Hiya.
If you're asking why 0!=1, then here is the answer.
Typically, at high school level, you are told that a factorial is simply the product of all integers down to 1 (e.g. 6!=6*5*4*3*2*1) and that 0! is "defined" to be equal to unity, right?
That's all well and good for the nonnegative integers; but what if we want to find, say, (-3)! or (1/2)!? The recursive definition doesn't work!
That's where GAMMA FUNCTIONS ([tex]\Gamma(n)[/tex]) come in! Basically, it can be shown that
[tex]\\\Gamma(n-1)=\int^{\infty}_{0}x^{n}\exp(-x)dx=n![/tex]
for ALL n larger than -1. (The integral is improper for -1<x<0, but can be shown to converge there.) So, punching n=0 to our new definition for n!, we get
[tex]\Gamma(-1)=\int^{\infty}_{0}\exp(-x)dx=1=0![/tex];
i.e.
0!=1. (QED)
In general,
[tex]\Gamma(n-1)=n![/tex];
it can also be shown that
[tex]\Gamma(n+1)=n\Gamma(n)[/tex], or [tex]\Gamma(n)=(1/n)\Gamma(n+1)[/tex];
i.e., we have a recursive definition for gamma functions. (Note that this means that the gamma function of any integer less than or equal to zero is
infinite; that is, the factorial of a negative
integer is infinite.) Also, it can be shown that
[tex]\Gamma(1/2)=\sqrt{\pi}[/tex].
Gamma functions are usually tabulated between 0 and 1; using the recursive formulas above, you can then find, say,
[tex](-3/2)!=\Gamma(-1/2)=2\Gamma(1/2)=2\sqrt{\pi}[/tex].
"Mathematical Methods In The Physical Sciences" (2nd ed., published by Wiley) by Mary Boas explains this beautifully. (I strongly recommend it: an absolute bible to me during my university years!)
Hope that helps!
dannyboy
