Question

Right ?

Answer 1

wha-?

Answer 2

??????

Answer 3

let him type lol

Answer 4

lol

Answer 5

\[\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi}\]\[\Gamma \left(\frac{n}{2}\right) \\~~ \\~~ \\ =\left(\frac{n-2}{2}\right)\left(\frac{n-4}{2}\right){\Huge....}\left(\frac{1}{2}\right)\Gamma \left(\frac{1}{2}\right) \\ =\left(\frac{n-2}{2}\right)\left(\frac{n-4}{2}\right){\Huge....}\left(\frac{1}{2}\right)\Gamma \left(\frac{1}{2}\right)\sqrt{\pi}\]

Answer 6

the first line is the given statement

Answer 7

oh boy, i'm outta here lol

Answer 8

I made a typo in last line

Answer 9

same, sorry, never studied this

Answer 10

@nincompoop @TuringTest @dan815

Answer 11

last line should say
\[\Gamma\left(\frac{n}{2}\right)=\left(\frac{n-2}{2}\right)\left(\frac{n-4}{2}\right){\Huge....}\left(\frac{1}{2}\right)\sqrt{\pi}\]

Answer 12

where n is an odd integer

Answer 13

gamma of n=(n-1)! so ...

Answer 14

yes, I know that, but this isn't a factorial of an integer.

Answer 15

sorry, my second didn't enter

Answer 16

So my question is, how do you know that n-(2k)=1 for some k?

Answer 17

yeah this is correct

Answer 18

@fbi2015

Answer 19

gamma(x+1)=xgamma(x) if x>0
comes directly from this formula