Chapter 11.10, Problem 23E

### 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.23. f(x) = e2x, a = 3

To determine

To find: The Taylor series for f(x) centered at 3 and radius of convergence.

Explanation

Given:

The function is, f(x)=e2x , a=3 .

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:

Consider the function f(x)=e2x centered at a=3 .

Since the function f(x)=e2x at 3 is f(3)=e6 .

Find the first derivative of f(x) at a=3 .

f(x)=ddx(e2x)=2e2x

f(x)=2e2x (1)

Obtain f(3) .

f(3)=2e6

Find the second derivative of f(x) at a=3 .

f(2)(x)=d2dx2(f(x))=ddx(f(x)) =ddx(2e2x)    (by equation(1))=2(2)e2x

That is, f(2)(x)=22e2x (2)

Compute f(2)(3) .

f(2)(3)=4e2(3)=4e6

That is, f(2)(3)=4e6 .

Find the third derivative of f(x) at a=3 as follows.

f(3)(x)=d3dx3(e2x)=ddx(f(2)(x))=ddx(22e2x)    (by equation(2))=22(2)e2x

That is, f(3)(x)=23e2x (3)

Compute f(3)(3) .

f(3)(3)=23e2(3)=8e6

Find the fourth derivative of f(x) at a=3 .

f(4)(x)=d4dx4(e2x)=ddx(f(3)(x))=ddx(23e2x)    (by equation(3))=23(2)e2x

That is, f(4)(x)=24e2x (4)

Compute f(4)(3)

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