From: "moth" Subject: Re: Gamma function as z->0 Date: Wed, 24 Jan 2001 11:39:19 +1100 Newsgroups: sci.math Summary: Introduction to Gamma function "Michael.Smith" wrote in message news:3A6DF8C4.FA8E9C9E@durham.ac.uk... > Can anyone tell me an expansion for the following expression involving > the gamma function (written G(z)) > > lim d->0 G(d)/A^d > > Where A is any denominator. > > I would have expected the expansion of G(z) at z->0 ie (2/d - Euler > Constant + terms regular) since 1/A^d --> 1 as d->0 but I have seen log > (A) terms in a text book and I'm wondering where they're from. > > Thanks in advance > > Michael E Smith The Log terms in this usual limit for the gamma function come about from the representation of... Gamma(z) = limit(n-->inf+) n!z^n / (z*(z+1)*....*(z+n)) so you take the log of both sides to get.... ln(Gamma(z)) = limit(n-->inf+) [ ln(n!) + z*ln(n) - ln(z) - ln(z+1) -....-ln(z+n) ] now take the limit at z-->0 so you have limit(z-->0) ln(Gamma(z)) = limit(n-->inf+) [ ln(n!) - limit(z-->0)ln(z) - ln(n!) ] now i have just taken the limit for z into the limit for n with out any justification, because the proof for doing that is quite hard. but if for the moment you are willing to accept it, you finaly get limit(z-->0)[ ln(Gamma(z))+ln(z) ] = 0 (*) of course there is away around the limits which makes everything much easier and this is limitz(z-->0) Gamma(z+1) = 1 and using the fact that Gamma(z+1) = z*Gamma(z) and taking the log of both sides you end up with (*) however i believe that the other derivation is the more fundamental and mathematicaly rigorous. either way, this should give you some indication of where the log factors come from when taking the limits you are after. all you need to do now is modifiy these results to obtain the limit you want. cheers moth ============================================================================== From: Doug Magnoli Subject: Re: Factorial stuff Date: Wed, 04 Jul 2001 07:15:47 GMT Newsgroups: sci.math,alt.math,alt.math.recreational What excellent questions! I remember going through this kind of stuff when I first learned about the gamma function, which I found (find) fascinating. You've got one little blunder in here, though.....You know that i! =/= i and especially that 2i! =/= 2i * i = -2. Remember that factorial means n(n-1)(n-2).... In the case of 2i, that means 2i(2i-1)(2i-2).... As far as exact answers for gamma(x), there aren't very many x's for which there are exact answers. I'm afraid that for your example of Gamma[1+i], you're stuck with an approximation. The Concise Encyclopedia of Mathematics, which used to be on the web as Eric's Mathworld, gives a few good gamma function expressions. I'll copy some of them here so you can see what they're like. Seems that a lot of them come from elliptic integrals, so some introduction. In that same encyclopedia under 'Elliptic Integral Singular Value' we find: "When the Modulus k has a singular value, the complete elliptic integrals may be computed in analytic form in terms of gamma functions. Abel...proved that whenever: K'(k) / K(k) = (a+b sqrt(n)) / (c + d sqrt(n)) where a,b,c,d, and n are integers, K(k) is a complete elliptic integral of the first kind, and K'(k)=K(sqrt(1-k^2)) is the complementary complete elliptic integral of the first kind, then the modulus k is the root of an algebraic equation with integer coefficients. A modulus k_r such that K'(k_r) / K(k_r) = sqrt(r) is called a singular value of the elliptic integral." You'll see that although there are exact expressions for the gamma function of a few arguments, they aren't quite exact expressions like 6 or sqrt(pi). Gamma[1/3] = 2^(7/9) 3^(-1/12) pi^(1/3) [K(k_3)]^(1/3) Gamma[1/4] = 2*pi^(1/4) [K(k_1)]^(1/2) Gamma[1/6] = 2^(-1/3) 3^(1/2) pi^(-1/2) [Gamma(1/3)]^2 Gamma(1/8) * Gamma(3/8) = sqrt(sqrt(2) - 1) sqrt(pi) 2^(13/4) K(k_2) Gamma(1/8) / Gamma(3/8) = 2 sqrt(sqrt(2)+1) pi^(-1/4) [K(k_1)]^(1/2) Gamma(1/12) = 2^(-1/4) 3^(3/8) (sqrt(3)+1)^(1/2) pi^(-1/2) Gamma(1/4) Gamma(1/3) Gamma(5/12) = 2^(1/4) 3^(-1/8) (sqrt(3)-1)^(1/2) pi^(1/2) Gamma(1/4) / Gamm(1/3) and then we get into a bunch of them that look like this: Gamma(1/15) * Gamma(2/15) * Gamma(7/15) / Gamma(2/15) = 2*3^(1/2) 5^(1/6) sin(2*pi/5) sin(4*pi/15) [Gamma(1/5)]^2 Under the elliptic integral heading I quoted earlier is a list of values, all based on gamma functions, of K(k_i), i=1=13,15-17,25. The most 'interesting' of them looks like this: K(k_13) = [ sqrt(sqrt(18+5 sqrt(13))) / sqrt(6656*pi^5) ] * sqrt (gamma(1/52)*gamma(7/52)*gamma(9/52)*gamma(11/52)* gamma(15/52)*gamma(17/52)*gamma(19/52)*gamma(25/52)* gamma(29/52)*gamma(31/52)*gamma(47/52)*gamma(49/52) And what anyone is supposed to do with that is beyond me. Just because they're interesting to look at, I'll add some more from the entry on the gamma function, where I've used P(n=1,m) f(n) to mean the product of f(n), n=1,m. The ones from Ramunjan are particularly unwieldy, and a body has to wonder what they're for, or how he stumbled across them, or why. But next time you run into something like cosh(n pi sqrt(3) - cos(n pi), you'll know to start thinking in terms of gamma functions. "A few curious identities include: P(n=1,8) Gamma(n/3) = (640 / 3^6) (pi / sqrt(3))^3 [Gamma(1/4)]^4 / (16 pi^2) = (3^2 / (3^2-1)) * ((5^2-1)/5^2) * (7^2 / (7^2-1)) ... Gamma'[1] / Gamma[1] - Gamma'[1/2] / Gamma[1/2] = 2 ln 2 Ramunjan also gave a number of fascinating identities: Gamma^2(n+1) / (Gamma(n+xi+1) * Gamma(n-xi+1) = P(k=1,inf) [1 + x^2 / (n+k)^2] phi(m,n)*phi(n,m) = Gamma^3(m+1)*Gamma^3(n+1) / (Gamma(2m+n+1)*Gamma(2n+m+1)) * (cosh[pi(m+n)*sqrt(3)] = cos[pi(m-n)]) / (2 pi^2 (m^2+mn+n^2)) where phi(m,n) = P(k=1,inf) [ 1 + ((m+n) / (k+m)(^3] P(k=1,inf) [ 1+(n/k)^3] * P(k=1,inf) [1 +3(n/(n+2k))^2] = [Gamma(n/2) / Gamma(n/2 + 1/2)] * (cosh(pi n sqrt(3) - cos(pi n)) / (2^n+2) pi^(3/2) n) -Doug Magnoli Jeremy Price wrote: > Are factorials defined for the imaginary and complex numbers, or just for > the reals? Given the definition of them I was taught in High School math, I > would think that i! = i, (2i)! = 2i*1i =-2, ect. following the pattern > n!=(n)(n-1)(n-2)...(1). But then I remembered that the gamma function also > defined factorials, x! = Gamma(x+1) = int[ e^-t t^(x-1) dt ] form 0 to inf. > I am unsure of how to get an exact answer for i! = Gamma(i+1), but I got an > approximation of i! ~ .498+.155i. Is there any way to get an exact answer > for this, if they are even defined? How would I go about finding the same > for complex factorials, if it is defined for complex numbers as well. > > I was also wondering if there could be found an "inverse gamma function" of > some sort where g(x!)=x. I would assume that some function exists to do > this, but how would I go about deriving it? > > Thank you, > Jeremy Price