214 BINOMIAL COEFFICIENTS
Now sin( x + y ) = sin x cos y + cos x sin y ; so this ratio of sines is cos 2n7t sin 2~ = (-qn(2 + O(e)) , cos n7t sin c7r by the methods of Chapter 9. Therefore, by (5.86), we have (-n-4!
=
2(-l),r(2n)
!‘_mo (-2n - 2e)!
=
,(-,),P-l)!
n Vn)!
(n-l)! = (-‘) 7’
r(n)
as desired. Let’s complete our survey by restating the other identities we’ve seen so far in this chapter, clothing them in hypergeometric garb. The triple-binomial sum in (5.29) can be written F
1 --a-2n, 1 -b-211, -2n , a, b
1)
(2n)! (a+b+2n-2)” = (-l)nn!ak’,‘i ’
integer n 3 0.
When this one is generalized to complex numbers, it is called Dixon’s formula: F
a, b, c 1 fc-a, 1 fc-b ,
= ( c / 2 ) ! (c-a)*(c-b)* c! (c-a-b)* ’
b6)
fla+Rb < 1 +Rc/2. One of the most general formulas we’ve encountered is the triple-binomial sum (5.28), which yields Saalschiitz’s identity: F
a, b, --n c, afb-c-n+1
= (c-a)K(c-b)” c”(c-a-b)K (a - c)n (b - c)E = (-c)s(a+b-c)n’
integer n 3 0.
This formula gives the value at z = 1 of the general hypergeometric series with three upper parameters and two lower parameters, provided that one of the upper parameters is a nonpositive integer and that bl + bz = al + a2 + a3 + 1. (If the sum of the lower parameters exceeds the sum of the upper parameters by 2 instead of by 1, the formula of exercise 25 can be used to express F(al , a2, as; bl , b2; 1) in terms of two hypergeometrics that satisfy Saalschiitz’s identity.) Our hard-won identity in Problem 8 of Section 5.2 reduces to 1 x+1, n+l, -n 1 = ---F 1+x ( 1, x+2 1)
(-‘)nX”X-n=l.
