How to show that $\frac{K(k')}{K(k)}=\sqrt{7} \implies k=k_{7}=\frac{\sqrt{2}}{8}(3-\sqrt{7})$?

Question

$$K(k)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}\tag{1}$$ is the complete elliptic of the first kind and $k'=\sqrt{1-k^2}$ the complementary modulus. We are interested in solving: $$\frac{K(k')}{K(k)}=\sqrt{n}\tag{2}$$ with $n$ a natural number. This question concerns the case: $$\frac{K(k')}{K(k)}=\sqrt{7}\tag{3} > \implies k=k_{7}=\frac{\sqrt{2}}{8}(3-\sqrt{7}).$$

Question

Is it possible to prove $(3)$ with elementary methods (no modular forms or elliptic functions) and without evaluating $K(k_{7})$? Thanks in advance for your efforts.

Updated

For esceptics. This formula is very useful: $$|K(k)(k+1)|=|-K(\frac{1-k}{1+k})+\frac{1+k}{1-k}K(\frac{1+k}{1-k})|\tag{4}$$ You have to choose the appropiate $k$. You can do calculations for the easier cases $k_{1}$,$k_{2}$,$k_{3}$,$k_{4}$,$k_{5}$ and $k_{6}$ first and you will see that $k_{3}$ and $k_{7}$ are the same in complexity.

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We show the cases $n=3,12$ for getting out doubts. We put $k=i^{1/3}=\frac{\sqrt{3}}{2}+\frac{i}{2}$ in $(4)$, then: $$|K(i^{1/3})|\frac{\sqrt{6}+\sqrt{2}}{2}=|-K(i(3-\sqrt{2})+i(3+\sqrt{2})K(i(3+\sqrt{2}))|.\tag{5}$$ Now we use: $$K(ik)=\frac{K(\frac{k}{\sqrt{1+k^2}})}{\sqrt{1+k^2}}\tag{6}.$$ We have: $$K(i(3-\sqrt{2}))=\frac{\sqrt{6}+\sqrt{2}}{4}K(\frac{\sqrt{6}-\sqrt{2}}{4})\tag{7}$$ and $$K(i(3+\sqrt{2}))=\frac{\sqrt{6}-\sqrt{2}}{4}K(\frac{\sqrt{6}+\sqrt{2}}{4})\tag{8}.$$ Replacing this two in $(5)$ we have:

$$\begin{align} |K(i^{1/3})| \frac{\sqrt{6} + \sqrt{2}}{2} &= \left| - \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) K \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right) \right. \nonumber \\ &\quad \left. + i \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) K \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) \right| \end{align}\tag{9} .$$

We use now Landen ascending transformation for $K(k)$: $$K(k)=\frac{2}{1+k'}K(\frac{1-k'}{1+k'})\tag{10}.$$ And we apply it to $K(i^{1/3})$ then: $$K(i^{1/3})=(1+i(2-\sqrt{3}))K(i(2-\sqrt{3}))\tag{11}.$$ And we use again $(6)$ in this last to obtain: $$K(i^{1/3})=(1+i(2-\sqrt{3}))(\frac{\sqrt{6}+\sqrt{2}}{4})K(\frac{\sqrt{6}-\sqrt{2}}{4})\tag{12}.$$ Replacing in $(9)$ we have:

$$\begin{align} \frac{\sqrt{2} (\sqrt{3} + 1) K \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right)}{2} &= \left| - \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) K \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right) \right. \nonumber \\ &\quad + i \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) K \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) \left. \right|. \tag{13} \end{align}$$

We can divide by $K(\frac{\sqrt{6}-\sqrt{2}}{4})$ and then: $$\frac{\sqrt{2}(\sqrt{3}+1)}{2}=|-\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)+i(\frac{\sqrt{6}+\sqrt{2}}{4})\frac{K(\frac{\sqrt{6}+\sqrt{2}}{4})}{K(\frac{\sqrt{6}-\sqrt{2}}{4})}| \\ \implies \frac{K(\frac{\sqrt{6}+\sqrt{2}}{4})}{K(\frac{\sqrt{6}-\sqrt{2}}{4})}=\sqrt{3}\hspace{.5cm}and{}\hspace{.5cm}k_{3}=\frac{\sqrt{6}-\sqrt{2}}{4}.$$ And again applying $(10)$ then $k_{12}=\frac{1-k'_{3}}{1+k'_{3}}$: $$\frac{K(k'_{12})}{K(k_{12})}=\sqrt{12} \implies k_{12}=(\sqrt{3}-\sqrt{2})^2(\sqrt{2}-1)^2.$$

Answer 1

This is not an answer (just for the fun of an approximate solution).

We can have a good approximation of the solution of $$\frac{K(k')}{K(k)}=\sqrt{n}$$ combining the series expansions of the elliptic integrals around $k=0$. This gives $$\frac{K(k')}{K(k)}=\frac{2 (2 \log (2)-\log (k))}{\pi}-\frac{k^2}{2 \pi }+O\left(k^4 \right)$$ and the solution is given in terms of Lambert function (let $k=\sqrt t$). This gives $$\large\color{blue}{k_n \sim \sqrt{2 W\left(8 e^{-\pi\sqrt{n} }\right)}}\tag 1$$ where $W(.)$ is the principal branch of Lambert function.

Expanding for "large" values of $n$, using the above estimate $$\frac{K(k')}{K(k)}-\sqrt{n}=-\frac{52}{\pi }\,e^{-2 \pi \sqrt{n}}\Big( 1-\frac{256}{39} e^{-\pi \sqrt{n}}+\cdots\Big)$$

Trying for a few values of $n$

$$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 2 & 0.41548680 & 0.41421356 \\ 3 & 0.25893793 & 0.25881902 \\ 4 & 0.17158802 & 0.17157287 \\ 5 & 0.11887936 & 0.11887695 \\ 6 & 0.08516469 & 0.08516423 \\ 7 & 0.06262301 & 0.06262301 \\ 8 & 0.04702192 & 0.04702192 \\ 9 & 0.03592157 & 0.03592157 \\ 10 & 0.02784245 & 0.02784245 \\ \end{array} \right)$$

Edit

We can improve the estimate using the first iterate of any Newton-like method of order $m$ starting with $(1)$. The formulae remain fully explicit.

For example, for $n=7$, we shall have

$$\left( \begin{array}{ccc} m & k_7^{(m)} & \text{method} \\ 2 & \color{red}{0.062622912543}0918265780130914496 & \text{Newton} \\ 3 & \color{red}{0.062622912543167970}0824424230563 & \text{Halley} \\ 4 & \color{red}{0.0626229125431679701026646}644880 & \text{Householder} \\ 5 & \color{red}{0.0626229125431679701026646800391} & \text{unnamed} \\ \end{array} \right)$$

Update

The series expansion is $$\frac{4 \log (2)}{\pi }-\frac{\log \left(k^2\right)}{\pi }-\frac 1{2\pi}\sum_{n=1}^\infty a_n\,k^{2n}$$ where the first coefficients are $$\left\{1,\frac{13}{32},\frac{23}{96},\frac{2701}{16384}, \frac{5057}{40960},\frac{76715}{786432},\frac{146749}{1835008},\frac{144644749}{2147483648},\frac{279805685}{4831838208}\right\}$$ which do not seem to correspond to any known sequence in OEIS.

Answer 2

We are going to consider: $l_{7}=\frac{3\sqrt{7}}{8}-\frac{i}{8}$. Then using formula $(4)$ of the question:

$$|K(\frac{3\sqrt{7}}{8}-\frac{i}{8})(1+\frac{3\sqrt{7}}{8}-\frac{i}{8})|=|-K(i(8-3\sqrt{7}))-i(3\sqrt{7}+8)K(i(3\sqrt{7}+8))|\tag{1}.$$ We consider now Landen ascending transformation: $$K(k)=\frac{2}{1+k'}K(\frac{1-k'}{1+k'})\tag{2},$$ where $k'=\sqrt{1-k^2}$ is the complementary modulus. And the imaginary transformation: $$K(ik)=\frac{1}{\sqrt{1+k^2}}K(\frac{k}{\sqrt{1+k^2}})\tag{3}.$$ Then we have: $$K(\frac{3\sqrt{7}}{8}-\frac{i}{8})\overset{(2)}=(\frac{11}{8}-\frac{\sqrt{7}i}{8})K(\frac{3}{8}-\frac{\sqrt{7}i}{8}) \overset{(2)}=(\frac{11}{8}-\frac{\sqrt{7}i}{8})(1+i(3\sqrt{7}-8))K(i(3\sqrt{7}-8))\\ \overset{(3)}=(\frac{5\sqrt{2}+\sqrt{14}}{8}+i(\frac{\sqrt{14}-5\sqrt{2}}{8}))K(\frac{3\sqrt{2}-\sqrt{14}}{8})\tag{4},$$ $$K(i(8-3\sqrt{7}))\overset{(3)}=\frac{3\sqrt{2}+\sqrt{14}}{8}K(\frac{3\sqrt{2}-\sqrt{14}}{8}),\tag{5}$$ and: $$K(i(3\sqrt{7}+8))\overset{(3)}=\frac{3\sqrt{2}-\sqrt{14}}{8}K(\frac{3\sqrt{2}+\sqrt{14}}{8}).\tag{6}$$ Replacing $(4)$, $(5)$ and $(6)$ in $(1)$: $$\frac{3+\sqrt{7}}{2}K(\frac{3\sqrt{2}-\sqrt{14}}{8})\\=|-\frac{3\sqrt{2}+\sqrt{14}}{8}K(\frac{3\sqrt{2}-\sqrt{14}}{8})-i(\frac{(3\sqrt{7}+8)(3\sqrt{2}-\sqrt{14})}{8}K(\frac{3\sqrt{2}+\sqrt{14}}{8}))|$$ And from this last it follows that: $$\frac{K(\frac{3\sqrt{2}+\sqrt{14}}{8})}{K(\frac{3\sqrt{2}-\sqrt{14}}{8})}=\sqrt{7}\hspace{.5cm}and\hspace{.5cm}k_{7}=\frac{3\sqrt{2}-\sqrt{14}}{8}.$$

Now we are going to derive: $$28E(k_{7})K(k_{7})-(4\sqrt{7}+14)K^2(k_{7})=\sqrt{7}\pi,\hspace{1cm}k_{7}=\frac{3\sqrt{2}-\sqrt{14}}{8}\tag{7}.$$

We follow the same approach as we have done for $k_{3}$: https://math.stackexchange.com/a/5080646/1187943.

We need also Landen ascending transformation and the imaginary transformation for $E$: $$\frac{E(k)}{1+k'}+\frac{k'K(k)}{1+k'}=E(\frac{1-k'}{1+k'}),\tag{8}$$ and $$E(ik)=\sqrt{1+k^2}E(\frac{k}{\sqrt{1+k^2}})\tag{9}.$$

Let be $s_{7}=\frac{3}{8}-\frac{\sqrt{7}i}{8}$ then we have to observe that: $$E(s_{7})K(s_{7})=2E(k_{7})K(k_{7})-K^2(k_{7}).\tag{10}$$ To confirm it we have: $$K(s_{7})=K(\frac{3}{8}-\frac{\sqrt{7}i}{8})\overset{(2)}=(1+i(3\sqrt{7}-8))K(i(3\sqrt{7}-8))\\ \overset{(3)}=(1+i(3\sqrt{7}-8))(\frac{3\sqrt{2}+\sqrt{14}}{8})K(k_{7}),\tag{11}$$ and $$E(s_{7})=E(\frac{3}{8}-\frac{\sqrt{7}i}{8})\overset{(8)}=(1+\frac{3\sqrt{7}}{8}+\frac{i}{8})E(i(3\sqrt{7}-8))-(\frac{3\sqrt{7}}{8}+\frac{i}{8})K(\frac{3}{8}-\frac{\sqrt{7}i}{8})\\ \overset{(9)}=(1+\frac{3\sqrt{7}}{8}+\frac{i}{8})(6\sqrt{2}-2\sqrt{14})E(k_{7})-(\frac{3\sqrt{7}}{8}+\frac{i}{8})K(\frac{3}{8}-\frac{\sqrt{7}i}{8}) \overset{(2)}=(1+\frac{3\sqrt{7}}{8}+\frac{i}{8})(6\sqrt{2}-2\sqrt{14})E(k_{7})-(\frac{3\sqrt{7}}{8}+\frac{i}{8})(1+i(3\sqrt{7}-8))K(i(3\sqrt{7}-8)) \\\overset{(3)} =(1+\frac{3\sqrt{7}}{8}+\frac{i}{8})(6\sqrt{2}-2\sqrt{14})E(k_{7})-(\frac{3\sqrt{7}}{8}+\frac{i}{8})(1+i(3\sqrt{7}-8))(\frac{3\sqrt{2}+\sqrt{14}}{8})K(k_{7}) \tag{12}.$$ And using $(11)$ and $(12)$ we confirm $(10)$. Also we have to note that: $$K(s'_{7})=K(\frac{3\sqrt{7}}{8}+\frac{i}{8})\overset{(2)}=(\frac{11}{8}+\frac{\sqrt{7}i}{8})K(\frac{3}{8}+\frac{\sqrt{7}i}{8})\overset{(2)}= (\frac{11}{8}+\frac{\sqrt{7}i}{8})(1+i(8-3\sqrt{7}))K(i(8-3\sqrt{7})) \\ \nonumber \overset{(3)}=(\frac{11}{8}+\frac{\sqrt{7}i}{8})(1+i(8-3\sqrt{7}))\left(\frac{3\sqrt{2}+\sqrt{14}}{8}\right)K(k_{7}).\tag{13}$$ Now we use: $$\frac{K(k')}{K(k)}\left(K^2(k)-\frac{K(k)E(k)}{1+k}\right)=-\frac{\pi}{2(k+1)}+K(k)E(\frac{1-k}{1+k})\tag{14}.$$ that is a consequence of Legendre's relation and Landen's descending transformations. We replace $s_{7}=\frac{3}{8}-\frac{\sqrt{7}i}{8}$ in $(14)$ and using $(10)$, $(11)$ and $(13)$ we have: $$\left(\frac{\sqrt{7}}{2}+\frac{i}{2}\right)\left(K^2(s_{7})-\frac{2E(k_{7})K(k_{7})-K^2(k_{7})}{1+s_{7}} \right)=-\frac{\pi}{2(1+s_{7})}+K(s_{7})E(\frac{1-s_{7}}{1+s_{7}}).\tag{15}$$ In the other hand: $$ E(\frac{1-s_{7}}{1+s_{7}})=E(\frac{3}{8}+\frac{\sqrt{7}i}{8}) \\ \overset{(8)}=\left( (1+\frac{3\sqrt{7}}{8})E(i(8-3\sqrt{7}))-\frac{3\sqrt{7}}{8}K(\frac{3}{8}+\frac{\sqrt{7}i}{8})\right)+ i\left(\frac{K(\frac{3}{8}+\frac{\sqrt{7}i}{8})}{8}-\frac{E(i(8-3\sqrt{7}))}{8} \right) \\ \overset{(9)}=\left( \frac{3\sqrt{2}+\sqrt{14}}{4}E(k_{7})-\frac{3\sqrt{7}}{8}K(\frac{3}{8}+\frac{\sqrt{7}i}{8})\right)+ i\left(\frac{K(\frac{3}{8}+\frac{\sqrt{7}i}{8})}{8}+\frac{\sqrt{14}-3\sqrt{2}}{4}E(k_{7}) \right) \\ \overset{(2)}=\left( \frac{3\sqrt{2}+\sqrt{14}}{4}E(k_{7})-K(i(8-3\sqrt{7}))\right)+ i\left((8-3\sqrt{7})K(i(8-3\sqrt{7}))+\frac{\sqrt{14}-3\sqrt{2}}{4}E(k_{7}) \right) \\ \overset{(3)}=\left( \frac{3\sqrt{2}+\sqrt{14}}{4}E(k_{7})-\frac{3\sqrt{2}+\sqrt{14}}{8}K(k_{7})\right)+ i\left(\frac{3\sqrt{2}-\sqrt{14}}{8}K(k_{7})+\frac{\sqrt{14}-3\sqrt{2}}{4}E(k_{7}) \right).\tag{16}$$ Finally using $(11)$, $(15)$ and $(16)$: $$ (\frac{5\sqrt{7}}{16}+\frac{11}{8})K^2(k_{7}) -\frac{5\sqrt{7}K(k_{7})E(k_{7})}{8}+i\left((\frac{\sqrt{7}}{8}+\frac{9}{16})K^2(k_{7}) -\frac{9K(k_{7})E(k_{7})}{8}\right) \\ =-\frac{3\sqrt{7}K^2(k_{7})}{8}+\frac{3\sqrt{7}K(k_{7})E(k_{7})}{4}-\frac{11\pi}{32}+i\left(\frac{K^2(k_{7})}{8}-\frac{K(k_{7})E(k_{7})}{4}-\frac{\sqrt{7}\pi}{32}\right) $$ We take real parts in this last equality (with the imaginary is the same) and: $$ E(k_{7})=(\frac{\sqrt{7}}{7}+\frac{1}{2})K(k_{7})+\frac{\sqrt{7}\pi}{28K(k_{7})}.$$ Or which is the same: $$28E(k_{7})K(k_{7})-(4\sqrt{7}+14)K^2(k_{7})=\sqrt{7}\pi.\tag{17}$$ For obtain the famous series by Ramanujan we invoke Clausen's formula: $$\frac{4K^2(s)}{\pi^2}=\sum_{n=0}^{\infty}\frac{(2n)!^3}{2^{4n}n!^6}(s^2(1-s^2))^{n}\hspace{.5cm}0<|s| \leq \frac{1}{\sqrt{2}}\tag{18},$$ and: $$\frac{4K(s)E(s)}{\pi^2}=\sum_{n=0}^{\infty}\frac{(2n)!^3}{2^{4n}n!^6}(s^2(1-s^2))^{n}(n(1-2s^2)+(1-s^2))\hspace{.5cm}0<|s|<\frac{1}{\sqrt{2}},\tag{19}$$ which is obtained from $(18)$ after derivating and using: $\frac{\mathrm{dK}}{\mathrm{d}k}=\frac{E(k)}{k(1-k^2)} - \frac{K(k)}{k}$.

As a corollary using $(17)$, $(18)$ and $(19)$: $$\sqrt{7}\pi=28E(k_{7})K(k_{7})-(4\sqrt{7}+14)K^2(k_{7})=\frac{\sqrt{7}\pi^2}{16}\sum_{n=0}^{\infty}\frac{(2n)!^3}{2^{12n}n!^6} (42n+5) \\ \implies \sum_{n=0}^{\infty}\frac{(2n)!^3}{2^{12n}n!^6} (42n+5)=\frac{16}{\pi}.$$

Wich is formula $(29)$ of Ramanujan's master paper.

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We have to note that using the hypergeometric approach and $(17)$ we can derive the following:

$$ \sum_{n=1}^{\infty}\binom{2n}{n}^3\frac{q(n)H_{n}}{2^{12n}(2n-1)^3}=\frac{16}{\pi}\tag{20} $$ where $q(n)=21168n^5-18816n^4+272n^3-180n^2+17n + 5$ and $H_{n}=\sum_{i=1}^{n}\frac{1}{i}$ is the n-th harmonic number. $$ \sum_{n=0}^{\infty}\frac{(-1)^n(4n)!}{63^{2n}n!^{4}}(65n+8)=\frac{9\sqrt{7}}{\pi}.\tag{21} $$ And: $$ \sum_{n=0}^{\infty}\frac{(-1)^n(6n)!}{15^{3n}n!^{3}(3n)!}(63n+8)=\frac{5\sqrt{15}}{\pi}.\tag{22} $$

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Here is some feedback of Zudilin.

You certainly master the techniques. This type of proof is based on the Legendre relation between periods and quasi-periods, it goes back to the Borweins' book "Pi and the AGM". I am impressed that you do such things by hand. You may feel that it looks new computationally, but mathematically there is exploitation of the very same ideas that have been already used for such derivations. These ideas are generalised in several directions; for example, James Wan had other extensions to 4F3 series in his work around 2013 in the Integral Transforms and Special Functions.

Several new ideas, also in executing modular proofs, originate from Jesús Guillera.

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Borwein's books is the seed of the work of many others but in their book this case is not showed in detail. I believe this proof has pedagogical value.

Answer 3

Using Mathematica notations, we need to use $$K(k^2)=\frac{\pi }{2\, \text{agm}\left(1,\sqrt{1-k^2}\right)} \quad \implies \quad \frac{K(1-k^2)}{K(k^2)}=\frac{\text{agm}\left(1,\sqrt{1-k^2}\right)}{\text {agm}(1,k)}$$ If $$k=\frac{3-\sqrt{7}}{4 \sqrt{2}}$$ then $$\frac{\text{agm}\left(1,\sqrt{1-k^2}\right)}{\text{agm}(1,k)}=\frac{\text{agm}\left(1,\frac{1}{4} \sqrt{8+3 \sqrt{7}}\right)}{\text{agm}\left(1,\frac{1}{4} \sqrt{8-3\sqrt{7}}\right)}$$

For sure, we could have use Jacobi theta functions to try to go further from a formal point of view.

Numerically, this is

$$2.6457513110645905905016157536392604257102591830825\cdots=\sqrt 7$$