## Posts Tagged ‘Ramanujan’

### Hypergeometric formulas for Ramanujan’s continued fractions 2

(continued from yesterday’s post)

III. Icosahedral group

Given the Rogers-Ramanujan identities (see also here),

\begin{aligned} G(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\H(q) &= \sum_{n=0}^\infty \frac{q^{n^2+n}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\end{aligned}

I observed that,

\begin{aligned}&q^{-1/60}G(q) = j^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\big) = (j-1728)^{1/60}\,_2F_1\big(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{1728-j}\big)\\[2.5mm]&q^{11/60}H(q) = j^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\big) = (j-1728)^{-11/60}\,_2F_1\big(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{1728-j}\big)\end{aligned}

where, as in the previous post, $j=j(\tau)$ is the j-function, $q = e^{2\pi i \tau} = \exp(2\pi i \tau)$, $\tau = \sqrt{-N}$, and $N>1$.  Since it is known that,

\begin{aligned}&r(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \cfrac{q^3}{1 + \ddots}}}} = \frac{q^{11/60}H(q)}{q^{-1/60}G(q)} = \frac{q^{11/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}}{q^{-1/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(q^{5n-4})}}\end{aligned}

this implies that,

\begin{aligned}r(q) &=\frac{j^{-11/60}\,_2F_1\big(\frac{31}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{j}\big) }{j^{1/60}\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{j}\big)}\\[3mm]&=\frac{(j-1728)^{-11/60}\,_2F_1\big(\frac{41}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{1728-j}\big) }{(j-1728)^{1/60}\,_2F_1\big(\frac{29}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{1728-j}\big)}\end{aligned}

Example. Let $\tau = \sqrt{-4}$, hence $j = j(\sqrt{-4}) = 66^3$. Then,

$1/r(q) -r(q) = \left(\frac{1+\sqrt{5}}{2}\right)^4+\left(\frac{1+\sqrt{5}}{2}\right)5^{3/4} = 12.2643\dots$

Furthermore, since Ramanujan established that,

$G(q^{11})H(q)-q^2G(q)H(q^{11}) = 1$

if we define the two functions,

\begin{aligned}U(\tau) &= \big(j(\tau)\big)^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j(\tau)}\big)\\V(\tau) &= \big(j(\tau)\big)^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j(\tau)}\big) \end{aligned}

then the counterpart hypergeometric identity is also beautifully simple and given by,

$U(11\tau)V(\tau)-U(\tau)V(11\tau)=1$

In the next post, we will use one of the hypergeometric formulas to solve the general quintic.

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### Hypergeometric formulas for Ramanujan’s continued fractions 1

There are five Platonic solids, two are duals to another two, while the tetrahedron is self-dual. As such, this gives rise to 3 polyhedral groups: the tetrahedral group of order 12, the octahedral group of order 24, and the icosahedral group of order 60.

Amazingly, Ramanujan found 3 continued fractions that can be associated with each group. See this article for more details.  It turns out there are also corresponding hypergeometric formulas, and the numbers 12, 24, and 60 naturally appears.

First though, define the j-function as,

$j = j(\tau) = \frac{1}{q} + 744 + 196884q + 21493760q^2 + \dots$

where,

$q = e^{2\pi i \tau} = \exp(2\pi i \tau)$

This can be conveniently calculated in Mathematica as,

$j(\tau) = 1728\text{KleinInvariantJ}(\tau)$

NOTE:  In the formulas below, it will be assumed that,

$\tau = \sqrt{-N},\;\; N > 1$

I. Tetrahedral group

Given,

\begin{aligned}&c=c(q)=\cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \cfrac{q^3+q^6}{1 + \ddots}}}} = q^{1/3}\prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})^2}\end{aligned}

and,

\begin{aligned}&d = j^{1/3}\,\frac{\,_2F_1\big(\frac{1}{4},\frac{-1}{12},\frac{2}{3},\frac{1728}{j}\big)}{\,_2F_1\big(\frac{1}{4},\frac{7}{12},\frac{4}{3},\frac{1728}{j}\big)}\end{aligned}

then we have the simple relationship,

$d = 4c^2+c^{-1}$

Example.  Let $\tau = \sqrt{-2}$, hence $j=j(\sqrt{-2}) = 20^3$, then,

$d = 3\sqrt{2(11+4\sqrt{6})} = 19.3484\dots$

and c(q) can then be easily solved for as a cubic equation.

II. Octahedral group

Let,

\begin{aligned}&u = u(q) = \cfrac{\sqrt{2}\,q^{1/8}}{1 + \cfrac{q}{1+q + \cfrac{q^2}{1+q^2 + \cfrac{q^3}{1+q^3 + \ddots}}}} = \sqrt{2}\,q^{1/8}\prod_{n=1}^\infty\frac{1-q^{2n-1}}{(1-q^{4n-2})^2} \end{aligned}

then,

\begin{aligned}&u = \frac{\sqrt{2}}{j^{1/8}}\left(\frac{\,_2F_1\big(\frac{5}{24},\frac{13}{24},\frac{5}{4},\frac{1728}{j}\big)}{\,_2F_1\big(\frac{7}{24},\frac{-1}{24},\frac{3}{4},\frac{1728}{j}\big)}\right)^{1/2}\end{aligned}

Example. Still using $\tau = \sqrt{-2}$, then,

$u = \sqrt{1+\sqrt{2}-\sqrt{2(1+\sqrt{2})}} = 0.4656\dots$

III. Icosahedral group

(To be discussed in the next post.)

### A new formula for Apery’s constant and other zeta(2n+1)?

I. Introduction

In Identities Inspired from Ramanujan’s Notebooks, Simon Plouffe recounts how, based on Ramanujan’s,

\begin{aligned}\sum_{k=1}^\infty \frac{\coth(\pi k)}{k^3} = \frac{7}{180}\pi^3\end{aligned}

he found,

\begin{aligned}\zeta(3) &= \frac{7\pi^3}{180}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{2\pi k}-1)}\\ \zeta(7) &= \frac{19\pi^7}{56700}-2\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k}-1)} \end{aligned}

and similar ones for other s = 4m+3.  On a hunch, and using Mathematica’s LatticeReduce function, I found that,

\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}

etc.

II. Functions

If we define,

\begin{aligned} &U_a(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{a\pi k}-1)}\end{aligned}

then Plouffe discovered integer relations between,

$\zeta(s), \pi^s, U_1(s),\, U_2(s),\, U_4(s)$

for odd s, with s = 3 being,

\begin{aligned} &\zeta(3) = 28U_1(3)-37U_2(3)+7U_4(3)\\ &\text{and,}\\ &\pi^3 =720U_1(3)-900U_2(3)+180U_4(3)\end{aligned}

Eliminating $U_1(s),\, U_4(s)$ leads to the 3-term equalities in the Introduction.  See Chamberland’s and Lopatto’s Formulas for Odd Zeta Values.  On the other hand, by defining the function,

\begin{aligned} &V_b(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{b\pi k\sqrt{2}}-1)}\end{aligned}

I observed integer relations between,

$\zeta(s), \pi^s \sqrt{2},\, V_1(s),\, V_2(s),\, V_3(s),\, V_6(s)$

also for odd s, with s = 3 as,

\begin{aligned} &\zeta(3)-102V_1(3)+99V_2(3)+10V_3(3)-5V_6(3) = 0\\ &\text{and,}\\ &\pi^3\,\sqrt{2}-3720V_1(3)+3540V_2(3)+360V_3(3)-180V_6(3) = 0\end{aligned}

and so on.  Eliminating $V_3(s),\, V_6(s)$ leads to the 4-term equalities in the Introduction.

III. Conjecture

The 4-term equalities have coefficients that are simple except for one term. Recall that,

\begin{aligned} V_1(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{\pi k \sqrt{2}}-1)}\\ V_2(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{2\pi k \sqrt{2}}-1)}\end{aligned}

Conjecture:

“Using the positive case of $\pm 1$ for s = 4m+3, and the negative for s = 4m+5, then in the equation,

\begin{aligned}&F(s)\zeta(s-1)\pi\sqrt{2} = \big(2^{(s-1)/2}\pm 1\big)\frac{\zeta(s)}{2}+2^{(s-1)/2}\,V_1(s) \pm V_2(s) \end{aligned}

$F(s)$ is a rational number.”

The first few for s = {3, 7, 11,…} are $F(s) = \frac{1}{4}, \frac{41}{40}, \frac{319}{84}, \dots$ while for s = {5, 9, 13,…} are $F(s) = \frac{1}{3}, \frac{19}{11}, \frac{5017}{691},\dots$  These rationals may have a closed-form expression in terms of Bernoulli numbers, but I do not yet know the exact formulation.

### Ramanujan’s continued fraction for Catalan’s constant

Ramanujan was a goldmine when it came to continued fractions (and many others).  In this post, two families will be given: they involve pi and Catalan’s constant as special cases.  However, versions will be given that roughly double the rate of convergence.

Recall that Catalan’s constant C is given by,

$C = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 0.915965\dots$

Ramanujan gave the beautiful pair of continued fractions.  Let $|x| > 1$, then,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{x^2-1 + \cfrac{2^2}{1 + \cfrac{2^2}{x^2-1 + \cfrac{4^2}{1 + \cfrac{4^2}{x^2-1 +\ddots}}}}}$

$g(x) \; = \;\frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4 (\frac{x+3}{4})}\; =\; \cfrac{16}{x^2-1 + \cfrac{1^2}{1 + \cfrac{1^2}{x^2-1 + \cfrac{3^2}{1 + \cfrac{3^2}{x^2-1 +\ddots}}}}}$

One can just admire Ramanujan’s artistry — one continued fraction uses even numerators, the other, odd numerators.  For even x > 0, it is easily seen the first one involves Catalan’s constant.  For example,

$f(2) = 2(1-C)$

On the other hand, for odd x > 1, both involve pi,

$f(3) = \frac{1}{24}(12-\pi^2)$

$g(3) = \pi^2$

Notice though that numerators are repeated. Thanks to the insight of J.M. from a mathstackexchange post about Apery’s constant, we can speed up the rate of convergence of this particular form by getting rid of every other level and extracting even convergents.  After some slightly tedious algebraic manipulation, given,

$y = \cfrac{a_1}{b_1 + \cfrac{a_2}{1 + \cfrac{a_3}{b_2 + \cfrac{a_4}{1 +\ddots}}}}$

then,

$y_{even} = \cfrac{a_1}{b_1+a_2 - \cfrac{a_2\, a_3}{b_2+a_3+a_4 - \cfrac{a_4\, a_5}{b_3+a_5+a_6 - \cfrac{a_6\, a_7}{b_4+a_7+a_8 -\ddots}}}}$

So Ramanujan’s continued fractions are now the more compact,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where, starting with n = 1,

$u_n = (2n-2)^2 + (2n)^2 + x^2-1$

and,

$g(x) \; = \; \frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4(\frac{x+3}{4})}\; =\; \cfrac{16}{-1+v_1 - \cfrac{1^4}{v_2 - \cfrac{3^4}{v_3 - \cfrac{5^4}{v_4 -\ddots}}}}$

$v_n = (2n-3)^2 + (2n-1)^2 + x^2-1$

Thus, we have the slightly faster continued fraction for Catalan’s constant C,

$f(2) = 2(1-C) = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where,

$u_n = 8n^2-8n+7$

(There is an even faster one by Zudilin given in An Apery-like difference equation for Catalan’s constant though he states this still does not prove C is irrational.)

### Continued fractions for Zeta(2) and Zeta(3)

It seems there is a nice “pattern” between the continued fractions for the Riemann zeta function,

$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\dots$

at s = 2, and s = 3.  First though, a little introduction.

The origins of this function go back to 1644 when, at the tender age of 18, the Italian mathematician Pietro Mengoli (1626-1686) first proposed what would be later known as the Basel Problem, namely to determine the exact value of the sum of the reciprocals of the squares.  Euler would later find that, for n a positive integer, then $\zeta(2n)$ is a rational multiple of $\pi^{2n}$, with the first case s = 2 as,

$\zeta(2) = \frac{\pi^2}{6}$

which obviously is irrational.  However, the status of odd s was not as easy to resolve.  It was only in 1979 that Apery proved that $\zeta(3)$ is irrational, which henceforth was called Apery’s constant.

I. Zeta(2)

One of its infinite number of continued fractions can be given as,

$m\,\zeta(2) = \cfrac{1}{u_1+\cfrac{1^4}{u_2+\cfrac{2^4}{u_3+\cfrac{3^4}{u_4+\ddots}}}}$

where $m = \frac{1}{2}$ and the $u_n$, starting with n = 1, are generated by,

$u_n = 2n-1 = 1, 3, 5, 7,\dots$

or simply the odd numbers.  The convergence is slow, but Apery found it can accelerated by using a quadratic function,

$u_n = 11n^2-11n+3 = 3, 25, 69, 135,\dots$

but now $m = \frac{1}{5}$.

II. Zeta(3)

Likewise, this also has an infinite number of continued fractions (see Ramanujan’s versions here), but one important form is,

$m\,\zeta(3) = \cfrac{1}{v_1-\cfrac{1^6}{v_2-\cfrac{2^6}{v_3-\cfrac{3^6}{v_4-\ddots}}}}$

where $m = 1$ and the $v_n$, again starting with n = 1, are,

$v_n = (n-1)^3+n^3 = (2n-1)(n^2-n+1) = 1, 9, 35, 91, \dots$

Apery again found an accelerated version,

$v_n = 34n^3-51n^2+27n-5 = (2n-1)(17n^2-17n+5) = 5, 117, 535, 1463, \dots$

where now $m = \frac{1}{6}$, and established that its rate of convergence was such that $\zeta(3)$ could not be a ratio of two integers.

III. Connection between Zeta(2) and Zeta(3)

Define the two sequences,

$B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251,\dots$

$A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, \dots$

then it was established that these Apery numbers have the recurrence relations,

$n^2 B_n = (11n^2-11n+3)B_{n-1}+(n-1)^2B_{n-2}$

$n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}-(n-1)^3A_{n-2}$

Interesting that the same polynomials pop up, isnt’ it?  Furthermore, their limiting ratios have the common form,

$\left(\frac{b+\sqrt{b^2+4}}{2}\right)^{6-b}$

thus for b = 1, 2,

$\lim_{n \to \infty} \frac{B_{n+1}}{B_n} = \left(\frac{1+\sqrt{5}}{2}\right)^5 = 11.0901\dots$

$\lim_{n \to \infty} \frac{A_{n+1}}{A_n} = \left(1+\sqrt{2}\right)^4 = 33.9705\dots$

hence the golden ratio and the silver ratio surprisingly turn up in the continued fractions for $\zeta(2)$ and $\zeta(3)$, respectively.  It is easy to check other sequences,

$C_n = \sum_{k=0}^n {\binom n k}^p {\binom {n+k}k}^q$

for some small {p, q}, but there are no other recurrence relations similar in form to the two above.

IV. Zeta(5)

There is an orderly non-simple continued fraction for all $\zeta(s)$, with the next odd s as,

$m \zeta(5) = \cfrac{1}{w_1-\cfrac{1^{10}}{w_2-\cfrac{2^{10}}{w_3-\cfrac{3^{10}}{w_4-\ddots}}}}$

where $m = 1$ and the $w_n$ are,

$w_n = (n-1)^5+n^5 = (2n-1)(n^4-2n^3+4n^2-3n+1) = 1, 33, 275, 1267, \dots$

Unfortunately, no one has yet found an accelerated version where the $w_n$ are generated by a 5th degree (or higher) polynomial, and m is rational.

Can you find one?

For further reading, refer to Alfred van der Poorten’s excellent, A Proof That Euler Missed.

### Some of Ramanujan’s continued fractions for pi

The digits of pi go on forever apparently with no discernible pattern. However, there are beautifully simple patterns in its (ironically) non-simple continued fraction expansions. Examples are,

$\pi = \cfrac{4}{1 + \cfrac{1^2}{2 + \cfrac{3^2}{2 + \cfrac{5^2}{2 + \ddots} }}}$

known by Lord Brouncker (1620-1684), and,

$\pi = 3 + \cfrac{1^2}{6 + \cfrac{3^2}{6 + \cfrac{5^2}{6 + \ddots} }}$

One can see the affinity between the two.  They in fact belong to the same family.  Given complex numbers {n, x} with $Re[x] > 0$, then,

$F(n,x) = \Large{\frac{\Gamma\left(\frac{x+n+1}{4} \right)\Gamma\left(\frac{x-n+1}{4} \right) }{\Gamma\left(\frac{x+n+3}{4} \right)\Gamma\left(\frac{x-n+3}{4} \right)}} = \cfrac{4}{x + \cfrac{1^2-n^2}{2x + \cfrac{3^2-n^2}{2x + \cfrac{5^2-n^2}{2x + \ddots} }}}$

where $\Gamma(\tau)$ is the gamma function.  This is Entry 25 in Ramanujan’s Second Notebook (Chapter 12) though this result was also known by Euler. For the case = 0, the continued fraction assumes the form of the examples and the function simplifies as,

$F(0,x) = \frac{\Gamma^2\left(\frac{x+1}{4}\right)}{\Gamma^2\left(\frac{x+3}{4}\right)}$

For x an odd integer, then (0, x) is a rational multiple of $\pi$ or $1/\pi$.  Specifically, for x = {1, 3, 5, 7, 9}, we have,

$F(0,x) = \pi,\; \frac{4}{\pi}, \frac{\pi}{4}, \frac{16}{9\pi}, \frac{9\pi}{64}$

More generally,

$F(0,4m+1) = \big(\frac{(2m)!}{m!^2}\big)^2\, \frac{\pi}{2^{4m}}$

$F(0,4m+3) = \big(\frac{m!^2}{(2m+1)!}\big)^2\, \frac{2^{4m+2}}{\pi}$

See p. 178 of Annie Cuyt’s HandBook of Continued Fractions for Special Functions.  For more examples, see also this article Ramanujan’s Continued Fractions, Apery’s Constant, and more.

### Ramanujan’s pi approximations and Pell equations

Ramanujan gave many fascinating formulas and approximations to pi. Using one of his examples, we can give its family. First, define the fundamental units,

$U_{2} = 1+\sqrt{2}$

$U_{29} = \frac{5+\sqrt{29}}{2}$

$U_{58} = 99+13\sqrt{58}$

$U_{174} = 1451+110\sqrt{174}$

These are involved in fundamental solutions to Pell equations.  For example, for $x^2-58y^2 = -1$, it is {x, y} = {99, 13}, (see the values above), while for $x^2-174y^2 = 1$ it is {x, y} = {1451, 110}. Using these solutions to Pell equations, then,

$\pi \approx \frac{1}{\sqrt{58}} \ln \Big[ 2^6 (U_{29})^{12} \Big]$

$\pi \approx \frac{1}{2\sqrt{58}} \ln \left[ 2^9 \left((U_2)^3 U_{29} \sqrt{U_{58}} \,\right)^6 \right]$

$\pi \approx \frac{1}{3\sqrt{58}} \ln \left[ 2^6 (U_{29})^{12} (U_{174})^2 \left( \sqrt{\frac{9+3\sqrt{6}}{4} } + \sqrt{\frac{5+3\sqrt{6}}{4}}\right)^{24}\right]$

$\pi \approx \frac{1}{4\sqrt{58}} \ln \Big[ 2^9 \left((U_2)^3 U_{29} \sqrt{2U_{58}} \,\right)^3 \left(\sqrt{v+1} +\sqrt{v}\right)^{12}\Big]$

where,

$v = 2^{-1/2}(U_2)^6(U_{29})^3$

Nice, isn’t it?  The second to the last approximation is by Ramanujan which is accurate to 31 digits, while the last is by this author and is accurate to 42 digits.  (Can anyone find a nice expression for the next step? )  The expression inside the log function is the exact value of,

$\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}$

where $\eta(\tau)$ is the Dedekind eta function, and $\tau = \frac{\sqrt{-58}}{2}$, $\tau = \frac{2\sqrt{-58}}{2}$$\tau = \frac{3\sqrt{-58}}{2}$$\tau = \frac{4\sqrt{-58}}{2}$, respectively.

The fundamental discriminant d = -4∙58 has class number h(d) = 2.  Another one with the same class number is d = -4∙37.  Hence, given,

$U_{37} = 6+\sqrt{37}$

$U_{111} = 295+28\sqrt{111}$

then,

$\pi \approx \frac{1}{\sqrt{37}} \ln \Big[ 2^6 (U_{37})^{6} \Big]$

$\pi \approx \frac{1}{3\sqrt{37}} \ln \left[ 2^6 (U_{37})^{6} (U_{111})^2 \left( \sqrt{\frac{37+20\sqrt{3}}{4} } + \sqrt{\frac{33+20\sqrt{3}}{4}}\right)^{12}\right]$

where the expression inside the log function is now the absolute value of the eta quotient at $\tau = \frac{1+\sqrt{-37}}{2}$ and $\tau = \frac{1+3\sqrt{-37}}{2}$.