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91Ó°ÊÓ

Find the indicated Maclaurin or Taylor series for the given function about the indicated point, and find the radius of convergence for the series.

x,x0=1

Short Answer

Expert verified

The radius of convergence for the series isPn(x)=1+12(x-1)+∑k=2∞(-1)k+11·3·5⋯(2k-3)2kk!(x-1)k

Step by step solution

01

Given information

The function isf(x)=x

02

Find the general of the Taylor series of the function 

The Taylor series at x=1for any function fwith a derivative of ordernis given by

Pn(x)=f(1)+f'(1)(x-1)+f''(1)2!(x-1)2+f'''(1)3!(x-1)3+f''''(1)4!(x-1)4+…

As a result, first determine the function's value as well as f'(x),f''(x),f'''(x)at x=1

Furthermore, the function's general Taylor series isPn(x)=∑k=0∞fkx0k!x-x0n

03

 Make a table of the Taylor series for the function f(x)=x at x=1 

0
x
1
1
1
12x
12
12
2
-122(x)-32
-122
-1222!
3
1·323(x)52
1·323
1·323
4
-1·3·524(x)-72
-1·3·524
-1·3·524
..
.
.

k
(-1)k+11·3·5⋯(2k-3)24(1-x)-2k-12
(-1)k+11·3·5⋯(2k-3)2k
(-1)k+11·3·5⋯(2k-3)2k
04

 Find the Taylor series for the function f(x)=x at x=1 

The Taylor series for the function f(x)=xat x=1is

1+12·(x-1)+-1222!(x-1)2+1·3233!(x-1)3+-1·3·5244!(x-1)4+.....+(-1)k+11·3·5.....(2k-3)2kk!(x-1)k

Or,

Pn(x)=1+12(x-1)+∑k=2∞(-1)k+11·3·5⋯(2k-3)2kk!(x-1)k

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