18.100pset9

18.100pset9 - 18.100B/C: Fall 2008 Solutions to Homework 9...

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Unformatted text preview: 18.100B/C: Fall 2008 Solutions to Homework 9 1. Let f ( x ) = sin x . Then f (0) = 0, f (0) = cos 0 = 1 and f (0) =- sin 0 = 0. Using Taylors Theorem for the degree-3 expansion for f ( x ) around the point x = 0, we have for any x R that f ( x ) = x 1! + f (3) ( t ) 3! x 3 for some t between 0 and x . We have f (3) ( x ) =- cos x , so we can rewrite this equation as f ( x )- x =- cos t 6 x 3 . Taking absolute values and using the fact that | cos t | 1 for all t R gives | sin( x )- x | | x | 3 6 , as needed. For the second part of the question, we always consider x and so also x > 0. From above we have | sin( x- c )- x- c | 1 6 x- 3 c and so | x a sin( x- c )- x a- c | 1 6 x a- 3 c . If a < c then x a- c 0 and so | x a sin( x- c ) | has the same limit as | x a sin( x- c )- x a- c | . This latter expression is bounded between 0 and 1 6 x a- 3 c 0, so in fact in this case x a sin x- c 0. If a = c then we have | x a sin( x- c )- 1 | 1 6 x- 2 a and so x a sin x- c 1. Finally, if a > c then x a sin x- c = x a- c ( x c sin x- c ). Since x a- c and x c sin x- c 1 we have x a sin x- c . 2. For the first part of the question, we proceed by induction on the degree of the polynomial. We know that lim x e x = and so lim x c e x = 0 for any c R . Suppose that lim x Q ( x ) e x = 0 1 for all polynomials Q of degree n . Let P be a polynomial of degree n + 1. Then as x...
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.

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18.100pset9 - 18.100B/C: Fall 2008 Solutions to Homework 9...

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