second order finite difference scheme

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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

Chad Greene
Chad Greene on 21 Feb 2017
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
Margaret Winding
Margaret Winding on 23 Feb 2017
Chad and Torsten,
Thank you so much for your help! I was able to get the correct answer :)
alburary daniel
alburary daniel on 3 Aug 2018
and how will be the code for using a 4-point first derivative?

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