Index in position 1 is invalid. Array indices must be positive integers or logical values.

Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in midpointfinal (line 21)

k1 = h.*f_m(t(1),y_m(:,1));

clc; clear all;
t=0:0.01:1;
h=0.01;
n=(t(2)-t(1))/h;
alpha=0.5;

%initials%
y_m(1)=2;
y_m(2)=exp(20.*h)+cos(h);
f_m(1)=20.*(y_m(1)-cos(t(1)))-sin(t(1));
f_m(2)=20.*(y_m(2)-cos(t(2)))-sin(t(2));


%Midpoint Two step method%
for i=3:n
    y_m(i)=y_m(i-2)+2.*h.*f_m(i-1);
    f_m(i)=20.*(y_m(i)-cos(t(i)))-sin(t(i));
end

% Initialization with second order Runge-Kutta method
k1 = h.*f_m(t(1),y_m(:,1));
    k2 = h.*f_m(t(1)+alpha.*h, y_m(:,1)+alpha.*k1);
    y_m(:,2) = y_m(:,1) + (1-1/2/alpha).*k1 + k2/2/alpha;

[Torsten]

f_m in your code is an array, not a function. Thus setting

k1 = h.*f_m(t(1),y_m(:,1));
k2 = h.*f_m(t(1)+alpha.*h, y_m(:,1)+alpha.*k1);

where f_m is used as a function makes no sense.

Answer 1

[James Tursa]

Looks like you changed the definition of what f_m is in your code. In these lines f_m appears to be an array intended to hold values:

f_m(1)=20.*(y_m(1)-cos(t(1)))-sin(t(1));
f_m(2)=20.*(y_m(2)-cos(t(2)))-sin(t(2));

But in these lines f_m appears to be a function handle intended to evaluate a derivative:

k1 = h.*f_m(t(1),y_m(:,1));
k2 = h.*f_m(t(1)+alpha.*h, y_m(:,1)+alpha.*k1);

You need to fix up your nomenclature.

Answer 2

[Voss]

t(1) is 0

t=0:0.01:1;
t(1)

ans = 0

and 0 is not a valid index in MATLAB

f_m = [1 2];
f_m(t(1))

Array indices must be positive integers or logical values.

It looks more like you're using f_m as if it were a function.