handles for functions of several variables
16 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Stina Ravdna Lorås
le 18 Mar 2021
Commenté : Stina Ravdna Lorås
le 19 Mar 2021
Im trying to make function handles that make it possible to calculate functions with different values for x1 and x2, but I cannot make it work. Please help?
(This is not the entire code, and it is split into two files)
syms x1 x2
x0 = [-1; -1];
f = @f_function;
J = @J_function;
X = NewtonsMethod(f, J, x0, lim, N)
function f = f_function(x)
f=[(x1*x2-2);
((x1.^4/4) + (x2.^3/3) -1)];
end
function J = J_function(x)
J = jacobian(f, [x1,x2]);
end
File:NewtonsMethod
xn = x0; % initial estimate
n = 1; % iteration number
fn = f(xn); % save calculation
iterate = norm(fn,Inf) > tol;
while iterate
xn = xn - J(xn)\fn;
n = n+1;
X(:,n) = xn;
fn = f(xn);
0 commentaires
Réponse acceptée
Walter Roberson
le 19 Mar 2021
syms x1 x2
x0 = [-1; -1];
fsym = f_function(x1,x2);
f = matlabFunction(fsym, 'vars', {[x1, x2]})
J = matlabFunction(jacobian(fsym), 'vars', {[x1, x2]})
X = NewtonsMethod(f, J, x0, lim, N)
function f = f_function(x1,x2)
f=[(x1*x2-2);
((x1.^4/4) + (x2.^3/3) -1)];
end
Plus de réponses (1)
Jan
le 18 Mar 2021
Modifié(e) : Jan
le 18 Mar 2021
Do you get an error message?
What is x1 and x2 in your functions "f_function" and "J_function"? Do you mean: x(1) and x(2) ?
Be careful with the unary minus in vectors:
f=[(x1*x2-2);
((x1.^4/4) + (x2.^3/3) -1)];
% ^
This can confuse the parser. See:
[1 1] % [1, -1]
[1 -1] % [1, -1]
[1 - 1] % [2]
2 commentaires
Voir également
Catégories
En savoir plus sur Interpolation dans Help Center et File Exchange
Produits
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!