Azzera filtri
Azzera filtri

Error calculating an integral: ""Input function must return 'double' or 'single' values. Found 'sym'."" How can i get the %% integral(F_potext,0,a) %% done? Thanks for the help :)

1 visualizzazione (ultimi 30 giorni)
a=3; b=(2/3)*a; h=0.01;
q0 = -1000;
syms x y z c
X_m = [0 0 (x/a)^2 (x/a)^3 (x/a)^4];
Y_n = [0 0 (y/b)^2 (y/b)^3 (y/b)^4];
fn = sym([0 0 0 0 0]);
for i=1:5
for j=1:5
fn(i) = fn(i) + X_m(i)*Y_n(j);
end
end
Pot_ext = [0 0 0 0 0];
for i=1:5
F_potext = @(x) ((q0.*x)./a)*subs(fn(i),y,2*b/3);
Pot_ext(i) = integral(F_potext,0,a)
end

Risposta accettata

Walter Roberson
Walter Roberson il 18 Dic 2020
a=3; b=(2/3)*a; h=0.01;
q0 = -1000;
syms x y z c
X_m = [0 0 (x/a)^2 (x/a)^3 (x/a)^4];
Y_n = [0 0 (y/b)^2 (y/b)^3 (y/b)^4];
fn = sym([0 0 0 0 0]);
for i=1:5
for j=1:5
fn(i) = fn(i) + X_m(i)*Y_n(j);
end
end
Pot_ext = [0 0 0 0 0];
for i=1:5
F_potext = matlabFunction(((q0.*x)./a)*subs(fn(i),y,2*b/3), 'vars', x);
Pot_ext(i) = integral(F_potext,0,a, 'arrayvalued', true);
end
Pot_ext
Pot_ext = 1×5
0 0 -703.7037 -562.9630 -469.1358
The reason for the 'arrayvalued', true is that the first two entries in fn come out as 0, so the code does a
matlabFunction(sym(0), 'vars', x)
which generates @(x) 0.0 as the anonymous code. But when you use that code in integral() or fplot() you have a problem because those pass in arrays of x values and require that you return back an array of the same size, but @(x) 0.0 returns back a single x not an array.

Più risposte (1)

Abhishek Gupta
Abhishek Gupta il 18 Dic 2020

Prodotti

Community Treasure Hunt

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

Start Hunting!

Translated by