Re: How to use GNU MP in octave?

[Ben Barrowes]

I have written a mex-based toolbox for ML which calls GMP and MPFR 
routines to achieve arbitrary precision in ML:
http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?objectId=6446&objectType=FILE
This toolbox defines a new mp data type and overloads common functions 
and operators using an @mp directory.
I have heard that there is mex functionality written into octave, but I 
have not experimented with it. If the mex wrapper in octave is 
substantially similar to ML's, perhaps this toolbox could be ported 
without too much trouble.
   Having the Mathematica toolbox available to octave users could also 
add symbolic capability to octave for those with a MMA installed on 
their system:
http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?objectId=6044&objectType=FILE
   Any comments on how easy/difficult a port looks by taking a quick 
once over of these toolboxes?

Ben Barrowes


Shai Ayal wrote:

> Javier Arantegui wrote:
>
> > Hello,
> >
> > El Martes, 24 de Mayo de 2005 05:28, Keith Goodman escribió:
> >
> > > Here's a quote from google
> >
> >
> > [...]
> >
> > > Let x = lg(8600!). Since lg(xy)=lg(x)+lg(y) you have
> > >
> > > x = sum(lg(k),k=1..86000)
> > >
> > > Define floor(x) as the integer part of x. Then
> > >
> > > 86000! = 10^(x-floor(x))*10^floor(x).
> > >
> > > You can evaluate x to the degree of accuracy you want, for example,
> > > using Maple I got:
> > >
> > > 86000! = 7.91222558*10^372239
> >
> >
> >
> > Unfortunately, Octave doesn't buy the trick :-(:-(
> >
> > octave:4> k=1:1:86000;
> > octave:5> x=sum(log10(k))
> > x =  3.8702e+05
> > octave:6> 10^(x-floor(x))*10^floor(x)
> > ans = Inf
>
> Octave doesn't buy this trick since the maximum double value is ~ 
> 10^300. I think Keith meant to say:
>
> use octave to calc 10^(x-floor(x)) which is hard to do by hand, and 
> multiply by 10^floor(x) in your head which is easy, (at least to me ...)
>
>
> octave:8>  k=1:1:86000;
> octave:9>  x=sum(log10(k));
> octave:10> format long
> octave:11> x
> x = 387020.407702849
> octave:12> floor(x)
> ans = 387020
> octave:13> 10^(x-floor(x))
> ans = 2.55683586378090
>
> so your answer is approximately 2.55683586378090*10^387020, (or 
> 2.55683586378090e+387020 in octaves output format). 16 decimal digits 
> is the limit of double precision floting point used in octave. If you 
> need more precision, than other programing tools, as suggested in 
> previous emails in this thread are appropriate.
>
> Note that this result is different from the maple result given above, 
> probably due to roundoff errors in the sum. I suppose this could be 
> improved with some more perusing, although the double precision output 
> of maple leads me to suspect that tin his case a a similar algorithm, 
> or some approximation was used, not real arbitrary precision arithmetic.
>
>
>
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--
---- Benjamin E. Barrowes ------- http://alum.mit.edu/www/barrowes
Los Alamos National Laboratory               address@hidden
Biophysics Group P-21, MS-D454                 Phone:(505)606-0105
Los Alamos, NM 87544                             FAX:(270)294-1268
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