Chapter 9.7, Problem 15E

### Calculus of a Single Variable

11th Edition
Ron Larson + 1 other
ISBN: 9781337275361

Chapter
Section

### Calculus of a Single Variable

11th Edition
Ron Larson + 1 other
ISBN: 9781337275361
Textbook Problem

# Conjecture Consider the function f ( x ) = cos x and its Maclaurin polynomials P 2 , P 4 . and P 6 (sec Example 5).(a) Use a graphing utility to graph f and the indicated polynomial approximations.(b) Evaluate and compare the values of f ( n ) ( 0 ) and P n ( n ) ( 0 ) for n = 2 , 4 , and 6.(c) Use the results in part (b) to make a conjecture about f ( n ) ( 0 ) and P n ( n ) ( 0 ) .

To determine

To graph: The function, f(x)=cosx and the polynomial approximations, P2,P4,P6 which are the Maclaurin Polynomials of the function.

Explanation

Given:

The function, f(x)=cosx and the polynomial approximations, P2,P4,P6 which are the Maclaurin Polynomials of the function.

Graph:

Consider the function,

f(x)=cosx

Differentiate both sides of the above equation with respect to x.

f(x)=ddx(cosx)

Apply the formula, ddx(cosx)=sinx in the derivative,

f(x)=sinx

Substitute 0 for x in the derivative,

f(0)=sinx=0

Substitute 0 for x in the function f(x)=cosx,

f(0)=cos0=1

Differentiate both sides of the function, f(x)=sinx with respect to x.

f(x)=ddx(sinx)

Apply the formula, ddx(sinx)=cosx to evaluate the derivative,

f(x)=ddx(sinx)=cosx

Substitute 0 for x in the derivative,

f(0)=cos0=1

Differentiate both sides of the function, f(x)=cosx with respect to x.

f(x)=ddx(cosx )

Apply the formula, ddx(cosx)=sinx in the above equation,

f(x)=ddx(cosx )=(sinx)=sinx

Substitute 0 for x in the derivative,

f(0)=sin0=0

Differentiate both sides of the equation, f(x)=sinx with respect to x.

f(4)(x)=ddx(sinx)

Apply the formula, ddx(sinx)=cosx to evaluate the derivative,

f(4)(x)=ddx(sinx)=cosx

Substitute 0 for x in the derivative,

f(4)(0)=cos0=1

Differentiate both sides of the equation, f(4)(x)=cosx with respect to x.

f(5)(x)=ddx(cosx )

Apply the formula, ddx(cosx)=sinx in the above equation,

f(5)(x)=ddx(cosx )=(sinx)

Substitute 0 for x in the derivative,

f(5)(0)=sin0=0

Differentiate both sides of the equation, f(5)(x)=(sinx) with respect to x,

f(6)(x)=ddx(sinx)

Apply the formula, ddx(sinx)=cosx to evaluate the above derivative,

f(6)(x)=ddx(sinx)=cosx

Substitute 0 for x in the derivative,

f(6)(0)=cos0=1

The definition of Taylor series for a nth degree polynomial is, if f has n derivatives at c, then the polynomial is known as the nth Taylor series for f at c. Here if c=0 then the polynomial is known as the Maclaurin series as shown below.

Pn(x)=f(0)+f(c)(x)+f(0)2(x)2++f(n)(0)n(x)n

The Maclaurin series for the function f(x)=cosx of degree two will be the series for n=2,

P2(x)=f(0)+f(0)(x)+f(0)2(x)2

Substitute the values for the functions and derivatives, f(0)=1 and f(0)=0,f(0)=1 in the above equation,

P2(x)=1+0(x)+12(x)2=112x2

The Maclaurin series for the function f(x)=cosx of degree four will be the series for n=4

(b)

To determine

To calculate: The value for fnn(0) and Pn(n)(0) for n=2,4,6, where f(x)=cosx and P2,P4,P6 Maclaurin series for function.

(c)

To determine

The conjecture about fnn(0) and Pn(n)(0) from the result determined in part (b) for n=2,4,6.

### 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