Re: Least squares

From: Robert Israel (israel_at_math.ubc.ca)
Date: 02/27/05 

Date: 27 Feb 2005 21:57:38 GMT

In article <KUmUd.57643$EH5.56507@blueberry.telenet-ops.be>,
John Dee <johndee@hotmail.com> wrote:
>To solve a least squares problem, one minimizes
>sum i=0:N | f(x_i)-P(x_i,theta) |^2
>in terms of the unknown parameters theta of a polynomial P, for all values of x.
>r=f(x_i)-P(x_i,theta) , represents the residual.

>When r is complex, this means that the magnitude of the residue is
>squared, but i don't understand why this must be squared, since the
>magnitude is always positive. When dropping the square, I think one is
>solving a different problem. Can someone point this out for me?

Yes, without squares it's not a least squares problem, it's a least
sum of absolute values problem. There's no reason why it "must" be
squared. However, squaring gives you a problem that is easier to solve.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

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