Index exceeds the number of array elements (1).

Errors from command window when calling function trapode

Error using fzero (line 306)

FZERO cannot continue because user-supplied function_handle ==> @(y)y(n)-y(n-1)-0.5*h*(f(t(n-1),y(n-1))+f(t(n),y(n))) failed with the error below.

Index exceeds the number of array elements (1).

Error in trapode (line 26)

Y = fzero(F,y(n-1));

I think the error is related to my function F has a single variable, y, but I'm creating it with y(n) and y(n-1). However, I don't know how to fix this one, and I couldn't get the other answers on this topic to work.

Thank you,

Rick

function [t,y] = trapode(f,a,b,alpha,N)
t = linspace(a,b,N+1);
y = zeros(size(t));
h = (b-a)/N;
y(1) = alpha;

for n = 2:N+1
    F = @(y) y(n) - y(n-1) - 0.5*h*(f(t(n-1),y(n-1)) + f(t(n),y(n)));
    Y = fzero(F,y(n-1));
    y(n) = Y;
end

end

Answer 1

[Walter Roberson]

fzero only ever passes a scalar as its argument. The "y" that reaches inside F will be a scalar, and so cannot be accessed as y(n) or y(n-1)

You are basically creating a conflict between the variable named y that you are storing values into, and the trial value that is being passed into F. Maybe

    F = @(ty) ty - y(n-1) - 0.5*h*(f(t(n-1),y(n-1)) + f(t(n),ty));
    

but I doubt it.

[Rick Sellers]

Thank you. That worked.

[Rick Sellers]

Made one change--the ty inside the function f I changed back to y(n). Still works, FYI.

[Walter Roberson]

Ummm, no. If changing that to y(n) "works" then you would have

ty - y(n-1) - 0.5*h*(f(t(n-1),y(n-1)) + f(t(n),y(n)))

and the - y(n-1) - 0.5*h*(f(t(n-1),y(n-1)) + f(t(n),y(n))) part would be constant in the input. When you fzero() input_variable minue constant then the only possible solution is that the input variable equals the constant. So if you believe that that code "works" then you should skip the fzero phase and simply have

Y = - y(n-1) - 0.5*h*(f(t(n-1),y(n-1)) + f(t(n),y(n)));