Ramanujan-type series for $\frac {1}{\pi }$, revisited

Abstract

In this note, we revisit Ramanujan-type series for $\frac {1}{\pi }$ and show how they arise from genus zero subgroups of $\mathrm {SL}_{2}(\mathbb {R})$ that are commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$. As illustrations, we reproduce a striking formula of Ramanujan for $\frac {1}{\pi }$ and a recent result of Cooper et al., as well as derive a new rational Ramanujan-type series for $\frac {1}{\pi }$. As a byproduct, we obtain a Clausen-type formula in some general sense and reproduce a Clausen-type quadratic transformation formula closely related to the aforementioned formula of Ramanujan.

1 Introduction

One of the best fortunes that Ramanujan left to his fellows might be the 17 extraordinary formulas for $\frac {1}{\pi }$ given in his seminal work [10], one of which is the following striking formula:

(1.1) $$ \begin{align} \frac{1}{\pi}=\frac{21}{8}\sum_{n=0}^{\infty}\binom{2n}{n}^{3}\left(n+\frac{5}{42}\right)\frac{1}{2^{12n}}, \end{align} $$

which was derived from Jacobi’s complete elliptic integral of the first kind

$$ \begin{align*}K(k)=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-k^{2}\sin(\theta)^{2}}}=\frac{\pi}{2}{}_{2}{F}_{1}\left(\frac{1}{2},\frac{1}{2};1;k^{2}\right) \end{align*} $$

and its modular transformation, and a Clausen-type quadratic transformation

(1.2) $$ \begin{align} {}_{2}{F}_{1}\left(\frac{1}{2},\frac{1}{2};1;k^{2}\right)^{2}={}_{3}F_{2}\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;4(1-k^{2})k^{2}\right). \end{align} $$

After Ramanujan’s work, series for $\frac {1}{\pi }$ of the form

(1.3) $$ \begin{align} C\sum_{n=0}^{\infty}A(n)(n-B)t^{n} \end{align} $$

with $A(n)$ , B, C algebraic numbers have been one of the central objects in related areas such as the theory of elliptic integrals, and have been extensively studied by many mathematicians, and now in the spirit of Ramanujan, we call a series convergent to $\frac {1}{\pi }$ of the form (1.3) a Ramanujan-type series for $\frac {1}{\pi }$ . In particular, when $A(n)$ , B, and C are all rational, it is called a rational Ramanujan-type series. The reader is referred to [3] and the references therein for some studies on Ramanujan-type series.

One of the notable works on Ramanujan-type series is due to Chan, Chan, and Liu [5], in which they extend Ramanujan’s modular approach and invent a method for producing Ramanujan-type series via modular transformations and modular equations. Building upon Chan et al.’s work, Cooper and/or his collaborators (see, e.g., [9] and the references therein) derived a number of families of explicit examples of Ramanujan-type series. Somewhat interestingly, if one takes a careful look at the derivations of Ramanujan’s series and Cooper’s work, one may note that they share some notable commonness, such as that:

• • they are all somehow connected to some subgroups of $\mathrm {SL}_{2}(\mathbb {R})$ commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ , whose associated modular curves are of genus zero,

• • the coefficients $A(n)$ all satisfy some recurrence relation of the form

  $$ \begin{align*}(n+1)^{3}A(n+1)+c_{1}(n)A(n)+\cdots+c_{\ell}(n)A(n+1-\ell)=0 \end{align*} $$

for some $\ell \geq 2$ and cubic polynomials $c_{i}(n)$ in n, from which a number of questions may arise, such as:

• • how does a genus zero $\mathrm {SL}_{2}(\mathbb {R})$ -subgroup commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ give rise to a Ramanujan-type series?

• • does the order $\ell $ of the recurrence relation encode any particular information?

• • why is the coefficient of $A(n+1)$ always $(n+1)^{3}$ ?

In this note, all of these questions shall be addressed, and more generally, we shall show that all those Ramanujan-type series can be uniformly formulated as follows (throughout the remainder of this note, when we write $t(z_{0})$ , it is assumed to be finite, i.e., $z_{0}$ is assumed to not be a pole).

Theorem 1.1 Let $\Gamma $ be a subgroup of $\mathrm {SL}_{2}(\mathbb {R})$ commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ such that the associated modular curve $X(\Gamma )=\overline {\Gamma \backslash \mathbb {H}}$ is of genus zero with a uniformizer $t=t(\tau )$ satisfying that:

• • its Fourier coefficients at the cusp $[i\infty ]$ are algebraic numbers,

• • it has a zero at the cusp $[i\infty ]$ ,

• • it has

1. – a pole at a cusp apart from $[i\infty ]$ , or at an elliptic point of period $h>2$ ,

   or

2. – a pole at an elliptic point of period $h=2$ if there are no cusps other than $[i\infty ]$ or elliptic points of period $h>2$ .

Then, for any imaginary quadratic point $\tau _{0}\in \mathbb {Q}[\sqrt {-d}]\cap \mathbb {H}$ such that $|t(\tau _{0})|<\mu (t)$ , where

$$ \begin{align*}\mu(t)=\min\{|t(z_{0})|:\, z_{0}\ \mbox{is a cusp apart from}\ [i\infty], \mbox{or an elliptic point}\}, \end{align*} $$

one has that

(1.4) $$ \begin{align} \frac{1}{\pi}=C\sum_{n=0}^{\infty}A(n)\left(n-B\right)t(\tau_{0})^{n}, \end{align} $$

where

$$ \begin{align*}B&=\left.t(\tau_{0})\left(\frac{1}{\pi i}\left.\frac{dy}{d\tau}\right|_{\tau_{0}}-\frac{1}{2\pi\mathrm{Im}(\tau_{0})}y(\tau_{0})\right)\right/\left(\frac{y(\tau_{0})}{2\pi i}\left.\frac{dt}{d\tau}\right|_{\tau_{0}}\right),\\ C&=\frac{2\mathrm{Im}(\tau_{0})}{t(\tau_{0})}\prod_{\substack{z_{0}\ \text{is a cusp}}}(t(\tau_{0})-t(z_0))\prod_{\substack{z_{0}\ \text{is an elliptic}\\ \text{ point of period}~h }}\left(t(\tau_{0})-t(z_{0})\right)^{\frac{h-1}{h}},\\ y(\tau)&=\prod_{\substack{z_{0}\ \text{is a cusp }}}(t-t(z_0))^{\frac{-1}{2}}\prod_{\substack{z_{0}\ \text{is an elliptic}\\ \text{ point of period}~h}}\left(t-t(z_{0})\right)^{\frac{1-h}{2h}}\left(\frac{1}{2\pi i}\frac{dt}{d\tau}\right)^{\frac{1}{2}}, \end{align*} $$

and the coefficients $A(n)$ are defined by

$$ \begin{align*}A(0)=\lim_{\tau\to i\infty}y(\tau)^{2} \end{align*} $$

and that $ \sum _{n=0}^{\infty }A(n)t^{n} $ is a solution of

(1.5) $$ \begin{align} \frac{d^{3}Y}{dt^{3}}\kern1.3pt{+}\kern1.3pt 3R_{1}(t)\frac{d^{2}Y}{dt^{2}}\kern1.3pt{+}\kern1.3pt\kern-1pt\left(2R_{1}(t)^{2}\kern1.3pt{+}\kern1.3pt 4R_{2}(t)\kern1.3pt{+}\kern1.3ptR_{1}'(t)\right)\frac{dY}{dt}\kern1.3pt{+}\kern1.3pt\kern-1pt\left(4R_{1}(t)R_{2}(t)\kern1.3pt{+}\kern1.3pt2R_{2}'(t)\right)\kern-1pt Y\kern1.3pt{=}\kern1.3pt 0, \end{align} $$

which, as one shall see in Section 2, is the symmetric square of

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}+R_{1}(t)\frac{dy}{dt}+R_{2}(t)y=0, \end{align*} $$

where

$$ \begin{align*}R_{1}(t)=\sum_{\substack{z_{0}\ \text{is a cusp}}}\frac{1}{t-t(z_{0})}+\sum_{\substack{z_{0}\ \text{is an elliptic }\\ \text{point of period}~h}}\frac{1-\frac{1}{h}}{t-t(z_{0})}, \end{align*} $$

and $R_{2}(t)$ is a rational function in t such that $R_{2}(t(\tau ))=\frac {3(t")^{2}-2t"'t'}{4(t')^{4}}+\frac {1}{2}R_{1}'(t(\tau ))+\frac {1}{4}R_{1}(t(\tau ))^{2}$ with $t'=\frac {dt}{d\tau }$ and $R_{1}'(t)=\frac {dR_{1}}{dt}$ .

In particular, the coefficients $A(n)$ satisfy a linear homogeneous recurrence relation of the form

$$ \begin{align*}(n+1)^{3}A(n+1)+c_{1}(n)A(n)+\cdots+c_{\nu(\Gamma)-2}(n)A(n+3-\nu(\Gamma))=0 \end{align*} $$

for some cubic polynomials $c_{i}(n)$ in n, where $\nu (\Gamma )$ is defined by

$$ \begin{align*}\nu(\Gamma)=\begin{cases}\displaystyle 2\left(\varepsilon_{\infty}+\sum_{h>2}\varepsilon_{h}-1\right)+\varepsilon_{2},&\mbox{if}\ \varepsilon_{\infty}+\sum_{h>2}\varepsilon_{h}\geq2,\\ \displaystyle \varepsilon_{2}+1,&\mbox{if}\ \varepsilon_{\infty}+\sum_{h>2}\varepsilon_{h}=1,\end{cases} \end{align*} $$

with $\varepsilon _{\infty }$ and $\varepsilon _{h}$ , respectively, denoting the numbers of cusps and elliptic points of period h of $\Gamma $ .

Remark 1.2 Since $\Gamma $ is commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ , then $t(\tau )$ is algebraically related with $j(\tau )$ . It is well known that for an imaginary quadratic point $\tau _{0}\in \mathbb {H}$ , $j(\tau _{0})$ can be rigorously computed by a modular equation. As a result, one can indeed algebraically obtain the value of t at $\tau _{0}$ , as well as the value of C. Similarly, the value of B can be algebraically computed using modular equations. See, e.g., [8, Section 5.10].

Remark 1.3 As is seen in Theorem 1.1, the order of the recurrence relation in fact encodes some geometric information of $\Gamma $ by the conjugation-invariant $\nu (\Gamma )$ . This answers one of the questions posed at the top of Theorem 1.1.

Remark 1.4 By [4, Section 10.4], under the given assumptions, it is not hard to show that $\nu (\Gamma )>2$ .

As an illustration, in the following example, we recover Ramanujan’s formula (1.1) from Theorem 1.1. Throughout the remainder of this note, we write $\eta (\tau )=q^{1/24}\prod _{n=1}^{\infty }(1-q^{n})$ with $q=\exp (2\pi i\tau )$ for the Dedekind eta function.

Example 1.5 (Ramanujan’s formula (1.1))

Let $\Gamma =\Gamma _{0}(4)+=\left \langle \Gamma _{0}(4),\begin {pmatrix}0&-1\\4&0\end {pmatrix}\right \rangle $ , the group generated by $\Gamma _{0}(4)$ and its Fricke involution $\begin {pmatrix}0&-1\\4&0\end {pmatrix}$ , with $\varepsilon _{\infty }=2$ and $\varepsilon _{2}=1$ and $\varepsilon _{h}=0$ for $h>2$ , and take the uniformizer

$$ \begin{align*}t=t(\tau)=\frac{\eta(\tau)^{24}\eta(4\tau)^{24}}{\eta(2\tau)^{48}} \end{align*} $$

with a zero at the cusp $[i\infty ]$ and a pole at the cusp $[\frac {1}{2}]$ and $\mu (t)=\frac {1}{64}$ . Then numerically, one can find that for $\tau _{0}=i\sqrt {7/4}$ , the coefficients $B=-\frac {5}{42}$ , $C=\frac {21i}{64}$ , and $t(\tau _{0})=\frac {1}{2^{12}}$ . Also, in this case, the coefficient $A(0)=-8i$ , and equation (1.5) can be specified as

$$ \begin{align*}\frac{d^{3}Y}{dt^{3}}+\frac{3(96t-1)}{t(64t-1)}\frac{d^{2}Y}{dt^{2}}+\frac{208t-1}{t^{2}(64t-1)}\frac{dY}{dt}+\frac{8}{t^{2}(64t-1)}Y=0. \end{align*} $$

Substituting in $Y=\sum _{n=0}^{\infty }A(n)t^{n}$ and equating the coefficient of $t^{n}$ of the resulting series to 0, one finds that $A(n)$ ’s satisfy the following $2$ -term recurrence relation:

$$ \begin{align*}(n+1)^{3}A(n+1)-(2(2n+1))^{3}A(n)=0, \end{align*} $$

and it is easy to show that

$$ \begin{align*}A(n)=A(0)\binom{2n}{n}^{3}=-8i\binom{2n}{n}^{3}. \end{align*} $$

Substituting all these into (1.4), we reproduce Ramanujan’s striking formula (1.1)

$$ \begin{align*}\frac{1}{\pi}=\frac{21}{8}\sum_{n=0}^{\infty}\binom{2n}{n}^{3}\left(n+\frac{5}{42}\right)\frac{1}{2^{12n}}. \end{align*} $$

We also obtain the following example of a rational Ramanujan-type series for $\frac {1}{\pi }$ , which, to the best of our knowledge, has not appeared before in the literature.

Example 1.6 Let $\Gamma =\gamma ^{-1}(\Gamma _{0}(4)\cap \Gamma (2))\gamma $ , where $\gamma =\begin {pmatrix}1&0\\1&1\end {pmatrix}$ , with $\varepsilon _{\infty }=4$ and ${\varepsilon _{h}=0}$ for $h\geq 2$ , and take the uniformizer

$$ \begin{align*}t=t(\tau)=4\frac{\eta\left({2\tau}\right)^{8}\eta\left(\frac{\tau}{2}\right)^{4}}{\eta(\tau)^{12}} \end{align*} $$

with a zero at the cusp $[i\infty ]$ and a pole at the cusp $[1]$ and $\mu (t)=1$ . Then numerically, one can find that for $\tau _{0}=i$ , the coefficients $B=0$ , $C=-1$ , and $t(\tau _{0})=\frac {\sqrt {2}}{2}$ . In this case, the coefficient $A(0)=-\frac {1}{4}$ , and equation (1.5) is given by

$$ \begin{align*}\frac{d^{3}Y}{dt^{3}}+\frac{3(3t^{2}-1)}{t^{3}-t}\frac{d^{2}Y}{dt^{2}}+\frac{19t^{4}-16t^{2}+1}{t^{2}(t^{2}-1)^{2}}\frac{dY}{dt}+\frac{4(2t^{2}-1)}{t(t^{2}-1)^{2}}Y=0. \end{align*} $$

Thus, the coefficients $A(n)$ satisfy the $3$ -term recurrence relation

$$ \begin{align*}(n+1)^{3}A(n+1)-2n(n^{2}+1)A(n-1)+(n-1)^{3}A(n-3)=0. \end{align*} $$

By Example 3.6, one can tell that $A(n)$ has the compact form

$$ \begin{align*}A(n)=\begin{cases}\displaystyle -\frac{1}{4}\left(\sum_{k}\binom{n-2k}{n/2-k}^{2}\binom{2k}{k}^{2}\right)4^{-n},&\mbox{if}\ n\ \mbox{is even,}\\ 0,&\mbox{otherwise.} \end{cases} \end{align*} $$

Substituting these into (1.4), one obtains a new, to the best of our knowledge, rational Ramanujan-type series

$$ \begin{align*}\frac{1}{\pi}=\frac{1}{2}\sum_{n=0}^{\infty}\left(\sum_{k}\binom{2n-2k}{n-k}^{2}\binom{2k}{k}^{2}\right)n\frac{1}{2^{5n}}. \end{align*} $$

Theorem 1.1 is mainly built upon properties of modular forms and a geometric notion called orbifold uniformization and relations between its attached differential forms called orbifold uniformizing differential equation, which is indeed a Fuchsian differential equation. In light of these, we organize the remainder of this note as follows. In Section 2, we review some basics of Fuchsian equations. In the first two subsections of Section 3, we briefly go over the notion of orbifold uniformization and relevant materials along with a few explicit examples, old and new, as illustrations, as well as establish a key lemma, Lemma 3.3, for Theorem 1.1. In the third subsection, we remark how a Clausen-type transformation formula in some general sense can be derived from the results of the preceding subsections as a byproduct, and recover the quadratic transformation (1.2) as an illustration. In the last section, we end this note with the proof of Theorem 1.1.

2 Fuchsian equations and their symmetric squares

In this section, we review Fuchsian equations and their basic properties, which we shall occasionally refer to. Most of the terminology and basic results are recalled from [12]. We start with a second-order linear differential equation

(2.1) $$ \begin{align} \frac{d^{2}y}{dx^{2}}+R_{1}(x)\frac{dy}{dx}+R_{2}(x)y=0 \end{align} $$

with $R_{i}(x)$ rational functions in x over $\mathbb {C}$ . A point $x_{0}$ is called a singular point for (2.1) if it is a pole of $R_{i}(x)$ for $i=1$ or $2$ . A singular point $x_{0}$ is called regular if $(x-x_{0})R_{1}(x)$ and $(x-x_{0})^{2}R_{2}(x)$ are holomorphic at $x_{0}$ for $x_{0}\ne \infty $ , and $R_{1}(1/x)/x$ and $R_{2}(1/x)/x^{2}$ are holomorphic at $x=0$ for $x_{0}=\infty $ . Equation (2.1) is called a Fuchsian equation if all of its singular points are regular. Moreover, a regular singular point $x_{0}$ is called apparent if the Fuchsian equation has two linearly independent holomorphic solutions near $x_{0}$ . For a Fuchsian equation of the form (2.1) with a regular singular point $x_{0}$ , the local solutions at $x_{0}$ to it may be characterized by a pair of so-called characteristic exponents $s_{1}(x_{0})$ and $s_{2}(x_{0})$ defined to be the roots of $s(s-1)+r_{1}s+r_{2}=0$ , where

$$ \begin{align*}r_{1}=\left.(x-x_{0})R_{1}(x)\right|_{x=x_{0}}\quad\mbox{and}\quad r_{2}=\left.(x-x_{0})^{2}R_{2}(x)\right|_{x=x_{0}}\quad\mbox{for}\ x_{0}\ne\infty, \end{align*} $$

and

$$ \begin{align*}r_{1}=-\left.\frac{1}{x}R_{1}\left(\frac{1}{x}\right)\right|_{x=0}\quad\mbox{and}\quad r_{2}=\left.\frac{1}{x^{2}}R_{2}\left(\frac{1}{x}\right)\right|_{x=0}\quad\mbox{for}\ x_{0}=\infty. \end{align*} $$

Several analytic properties [12, Section 2.5] of the solutions of (2.1) are characterized by the difference $s_{2}(x_{0})-s_{1}(x_{0})$ of the characteristic exponents. Moreover, the difference $s_{2}(x_{0})-s_{1}(x_{0})$ of the characteristic exponents is a “projective” invariant, up to a sign, of a Fuchsian equation. One can actually transform (2.1) into a uniquely defined equation of the form

$$ \begin{align*}\frac{d^{2}y}{dx^{2}}+R(x)y=0, \end{align*} $$

so that the ratio of two of its linearly independent solutions remains, and such an equation is called the projective normal form of a Fuchsian equation of the form (2.1) defined by setting $R(x)=R_{2}(x)-\frac {1}{2}R_{1}'(x)-\frac {1}{4}R_{1}(x)^{2}$ . Such a differential equation has the same singular behavior as that of (2.1), i.e., they have the same (regular) singular points, whose solutions are projectively equivalent to that of (2.1) by a multiplier of $\exp \left (\int \frac {1}{2}R_{1}(x)dx\right )$ . As is just noted, the difference of characteristic exponents of the projective normal form differs from $s_{2}(x_{0})-s_{1}(x_{0})$ by a sign.

Finally, for a Fuchsian equation (2.1) with two linearly independent solutions $y_{1}=y_{1}(x)$ and $y_{2}=y_{2}(x)$ , its symmetric square is defined by

$$ \begin{align*}\frac{d^{3}Y}{dx^{3}}+3R_{1}(x)\frac{d^{2}Y}{dx^{2}}+\left(2R_{1}(x)^{2}+4R_{2}(x)+R_{1}'(x)\right)\frac{dY}{dx}+\left(4R_{1}(x)R_{2}(x)+2R_{2}'(x)\right)Y=0, \end{align*} $$

and a basis for the solutions to such a third-order linear differential equation consists of $y_{1}^{2},\, y_{1}y_{2},\, y_{2}^{2}$ . See, e.g., [11, Theorem 5.5]. So (1.5) defined in Theorem 1.1 is literally the symmetric square of some Fuchsian equation.

3 Orbifold uniformizations

In this section, we shall review the notion of orbifold uniformization and orbifold uniformizing differential equation used by Yoshida [12]. In particular, in Section 3, we establish a key lemma to Theorem 1.1 (see Lemma 3.3), wherein the question regarding the coefficient of $A(n+1)$ in the recurrence relation is answered. At the end, we remark how one may obtain Clausen-type transformation formulas via orbifold uniformizations.

3.1 Orbifolds

Recall that an orbifold is a triple $(X,Y,(b_{j}))$ consisting of a complex manifold X, a hypersurface $Y\subset X$ which can be decomposed into irreducible components $\cup _{j}Y_{j}$ , each of which is assigned with a number $b_{j}$ , either $\infty $ or an integer greater than or equal to $2$ , called weight, so that for any point in $X-\cup _{j}\{Y_{j}|\,b_{j}=\infty \}$ , there is an open neighborhood U and a covering manifold which ramifies along $U\cap Y$ with the given weights $b_{j}$ . Moreover, an orbifold $(X,Y,(b_{j}))$ is called uniformizable if there is a simply connected global covering manifold of X with the given ramification data $(Y,(b_{j}))$ . For example, the triple $(X,Y,(b_{j}))$ with $X=\mathbb {P}^{1}(\mathbb {C})$ , $Y=\{\infty \}\cup \{0\}\cup \{1\}$ and $(b_{j})=(\infty ,3,2)$ is a uniformizable orbifold with the global covering map $J:\mathbb {H}\to \mathbb {P}^{1}(\mathbb {C})$ , where J is the modular j-invariant of the form $\frac {1}{1728q}+O(1)$ near $i\infty $ .

Note that for a modular curve $X(\Gamma )$ associated with an $\mathrm {SL}_{2}(\mathbb {R})$ -subgroup of genus zero commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ , $(X(\Gamma ),S,(b_{j}))$ with S the set of cusps and elliptic points and $b_{j}$ being either $\infty $ or the period of an elliptic point depending on its attached component being a cusp or an elliptic point is a uniformizable orbifold with the global covering manifold $\pi _{\Gamma }:\mathbb {H}\to X(\Gamma )$ . This can also be identified with the orbifold $(\mathbb {P}^{1}(\mathbb {C}),Y_{\Gamma },(b_{j}))$ through the uniformizer $t:X(\Gamma )\to \mathbb {P}^{1}(\mathbb {C})$ , where $Y_{\Gamma }=t(S)$ .

3.2 Orbifold uniformizing differential equations

Now, let $(X,Y,(b_{j}))$ be a uniformizable orbifold with $X=\mathbb {P}^{1}(\mathbb {C})$ whose global covering manifold is $\mathbb {H}$ , and covering map is denoted by t. By the Galois theory of covering maps, $X-\cup _{j}\{Y_{j}|\,b_{j}=\infty \}\cong \Gamma \backslash \mathbb {H}$ for some $\Gamma \subset \mathrm {SL}_{2}(\mathbb {R})$ . Letting w be any local coordinate on $\mathbb {H}$ , one can associate the orbifold with a differential equation called the orbifold uniformizing differential equation for X defined by

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}-\{w,t\}y=0, \end{align*} $$

where

$$ \begin{align*}\{w,t\}=\frac{3(w")^{2}-2w"'w'}{4(w')^{2}} \end{align*} $$

with $w^{(n)}=\frac {d^{n}w}{dt^{n}}$ called the Schwarzian derivative. It is known [12, p. 50] that an orbifold uniformizing differential equation is a Fuchsian equation. Moreover, it is noteworthy to remark that the Schwarzian derivative is $\mathrm {PSL}_{2}(\mathbb {C})$ invariant in w, i.e., for any $\gamma \in \mathrm {PSL}_{2}(\mathbb {C})$ , $ \{w,t\}=\{\gamma \cdot w,t\}. $

Lemma 3.1 A solution to the orbifold uniformizing differential equation is given by $y(t)=\left (\frac {1}{dw/dt}\right )^{\frac {1}{2}}$ for a suitable choice of branch.

Proof This follows from the Inverse Function Theorem and the simple relation $\{w,t\}=-\{t,w\}/(t')^{2}$ with $t^{(n)}=\frac {d^{n}t}{dw^{n}}$ .

Suppose that $(X,Y,(b_{j}))$ is a uniformizable orbifold with $X=\mathbb {P}^{1}(\mathbb {C})\cong X(\Gamma )$ for some $\mathrm {SL}_{2}(\mathbb {R})$ -subgroup $\Gamma $ of genus zero commensurable with $\mathrm { SL}_{2}(\mathbb {Z})$ . Clearly, every meromorphic function on $X(\Gamma )$ is a rational function in t. With respect to the canonical local coordinate $\tau $ on $\mathbb {H}$ , it is easy to check that $-\{t,\tau \}/(t')^{2}$ is a meromorphic function on $X(\Gamma )$ by the fact that

$$ \begin{align*}\left.\frac{dt}{d\tau}\right|_{\tau\to\gamma\cdot\tau}=(c\tau+d)^{2}\frac{dt}{d\tau}, \end{align*} $$

where $\gamma =\begin {pmatrix}a&b\\c&d\end {pmatrix}\in \Gamma $ , and thus the orbifold uniformizing differential equation associated with $(\mathbb {P}^{1}(\mathbb {C}),Y_{\Gamma },(b_{j}))$ can be indeed rewritten as

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}+R(t)y=0, \end{align*} $$

for some rational function in t which can be determined when t in terms of $\tau $ is explicit.

Example 3.2 For $\Gamma =\Gamma _{0}(4)+=\left \langle \Gamma _{0}(4),\begin {pmatrix}0&-1\\4&0\end {pmatrix}\right \rangle $ , the group generated by $\Gamma _{0}(4)$ and its Fricke involution $\begin {pmatrix}0&-1\\4&0\end {pmatrix}$ , as is seen in Example 1.5, a uniformizer, is given by

$$ \begin{align*}t(\tau)=\frac{\eta(4\tau)^{24}\eta(\tau)^{24}}{\eta(2\tau)^{48}}, \end{align*} $$

which has a simple zero at the cusp $[i\infty ]$ and a simple pole at the cusp $[\frac {1}{2}]$ . It is easy to check (or see Lemma 3.3) that

$$ \begin{align*}\frac{\{t,\tau\}}{(t')^{2}}=\frac{4096t^2-80t+1}{4t^2(64t-1)^2}, \end{align*} $$

and thus by the remarks in Section 2, one can tell that a solution in terms of $\tau $ to the orbifold uniformizing differential equation

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}-\{\tau,t\}y=0, \end{align*} $$

i.e.,

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}+\frac{4096t^2-80t+1}{4t^2(64t-1)^2}y=0, \end{align*} $$

is $y=\left (\frac {1}{2\pi i}\frac {d}{d\tau }\left (\frac {\eta (4\tau )^{24}\eta (\tau )^{24}}{\eta (2\tau )^{48}}\right )\right )^{\frac {1}{2}}$ for some choice of branch. Moreover, one can also check that the Fuchsian equation above is indeed the projective normal form of

(3.1) $$ \begin{align} \frac{d^{2}y}{dt^{2}}+\frac{96t-1}{t(64t-1)}\frac{dy}{dt}+\frac{4}{t(64t-1)}y=0, \end{align} $$

and thus, in terms of $\tau $ , a solution to this equation is given by

$$ \begin{align*}y=\left(\left(\frac{1}{64t^{3}-t^{2}}\right)^{\frac{1}{2}}\frac{1}{2\pi i}\frac{d}{d\tau}\left(\frac{\eta(4\tau)^{24}\eta(\tau)^{24}}{\eta(2\tau)^{48}}\right)\right)^{\frac{1}{2}}. \end{align*} $$

In the example above, we mention that the projective normal form of the Fuchsian equation (3.1) is exactly the orbifold uniformizing differential equation associated with $X(\Gamma _{0}(4)+)$ . Moreover, one may note that the denominators of its coefficients have lower degrees than that of the orbifold uniformizing differential equation does. In fact, for an orbifold uniformizing differential equation associated with an $\mathrm { SL}_{2}(\mathbb {R})$ -subgroup commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ , one can always “anti-projectivize” its orbifold uniformizing differential equation

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}+R(t)y=0 \end{align*} $$

to obtain a Fuchsian equation whose coefficients have denominators of lower degrees than that of $R(t)$ . We elaborate this in the following lemma.

Lemma 3.3 For a given genus zero modular curve $X(\Gamma )$ associated with an $\mathrm {SL}_{2}(\mathbb {R})$ -subgroup $\Gamma $ commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ , choose a uniformizer $t=t(\tau )$

• • having a zero at the cusp $[i\infty ]$

• • and

1. – a pole at a cusp apart from $[i\infty ]$ , or at an elliptic point of period $h>2$ ,

   or

2. – a pole at an elliptic point of period $h=2$ if there are no cusps other than $[i\infty ]$ or elliptic points of period $h>2$ .

Let $\nu (\Gamma )$ be defined as in Theorem 1.1 and denote by $\omega (\Gamma )$ the total number of cusps and elliptic points of $X(\Gamma )$ . Suppose that the orbifold uniformizing differential equation of $X(\Gamma )$ is given by

(3.2) $$ \begin{align} \frac{d^{2}y}{dt^{2}}+R(t)y=0. \end{align} $$

Then:

1. (1) this is the projective normal form of

   (3.3) $$ \begin{align} \frac{d^{2}y}{dt^{2}}+R_{1}(t)\frac{dy}{dt}+R_{2}(t)y=0, \end{align} $$
   where
   $$ \begin{align*}R_{1}(t)=\sum_{\substack{z_{0}\ \text{is a cusp}}}\frac{1}{t-t(z_{0})}+\sum_{\substack{z_{0}\ \text{is an elliptic }\\\text{point of period}~h}}\frac{1-\frac{1}{h}}{t-t(z_{0})} \end{align*} $$
   is of the form
   $$ \begin{align*}P_{1}(t)\left/\left(\prod_{\substack{z_{0}\ \text{is a cusp or}\\ \text{an elliptic point}}}(t-t(z_{0}))\right)\right. \end{align*} $$
   for some polynomial $P_{1}(t)$ of degree $\omega (\Gamma )-2$ having no common factor with the denominator, and $R_{2}(t)=R(t)+\frac {1}{2}R_{1}'(t)+\frac {1}{4}R_{1}(t)^{2}$ ,

2. (2) for the symmetric square of (3.3)

   (3.4) $$ \begin{align} \frac{d^{3}Y}{dt^{3}}+3R_{1}(t)\frac{d^{2}Y}{dt^{2}}+\left(2R_{1}(t)^{2}+4R_{2}(t)+R_{1}'(t)\right)\frac{dY}{dt}+\left(4R_{1}(t)R_{2}(t)+2R_{2}'(t)\right)Y=0, \end{align} $$
   the coefficient $2R_{1}(t)^{2}+4R_{2}(t)+R_{1}'(t)$ satisfies that
   $$ \begin{align*}\left(2R_{1}(t)^{2}+4R_{2}(t)+R_{1}'(t)\right)\times\left({\prod_{\substack{z_{0}\ \text{is a cusp or }\\\text{an elliptic point }\\\text{of period}~h>2}}}(t-t(z_{0}))^{2}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~2}}(t-t(z_{0}))\right) \end{align*} $$
   is a polynomial in t of degree $\nu (\Gamma )-2$ coprime to
   $$ \begin{align*}\prod_{\substack{z_{0}\ \text{is a cusp or }\\\text{an elliptic point }\\\text{of period}~h>2}}(t-t(z_{0}))^{2}, \end{align*} $$
   and similarly the coefficient $4R_{1}(t)R_{2}(t)+2R_{2}'(t)$ satisfies that
   $$ \begin{align*}\left(4R_{1}(t)R_{2}(t)+2R_{2}'(t)\right)\times\left({\prod_{\substack{z_{0}\ \text{is a cusp or }\\\text{an elliptic point }\\\text{of period}~h>2}}}(t-t(z_{0}))^{2}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~2}}(t-t(z_{0}))\right) \end{align*} $$
   is a polynomial in t of degree $\nu (\Gamma )-3$ ,

3. (3) a solution to (3.3) is given by

   $$ \begin{align*}y=\prod_{\substack{z_{0}\ \text{is a cusp}}}(t-t(z_0))^{\frac{-1}{2}}\prod_{\substack{z_{0}\ \text{is an elliptic }\\\text{point of period}~h}}\left(t-t(z_{0})\right)^{\frac{1-h}{2h}}\left(\frac{1}{2\pi i}\frac{1}{d\tau/dt}\right)^{\frac{1}{2}}, \end{align*} $$
   and a solution to (3.4) is given by
   $$ \begin{align*}Y=\prod_{\substack{z_{0}\ \text{is a cusp}}}(t-t(z_0))^{-1}\prod_{\substack{z_{0}\ \text{is an elliptic }\\\text{point of period}~h}}\left(t-t(z_{0})\right)^{\frac{1-h}{h}}\frac{1}{2\pi i}\frac{1}{d\tau/dt}, \end{align*} $$
   and as functions in $\tau $ , they are, respectively, modular forms of weights 1 and 2,

4. (4) as a function in t, the solution $Y=Y(t)$ has a local holomorphic representation $\sum _{n=0}^{\infty }A(n)t^{n}$ at $t=0$ converges for

   $$ \begin{align*}|t|<\mu(t)=\min\{|t(z_{0})|:\, z_{0}\ \mbox{is a cusp apart from}\ [i\infty],\ \mbox{or an elliptic point}\}, \end{align*} $$
   and the coefficients $A(n)$ satisfy a linear homogeneous recurrence relation of the form
   $$ \begin{align*}(n+1)^{3}A(n+1)+c_{1}(n)A(n)+\cdots+c_{\nu(\Gamma)-2}(n)A(n+3-\nu(\Gamma))=0 \end{align*} $$
   for some cubic polynomials $c_{i}(n)$ in n.

Proof Throughout the proof, we write

$$ \begin{align*}\alpha=\sum_{\substack{z\ \text{is a cusp}\\\text{not a pole}}}1+\sum_{\substack{z\ \text{is an elliptic}\\\text{point of period}~h}}\left(1-\frac{1}{h}\right). \end{align*} $$

It is known by [4, Section 10.4] together with the assumption that $X(\Gamma )$ is of genus zero and $\Gamma $ is commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ that $\alpha>1$ .

Now, recall that

$$ \begin{align*}R(t)=\frac{\{t,\tau\}}{(t')^{2}}=\frac{3(t")^{2}-2t"'t'}{4(t')^{4}}. \end{align*} $$

Since the argument under other circumstance is similar, without loss of generality, suppose that

$$ \begin{align*}t=\begin{cases}t(z_{0})+c_{1}(\tau-z_{0})+\cdots,&\mbox{if}\ z_{0}\ \mbox{is not a cusp or an elliptic point},\\\\ t(z_{0})+c_{1}(\tau-z_{0})^{h}+c_{2}(\tau-z_{0})^{h+1}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is an elliptic point of period}~h,\\\\ t(z_{0})+c_{1}q_{e}+c_{2}q_{e}^{2}+c_{3}q_{e}^{3}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{not a pole,}\\\\ \frac{1}{q_{e}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{that is a pole,} \end{cases} \end{align*} $$

where $c_{1}$ are all nonzero. One, therefore, can find that

$$ \begin{align*} &R(t(\tau))=\frac{\{t,\tau\}}{(t')^{2}}\\ &\quad=\begin{cases}O(1),&\mbox{if}\ z_{0}\ \mbox{is not a cusp or an elliptic point,}\\\\ \frac{h^{2}-1}{4h^{2}c_{1}^{2}}\frac{1}{(\tau-z_{0})^{2h}}-\frac{(h^{2}-1)c_{2}}{2h^{2}c_{1}^{3}}\frac{1}{(\tau-z_{0})^{2h-1}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is an elliptic point of period}~h,\\\\ \frac{1}{4c_{1}^{2}}\frac{1}{q_{e}^{2}}-\frac{c_{2}}{c_{1}^{3}}\frac{1}{q_{e}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{not a pole of}~t,\\\\ \frac{q_{e}^{2}}{4}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{that is a pole of}~t, \end{cases} \end{align*} $$

and

$$ \begin{align*} &R_{1}(t(\tau))=\begin{cases}O(1),&\mbox{if}\ z_{0}\ \mbox{is not a cusp or an elliptic point,}\\\\ \frac{h-1}{hc_{1}(\tau-z_{0})^{h}}-\frac{(h-1)c_{2}}{hc_{1}^{2}(\tau-z_0)^{h-1}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is an elliptic point of period}~h,\\\\ \frac{1}{c_{1}q_{e}}-\frac{c_{2}}{c_{1}^{2}}+\cdots,&\mbox{if} z_{0}\ \mbox{is a cusp of width}~e\ \mbox{not a pole of}~t,\\\\ \alpha q_{e}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{that is a pole of}~t, \end{cases} \end{align*} $$

and

$$ \begin{align*} &R_{2}(t(\tau))=\begin{cases}O(1),&\mbox{if}\ z_{0}\ \mbox{is not a cusp or an elliptic point,}\\\\ -\frac{3(h^{2}-1)c_{2}}{2h^{2}c_{1}^{3}(\tau-z_{0})^{h}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is an elliptic point of period}~h,\\\\ -\frac{c_{2}}{2c_{1}^{3}q_{e}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{not a pole of}~t,\\\\ \frac{1}{4}(\alpha-1)^{2} q_{e}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{that is a pole of}~t, \end{cases} \end{align*} $$

which has at most a simple pole at a cusp that is not a pole of t and always has a double zero at the cusp $z_{0}$ that is a pole of t since $\alpha>1$ . Building upon these, we are now ready to give proofs to our assertions.

For (i), first, it is clear by definition that (3.2) is the projective normal form of (3.1). For the rational form of $R_{1}(t)$ , note that $R_{1}(t)$ has simple poles at exactly that cusps and elliptic points that is not a pole of t, and a simple zero at the pole of t, and thus

$$ \begin{align*}R_{1}(t)\times \prod_{\substack{z_{0}\ \text{is a cusp or}\\ \text{an elliptic point}}}(t-t(z_{0})) \end{align*} $$

has a pole only at the cusp that is a pole of t, which is of order $\omega (\Gamma )-1$ , and no zeros at other cusps or elliptic points. As such, this must be a polynomial of degree $\omega (\Gamma )-1$ with no common factor with the multiplier. It is worth remarking that $R(t)$ is a rational function in t whose denominator is of degree $2\omega (\Gamma )-2>\omega (\Gamma )-1$ by [4, Section 10.4].

For (ii), similarly, working through the computations, one can deduce that

$$ \begin{align*} &2R_{1}(t)^{2}+4R_{2}(t)+R_{1}'(t)\\ &= \begin{cases}O(1),&\mbox{if}\ z_{0}\ \mbox{is not a cusp or an elliptic point,}\\\\ \frac{(h-2)(h-1)}{h^{2}c_{1}^{2}}\frac{1}{(\tau-z_{0})^{2h}}+\frac{(h-1)(2-4h)c_{2}}{h^{2}c_{1}^{3}}\frac{1}{(\tau-z_{0})^{2h-1}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is an elliptic point of period}~h,\\\\ \frac{1}{c_{1}^{2}q_{e}^{2}}-\frac{4c_{2}}{c_{1}^{3}}\frac{1}{q_{e}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{not a pole of}~t,\\\\ (\alpha^{2}-\alpha+1)q_{e}^{2}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{that is a pole of}~t, \end{cases} \end{align*} $$

which has at most a simple pole at an elliptic point of period $2$ , and always has a double pole at an elliptic point of period $h>2$ and at a cusp that is not a pole of t, as well as a double zero at the cusp that is a pole of t since $\alpha>1$ , and

$$ \begin{align*} &4R_{1}(t)R_{2}(t)+2R_{2}'(t)\\ &= \begin{cases}O(1),&\mbox{if}\ z_{0}\ \mbox{is not a cusp or an elliptic point,}\\\\ -\frac{3(h^{2}-1)(h-2)c_{2}}{h^{3}c_{1}^{4}}\frac{1}{(\tau-z_{0})^{2h}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is an elliptic point of period}~h,\\\\ -\frac{c_{2}}{c_{1}^{4}}\frac{1}{q_{e}^{2}}+\frac{21c_{2}^{2}-16c_{1}c_{3}}{c_{1}^{5}}\frac{1}{q_{e}}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{not a pole,}\\\\ (\alpha-1)^{3}q_{e}^{3}+\cdots,&\mbox{if}\ z_{0}\ \mbox{is a cusp of width}~e\ \mbox{that is a pole,} \end{cases} \end{align*} $$

which has at most a simple pole at an elliptic point of period $2$ and at most a double pole at a cusp that is not a pole of t, and always has a double pole at an elliptic point of period $h>2$ and a zero of order 3 at the cusp $z_{0}$ that is a pole of t.

Then, clearly,

$$ \begin{align*}\left(2R_{1}(t)^{2}+4R_{2}(t)+R_{1}'(t)\right)\times\left({\prod_{\substack{z_{0}\ \text{is a cusp or }\\\text{an elliptic point }\\\text{of period}~h>2}}}(t-t(z_{0}))^{2}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~2}}(t-t(z_{0}))\right) \end{align*} $$

has a pole only at the cusp that is a pole of t, which is order $\nu (\Gamma )-2$ , and thus a polynomial in t of degree $\nu (\Gamma )-2$ , which can possibly have a common factor with the multiplier in the case that $t=t(z_{0})+c_{1}(\tau -z_{0})^{h}+O((\tau -z_{0})^{3h})$ at some elliptic point of period $h=2$ . Argument for the case of $4R_{1}(t)R_{2}(t)+2R_{2}'(t)$ is similar and thus omitted.

Item (iii) follows from the remarks given in Section 2 and Lemma 3.1.

For (iv), first, by items (i) and (ii), one can tell that the least common multiple of the denominators of $3R_{1}(t)$ , $2R_{1}(t)^{2}+4R_{2}(t)+R_{1}'(t)$ , and $4R_{1}(t)R_{2}(t)+2R_{2}'(t)$ is exactly

$$ \begin{align*}\mathcal{P}_{3}(t):= t^{2}\prod_{\substack{\tau_{0}\ \text{is a cusp apart}\\\text{from}\ [i\infty],\ \text{or an}\\\text{ elliptic point}\\\text{ of period}~h>2}}(t-t(z_{0}))^{2}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~2}}(t-t(z_{0})). \end{align*} $$

So clearing denominators, the symmetric square (3.4) is of the form

(3.5) $$ \begin{align} \mathcal{P}_{3}(t)\frac{d^{3}Y}{dt^{3}}+\mathcal{P}_{2}(t)\frac{d^{2}Y}{dt^{2}}+\mathcal{P}_{1}(t)\frac{dY}{dt}+\mathcal{P}_{0}(t)Y=0, \end{align} $$

where $\mathcal {P}_{2}(t)$ is a polynomial of the form $t\tilde {\mathcal {P}}_{2}(t)$ with $\tilde {\mathcal {P}}_{2}(t)$ of degree $\nu (\Gamma )-2$ coprime to

$$ \begin{align*}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~2}}(t-t(z_{0})), \end{align*} $$

$\mathcal {P}_{1}(t)$ is a polynomial of degree $\nu (\Gamma )-2$ coprime to

$$ \begin{align*}\prod_{\substack{\tau_{0}\ \text{is a cusp apart}\\ \text{from}\ [i\infty],\ \text{or an}\\\text{ elliptic point}\\\text{ of period}~h>2}}(t-t(z_{0}))^{2}, \end{align*} $$

$\mathcal {P}_{0}(t)$ is a polynomial of degree $\nu (\Gamma )-3$ , and $\mathcal {P}_{3}(t)/t^{2}$ is a polynomial in t of degree $\nu (\Gamma )-2$ with nonzero constant, that is to say, the greatest common divisor of $\mathcal {P}_{3}(t)/t^{2}$ and $\mathcal {P}_{i}(t)$ for $i\in \{0,1,2\}$ is 1. As a result, substituting $\sum _{n=0}^{\infty }A(n)t^{n}$ into the left-hand side of the equation, one can deduce that for any m, the coefficient of $t^{m}$ is exactly a linear combination of $A(m+1),\ldots ,A(m+3-\nu (\Gamma ))$ over $\overline {\mathbb {Q}}[m]$ , and the coefficient of $A(m+1)$ is

$$ \begin{align*}\left(\lim_{t\to0}\frac{\mathcal{P}_{3}(t)}{t^{2}}\right) (m+1)m(m-1)+\tilde{\mathcal{P}}_{2}(0)(m+1)m+\mathcal{P}_{1}(0)(m+1), \end{align*} $$

where

$$ \begin{align*}\lim_{t\to0}\frac{\mathcal{P}_{3}(t)}{t^{2}}=\prod_{\substack{\tau_{0}\ \text{is a cusp apart}\\\text{from }[i\infty],\ \text{or an}\\\text{ elliptic point}\\\text{ of period}~h>2}}t(z_{0})^{2}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~2}}(-t(z_{0}))\ne0. \end{align*} $$

Note that by the proofs of (i) and (ii),

$$ \begin{align*} \tilde{\mathcal{P}}_{2}(0)&=\lim_{t\to0}3R_{1}(t)t\frac{\mathcal{P}_{3}(t)}{t^{2}}=3\left(\lim_{t\to0}\frac{\mathcal{P}_{3}(t)}{t^{2}}\right) \end{align*} $$

and

$$ \begin{align*} \mathcal{P}_{1}(0)&=\lim_{t\to0}\left(2R_{1}(t)^{2}+4R_{2}(t)+R_{1}'(t)\right)t^{2}\frac{\mathcal{P}_{3}(t)}{t^{2}}=\lim_{t\to0}\frac{\mathcal{P}_{3}(t)}{t^{2}}. \end{align*} $$

Therefore, up to a multiplier of $\lim _{t\to 0}\frac {\mathcal {P}_{3}(t)}{t^{2}}$ , the coefficient of $A(m+1)$ is exactly

$$ \begin{align*}(m+1)m(m-1)+3(m+1)m+m+1=(m+1)^{3}. \end{align*} $$

Finally, setting the coefficient of $t^{m}$ to be 0 and cancelling the multiplier highlighted above, one obtains the desired linear homogeneous recurrence relation.

The following example is recently discovered by Cooper et al. [2, Theorem 6.2] using various q-series identities. We hereby provide a much simpler and more systematic derivation.

Example 3.4 For $\Gamma =\Gamma _{0}(21)+$ , the Fuchsian group generated by $\Gamma _{0}(21)$ and all of its Atkin–Lehner involutions, with

$$ \begin{align*}t=\frac{u}{(u-1)^{2}}, \end{align*} $$

where $u=\left (\frac {\eta (\tau )\eta (21\tau )}{\eta (3\tau )\eta (7\tau )}\right )^{2}$ , one can compute and find the associated orbifold uniformizing differential equation

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}+\frac{11,340t^6+5,916t^5+1,705t^4+352t^3+30t^2+4t+1}{4t^2(4t+1)^2(27t^2+2t-1)^2}y=0, \end{align*} $$

whose “anti-projectivized” form is given by

$$ \begin{align*}\frac{d^{2}y}{dt^{2}}+\frac{270t^3+70t^2-3t-1}{t(4t+1)(27t^{2}+2t-1)}\frac{dy}{dt}+\frac{240t^2+47t-2}{4t(4t+1)(27t^2+2t-1)}y=0. \end{align*} $$

And thus, taking symmetric square, one obtains the third-order differential equation

$$ \begin{align*}\frac{d^{3}Y}{dt^{3}}&+\frac{3(270t^3+70t^2-3t-1)}{t(4t+1)(27t^{2}+2t-1)}\frac{d^{2}Y}{dt^{2}}\\&+\frac{(15t-1)(88t^2+23t+1)}{t^2(4t+1)(27t^2+2t-1)}\frac{dY}{dt}+\frac{360t^2+47t-1}{t^2(4t+1)(27t^2+2t-1)}Y=0 \end{align*} $$

satisfied by

$$ \begin{align*}Y=\frac{1}{t(4t+1)^{\frac{1}{2}}(27t^{2}+2t+1)^{\frac{1}{2}}}\frac{1}{2\pi i}\frac{dt}{d\tau}. \end{align*} $$

Using the theory of modular forms or Sturm’s bound, one can check that

$$ \begin{align*}\frac{1}{t(4t+1)^{\frac{1}{2}}(27t^{2}+2t+1)^{\frac{1}{2}}}\frac{1}{2\pi i}\frac{dt}{d\tau}=-i\frac{\eta(\tau)\eta(3\tau)\eta(7\tau)\eta(21\tau)}{t^{\frac{4}{3}}}, \end{align*} $$

which recovers [2, Theorem 6.2].

3.3 Remark: Clausen-type transformations

As is noted at the beginning, a key ingredient of the derivation of Ramanujan’s formula (1.1) is the Clausen-type quadratic transformation formula

(3.6) $$ \begin{align} {}_{2}{F}_{1}\left(\frac{1}{2},\frac{1}{2};1;t\right)^{2}={}_{3}F_{2}\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;4(1-t)t\right). \end{align} $$

In this subsection, one shall see that this can be regarded as a special case of a general Clausen-type transformation formula obtained by modular transformations between orbifold uniformizations associated with genus zero $\mathrm {SL}_{2}(\mathbb {R})$ -subgroups.

For any genus zero $\mathrm {SL}_{2}(\mathbb {R})$ -subgroups $\Gamma _{1},\Gamma _{2}$ commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ with $\Gamma _{1}<\Gamma _{2}$ , suppose that for $i=1,2$ , $t_{i}$ is a uniformizer for $X(\Gamma _{i})$ . Then there is a rational function $r(t_{1})$ in $t_{1}$ such that $r(t_{1})=t_{2}$ . In addition, we have seen that

$$ \begin{align*}y_{i}=\prod_{\substack{z_{0}\ \text{is a cusp}}}(t_{i}-t_{i}(z_0))^{\frac{-1}{2}}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~h}}\left(t_{i}-t_{i}(z_{0})\right)^{\frac{1-h}{2h}}\left(\frac{1}{2\pi i}\frac{dt_{i}}{d\tau}\right)^{\frac{1}{2}} \end{align*} $$

is a solution to

$$ \begin{align*}\frac{d^{2}y}{dt_{i}^{2}}+R_{1}(t_{i})\frac{dy}{dt_{i}}+R_{2}(t_{i})y=0, \end{align*} $$

where

$$ \begin{align*}R_{1}(t_{i})=\sum_{\substack{z_{0}\ \text{is a cusp }}}\frac{1}{t_{i}-t_{i}(z_{0})}+\sum_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~h}}\frac{1-\frac{1}{h}}{t_{i}-t_{i}(z_{0})}, \end{align*} $$

and $R_{2}(t)=\{t_{i},\tau \}/(t_{i}')^{2}+\frac {1}{2}R_{1}'(t_{i})+\frac {1}{4}R_{1}(t_{i})^{2}$ , and that

$$ \begin{align*}Y_{i}=\prod_{\substack{z_{0}\ \text{is a cusp }}}(t_{i}-t_{i}(z_0))^{-1}\prod_{\substack{z_{0}\ \text{is an elliptic}\\\text{point of period}~h}}\left(t_{i}-t_{i}(z_{0})\right)^{\frac{1-h}{h}}\frac{1}{2\pi i}\frac{dt_{i}}{d\tau} \end{align*} $$

as a modular form for $\Gamma _{i}$ of weight 2 is a solution to the symmetric square of the Fuchsian equation above. Since $Y_{2}/Y_{1}$ is a modular function for $\Gamma _{1}$ , there is a rational function in $t_{1}$ such that $Y_{2}=R(t_{1})Y_{1}$ . Suppose that near $t_{i}=0$ , one has the following local expansions:

$$ \begin{align*}y_{i}=\sum_{n=0}^{\infty}a_{i}(n)t_{i}^{n}\quad\mbox{and}\quad Y_{i}=\sum_{n=0}^{\infty}A_{i}(n)t_{i}^{n} \end{align*} $$

for $y_{i}$ and $Y_{i}$ . Then near $t_{1}=0$ such that $r(t_{1})$ is also near 0, in this way, one obtains a Clausen-type transformation formula

(3.7) $$ \begin{align} \sum_{n=0}^{\infty}A_{2}(n)r(t_{1})^{n}=R(t_{1})\left(\sum_{n=0}^{\infty}a_{1}(n)t_{1}^{n}\right)^{2}. \end{align} $$

In what follows, we exemplify this Clausen-type transformation by reproducing the quadratic transformation (3.6).

Example 3.5 Take $\Gamma _{1}=\gamma \Gamma (2)\gamma ^{-1}$ with $\gamma =\begin {pmatrix}1&0\\0&2\end {pmatrix}$ , i.e., $\Gamma _{1}=\Gamma _{0}(4)$ , and $\Gamma _{2}=\Gamma _{0}(4)+$ . Then correspondingly,

$$ \begin{align*}t_{1}=16\frac{\eta(4\tau)^{16}\eta\left({\tau}\right)^{8}}{\eta(2\tau)^{24}}\quad\mbox{and}\quad t_{2}=\frac{\eta(4\tau)^{24}\eta(\tau)^{24}}{\eta(2\tau)^{48}}=\frac{t_{1}}{16}(1-t_{1}), \end{align*} $$

and by properties of modular forms or Sturm’s bound, one can check that

$$ \begin{align*}Y_{2}=\frac{1}{t_{2}\left(64t_{2}+1\right)^{\frac{1}{2}}}\frac{1}{2\pi i}\frac{dt_{2}}{d\tau}=\frac{1}{t_{1}(t_{1}-1)}\frac{1}{2\pi i}\frac{dt_{1}}{d\tau}=Y_{1}=y_{1}^{2}. \end{align*} $$

One can also deduce from the orbifold uniformizing differential equation satisfied by $y_{1}$ with respect to $t_{1}$ and the symmetric square satisfied by $Y_{2}$ with respect to $t_{2}$ that

$$ \begin{align*}y_{1}=\sum_{n=0}^{\infty}\binom{2n}{n}^{2}\left(\frac{t_{1}}{4^{2}}\right)^{n}={}_{2}F_{1}\left(\frac{1}{2},\frac{1}{2};1;t_{1}\right) \end{align*} $$

and

$$ \begin{align*}Y_{2}=\sum_{n=0}^{\infty}\binom{2n}{n}^{3}t_{2}^{n}=\sum_{n=0}^{\infty}\binom{2n}{n}^{3}\left(\frac{4t_{1}(1-t_{1})}{4^{3}}\right)^{n}={}_{3}F_{2}\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;4t_{1}(1-t_{1})\right). \end{align*} $$

Therefore, near $t_{1}=0$ ,

$$ \begin{align*}{}_{2}F_{1}\left(\frac{1}{2},\frac{1}{2};1;t_{1}\right)^{2}={}_{3}F_{2}\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;4t_{1}(1-t_{1})\right). \end{align*} $$

Another example given as follows plays an important role in the derivation of Example 1.6.

Example 3.6 Take $\Gamma _{1}=\gamma ^{-1}\left (\Gamma _{0}(4) \cap \Gamma (2)\right )\gamma $ with $\gamma =\begin {pmatrix}1&0\\1&1\end {pmatrix}$ and $\Gamma _{2}=\Gamma (2)$ , and correspondingly

$$ \begin{align*}t_{1}=4\frac{\eta(2\tau)^{8}\eta\left(\frac{\tau}{2}\right)^{4}}{\eta(\tau)^{12}}\quad\mbox{and}\quad t_{2}=t_{1}^{2}, \end{align*} $$

where $t_{2}$ is the well-known Lambda modular invariant. Following the remarks above, one first finds that

$$ \begin{align*}Y_{2}=\frac{1}{t_{2}^{2}-t_{2}}\frac{1}{2\pi i}\frac{dt_{2}}{d\tau}=\frac{2}{t_{1}^{3}-t_{1}}\frac{1}{2\pi i}\frac{dt_{1}}{d\tau}=2Y_{1}=2y_{1}^{2}. \end{align*} $$

By [1, equation (n)] and [6, p. 373], one can deduce by changes of variables that

$$ \begin{align*}y_{1}^{2}=-\frac{1}{4}\left(\sum_{n=0}^{\infty}\left(\sum_{k}(-1)^{k}\binom{-1/2}{k}\binom{-1/2}{n-k}\binom{n}{k}\right)t_{1}^{n}\right)^{2} \end{align*} $$

and

$$ \begin{align*}Y_{2}=-\frac{1}{2}\sum_{n=0}^{\infty}\left(\sum_{k}\binom{2n-2k}{n-k}^{2}\binom{2k}{k}^{2}\right)16^{-n}t_{2}^{n}, \end{align*} $$

and thus one has the following transformation identity near $t_{1}=0$ :

$$ \begin{align*}\sum_{n=0}^{\infty}\left(\sum_{k}\binom{2n-2k}{n-k}^{2}\binom{2k}{k}^{2}\right)16^{-n}t_{1}^{2n} =\left(\sum_{n=0}^{\infty}\left(\sum_{k}(-1)^{k}\binom{-1/2}{k}\binom{-1/2}{n-k}\binom{n}{k}\right)t_{1}^{n}\right)^{2}. \end{align*} $$

Remark 3.7 The notion of Clausen-type transformation formulas arising from modular forms was first used in [7] by Chan et al., in which they establish three examples that can be viewed as the specialization of (3.7) to the cases of $\Gamma _{1}=\Gamma _{0}(6)$ and $\Gamma _{2}=\Gamma _{0}(6)+$ , where $\Gamma _{0}(6)+$ is the group generated by $\Gamma _{0}(6)$ and all its Atkin–Lehner involutions.

4 Proof of Theorem 1.1

We are now in a position to give a proof to our main theorem as a closing of this note.

Proof of Theorem 1.1

Under the given assumptions, write

$$ \begin{align*}Y=Y(\tau)=\prod_{\substack{z_{0}\ \text{is a cusp}}}(t-t(z_0))^{-1}\prod_{\substack{z_{0}\ \text{is an elliptic }\\\text{point of period}~h}}\left(t-t(z_{0})\right)^{\frac{1-h}{h}}\frac{1}{2\pi i}\frac{dt}{d\tau}, \end{align*} $$

and start with the simple manipulation

$$ \begin{align*}\frac{1}{2\pi i}\frac{dY}{d\tau}=\left(\frac{1}{2\pi i}\frac{dY}{d\tau}-\frac{2Y}{4\pi\mathrm{Im}(\tau)}\right)+\frac{2Y}{4\pi\mathrm{Im}(\tau)}. \end{align*} $$

Writing $Y=y^{2}$ and multiplying both sides by y gives

(4.1) $$ \begin{align} y\frac{1}{2\pi i}\frac{dY}{d\tau}&=y\left(\frac{2y}{2\pi i}\frac{dy}{d\tau}-\frac{2y^{2}}{4\pi\mathrm{Im}(\tau)}\right)+y\frac{2Y}{4\pi\mathrm{Im}(\tau)}\nonumber\\ &= 2Y\left(\frac{1}{2\pi i}\frac{dy}{d\tau}-\frac{y}{4\pi\mathrm{Im}(\tau)}\right)+y\frac{2Y}{4\pi\mathrm{Im}(\tau)}. \end{align} $$

Recall by Lemma 3.3(iv) that $Y(\tau )=\sum _{n=0}^{\infty }A(n)t(\tau )^{n}$ with coefficients $A(n)$ satisfying a linear homogeneous recurrence relation of at most $(\nu (\Gamma )-1)$ terms induced by (3.4) for

$$ \begin{align*}|t|<\mu(t)=\min\{|t(z_{0})|:\, z_{0}\ \mbox{is a cusp apart from}\ [i\infty],\ \mbox{or an elliptic point}\}. \end{align*} $$

Substitute this for the Y’s attached to the first two terms from left in (4.1) to get

$$ \begin{align*} y\frac{1}{2\pi i}\frac{dt}{d\tau}\sum_{n=0}^{\infty}A(n)nt^{n-1}&=2\left(\frac{1}{2\pi i}\frac{dy}{d\tau}-\frac{y}{4\pi \mathrm{Im}(\tau)}\right)\sum_{n=0}^{\infty}A(n)t^{n}+y\frac{Y}{2\pi \mathrm{Im}(\tau)}. \end{align*} $$

A simple manipulation gives the desired formula (1.4) for $\frac {1}{\pi }$ ,

$$ \begin{align*} \frac{1}{\pi}={C}\sum_{n=0}^{\infty}A(n)(n-{B})t^{n}, \end{align*} $$

where

$$ \begin{align*} {B}&=t\left(\frac{1}{\pi i}\frac{dy}{d\tau}-\frac{y}{2\pi \mathrm{Im}(\tau)}\right)\left/\left(y\frac{1}{2\pi i}\frac{dt}{d\tau}\right)\right. \end{align*} $$

and

$$ \begin{align*} {C}&=2\mathrm{Im}(\tau)\left.\frac{1}{2\pi i}\frac{dt}{d\tau}\right/\left(tY\right)\\ &=\frac{2\mathrm{Im}(\tau)}{t}\prod_{\substack{z_{0}\ \text{is a cusp}}}(t-t(z_0))\prod_{\substack{z_{0}\ \text{is an elliptic }\\\text{point of period}~h}}\left(t-t(z_{0})\right)^{\frac{h-1}{h}}. \end{align*} $$

Finally, since $y=y(\tau )$ is a meromorphic modular form of weight 1 and $t=t(\tau )$ is a modular function for an $\mathrm {SL}_{2}(\mathbb {R})$ -subgroup commensurable with $\mathrm {SL}_{2}(\mathbb {Z})$ , and both y and t have algebraic coefficients at $[i\infty ]$ as assumed, by [8, Corollary 5.10.7], one can conclude that both B and C are algebraic numbers for $\tau $ an imaginary quadratic point in $\mathbb {H}$ .