Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series f(x) Ο + Σε η = 1 8 ο+Σ n = 0 1 Οξ+ Σι-11 M=1 00 (2η − 1)! 20.52n +1 (-1)^ ο+Σ (1)n1 n = 1 a = 25 .3.5. 20.52n + 1. n! οξ+ Σ (1) + 1 n = 1 (x – 25)n · (2n – 1) 2.52n +1.n! (x – 25)n 1 3 5 (2n-1) . .... . 20.52n-1.n! 1 27.52m.n! (x – 25) (x – 25)n - (x – 25)n

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R(x)→ 0.]
√
5
+
f(x) =
n = 1
00
n=1
n = 0
(2n-1)!
20.52n +1 (x - 25)
0 + (-1)0¹-3-5
a = 25
1.3.5 (2n - 1)
2n. 52n + 1. n!
1
20.52n + 1. n!
0² + (-1)-²
n = 1
οξέων
0 / + (-1)² + 1-
n = 1
(x - 25)
1.3.5 (2n-1) (x - 25)"
20.52n-1.n!
-(x - 25)"
1
2.52n.n!
(x - 25)
Transcribed Image Text:Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R(x)→ 0.] √ 5 + f(x) = n = 1 00 n=1 n = 0 (2n-1)! 20.52n +1 (x - 25) 0 + (-1)0¹-3-5 a = 25 1.3.5 (2n - 1) 2n. 52n + 1. n! 1 20.52n + 1. n! 0² + (-1)-² n = 1 οξέων 0 / + (-1)² + 1- n = 1 (x - 25) 1.3.5 (2n-1) (x - 25)" 20.52n-1.n! -(x - 25)" 1 2.52n.n! (x - 25)
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