar data points. 

But my project is already complex. I have 3D convolutions, my function calculates seconds, so I would prefer "elegant" solution.
So, any help or suggestion is appreciated.
 
With best regards, Kristof
-- 
Kristof M. Lebecki, Dr,
  Fachbereich Physik
  Postfach 674
  Universitat Konstanz
  D-78457 Konstanz, Germany
voice: +49 (0)7531 88 - 3796
  fax: +49 (0)7531 88 - 5325
  web: theorie.physik.uni-konstanz.de/lebecki



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From: DrMajorBob <btreat1 at austin.rr.com>
Subject: [mg120557] FinancialData errors
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FUNNY STORY: Here is Tech Support's non-response to my problem, described  
farther down:

Hello,

Thank you for taking the time to send in this report to us. The  accuracy
of the financial data depends on how quickly our sources  of data provide
updates. A 15 minutes delay on current prices is common. Longer delays are
possible depending on the data. A complete list of the sources queried by
FinancialData is available at:

http://reference.wolfram.com/mathematica/note/FinancialDataSourceInformatio
n.html

We are always looking for alternative sources of financial information to
query. If you have any suggestions on alternate sources of information we
should consider, please let us know.

Sincerely,

Sean Clarke


========================================================================

Now, isn't that special? The data is 5 months old, and TS tells me there's  
a 15-minute delay?

========================================================================


I've found a new (to me) species of error in FinancialData.

One example is the price of the following stock on Feb 10, 2011:

s = "JANFX";
FinancialData["JANFX", "Name"]
FinancialData["JANFX", {{2011, 2, 9}, {2011, 2, 11}}]

"Janus Flexible Bond Fund D Shar"

{{{2011, 2, 9}, 10.25}, {{2011, 2, 10}, 10.23}, {{2011, 2, 11},
     10.24}}

As you see, FinancialData says the price was 10.23.

But I BOUGHT the stock through Janus Funds on that day, and the price was
10.36.

Moreover, if I go to Yahoo's interactive price chart for the stock at the
following page, 10.36 is the price I find:

http://finance.yahoo.com/echarts?s=JANFX+Interactive#chart3:symbol=janfx;range=ytd;indicator=volume;charttype=line;crosshair=on;ohlcvalues=0;logscale=on;source=undefined

So... where are the errors really coming from?

This is not an astronomically large price or a Missing["Not Available"]
price; it's within 2% of the correct price. But it's wrong.

These errors make it very difficult to trust FinancialData for any useful
purpose.

Bobby

-- 
DrMajorBob at yahoo.com



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From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
Subject: [mg120555] Re: how to ListPlot3D large data sets
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On Thu, 28 Jul 2011, Oliver Ruebenkoenig wrote:

> On Wed, 27 Jul 2011, Ted Sariyski wrote:
>
>> Hi,
>>
>> In one attempt MaxPlotPoints did not help, but I was not persistent
>> because I have to resolve other issues before I return to ListPlot3D.
>> One issue is that I cannot pack data. On the big data set I verify that
>> it is numeric array and still  I cannot get it packed. Here is an example:
>>
>> mydata={{0.00100,10.,0.},{0.00100,10.,0.00100},{0.00100,10.,0.00200},{0.00100,10.,0.00300},{0.00100,10.,0.00400}};
>>
>> MatrixQ[mydata,NumericQ]
>> True
>>
>> myPackedData=ToPackedArray[mydata,Real];
>>
>> MatrixQ[myPackedData, NumericQ]
>> True
>>
>> PackedArrayQ[myPackedData]
>> False
>>
>> What I am doing wrong?
>> Thanks,
>> --Ted
>
>
> Ted, possibly ToPackedArray are not on you context path
>
> either try
>
> <<Developer`
>
> or
>
> mydata = {{0.00100, 10., 0.}, {0.00100, 10., 0.00100}, {0.00100, 10.,
>     0.00200}, {0.00100, 10., 0.00300}, {0.00100, 10., 0.00400}};
> myPackedData = Developer`ToPackedArray[mydata, Real];
> Developer`PackedArrayQ[myPackedData]
>
> Oliver


Daniel Lichtblau actually pointed out to me that I might have 
misunderstood the implications of you result.

Some comments: Entering a list like mydata by "hand" (tying, 
cut&copy&paste) will always be an unpacked list.

You'd have to check PackedArrayQ directly on the import of your large data 
set, assuming that returned False, you could wrap a 
N[yourLargeDataImport] around it and see if it then is packed. Should that 
still not be the case, there might be non numerical stuff in your imported 
list and you could give it a shot at removing that and the see if and that 
affects performance for you plot.

I hope this helps in a better way.

Oliver



>
>>
>> On 7/25/2011 7:30 AM, Oliver Ruebenkoenig wrote:
>>> On Sat, 23 Jul 2011, Ted Sariyski wrote:
>>>
>>>> Hi,
>>>> I guess I am pushing the limits of ListPlot3D. I try to ListPlot3d a data
>>>> set with many millions of records. I was not able to get an image from the
>>>> full dataset, it takes forever. If I use e.g. every fifth record, although
>>>> slow, I get an image. The machine running Mathematica is Windows 7 (64 bit),
>>>> has 24 GB RAM and there was no swapping. I wonder what is considered as a
>>>> reasonable data size for ListPlot3D and are there other tools in Mathematica
>>>> for visualization of large data sets?
>>>> Thanks in advance,
>>>> --Ted
>>>>
>>>>
>>>>
>>> Ted,
>>>
>>> is your data packed?
>>>
>>> Developer`PackedArrayQ[yourData]
>>>
>>> Oliver
>>>
>>
>>
>>
>
>



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Date: Fri, 29 Jul 2011 04:43:47 -0400 (EDT)
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From: Richard Kendall-Smith <richpks at gmail.com>
Subject: [mg120569] Re: Roots of a Jacobi polynomial
References: <201107261107.HAA09296 at smc.vnet.net> <4E2EFAAD.60403 at wolfram.com> <201107281154.HAA04077 at smc.vnet.net>
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Hi,

I'm trying to solve a set of non-linear equations by doing the following:

eqlist = {101.74 == a + ((0.055/(479248/c))*(352343 + ((479248/c)*a) -
(158194*b/c))),
 47205 + ((158194/c)*a) - ((61842/c)*b) == 68043.98 + (299.73*b),
 460584 + (479248*(a^2)/(c^2)) - (158194*b*a/(c^2)) + (61842*(b^2)/(c^2))
== 21986 + (3576*c)}

NSolve[eqlist, {a, b, c}, Reals]

But it interprets c as Musical Notation. I have tried altering the variables
to x y and z but then it comes up with a Visual Form error. I'm new to
Mathematica but I don't see why this is happening because aren't we
specifying a, b, c as variables to be solved?

Thanks,

Richard

On Thu, Jul 28, 2011 at 5:54 AM, Jacopo Bertolotti
<J.Bertolotti at utwente.nl>wrote:

> On 07/26/2011 07:34 PM, Daniel Lichtblau wrote:
> > On 07/26/2011 06:07 AM, Jacopo Bertolotti wrote:
> >> Dear MathGroup,
> >> Mathematica implements Jacobi polynomials as JacobiP[n,a,b,x] where n is
> >> the order of the polynomial. As it can be checked plotting it a Jacobi
> >> polynomial has n real roots in the interval [-1,1] and it goes rapidly
> >> to infinity outside this interval (at least when both a and b are>-1).
> >> The problem arise when you try to find the roots of such a polynomial
> >> for a relatively high value of n. As an example the command
> >> NSolve[JacobiP[20, -0.5, -0.5, x] == 0, x] correctly returns
> >>
> >> {{x ->  -0.99702}, {x ->  -0.972111}, {x ->  -0.92418}, {x ->
> >> -0.852398},
> >> {x ->  -0.760555}, {x ->  -0.649375}, {x ->  -0.522529}, {x ->
> >> -0.382672},
> >> {x ->  -0.233449}, {x ->  -0.0784582}, {x ->  0.0784591}, {x ->
> >> 0.233445},
> >> {x ->  0.382684}, {x ->  0.522504}, {x ->  0.649423}, {x ->
> >> 0.760466}, {x
> >> ->  0.852539}, {x ->  0.924002}, {x ->  0.972267}, {x ->  0.996958}}
> >>
> >> while NSolve[JacobiP[25, -0.5, -0.5, x] == 0, x] gives
> >>
> >> {{x ->  -1.01869}, {x ->  -0.979859 - 0.0479527 I}, {x ->  -0.979859 +
> >> 0.0479527 I}, {x ->  -0.870962 - 0.070991 I}, {x ->  -0.870962 +
> >> 0.070991
> >> I}, {x ->  -0.71378 - 0.0505783 I}, {x ->  -0.71378 + 0.0505783 I},
> >> {x ->
> >> -0.571283}, {x ->  -0.486396}, {x ->  -0.367829}, {x ->  -0.248377},
> >> {x ->
> >> -0.125513}, {x ->  -0.0000434329}, {x ->  0.125442}, {x ->  0.2489},
> >> {x ->
> >> 0.365644}, {x ->  0.496977}, {x ->  0.555743}, {x ->  0.717741-
> >> 0.0573399
> >> I}, {x ->  0.717741+ 0.0573399 I}, {x ->  0.87423- 0.0652273 I}, {x ->
> >> 0.87423+ 0.0652273 I}, {x ->  0.977876- 0.0422422 I}, {x ->  0.977876+
> >> 0.0422422 I}, {x ->  1.01494}}
> >>
> >> i.e. both complex roots and roots outside the [-1,1] interval.
> >> Substituting any of these values back into the polynomial easily show
> >> that these values are not roots at all. Also notice that using the
> >> command Root (e.g. Sort@Table[Root[JacobiP[25, -0.5, -0.5, x], i], {i,
> >> 1, 25}]) gives different but still wrong results.
> >> On a related note: NIntegrate sometimes gives wrong results when
> >> integrating function of the form (1-x)^a (1+x)^b f[x] and I feel this
> >> might related to the Gauss-Jacobi quadrature failing to retrieve the
> >> correct roots of a Jacobi polynomial.
> >>
> >> Do anyone have a solution for that?
> >>
> >> Thank you
> >>
> >> Jacopo
> >
> > This is not a well conditioned problem. To do better you'll need to
> > allow NSolve to use higher precision by not giving machine numbers in
> > the input.
> >
> > In[102]:= solns = Sort[NSolve[JacobiP[25, -1/2, -1/2, x] == 0, x]]
> >
> > Out[102]= {{x -> -0.998027}, {x -> -0.982287}, {x -> -0.951057}, {x \
> > -> -0.904827}, {x -> -0.844328}, {x -> -0.770513}, {x -> -0.684547}, \
> > {x -> -0.587785}, {x -> -0.481754}, {x -> -0.368125}, {x -> \
> > -0.24869}, {x -> -0.125333}, {x -> 0.}, {x -> 0.125333}, {x ->
> >    0.24869}, {x -> 0.368125}, {x -> 0.481754}, {x -> 0.587785}, {x ->
> >    0.684547}, {x -> 0.770513}, {x -> 0.844328}, {x -> 0.904827}, {x ->
> >     0.951057}, {x -> 0.982287}, {x -> 0.998027}}
> >
> > These might seem imperfect when you check residuals (see below). In
> > fact they are not at all bad. We first show that by comparing to
> > higher precision versions that do validate fairly well as giving
> > smallish residuals.
> >
> > In[104]:= solns30 =
> >  Sort[NSolve[JacobiP[25, -1/2, -1/2, x] == 0, x,
> >    WorkingPrecision -> 30]]
> >
> > Out[104]= {{x -> -0.998026728428271561952336806863}, {x -> \
> > -0.982287250728688681085641742865}, {x -> \
> > -0.951056516295153572116439333379}, {x -> \
> > -0.904827052466019527713668647933}, {x -> \
> > -0.844327925502015078548558063967}, {x -> \
> > -0.770513242775789230803009636396}, {x -> \
> > -0.684547105928688673732283357621}, {x -> \
> > -0.587785252292473129168705954639}, {x -> \
> > -0.481753674101715274987191502872}, {x -> \
> > -0.368124552684677959156947147493}, {x -> \
> > -0.248689887164854788242283746006}, {x -> \
> > -0.125333233564304245373118759817}, {x -> 0}, {x ->
> >    0.125333233564304245373118759817}, {x ->
> >    0.248689887164854788242283746006}, {x ->
> >    0.368124552684677959156947147493}, {x ->
> >    0.481753674101715274987191502872}, {x ->
> >    0.587785252292473129168705954639}, {x ->
> >    0.684547105928688673732283357621}, {x ->
> >    0.770513242775789230803009636396}, {x ->
> >    0.844327925502015078548558063967}, {x ->
> >    0.904827052466019527713668647933}, {x ->
> >    0.951056516295153572116439333379}, {x ->
> >    0.982287250728688681085641742865}, {x ->
> >    0.998026728428271561952336806863}}
> >
> > This shows the two results are quite close.
> >
> > In[106]:= Max[Abs[(x /. solns) - (x /. solns30)]]
> > Out[106]= 4.47615*10^-10
> >
> > This shows the high precision solutions give reasonable residuals.
> >
> > In[108]:= Max[Abs[JacobiP[25, -1/2, -1/2, x] /. solns30]]
> > Out[108]= 0.*10^-12
> >
> > The machine precision results will not give good residuals. There are
> > two possible (and related) reasons. One is that the conditioning is
> > such that we are bound to lose many digits in assessing residuals (as
> > seems to happen above). The other is that the form of JacobiP
> > polynomials might just be inducing cancellation error.
> >
> > Certainly this second item is part of the problem:
> >
> > In[109]:= Max[Abs[Expand[JacobiP[25, -1/2, -1/2, x]] /. solns30]]
> > Out[109]= 0.*10^-21
> >
> > Here are the corresponding machine precision evaluations of residuals.
> >
> > In[112]:= Max[Abs[JacobiP[25, -1/2, -1/2, x] /. solns]]
> > Out[112]= 17.
> >
> > In[113]:= Max[Abs[Expand[JacobiP[25, -1/2, -1/2, x]] /. solns]]
> > Out[113]= 9.31323*10^-9
> >
> > Daniel Lichtblau
> > Wolfram Research
> >
> Thank you Daniel. And thank you also to everyone who wrote me
> suggestions (both here and in private). The combination of not using
> machine number in the input and increasing the WorkingPrecision seems to
> work fine. Apparently just using one of the two tricks do not yield any
> sensible result.
>
> Jacopo
>
>


From mathgroup-adm at smc.vnet.net  Fri Jul 29 06:09:14 2011
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From: Alexey Popkov <lehin.p at gmail.com>
Subject: [mg120571] Is it possible to delete "context" from the list of loaded Contexts[]?
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Hello,

We can remove all symbols in a particular context by using
Remove["context`*"]. But is it possible to remove "context`" itself
from the system so that it will no longer be listed in Contexts[]?



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From: gopher <gophergoon at gmail.com>
Subject: [mg120570] NIntegrate issue with symbolic parameters
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The docs say that NIntegrate evaluates the integrand with the
variables being symbolic, and then repeatedly evaluates the result
numerically. This is a problem for me as my integrand calls
Eigenvectors which results in a symbolic calculation of the
eigenvectors. If the variables were passed numerically instead, there
would be no problem.

In the end, I get errors about oscillation and convergence, 'slwcon'
and 'eincr', which should never occur since the integrand is actually
constant! (theoretically and also confirmed by Plot3D).

Is there a way out?  Since the details are a bit long, I am putting a
link to the file below:

https://netfiles.uiuc.edu/aroy2/www/nintegrate-issue.nb

Thanks for any help,
Abhishek



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From: A Retey <awnl at gmx-topmail.de>
Subject: [mg120574] Re: CompressedData in Buttons
References: <j0trvu$d59$1 at smc.vnet.net>
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Hi,

> I have some Buttons that display various data, e.g., area,
> population, density, etc. of U.S. states, and would like to update
> the data. I have only the buttons, not the Input code that produced
> them long ago. Using Show Expression on the button reveals that,
> instead of being in List form, much of the data appears as
> "CompressedData[ ... (gibberish) ...].
>
> How can I recover the original, uncompressed, data so that I can
> revise it? I found nothing in the Documentation Center or Virtual
> Book, and none of the  74 hits on wolfram.com appear to shed any
> relevant light.

I think you will just need to apply Uncompress to the "glibberish", 
which should be a string. If you have the CompressedData[..] as an 
expression, you can convert the compressed part with:

CompressedData["..."] /. Verbatim[CompressedData][s_String] :> Uncompress[s]

hth,

albert



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From: gal bevc <gal.bevc at gmail.com>
Subject: [mg120573] Poincare section for double pendulum
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Hello,

I'm a relatively new user of Mathematica, who doesn't have much of
programming skills. For my undergraduate assingment I must analyze chaotic
motion of double pendulum.

Until now i have got system of differential equations for equations of
motion for double pendulum(i have x''[t]=function(t) and
y''[t]=function(t)). System of differential equations can be solved for 4
inital conditions, x[0],y[0],x'[0] and y'[0]. With using function NDSolve i
got functions of angles and angular velocities for upper and lower pendulum
with respect to time, x[t],x'[t],y[t] and y'[t].

To get a poincare section of double pendulum, i have to record position of
y[t] and y'[t] whenever x[t] is equal to zero and the velocity of x'[t] is a
positive number. In the end I must get some sort of phase diagram y[t] and
y'[t].
Because this is a Hamilton non-dissipative system, inital energy of the
system is a constant of time and initial energy is a function of initial
conditions. To get a real poincare diagram i must repeat the procedure
described above for different initial conditions, but for the same energy
level. I need mathematica to use some random numbers for initial conditions
in a way that the initial energy of the system stays the same. So i must
repeat procedure for poincare section(surface of section) for let's say 50
different initial conditions and then display all results in one y[t],y'[t]
diagram.
Hope that someone can help me.

Thank you,
Gal Bevc


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From: Pierpaolo Bernardi <olopierpa at gmail.com>
Subject: [mg120572] Re: FinancialData errors
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On Fri, Jul 29, 2011 at 10:41, DrMajorBob <btreat1 at austin.rr.com> wrote:
> FUNNY STORY: Here is Tech Support's non-response to my problem, described
> farther down:

> I've found a new (to me) species of error in FinancialData.
>
> One example is the price of the following stock on Feb 10, 2011:
>
> s = "JANFX";
> FinancialData["JANFX", "Name"]
> FinancialData["JANFX", {{2011, 2, 9}, {2011, 2, 11}}]
>
> "Janus Flexible Bond Fund D Shar"
>
> {{{2011, 2, 9}, 10.25}, {{2011, 2, 10}, 10.23}, {{2011, 2, 11},
> =C2  =C2  10.24}}
>
> As you see, FinancialData says the price was 10.23.
>
> But I BOUGHT the stoc