What is the definition of a gamma function?
I know that the factorial is the integral from 0 to infinity of a gamma function, but aren't there lots of gamma functions?

for n > 0 - the integral doesn't exist for n £ 0. (The limits of the integral are 0 and infinity, in case it doesn't look very clear).

I'm not sure why it is defined with an n-1 instead of n, but it means that it's G(n+1) rather than G(n) that is equal to n!. This isn't very difficult to prove: if you know how to do integration by parts try to show that

G(n+1)=n×G(n),

and this, together with the fact that

G(1)=ò₀^¥ e^-x dx=1(=0!)

proves that G(n+1)=n! Other results like

(-1/2)!=

Ö

p

can be derived, and used to baffle people who haven't heard of the alternate definition.

Pras

Ah but if you write n!, then doesn't n automatically have to be an integer in that notation?

You could write

G(0.5)=

Ö

p

.
Presumably your proof by integration by parts is really a proof by induction.

So there is only one gamma functions. Does it have any other uses/properties other than to bemuse people who only know use !

Thanks,
Neil M

PS: never done a pun with punctuation before

BTW - I made a mistake when I said the gamma function isn't defined for n < 0 - having looked at a graph of G(n), it seems it only doesn't exist at the negative integers and 0. Pras

It does have a very interesting property which I can neither prove nor understand (I haven't thought about it as much as I should though.). This may very well may be an unhelpful way to go about it, but it does seem unusual. This came out of the discussion in Statue of the Anonymous: half derivatives.

If you let f(x)=xⁿ then the kth derivative of f(x) is n!/(n-k)! x^n-k for integral k. So can we perhaps use this result to generalise the factorial??

Well the idea is to let n be natural and k be rational. Then we have

(n-k)! = x^n-k n!/kth derivative of xⁿ

So the left hand side is the kind of thing we are looking for. The RHS is easy apart from the kth derivative of xⁿ for non-integral k. We somehow want to be able to partially differentiate...

Well one sensible way of doing this is looking at the limiting definition of differentiation.

f'(x) = lim(d-> 0) [f(x)-f(x-d)]/d
f''(x) = lim(d-> 0) [f(x)-2f(x-d)+f(x-2d]/d²
f'''(x) = lim(d-> 0) [f(x)-3f(x-d)+3f(x-2d)-f(x-3d)]/d³

and you get the binomial co-efficients

Now to evaluate a non-integral derivative you must use the infinite binomial expansion. To make it converge, you can let the number of terms increase as proportional to 1/d. This is a self-consistent definition for a fractional derivative for exactly the same reasons that (1-x)ⁿ can be expanded out in an infinite series. (I'll explain the analogy if you like).

So using this method we can define n! for all rational n. If we let it be continuous we can define it for all real n.

And lo and behold, defining everything like this we do indeed appear to get the gamma function! (At least for all the values I've tried. It predicted for example that (1/2)! = sqrt(pi)/2.)

Now the infinite binomial expansion is based on the generalisation of powers (to non-integral indices). So anyway I believe that there is some deep analogy between the way factorials are generalised and the way the powers and indices are generalised, for non-integers.

But then I could be wrong.

Thanks,

Michael

-Dave

Pras-

You said you had looked at a graph for gamma function. It obviously is a very exponentially shaped graph but does it have any bizarre parts?

Nothing very bizarre - here's a link to a (reasonable) graph of the gamma function: click here .

Pras

Just to quickly add something to this discussion.

The integral which was given to define the gamma function only converges for n> 0 (as was mentioned) however the gamma function is actually defined on the real line (except for 0,-1,-2,...). Infact it can be defined on the complex plane apart from these points. This is an example of what is called analytic continuation and is basically a way to define function on as large a domain as possible.

How do we do this for the gamma function? Well one of the properties was that G(n+1) = n*G(n) so if we assume that this is a general property of the function we can write it as G(n-1) = G(n)/(n-1). And now we can use this identity to start defining the gamma function on places where the original definition failed. Clearly we can't stick n=1 into this definition which is why the function is ill-defined at the origin. And then when we repeatedly use this we see that the problem at 0 propogates backwards to the points -1,-2,-3,...

Analytic continuation is a very important part of mathematics. Normally it is impossible for a function to be defined everywhere in one formula (even the series 1+x+x² +x³ +... doesn't work for x> 1 even though it clearly is trying to define the function 1/(1-x) which does exist for x> 1) so it is vital to obtain properties of the function and then use these to claw one's way up to the whole function.

AlexB.