0 x4n1 4n 12n this is valid for any x r moreover

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Unformatted text preview: f convergence of this series is 1. At x = 1, the series becomes ∞ (−1)n n=0 1 2n + 1 which converges by the Alternating Series Test. When x = −1, the series is ∞ ∞ 2n+1 n (−1) (−1) 2n + 1 n=0 (−1)3n+1 = n=0 1 . 2n + 1 Since 1 if n is odd −1 if n is even (−1)3n+1 = the last series is the same as ∞ (−1)n+1 n=0 1 2n + 1 which also converges by the Alternating Series Test. Therefore, the Continuity Theorem for Power Series shows that ∞ (−1)n arctan(x) = n=0 x2n+1 2n + 1 for all x ∈ [−1, 1] This last statement has an interesting application. We get that π 4 = arctan(1) ∞ (−1)n = n=0 DE Math 128 339 1 2n + 1 (B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series Multiplying both sides of this equation by 4 gives 1111 + − + − ···) 3579 4444 = 4 − + − + − ··· 3579 π = 4(1 − The series representation for arctan(x) given above is called Gregory’s series after the Scottish mathematician of the same name. The famous series expansion for π which we derived from Gregory’s series is called Leibniz’s formula...
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This document was uploaded on 03/23/2013.

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