Vandermonde Matrix and an Error Vector
4 views (last 30 days)
Show older comments
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.
0 Comments
Answers (2)
Matt J
on 25 Apr 2013
Edited: Matt J
on 25 Apr 2013
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;
0 Comments
See Also
Categories
Find more on Logical in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!