Some double-angle formulas related to a generalized lemniscate function

Abstract

In this paper, we will establish some double-angle formulas related to the inverse function of $\int _0^x \text {d}t/\sqrt{1-t^6}$. This function appears in Ramanujan’s Notebooks and is regarded as a generalized version of the lemniscate function.

1 Introduction

Let $1<p,\ q<\infty$ and

$$\begin{aligned} F_{p,q}(x):=\int _0^x \frac{\text {d}t}{(1-t^q)^{1/p}}, \quad x \in [0,1]. \end{aligned}$$

We will denote by $\sin _{p,q}$ the inverse function of $F_{p,q}$, i.e.,

$$\begin{aligned} \sin _{p,q}{x}:=F_{p,q}^{-1}(x). \end{aligned}$$

Clearly, $\sin _{p,q}{x}$ is an increasing function mapping $[0,\pi _{p,q}/2]$ to [0, 1], where

$$\begin{aligned} \pi _{p,q}:=2F_{p,q}(1)=2\int _0^1 \frac{\text {d}t}{(1-t^q)^{1/p}}. \end{aligned}$$

We extend $\sin _{p,q}{x}$ to $(\pi _{p,q}/2,\pi _{p,q}]$ by $\sin _{p,q}{(\pi _{p,q}-x)}$ and to the whole real line $\mathbb {R}$ as the odd $2\pi _{p,q}$-periodic continuation of the function. Since $\sin _{p,q}{x} \in C^1(\mathbb {R})$, we also define $\cos _{p,q}{x}$ by $\cos _{p,q}{x}:=(\text {d}/\text {d}x)(\sin _{p,q}{x})$. Then, it follows that

$$\begin{aligned} |\cos _{p,q}{x}|^p+|\sin _{p,q}{x}|^q=1. \end{aligned}$$

In case $(p,q)=(2,2)$, it is obvious that $\sin _{p,q}{x},\ \cos _{p,q}{x}$ and $\pi _{p,q}$ are reduced to the ordinary $\sin {x},\ \cos {x}$ and $\pi$, respectively. This is a reason why these functions and the constant are called generalized trigonometric functions (with parameter (p, q)) and the generalized $\pi$, respectively.

The generalized trigonometric functions are well studied in the context of nonlinear differential equations (see [4, 6, 7] and the references given there). Suppose that u is a solution of the initial value problem of the p-Laplacian

$$\begin{aligned} -(|u'|^{p-2}u')'=\frac{(p-1)q}{p} |u|^{q-2}u, \quad u(0)=0,\ u'(0)=1, \end{aligned}$$

which is reduced to the equation $-u''=u$ of simple harmonic motion for $u=\sin {x}$ in case $(p,q)=(2,2)$. Then,

$$\begin{aligned} \frac{d}{dx}(|u'|^p+|u|^q)=\left( \frac{p}{p-1}(|u'|^{p-2}u')'+q|u|^{q-2}u\right) u'=0. \end{aligned}$$

Therefore, $|u'|^p+|u|^q=1$. It is possible to show that u coincides with $\sin _{p,q}$ defined as above. The generalized trigonometric functions are often applied to the eigenvalue problem of the p-Laplacian.

Now, we are interested in finding double-angle formulas for generalized trigonometric functions. It is possible to discuss addition formulas for these functions: for instance $\sin _{2,6}$ has the addition formula (3) with (2) below (see also [5] for $\sin _{4/3,4}$), but for simplicity we will not develop this point here.

We have known the double-angle formulas of $\sin _{2,q},\ \sin _{q^*,q}$, and $\sin _{q^*,2}$ for $q=2,3,4$ except for $\sin _{3/2,2}$, where $q^*:=q/(q-1)$ (Table 1). For details for each formula, we refer the reader to [8] (after having proved the formula for $\sin _{2,3}$ in the co-authored paper [8], the author noticed that the formula has already been obtained as “$\varphi (2s)$” by Cox and Shurman [3, p. 697]). It is worth pointing out that Lemma 3.1 (resp. Lemma 3.2) below connects the parameter (2, q) to $(q^*,q)$ (resp. $(q^*,2)$) and yields the possibility to obtain the other formula from one formula. Indeed, in this way, the formulas of $\sin _{4/3,4}$ and $\sin _{4/3,2}$ follow from that of $\sin _{2,4}$ ([10, Subsect. 3.1] and [8, Theorem 1.1], respectively), and the formula of $\sin _{2,3}$ follows from that of $\sin _{3/2,3}$ ([8, Theorem 1.2]). Nevertheless, the parameter (3/2, 2) is still open because of the difficulty of the inverse problem corresponding to (10).

In this paper, we wish to investigate the double-angle formula of the function $\sin _{2,6}{x}$, whose inverse function is defined as

$$\begin{aligned} \sin _{2,6}^{-1}{x}=\int _0^x \frac{\text {d}t}{\sqrt{1-t^6}}. \end{aligned}$$

The function $\sin _{2,6}{x}$ appears as the inverse of “H(v)” in Ramanujan’s Notebooks [1, p. 246] and is regarded as a generalized version of the lemniscate function $\sin _{2,4}{x}$. For the function, Shinohara [9] gives the novel double-angle formula

$$\begin{aligned} \sin _{2,6}{(2x)}=\frac{2\sin _{2,6}{x}\cos _{2,6}{x}}{\sqrt{1+8\sin _{2,6}^6{x}}}, \quad x \in [0,\pi _{2,6}/2]. \end{aligned}$$

(1)

In fact, he found (1) in “trial and error calculations” (according to private communication), but instead we will give a proof of (1) in Sect. 2. Moreover, as mentioned above, we can show the following counterparts of (1) for $\sin _{6/5,6}$ and $\sin _{6/5,2}$, respectively.

Theorem 1.1

Let $(p,q)=(6/5,6)$. Then, for $x \in [0,\pi _{6/5,6}/4]$,

$$\begin{aligned}&\sin _{6/5,6}{(2x)}\\&\quad =\frac{2^{1/6}\sin _{6/5,6}{x}\cos _{6/5,6}^{1/5}{x}\left( 3+\sqrt{1+32\sin _{6/5,6}^6{x} \cos _{6/5,6}^{6/5}{x}}\right) ^{1/2}}{\left( 1+32\sin _{6/5,6}^6{x}\cos _{6/5,6}^{6/5}{x}\right) ^{1/4} \left( 1+\sqrt{1+32\sin _{6/5,6}^6{x}\cos _{6/5,6}^{6/5}{x}}\right) ^{1/6}}. \end{aligned}$$

Theorem 1.2

Let $(p,q)=(6/5,2)$. Then, for $x \in [0,\pi _{6/5,2}/2]$,

$$\begin{aligned} \sin _{6/5,2}{(2x)} =\sqrt{1-\left( \frac{9-8\sin _{6/5,2}^2{x}-4\sin _{6/5,2}^2{x}\cos _{6/5,2}^{2/5}{x}}{9-8\sin _{6/5,2}^2{x}+8\sin _{6/5,2}^2{x}\cos _{6/5,2}^{2/5}{x}}\right) ^3}. \end{aligned}$$

Table 1 The parameters for which the double-angle formulas have been obtained

2 Proof of (1)

The change of variable $s=t^2$ leads to the representation

$$\begin{aligned} \sin _{2,6}^{-1}{x}=\frac{1}{2}\int _0^{x^2}\frac{\text {d}s}{\sqrt{s(1-s^3)}},\quad 0 \le x \le 1. \end{aligned}$$

The furthermore change of variable ([2, 576.00 in p. 256])

$$\begin{aligned} {\text {cn}}{u}=\frac{1-(\sqrt{3}+1)s}{1+(\sqrt{3}-1)s}, \quad k^2=\frac{2-\sqrt{3}}{4} \end{aligned}$$

gives

$$\begin{aligned} \sin _{2,6}^{-1}{x}&=\frac{1}{2}\int _0^{{\text {cn}}^{-1}{\phi (x)}} \frac{((\sqrt{3}+1)+(\sqrt{3}-1){\text {cn}}{u})^2}{2\cdot 3^{3/4}{\text {sn}}{u}{\text {dn}}{u}}\\&\qquad \times \frac{2\sqrt{3}{\text {sn}}{u}{\text {dn}}{u}}{((\sqrt{3}+1)+(\sqrt{3}-1){\text {cn}}{u})^2}\,\text {d}u\\&=\frac{1}{2\cdot 3^{1/4}}\int _0^{{\text {cn}}^{-1}{\phi (x)}}\,\text {d}u\\&=\frac{1}{2\cdot 3^{1/4}} {\text {cn}}^{-1}{\phi (x)}, \end{aligned}$$

where ${\text {sn}}{u}={\text {sn}}{(u,k)},\ {\text {cn}}{u}={\text {cn}}{(u,k)}$, and ${\text {dn}}{u}={\text {dn}}{(u,k)}$ are the Jacobian elliptic functions (see e.g., [11, Chap. XXII] for more details), and

$$\begin{aligned} \phi (x)=\frac{1-(\sqrt{3}+1)x^2}{1+(\sqrt{3}-1)x^2}. \end{aligned}$$

(2)

Thus,

$$\begin{aligned} \sin _{2,6}{u}=\phi ^{-1}{({\text {cn}}{(2\cdot 3^{1/4}u)})}, \quad 0 \le u \le \pi _{2,6}/2=K/(3^{1/4}), \end{aligned}$$

where $K=K(k)$ is the complete elliptic integral of the first kind and

$$\begin{aligned} \phi ^{-1}(x)=\sqrt{\frac{1-x}{(\sqrt{3}+1)+(\sqrt{3}-1)x}}. \end{aligned}$$

Now, we use the addition formula of ${\text {cn}}$. For $u,\ v,\ u \pm v \in [0,K/(3^{1/4})]$,

$$\begin{aligned} \sin _{2,6}{(u \pm v)}=\phi ^{-1}{({\text {cn}}{(\tilde{u} \pm \tilde{v})})}=\phi ^{-1}{\left( \frac{{\text {cn}}{\tilde{u}}{\text {cn}}{\tilde{v}} \mp {\text {sn}}{\tilde{u}}{\text {sn}}{\tilde{v}}{\text {dn}}{\tilde{u}}{\text {dn}}{\tilde{v}}}{1-k^2{\text {sn}}^2{\tilde{u}}{\text {sn}}^2{\tilde{v}}}\right) } \end{aligned}$$

where $\tilde{u}:=2\cdot 3^{1/4}u$ and $\tilde{v}:=2\cdot 3^{1/4}v$. Recall that ${\text {sn}}^2{x}+{\text {cn}}^2{x}=1$ and $k^2{\text {sn}}^2{x}+{\text {dn}}^2{x}=1$; then the last equality gives

$$\begin{aligned}&\sin _{2,6}{(u \pm v)}\nonumber \\&\quad =\phi ^{-1}{\left( \frac{\phi (U)\phi (V) \mp \sqrt{(1-\phi (U)^2)(1-\phi (V)^2)(1-k^2(1-\phi (U)^2))(1-k^2(1-\phi (V)^2))}}{1-k^2(1-\phi (U)^2)(1-\phi (V)^2)}\right) },\nonumber \\ \end{aligned}$$

(3)

where $U:=\sin _{2,6}{u}$ and $V:=\sin _{2,6}{v}$.

With $u=v$ and the observation that

$$\begin{aligned} 1-\phi (U)^2=\frac{4\sqrt{3}U^2(1-U^2)}{(1+(\sqrt{3}-1)U^2)^2},\\ 1-k^2(1-\phi (U)^2)=\frac{1+U^2+U^4}{(1+(\sqrt{3}-1)U^2)^2}, \end{aligned}$$

this implies that

$$\begin{aligned} \sin _{2,6}{(2u)}&=\phi ^{-1}{\left( \frac{\phi (U)^2-(1-\phi (U)^2)(1-k^2(1-\phi (U)^2))}{1-k^2(1-\phi (U)^2)^2}\right) }\\&=\phi ^{-1}{\left( \frac{1-4(\sqrt{3}+1)U^2+8U^6+4(\sqrt{3}+1)U^8}{1+4(\sqrt{3}-1)U^2+8U^6-4(\sqrt{3}-1)U^8}\right) }. \end{aligned}$$

Routine simplification now results in the formula

$$\begin{aligned} \sin _{2,6}{(2u)} =\frac{2U\sqrt{1-U^6}}{\sqrt{1+8U^6}} =\frac{2\sin _{2,6}{u}\cos _{2,6}{u}}{\sqrt{1+8\sin _{2,6}^6{u}}}, \end{aligned}$$

and the proof is complete.

3 Proofs of theorems

To prove Theorem 1.1, we use the following multiple-angle formulas.

Lemma 3.1

([10]) Let $1<q<\infty$ and $q^*:=q/(q-1)$. If $x \in [0,\pi _{2,q}/(2^{2/q})]=[0,\pi _{q^*,q}/2]$, then

$$\begin{aligned} \sin _{2,q}{(2^{2/q}x)}&=2^{2/q}\sin _{q^*,q}{x}\cos _{q^*,q}^{q^*-1}{x}, \end{aligned}$$

(4)

$$\begin{aligned} \cos _{2,q}{(2^{2/q} x)}&=\cos _{q^*,q}^{q^*}{x}-\sin _{q^*,q}^q{x} \nonumber \\&=1-2\sin _{q^*,q}^q{x}=2\cos _{q^*,q}^{q^*}{x}-1. \end{aligned}$$

(5)

Proof of Theorem 1.1

Let $x \in [0,\pi _{6/5,6}/4]$. Applying (4) of Lemma 3.1 in case $q=6$ with x replaced by $2x \in [0,\pi _{6/5,6}/2]$, we get

$$\begin{aligned} \sin _{2,6}{(2\cdot 2^{1/3}x)}=2^{1/3}\sin _{6/5,6}{(2x)}(1-\sin _{6/5,6}^6{(2x)})^{1/6}. \end{aligned}$$

(6)

First, we consider the case

$$\begin{aligned} 0 \le x <\frac{\pi _{6/5,6}}{8}. \end{aligned}$$

Then, since $0 \le 2\sin _{6/5,6}^6{(2x)}<1$ by [10, Lemma 2.1], Eq. (6) gives

$$\begin{aligned} 2\sin _{6/5,6}^6{(2x)} =1-\sqrt{1-\sin _{2,6}^6{(2\cdot 2^{1/3}x)}}. \end{aligned}$$

Set $S=S(x):=\sin _{2,6}{(2^{1/3}x)}$. Using the double-angle formula (1) for $\sin _{2,6}{x}$, we have

$$\begin{aligned} 2\sin _{6/5,6}^6{(2x)}&=1-\sqrt{1-\left( \frac{2S\sqrt{1-S^6}}{\sqrt{1+8S^6}}\right) ^6}\\&=1-\frac{\sqrt{1-40S^6+384S^{12}+320S^{18}+64S^{24}}}{(1+8S^6)^{3/2}}\\&=1-\frac{|1-20S^6-8S^{12}|}{(1+8S^6)^{3/2}}. \end{aligned}$$

Since $0 \le S^6 <\sin _{2,6}^6{(\pi _{2,6}/4)}=(3\sqrt{3}-5)/4$, evaluated by (1), we see that $1-20S^6-8S^{12}>0$. Thus,

$$\begin{aligned} 2\sin _{6/5,6}^6{(2x)}&=1-\frac{1-20S^6-8S^{12}}{(1+8S^6)^{3/2}} \nonumber \\&=\frac{(\sqrt{1+8S^6}-1)(\sqrt{1+8S^6}+3)^3}{8(1+8S^6)^{3/2}} \nonumber \\&=\frac{S^6(3+\sqrt{1+8S^6})^3}{(1+8S^6)^{3/2}(1+\sqrt{1+8S^6})}. \end{aligned}$$

(7)

Therefore, by (4),

$$\begin{aligned} \sin _{6/5,6}{(2x)} =\frac{2^{1/6}\sin _{6/5,6}{x}\cos _{6/5,6}^{1/5}{x}\left( 3+\sqrt{1+32\sin _{6/5,6}^6{x} \cos _{6/5,6}^{6/5}{x}}\right) ^{1/2}}{\left( 1+32\sin _{6/5,6}^6{x}\cos _{6/5,6}^{6/5}{x}\right) ^{1/4} \left( 1+\sqrt{1+32\sin _{6/5,6}^6{x}\cos _{6/5,6}^{6/5}{x}}\right) ^{1/6}}. \end{aligned}$$

In the remaining case

$$\begin{aligned} \frac{\pi _{6/5,6}}{8} \le x \le \frac{\pi _{6/5,6}}{4}, \end{aligned}$$

it follows easily that $1 \le 2\sin _{6/5,6}^6{(2x)}<2$ and $1-20S^6-8S^{12} \le 0$, hence we obtain (7) again. The proof is complete.$\square$

To show Theorem 1.2, the following lemma is useful.

Lemma 3.2

([5, 6]) Let $1<p,\ q <\infty$. For $x \in [0,2]$,

$$\begin{aligned} q\pi _{p,q}= & {} p^*\pi _{q^*,p^*},\\ \sin _{p,q}{\left( \frac{\pi _{p,q}}{2}x\right) }= & {} \cos _{q^*,p^*}^{q^*-1}{\left( \frac{\pi _{q^*,p^*}}{2}(1-x)\right) }. \end{aligned}$$

Proof of Theorem 1.2

Let $x \in [0,\pi _{6/5,2}/2]$. Then, since $4x/\pi _{6/5,2} \in [0,2]$, it follows from Lemma 3.2 that

$$\begin{aligned} \sin _{6/5,2}{(2x)} =\cos _{2,6}{\left( \frac{\pi _{2,6}}{2}\left( 1-\frac{4x}{\pi _{6/5,2}}\right) \right) } =\cos _{2,6}{\left( \frac{\pi _{2,6}}{2}-\frac{2x}{3}\right) }. \end{aligned}$$

Thus,

$$\begin{aligned} \sin _{6/5,2}{2x}=\sqrt{1-\sin _{2,6}^6{\left( \frac{\pi _{2,6}}{2}-\frac{2x}{3}\right) }}. \end{aligned}$$

(8)

The function $\sin _{2,6}$ has the addition formula (3). Letting $u=\pi _{2,6}/2$ and $v=2x/3$, we have

$$\begin{aligned} \sin _{2,6}{\left( \frac{\pi _{2,6}}{2}-\frac{2x}{3}\right) } =\phi ^{-1}{\left( -\phi (V)\right) }=\sqrt{\frac{1-V^2}{1+2V^2}}, \end{aligned}$$

(9)

where $V:=\sin _{2,6}{(2x/3)}$. Applying (9) to the right-hand side of (8), we obtain

$$\begin{aligned} \sin _{6/5,2}{2x}=\sqrt{1-\left( \frac{1-\sin _{2,6}^2{(2x/3)}}{1+2\sin _{2,6}^2{(2x/3)}}\right) ^3}. \end{aligned}$$

Let $f(x):=\sin _{6/5,2}{x}$ and $g(x):=\sin _{2,6}{(2x/3)}$. Then

$$\begin{aligned} f(2x)=\sqrt{1-\left( \frac{1-g(x)^2}{1+2g(x)^2}\right) ^3}. \end{aligned}$$

(10)

Therefore, it is easy to see that

$$\begin{aligned} g(x)=\sqrt{\frac{1-(1-f(2x)^2)^{1/3}}{1+2(1-f(2x)^2)^{1/3}}}. \end{aligned}$$

(11)

On the other hand, by (1) with x replaced with x/2, we see that g(x) satisfies

$$\begin{aligned} g(x)=\frac{2g(x/2)\sqrt{1-g(x/2)^6}}{\sqrt{1+8g(x/2)^6}}. \end{aligned}$$

Applying (11) with x replaced with x/2 to the right-hand side, we obtain

$$\begin{aligned} g(x)=\frac{2f(x)(1-f(x)^2)^{1/6}}{\sqrt{9-8f(x)^2}}. \end{aligned}$$

(12)

Substituting (12) into (10), we can express f(2x) in terms of f(x), i.e.,

$$\begin{aligned} f(2x)=\sqrt{1-\left( \frac{9-8f(x)^2-4f(x)^2(1-f(x)^2)^{1/3}}{9-8f(x)^2+8f(x)^2(1-f(x)^2)^{1/3}}\right) ^3}. \end{aligned}$$

Since $1-f(x)^2=\cos _{6/5,2}^{6/5}{x}$, the proof is complete. $\square$