   Chapter 11.10, Problem 7E

Chapter
Section
Textbook Problem

# Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of a.7. f ( x ) = x 3 , a = 8

To determine

To find: The first four nonzero terms of the series for f(x) centered at 8.

Explanation

Result used:

If f has a power series expansion at a , f(x)=n=0f(n)(a)n!(xa)n , f(x)=f(a)+f(a)1!(xa)+f(a)2!(xa)2+f(a)3!(xa)3+

Calculation:

Consider the function f(x)=x3 centered at a=8 .

Obtain the first four nonzero terms of the series as follows,

The function f(x) at a=8 is computed as follows,

f(8)=(8)13=2

The first derivative of f(x) at a=8 is computed as follows,

f(x)=ddx(f(x))=13(x)131

f(x)=13(x)23 (1)

Substitute 8 for x,

f(8)=13(8)23=13(8)23=13(82)13=13(64)13

That is, f(8)=112

The second derivative of f(x) at a=8 is computed as follows,

f(x)=d2dx2(f(x))=ddx(f(x))=ddx(13(x)23)    (by equation(1))=13ddx((x)23)

Simplify further and compute f(2)(x) ,

f(2)(x)=13(23)(x)231=29(x)53

f(2)(x)=29(x)53 (2)

Substitute 8 for x,

f(2)(8)=29(8)53=29(85)13=29(32768)13=29(32)

That is, f(2)(8)=2288

The third derivative of f(x) at a=8 is computed as follows,

f(3)(x)=d3dx3(f(x)

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