Of French Gothic architecture and Ramanujan's $\pi\approx \frac{9}{5}+ \sqrt{ \frac{9}{5} } = 3.1416\dots $?

Question

I. Data

Ramanujan found,

$$\pi\approx \frac{9}{5}+ \sqrt{ \frac{9}{5} } = 3.1416\dots $$

As John Alexiou stated in this 2012 post, the golden ratio $\phi$ is exactly,

$$ \phi = \frac{5}{6} \left( \frac{9}{5}+\sqrt{\frac{9}{5}}\right) - 1 $$

and Jaume Oliver Lafont pointed out in his answer that this is equivalent to,

$$ \phi^2 = \frac{5}{6} \left( \frac{9}{5}+\sqrt{\frac{9}{5}}\right)\qquad$$

since $\phi^2=\phi+1$. Of course, this immediately implies,

$$ \frac{6\,\phi^2}{5} \approx \pi\qquad$$

Interestingly, the approximation seems long known by the French from the 12th century. (See Frédéric Beatrix's "Squaring the circle like a medieval master mason", Section 5, Proposed dating for the approximation $\frac{6\,\phi^2}{5} \approx \pi$.)

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II. Squaring the circle

From this answer by Tankut Begyu on a post about geometric constructions, taking the square root of both sides, we get,

$$\overline{FM}= \phi\sqrt{\frac65}\approx\sqrt{\pi}$$

and this is the red line segment in Beatrix's construction below,

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III. Pi formulas

In Ramanujan's 1914 paper, "Modular Equations and Approximations to $\pi$", one formula is,

$$\frac1{\pi}=\frac{-1+5\sqrt5}{2^5}+\frac1{2^5}\sum_{n=1}^\infty\frac{(2n)!^3}{n!^6} \frac{(30n-1)+(42n+5)\sqrt5}{(2^{12}\phi^8)^n}$$

If we truncate this, then,

$$\frac1{\pi}\approx\frac{-1+5\sqrt5}{2^5}$$

but is not Ramanujan's other approximation,

$$\pi\approx \frac{9}{5}+ \sqrt{ \frac{9}{5} }$$

even if we flip it over.

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IV. Question

Is there an unknown Ramanujan-type pi formula such that if we trucate it, then,

$$\frac1{\pi}\approx \left(\frac{9}{5}+ \sqrt{ \frac{9}{5} }\right)^{-1}$$

is the first approximation?

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Edit: In response to a comment, a Ramanujan-type pi formula has the simple form,

$$\frac1{\pi}=\sum_{n=0}^\infty S(n) \frac{An+B}{C^n}$$

where $(A,B,C)$ are algebraic numbers and $S(n)$ is one of four well-defined integer sequences. Given the Pochhammer symbol $(a)_n$ or binomial coefficients $\binom{n}{k}$, these are,

\begin{align} S_1(n) &= \frac{(1/2)_n(1/6)_n(5/6)_n\, 1728^n }{n!^3} = \binom{2n}{n}\binom{3n}{n}\binom{6n}{3n}= \frac{(6n)!}{(3n)!\,n!^3}\\ S_2(n) &= \frac{(1/2)_n(1/4)_n(3/4)_n\, 256^n}{n!^3} \, = \, \binom{2n}{n}\binom{2n}{n}\binom{4n}{2n} = \frac{(4n)!}{n!^4}\\ S_3(n) &= \frac{(1/2)_n(1/3)_n(2/3)_n\, 108^n}{n!^3} \, = \, \binom{2n}{n}\binom{2n}{n}\binom{3n}{n} = \frac{(2n)!\,(3n)!}{n!^5}\\ S_4(n) &= \frac{(1/2)_n(1/2)_n(1/2)_n\, 64^n}{n!^3} \; = \; \binom{2n}{n}\binom{2n}{n}\binom{2n}{n} \; = \;\frac{(2n)!^3}{n!^6} \end{align}

or explicitly,

$S_1(n) = 1, 120, 83160, 81681600,\dots$ (A001421)

$S_2(n) = 1, 24, 2520, 369600,\dots$ (A008977)

$S_3(n) = 1, 12, 540, 33600,\dots$ (A184423)

$S_4(n) = 1, 8, 216, 8000,\dots$ (A002897)

P.S. Ramanujan gave 17 such formulas in his 1914 paper, but there are actually 36 formulas where $(A,B,C)^2$ are integers (he missed half, mostly from $S_1$), and infinitely many if they are algebraic numbers. A Ramanujan-Sato formula for $\dfrac1{\pi}$ of level 5 and above uses different $S(n)$ and $C$ is a value of a modular form.

Answer 1

I. First search

I checked my old notes to see if the answer was there. Given the golden ratio $\phi$, then,

\begin{align} e^{\pi\sqrt{5}}&= 2^6\phi^6-24.247\dots\\[4pt] e^{\pi\sqrt{10}}&= 2^6\phi^{12}+23.987\dots\\[4pt] e^{\pi\sqrt{15}}&= 2^{12}\phi^8-24.001\dots\\[4pt] e^{\pi\sqrt{25}}&= 2^6\phi^{24}-24.00004\dots \end{align}

We can use these four radicals in Ramanujan-type pi formulas,

\begin{align} \qquad\frac1{\pi} &= \frac1{\phi^{5/2}}\sum_{n=0}^\infty\frac{(2n)!^3}{n!^6}\frac{1+2n\phi\sqrt5}{(2^6\phi^6)^n}\\[6pt] \frac1{\pi} &= \frac1{2\,\phi^{3}}\sum_{n=0}^\infty\frac{(2n)!^3}{n!^6}\frac{-4+3(4n+1)\sqrt5}{(-2^6\phi^{12})^n}\\[6pt] \color{red}{\frac1{\pi}} &= \frac1{32}\sum_{n=0}^\infty\frac{(2n)!^3}{n!^6}\frac{(30n-1)+(42n+5)\sqrt5}{(2^{12}\phi^8)^n}\\[6pt] \frac1{\pi} &= \frac{5^{5/4}}{\phi^{8}}\sum_{n=0}^\infty\frac{(2n)!^3}{n!^6}\frac{2+12n\phi^2}{(2^6\phi^{24})^n} \end{align}

where only the red one was found by Ramanujan. Truncating to the first term,

\begin{align} \frac1{\pi} &\approx \frac{1}{\phi^{5/2}}\\[5pt] \frac1{\pi} &\approx \frac{-4+3\sqrt5}{2\phi^{3}}\\[5pt] \frac1{\pi} &\approx \frac{-1+5\sqrt5}{32}\\[5pt] \frac1{\pi} &\approx \frac{2\cdot5^{5/4}}{\phi^{8}} \end{align}

So none (after manipulation) is Ramanujan's $\frac95+\sqrt{\frac95}$, and only the last one has comparable accuracy.

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II. Second search

So I looked at my copy of Ramanujan's 1914 paper. It turned out it was there all along, a special case of eq. $21$. Let $\color{blue}{q=e^{\pi\sqrt{n}}}$ and define a function $R_n$ as,

$$R_n = \left(1-\frac3{\pi\sqrt{n}}-24 \sum_{k=1}^\infty \frac{k}{q^{2k}-1}\right) \left(1-24\sum_{k=1}^\infty \frac{2k-1}{q^{2k-1}+1}\right)^{-1} $$

then Ramanujan states that $R_n$ is a radical for rational $n$ and that,

$$\pi \approx \frac3{(1-R_n)\sqrt{n}}$$

where the approximation gets better as $n$ increases. Thus, if $n=25$ and using the formula for $R_n$, one can indeed verify that,

$$\frac3{(1-R_{25})\sqrt{25}} = \frac95+\sqrt{\frac95}$$

Thus, it is not a truncated pi formula but originates from the different function $R_n$ whose output is a radical, though Ramanujan knew it could also be used to approximate pi.

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P.S. One can also notice that $R_n$ for $n=(11,19,43,67,163)$ are roots of cubics. For example,

$$11R_{11}=-\frac{5T+9}{2T-6}$$

where $T$ is the tribonacci constant, a root of $T^3-T^2-T-1=0$.

Answer 2

This is not an answer but a curiosity.

When I was working in quantum physics, just for the fun of playing with fundamental constants, I proposed as an approximation $$\pi \sim 3+\frac 3{10}\Big(\frac{360}{\varphi ^2}-\frac{1}{\alpha} \Big)$$ where $\alpha$ is Sommerfeld constant (the fine structure constant) defined as

$$\alpha=\frac{e^2}{2\, \varepsilon_0\, h \,c}$$ where

• $e$ is the elementary charge
• $\varepsilon_0$ is absolute dielectric permittivity of classical vacuum.
• $h$ is Planck constant
• $c$ is the speed of light

Using the most recent value (february 2023) $$\frac 1 \alpha=137.035 999 166 (02)$$ then

$$3+\frac 3{10}\Big(\frac{360}{\varphi ^2}-\frac{1}{\alpha} \Big)=\color{red}{3.1415}29$$

Answer 3

This is not an answer. Just out of personal curiosity.

Being just fascinated by all results shown in the question, in @Tito Piezas III's answer and the links in comments, I wondered what would happen if we replace in $R_n$ the summations by the corresponding integrals. $$\alpha=\int_1^\infty \frac{k}{q^{2 k}-1}\,dk=\frac{1}{4 \log ^2(q)} \left(\text{Li}_2\left(\frac{1}{q^2}\right)-2 \log \left(1-\frac{1}{q^2}\right) \log(q)\right)$$ $$\beta=\int_1^\infty \frac{2 k-1}{q^{2 k-1}+1}\,dk=\frac{1}{2 \log ^2(q)}\left(\log \left(1+\frac{1}{q}\right) \log(q)-\text{Li}_2\left(-\frac{1}{q}\right)\right)$$ I did not change anything else and computed the value $$\color{blue}{\large\tilde \pi_{(n)}=\frac3{(1-R_n)\sqrt{n}}}$$ Converting the results to decimals $$\left( \begin{array}{cc} n & \tilde \pi_{(n)}-\pi \\ 10 & 4.6801\times 10^{-04} \\ 20 & 8.3768\times 10^{-06} \\ 30 & 3.6944\times 10^{-07} \\ 40 & 2.6326\times 10^{-08} \\ 50 & 2.5570\times 10^{-09} \\ 60 & 3.0990\times 10^{-10} \\ 70 & 4.4430\times 10^{-11} \\ 80 & 7.3590\times 10^{-12} \\ 90 & 1.3611\times 10^{-12} \\ 100 & 2.1094\times 10^{-13} \\ \end{array} \right)$$