1 / 13

Tutorial 4. Least Squares.

Tutorial 4. Least Squares. 1. Reminder: Least squares solution. Taking the derivative we obtain:. The normal equation:. If then there is a unique global solution:. If A is square and invertible, then the unique solution is.

gilead
Download Presentation

Tutorial 4. Least Squares.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Tutorial 4. Least Squares.

  2. 1. Reminder: Least squares solution Taking the derivative we obtain: The normal equation: If then there is a unique global solution: If A is square and invertible, then the unique solution is

  3. General Quadratic Equation (K positive semi-definite): The derivative this time is given by: If then there is a unique global solution: If K is singular, any solution satisfying leads to the minimal value, and there are infinitely many of those.

  4. 2. Derivatives of Matrix-Vector Expressions Given the function: With symmetric matrices: The minimizer of f(x) should satisfy the cubic equation:

  5. (continue) For the special case: We obtain the requirement: Thus, the possible solutions (zeroing the gradient of the function) are: If , the first solution is meaningless, and second solution is the correct one. If , the first solution is the correct one, and the second represents a local maximum!!!!!!

  6. Example: A=[4 1; 1 2]; b=[2;1]; c=-50; for ind1=1:100 x1=ind1/10-5; for ind2=1:100 x2=ind2/10-5; h(ind1,ind2)=[x1,x2]*A*[x1;x2]-b'*[x1;x2]+c; end; end; figure(1); meshc(h); figure(2); meshc(h.*h); axis([0 100 0 100 0 3000]); caxis([0 3000]); Note: In this case, the value of (x) can get below zero, as the next figures show

  7. 3. Perform: Polynomial fitting x = (-3:.06:3)'; PointNum=size(x,1); C=[-3 1 2 -1]; y0=C(1)+C(2)*x+C(3)*x.^2+C(4)*x.^3; y=y0+10*randn(PointNum,1); figure(1); clf; plot(x,y0,'r'); hold on; plot(x,y,'.b'); A=[ones(PointNum,1),x,x.^2,x.^3,x.^4,x.^5,x.^6,x.^7,x.^8,x.^9,x.^10]; c10=inv(A'*A)*A'*y; c8=inv(A(:,1:8)'*A(:,1:8))*A(:,1:8)'*y; c6=inv(A(:,1:6)'*A(:,1:6))*A(:,1:6)'*y; c4=inv(A(:,1:4)'*A(:,1:4))*A(:,1:4)'*y; c2=inv(A(:,1:2)'*A(:,1:2))*A(:,1:2)'*y; Lets search by curve fitting the best Polynomial of orders 2-10, and see how they perform (visually and by evaluating the error):

  8. (continue) disp(sqrt(mean((y0-A*c10).^2))); figure(2); clf; plot(x,y0,'r'); hold on; plot(x,y,'.b'); plot(x,A*c10,'g'); 2.2604

  9. (continue) disp(sqrt(mean((y0-A*c8).^2))); figure(3); clf; plot(x,y0,'r'); hold on; plot(x,y,'.b'); plot(x,A(:,1:8)*c8,'g'); 1.9877

  10. (continue) disp(sqrt(mean((y0-A*c6).^2))); figure(4); clf; plot(x,y0,'r'); hold on; plot(x,y,'.b'); plot(x,A(:,1:6)*c6,'g'); 1.8918

  11. (continue) disp(sqrt(mean((y0-A*c4).^2))); figure(5); clf; plot(x,y0,'r'); hold on; plot(x,y,'.b'); plot(x,A(:,1:4)*c4,'g'); 0.8114

  12. (continue) disp(sqrt(mean((y0-A*c2).^2))); figure(6); clf; plot(x,y0,'r'); hold on; plot(x,y,'.b'); plot(x,A(:,1:2)*c2,'g'); 6.9345

More Related