Vandermonde Matrix and an Error Vector

format long 

n = inputdlg('Please enter a series of numbers seperated by spaces/commas: ')
numbers = str2num(n{1});
X = 0
V = fliplr(vander(numbers))
r_size = size(V,2);
c_size = size(V,1);
B = 1+(r_size./numbers)
X = B/V

Here's what I have so far. I've been running the program with n = 5, 15, 20, and basically I want to implement an a way of calculating an error vector using Error = (X - Dt), where Dt = [1, 1, 0, 0, 0, . . . , 0]T.

I'm not really quite sure how to go about this, and is this a good way to solve VX = B for X? Cheers for any advice guys.

Answer 1

[bym]

X = V\B

[Joseph Percsy]

No, the system is X = B/V. That's not going to change.

[Matt J]

That's not what you posted. You posted that you wanted to solve V*X=B. The solution for that will be

X=V\B

Answer 2

[Matt J]

Why not just use POLYFIT?

X=polyfit(numbers,B,length(B)-1);

It is more numerically stable than using the Vandermonde matrix directly.

I want to implement an a way of calculating an error vector using Error = (X - Dt), where Dt = [1, 1, 0, 0, 0, . . . , 0]T.

I don't really understand where Dt comes from, but why not implement it directly as you've written it?

   Dt=[1 1, zeros(1,length(X)-2)].';
   Error=X-Dt;