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

  1. 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

    [tex] \Gamma\left(\frac{m_1}{m_2}+1\right)= \frac{m_1}{m_2}\ ![/tex]


    [tex] m_1 , m_2 \in \mathbf{N} [/tex]


    [tex] 1<m_2 [/tex]

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

    [tex] L(q,x)=\prod^{\infty}_{k=0}\frac{(k+q+x) (k+1)}{(k+q)(k+1+x)}[/tex]


    [tex] q\neq [/tex]

    0,-1,-2,-3, …… and

    [tex] x\neq [/tex]

    –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.

    [tex] \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} [/tex]

    [tex] m_1 , m_2 \in\mathbf{N} [/tex]


    [tex] 1<m_2 [/tex]

    Please feel free to react.
  2. jcsd
  3. 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).
  4. @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:

  5. You did write that (sorry, was too focused elsewhere).
  6. I made a mistake in the last equation. It must be:
    This formula compares to Euler's reflection formula.
  7. This is a factorial summation formula.

    For example if y=0
    If y=1
    (x+1)!L(x+1,1)= (x+1)!/(x+1)=x!1!=x!
  8. 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'?
  9. 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
  10. 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)
  11. To complete the formula about the Euler-Mascheroni constant I'll add this one:

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