Solution to HW4_S09

Solution to HW4_S09 - HW#4 From the textbook by Atkinson pp...

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HW#4 EGM6341 From the textbook by Atkinson : pp 496-503: #6a, (just expand the determinant using cofactors and minor) Solution: = x x x x x x f n 1 0 0 ... 0 1 1 0 ... 1 1 0 0 ... 0 1 1 0 ... 0 1 det ) ( Expand the determinant using first row, we obtain - = x x x x x x x x x x x x f n 1 0 0 ... 0 1 1 0 ... 1 1 0 0 ... 0 1 1 0 ... 0 1 * 1 1 0 0 ... 0 1 1 0 ... 1 1 0 0 ... 0 1 1 0 ... 0 1 det ) ( = ) ( ) ( 2 1 x f x xf n n - - - Which is equivalent to ) ( ) ( ) ( 1 1 x f x xf x f n n n - + - = #21 Probe the following: for n xf C (a) 1 x x n x f f (b) 2 x x n x f f (c) 2 1 2 x x n x f Solution. (a) 1 x x n x f f (1) 1 1 1 1 0 Max n n j k j k n j j j k x x x x x f f = = - = - =
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where k denotes index for maximum component of x vector 1 x x f (2) 1 n x x f - 1 1 1 1 1 1 1 Max Max n n n j j j k j n j n j k k n x x n x x x x f f = = = - = - = - 1 1 1 1 1 0 Max Max n n n j k j k j n j n k k k x x x x f = = = = - = - 1 n x x f From (1) and (2) 1 x x n x f f (b) 2 x x n x f f (1) 1 1 2 2 2 2 2 1 1 Max n j j j n j x x x x f f = - = - & consider 2 2 2 1 1 1 0 Max n n j k j k n j j j k x x x f = = - = where k denotes index for maximum component of x vector 0 a b - if a b f , for positive a and b 1 1 2 2 2 2 1 1 0 Max n j j j n j x x f = - & 2 x x f (2) 1 1 2 2 2 2 2 1 1 Max n j j j n j n x x n x x f f = - = - & consider 2 2 1 1 Max n j j j n j n x x f = - & 2 2 2 2 2 2 1 1 1 1 1 1 1 Max Max Max n n n n k j k j k j k n k n k n j j j j n x x x x x x = = = = - = - = -
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2 2 1 0 Max k j k n x x f - 2 2 2 2 1 1 1 1 0 0 Max Max n n k j j j k n j n j j x x n x x f = = - - 1 1 2 2 2 2 2 1 1 0 Max n j j j n j n x x n x x f f = - = - & 2 n x x f From (1) and (2) 2 x x n x f f (c) 2 1 2 x x n x f (1) ( 29 1 1 2 2 2 2 1 2 1 1 n n j j j j x x x x = = - = - consider ( 29 ( 29 2 2 2 1 1 1 1 1 n n n n n j j i j j j j j i j x x x x x = = = = = - = - ( 29 2 1 1 1 1 0 n n n n i j j i j j i j i i j x x x x x = = = = = - = ( 29 ( 29 1 1 2 2 2 2 2 2 1 1 1 1 0 0 n n n n j j j j j j j j x x x x = = = = - - ( 29 1 1 2 2 2 2 1 2 1 1 0 n n j j j j x x x x = = - = - 1 2 x x (2) 2 1 n x x - v Use Hölder’s inequality: 1 2 1 2 1 2 1 1 1 1 1 1 1 n n n x x x x x x x x x x = + + + = + + + = + + + L L L ( 29 ( 29 1 1 2 2 2 2 2 2 2 2 1 2 2 1 1 1 n x x x n x + + + + = L L 2 1 n x x f
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From (1) and (2) 2 1 2 x x n x f Problem 23 Show 1 1 1 lim Max p n p j i p i n j x x f f = = & Solution. Let B= | | 1 i n i x Max . Note that there may be several (k 1) elements of x i having the same maximum magnitude of B. Since B x Max x i n i j = | | | | 1 p p j B x | | for 1 j n For p j n j p p j n j B x B x ) / | (| | | 1 1 = = = Inside the summation of p j n j B x ) / | (| 1 = , there are k terms of 1 and the rest of the terms are less than 1 due to the definition of B. As p →∞ , p j n j p B x ) / | (| lim 1 = = k 1 since 1. a for 0 lim < = p p a Thus, p p p j n j p B k x = = lim | | lim 1 , Hence, ( 29 ( 29 B k B kB x p p p p p p p j n j p = = = = / 1 / 1 / 1 1 lim lim | | lim (because 1. b for 1 lim / 1 = p p b Finally, | | | | lim 1 / 1 1 i n i p p j n j p x Max x = = pp 574-583: #2
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0 2 1 3 1 3 2 2 2 2 2 6 3 2 1 3 2 1 3 2 1 = - + = + + - = + + x x x x x x x x x Soln:
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This note was uploaded on 04/23/2009 for the course EGM 6341 taught by Professor Mei during the Spring '09 term at University of Florida.

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Solution to HW4_S09 - HW#4 From the textbook by Atkinson pp...

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