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Unformatted text preview: Math 301/ENAS 513: Homework 12 Solution Triet M. Le Page 116: 2: Let f ( x ) = , x e 1 /x , x > . To show f is infinitely differentiable at x = 0. Clearly, f is infinitely differentiable on ( , 0) (0 , ), with f ( n ) ( x ) = , x < P 2 n (1 /x ) e 1 /x , x > , where P 2 n ( x ) is some polynomial of degree 2 n . (This can be shown by induction.) Note that for any m N , y m e y 0 as y . Hence, for all integer n 0, P 2 n (1 /x ) e 1 /x , as x . If x n is any sequence converging to 0, then  f ( x n )  e 1 /x n 0 = f ( x ) , as n . Hence f is continuous at 0. Now suppose f ( n ) (0) exists and is = 0 for some n 0, then lim x  f ( n ) (0) f ( n ) ( x ) x = 0 , and lim x + f ( n ) (0) f ( n ) ( x ) x = 1 x P 2 n (1 /x ) e 1 /x = P 2( n +1) (1 /x ) e 1 /x , as x . This implies, f ( n +1) (0) = lim x f ( n ) (0) f ( n ) ( x ) x = 0. So by induction f ( n ) exists and is equal to 0 for all n 0. The Taylor polynomial T n at a = 0 is identically 0 for all n . Page 116: 3: Suppose f is a bounded realvalued on R , and that f and f 00 are also bounded and continuous. To show sup x R  f ( x )  4 sup x R  f ( x )  sup x R  f 00 ( x )  . (1) Pick any x, a R , x 6 = a , Taylor theorem implies f ( x ) = f ( a ) + f ( a )( x a ) + f 00 ( c ) 2 ( x a ) 2 f ( a ) = f ( x ) f ( a ) x a f 00 ( c ) 2 ( x a ) , 1 for some c between a and x . Now, let h > 0 and x = a + h . We have f ( a ) = f ( a + h ) f ( a ) h f 00 ( c ) 2 ( h )  f ( a )  2 h sup x R  f ( x )  + h 2 sup x R  f 00 ( x )  ....
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 Fall '07
 TrietLe
 Math

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