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m425b-ex2soln-s08

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

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MATH 425b MIDTERM 2 SOLUTIONS SPRING 2008 Prof. Alexander (1) Let A = f 0 ( 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 0 = I, which says that I - A is the inverse of n =0 A n . (3) We apply the Implicit Function Theorem to ˜ f ( x, y ) = f ( x, y ) - T ( x ). Since T is a linear map on R n , it is differentiable with derivative T . Therefore ˜ f 0 ( x, y ) = f 0 ( x, y ) - T ( x ). If A x and A y are the two pieces of the matrix of f 0 ( x, y ), then the corresponding pieces of ˜ f 0 ( x, y ) are ˜ A x = A x - T and ˜ A y = A y . Since by assumption A x is invertible, from Theorem 9.8 (applied to A x and to B = A x - T ) we get that ˜ A x = A x - T is invertible for sufficiently small T , say for || T || < 0 . For such

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