34 Georgia Tech ECE 6250 Fall 2019 Notes by J Romberg and M Davenport Last

34 georgia tech ece 6250 fall 2019 notes by j romberg

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34 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 2:06, November 18, 2019
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By the chain rule, d d α f ( x k +1 ) = f ( x k +1 ) T d d α x k +1 = f ( x k +1 ) T r k . So we need to choose α k such that f ( x k +1 ) r k , or more concisely r k +1 r k ( r T k +1 r k = 0) . So let’s do this r T k +1 r k = 0 ( b - Hx k +1 ) T r k = 0 ( b - H ( x k + α k r k )) T r k = 0 ( b - Hx k ) T r k - α k r T k Hr k = 0 r T k r k - α k r T k Hr k = 0 and so the optimal step size is α k = r T k r k r T k Hr k . The steepest descent algorithm performs this iteration until k Hx k - b k 2 is below some tolerance δ : 35 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 2:06, November 18, 2019
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Steepest Descent, version 1 Initialize: x 0 = some guess, k = 0, r 0 = b - Hx 0 . while not converged, k r k k 2 δ do α k = r T k r k / r T k Hr k x k +1 = x k + α k r k r k +1 = b - Hx k +1 k = k + 1 end while There is a nice trick that can save us one of two applications of H needed in each iteration above. Notice that r k +1 = b - Hx k +1 = b - H ( x k + α k r k ) = r k - α k Hr k . So we can save an application of H by updating the residual rather than recomputing it at each stage. Steepest Descent, more efficient version 2 Initialize: x 0 = some guess, k = 0, r 0 = b - Hx 0 . while not converged, k r k k 2 δ do q = Hr k α k = r T k r k / r T k q x k +1 = x k + α k r k r k +1 = r k - α k q k = k + 1 end while 36 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 2:06, November 18, 2019
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The effectiveness of SD depends critically on how H is conditioned and the starting point. Consider the two examples on the next page. -4 -2 2 4 -4 -2 2 4 6 -4 -2 2 4 -4 -2 2 4 6 -4 -2 2 4 -4 -2 2 4 6 -4 -2 2 4 -4 -2 2 4 6 1 (c) 2 1 (d) 2 1 (a) 2 1 (b) 2 (from Shewchuk, “... without the agonizing pain”) When the conditioning of H is poor and we choose a bad starting point, convergence can take many iterations even in simple cases. 37 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 2:06, November 18, 2019
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The method of conjugate gradients (CG) An excellent companion resource for this section is the manuscript: J. Shewchuk: “An introduction to the conjugate gradient method without the agonizing pain”. We can see from the example on the last page that steepest descent can be inefficient because it can move in essentially the same direction many times. CG avoids this by ensuring that each step is orthogonal (in an ap- propriate inner product) to all of the previous steps that have been taken. Miraculously, this can be done with very little overhead. Suppose for a moment that we pre-determine N step directions d 0 , . . . , d N - 1 that are orthogonal (but not necessarily normalized), d T j d i = 0 for i 6 = j . This means that { d k / k d k k 2 , k = 0 , . . . , N - 1 } is an orthobasis for R N . Then given a starting point x 0 , the initial error e 0 = x 0 - b x , where b x is the solution that satisfies H b x = b , can be expanded as e 0 = N - 1 X =0 c d k d k 2 , where c = d T e 0 k d k 2 . (3) Given step sizes α 0 , α 1 , . . . , the error after the k th step is e k = x k - b x = x k - 1 + α k - 1 d k - 1 - b x = e k - 1 + α k - 1 d k - 1 = e k - 2 + α k - 1 d k - 1 + α k - 2 d k - 2 .
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