Calculus AB and Calculus BC
CHAPTER 10 Sequences and Series
C. POWER SERIES
C1. Definitions; Convergence.
An expression of the form
where the a’s are constants, is called a power series in x; and
is called a power series in (x − a).
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 |x − a| < r and diverges when |x − a| > 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.
Find all x for which the following series converges:
SOLUTION: By the Ratio Test, the series converges if
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.
For what x does converge?
The radius of convergence is 1. When x = 1, we have an alternating convergent series; when x = −1, the series is which diverges. Thus, the series converges if − 1 < x 1.
For what values of x does converge?
which is always less than 1. Thus the series converges for all x.
Find all x for which the following series converges:
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.
Find all x for which the series converges.
SOLUTION: converges only at x = 0, since
unless x = 0.
C2. Functions Defined by Power Series.
Let the function f be defined by
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,
Note that power series (1) and its derived series (2) have the same radius of convergence but not necessarily the same interval of convergence.
Find the intervals of convergence of the power series for f (x) and f ′(x).
Hence, the power series for f converges if −1 x 1.
For the derivative
Hence, the power series for f ′ converges if −1 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 for each x within the interval of convergence of (1); that is,
Let Show that the power series for converges for all values of x in the interval of convergence of the power series for f (x).
SOLUTION: Obtain a series for by long division.
It can be shown that the interval of convergence is −1 < x < 1.
Then by Property 2c
Since when x = 0 we see that c = 1, we have
Note that this is a geometric series with ratio r = x and with a = 1; if |x| < 1, its sum is
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 (x − a) + c2 (x − a)2 + ··· + cn (x − a)n +···
on an interval |x − a| < r, then the coefficients are given by
is called the Taylor series of the function f about the number a. There is never more than one power series in (x − a) 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
called the Maclaurin series of the function f; this is the expansion of f about x = 0.
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
Find the Maclaurin expansion for f (x) = sin x.
Find the Maclaurin series for
Note that this agrees exactly with the power series in x obtained by different methods in Example 39.
Find the Taylor series for the function f (x) = ln x about x = 1.
COMMON MACLAURIN SERIES
We list here for reference some frequently used Maclaurin series expansions together with their intervals of convergence:
FUNCTIONS THAT GENERATE NO SERIES.
Note that the following functions are among those that fail to generate a specific series in (x − a) because the function and/or one or more derivatives do not exist at x = a:
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:
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
The Taylor polynomial P1 (x) at x = 0 is the tangent-line approximation to f (x) near zero given by
f (x) 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 (x − a) 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 or just x, which has degree 1, not 2.
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
With x = 0.25 we have
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:
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, at x = 1. In fact, P4 (1) = 0.375 to three decimal places, close to e−1 ≈ 0.368.
(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 using each of the four polynomials.
(a) The derivatives of the sine function at 0 are given by the following table:
From the table we know that
(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.
(c) To four decimal places, sin = 0.8660. Evaluating the polynomials at we get
We see that P7 is correct to four decimal places.
(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.
(a) The derivatives of ln x at x = 1 are given in the table:
From the table we have
(b) Figure N10–4 shows the graphs of ln x and the four Taylor polynomials above, in [0,2.5] × [−1,1].
(c) ln 1.3 ≈ P2 (1.3) = (1.3 − 1) − − 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:
For what positive values of x is the approximate formula
ln (1 + x) =
correct to three decimal places?
SOLUTION: We can use series (4) of Common Maclaurin Series:
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 If 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 |x − a| < r, then for each x in this interval
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:
the Lagrange error bound.
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
The Lagrange error bound is
To estimate e, we use x = 1. For 0 < c < 1, the maximum value of ec is e. Thus:
Find the Maclaurin series for ln (1 + x) and the associated Lagrange error bound.
where the Lagrange error bound is
NOTE: For 0 < x < 1 the Maclaurin series is alternating, and the error bound simplifies to the first omitted term. The more difficult Lagrange error bound applies for −1 < x <0.
Find the third-degree Maclaurin polynomial for 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.
Since this is not an alternating series for x = 0.1, we must use the Lagrange error bound:
where x = 0.1 and 0 < c < 0.1.
Note that is decreasing on the interval 0 < c < 0.1, so its maximum value occurs at c = 0. Hence:
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.
Compute to four decimal places.
SOLUTION: We can use the Maclaurin series,
and let to get
Note that, since this series converges by the Alternating Series Test, R5 is less than the first term dropped:
so correct to four decimal places.
Estimate the error if the approximate formula
is used and |x| < 0.02.
SOLUTION: We obtain the first few terms of the Maclaurin series generated by
Note that for x < 0, the series is not alternating, so we must use the Lagrange error bound. Here where −0.02 < c < 0.02. With |x| < 0.02, we see that the upper bound uses c = −0.02:
Use a series to evaluate
SOLUTION: From series (1) in Common Maclaurin Series,
a well-established result obtained previously.
Use a series to evaluate
SOLUTION: We can use series (4) in Common Maclaurin Series, and write
Show how a series may be used to evaluate π.
SOLUTION: Since a series for tan−1 x may prove helpful. Note that
and that a series for is obtainable easily by long division to yield
If we integrate this series term by term and then evaluate the definite integral, we get
(Compare with series (5) in Common Maclaurin Series and note especially that this series converges on −1 ≤ x ≤ 1.)
For x = 1 we have:
Here are some approximations for π using this series:
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.)
Use a series to evaluate to four decimal places.
SOLUTION: Although 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,
Since this is a convergent alternating series (with terms decreasing in magnitude and approaching 0), which will not affect the fourth decimal place. Then, correct to four decimal places,
†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
When x = yi, then (1) becomes
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.
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.