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,

$$u = (a+\sqrt{a^2-1})^2(b+\sqrt{b^2-1})^2(c+\sqrt{c^2-1})(d+\sqrt{d^2-1}) \approx 1.43\times10^{13}$$


$$\begin{aligned} a &= \tfrac{1}{2}(23+4\sqrt{34})\\ b &= \tfrac{1}{2}(19\sqrt{2}+7\sqrt{17})\\ c &= (429+304\sqrt{2})\\ d &= \tfrac{1}{2}(627+442\sqrt{2}) \end{aligned}$$

(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?

  • $\begingroup$ Shanks' cited paper 'Dihedral...' (accessible for guests) and a follow up with Newman ’On a sequence arising in series for pi'. $\endgroup$ – Raymond Manzoni Oct 5 '13 at 9:21
  • $\begingroup$ Thanks. This paper is longer than I remembered and I just realized what I found before was the shorter Quartic Approximations for Pi. Curious that Shanks does not mention the connection to the eta quotient. $\endgroup$ – Tito Piezas III Oct 5 '13 at 17:19

Let $a = \displaystyle{ \frac{11 + \sqrt{106}}{2}}$ and $b = \displaystyle{ \frac{21 + 2 \sqrt{106}}{2}}.$ Then

$$x = (a + \sqrt{a^2 - 1}) (b + \sqrt{b^2 + 1}).$$

As requested, this exhibits $x$ as a product of two quartic units. (For the purists, note that $a$ and $b$ are only half algebraic integers, but the expressions above are genuinely units. I wrote them in this form to conform with the previous example of the OP) The first unit (involving $a$) generates a degree four extension whose Galois closure has Galois group $D_8$, but the latter generates a Galois extension with Galois group $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$.

For those playing at home, this allows for the extra simplification

$$b + \sqrt{b^2 + 1} = (1 + \sqrt{2})^2 \cdot \frac{7 + \sqrt{53}}{2} .$$

  • $\begingroup$ Beautiful! It took two years for someone to answer, but this implies $$\pi \approx \frac{1}{\sqrt{742}} \ln\Big( \big(\sqrt{2}\,u \big)^{12}+24\Big) $$ which differs by a mere $10^{-74}$ and where $u=(a+\sqrt{a^2-1})\, U_2^2\, U_{53}$ with "$a$" as in the answer, and $U_2$ and $U_{53}$ are fundamental units. $\endgroup$ – Tito Piezas III May 28 '15 at 6:33
  • $\begingroup$ A quick look at a table of class numbers shows that $x_d$ as an eta quotient has that form for both $$x_{742} = (a+\sqrt{a^2-1})\,U_2^2\,U_{53}$$ $$x_{598} = (c+\sqrt{c^2-1})\,U_2^2\,U_{13}$$ where $c=6+\sqrt{26}$. $\endgroup$ – Tito Piezas III May 28 '15 at 7:09

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 $\pi$".

  • $\begingroup$ Yes, it was to be expected that Ramanujan's $g_n$ and $G_n$ functions would not be that far away. However, sometimes I prefer the Dedekind eta function since Mathematica can calculate it efficiently. And we also have to give credit to Shanks for factoring a 16-deg algebraic number into four quartic units with a common form, which is no easy feat! (I'm not sure if all $d$ with $h(-d)=16$ can be factored so prettily.) $\endgroup$ – Tito Piezas III Jun 19 '15 at 13:34

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