second order finite difference scheme

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Margaret Winding
Margaret Winding 2017년 2월 21일
댓글: Rena Berman 2020년 5월 14일
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?

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Chad Greene
Chad Greene 2017년 2월 21일
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));
  댓글 수: 6
Margaret Winding
Margaret Winding 2017년 2월 23일
Chad and Torsten,
Thank you so much for your help! I was able to get the correct answer :)
alburary daniel
alburary daniel 2018년 8월 3일
and how will be the code for using a 4-point first derivative?

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