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MathGroup Archive 2007

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Re: Fw: Re: Fw: 2

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73724] Re: Fw: Re: Fw: 2
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Mon, 26 Feb 2007 06:17:37 -0500 (EST)
  • References: <errl49$80t$1@smc.vnet.net>

Hi Jerry.

Someone could say this conversation is a little pointless since I
don't have version 4.2.

As I said I could find many occasions that 5.2 shows superior
performance than
4=2E0. But as I understood is a matter of what applications someone is
interested in.

Anyway...
I want you just to execute the following in your version 4.2:

In[183]:=
Integrate[Abs[Cos[u]], {u, 0, x*Pi}]

Out[183]=
Integrate[Abs[Cos[u]], {u, 0, Pi*x}]

(*in version 4.0 you get the incorrect result Sqrt[Cos[Pi x]^2] Tan[Pi
x]; the correct symbolic result
is Sqrt[Cos[Pi x]^2] Tan[Pi x] + 2 Floor[x + 1/2]  see here

http://groups.google.gr/group/comp.soft-sys.math.mathematica/browse_thread/thread/e95bb3767782b1e0/981567b0d6c1e7dc?lnk=gst&q=Integrate&rnum=722&hl=el#981567b0d6c1e7dc
)

So comparing 4.0 and 5.2 (and 4.2 by your own...) what performance do
you prefer?

Incorrect evaluation or no evaluation at all?

Also try the following in 4.2...

In[202]:=
Integrate[Abs[Cos[u]], {u, 0, x*Pi}, Assumptions -> 0 <= x <= 10]
(% /. x -> #1 & ) /@ Range[10]
Out[202]=
Boole[1/2 < x < 3/2] + Boole[Inequality[1/2, Less, x, LessEqual, 10]]
+ Boole[3/2 < x < 5/2] + 2*Boole[3/2 <= x <= 10] +
  Boole[5/2 < x < 7/2] + Boole[7/2 < x < 9/2] + 2*Boole[7/2 <= x <=
10] + Boole[9/2 < x < 11/2] + Boole[11/2 < x < 13/2] +
  2*Boole[11/2 <= x <= 10] + Boole[13/2 < x < 15/2] + Boole[15/2 < x <
17/2] + 2*Boole[15/2 <= x <= 10] +
  Boole[17/2 < x < 19/2] + Boole[Inequality[19/2, Less, x, LessEqual,
10]] + 2*Boole[19/2 <= x <= 10] +
  2*Boole[x == 5/2 || Inequality[5/2, Less, x, LessEqual, 10]] +
2*Boole[x == 9/2 || Inequality[9/2, Less, x, LessEqual, 10]] +
  2*Boole[x == 13/2 || Inequality[13/2, Less, x, LessEqual, 10]] +
  2*Boole[x == 17/2 || Inequality[17/2, Less, x, LessEqual, 10]] +
Boole[Inequality[0, Less, x, LessEqual, 1/2]]*Sin[Pi*x] -
  Boole[1/2 < x < 3/2]*Sin[Pi*x] + Boole[3/2 < x < 5/2]*Sin[Pi*x] -
Boole[5/2 < x < 7/2]*Sin[Pi*x] +
  Boole[7/2 < x < 9/2]*Sin[Pi*x] - Boole[9/2 < x < 11/2]*Sin[Pi*x] +
Boole[11/2 < x < 13/2]*Sin[Pi*x] -
  Boole[13/2 < x < 15/2]*Sin[Pi*x] + Boole[15/2 < x < 17/2]*Sin[Pi*x]
- Boole[17/2 < x < 19/2]*Sin[Pi*x] +
  Boole[Inequality[19/2, Less, x, LessEqual, 10]]*Sin[Pi*x]
Out[203]=
{2, 4, 6, 8, 10, 12, 14, 16, 18, 20}


(*check*)

In[204]:=
(Integrate[Abs[Cos[x]], {x, 0, #1*Pi}] & ) /@ Range[10]
Out[204]=
{2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

In[214]:=
(NIntegrate[Abs[Cos[x]], {x, 0, #1*Pi}] & ) /@ Range[10]
Out[214]=
{2.,4.,6.,8.,10.,12.,14.,16.,18.,20.}

Try also

In[218]:=
Integrate[Abs[Cos[u]], {u, 0, x*Pi}, Assumptions -> 0 <= x <= 10];
FullSimplify[%, x =E2=88=88 Integers]
Out[219]=
Piecewise[{{2, 1/2 < x < 3/2}, {4, 3/2 < x < 5/2}, {6, 5/2 < x < 7/2},
{8, 7/2 < x < 9/2}, {10, 9/2 < x < 11/2},
   {12, 11/2 < x < 13/2}, {14, 13/2 < x < 15/2}, {16, 15/2 < x <
17/2}, {18, 17/2 < x < 19/2},
   {20, Inequality[19/2, Less, x, LessEqual, 10]}}]

Of course nobody is perfect:

In[211]:=
Integrate[Abs[Cos[u]], {u, 0, x*Pi}, Assumptions -> 0 <= x <= 10 && x
=E2=88=88 Integers]
(% /. x -> #1 & ) /@ Range[10]
Out[211]=
1 - 1/Sign[1 - 2*x]
Out[212]=
{2, 2, 2, 2, 2, 2, 2, 2, 2, 2}


Kind Regards
Dimitris







=CE=9F/=CE=97 fizzy =CE=AD=CE=B3=CF=81=CE=B1=CF=88=CE=B5:
> Perhaps I should have added this.....I have upgraded to 5.2 and I realize
> that there are many additions that 4.2 doesnt have, etc.etc.etc.....that =
was
> hardly my point!!!!!.......part of what I was trying to say is that I sti=
ll
> use 4.2 just in case and have found it, at least for some of the things I=
've
> worked on, to be quite helpful to me in  a way that the upgrades were
> not........naturally, 4.2 doesnt work on every ocassion but I have found
> many instances where 4.2 will work although 5.2 wont or the answers are
> completely different......also, in terms of using , for example, a
> subscripted definition , such as
>
> E_subscript_x[x, y, z] , I have also found that 5.2 handles this quite
> differently then 4.2 especially if you take the derivative of the
> function.....(the subscript is gotten from the appropriate Palette).....in
> fact, I can't use this definition in 5.2....
>
> When I sent these comments, I thought that someone, at least at Wolfram,
> would care and would want to investigate this......if something doesnt wo=
rk
> in a New Version but does in an Old Version, strikes me as a Bug in the n=
ew
> software assuming that the earlier Version's answer was, in fact,
> correct....or, at least, it's food for thought...
>
> or I hope it is....Jerry
>
> ----- Original Message -----
> From: "dimitris" <dimmechan at yahoo.com>
> To: <mathgroup at smc.vnet.net>
> Sent: Saturday, February 24, 2007 1:11 AM
> Subject:  Re: Fw: 2
>
>
> > Try also the following links
> >
> > http://documents.wolfram.com/mathematica/Built-inFunctions/NewInVersion=
5=2E0/
> > http://www.wolfram.com/products/mathematica/newin51/
> > http://www.google.com/search?hl=en&q=mathematica+version+5.2
> >
> > May be this material will convince for upgrading!
> >
> > Regards
> > Dimitris
> >
> > =CF/=C7 fizzy =DD=E3=F1=E1=F8=E5:
> >> This is an interesting question for me for another reason.....my
> >> 'favorit=
> > e'
> >> version of Mathematica is still 4.2....I have found many things that
> >> work=
> > in
> >> 4.2 but do not work in 5.2....I have always been meaning to send them =
to
> >> mathgroup but, must admit, I'm too lazy for it.....however, this examp=
le
> >> here is perfect for the ocassion....
> >>
> >> It works in 4.2......interesting also, that the Output in 5.2 for  the
> >> Series[ ] question, the answer seems different....I havent investigated
> >> it
> >> further, just want to point out the interesting differences.....
> >>
> >> Jerry Blimbaum
> >>
> >> ----- Original Message -----
> >> From: "Andrzej Kozlowski" <akoz at mimuw.edu.pl>
> >> To: <mathgroup at smc.vnet.net>
> >> Sent: Thursday, February 22, 2007 3:35 AM
> >> Subject:
> >>
> >>
> >> > Try:
> >> >
> >> > Series[ArcCosh[x], {x, 0, 11}]
> >> >
> >> > and now try
> >> >
> >> > ArcCosh[x] + O[x]^12
> >> >
> >> > At least with my version of Mathematica:
> >> >
> >> > $Version
> >> > 5.2 for Mac OS X (February 24, 2006)
> >> >
> >> >
> >> > I do not get the same answer (in fact in the latter case the input is
> >> > returned unevaluated). With ArcSinh and any other function that I
> >> > have tried in place of ArcCosh  the outputs are always the same.
> >> >
> >> > Andrzej Kozlowski
> >> >
> >> >
> >> >
> >> >
> >> >
> >> >
> >
> >
> >
> >



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