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MathGroup Archive 2005

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Re: Re: [Mathematica 5.1] Bug Report - Two numerical values for a same variable

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54219] Re: [mg54195] Re: [Mathematica 5.1] Bug Report - Two numerical values for a same variable
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sun, 13 Feb 2005 22:17:06 -0500 (EST)
  • References: <cuhr4q$978$1@smc.vnet.net> <cukc65$lup$1@smc.vnet.net> <200502130521.AAA03676@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

>> But wasn't Ling _already_ using arbitrary-precision numbers in the input?

No, he was using exact numbers, and got an answer that was still exact. The substitution

-529/20000 -> -529/20000`20

made the expression approximate, but more than machine-precision. Numbers like 20000`20 are called "arbitrary precision" for some unfathomable reason.

Bobby

On Sun, 13 Feb 2005 00:21:26 -0500 (EST), David W. Cantrell <DWCantrell at sigmaxi.org> wrote:

> Maxim <ab_def at prontomail.com> wrote:
>> On Fri, 11 Feb 2005 08:42:02 +0000 (UTC), Ling Li <ling at caltech.edu>
>> wrote:
>>
>> > Hi,
>> >
>> > Mathematica v5.1 is much better than v5.0 on the accuracy of
>> > integration. I am not talking about numerical accuracy---v5.0 sometimes
>> > gives out two answers that are off by a constant for the same input.
>> >
>> > However, I still get two different numerical values for a same variable
>> > in v5.1
>> >
>> > a=Exp[Integrate[(Cos[w]-Exp[-w^2/2]*Cos[23/100*w])/w, {w,0,Infinity}]]
>> > N[a]
>> > N[a,20]
>> >
>> > Thanks for looking into this problem.
>> > --Ling
>
> I apologize for not having checked the result given in my previous post
> via numerical integration. Had I done so, I would have realized that the
> result was incorrect. Your result of approx 0.5439146 indeed appears to be
> correct.
>
> Problems may be compounded by the fact that I'm still using version 5.0.
>
>> The expression which is evaluated incorrectly is N[a, 20] (I get
>> Divide::infy warnings and 0.52983935469483822112 as the output). This is
>> not so unusual when Hypergeometric and MeijerG are involved; for example,
>> see http://forums.wolfram.com/mathgroup/archive/2004/Jul/msg00185.html .
>> If we start with arbitrary-precision numbers in the input, we obtain the
>> correct result without any problems:
>
> But wasn't Ling _already_ using arbitrary-precision numbers in the input?
>
> David
>
>> In[1]:=
>> f[w_] = (Cos[w] - E^(-w^2/2)*Cos[23/100*w])/w;
>> a = E^Integrate[f[w], {w, 0, Infinity}];
>> a /. -529/20000 -> -529/20000`20
>>
>> Out[3]=
>> 0.54391462270509428
>>
>> Here are two ways to verify the result. First, you can use the lim
>> function from
>> http://forums.wolfram.com/mathgroup/archive/2003/Aug/msg00233.html to
>> simplify the derivative of Hypergeometric1F1:
>>
>> In[7]:=
>> a /. Derivative[1, 0, 0][Hypergeometric1F1][a_, b_, z_] :>
>>    lim[(Hypergeometric1F1[a + eps, b, z] -
>>          Hypergeometric1F1[a, b, z])/eps,
>>      eps -> 0]
>> N[%, 20]
>>
>> Out[7]=
>> E^((-EulerGamma + (529*HypergeometricPFQ[{1, 1}, {3/2, 2},
>> -529/20000])/10000 - Log[2])/2)
>>
>> Out[8]=
>> 0.54391462270509428214
>>
>> Another way is to evaluate the integral numerically (well, almost). To do
>> that, we will split the integration range into [0, 1] and [1, Infinity)
>> and then integrate the term with E^(-w^2/2) using Method ->
>> DoubleExponential and integrate Cos[w]/w symbolically:
>>
>> In[9]:=
>> E^(NIntegrate[f[w], {w, 0, 1}, WorkingPrecision -> 30] +
>>      (If[FreeQ[#, E],
>>         Integrate[#, {w, 1, Infinity}],
>>         NIntegrate[#, {w, 1, Infinity},
>>           Method -> DoubleExponential, WorkingPrecision -> 30]
>>       ]& /@ Expand[f[w]]))
>>
>> Out[9]=
>> 0.5439146227050942821365022967
>>
>> We could also do it in a purely numerical way, integrating the
>> oscillatory terms on [1, Infinity) separately by using Method ->
>> Oscillatory and increasing the value of MaxRecursion, but I wasn't able
>> to get the result with more than 15 digits of precision then.
>>
>> Mathematica has trouble with high-precision evaluation of MeijerG as
>> well: N[MeijerG[{{0, 1}, {}}, {{1/2}, {}}, 1/100], 20] seems unable to
>> find even one significant digit with any setting of $MaxExtraPrecision,
>> and gives an incorrect result if we increase $MinPrecision (MeijerG[{{0,
>> 1}, {}}, {{1/2}, {}}, z] is O[Sqrt[z]] when z->0). Perhaps there are no
>> general enough methods to evaluate MeijerG in such cases, but the claim
>> that Mathematica "can evaluate special functions with any parameters to
>> any precision" seems a little too strong anyway.
>>
>> Maxim Rytin
>> m.r at inbox.ru
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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