Replace gamma(x) with the \(O(1/x^(2n-1))\) Stirling formula. When
n isn’t a nonnegative integer, signal an error. With the optional third
argument pred, the Stirling formula is applied only when pred is
true.
To use this function write first load("stirling").
Reference: Abramowitz & Stegun, "Handbook of mathematical functions", 6.1.40.
Examples:
(%i1) load ("stirling")$
(%i2) stirling(gamma(%alpha+x)/gamma(x),1);
1/2 - x x + %alpha - 1/2
(%o2) x (x + %alpha)
1 1
--------------- - ---- - %alpha
12 (x + %alpha) 12 x
%e
(%i3) taylor(%,x,inf,1);
%alpha 2 %alpha
%alpha x %alpha - x %alpha
(%o3)/T/ x + -------------------------------- + . . .
2 x
(%i4) map('factor,%);
%alpha - 1
%alpha (%alpha - 1) %alpha x
(%o4) x + -------------------------------
2
The function stirling knows the difference between the variable ’gamma’
and the function gamma:
(%i5) stirling(gamma + gamma(x),0);
x - 1/2 - x
(%o5) gamma + sqrt(2) sqrt(%pi) x %e
(%i6) stirling(gamma(y) + gamma(x),0);
y - 1/2 - y
(%o6) sqrt(2) sqrt(%pi) y %e
x - 1/2 - x
+ sqrt(2) sqrt(%pi) x %e
To apply the Stirling formula only to terms that involve the variable k,
use an optional third argument; for example
(%i7) makegamma(pochhammer(a,k)/pochhammer(b,k));
gamma(b) gamma(k + a)
(%o7) ---------------------
gamma(a) gamma(k + b)
(%i8) stirling(%,1, lambda([s], not(freeof(k,s))));
b - a k + a - 1/2 - k - b + 1/2
%e gamma(b) (k + a) (k + b)
(%o8) --------------------------------------------------------
gamma(a)
The terms gamma(a) and gamma(b) are free of k, so the
Stirling formula was not applied to these two terms.