What is the formula for factorial (!) .

Recommended Posts

It's ok with me' date=' in fact the structure of your proof is good, its just that a lemma would be nice, which goes into the gamma function.

in other words prove the following statement

[b']Lemma[/b]:

$\Gamma (n+1) = n!$

well, i know you're gonna hate this explanation, but the Gamma function is defined as

$$\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} \,dt$$ for x > 0.

although, the exact proof that you're looking for, which is a simple expansion on this definition can be found here:

http://mathworld.wolfram.com/GammaFunction.html

i'm just honestly too lazy to redo the LaTeX to make it clear.

just scroll a little down the page right under those pretty little colored graphs. there lies your answer.

if you need any furthur proofs of why this and that, you can email wolfram.

Share on other sites

I don't hate it, it's just not what I'm thinking of.

I'll come up with something, then post it. Basically, what I have in mind uses integration by parts in the proof, but its been so long since i formally proved it, i was hoping you knew the argument.

Regards

Share on other sites
I don't hate it' date=' it's just not what I'm thinking of.

[/quote']

of course its not.

I'll come up with something, then post it.

i'm sure you will. you always do.

Share on other sites

were you able to invent something over the weekend, johnny?

Share on other sites
were you able to invent something over the weekend, johnny?

There's nothing to invent, I already know the basis of the argument. Let me see what the heck this thread was about again though.

Ah yes this:

Prove the following:

$\Gamma (n+1) = n!$

By definition, the following statement is true:

$\Gamma(n+1) \equiv \int_{x=0}^{x=\infty} t^n e^{-t} dt$

So we can focus on the integral.

Here is the integration by parts formula:

$\int u dv = uv- \int v du$

let dv = e^-t dt, whence it follows that v = -e^-t

let u = t^n, thus du = n t^(n-1) dt

Substituting we have:

$\int_{t=0}^{t=\infty} t^n e^{-t} dt = -t^n e^{-t} |_{t=0}^{t=\infty} + n\int_{t=0}^{t=\infty} e^{-t} t^{n-1} dt$

Therefore:

$\int_{t=0}^{t=\infty} t^n e^{-t} dt = 0 + n\int_{t=0}^{t=\infty} e^{-t} t^{n-1} dt$

Now, replace n by n-1 in the statement above, to obtain:

$\int_{t=0}^{t=\infty} t^{n-1} e^{-t} dt = (n-1) \int_{t=0}^{t=\infty} e^{-t} t^{n-2} dt$

Thus, we have:

$\int_{t=0}^{t=\infty} t^n e^{-t} dt = n(n-1) \int_{t=0}^{t=\infty} e^{-t} t^{n-2} dt$

Continuing on in this manner, eventually, you will reach the following integral:

$\int_{t=0}^{t=\infty} e^{-t} t dt$

This integral will be reached when n-k=1, hence when n=k+1.

When k=1, the coefficient was n. When k=2, the coefficient was n(n-1). When k=3, the coefficient would be n(n-1)(n-2). Thus, when k=(n-1), the coefficient would be n(n-1)(n-2)(n-3)... (n-(n-1-1)) or rather

n(n-1)(n-2)...(n-n+1+1)

Therefore:

$\int_{t=0}^{t=\infty} t^n e^{-t} dt = n(n-1)(n-2)...(2)\int_{t=0}^{t=\infty} e^{-t} t dt$

Now, evaluate the following integral:

$\int_{t=0}^{t=\infty} t e^{-t} dt$

Using the integration by parts formula, with u=t, and dv = e^-t dt we have:

$\int_{t=0}^{t=\infty} t e^{-t} dt = -te^{-t} |_{t=0}^{t=\infty} + \int_{t=0}^{t=\infty} e^{-t}dt$

Whence it follows that:

$\int_{t=0}^{t=\infty} t^n e^{-t} dt = n(n-1)(n-2)...(2)\int_{t=0}^{t=\infty} e^{-t} dt$

Now, evaluate the following integral:

$\int_{t=0}^{t=\infty} e^{-t} dt$

And this is easy.

$\int_{t=0}^{t=\infty} e^{-t} dt = -e^-t |_{t=0}^{t=\infty} = e^-t |_{t=\infty}^{t=0} = 1-0$

Hence:

$\int_{t=0}^{t=\infty} t^n e^{-t} dt = n(n-1)(n-2)...(2)1 = n!$

The LHS is the gamma function of n+1, therefore:

$\Gamma (n+1) = n!$

QED

Oh wait, that is the proof you cited, which was used at wolfram, i just looked down far enough, and saw it.

Man, I feel like i did all that for nothing.

Regards