To resolve issues starting MATLAB on Mac OS X 10.10 (Yosemite) visit: http://www.mathworks.com/matlabcentral/answers/159016
I'm trying to solve an implicit function for x that looks like this:
A*x^q + Bx - C = 0
The solution is within (0,1). Speed is a huge issue. The solution is used to write down a new problem. (A and B both depend on the previous value of x). Both fsolve and fzero are too slow. Any ideas?
No products are associated with this question.
You can try the following approach:
abc=rand(3,1); % A, B, C parameters q=8; % order of the polynomial
Nmax=1E3; % maximum number of iterations f_tol=1E-6; % convergence tolerance k=0.9; % relaxation parameter, must be in the range (0,1]
x=0.5; % starting point f=abc(1)*x^q + abc(2)*x - abc(3); % initial function value
% search for the zero for i=1:Nmax x=k*x+(1-k)*(x-f); % update x f=abc(1)*x^q + abc(2)*x - abc(3); disp([i f x]) if abs(f)<f_tol, break; end end clear q Nmax f_tol k i
I tested the code given above for integer value of q in the range [1,10], and was able to find a solution in at most ~200 iterations. You can also play around with the k parameter to tune the rate of convergence. Note that setting k too low will cause the search to become unstable.
How about doing a fixed number of Newton steps for each problem? (not necessarily doing a convergence check)
A, B, c, q = ... nSteps = 5; x = 0.5
for i = 1:nProblems % Use x from previous problem to update for i = 1:nSteps x = x - (A*x^q + B*x + c) / (q*A*x^(q-1) + B); end
% Update A, B, c, q to next problem
You could always add in a convergence check, or track the errors if you need to. Of course whether or not this works depends on the parameters of your problem, but with a bit of luck it will.