624
C Stirling Approximations
Fig. C.1 Illustration of the
Stirling approximation for
ln n!
0
5
10
15
20
25
30
0.0
0.5
1.0
1.5
2.0
2.5
3.0
ln x
x
which is Stirling’s approximation for ln n!. If we recognize that n is large enough
that we can neglect 1 in comparison to n, then we can also rewrite the Stirling
approximation as
n! !
n
e
n
.
(C.1.5)
We shall refer to this as the lowest-order Stirling approximation to n!. Moreover, it
is the one that we shall employ for the most part in this course.
We can also obtain directly an expression in the same spirit for the derivative of
ln n!: if m is sufficiently small relative to n, we can approximate the derivative of
ln n! by the finite-difference expression
d
dn
ln n! !
ln(n + m)! − ln n!
m
which becomes, using the properties of the (natural) logarithm
d
dn
ln n! =
1
m
ln
(n + m)!
n!
=
1
m
ln
(n + m)(n + m − 1) · · · (n + 1)n!
n!
.
Now, recalling that the spirit of Stirling’s approximation is for n 1, and recalling
further that m ≈ 1, we obtain
C Stirling Approximations
Fig. C.1 Illustration of the
Stirling approximation for
ln n!
0
5
10
15
20
25
30
0.0
0.5
1.0
1.5
2.0
2.5
3.0
ln x
x
which is Stirling’s approximation for ln n!. If we recognize that n is large enough
that we can neglect 1 in comparison to n, then we can also rewrite the Stirling
approximation as
n! !
n
e
n
.
(C.1.5)
We shall refer to this as the lowest-order Stirling approximation to n!. Moreover, it
is the one that we shall employ for the most part in this course.
We can also obtain directly an expression in the same spirit for the derivative of
ln n!: if m is sufficiently small relative to n, we can approximate the derivative of
ln n! by the finite-difference expression
d
dn
ln n! !
ln(n + m)! − ln n!
m
which becomes, using the properties of the (natural) logarithm
d
dn
ln n! =
1
m
ln
(n + m)!
n!
=
1
m
ln
(n + m)(n + m − 1) · · · (n + 1)n!
n!
.
Now, recalling that the spirit of Stirling’s approximation is for n 1, and recalling
further that m ≈ 1, we obtain