Efficient way to solve an equation in MATLAB

20 vues (au cours des 30 derniers jours)
Ali Almakhmari
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

Réponse acceptée

Steven Lord
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.
  3 commentaires
Steven Lord
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
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

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Produits


Version

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by