   Chapter 11.10, Problem 4E

Chapter
Section
Textbook Problem

# Find the Taylor series for f centered at 4 if f ( n ) ( 4 ) = ( − 1 ) n n ! 3 n ( n + 1 ) What is the radius of convergence of the Taylor series?

To determine

To find: The Taylor series for f and its radius of the convergence.

Explanation

Given:

The nth derivative of the function f(x) at the point 4 is, fn(4)=n=0n!3n(n+1) (1)

Result used:

(1) 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+ .

(2)The Ratio Test:

(i) If limn|an+1an|=L<1 , then the series n=1an is absolutely convergent (and therefore convergent.)

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

(ii) If limn|an+1an|=1 , the Ratio Test inconclusive; that is, no conclusion can be drawn about the convergence or divergence of n=1an .

Calculation:

The function f has a power series expansion at 4 from equation (1) is f(x)=n=0f(4)(4)n!(x4)n . (2)

Substitute equation (1) in equation (2),

f(x)=n=0(1)nn!3n(n+1)n!(x4)n   (by equation(1))=n=0(1)nn!3n(n+1)n!(x4)n=n=0(1)n3n(n+1)(x4)n

Hence, the Taylor series for f(x) is n=0(1)n3n(n+1)(x4)n

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 