1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Aug 18, 2014 #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]

    with

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

    and

    [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]

    with

    [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]
     
     
    with

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

    and

    [tex] 1<m_2 [/tex]

    Please feel free to react.
     
  2. jcsd
  3. Aug 19, 2014 #2
    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. Aug 21, 2014 #3
    @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
     
  5. Aug 21, 2014 #4
    You did write that (sorry, was too focused elsewhere).
     
  6. Aug 30, 2014 #5
    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.
     
  7. Oct 18, 2014 #6
    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!
     
  8. Oct 26, 2014 #7
    The next formula is about binomial coefficients.
    If I use my definition of the factorial in this formula:
    [tex]
    {x \choose y}=\frac{x\ !}{y\ !(x-y)\ !}
    [/tex]
    (x, y are rational numbers) then the next equation appears :
    [tex]
    {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
    [/tex]
    For example if y=0:
    [tex]
    {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
    [/tex]
    Or if x=1 and y=1/2
    [tex]
    {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
    [/tex]
    Does this give the same results when using the 'gamma function'?
     
  9. Oct 26, 2014 #8
    The remaining text (x=1 and y=1/2):

    [tex]
    ={1 \choose \frac{1}{2} }L(1+1-\frac{1}{2},\frac{1}{2})={1 \choose \frac{1}{2}}\frac {\pi}{4}=1
    [/tex]
     
  10. Feb 24, 2015 #9
    I think I can approach the Euler–Mascheroni constant with the next formula:

    [tex]
    - \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)
    [/tex]
     
  11. Mar 22, 2015 #10
    To complete the formula about the Euler-Mascheroni constant I'll add this one:

    [tex]
    \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)
    [/tex]
     
  12. Apr 5, 2015 #11
    I also think the digamma function for
    [tex]
    x\in\mathbb{Q}
    [/tex] can be expressed in the next formula:

    [tex]
    \Psi(x)=-\gamma+\lim_{n\rightarrow\pm\infty}(n(1-L(x,\frac{1}{n}))
    [/tex]
     
  13. Oct 6, 2015 #12
    Of course you have already noticed that a multiplication formula can easely be derivered from the summation formula.


    For [tex] n \in \mathbf{N}, 1<n [/tex] and [tex] x \in \mathbf{Q} [/tex]

    [tex]
    (nx) ! = \frac {{x !}^n}{\prod\limits_{k=1}^{n-1}L\left(1+ kx,x\right)}
    [/tex]
     
  14. Oct 20, 2015 #13
    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:



    [tex]

    n, m \in \mathbf{N} \ (starting\ with\ zero),\ x \in \mathbf{C}

    [/tex]


    [tex]
    X1: \ Pochhammer(x,0)=1 \ (definition\ one)\\
    [/tex]
    [tex]
    X2: \ Pochhammer(x,n+1)=Pochhammer(x,n)*(x+n)\ (definition\ two)
    [/tex]



    My idea is to use the symbol ! as a binary operator like this:



    [tex]
    X1: \ x!0=1\ (definition\ one)\\
    [/tex]
    [tex]
    X2: \ x!(n+1)=x!n*(x+n)\ (definition\ two)
    [/tex]


    Now a theorem can be proved bij these definitions and mathematical induction.



    [tex] Pochhammer(x,n+m)=Pochhammer(x,n)*Pochhammer(x+n,m)

    [/tex]


    This is my formula, I prefer to use [tex]x!y[/tex] instead of [tex]Pochhammer(x,y)[/tex]:



    [tex]

    for\ x,y \in \mathbf{Q}

    [/tex]



    [tex]

    x!y=Pochhammer(x,y)=\frac{1!y}{L(x,y)}

    [/tex]
     
  15. Apr 22, 2016 #14
    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 [itex] \Psi(x)[/itex] because I think, and maybe somebody can prove this, it is the same as my formula.





    [tex] \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}} [/tex]
     
  16. May 10, 2016 #15
    I Want to make a few more steps. Hopefully mathematically correct.





    [tex]



    \lim_{h\rightarrow 0}\ln{h!^\frac{1}{h}}=\lim_{h\rightarrow 0}\frac{\ln{(0+h)!}-\ln{0!}}{h}=\Psi(1)=-\gamma



    [/tex]





    and



    [tex]



    \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}}



    [/tex]
     
  17. May 16, 2016 #16
    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.



    [tex]

    \Psi(x)=-\gamma +\sum^{\infty}_{k=0}{\left(\frac{1}{k+1}-\frac{1}{k+x}\right)}

    \Rightarrow

    [/tex]



    [tex]

    \ln\Gamma(x)=-\gamma x+\sum^{\infty}_{k=0}{\left(\frac{x-1}{k+1}-\ln\frac{k+x}{k+1}\right)}+c

    \Rightarrow

    [/tex]



    [tex]

    \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=

    [/tex]



    [tex]
    e^{-\gamma x+c}\prod^{\infty}_{k=0}{e^{\left(\frac{x-1}{k+1}\right)}\left(\frac{k+1}{k+x}\right)}=

    [/tex]



    [tex]

    e^{-\gamma x+c}\prod^{\infty}_{k=1}{e^{\left(\frac{x-1}{k}\right)}\left(\frac{k}{k+x-1}\right)}=

    [/tex]



    According to the definition



    [tex]

    \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}}

    [/tex]



    [tex]

    \Gamma(x+1)=

    e^{-\gamma x}\prod^{\infty}_{k=1}{e^{\frac{x}{k}}\frac{k}{k+x}}

    [/tex]



    According to the definition



    [tex]

    \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}}

    [/tex]



    [tex]

    \Gamma(x)=\frac{e^{-\gamma x}}{x}\prod^{\infty}_{k=1}{e^{\frac{x}{k}}(1+\frac{x}{k})^{-1}}

    [/tex]
     
  18. Jun 2, 2016 #17

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The factorial of a rational number, the gamma function not used
  1. Rational Numbers (Replies: 6)

Loading...