- #1
H.B.
- 20
- 0
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.
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.