Chapter 11.10, Problem 20E

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

Chapter
Section

### Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
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 Rn(x) → 0.] Also find the associated radius of convergence.20. f(x) = x6 − x4 + 2, a = −2

To determine

To find: The Taylor series for f(x) centered at −2 and radius of convergence.

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)=x6x4+2 centered at a=2 .

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

The function f(x)=x6x4+2 and obtain f(2) .

f(2)=(2)6(2)4+2=6416+2=50

That is, f(2)=50 .

The first derivative of f(x) at a=2 as follows,

f(x)=ddx(x6x4+2)=ddx(x6)ddx(x4)+ddx(2)=6x5(4x3)+0=6x54x3

f(x)=6x54x3 (1)

Obtain f(2) .

f(2)=6(2)54(2)3=6(32)4(8)=192+32=160

That is, f(2)=160

The second derivative of f(x) at a=2 as follows,

f(2)(x)=d2dx2(f(x))=ddx(f(x))=ddx(6x54x3)    (by equation(1))=6ddx(x5)4ddx(x3)

Simplify further and obtain f(2)(2) ,

f(x)=6(5x4)4(3x2)=30x412x2

f(2)(x)=30x412x2 (2)

Compute f(2)(2) .

f(2)(2)=30(2)412(2)2=30(16)12(4)=48048=432

That is, f(2)(2)=432

The third derivative of f(x) at a=2 as follows,

f(3)(x)=d3dx3(f(x))=ddx(f(2)(x))=ddx(30x412x2)    (by equation(2))=30ddx(x4)12ddx(x2)

Obtain f(3)(2) .

f(3)(x)=30(4x3)12(2x)=120x324x

f(3)(x)=120x324x (3)

Compute f(3)(2) .

f(3)(2)=120(2)324(2)=120(8)+48=960+48=912

That is, f(3)(2)=912 .

The fourth derivative of f(x) at a=2 as follows,

f(4)(x)=d4dx4(f(x))=ddx(f(3)(x))=ddx(120x324x)    (by equation(3))=120ddx(x3)24ddx(x)

f(4)(x)=120(3x2)24(1)=360x224

f(4)(x)=360x224 (4)

Compute f(4)(2)

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