Taylor Series

# At least theoretically this should work however in

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Unformatted text preview: for π . Example. Find the Taylor series centered at x = 0 for the integral function x cos(t2 ) dt. F (x) = 0 Also ﬁnd F (9) (0) and F (16) (0). We know that for any u ∈ R, ∞ (−1)n cos(u) = n=0 x 2n . (2n)! If we let u = t2 , we get that for any t ∈ R, ∞ 2 2n n (t ) (−1) cos(u) = n=0 (2n)! ∞ (−1)n = n=0 t4n . (2n)! The Integration Theorem for Power Series gives us that x cos(t2 ) dt F (x) = 0 x∞ (−1)n 0 t4n dt (2n)! n=0 ∞ t4n dt (2n)! (−1)n = x = n=0 ∞ 0 [(−1)n = n=0 ∞ (−1)n = n=0 t4n+1 |x ] (4n + 1)(2n)! 0 x4n+1 . (4n + 1)(2n)! This is valid for any x ∈ R. Moreover, by the Uniqueness Theorem for Power Series Representations, this must be the Taylor series for F (x). DE Math 128 340 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series To ﬁnd F (9) (0), we recall that if ∞ ak x k , F (x) = k=0 then a9 = F (9) (0) . 9! This tells us that to ﬁnd F (9) (0) we must ﬁrst identify the coeﬃcient of x9 in ∞ [(−1)n n=0 x4n+1 . (4n + 1)(2n)! It is an easy observation that to get x9 we let n = 2...
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