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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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This note was uploaded on 07/18/2008 for the course MATH 301 taught by Professor Trietle during the Fall '07 term at Yale.
 Fall '07
 TrietLe
 Math

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