Find the Taylor series, centered at c = 3, for the function 1 1 - x² f(x) = (-1)^(4n-3) 8hn! The interval of convergence is: = f(x): Σ n=0 = (x − 3)"
Find the Taylor series, centered at c = 3, for the function 1 1 - x² f(x) = (-1)^(4n-3) 8hn! The interval of convergence is: = f(x): Σ n=0 = (x − 3)"
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 50E
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Transcribed Image Text:Find the Taylor series, centered at c = 3, for the function
1
1 - x²
f(x) =
(-1)^(4n − 3)
8hn!
The interval of convergence is:
=
f(x):
Σ
n=0
=
(x-3)"
Transcribed Image Text:Problem 71
n
O
1
2
F"(x)
(1-x²)-0
2-6x2
(1-x2)3
21/8
-2x(1-x²)-2 3/32
그
64
Ge
-(x-3)
Find the taylor sedes cender at c=3 for the
function
f(x) = 1 = x ²
f^(3)
8 доп
↓
64 (x-3) ²
2%
- 1/2+1/31/20
- 3 + 3² (193) 5 ² ₂ (x-3) ²
129
(-1)^(1+y)(x-3)
5390
n
% (-1)^(-3)) (x-3)^
1²0
8onn
(1-x²))
(2-(3)²) = ¹ =
- (1-x²)-²-2x) = +2×(1-x²)
= 28
2x
(1-x²/2 =>
-2
(1-x²) ²
(2(1-x2) ²-2 (1-x²) (-2) (20)
(1-x²)4
2(1-x²) +8x
2-282 18x2
(1-7²)3
2(-6131²)
(1-(37²)3
(1-x²) 3
|X-3121
-14X-3 <1
+3
13 d3
22x24
Fuctor by C-X²1
2 | (₁-x²1²58²)
(1-x²)
2-68²
(1-x²)3
64
(2,4)
1
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