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I'm trying to find an answer to this question. Let $K(k)$ be the elliptic integral of the first kind and $K'=K(\sqrt{1-k^2})$. According to Abel's theorem (see this link) we know that if $\frac{K'}{K}=\frac{a+b\sqrt{n}}{c+d\sqrt{n}}$ where $a,b,c,d,n$ are integers, then $k$ is the root of an algebraic equation with integer coefficients. (As I understand it, by algebraic equation with integer coefficients they mean a polynomial with integer coefficients?) So I solved the equation $$ \displaystyle \frac{K(\sqrt{1-k^2})}{K(k)}=\sqrt{2}-1 $$ numerically with 200 digit accuracy and found the corresponding $k$:

$$ k=0.995942004483485267626221391206685115425856878468293704688032877263852669\\384783141641717390148240933985687938743309287701383005201421919573984016706\\406957193055047393475454459046127151355572762924 $$

Here is the Mathematica code I used:

k:=Sqrt[m/.FindRoot[EllipticK[1 - m]/EllipticK[m] - Sqrt[ 2 ] +1==0,{m,1},WorkingPrecision -> 200]]; N[k,200]

(As was pointed out below by ccorn EllipticK(k^2)$=K(k)$.)

Then I tried to identify this algebraic constant by Inverse symbolic calculator but it couldn't identify it. Is there some other way to identify this equation with integer coefficients?

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  • The Recognize[...] command does not yield any likely candidates. – Lucian Dec 03 '15 at 14:40
  • Isn't $m=k^2$ in Mathematica? Then you should use 1-m instead of Sqrt[1-m^2] and take the squareroot of the $m$ you found. – ccorn Dec 04 '15 at 00:15
  • Cf. Wolfram's reference where they state that $K'(m) = K(1-m)$. – ccorn Dec 04 '15 at 00:45
  • @ccorn thanks, I edited the question. But still inverse symbolic calculator couldn't recognize the corrected value of $k$. –  Dec 04 '15 at 11:59
  • @ccorn: From what you recall about the eta quotient $\frac{\eta(2\tau)}{\eta(\tau)}$, does Part II of my answer hold true? – Tito Piezas III Dec 14 '15 at 04:59
  • @TitoPiezasIII: Yes, since $\frac{k^4}{1-k^2} = 256\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{24}$. I also doubt that $k$ is algebraic, and that Whittaker & Watson have forgotten some $\mathrm{i}$, but first I'd like to see the article by Abel which Whittaker & Watson seem to have referenced. – ccorn Dec 14 '15 at 08:05
  • Mathworld references Whitaker & Watson, which archive.org has online. It's just a remark with an opaque reference to Abels oeuvres vol. I (published papers), the edition by Sophus Lie et al. Archive.org seems to only have the earlier collection, and I have not been able to identify Abel's article there. – ccorn Dec 14 '15 at 08:29
  • If the reference is to Crelle, this article might be it, but my french has been out of use for some decades, and I have difficulties identifying the symbols. – ccorn Dec 14 '15 at 08:41
  • There is an 1881 article by Hurwitz online, where he recalls that, given period ratios $\omega$ and $\omega_1$, $k(\omega), k(\omega_1)$ are algebraically related if $\omega$ and $\omega_1$ satisfy $a\omega\omega_1+b\omega+c\omega_1+d=0$ with $a,b,c,d\in\mathbb{Z}$ and $ad-bc>0$. No surprise there. Hurwitz essentially proves that this bilinear form in $\omega$ and $\omega_1$ is the only possible form of an algebraic relation between $\omega$ and $\omega_1$ whenever $k(\omega), k(\omega_1)$ are algebraically related. – ccorn Dec 14 '15 at 08:57
  • @ccorn: Thanks for finding those references. I normally deal with $\tau$ that is a complex root of a quadratic, and I haven't really explored when it is a root of higher deg eqns. – Tito Piezas III Dec 14 '15 at 09:34
  • @TitoPiezasIII: I'm not into transcendence matters, but we had such questions discussed in comments before and found that when $\tau$ is algebraic of degree greater than 2, then $j$, and hence $k$, must be transcendental. Which means your answer here is right, and W&W must have got it slightly wrong. This time Mathworld is not to blame, they just copied the statement almost verbatim from W&W. – ccorn Dec 14 '15 at 10:01
  • @ccorn: Oh, I remember that post now. Thanks for the clarification re Whittaker and Watson. – Tito Piezas III Dec 14 '15 at 13:12

1 Answers1

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I think Mathworld's explanation of Abel's Theorem is incomplete.

Part I. The Mathematica command is,

$$\frac{K'(k)}{K(k)}=\frac{\text{EllipticK[1-ModularLambda[}\tau]]}{\text{EllipticK[ModularLambda[}\tau]]}=\sqrt{2}-1\tag1$$

where the argument $\tau$ is,

$$\color{brown}{\tau = \sqrt{-2}-\sqrt{-1}}\tag2$$

Thus, we get your constant,

$$k = \sqrt{\lambda(\tau)}=\sqrt{\text{ModularLambda[}\tau]}=0.9959420044834\dots\tag3$$

Other expressions are,

$$\big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^{1/2} = \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$

with the nome $q = \exp(i \pi \tau)$, Dedekind eta function $\eta(\tau)$, and Jacobi theta function $\vartheta_n(0,q)$, with both functions built in Mathematica. See also this post.

Part II. Now as to whether $k$ given by $(3)$ is algebraic, your question is equivalent to asking if the eta quotient,

$$x =\frac{\eta(2\tau)}{\eta(\tau)}\tag4$$

is algebraic. If $\tau$ is an imaginary quadratic $\tau_2$, then it is yes. However, your $\tau$ is an imaginary quartic root $\tau_4$, a root of,

$$\color{brown}{\tau^4+6\tau^2+1 = 0}\tag5$$

Conclusion: I think Abel's Theorem covers only at most $\tau_2$, not $\tau_4$, and Mathworld forgot(?) that crucial detail. Re this comment in an MO post.

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