Calculus AB and Calculus BC

CHAPTER 10 Sequences and Series

C. POWER SERIES

C1. Definitions; Convergence.

An expression of the form

Image

where the a’s are constants, is called a power series in x; and

Image

is called a power series in (xa).

If in (1) or (2) x is replaced by a specific real number, then the power series becomes a series of constants that either converges or diverges. Note that series (1) converges if x = 0 and series (2) converges if x = a.

RADIUS AND INTERVAL OF CONVERGENCE

If power series (1) converges when |x| < r and diverges when |x| > r, then r is called the radius of convergence. Similarly, r is the radius of convergence of power series (2) if (2) converges when |xa| < r and diverges when |xa| > r.

The set of all values of x for which a power series converges is called its interval of convergence. To find the interval of convergence, first determine the radius of convergence by applying the Ratio Test to the series of absolute values. Then check each endpoint to determine whether the series converges or diverges there.

EXAMPLE 33

Find all x for which the following series converges:

Image

SOLUTION: By the Ratio Test, the series converges if

Image

Thus, the radius of convergence is 1. The endpoints must be tested separately since the Ratio Test fails when the limit equals 1. When x = 1, (3) becomes 1 + 1 + 1 + · · · and diverges; when x = −1, (3) becomes 1−1 + 1−1 + ··· and diverges. Thus the interval of convergence is −1 < x < 1.

EXAMPLE 34

For what x does Image converge?

SOLUTION: Image

The radius of convergence is 1. When x = 1, we have Image an alternating convergent series; when x = −1, the series is Image which diverges. Thus, the series converges if − 1 < x Image 1.

EXAMPLE 35

For what values of x does Image converge?

SOLUTION: Image

which is always less than 1. Thus the series converges for all x.

EXAMPLE 36

Find all x for which the following series converges:

Image

SOLUTION: Image

which is less than 1 if |x − 2| < 2, that is, if 0 < x < 4. Series (4) converges on this interval and diverges if |x − 2| > 2, that is, if x < 0 or x > 4.

When x = 0, (4) is 1 − 1 + 1 − 1 + · · · and diverges. When x = 4, (4) is 1 + 1 + 1 + · · · and diverges. Thus, the interval of convergence is 0 < x < 4.

EXAMPLE 37

Find all x for which the series Image converges.

SOLUTION: Image converges only at x = 0, since

Image

unless x = 0.

C2. Functions Defined by Power Series.

Let the function f be defined by

Image

its domain is the interval of convergence of the series.

Functions defined by power series behave very much like polynomials, as indicated by the following properties:

PROPERTY 2a. The function defined by (1) is continuous for each x in the interval of convergence of the series.

PROPERTY 2b. The series formed by differentiating the terms of series (1) converges to f (x) for each x within the radius of convergence of (1); that is,

Image

Note that power series (1) and its derived series (2) have the same radius of convergence but not necessarily the same interval of convergence.

EXAMPLE 38

Let Image

Find the intervals of convergence of the power series for f (x) and f (x).

SOLUTION: Image

also,

Image

and

Image

Hence, the power series for f converges if −1 Image x Image 1.

For the derivative Image

Image

also,

Image

and

Image

Hence, the power series for f converges if −1 Image x < 1.

Thus, the series given for f (x) and f (x) have the same radius of convergence, but their intervals of convergence differ.

PROPERTY 2c. The series obtained by integrating the terms of the given series (1) converges to Image for each x within the interval of convergence of (1); that is,

Image

EXAMPLE 39

Let Image Show that the power series for Image converges for all values of x in the interval of convergence of the power series for f (x).

SOLUTION: Obtain a series for Image by long division.

Image

Then,

Image

It can be shown that the interval of convergence is −1 < x < 1.

Then by Property 2c

Image

Since when x = 0 we see that c = 1, we have

Image

Note that this is a geometric series with ratio r = x and with a = 1; if |x| < 1, its sum is Image

C3. Finding a Power Series for a Function: Taylor and Maclaurin Series.

If a function f (x) is representable by a power series of the form

c0 + c1 (xa) + c2 (xa)2 + ··· + cn (xa)n +···

on an interval |xa| < r, then the coefficients are given by

Image

Taylor series

The series

Image

is called the Taylor series of the function f about the number a. There is never more than one power series in (xa) for f (x). It is required that the function and all its derivatives exist at x = a if the function f (x) is to generate a Taylor series expansion.

When a = 0 we have the special series

Image

called the Maclaurin series of the function f; this is the expansion of f about x = 0.

EXAMPLE 40

Find the Maclaurin series for f (x) = ex.

SOLUTION: Here f (x) = ex, …, f (n) (x) = ex, …, for all n. Then

f (0) = 1, …, f (n) (0) = 1, …,

for all n, making the coefficients Image

Image

EXAMPLE 41

Find the Maclaurin expansion for f (x) = sin x.

SOLUTION: Image

Thus,

Image

EXAMPLE 42

Find the Maclaurin series for Image

SOLUTION:

Image

Then

Image

Note that this agrees exactly with the power series in x obtained by different methods in Example 39.

EXAMPLE 43

Find the Taylor series for the function f (x) = ln x about x = 1.

SOLUTION:

Image

Then

Image

COMMON MACLAURIN SERIES

We list here for reference some frequently used Maclaurin series expansions together with their intervals of convergence:

Image

FUNCTIONS THAT GENERATE NO SERIES.

Note that the following functions are among those that fail to generate a specific series in (xa) because the function and/or one or more derivatives do not exist at x = a:

Image

C4. Approximating Functions with Taylor and Maclaurin Polynomials.

The function f (x) at the point x = a is approximated by a Taylor polynomial Pn (x) of order n:

Image

The Taylor polynomial Pn (x) and its first n derivatives all agree at a with f and its first n derivatives. The order of a Taylor polynomial is the order of the highest derivative, which is also the polynomial’s last term.

In the special case where a = 0, the Maclaurin polynomial of order n that approximates f (x) is

Image

The Taylor polynomial P1 (x) at x = 0 is the tangent-line approximation to f (x) near zero given by

f (x) Image P1 (x) = f (0) + f (0)x.

It is the “best” linear approximation to f at 0, discussed at length in Chapter 4 §L.

A NOTE ON ORDER AND DEGREE

A Taylor polynomial has degree n if it has powers of (xa) up through the nth. If f (n) (a) = 0, then the degree of Pn (x) is less than n. Note, for instance, in Example 45, that the second-order polynomial P2 (x) for the function sin x (which is identical with P1 (x)) is Image or just x, which has degree 1, not 2.

EXAMPLE 44

Find the Taylor polynomial of order 4 at 0 for f (x) = e−x. Use this to approximate f (0.25).

SOLUTIONS: The first four derivatives are −e−x, e−x, −e−x and e−x ; at a = 0, these equal −1, 1,−1, and 1 respectively. The approximating Taylor polynomial of order 4 is therefore

Image

With x = 0.25 we have

Image

This approximation of e−0.25 is correct to four places.

In Figure N10–2 we see the graphs of f (x) and of the Taylor polynomials:

Image

Image

FIGURE N10–2

Notice how closely P4 (x) hugs f (x) even as x approaches 1. Since the series can be shown to converge for x > 0 by the Alternating Series Test, the error in P4 (x) is less than the magnitude of the first omitted term, Image at x = 1. In fact, P4 (1) = 0.375 to three decimal places, close to e−1 ≈ 0.368.

EXAMPLE 45

(a) Find the Taylor polynomials P1, P3, P5, and P7 at x = 0 for f (x) = sin x.

(b) Graph f and all four polynomials in [−2π,2π] × [−2,2].

(c) Approximate sin Image using each of the four polynomials.

SOLUTIONS:

(a) The derivatives of the sine function at 0 are given by the following table:

Image

From the table we know that

Image

(b) Figure N10–3a shows the graphs of sin x and the four polynomials. In Figure N10–3b we see graphs only of sin x and P7 (x), to exhibit how closely P7 “follows” the sine curve.

Image

FIGURE N10–3a

Image

FIGURE N10–3b

(c) To four decimal places, sin Image = 0.8660. Evaluating the polynomials at Image we get

Image

We see that P7 is correct to four decimal places.

EXAMPLE 46

(a) Find the Taylor polynomials of degrees 0, 1, 2, and 3 generated by f (x) = ln x at x = 1.

(b) Graph f and the four polynomials on the same set of axes.

(c) Using P2, approximate ln 1.3, and find a bound on the error.

SOLUTIONS:

(a) The derivatives of ln x at x = 1 are given in the table:

Image

From the table we have

Image

(b) Figure N10–4 shows the graphs of ln x and the four Taylor polynomials above, in [0,2.5] × [−1,1].

Image

FIGURE N10–4

(c) ln 1.3 ≈ P2 (1.3) = (1.3 − 1) − Image − 0.045 = 0.255.

For x = 1.3 the Taylor series converges by the Alternating Series Test, so the error is less than the magnitude of the first omitted term:

Image

EXAMPLE 47

For what positive values of x is the approximate formula

ln (1 + x) = Image

correct to three decimal places?

SOLUTION: We can use series (4) of Common Maclaurin Series:

Image

For x > 0, this is an alternating series with terms decreasing in magnitude and approaching 0, so the error committed by using the first two terms is less than Image If Image then the given approximation formula will yield accuracy to three decimal places. We therefore require that |x|3 < 0.0015 or that |x| < 0.114.

C5. Taylor’s Formula with Remainder; Lagrange Error Bound.

When we approximate a function using a Taylor polynomial, it is important to know how large the remainder (error) may be. If at the desired value of x the Taylor series is alternating, this issue is easily resolved: the first omitted term serves as an upper bound on the error. However, when the approximation involves a nonnegative Taylor series, placing an upper bound on the error is more difficult. This issue is resolved by the Lagrange remainder.

TAYLOR’S THEOREM. If a function f and its first (n + 1) derivatives are continuous on the interval |xa| < r, then for each x in this interval

Image

where

Image

and c is some number between a and x. Rn (x) is called the Lagrange remainder.

Note that the equation above expresses f (x) as the sum of the Taylor polynomial Pn (x) and the error that results when that polynomial is used as an approximation for f (x).

When we truncate a series after the (n + 1)st term, we can compute the error bound Rn, according to Lagrange, if we know what to substitute for c. In practice we find, not Rn exactly, but only an upper bound for it by assigning to c the value between a and x that determines the largest possible value of Rn. Hence:

Image

the Lagrange error bound.

EXAMPLE 48

Estimate the error in using the Maclaurin series generated by ex to approximate the value of e.

SOLUTION: From Example 40 we know that f (x) = ex generates the Maclaurin series

Image

The Lagrange error bound is

Image

To estimate e, we use x = 1. For 0 < c < 1, the maximum value of ec is e. Thus:

Image

EXAMPLE 49

Find the Maclaurin series for ln (1 + x) and the associated Lagrange error bound.

SOLUTION:

Image

Then

Image

where the Lagrange error bound is

Image

NOTE: For 0 < x < 1 the Maclaurin series is alternating, and the error bound simplifies to Image the first omitted term. The more difficult Lagrange error bound applies for −1 < x <0.

EXAMPLE 50

Find the third-degree Maclaurin polynomial for Image and determine the upper bound on the error in estimating f (0.1).

SOLUTION: We first make a table of the derivatives, evaluated at x = 0 and giving us the coefficients.

Image

Thus Image

Since this is not an alternating series for x = 0.1, we must use the Lagrange error bound:

Image where x = 0.1 and 0 < c < 0.1.

Note that Image is decreasing on the interval 0 < c < 0.1, so its maximum value occurs at c = 0. Hence:

Image

C6. Computations with Power Series.

The power series expansions of functions may be treated as any other functions for values of x that lie within their intervals of convergence. They may be added, subtracted, multiplied, divided (with division by zero to be avoided), differentiated, or integrated. These properties provide a valuable approach for many otherwise difficult computations. Indeed, power series are often very useful for approximating values of functions, evaluating indeterminate forms of limits, and estimating definite integrals.

EXAMPLE 51

Compute Image to four decimal places.

SOLUTION: We can use the Maclaurin series,

Image

and let Image to get

Image

Note that, since this series converges by the Alternating Series Test, R5 is less than the first term dropped:

Image

so Image correct to four decimal places.

EXAMPLE 52

Estimate the error if the approximate formula

Image

is used and |x| < 0.02.

SOLUTION: We obtain the first few terms of the Maclaurin series generated by Image

Image

Then

Image

Note that for x < 0, the series is not alternating, so we must use the Lagrange error bound. Here Image where −0.02 < c < 0.02. With |x| < 0.02, we see that the upper bound uses c = −0.02:

Image

EXAMPLE 53

Use a series to evaluate Image

SOLUTION: From series (1) in Common Maclaurin Series,

Image

Then

Image

a well-established result obtained previously.

EXAMPLE 54

Use a series to evaluate Image

SOLUTION: We can use series (4) in Common Maclaurin Series, and write

Image

EXAMPLE 55

Image

EXAMPLE 56

Show how a series may be used to evaluate π.

SOLUTION: Since Image a series for tan−1 x may prove helpful. Note that

Image

and that a series for Image is obtainable easily by long division to yield

Image

If we integrate this series term by term and then evaluate the definite integral, we get

Image

(Compare with series (5) in Common Maclaurin Series and note especially that this series converges on −1 x 1.)

For x = 1 we have:

Image

Here are some approximations for π using this series:

Image

Since the series is alternating, the odd sums are greater, the even ones less, than the value of π. It is clear that several hundred terms may be required to get even two-place accuracy. There are series expressions for π that converge much more rapidly. (See Miscellaneous Free-Response Practice, Problem 12.)

EXAMPLE 57

Use a series to evaluate Image to four decimal places.

SOLUTION: Although Image cannot be expressed in terms of elementary functions, we can write a series for eu, replace u by (−x2), and integrate term by term. Thus,

Image

Image

Since this is a convergent alternating series (with terms decreasing in magnitude and approaching 0), Image which will not affect the fourth decimal place. Then, correct to four decimal places,

Image

C7. Power Series over Complex Numbers.

A complex number is one of the form a + bi, where a and b are real and i2 = −1. If we allow complex numbers as replacements for x in power series, we obtain some interesting results.

Consider, for instance, the series

Image

When x = yi, then (1) becomes

Image

Then

Image

since the series within the parentheses of equation (2) converge respectively to cos y and sin y. Equation (3) is called Euler’s formula. It follows from (3) that

ei π = − 1,

and thus that

ei π + 1 = 0,

sometimes referred to as Euler’s magic formula.

This is an optional topic not in the BC Course Description. We include it here because of the dramatic result.

Chapter Summary

In this chapter, we have reviewed an important BC Calculus topic, infinite series. We have looked at a variety of tests to determine whether a series converges or diverges. We have worked with functions defined as power series, reviewed how to derive Taylor series, and looked at the Maclaurin series expansions for many commonly used functions. Finally, we have reviewed how to find bounds on the errors that arise when series are used for approximations.