Efficient way to solve an equation in MATLAB

I have a nested for loop system that will run 251658240...you heard me correctly. There isn't much in this for loop that is time-consuming, except solving this equation: where "x" is a costant that changes with each iteration. The method I am using right now is

syms theta
theta = vpasolve((2*theta + sin(2*theta))==(pi*sin(x)));

Is there a way to make this solving process faster? Cause its soooo time-consuming doing it this way

Answer 1

[Steven Lord]

Solve numerically using fzero. Here I've written a function handle that itself makes function handles. I can pass that generated function handle into fzero to get a solution.

f = @(x) @(theta) 2*theta + sin(2*theta) - (pi*sin(x));
h = f(1) % h "remembers" that x is 1

h = function_handle with value:
    @(theta)2*theta+sin(2*theta)-(pi*sin(x))

sol = fzero(h, 1)

sol = 0.8232

Check the solution

h(sol) % Should be close to 0

ans = 0

Or check explicitly, if the way f creates a function handle looks like "magic".

2*sol + sin(2*sol) - pi*sin(1)

ans = 0

If you're going to solve this repeatedly for potentially the same value of x, you may also want to memoize h.

[Dyuman Joshi]

You can call fzero directly though as well

x=1;
f=fzero(@(theta) 2*theta+sin(2*theta)-pi*sin(x),0)

f = 0.8232

%or in a for loop
for x=1:10
    f(x)=fzero(@(theta) 2*theta+sin(2*theta)-pi*sin(x),0);
end
f

f = 1×10

0.8232 0.9577 0.1113 -0.6966 -1.1050 -0.2231 0.5754 1.2764 0.3360 -0.4580

[Steven Lord]

That works if x has been defined before you create the function handle, but note that changing the value of x after the function handle has been created does not change the function handle.

x = 1;
f = @(y) x+y;
f(2) % 3

ans = 3

x = 999;
f(2) % still 3 not 1002

ans = 3

[Dyuman Joshi]

Yes, I am aware of that. However, what would be the difference between -

f = @(x) @(theta) 2*theta + sin(2*theta) - (pi*sin(x));
h = f(1) % h "remembers" that x is 1

h = function_handle with value:
    @(theta)2*theta+sin(2*theta)-(pi*sin(x))

sol = fzero(h, 1)

sol = 0.8232

F = @(x,theta) 2*theta + sin(2*theta) - (pi*sin(x));
fzero(@(theta) F(1,theta),1)

ans = 0.8232