Concrete mathematics : a foundation for computer science

9.3 0 MANIPULATION 441
(We are expanding everything out until we get a relative error of O(nP3),
because the relative error of a product is the sum of the relative errors of the
individual factors. All of the O(nP3) terms will coalesce.)
In order to expand (1 - nP’)n, we first compute ln(1 - nP’ ) and then
form the exponential, enln(‘Pnm’l:
(1 - nP’)n = exp(nln(1 -n-l))
= exp(n(-nP’ - in-’ - in3 + O(nP4)))
= exp(-1 - in-’ - in2 + O(nP3))
= exp(-1) . exp(-in-‘) . exp(-$n2) . exp(O(nP3))
= exp(-1) . (1 - in-’ + in2 + O(nP3))
. (1 - in2 + O(nP4)) . (1 + O(nP3))
= e-l (1 - in-’ - $ne2 + O(nP3)) .
Here we use the notation expz instead of e’, since it allows us to work with
a complicated exponent on the main line of the formula instead of in the
superscript position. We must expand ln(1 -n’) with absolute error O(ne4)
in order to end with a relative error of O(nP3), because the logarithm is being
multiplied by n.
The right-hand side of (9.41) has now been reduced to fi times
n+‘/e” times a product of several factors:
(1 - in-’ - AnP2 + O(nP3))
. (1 + n-l -t nP2 + O(nP3))
. (1 - in-’ - &nP2 + O(nP3))
. (1 + an-’ + (a + b)nP2 + O(nP3)) .
Multiplying these out and absorbing all asymptotic terms into one O(n-3)
yields
l+an’+(a$-b-&)nP2+O(nP3).
Hmmm; we were hoping to get 1 + an’ + bn2 + O(nP3), since that’s what
we need to match the right-hand side of (9.40). Has something gone awry?
No, everything is fine; Table 438 tells us that a = A, hence a + b - & = b.
This perturbation argument doesn’t prove the validity of Stirling’s approximation,
but it does prove something: It proves that formula (9.40) cannot
be valid unless a = A. If we had replaced the O(nA3) in (9.40) by
cne3 + O(nP4) and carried out our calculations to a relative error of O(nP4),
we could have deduced that b = A. (This is not the easiest way to determine
the values of a and b, but it works.)