Dilogarithm identities for solutions to Pell's equation in terms of continued fraction convergents

TL;DR

This paper establishes a novel link between the dilogarithm function and solutions to Pell's equation, using continued fraction expansions, and connects Ramanujan's identities to hyperbolic geometry.

Contribution

It introduces a new method to derive dilogarithm identities from Pell's equation solutions via continued fractions, and relates Ramanujan's identities to hyperbolic geometry.

Findings

Dilogarithm identities associated with Pell's solutions are derived.

Ramanujan's dilogarithm identities are interpreted through hyperbolic geometry.

Continued fraction expansions of units in quadratic fields are central to the identities.

Abstract

In this paper we give describe a new connection between the dilogarithm function and solutions to Pell's equation $x^2-ny^2 = \pm 1$. For each solution $x,y$ to Pell's equation we obtain a dilogarithm identity whose terms are given by the continued fraction expansion of the associated unit $x+y\sqrt{n} \in \Z[\sqrt{n}]$. We further show that Ramanujan's dilogarithm value-identities correspond to an identity for the regular ideal hyperbolic hexagon.

Equations

$$L{i}_2(z)=\sum _{n=1}^{\infty }\frac{z^n}{n^2}|z|\le 1$$

$$L{i}_2(z)=-{\int }_0^z\frac{\log (1-t)}{t}dt.$$

$$\mathcal{L}(z)=L{i}_2(z)+\frac{1}{2}\log |z|\log (1-z).$$

$$Vol(T)=\mathrm{Im}(\mathcal{L}([z_1,z_2,z_3,z_4]))$$

$$L{i}_2(z)+L(-z)=\frac{1}{2}L{i}_2(z^2).$$

$$\mathcal{L}(z)+\mathcal{L}(-z)=\frac{1}{2}\mathcal{L}(z^2)\text{mbox}(SquaringIdentity).$$

$$\mathcal{L}(x)+\mathcal{L}(1-x)=\frac{{\pi }^2}{6}\mathcal{L}(x)+\mathcal{L}(x^{-1})=\frac{{\pi }^2}{6}\text{mbox}(ReflectionIdentity)$$

$$\mathcal{L}(x)+\mathcal{L}(y)=\mathcal{L}(xy)+\mathcal{L}\left(\frac{x(1-y)}{1-xy}\right)+\mathcal{L}\left(\frac{y(1-x)}{1-xy}\right)\text{mbox}(AbelIdentity)$$

$$\mathcal{L}\left(-\frac{1}{x}\right)=-\mathcal{L}\left(\frac{1}{x+1}\right)x>0\text{mbox}(LandauIdentity).$$

$$\mathcal{L}(0)=0\mathcal{L}(1)=\frac{{\pi }^2}{6}\mathcal{L}\left(\frac{1}{2}\right)=\frac{{\pi }^2}{12}\mathcal{L}({\varphi }^{-1})=\frac{{\pi }^2}{10}\mathcal{L}({\varphi }^{-2})=\frac{{\pi }^2}{15}$$

$$\sum _{k=2}^{\infty }\mathcal{L}\left(\frac{1}{k^2}\right)=\frac{{\pi }^2}{6}.$$

$$\mathcal{L}(e^{-L})=\sum _{k=2}^{\infty }\mathcal{L}\left(\frac{{\sinh }^2\left(\frac{L}{2}\right)}{{\sinh }^2\left(\frac{kL}{2}\right)}\right)$$

$$\mathcal{L}\left(\frac{1}{u^2}\right)=\sum _{k=1}^{\infty }\mathcal{L}\left(\frac{1}{(h_{2k-1}{)}^2}\right).$$

$$\mathcal{L}\left(\frac{1}{u^2}\right)=\sum _{k=0}^{\infty }\mathcal{L}\left(\frac{1}{b^{2}n(2H_{2k-1}{)}^2}\right)+\mathcal{L}\left(\frac{1}{(2H_{2k+1}-H_{2k}{)}^2}\right).$$

$$\mathcal{L}\left(\frac{1}{(3+2\sqrt{2}{)}^2}\right)=\mathcal{L}\left(\frac{1}{6^2}\right)+\mathcal{L}\left(\frac{1}{35^2}\right)+\mathcal{L}\left(\frac{1}{204^2}\right)+\mathcal{L}\left(\frac{1}{1189^2}\right)+\dots .$$

$$\frac{5}{1},\frac{6}{1},\frac{29}{5},\frac{35}{6},\frac{169}{29},\frac{204}{35},\frac{985}{169},\frac{1189}{204}.$$

$$\mathcal{L}\left(\frac{1}{3+2\sqrt{2}}\right)=\mathcal{L}\left(\frac{1}{2(2{)}^2}\right)+\mathcal{L}\left(\frac{1}{7^2}\right)+\mathcal{L}\left(\frac{1}{2(12{)}^2}\right)+\mathcal{L}\left(\frac{1}{41^2}\right)+\mathcal{L}\left(\frac{1}{2(70{)}^2}\right)+\mathcal{L}\left(\frac{1}{239^2}\right)+\mathcal{L}\left(\frac{1}{2(408{)}^2}\right)+\dots .$$

$$\mathcal{L}\left(\frac{1}{842401+233640\sqrt{13}}\right)=\mathcal{L}\left(\frac{1}{1298^2}\right)+\mathcal{L}\left(\frac{1}{1684803^2}\right)+\mathcal{L}\left(\frac{1}{2186872996^2}\right)\dots .$$

$$\frac{1297}{1},\frac{1298}{1},\frac{1683505}{1297},\frac{1684803}{1298},\frac{2185188193}{1683505},\frac{2186872996}{1684803}\dots$$

$$\mathcal{L}\left(\frac{1}{u^2}\right)=\sum _{k=2}^{\infty }\mathcal{L}\left(\frac{1}{(b_k/b{)}^2}\right).$$

$$\mathcal{L}\left(\frac{1}{u^2}\right)=\sum _{k=1}^{\infty }\mathcal{L}\left(\frac{1}{n(b_{2k}/a{)}^2}\right)+\mathcal{L}\left(\frac{1}{(a_{2k+1}/a{)}^2}\right)$$

$${\varphi }^k=\frac{g_k+f_k\sqrt{5}}{2}$$

$$\sum _{k=1}^{\infty }\left(\mathcal{L}\left(\frac{1}{5f_{2n}^2}\right)+\mathcal{L}\left(\frac{1}{g_{2n+1}^2}\right)\right)=\frac{{\pi }^2}{15}$$

$$U_n(\cos (\theta ))=\frac{\sin ((n+1)\theta )}{\sin (\theta )}$$

$$\mathcal{L}\left(\frac{1}{{\left(x+\sqrt{x^2-1}\right)}^2}\right)=\sum _{n=1}^{\infty }\mathcal{L}\left(\frac{1}{U_n(x{)}^2}\right).$$

$$a^2-n{b}^2=\pm 1.$$

$$u=c_0+\frac{1}{c_1+\frac{1}{c_2+\frac{1}{c_3+\dots }}}$$

$$h_i=c_i{h}_{i-1}+h_{i-2}{k}_i=c_i{k}_{i-1}+k_{i-2}i\ge 0$$

$$r_n=[c_0,c_1,c_2,\dots ,c_n]=\frac{h_n}{k_n}$$

$$a_k+b_k\sqrt{n}=(a+b\sqrt{n}{)}^k.$$

Full text

1 Dilogarithm and Pell’s Equation

Dilogarithm: The dilogarithm function $Li_{2}(z)$ is given by the power series

[TABLE]

which has integral form

[TABLE]

In [13], Rogers introduced the following normalization.

[TABLE]

The dilogarithm function arises in many areas of mathematics, especially hyperbolic geometry and number theory (see [14]). In particular the volume of an ideal hyperbolic tetrahedron $T$ in ${\mathbb{H}}^{3}$ with ideal vertices $z_{1},z_{2},z_{3},z_{4}\in\hat{\mathbb{C}}$ is

[TABLE]

where $[z_{1},z_{2},z_{3},z_{4}]$ is the cross-ratio.

Pell’s Equation: Pell’s equation for $n\in\mathbb{N}$ is the Diophantine equation $x^{2}-ny^{2}=\pm 1$ over $\mathbb{Z}$. Pell’s equation has a long and interesting history going back to Archimedes’ cattle problem (see [7]). The equation only has solutions for $n$ square-free, so we assume $n$ is square-free. Also by symmetry, we only need consider solutions with $x,y>0$. A solution is positive/negative depending on whether $x^{2}-ny^{2}=1$ or $x^{2}-ny^{2}=-1$. For all square-free $n$ there is always a positive solution but not necessarily a negative solution. Solutions to Pell’s equation correspond to units in $\mathbb{Z}[\sqrt{n}]$ by identifying $x,y$ with $x+y\sqrt{n}$ and it is natural to identify the two. The smallest positive unit $u=x+y\sqrt{n}$ is called the fundamental unit and a well-known result gives that the set of positive units is exactly $\{u^{k}\},k\in\mathbb{N}$ (see [12, Theorem 7.26])).

In this paper we prove a new and surprising connection between the dilogarithm and solutions to Pell’s equation. Using earlier work of the author which gave a dilogarithm identity associated to a hyperbolic surface, we obtain a dilogarithm identity for each solution $x,y$ to Pell’s equation whose terms are given by the continued fraction expansion of $x+y\sqrt{n}$.

1.1 Dilogarithm Identities

The dilogarithm satisfies a number of classical identities. In particular by adding power series we have the squaring identity

[TABLE]

It follows by direct computation that this identity holds for the Rogers dilogarithm with

[TABLE]

The other classic identities are the reflection identity of Euler

[TABLE]

Abel’s well-known 5-term identity

[TABLE]

and Landau’s identity

[TABLE]

A closed form for values of $\mathcal{L}$ is only known for a small set of values. These are

[TABLE]

where $\phi$ is the golden ratio. In [8] Lewin gave the following remarkable infinite identity.

[TABLE]

2 Results

Using earlier work of the author we first prove the below new infinite identity for $\mathcal{L}$. We prove

Theorem 2.1

If $L>0$ then

[TABLE]

One immediate observation is if we let $L\rightarrow 0$ we recover the formula of Lewin in equation 1.1 above.

We now apply the above identity to solutions of Pell’s equation and units in the ring $\mathbb{Z}[\sqrt{n}]$.

Dilogarithm identity for Solution to Pell’s Equation

In order to obtain our identity associated to a given solution $a^{2}-nb^{2}=\pm 1$ of Pell’s equation, we will choose $e^{L/2}=a+b\sqrt{n}$ in the above. We then show that the righthandside is given in terms of the continued fraction expansion of $a+b\sqrt{n}$. We obtain:

Theorem 2.2

Let $u=a+b\sqrt{n}\in\mathbb{Z}[\sqrt{n}]$ be a solution to Pell’s equation.

• •

If $u$ is a positive solution with continued fraction convergents $r_{j}=h_{j}/k_{j}$ then

[TABLE]

• •

If $u$ is a negative solution and $u^{2}$ has convergents $R_{j}=H_{j}/K_{j}$ then

[TABLE]

Examples

We now consider some examples:

Case of $\mathbb{Z}[\sqrt{2}]$: For $\mathbb{Z}[\sqrt{2}]$ the fundamental unit is $3+2\sqrt{2}$ giving

[TABLE]

We note that $3+2\sqrt{2}$ has convergents $r_{k}$ given by

[TABLE]

It can be further shown that the units of $\mathbb{Z}[\sqrt{2}]$ are given by $(1+\sqrt{2})^{k}$. As $u$ is a negative solution to Pell’s equation and $u^{2}=3+2\sqrt{2}$ we get

[TABLE]

Case of $\mathbb{Z}[\sqrt{13}]$: An interesting case of a large fundamental solution occurs for $\mathbb{Z}[\sqrt{13}]$. Here $u=649+180\sqrt{13}$ is the fundamental unit giving

[TABLE]

The continued fraction convergents of $u$ are

[TABLE]

Pell’s equation over $\mathbb{Q}$

Similarly we consider Pell’s equation over $\mathbb{Q}$. If $a,b\in\mathbb{Q}$ satisfy Pell’s equation $a^{2}-nb^{2}=\pm 1$ we will identify this with the element $a+b\sqrt{n}\in\mathbb{Q}[\sqrt{n}]$. Applying the identity in Theorem 2.1 we get the following.

Theorem 2.3

Let $u=a+b\sqrt{n}\in\mathbb{Q}[\sqrt{n}],a,b>0$ satisfy Pell’s equation and let $u^{k}=a_{k}+b_{k}\sqrt{n}$.

If $u$ is a positive solution, then

[TABLE]

Further if $u\in\mathbb{Z}[\sqrt{n}]$ then $b_{k}/b\in\mathbb{Z}$ for all $k$.

If $u$ is a negative solution then

[TABLE]

Further if $u\in\mathbb{Z}[\sqrt{n}]$ then $b_{2k}/a,a_{2k+1}/a\in\mathbb{Z}$ for all $k$.

Fibonacci Numbers: The golden mean $\phi\in\mathbb{Q}[\sqrt{5}]$ corresponds to a negative solution to Pell’s equation over $\mathbb{Q}$. Also we have

[TABLE]

where $f_{k}$ is the classic Fibonacci sequence $1,1,2,3,5\ldots$ and $g_{k}$ is the Fibonacci sequence $1,3,4,7,11,\ldots$.

As $\mathcal{L}(\phi^{-2})=\pi^{2}/15$ we get the identity

[TABLE]

Chebyshev Polynomials, Pell’s Equation and Dilogarithms

Chebyshev polynomials arise in numerous areas of mathematics and have a natural interpretation in terms of Pell’s equation. The Chebyshev polynomials of the first kind $T_{n}$ are the unique polynomials satisfying $T_{n}(\cos(\theta))=\cos(n\theta)$ and the Chebyshev polynomials of the second kind $U_{n}$ are given by

[TABLE]

We obtain the following corollary.

Corollary 2.4

Let $x>1$ then

[TABLE]

The reader interested in knowing more about the dilogarithm function and its generalizations we refer to the book [8], Structural Properties of Polylogarithms, by L. Lewin and the aforementioned article [14], The dilogarithm function, by D. Zagier.

3 Units in $\mathbb{Z}[\sqrt{n}]$, Pell’s equation

We assume $n$ is not a perfect square. If $a+b\sqrt{n}\in\mathbb{Z}[\sqrt{n}]$ is a unit, then so are $\pm a\pm b\sqrt{n}$ and therefore we only need to consider solutions $(a,b)\in\mathbb{N}^{2}$. It follows easily that $a\pm b\sqrt{n}\in\mathbb{Z}[\sqrt{n}]$ is a unit if and only if $(a,b)$ satisfy Pell’s equation over $\mathbb{Z}$

[TABLE]

We call a solution $(a,b)$ (or the unit $a+b\sqrt{n}$) positive/negative depending on if the righthandside of the Pell equation is positive/negative. Whereas there is always a solution to the positive Pell equation $x^{2}-ny^{2}=1$, it can be shown that there are no solutions to $x^{2}-ny^{2}=-1$ for certain $n$ (see [12, Chapter 7]).

Continued Fraction Convergents

If $u\in\mathbb{R}_{+}$ we say $u$ has continued fraction expansion $u=[c_{0},c_{1},c_{2},c_{3},\ldots]$ if $c_{i}\in\mathbb{Z}$ and

[TABLE]

This means that if we define $r_{n}=[c_{0},c_{1},c_{2},\ldots,c_{n}]\in\mathbb{Q}$ to be the $n^{th}$ convergent, then $r_{n}\rightarrow u$ as $n\rightarrow\infty$. If the continued fraction coefficients satisfy $c_{n+r}=c_{n}$ for $n>k$ we say $u$ is periodic with period $r$ and write $u=[c_{0},c_{1},\ldots,c_{k},\overline{c_{k+1},\ldots,c_{k+r}}]$. We have the following standard description of $r_{n}$;

Theorem 3.1

([12, Theorems 7.4, 7.5]) Let $u\in\mathbb{R}_{+}$ with $u=[c_{0},c_{1},c_{2},\ldots]$ and define $h_{n},k_{n}$ by

[TABLE]

with $(h_{-2},k_{-2})=(0,1),(h_{-1},k_{-1})=(1,0)$. Then $\gcd(h_{i},k_{i})=1$ and

[TABLE]

The positive units in $\mathbb{Z}[\sqrt{n}]$ have the following elegant description.

Theorem 3.2

([12, Theorem 7.26]) Let $n\in\mathbb{N}$ not be a perfect square. Then there is a unique solution $(a,b)\in\mathbb{N}^{2}$ of Pell’s equation $x^{2}-ny^{2}=1$ such that the set of solutions to $x^{2}-ny^{2}=1$ in $\mathbb{N}^{2}$ is $\{(a_{k},b_{k})\}_{k=1}^{\infty}$ where

[TABLE]

The pair $(a,b)$ is called the fundamental solution of $x^{2}-ny^{2}=1$. Thus one consequence of the above is if we let $u$ be the fundamental unit then $\{u^{k}\}$ gives the set of all positive solutions to Pell’s equation and the dilogarithm identity in Theorem 2.2 can be interpreted as a sum over all solutions to Pell’s equation.

4 The Orthospectrum Identity

In a prior paper, the author proved a dilogarithm identity for a hyperbolic surface with geodesic boundary. The identity was generalized to hyperbolic manifolds by the author and Kahn in [5]. The relation to other identities on hyperbolic manifolds such as the Basmajian identity (see [3]), McShane-Mirzakhani identity (see [10], [11]) and Luo-Tan identity (see [9]) is discussed in [6].

In order to state the orthospectrum identity, we recall some basic terms.

If $S$ is a hyperbolic surface with totally geodesic boundary, an orthogeodesic $\alpha$ is a proper geodesic arc which is perpendicular to the boundary $\partial S$ at its endpoints. The set of orthogeodesics of $S$ is denoted $O(S)$. Each boundary component is either a closed geodesic or an infinite geodesic whose endpoints are boundary cusps of $S$. We let $N(S)$ be the number of boundary cusps of $S$. Further let $\chi(S)$ be given by $Area(S)=2\pi|\chi(S)|$.

One elementary example of a surface will be an ideal n-gon which has $N(S)=n$ and $O(S)$ a finite set and $\chi(S)=1-n/2$. In fact these are the only surfaces with $O(S)$ finite.

The dilogarithm orthospectrum identity is as follows;

Theorem 4.1

(Dilogarithm Orthospectrum Identity, [4]) Let $S$ be a finite area hyperbolic surface with totally geodesic boundary $\partial S\neq 0$. Then

[TABLE]

and equivalently

[TABLE]

In the original paper [4], we showed that the above identity recovers the reflection identity, Abels identity and Landau’s identity by considering the elementary cases of the ideal quadrilateral and ideal pentagon respectively.

5 An infinite dilogarithm identity

We define the cross-ratio of 4 distinct points in $\hat{\mathbb{C}}$ by

[TABLE]

We let $x_{1},x_{2},x_{3},x_{4}\in\partial{\mathbb{H}}^{2}$ be distinct points ordered counterclockwise on $\partial{\mathbb{H}}^{2}$. If $g$ is the geodesic with endpoints $x_{1},x_{2}$ and $h$ is the geodesic with endpoints $x_{3},x_{4}$ then $g,h$ are disjoint. A simple calculation shows that the perpendicular distance between them $l$ is given by

[TABLE]

We now prove Theorem 2.1 which we now restate.

Theorem 2.1 If $L>0$ then

[TABLE]

Proof: We consider the hyperbolic surface $S$ which is topologically an annulus with one boundary component being a closed geodesic $g$ of length $L$ and the other $h$ an infinite geodesic with a single boundary cusp (see figure 1).

We lift $S$ to the upper half plane with $g$ lifted to the y-axis and let $\lambda=e^{L}$. Then $\tilde{S}$ is an infinite sided ideal polygon, which is invariant under multiplication by $\lambda$ (see figure 2).

We normalize so that one of the ideal vertices is at $1$. Then the vertices of $\tilde{S}$ are $\lambda^{k}$ for $k\in\mathbb{Z}$. There is a single orthogeodesic with endpoint on $g$ and it has length $l$ satisfying

[TABLE]

The other ortholengths are given by $l_{k}$ where

[TABLE]

[TABLE]

Thus the dilogarithm identity for $S$ is

[TABLE]

Using the reflection identity $\mathcal{L}(1-x)+\mathcal{L}(x)=\pi^{2}/6$ we get

[TABLE]

$\Box$

6 Proof of identity for Solutions to Pell’s Equation over $\mathbb{Q}$

We now prove the dilogarithm identity for solutions to Pell’s equation over $\mathbb{Q}$ given in Theorem 2.3.

Proof of Theorem 2.3:

Let $e^{L/2}=u=a+b\sqrt{n}$, then $e^{-L/2}=u^{-1}=\pm(a-b\sqrt{n})$ with the sign depending on if $u$ is a positive unit or negative. If $u$ is a positive unit then

[TABLE]

and if $u$ is a negative unit

[TABLE]

Either way we have

[TABLE]

We let $m_{k}=\sinh(kL/2)$ and $n_{k}=\cosh(kL/2)$. The dilogarithm identity gives

[TABLE]

If $u$ is a positive root, then $m_{1}=\sinh(L/2)=b\sqrt{n}$ and $n_{1}=\cosh(L/2)=a$. Then by the addition formulae we have

[TABLE]

Then by induction we have $n_{k}=a_{k}$ and $m_{k}=b_{k}\sqrt{n}$ and

[TABLE]

[TABLE]

If $u$ is a negative solution, then $m_{1}=\sinh(L/2)=a$ and $n_{1}=\cosh(L/2)=b\sqrt{n}$. Then by the addition formulae we have

[TABLE]

Therefore

[TABLE]

Therefore

[TABLE]

Therefore

[TABLE]

$\Box$

We now prove Corollary 2.4 relating the identity to the Chebyshev polynomials $U_{n}$ of the second kind.

Proof of Corollary 2.4: We have the Chebyshev polynomials $T_{n}(x),U_{n}(x)\in\mathbb{R}[x]$. We let $x=\cos(\theta)$, then $\sin(\theta)=\sqrt{1-x^{2}}$. Therefore

[TABLE]

and

[TABLE]

Thus

[TABLE]

and

[TABLE]

As this holds for $|x|<1$, it also holds for all $x\in\mathbb{R}$. Now if $x>1$ then we let $x=\cosh(L/2)$ then $\sqrt{x^{2}-1}=\sinh(L/2)$ giving

[TABLE]

Therefore by the above formulae

[TABLE]

Thus

[TABLE]

$\Box$

7 Identity for continued fraction convergents

We now consider the case where $u\in\mathbb{Z}[\sqrt{n}]$ and prove Theorem 2.2 expressing the above in terms of the convergents $r_{j}=h_{j}/k_{j}$ of their continued fractions expansion. First we have the following lemma.

Lemma 7.1

Let $u=a+b\sqrt{n}\in\mathbb{Z}[\sqrt{n}]$ be a solution to Pell’s equation with $a,b\in\mathbb{N}$. If $u$ is a positive solution then $u=[2a-1,\overline{1,2a-2}]$ and if $u$ is a negative solution then $u=[\overline{2a}]$.

Proof: if $u$ is a negative solution. then $u=a+\sqrt{a^{2}+1}$ and $u^{2}-2au-1=0$. Then

[TABLE]

giving $u=[\overline{2a}]$.

If $u$ is a positive solution then $u=a+\sqrt{a^{2}-1}$. Therefore $u$ satisfies the quadratic $u^{2}-2au+1=0$. Rewriting we have

[TABLE]

Now we have

[TABLE]

Therefore $u=[2a-1,\overline{1,2a-2}]$.

$\Box$

Using the above description of the continued fraction, we will show the relation between the approximates $r_{j}=h_{j}/k_{j}$ for $u$ and the coefficients $a_{j},b_{j}$ given by $u^{j}=a_{j}+b_{j}\sqrt{n}$. This will allow us to prove Theorem 2.2 giving the dilogarithm identity in terms of the convergents of the the continued fraction convergents.

Lemma 7.2

Let $u=a+b\sqrt{n}\in\mathbb{Z}[\sqrt{n}]$ be a solution to Pell’s equation.

If $u$ is a positive solution and $u$ has continued fraction convergents $r_{j}=h_{j}/k_{j}$ then $k_{j}=h_{j-2}$ and

[TABLE]

If $u$ is a negative solution and $u^{2}$ has continued fraction convergents $R_{j}=H_{j}/K_{j}$ then

[TABLE]

Proof: Let $u=a+b\sqrt{n}=e^{L/2}$, then $u^{k}=a_{k}+b_{k}\sqrt{n}=\cosh(kL/2)+\sinh(kL/2)$. As $u$ is a positive solution $u=[2a-1,\overline{1,2a-2}]$. Therefore we have $(h_{0},h_{-1})=(2a-1,1)$ and for $k>0$

[TABLE]

The matrix $A$ has characteristic polynomial $x^{2}-2ax+1$ giving eigenvalues $u,1/u$ and eigenvectors $(u-1,1),(1-u,u)$. Therefore diagonalizing we get

[TABLE]

Multiplying out and noting that $u=e^{L/2}$ we have

[TABLE]

It follows that for $k\geq 1$

[TABLE]

Therefore

[TABLE]

Similarly we note that as $(k_{0},k_{1})=(1,0)$ then applying the above analysis we get

[TABLE]

and

[TABLE]

giving $k_{j}=h_{j-2}$.

If $u$ is a negative solution, then for $k$ odd $a_{k}=\sinh(kL/2),b_{k}\sqrt{n}=\cosh(kL/2)$ and for $k$ even $b_{k}\sqrt{n}=\sinh(kL/2),a_{k}=\cosh(kL/2)$.

As $u=[\overline{2a}]$ we have the formula

[TABLE]

with $(h_{-2},k_{-2})=(0,1)$ and $(h_{-1},k_{-1})=(1,0)$. Iterating we get $h_{j}=0,1,2a,\ldots$ and $k_{j}=1,0,1,2a,\ldots$. Therefore $k_{j}=h_{j-1}$ for $j\geq-1$. We focus on calculating $h_{k}$. As $(h_{-1},h_{-2})=(1,0)$ we have the recursion

[TABLE]

The matrix $A$ has characteristic polynomial $x^{2}-2ax-1=0$ giving eigenvalues $u,-u^{-1}$ and eigenvectors $(u,1),(1,-u)$. Thus

[TABLE]

Multiplying we get

[TABLE]

For $k$ odd we have

[TABLE]

Similarly for $k$ even we have

[TABLE]

Thus for all $k\geq 0$

[TABLE]

We let $H_{j},K_{j}$ be the convergents for the continued fraction expansion of $u^{2}$. Then as $u^{2}=e^{L}$ is a positive solution to Pell’s equation. Applying equations 7.2 and 7.3 above we have,

[TABLE]

[TABLE]

Also if $(u^{2})^{k}=A_{k}+B_{k}\sqrt{n}$ then $A_{k}=a_{2k},B_{k}=b_{2k}$ and by equation 7.4

[TABLE]

giving

[TABLE]

Also

[TABLE]

Thus if $u$ is a negative solution to Pell’s equation

[TABLE]

$\Box$

8 Ideal n-gon identities

Ramanujan gave a number of value-identities for linear combinations of specific values of $\mathcal{L}$ as follows (see [1, Entry 39]);

[TABLE] 2. 2.

[TABLE] 3. 3.

[TABLE] 4. 4.

[TABLE] 5. 5.

[TABLE]

More recently in their article [2], Bailey, Borwein, and Plouffe gave the identity

[TABLE]

Applying Landau’s identity gives $\mathcal{L}(-1/3)=-\mathcal{L}(1/4)$ and $\mathcal{L}(-1/8)=-\mathcal{L}(1/9)$, which reduces the value-identities of Ramanujan to the two equations

[TABLE]

In this section we will show that these value-identities follow from considering ideal hyperbolic hexagons.

We recall the dilogarithm identity in [4] for ideal hyperbolic polygons. Let $P$ be an ideal polygon in ${\mathbb{H}}^{2}$ with vertices in counterclockwise order $x_{1},\ldots,x_{n}$ about $\partial{\mathbb{H}}^{2}$. If $l_{ij}$ is the length of the orthogeodesic joining side $[x_{i},x_{i+1}]$ to $[x_{j},x_{j+1}]$ then a simple calculation gives

[TABLE]

Applying the orthospectrum identity in Theorem 4.1 to $P$ we obtain the equation

[TABLE]

If $P$ is the regular ideal n-gon then we obtain the equation

[TABLE]

where $e_{n}=0$ if n is odd and $e_{n}=1$ if n is even.

9 Ideal Hexagons and Ramanujan’s value identities

We now show that Ramanujan’s value identities 1-5 and identity 8.5 of Bailey, Borwein, Plouffe, correspond to identities for the regular ideal hexagons.

For the regular 6-gon $H_{reg}$ the orthospectrum identity gives

[TABLE]

We also have from Landau’s identity that $\mathcal{L}(-1/3)=-\mathcal{L}(1/4)$. Therefore applying the squaring identity we get

[TABLE]

Thus we obtain

[TABLE]

Combining this and the identity above for the regular hexagon, we obtain Ramanujan’s value-identities

[TABLE]

To recover the identity 8.5 of Bailey, Borwein, Plouffe we note that by Landau $\mathcal{L}(-1/8)=-\mathcal{L}(1/9)$ then by the squaring identity we have

[TABLE]

Therefore substituting for $\mathcal{L}(1/8)$ we get

[TABLE]

As $\mathcal{L}(1/2)=\pi^{2}/12$ and by the hexagon identity $3\mathcal{L}(1/4)+\mathcal{L}(1/9)=\pi^{2}/6$ we recover identity 8.5.