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

[H.B.]

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

$$1<m_2$$

Please feel free to react.

 

[mesa]

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).

 

[H.B.]

@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

 

[mesa]

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

 

[H.B.]

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.

 

[H.B.]

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!

 

[H.B.]

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'?

 

[H.B.]

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

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

 

[H.B.]

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

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

 

[H.B.]

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

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

 

[H.B.]

I also think the digamma function for
$$x\in\mathbb{Q}$$ can be expressed in the next formula:

$$\Psi(x)=-\gamma+\lim_{n\rightarrow\pm\infty}(n(1-L(x,\frac{1}{n}))$$

 

[H.B.]

Of course you have already noticed that a multiplication formula can easely be derivered from the summation formula.For $$n \in \mathbf{N}, 1<n$$ and $$x \in \mathbf{Q}$$

$$(nx) ! = \frac {{x !}^n}{\prod\limits_{k=1}^{n-1}L\left(1+ kx,x\right)}$$

 

[H.B.]

I think Leo Pochhammer did a good thing by changing the factorial function from a unary operation into a binary operation.

As I see it this is the definition of the rising factorial:
$$n, m \in \mathbf{N} \ (starting\ with\ zero),\ x \in \mathbf{C}$$$$X1: \ Pochhammer(x,0)=1 \ (definition\ one)\\$$
$$X2: \ Pochhammer(x,n+1)=Pochhammer(x,n)*(x+n)\ (definition\ two)$$
My idea is to use the symbol ! as a binary operator like this:
$$X1: \ x!0=1\ (definition\ one)\\$$
$$X2: \ x!(n+1)=x!n*(x+n)\ (definition\ two)$$Now a theorem can be proved bij these definitions and mathematical induction.
$$Pochhammer(x,n+m)=Pochhammer(x,n)*Pochhammer(x+n,m)$$This is my formula, I prefer to use $$x!y$$ instead of $$Pochhammer(x,y)$$:
$$for\ x,y \in \mathbf{Q}$$
$$x!y=Pochhammer(x,y)=\frac{1!y}{L(x,y)}$$

 

[H.B.]

My next formula is also about the digamma function or more precisely the derivative of the natural logarithm of the factorial function as I defined this, the input x shift by 1. I use the notation of the digamma function $\Psi(x)$ because I think, and maybe somebody can prove this, it is the same as my formula.$$\Psi(x)=\lim_{h\rightarrow 0}\ln{h!^\frac{1}{h}}-\lim_{h\rightarrow 0}\ln{L(x,h)^\frac{1}{h}}=-\gamma-\lim_{h\rightarrow 0}\ln{L(x,h)^\frac{1}{h}}$$

 

[H.B.]

I Want to make a few more steps. Hopefully mathematically correct.$$\lim_{h\rightarrow 0}\ln{h!^\frac{1}{h}}=\lim_{h\rightarrow 0}\frac{\ln{(0+h)!}-\ln{0!}}{h}=\Psi(1)=-\gamma$$and
$$\lim_{h\rightarrow 0}\ln{L(x,h)^\frac{1}{h}}= \lim_{h\rightarrow 0}\ln\prod^{\infty}_{k=0}{\left(\frac{(k+x+h) (k+1)}{(k+x)(k+1+h)}\right)^\frac{1}{h}}=\lim_{h\rightarrow 0}\sum^{\infty}_{k=0}{\ln\left(\frac{(k+x+h) (k+1)}{(k+x)(k+1+h)}\right)^\frac{1}{h}}=\lim_{h\rightarrow 0}\sum^{\infty}_{k=0}{\frac{\ln(k+x+h)-\ln(k+x)}{h}-\frac{\ln(k+1+h)-\ln(k+1)}{h}}=\sum^{\infty}_{k=0}{\lim_{h\rightarrow 0}\frac{\ln(k+x+h)-\ln(k+x)}{h}-\frac{\ln(k+1+h)-\ln(k+1)}{h}}=\sum^{\infty}_{k=0}{\frac{1}{k+x}-\frac{1}{k+1}}$$

 

[H.B.]

I started with a definition of the factorial function for x is a rational number. Then I could derive a formula that gives the derivative of the natural logarithm of the factorial function (shifted by one this match with the digamma function). I think and tell me if I am wrong, it is easy to derive 'the Weierstrass representation‏ of the gamma function' by integrating this function and so on.
$$\Psi(x)=-\gamma +\sum^{\infty}_{k=0}{\left(\frac{1}{k+1}-\frac{1}{k+x}\right)} \Rightarrow$$
$$\ln\Gamma(x)=-\gamma x+\sum^{\infty}_{k=0}{\left(\frac{x-1}{k+1}-\ln\frac{k+x}{k+1}\right)}+c \Rightarrow$$
$$\Gamma(x)=e^{-\gamma x}e^{\sum\limits^{\infty}_{k=0}{\left(\frac{x-1}{k+1}-\ln\frac{k+x}{k+1}\right)}}e^c=$$
$$e^{-\gamma x+c}\prod^{\infty}_{k=0}{e^{\left(\frac{x-1}{k+1}\right)}\left(\frac{k+1}{k+x}\right)}=$$
$$e^{-\gamma x+c}\prod^{\infty}_{k=1}{e^{\left(\frac{x-1}{k}\right)}\left(\frac{k}{k+x-1}\right)}=$$
According to the definition
$$\Gamma(1)=1\\ c=\gamma\\ \Gamma(x)=e^{-\gamma x+\gamma}\prod^{\infty}_{k=1}{e^{\frac{x-1}{k}}\frac{k}{k+x-1}}$$
$$\Gamma(x+1)= e^{-\gamma x}\prod^{\infty}_{k=1}{e^{\frac{x}{k}}\frac{k}{k+x}}$$
According to the definition
$$\Gamma(x+1)=\Gamma(x)x\\ \Gamma(x)=\frac{e^{-\gamma x}}{x}\prod^{\infty}_{k=1}{e^{\frac{x}{k}}\frac{k}{k+x}}$$
$$\Gamma(x)=\frac{e^{-\gamma x}}{x}\prod^{\infty}_{k=1}{e^{\frac{x}{k}}(1+\frac{x}{k})^{-1}}$$

 

[mfb]

I don't see where this is leading to - we are not a website to host a bunch of equations, especially without proof.
I closed the thread.