       Re: Re: Accuracy and Precision

• To: mathgroup at smc.vnet.net
• Subject: [mg37119] Re: Re: Accuracy and Precision
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Thu, 10 Oct 2002 03:21:01 -0400 (EDT)
• References: <ao0t54\$gqo\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Bobby,

You note some important points but there are situations in which raising
accuracy is a practical necessity.

1) It may be that a calculation (perhaps describing a real world phenonenon)
necessitates our raising the precision of inuts so as to work to high
precision internally even when the output is not very sensitive to changes
in input.

Compare the following graphs

pts = Table[({#1, Sin[#1]*10^#1*Log[1 + 10^(-#1)]} & )[x], {x, 15., 20.,
0.1}];
ListPlot[pts, PlotJoined -> True]

pts = Table[({#1, Sin[#1]*10^#1*Log[1 + 10^(-#1)]} & )[
SetAccuracy[x, 20]], {x, 15., 20., 0.1}];
ListPlot[pts, PlotJoined -> True]

2) Another situation in which high precision numbers are needed is when a
computation with exact numbers  would be slow or might use up all the memory
available. We then need to replace the exact numbers with high precision
bigfloats  which causes the number of digits used internally to be
restricted and the maximum error in the output to be reported.
The N function will raise the precision of the exact numbers to try and
reach the requested precision.

There are also concerns with, for example, how the replacing bigfloats are
related to the original numbers.
How important this depends on the use to which the calculation is being put
and how sensitive the output is to changes in inputs - in the plotting above
it is not important.

--
Allan

---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"DrBob" <drbob at bigfoot.com> wrote in message
news:ao0t54\$gqo\$1 at smc.vnet.net...
> Let's be realistic.  If you want 60 digits of precision, too bad! -- in
> the real world.  There's nothing we can measure that closely.  Drug
> concentrations in clinical trials are generally measured within 15%, for
> instance.  Even machine precision is more than can be realistically
> expected in any application I can think of.  Even getting a satellite to
> Jupiter probably involves more error in the final result than machine
> precision.  (If not, it's because we rely on ongoing corrections and
> natural factors that put the satellite where it should be, such as
> gravity drawing it toward each rendezvous -- not on that kind of
> precision in propulsion or guidance.)
>
> So... unless all numerics in a problem have a theoretical origin, and
> could be represented in Mathematica as Infinite precision expressions...
> all this talk of higher-precision computation seems futile.
>
> The realistic question is this: given that I have confidence, say, in 6
> digits of precision for the inputs of an expression, how many digits can
> I trust in the end result?  Giving inputs MORE precision than they
> deserve isn't the best way to answer that question.  Here are two
> methods of answering it in Mathematica.  One uses "bignums" and the
> other uses Intervals.
>
> Repetitious trial and error are NOT required either way.
>
> BIGNUMS:
>
> ClearAll[a, b, f, x, y]
> f = x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b);
> a = SetPrecision[77617, 6];
> b = SetPrecision[33096, 6];
> x = SetPrecision[33375/100, 6];
> y = SetPrecision[55/10, 6];
> InputForm[f]
>
> -4.1396`-12.5121*^19
>
> Several previous solutions have set the precision or accuracy of f
> before giving a and b (and possibly x and y) values.  That results in
> making the exponents imprecise along with all coefficients (not just x
> and y), which may or may not be what we want.
>
> INTERVALS:
>
> I'll first digress to figure out what Interval is equivalent to 6-digit
> precision.  You might not actually do this if you like the Interval
> method, but you have to decide SOMEHOW what Interval width to use.
>
> nums = {77617, 33096, 33375/100, 55/10};
> (Interval[SetPrecision[#1, 6]] & ) /@ nums /. Interval[{a_, b_}] :>
> 2*((b - a)/(b + a));
> InputForm[%]
>
> {3.2209438653903`0.207*^-6,
>   3.7768914672467`0.2761*^-6,
>   2.9260299625467`0.1652*^-6,
>   2.7743252840909`0.1421*^-6}
>
> For these numbers, # + Interval[{-1,1}]*#/630000& gets us very close, so
> I'll use that.  The second method therefore is:
>
> g = #1 + Interval[{-1, 1}]*(#1/630000) & ;
> a = g;
> b = g;
> x = g[333.75];
> y = g[5.5];
> f = x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b)
> (Min[f] + Max[f])/2
>
> Interval[{-2.136361928005054*^32, 2.1363651195928296*^32}]
>
> 1.5957938878075177*^26
>
> Either method shows the answer has no trustworthy digits, but I think
> the second result is far easier to interpret.
>
> Here's another example, using the Sin function, whose derivative is Cos,
> whose magnitude is bounded by one.  The precision of the Sin of a number
> should be GREATER than the precision of the number itself, especially
> when Cos is small.
>
> a = SetPrecision[Pi/2, 6];
> InputForm[Sin[a]]
>
> 0.9999999999990905052982256654`11.6078
>
> a = g[N[Pi/2]];
> Sin[a] - 1
>
> Interval[{-3.1084024243455137*^-12, 0}]
>
> I hope this was worthwhile to someone.
>
> DrBob
>
> -----Original Message-----
> From: Peter Kosta [mailto:pkosta2002 at yahoo.com]
To: mathgroup at smc.vnet.net
> Subject: [mg37119]  Re: Accuracy and Precision
>
>
>
> --- Daniel Lichtblau <danl at wolfram.com> wrote:
> > Peter Kosta wrote:
> > >
> > > Andrzej Kozlowski <andrzej at tuins.ac.jp> wrote in
> > message news:<anp065\$qtb\$1 at smc.vnet.net>...
> > > > On Friday, October 4, 2002, at 06:01 PM, DrBob
> > wrote:
> > > >
> > > >[...]
> > > >
> > > > I would say this is correct and show that
> > SetPrecision is very useful
> > > > indeed. It tells you (what of course you ought
> > to already know in this
> > > > case anyway) that machine precision will not
> > give you a realiable
> > > > answer in this case. If you know your numbers
> > with a great deal of
> > > > accuracy you can get an accurate answer:
> > > >
> > > > In:=
> > > > f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 -
> > b^6 -
> > > >          121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100];
> > > > a=SetPrecision[77617.,100];  b =
> > SetPrecision[33096.,100];
> > > >
> > > >
> > > > In:=
> > > > {f, Precision[f]}
> > > >
> > > > Out=
> > > >
> >
> {-0.82739605994682136814116509547981629199903311578438481991\
> > > > 781484167246798617832`61.2597, 61}
> > > >
> > >
> > > Congratulations! You just requested accuracy of
> > 100 for f and got 61 (
> > > to convince yourself add Accuracy[f] to In).
> > If In one
> > > replaces SetAccuracy by SetPrecision the result is
> > similar.
> > >
> > > PK
> > > [...]
> >
> > One has (initially) an accuracy of 100 for an
> > expression that contains
> > variables.
> >
> > In:= Clear[a,b,f]
> >
> > In:= f = SetAccuracy[333.75*b^6 +
> > a^2*(11*a^2*b^2 - b^6 -
> > 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 100];
> >
> > In:= Accuracy[f]
> > Out= 100.
> >
> > Now we assign values to some indeterminants in f.
> >
> > In:= a = SetPrecision[77617.,100]; b =
> > SetPrecision[33096.,100];
> >
> > In:= {f, Precision[f], Accuracy[f]}
> > Out=
> >
> {-0.8273960599468213681411650954798162919990331157843848199178148,
> >
> > 61.2599, 61.3422}
> >
> > The precision and accuracy has dropped. This is all
> > according to
> > standard numerical analysis regarding cancellation
> > error. You'll find it
> > in any textbook on the topic.
> >
>
> Assume that I want accuracy and precision of 100 for
> f. You advice me to make experiments to find out, what
> should be the initial precision and accuracy of a and
> b to reach the requested accuracy and precision for f.
> Notice, that you cannot just repeat I, we saw
> already what happens. I have to re-type I, I,
> I, I, I, and I as many times as needed
> to get f with accuracy and precision 100.
>
> Dan, you simply advocate to do MANUAL WORK that should
> be done by machine.
>
> Let's suppose that in the above example I just want 60
> digits not 61. Precisely, I want 60 digits and nothing
> or zeros afterwards. Let's see if I could use
> SetAccuracy.
>
> In:=
> SetAccuracy[%, 60]
>
> Out=
> -0.82739605994682136814116509547981629199903311578438481991781
>
> In:=
> % // FullForm
>
> Out//FullForm=
> -0.827396059946821368141165095479816291999033115784384819917814841672467
> 988`\
> 59.9177
>
> Oops, it did not work (as expected). Let's highlight
> with mouse the expression in Out and copy to a new
> cell. Oops, we got
> -0.827396059946821368141165095479816291999033115784384819917814841672467
> 988`\
> 59.9177
> again. Let's change Out to a text cell and then
> copy.
>
> In:=
> -0.82739605994682136814116509547981629199903311578438481991781
>
> Out=
> -0.82739605994682136814116509547981629199903311578438481991781
>
> Success? Not so fast.
>
> In:=
> % // FullForm
>
> Out//FullForm=
> -0.827396059946821368141165095479816291999033115784384819917809999999999
> 998635\
> 08`59.2041
>
> Dan, is there any simple way to get what I want?
>
> As I repeated already number of times, at this stage
> of the development of computer technology, software
> should do it for me (!). We both know that this is
> doable. Some of the textbooks that you just advised me
> to read describe it. As a developer of Mathematica,
> tell us why do you consider this to be a bad idea?
>
> Peter Kosta
>
> > As for what happens when you artificially raise
> > precision (or accuracy)
> > of machine numbers far beyond that guaranteed by
> > their internal
> > representation, that falls into to category of
> > garbage in, garbage out.
> > It is, howoever, valid to use SetPrecision to raise
> > precision in
> > (typically iterative) algorithms where significance
> > arithmetic might be
> > unduly pessimistic due to incorrect assumptions
> > of numerical error. Examples of such usage have
> > appeared in this news
> > group.
> >
> >
> > Daniel Lichtblau
> > Wolfram Research
>
>
> __________________________________________________
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>
>

```

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