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The following sequence converging to Gelfond's constant ($e^\pi$) is apparently mentioned (and originates?) in the book "Mathematics by Experiment: Plausible Reasoning in the 21st Century.", which I unfortunately do not have access to. $$k_0 = 1/\sqrt{2},\quad k_{n+1}={\frac {1-{\sqrt {1-k_{n}^{2}}}} {1+{\sqrt {1-k_{n}^{2}}}}}$$ $$e^\pi =\lim_{n\to\infty} \left(\frac{4}{k_{n+1}}\right)^{2^{-n}}$$ How is this sequence and similar ones derived?

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    $\begingroup$ The algorithm above appears at page 137 of Mathematics by Experiment, but no proof is given there either. $\endgroup$ – Jack D'Aurizio Oct 25 '17 at 23:45
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    $\begingroup$ Anyway, it is not difficult to guess that such algorithm is related to $e^{\pi K'/K}$ and to the fact that $\frac{1}{\sqrt{2}}$ is an elliptic singular value, hence it turns out to be a minor twist on the classical AGM algorithm for the computation of $K(k)$. I will write a detailed answer tomorrow, if no one comes before me. Time to sleep in Italy, have a good night. $\endgroup$ – Jack D'Aurizio Oct 25 '17 at 23:59
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As mentioned in comments (to the question) from Jack d'Aurizio this is related to elliptic integrals. And based on the feedback received in comments to this answer I provide some background in the theory of elliptic integrals and theta functions. The details along with proofs are available in the linked blog posts.


Let $k\in(0,1)$ be the elliptic modulus and then we define complementary modulus $k'=\sqrt{1-k^{2}}$. Further the complete elliptic integrals of the first kind are defined by $$K(k)=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^{2}\sin^{2}x}},\, K'(k)=K(k')\tag{1}$$ By the way the use of prime $(') $ here does not denote derivative.

Starting with a modulus $k$ we define a new modulus $l$ via ascending Landen transformation $$l=\frac{2\sqrt{k}}{1+k},\,k=\frac{1-l'}{1+l'},\, l>k\tag{2}$$ where $l'$ is complementary to $l$ ie $l'=\sqrt{1-l^{2}}$. It can be proved with some effort that $$K(l)=(1+k) K(k)$$ Using a little amount of algebra it can be seen that the relation between $l$ and $k$ is same as the relation between $k'$ and $l'$ and hence from the above equation we get $$K'(k) =(1+l') K'(l)=\frac{2K'(l)}{1+k}$$ and therefore from the last two equations we get $$\frac{K'(k)} {K(k)} =2\cdot\frac{K'(l)}{K(l)}\tag{3}$$ Corresponding to elliptic modulus $k$ we have another variable $q$ called nome defined by $q=\exp(-\pi K'(k) /K(k)) $ and it is possible to get an expression for $k$ in terms of $q$ via the use of theta functions. We have $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}\tag{4}$$ where $$\vartheta_{2}(q)=2q^{1/4}\sum_{n=0}^{\infty}q^{n(n+1)},\,\vartheta_{3}(q)=1+2\sum_{n=1}^{\infty}q^{n^{2}}\tag{5}$$


The sequence $k_{n} $ in question is descending Landen sequence of elliptic moduli and using equation $(3)$ we have $$\frac{K'(k_{n+1})}{K(k_{n+1})}=2\cdot\frac{K'(k_{n})}{K(k_{n})},\, K'(k_{0})=K(k_{0})\tag{6}$$ The corresponding nomes $q_{n} =\exp(-\pi K'(k_{n}) /K(k_{n})) $ have a relation between them as $q_{n+1}=q_{n}^{2}$. Thus we can see that $$e^{-\pi}=q_{0}=(q_{n})^{2^{-n}}\tag{7}$$ Next also note that $q_{n} \to 0, k_{n} \to 0$ as $n\to \infty$ and using $(4)$ we have the relation $$\lim_{n\to\infty}\frac{k_{n}^{2}}{16q_{n}}= 1\tag{8}$$ (the above equation can also be proved without using the theory of theta function via the fundamental asymptotic $K'(k) =\log(4/k)+o(1)$ as $k\to 0^{+}$). The desired relation follows easily from $(7)$ and $(8)$ and the relation $k_{n} ^{2}=4k_{n+1}/(1+k_{n+1})^{2}$.

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  • $\begingroup$ Do you think I understand anything of what you wrote ? $\endgroup$ – reuns Oct 29 '17 at 13:50
  • $\begingroup$ Thank you! Could add some more detail for those who aren't familiar with the topic? (e.g. What's a Landen sequence? How do we get the value of the limit?) $\endgroup$ – nbubis Oct 29 '17 at 16:29
  • $\begingroup$ @reuns : I will add some more details, but let me know specific areas which you find need explanation. $\endgroup$ – Paramanand Singh Oct 29 '17 at 16:43
  • $\begingroup$ @nbubis: will add more details, wait for sometime. $\endgroup$ – Paramanand Singh Oct 29 '17 at 16:46
  • $\begingroup$ I can't even explain what I want explanations on, honestly this is just a mess of symbols without any explanation. In the theory of elliptic integrals and functions there is a lattice and an associated cubic polynomial with some special functions (and their inverse functions) attached to it. $\endgroup$ – reuns Oct 29 '17 at 16:52

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