(1/2)!

rebghb

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
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??
$$\displaystyle \left(n+\tfrac{1}{2}\right)!=(n+\tfrac{1}{2})(n-1+\tfrac{1}{2})!=\cdots=(n+\tfrac{1}{2})\cdot(n-1+\tfrac{1}{2})\cdots(\tfrac{1}{2})!=\sqrt{\pi}\frac{(2n-1)!!}{2^n}$$

rebghb

$$\displaystyle \left(n+\tfrac{1}{2}\right)!=(n+\tfrac{1}{2})(n-1+\tfrac{1}{2})!=\cdots=(n+\tfrac{1}{2})\cdot(n-1+\tfrac{1}{2})\cdots(\tfrac{1}{2})!=\sqrt{\pi}\frac{(2n-1)!!}{2^n}$$
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
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

rebghb

ok ok i thought the double factorial was completely a different thing, thanks for clearing that matter out.