We get that 4 arctan1 1n n0 de math 128 339 1 2n

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Unformatted text preview: f (x) = arctan(x) and determine the interval on which the representation is valid. d 1 We begin with the observation that dx (arctan(x)) = 1+x2 . Therefore, if we can find a power 1 series representation for 1+x2 , we can use our integration techniques to find a representation for arctan(x). We know that for any u ∈ (−1, 1), 1 = 1−u ∞ un . n=0 Let x ∈ (−1, 1). If we let u = −x2 , then u ∈ (−1, 1). It follows that 1 1 + x2 = 1 1 − (−x2 ) ∞ (−x2 )n = n=0 ∞ (−1)n x2n = n=0 1 Since arctan(x) is an antiderivative of 1+x2 , the Integration of Power Series Theorem shows that there exists a constant C such that ∞ (−1)n x2n dx arctan(x) = C + n=0 ∞ (−1)n = C+ n=0 DE Math 128 338 x2n+1 2n + 1 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series To find C , note that arctan(0) = 0, so 0 = arctan(0) ∞ (−1)n = C+ n=0 02n+1 2n + 1 =C Therefore, we have shown that if x ∈ (−1, 1) ∞ (−1)n arctan(x) = n=0 x2n+1 2n + 1 The radius o...
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This document was uploaded on 03/23/2013.

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