Hello,
Thanks for replying!
However I don't understand why mathematica 8 student version would differ
from the online mathematica trial version? Because according to below
results from both mathematica 8 and maple 12, the results are the same with
python mpmath, that is after the 186th digit after the digits "5016." the
result shows "77360673..", instead of "7973360555..", as in the trial
version result in my attached file!
I guess the result from the trial version is a tentative result and not the
real thing!
I would not mind a fourth confirmation from anyone else using a bought
mathematica package!
Thanks
Vick
_____
From: Richard Hennessy [mailto:rich.hennessy at verizon.net]
Sent: Sunday, December 09, 2012 4:53 AM
To: Vickram
Cc: maxima mailing list
Subject: Re: [Maxima] Hello
Hi Vickram:
I have access to Maple 12 and Mathematica 8 Student editions. I have
computed this integral with both.
Mathematica
f[x] := 1/ Sqrt[(812/10^7 (1 + x)^4) + (2719188/10^7 (1 + x)^3) +
728/1000];
NIntegrate[f[x], {x, 0, 10^13}, WorkingPrecision -> 2000, AccuracyGoal ->
350, MaxRecursion -> 12];
N[%, 350]
->
3.37384610380011411639276017299071962336315208512187429308244524692045\
4083831592619860587001647852397157609808820051137444090274914547366982\
6567770689784941030012023564186263482078797150167736067372209208550072\
9400684840196362348221869360897821068576910702514214008864442443503717\
6845021973313727527210853511703545228534630327760394887402637131280984\
3
Maple 12,
Digits:=3000;
evalf[350](int(1/sqrt((812/10^7 * (1 + x)^4) + (2719188/10^7 * (1 + x)^3) +
728/1000), x = 0..10^13));
-> 3.37384610380011411639276017299071962336315208512187429308244524692045
4083831592619860587001647852397157609808820051137444090274914547366982
6567770689784941030012023564186263482078797150167736067372209208550072
9400684840196362348221869360897821068576910702514214008864442443503717
68450219733137275272108535117035452285346303277603948874026371313161363
[Mathematica 8 answer] - [Maple 12 answer] = 3.5152*10^-344
My 2 cents worth, I hope this helps. Disclaimer, I am just a
amateur/student. (Maple 12 is faster).
Richard Hennessy
From: Stavros Macrakis <mailto:macrakis at gmail.com>
Sent: Saturday, December 08, 2012 3:17 PM
To: Vickram <mailto:vick1975 at intnet.mu>
Cc: maxima <mailto:maxima at math.utexas.edu> mailing list
Subject: Re: [Maxima] Hello
Vickram,
Thanks for the additional information. I am forwarding your reply to the
whole mailing list, as I am not a specialist in numerical quadrature, but
others on the list are.
-s
On Sat, Dec 8, 2012 at 6:34 AM, Vickram <vick1975 at intnet.mu> wrote:
I am trying to do so for two main reasons:
1) This definite integral is a computation for the Hubble parameter. I use
the Hubble parameter in a python code to calculate cosmological distances
from a given redshift. In order to calculate the conditions when the
universe was several seconds old, the redshift must be in the order of 1e+13
and above, up to 1e+43! However, even using such large redshift the
conditions become negative or meaningless, if higher precision is not
ordered. For me to get a meaningful or positive figure for such large
redshifts, I need a minimum of 200 digits accuracy.
2) My main program to perform these calculations is python with mpmath.
Python mpmath has always been very accurate and allow for the computation of
larger digits accuracy. However, I'm not so sure with regard to its
quadrature module. And so I checked the answer against mathematica (trial
version), and the result diverged at about the 186th digit. So I needed a
third source to know which one is correct!!!
Yeah I checked maxima computation of pi, it worked very well for a 1000
digits. But I just don't know how to use the same precision with regard to
any in-built maxima quadrature module.
I've attached the result from mathematica in the file resultmath.docx
And here is a reproduction of sage or python mpmath result:
mpf('3.37384610380011411639276017299071962336315208512187429308244524692\
045408383159261986058700164785239715760980882005113744409027491454736698\
265677706897849410300120235641862634820787971501677360673722092085500729\
400684840196362348221869360897821068576910702514214008864442443503717684\
502197331334033546973')
The results diverge after the red digits in the above result!
I'd really appreciate if you could just show me how to compute the tanh-sinh
quadrature definite integral of my function with a precision of above 200
digits please!
Thanks
Vick
_____
From: macrakis at gmail.com [mailto:macrakis at gmail.com] On Behalf Of Stavros
Macrakis
Sent: Saturday, December 08, 2012 3:52 AM
To: Vickram Ratnam
Cc: maxima at math.utexas.edu
Subject: Re: [Maxima] Hello
Vickram,
Thanks for your interest in Maxima.
What problem are you trying to solve? i.e. why do you want to calculate a
numeric integral to 200 digits of precision in the first place?
Also, what do you mean by "Maxima cannot compute up to 1e+13!"? Maxima
floating-point numbers have standard IEEE float ranges (up to roughly
1e+308). Maxima bfloat's have essentially unlimited range (10^10^100b0 is
unproblematic).
Perhaps if you could show us a self-contained minimal example of how to
generate the error you're seeing, we could help find the problem.
-s
--
On Fri, Dec 7, 2012 at 12:07 AM, Vickram Ratnam <accts at goldox.mu> wrote:
Hi,
I'm new to Maxima. I am using Maxima for windows!
I'm trying to compute a definite integral using Woollett's tanh-sinh maxima
code and the maxima built-in integration methods but with no success!
My function is the following:
1/sqrt((.0000812*(1+x)^4)+(.2719188*(1+x)^3)+(0*(1+x)^2)+.728)
>From 0 to 1e+13
Using preferably tanh-sinh quadrature and to over 300 digits accuracy!
I can do the same with python mpmath and on mathematica. I wanted to check
the answers from a third source.
But Maxima cannot compute up to 1e+13!
Can I do that or is Maxima limited to certain accuracy?
Thanks
Vick
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