Is it possible to get several solutions to an arbitrary equation in Matlab?

7
votes

Suppose I have a function f = @(x) myfun(x);

I can use fzero to get the solution closest to a given x0, but can I get all solutions in a specific area, for instance: -5 < x < 5?

I.e. Is it possible to get a solution similar to the result of roots, but for non-polynomials?

2
matlabequation

2 Answers

6
votes

Yes, you can.

There's a nice submission on the file exchange that allows you to do exactly that. It works by approximating your curve by a Chebychev polynomial, and then finding all real roots of that polynomial.

If you want you can use these estimates for the roots as initial values for fzero, but often (at least for smooth and otherwise well-behaved curves) the accuracy demands can already be met by using a higher-order Chebychev approximation.

For your example, using only 18 function evaluations (I have a slightly modified version of the file, but the essence is the same): 

>> f = @(A) 17.7*sin(A).*cos(A)+87*sin(A).^2-9.65*cos(A)-47*sin(A);
>> R = FindRealRoots(f, -5,5, 17)
R =
  -3.709993256346244
  -3.345207732130925
  -0.201929737187637
   0.572382702285053
   2.573423209113534
   2.937157987217741

>> R2 = R; 
>> funcCount = 0;
>> for ii = 1:numel(R) 
        [R2(ii), ~,~, output]  = fzero(f,R2(ii)); 
        funcCount = funcCount + output.funcCount;
   end
>> max(abs(R2(:)-R(:)))
ans =
    8.564253235401331e-004
>> funcCount
ans = 
    46

e.g., there is only 8 parts per ten thousand improvement for no less than 46 additional function evaluations.

2
votes

First there is Matlab's built in option that uses the symbolic math toolbox: mathworks.com/help/symbolic/mupad_ref/numeric-realroots.html

Another option, if your function is well behaved, is just to feed fsolve with the right guesses, so even though it is using a loop, it is an efficient computation. For example:

A=linspace(-5,5,1000);
f=@(A) 17.7.*sin(A).*cos(A)+87.*sin(A).^2-9.65*cos(A)-47*sin(A)

idx = find(diff(sign(f(A))));
for n=1:numel(idx)
    r(n)=fzero(f,A(idx(n)))
end

r=
  -3.709541990613713
  -3.345170894638306
  -0.202018624930518
   0.572128202319968
   2.573643316565874
   2.938014412541281