Hopefully the reader is familiar with the counting properties of the factorial that follows the recurrence relation of

  • n!=n (n-1)!
  • 1!=1.

From the above it is not hard to show that:

  • 2!=2(1!)=2(1)=2
  • 3!=3(2!)=3(2)=6
  • 4!=4(3!)=24
  • 5!=5(4!)=120
  • 6!=6(5!)=720

Perhaps a bit complexing is the value of the factorial at 0, this is best viewed as

  • 1!=1(0!)
  • 1=1(0!).

It follows naturally that 0!=1. {author note, ref linked.com group posting via India.}

As gleanned from the unit shifted gamma funciton, .i.e.\Gamma(x+1)=x!, the factorial is never zero, but has poles at the negative integers {-1,-2,-3, … }.

It can be shown that the reciporical of the factorial function is entire. In addition, it has zeros at the negative integers. It is well known that

\Gamma(s) = \frac{1}{s}e^{-\gamma s}\prod_n \exp(s/n)\frac{s}{s+n}

The reciporical factioral (1/s!)! can be shown to satisfy the recurrence of

\frac{1}{(s-1)!} = s \frac{1}{s!}.

  • \frac{1}{(-1)!} = 0 \frac{1}{0!} =0
  • \frac{1}{(-2)!} = -1 \frac{1}{(-1)!} =0
  • \frac{1}{(-3)!} = -2 \frac{1}{(-2)!} =0

ad infinitium. Therefore,

\frac{1}{s!} = e^{\gamma s} \prod_n e^{\frac{-s}{n}}\frac{s+n}{n},

where the product is taken over the naturals.


Binomial Coefficients

\binom{s}{t} = \prod_n \frac{t+n}{s+n}\frac{s-t+n}{n} =  \prod_{n=1}^\infty \frac{t+n}{s+n}\frac{s-t+n}{n},

where the product is taken over the naturals, for s \notin \{-1, -2, -3, \dots\}, not a negative integer. Therefore,

\binom{s}{t} = \frac{t+1}{s+1}\frac{s-t+1}{1}\frac{t+2}{s+2}\frac{s-t+2}{2}\frac{t+3}{s+3}\frac{s-t+3}{3}\cdots

Power of a (Ordinary) Generating Function

Consider a non zeroth cusp (\alpha_0 \neq 0) (ordinary) generating function raised to a power, then
\left(\sum_{m=0}^\infty \alpha_m x^m\right)^z =  	\left(\sum_m \alpha_m x^m\right)^z  	=\alpha_0^z\sum_{m=0}^\infty \left(\sum_{n=0}^m \frac{z^{\underline{n}}}{\alpha_0^{n}} \beta_{m,n}(\underline{\alpha})\right) x^m\\  	 =\sum_{m,n} \alpha_0^{z-n} z^{\underline{n}} \beta_{m,n}(\underline{\alpha}) x^m,

  • m \in \mathbb{W} = \left\{0\right\} \cup \mathbb{N} = \left\{0, 1, 2, 3, \dots \right\}
  • n \in \left\{0, 1, 2, \dots, m\right\}
  • falling factorial: z^{\underline{1+n}}=(z-n)^{\underline{1}}z^{\underline{n}}=(z-n)z^{\underline{n}},


  • z^{\underline{1}}=z, z^{\underline{2}}=z(z-1),\dots
  • \left(\underline{\alpha}\right) = \left(\alpha_1,\alpha_2,\alpha_3,\dots\right)
  • \forall \underline{\alpha}:  \beta_{m,0}(\underline{\alpha})=\left[m=0\right]=  	\begin{cases}  	1,\ m=0\,\\ 0, m \neq 0.   	\end{cases}
  • n \beta_{m,n}(\underline{\alpha}) = \sum_j \beta_{j,n-1}(\underline{\alpha}) \alpha_{m-j}, j \in \left\{n-1,n,\dots,m-1\right\}

The coefficents can be viewed as the result of a number triangle product with a column vector, \ie,

\begin{bmatrix}  1 &0 &\cdots\\  0 &\alpha_{1}&0&0&0&0\\   0 &\alpha_{2}&\alpha_1^2/2!&0&0&0\\  0 &\alpha_{3}&\alpha_{2}\,\alpha_{1}&\alpha_1^3/{3!}&0&0\\  0 &\alpha_{4}&\alpha_3\,\alpha_1+\alpha_2^2/2&\alpha_2\,\alpha_1^2/2&\alpha_1^4/4!&0 \\  0 &\alpha_{5} &\alpha_{4}\,\alpha_{1}+\alpha_{3} \,\alpha_{2} &\left(\alpha_2^2/2\right)\,\alpha_1+\alpha_3\,\alpha_1^2/2&\alpha_2\,\alpha_1^3/3! &\alpha_1^5/5!\\  \vdots &\vdots & & & & &\ddots  \end{bmatrix}  \begin{bmatrix}  \alpha_0^z \\  \alpha_0^z \frac{z}{\alpha_0}\\   \alpha_0^z \frac{z(z-1)}{\alpha_0^2}\\  \alpha_0^z \frac{z(z-1)(z-2)}{\alpha_0^3}\\  \alpha_0^z \frac{z}{\alpha_0} \frac{z-1}{\alpha_0} \frac{z-2}{\alpha_0} \frac{z-3}{\alpha_0}\\  \vdots   \end{bmatrix}

latex test

Theorem Let f\in C^{\infty}(a,b), then


where S_k(n)=\sum_{j=1}^{n}j^k, f^{(k)} is the k-th derivative of f, and a<x<x+jt<b.



Proof Let f(x)=e^x in the above Theorem.