second order finite difference scheme
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Margaret Winding
am 21 Feb. 2017
Kommentiert: Rena Berman
am 14 Mai 2020
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(t−h))/2h =y′(t)+O(h^2)
(−y(t+2h)+4y(t+h)−3y(t))/2h =y′(t)+O(h^2)
(y(t−2h)−4y(t−h)+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
am 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));
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