chapter  13
4 Pages

## Error Bounds for Hankel's Expansions

We may view this result in the following way. Let lzl have a given large value and n be fixed. As phz increases continuously from fn, the right-hand side of (13.01), without the error term, gives a good approximation to Hil)(z) up to and including ph z = $n. To achieve comparable numerical accuracy when$n < ph z < 2n it is necessary to add to this approximation the second series on the right of (13.06), even though in this region e-'c is exponentially small compared with elz and therefore negligible in PoincarP's sense. As the value ph z = 2n is passed, e-Iz becomes large compared with elz, causing the roles of the two series in (13.06) to interchange; the inclusion of the second is mandatory, and the first cannot be discarded without some loss of accuracy. Beyond ph z = Qn the error bound for q,,, l(ze-ni) in (13.06) becomes large, and to maintain accuracy a new multiple of the first series (obtainable from (4.13) with m = 3) has to be used. And so on. Thus it is possible to compute H,")(z) for any value of phz by one or two applications of (13.01), with phz confined to the numerically acceptable range [-fn, 3x1. Similarly for H,!2)(z). 13.3 Another way of obtaining error bounds for the remainder terms in Hankel's expansions, which is particularly valuable when the variables are real, is to apply the methods of Chapters 3 and 4 to Hankel's integrals, as followst:

Let us assume that v > -f and z > 0. Then the path in (4.20) may be collapsed onto the two sides of the join of 1 and 1 + i a . Taking a new integration variable 7 = (t-l)/i and using the reflection formula for the Gamma function, we arrive at

the factors in the integrand having their principal values. For any positive integer n,

the form of Taylor's theorem obtained by repeated partial integrations shows that

where

Substitution of the sum on the right-hand side of (13.08) into (13.07) produces the first n terms of (13.01). For the remainder term, assume that n 2 v-+. Then )(1 ++im)"-"-(112)1 <1. Hence