Vandermonde Matrix and an Error Vector

6 次查看（过去 30 天）
Joseph Percsy
Joseph Percsy 2013-4-25
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.

请先登录，再回答此问题。

回答（2 个）

bym
bym 2013-4-25
X = V\B
  2 个评论
Joseph Percsy
Joseph Percsy 2013-4-25
No, the system is X = B/V. That's not going to change.
Matt J
Matt J 2013-4-25
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

请先登录，再进行评论。

Matt J
Matt J 2013-4-25
编辑：Matt J 2013-4-25
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;

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by