Factorials

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

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

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

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

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
$\left(\sum_{m=0}^\infty \alpha_m x^m\right)^z = \left(\sum_m \alpha_m x^m\right)^z =\alpha_0^z\sum_{m=0}^\infty \left(\sum_{n=0}^m \frac{z^{\underline{n}}}{\alpha_0^{n}} \beta_{m,n}(\underline{\alpha})\right) x^m\\ =\sum_{m,n} \alpha_0^{z-n} z^{\underline{n}} \beta_{m,n}(\underline{\alpha}) x^m,$
where

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

therefore,

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

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

$\sum_{j=1}^{n}f(x+jt)=\sum_{k=0}^{\infty}S_k(n)\frac{t^kf^{(k)}(x)}{k!}$,

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

Lemma

$e^t\frac{e^{nt}-1}{e^t-1}=\sum_{k=0}^{\infty}S_k(n)\frac{t^k}{k!}$

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