10. A commercial for q-trigonometry

In the preceding section, we saw both the factorial and q^{2}-factorial of −1/2. The former is \sqrt{π}; why not call the latter \sqrt{π}_{q}? More generally, why don’t we have a q-version of the factorial reflection formula z! (−z)! = πz/ sin πz? I propose

Π_{q} := \frac{π_{q}}{1 − q^{2}} = q^{\frac{1}{4}} \frac{(−\frac{1}{2})!_{q^{2}}^{2}}{1 − q^{2}} = q^{\frac{1}{4}} ∏_{n≥1} \frac{(1 − q^{2n})^{2}}{(1 − q^{2n−1})^{2}} = \frac{q-Wallis product}{1 − q}

= q^{\frac{1}{4}} (1 + q + q^{3} + q^{6} + q^{10} + ⋯)^{2} = \frac{ϑ_{2}ϑ_{3}}{2} = \frac{ϑ_{2}^{2}(0, q^{\frac{1}{2}} )}{4}

= lim_{z→0} \frac{sin_{q} πz}{q^{−z} − q^{z}} = −\frac{π}{2} \frac{sin′_{q} 0}{ln q} = \frac{η(q^{2})^{4}}{η(q)^{2}}

where

sin_{q} πz := \frac{q^{z(z−1)}π_{q}}{(z − 1)!_{q^{2}} (−z)!_{q^{2}}} = q^{(z−1/2)^{2}} ∏_{n≥1} \frac{(1 − q^{2n−2z})(1 − q^{2n+2z−2})}{(1 − q^{2n−1})^{2}}

= iq^{z^{2}} \frac{ϑ_{1}(iz ln q)}{ϑ_{4}} = − sin_{q} π(z + 1)

= \frac{∑_{−∞<n<∞}(−1)^{n}q^{(n−z+ \frac{1}{2} )^{2}}}{∑_{−∞<n<∞}(−q)^{n^{2}}}

and

cos_{q} πz := sin_{q} π(\frac{1}{2} − z) = q^{z^{2}} ∏_{n≥1} \frac{(1 − q^{2n−2z−1})(1 − q^{2n+2z−1})}{(1 − q^{2n−1})^{2}}

= q^{z^{2}} \frac{ϑ_{4}(iz ln q)}{ϑ_{4}} = \frac{∑_{−∞<n<∞}(−1)^{n}q^{(n−z)^{2}}}{∑_{−∞<n<∞}(−q)^{n^{2}}} .

While Π_{q} is expressible with two Dedekind η functions, Jacobi’s æquatio identica satis abstrusa ([5]) returns the favor:

η(q) := q^{1/24} ∏_{n≥1} 1 − q^{n} = (\begin{matrix}Π & 5 q \\ Π & {2 \atop q^{2}} \end{matrix} − 16 Π_{q} Π_{q^{2}}^{2})^{1/6} .

So we see that sin_{q} πz and cos_{q} πz are period 2, unit amplitude functions with many of the properties of their q → 1 ancestors:

sin_{q}(x + y) = \frac{sin_{q}(x − y)}{sin_{q^{2}}(x − y)}(sin_{q^{2}} x cos_{q^{2}} y + cos_{q^{2}} x sin_{q^{2}} y)

cos_{q}(x + y) = \frac{cos_{q}(x − y)}{cos_{q^{2}}(x − y)}(cos_{q^{2}} x cos_{q^{2}} y − sin_{q^{2}} x sin_{q^{2}} y)
cos_{q} 2z = (cos_{q^{2}} z)^{2} − (sin_{q^{2}} z)^{2} = cos 2z ∏_{n≥0}(sin_{q^{2^{−n}}}^{2} z + cos_{q^{2^{−n}}}^{2} z)

= (cos_{q} z)^{4} − (sin_{q} z)^{4}

sin_{q} 2z = \begin{matrix}Π & q \\ Π & q^{2} \end{matrix} sin_{q^{2}} z cos_{q^{2}} z

= \frac{1}{2} \begin{matrix}Π & q \\ Π & q^{4} \end{matrix} \sqrt{(sin_{q^{4}} z)^{2} − (sin_{q^{2}} z)^{4}} = \frac{1}{2} \begin{matrix}Π & q \\ Π & q^{4} \end{matrix} \sqrt{(cos_{q^{4}} z)^{2} − (cos_{q^{2}} z)^{4}}

sin_{q} 3z = \begin{matrix}Π & q \\ Π & q^{3} \end{matrix} (cos_{q^{3}} z)^{2} sin_{q^{3}} z − (sin_{q^{3}} z)^{3}

= \frac{1}{3} \begin{matrix}Π & q \\ Π & q^{9} \end{matrix} sin_{q^{9}} z − (1 + \frac{1}{3} \begin{matrix}Π & q \\ Π & q^{9} \end{matrix} )(sin_{q^{3}} z)^{3}

sin_{q} 5z = \begin{matrix}Π & q \\ Π & q^{5} \end{matrix} (cos_{q^{5}} z)^{4} sin_{q^{5}} z − \sqrt{\begin{matrix}Π & 3 q \\ Π & {3 \atop q^{5}} \end{matrix} − 2\begin{matrix}Π & 2 q \\ Π & {2 \atop q^{5}} \end{matrix} + 5\begin{matrix}Π & q \\ Π & q^{5} \end{matrix}} (cos_{q^{5}} z)^{2}(sin_{q^{5}} z)^{3}

+ (sin_{q^{5}} z)^{5}

sin_{q}(x − w) sin_{q}(x − y) sin_{q}(y − w) sin_{q}(y − z) sin_{q}(z − w) sin_{q}(z − x)
+ cos_{q}(x − w) cos_{q}(x − y) sin_{q}(y − w) cos_{q}(y − z) cos_{q}(z − w) sin_{q}(z − x)
+ cos_{q}(x − w) sin_{q}(x − y) cos_{q}(y − w) cos_{q}(y − z) sin_{q}(z − w) cos_{q}(z − x)
+ sin_{q}(x − w) cos_{q}(x − y) cos_{q}(y − w) sin_{q}(y − z) cos_{q}(z − w) cos_{q}(z − x) = 0

I actually discovered this last formula in q-land, via a sequence of symmetrizing generalizations guided by the additional structure imposed by the qs.

Strip mining over undetermined linear combinations of extensive Taylor expansions turned up the empirical relations

∑_{n≥1} \frac{q^{n}}{(1 − q^{n})^{2}} − 2 ∑_{n≥1} \frac{q^{2n}}{(1 − q^{2n})^{2}} = \frac{1}{24}(\begin{matrix}Π & 4 q \\ Π & {2 \atop q^{2}} \end{matrix} − 1) + \frac{2}{3} Π_{q^{2}}^{2}

∑_{n≥1} \frac{q^{n}}{(1 − q^{n})^{2}} − 3 ∑_{n≥1} \frac{q^{3n}}{(1 − q^{3n})^{2}} = \frac{(Π_{q}^{2} +3 Π_{q^{3}}^{2})^{2}}{12 Π_{q} Π_{q^{3}}} − \frac{1}{12}

∑_{n≥1} \frac{q^{n}}{(1 − q^{n})^{2}} − 4 ∑_{n≥1} \frac{q^{4n}}{(1 − q^{4n})^{2}} = \frac{1}{8}(\begin{matrix}Π & 4 q \\ Π & {2 \atop q^{2}} \end{matrix} − 1)

∑_{n≥1} \frac{q^{2n}}{(1 − q^{2n})^{2}} − 9 ∑_{n≥1} \frac{q^{18n}}{(1 − q^{18n})^{2}} = \begin{matrix}Π & {3 \atop q^{3}} \\ Π & q \end{matrix} + \frac{1}{3}(\begin{matrix}Π & {3 \atop q^{3}} \\ Π & q^{9} \end{matrix} − 1)

∑_{n≥1} \frac{q^{2n−1}}{(1 − q^{2n−1})^{2}} − 2 ∑_{n≥1} \frac{q^{4n−2}}{(1 − q^{4n−2})^{2}} = Π_{q^{2}}^{2} = ∑_{n≥1} \frac{(2n − 1)q^{2n−1}}{1 − q^{4n−2}}

∑_{n≥1} \frac{q^{2n−1}}{(1 − q^{2n−1})^{2}} − 5 ∑_{n≥1} \frac{q^{10n−5}}{(1 − q^{10n−5})^{2}} = Π_{q^{5}}^{2} \frac{\begin{matrix}Π & {2 \atop q^{5}} \\ Π & {2 \atop q^{10}} \end{matrix} + 16\begin{matrix}Π & {2 \atop q^{10}} \\ Π & {2 \atop q^{5}} \end{matrix}}{\begin{matrix}Π & q \\ Π & q^{5} \end{matrix} − 4 − \begin{matrix}Π & q^{5} \\ Π & q \end{matrix}} .

∑_{n≥0} \frac{1 − q^{12n+1}}{1 − q^{2}} \frac{(2n − \frac{3}{2})!_{q^{2}}(2n + \frac{1}{2})!_{q^{2}}}{(4n + 1)!_{q^{2}}} q^{4n^{2}} = −q^{1/4}\sqrt{π_{q}π_{q^{5}} \frac{1 − q^{2}}{1 − q^{10}}}

This last is the a → 0 case of the summation identity of Section 1. Notice how much easier it is to take the q → 1 limit in this form than in that.

A limiting case of a certain two parameter path invariance result (omitted) can be written

∑_{k≥0} \frac{(a + k − 1)!_{q}(k − a)!_{q}}{(2k)!_{q}} q^{k^{2}} = \frac{π_{q^{1/2}}}{sin_{q^{1/2}} πa} \frac{sin_{q^{3/2}} π \frac{a+1}{3}}{sin_{q^{3/2}} \frac{π}{3}} q^{(a−1)a/3}.

Although Professor Andrews has discovered an equivalent result in Ramanujan’s “lost” “notebook” ([6]), it is worth noting how the q-trigonometric version clearly reveals this f(a) as period 2 poles times period 6 zeros times a quadratic power of q.

Professors Andrews and Berndt also inform me that J. W. L. Glaisher had at least the first half of the identity

(Glaisher)  Π_{q}^{4} = ∑_{n≥1} \frac{n^{3}q^{n}}{1 − q^{2n}} = ∑_{n≥1} Li_{−3}(q^{2n−1})

by 1905 ([7]). Combinatorially, it says that the number of ways to express n − 1 as the sum of 8 triangular numbers equals n^{3} times the sum of the cubes of the reciprocals of the odd divisors of n. For an explanation of the Li_{−3}, see section 12.

11. From the q-Land Chamber of Commerce

I found the (Glaisher) identity simply by letting the angles vanish in a q-trig identity. This illustrates one of the several advantages of q-identities: even after giving back all the conventional parameters, you may still have a neat result.

I have already mentioned that q-structures led me to the four angle trig identity.

Yet another advantage is that you can test q-identities merely by Taylor expanding with respect to q, whereas in q → 1 land, you are unlikely to be able to expand with respect to any of the remaining parameters.

Computer algebra has been of great help in exploring q-land, but now perhaps q-land can return the favor, by deflating the intermediate expression swell that burdens many rational function manipulations.

For example, the path-invariance check for the matrices underlying Dougall’s theorem requires the expansion of the numerator of expression (D2), below,

(D1) \begin{matrix}q^{−n+k−i−h−g+1} & (1 − q^{n+g}) & (1 − q^{n+h}) & (1 − q^{n+i}) & (1 − q^{n+2k−i−h−g+2}) \\ (1 − q^{k−i−h−g}) & (1 − q^{n+k−g+1}) & (1 − q^{n+k−h+1}) & (1 − q^{n+k−i+1}) \end{matrix}

−\begin{matrix}(1 − q^{k−h−g+1}) & (1 − q^{k−i−g+1}) & (1 − q^{k−i−h+1}) & (1 − q^{2n+k+1}) \\ (1 − q^{k−i−h−g}) & (1 − q^{n+k−g+1}) & (1 − q^{n+k−h+1}) & (1 − q^{n+k−i+1}) \end{matrix}

(C2) COMBINE(QONE(D1)); /* lim_{q→1} of above */

(D2)  \frac{(n+g)(n+h)(n+i)(n+2k−i−h−g+2)−(k−h−g+1)(k−i−g+1)(k−i−h+1)(2n+k+1)}{(k−i−h−g)(n+k−g+1)(n+k−h+1)(n+k−i+1)}

which results in the 99 term expression (D3).

(C3) EXPAND(NUM(D2));

(D3) n^{4}+2kn^{3}+2n^{3}+2ikn^{2}+2hkn^{2}+2gkn^{2}−i^{2}n^{2}−hin^{2}−gin^{2}+2in^{2}−h^{2}n^{2}−ghn^{2}+2hn^{2}− g^{2}n^{2}+2gn^{2}−2k^{3}n+4ik^{2}n+4hk^{2}n+4gk^{2}n−6k^{2}n−2i^{2}kn−4hikn−4gikn+8ikn−2h^{2}kn− 4ghkn+8hkn−2g^{2}kn+8gkn−6kn+hi^{2}n+gi^{2}n−2i^{2}n+h^{2}in+2ghin−4hin+g^{2}in−4gin+ 4in+gh^{2}n−2h^{2}n+g^{2}hn−4ghn+4hn−2g^{2}n+4gn−2n−k^{4}+2ik^{3}+2hk^{3}+2gk^{3}−4k^{3}−i^{2}k^{2}− 3hik^{2} −3gik^{2} +6ik^{2} −h^{2}k^{2} −3ghk^{2} +6hk^{2} −g^{2}k^{2} +6gk^{2} −6k^{2} +hi^{2}k +gi^{2}k −2i^{2}k +h^{2}ik + 4ghik−6hik+g^{2}ik−6gik+6ik+gh^{2}k−2h^{2}k+g^{2}hk−6ghk+6hk−2g^{2}k+6gk−4k−ghi^{2}+ hi^{2}+gi^{2}−i^{2}−gh^{2}i+h^{2}i−g^{2}hi+4ghi−3hi+g^{2}i−3gi+2i+gh^{2}−h^{2}+g^{2}h−3gh+2h−g^{2}+2g−1

(C4) LENGTH(D3);

(D4)  99

But (D2) is the q → 1 limit of (D1), from the q version of Dougall’s theorem, known as Jackson’s. Even though (D1) is more complicated than (D2), its expansion, (D5), has fewer than 1/6 as many terms!

(C5) EXPAND(NUM(COMBINE(D1)));

(D5) q^{3n+3k−i−h−g+3} −q^{2n+4k−2i−2h−2g+4} −q^{2n+2k−i−h+2} −q^{2n+2k−i−g+2} −q^{2n+2k−h−g+2} +q^{n+3k−i−2h−2g+3} + q^{n+3k−2i−h−2g+3} + q^{n+3k−2i−2h−g+3} + q^{n+k−i+1} + q^{n+k−h+1} +q^{n+k−g+1} + q^{−n+k−i−h−g+1} − q^{2k−i−h−2g+2} − q^{2k−i−2h−g+2} − q^{2k−2i−h−g+2} − 1

(C6) LENGTH(D5);

(D6)  16

This is because the parameters in the exponents can only combine additively, and are therefore immune to the profusion of cross-terms plaguing ordinary polynomial multiplication.

And yet, if you needed it, the 99 term expansion is recoverable from the 16 q-terms via four applications of l’Hospital’s rule. But fortunately, (D5) is only an intermediate result, whose conversion can be postponed the until it’s trivial.

12. Commercial for negapolylogs

(If Sesame Street can run commercials for letters of the alphabet, I demand equal time for my favorite special functions.) Running the polylog recurrence in reverse, we differentiate our way back through a sequence of rational functions which generate the Eulerian number triangle:

Li_{0}(x) = x\frac{1}{1 − x}

Li_{−1}(x) = x\frac{1}{(1 − x)^{2}}

Li_{−2}(x) = x\frac{x + 1}{(1 − x)^{3}}

Li_{−3}(x) = x\frac{x^{2} + 4x + 1}{(1 − x)^{4}}

Li_{−4}(x) = x\frac{x^{3} + 11x^{2} + 11x + 1}{(1 − x)^{5}}

⋮

These are useful rational functions. While yet retaining certain loggish tendencies:

Li_{−k}(e^{z}) = \frac{d^{k}}{dz^{k}} Li_{0}(e^{z})  k ≥ 0

= \frac{k!}{(−z)^{k+1}} − ∑_{n≥0} \frac{B_{n+k+1}(1)}{n + k + 1} \frac{z^{n}}{n!} .

(Note the gap of k powers of z.) The generating function is

∑_{k≥0} \frac{Li_{−k}(e^{−z})}{k!}t^{k} = \frac{1}{e^{z−t} − 1}

A sample application is the expansion of tan (or sec or cot or …) about an arbitrary point,

\frac{d^{k}}{dx^{k}} tan x = (2i)^{k+1} Li_{−k}(−e^{2ix}),

which would otherwise invoke misleadingly transcendental polygammas. And here is the nub of the “q-Stirling” formula

log ∏_{n≥z+y} 1 − q^{n} ∼_{z} ∑_{k≥0} \frac{B_{k}(y)}{k!}(ln q)^{k−1} Li_{2−k}(q^{z}).

And we can use negapolylogs to compute Bernoulli numbers and special values of the Euler polynomials.

Li_{−k}(−1) = (−1)^{k}(2^{k+1} − 1)\frac{B_{k+1}}{k + 1} = \frac{(−1)^{k+1}}{2} E_{k}(0) = −\frac{E_{k}(1)}{2}
Li_{−k}(i) = −2^{k−1}E_{k}(1) + \frac{i}{2} E_{k}
Li_{−k}(e^{\frac{πi}{3}} ) = \frac{1}{4}((1 − 3^{k+1})E_{k}(1) + i 3^{k+ \frac{1}{2}} (1 + (−1)^{k})E_{k}(\frac{1}{3}))

Li_{−k}(e^{\frac{2πi}{3}} ) = \frac{1}{4}(\frac{1 − 3^{k+1}}{2^{k+1} − 1}E_{k}(1) + i 3^{k+ \frac{1}{2}} \frac{1 + (−1)^{k}}{2^{k−1} + 1} E_{k}(\frac{1}{3}))

Finally, do not confuse these with negapolygammas, ψ_{−k}, which become “more transcendental” (deeply nested integrals) with increasing k, and are effectively the logarithmic derivatives of the “higher factorials” 1^{1^{k}}2^{2^{k}} . . .. But that’s another story.

ω. References

[1] R. Wm. Gosper, Decision procedure for indefinite hypergeometric summation, Proceedings of the National Academy of Sciences (USA) 75, No. 1 (January 1978), 40-42.

[2] W. Heymann, Ueber hypergeometrische Functionen, deren letztes Element speciell ist, nebst einer Anwendung auf Algebra, Jahresbericht der Technischen Staatslehrenstalten in Chemnitz 1897-1898, (1898), 3-44.

[3] W. Heymann, Über hypergeometrische Funktionen, deren letztes Element speziell ist, Z. Math. Phys. 44, (1899), 280-288.

[4] G. Andrews, On the q-Analog of Kummer’s Theorem and Applications, Duke Mathematical Journal 40, No. 3 (September 1973), 525-528.

[5] E. T. Whittaker and G. N. Watson, (A Course of) Modern Analysis, (1902-1984), 470.

[6] G. Andrews, Ramanujan’s “Lost” Notebook II, Adv. in Math. 41, (1981), 173-185 (Eq. (1.10.)_{R} of Thm. 1).

[7] J. W. L. Glaisher, Representations of a number as a sum of four squares, Quarterly J. of Pure and Applied Math. 36 (1905), 355.