# Cherokee Ridge Homes

Been updating Cherokee Ridge Homes.  About mastered LESS, now moving on to SASS for CSS pre-processing.

Seems tougher these days to stay up to date on programming trends. 😦

# Why 0!=1

From the factorial recurrence relationship of n!=n×(n-1)! and 1!=1:

1=1!=1×0! imples 0!=1 (since identically 1×x =x for all numbers x).

# 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

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

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

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