Forward, Backwawrd, Central, and Perfect Difference

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]

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

[Jan]

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.

Is this your homework?

[Bailey Smith]

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.

Answer 1

[Ankita Bansal]

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.

Change your code to

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

Answer 2

[Jan]

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.

See also the diff command and gradient.