{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lect04 - EE553 LECTURE 4 LECTURE OUTLINE Principal Gradient...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
EE553 LECTURE 4 LECTURE OUTLINE Principal Gradient Methods Gradient Methods - Choices of Direction Gradient Methods - Choice of Stepsize Gradient Methods - Convergence Issues
Image of page 1

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

View Full Document Right Arrow Icon
CONVERGENCE RESULTS CONSTANT STEPSIZE Let { x k } be a sequence generated by a gradient method x k +1 = x k + α k d k , where { d k } is gradient related. Assume that for some constant L > 0 , we have f ( x ) - ∇ f ( y ) L x - y , x, y n , Also, assume that there exists a scalar such that for all k 0 < α k (2 - ) |∇ f ( x k ) d k | L d k 2 . Then either f ( x k ) → -∞ or else { f ( x k ) } converges and every limit point of { x k } is stationary.
Image of page 2
MAIN PROOF IDEA 0 α α∇ f(x k )'d k + (1/2) α 2 L||d k || 2 × α∇ f(x k )'d k α = | f(x k )'d k | L||d k || |2 f(x k + α d k ) - f(x k ) The idea of the convergence proof for a constant stepsize. Given x k and the descent direction d k , the cost difference f ( x k + αd k ) - f ( x k ) is ma- jorized by α f ( x k ) d k + 1 2 α 2 L d k 2 (based on the Lipschitz assumption; see descent lemma). Mini- mization of this function over α yields the stepsize α = |∇ f ( x k ) d k | L d k 2 This stepsize reduces the cost function f as well.
Image of page 3

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

View Full Document Right Arrow Icon
PRINCIPAL GRADIENT METHODS x k +1 = x k + α k d k , k = 0 , 1 , . . . where, if f ( x k ) = 0 , the direction d k satisfies f ( x k ) d k < 0 , and α k is a positive stepsize. Principal example: x k +1 = x k - α k D k f ( x k ) , where D k is a positive definite symmetric matrix
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern