TITLE: On Shanks' quartic approximation $\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)$
QUESTION [37 upvotes]: In Mathworld's "Pi Approximations", (line 58), Weisstein mentions one by the mathematician Daniel Shanks that differs by a mere $10^{-82}$,
$$\pi \approx \frac{6}{\sqrt{3502}}\ln(2u)\color{blue}{+10^{-82}}\tag{1}$$
and says that $u$ is a product of four simple quartic units, but does not explicitly give the expressions. I managed to locate Shanks' Dihedral Quartic Approximations and Series for Pi (1982) Quartic Approximations for Pi (1980) online before so,
$$u = (a+\sqrt{a^2-1})^2(b+\sqrt{b^2-1})^2(c+\sqrt{c^2-1})(d+\sqrt{d^2-1}) \approx 1.43\times10^{13}$$
where,
$$\begin{aligned}
a &= \tfrac{1}{2}(23+4\sqrt{34})\\
b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\
c &= (429+304\sqrt{2})\\
d &= \tfrac{1}{2}(627+442\sqrt{2})
\end{aligned}$$
(Remark: In Shank's paper, the expressions for $a,b$ are different since he didn't express them as squares.)  
A small tweak to $(1)$ can vastly increase its accuracy to $10^{-161}$,
$$\pi \approx \frac{1}{\sqrt{3502}}\ln\big((2u)^6+24\big)\color{blue}{+10^{-161}}\tag{2}$$
I noticed the constant $u$ can also be expressed in terms of the Dedekind eta function $\eta(\tau)$ as,
$$u = \frac{1}{2}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-3502})}{\eta(\sqrt{-3502})}\right)^4\approx 1.43\times 10^{13}$$
which explains why $24$ improves the accuracy. Note that the class number of $d = 4\cdot3502$ is $h(-d)=16$, and $u$ is an algebraic number of deg $16$. Mathworld has a list of class numbers of d. However, we can also use those with $h(-d)=8$ such as,
$$x = \frac{1}{\sqrt{2}}\left(\frac{\eta(\tfrac{1}{2}\sqrt{-742})}{\eta(\sqrt{-742})}\right)^2 \approx 884.2653\dots$$
which is a root of the 8th deg,
$$1 + 886 x + 1535 x^2 + 962 x^3 + 1628 x^4 - 962 x^5 + 1535 x^6 - 886 x^7 + x^8 = 0$$
Question: Analogous to $u$, how do we express $x$ as a product of two quartic units?

REPLY [6 votes]: The following is not a direct answer to your question, but is rather too long for a comment.
This is basically Ramanujan's approximation $$\pi \approx \frac{24}{\sqrt{n}}\log(2^{1/4}g_{n}) = \frac{6}{\sqrt{n}}\log (2g_{n}^{4}) = \frac{6}{\sqrt{n}}\log (2u)\tag{1}$$ where $g_{n}$ is Ramanujan's class invariant given by $$g_{n} = 2^{-1/4}e^{\pi\sqrt{n}/24}(1 - e^{-\pi\sqrt{n}})(1 - e^{-3\pi\sqrt{n}})(1 - e^{-5\pi\sqrt{n}})\cdots\tag{2}$$ Here $n = 3502 = 2 \times 17 \times 103$ and it is known that for many values of $n$ the class invariant $g_{n}$ is a unit so that that $u = g_{n}^{4}$ is also a unit. The value of $u$ given in the question is clearly a unit.
Also raising both sides of $(2)$ to 24th power and doing some simpification we get $$(2u)^{6} + 24 = 64g_{n}^{24} + 24 = e^{\pi\sqrt{n}} + 276e^{-\pi\sqrt{n}} + \cdots = e^{\pi\sqrt{n}}(1 + 276e^{-2\pi\sqrt{n}} + \cdots)\tag{3}$$ and hence on taking logs we get $$\log\{(2u)^{6} + 24\} = \pi\sqrt{n} + \log(1 + 276e^{-2\pi\sqrt{n}}) \approx \pi\sqrt{n} + 276e^{-2\pi\sqrt{n}}$$ and thus $$\pi \approx \frac{1}{\sqrt{n}}\log\{(2u)^{6} + 24\} - \frac{276}{\sqrt{n}}e^{-2\pi\sqrt{n}}\tag{4}$$ The error is of the order of $e^{-2\pi\sqrt{n}}$ and in formula $(1)$ it is of the order $e^{-\pi\sqrt{n}}$ (as evident from infinite product $(2)$). Putting $n = 3502$ we see that error in formula $(1)$ is of the order $10^{-81}$ and that in formula $(4)$ is of the order $10^{-161}$. See Ramanujan's paper "Modular Equations and Approximations to $\pi$".