   Chapter 11.10, Problem 26E

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 ) = x , a = 16

To determine

To find:

The first four nonzero terms of Taylor series

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=x,   a=16

3) Calculation:

As   a=16, the Taylor series is given by:

fx=n=0fn16n!(x-16)n=f16+f'161!x-16+f''162!x-162+f'''163!x-163+

Let’s find the coefficients of this series

fx=x

So,

f16=16=4

Differentiate   f(x) with respect to x

f(x) can be written as   fx=x12

f'x=12x12-1=12x-12=12x12

So,

f16=12(16)12=12·14

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

f''x=12·-12·x-12-1

f''x=-14x-32=-14x32

So,

f''16=-141632=-14·143

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

f'''x=-14·-32·x-32-1

f'''x=38x-52=38x52

So,

f'''16=38(16)52=38·145

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

f''''x=38·-52·x-52-1

f''''x=-1516·x-72=-1516x72

So,

f''''16=-15161672=-1516·147

From this Taylor series becomes:

fx=x=n=0fn16n!(x-16)n=f16+f'161!x-16+f''162!x-162+f'''163!x-163+

x=4+12·141!x-16+-14·1432!x-162+38·1453!x-163+-1516·1474!x-164+

x=4+12·14·

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