%revenge.tex: 
%The Revenge of the Plain Mathematician:
%How I Answered
%Two Questions by Two Fancy Jerusalem Mathematicians In One Day
%a Plain TeX file by Doron Zeilberger (1 page)
 
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\bf
\centerline
{The Revenge of the Plain Mathematician:}
\centerline
{How I Answered
Two Questions by Two Fancy Jerusalem Mathematicians In One Day
}
\rm
\bigskip
\centerline{ {\it 
Doron ZEILBERGER
}\footnote{$^1$}
{\eightrm  \raggedright
Department of Mathematics, Temple University,
Philadelphia, PA 19122, USA. 
%\break
{\eighttt zeilberg@math.temple.edu} \hfill \break
{\eighttt http://www.math.temple.edu/\Tilde zeilberg/   .}
Oct. 5, 1999.  Supported in part by the NSF.
Exclusive to the Personal Journal of Shalosh B. Ekhad
and D. Zeilberger:
{\eighttt http://www.math.temple.edu/\Tilde zeilberg/pj.html   .}
} 
}
 
Like all mathematicians who grew up in Israel, I was always in awe
of the members of the faculty of the Hebrew University at Jerusalem,
roughly the Israeli analog of Princeton. Hence I felt a huge
personal satisfaction, when, on the very same day (yesterday, Oct. 4, 1999), I
was able to answer, {\it on the spot}, two questions posed by two 
{\it distinct} Jerusalem mathematicians, that furthermore, were utterly trivial
to me, and showed me that at least I know something that these
fancy guys do not seem to know. I was also amused that, roughly,
I am the average of these two, at least in age and in piety.
 
The first question came from the famous Knotter Dror Bar-Natan
(15 years my junior), who asked me to prove the identity:
$$
{{(q)_{a+b} } \over {(q)_a (q)_b}} =
\sum_{c=0}^{b} {{q^{ac}} \over {(q^{-1})_c (q)_{b-c}}} \quad,
$$
where $(q)_n:=(1-q)(1-q^2)...(1-q^n)$. After a few minutes
of blabbing that these identities are all shaloshable
(using my Maple package qEKHAD), and him retorting that he
meant a {\it real} proof, not a {\it bogus} (computer-generated) one,
I realized that it would be a waste of Shalosh's precious time to
give this trifle to it. All one has to do is set
$y:=-q^a$ in Cauchy's celebrated (and trivial) q-binomial theorem
(e.g. [W])
$$
\prod_{k=1}^{b} (1+y q^k) =
\sum_{c=0}^{b} y^c q^{c(c+1)/2} {{(q)_b } \over {(q)_c (q)_{b-c}}} \quad ,
$$
and do some trivial manipulations. \halmos
 
The second question was posed by the famous Riemann-Surfacer, Herschel
Farkas (11 years my senior), later the same day, in a Temple University
math colloquium talk. He asked to give an `elementary' proof of
the following identity, that he and collaborators derived using
fancy, theta function methods: If $k$ is odd (for some reason Herschel
said `prime', but he probably meant odd, since it is true for all odd
$k$)
$$
\sum_{r=0}^{(k-3)/2} \tan^2 ({{(2r+1)\pi} \over {2k}})= (k-1)(k-2)/6 \quad.
\eqno(Herschel)
$$
I noticed right away, that this is (using $\tan^2=\sec^2-1$),\quad
$-(k-1)/2$ plus
the sum of the reciprocals of the roots of the $(k-1)/2$-degree
polynomial $p(x)$ , where
$p(x^2)=T_{k}(x)/x$,
$T_k(x)$ being the venerable Chebychev polynomial (see. e.g. [W])
(when $k$ is odd): $x(1-((k+1)(k-1)/6)x^2+...)$.
Hence $p(x)=1-((k+1)(k-1)/6)x+... $, and the sum of the reciprocals
of its roots is $(k+1)(k-1)/6$, hence the left side of $(Herschel)$
equals $(k+1)(k-1)/6-(k-1)/2=(k-1)(k-2)/6$. \quad \halmos.
 
{\bf Moral:} You don't have to be a Zeilberger to solve the above two
`problems', they are both trivial, once you know the relevant facts.
Hence it is good to assemble as many facts as possible from 
{\it all over mathematics}. A painless, indeed extremely enjoyable,
way to increase your `mathematical facts power' is to browse,
at least an hour a day, in Eric Weisstein's[W] magnificent opus,
either in its paper version (if your age is $>60$),
or the web version (if your age is $<40$), or both (if you are
somewhere in between, like me).
 
{\bf References}
 
[W] Eric W. Weisstein, {\it CRC Concise Encylopedia of Mathematics},
web version:\hfill\break
{\tt http://www.astro.virginia.edu/\Tilde eww6n/\#TreasureTroves} .
 
 
\bye
 
 
 
 
 
 
