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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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