3

$$ a_0 = \frac{1}{\sqrt 2} $$

$$ a_{n+1} = \frac{( \sqrt {1 - a_n^2} -1)^2}{a_n^2} $$

$$ \lim_{n \to \infty} \frac{4^{\frac{1}{2^n}}}{a_{n+1}^{\frac{1}{2^n}}} = \exp(\pi) $$

How did Gelfond find this nice result ? And how to prove it ? The 4 is trivial , but the rest is not. Is this related to trigonometry ? Is this related to continued fractions ?

Are there analogues known for cube roots ?

Notice a proof alone might not explain how he found the result.

Xander Henderson
  • 29,772
mick
  • 15,946
  • mick Please check my edits, fairly minor, but primarily check those for the for $a_{n+1}$. It seemed as though you intended for the square root to extend over all of $1 -a_n^2$. – amWhy Jun 29 '18 at 11:31
  • Yes I did amwhy. – mick Jun 29 '18 at 11:33
  • The theory behind the result is difficult and you may get a brief explanation from my answer to the original question. The details would require some study of elliptic integrals and theta functions which are covered nicely in my blog. – Paramanand Singh Jun 30 '18 at 04:09
  • Also Gelfond did not find this result. Rather the constant $e^\pi$ goes by the name Gelfond's constant. – Paramanand Singh Jun 30 '18 at 04:33
  • A similar iteration can be given using same approach involving cube roots (or nth roots in general) but the iteration would be clumsy and hard to follow. – Paramanand Singh Jun 30 '18 at 08:38

1 Answers1

4

This is the Gauss-Legendre algorithm non homogeneous version for the elliptic modulus k. That is, let $\, a_n = k(q_n) = (\theta_2(q_n)/\theta_3(q_n))^2 \,$ where $\, q_n = \exp(-\pi 2^n). \,$ We have $\, a_{n+1} = 4 q_n + O(q_n^3). \,$

Legendre used the following: $\, k_{n+1} = (1 - k'_n)/(1 + k'_n) = k_n^2/(1 + k'_n)^2 = (1 - k'_n)^2/k_n^2 \,$ where $\, 1 = k_n^2 + k_n'^2. \,$ The AGM of Gauss is $\, a_{n+1} \!=\! \frac12(a_n \!+\! b_n), $ $ b_{n+1} \!=\! \sqrt{a_n b_n}, \, $ $ c_{n+1} \!=\! \frac12(a_n\!-\!b_n). \,$ The connection is $\, k_n = c_n/a_n, \, k_n' = b_n/a_n. \,$ For more details, you can read the standard book Pi and the AGM by Borwein and Borwein.

By the way, the recursion as written $\, a_{n+1} = (\sqrt {1 - a_n^2} -1)^2/a_n^2, \,$ is not good from a numerical point of view. The subtraction of two numbers nearly equal causes drastic loss of significance. The equivalent recursion $\, a_{n+1} = a_n^2 / (1 + \sqrt {1 - a_n^2})^2 \,$ does not suffer from significance loss.

Somos
  • 35,251
  • 3
  • 30
  • 76
  • Plz Explain more. I do not see a geometric average nor an arithmetic explicitly , I do not know What your thetas are. And are saying it is actually gauss result ? Even If it is the AGM , How was the result reached and found ? Integral representation of AGM from Gauss or What ? Thank you. – mick Jun 29 '18 at 13:11