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 :)

3 views (last 30 days)
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

Accepted Answer

Walter Roberson
Walter Roberson on 18 Dec 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.

More Answers (1)

Abhishek Gupta
Abhishek Gupta on 18 Dec 2020

Products

Community Treasure Hunt

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

Start Hunting!