m425b-ex2soln-s08

m425b-ex2soln-s08 - MATH 425b MIDTERM 2 SOLUTIONS SPRING...

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Unformatted text preview: MATH 425b MIDTERM 2 SOLUTIONS SPRING 2008 Prof. Alexander (1) Let A = f ( x ). Then f ( x + h n + k n )- f ( x ) = A ( h n + k n ) + o ( | h n + k n | ) = Ah n + Ak n + o ( | h n | + | k n | ) = [ f ( x + h n )- f ( x ) + o ( | h n | )] + [ f ( x + k n )- f ( x ) + o ( | k n | ) = [ f ( x + h n )- f ( x )] + [ f ( x + k n )- f ( x )] + o ( | h n | + | k n | ) . (2)(a) For N > M , || S N- S M || = || N X n = M +1 A n || ≤ N X n = M +1 || A n || ≤ N X n = M +1 || A || n and this approaches 0 as M,N → ∞ since the series ∑ ∞ n =1 || A | n converges. Therefore { S N } is Cauchy (in the matrix norm.) Now A is k × k for some k , and convergence in the matrix norm is the same as convergence in R k 2 which is complete. Therefore { S N } , being Cauchy, must be converent, that is, ∑ ∞ n =0 A n converges. (b) Since by (a) the series converges, we have ( I- A ) ∞ X n =0 A n = ∞ X n =0 A n- ∞ X n =0 A n +1 = ∞ X n =0 A n- ∞ X n =1 A n = A = I, which says that I- A is the inverse of...
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This note was uploaded on 06/23/2008 for the course MATH 425B taught by Professor Alexander during the Fall '07 term at USC.

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m425b-ex2soln-s08 - MATH 425b MIDTERM 2 SOLUTIONS SPRING...

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