Homework_Set_7_Solution

# Homework_Set_7_Solution - MA 1A FALL 2005 HOMEWORK 7...

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Unformatted text preview: MA 1A FALL 2005 HOMEWORK 7 SOLUTIONS Problem 1. The purpose of this problem is to provide an alternative approach to the definitions of log( x ) and e x used in class. (a) Prove that the radius of convergence of the power series ∞ X n =0 x n n ! is infinite. Define the limit to be e x . Proof. To find the radius of convergence of the power series ∞ X n =0 x n n ! , we apply the ratio test. If x 6 = 0, the ratio of consecutive terms has absolute value fl fl fl fl x n +1 n ! ( n + 1)! x n fl fl fl fl = | x | n + 1 . Since this ratio tends to 0 as n → ∞ , we conclude that the series converges absolutely for all real numbers x 6 = 0. It also converges for x = 0, so the radius of convergence is + ∞ . (b) Define P n ( x ) = n X j =0 x j j ! , E n ( x ) = ∞ X j = n +1 x j j ! the approximation and error. Prove that if | x | < n , then | E n ( x ) | ≤ | x | n +1 ( n + 1)! 1- | x | n ¶- 1 . Proof. By definition of E n ( x ), | E n ( x ) | = fl fl fl fl fl fl ∞ X j = n +1 x j j ! fl fl fl fl fl fl ≤ ∞ X j = n +1 | x | j j ! = | x | n +1 ( n + 1)! ∞ X j = n +1 | x | j- n- 1 j ! / ( n + 1)! ≤ | x | n +1 ( n + 1)! ∞ X j =0 | x | j n j . Since | x | < n , we have ∞ X j =0 | x | j n j = 1- | x | n ¶- 1 . Therefore, | E n ( x ) | ≤ | x | n +1 ( n + 1)! 1- | x | n ¶- 1 . (c) Use the binomial theorem to prove that | P 2 n ( x + y )- P 2 n ( x ) P 2 n ( y ) | ≤ e | x | E n ( | y | ) + e | y | E n ( | x | ) . Proof. By definition of P n ( x ), | P 2 n ( x + y )- P 2 n ( x ) P 2 n ( y ) | = fl fl fl fl fl fl 2 n X j =0 ( x + y ) j j !- ˆ 2 n X i =0 x i i ! ! 2 n X j =0 y j j ! fl fl fl fl fl fl . 1 2 Use the binomial theorem to the first summation, we have | P 2 n ( x + y )- P 2 n ( x ) P 2 n ( y ) | = fl fl fl fl fl fl 2 n X j =0 1 j ! j X i =0 j i ¶ x i y j- i- ˆ 2 n X i =0 x i i ! ! 2 n X j =0 y j j ! fl fl fl fl fl fl = fl fl fl fl fl fl 2 n X j =0 j X i =0 x i y j- i i !( j- i )!- 2 n X i =0 2 n X j =0 x i y j i ! j ! fl fl fl fl fl fl = fl fl fl fl fl fl 2 n X i =0 2 n X j = i x i y j- i i !( j- i )!- 2 n X i =0 2 n X j =0 x i y j i ! j ! fl fl fl fl fl fl = fl fl fl fl fl fl 2 n X i =0 x i i ! 2 n X j = i y j- i ( j- i )!- 2 n X j =0 y j j ! fl fl fl fl fl fl = fl fl fl fl fl fl 2 n X i =0 x i i !...
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## This note was uploaded on 03/30/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.

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Homework_Set_7_Solution - MA 1A FALL 2005 HOMEWORK 7...

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