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It is well known that $$\operatorname{R}(-e^{-\pi})=-\cfrac{e^{-\frac{\pi}{5}}}{1-\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1-\cfrac{e^{-3\pi}}{1+\ddots}}}}=\frac{\sqrt{5}-1}{2}-\sqrt{\frac{5-\sqrt{5}}{2}}$$ where $\operatorname{R}$ is the Rogers-Ramanujan continued fraction: $$\operatorname{R}(q)=\cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}},\, q=e^{\pi i\tau}.$$

But I'm interested in $\operatorname{R}(e^{-\pi})$. Numerically, I checked that it agrees with the root of the following octic equation near $x=\frac{1}{2}$ to at least $16$ decimal places: $$x^8+14x^7+22x^6+22x^5+30x^4-22 x^3+22 x^2-14x+1=0;$$ the root turns out to be equal to $$\frac{\sqrt{5}-1}{2}\frac{\sqrt[4]{5}+\sqrt{2+\sqrt{5}}}{\sqrt{5}+\sqrt{2+\sqrt{5}}}.$$

So is it true that $$\frac{e^{-\frac{\pi}{5}}}{1+\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1+\cfrac{e^{-3\pi}}{1+\ddots}}}}=\frac{\sqrt{5}-1}{2}\frac{\sqrt[4]{5}+\sqrt{2+\sqrt{5}}}{\sqrt{5}+\sqrt{2+\sqrt{5}}}?$$ The $2$'s and $5$'s under the square roots seem very suggestive of the nature of $\operatorname{R}$.

Also, how could it be proved in that case?

Using $$\frac{1}{\operatorname{R}(q)}-\operatorname{R}(q)=\frac{\left(q^{\frac{1}{5}};q^{\frac{1}{5}}\right)_{\infty}}{q^{\frac{1}{5}}(q^5;q^5)_{\infty}}+1$$ and $$\frac{\eta (e^{-\pi\sqrt{n}})}{\eta \left(e^{-\frac{\pi}{\sqrt{n}}}\right)}=n^{-\frac{1}{4}},\, n\gt 0$$ (where $\eta (q)=q^{\frac{1}{12}}\prod_{n\ge 1}(1-q^{2n})$ is the Dedekind eta function), I've been able to evaluate $\operatorname{R}(-e^{-\pi})$ and $\operatorname{R}(e^{-2\pi})$, but I don't know how could it be used to evaluate $\operatorname{R}(e^{-\pi})$. Perhaps something else is necessary.

I was inspired by Ramanujan's first letter to Hardy, where Ramanujan's $7\text{th}$ theorem states that $$\cfrac{1}{1+\cfrac{e^{-\pi\sqrt{n}}}{1+\cfrac{e^{-2\pi\sqrt{n}}}{1+\cfrac{e^{-3\pi\sqrt{n}}}{1+\ddots}}}}$$ can be exactly found for any $n\in\mathbb{Q}^{+}$.

Poder Rac
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    How did you get that octic equation? – eyeballfrog Sep 21 '20 at 19:42
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    https://faculty.math.illinois.edu/~berndt/articles/rrcf.pdf seems to be relevant, in particular, Eq. (4.5), Th. 4.5 and Ref. [36]. – Start wearing purple Sep 21 '20 at 20:03
  • $R(q)$ is only defined if $|q|<1$ which implies that $R(e^{2\pi})$ probably is a typo for $R(e^{-2\pi})$. You also have to be careful in $q^{\frac15}$ to use the correct branch. – Somos Sep 21 '20 at 21:43
  • There were roots of polynomials of lower degrees as well but the coefficients were too large (e.g. $\gt 500$). I was trying to find a polynomial with coefficients as small as possible. – Poder Rac Sep 22 '20 at 00:14
  • @Somos You're right, it was a typo. Corrected. – Poder Rac Sep 22 '20 at 00:15
  • @Somos Here, $q^{\frac{1}{5}}$ is simply $e^{\frac{\pi i\tau}{5}}$. – Poder Rac Sep 22 '20 at 00:27
  • @eyeballfrog This was the first "nice" polynomial I found. In the case of lower degree polynomials & their respective roots, either the coefficients were too large, or they were not solvable in radicals, or both. – Poder Rac Sep 22 '20 at 01:04
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    Probably irrelevant to the main question, but due to its skew-reciprocal (is that a term?) nature the octic, call it $f(x)$, can be rewritten in terms of the variable $u=x-\dfrac1x$ as $$f(x)=x^4(u^4+14u^3+26u^2+64u+76).$$ Only relevant if you want to find all the zeros - and identify them in terms of continued fractions and/or modular functions :-). Anyway the Galois group of that quartic is dihedral. – Jyrki Lahtonen Sep 22 '20 at 05:21
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    Tidied up a bit as: $$\frac{e^{-\pi/5}}{1+\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1+\cfrac{e^{-3\pi}}{1+\ddots}}}}\stackrel{?}{=}\frac{\sqrt{5}-1}{2}\cdot\frac{\sqrt[4]{5}+\sqrt{2+\sqrt{5}}}{\sqrt{5}+\sqrt{2+\sqrt{5}}}$$ – Mourad Sep 22 '20 at 05:27

2 Answers2

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$R(q)$ is expressible in terms of radicals for every $q=e^{2\pi i \tau}$, where $\tau$ is an imaginary quadratic irrational in the upper half plane. OP already observed it for $R(e^{-\pi\sqrt{n}})$. It is easy to obtain an algebraic equation satisfied by $R(q)$, but unwinding into radical form is more difficult.


The Roger-Ramanujan continued fraction has many equivalent forms:$$R(q) = \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}} = q^{1/5} \prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}.$$

Writing $q=e^{2\pi i \tau}$. From the latter expression, it can be shown that $R$ is modular of level $5$, therefore $R$ and the $j$-invariant has an algebraic relation of degree $[\bar{\Gamma}(1):\bar{\Gamma}(5)] = 60$ (bar means quotienting out center): $$\tag{*} R^5 (R^{10}+11 R^5-1)^5j+(R^{20}-228 R^{15}+494 R^{10}+228 R^5+1)^3 = 0.$$

You are concerned with $R(e^{-\pi})$. Since $j(i/2) = 287496$, $R$ is a root of degree $60$ equations over $\mathbb{Z}$. $(*)$ factors over $\mathbb{Q}$, the octic $$R^8+14R^7+22R^6+22R^5+30R^4-22 R^3+22 R^2-14R+1$$ appears as one of the factors. Numerical evaluation will tell you that $R(e^{-\pi})$ is indeed a root of this factor, completing the proof.

The other $59$ roots of $(*)$ are values of $R$ at $\Gamma(1)$-orbit of $i/2$ which are $\bar{\Gamma}(5)$-inequivalent.

pisco
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  • Is there an elementary proof of the relation between $R$ and $j$ available? – Paramanand Singh Jan 25 '21 at 03:17
  • @ParamanandSingh I am not sure what you mean by "elementary". The most mechanical way is to note that $R^5 j$ is modular of level $5$, and has $q$-expansion: $1+O(q)$. Then $R^5 j = f(R^5)/g(R^5)$ for some polynomial $f,g$ of degree $\leq 12$. They can be then found via for example Hensel's lemma by looking at $q$-expansion. There is also smarter but *ad hoc* argument using geometry of icosahedron. – pisco Jan 26 '21 at 08:27
  • Well I meant a proof which just uses algebraic manipulation. For example I know how the j invariant is related to Eisenstein series. And I know how the $R$ is related to the eta functions (see my answer here). There should be some way to use a little or more of algebra and get the relation between $j$ and $R$. I am not well versed in complex analysis or the algebra of groups/rings etc. – Paramanand Singh Jan 26 '21 at 08:33
  • I also checked Wikipedia which gives the following $$R^{-5}-R^5-11=\left(\frac{\eta(q)} {\eta(q^5)}\right)^6$$ and $$j=\frac{(x^2+10x+5)^3}{x}$$ where $$x=125\left(\frac{\eta(q^5)}{\eta(q)}\right)^6$$ So I guess my problem is reduced to a proof of the relation between $j$ and $x$. – Paramanand Singh Jan 26 '21 at 08:39
  • Maybe I will ask a separate question with all the details. – Paramanand Singh Jan 26 '21 at 08:41
  • @ParamanandSingh Given $j$, how can I solve $j=\frac{(x^2+10x+5)^3}{x}$ for $x$? – Nomas Aug 14 '21 at 19:38
  • @ParamanandSingh For example, for $j$ with $\tau =2i$ I get $x^6+30x^5+315x^4+1300x^3+1575x^2-286746x+125=0$... How can this sextic be solved algebraically? – Nomas Aug 14 '21 at 20:05
  • @Nomas: usually some computer algebra packages are needed to get explicit solutions. But the approach by Ramanujan is more direct. If $\tau=2i$ then $q=e^{-4\pi}$ and values of $R(e^{-\pi}), R(e^{-4\pi})$ have a simple algebraic relation between. Using that we can find the desired value of $R(e^{-4\pi})$. See details in my answer. – Paramanand Singh Aug 15 '21 at 03:26
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Another approach is to use the identity $$\frac{1}{R(q)}-1-R(q)=\frac{f(-q^{1/5})}{q^{1/5}f(-q^5)}\tag {1}$$ where $$f(-q) =\prod_{n=1}^{\infty} (1-q^n)\tag{2}$$ Also note that the above is essentially Dedekind eta function $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty}(1-q^n)\tag{3}$$ Thus in order to evaluate $R(q) $ for $q=e^{-\pi} $ we need the values of $\eta(e^{-\pi/5})$ and $\eta(e^{-5\pi})$. In fact what is needed is the ratio of their values and not the individual values.

We can note that if $k$ is the elliptic modulus corresponding to $q$ and $K$ is the corresponding complete elliptic integral of first kind then $$\eta(q) =2^{-1/6}\sqrt {\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{4}$$ Also it should be noted that if $k$ is replaced by $k'$ then $K$ is replaced by $K'$ and $q$ is replaced by $q'$ where $(\log q) (\log q') =\pi^2$. Clearly the values $q=e^{-\pi/5},q'=e^{-5\pi}$ satisfy this equation and hence using a similar equation for $\eta(q') $ we get $$\frac{\eta(q)} {\eta(q')} =\sqrt{\frac{K} {K'}} (k'/k) ^{1/4}$$ and $\pi K'/K=-\log q=\pi/5$. Thus we get $$\frac{\eta(q)} {\eta(q')} =\sqrt{5}\cdot(k'/k)^{1/4}$$ where $k$ is the modulus corresponding to $q=e^{-\pi/5}$. It is well known from the value of Ramanujan class invariant $G_{25}=\phi$ (golden ratio) that $$2k'^2=1-\sqrt{1-\phi^{-24}}=1-12\sqrt{161\sqrt{5}-360}$$ and $$2k^2=1+\sqrt {1-\phi^{-24}}=1+12\sqrt{161\sqrt{5}-360}$$ and hence $$\frac{\eta(e^{-\pi/5})} {\eta(e^{-5\pi})} =\sqrt{5}\left(\frac{1-12\sqrt{161\sqrt{5}-360}}{1+12\sqrt{161\sqrt{5}-360}}\right)^{1/8}=A\text{(say)}\tag{5}$$ It should now be observed that if $q=e^{-\pi}$ then the value of right hand side of equation $(1)$ is $A$. Thus $R(e^{-\pi}) $ is the root of the equation $$\frac{1}{x}-1-x=A$$ or $$x^2+(A+1)x-1=0$$ so that $$R(e^{-\pi}) =x=\frac{-(A+1)+\sqrt{(A+1)^2+4}}{2}$$ where $A$ is given by equation $(5)$.

It requires some patience to get a simple expression for the above root though. Fortunately Ramanujan provided a denesting formula which helps us here. I reproduce it from one of my previous questions here: $$ \sqrt[8]{1\pm\sqrt{1-\phi^{-24}}} = \frac{-1+\sqrt{5}}{2}\cdot \frac{\sqrt[4]{5}\pm 1}{\sqrt{2}}\tag{6}$$ and thus $$A=\sqrt{5}\cdot\frac{\sqrt[4]{5}-1}{\sqrt[4]{5}+1}$$ It is now easy to verify the desired form for $x$ using a little symbolic algebra with the symbol $a$ given $a>0,a^4=5$.

Another curious fact is that if we change the value of $A$ to $A'=5/A$ then the corresponding $x$ gives the value of $R(e^{-4\pi})$. This is generalized by Ramanujan as a relation between the values of $R(e^{-2\alpha}) $ and $R(e^{-2\beta})$ where $\alpha, \beta$ are positive with $\alpha\beta=\pi^2$: $$\left(\frac{\sqrt {5}+1}{2}+R(e^{-2\alpha}) \right) \left(\frac{\sqrt{5}+1}{2}+R(e^{-2\beta})\right)=\frac{5+\sqrt{5}}{2}\tag{7}$$

Paramanand Singh
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