 # What is the formula for factorial (!) .

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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.

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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

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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

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