fsolve
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Is there a way to accelerate the fsolve function, with the least lost of precision possible. In:
beta(n+1)=fsolve(F,beta(n))
6 Comments
Accepted Answer
Sean de Wolski
on 17 Jun 2011
preallocate beta
beta = zeros(nmax+1,1);
beta(1) = beta_of_1;
for ii = 1:nmax
beta(ii+1) = fsolve(F,beta(ii));
end
EDIT more stuff:
You calculate:
- 'sqrt((Ko^2-(x)^2))*b': 4x
- 'sqrt((Ko^2*Ed-(x)^2))*a': 4x
- the bessel functions multiple times a pop.
Turn your function handle into a function. Make each of these calculations once, then use them multiple times.
More Answers (1)
Walter Roberson
on 17 Jun 2011
fsolve() can be much faster if you can constrain the range to search in.
2 Comments
Walter Roberson
on 20 Jun 2011
Sorry it turns out that fsolve() has no way of constraining ranges. fzero() can operate over an interval, if your function has only one independent variable.
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