EXAMPLE 1 Convergent and Divergent Sequences

The sequence is convergent with limit 0.

The sequence {iⁿ} = {i, −1, −i, 1, …} is divergent, and so is {z_n} with z_n = (1 + i)ⁿ.

EXAMPLE 2 Sequences of the Real and the Imaginary Parts

The sequence {z_n} with z_n = x_n + iy_n = 1 − 1/n² + i(2 + 4/n) is (Sketch it.) It converges with the limit c = 1 + 2i. Observe that {x_n} has the limit 1 = Re c and {y_n} has the limit 2 = Im c. This is typical. It illustrates the following theorem by which the convergence of a complex sequence can be referred back to that of the two real sequences of the real parts and the imaginary parts.

THEOREM 1 Sequences of the Real and the Imaginary Parts

A sequence z₁, z₂, …, z_n, … of complex numbers z_n = x_n + y_n (where n = 1, 2, …) converges to c = a + ib if and only if the sequence of the real parts x₁, x₂, … converges to a and the sequence of the imaginary parts y₁, y₂, … converges to b.

PROOF

Convergence z_n → c = a + ib implies convergence x_n → a and y_n → b because if |z_n − c| < then z_n lies within the circle of radius about c = a + ib so that (Fig. 363a)

Conversely, if x_n → a and y_n → b as n → ∞, then for a given > 0 we can choose N so large that, for every n > N,

Fig. 363. Proof of Theorem 1

These two inequalities imply that z_n = x_n + iy_n lies in a square with center c and side . Hence, z_n must lie within a circle of radius with center c (Fig. 363b).

THEOREM 2 Real and Imaginary Parts

A series (3) with z_m = x_m + iy_m converges and has the sum s = u + iv if and only if x₁ + x₂ + … converges and has the sum u and y₁ + y₂ + … converges and has the sum v.

THEOREM 3 Divergence

If a series z₁ + z₂ + … converges, then Hence if this does not hold, the series diverges.

PROOF

If z₁ + z₂ + … converges, with the sum s, then, since z_m = s_m − s_m−1,

THEOREM 4 Cauchy's Convergence Principle for Series

A series z₁ + z₂ + … is convergent if and only if for every given > 0 (no matter how small) we can find an N (which depends on , in general) such that

EXAMPLE 3 A Conditionally Convergent Series

The series converges, but only conditionally since the harmonic series diverges, as mentioned above (after Theorem 3).

THEOREM 5 Comparison Test

If a series z₁ + z₂ + … is given and we can find a convergent series b₁ + b₂ + … with nonnegative real terms such that |z₁| b₁, |z₂| b₂, …, then the given series converges, even absolutely.

PROOF

By Cauchy's principle, since b₁ + b₂ + … converges, for any given > 0 we can find an N such that

From this and |z₁| b₁, |z₂| b₂, … we conclude that for those n and p,

Hence, again by Cauchy's principle, |z₁| + |z₂| + … converges, so that z₁ + z₂ + … is absolutely convergent.

THEOREM 6 Geometric Series

The geometric series

converges with the sum 1/(1 − q) if |q| < 1 and diverges if |q| 1.

PROOF

If |q| 1, then |q^m| 1 and Theorem 3 implies divergence.

Now let |q| < 1. The nth partial sum is

From this,

On subtraction, most terms on the right cancel in pairs, and we are left with

Now 1 − q ≠ 0 since q ≠ 1, and we may solve for s_n finding

Since |q| < 1, the last term approaches zero as n → ∞. Hence if |q| < 1. the series is convergent and has the sum 1/(1 − q). This completes the proof.

THEOREM 7 Ratio Test

If a series z₁ + z₂ + … with z_n ≠ 0 (n = 1, 2, …) has the property that for every n greater than some N,

(where q < 1 is fixed), this series converges absolutely. If for every n > N,

the series diverges.

PROOF

If (8) holds, then |z_{n + 1}| |z_n| for n > N, so that divergence of the series follows from Theorem 3.

If (7) holds, then |z_{n + 1}| |z_n| q for n > N, in particular,

and in general, |z_{N + p}| |z_{N + 1}|q^p−1. Since q < 1, we obtain from this and Theorem 6

Absolute convergence of z₁ + z₂ + … now follows from Theorem 5.

THEOREM 8 Ratio Test

If a series z₁ + z₂ + … with z_n ≠ 0(n = 1, 2, …) is such that then:

1. If L < 1, the series converges absolutely.
2. If L > 1, the series diverges.
3. If L = 1, the series may converge or diverge, so that the test fails and permits no conclusion.

PROOF

(a) We write k_n = |z_{n + 1}/z_n| and let L = 1 − b < 1. Then by the definition of limit, the k_n must eventually get close to 1 − b, say, for all n greater than some N. Convergence of z₁ + z₂ + … now follows from Theorem 7.

(b) Similarly, for L = 1 + c > 1 we have for all n > N* (sufficiently large), which implies divergence of z₁ + z₂ + … by Theorem 7.

(c) The harmonic series has z_{n + 1}/z_n = n/(n + 1), hence L = 1, and diverges. The series

hence also L = 1, but it converges. Convergence follows from (Fig. 364)

so that s₁, s₂, … is a bounded sequence and is monotone increasing (since the terms of the series are all positive); both properties together are sufficient for the convergence of the real sequence s₁, s₂, …. (In calculus this is proved by the so-called integral test, whose idea we have used.)

EXAMPLE 4 Ratio Test

Is the following series convergent or divergent? (First guess, then calculate.)

Solution. By Theorem 8, the series is convergent, since

EXAMPLE 5 Theorem 7 More General Than Theorem 8

Let a_n = i/2³ⁿ and b_n = 1/2^{3n + 1}. Is the following series convergent or divergent?

Solution. The ratios of the absolute values of successive terms are Hence convergence follows from Theorem 7. Since the sequence of these ratios has no limit, Theorem 8 is not applicable.

THEOREM 9 Root Test

If a series z₁ + z₂ + … is such that for every n greater than some N,

(where q < 1 is fixed), this series converges absolutely. If for infinitely many n,

the series diverges.

PROOF

If (9) holds, then |z_n| qⁿ < 1 for all n > N. Hence the series |z₁| + |z₂| + … converges by comparison with the geometric series, so that the series z₁ + z₂ + … converges absolutely. If (10) holds, then |z_n| 1 for infinitely many n. Divergence of z₁ + z₂ + … now follows from Theorem 3.

THEOREM 10 Root Test

If a series z₁ + z₂ + … is such that then:

1. The series converges absolutely if L < 1.
2. The series diverges if L > 1.
3. If L = 1, the test fails; that is, no conclusion is possible.

PROBLEM SET 15.1

1–10 SEQUENCES

Is the given sequence z₁, z₂, …, z_n, … bounded? Convergent? Find its limit points. Show your work in detail.

1. z_n = (1 + i)²ⁿ/2ⁿ
2. z_n = (3 + 4i)ⁿ/n!
3. z_n = nπ/(4 + 2ni)
4. z_n = (1 + 2i)ⁿ
5. z_n = (−1)ⁿ + 10i
6. z_n = (cos nπi)/n
7. z_n = n² + i/n²
8. 
9. z_n = (3 + 3i)⁻ⁿ
10. 
11. CAS EXPERIMENT. Sequences. Write a program for graphing complex sequences. Use the program to discover sequences that have interesting “geometric” properties, e.g., lying on an ellipse, spiraling to its limit, having infinitely many limit points, etc.
12. Addition of sequences. If z₁, z₂, … converges with the limit l and converges with the limit l*, show that is convergent with the limit l + l*.
13. Bounded sequence. Show that a complex sequence is bounded if and only if the two corresponding sequences of the real parts and of the imaginary parts are bounded.
14. On Theorem 1. Illustrate Theorem 1 by an example of your own.
15. On Theorem 2. Give another example illustrating Theorem 2.

16–25 SERIES

Is the given series convergent or divergent? Give a reason. Show details.

• 16.
• 17.
• 18.
• 19.
• 20.
• 21.
• 22.
• 23.
• 24.
• 25.
• 26. Significance of (7). What is the difference between (7) and just stating |z_{n + 1}/z_n| < 1?
• 27. On Theorems 7 and 8. Give another example showing that Theorem 7 is more general than Theorem 8.
• 28. CAS EXPERIMENT. Series. Write a program for computing and graphing numeric values of the first n partial sums of a series of complex numbers. Use the program to experiment with the rapidity of convergence of series of your choice.
• 29. Absolute convergence. Show that if a series converges absolutely, it is convergent.
• 30. Estimate of remainder. Let |z_{n + 1}/z_n| q < 1, so that the series z₁ + z₂ + … converges by the ratio test. Show that the remainder R_n = z_{n + 1} + z_{n + 2} + … satisfies the inequality |R_n| |z_{n + 1}|/(1 − q). Using this, find how many terms suffice for computing the sum s of the series

  

  with an error not exceeding 0.05 and compute s to this accuracy.

EXAMPLE 1 Convergence in a Disk. Geometric Series

The geometric series

converges absolutely if |z| < 1 and diverges if |z| 1 (see Theorem 6 in Sec. 15.1).

EXAMPLE 2 Convergence for Every z

The power series (which will be the Maclaurin series of e^z in Sec. 15.4)

is absolutely convergent for every z. In fact, by the ratio test, for any fixed z,

EXAMPLE 3 Convergence Only at the Center. (Useless Series)

The following power series converges only at z = 0, but diverges for every z ≠ 0, as we shall show.

In fact, from the ratio test we have

THEOREM 1 Convergence of a Power Series

(a) Every power series (1) converges at the center z₀

(b) If (1) converges at a point z = z₁ ≠ z₀, it converges absolutely for every z closer to z₀ than z₁, that is, |z − z₀| < |z₁ − z₀|. See Fig. 365.

(c) If (1) diverges at z = z₂, it diverges for every z farther away from z₀ than z₂. See Fig. 365.

PROOF

(a) For z = z₀ the series reduces to the single term a₀.

(b) Convergence at z = z₁ gives by Theorem 3 in Sec. 15.1 a_n(z₁ − z₀)ⁿ → 0 as n → ∞. This implies boundedness in absolute value,

Multiplying and dividing a_n(z − z₀)ⁿ by (z₁ − z₀)ⁿ we obtain from this

Summation over n gives

Now our assumption |z − z₀| < |z₁ − z₀| implies that |(z − z₀)/(z₁ − z₀)| < 1. Hence the series on the right side of (3) is a converging geometric series (see Theorem 6 in Sec. 15.1). Absolute convergence of (1) as stated in (b) now follows by the comparison test in Sec. 15.1.

(c) If this were false, we would have convergence at a z₃ farther away from z₀ than z₂. This would imply convergence at z₂, by (b), a contradiction to our assumption of divergence at z₂.

EXAMPLE 4 Behavior on the Circle of Convergence

On the circle of convergence (radius R = 1 in all three series),

Σ zⁿ/n² converges everywhere since Σ 1/n² converges,

Σ zⁿ/n converges at −1 (by Leibniz's test) but diverges at 1,

Σ zⁿ diverges everywhere.

THEOREM 2 Radius of Convergence R

Suppose that the sequence |a_{n + 1}/a_n|, n = 1, 2, …, converges with limit L*. If L* = 0, then R = ∞; that is, the power series (1) converges for all z. If L* ≠ 0 (hence L* > 0), then

If |a_{n + 1}/a_n| → ∞, then R = 0 (convergence only at the center z₀).

PROOF

For (1) the ratio of the terms in the ratio test (Sec. 15.1) is

Let L* ≠ 0, thus L* > 0. We have convergence if L = L*|z − z₀| < 1, thus |z − z₀| < 1/L*, and divergence if |z − z₀| > 1/L*. By (4) and (5) this shows that 1/L* is the convergence radius and proves (6).

If L* = 0, then L = 0 for every z, which gives convergence for all z by the ratio test. If |a_{n + 1}/a_n| → ∞, then |a_{n + 1}/a_n||z − z₀| > 1 for any z ≠ z₀ and all sufficiently large n. This implies divergence for all z ≠ z₀ by the ratio test (Theorem 7, Sec. 15.1).

EXAMPLE 5 Radius of Convergence

By (6) the radius of convergence of the power series is

The series converges in the open disk of radius and center 3i.

EXAMPLE 6 Extension of Theorem 2

Find the radius of convergence R of the power series

Solution. The sequence of the ratios does not converge, so that Theorem 2 is of no help. It can be shown that

This still does not help here, since does not converge because for odd n, whereas for even n we have

so that has the two limit points and 1. It can further be shown that

Here = I, so that R = 1. Answer. The series converges for |z| < 1.

PROBLEM SET 15.2

1. Power series. Are 1/z + z + z² + … and z + z^3/2 + z² + z³ + … power series? Explain.
2. Radius of convergence. What is it? Its role? What motivates its name? How can you find it?
3. Convergence. What are the only basically different possibilities for the convergence of a power series?
4. On Examples 1–3. Extend them to power series in powers of z − 4 + 3πi. Extend Example 1 to the case of radius of convergence 6.
5. Powers z²ⁿ. Show that if Σa_nzⁿ has radius of convergence R (assumed finite), then Σa_nz²ⁿ has radius of convergence

6–18 RADIUS OF CONVERGENCE

Find the center and the radius of convergence.

• 6.
• 7.
• 8.
• 9.
• 10.
• 11.
• 12.
• 13.
• 14.
• 15.
• 16.
• 17.
• 18.
• 19. CAS PROJECT. Radius of Convergence. Write a program for computing R from (6), (6*), or (6**), in this order, depending on the existence of the limits needed. Test the program on some series of your choice such that all three formulas (6), (6*), and (6**) will come up.
• 20. TEAM PROJECT. Radius of Convergence. (a) Understanding (6). Formula (6) for R contains |a_n/a_{n + 1}|, not |a_{n + 1}/a_n|. How could you memorize this by using a qualitative argument?

  (b) Change of coefficients. What happens to R (0 < R < ∞) if you (i) multiply all a_n by, k ≠ 0, (ii) multiply all a_n by Kⁿ ≠ 0, (iii) replace a_n by 1/a_n? Can you think of an application of this?

  (c) Understanding Example 6, which extends Theorem 2 to nonconvergent cases of a_n/a_{n + 1}. Do you understand the principle of “mixing” by which Example 6 was obtained? Make up further examples.

  (d) Understanding (b) and (c) in Theorem 1. Does there exist a power series in powers of z that converges at z = 30 + 10i and diverges at z = 31 − 6i? Give reason.

THEROEM 1 Continuity of the Sum of a Power Series

If a function f(z) can be represented by a power series (2) with radius of convergence R > 0, then f(z) is continuous at z = 0.

PROOF

From (2) with z = 0 we have f(0) = a₀. Hence by the definition of continuity we must show that lim_z→0 f(z) = f(0) = a₀. That is, we must show that for a given > 0 there is a δ > 0 such that |z| < δ implies |f(z) − a₀| < . Now (2) converges absolutely for |z| r with any r such that 0 < r < R, by Theorem 1 in Sec. 15.2. Hence the series

converges. Let S ≠ 0 be its sum. (S = 0 is trivial.) Then for 0 < |z| r,

and |z|S < when |z| < δ, where δ > 0 is less than r and less than /S. Hence |z|S < δS < (/S)S = . This proves the theorem.

THEOREM 2 Identity Theorem for Power Series. Uniqueness

Let the power series a₀ + a₁z + a₂z² + … and b₀ + b₁z + b₂z² + … both be convergent for |z| < R, where R is positive, and let them both have the same sum for all these z. Then the series are identical, that is, a₀ = b₀, a₁ = b₁, a₂ = b₂, ….

Hence if a function f(z) can be represented by a power series with any center z₀ this representation is unique.

PROOF

We proceed by induction. By assumption,

The sums of these two power series are continuous at z = 0, by Theorem 1. Hence if we consider |z| > 0 and let z → 0 on both sides, we see that a₀ = b₀: the assertion is true for n = 0. Now assume that a_n = b_n for n = 0, 1, …, m. Then on both sides we may omit the terms that are equal and divide the result by z^{m + 1} (≠ 0); this gives

Similarly as before by letting z → 0 we conclude from this that a_{m + 1} = b_{m + 1}. This completes the proof.

THEOREM 3 Termwise Differentiation of a Power Series

The derived series of a power series has the same radius of convergence as the original series.

PROOF

This follows from (6) in Sec. 15.2 because

or, if the limit does not exist, from (6**) in Sec. 15.2 by noting that as n → ∞.

EXAMPLE 1 Application of Theorem 3

Find the radius of convergence R of the following series by applying Theorem 3.

Solution. Differentiate the geometric series twice term by term and multiply the result by z²/2. This yields the given series. Hence R = 1 by Theorem 3.

THEOREM 4 Termwise Integration of Power Series

The power series

obtained by integrating the series a₀ + a₁z + a₂z² + … term by term has the same radius of convergence as the original series.

THEOREM 5 Analytic Functions. Their Derivatives

A power series with a nonzero radius of convergence R represents an analytic function at every point interior to its circle of convergence. The derivatives of this function are obtained by differentiating the original series term by term. All the series thus obtained have the same radius of convergence as the original series. Hence, by the first statement, each of them represents an analytic function.

PROOF

(a) We consider any power series (1) with positive radius of convergence R. Let f(z) be its sum and f₁(z) the sum of its derived series; thus

We show that f(z) is analytic and has the derivative f₁(z) in the interior of the circle of convergence. We do this by proving that for any fixed z with |z| < R and Δz → 0 the difference quotient [f(z + Δz) − f(z)]/Δz approaches f₁(z). By termwise addition we first have from (4)

Note that the summation starts with 2, since the constant term drops out in taking the difference f(z + Δz) − f(z), and so does the linear term when we subtract f₁(z) from the difference quotient.

(b) We claim that the series in (5) can be written

The somewhat technical proof of this is given in App. 4.

(c) We consider (6). The brackets contain n − 1 terms, and the largest coefficient is n − 1. Since (n − 1)² n(n − 1), we see that for |z| R₀ and |z + Δz| R₀, R₀ < R, the absolute value of this series (6) cannot exceed

This series with a_n instead of |a_n| is the second derived series of (2) at z = R₀ and converges absolutely by Theorem 3 of this section and Theorem 1 of Sec. 15.2. Hence our present series (7) converges. Let the sum of (7) (without the factor |Δz|) be K(R₀). Since (6) is the right side of (5), our present result is

Letting Δz → 0 and noting that R₀(< R) is arbitrary, we conclude that f(z) is analytic at any point interior to the circle of convergence and its derivative is represented by the derived series. From this the statements about the higher derivatives follow by induction.

PROBLEM SET 15.3

1. Relation to Calculus. Material in this section generalizes calculus. Give details.
2. Termwise addition. Write out the details of the proof on termwise addition and subtraction of power series.
3. On Theorem 3. Prove that as n → ∞, as claimed.
4. Cauchy product. Show that

   (a) by using the Cauchy product, (b) by differentiating a suitable series.

5–15 RADIUS OF CONVERGENCE BY DIFFERENTIATION OR INTEGRATION

Find the radius of convergence in two ways: (a) directly by the Cauchy–Hadamard formula in Sec. 15.2, and (b) from a series of simpler terms by using Theorem 3 or Theorem 4.

• 5.
• 6.
• 7.
• 8.
• 9.
• 10.
• 11.
• 12.
• 13.
• 14.
• 15.

16–20 APPLICATIONS OF THE IDENTITY THEOREM

State clearly and explicitly where and how you are using Theorem 2.

• 16. Even functions. If f(z) in (2) is even (i.e., f(− z) = f(z)), show that a_{n = 0 for odd n. Give examples.}
• 17. Odd function. If f(z) in (2) is odd (i.e., f(−z) = −f(z)), show that a_n = 0 for even n. Give examples.
• 18. Binomial coefficients. Using (1 + z)^p(1 + z)^q = (1 + z)^{p + q}, obtain the basic relation

  

• 19. Find applications of Theorem 2 in differential equations and elsewhere.
• 20. TEAM PROJECT. Fibonacci numbers.² (a) The Fibonacci numbers are recursively defined by a₀ = a₁ = 1, a_{n + 1} = a_n + a_n−1 if n = 1, 2, …. Find the limit of the sequence (a_{n + 1}/a_n).

  (b) Fibonacci's rabbit problem. Compute a list of a₁, …, a₁₂. Show that a₁₂ = 233 is the number of pairs of rabbits after 12 months if initially there is 1 pair and each pair generates 1 pair per month, beginning in the second month of existence (no deaths occurring).

  (c) Generating function. Show that the generating function of the Fibonacci numbers is f(z) = 1/(1 − z − z²); that is, if a power series (1) represents this f(z), its coefficients must be the Fibonacci numbers and conversely. Hint. Start from f(z)(1 − z − z²) = 1 and use Theorem 2.

THEOREM 1 Taylor's Theorem

Let f(z) be analytic in a domain D, and let z = z₀ be any point in D. Then there exists precisely one Taylor series (1) with center z₀ that represents f(z). This representation is valid in the largest open disk with center z₀ in which f(z) is analytic. The remainders R_n(z) of (1) can be represented in the form (3). The coefficients satisfy the inequality

where M is the maximum of |f(z)| on a circle |z − z₀| = r in D whose interior is also in D.

PROOF

The key tool is Cauchy's integral formula in Sec. 14.3; writing z and z* instead of z₀ and z (so that z* is the variable of integration), we have

z lies inside C, for which we take a circle of radius r with center z₀ and interior in D (Fig. 367). We develop 1/(z* − z) in (6) in powers of z − z₀. By a standard algebraic manipulation (worth remembering!) we first have

Fig. 367. Cauchy formula (6)

For later use we note that since z* is on C while z is inside C, we have

To (7) we now apply the sum formula for a finite geometric sum

which we use in the form (take the last term to the other side and interchange sides)

Applying this with q = (z − z₀)/(z* − z₀) to the right side of (7), we get

We insert this into (6). Powers of z − z₀ do not depend on the variable of integration z*, so that we may take them out from under the integral sign. This yields

with R_n(z) given by (3). The integrals are those in (2) related to the derivatives, so that we have proved the Taylor formula (4).

Since analytic functions have derivatives of all orders, we can take n in (4) as large as we please. If we let n approach infinity, we obtain (1). Clearly, (1) will converge and represent f(z) if and only if

We prove (9) as follows. Since z* lies on C, whereas z lies inside C (Fig. 367), we have |z* − z| > 0. Since f(z) is analytic inside and on C, it is bounded, and so is the function f(z*)/(z* − z), say,

for all z* on C. Also, C has the radius r = |z* − z₀| and the length 2πr. Hence by the ML-inequality (Sec. 14.1) we obtain from (3)

Now |z − z₀| < r because z lies inside C. Thus |z − z₀|/r < 1, so that the right side approaches 0 as n → ∞. This proves that the Taylor series converges and has the sum f(z). Uniqueness follows from Theorem 2 in the last section. Finally, (5) follows from a_n in (1) and the Cauchy inequality in Sec. 14.4. This proves Taylor's theorem.

THEOREM 2 Relation to the Previous Section

A power series with a nonzero radius of convergence is thes Taylor series of its sum.

PROOF

Given the power series

Then f(z₀) = a₀. By Theorem 5 in Sec. 15.3 we obtain

and in general f⁽ⁿ⁾(z₀) = n!a_n. With these coefficients the given series becomes the Taylor series of f(z) with center z₀.

EXAMPLE 1 Geometric Series

Let f(z) = 1/(1 − z). Then we have f⁽ⁿ⁾(z) = n!/(1 − z)^{n + 1}, f⁽ⁿ⁾(0) = n!. Hence the Maclaurin expansion of 1/(1 − z) is the geometric series

f(z) is singular at z = 1; this point lies on the circle of convergence.

EXAMPLE 2 Exponential Function

We know that the exponential function e^z (Sec. 13.5) is analytic for all z, and (e^z)′ = e^z. Hence from (1) with z₀ = 0 we obtain the Maclaurin series

This series is also obtained if we replace x in the familiar Maclaurin series of e^x by z.

Furthermore, by setting z = iy in (12) and separating the series into the real and imaginary parts (see Theorem 2, Sec. 15.1) we obtain

Since the series on the right are the familiar Maclaurin series of the real functions cos y and sin y, this shows that we have rediscovered the Euler formula

Indeed, one may use (12) for defining e^z and derive from (12) the basic properties of e^z. For instance, the differentiation formula (e^z)′ = e^z follows readily from (12) by termwise differentiation.

EXAMPLE 3 Trigonometric and Hyperbolic Functions

By substituting (12) into (1) of Sec. 13.6 we obtain

When z = x these are the familiar Maclaurin series of the real functions cos x and sin x. Similarly, by substituting (12) into (11), Sec. 13.6, we obtain

EXAMPLE 4 Logarithm

From (1) it follows that

Replacing z by −z and multiplying both sides by −1, we get

By adding both series we obtain

EXAMPLE 5 Substitution

Find the Maclaurin series of f(z) = 1/(1 + z²).

Solution. By substituting −z² for z in (11) we obtain

EXAMPLE 6 Integration

Find the Maclaurin series of f(z) = arctan z.

Solution. We have f′(z) = 1/(1 + z²). Integrating (19) term by term and using f(0) = 0 we get

this series represents the principal value of w = u + iv = arctan z defined as that value for which |u| < π/2.

EXAMPLE 7 Development by Using the Geometric Series

Develop 1/(c − z) in powers of z − z₀, where c − z₀ ≠ 0.

Solution. This was done in the proof of Theorem 1, where c = z*. The beginning was simple algebra and then the use of (11) with z replaced by (z − z₀)/(c − z₀):

This series converges for

EXAMPLE 8 Binomial Series, Reduction by Partial Fractions

Find the Taylor series of the following function with center z₀ = 1.

Solution. We develop f(z) in partial fractions and the first fraction in a binomial series

with m = 2 and the second fraction in a geometric series, and then add the two series term by term. This gives

We see that the first series converges for |z − 1| < 3 and the second for |z − 1| < 2. This had to be expected because 1/(z + 2)² is singular at −2 and 2/(z − 3) at 3, and these points have distance 3 and 2, respectively, from the center z₀ = 1. Hence the whole series converges for |z − 1| < 2.

PROBLEM SET 15.4

1. Calculus. Which of the series in this section have you discussed in calculus? What is new?
2. On Examples 5 and 6. Give all the details in the derivation of the series in those examples.

3–10 MACLAURIN SERIES

Find the Maclaurin series and its radius of convergence.

• 3. sin 2z²
• 4.
• 5.
• 6.
• 7.
• 8. sin²z
• 9.
• 10.

11–14 HIGHER TRANSCENDENTAL FUNCTIONS

Find the Maclaurin series by termwise integrating the integrand. (The integrals cannot be evaluated by the usual methods of calculus. They define the error function erf z, sine integral Si(z), and Fresnel integrals⁴ S(z) and C(z), which occur in statistics, heat conduction, optics, and other applications. These are special so-called higher transcendental functions.)

• 11.
• 12.
• 13.
• 14.
• 15. CAS Project. sec, tan. (a) Euler numbers. The Maclaurin series

  

  defines the Euler numbers E_2n. Show that E₀ = 1, E₂ = −1, E₄ = 5, E₆ = −61. Write a program that computes the E_2n from the coefficient formula in (1) or extracts them as a list from the series. (For tables see Ref. [GenRef1], p. 810, listed in App. 1.)

  • (b) Bernoulli numbers. The Maclaurin series

    

    defines the Bernoulli numbers B_n. Using undetermined coefficients, show that

    

    Write a program for computing B_n.

  • (c) Tangent. Using (1), (2), Sec. 13.6, and (22), show that tan z has the following Maclaurin series and calculate from it a table of B₀, …, B₂₀:

    

• 16. Inverse sine. Developing and integrating, show that

  

  Show that this series represents the principal value of arcsin z (defined in Team Project 30, Sec. 13.7).

• 17. TEAM PROJECT. Properties from Maclaurin Series. Clearly, from series we can compute function values. In this project we show that properties of functions can often be discovered from their Taylor or Maclaurin series. Using suitable series, prove the following.

  (a) The formulas for the derivatives of and e^z, cos z, sin z, cosh z, sinh z. and Ln(1 + z)

  (b)

  (c) sin z ≠ 0 for all pure imaginary z = iy ≠ 0

18–25 TAYLOR SERIES

Find the Taylor series with center z₀ and its radius of convergence.

• 18. 1/z, z₀ = i
• 19. 1/(1 − z), z₀ = i
• 20. cos²z, z₀ = π/2
• 21. sin z, z₀ = π/2
• 22. cosh (z − πi), z₀ = πi
• 23. 1/(z + i)², z₀ = i
• 24. e^z(z−2), z₀ = 1
• 25. sinh (2z − i), z₀ =i/2

DEFINITION Uniform Convergence

A series (1) with sum s(z) is called uniformly convergent in a region G if for every > 0 we can find an N = N(), not depending on z, such that

Uniformity of convergence is thus a property that always refers to an infinite set in the z-plane, that is, a set consisting of infinitely many points.

EXAMPLE 1 Geometric Series

Show that the geometric series 1 + z + z² + … is (a) uniformly convergent in any closed disk |z| r < 1, (b) not uniformly convergent in its whole disk of convergence |z| < 1.

Solution. (a) For z in that closed disk we have |1 − z| 1 − r (sketch it). This implies that 1/|1 − z| 1/(1 − r). Hence (remember (8) in Sec. 15.4 with q = z)

Since r < 1, we can make the right side as small as we want by choosing n large enough, and since the right side does not depend on z (in the closed disk considered), this means that the convergence is uniform.

(b) For given real K (no matter how large) and n we can always find a z in the disk |z| < 1 such that

simply by taking z close enough to 1. Hence no single N() will suffice to make |s(z) − s_n(z)| smaller than a given > 0 throughout the whole disk. By definition, this shows that the convergence of the geometric series in |z| < 1 is not uniform.

THEOREM 1 Uniform Convergence of Power Series

A power series

with a nonzero radius of convergence R is uniformly convergent in every circular disk |z − z₀| r of radius r < R.

PROOF

For |z − z₀| r and any positive integers n and p we have

Now (2) converges absolutely if |z − z₀| = r < R (by Theorem 1 in Sec. 15.2). Hence it follows from the Cauchy convergence principle (Sec. 15.1) that, an > 0 being given, we can find an N() such that

From this and (3) we obtain

for all z in the disk |z − z₀| r, every n > N(), and every p = 1, 2, …. Since N() is independent of z, this shows uniform convergence, and the theorem is proved.

THEOREM 2 Continuity of the Sum

Let the series

be uniformly convergent in a region G. Let F(z) be its sum. Then if each term f_m(z) is continuous at a point z₁ in G, the function F(z) is continuous at z₁.

PROOF

Let s_n(z) be the nth partial sum of the series and R_n(z) the corresponding remainder:

Since the series converges uniformly, for a given > 0 we can find an N = N() such that

Since s_N(z) is a sum of finitely many functions that are continuous at z₁, this sum is continuous at z₁. Therefore, we can find δ > 0 a such that

Using F = s_N + R_N and the triangle inequality (Sec. 13.2), for these z we thus obtain

This implies that F(z) is continuous at z₁, and the theorem is proved.

EXAMPLE 2 Series of Continuous Terms with a Discontinuous Sum

Consider the series

This is a geometric series with q = 1/(1 + x²) times a factor x². Its nth partial sum is

We now use the trick by which one finds the sum of a geometric series, namely, we multiply s_n(x) by −q = −1/(1 + x²),

Adding this to the previous formula, simplifying on the left, and canceling most terms on the right, we obtain

thus

The exciting Fig. 368 “explains” what is going on. We see that if x ≠ 0, the sum is

but for x = 0 we have s_n(0) = 1 − 1 = 0 for all n, hence s(0) = 0. So we have the surprising fact that the sum is discontinuous (at x = 0), although all the terms are continuous and the series converges even absolutely (its terms are nonnegative, thus equal to their absolute value!).

Theorem 2 now tells us that the convergence cannot be uniform in an interval containing x = 0. We can also verify this directly. Indeed, for x ≠ 0 the remainder has the absolute value

and we see that for a given (< 1) we cannot find an N depending only on such that |R_n| < for all n > N() and all x, say, in the interval 0 x 1.

EXAMPLE 3 Series for Which Termwise Integration Is Not Permissible

Let u_m(x) = mxe^−mx2 and consider the series

in the interval 0 x 1. The nth partial sum is

Hence the series has the sum (0 x 1). From this we obtain

On the other hand, by integrating term by term and using f₁ + f₂ + … + f_n = s_n, we have

Now s_n = u_n and the expression on the right becomes

but not 0. This shows that the series under consideration cannot be integrated term by term from x = 0 to x = 1.

THEOREM 3 Termwise Integration

Let

be a uniformly convergent series of continuous functions in a region G. Let C be any path in G. Then the series

is convergent and has the sum ∫_C F(z) dz.

PROOF

From Theorem 2 it follows that F(z) is continuous. Let s_n(z) be the nth partial sum of the given series and R_n(z) the corresponding remainder. Then F = s_n + R_n and by integration,

Let L be the length of C. Since the given series converges uniformly, for every given > 0 we can find a number N such that |R_n(z)| < /L for all n > N and all z in G. By applying the ML-inequality (Sec. 14.1) we thus obtain

Since R_n = F − s_n, this means that

Hence, the series (4) converges and has the sum indicated in the theorem.

THEOREM 4 Termwise Differentiation

Let the series f₀(z) + f₁(z) + f₂(z) + … be convergent in a region G and let F(z) be its sum. Suppose that the series converges uniformly in G and its terms are continuous in G. Then

THEOREM 5 Weierstrass⁵ M-Test for Uniform Convergence

Consider a series of the form (1) in a region G of the z-plane. Suppose that one can find a convergent series of constant terms,

such that |f_m(z)| M_m for all z in G and every m = 0, 1, …. Then (1) is uniformly convergent in G.

EXAMPLE 4 Weierstrass M-Test

Does the following series converge uniformly in the disk |z| 1?

Solution. Uniform convergence follows by the Weierstrass M-test and the convergence of Σ1/m² (see Sec. 15.1, in the proof of Theorem 8) because

EXAMPLE 5 No Relation Between Absolute and Uniform Convergence

The series in Example 2 converges absolutely but not uniformly, as we have shown. On the other hand, the series

converges uniformly on the whole real line but not absolutely.

Proof. By the familiar Leibniz test of calculus (see App. A3.3) the remainder R_n does not exceed its first term in absolute value, since we have a series of alternating terms whose absolute values form a monotone decreasing sequence with limit zero. Hence given > 0, for all x we have

This proves uniform convergence, since N() does not depend on x.

The convergence is not absolute because for any fixed x we have

where k is a suitable constant, and kΣ1/m diverges.

PROBLEM SET 15.5

1. CAS EXPERIMENT. Graphs of Partial Sums. (a) Fig. 368. Produce this exciting figure using your CAS. Add further curves, say, those of s₂₅₆, s₁₀₂₄, etc. on the same screen.
   • (b) Power series. Study the nonuniformity of convergence experimentally by graphing partial sums near the endpoints of the convergence interval for real z = x.

2–9 POWER SERIES

Where does the power series converge uniformly? Give reason.

• 2.
• 3.
• 4.
• 5.
• 6.
• 7.
• 8.
• 9.

10–17 UNIFORM CONVERGENCE

Prove that the series converges uniformly in the indicated region.

• 10.
• 11.
• 12.
• 13.
• 14.
• 15.
• 16.
• 17.
• 18. TEAM PROJECT. Uniform Convergence.

  (a) Weierstrass M-test. Give a proof.

  (b) Termwise differentiation. Derive Theorem 4 from Theorem 3.

  (c) Subregions. Prove that uniform convergence of a series in a region G implies uniform convergence in any portion of G. Is the converse true?

  (d) Example 2. Find the precise region of convergence of the series in Example 2 with x replaced by a complex variable z.

  (e) Figure 369. Show that if x ≠ 0 and 0 if x = 0. Verify by computation that the partial sums s₁, s₂, s₃ look as shown in Fig. 369.

  

  Fig. 369. Sum s and partial sums in Team Project 18(e)

19–20 HEAT EQUATION

Show that (9) in Sec. 12.6 with coefficients (10) is a solution of the heat equation for t > 0, assuming that f(x) is continuous on the interval 0 x L and has one-sided derivatives at all interior points of that interval. Proceed as follows.

• 19. Show that |B_n| is bounded, say |B_n| < K for all n. Conclude that

  

  and, by the Weierstrass test, the series (9) converges uniformly with respect to x and t for t t₀, 0 x L. Using Theorem 2, show that u(x, t) is continuous for t t₀ and thus satisfies the boundary conditions (2) for t t₀.

• 20. Show that if t t₀ and the series of the expressions on the right converges, by the ratio test. Conclude from this, the Weierstrass test, and Theorem 4 that the series (9) can be differentiated term by term with respect to t and the resulting series has the sum ∂u/∂t. Show that (9) can be differentiated twice with respect to x and the resulting series has the sum ∂²u/∂x². Conclude from this and the result to Prob. 19 that (9) is a solution of the heat equation for all t t₀. (The proof that (9) satisfies the given initial condition can be found in Ref. [C10] listed in App. 1.)