# Efficient way to solve an equation in MATLAB

20 vues (au cours des 30 derniers jours)
Ali Almakhmari le 25 Sep 2022
Commenté : Dyuman Joshi le 26 Sep 2022
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
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### Réponse acceptée

Steven Lord le 25 Sep 2022
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.
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Steven Lord le 25 Sep 2022
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 le 26 Sep 2022
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

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