Why is 0! = 1?

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- Sep 1st 2009, 07:19 AMpacman0!
Why is 0! = 1?

- Sep 1st 2009, 07:24 AMynj
it is specially defined.

Since 1!=1*0!. - Sep 1st 2009, 07:45 AMSoroban

A silly poem I wrote while in college:

Man has wondered

Since time immemorial

Why 1 is the value

Of 0!.

- Sep 1st 2009, 07:57 AMdhiab
**Hello 0! is a convention without demonstration.**

**Same$\displaystyle 2^0 = 1$**$\displaystyle

$ - Sep 1st 2009, 09:17 AMChris L T521
We can show this is the case using calculus (if this is a bit advanced for you, please bear with me (Nod) )

The Gamma Function is defined by $\displaystyle \Gamma\left(x\right)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$ for $\displaystyle x>0$

If $\displaystyle n\in\mathbb{N}$, the Gamma Function has a special property: $\displaystyle \Gamma\left(n\right)=\left(n-1\right)!$.

So it follows that $\displaystyle {\color{blue}\Gamma\left(1\right)}=\left(1-1\right)!=\color{red}\boxed{0!}$

By definition, $\displaystyle {\color{blue}\Gamma\left(1\right)}=\int_0^{\infty} e^{-t}t^{1-1}\,dt=\int_0^{\infty}e^{-t}\,dt=\lim_{b\to\infty}\left.\left[-e^{-t}\right]\right|_0^{b}$ $\displaystyle =-\lim_{b\to\infty}e^{-b}-\left(-e^{-0}\right)=0-(-1)=\color{red}\boxed{1}$.

Thus, $\displaystyle {\color{blue}\Gamma\left(1\right)}=\color{red}\box ed{0!=1}$. - Sep 1st 2009, 09:35 AMRobLikesBrunch
I tried figuring out the Gamma function from Wikipedia a few weeks ago...but didn't understand it that well. Do you know any website where I can find a proper introduction to it? Or should I just wait until they cover it in one of my future college courses....once I start college?

Thanks. - Sep 1st 2009, 09:43 AMmasters
Hi pacman,

I posted this in another thread, but I'll repost it here.

See if this helps.

n! is defined as 1 * 2 * 3 * . . . * n

And with a little manipulation, we can show that 0! = 1 by demonstrating that...

1! = 1

2! = 1! * 2

3! = 2! * 3

4! = 3! * 4

etc.

Reversing that, we can achieve...

3! = 4!/4

2! = 3!/3

1! = 2!/2

0! = 1!/1 = 1 - Sep 1st 2009, 09:44 AMChris L T521
See if this helps you: The Gamma Function

- Sep 1st 2009, 09:44 AMRandom Variable
Let's say that 0! is equal to something else, say 2.

then the taylor series of $\displaystyle \cos x $ expanded about $\displaystyle x=0$ would be $\displaystyle \frac{1}{2} - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + ... $

which would imply that $\displaystyle \cos (0) = \frac{1}{2} $

and all hell would break loose (Giggle) - Sep 1st 2009, 06:22 PMpacman
**random variable**, can you elaborate further? (Rock)(Rock)(Rock)(Rock)(Rock) thanks! - Sep 1st 2009, 06:35 PMChris L T521
When you evaluate the taylor expansion of cosine at zero, you get $\displaystyle \cos\left(0\right)=\frac{1}{0!}$.

When he said all hell would break loose, he was commenting on the fact that if $\displaystyle 0!$ was some other value than 1, then $\displaystyle \cos\left(0\right)=\frac{1}{\text{other value}}$, which is absurd since we know from trig that $\displaystyle \cos\left(0\right)={\color{red}1}$. This would then force $\displaystyle 0!=1$. - Sep 1st 2009, 06:45 PMpacman
ahhh, that is much CLEARER now. Thanks

**Cris** - Sep 1st 2009, 07:31 PMpomp
So many different ways! I like this.

The way I was taught was using the fact that that there are $\displaystyle ^{n}C_k$ ways of picking k items from a set of n, where

$\displaystyle ^{n}C_k = \frac{n!}{(n-k)!k!}$ .

If we think about it, it should be that there is only 1 way of choosing no items from a set of n items, therefore $\displaystyle ^{n}C_0 = 1$ and so,

$\displaystyle \frac{n!}{(n-0)!0!} = 1$ $\displaystyle \Rightarrow$ $\displaystyle \frac{1}{0!} = 1$ $\displaystyle \Rightarrow$ $\displaystyle 0! = 1$

Some of the other explanations are nicer I think, but there's my two cents! - Sep 1st 2009, 07:42 PMpacman
**Pomp**, i like your way of obtaining 0! = 1 through Combination. - Sep 1st 2009, 08:32 PMWilmer
If 0! was equal to 0, then n! would = 0,

since we'd have to assume that ! represents 0*1*2......*n