   Chapter 11.10, Problem 16E

Chapter
Section
Textbook Problem

# Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.] Also find the associated radius of convergence.16. f(x) = x cos x

To determine

To find: The Maclaurin series for f(x) by using the definition of a Maclaurin series and also the radius of the convergence.

Explanation

Given:

The given function is f(x)=xcosx .

Result used:

(1) "The expansion of the Maclaurin series f(x)=n=0f(n)(0)n! is f(0)+f01!x+f(0)2!x2+f(0)3!x3+

(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:

Obtain f(0) .

Substitute 0 for x in f(x) .

f(0)=0(cos0)=0

Find the first derivative of f(x) at a=0 .

f(x)=ddx(xcosx)=cosxddx(x)+xddx(cosx)=cosx(1)xsinx=cosxxsinx

f(x)=cosxxsinx (1)

Substitute 0 for x in f(x) ,

f(0)=cos(0)(0)sin(0)=(1)0=1

Find the second derivative of f(x) at a=0 .

f(2)(x)=d2dx2(f(x))=ddx(f(x)) =ddx(cosxxsinx)    (by equation(1))=ddx(cosx)ddx(xsinx)

Simplify further and obtain f(2)(0) as shown below.

f(2)(x)=sinx(sinxddx(x)+xddx(sinx))=sinxsinxxcosx=xcosx2sinx

f(2)(x)=xcosx2sinx (2)

Substitute 0 for x in f(2)(x) ,

f(2)(0)=0cos02sin0=0

Find the third derivative of f(x) at a=0 .

f(3)(x)=d3dx3(f(x))=ddx(f(2)(x))=ddx(xcosx2sinx)    (by equation(2))f(3)(x)=ddx(xcosx)2ddx(sinx)

Simplify further and obtain f(3)(0) as shown below.

f(3)(x)=(xsinx+cosx)2cosx   [ddx(xcosx)=xsinx+cosx]=xsinx3cosx

That is, f(3)(x)=xsinx3cosx (3)

Substitute 0 for x in f(3)(x) ,

f(3)(0)=0sin(0)3cos(0)=3(1)=3

Find the fourth derivative of f(x) at a=0 .

f(4)(x)=d4dx4(f(x))=ddx(f(3)(x))=ddx(xsinx3cosx)    (by equation(3))=ddx(xsinx)3ddx(cosx)

Simplify further and obtain f(4)(0) as shown below.

f(4)(x)=sinx+xcosx3(sinx)    [ddx(xsinx)=sinx+xcosx]=xcosx+4sinx

That is, f(4)(x)=xcosx+4sinx (4)

Substitute 0 for x in f(4)(x) ,

f(4)(0)=0cos(0)+4sin(0)=0+0=0

Find the fifth derivative of f(x) at a=0

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