The Revenge of the Plain Mathematician:

How I Answered Two Questions by Two Fancy Jerusalem Mathematicians In One Day

Doron ZEILBERGER ^{1}

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, on the spot, two questions posed by two 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:

\frac{(q)_{a+b}}{(q)_{a}(q)_{b}} = ∑_{c=0}^{b} \frac{q^{ac}}{(q^{−1})_{c}(q)_{b−c}} ,

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 real proof, not a 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])

∏_{k=1}^{b}(1 + yq^{k}) = ∑_{c=0}^{b} y^{c}q^{c(c+1)/2} \frac{(q)_{b}}{(q)_{c}(q)_{b−c}} ,

and do some trivial manipulations.

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)

∑_{r=0}^{(k−3)/2} tan^{2}(\frac{(2r + 1)π}{2k}) = (k − 1)(k − 2)/6 .  (Herschel)

I noticed right away, that this is (using tan^{2} = sec^{2} −1), −(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. .

1 Department of Mathematics, Temple University, Philadelphia, PA 19122, USA. zeilberg@math.temple.edu

http://www.math.temple.edu/~zeilberg/ . Oct. 5, 1999. Supported in part by the NSF. Exclusive to the Personal Journal of Shalosh B. Ekhad and D. Zeilberger: http://www.math.temple.edu/~zeilberg/pj.html .

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 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).

References

[W] Eric W. Weisstein, CRC Concise Encylopedia of Mathematics, web version:

http://www.astro.virginia.edu/~eww6n/#TreasureTroves .