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In this paper, it says that the three Borwein cubic theta functions obey the identity $a(q)^{3}=b(q)^{3}+c(q)^{3}$, which is strongly reminiscent of the identity that Dixonian elliptic functions obey $\mathrm{sm}^{3}(z)+\mathrm{cm}^{3}(z)=1$. What relationship (if any) exists between the Dixonian elliptic functions and the Borwein cubic theta functions?

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    $\begingroup$ Hmm, you may be on to something. ${}_2 F_1\left({{\frac13}\atop{}}{{}\atop{\frac43}}{{\frac23}\atop{}}\mid z\right)$ crops up in Dixonian theory, while ${}_2 F_1\left({{\frac13}\atop{}}{{}\atop{1}}{{\frac23}\atop{}}\mid z\right)$ crops up in relating the Borwein theta functions... it might take some tedious algebra to entirely display the connection, though. $\endgroup$ – J. M. is a poor mathematician May 3 '11 at 19:41
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    $\begingroup$ This is a good observation. My student Ng Say Tiong and I have looked at it a few years ago and come up with some results. These are contained in Say Tiong's Master thesis at National University of Singapore. $\endgroup$ – user370914 Sep 21 '16 at 7:06
  • $\begingroup$ Related post. $f^3+g^3=1$ for two meromorphic functions $\endgroup$ – Tito Piezas III Jan 7 '17 at 8:25
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(Too long for a comment.)

And to add to the mystery, there's also a $x^3+y^3=1$ relation involving Ramanujan's beautiful cubic continued fraction, $$C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}=\cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}$$

where $q=e^{2\pi i \tau}$ and Dedekind eta function $\eta(\tau)$. Define, $$\alpha =\left(\left(\frac{4C^3(q)+1}{C(q)}\right)^3-27\right)^{1/4}$$ $$\beta=\left(\left(\frac{4C^3(q^3)+1}{C(q^3)}\right)^3-27\right)^{1/4}$$ then we have the relation,

$$\alpha^3+\left(\frac{3\eta(3\tau)}{\eta(\tau)}\right)^3 =\left(\alpha+\frac{3^2}{\beta}\right)^3\tag1$$

Equivalently, and more simply in terms of eta quotients,

$$\left(\frac{\eta^3(\tau)}{\eta^3(3\tau)}\right)^3+\left(\frac{3\eta(3\tau)}{\eta(\tau)}\right)^3 =\left(\frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta^3(3\tau)}\right)^3\tag2$$ See also this MO post.

$\color{blue}{Update:}$ I just realized that if we multiply $(2)$ by $\eta^6(3\tau)$ to get, $$\left(\frac{\eta^3(\tau)}{\eta(3\tau)}\right)^3+\left(\frac{3\eta^3(3\tau)}{\eta(\tau)}\right)^3 =\left(\frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}\right)^3$$ this is in fact the Borwein's, $$b^3(q)+c^3(q) = a^3(q)$$ where, $$a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$ and similarly for $b(q),\,c(q)$.

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In a Master's thesis [1], the author makes the insightful observation in Section 5.2 that if $smu$ and $cmu$ are the standard Dixonian elliptic functions, where $$ cm^3u+sm^3u-3{\alpha}*cmu*smu=1$$ with $\alpha\neq-1$, and if the following are defined as

\begin{align} & z=\vartheta_1(\pi/3)*u/\vartheta_1'(0) \\ & s(u)=\vartheta_1(z)/\vartheta_1(z+\pi/3) \\ & c(u)=-\vartheta_1(z-\pi/3)/\vartheta_1(z+\pi/3) \\ & \alpha=-a(q)*b(q) \end{align} then the functions $s(u)$ and $c(u)$ have the same Taylor expansions as $smu$ and $cmu$ at 0, where $a(q)$ and $b(q)$ are the BBG cubic theta functions (as above).

The author states this as a conjecture. But since the Taylor expansion is unique within it's radius of convergence, he has essentially outlined a proof of the result without providing the specific details. This is entirely based on the author's Theorem 5.1.1 which states a formula that is equivalent to a Jacobian theta function identity proved by Liu.

The paper is worth reading even though it contains some errors. For example on the final page the author fails to observe that $\vartheta_1(z,q)$ is an odd function of $z$.

I can't assert the full accuracy of the claims made in the paper as I have only made a cursory reading of it.

References:
[1] Elliptic Functions, Theta Functions and Identities, Ng Say Tiong

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