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Topic: calculation of exp(200)
Replies: 5   Last Post: Jul 10, 2014 12:51 AM

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kumar vishwajeet

Posts: 89
Registered: 4/6/10
Re: calculation of exp(200)
Posted: Jul 10, 2014 12:51 AM
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"Steven Lord" <Steven_Lord@mathworks.com> wrote in message <lpjplt$ler$1@newscl01ah.mathworks.com>...
>
> "kumar vishwajeet" <kwzeet@gmail.com> wrote in message
> news:lphu22$5i$1@newscl01ah.mathworks.com...

> > "Nasser M. Abbasi" wrote in message <lpfj5e$4c4$1@speranza.aioe.org>...
> >> On 7/7/2014 8:26 PM, kumar vishwajeet wrote:
> >> > I want to find exp(N) where N can range from 0 to 200.
> >> >Is there any method to calculate it with good accuracy?

> >>
> >> as James said, you can use vpa
> >>
> >> vpa(exp(sym(200)),500)
> >> 7225973768125749258177477042189305697356874428527
> >> 31928403269789123221909361473891661561.9265890625
> >> 7055746840204310142941817711067711936822648098307
> >> 7273278800877934252667473057807294372135876617806
> >> 9702350324220401483115192088442120622525378469924
> >> 9272881421981093552839024711140014485905504285329
> >> 2850053281888896583514044902234770562940736477036
> >> 9022153799888245922895391403712478563813079673045
> >> 5316370477078232246232166391756343572294659379732
> >> 9550687822348546666476886365318979823972170480437
> >> 40943587594
> >>
> >> But you have to do all the rest of the computation
> >> is syms land, otherwise what is the point of
> >> getting this accuracy if you can't use it in
> >> numerical matlab.
> >>
> >>
> >>

> > Actually I have a matrix whose elements range from exp(200) to exp(-200).
>
> Is your matrix a double precision matrix or a symbolic matrix?
>

> >The matrix is ill conditioned with conditioning number of 10^167.
>
> So ... if you're doing ANYTHING in double precision with this matrix, you're
> getting garbage.
>
> http://en.wikipedia.org/wiki/Condition_number
>
> "As a general rule of thumb, if the condition number is 10^k , then you may
> lose up to k digits of accuracy on top of what would be lost to the
> numerical method due to loss of precision from arithmetic methods."
> [replaced images with text.]
>
> By this guideline, you're losing roughly 167 digits of accuracy for the
> numbers you know to roughly 16 digits.
>

> > I am calculating determinant of this matrix. MATLAB gives it in the
> > 10^-13. The matrix is posted in the following link.
> > http://math.stackexchange.com/questions/859597/determinant-of-an-ill-conditioned-matrix/859614#859614
> > So, I was just wondering whether to rely on this calculation of
> > determinant or not.

>
> That looks like you're trying to perform the computations in double
> precision, in which case I would try not to have anything to do with this
> matrix at all.
>
> What is the underlying problem that you're trying to use this problem to
> solve? There may be a way to solve that problem that doesn't require
> creating a matrix whose elements span 200+ orders of magnitude.
>
> --
> Steve Lord
> slord@mathworks.com
> To contact Technical Support use the Contact Us link on
> http://www.mathworks.com


Here is the link to the calculations I am doing
http://www.docdroid.net/ej8l/mathworks.pdf.html
Depending upon different conditions I am calculating f_ij, g_ij and h_ij, where, i,j indicate the row and column of matrix. Instead of using \textit{hypergeom} command of MATLAB, I am computing hypergeometric series term by term and checking for convergence. Because of very large argument of hypergeometric function, I am making all calculations using log and then I convert those values back to original. If you want I can upload the code as well.
Thanks.



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