## Forward, Backwawrd, Central, and Perfect Difference

### Bailey Smith (view profile)

on 14 Jun 2018
Latest activity Edited by Jan

on 18 Jun 2018

### Ankita Bansal (view profile)

Could someone please explain how to use the differences, especially with vectors? I would like to make a chart of the values of the four differences types, along with the corresponding x value. Say x=[0:0.2:0.6], y=sin(x), and yperfect=cos(x), how would I go about this?
UPDATE: This is the code I have to take the forward difference (first order).
function [out] = forwarddiff(x,y)
n=1;
L=length(x);
while n < L
out(n,1)=x(n);
out(n,2)=(y(n+1)-y(n))/(x(n+1)-x(n));
n=n+1;
end
out(L,1)=x(L);
out(L,2)=NaN;
end
X and Y values can be input by the user. My new question is: Would the second order forward difference look something like this?
function [out] = forwarddiff(x,y)
n=1;
L=length(x);
while n < L
out(n,1)=x(n);
out(n,2)=(y(n+2)-2*y(n+1)+y(n))/(x(n+1)-x(n))^2;
n=n+1;
end
out(L,1)=x(L);
out(L,2)=NaN;
end
I get an error about exceeding matrix dimensions when I run this.

John D'Errico

### John D'Errico (view profile)

on 15 Jun 2018
Why not try something, then ask for help when you get stuck? Show what you did. Make an effort.
Jan

### Jan (view profile)

on 15 Jun 2018
I agree with John. Please try it, post the code and ask a specific question. It is unlikely that the forum will do your work in exactly the way you need it.
Bailey Smith

### Bailey Smith (view profile)

on 15 Jun 2018
An old homework, but no Matlab required. I've been wanting to take my old homeworks and code them into Matlab so that I can better understand the program before I take any advanced computing class. If I have time, I will try to get a code posted tonight or tomorrow.

### Ankita Bansal (view profile)

on 18 Jun 2018

Hi, in line
out(n,2)=(y(n+2)-2*y(n+1)+y(n))/(x(n+1)-x(n))^2;
you are getting error because you are trying to access y(n+2) at n=L-1.
function [out] = forwarddiff(x,y)
n=1;
L=length(x);
while n < L-1
out(n,1)=x(n);
out(n,2)=(y(n+2)-2*y(n+1)+y(n))/(x(n+1)-x(n))^2;
n=n+1;
end
out(L-1,1)=x(L-1);
out(L-1,2)=NaN;
end

on 18 Jun 2018
Edited by Jan

### Jan (view profile)

on 18 Jun 2018

A shorter form of your 1st order forward difference:
function [out] = forwarddiff(x,y)
x = x(:); % Consider row vectors
y = y(:);
dy = (y(2:end) - y(1:end-1)) / (x(2:end) - x(1:end-1));
out = [x, [dy; NaN]];
end
If you really want to do this with a loop, care for a pre-allocation. Letting the output grow iteratively wastes a lot of resources.