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\line{\sevenrm a18.tex[162,phy] \today\hfill}

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\line{\bf Factorials and Binomial Coefficients as Formal Products\hfil}

\bigskip
Let $f\colon\, Z→Z$. Define $\pi f$ as the formal product 
$\cdots (-1)↑{f(-1)}\,0↑{f(0)}\,1↑{f(1)}\,2↑{f(2)}\,\ldots\,$,
or $\prod↓i i↑{f(i)}$; the product is a real number only if
$\sum↓i|f(i)|$ is finite and $f(0)≥0$. Then $\pi(f+g)=\pi f\cdot\pi g$,
$\pi(-f)=1/\pi f$, $\pi 0=1$, $\pi(b\delta↓{ia})=a↑b$.

Define $a\!\!\downarrow$ as $\pi[i≤a]$, $a\!\!\uparrow$ as $\pi[i>a]$,
where the square brackets indicate a characteristic function. Since
$[i≤a]+[i>a]=1$, $a\!\!\downarrow\cdot a\!\!\uparrow=\pi1$. Therefore
$a\!\!\downarrow/b\!\!\downarrow=b\!\!\uparrow/a\!\!\uparrow$. We define
$a!=a\!\!\downarrow/0\!\!\downarrow$, $a↑{\underline{b}}=
a\!\!\downarrow/(a-b)\!\!\downarrow=a!/(a-b)!$,
$a↑{\overline{b}}=(a-1)\!\!\uparrow/(a+b-1)\!\!\uparrow
=(a+b-1)\!\!\downarrow/(a-1)\!\!\downarrow
=(a+b-1)!/(a-1)!=(a+b-1)↑{\underline{b}}$.
These are formally defined for all $a$ and~$b$.
Some easy identities:
$$\eqalign{a↑{\underline{b}}(a-b)↑{\underline{c}}&={a\!\!\downarrow\over
(a-b)\!\!\downarrow}\;{(a-b)\!\!\downarrow\over (a-b-c)\!\!\downarrow}
=a↑{\underline{a+c}}\cr
\noalign{\smallskip}
a↑{\underline{-b}}&={a\!\!\downarrow\over (a+b)\!\!\downarrow}
=1/(a+b)↑{\underline{b}}\cr
\noalign{\smallskip}
{1\over a↑{\underline{b}}}&={(a-b)\!\!\downarrow\over a\!\!\downarrow}
=(a-b)↑{\underline{-b}}\cr
\noalign{\smallskip}
a!&=a↑{\underline{a}}\,,\quad a↑{\underline{0}}=1\,,\quad
a↑{\underline{1}}=a\,,\quad
a↑{\underline{-1}}={1\over a+1}\cr}$$
{}
$$\pi f(-i)=\prod↓ii↑{f(-1)}=\prod↓j(-j)↑{f(j)}=(-1)↑{\sum↓jf(j)}\pi f(i)$$
whenever $\sum↓jf(j)$ is defined.
$$\eqalign{{a\!\!\downarrow\over b\!\!\downarrow}&={b\!\!\uparrow\over a\!\!\uparrow}
=\pi([i≥b+1]-[i≥a+1])\cr
\noalign{\smallskip}
&=\pi([-i≤-b-1]-[-i≤-a-1])\cr
\noalign{\smallskip}
&=(-1)↑{a-b}\pi([i≤-b-1]-[i≤-a-1])\cr
\noalign{\smallskip}
&=(-1)↑{a-b}\,{(-b-1)\!\!\downarrow\over (-a-1)\!\!\downarrow}\,.\cr}$$
Applying this result to factorial powers,
$$\eqalign{a↑{\underline{b}}&={a\!\!\downarrow\over (a-b)\!\!\downarrow}=
(-1)↑b\,{(b-a-1)\!\!\downarrow\over 
(-a-1)\!\!\downarrow}=(-1)↑b(b-a-1)↑{\underline{b}}\cr
\noalign{\smallskip}
{1\over a↑{\underline{b}}}&={(a-b)\!\!\downarrow\over a\!\!\downarrow}=(-1)↑b\,
{(-a-1)\!\!\downarrow\over (b-a-1)\!\!\downarrow}=(-1)↑b(-a-1)↑{\underline{-b}}\,.\cr}$$
When $a≥0$ and $b>a$, the above gives the useful
$$a↑{\underline{b}}=a↑{\underline{a}}\,0↑{\underline{1}}\,(-1)↑{\underline{b-a-1}}
=a!\,0↑1(-1)↑{b-a-1}(b-a-1)↑{\underline{b-a-1}}
=(-1)↑{b-a-1}a!\,(b-a-1)!\,0↑1\,.$$
All falling factorial powers $a↑{\underline{b}}$ can now be expressed
as products of ordinary factorials and parity functions, with a
factor of $0↑{-1}$, $0↑0$, or $0↑1$. If $b<0$, use
$a↑{\underline{-b}}=1/(a+b)↑{\underline{b}}$. If $0≤b≤a$,
$a↑{\underline{b}}=a!/(a-b)!$. If $0≤a<b$, $a↑{\underline{b}}=(-1)↑{b-a-1}a!\,
(b-a-1)!\,0↑1$.
If $a<0≤b$, $a↑{\underline{b}}=(-1)↑b(b-a-1)↑{\underline{b}}=
(-1)↑b(b-a-1)!/(-a-1)!$.
Factorials of negative numbers are now $(-a)!={(-1)↑{a-1}0↑{-1}\over (a-1)!}$.

Naturally, binomial coefficients fit into this scheme;
${a\choose b}={a!\over b!\,(a-b)!}={a\!\downarrow/0\!\downarrow\over
(b\!\downarrow/0\!\downarrow)(a-b)\!\downarrow/0\!\downarrow}=
{a\!\downarrow0\!\downarrow\over b\!\downarrow(a-b)\!\downarrow}$. 
This is zero if
$[a≥0]+[0≥0]-[b≥0]-[a-b≥0]≥1$. For $a≥0$, ${a\choose b}=0$ unless
$0≤b≤a$. For $a<0$, ${a\choose b}=0$ requires $b<0$ and $a-b<0$, or
$a<b<0$.

\bigskip
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\bigskip
As the figure shows, if $[a≥0]+[b≥0]+[a≥b]$ is even, ${a\choose b}=0$.
To have ${a\choose b}$ infinite requires $[a≥0]-[b≥0]-[a≥b]≤-2$,
so $0≤b≤a<0$, impossible. If $a<0$, $b≥0$, $a-b<0$,
$${a\choose b}={a!\over b!\,(a-b)!}={(-1)↑{-a-1}\,0↑{-1}\,(b-a-1)!\over
(-a-1)!\,b!\,(-1)↑{b-a-1}\,0↑{-1}}= {(-1)↑b(b-a-1)!\over (-a-1)!\,b!}
=(-1)↑b{b-a-1\choose b}\,.$$



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\copyright 1988 Robert W. Floyd.
First draft (not published)
April 1, 1988.
%revised date;
%subsequently revised.

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