Divergent or Semiconvergent Series

3.2.6 Divergent or Semiconvergent Series

That a series is convergent is no guarantee that it is numerically useful. In this section, we shall see examples of the reverse situation: a divergent series can be of use in numerical computations. This sounds strange, but it refers to series where the size of the terms decreases rapidly at first and increases later, and where an error bound (see Figure 3.2.4), can be obtained in terms of the first neglected term. Such series are sometimes called

semiconvergent . 66 An important subclass are the asymptotic series; see below.

Example 3.2.9.

We shall derive a semiconvergent series for the computation of Euler’s function

for large values of x. (The second integral was obtained from the first by the substitution t = u + x.) The expression (u + x) −1 should first be expanded in a geometric series with remainder term, valid even for u > x,

−1) n x −j u + (−1) (u + x) −1 (x −1 u) .

j =0

We shall frequently use the well-known formula

u j e −u du = j! = Ŵ(j + 1).

We write f (x) = S n (x) +R n (x) , where

x 3 − · · · + (−1) x n

j =0

R n (x) = (−1) n (u + x) −1

e −u du.

The terms in S n (x) qualitatively behave as in Figure 3.2.4. The ratio between the last term in S n +1 and the last term in S n is

65 For hypergeometric or confluent hypergeometric series see Lebedev [240, Secs. 9.1 and 9.11] or [1, Secs. 15.3 and 13.2].

66 A rigorous theory of semiconvergent series was developed by Stieltjes and Poincaré in 1886.

3.2. More about Series 213

x 10 − 1.5 3

Error estimate − 0.5

3.2.9 for x = 10; see (3.2.43).

Figure 3.2.4. Error estimates of the semiconvergent series of Example

and since the absolute value of that ratio for fixed x is unbounded as n → ∞, the sequence {S n (x) } ∞ n =1 diverges for every positive x . But since sign R n (x)

= (−1) n for x > 0, it follows fromTheorem 3.1.4 that

The idea is now to choose n so that the estimate of the remainder is as small as possible. According to (3.2.42), this happens when n is equal to the integer part of x. For x = 5 we

choose n = 5,

S 5 ( 5) = 0.2 − 0.04 + 0.016 − 0.0096 + 0.00768 = 0.17408, S 6 ( 5) = S 5 ( 5) − 0.00768 = 0.16640,

which gives f (5) = 0.17024 ± 0.00384. The correct value is 0.17042, so the actual error is only 5% of the error bound. For n = x = 10, the error estimate is 1.0144 · 10 −5 .

For larger values of x the accuracy attainable increases. One can show that the bound for the relative error using the above computational scheme decreases approximately as (π

·x/2) 1/2 e −x , an extremely good accuracy for large values of x, if one stops at the smallest term. It can even be improved further, by the use of the convergence acceleration techniques

presented in Sec. 3.4, notably the repeated averages algorithm, also known as the Euler transformation ; see Sec. 3.4.3. The algorithms for the transformation of a power series into

a rapidly convergent continued fraction, mentioned in Sec. 3.5.1, can also be successfully applied to this example and to many other divergent expansions.

One can derive the same series expansion as above by repeated integration by parts. This is often a good way to derive numerically useful expansions, convergent or semi-

214 Chapter 3. Series, Operators, and Continued Fractions convergent, with a remainder in the form of an integral. For convenient reference, we

formulate this as a lemma that is easily proved by induction and the mean value theorem of integral calculus. See Problem 3.2.10 for applications.

Lemma 3.2.6 (Repeated Integration by Parts). Let F ∈ C p (a, b), let G 0 be a piecewise continuous function, and let G 0 ,G 1 ,...

be a sequence of functions such that G ′ j +1 (x) =G j (x) with suitably chosen constants of integration. Then

The sum is the “expansion,” and the last integral is the “remainder.” If G p (t ) has a constant sign in (a, b), the remainder term can also be written in the form

( −1) p F (p) (ξ )(G p +1 (b) −G p +1 (a)), ξ ∈ (a, b). The expansion in Lemma 3.2.6 is valid as an infinite series, if and only if the remainder

tends to 0 as p → ∞. Even if the sum converges as p → ∞, it may converge to the wrong result.

The series in Example 3.2.9 is an expansion in negative powers of x, with the property that for all n, the remainder, when x → ∞, approaches zero faster than the last included

term. Such an expansion is said to represent f (x) asymptotically as x → ∞. Such an asymptotic series can be either convergent or divergent (semiconvergent). In many

branches of applied mathematics, divergent asymptotic series are an important aid, though they are often needlessly surrounded by an air of mysticism.

It is important to appreciate that an asymptotic series does not define a sum uniquely. For example f (x) = e −x is asymptotically represented by the series

j =0 ∞ 0·x −j , as x → ∞. Thus e −x (and many other functions) can be added to the function for which the expansion was originally obtained.

Asymptotic expansions are not necessarily expansions into negative powers of x. An expansion into positive powers of x − a,

n −1

f (x)

c ∼ ν ν (x − a) +R n (x),

represents f (x) asymptotically when x → a if

lim (x − a) −(n−1) x R n (x)

→a

Asymptotic expansions of the error of a numerical method into positive powers of a step length h are of great importance in the more advanced study of numerical methods. Such expansions form the basis of simple and effective acceleration methods for improving nu- merical results; see Sec. 3.4.

Problems and Computer Exercises 215

ReviewQuestions

3.2.1 Give the Cauchy formula for the coefficients of Taylor and Laurent series, and describe the Cauchy–FFT method. Give the formula for the coefficients of a Fourier series. For which of the functions in Table 3.1.1 does another Laurent expansion also exist?

3.2.2 Describe by an example the balancing procedure that was mentioned in the subsection about perturbation expansions.

3.2.3 Define the Chebyshev polynomials, and tell some interesting properties of these and of Chebyshev expansions. For example, what do you know about the speed of con- vergence of a Chebyshev expansion for various classes of functions? (The detailed expressions are not needed.)

3.2.4 Describe and exemplify what is meant by an ill-conditioned power series and a pre- conditioner for such a series.

3.2.5 (a) Define what is meant when one says that the series ∞ 0 a n x −n • converges to a function f (x) for x ≥ R;

• represents a function f (x) asymptotically as x → ∞. (b) Give an example of a series that represents a function asymptotically as x → ∞,

although it diverges for every finite positive x. (c) What is meant by semiconvergence? Say a few words about termination criteria

and error estimation.

Problems and Computer Exercises

3.2.1 Some of the functions appearing in Table 3.1.1 and in other examples and problems are not single-valued in the complex plane. Brush up your complex analysis and find out how to define the branches, where these expansions are valid, and (if nec- essary) define cuts in the complex plane that must not be crossed. It turns out not to be necessary for these expansions. Why?

(a) If you have access to programs for functions of complex variables (or to com- mands in some package for interactive computation), find out the conventions used for functions like square root, logarithm, powers, arc tangent, etc. If the manual does not give enough detail, invent numerical tests, both with strategically chosen values of z and with random complex numbers in some appropriate domain around the origin. For example, do you obtain

z +1

ln

− ln(z + 1) + ln(z − 1) = 0 ∀z?

z −1 √

Or, what values of z 2 − 1 do you obtain for z = ±i? What values should you obtain, if you want the branch which is positive for z > 1?

216Chapter 3. Series, Operators, and Continued Fractions (b) What do you obtain if you apply Cauchy’s coefficient formula or the Cauchy–FFT

method to find a Laurent expansion for √z? Note that √z is analytic everywhere in an annulus, but that does not help. The expansion is likely to become weird. Why?

3.2.2 Apply (on a computer) the Cauchy–FFT method to find the Maclaurin coefficients

a n of (say) e z , ln(1 − z), and (1 + z) 1/2 . Conduct experiments with different values of r and N, and compare with the exact coefficients. This presupposes that you

have access to good programs for complex arithmetic and FFT. Try to summarize your experiences of how the error of a n depends on r, N. You may find some guidance in Example 3.2.2.

3.2.3 (a) Suppose that r is located inside the unit circle; t is real. Show that

2 1−r ∞

1 − 2r cos t + r 2 =1+2

Hint: First suppose that r is real. Set z = re it . Show that the two series are the real and imaginary parts of (1 + z)/(1 − z). Finally, make an analytic continuation of

the results. (b) Let a be positive, x ∈ [−a, a], while w is complex, w /∈ [−a, a]. Let r = r(w),

|r| < 1 be a root of the quadratic r 2 − (2w/a)r + 1 = 0. Show that (with an appropriate definition of the square root)

, (w / ∈ [−a, a], x ∈ [−a, a]). w −x

w 2 −a 2 n =1

(c) Find the expansion of 1/(1 + x 2 ) for x ∈ [−1.5, 1.5] into the polynomials T n (x/ 1.5). Explain the order of magnitude of the error and the main features of the

error curve in Figure 3.2.5. Hint: Set w = i, and take the imaginary part. Note that r becomes imaginary.

4 3 x 10 −

The error of the expansion of f (x) = 1/(1 + x 2 ) in a sum of Chebyshev polynomials {T n (x/ 1.5)}, n ≤ 12.

Figure 3.2.5.

Problems and Computer Exercises 217

3.2.4 (a) Find the Laurent expansions for

f (z) = 1/(z − 1) + 1/(z − 2).

(b) How do you use the Cauchy–FFT method for finding Laurent expansions? Test your ideas on the function in the previous subproblem (and on a few other functions). There may be some pitfalls with the interpretation of the output from the FFT program, related to so-called aliasing; see Sec. 4.6.6 and Strang [339].

(c) As in Sec. 3.2.1, suppose that F (p) is of bounded variation in [−π, π] and denote the Fourier coefficients of F (p) (p) by c

n . Derive the following generalization of (3.2.7):

+ (in) p .

(in) j +1

j =0

Show that if we add the condition that F ∈ C j [−∞, ∞], j < p, then the asymptotic results given in (and after) (3.2.7) hold.

(d) Let z = 1 (w +w −1

2 ) . Show that |z − 1| + |z + 1| = |w| + |w| −1 . Hint:

Use the parallelogram law, |p − q| 2 + |p + q| 2 = 2(|p| 2 + |q| 2 ) .

3.2.5 (a) The expansion of arcsinh t into powers of t, truncated after t 7 , is obtained from Problem 3.1.6 (b). Using economization of a power series, construct from this a

polynomial approximation of the form c

1 t +c 3 t 3 for the interval t ∈ [− 2 , 2 ]. Give bounds for the truncation error for the original truncated expansion and for the

economized expansion. (b) The graph of T 20 (x) for x ∈ [−1, 1] is shown in Figure 3.2.1. Draw the graph

of T 20 (x) for (say) x ∈ [−1.1, 1.1].

3.2.6 Compute a few terms of the expansions into powers of ǫ or k of each of the roots of the following equations, so that the error is O(ǫ 2 ) or O(k −2 ) (ǫ is small and positive; k is large and positive). Note that some terms may have fractional or negative exponents. Also try to fit an expansion of the wrong form in some of these examples, and see what happens.

2 (b) ǫz 3 2 (a) (1 + ǫ)z 3 − ǫ = 0; −z + 1 = 0; (c) ǫz − z + 1 = 0; (d) z 4 2 2 2 − (k 2 + 1)z −k = 0, (k ≫ 1).

3.2.7 The solution of the boundary value problem

1 + ǫ)y ′′ − ǫy = 0, y(

0) = 0, y(1) = 1, has an expansion of the form y(t; ǫ) = y 0 (t ) +y 1 (t )ǫ +y 2 (t )ǫ 2 +···.

(a) By coefficient matching, set up differential equations and boundary conditions for y 0 ,y 1 ,y 2 , and solve them. You naturally use the boundary conditions of the original problem for y 0 . Make sure you use the right boundary conditions for y 1 ,y 2 .

(b) Set R(t) = y 0 (t ) + ǫy 1 (t ) − y(t; ǫ). Show that R(t) satisfies the (modified) differential equation

( 1 + ǫ)R ′′

2 − ǫR = ǫ 3 ( 7t − t )/ 6, R( 0) = 0, R(1) = 0.

218 Chapter 3. Series, Operators, and Continued Fractions

3.2.8 (a) Apply Kummer’s first identity (3.2.39) to the error function erf(x), to show that 2x −x 2 3 2 2x −x 2 2x 2 ( 2x 2 ) 2 ( 2x 2 ) 3

erf(x) = √ e M 1, ,x = √ e + +··· . π

3·5 3·5·7 Why is this series well-conditioned? (Note that it is a bell sum; compare Fig-

ure 3.2.3.) Investigate the largest term, rounding errors, truncation errors, and termination criterion.

(b) erfc(x) has a semiconvergent expansion for x ≫ 1 that begins

1 3 15 erfc(x) = 1 − erf(x) = √

2 e −x

2x 2 + 4x 4 − 8x 6 +··· . Give an explicit expression for the coefficients, and show that the series diverges

e −t dt

= x √ π 1−

for every x. Where is the smallest term? Estimate its size. Hint: Set t 2 =x 2 + u, and proceed analogously to Example 3.2.8. See Prob-

lem 3.1.7 (c), α = 1

2 , about the remainder term. Alternatively, apply repeated integration by parts; it may be easier to find the remainder in this way.

3.2.9 Other notations for series, with application to Bessel functions . (a) Set

f (x) =

g(x) =

h(x) = ,

!n! Let h(x) = f (x) · g(x), χ(w) = φ(w) · ψ(w). Show that

Derive analogous formulas for series of the form ∞ n =0 a n w n /( 2n)!. Suggest how to divide two power series in these notations. (b) Let a j

= (−1) x

j , g(x) = e . Show that

−1) j a ′ j .

Comment: By (3.2.1), this can also be written c n = (−1) n 4 n a 0 . This proves the mathematical equivalence of the preconditioners (3.1.55) and (3.1.59) if P (x) = e x . (c) Set, according to Example 3.2.8 and part (a) of this problem, w = −x 2 / 4,

I J (x) ≡I 0 (x)J 0 (x) = .

n !n!

n =0 n !n!

n =0 n !n!

n =0

Problems and Computer Exercises 219 Show that

0 if n = 2m + 1. Hint: The first expression for γ n follows from (a). It can be interpreted as the

−j

coefficient of t n in the product (1 − t) n (

1 + t) n . The second expression for γ n is the same coefficient in (1 − t 2 ) n .

(d) The second expression for γ n in (c) is used in Example 3.2.8. 67 Reconstruct and extend the results of that example. Design a termination criterion. Where is the largest modulus of a term of the preconditioned series, and how large is it approximately? Make a crude guess in advance of the rounding error in the preconditioned series.

(e) Show that the power series of J 0 (x) can be written in the form

where a n is positive and decreases slowly and smoothly. Hint: Compute a n +1 /a n . (f) It is known (see Lebedev [240, eq. (9.13.11)]) that

J 0 (x) =e −ix M

2 , 1; 2ix ,

where M(a, b, c) is Kummer’s confluent hypergeometric function, this time with an imaginary argument. Show that Kummer’s first identity is unfortunately of no use here for preconditioning the power series.

Comment: Most of the formulas and procedures in this problem can be generalized to the series for the Bessel functions of the first kind of general integer order, (z/ 2) −n J n (x) . These belong to the most studied functions of applied mathematics, and there exist more efficient methods for computing them; see, e.g., Press et al. [294, Chapter 6]. This problem shows, however, that preconditioning can work well for

a nontrivial power series, and it is worth being tried.

3.2.10. (a) Derive the expansion of Example 3.2.5 by repeated integration by parts. (b) Derive the Maclaurin expansion with the remainder according to (3.1.5) by the

application of repeated integration by parts to the equation

f (z) − f (0) = z

f ′ (zt ) d(t − 1).

3.2.11. Show the following generalization of Theorem 3.2.5. Assume that |f (z)| ≤ M for z ∈E R . Let |ζ | ∈ E ρ , 1 < ρ < r ≤ R − ǫ. Then the Chebyshev expansion of f (ζ )

67 It is much better conditioned than the first expression. This may be one reason why multiple precision is not needed here.

220 Chapter 3. Series, Operators, and Continued Fractions satisfies the inequality

( n −1

( ( n 2M(ρ/R)

(f (ζ ) − .

c j T j (ζ ) (

1 − ρ/R Hint:

Set ω = ζ + 1 ζ 2 − 1, and show that |T j j (ζ ) |=|

2 (ω +ω −j )

|≤ρ j .