Recent content by 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...

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)...

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...

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\...

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)}

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}))

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)

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)

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

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)...

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!

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.

@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

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...

Tell me if I’m wrong. The formula f=ma is a definition of force. You can’t prove a definition. You can ask, is it meaningful? I think it is not. Can you measure force? Well, if you can measure mass and if you can measure the acceleration then you can measure the force. Otherwise I don’t think...