Linear Least Squares Approximation - PowerPoint PPT Presentation

1 / 26
About This Presentation
Title:

Linear Least Squares Approximation

Description:

Linear Least Squares Approximation By Kristen Bauer, Renee Metzger, Holly Soper, Amanda Unklesbay Linear Least Squares Is the line of best fit for a group of points ... – PowerPoint PPT presentation

Number of Views:421
Avg rating:3.0/5.0
Slides: 27
Provided by: EricTh
Category:

less

Transcript and Presenter's Notes

Title: Linear Least Squares Approximation


1
(No Transcript)
2
Linear Least SquaresApproximation
  • By
  • Kristen Bauer, Renee Metzger,
  • Holly Soper, Amanda Unklesbay

3
Linear Least Squares
  • Is the line of best fit for a group of points
  • It seeks to minimize the sum of all data points
    of the square differences between the function
    value and data value.
  • It is the earliest form of linear regression

4
Gauss and Legendre
  • The method of least squares was first published
    by Legendre in 1805 and by Gauss in 1809.
  • Although Legendres work was published earlier,
    Gauss claims he had the method since 1795.
  • Both mathematicians applied the method to
    determine the orbits of bodies about the sun.
  • Gauss went on to publish further development of
    the method in 1821.

5
Example
  • Consider the points (1,2.1), (2,2.9), (5,6.1),
    and (7,8.3)
  • with the best fit line f(x) 0.9x 1.4
  • The squared errors are
  • x11 f(1)2.3 y12.1 e1 (2.3 2.1)² .04
  • x22 f(2)3.2 y22.9 e2 (3.2 2.9)² . 09
  • x35 f(5)5.9 y36.1 e3 (5.9 6.1)² .04
  • x47 f(7)7.7 y48.3 e4 (7.7 8.3)² .36
  • So the total squared error is .04 .09 .04
    .36 .53
  • By finding better coefficients of the best fit
    line, we can make this error smaller

6
We want to minimize the vertical distance between
the point and the line.
  • E (d1)² (d2)² (d3)² (dn)² for n data
    points
  • E f(x1) y1² f(x2) y2² f(xn)
    yn²
  • E mx1 b y1² mx2 b y2² mxn
    b yn²
  • E ?(mxi b yi )²

7
E must be MINIMIZED!
  • How do we do this?
  • E ?(mxi b yi )²
  • Treat x and y as constants, since we are trying
    to find m and b.
  • SoPARTIALS!
  • ?E/?m 0 and ?E/?b 0
  • But how do we know if this will yield maximums,
    minimums, or saddle points?

8
Minimum Point
Maximum Point
Saddle Point
9
Minimum!
  • Since the expression E is a sum of squares and is
    therefore positive (i.e. it looks like an upward
    paraboloid), we know the solution must be a
    minimum.
  • We can prove this by using the 2nd Partials
    Derivative Test.

10
2nd Partials Test
Suppose the gradient of f(x0,y0) 0. (An
instance of this is ?E/?m ?E/?b 0.) We set
  • And form the discriminant D AC B2
  • If D lt 0, then (x0,y0) is a saddle point.
  • If D gt 0, then f takes on
  • A local minimum at (x0,y0) if A gt 0
  • A local maximum at (x0,y0) if A lt 0

11
Calculating the Discriminant
12
  • If D lt 0, then (x0,y0) is a saddle point.
  • If D gt 0, then f takes on
  • A local minimum at (x0,y0) if A gt 0
  • A local maximum at (x0,y0) if A lt 0
  • Now D gt 0 by an inductive proof showing that
  • Those details are not covered in this
    presentation.
  • We know A gt 0 since A 2 ? x2 is always positive
    (when not all xs have the same value).

13
Therefore
Setting ?E/?m and ?E/?b equal to zero will yield
two minimizing equations of E, the sum of the
squares of the error.
Thus, the linear least squares algorithm (as
presented) is valid and we can continue.
14
  • E ?(mxi b yi)² is minimized (as just shown)
    when the partial derivatives with respect to each
    of the variables is zero. ie ?E/?m 0 and ?E/?b
    0

?E/?b ?2(mxi b yi) 0 set equal to 0
m?xi ?b ?yi
mSx bn Sy ?E/?m ?2xi (mxi b yi)
2?(mxi² bxi xiyi) 0
m?xi² b?xi
?xiyi mSxx bSx Sxy
NOTE ?xi Sx ?yi Sy ?xi² Sxx ?xiyi
SxSy
15
Next we will solve the system of equations for
unknowns m and b
Solving for m
nmSxx bnSx nSxy Multiply by n mSxSx bnSx
SySx Multiply by Sx nmSxx mSxSx nSxy
SySx Subtract
m(nSxx SxSx) nSxy SySx Factor m
16
Next we will solve the system of equations for
unknowns m and b
Solving for b
mSxSxx bSxSx SxSxy Multiply by Sx mSxSxx
bnSxx SySxx Multiply by Sxx bSxSx bnSxx
SxySx SySxx Subtract
b(SxSx nSxx) SxySx SySxx Solve for b
17
  • Example Find the linear least squares
    approximation to the data (1,1), (2,4), (3,8)

Use these formulas
Sx 123 6 Sxx 1²2²3² 14 Sy 148
13 Sxy 1(1)2(4)3(8) 33 n number of points
3
The line of best fit is y 3.5x 2.667
18
Line of best fit y 3.5x 2.667
19
THE ALGORITHMin Mathematica
20
(No Transcript)
21
(No Transcript)
22
Activity
  • For this activity we are going to use the linear
    least squares approximation in a real life
    situation.
  • You are going to be given a box score from either
    a baseball or softball game.
  • With the box score you are given you are going to
    write out the points (with the x coordinate being
    the number of hits that player had in the game
    and the y coordinate being the number of at-bats
    that player had in the game).
  • After doing that you are going to use the linear
    least squares approximation to find the best
    fitting line.
  • The slope of the besting fitting line you find
    will be the teams batting average for that game.

23
In Conclusion
  • E ?(mxi b yi )² is the sum of the squared
    error between the set of data points
    (x1,y1),,(xi,yi),,(xn,yn) and the line
    approximating the data f(x) mx b.
  • By minimizing the error by calculus methods, we
    get equations for m and b that yield the least
    squared error

24
Advantages
  • Many common methods of approximating data seek to
    minimize the measure of difference between the
    approximating function and given data points.
  • Advantages for using the squares of differences
    at each point rather than just the difference,
    absolute value of difference, or other measures
    of error include
  • Positive differences do not cancel negative
    differences
  • Differentiation is not difficult
  • Small differences become smaller and large
    differences become larger

25
Disadvantages
  • Algorithm will fail if data points fall in a
    vertical line.
  • Linear Least Squares will not be the best fit for
    data that is not linear.

26
The End
Write a Comment
User Comments (0)
About PowerShow.com