ASYMPTOTIC SERIES

10. ASYMPTOTIC SERIES

Since you have spent some time learning to test series for convergence, it may surprise you to learn that there are divergent series which can be of practical use. We can show this best by an example.

Example 1. From (9.3a)

2 ∞ −t 2

erfc(x) = 1 − erf(x) = √

e dt.

We are going to expand the integral in (10.1) in a series of inverse powers of x. To do this we write

and integrate by parts as follows:

Now in the last integral in (10.3), write (1/t 2 )e −t 2 = (1/t 3 )(d/dt)(− 1 e −t 2 2 ), and again integrate by parts:

Continue this process, and substitute (10.3) and the following steps into (10.1) to get (Problem 1)

. 2x 2 (2x 2 ) 2 − (2x 2 ) 3 +··· [Use this when |x| ≫ 1. Compare (9.6).]

erfc(x) = 1 − erf(x) ∼ √

x π 1−

550 Special Functions Chapter 11

(We shall explain the exact meaning of the symbol ∼ shortly.) This series diverges for every x because of the factors in the numerator. However, suppose we stop after

a few terms and keep the integral at the end so that we have an exact equation. If we stop after the second term, we have

−x 2 e 1 3 ∞

(10.5) 2 erfc(x) = √ 1−

t −4 e −t dt.

2x 2

There is no approximation here. This is not an infinite series so there is no question of convergence. However, we shall show that the integral at the end is negligible for large enough x; this will then make it possible for us to use the rest of (10.5) [that is, the first two terms of (10.4)] as a good approximation for erfc(x) for large x. This is the meaning of an asymptotic series. As an infinite series it may diverge, but we do not use the infinite series. Instead, using an exact equation [like (10.5) for this example], we show that the first few terms which we do use give a good approximation if x is large.

Example 2. Now let’s look at the integral in (10.5); we want to estimate its size for large x. The t in the integrand takes values from x to ∞; therefore t ≥ x or 1/x ≥ 1/t

for all values of t from x to ∞. Let us write the integral as

We increase the value of this integral if we replace 1/t 5 by 1/x 5 since 1/x ≥ 1/t. Thus

When we stop in (10.5) with the term in e 2 −x 3

/x , the error is of the order of

e /x 5 , which becomes much smaller than e −x 2 /x 3 as x increases. Thus we have shown that two terms of (10.4) give a good approximation for erfc(x) when x ≫ 1.

−x 2

A similar result can be shown for an approximation using any number of terms of the asymptotic series (10.4) with the error depending on the “left-over” integral and the value of x.

We can make the above discussion more precise. For (10.4), we have seen that if we stop after the term in x −3 e −x 2 , the error is of the order of x −5 −x 2 e 2 . Then the

ratio of the error to the last term kept (namely x −5 e −x 2 ÷x −3 e −x =x −2 ) tends to zero as x tends to infinity, that is, the approximation becomes increasingly good for larger x as we have said. The “error” in an asymptotic expansion means in general the difference between the function being expanded and a partial sum (first N terms) of the series. A series is called an asymptotic expansion (about ∞) of a function f (x) if, for each fixed N , the ratio of the error to the last (nonzero) term

Section 10 Asymptotic Series 551

kept, tends to zero as x → ∞. In symbols

(10.6) read φ n (x) is an asymptotic expansion of f (x)

n=0

if for each fixed N

Frequently, the terms of an asymptotic series (about ∞) are inverse powers of x. [We could write (10.4) this way by multiplying through by e x 2 .] Then (10.6) becomes

if for each fixed N

We can also have asymptotic series about the origin (or any point—compare Taylor series). We say that

if for each fixed N

f (x) − N a n x n ÷x → 0 as x → 0.

n=0

Although we have discussed the particularly interesting case of divergent asymp- totic series, it is not necessary for such series to diverge. Note that to test a series for convergence, we fix x and let n tend to infinity; to see if a series is asymptotic, we fix n and let x tend to a limit. A given series may meet both tests, or only one or the other (or neither).

PROBLEMS, SECTION 10

1. Carry through the algebra to get equation (10.4). 2. R The integral ∞

t p−1 e −t dt = Γ(p, x) is called an incomplete Γ function. [Note that if x = 0, this integral is Γ(p).] By repeated integration by parts, find several terms of the asymptotic series for Γ(p, x).

3. Express the complementary error function erfc(x) as an incomplete Γ function (see Problem 2) and use your result in Problem 2 to obtain (again) the asymptotic expansion of erfc(x) as in (10.4).

552 Special Functions Chapter 11

Z ∞ e −xt Z x e t 4. E n (x) =

dt, and other similar 1 t

dt, n = 0, 1, 2, · · ·, and Ei(x) =

−∞ t

integrals are called exponential integrals. By making appropriate changes of variable, show that

(b) Ei(x) = −

(−1/x) (Caution: Various notations are used; check carefully the notation in references you

(c) E

1 (x) = − Ei(−x)

(d)

dt = E 1

are using.) 5. (a)

Express E 1 (x) as an incomplete Γ function. (b)

Find the asymptotic series for E 1 (x).

6. The logarithmic integral is li(x) = dt . Express as exponential integrals 0 ln t

(c) dt 0 ln(1/t) 7. Computer plot graphs of

Z x (a) li(x)

(b) li(e x )

(a) E n (x) for n = 0 to 10 and x = 0 to 2; (b)

E 1 (x) and Ei(x) for x = 0 to 2;

Z ∞ cos t (c)

Z x sin t

the sine integral Si(x) = dt

dt and the cosine integral Ci(x) = −

t for x = 0 to 4π.