# Why do Pell equations appear in Ramanujan's pi formulas?

While answering this MSE question about the Pell equation $x^2-29y^2=1$, I noticed that certain fundamental solutions appeared in Ramanujan's famous pi formula.

I. Given the fundamental unit $\displaystyle U_{29} =\tfrac{5+\sqrt{29}}{2}$ and,

$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$

$$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$

$$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$

we find those integers all over Ramanujan's,

$$\frac{1}{\pi} = \frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k}$$

The same integers appear in related pi formulas in this post.

II. A similar thing happens with the Chudnovsky formula. I knew that H. Chan explored $x^2-3dy^2 = 1$ in relation to $j(\tau)$. Given the fundamental unit for $n=3d=3\times163=489$,

$$U_n =u+v\sqrt{489} =7592629975+343350596\sqrt{489} = \big(35573\sqrt{3}+4826\sqrt{163}\big)^2$$

so $u^2-489v^2=1$. The units had a common form for $d=19,43,67,163$, for the last being,

$$U_n = \left(\tfrac{1}{18}(\color{brown}{640320}-6)\sqrt{3}+4826\sqrt{163}\right)^2$$

and some experimentation yielded,

$$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+6 = \color{brown}{640320}$$

$$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+18 = \color{brown}{2^2\cdot3^3\cdot7^2\cdot11^2}$$

$$\sqrt{163}\big(U_n^{1/2}+U_n^{-1/2}\big) = \color{brown}{2^2\cdot19\cdot127\cdot163}$$

and the Chudnovsky formula,

$$12\sum_{k=0}^\infty (-1)^k \frac{(6k)!}{k!^3(3k)!} \frac{\color{brown}{2\cdot3^2\cdot7\cdot11\cdot19\cdot127\cdot163}\,k+13591409}{(\color{brown}{640320}^3)^{k+1/2}} = \frac{1}{\pi}$$

The same thing happens with the other $d$, for example,

$$U_{201} = \left(\tfrac{1}{18}(\color{brown}{5280}-6)\sqrt{3}+62\sqrt{67}\right)^2$$

so it is not a fluke. Plus, these $U_n$ are denominators in Ramanujan-Sato pi formulas of level 9.

Q. Anyone knows the reason why a fundamental unit $U_{n}$ would appear in a pi formula?

I happened to read the wonderful book "Pi and the AGM" written by Borwein brothers several months ago, and I wondered how they proved the famous formula

$$\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\cdots$$

After reading their proof, I realized that the theory of special value of modular forms is closely related to your question.

Modular form of level 2

$$8\left(\frac{\eta(2\tau)}{\eta({\tau})}\right)^{12}$$ appears in the formula given by Ramanujan. I hear that H. M. Weber computed some special values of this modular form with Kronecker limit formula, where some Eisenstein series appears in it. Proper linear combination(which is closely related to the ideal class of some imaginary quadratic field) of Eisenstein series will give logarithmic of the modular form on the RHS, while the linear combination itself is a Hecke L-function which can be decomposed to the product of Dirichlet L-functions. Then the fundamental unit naturally appears in the Dirichlet class number formula.

More details can be found in the book of C.L.Siegel(Chapter II).

After staring at my question for a while, I think I can answer it partially. But not rigorously, so perhaps someone can improve it. Ramanujan's 4 kinds of pi formulas,

$$\frac{1}{\pi} = \sum_{n=0}^\infty s(n) \frac{An+B}{C^{n+1/2}}$$

for a defined integer sequence $s(n)$ have a general closed-form in terms of the hypergeometric function and Dedekind eta function. See this short article.

After some fiddling, it turns out $A$ can be nicely factored. For level $p=2$, define the eta quotient,

$$v:=v(\tau)=\frac{1}{2\sqrt{2}}\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^6\tag1$$

(similar to zy_'s answer), then,

$$A:= A(\tau) = 4\sqrt{d}\left(v-\frac{1}{v}\right)\left(v+\frac{1}{v}\right)\tag2$$

For example:

1. For $\tau=\tfrac{1}{2}\sqrt{-d},\;d=58$: $$v(\tau)=\Big(\tfrac{5+\sqrt{29}}{2}\Big)^3=\color{blue}{70}+\color{blue}{13}\sqrt{29}\tag3$$

so the two factors of $(2)$ explain,

$$A(\tau)=4\sqrt{58}\left(v-\frac{1}{v}\right)\left(v+\frac{1}{v}\right) = 16\sqrt{2}\cdot29\color{blue}{\cdot70\cdot13}$$

2. However, for $\tau=\tfrac{1}{2}(1+\sqrt{-d}),\;d=37:$

$$v(\tau)=\zeta^3\big(6+\sqrt{37}\big)^{3/2}\tag4$$

with root of unity $\zeta = e^{2\pi i/8}$, and cannot be expanded out like $(3)$, though $A(\tau)$ is still an integer (with the imaginary unit affixed),

$$A(\tau)=2^3\cdot5\cdot29\cdot37\,i$$

So while $(4)$ also involved a unit, namely $U_{37}$, I guess it was because $(3)$ was an integer power of the unit $U_{29}$ that made it different.

Not the complete answer, but at least a try.

I think is simple. All are due to algebraic units. If $$z=a+b\sqrt{d}$$ is a unit and $$N(z)=a^2-d\cdot b^2$$, then $$N(z)=1$$ (unit). Hence if we deal with algebraic units, then $$N(z)=1=a^2-d\cdot b^2$$, is a Pell equation. See paper [1] (the knowledge of elliptic singular modulus $$k_r$$ requires the knowledge of algebraic units).

It have been proven that most of these formulas (Ramanujan formulas for $$1/\pi$$, $$1/\pi^2$$, etc) rise from elliptic functions (see [3],[5],[6]). For example:

Set $$\phi(z):={}_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,z\right)=\frac{4K^2\left(\sqrt{\frac{1}{2}\left(1-\sqrt{1-z}\right)}\right)}{\pi^2}$$ Then we ask whether exist constants $$a_1,b_1$$ and $$g$$ such that $$\sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}z^n(a_1n+b_1)=\frac{g}{\pi}\leftrightarrow b_1\phi(z)+a_1\phi'(z)=\frac{g}{\pi}.$$ Setting $$w=\sqrt{\frac{1}{2}\left(1-\sqrt{1-z}\right)}$$, then $$1-2w^2=\sqrt{1-z}$$. Also by direct calculation $$\sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}4^n(w-w^2)^n(a_1n+b_1)=\frac{4K(w)\left[a_1E(w)+(b_1-a_1(1-w)-2b_1w)K(w)\right]}{\pi^2(1-2w)}$$ Here ofcourse $$K(w)$$ and $$E(w)$$ are the complete elliptic integrals of the first and second kind. Also $$\alpha(r)=\frac{\pi}{4K^2(k_r)}-\sqrt{r}\left(\frac{E(k_r)}{K(k_r)}-1\right)$$ where $$k_r$$ is the elliptic singular modulus i.e the solution of $$\frac{K(\sqrt{1-k_r^2})}{K(k_r)}=\sqrt{r}, r>0.$$ Also $$\frac{dK(x)}{dx}=\frac{E(x)}{x(1-x^2)}-\frac{K(x)}{x}$$ Hence setting $$w=k_r$$, $$a_1=1$$ and $$b_1=\left(\frac{\alpha(r)}{\sqrt{r}}-k_r^2\right)\left(1-2k_r^2\right)^{-1}$$ we arrive to the general formula $$\sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}4^n(k_rk'_r)^{2n}\left(n+\frac{\alpha(r)-k_r^2\sqrt{r}}{\sqrt{r}(1-2k_r^2)}\right)=\frac{1}{\pi\sqrt{r}(1-2k_r^2)}$$ Evaluations of $$k_r$$ and $$\alpha(r)$$ can given for positive rationals $$r$$. For example with $$r=2$$, we have $$k_2=\sqrt{2}-1$$, $$\alpha(2)=\sqrt{2}-1$$. Hence we get the next formula $$\sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}(40\sqrt{2}-56)^{n}\left(n+\frac{2}{7}-\frac{1}{7\sqrt{2}}\right)=\frac{8+5\sqrt{2}}{14\pi}$$ For $$r$$ positive integer we can find $$k_r$$ and $$\alpha(r)$$ using methods of [1],[2]. For a table of values of $$k_r$$ from r=1 to 100 see [4]. For evaluations of $$\alpha(r)$$ see[5]

[1]: Mark B. Villarino: "Ramanujan most singular modulus". arXiv:math/0308028v4 [math.HO] 2005.

[2]: D. Broadhurst. "Solutions by radicals at Singular Values $$k_N$$ from New Class Invariants for $$N\equiv3(\textrm{mod})8$$". arXiv:0807.2976 [math-ph], (2008).

[3]: N.D. Bagis, M.L. Glasser. "Ramanujan type $$1/\pi$$ approximation formulas". Journal of Number Theory, Elsevier., (2013)

[4]: J.M. Borwein, M.L. Glasser, R.C. McPhedran, J.G. Wan, I.J. Zucker. "Lattice Sums Then and Now". Cambridge University Press. New York, (2013).

[5]: J.M. Borwein and P.B. Borwein. "Pi and the AGM". 1987 ed., New York: John Wiley and Sons Inc., (1987)

[6]: N.D. Bagis. "A General Method for Constructing Ramanujan-Type Formals for Powers of $$1/\pi$$". The Mathematica Journal. Vol. 15., (2013)

...CONTINUED

From Chan Huang's Unit Theorem we have that: If $$n\equiv 2(mod4)$$, then $$k_n$$ is unit.

From relation $$\frac{(k'_n)^2}{k_n}=2g_n^{12}\textrm{, }k'_n=\sqrt{1-k_n^2}$$ where $$g_n$$ is Weber's invariant: $$g_n=2^{-1/4}q^{-1/24}\prod^{\infty}_{m=0}(1-q^{2m+1})\textrm{, }q=e^{-\pi\sqrt{n}}.$$ Now if $$\delta_1,\delta_2,\ldots,\delta_{\nu(-8n)}$$ are the distinct odd divisors of $$-8n$$ and $$\delta'_l$$ are the complimentary divisors of $$\delta_l$$, such $$\delta'_l\delta_l=-8n$$ $$g_{2n}=\prod^{\nu(-8n)}_{l=1}\left(\frac{T_l+U_l\sqrt{\delta_l}}{2}\right)^{\nu(\delta_l)\nu(\delta'_l)/\nu(-8n)},$$ where $$\nu(n)$$ is the number of properly primitive classes of discriminant of $$n$$ and $$(T_l,U_l)$$ is the minimal solution of the Pell equation $$x^2-\delta_l y^2=4$$

NOTES.

LEMMA 1. If

1. $$uv:=g_n^6$$

2. $$2U:=u^2+u^{-2}$$, $$2V=v^2+v^{-2}$$

3. $$W=\sqrt{U^2+V^2-1}$$,

4. $$2S=U+V+W+1$$, then

$$k_n^2=(\sqrt{S}-\sqrt{S-1})^2(\sqrt{S-U}-\sqrt{S-U-1})^2(\sqrt{S-V}-\sqrt{S-V-1})^2\times$$ $$\times(\sqrt{S-W}-\sqrt{S-W-1})^2.$$

LEMMA 2. If

$$\sqrt{\alpha}:=\sqrt{ab}+\sqrt{(a+1)(b-1)}$$

$$\sqrt{\beta}:=\sqrt{cd}+\sqrt{(c-1)(d-1)}$$, then $$x_1:=(\sqrt{a+1}-\sqrt{a})(\sqrt{b}-\sqrt{b-1})(\sqrt{c}-\sqrt{c-1})(\sqrt{d}-\sqrt{d-1})$$ $$x_2:=-(\sqrt{a+1}+\sqrt{a})(\sqrt{b}+\sqrt{b-1})(\sqrt{c}+\sqrt{c-1})(\sqrt{d}+\sqrt{d-1})$$ are roots of $$x-x^{-1}=2(\sqrt{\alpha\beta}+\sqrt{(\alpha+1)(\beta-1)})$$

This is discussed on both the Wikipedia page on computing $\pi$ and in Eymard and Lafon's The number $\pi$ (where they actually refer to a book of Weber's, which I can't locate since google books is not showing the table of contents).