# The factorial of a rational number, the gamma function not used

1. ### H.B.

14
My first question is: is this formula (at the bottom) a known formula?
In this subject i haven't explained how i build up the formula.
So far i think it is equal to the gamma function of Euler with

$$\Gamma\left(\frac{m_1}{m_2}+1\right)= \frac{m_1}{m_2}\ !$$

with

$$m_1 , m_2 \in \mathbf{N}$$

and

$$1<m_2$$

This gamma function however I didn't use.
The next limit i write like L(q,x) in de last formula.

$$L(q,x)=\prod^{\infty}_{k=0}\frac{(k+q+x) (k+1)}{(k+q)(k+1+x)}$$

with

$$q\neq$$

0,-1,-2,-3, …… and

$$x\neq$$

–1,-2,-3,-4, ……..

This is the formula where i use the factorial symbol ! because i think that it gives no problems with arguments with real values.

$$\frac{ m_1}{m_2}\ !=\left(m_1\ ! \prod^{ m_2-1}_{ i=1}L\left(1+ i\frac{ m_1}{m_2}, \frac{ m_1}{m_2}\right) \right)^\frac{1}{m_2}$$

with

$$m_1 , m_2 \in\mathbf{N}$$

and

$$1<m_2$$

2. ### mesa

583
Does it only work for positive natural numbers? Even if limited by those inputs I still like your 'q' and 'x' general form (assuming this identity is correct).

3. ### H.B.

14
@mesa,
The input is a positive rational number m1/m2.
For an input of a negative rational number the relation between x! and (-x)! according to my definition (which i didn't show here) of the factorial (!) is:

x!(-x)!L(1-x,x)=1

4. ### mesa

583
You did write that (sorry, was too focused elsewhere).

5. ### H.B.

14
I made a mistake in the last equation. It must be:
x!(-x)!=L(1-x,x)
This formula compares to Euler's reflection formula.

6. ### H.B.

14
This is a factorial summation formula.

(x+y)!L(x+1,y)=x!y!
For example if y=0
(x+0)!L(x+1,0)=x!L(x+1,0)=x!=x!0!
If y=1
(x+1)!L(x+1,1)= (x+1)!/(x+1)=x!1!=x!

7. ### H.B.

14
The next formula is about binomial coefficients.
If I use my definition of the factorial in this formula:
$${x \choose y}=\frac{x\ !}{y\ !(x-y)\ !}$$
(x, y are rational numbers) then the next equation appears :
$${x \choose y}L(1+x-y,y)={x \choose y}\prod^{\infty}_{k=0}\frac{(k+x+1) (k+1)}{(k+1+x-y)(k+1+y)}={x \choose y}\prod^{\infty}_{k=1}\frac{(k+x) k}{(k+x-y)(k+y)}=1$$
For example if y=0:
$${x \choose 0}L(1+x,0)={x \choose 0}\prod^{\infty}_{k=0}\frac{(k+x+1) (k+1)}{(k+1+x)(k+1)}=1$$
Or if x=1 and y=1/2
$${1 \choose \frac{1}{2} }L(1+1-\frac{1}{2},\frac{1}{2})={1 \choose \frac{1}{2}}\prod^{\infty}_{k=0}\frac{(k+1+1) (k+1)}{(k+1+1-\frac{1}{2})(k+1+\frac{1}{2})}={1 \choose \frac{1}{2}}\prod^{\infty}_{k=1}\frac{(k+1) k}{(k+1-\frac{1}{2})(k+\frac{1}{2})}={1 \choose \frac{1}{2}}\prod^{\infty}_{k=1}\frac{(k+1) k}{(k+\frac{1}{2})^2}={1 \choose \frac{1}{2}}\frac {\pi}{4}=1$$
Does this give the same results when using the 'gamma function'?

8. ### H.B.

14
The remaining text (x=1 and y=1/2):

$$={1 \choose \frac{1}{2} }L(1+1-\frac{1}{2},\frac{1}{2})={1 \choose \frac{1}{2}}\frac {\pi}{4}=1$$

9. ### H.B.

14
I think I can approach the Euler–Mascheroni constant with the next formula:

$$- \gamma =\lim_{n \rightarrow \infty}((\prod^{\infty}_{k=1}(\frac{(k+1)^\frac{1}{n} k^\frac{n-1}{n}}{(k+\frac{1}{n})})-1)n)$$

10. ### H.B.

14
To complete the formula about the Euler-Mascheroni constant I'll add this one:

$$\gamma = \lim_{n \rightarrow \infty} ((\prod^{\infty}_{k=1} (\frac{(k+1)^\frac{-1}{n} k^\frac{n+1}{n}}{(k-\frac{1}{n})})-1)n)$$