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Unformatted text preview: Lecture 13 Differentiability (of inversion) First suppose F is differentiable at A ∈ G ( X ). From F ( A ) A = I we differentiate by the Leibniz rule and obtain F A ( H ) A + F ( A ) H = 0 F A ( H ) = F ( A ) HA 1 = A 1 HA 1 . This shows what the inverse is, assuming its existence. Now calculate, F ( A + H ) F ( A ) { A 1 HA 1 } = ( A + H ) 1 A 1 + A 1 HA 1 = (( I + HA 1 ) A ) 1 A 1 + A 1 HA 1 = A 1 { ( I + HA 1 ) 1 I + HA 1 . } Now recall, ( I + K ) 1 = ∞ X n =0 ( K ) n = I K + ∞ X n =2 ( K ) n So k ( I + K ) 1 I + K k ≤ k K k 2 ∞ X n =0 k K k n 1 = k K k 2 1 k K k ( k K k < 1) = k K k 2 1 k K k ≤ 2 k K k 2 ( k K k < 1 2 ) . Therefore, k F ( A + H ) F ( A ) + A 1 HA 1 k ≤ k A 1 kk ( I + HA 1 ) 1 I + HA 1 k ≤ k A 1 k 2 k HA 1 k 2 (when k HA 1 ≤ 1 2 ; e . g . when k H k ≤ 1 2 k A 1 k 1 k ) Thus, if k H k < 1 2 k A 1 k 1 then F ( A + H ) F ( A ) + A 1 HA 1 k ≤ 2 k A 1 k 3 k H k 2 and so k F ( A + H ) F ( A ) + A 1 HA 1 k...
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This note was uploaded on 07/08/2011 for the course MAT 6932 taught by Professor Staff during the Summer '10 term at University of Florida.
 Summer '10
 Staff
 Algebra

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