I was trying to slove this question and I don't know if I am doing right? Find the Taylor series, centered at c = 3, for the function11 - x²f(x) =(-1)^(4n − 3)8hn!The interval of convergence is:=f(x):Σn=0=(x-3)" Problem 71nO12F"(x)(1-x²)-02-6x2(1-x2)321/8-2x(1-x²)-2 3/32그64Ge-(x-3)Find the taylor sedes cender at c=3 for thefunctionf(x) = 1 = x ²f^(3)8 доп↓64 (x-3) ²2%- 1/2+1/31/20- 3 + 3² (193) 5 ² ₂ (x-3) ²129(-1)^(1+y)(x-3)5390n% (-1)^(-3)) (x-3)^1²08onn(1-x²))(2-(3)²) = ¹ =- (1-x²)-²-2x) = +2×(1-x²)= 282x(1-x²/2 =>-2(1-x²) ²(2(1-x2) ²-2 (1-x²) (-2) (20)(1-x²)42(1-x²) +8x2-282 18x2(1-7²)32(-6131²)(1-(37²)3(1-x²) 3|X-3121-14X-3 <1+313 d322x24Fuctor by C-X²12 | (₁-x²1²58²)(1-x²)2-68²(1-x²)364(2,4)1

The given function isCentered at c = 3.

Answered: Find the Taylor series, centered at c =…

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)"

Calculus: Early Transcendentals

8th Edition

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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Question

I was trying to slove this question and I don't know if I am doing right?

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)"

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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