How do I create variable vectors with a new value for each iteration of a loop?

I wrote the following function, where t and y are supposed to give values for each iteration of the Euler method. However, I don't know how to properly allocate space for the values, nor ensure that the values change and are added to a 1Xn array for each iteration of the loop. Thanks!

function [t,y]=eulerMethod(f3, dt, Tf, t0, y0)
n=(Tf-t0)/dt;
nf=round(n,1);
y= zeros(1, nf);
yp = zeros(1, nf);
yn = zeros(1, nf);
tp= zeros(1, nf);
yp=y0;
tp(n+1,:);
for n=0:nf
    tp(n+1)=t0+n*dt;
    f3p(n+1)=f3(tp(n+1),yp(n+1));
    yn(n+1)=yp(n+1)+dt*f3p(n+1);
    yp(n+1) = yn(n+1);
end
t=tp;
y=yp;
end

Answer 1

[darova]

Here are some corrections

% yp=y0;      % you are replacing variable yp with y0
yp(1) = y0;   % you need to replace first value only

corrections2

tp= zeros(1, nf);   % length of array nf
for n=1:nf-1        % start from '1' index
    tp(n)=t0+n*dt;  % you can access tp(n+1). It doesn't exist
    f3p(n)=f3(tp(n),yp(n));
    yp(n+1) = yp(n) + dt*f3p(n);
    %yp(n+1) = yn(n+1); % you don't need yn variable at all
end

[Ragini Ravichandren]

Hi!

Thanks so much! So, I ran the code as follows:

function [t,y]=eulerMethod(f3, dt, Tf, t0, y0)
n=(Tf-t0)/dt;
nf=round(n,1);
y= zeros(1, nf);
yp = zeros(1, nf);
tp= zeros(1, nf);
yp(1)=y0;
for n=1:nf
    tp(n)=t0+n*dt;
    f3p(n)=f3(tp(n),yp(n));
    yp(n+1)=yp(n)+dt*f3p(n);
end
t=tp;
y=yp;
end

However, the final y value (2.310559058101177e4) is different from the one I got using the previous code: 2.198756668645101e+04, probably since the for loop starts at 1, rather than 0, and the loop is supposed to build on the inital tn value which should be t0+n*dt, where n=0. Is ther any way I could start the loop at 0 vs 1?

[darova]

Increase size of tp

tp= zeros(1, nf+1);   % length of array nf
for n=1:nf            % till the end

Modify tp a bit

tp(n)=t0+(n-1)*dt;  % you can access tp(n+1). It doesn't exist