asst10soln

# asst10soln - MATH 138 Assignment 10 Solutions Problem 1 You...

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Unformatted text preview: MATH 138 Assignment 10 Solutions Problem 1: You may have wondered why odd functions are called “odd”, and even functions are called “even”. Here is a possible reason. Suppose f ( x ) = a + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + . .. for x in an interval (- R,R ). If f is even, prove that 0 = a 1 = a 3 = a 5 = a 7 = . .. . Thus only the even coefficients appear in the power series expansion of f . Likewise, prove that if f is odd, then 0 = a = a 2 = a 4 = a 6 = . .. . Hint. The coefficients of a power series converging on an interval have to be the Taylor coefficients of the function that the series converges to. Thus it is not possible for different power series to converge to the same function on an interval (- R,R ). Solution: If f is even, then f ( x ) = f (- x ). Thus a + a 1 x + a 2 x 2 + a 3 x 3 + . .. = a- a 1 x + a 2 x 2- a 3 x 3 + . .. Since the power series representation of a function is unique, the coefficients must match. ∴ a 1 =- a 1 , a 3 =- a 3 , a 5 =- a 5 , .. . This is only possible if 0 = a 1 = a 3 = a 5 = . .. . If f is odd, then- f ( x ) = f (- x ). Thus- a- a 1 x- a 2 x 2- a 3 x 3- . .. = a- a 1 x + a 2 x 2- a 3 x 3 + . .. Since the power series representation of a function is unique, the coefficients must match. ∴- a = a ,- a 2 = a 2 ,- a 4 = a 4 , .. . This is only possible if 0 = a = a 2 = a 4 = . .. . Problem 2: Find the Taylor polynomial of degree up to 2 expanded about 0 for the function y = sec x . Solution: y = tan x sec x, y 00 = sec 2 x (sec x ) + tan x (tan x sec x ) = sec 3 x + tan 2 x sec x y (0) = 1 , y (0) = 0 , y 00 (0) = 1 The Taylor polynomial of degree 2 centered at 0 is T 2 ( x ) = 2 X n =0 f ( n ) (0) n ! x n = 1 + x 2 2 Problem 3: Let T n ( x ) be the Taylor polynomial of degree up to n for y = e x expanded about 0. Find an n such that | 3 √ e- T n (1 / 3) | ≤ 10- 6 , and show that your n does the desired job. Solution: Let f ( x ) = e x . By the Lagrange Remainder Formula, 3 √ e- T n (1 / 3) = f ( n +1) ( z )(1 / 3) n +1 ( n + 1)! where z ∈ [0 , 1 / 3] f ( n +1) ( z ) = e z for all n and 0 ≤ e z ≤ 3 for z ∈ [0 , 1 / 3]. ∴ | 3 √ e- T n (1 / 3) | ≤ 3(1 / 3) n +1 ( n + 1)!...
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## This note was uploaded on 03/16/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.

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asst10soln - MATH 138 Assignment 10 Solutions Problem 1 You...

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