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In a q-continued fraction related to the octahedral group I defined a new q-continued fraction for the square of ramanujan's octic continued fraction which I discovered using certain three term relations and algebraic manipulations.

Given $$\big(u(2\tau)\big)^2=\cfrac{2\,q^{1/2}}{1-q+\cfrac{q(1+q)^2}{1-q^3+\cfrac{q^2(1+q^2)^2}{1-q^5+\cfrac{q^3(1+q^3)^2}{1-q^7+\ddots}}}}$$

then by using the well known special value $$\big(u(i)\big)^2= \sqrt{2}-1$$ which was first found by Srinivasa Ramanujan in his first letter to GH Hardy ,leads immediately to the following new continued fraction of square root 2

$$\sqrt{2}=1+\cfrac{2\,e^{-\pi/2}}{1-e^{-\pi}+\cfrac{e^{-\pi}(1+e^{-\pi})^2}{1-e^{-3\pi}+\cfrac{e^{-2\pi}(1+e^{-2\pi})^2}{1-e^{-5\pi}+\cfrac{e^{-3\pi}(1+e^{-3\pi})^2}{1-e^{-7\pi}+\ddots}}}}$$

Can anyone verify the identity, either by algebraic or numerical methods?

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Nicco
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  • Should $e^{-3\pi}$ in the bottom display line be $e^{-5\pi}$? – John Bentin Aug 01 '15 at 11:40
  • @Nicco: Yes, it is true. The general case is proved in your other post. Note the incidental fact that Mathematica command ModularLambda[Sqrt[-4]] yields $a=(\sqrt{2}-1)^4$ and, $$\frac{K'(k)}{K(k)} = \frac{\text{EllipticK[1-a]}} {\text{EllipticK[a]}} = \sqrt{4}$$ The other post explains the connection of this cfrac to $K(k)$. – Tito Piezas III Aug 22 '15 at 16:04

2 Answers2

8

I will stick to verifying the identity numerically. A backwards recursion formula for the $n$'th partial quotient of the continued fraction is

$$s_{k-1} = 1 - e^{-(2k-1)\pi} + \frac{e^{-k\pi}(1+e^{-k\pi})^2}{s_{k}}$$

for $k=n,n-1,\ldots,3,2,1$ and $s_n = 1$. Having calculated $s_0$ the $n$'th partial quotient is then given as $1 + \frac{2e^{-\pi/2}}{s_0}$.

Here is a numerical implementation of this recursion in Mathematica:

n = 5;
s = 1;
Do[
 s = 1 - Exp[-(2 j + 1) \[Pi]] + (Exp[-\[Pi] (j + 1)] (1 + Exp[-\[Pi] (j + 1)])^2)/s;
, {j, n, 0, -1}]
(1 + (2 Exp[-\[Pi]/2])/s) - Sqrt[2]

Below is a plot of the difference between the $n$'th partial quotient of the continued fraction and $\sqrt{2}$. Even for $n=3$ the error is found to be smaller than double precision $\sim 10^{-16}$ and one needs to use higher precision in the calculation to be able to find any difference. With higher precision I find that at $n=25$ the agreement is better than $500$ digits.

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
(source: folk.uio.no)

For completeness here is the code used to calculate to arbitrary precision (here for $500$ digits):

Block[{$MinPrecision = 500, $MaxPrecision = 500},
n = 25;
s = 1;
Do[
  s = 1 - Exp[-\[Pi] (2 j + 1)] + (Exp[-\[Pi] (j + 1)] (1 + Exp[-\[Pi] (j + 1)])^2)/s;
, {j, n, 0, -1}];
N[(1 + (2 Exp[-\[Pi]/2])/s) - Sqrt[2], 500]
]
Glorfindel
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Winther
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  • I guess this numerical analysis provides evidence that the continued fraction converges very fast. – Nicco Aug 04 '15 at 09:21
  • @Nicco Yes it does. One can compare it with $\sqrt{2} = \cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\ldots}}} - 1$. In this case one needs to go down to $n=130$ to get $100$ digits of $\sqrt{2}$. – Winther Aug 04 '15 at 12:37
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If $|q|<1$ and we set $$ (a;q)_{\infty}=\prod^{\infty}_{n=0}\left(1-aq^n\right), $$ then (see [1]): $$ P(a,q):=\left(\frac{(-a;q)_{\infty}}{(a;q)_{\infty}}\right)^2= $$ $$ =-1+\frac{2}{1-}\frac{2a}{1-q+}\frac{a^2(1+q)^2}{1-q^3+}\frac{a^2q(1+q^2)^2}{1-q^5+}\frac{a^2q^2(1+q^3)^2}{1-q^7+}\ldots,\tag 1 $$ where $a$ is a complex number.

Take the logarithms in (1) and expand the two products in Taylor series (of $\log(1-x)$). Then the double sums are easily rearranged and we get $$ 4\sum^{\infty}_{n=0}\frac{a^{2n+1}}{(2n+1)(1-q^{2n+1})}=\log P(a,q). \tag 2 $$ Set also $$ u_0(a,q):=\frac{P(a,q)-1}{P(a,q)+1}. \tag 3 $$ Then we have $$ u_0(a,q):=\frac{2a}{1-q+}\frac{a^2(1+q)^2}{1-q^3+}\frac{a^2q(1+q^2)^2}{1-q^5+}\frac{a^2q^2(1+q^3)^2}{1-q^7+}\ldots,\tag 4 $$ Set now the value $a=q^{\nu+1/2}$, $\nu\in\{0,1,2,\ldots\}$ in (2) to get $$ P\left(q^{\nu+1/2},q\right)=\exp\left(-4\sum^{\infty}_{n=0}\frac{q^{(2n+1)(\nu+1/2)}}{(2n+1)(1-q^{2n+1})}\right)= $$ $$ =\exp\left(-4\sum^{\nu-1}_{j=0}\textrm{arctanh}(q^{j+1/2})+\textrm{arctanh}(k_r)\right) \tag 5 $$

( For many applications and very interested notes see:

http://www-elsa.physik.uni-bonn.de/~dieckman/InfProd/InfProd.html )

Here the function $k_r$ is the elliptic singular modulus and $k'_r=\sqrt{1-k_r^2}$.

For $\nu=0$ in (5) we get $$ P\left(q^{1/2},q\right)=\frac{k'_r}{1-k_r}. $$ Hence your continued fraction $u_0(q^{1/2},q)$ is $$ u_0(q^{1/2},q)=\frac{k'_r+k_r-1}{k'_r-(k_r-1)}=\frac{2q^{1/2}}{1-q+}\frac{q(1+q)^2}{1-q^3+}\frac{q^2(1+q^2)^2}{1-q^5+}\frac{q^3(1+q^3)^2}{1-q^7+}\ldots,\tag 6 $$ where $q=e^{-\pi\sqrt{r}}$, $r>0$.

Relation (6) with $r=1$, $k_1=k'_1=\frac{1}{\sqrt{2}}$ gives: $$ u_0\left(e^{-\pi/2},e^{-\pi}\right)=\sqrt{2}-1. $$

References

[1]: N.D. Bagis and M.L. Glasser. "Evaluations of a Continued Fraction of Ramanujan". Rend. Sem. Mat. Univ. Padova, Vol. 133 (2015). (submited 2013)

Nikos Bagis
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