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About LinkBacksWhy is 0! = 1?
We can show this is the case using calculus (if this is a bit advanced for you, please bear with meOriginally Posted by pacman
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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, $\displaystyle {\color{blue}\Gamma\left(1\right)}=\color{red}\box ed{0!=1}$.
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.
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
See if this helps you: The Gamma Function
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
When you evaluate the taylor expansion of cosine at zero, you get $\displaystyle \cos\left(0\right)=\frac{1}{0!}$.random variable, can you elaborate further?thanks!
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 $\displaystyle 0!=1$.
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!
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