Differential Equations Solutions 38

Differential Equations Solutions 38 - p T p = h 2 ....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
48 Chapter 9. Solutions: Numerical Methods for Unconstrained Optimization CHALLENGE 9.3. x = [1;2]; for i=1:5, g = [4*(x(1) - 5)^3 - x(2); 4*(x(2) + 1)^3 - x(1)]; H = [12*(x(1) - 5)^2, -1; -1, 12*(x(2) + 1)^2]; p=-H\g; x=x+p; end CHALLENGE 9.4. The Lagrangian function is L ( p )= f ( x )+ p T g + 1 2 p T Hp + λ 2 ( p T p h 2 ) . Setting the partial derivatives to zero yields g + Hp + λ p = 0 , 1 2 ( p T p h 2 )=0 . Thus the Frst equation is equivalent to ( H + λ I ) p = g ,o r b Hp = g ,a sinthe Levenberg-Marquardt algorithm. The parameter
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: p T p = h 2 . CHALLENGE 9.5. If f is quadratic, then H ( k ) s ( k ) = g ( x ( k +1) ) g ( x ( k ) ) , where H is the Hessian matrix of f . Close to x ( k +1) , a quadratic model is a close Ft to any function, so we demand this property to hold for our approximation to H . CHALLENGE 9.6. The formula for Broydens good method is B ( k +1) = B ( k ) ( B ( k ) s ( k ) y ( k ) ) s ( k ) T s ( k ) T s ( k ) ....
View Full Document

This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

Ask a homework question - tutors are online