least_squares - Least Squares and matrices Jonathan L.F...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Least Squares and matrices Jonathan L.F. King University of Florida, Gainesville FL 32611-2082, USA [email protected] Webpage http://www.math.ufl.edu/ ∼ squash/ The Problem Suppose we have a collection K of N points Q 1 ,...,Q j ,...Q N in the plane. Consider now the line L with equation y = βx + α . It has slope β and y-intercept α . At a given point Q = ( x,y ), the vertical ( signed ) distance to L is v := [ α + βx ]- y . 1 : Letting v j denote the vertical distance at Q j , de- fine the least-square distance from K to L by g ( α,β ) := X N j =1 [ v j ] 2 . 1 : Our goal is to find all pairs ( α,β ) which mini- mize g . It will turn out there is a unique min- imum, except in the silly case that all the given points lie on one vertical line. That is, writing Q j as ( x j ,y j ), except when x 1 = ··· = x N . The quantities that we will need are X := ∑ N j =1 x j , Y := ∑ N j =1 y j , S := ∑ N j =1 x j 2 , P := ∑ N j =1 y j x j . ( “ S ” is for Squares and “ P ” is for Product. ) Using Calculus Evidently in computing the first-partials of g we will want to compute them for each v j . From (1) we compute that d v d α = 1, so d d α ( v 2 ) = 2 v · 1 and d v d β = x , so d d β ( v 2 ) = 2 v · x, by the Chain Rule. Consequently d g d α = X N j =1 2 v j and d g d β = X N j =1 2 v j x j . Thus, the pair ( α,β ) is a critical point of g IFF at ( α,β ) we have that 0 = X N j =1 v j and 0 = X N j =1 v j x j . 2 : Recall that v j is α + x j β- y j . So multiplying out and distributing the summations in (2) yields that 0 = Nα + Xβ- Y , 0 = Xα + Sβ- P . 2 : We can rewrite this to say that ( α,β ) is a critical point of g IFF Y = Nα + Xβ , P = Xα + Sβ . 3 : Matrices. Let M denote the matrix [ N X X S ] and let D := Det( M ) note === NS- X 2 . It follows from a standard ♥ 1 inequality that: All the points x 1 ,...,x N are equal IFF D = 0. We henceforth assume that our scatterplot has at least two distinct x-values. Bare-hands computation [ or matrix algebra ] shows that (3) has a unique solution, which is α = 1 D h S Y- XP i , β = 1 D h XY + NP i ....
View Full Document

This note was uploaded on 01/26/2012 for the course MAC 3472 taught by Professor Jury during the Fall '07 term at University of Florida.

Page1 / 3

least_squares - Least Squares and matrices Jonathan L.F...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online