Then we will also have the following 1 h r k 1 r i r T r k 1 0 for all k To see

Then we will also have the following 1 h r k 1 r i r

This preview shows page 12 - 16 out of 16 pages.

Then we will also have the following: 1. h r k +1 , r i = r T r k +1 = 0 for all k . To see this, notice that r T Hd k = ( d - β d - 1 ) T Hd k (5) = ( d T k Hd k = k 0 ‘ < k, (6) where the second step follows directly from the fact that h d k , d i H = 0 for ‘ < k . As a result r T r k +1 = r T r k - r T k r k d T k Hd k r T Hd k = 0 for all k. (7) 2. h d k +1 , d i H = d T Hd k +1 = 0 for all k . This follows from the expansion d T Hd k +1 = d T Hr k +1 + β k +1 d T Hd k . Notice that r T i r k +1 = r T i r k - α k r T i Hd k r T i Hd k = 1 α k r T k r k i = k - 1 α k r T k +1 r k +1 i = k + 1 0 i < k. (8) 42 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 2:06, November 18, 2019
Image of page 12

Subscribe to view the full document.

Then for = k d T k Hd k +1 = - 1 α k r T k +1 r k +1 + β k +1 d T k Hd k = - r T k +1 r k +1 r T k r k d T k Hd k + r T k +1 r k +1 r T k r k d T k Hd k = 0 . For ‘ < k , d T Hd k +1 = d T Hr k +1 + β k +1 d T Hd k . For the first term d T Hr k +1 = 0 , since Hd = α - 1 ( r - r +1 ) and we have ( 7 ); for the second term β k +1 d T Hd k = 0 , since the d 0 , d 1 , . . . , d k are H -orthogonal already. We have established that the direction d k that CG moves on iteration k is H -orthogonal to all previous directions. Now let’s look at the step sizes, where we want to establish that α k = - c k / k d k k H = - d T k He 0 / k d k k 2 H . Start by noting ( 6 ) above, and recall that r k = b - Hx k = H ( b x - x k ) = - He k . At the first step, we have d 0 = r 0 , and so α 0 = r T 0 r 0 d T 0 Hd 0 = d T 0 r 0 d T 0 Hd 0 = d T 0 H ( b x - x 0 ) d T 0 Hd 0 = - d T 0 He 0 d T 0 Hd 0 . 43 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 2:06, November 18, 2019
Image of page 13
At subsequent steps, since d k = r k + k - 1 X i =0 γ i r i for some γ i R , by Fact 1, we have r T k r k = d T k r k , and so α k = d T k r k d T k Hd k = - d T k H e 0 + k - 1 =0 α d d T k Hd k = - d T k He 0 d T k Hd k . So finally, this means that for the method of conjugate gradients, e k = N - 1 X = k d T r d T Hd ! d , k e k k 2 H = N - 1 X = k | d T r | 2 d T Hd . As k increases, the number of (positive) terms in the sum above gets smaller and smaller, until finally e N = 0 . Thus CG is guaranteed to converge exactly in N steps . Since each iteration of CG involves a vector-matrix multiply, each of which are O ( N 2 ), and we converge in O ( N ) iterations, CG solves Hx = b in O ( N 3 ) computations in general, the same as other solvers. 44 Georgia Tech ECE 6250 Fall 2019; Notes by J. Romberg and M. Davenport. Last updated 2:06, November 18, 2019
Image of page 14

Subscribe to view the full document.

But there are two important things to realize : 1. If H is specially structured so that it takes O ( N 2 ) com- putations to apply, then CG takes advantage of this. The real cost is N applications of H . 2. It is often the case that k e k k 2 H is acceptably small for relatively modest values of k . This is particularly true if H is well- conditioned. Each iteration (application of H ) gets us closer, in a measurable way, to the solution. CG can get an approximate (but still potentially very good) solu- tion using much less computation than solving the system directly.
Image of page 15
Image of page 16
  • Fall '08
  • Staff

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

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes