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Find the value of the following expression $$ \prod^{\infty}_{n=0}\left(1-\frac{1}{2^{3n}}+\frac{1}{2^{6n}}\right)$$

What I tried:

If $\frac{1}{2^3}=x$, then we can write expression as

$$\prod^{\infty}_{n=0}(1-x^n+x^{2n})=(1-1+1)(1-x+x^2)(1-x^2+x^4)(1-x^3+x^6)\cdots$$

How do I find that infinite product

jacky
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  • Use distributive law and show by induction, what $\prod_{n=0}^k$ equals for all arbitrary $k \in \mathbb{N}$. Then take the limit. It should be 1, I guess. – Mathy Mar 07 '20 at 11:33
  • There is no closed form in terms of usual constants and functions. If you allow theta functions and their friends then there is a closed form (see the answer by Nikos Bagis). If on the other hand $x=\exp(-\pi\sqrt{r}), r>0,r\in\mathbb {Q} $ then in principle one can get a closed using elementary functions and Gamma function values at rational numbers. And even this case is easily said than done unless one has Ramanujan type powers of algebraic manipulation. – Paramanand Singh Mar 14 '20 at 11:47

3 Answers3

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Almost all known information about the infinite product $\prod_{k=1}^\infty(1-x^k+x^{2k})$ is found at the Online Encylopedia of Integer Sequences, sequence A109389.

For example: Expansion of $\displaystyle \frac{q^{-1/12}\eta(q)\eta(q^6)}{\eta(q^2)\eta(q^3)}$ in powers of $q$. Here, $\eta$ is the Dedekind eta function.

I suppose an expert could tell if that eta-quotient is Hauptmodul for a certain subgroup of $SL(2,\mathbb Z)$, and then if $q=1/8$ turns out to be a cusp of the fundamental domain, we could evaluate this at $q=1/8$.

This is $$ j_{6B}(\tau) = \left(\frac{\eta(\tau)\eta(6\tau)}{\eta(2\tau)\eta(3\tau)}\right)^{-12} $$ where $q=e^{2\pi i\tau}$. If $q=1/8$ then $\tau = i(3\log 2)/(2\pi)$. The known special values have $\tau$ algebraic, so this does not work.

GEdgar
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This is not an answer but just the result from a CAS.

Using Pochhammer symbols, the infinite product corresponds to $$L=\lim_{p\to \infty } \, 2^{-3 p (p+1)} \left(\frac{1-i \sqrt{3}}{2} ;8\right)_{p+1} \left(\frac{1+i \sqrt{3}}{2} ;8\right)_{p+1}$$ What is really amazing (at least to me) is the approximation $$L=64\, \sqrt[31] {2\times 3^{10}\times e^{74}\times \log ^{759}(2)\times \log ^{632}(3)}$$ which is correct for $18$ significant figures.

Claude Leibovici
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We will use (see [3]): If $q=e^{-\pi\sqrt{r}}$, $r>0$, $k^{*}_r=\sqrt{1-k_r^2}$: $$ \prod^{\infty}_{n=1}(1+q^n)=q^{-1/24}2^{-1/6}\left(\frac{k_r}{(k^{*}_r)^2}\right)^{1/12}. $$ Hence if $$ P=\prod_{n=0}^{\infty}\left(1-q^{n}+q^{2n}\right),\tag 1 $$ we have $$ P^{12}=\left(\prod_{n=1}^{\infty}\frac{(1+q^{3n})}{(1+q^n)}\right)^{12}=q^{-1}\frac{k_{9r}}{k_r}\left(\frac{k^{*}_{r}}{k^{*}_{9r}}\right)^2. $$ Hence in view of [1] pg.229-238 with $\alpha=k_r^2$, $\beta=k_{9r}^2$ and $$ m=\frac{_{2}F_1\left(\frac{1}{2},\frac{1}{2};1;k_{9r}^2\right)}{{}_2F_1\left(\frac{1}{2},\frac{1}{2};1;k_r^2\right)}=\frac{z_9}{z_1}, $$ instead of $m=\frac{z_1}{z_9}$, we have $$ P=\sqrt[12]{8\frac{(1-3m)^2(1-m) m}{(1+m)^2(1+3m)}}.\tag 2 $$ The $m$ is the solution of ([2] chapter 5): $$ 27m^4-18m^2-8(1-2k^2)m-1=0\tag 3 $$ and $$ k=k_r=\left(\frac{\vartheta_2(0;1/8)}{\vartheta_3(0;1/8)}\right)^2\textrm{, }r=9\log^2(2)/\pi^2.\tag 4 $$ The theta functions are $$ \vartheta_2(0,q):=\theta_2(q)=\sum^{\infty}_{n=-\infty}q^{(n+1/2)^2}\textrm{, }\vartheta_3(0,q):=\theta_3(q)=\sum^{\infty}_{n=-\infty}q^{n^2}. $$

REFERENCES.

[1]: B.C. Berndt, 'Ramanujan`s Notebooks Part III'. Springer Verlang, New York (1991)

[2]: J.M. Borwein and P.B. Borwein. 'Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity', Wiley, New York, 1987.

[3]: E.T.Whittaker and G.N.Watson, 'A course on Modern Analysis'. Cambridge U.P. (1927)

Nikos Bagis
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