# Solving an implicit function, fsolve vs. fzero

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darius on 4 Jul 2012
Hi,
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?
Thanks.
darius on 5 Jul 2012
Thank you for the comments. fminsearch is also too slow.
x is the only unknown. q is a fraction for sure. The solution to the subsequent problem is known to be greater than the solution tot he current one.

Anton Semechko on 5 Jul 2012
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.
Richard Brown on 5 Jul 2012
Actually, what I said doesn't work anyway! Too early in the morning, no coffee yet :)

Richard Brown on 5 Jul 2012
Edited: Richard Brown on 5 Jul 2012
How about doing a fixed number of Newton steps for each problem? (not necessarily doing a convergence check)
For example:
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
end
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.