On Ramanujan's Question 359

In JIMS 4, p.78, Question 359 was asked by Ramanujan. (See The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society, p. 9, by Bruce Berndt, et al.) If,

$$\sin(x+y) = 2\sin\big(\tfrac{1}{2}(x-y)\big)\tag1$$

$$\sin(y+z) = 2\sin\big(\tfrac{1}{2}(y-z)\big)\tag2$$

prove that,

$$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}\tag3$$

The example by Ramanujan was,

$$\begin{aligned} x &= \frac{\pi-\arcsin\big((\sqrt{5}-2)^3(4+\sqrt{15})^2\big)}{2}=1.094\dots\\ y &=\frac{\arcsin\big(\sqrt{5}-2\big)}{2}=0.119\dots\\ z &=\frac{\arcsin\big((\sqrt{5}-2)^3(4-\sqrt{15})^2\big)}{2}=0.0001\dots \end{aligned}$$

Ten years later, a 3-page proof was given in JIMS 15, p.114-117.

I got an email asking if there was a shorter proof. Considering the problem, I observed the following. Given the quartic,

$$a^3w^4+(1-3a^2)w^3+3a(1-a^2)w^2+a^2(3-a^2)w-a=0\tag4$$ with $a=\tan(\color{blue}y/4)$. Define, $$x=4\tan^{-1} u\\z=4\tan^{-1} v$$ where $u,v$ are two appropriate roots of the quartic, then we get the same bizarre relation as Ramanujan,

$$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2\color{blue}y)^{1/12}\tag5$$

Equivalently, those two roots $u,v$ obey,

$$\left(\frac{1}{2}\,\frac{4(u-u^3)}{(u^2+1)^2}\frac{v^4-6v^2+1}{(v^2+1)^2} \right)^{1/4}+\left(\frac{1}{2}\,\frac{4(v-v^3)}{(v^2+1)^2}\frac{u^4-6u^2+1}{(u^2+1)^2} \right)^{1/4}=\big(\sin 2\color{blue}y\big)^{1/12}\tag6$$

For example, let $y=1$, so $a=\tan(1/4)$, then $u,v$ are the two real roots of (4).

Question: Anyone knows a short proof for (3) and (6)?

Solution 1:

Thanks to a deleted answer by G. Manco, we finally have a partial answer after four years to a closed-form solution $x,y,z$ to, $$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}\tag3$$ in terms of the Ramanujan G function. Using the Dedekind eta function $\eta(\tau)$, this can be calculated in Mathematica as, $$G_n=\frac{2^{-1/4}\,\eta^2(\sqrt{-n})}{\eta\big(\tfrac{\sqrt{-n}}{2}\big)\,\eta(2\sqrt{-n})}\tag4$$ However, because of the constraints $(1),(2)$ in the post, there is a difference when $1\leq n\leq9$ and $n\geq9$. Thus,

If $1\leq n\leq9$: $$\begin{aligned} x &=\pi/2-\tfrac12\arcsin G_{n/9}^{-12}\\ y &=\tfrac12\arcsin G_{n}^{-12}\\ z &=\tfrac12\arcsin G_{9n}^{-12} \end{aligned}$$ If $n\geq9$: $$\begin{aligned} x &=\tfrac12\arcsin G_{n/9}^{-12}\\ y &=\tfrac12\arcsin G_{n}^{-12}\\ z &=\tfrac12\arcsin G_{9n}^{-12} \end{aligned}$$ We recover Ramanujan's solution $x,y,z \approx 1.094,\; 0.119,\; 0.0001$ by plugging in $n=5$ to the first set of formulas. Also, since,

$$\sin(\tfrac12\arcsin \lambda)=\tfrac12\big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\big)$$ $$\cos(\tfrac12\arcsin \lambda)=\tfrac12\big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\big)$$ then $(3)$ can be expressed in radical form,

$$\left(\tfrac12\Big(\tfrac{\sqrt{1+\alpha}+\sqrt{1-\alpha}}2\Big)\Big(\tfrac{\sqrt{1+\beta}\,\color{blue}\pm\,\sqrt{1-\beta}}2\Big)\right)^{1/4}+\left(\tfrac12\Big(\tfrac{\sqrt{1+\alpha}-\sqrt{1-\alpha}}2\Big)\Big(\tfrac{\sqrt{1+\beta}\,\color{blue}\mp\,\sqrt{1-\beta}}2\Big)\right)^{1/4}=\gamma^{1/12}$$ where $\alpha = G_{n/9}^{-12},\;\beta= G_{9n}^{-12},\;\gamma= G_{n}^{-12}$ and the $+,-$ case for $n\leq9$ and $-,+$ case for $n\geq9$.

P.S. The second part of the question is still a bit tricky to prove.

Solution 2:

An elegant solution to question 359 submitted by Ramanujan to Journal of the Indian Mathematical Society.

The three equations (1)-(3) are three modular equations: the first two are of third degree and the last one of ninth degree. These equations are in Ramanujan's notebooks: http://www.imsc.res.in/~rao/ramanujan/index.html.

If $\alpha$ be of the $1^{st}$, $\beta$ the $3^{rd}$ and $\gamma$ the $9^{th}$ degree then.

$\sqrt{\alpha(1-\beta)}$+$\sqrt{\beta(1-\alpha)}$=$2(\alpha\beta(1-\alpha)(1-\beta))^\frac {1} {8}$$\tag1$

$\sqrt{\gamma(1-\beta)}$+$\sqrt{\beta(1-\gamma)}$=$2(\beta \gamma (1-\beta)(1-\gamma))^\frac{1} {8}$$\tag2$

and

$(\alpha(1-\gamma))^\frac{1}{8}$+ $(\gamma(1-\alpha))^\frac{1}{8}$=$2^\frac{1}{ 3}$($\beta(1-\beta))^\frac{1}{24}$ $\tag3$

Here is my interpretation. The reading key is:

$\sqrt{\alpha}=\sin{x}$, $\sqrt{\beta}=\sin{y}$, and $\sqrt{\gamma}=\sin{z}$

moreover

$\sqrt{1-\alpha}=\cos x$, $\sqrt{1-\beta}=\cos y$ and $\sqrt{1-\gamma}=\cos z$.

and with class invariants

$4\alpha(1-\alpha)G_{\frac n 9}^{24}=1$, $4\beta(1-\beta)G_{n}^{24}=1$, and $4\gamma(1-\gamma)G_{9n}^{24}=1$

or

$\sin{2x}.G_{\frac {n} {9}}^{12}=1$ , $\sin{2y}.G_{n}^{12}=1$ , and $\sin{2z}.G_{9n}^{12}=1$.

For $n=5$ we have

$\sin 2x=2\sqrt{\alpha(1-\alpha)} =G_{\frac {5} {9}}^{-12}$=$((\sqrt 5+2)^{\frac 1 4}(4-\sqrt{15})^{\frac 1 6})^{-12}=(\sqrt 5-2)^3(4+\sqrt 15)^2$

$\sin 2y=2\sqrt{\beta(1-\beta)}=G_{5}^{-12}=\big((\frac {\sqrt 5+1}{2})^{\frac1 4}\big)^{-12}=(\sqrt 5-2)$

$\sin 2z=2\sqrt{\gamma(1-\gamma)}=G_{45}^{-12}=((\sqrt 5+2)^{\frac 1 4}(4+\sqrt{15})^{\frac 1 6})^{-12}=(\sqrt 5-2)^3(4-\sqrt 15)^2$

$(\frac 1 2 \sin x \cos z)^{\frac 1 4}=\frac{ 1} {\sqrt 2}(\sqrt 5-2)^{\frac 3 4}\big(\sqrt \frac {7+3\sqrt 5}{4}+\sqrt \frac {3+3\sqrt 5}{4}\big)$

$(\frac 1 2 \sin z \cos x)^{\frac 1 4}=\frac{ 1} {\sqrt 2}(\sqrt 5-2)^{\frac 3 4}\big(\sqrt \frac {7+3\sqrt 5}{4}-\sqrt \frac {3+3\sqrt 5}{4}\big)$

adding we have

$\frac{ 1} {\sqrt 2}(\sqrt 5-2)^{\frac 3 4}\big(2\sqrt \frac {7+3\sqrt 5}{4}\big)=(\sqrt 5-2)^{\frac 1 {12}}=(\frac {\sqrt 5-1}{2})^{\frac1 4}$

This result is also equal to

$(\sin 2y )^{\frac {1} {12}}=\frac {1} {G_{5}}=(\frac {\sqrt 5-1}{2})^{\frac{1} {4}}$

I calculated several solutions for $n=1,1/2,2,3,1/3,4,1/4,5,1/5,6,1/6,7,1/7,8,1/8,9,1/9,10,1/10,15,1/15.$

For $n=12$ we have

$\sin x$=$(\sqrt{2}+1)^2(\sqrt{3}-\sqrt{2})^2$

$\sin y$=$(\sqrt{2}-1)^2(\sqrt{3}-\sqrt{2})^2$

$\sin z$=$\frac{2-\sqrt{2+\sqrt{\delta}}} {2+\sqrt{2+\sqrt{\delta}}}$

with

$\delta=3-100.2^{\frac{1} {3}}+80.2^{\frac{2} {3}}$.