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Re: calculation of exp(200)
Posted:
Jul 10, 2014 12:51 AM


"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/determinantofanillconditionedmatrix/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.



