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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 Fe component by component, we find that F is a Toeplitz matrix of size m × ( m + n 1) with first row equal to [ f n , f n 1 , . . . , f 1 , 0 , . . . , 0] and first 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 ( e , f ) 1 2 E 2 F + 1 2 r 2 2 = 1 2 m + n 1 i =1 d 2 i e 2 i + 1 2 m i =1 r 2 i = 1 2 m + n 1 i =1 d 2 i e 2 i + 1 2 m i =1 ( g i n j =1 ( k ij + e ij ) f j ) 2 . We need the gradient and Hessian matrix of this function. Noting that
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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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