fsolve

Is there a way to accelerate the fsolve function, with the least lost of precision possible. In:

beta(n+1)=fsolve(F,beta(n))

[Andrew Newell]

Why are you re-solving the same problem? If beta(1) is your initial guess, beta(2) should already be accurate to the default tolerance. Perhaps you really want to reduce the tolerance?

[Sean de Wolski]

Unless F is dependent on persistent variables.

[Liber-T]

It is because my beta(n+1) is used to solve an ODE, that give us some number we enter in a matrix, than we the det of the matrix give us the Function F, and we continue until beta(n+1)=beta(n).

[Sean de Wolski]

Why don't you show us some of the code in F to see if that can be further optimized.

[Liber-T]

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]

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

Answer 1

[Sean de Wolski]

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.

[Liber-T]

Thnks, but I already know that trick, is there something else for fsolve?

Answer 2

[Walter Roberson]

fsolve() can be much faster if you can constrain the range to search in.

[Liber-T]

how do I constrain the range

[Walter Roberson]

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.