Page 385 - Schaum's Outline of Theory and Problems of Advanced Calculus

376 GAMMA AND BETA FUNCTIONS [CHAP. 15

1.40 0.8873
1.50 0.8862
1.60 0.8935
1.70 0.9086
1.80 0.9314
1.90 0.9618
2.00 1.0000
The equation (2)isa recurrence relationship that leads to the factorial concept. First observe that if
x ¼ 1, then (1) can be evaluated, and in particular,
ð1Þ¼ 1:
From (2)

ðx þ 1Þ¼ x ðxÞ¼ xðx 1Þ ðx 1Þ¼ xðx 1Þðx 2Þ ðx kÞ ðx kÞ
If x ¼ n, where n is a positive integer, then
ðn þ 1Þ¼ nðn 1Þðn 2Þ ... 1 ¼ n! ð3Þ

If x is a real number, then x! ¼ ðx þ 1Þ is defined by ðx þ 1Þ. The value of this identification is in
intuitive guidance.
If the recurrence relation (2)is characterized as a differential equation, then the definition of ðxÞ
can be extended to negative real numbers by a process called analytic continuation. The key idea is that
1
ðx þ 1Þ allows the
x
even though ðxÞ is defined in (1)is not convergent for x < 0, the relation ðxÞ¼
meaning to be extended to the interval 1 < x < 0, and from there to 2 < x < 1, and so on. A
general development of this concept is beyond the scope of this presentation; however, some information
is presented in Problem 15.7.
The factorial notion guides us to information about ðx þ 1Þ in more than one way. In the
eighteenth century, Sterling introduced the formula (for positive integer values n)
p ffiffiffiffiffiffi nþ1 n
2 n e
lim ¼ 1 ð4Þ
n!
n!1
p ffiffiffiffiffiffi nþ1 n
This is called Sterling’s formula and it indicates that n! asymptotically approaches 2 n e for large
values of n. This information has proved useful, since n! is difficult to calculate for large values of n.
There is another consequence of Sterling’s formula. It suggests the possibility that for sufficiently
large values of x,
p ffiffiffiffiffiffi xþ1 x
2 x e
x! ¼ ðx þ 1Þ ð5aÞ
(An argument supporting this is made in Problem 15.20.)
It is known that ðx þ 1Þ satisfies the inequality
ffiffiffiffiffiffi ffiffiffiffiffiffi
p xþ1 x p xþ1 x 1
2 x e < ðx þ 1Þ < 2 x e e 12ðxþ1Þ ð5bÞ
1
Since the factor e 12ðxþ1Þ ! 0 for large values of x, the suggested value (5a)of ðx þ 1Þ is consistent
with (5b).
An exact representation of ðx þ 1Þ is suggested by the following manipulation of n!. (It depends on
ðn þ kÞ! ¼ðk þ nÞ!.)
k! k n
n! ¼ lim 12 ... nðn þ 1Þþðn þ 2Þ ... ðn þ kÞ ¼ lim lim ðk þ 1Þðk þ 2Þ ... ðk þ nÞ :
k n
k!1 k!1 ðn þ 1Þ ... ðn þ kÞ k!1
ðn þ 1Þðn þ 2Þ ... ðn þ kÞ
k! k n
Since n is fixed the second limit is one, therefore, n! ¼ lim : (This must be read as an
infinite product.) k!1 ðn þ 1Þ ... ðn þ kÞ