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and the term is included because that was obtained for (2.65)—although it had a
zero coefficient. In order to match (2.71) and (2.72), we require and
i.e.

(It is left as an exercise to show that (2.65) and (2.73) are recovered from suitable
expansions of (1.22).) We should note that (2.65) exhibits no breakdown—there is
only one term here, after all—but (2.73) does break down where i.e.
x = O(1), as we would expect. Any indication of a breakdown in the asymptotic
expansion valid for x = O(1) will come from the exponentially small terms; let us
briefly address this aspect of the problem.
The first point to note is that, from (2.70) and (2.73), we would require not only
O(1) and but also terms (and others), in order to complete the matching
procedure; this is simply because we need, at least in principle, to match to all the terms
(cf. (2.72))

Thus the expansion valid for x = O(1) must include a term to allow matching
to this order (and then it is not too difficult to see that a complete asymptotic expansion
requires all the terms in the sequence n = 0, 1, 2,…, m = 0, 1, 2,…).
In passing, we observe that this use of the matching principle is new in the context
of our presentation here. We are using it, first, in a general sense, to determine the
type(s) of term(s) required in the expansion valid in an adjacent region in order to
allow matching. Then, with these terms included, the ‘full’ matching procedure may
be employed to check the details and fix the values of any arbitrary constants left
undetermined. We will find the first of the exponentially small terms here.
For x = O(1), we seek a solution

where in this example); thus equation (2.63) be-
comes

(Again ‘= 0’ means zero to all orders in We already have that satisfies the
original equation, and so must satisfy