Question: How would you prove $$\dfrac 1{R(q)}-1-R(q)=\dfrac {f(-q^{1/5})}{q^{1/5} f(-q^5)}\tag{1}$$ where $f(-q)=(q;q^5)_\infty$ and $R(q)$ is defined as $$R(q)=q^{1/5}\dfrac {(q;q^5)_\infty (q^4;q^5)_\infty}{(q^2;q^5)_\infty (q^3;q^5)_\infty}\tag{2}$$

I'm not sure what to do. I started with Euler's Pentagonal Theorem $$f(-q)=\sum\limits_{n=-\infty}^{\infty} (-1)^n q^{n(3n+1)/2}\tag3$$ and got the following: $$\begin{align*} & \color{red}{f(-q^{1/5})}=\sum\limits_{n=-\infty}^{\infty} (-1)^n q^{n(3n+1)/10}\tag4\\ & \color{blue}{f(-q^5)}=\sum\limits_{n=-\infty}^{\infty} (-1)^n q^{5n(3n+1)/2}\tag5\end{align*}$$ Thus, the RHS can be expressed as $$\dfrac {f(-q^{1/5})}{q^{1/5} f(-q^5)}=\dfrac {\color{red}{\sum\limits_{n=-\infty}^\infty (-1)^n q^{n(3n+1)/10}}}{q^{1/5}\color{blue}{\sum\limits_{n=-\infty}^\infty (-1)^n q^{5n(3n+1)/2}}}\tag6$$ However, I'm not sure how to further reduce down $(6)$. And "even furthermore," I don't know how to simplify $R(q)$ to get $$R(q)=q^{1/5}\dfrac {(q;q^5)_\infty (q^4;q^5)_\infty}{(q^2;q^5)_\infty (q^3;q^5)_\infty}\tag7$$

  • 1
    $\begingroup$ $(q^a;q^b)_\infty = \prod_{n=1}^\infty (1-q^a q^{nb})$](en.wikipedia.org/wiki/Q-Pochhammer_symbol). Did you show they are modular : congruence group, weight, order of vanishing at the cusp ? $\endgroup$ – reuns Jan 28 '17 at 23:21
  • 1
    $\begingroup$ See this blog post paramanands.blogspot.com/2013/09/… $\endgroup$ – Paramanand Singh Feb 28 '17 at 16:51
  • 1
    $\begingroup$ @ParamanandSingh Cool! Is it possible for you to compress what you wrote and answer the question? $\endgroup$ – Crescendo Feb 28 '17 at 18:17
  • 1
    $\begingroup$ I can only provide an outline of the proof as the detailed proof is long enough. Of if you prefer I can copy paste from blog with some editing. But this will be possible tomorrow only. $\endgroup$ – Paramanand Singh Feb 28 '17 at 20:23

I checked some of your questions which deal with the work of Ramanujan (like class invariants $G_{n}, g_{n} $, series for $1/\pi$, Rogers-Ramanujan function $R(q) $ of the current question). A lot of Ramanujan's work has been decoded and proved by Dr. Bruce C. Berndt in his five volume work Ramanujan Notebooks and you can get answers to your queries with more details and references there.

On the other hand I have myself studied some material on Ramanujan from various sources (including the Notebooks mentioned above) and have described the same in my blog posts. Material in the blog is self contained and presented in a systematic fashion so that pre-requisites for a result are established before proving the result. I would advise you to have a look at the archives page and check the posts related to "Modular Equations", "Rogers-Ramanujan function", "Class Invariants", "Lambert series", and "Partitions". It also helps if you are aware of the classical theory of elliptic and theta functions as given by Jacobi (so that you can compare it with Ramanujan's work) and some material on these topics is also available on the blog.

I earlier tried to provide an answer here by copy pasting from the blog, but it appears that there are many interconnected results which need to be established first and a coherent answer with all the details would be a bit long and therefore I suggest you take time to study the blog posts mentioned earlier.


The proof I present has been considered in Duke's paper [Du]. Another proof can be found in Berndt's book [Ber] but it is much more involved (it requires quite complicated transformations of theta series) and is still lengthy.

If $|q|<1$, then $$ R(q)=q^{1/5}\frac{\sum^{\infty}_{n=-\infty}(-1)^nq^{5n^2/2+3n/2}}{\sum^{\infty}_{n=-\infty}(-1)^nq^{5n^2/2+n/2}}=q^{1/5}\prod^{\infty}_{n=1}(1-q^n)^{(n|5)},\tag A $$ where $(n|m)$ is the usual Jacobi symbol.

We define the theta functions with parameters $\alpha,\beta$ in $\textbf{R}$ by $$ \theta(\alpha,\beta;z):=\sum^{\infty}_{n=-\infty}e\left(\frac{1}{2}\left(n+\frac{\alpha}{2}\right)^2z+\frac{\beta}{2}\left(n+\frac{\alpha}{2}\right)\right)\textrm{, }Im(z)>0,\tag 1 $$ where $e(z):=e^{2\pi i z}$.

These theta functions we have the following properties:

Theorem 1. (see [FK])

If $l,m\in\textbf{Z}$ and $N$ positive integer, then $$ \theta(\alpha,\beta;z)=e\left(\mp \frac{\alpha m}{2}\right)\theta\left(\pm \alpha+2l,\pm \beta+2m;z\right)\tag 2 $$ $$ \theta(\alpha,\beta;z)=\sum^{N-1}_{k=0}\theta\left(\frac{\alpha+2k}{N},N\beta;N^2z\right)\tag 3 $$ The theta function (1) also satisfies for $a,b,c,d\in\textbf{Z}$, with $ad-bc=1$ the following functional (modular) relation $$ \theta\left(\alpha,\beta;\frac{az+b}{cz+d}\right)=\kappa\sqrt{cz+d}\cdot\theta\left(a\alpha+c\beta-ac,b\alpha+d\beta+bd;z\right),\tag 4 $$
where $$ \kappa=e\left(-\frac{1}{4}(a\alpha+c\beta)bd-\frac{1}{8}(ad\alpha^2+cd\beta^2+2bc\alpha\beta)\right)\kappa_0,\tag 5 $$ with $\kappa_0$ beeing an eighth root of unity, depending only on $a,b,c,d$.

In particular we have $$ \theta(\alpha,\beta;z+1)=e\left(-\frac{\alpha}{4}\left(1+\frac{\alpha}{2}\right)\right)\theta(\alpha,\beta;z)\tag 6 $$ and $$ \theta\left(\alpha,\beta;\frac{-1}{z}\right)=e\left(-\frac{1}{8}\right)\sqrt{z}\cdot e\left(\frac{\alpha\beta}{4}\right)\theta\left(\beta,-\alpha;z\right).\tag 7 $$ Also holds the following product formula $$ \theta(\alpha,\beta;z)= $$ $$ e\left(\frac{\alpha\beta}{4}\right)q^{\alpha^2/8}f(-q)\prod^{\infty}_{n=1}\left(1+e\left(\beta/2\right)q^{n-\frac{\alpha+1}{2}}\right)\left(1+e\left(-\beta/2\right)q^{n+\frac{\alpha-1}{2}}\right),\tag B $$ where $q=e(z)$, $Im(z)>0$.

Theorem 2

If $q_1=e(-1/z)$ and $q=e(z)$, $Im(z)>0$, then exists a fifth root of unity $\zeta=\zeta(q)=e\left(\nu(q)/5\right)$ such that $\nu(q)\in\{0,1,2,3,4\}$ and $$ R(q_1)=\overline{\zeta\left(q_1\right)}\cdot\frac{1-\phi \zeta(q) R(q)}{\phi+\zeta(q) R(q)},\tag 8 $$ where $\phi:=\frac{1+\sqrt{5}}{2}$ and where $\overline{z}$ denotes the conjugate complex number of $z$.


If $z$ is in the upper half plane, then relation (A) can be written, with the help of (1) as $$ R(q)^5=\left(e\left(-1/10\right)\frac{\theta(3/5,1;5z)}{\theta(1/5,1;5z)}\right)^5.\tag 9 $$ The 5th power in both sides of (9) is for avoiding the factor $q^{1/5}$ of (A). Hence we can write $$ \zeta(q)\cdot R(q)=e\left(-1/10\right)\frac{\theta(3/5,1;5z)}{\theta(1/5,1;5z)},\tag {10} $$ where $\zeta(q)$ is a 5th root of unity. Using (7),(2),(3) we get $$ \zeta(q_1)R(q_1)= e\left(-1/10\right)\frac{\theta(3/5,1;-5/z)}{\theta(1/5,1;-5/z)}=\frac{\theta(1,3/5;z/5)}{\theta(1,1/5;z/5)}= $$ $$ =\frac{\sum^{4}_{\nu=0}\theta(\frac{2\nu+1}{5},3;5z)}{\sum^{4}_{\nu=0}\theta(\frac{2\nu+1}{5},1;5z)}.\tag{11} $$ But $$ \sum^{4}_{\nu=0}\theta\left(\frac{2\nu+1}{5},3;5z\right)=\theta(1/5,3;5z)+\theta(3/5,3;5z)+\theta(1,3;5z)+ $$ $$ +\theta(7/5,3;5z)+\theta(9/5,3;5z)= $$ $$ =e(1/10)\theta(1/5,1;5z)+e(3/10)\theta(3/5,1;5z)+e(1)\theta(1,1;5z)+ $$ $$ +e(7/10)\theta(2-7/5,1,5z)+e(9/10)\theta(2-9/5,1,5z)= $$ $$ =\left(e^{\frac{i\pi}{5}}+e^{\frac{18i\pi}{5}}\right)\theta(1/5,1;5z)+\left(e^{\frac{3i\pi}{5}}+e^{\frac{4i\pi}{5}}\right)\theta(3/5,1,5z).\tag{12} $$ In the same way we get $$ \sum^{4}_{\nu=0}\theta\left(\frac{2\nu+1}{5},1,5z\right)=\left(1+e^{\frac{9\pi i}{5}}\right)\theta(1/5,1;5z)+\left(1+e^{\frac{7\pi i}{5}}\right)\theta(3/5,1;5z).\tag{13} $$ From (10),(11),(12) and (13) we get the result. $qed$

Notes 1

Actualy the ''constant'' term $\zeta=\zeta(q)=e^{2\pi i\nu(q)/5}$ in (8) and (10), is chainging along with the value of $Re(z)$. If $-1/2<Re(z)\leq 1/2$, then $\nu(q)=0$. When $1/2<Re(z)\leq 3/2$, $\nu(q)=1$, etc $\ldots$. In general if $k-1/2< Re(z)\leq k+1/2$, $k\in\textbf{Z}$, then $\nu(q)=k$. Hence $\nu(q)$ is the nearest integer to $Re(z)$ (with the further assuption that if $k\in\textbf{Z}$ and $Re(z)=k+\frac{1}{2}$, then $\nu(q)=k$).

Theorem 3

If $q_1=e(-1/z)$ and $q=e(z)$, $Im(z)>0$ and define $r_0(z):=\zeta(q) R(q)$ (according to (10)), then $$ r_0(z+1)=e^{2\pi i/5}r_0(z)\tag{14} $$ and $$ r_0(-1/z)=\frac{1-\phi r_0(z)}{\phi+r_0(z)}\tag{15} $$


For to prove (14) we use (6) in (10). The result follows easily. The second relation is immediate consequence of the definition of $r_0(z)$ and Theorem 2. $qed$

Deffinition 1

i) We denote with $\textbf{H}$ the upper half complex plane.

ii) We call fundamental domain $D$ the set $$ D=\{z\in\textbf{H}\textrm{ : }|Re(z)|\leq\frac{1}{2}\textrm{, }|z|\geq 1\}\tag{16} $$ iii) $$ \Gamma=\left\{\left( \begin{array}{cc} a\textrm{ }b\\ c\textrm{ }d \end{array} \right)\textrm{ : }a,b,c,d\in\textbf{Z}\textrm{, }ad-bc=1\right\}\tag{17} $$ iv) We call Moebious transformation$-\sigma$ the map $\sigma:\textbf{C}\rightarrow\textbf{C}$ such that $$ \sigma(z)=\frac{az+b}{cz+d}\textrm{, }ad-bc\neq0.\tag{18} $$

Notes 2 (see [Mor,Wag] pg.134-135)

The main properties of the fundamental domain $D$ are

1) For every $z\in\textbf{H}$, there is a $\sigma$ Moebious transform in $\Gamma$ such that $\sigma(z)\in D$,


2) If $z\in D$ and $\sigma\in \Gamma$ are such that $\sigma(z)\in D$ and $\sigma$ is not the identity, then either $Re(z)=\pm\frac{1}{2}$ and $\sigma(z)=z\pm 1$, or $|z|=1$ and $\sigma(z)=-1/z$.

Lemma 1

If $h$ is bounded and holomorphic function in $\textbf{H}$ such that $h(z+1)=h(z)$ and $h(-1/z)=h(z)$, for all $z$ in $\textbf{H}$, then $h$ is constant.


Since the function $h(z)$ is $1-$periodic in $\textbf{H}$, then exists function $g:\textbf{K}\rightarrow\textbf{C}$, where $\textbf{K}=\{z\in\textbf{C}:|z|<1\}$, such that $h(z)=g(q)$, $q=e(z)$, $Im(z)>0$. But $h(z)$ is bounded and holomorphic in $\textbf{H}$. Hence we conclude that $h$ it is regular at the origin. By subtracting the value $g(0)$ we may assume that $h(z)\rightarrow 0$ as $Im(z)\rightarrow\infty$. Now for any point $z\in\textbf{H}$ there is a $\sigma$ Moebius transformation in $\Gamma$ with $\sigma(z)\in D$, where $D$ is the fundamental domain. Assume that $|h(z)|$ attains its maximum at some point $z_0$ in $D$. This point is in $\partial D$ but not at infinity. Suppose $|h(z_0)|=M$ is the maximum where $z_0\in\partial D$. If $h(z)$ where not a constant, we could find another point $z_1$ in a neighborhood of $z$ and not in $D$ with $|h(z_1)|>M$. Hence for the Moebius map $\sigma$ we have $\sigma(z_1)\in D$ and $|h(z_1)|>M\geq |h(\sigma(z_1))|=|h(z_1)|$, which is a contradiction. This proves the lemma. $qed$

Lemma 2

If $h$ is bounded and holomorphic function in $\textbf{H}$, such that $h(z+T)=h(z)$, $T\in\textbf{N}:=\{1,2,3,\ldots\}$ and $h(-1/z)=1/h(z)$, for all $z\in \textbf{H}$, then $h$ is constant.


Suppose that $h(z)$ is $T-$periodic bounded not constant and holomorphic in $\textbf{H}$. Then exists function $g:\textbf{K}\rightarrow\textbf{C}$, such that $h(z)=g(q)=g\left(e^{2\pi i z/T}\right)$, $Im(z)>0$. Since $h(z)$ is bounded in $\textbf{H}$, then $h$ is bounded in $D$.
Also exists $M>0$ and $z_0$ in $\partial D$ ($z_0$ is the point in which $h(z)$ attends its maximum in $D$), such that $\frac{1}{M}=|h\left(\frac{-1}{z_0}\right)|\leq|h(z)|\leq |h(z_0)|= M$ for all $z\in D$, (since $h(-1/z)=1/h(z)$). Obviously the point $z_0$ is not at infinity. Also in the neighborhood of $z_0$ in the upper halph plane (not in $D$) exists a $z_1$ such that $|h(z_1)|>M$. But for the Moebious map $\sigma\in\Gamma$, $\sigma(z):\textbf{H}\rightarrow D$, we have $h\left(\sigma(z_1)\right)=\frac{1}{h(z_1)}$ and $|h(\sigma(z_1))|\geq\frac{1}{M}$, or equivalently $\frac{1}{|h(z_1)|}\geq\frac{1}{M}$, or equivalently $M<|h(z_1)|\leq M$. But this is contradiction. $qed$

Definition 2

We define the Dedekind eta function $\eta(z)$ from $\textbf{H}$ to $\textbf{C}$ as $$ \eta(z):=q^{1/24}f(-q)\textrm{, }q=e(z)\textrm{, }Im(z)>0,\tag{19} $$ where $$ f(-q):=\prod^{\infty}_{n=1}(1-q^n) $$

For $\eta$ function holds the following modular relation:

Theorem 4

$$ \eta(-1/z)=\sqrt{-iz}\cdot\eta(z)\textrm{, }Im(z)>0.\tag{20} $$

Theorem 5

Let $q=e(z)$, $Im(z)>0$, then $$ r_0(z)^{-1}-1-r_0(z)=\frac{\eta(z/5)}{\eta(5z)}\tag{21} $$


For $q_1=e(-1/z)$, $q=e(z)$, with $Im(z)>0$ we set $$ h(z):=\left(r_0(z)^{-1}-1-r_0(z)\right)\frac{\eta(5z)}{\eta(z/5)}. $$ Then $$ h(-1/z)=\left(r_0(-1/z)^{-1}-1-r_0(-1/z)\right)\frac{\eta(-5/z)}{\eta(-1/(5z))} $$ Using (15) and (20) we get after elementary algebraic evaluations $$ h(-1/z)=\frac{5r_0(z)}{1-r_0(z)-r_0(z)^2}\cdot\frac{\sqrt{-iz/5}\cdot\eta(z/5)}{\sqrt{-i5z}\cdot\eta(5z)}=1/h(z). $$ From (14) $h(z)$ is $5-$periodic i.e. $h(z+5)=h(z)$. The boundedness of $h$ follows from the fact that $r_0\left(x+it\right)$ has no poles or roots when $x\in\textbf{R}$, $0<t<\infty$ (this is from $|q|<1$ and (A)). The equation $r_0(z)^{-1}-1-r_0(z)=0$, $z=x+it$ is equivalent to $r_0(z)=\frac{-1\pm\sqrt{5}}{2}$ and this last has no roots. The function $\frac{\eta((x+it)/5)}{\eta(5(x+it))}$ for $x\in\textbf{R}$ and $t>0$, is continuous and have no roots and poles. Also we have $$ R(q)=q^{1/5}\left(1-q+O\left(q^2\right)\right)\textrm{, }q\rightarrow 0 $$ $$ R(q)^{-1}=q^{-1/5}\left(1+q+O\left(q^3\right)\right)\textrm{, }q\rightarrow 0 $$ and $$ \frac{f(-q^{1/5})}{q^{1/5}f(-q^5)}=q^{-1/5}-1-q^{1/5}+q^{4/5}+q^{6/5}+O\left(q^{11/5}\right)\textrm{, }q\rightarrow 0. $$ From these arguments we have the limit $$ \lim_{t\rightarrow +\infty}\left(r_0(x+it)^{-1}-1-r_0(x+it)\right)\frac{\eta\left(5(x+it)\right)}{\eta\left((x+it)/5\right)}=1. $$ Hence we conclude that $h(x+it)$ for $x\in\textbf{R}$ and $0<t<\infty$ is continuous and bounded. Using Lemma 2 we get that $h(z)$ is constant and $h(z)=1$. $qed$

Theorem 6

If $q=e(z)$, $Im(z)>0$, then $$ R(q)^{-5}-11-R(q)^5=r_0(z)^{-5}-11-r_0(z)^5=\left(\frac{\eta(z)}{\eta(5z)}\right)^6\tag{22} $$


Suppose that $q_1=e(-1/z)$, $q=e(z)$, $Im(z)>0$. Then set $$ A_1(z):=\left(r_0(z)^{-5}-11-r_0(z)^5\right)\left(\frac{\eta(5z)}{\eta(z)}\right)^6.\tag{23} $$ and $$ h_1(z):=r_0(z)^{-5}-11-r_0(z)^5.\tag{24} $$ Hence $$ A_1(z)=h_1(z)\left(\frac{\eta(5z)}{\eta(z)}\right)^6. $$ A simple algebraic calculation using (15) can show us that $$ h_1(-1/z)=h_1(z)\frac{125}{h_2(z)^6},\tag 25 $$ where from Theorem 5: $$ h_2(z)=r_0(z)^{-1}-1-r_0(z)=\frac{\eta(z/5)}{\eta(5z)}.\tag{26} $$ Hence from (20),(23),(25),(26) we get $$ \frac{A_1(-1/z)}{A_1(z)}=\frac{h_1(-1/z)}{h_1(z)}\frac{\eta(-1/(z/5))^6}{\eta(-1/z)^6}\frac{\eta(z)^6}{\eta(5z)^6}= $$ $$ =\frac{h_1(z)125}{h_1(z)h_2(z)^6}\frac{\eta(-1/(z/5))^6}{\eta(-1/z)^6}\frac{\eta(z)^6}{\eta(5z)^6}= $$ $$ =\frac{125}{h_2(z)^6}\frac{\left(\sqrt{-iz/5}\cdot\eta(z/5)\right)^6}{\left(\sqrt{-iz}\cdot\eta(z)\right)^6}\frac{\eta(z)^6}{\eta(5z)^6} =\frac{125\eta(5z)^6}{\eta(z/5)^6}\frac{\eta(z/5)^6}{125\eta(z)^6}\frac{\eta(z)^6}{\eta(5z)^6}=1. $$ Hence $$ A_1(-1/z)=A_1(z).\tag{27} $$ Observe also (easily) that $A_1(z)$ is $1-$periodic. Hence using Lemma 1 we get that $A_1(z)$ is constant and from this we easily get that $A_1(z)=1$, for all $z$ in the upper half plane. $qed$

Notes 3

Actualy there hold the following more general identities:

For all $|q|<1$, we have $$ \frac{1}{R(q)}-1-R(q)=\frac{f(-q^{1/5})}{q^{1/5}f(-q^5)} $$ and $$ \frac{1}{R(q)^5}-11-R(q)^5=\frac{f^6(-q)}{q f^6(-q^5)}. $$

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

[Du]: W. Duke. 'Continued fractions and Modular functions'. Bull. Amer. Math. Soc. (N.S.), 42 (2005), 137-162.

[Mor,Wag]: Carlos J. Moreno, Samuel S. Wagstaff. Jr. 'Sums of Quares of Integers'. Chapman and Hall/CRC, Taylor and Francis Group, (2006).

[FK]: H. Farkas and I. Kra, 'Theta Constants, Riemann surfaces and the modular group'. Amer. Math. Soc., Providence, RI, 2001.

Note. If there are any problems or gaps in the proof please let me know.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.