second order finite difference scheme

I am given data t=[0 1 2 3 4 5] and y(t)=[1 2.7 5.8 6.6 7.5 9.9] and have to evaluate the derivative of y at each given t value using the following finite difference schemes.
(y(t+h)y(th))/2h =y(t)+O(h^2)
(y(t+2h)+4y(t+h)3y(t))/2h =y(t)+O(h^2)
(y(t2h)4y(th)+3y(t))/2h =y(t)+O(h^2)
I started the code, but I haven't learned what to do in the second order case. This what I have so far for the first given equation:
t= 0: 1: 5;
y(t)= [1 2.7 5.8 6.6 7.5 9.9];
n=length(y);
dfdx=zeros(n,1);
dfdx(t)=(y(2)-y(1))/(t(2)-t(1));
for i=2:n-1
dfdx(1)=(y(i+1)-y(i-1))/(t(i+1)-t(i-1));
end
dfdx(n)=(y(n)-y(n-1))/(t(n)-t(n-1));
the error that returns is "Subscript indices must either be real positive integers or logicals." referencing my use of y(t). How do I fix this to make my code correct?

 Accepted Answer

There's no need for the (t) when you define y(t). Same with dfdx. Also, make sure you change dfdx(1) in the loop to dfdx(i).
t= 0: 1: 5;
y= [1 2.7 5.8 6.6 7.5 9.9];
n=length(y);
dfdx=zeros(n,1);
dfdx=(y(2)-y(1))/(t(2)-t(1));
for i=2:n-1
dfdx(i)=(y(i+1)-y(i-1))/(t(i+1)-t(i-1));
end
dfdx(n)=(y(n)-y(n-1))/(t(n)-t(n-1));

6 Comments

By the way, the gradient function gives the same results.
t= 0: 1: 5;
y= [1 2.7 5.8 6.6 7.5 9.9];
plot(t,gradient(y,t))
Thank you very much! I learned about the gradient function but my professor does not want us to use that for this problem.
I fixed the errors you mentioned, but it now gives me the same error "Subscript indices must either be real positive integers or logicals" but for dfdx=(y(2)-y(1))/(t(2)-t(1))
I don't understand what formatting error is occurring at this point, any ideas? Thank you so much!!
dfdx(1)=(y(2)-y(1))/(t(2)-t(1))
Best wishes
Torsten.
Ah, yes, sorry Margaret; thanks Torsten.
Chad and Torsten,
Thank you so much for your help! I was able to get the correct answer :)
and how will be the code for using a 4-point first derivative?

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