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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 byfor
If, the Gamma Function has a special property:
.
So it follows that
By definition,![{\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}](http://latex.codecogs.com/png.latex?{\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})
.
Thus,.
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
When you evaluate the taylor expansion of cosine at zero, you getrandom variable, can you elaborate further?thanks!
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When he said all hell would break loose, he was commenting on the fact that ifwas some other value than 1, then
, which is absurd since we know from trig that
. This would then force
.
So many different ways! I like this.
The way I was taught was using the fact that that there areways of picking k items from a set of n, where
.
If we think about it, it should be that there is only 1 way of choosing no items from a set of n items, thereforeand so,
!0!} = 1)
Some of the other explanations are nicer I think, but there's my two cents!
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