Differential Equations Solutions 79

# Differential Equations Solutions 79 - and 1 iF i = j we...

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Chapter 15 Solutions: Case Study: Blind Deconvolution: A Matter of Norm CHALLENGE 15.1. Writing out the expressions Ef and F b e component by component, we fnd that F is a Toeplitz matrix oF size m × ( m + n 1) with frst row equal to [ f n ,f n 1 ,...,f 1 , 0 ,..., 0] and frst column equal to [ f n , 0 ,..., 0]. CHALLENGE 15.2. Let d =[1 , 2 ,..., n,. .., n, n 1 ,..., 1] be a vector oF length m + n 1 and let D be the diagonal matrix with entries d . Then F ( b e , f ) 1 2 ± E ± 2 F + 1 2 ± r ± 2 2 = 1 2 m + n 1 X i =1 d 2 i b e 2 i + 1 2 m X i =1 r 2 i = 1 2 m + n 1 X i =1 d 2 i b e 2 i + 1 2 m X i =1 ( g i n X j =1 ( k ij + e ij ) f j ) 2 . We need the gradient and Hessian matrix oF this Function. Noting that E ij = b e n + i j , and letting δ ij =0iF i ² = j
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Unformatted text preview: and 1 iF i = j , we compute ∂F ( b e , f ) ∂ b e ± = d 2 ± b e ± − m X i =1 r i f n + i − ± = ( D 2 b e − F T r ) ± , ∂F ( b e , f ) ∂f ± = − m X i =1 r i ( k i± + b e n + i − ± ) = − (( K + E ) T r ) ± , ∂ 2 F ( b e , f ) ∂ b e ± ∂ b e q = δ ±,q d 2 ± + m X i =1 f n + i − ± f n + i − q = ( D 2 + F T F ) ±q , ∂ 2 F ( b e , f ) ∂ b e ± ∂f q = r ± + q + m X i =1 ( k iq + e iq ) f n + i − ± = ( R + ( K + E ) T F ) ±q , 89...
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