matlab precision error in comparison with fortran's code
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Hi everyone
I need your help with the problem am dealing with. I have converted fortran code to matlab and the matlab's result is not exactly the same with fortran's code.
for i = 1 : n - 1;
sumlv = sym(sumlv) + sym(conj(q(i))).*sym(hq(i));
end; i = fix(n);
aa=vpa(sumlv);
cen = rdx.*aa;
--------------------------
The n=2048. The matrices q and hq have exactly the same values with fortran's code.
Matlab's cen= 1.110535470488675 .Fortran's cen= .110535470488684d+01.
When i use the vpa i have cen=1.1105354704886762457328491316803 which still is not the same with fortran's result.Does anyone know what to do with the specific problem?
As i saw in Matlab's code the higher the number of values in metrices the higher the presicion error is occured in comparison with fortran's code.
Thanks in advance
5 件のコメント
Walter Roberson
2014 年 3 月 3 日
what value are you using for digits() ?
Is it possible that Fortran is the one that is being less precise?
James Tursa
2014 年 3 月 3 日
How do you know that the inputs are "exactly" the same between MATLAB and Fortran? Did you write out a binary file on one side and read it in on the other side? Did you pass inputs directly via mex function? Or what? What does your Fortran code look like?
Strat Mpi
2014 年 3 月 4 日
Strat Mpi
2014 年 3 月 4 日
James Tursa
2014 年 3 月 4 日
"... because I checked ..." doesn't tell me anything. HOW did you check? By comparing the printed digits on the screen of both versions (inadequate)? Or by comparing bit-for-bit the underlying floating point bit patterns (adequate)? Once that is settled, then we can talk about other differences such as compiled Fortran code perhaps holding intermediate results in 80-bit registers but the MATLAB code not doing this, etc.
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その他の回答 (1 件)
Walter Roberson
2014 年 3 月 3 日
0 投票
Consider using the built-in symbolic dot product function. The following post shows how you would access it: http://www.mathworks.com/matlabcentral/answers/15949
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