(1/2)!

Jan 2010
133
7
Hello everyone,

I am familiar with the gamma function, but When I needed to simplify the bessel function of the 1st kind \(\displaystyle J_{\frac{1}{2}}\) I stuck on \(\displaystyle \Gamma (n+\frac{3}{2})=(n+\frac{1}{2}!)\).

Isn't \(\displaystyle n! = n \cdot (n-1) \cdot (n-2)\cdot \dots \cdot 2 \cdot 1\) where \(\displaystyle n\) is an integer. But what if \(\displaystyle n\) is not, namely \(\displaystyle n=1/2\).
I started with \(\displaystyle (n+\frac{1}{2}!)=(n+\frac{1}{2})(n+\frac{-1}{2})\dots(\frac{1}{2}!)\) but wasn't sure if I must stop at 1 or at \(\displaystyle (\frac{1}{2}!)\). But then I multiplied each term by 2 so I got \(\displaystyle 2^n(2n+1)(2n-1)\dots(\frac{1}{2}!)\)

WHAT NEXT??
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Hello everyone,

I am familiar with the gamma function, but When I needed to simplify the bessel function of the 1st kind \(\displaystyle J_{\frac{1}{2}}\) I stuck on \(\displaystyle \Gamma (n+\frac{3}{2})=(n+\frac{1}{2}!)\).

Isn't \(\displaystyle n! = n \codt (n-1) \cdot (n-2) \cdot \dots \cdot 2 \cdot 1\) where \(\displaystyle n\) is an integer. But what if \(\displaystyle n\) is not, namely \(\displaystyle n=1/2\).
I started with \(\displaystyle (n+\frac{1}{2}!)=(n+\frac{1}{2})(n+\frac{-1}{2})\dots(\frac{1}{2}!)\) but wasn't sure if I must stop at 1 or at \(\displaystyle (\frac{1}{2}!)\). But then I multiplied each term by 2 so I got \(\displaystyle 2^n(2n+1)(2n-1)\dots(\frac{1}{2}!)\)

WHAT NEXT??
\(\displaystyle \left(n+\tfrac{1}{2}\right)!=(n+\tfrac{1}{2})(n-1+\tfrac{1}{2})!=\cdots=(n+\tfrac{1}{2})\cdot(n-1+\tfrac{1}{2})\cdots(\tfrac{1}{2})!=\sqrt{\pi}\frac{(2n-1)!!}{2^n}\)
 
Jan 2010
133
7
\(\displaystyle \left(n+\tfrac{1}{2}\right)!=(n+\tfrac{1}{2})(n-1+\tfrac{1}{2})!=\cdots=(n+\tfrac{1}{2})\cdot(n-1+\tfrac{1}{2})\cdots(\tfrac{1}{2})!=\sqrt{\pi}\frac{(2n-1)!!}{2^n}\)
Well, I saw something like that somewhere, but what do you mean by double factorial , what is \(\displaystyle (2n-1)!!\) ?? I mean where did that come from (the double factorial), \(\displaystyle (2n-1)! \) must stop at 1, but here it stops at \(\displaystyle \sqrt{\pi}\) so how can we compensate for that?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
Well, I saw something like that somewhere, but what do you mean by double factorial , what is \(\displaystyle (2n-1)!!\) ?? I mean where did that come from (the double factorial), \(\displaystyle (2n-1)! \) must stop at 1, but here it stops at \(\displaystyle \sqrt{\pi}\) so how can we compensate for that?
What do you mean by compensate? \(\displaystyle (2n-1)!!=\frac{(2n)!}{2^nn!}\) more info
 
Jan 2010
133
7
ok ok i thought the double factorial was completely a different thing, thanks for clearing that matter out.