19.2 Taylor Series

INTRODUCTION

The correspondence between a complex number z within the circle of convergence and the number to which the series converges is single-valued. In this sense, a power series defines or represents a function f; for a specified z within the circle of convergence, the number L to which the power series converges is defined to be the value of f at z; that is, f(z) = L. In this section we present some important facts about the nature of this function f.

In the preceding section we saw that every power series has a radius of convergence R. Throughout the discussion in this section, we will assume that a power series has either a positive or an infinite radius R of convergence. The next three theorems will give some important facts about the nature of a power series within its circle of convergence |z − z0| = R, R ≠ 0.

Theorem 19.2.1 Continuity

A power series ak(z − z0)k represents a continuous function f within its circle of convergence |z − z0| = R, R ≠ 0.

Theorem 19.2.2 Term-by-Term Integration

A power series ak(z − z0)k can be integrated term by term within its circle of convergence |z − z0| = R, R ≠ 0, for every contour C lying entirely within the circle of convergence.

Theorem 19.2.3 Term-by-Term Differentiation

A power series ak(z − z0)k can be differentiated term by term within its circle of convergence |z − z0| = R, R ≠ 0.

Taylor Series

Suppose a power series represents a function f for |z − z0| < R, R ≠ 0; that is,

(1)

It follows from Theorem 19.2.3 that the derivatives of f are

(2)

(3)

(4)

and so on. Each of the differentiated series has the same radius of convergence as the original series. Moreover, since the original power series represents a differentiable function f within its circle of convergence, we conclude that when R ≠ 0:

A power series represents an analytic function within its circle of convergence.

There is a relationship between the coefficients ak and the derivatives of f. Evaluating (1), (2), (3), and (4) at z = z0 gives

respectively. In general, f(n)(z0) = n!an or

(5)

When n = 0, we interpret the zeroth derivative as f(z0) and 0! = 1. Substituting (5) into (1) yields

(6)

This series is called the Taylor series for f centered at Although the English mathematician Brook Taylor (1685–1731) did not invent this series, he did use it extensively in his research on finite differences. Because the Scottish mathematician Colin Maclaurin (1698–1746) frequently used Taylor series centered at 0 in his research, the series

(7)

was named the Maclaurin series in his honor.

We have just seen that a power series with a nonzero radius of convergence represents an analytic function. On the other hand, if we are given a function f that is analytic in some domain D, can we represent it by a power series of the form (6) and (7)? Since a power series converges in a circular domain, and a domain D is generally not circular, the question becomes: Can we expand f in one or more power series that are valid in circular domains that are all contained in D? The question will be answered in the affirmative by the next theorem.

Theorem 19.2.4 Taylor’s Theorem

Let f be analytic within a domain D and let z0 be a point in D. Then f has the series representation

(8)

valid for the largest circle C with center at z0 and radius R that lies entirely within D.

PROOF:

Let z be a fixed point within the circle C and let s denote the variable of integration. The circle C is then described by |s − z0| = R. See FIGURE 19.2.1. To begin, we use the Cauchy integral formula to obtain the value of f at z:

(9)

A circle C with radius R and center z subscript 0 is shown in domain D. A point labeled z is inside the circle and a point labeled S is on the circumference of the circle. The circle has anticlockwise arrows on it.

FIGURE 19.2.1 Circular contour C used in proof of Theorem 19.2.4

By replacing z by (z − z0)/(s − z0) in (8) of Section 19.1, we have

and so (9) becomes

(10)

Utilizing Cauchy’s integral formula for derivatives, we can write (10) as

(11)

where

Equation (11) is called Taylor’s formula with remainder Rn. We now wish to show that Rn(z) → 0 as n → ∞. Since f is analytic in D, |f(z)| has a maximum value M on the contour C. In addition, since z is inside C, we have |z − z0| < R, and, consequently,

|s − z| = |s − z0 − (z − z0)| ≥ |s − z0| − |z − z0| = R − d,

where d = |z − z0| is the distance from z and z0. The ML-inequality then gives

Because d < R, (d/R)n → 0 as n → ∞, we conclude that |Rn(z)| → 0 as n → ∞. It follows that the infinite series

converges to f(z). In other words, the result in (8) is valid for any point z interior to C. ≡

We can find the radius of convergence of a Taylor series in exactly the same manner illustrated in Examples 5–7 of Section 19.1. However, we can simplify matters even further by noting that the radius of convergence is the distance from the center z0 of the series to the nearest isolated singularity of f. We shall elaborate more on this concept in the next section, but an isolated singularity is a point at which f fails to be analytic but is, nonetheless, analytic at all other points throughout some neighborhood of the point. For example, z = 5i is an isolated singularity of f(z) = 1/(z − 5i). If the function f is entire, then the radius of convergence of a Taylor series centered at any point z0 is necessarily infinite. Using (8) and the last fact, we can say that the Maclaurin series representations

(12)

(13)

(14)

are valid for all z.

If two power series with center z0,

,

represent the same function and have the same nonzero radius of convergence, then ak = bk , k = 0, 1, 2, . . . . Stated in another way, the power series expansion of a function with center z0 is unique. On a practical level, this means that a power series expansion of an analytic function f centered at z0, irrespective of the method used to obtain it, is the Taylor series expansion of the function. For example, we can obtain (14) by simply differentiating (13) term by term. The Maclaurin series for ez2 can be obtained by replacing the symbol z in (12) by z2.

EXAMPLE 1 Maclaurin Series

Find the Maclaurin expansion of f(z) = .

SOLUTION

We could, of course, begin by computing the coefficients using (8). However, we know from (5) of Section 19.1 that for |z| < 1,

(15)

Differentiating both sides of the last result with respect to z then yields

Since we are using Theorem 19.2.3, the radius of convergence of this last series is the same as the original series, R = 1. ≡

EXAMPLE 2 Taylor Series

Expand f(z) = in a Taylor series with center z0 = 2i.

SOLUTION

We shall solve this problem in two ways. We begin by using (8). From the first several derivatives,

we conclude that f(n)(z) = n!/(1 − z)n + 1 and so f(n)(2i) = n!/(1 − 2i)n + 1. Thus from (8) we obtain the Taylor series

(16)

Since the distance from the center z0 = 2i to the nearest singularity z = 1 is , we conclude that the circle of convergence for the power series in (16) is |z − 2i| = . This can be verified by the ratio test of the preceding section.

ALTERNATIVE SOLUTION

In this solution we again use the geometric series (15). By adding and subtracting 2i in the denominator of 1/(1 − z), we can write

We now write as a power series by using (15) with the symbol z replaced by (z − 2i)/(1 − 2i):

The reader should verify that this last series is exactly the same as that in (16). ≡

In (15) and (16) we represented the same function 1/(1 − z) by two different power series. The first series

has center zero and radius of convergence one. The second series

has center 2i and radius of convergence . The two different circles of convergence are illustrated in FIGURE 19.2.2. The interior of the intersection of the two circles (shaded) is the region where both series converge; in other words, at a specified point z* in this region, both series converge to the same value f(z*) = 1/(1 − z*). Outside the shaded region, at least one of the two series must diverge.

A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled z asterisk in the first quadrant close to the origin. The graph also shows a big circle labeled mod (z minus 2 i) = under root 5, which is majorly in the second and the first quadrant. It also shows a small circle, labeled mod z = 1, centered at the origin and intersecting the bigger circle. The intersecting region between the two circles is shaded.

FIGURE 19.2.2 Series (15) and (16) both converge within the shaded region

19.2 Exercises Answers to selected odd-numbered problems begin on page ANS-47.

In Problems 1–12, expand the given function in a Maclaurin series. Give the radius of convergence of each series.

  1. f(z) =
  2. f(z) =
  3. f(z) =
  4. f(z) =
  5. f(z) = e−2z
  6. f(z) = ze−z2
  7. f(z) = sinh z
  8. f(z) = cosh z
  9. f(z) = cos
  10. f(z) = sin 3z
  11. f(z) = sin z2
  12. f(z) = cos2z [Hint: Use a trigonometric identity.]

In Problems 13–22, expand the given function in a Taylor series centered at the indicated point. Give the radius of convergence of each series.

  1. f(z) = 1/z, z0 = 1
  2. f(z) = 1/z, z0 = 1 + i
  3. f(z) = , z0 = 2i
  4. f(z) = , z0 = −i
  5. f(z) = , z0 = 1
  6. f(z) = , z0 = i
  7. f(z) = cos z, z0 = π/4
  8. f(z) = sin z, z0 = π/2
  9. f(z) = ez, z0 = 3i
  10. f(z) = (z − 1)e−2z, z0 = 1

In Problems 23 and 24, use (7) to find the first three nonzero terms of the Maclaurin series of the given function.

  1. f(z) = tan z
  2. f(z) = e1/(1+z)

In Problems 25 and 26, use partial fractions as an aid in obtaining the Maclaurin series for the given function. Give the radius of convergence of the series.

  1. f(z) =
  2. f(z) =

In Problems 27 and 28, without actually expanding, determine the radius of convergence of the Taylor series of the given function centered at the indicated point.

  1. f(z) = , z0 = 2 + 5i
  2. f(z) = cot z, z0 = πi

In Problems 29 and 30, expand the given function in the Taylor series centered at the indicated points. Give the radius of convergence of each series. Sketch the region within which both series converge.

  1. f(z) = , z0 = −1, z0 = i
  2. f(z) = , z0 = 1 + i, z0 = 3
    1. Suppose the principal branch of the logarithm f(z) = Ln z = loge|z| + i Arg z is expanded in a Taylor series with center z0 = −1 + i. Explain why R = 1 is the radius of the largest circle centered at z0 = −1 + i within which f is analytic.
    2. Show that within the circle |z − (−1 + i)| = 1 the Taylor series for f is
    3. Show that the radius of convergence for the power series in part (b) is R = . Explain why this does not contradict the result in part (a).
    1. Consider the function f(z) = Ln(1 + z). What is the radius of the largest circle centered at the origin within which f is analytic?
    2. Expand f in a Maclaurin series. What is the radius of convergence of this series?
    3. Use the result in part (b) to find a Maclaurin series for Ln(1 − z).

    (d) Find a Maclaurin series for Ln.

In Problems 33 and 34, approximate the value of the given expression using the indicated number of terms of a Maclaurin series.

  1. e(1+i)/10, three terms
  2. sin, two terms
  3. In Section 15.1 we defined the error function as

    Find a Maclaurin series for erf (z).

  4. Use the Maclaurin series for eiz to prove Euler’s formula for complex z:

    eiz = cos z + i sin z.