   Chapter 11.10, Problem 21E

Chapter
Section
Textbook Problem

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 n ( x ) → 0 .] Also find the associated radius of convergence. f ( x ) = ln x , a = 2

To determine

To find:

Taylor series for   f(x), and find its associated radius of convergence

Explanation

1) Concept:

Taylor series of the function   f at   a is,

fx=n=0fnan!(x-a)n=fa+f'a1!(x-a)+f''a2!(x-a)2+f'''a3!(x-a)3

The ratio test

(i) If limnan+1an=L<1, then the series n=1an is absolutely convergent.

(ii) If limnan+1an=L>1 or limnan+1an=, then the series n=1an is divergent.

(iii) If limnan+1an=1, the ratio test is inconclusive; that is, no conclusion can be draw about the convergence or divergence of an.

2) Given:

fx=lnx,   a=2

3) Calculation:

As   a=2, Taylor series is given by:

fx=n=0fn2n!(x-2)n=f2+f'21!x-2+f''22!x-22+f'''23!x-23+

Let’s find the coefficients of this series

fx=lnx

So,

f2=ln2

Differentiate   f(x) with respect to x.

f'x=1x

So,

f2=12

Now, differentiate   f'(x) with respect to x to get f''x

f'(x) can be written as f'x=x-1

f''x=-1x-1-1=-x-2=-1x2

So,

f''2=-122

Now, differentiate f''x with respect to x to get f'''x

f'''x=--2x-2-1=2x-3

f'''x=2x3

So,

f'''2=223

Now, differentiate f'''x with respect to x to get f''''x.

f''''x=2·-3x-3-1=-6x-4

f''''x=-6x4

So,

f''''2=-624

From this, Taylor series becomes:

fx=lnx=n=0fn2n!(x-2)n=f2+f'21!x-2+f''22!x-22+f'''23!x-23+

lnx=ln2+1

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