{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework_Set_8_Solution

Homework_Set_8_Solution - 2005 MA 1A HOMEWORK 8 SOLUTIONS(1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
2005 MA 1A HOMEWORK 8 SOLUTIONS (1) Suppose a n > a n +1 0 for all n . Define f ( x ) = X n =0 ( - 1) n a n x n . (a) Prove that this series is absolutely convergent for | x | < 1. Proof. X n =0 | ( - 1) n a n x n | = X n =0 a n | x n | = X n =0 a n | x | n . Since a n a 0 for all n , we get that a n | x | n a 0 | x | n for all n . Therefore we may apply the comparison test, comparing n =0 a n | x | n to n =0 a 0 | x | n . But the latter series is just a geometric series, with ratio | x | . Since | x | < 1, this series is convergent. Thus by the comparison test, we get that n =0 a n | x | n is convergent, so X n =0 ( - 1) n a n x n is absolutely convergent. (b) Suppose from now on that a n 0 as n → ∞ . This means that n =0 ( - 1) n a n is convergent. Define L = n =0 ( - 1) n a n . Let f k ( x ) be the partial sum used to define f ( x ) where n runs from 0 to k . Prove that, for 0 < x < 1, f 2 m ( x ) f ( x ) f 2 m +1 . Proof. Since the series is absolutely convergent, we get f ( x ) = X n =0 ( - 1) n a n x n = 2 m X n =0 ( - 1) n a n x n + X n =2 m +1 ( - 1) n a n x n = 2 m X n =0 ( - 1) n a n x n + X j = m +1 ( - a 2 j - 1 x 2 j - 1 + a 2 j x 2 j ) = 2 m X n =0 ( - 1) n a n x n - X j = m +1 ( a 2 j - 1 x 2 j - 1 - a 2 j x 2 j ) 2 m X n =0 ( - 1) n a n x n = f 2 m ( x ), because a 2 j - 1 x 2 j - 1 a 2 j x 2 j for all j .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}