fsolve

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Liber-T
Liber-T 2011-6-17
Is there a way to accelerate the fsolve function, with the least lost of precision possible. In:
beta(n+1)=fsolve(F,beta(n))
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Liber-T
Liber-T 2011-6-20
F=@(x)10000000000000*det([0 besselj(0,(sqrt(Ko^2*Ed-(x)^2))*b) (i*bessely(0,sqrt((Ko^2*Ed-(x)^2))*b)+besselj(0,sqrt((Ko^2*Ed-(x)^2))*b)) -(i*bessely(0,sqrt((Ko^2-(x)^2))*b)+besselj(0,sqrt((Ko^2-(x)^2))*b));y(length(t),1) -besselj(0,(sqrt(Ko^2*Ed-(x)^2))*a) -(i*bessely(0,sqrt((Ko^2*Ed-(x)^2))*a)+besselj(0,sqrt((Ko^2*Ed-(x)^2))*a)) 0;0 -Ed*besselj(1,(sqrt(Ko^2*Ed-(x)^2))*b)/((sqrt(Ko^2*Ed-(x)^2))) -Ed*(i*bessely(1,sqrt((Ko^2*Ed-(x)^2))*b)+besselj(1,sqrt((Ko^2*Ed-(x)^2))*b))/((sqrt(Ko^2*Ed-(x)^2))) (i*bessely(1,sqrt((Ko^2-(x)^2))*b)+besselj(1,sqrt((Ko^2-(x)^2))*b))/((sqrt(Ko^2-(x)^2)));-(1-((e^2*ne0*besselj(0,mu1)/(me*Eo))/(omega*(omega-i*v))))*-y(length(t),2)/(((Ko^2*(1-((e^2*ne0*besselj(0,mu1)/(me*Eo))/(omega*(omega-i*v))))-(x)^2))) Ed*besselj(1,(sqrt(Ko^2*Ed-(x)^2))*a)/((sqrt(Ko^2*Ed-(x)^2))) Ed*(i*bessely(1,sqrt((Ko^2*Ed-(x)^2))*a)+besselj(1,sqrt((Ko^2*Ed-(x)^2))*a))/((sqrt(Ko^2*Ed-(x)^2))) 0]);
Liber-T
Liber-T 2011-6-20
s=0.1
t=0.001
f=200000000
%a=0.013;
%b=0.015;
%Ed=4.52;
omega=f*2*pi;
%v/omega=t
v=t*omega;
omegap=omega/s;
Eo=8.85418782*10^-12;
muo=1.25663706*10^-6;
Ko=sqrt((omega^2)*Eo*muo);
Ep=1-((omegap^2)/(omega*(omega-i*v)));
The answer here is 8.4049+0.0038*i

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采纳的回答

Sean de Wolski
Sean de Wolski 2011-6-17
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.
  1 个评论
Liber-T
Liber-T 2011-6-17
Thnks, but I already know that trick, is there something else for fsolve?

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更多回答（1 个）

Walter Roberson
Walter Roberson 2011-6-17
fsolve() can be much faster if you can constrain the range to search in.
  2 个评论
Liber-T
Liber-T 2011-6-17
how do I constrain the range
Walter Roberson
Walter Roberson 2011-6-20
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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