4

Let $\operatorname{ns}\left(z,k\right)$ be one of the Jacobi's elliptic functions, and $K(k)$ the complete elliptic integral of the first kind. It's well-known that $\operatorname{ns}(mK(k),k)$ where $m\in\mathbb{Q}/2\mathbb{Z}$ is always an algebraic function with respect to $k$. With its addition formula, $$ \operatorname{ns}\left ( u+v ,k\right )\operatorname{ns} \left ( u-v,k \right ) =\frac{k^2-\operatorname{ns}(u,k)^2\operatorname{ns}(v,k)^2}{ \operatorname{ns}(u,k)^2-\operatorname{ns}(v,k)^2}, $$ place $(u,v)=\left ( \frac{2}{3}K(k), \frac{1}{3}K(k) \right )$ and $(u,v)=\left ( K(k), \frac13K(k) \right )$ respectively, giving \begin{aligned} &\operatorname{ns}\left ( \frac{2}{3}K(k),k \right )^2 =\frac{k^2-\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2 }{ 1-\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2} \\ &\operatorname{ns}\left ( \frac{1}{3}K(k),k \right ) =\frac{k^2-\operatorname{ns}\left ( \frac{2}{3}K(k),k \right )^2\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2}{ \operatorname{ns}\left ( \frac{2}{3}K(k),k \right )^2-\operatorname{ns}\left ( \frac{1}{3}K(k),k \right )^2} \end{aligned} The two equations are linear independent with each other, producing exact expressions of $\operatorname{ns}\left ( \frac{1}{3}K(k),k \right ),\operatorname{ns}\left ( \frac{2}{3}K(k),k \right )$. But the nasty appearances make it hard to do for some calculation. So is there a concise algebraic relation between the two? Thanks for your help.

My ultimate goal is to develop another way to calculate some algebraic number $k_n$ which satisfying $K^\prime(k)/K(k)=\sqrt{n}$. And $n=1,2,3,4,5,6,8,9,10,12,16,18...$ can be evaluated, whereas $k_{7},k_{15}...$ can't because the restriction of Wolfram Cloud(I use it to solve equations). Basic ideas are recorded here. It signifies the need of relations among $\operatorname{ns}\left (mK(k),k \right )$ having the same denominator.

Setness Ramesory
  • 6,789
  • I'm working on the same problem. Being: \begin{align} P(q)=1-24\sum_{n=1}^{\infty}\frac{nq^{n}}{1-q^{n}}. \end{align} It can be shown that: $$1+k^2=\frac{6P(q^2)-3P(q)}{4P(q^4)-P(q)},$$ when $q=e^{-\pi\frac{K(k')}{K(k)}}$. The a recurrence relation is derived. – User-Refolio Jun 11 '23 at 10:14
  • 1
    Wikipedia gives the relation $\text{cn} (2K/3,k)=1-\text{sn}(K/3,k)$ which translates to $\text{ns}(2K/3,k)=\dfrac{\text{ns}(K/3,k)}{\sqrt{2, \text{ns}(K/3,k)-1}}$ – Paramanand Singh Jun 12 '23 at 03:41
  • @User: just curious, how does your formula relate to evaluation of singular moduli $k_n$? – Paramanand Singh Jun 12 '23 at 03:46
  • @ParamanandSingh sure, I will give full details as soon as possible. – User-Refolio Jun 12 '23 at 12:35
  • @ParamanandSingh it can be shown only knowing the first singular moduli ($k=\frac{1}{\sqrt{2}}$) that: $$P(e^{-\pi/2})=\frac{12}{\pi}-\frac{\Gamma({1/4})^4\left(\frac{3\sqrt{2}}{4}+\frac{9}{8} \right)}{\pi^3},$$ $$P(e^{-\pi})=\frac{6}{\pi}-\frac{12\Gamma({1/4})^4}{32\pi^3},$$ $$P(e^{-2\pi})=\frac{3}{\pi},$$ $$P(e^{-4\pi})=\frac{3\Gamma({1/4})^4}{32\pi^3}+\frac{3}{2\pi},$$ $$P(e^{-8\pi})=\frac{\Gamma({1/4})^4\left(3\sqrt{2}/64+9/128\right)}{\pi^3}+\frac{3}{4\pi},$$ $$P(e^{-16\pi})=\frac{\Gamma({1/4})^4\left(3(2^{1/4}+2^{3/4})/128+9\sqrt{2}/256+21/512\right)}{\pi^3}+\frac{3}{8\pi}.$$ – User-Refolio Jun 17 '23 at 10:07
  • @ParamanandSingh inserting these values on: $$1+k^2=\frac{6P(q^2)-3P(q)}{4P(q^4)-P(q)},$$ when $q=e^{-\pi\frac{K(k')}{K(k)}}$. Gives some singular moduli. As I said the process can be iterated. – User-Refolio Jun 17 '23 at 10:09
  • 1
    @User: If your objective is to find the values of $k_{2^n}$ starting from $k_1=2^{-1/2}$ then one can directly use the Landen transformation without much hassle. I won't comment further here as this may seem like a digression with respect to current question. – Paramanand Singh Jun 17 '23 at 10:51

1 Answers1

3

Define the two quantities

$$ a_1 := \text{ns}\left(\frac{K(k)}{3},k\right) \quad\text{ and } \quad a_2 := \text{ns}\left(\frac{2K(k)}{3},k\right). \tag1 $$

The power series for these using the notation $\,k^2 = 16x\,$ is

$$ a_1 = 2 - 6 x - 30 x^2 - 258 x^3 - 2694 x^4 - 31128 x^5 - \dots,\\ a_2 = (2 - 2 x - 10 x^2 - 82 x^3 - 818 x^4 - 9076 x^5 -\dots)/\sqrt{3}. \tag2 $$

The question asked is

So is there a concise algebraic relation between the two?

The answer is $$ a_2 = \frac{a_1}{\sqrt{2a_1-1}} \quad \text{ and } \quad a_1 = a_2^2 + \sqrt{a_2^4-a_2^2}. \tag3 $$

Other algebraic relations are $$ 0 = a_1^4 - 2a_1^3 + 2k^2a_1 - k^2, \\ 0 = 3a_2^8 - 4(1+k^2)a_2^6 + 6k^2a_2^4 - k^4. \tag4 $$

Somos
  • 37,457
  • 3
  • 35
  • 85