TITLE: Use the remainder theorem to find $P(2)$ where $P(x)=-x^4+3x^3-4x+7$
QUESTION [0 upvotes]: Use the remainder theorem to find $P(2)$ where $P(x)=-x^4+3x^3-4x+7$
Quotient = ?
Remainder = $P(2)$ = ?

REPLY [0 votes]: The following is the scheme: pull down the first coefficient; then multiply it by $2$, writing the result in the free space under $3$; do the sum, repeat.
$$
\begin{array}{c|cccc|c}
& -1 & 3 & 0 & -4 & 7 \\
2& \downarrow & -2  & 2  &  4  & 0 \\
\hline
 &-1 & 1 & 2 & 0 & 7
\end{array}
$$
Quotient of the division by $x-2$: $-x^3+x^2+2x$, remainder $7$.


TITLE: Gauss measure is not a pushforward of product measure
QUESTION [3 upvotes]: Let $ N = \{1,2,3, \ldots \}$. We define $\varphi : N^N \mapsto [0,1]$ as
$$ \varphi \left( (a_n)_{n \in_N} \right) = [0;a_1, a_2, \ldots ]$$
Where the expression on the right is a infinite continued fraction:
$$[0;a_1,a_2,a_3,\dotsc] = \cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}}$$
 It is standard exercise that this is borel mapping (continuous in fact, see here).
We define the Gauss measure on $[0,1]$ as:
$$ d\nu = \frac{1}{\ln(2)}\,\frac{dx}{1+x} $$
Let $\mu$ be a probability measure on $N$. By a product measure on $N^N$ i mean probabilistic measure $\widetilde { \mu }$ such that for any set of the form $F = F_1 \times \cdots \times F_n \times N \times N \times \cdots$ holds equality
$$ \widetilde {\mu } (F) = \prod_{n=1}^n \mu (F_k) $$
Such a measure exists and is unique due to teorem about extending funtion on ring to a measure.
I am to prove that the Gauss measure $\nu$ is not a pushforward of any such measure $\mu$, meaning that for any $\mu$ the equality
$$ \nu( A ) = \mu ( \varphi^{-1} [A])$$
cannot hold for every borel $A \subset [0,1]$.
I am looking for any help. I have no experience with infinite measure products. This is an exercise for my ergodic theory course.

I had obtained a tip that i don't understand:

Product measure has the property that basis clopen sets with disjoint supports are stochastically independent. Gauss measure does not have this property.

REPLY [0 votes]: Thanks to @Jonas i came up with a solution and my supervisor accepted it. Hence, i am posting it for completeness. Don't judge me if it is poorly written, i am translating it from Polish. ;)
From the source given in question we know that for any irrational  $x \in [0,1]$ it correspond to some continued fraction $[0; a_1, a_2, \ldots ]$. Another thing that we know is that it is between points
$$ x_n = [ 0; a_1, a_2, \ldots , a_n] \qquad x_{n+1} = [0; a_1, a_2, \ldots, a_{n+1} ] $$
To ease the notation lets denote for any $ \sigma \in N^{ < N}$ the set $ [[ \sigma ]] = \{ f \in N^N : \sigma \prec f \}$ and by $| \sigma|$ the length of $\sigma$.
Assume that for some product measure $\widetilde{ \mu}$ we have for all borel sets $A$
$$ \nu (A) = \widetilde{ \mu} \left( \varphi^{-1} (A) \right) $$
From the fact that $\varphi$ is a bijection  from $N^N$ to  $[0,1] \setminus Q$ and easy-to-check equalities
$$[0;1] = 1 \qquad [0;2] = 1/2 \qquad [0;3] = 1/3 \qquad [0;1,2] = 2/3 \qquad [0;1,3] = 3/4 $$
we get the following equalities:
$$ \varphi^{-1} [1/2, 1] = [[(1) ]] \qquad \varphi^{-1} [1/3 ,1/2] = [[(2) ]] \qquad \varphi^{-1} [2/3, 3/4] = [[(1,2) ]] $$
Hence, from the definition of product measure we have that
$$ \widetilde{\mu} \left( [[(1,2)]] \right)= \widetilde{\mu} \left( [[(1)]] \right) \cdot \widetilde{\mu} \left( [[(2)]] \right)$$
By assertion that $\nu$ is a pushforward measure we should have $ \nu (2/3, 3/4) = \nu (1/2,1) \nu (1/3, 1/2)$. However, with the help of almighty Wolfram we can tell that this is not the case because:
$$\nu (2/3, 3/4) = (\log 2)^{-1} ( \log 21 -\log 20) \qquad \nu (1/2,1) =  (\log 2)^{-1} ( \log 4 -\log 3) \qquad  \nu (1/3, 1/2) = (\log 2)^{-1} ( \log 9 -\log 8)$$
and
$$( \log 4- \log 3)(\log 9 - \log 8) \neq \log 2 ( \log 21 - \log 20) $$
So we have contradiction.


text:
\begin{document}

\title{Hypergeometric formulas for lattice sums and Mahler measures}
\author{Mathew D. Rogers\footnote{The author is supported by NSF award DMS-0803107} \\
        \small{\textit{Department of Mathematics, University of
        Illinois}}\\
        \small{\textit{Urbana, IL 61801, USA}}}

\maketitle

\abstract{We prove a variety of explicit formulas relating special
values of generalized hypergeometric functions to lattice sums with
four indices of summation.  These results are related to Boyd's
conjectured identities between Mahler measures and special values of
$L$-series of elliptic curves.}

\section{Introduction}
\label{proofs} \init

In this paper we will prove a number of formulas relating lattice
sums to
hypergeometric functions. This research was inspired by the work of
Boyd, who used numerical methods to conjecture hundreds of relations
between the $L$-series of elliptic curves and special values of
Mahler's measure \cite{Bo1}. The first example of such an identity
was due to Deninger \cite{De}, who hypothesized that
\begin{equation}\label{deninger's formula}
\m\left(1+y+y^{-1}+z+z^{-1}\right)\stackrel{?}{=}\frac{15}{4\pi^2}L(E,2),
\end{equation}
where $E$ is a conductor $15$ elliptic curve.  As usual,
``$\stackrel{?}{=}$'' denotes a conjectured equality which holds to
at least $50$ decimal places. The Mahler measure of a
$n$-dimensional polynomial is defined in equation \eqref{Mahler
definition}. Boyd observed that since every elliptic curve is
modular, this identity can be translated into a completely explicit
formula:
\begin{equation}\label{deninger's formula after Boyds remark}
\int_{0}^{1}\int_{0}^{1}\log\left|1+2\cos(2\pi t)+2\cos(2\pi
s)\right|\d s\d
t\stackrel{?}{=}\frac{15}{4\pi^2}\sum_{n=1}^{\infty}\frac{a_n}{n^2},
\end{equation}
where
\begin{equation*}
\sum_{n=1}^{\infty}a_n
q^n=q\prod_{n=1}^{\infty}\left(1-q^n\right)\left(1-q^{3n}\right)\left(1-q^{5n}\right)\left(1-q^{15n}\right).
\end{equation*}
If this line of thought is pushed somewhat further (see Theorem
\ref{F(b,c) related to L function}), the following conjecture
arises:
\begin{equation}\label{Deninger intro conjecture latticized}
\begin{split}
\sum_{n=0}^{\infty}&{2n\choose
n}^2\frac{(1/16)^{2n+1}}{2n+1}\\
&\stackrel{?}{=}\frac{540}{\pi^2}\sum_{n_i=-\infty}^{\infty}\frac{(-1)^{n_1+n_2+n_3+n_4}}{\left((6n_1+1)^2+3
(6n_2+1)^2+5(6n_3+1)^2+15(6n_4+1)^2\right)^2}.
\end{split}
\end{equation}
Equation \eqref{Deninger intro conjecture latticized} is an
interesting conjecture, because it relates a complicated lattice sum
to the $_3F_2$ hypergeometric function. Lattice sums have been
extensively studied in physics, since they often arise when
calculating electrostatic potentials of crystal lattices (for
instance see \cite{Gl1}, \cite{Zu} and \cite{FG}).  It is often
difficult to calculate lattice sums numerically, and it is quite
unusual to be able to reduce them to known special functions
\cite{Bor}. The problem of finding a closed form for Madelung's
constant is probably the most famous open problem in this area
\cite{Cr}.
\begin{definition} Let us define $F(a,b,c,d)$ by
\begin{equation*}
\begin{split}
F(a,b,c,d):=&(a+b+c+d)^2\\
&\times\sum_{n_i=-\infty}^{\infty}\frac{(-1)^{n_1+n_2+n_3+n_4}}{\left(
a(6n_1+1)^2+b (6n_2+1)^2+c (6n_3+1)^2+d (6n_4+1)^2\right)^2}.
\end{split}
\end{equation*}
We will also fix the following shorthand notation:
\begin{equation}
F(b,c):=F(1,b,c,b c).
\end{equation}
Finally, the default method of summation will be
$\displaystyle\sum_{n_i=-\infty}^{\infty}=\lim_{v\rightarrow\infty}\sum_{n_1=-v}^{v}\dots\sum_{n_4=-v}^{v}.$
\end{definition}

In this paper, we will prove many new formulas for special values of
$F(a,b,c,d)$.  While we believe that it should be possible to find a
general formula for the function, we have not managed to accomplish
that goal yet.  The second and third sections of the paper summarize
$18$ formulas relating $F(b,c)$ to rational hypergeometric
functions. Virtually all of those identities were extracted directly
from Boyd's tables \cite{Bo1}. For example, Boyd conjectured a
formula for a conductor $20$ elliptic curve, which is equivalent to:
\begin{equation}\label{intro F(1,5) example}
\frac{25}{6\pi^2}F(1,5)\stackrel{?}{=}\sqrt[3]{2} A {_3F_2}\left(\substack{\frac{1}{3},\frac{1}{3},\frac{1}{3}\\
\frac{2}{3},\frac{4}{3}}; \frac{2}{27}\right)
+\sqrt[3]{4} B {_3F_2}\left(\substack{\frac{2}{3},\frac{2}{3},\frac{2}{3}\\
\frac{4}{3},\frac{5}{3}}; \frac{2}{27}\right),
\end{equation}
where $A$ and $B$ are given in terms of gamma functions (see Theorem
\ref{theorem: hypergeometric reduction}).  When more general lattice
sums are considered, hypergeometric functions with irrational
arguments frequently appear. For instance, if
$\phi=\frac{1+\sqrt{5}}{2}$, we have
\begin{equation}\label{intro lattic example}
\frac{225}{32\sqrt{5}\pi^2}F\left(1,5,5,5\right)=\frac{A}{\sqrt[3]{\phi}}{_3F_2}\left(\substack{\frac{1}{3},\frac{1}{3},\frac{1}{3}\\
\frac{2}{3},\frac{4}{3}};\frac{1}{\phi}\right)
+\frac{3B}{\sqrt[3]{\phi^2}}{_3F_2}\left(\substack{\frac{2}{3},\frac{2}{3},\frac{2}{3}\\
\frac{4}{3},\frac{5}{3}};\frac{1}{\phi}\right).
\end{equation}
Formula \eqref{intro lattic example} closely resembles identities
that Forrester and Glasser established for three-dimensional sums
associated with NaCl lattices \cite{FG}.  While it seems likely that
equations \eqref{intro lattic example} and \eqref{intro F(1,5)
example} both arise as special cases of formulas for $F(a,b,c,d)$,
the exact nature of those formulas remains unclear. From our
calculations, we have determined that \eqref{intro lattic example}
is not equivalent to any of Boyd's identities, and it also does not
appear to reduce to a Mahler measure formula.

Based on the computations in this paper, it is probably safe to
conjecture that \textit{many} values of $F(b,c)$ reduce to
generalized hypergeometric functions and Meijer $G$-functions. For
instance, by \eqref{F(1,4) final formula}, we have
\begin{equation*}
\frac{144}{25\pi^2}F(1,4)=m\left(\frac{4}{\theta}\right)+\frac{1}{4\pi^2}\Im\left(G_{3,3}^{3,2}\left(\theta^2\big|\substack{\frac{1}{2},\frac{1}{2},1\\0,0,0}\right)\right),
\end{equation*}
where $m(k)$ is defined in \eqref{m(k) 4F3}, and $\theta\approx
1.93+1.79 i$.  We will briefly describe why $G$-functions appear in
this context.

It is often possible to show that lattice sums satisfy
hypergeometric differential equations with respect to modular
parameters.  If the standard solution to the equation (typically a
$_4F_3$ function), is analytic over the same domain as the lattice
sum, then they can be equated (see \eqref{F(1,2)(x) as a Mahler
measure}).  On the other hand, if the two functions have different
domains of analyticity, then the lattice sum will equate to a
piecewise-defined function, which incorporates a second solution of
the differential equation (sometimes a Meijer G-function). One of
our main results is such a formula for $F(1,1,1,x)$, which holds
whenever $x>0$ (see \eqref{F(1,1) hypergeometric form}). We will
also present explicit formulas for $F(1,4)$, $F(2,2)$, $F(1,1,2,4)$,
and $F(1,2,4,4)$. Additionally, we will recover formulas for
$F(1,1)$, $F(1,2)$, and $F(1,3)$, which were first proved in
\cite{RV}.

Therefore, we will briefly outline the approach contained in
Sections \ref{section:hypergeometric reductions}, \ref{section:
F(1,1,2,4) and F(1,2,4,4)}, and \ref{section: F(1,3) section}.
Identities such as those presented in Sections \ref{section : Boyds
conjectures} and \ref{section : summary of hypergeometric formulas}
require the reduction of four-dimensional sums to one-dimensional
sums.  If possible, the first step is a reduction to a
two-dimensional sum. For instance, the left-hand side of
\eqref{intro lattic example} becomes
\begin{equation}\label{intro conductor 32 exam cont}
\begin{split}
F(1,5,5,5)
=16^2\sum_{\substack{n=-\infty\\
k=0}}^{\infty}\frac{(-1)^{n+k}(2k+1)}{\left((6n+1)^2+15(2k+1)^2\right)^2}.
\end{split}
\end{equation}
These sorts of transformations follow from well-known $q$-series
results, and are quite rare. Equation \eqref{intro conductor 32 exam
cont} is a consequence of the following corollary to the Jacobi
triple product:
\begin{equation}\label{intro cm q expansion}
q^2\prod_{n=1}^{\infty}\left(1-q^{3n}\right)\left(1-q^{15n}\right)^3=\sum_{\substack{n=-\infty\\
k=0}}^{\infty}(-1)^{n+k}(2k+1)q^{\frac{15(2k+1)^2+(6n+1)^2}{8}}.
\end{equation}
Notice that equation \eqref{intro cm q expansion} gives an example
of a \textit{lacunary} modular form \cite{Fi2}. The two-dimensional
lattice sums are then evaluated using Ramanujan's theories of
elliptic functions and modular equations. Several of these
calculations are quite involved.

While the main goal of this research was to find formulas for
lattice sums, we have also proved several new relations between
modular forms and Mahler measures. Section \ref{section: more
mahlers} summarizes identities between Mahler measures and Mellin
transforms of non-multiplicative modular forms. Additionally, in
Section \ref{section: conclusion}, we have reformulated a conjecture
concerning a higher Mahler measure.

\section{Summary of Boyd's conjectures for $F(b,c)$}\label{section : Boyds conjectures}\init In this section we
will summarize a variety of explicit formulas relating
four-dimensional lattice sums to Mahler measures of polynomials.
Most of these results are only conjectures, although numerical
calculations can be used to verify them to any degree of accuracy.
Our first step will be to invoke the modularity theorem to find
explicit formulas for $L$-functions of elliptic curves with
conductors $N\in\{11, 14, 15, 20, 24, 27, 32, 36\}$.  The following
theorem is an easy consequence of a paper due to Martin and Ono
\cite{Ono}:

\begin{theorem} Suppose that $E_N$ is an elliptic curve of conductor
$N$, then
\begin{equation}\label{F(b,c) related to L function}
L(E_N,2)=F(b,c)
\end{equation}
for the following values of $N$ and $(b,c)$:
\begin{equation*}

    \begin{tabular}{|c|c|p{6 in}|}
        \hline
        $N$ & $(b,c)$\\
        \hline
        $11$ & $(1,11)$\\
        $14$ & $(2,7)$\\
        $15$ & $(3,5)$\\
        $20$ & $(1,5)$\\
        $24$ & $(2,3)$\\
        $27$ & $(1,3)$\\
        $32$ & $(1,2)$\\
        $36$ & $(1,1)$\\
        \hline
    \end{tabular}

\end{equation*}
\end{theorem}
\begin{proof} We are interested in cases where cusp forms of elliptic
curves equal the product of four eta functions.  Such equalities are
consequences of the modularity theorem. An exhaustive list of all
such cusp forms is provided in \cite{Ono}. By inspection of that
list, the eta product associated with $E_N$ will have the form
\begin{equation*}
g(q):=q\prod_{n=1}^{\infty}\left(1-q^{A n}\right)\left(1-q^{A b
n}\right)\left(1-q^{A c n}\right)\left(1-q^{A b c n}\right),
\end{equation*}
where $(1+b)(1+c)A=24$.  Recalling Euler's pentagonal number theorem
\begin{equation*}
\prod_{n=1}^{\infty}\left(1-q^n\right)=\sum_{n=-\infty}^{\infty}(-1)^n
q^{n (3n+1)/2},
\end{equation*}
this becomes
\begin{equation*}
g(q)=\sum_{n_i=-\infty}^{\infty}(-1)^{n_1+n_2+n_3+n_4}q^{\frac{A
(6n_1+1)^2+A b (6n_2+1)^2+A c (6n_3+1)^2+A b c (6n_4+1)^2}{24}},
\end{equation*}
and it follows immediately that
\begin{equation*}
L(E_N,2)=\frac{24^2}{A^2}\sum_{n_i=-\infty}^{\infty}\frac{(-1)^{n_1+n_2+n_3+n_4}}{\left(
(6n_1+1)^2+b (6n_2+1)^2+c (6n_3+1)^2+b c (6n_4+1)^2\right)^2}.
\end{equation*}
Since $(1+b)(1+c)=24/A$, the theorem follows. $\blacksquare$
\end{proof}

Since we now have expressed several different $L$-values in terms of
$F(b,c)$, it seems logical to list all of the known Mahler measures
which reduce to values of that function.
\begin{definition}  The Mahler measure of
an $n$-dimensional Laurent polynomial, $P(z_1,\dots, z_n)$, is
defined by
\begin{equation}\label{Mahler definition}
m(P):=\int_{0}^{1}\dots\int_{0}^{1}\log\left|P\left(e^{2\pi i
t_1},\dots, e^{2\pi i t_n}\right)\right| \d t_1\dots \d t_n.
\end{equation}
Furthermore, we will use the following notation:
\begin{align}
m(k):=&\m\left(k+y+y^{-1}+z+z^{-1}\right),\label{m(k) def}\\
n(k):=&\m\left(y^3+z^3+1-k y z\right),\label{n(k) def}\\
g(k):=&\m\left((1+y)(1+z)(y+z)-k y z\right),\label{g(k) def}\\
r(k):=&\m\left((1+y)(1+z)(1+y+z)-k y z\right).\label{r(k) def}
\end{align}
\end{definition}
For convenience we have slightly altered the definitions of $n(k)$,
$g(k)$ and $r(k)$ that appeared in \cite{LR}.  All of the following
examples were either extracted from Boyd's paper \cite{Bo1}, or were
deduced by combining Boyd's conjectures with functional equations in
\cite{LR}. While Boyd's minimal Weierstrass models often do not
coincide with the minimal Weierstrass models in \cite{Ono}, the
elliptic curves are isogenous, and the following results are all
numerically true: \allowdisplaybreaks{
\begin{align}
n(3\sqrt[3]{2})=&\frac{27}{2\pi^2}F(1,1)\\
g(2)=&\frac{9}{2\pi^2}F(1,1)\\%=\frac{1}{3}n(2\times3^{1/3})
g(-4)=&\frac{18}{\pi^2}F(1,1)\\%=\frac{2}{3}n(2\times3^{1/3})
m(4 i)=&\frac{16}{\pi^2}F(1,2)\label{m(4i) formula}\\
m(2\sqrt{2})=&\frac{8}{\pi^2}F(1,2)\\
n(-6)=&\frac{81}{4\pi^2}F(1,3)\\
n(\sqrt[3]{2})\stackrel{?}{=}&\frac{25}{6\pi^2}F(1,5)\\
n(\sqrt[3]{32})\stackrel{?}{=}&\frac{40}{3\pi^2}F(1,5)\\
g(-2)\stackrel{?}{=}&\frac{15}{\pi^2}F(1,5)\\%=\frac{1}{3}n(2^{1/3})+\frac{4}{3}n(2^{5/3})
g(4)\stackrel{?}{=}&\frac{10}{\pi^2}F(1,5)\\%=\frac{4}{3}n(2^{1/3})+\frac{1}{3}n(2^{5/3})
r(-1)=&\frac{77}{4\pi^2}F(1,11)\\
m(2)\stackrel{?}{=}&\frac{6}{\pi^2}F(2,3)\\
m(8)\stackrel{?}{=}&\frac{24}{\pi^2}F(2,3)\\
m(3\sqrt{2})\stackrel{?}{=}&\frac{15}{\pi^2}F(2,3)\\
m( i\sqrt{2})\stackrel{?}{=}&\frac{9}{\pi^2}F(2,3)\\%=m(8)-m(3\sqrt{2})
n(-1)\stackrel{?}{=}&\frac{7}{\pi^2}F(2,7)\label{n(-1), F(2,7) conjecture}\\
n(5)\stackrel{?}{=}&\frac{49}{2\pi^2}F(2,7)\\
g(1)\stackrel{?}{=}&\frac{7}{2\pi^2}F(2,7)\\%=\frac{4}{3}n(-1)+\frac{1}{3}n(5)
g(7)\stackrel{?}{=}&\frac{21}{\pi^2}F(2,7)\\
g(-8)\stackrel{?}{=}&\frac{35}{\pi^2}F(2,7)\\%=\frac{1}{3}n(-1)+\frac{4}{3}n(5)
m(1)\stackrel{?}{=}&\frac{15}{4\pi^2}F(3,5)\label{m(1), F(3,5) conjecture}\\
m(3 i)\stackrel{?}{=}&\frac{75}{4\pi^2}F(3,5)\\
m(5)\stackrel{?}{=}&\frac{45}{2\pi^2}F(3,5)\\
m(16)\stackrel{?}{=}&\frac{165}{4\pi^2}F(3,5)
\end{align}
All of the results involving $F(1,1)$, $F(1,2)$, and $F(1,3)$ can be
deduced from Rodriguez-Villegas's paper \cite{RV}. In particular,
those Mahler measures can be written in terms of two-dimensional
Eisenstein-Kronecker series, and then the results follow from
Deuring's theorem.

\section{Summary of rational hypergeometric formulas for
$F(b,c)$}\label{section : summary of hypergeometric formulas}\init

In this section we will translate almost all of the
known Mahler measures for $F(b,c)$ into hypergeometric functions. In
Corollary \ref{final reduction cor} we will also prove that similar
expressions exist for both $F(2,2)$ and $F(1,4)$, even though those
sums are apparently unrelated to the theory of elliptic curves.

\begin{theorem}\label{theorem: hypergeometric reduction} We can express $m(k)$, $n(k)$, and $g(k)$ in terms
of generalized hypergeometric functions for most values of $k$:
\begin{align}
m(k)=&\Re\left(\log(k)-\frac{2}{k^2}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; \frac{16}{k^2}\right)\right)\label{m(k) 4F3},\\
n(k)=&\Re\left(\log(k)-\frac{2}{k^3}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2};\frac{27}{k^3}\right)\right),\label{n(k) 4F3}\\
g(k)=&\Re\left(\log\left(\frac{(4+k)(k-2)^4}{k^2}\right)-\frac{2k^2}{(4+k)^3}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2};\frac{27k^2}{(4+k)^3}\right)\right.\label{g(k) 4F3}\\
&\qquad\left.-\frac{8k}{(k-2)^3}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2};\frac{27k}{(k-2)^3}\right)\right)\notag.
\end{align}
Equation \eqref{m(k) 4F3} is valid in $\mathbb{C}\setminus\{0\}$,
while \eqref{g(k) 4F3} holds in $\mathbb{C}\setminus[-4,2]$, and
\eqref{n(k) 4F3} is true for $|k|$ is sufficiently large.

In certain cases we can reduce these hypergeometric functions
further. Suppose that $k\in\mathbb{R}\setminus\{0\}$, then
\begin{equation}\label{m(k) 4F3 to 3F2}
\Re\left(\log(k)-\frac{2}{k^2}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; \frac{16}{k^2}\right)\right)=\Re\left(\frac{|k|}{4}{_3F_2}\left(\substack{\frac{1}{2},\frac{1}{2},\frac{1}{2}\\
1,\frac{3}{2}}; \frac{k^2}{16}\right)\right),
\end{equation}
and
\begin{equation}\label{n(k) 4F3 to 3F2}
\begin{split}
Re\left(\log(k)-\frac{2}{k^3}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2};\frac{27}{k^3}\right)\right)=&s(k)\Re\left(A k {_3F_2}\left(\substack{\frac{1}{3},\frac{1}{3},\frac{1}{3}\\
\frac{2}{3},\frac{4}{3}}; \frac{k^3}{27}\right)\right.\\
&\qquad\qquad\left.+B k^2 {_3F_2}\left(\substack{\frac{2}{3},\frac{2}{3},\frac{2}{3}\\
\frac{4}{3},\frac{5}{3}}; \frac{k^3}{27}\right)\right),
\end{split}
\end{equation}
where
$A=\frac{\sqrt[3]{2}\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{2}\right)}{8\sqrt{3}\pi^2}$,
$B=\frac{\Gamma^3\left(\frac{2}{3}\right)}{16\pi^2}$, and
$s(k)=\frac{1+3\sgn(k)}{4}$.
\end{theorem}
\begin{proof} Equations \eqref{m(k) 4F3} and \eqref{n(k) 4F3} are
due to Rodriguez-Villegas \cite{RV}, while \eqref{g(k) 4F3} was
proved in \cite{LR}.  Kurokawa and Ochiai have examined a version of
\eqref{m(k) 4F3 to 3F2} in \cite{KO}, although it can also be proved
by integrating the following identity:
\begin{equation*}
\Re\left({_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};
u\right)\right)=\Re\left(\frac{1}{\sqrt{u}}{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1}; \frac{1}{u}\right)\right),
\end{equation*}
which holds for $u\in \left(0,\infty\right)$.  A similar argument
can be used to establish \eqref{n(k) 4F3 to 3F2}. For example, when
$u\in \left(0,\infty\right)$ we can integrate the following
transformation:
\begin{equation*}
\Re\left({_2F_1}\left(\substack{\frac{1}{3},\frac{2}{3}\\ 1};
\frac{1}{u}\right)\right)=\Re\left(\frac{9\Gamma^3\left(2/3\right)}{4\pi^2}
u^{2/3}{_2F_1}\left(\substack{\frac{2}{3},\frac{2}{3}\\
\frac{4}{3}}; u \right)
 +\frac{1}{2} u^{1/3}{_2F_1}\left(\substack{\frac{1}{3},\frac{1}{3}\\
 1};1-u\right)\right),
\end{equation*}
with respect to $u$. $\blacksquare$
\end{proof}

Equations \eqref{m(k) 4F3 to 3F2} and \eqref{n(k) 4F3 to 3F2} will
often allow us to obtain convergent series expansions from divergent
hypergeometric formulas.  For example, applying the results of the
last theorem to conjecture \eqref{m(1), F(3,5) conjecture}, we
obtain formula \eqref{Deninger intro conjecture latticized}. It is
hardly coincidental that \eqref{Deninger intro conjecture
latticized} bears a striking resemblance to a famous formula that
Ramanujan discovered for Catalan's constant \cite{Ad}:
\begin{equation*}
L\left(\chi_{-4},2\right)=\pi\sum_{n=0}^{\infty}{2n\choose
n}^2\frac{\left(1/4\right)^{2n+1}}{2n+1}.
\end{equation*}
Ramanujan's formula follows easily from Boyd's evaluation of the
degenerate Mahler measure $m(4)$.

 The following list summarizes the identities that can be obtained
by reducing the Mahler measures in the previous section to
hypergeometric functions.  Whenever possible, we have used equations
\eqref{m(k) 4F3 to 3F2} and \eqref{n(k) 4F3 to 3F2} to obtain
hypergeometric functions with convergent arguments. Since no such
expression is known for $r(-1)$, we have simply retained that Mahler
measure in our list.  Finally, because several Mahler measures such
as $g(2)$ and $n\left(3\sqrt[3]{2}\right)$ are equivalent, this list
contains fewer entries than we might expect.  Once again, define
\begin{align*}
A:=&\frac{\sqrt[3]{2}\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{2}\right)}{8\sqrt{3}\pi^2},
&B&:=\frac{\Gamma^3\left(\frac{2}{3}\right)}{16\pi^2},
\end{align*}
then the following results are numerically true:
\begin{align}
\frac{9}{2\pi^2}F(1,1)=&\frac{1}{9}\log(54)-\frac{1}{81}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2}; \frac{1}{2}\right),\label{hypergeo F(1,1)}\\
\frac{16}{\pi^2}F(1,2)=&2\log(2)+\frac{1}{8}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; -\frac{1}{4}\right),\\
\frac{8}{\pi^2}F(1,2)=&\frac{1}{\sqrt{2}}{_3F_2}\left(\substack{\frac{1}{2},\frac{1}{2},\frac{1}{2}\\
1,\frac{3}{2}}; \frac{1}{2}\right),\\
\frac{81}{4\pi^2}F(1,3)=&\log(6)+\frac{1}{108}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2}; -\frac{1}{8}\right),\label{hypergeo F(1,3)}\\
\frac{25}{6\pi^2}F(1,5)\stackrel{?}{=}&\sqrt[3]{2} A {_3F_2}\left(\substack{\frac{1}{3},\frac{1}{3},\frac{1}{3}\\
\frac{2}{3},\frac{4}{3}}; \frac{2}{27}\right)
+\sqrt[3]{4} B {_3F_2}\left(\substack{\frac{2}{3},\frac{2}{3},\frac{2}{3}\\
\frac{4}{3},\frac{5}{3}}; \frac{2}{27}\right),\\
\frac{40}{3\pi^2}F(1,5)\stackrel{?}{=}&\frac{5}{3}\log(2)-\frac{1}{16}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2}; \frac{27}{32}\right),\\
\frac{77}{4\pi^2}F(1,11)=&r(-1),\label{hypergeo F(1,11)}\\
\frac{6}{\pi^2}F(2,3)\stackrel{?}{=}&\frac{1}{2}{_3F_2}\left(\substack{\frac{1}{2},\frac{1}{2},\frac{1}{2}\\
1,\frac{3}{2}}; \frac{1}{4}\right),\label{hypergeo F(2,3) first 0}\\
\frac{24}{\pi^2}F(2,3)\stackrel{?}{=}&3\log(2)-\frac{1}{32}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; \frac{1}{4}\right),\label{hypergeo F(2,3) first}\\
\frac{15}{\pi^2}F(2,3)\stackrel{?}{=}&\frac{1}{2}\log(18)-\frac{1}{9}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; \frac{8}{9}\right),\\
\frac{9}{\pi^2}F(2,3)\stackrel{?}{=}&\frac{1}{2}\log(2)+{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\{2,2,2}},-8\right),\label{hypergeo F(2,3) last}\\
\frac{7}{\pi^2}F(2,7)\stackrel{?}{=}&\frac{A}{2} {_3F_2}\left(\substack{\frac{1}{3},\frac{1}{3},\frac{1}{3}\\
\frac{2}{3},\frac{4}{3}}; -\frac{1}{27}\right)
-\frac{B}{2} {_3F_2}\left(\substack{\frac{2}{3},\frac{2}{3},\frac{2}{3}\\
\frac{4}{3},\frac{5}{3}}; -\frac{1}{27}\right),\\
\frac{49}{2\pi^2}F(2,7)\stackrel{?}{=}&\log(5)-\frac{2}{125}{_4F_3}\left(\substack{\frac{4}{3},\frac{5}{3},1,1\\
2,2,2}; \frac{27}{125}\right),\\
\frac{21}{\pi^2}F(2,7)\stackrel{?}{=}&g(7),\\
\frac{15}{4\pi^2}F(3,5)\stackrel{?}{=}&\frac{1}{4}{_3F_2}\left(\substack{\frac{1}{2},\frac{1}{2},\frac{1}{2}\\
1,\frac{3}{2}}; \frac{1}{16}\right),\label{f(3,5) first}\\
\frac{45}{2\pi^2}F(3,5)\stackrel{?}{=}&\log(5)-\frac{2}{25}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; \frac{16}{25}\right),\\
\frac{165}{4\pi^2}F(3,5)\stackrel{?}{=}&4\log(2)-\frac{1}{128}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; \frac{1}{16}\right),\\
\frac{75}{4\pi^2}F(3,5)\stackrel{?}{=}&\log(3)+\frac{2}{9}{_4F_3}\left(\substack{\frac{3}{2},\frac{3}{2},1,1\\
2,2,2}; -\frac{16}{9}\right).\label{hypergeo very last}
\end{align}
While most of these formulas remain unproven, a variety of partial
results exist.  For instance, identities \eqref{hypergeo F(2,3)
first 0}} through \eqref{hypergeo F(2,3) last} are equivalent to one
another \cite{LR}, formulas \eqref{f(3,5) first} through
\eqref{hypergeo very last} are equivalent to one another \cite{Ml2},
and formulas \eqref{hypergeo F(1,1)} through \eqref{hypergeo F(1,3)}
follow from \cite{RV}.  Additionally, Mellit and Brunault have given
$K$-theoretic proofs of the formulas for $F(2,7)$ and $F(1,11)$ (see
\cite{Me} and \cite{Br}).

\section{Proofs of formulas for $F(1,1)$, $F(1,2)$, $F(1,4)$, $F(2,2)$, and $F(1,1,1,x)$}\label{section:hypergeometric reductions}\init

In the previous section we translated many of Boyd's conjectures
into explicit identities between hypergeometric functions and
lattice sums.  This approach has two essential consequences. Not
only does it eliminate any obvious connection with elliptic curves,
but it also allows for the construction of proofs based upon series
manipulation.  In this section we will discuss the cases that occur
when $(b,c)\in\{(1,1),(1,2),(1,4),(2,2)\}$.  We will rely heavily on
the $q$-series theorems contained in Ramanujan's notebooks (see
\cite{Be3} and \cite{Be5}).

\begin{definition} Let us recall Ramanujan's $q$-series notation:
\begin{align*}
\varphi(q):=&\sum_{n=-\infty}^{\infty}q^{n^2},&
\psi(q):=&\sum_{n=0}^{\infty}q^{\frac{n(n+1)}{2}},\\
 f(-q):=&\prod_{n=1}^{\infty}(1-q^n),&
(x;q)_{\infty}:=&\prod_{n=0}^{\infty}\left(1-x q^n\right).
\end{align*}
When convenient, we will also employ the following notation:
\begin{equation*}
e_j:=q^{j/24}\prod_{n=1}^{\infty}\left(1-q^{j n}\right).
\end{equation*}
\end{definition}

Proposition \ref{lemma: cm reduction} reduces the aforementioned
cases of $F(b,c)$ to two-dimensional sums.  Such identities exist
because various eta-quotients can be written in terms theta
functions. Euler's pentagonal number formula is probably the
simplest such identity:
\begin{equation*}
e_j=\sum_{n=-\infty}^{\infty}(-1)^n q^{\frac{j(6n+1)^2}{24}}.
\end{equation*}
Unfortunately, similar formulas are not known (and probably do not
exist) for $e_1^2$, $e_1e_2$, or $e_1e_3$ \cite{Fi}. This fact
represents the main obstruction to proving Boyd's conjectures for
$F(1,5)$, $F(2,3)$, $F(2,7)$, and $F(3,5)$.

\begin{definition} We will use the following notation:
\begin{align}
F_{(1,2)}(x):=&\sum_{\substack{n=-\infty\\ k=0}}^{\infty}\frac{(-1)^{n+k}(2k+1)}{\left((2k+1)^2+x^2(2n)^2\right)^2},\label{F(1,2) 2D sum}\\

F_{(1,4)}(x):=&25\sum_{n,k=-\infty}^{\infty}\frac{(-1)^n
(3k+1)}{\left(4(3k+1)^2+x^2(6n+1)^2\right)^2},\label{F(1,4) 2D sum}\\
F_{(2,2)}(x):=&9\sum_{n,k=0}^{\infty}\frac{(-1)^{\frac{n(n+1)}{2}+k}(2k+1)}{\left(2(2k+1)^2+x^2(2n+1)^2\right)^2}.\label{F(2,2)
2D sum}
\end{align}
\end{definition}

In the next proposition we will show that each of these functions
equals a value of $F(b,c)$ when $x\rightarrow 1$. Furthermore, we
will also demonstrate that a two-dimensional series exists for
$F(1,1,1,x)$.

\begin{proposition}\label{lemma: cm reduction} Suppose that
$(b,c)\in\{(1,2),(1,4),(2,2)\}$, then
\begin{equation}\label{4d to 2d sum reduction}
F_{(b,c)}(1)=F(b,c).
\end{equation}
Furthermore, the following series expansion is true for any $x>0$:
\begin{equation}
\frac{F\left(1,1,1,x\right)}{\left(3+x\right)^2}=\sum_{\substack{n=-\infty\\
k=0}}^{\infty}\frac{(-1)^{n+k}(2k+1)}{\left(3(2k+1)^2+x
(6n+1)^2\right)^2}.\label{F(1,1) 2D sum}
\end{equation}
\end{proposition}
\begin{proof}
First notice that $F(a,b,c,d)$ has the following integral
representation for all positive values of $a$, $b$, $c$, and $d$:
\begin{equation*}
\frac{24^2 F(a,b,c,d)}{(a+b+c+d)^2} =\int_{0}^{1}\int_{0}^{q_1}e_a
e_b e_c e_d\frac{\d q}{q}\frac{\d q_1}{q_1}.
\end{equation*}
Taking note of the following identities:
\begin{align*}
e_x e_1^3=&(e_x)e_1^3,\\
e_1^2 e_2^2=&\left(\frac{e_1^2}{e_2}\right)e_2^3,\\
e_1^2 e_4^2=&\left(\frac{e_1^2e_4^2}{e_2}\right)e_2,\\
e_1 e_2^2 e_4=&\left(\frac{e_1e_4}{e_2}\right)e_2^3,
\end{align*}
and then employing well known series expansions:
\begin{align}
\frac{e_1^2}{e_2}=&\sum_{n=-\infty}^{\infty}(-1)^n q^{n^2},\\
\frac{e_1e_4}{e_2}=&\sum_{n=0}^{\infty}(-1)^{\frac{n(n+1)}{2}}q^{\frac{(2n+1)^2}{8}},\label{weight 1/2 series}\\
e_j=&\sum_{n=-\infty}^{\infty}(-1)^n q^{\frac{j(6n+1)^2}{24}},\\
e_j^3=&\sum_{n=0}^{\infty}(-1)^n (2n+1)q^{\frac{j(2n+1)^2}{8}},\\
\frac{e_1^2e_4^2}{e_2}=&\sum_{n=-\infty}^{\infty}(3n+1)q^{\frac{(3n+1)^2}{3}}\label{weight
3/2 series},
\end{align}
we recover equations \eqref{F(1,1) 2D sum} and \eqref{4d to 2d sum
reduction} in every case. $\blacksquare$
\end{proof}

We will use the next proposition to reduce each of the
two-dimensional sums to a $q$-series.  Then, in Theorem
\ref{hypergeometric theorem}, we will reduce each $q$-series to an
integral of hypergeometric functions. In certain special cases those
integrals translate into identities involving Mahler measures.

\begin{proposition}\label{q-series proposition} Let $\chi_{-3}(k)$ and $\chi_{-4}(k)$ denote Legendre
symbols modulo three and four, and assume that $x>0$.

If $q=e^{-\pi x/\sqrt{12}}$ and $\omega=e^{\pi i/6}$, then
\begin{align}\label{F(1,1)(x) q series sum}
\frac{F\left(1,1,1,x^2\right)}{\left(3+x^2\right)^2}&=\frac{\pi^2}{72x}\sum_{k=1}^{\infty}k\chi_{-4}(k)
\log\left|\frac{1+\omega q^{k}}{1-\omega q^{k}}\right|.
\end{align}
If $q=e^{-\pi x}$, then
\begin{equation}\label{F(1,2)(x) q series sum}
F_{(1,2)}(x)=-\frac{\pi^2}{32x}\left(\log(q)+4\sum_{k=1}^{\infty}k\chi_{-4}(k)\log\left(1+q^{k}\right)\right).
\end{equation}
If $q=e^{-\pi x/3}$ and $\omega=e^{\pi  i/6}$, then
\begin{align}\label{F(1,4)(x) q-series sum}
F_{(1,4)}(x)&=\frac{25\pi^2}{72x}\sum_{k=1}^{\infty}k\chi_{-3}(k)\log\left|\frac{1+\omega
q^k}{1-\omega q^k}\right|.
\end{align}
If $q=e^{-\pi x/\sqrt{8}}$ and $\omega=e^{\pi i/4}$, then
\begin{align}\label{F(2,2)(x) q-series sum}
F_{(2,2)}(x)&=\frac{9\pi^2}{32x}\sum_{k=1}^{\infty}
k\chi_{-4}(k)\log\left|\frac{1+\omega q^{k}}{1-\omega q^{k}}\right|.
\end{align}
\end{proposition}
\begin{proof} All of the proofs are very similar, so we
will only prove \eqref{F(1,1)(x) q series sum} in detail. First
notice that
\begin{align*}
\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}=&\frac{1}{144}\sum_{\substack{n=-\infty\\
k=0}}^{\infty}\frac{(-1)^{n+k}(2k+1)}{\left((k+1/2)^2+\frac{x^2}{12}
(6n+1)^2\right)^2},\\
=&\frac{\pi^2}{144}\int_{0}^{\infty}u\left(\sum_{
k=0}^{\infty}(-1)^{k}(2k+1)e^{-\pi(k+1/2)^2
u}\right)\\
&\qquad\qquad\times\left(\sum_{n=-\infty}^{\infty}(-1)^n e^{-\pi
(6n+1)^2 x^2 u/12}\right)\d u.
\end{align*}
Next, by the involution for the weight $3/2$ theta function:
\begin{equation*}
\sum_{k=0}^{\infty}(-1)^k (2k+1)e^{-\pi(k+1/2)^2
u}=\frac{1}{u^{3/2}}\sum_{k=0}^{\infty}(-1)^k (2k+1)e^{-\pi
(k+1/2)^2\frac{1}{u}},
\end{equation*}
this becomes
\begin{align*}
\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}=&\frac{\pi^2}{144}\sum_{\substack{n=-\infty\\
k=0}}^{\infty}(-1)^{n+k}(2k+1)\int_{0}^{\infty}u^{-1/2}e^{-\frac{\pi(k+1/2)^2}{
u}- \frac{\pi(6n+1)^2 x^2}{12}u}\d u.
\end{align*}
The following integral holds whenever $A,B\in\mathbb{R}$ and
$|A|\not=0$:
\begin{equation*}
\int_{0}^{\infty}u^{-1/2}e^{-\pi \left(A^2 u+\frac{B^2}{u}\right)}\d
u=\frac{e^{-2\pi|AB|}}{|A|},
\end{equation*}
and therefore
\begin{align*}
\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}=&\frac{\pi^2}{24\sqrt{3}x}\sum_{\substack{n=-\infty\\
k=0}}^{\infty}(-1)^{n+k}\frac{(2k+1)}{|6n+1|}e^{-\pi(2k+1)|6n+1|x/\sqrt{12}}\\
=&\frac{\pi^2}{24\sqrt{3}x}\sum_{k=1}^{\infty}k\chi_{-4}(k)\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}}{|6n+1|}q^{k|6n+1|}\\
=&\frac{\pi^2}{72x}\sum_{k=1}^{\infty}k\chi_{-4}(k)\log\left|\frac{1+\omega
q^{k}}{1-\omega q^{k}}\right|,
\end{align*}
where $\omega=e^{\pi i/6}$ and $q=e^{-\pi x/\sqrt{12}}$.
$\blacksquare$
\end{proof}

At this point, a hypergeometric formula for $F_{(1,2)}(x)$ can be
recovered.  By combining equation \eqref{F(1,2)(x) q series sum}
with formulas (2-9) and (2-16) in \cite{LR}, it is possible to show
that if $q=e^{-\pi x}$, then
\begin{equation}\label{F(1,2)(x) as a Mahler measure}
F_{(1,2)}(x)=\frac{\pi^2}{16x}m\left(\frac{ i
f^4(-q)}{\sqrt{q}f^4\left(-q^4\right)}\right).
\end{equation}
In the next section we will use values of class invariants to deduce
explicit examples from \eqref{F(1,2)(x) as a Mahler measure}.
Unfortunately, we will require another theorem to obtain useful
results on the other lattice sums.

\begin{theorem}\label{q-series integrals theorem} In this theorem we will
always assume that $x>0$.  If $q=e^{-\pi x/\sqrt{12}}$, then
\begin{align}\label{F(1,1)(x) q-series integral}
\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}&=\frac{\pi^2}{24\sqrt{3}x}\Im\left(\int_{0}^{
i q}\frac{f^9\left(-u^3\right)}{f^3\left(-u\right)}\d u\right).
\end{align}
If $q=e^{-\pi x}$, then
\begin{equation}\label{F(1,2)(x) q-series integral}
F_{(1,2)}(x)=\frac{\pi^3}{32}-\frac{\pi^2}{16x}\int_{0}^{q}\frac{\varphi^2(-u)\varphi^4(u)-1}{u}\d
u.
\end{equation}
If $q=e^{-\pi x/3}$ and $\rho=e^{2\pi i/3}$, then
\begin{align}\label{F(1,4)(x) q-series integral}
F_{(1,4)}(x)=\frac{25\pi^2}{36 x}\Im\left(\int_{0}^{\rho
q}\varphi^2(u)\psi^4\left(u^2\right)\d u\right).
\end{align}
If $q=e^{-\pi x/\sqrt{8}}$, then
\begin{align}\label{F(2,2)(x) q-series integral}

F_{(2,2)}(x)&=\frac{9\pi^2}{32\sqrt{2} x}\int_{0}^{q}
\varphi(-u^2)\varphi(u^4)\left(3\psi^4(-u^2)-\psi^4(u^2)\right)\d u.
\end{align}
\end{theorem}
\begin{proof}  Equations \eqref{F(1,2)(x) q-series integral} and \eqref{F(1,4)(x)
q-series integral} have similar proofs, so we will only prove the
latter identity. Notice that \eqref{F(1,4)(x) q-series sum} can be
rearranged to obtain
\begin{align*}
\frac{72x}{25\pi^2}F_{(1,4)}(x)=&\Re\left(\int_{0}^{q}
\sum_{k=1}^{\infty}k^2\chi_{-3}(k)\frac{2\omega u^k}{1-\omega^2
u^{2k}}\frac{\d u}{u}\right)\\
=&\sqrt{3}\int_{0}^{q}\sum_{k=1}^{\infty}k^2\chi_{-3}(k)\left(\frac{u^k-u^{5k}}{1+u^{6k}}\right)\frac{\d
u}{u}\\
=&\Im\left(2\int_{0}^{\rho
q}\sum_{k=1}^{\infty}k^2\left(\frac{u^k-u^{3k}+u^{5k}}{1+u^{6k}}\right)\frac{\d
u}{u}\right)\\
=&\Im\left(2\int_{0}^{\rho q}\sum_{k=1}^{\infty}\frac{k^2
u^k}{1+u^{2k}}\frac{\d u}{u}\right),
\end{align*}
where $\rho=e^{2\pi i/3}$. Combining entries 10.1, 11.3, and 17.2 in
Chapter 17 of \cite{Be3}, we deduce that for $|u|<1$:
\begin{equation*}
\sum_{k=1}^{\infty}\frac{k^2
u^k}{1+u^{2k}}=u\varphi^2(u)\psi^4\left(u^2\right),
\end{equation*}
which completes the proof of \eqref{F(1,4)(x) q-series integral}.

The proofs of equations \eqref{F(1,1)(x) q-series integral} and
\eqref{F(2,2)(x) q-series integral} will require the following
formula:
\begin{equation}\label{fine identity}
\Im\left(g\left( i
u,t\right)\right)=\frac{1}{t}\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^2
u^{2k+1}}{1+u^{2(2k+1)}}\left(t^{2k+1}-t^{-(2k+1)}\right),
\end{equation}
where
\begin{equation*}
g(u,t)=\frac{(u;u)_{\infty}^6\left(t^{-2}u;u\right)_{\infty}
\left(t^2;u\right)_{\infty}}{\left(t^{-1}u;u\right)_\infty^4
\left(t;u\right)_{\infty}^4}.
\end{equation*}
Equation \eqref{fine identity} is a direct consequence of identity
(14.2.9) in \cite{BA}, which follows from product expansions for the
Weierstrass $\wp$-function \cite{Ca}.

Rearranging equation \eqref{F(1,1)(x) q series sum} we have
\begin{align*}
\frac{72x}{\pi^2}\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}=&\Re\left(\int_{0}^{q}
\sum_{k=0}^{\infty}(-1)^k (2k+1)^2\frac{2\omega u^{2k+1}}{1-\omega^2
u^{2(2k+1)}}\frac{\d u}{u}\right)\\
=&\sqrt{3}\int_{0}^{q} \sum_{k=0}^{\infty}(-1)^k
(2k+1)^2\frac{u^{2k+1}-u^{5(2k+1)}}{1+u^{6(2k+1)}}\frac{\d u}{u}.
\end{align*}
Applying \eqref{fine identity} after letting $u\rightarrow u^3$ and
$t\rightarrow u^{-2}$, transforms this last integral into
\begin{equation*}
\frac{24\sqrt{3}x}{\pi^2}\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}=\Im\left(\int_{0}^{
i
q}\frac{f^5\left(-u^2\right)f^4\left(-u^3\right)f\left(-u^6\right)}{f^4(-u)}\d
u\right).
\end{equation*}
By the following eta function identity:
\begin{equation}\label{eta function identity in signature three}
\frac{f^5\left(-u^2\right)f^4\left(-u^3\right)f\left(-u^6\right)}{f^4(-u)}=\frac{f^9\left(-u^3\right)}{f^3\left(-u\right)}
+u\frac{f^9\left(-u^6\right)}{f^3\left(-u^2\right)},
\end{equation}
this becomes
\begin{align*}
\frac{24\sqrt{3}x}{\pi^2}\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}=&\Im\left(\int_{0}^{
i q}\frac{f^9\left(-u^3\right)}{f^3\left(-u\right)}\d
u\right)-\Im\left(\int_{0}^{q}
u\frac{f^9\left(u^6\right)}{f^3\left(u^2\right)}\d u\right)\\
=&\Im\left(\int_{0}^{ i
q}\frac{f^9\left(-u^3\right)}{f^3\left(-u\right)}\d u\right)-0,
\end{align*}
which completes the proof of \eqref{F(1,1)(x) q-series integral}.
Although we will not elaborate on the proof of \eqref{eta function
identity in signature three} here, it suffices to say that it
follows from algebraic transformations for the hypergeometric
function.

The proof of \eqref{F(2,2)(x) q-series integral} follows the same
lines, but requires a few extra steps.  Proceeding as before, we
find that
\begin{align*}
\frac{32x}{9\pi^2}F_{(2,2)}(x)=&\Re\left(\int_{0}^{q}
\sum_{k=0}^{\infty}(-1)^k (2k+1)^2\frac{2\omega u^{2k+1}}{1-\omega^2
u^{2(2k+1)}}\frac{\d u}{u}\right)\\
 =&\int_{0}^{q}
\sum_{k=0}^{\infty}(-1)^k (2k+1)^2\frac{u^{2k+1}-u^{3(2k+1)}}{1+
u^{4(2k+1)}}\frac{\d u}{u}.
\end{align*}
Applying equation \eqref{fine identity} after letting $u\rightarrow
u^2$ and $t\rightarrow u^{-1}$, this becomes
\begin{align*}
\frac{32x}{9\pi^2}F_{(2,2)}(x)=\Re\left(2\int_{0}^{\omega q}
\frac{f^6\left(-u^2\right)f\left(-u^8\right)}{f\left(-u^4\right)}\frac{1}{\left(\omega
u, u^2\right)_{\infty}^4\left(\bar{\omega}u,u^2\right)_{\infty}^4}\d
u\right),
\end{align*}
where $\omega=e^{\pi i/4}$.  For brevity of notation let us define a
new function
\begin{align}
g(u):=&\left(\omega u,
u^2\right)_{\infty}\left(\bar{\omega}u,u^2\right)_{\infty}\notag\\
=&\prod_{n=0}^{\infty}\left(1-\sqrt{2}
u^{2n+1}+u^{2(2n+1)}\right)\label{g(u) product}.
\end{align}
Since \eqref{F(2,2)(x) q-series sum} is odd with respect to $q$, our
integral can be transformed into
\begin{align}\label{F(2,2)(x) qintegral intermed}
\frac{32x}{9\pi^2}F_{(2,2)}(x)=\Re\left(\int_{0}^{\omega q}
\frac{f^6\left(-u^2\right)f\left(-u^8\right)}{f\left(-u^4\right)}\frac{g^4(u)+g^4(-u)}{g^4(u)g^4(-u)}\d
u\right).
\end{align}
Next we will reduce
$\left(g^4(u)+g^4(-u)\right)/\left(g(u)g(-u)\right)^4$ to theta
functions.  Observe by equation \eqref{g(u) product} that
\begin{align}\label{g(u)g(-u) product}
g(u)g(-u)=&\prod_{n=0}^{\infty}\left(1+u^{4(2n+1)}\right)
=\frac{\varphi\left(-u^8\right)}{f\left(-u^4\right)}.
\end{align}
With two applications of the Jacobi triple product \cite{Be3}, we
also have
\begin{align}
g(u)+g(-u)=&\frac{1}{f\left(-u^2\right)}\left(\sum_{n=-\infty}^{\infty}(-\omega)^n
u^{n^2}+\sum_{n=-\infty}^{\infty}\omega^n u^{n^2}\right)\notag\\
=&\frac{2}{f\left(-u^2\right)}\sum_{n=-\infty}^{\infty} (-1)^{n}
u^{16n^2}\notag\\
=&\frac{2\varphi\left(-u^{16}\right)}{f\left(-u^2\right)}
\label{g(u)+g(-u) sum}.
\end{align}
So finally, combining \eqref{g(u)g(-u) product} and
\eqref{g(u)+g(-u) sum}, we find that
\begin{equation*}
\frac{g^4(u)+g^4(-u)}{g^4(u)g^4(-u)}
=2\frac{f^4\left(-u^4\right)}{f^4\left(-u^2\right)}
\left(8\frac{\varphi^4\left(-u^{16}\right)}{\varphi^4\left(-u^8\right)}-8\frac{\varphi^2\left(-u^{16}\right)\varphi\left(-u^2\right)}{\varphi^3\left(-u^8\right)}+\frac{\varphi^2\left(-u^2\right)}{\varphi^2\left(-u^8\right)}\right).
\end{equation*}
Recalling that
$\varphi^2\left(-q^{16}\right)=\varphi\left(-q^{8}\right)\varphi\left(q^8\right)$,

this becomes
\begin{align*}
\frac{g^4(u)+g^4(-u)}{g^4(u)g^4(-u)}
=&2\frac{f^4\left(-u^4\right)}{f^4\left(-u^2\right)}
\left(\frac{8\varphi^2\left(u^{8}\right)-8\varphi\left(u^8\right)\varphi\left(-u^2\right)+\varphi^2\left(-u^2\right)}{\varphi^2\left(-u^8\right)}\right),\\
=&2\frac{f^5\left(-u^4\right)}{f^6\left(-u^2\right)}
\left(\frac{8\varphi^2\left(u^{8}\right)\varphi\left(-u^2\right)-8\varphi\left(u^8\right)\varphi^2\left(-u^2\right)+\varphi^3\left(-u^2\right)}{\varphi^2\left(-u^8\right)}\right),
\end{align*}
and therefore \eqref{F(2,2)(x) qintegral intermed} simplifies to

\begin{equation*}
\begin{split}
\frac{32x}{9\pi^2}F_{(2,2)}(x)=&\Re\left(2\int_{0}^{\omega q}
f^4\left(-u^4\right)f\left(-u^8\right)\right.\\
&\qquad\qquad\left.\times\frac{8\varphi^2\left(u^8\right)\varphi\left(-u^2\right)-8\varphi\left(u^8\right)\varphi^2\left(-u^2\right)+\varphi^3\left(-u^2\right)}{\varphi^2\left(-u^8\right)}\d
u\right).
\end{split}
\end{equation*}
Next, let $u\rightarrow \omega u$ to obtain
\begin{equation*}
\begin{split}
=&2\int_{0}^{q}
\frac{f^4\left(u^4\right)f\left(-u^8\right)}{\varphi^2\left(-u^8\right)}\\
&\qquad\times\Re\left(\omega\left(8\varphi^2\left(u^8\right)\varphi\left(-
i u^2\right)-8\varphi\left(u^8\right)\varphi^2\left(- i
u^2\right)+\varphi^3\left(- i u^2\right)\right)\right)\d u.
\end{split}
\end{equation*}
If we recall that $\varphi\left(- i
u^2\right)=\varphi\left(u^8\right)-2 i u^2\psi\left(u^{16}\right)$,
then we are left with
\begin{align*}
\frac{32x}{9\pi^2}F_{(2,2)}(x) =&\sqrt{2}\int_{0}^{q}
\frac{f^4\left(u^4\right)f\left(-u^8\right)}{\varphi^2\left(-u^8\right)}\\
&\qquad\times\left(\left(\varphi\left(u^8\right)-2u^2\psi\left(u^{16}\right)\right)^3
-4u^2\varphi\left(u^8\right)\psi\left(u^{16}\right)\left(\varphi\left(u^8\right)-2u^2\psi\left(u^{16}\right)\right)\right)\d
u.
\end{align*}
In order to simplify this last formula, we will freely apply theta
function identities on pages 34 and 40 of \cite{Be3}.  Therefore, we
find that
\begin{align*}
=&\sqrt{2}\int_{0}^{q}\frac{f^4\left(u^4\right)f\left(-u^8\right)}{\varphi^2\left(-u^8\right)}\left(\varphi^3\left(-u^2\right)-4u^2\varphi\left(u^8\right)\psi\left(u^{16}\right)\varphi\left(-u^2\right)\right)\d
u\\
=&\sqrt{2}\int_{0}^{q}\frac{f^4\left(u^4\right)f\left(-u^8\right)}{\varphi^2\left(-u^8\right)}\varphi\left(-u^2\right)\left(\varphi^2\left(-u^2\right)-4u^2\psi^2\left(u^{8}\right)\right)\d
u\\
=&\frac{1}{\sqrt{2}}\int_{0}^{q}\frac{f^4\left(u^4\right)f\left(-u^8\right)}{\varphi^2\left(-u^8\right)}\varphi\left(-u^2\right)\left(3\varphi^2\left(-u^2\right)-\varphi^2\left(u^{2}\right)\right)\d
u\\
=&\frac{1}{\sqrt{2}}\int_{0}^{q}\frac{f^4\left(u^4\right)f\left(-u^8\right)}{\varphi^2\left(-u^8\right)\psi^2\left(u^4\right)}
\varphi\left(-u^2\right)\left(3\psi^4\left(-u^2\right)-\psi^4\left(u^{2}\right)\right)\d
u\\
=&\frac{1}{\sqrt{2}}\int_{0}^{q}\varphi\left(u^4\right)
\varphi\left(-u^2\right)\left(3\psi^4\left(-u^2\right)-\psi^4\left(u^{2}\right)\right)\d
u,
\end{align*}
which completes the proof of \eqref{F(2,2)(x) q-series integral}.
$\blacksquare$
\end{proof}

The next theorem requires the signature-three theta functions.
Recall that if $\omega=e^{2\pi i/3}$, then the signature-three theta
functions are defined by:
\begin{align*}
a(q):=&\sum_{n,m=-\infty}^{\infty}q^{m^2+m n+n^2},\\
b(q):=&\sum_{n,m=-\infty}^{\infty}\omega^{m-n}q^{m^2+m n+n^2},\\
c(q):=&\sum_{n,m=-\infty}^{\infty}q^{(m+1/3)^2+(m+1/3)(n+1/3)+(n+1/3)^2}.
\end{align*}
The signature-three theta functions satisfy many interesting
formulas, including the following cubic relation:
\begin{equation*}
a^3(q)=b^3(q)+c^3(q).
\end{equation*}
Various other properties of $a(q)$, $b(q)$, and $c(q)$ have been
catalogued in \cite{Be5}.

\begin{theorem}\label{hypergeometric theorem} We can reduce the two-dimensional lattice sums to
integrals of hypergeometric functions.

Suppose that $q=e^{-\pi x/\sqrt{12}}$, then
\begin{equation}\label{F(1,1) hypergeometric form}
\begin{split}
\frac{648x}{\pi^2}\frac{F(1,1,1,x^2)}{\left(3+x^2\right)^2}
        &=\left\{\begin{array}{ll}
                     3\tilde{n}\left(3\frac{a( i q)}{b( i q)}\right)
                        +\frac{4}{\sqrt{3}}n_2\left(\frac{b^3( i q)}{a^3( i
                        q)}\right) &\text{if $x\in\left(0,\frac{1}{\sqrt{5}}\right)$},\\
                     3\tilde{n}\left(3\frac{a( i q)}{b( i q)}\right)
                        +\frac{1}{\sqrt{3}}n_2\left(\frac{b^3( i q)}{a^3( i
                        q)}\right) &\text{if $x\in\left(\frac{1}{\sqrt{5}},\sqrt{5}\right)$},\\
                        \frac{1}{\sqrt{3}}n_2\left(\frac{b^3( i q)}{a^3( i
                        q)}\right)  &\text{if $x\in\left(\sqrt{5},\infty\right)$},
                \end{array}\right.
\end{split}
\end{equation}
where
\begin{equation*}
\tilde{n}(k)=\Re\left(\log(k)-\frac{2}{k^3}{_4F_3}\left(\substack{1,1,\frac{4}{3},\frac{5}{3}\\
2,2,2};\frac{27}{k^3}\right)\right),
\end{equation*}
and
\begin{equation*}
n_2(k)=\Im\left(\int_{k}^{1}\frac{{_2F_1}\left(\substack{\frac{1}{3},\frac{2}{3}\\
1}; 1-u\right)}{u}\d u\right).
\end{equation*}
Notice that $\tilde{n}(k)=n(k)$ whenever $|k|$ is sufficiently
large.

Suppose that $q=e^{-\pi x}$ and $x>0$, then
\begin{equation}\label{F(1,2) hypergeometric form}
\frac{16x}{\pi^2}F_{(1,2)}(x)=m\left(\frac{ i
f^4(-q)}{\sqrt{q}f^4\left(-q^4\right)}\right),
\end{equation}
where $m(k)$ is defined in \eqref{m(k) def}.

Suppose that $q=e^{-\pi x/3}$ and $\omega=e^{2\pi  i/3}$, then
\begin{equation}\label{F(1,4)(x) hypergeometric integral}
\begin{split}
\frac{144x}{25\pi^2}F_{(1,4)}(x)
        &=\left\{\begin{array}{ll}
                     m\left(4\frac{\varphi^2(\omega q)}{\varphi^2(-\omega q)}\right)
                        -\frac{3}{4}m_2\left(\frac{\varphi^4(-\omega
                        q)}{\varphi^4(\omega
                        q)}\right) &\text{if $x\in\left(0,\frac{1}{\sqrt{2}}\right)$},\\
                     m\left(4\frac{\varphi^2(\omega q)}{\varphi^2(-\omega q)}\right)
                        +\frac{1}{4}m_2\left(\frac{\varphi^4(-\omega q)}{\varphi^4(\omega
                        q)}\right) &\text{if $x\in\left(\frac{1}{\sqrt{2}},\sqrt{2}\right)$},\\
                        \frac{1}{4} m_2\left(\frac{\varphi^4(-\omega q)}{\varphi^4(\omega
                        q)}\right)  &\text{if $x\in\left(\sqrt{2},\infty\right)$},
                \end{array}\right.
\end{split}
\end{equation}
where $m(k)$ is defined in \eqref{m(k) def}, and
\begin{equation*}
m_2(k):=\Im\left(\int_{k}^{1}
                        \frac{_2F_1\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1}; 1-u\right)}{u}\d
                        u\right).
\end{equation*}

If $q=e^{-\pi x/\sqrt{8}}$, then
\begin{equation}\label{F(2,2)(x) hypergeometric integral}
F_{(2,2)}(x)=\frac{9\pi^2}{256x}\int_{\frac{\varphi^2\left(-q^2\right)}{\varphi^2\left(q^2\right)}}^{1}\frac{\left(3u-1\right)}
{\sqrt{u(1-u)}}{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1}; 1-u^2\right)\d u.
\end{equation}
\end{theorem}
\begin{proof}  The proof of this theorem follows from our ability to
invert theta functions.  First recall the classical inversion
formula for the theta function:
\begin{equation}\label{theta inversion formula}
\varphi^2(q)={_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};
1-\frac{\varphi^4(-q)}{\varphi^4(q)}\right),
\end{equation}
which holds whenever $q\in(-1,1)$ \cite{Be3}.  If we use the
notation $\alpha=1-\varphi^4(-q)/\varphi^4(q)$ and $z=\varphi^2(q)$,
then many different theta functions can be expressed in terms of
these two parameters.  The following identities are true whenever
$|q|<1$:
\begin{align*}
\varphi(q)=&\sqrt{z},\\
\varphi(-q)=&(1-\alpha)^{1/4}\sqrt{z},\\
\varphi\left(q^2\right)=&\left(1+\sqrt{1-\alpha}\right)^{1/2}\sqrt{\frac{z}{2}},\\
\psi(-q)=&q^{-1/8}\left\{\alpha(1-\alpha)\right\}^{1/8}\sqrt{\frac{z}{2}},\\
\psi(q)=&q^{-1/8}\alpha^{1/8}\sqrt{\frac{z}{2}},
\end{align*}
and it is also well known that
\begin{equation*}
\frac{\d \alpha}{\d q}=\frac{\alpha(1-\alpha)z^2}{q}.
\end{equation*}
Both \eqref{F(1,2) hypergeometric form} and \eqref{F(2,2)(x)
hypergeometric integral} follow from applying these
parameterizations to equations \eqref{F(1,2)(x) q-series integral}
and \eqref{F(2,2)(x) q-series integral} respectively.

Since equation \eqref{theta inversion formula} does not hold in the
entire open unit disk, we will need to generalize that result. First
notice that $z$ satisfies the hypergeometric differential equation
with respect to $\alpha$:
\begin{equation}\label{hypergeometric diff equation}
\alpha(1-\alpha)\frac{\d^2 z}{\d\alpha^2}+(1-2\alpha)\frac{\d
z}{\d\alpha}-\frac{z}{4}=0.
\end{equation}
We can use the relation
$\frac{\d}{\d\alpha}=\frac{1}{\frac{\d\alpha}{\d
q}}\times\frac{\d}{\d q}$, to show that \eqref{hypergeometric diff
equation} holds (excluding possible poles) for $|q|<1$.  The most
general solution of this differential equation has the form
\begin{equation*}
z=C {_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};
\alpha\right)+D {_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1} ; 1-\alpha\right),
\end{equation*}
where $C$ and $D$ are undetermined constants.  When $q$ lies in a
neighborhood of zero, \eqref{theta inversion formula} shows that
$(C,D)=(1,0)$.  We can analytically continue that solution to a
larger connected $q$-domain, provided that $\alpha$ (and $1-\alpha$
if $D\not=0$) does not intersect the line
$[1,\infty)$.  In particular, the function ${_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1}; \alpha\right)$ has a branch cut running along $[1,\infty)$.

If we consider values of $q\in(0,\omega)$ with $\omega=e^{2\pi
i/3}$, then $\alpha$ crosses $[1,\infty)$ at the point $q=\omega
e^{-\pi\sqrt{2}/3}$. Similarly, $1-\alpha$ intersects the branch cut
at $q=\omega e^{-\pi/3\sqrt{2}}$. It follows that we will have to
solve the hypergeometric differential equation separately on each of
the three line segments. If $u=\omega e^{-\pi x/3}$, then
\begin{equation}\label{z inversion on a ray}
\begin{split}
\varphi^2(u)    &=\left\{\begin{array}{ll}
                      -3{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};1-\frac{\varphi^4(-u)}{\varphi^4(u)}\right)+2 i{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};\frac{\varphi^4(-u)}{\varphi^4(u)}\right) &\text{if $x\in\left(0,\frac{1}{\sqrt{2}}\right)$},\\
                      {_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
                        1};1-\frac{\varphi^4(-u)}{\varphi^4(u)}\right)+2 i
                        {_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};\frac{\varphi^4(-u)}{\varphi^4(u)}\right) &\text{if $x\in\left(\frac{1}{\sqrt{2}},\sqrt{2}\right)$},\\
                        {_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};1-\frac{\varphi^4(-u)}{\varphi^4(u)}\right)  &\text{if
                        $x\in\left(\sqrt{2},\infty\right)$}.
                \end{array}\right.
\end{split}
\end{equation}
The coefficients in \eqref{z inversion on a ray} can be verified
from the fact that $\varphi^2(u)$ is analytic when $x\in
(0,\infty)$. For example, we can check the continuity of the
right-hand side of \eqref{z inversion on a ray} by letting
$u\rightarrow\omega e^{-\pi\sqrt{2}/3}$. In that case
$\alpha=1-\frac{\varphi^4(-u)}{\varphi^4(u)}\approx 5.828\dots$, and
we have:
\begin{equation*}
\begin{split}
0=&\varphi^2\left(\omega e^{-\pi\frac{\sqrt{2}+0}{3}}\right)-\varphi^2\left(\omega e^{-\pi\frac{\sqrt{2}-0}{3}}\right)\\
=&{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};\alpha+ i 0\right)-{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};\alpha- i 0\right)-2 i{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1};1-\alpha\right).
\end{split}
\end{equation*}
This vanishing of this last expression follows from basic properties
of the hypergeometric function (see problem $1$ on page 276 of
\cite{Le}), and therefore the right-hand side of \eqref{z inversion
on a ray} is indeed continuous at $x=\sqrt{2}$. In practice, we
simply discovered \eqref{z inversion on a ray} numerically.

We will use the theory of signature-three theta functions to prove
equation \eqref{F(1,1) hypergeometric form}. Recall that $c(q)$ can
be expressed as an infinite product:
\begin{equation*}
\frac{c^3(q)}{27 q}=\frac{f^9(-q^3)}{f^3(-q)},
\end{equation*}
and that the signature-three theta functions obey a differentiation
formula:
\begin{equation*}
\frac{c^3(q)}{q}=\frac{a(q)}{1-\frac{c^3(q)}{a^3(q)}}\frac{\d}{\d
q}\left(\frac{c^3(q)}{a^3(q)}\right).
\end{equation*}
It follows immediately that equation \eqref{F(1,1)(x) q-series
integral} reduces to
\begin{equation}\label{F(1,1)(x) q integral reduced}
\frac{F(1,1,1,x^2)}{(3+x^2)^2}=\frac{\pi^2}{648\sqrt{3}x}\Im\left(\int_{0}^{
i q}\frac{a(u)}{1-\frac{c^3(u)}{a^3(u)}}\frac{\d}{\d
u}\left(\frac{c^3(u)}{a^3(u)}\right)\d u\right).
\end{equation}
Next recall that for $|u|$ sufficiently small:
\begin{equation}\label{a(q) inversion formula}
a(u)={_2F_1}\left(\substack{\frac{1}{3}, \frac{2}{3}\\
1};\frac{c^3(u)}{a^3(u)}\right).
\end{equation}
In order to apply \eqref{a(q) inversion formula} to our integral, we
will need to establish a generalized inversion formula which holds
for $u\in (0, i)$.  The reasoning closely follows the proof of
\eqref{z inversion on a ray}, except that
$c^3(u)/a^3(u)\in[1,\infty)$ when $u= i e^{-\pi\sqrt{5/12}}$, and
$1-c^3(u)/a^3(u)\in[1,\infty)$ when $u= i e^{-\pi/\sqrt{60}}$.
Suppose that $u= i e^{-\pi x/\sqrt{12}}$, then we obtain
\begin{equation}\label{a(q) inversion on a ray}
\begin{split}
a(u)        &=\left\{\begin{array}{ll}
                      4{_2F_1}\left(\substack{\frac{1}{3},\frac{2}{3}\\ 1};\frac{c^3(u)}{a^3(u)}\right)
                      +\sqrt{3} i{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};1-\frac{c^3(u)}{a^3(u)}\right) &\text{if $x\in\left(0,\frac{1}{\sqrt{5}}\right)$},\\
                      2{_2F_1}\left(\substack{\frac{1}{3},\frac{2}{3}\\ 1};\frac{c^3(u)}{a^3(u)}\right)
                      +\sqrt{3} i{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\ 1};1-\frac{c^3(u)}{a^3(u)}\right) &\text{if $x\in\left(\frac{1}{\sqrt{5}},\sqrt{5}\right)$},\\
                      {_2F_1}\left(\substack{\frac{1}{3},\frac{2}{3}\\ 1};\frac{c^3(u)}{a^3(u)}\right) &\text{if $x\in\left(\sqrt{5},\infty\right)$}.
                \end{array}\right.
\end{split}
\end{equation}
Finally, \eqref{F(1,1) hypergeometric form} follows from
substituting \eqref{a(q) inversion on a ray} into \eqref{F(1,1)(x) q
integral reduced} and simplifying. $\blacksquare$
\end{proof}

Finally, we will conclude this section by summarizing the formulas
that follow from setting $x=1$ in Theorem \ref{hypergeometric
theorem}.

\begin{corollary}\label{final reduction cor} The following identities are true:
\begin{align}
\frac{27}{2\pi^2}F(1,1)=&n\left(3\sqrt[3]{2}\right),\\
\frac{16}{\pi^2}F(1,2)=&m\left(4 i\right),\\
\frac{144}{25\pi^2}F(1,4)=&m\left(\frac{4}{\theta}\right)
+\frac{1}{4}\Im\left(\int_{\theta^2}^{1}\frac{{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1}; 1-u\right)}{u}\d u\right)\label{F(1,4) final formula},\\
\frac{256}{9\pi^2}F(2,2)=&\int_{\sqrt{2}-1}^{1}\frac{3u-1}{\sqrt{u(1-u)}}{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1}; 1-u^2\right)\d u,\label{F(2,2) final formula}
\end{align}
where $\theta=\frac{(4-2t-2t^2+t^3)}{4\sqrt{2}}$, and $t=-
i\sqrt[4]{12}$.
\end{corollary}
Notice that equation \eqref{F(1,4) final formula} involves Meijer's
$G$-function disguised as a hypergeometric integral:
\begin{equation*}
\Im\left(\int_{k}^{1}\frac{{_2F_1}\left(\substack{\frac{1}{2},\frac{1}{2}\\
1};1-u\right)}{u}\d
u\right)=\frac{1}{\pi^2}\Im\left({G_{3,3}^{3,2}}\left(k\big\vert\substack{\frac{1}{2},\frac{1}{2},1\\
0,0,0}\right)\right).
\end{equation*}
This identity probably rules out the possibility of expressing
$F(1,4)$ as a Mahler measure, and it also indicates that any
explicit formula for $F(b,c)$ should reduce to Meijer $G$-functions
in certain instances.

\section{Formulas for $F(1,1,2,4)$, $F(1,2,4,4)$, and additional explicit examples}\label{section: F(1,1,2,4) and F(1,2,4,4)}\init

In this section, we will present several additional formulas for
$F(a,b,c,d)$.  Since the ideas are the same as in the previous
section, we will only sketch brief details of each proof.

\begin{theorem} The following identities are true:
\begin{align}
\frac{144\sqrt{2}}{121\pi^2}F(1,2,4,4)&=\frac{1}{3}n\left(3\left(\frac{4}{2+17\sqrt{2}-9\sqrt{6}}\right)^{1/3}\right)-\frac{1}{6}n\left(3\left(\frac{4}{2+\sqrt{2}}\right)^{1/3}\right),\label{F(1,1,2,4) identity}\\
\frac{27}{\sqrt{2}\pi^2}F(1,1,2,4)&=n\left(3\left(\frac{4}{2-17\sqrt{2}+9\sqrt{6}}\right)^{1/3}\right)-n\left(3\left(\frac{4}{2-\sqrt{2}}\right)^{1/3}\right).\label{F(1,2,4,4)
identity}
\end{align}
\end{theorem}
\begin{proof} The proof of this theorem requires the following identities:
\begin{align*}
e_1^2 e_2 e_4=&\left(\frac{e_1^2
e_4^2}{e_2}\right)\left(\frac{e_2^2}{e_4}\right)=\sum_{n,k=-\infty}^{\infty}(-1)^k(3n+1)q^{\frac{(3n+1)^2+6k^2}{3}},\\
e_1 e_2 e_4^2=&\left(\frac{e_1^2
e_4^2}{e_2}\right)\left(\frac{e_2^2}{e_1}\right)=\sum_{\substack{n=-\infty\\
k=0}}^{\infty}(3n+1)q^{\frac{8(3n+1)^2+3(2k+1)^2}{24}}.
\end{align*}
By an argument similar to the one in Proposition \ref{q-series
proposition}, we can prove that
\begin{equation*}
\begin{split}
F(1,1,2,4)=&\sum_{n,k=-\infty}^{\infty}\frac{(-1)^k(3n+1)}{\left((3n+1)^2+6k^2\right)^2}\\
=&-\frac{\sqrt{2}\pi^2}{81}\left(\log(q^4)+9\sum_{n=1}^{\infty}n\chi_{-3}(n)\log\left(1+q^{4n}\right)\right),
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
F(1,2,4,4)=&\frac{121}{2}\sum_{n,k=-\infty}^{\infty}\frac{(3n+1)}{\left(8(3n+1)^2+3(2k+1)^2\right)^2}\\
=&\frac{121\pi^2}{288\sqrt{2}}\sum_{n=1}^{\infty}n\chi_{-3}(n)\log\left(\frac{1+q^n}{1-q^n}\right),
\end{split}
\end{equation*}
 where $q=e^{-\pi/\sqrt{6}}$.  Both of these $q$-series can be evaluated in terms of $n(k)$ by the results
 of Rodriguez-Villegas (see \cite{LR} for details). $\blacksquare$
\end{proof}

If we consider additional examples involving $F(1,1,1,x^2)$, then we
can establish many interesting formulas by setting $x=\sqrt{a/b}$.
For instance, if $x\in\left\{1,\frac{1}{\sqrt{5}}\right\}$, we have
$\Im\left(b^3( i q)/a^3( i q)\right)=0$, and as a result it is
possible to show that the $n_2$ term in \eqref{F(1,1) hypergeometric
form} vanishes.

\begin{theorem}Let $\phi=\frac{1+\sqrt{5}}{2}$, $y=\sqrt[3]{2}e^{\pi i/3}$, and suppose that $z\approx-.58+.56 i$ is a root of the equation $(z^2+3z+1)^3-2(z^6+1)=0$, then
\begin{align}
F(1,1)=&\frac{2\pi^2}{27}n\left(3\sqrt[3]{2}\right),\label{F(1,1)(1) to ntilde}\\
F\left(1,1,1,\frac{1}{5}\right)=&\frac{32\pi^2}{135\sqrt{5}}\tilde{n}\left(\frac{3}{\sqrt[3]{\phi}}\right),\label{F(1,1)sqrt5}\\
F\left(1,1,1,5\right)\stackrel{?}{=}&\frac{4\pi^2}{27\sqrt{5}}n\left(-3\sqrt[3]{\phi}\right),\label{F(1,1)sqrt5b}\\
F(1,1,1,9)=&
\frac{2\pi^2}{27\sqrt{3}}n_2\left(\frac{9}{2(1+y+3y^2)}\right),\\
F(1,1,1,25)=&
\frac{98\pi^2}{405\sqrt{3}}n_2\left(\frac{1}{1+z^6}\right).
\end{align}
\end{theorem}

Notice that the result in the introduction, \eqref{intro lattic
example}, follows from combining \eqref{F(1,1)sqrt5} with
\eqref{n(k) 4F3 to 3F2}.  We have also stated \eqref{F(1,1)sqrt5b}
as a conjecture, because the result presumably follows from
\eqref{F(1,1) hypergeometric form} when $x\rightarrow{\sqrt{5}}$,
but we still have not found a completely rigorous proof.
 Finally, we will conclude this section with two additional identities.
If we set $q=e^{-\frac{\pi}{2\sqrt{3}}}$ and $\omega=e^{\pi i/4}$,
then
\begin{align}
\frac{24^2\sqrt{2}}{19^2
\pi^2}F(1,2,8,8)=&\sum_{n=1}^{\infty}n\chi_{-3}(n)\log\left|\frac{1+\omega
q^n}{1-\omega q^n}\right|,\label{F(1,2,8,8) formula}\\
\frac{72\sqrt{2}}{49\pi^2}\F(1,1,4,8)=&\sum_{n=1}^{\infty}n\chi_{-3}(n)\log\left|\frac{1+\omega
q^{2n}}{1-\omega q^{2n}}\right|.\label{F(1,1,2,8) formula}
\end{align}
Unfortunately, we have not been able to reduce these last two
$q$-series to known functions.  These $q$-expansions follow from
combining $e_1 e_2 e_8^2=\left(\frac{e_2^2
e_8^2}{e_4}\right)\left(\frac{e_1 e_4}{e_2}\right)$ and $e_1^2 e_4
e_8=\left(\frac{e_1^2 e_4^2}{e_2}\right)\left(\frac{e_2
e_8}{e_4}\right)$, with equations \eqref{weight 1/2 series} and
\eqref{weight 3/2 series}.

\section{Remarks on $F(1,3)$, $F(2,9)$, $F(5,9)$ and higher values of
$F(a,b,c,d)$}\label{section: F(1,3) section}\init

In this section, we will briefly demonstrate how to reduce several
higher values of $F(a,b,c,d)$ to known functions.  Our proof of the
$F(1,3)$ formula will be instructive. Recall that Rodriguez-Villegas
showed that
\begin{equation}\label{F(1,3) RV formula}
\frac{4\pi^2}{81}n(-6)
=\Re\left(\frac{1}{2}\sum_{\substack{m,n\in\mathbb{Z}\\
(m,n)\not=(0,0)}}\frac{\chi_{-3}(n)} {\left(3\left(\frac{1+
i\sqrt{3}}{2}\right)m+n\right)^2\left(3\left(\frac{1-
i\sqrt{3}}{2}\right)m+n\right)}\right),
\end{equation}
and then used Deuring's theorem to equate this Eisenstein series to
the $L$ series of a CM elliptic curve of conductor $27$
\cite[p.~32]{RV}. A different proof could have been constructed from
numerically observing that
\begin{equation}\label{F(1,3) essential q-series identity}
\begin{split}
q\prod_{n=1}^{\infty}&\left(1-q^{3n}\right)^2\left(1-q^{9n}\right)^2\\
&=\frac{1}{4}\sum_{j=1}^{2}\chi_{-3}(j)
\sum_{n,m=-\infty}^{\infty}\left((6m+j)+3(6n+j)\right)q^{\frac{(6m+j)^2+3(6n+j)^2}{4}}.
\end{split}
\end{equation}
The modularity theorem implies that $e_3^2 e_9^2$ is associated to
the correct elliptic curve, hence the Mellin transform of the
left-hand side of equation \eqref{F(1,3) essential q-series
identity} will equal $L(E,s)$.  Since the Mellin transform (at
$s=2$) of the right-hand side trivially equals the right-hand side
of equation \eqref{F(1,3) RV formula}, it just remains to prove
\eqref{F(1,3) essential q-series identity}.  By applying limiting
cases of the triple and quintuple product identities, we can show
that equation \eqref{F(1,3) essential q-series identity} is
equivalent to an identity between eta functions:
\begin{equation}\label{F(1,3) equivalent eta function identity}
\begin{split}
4e_3^2e_9^2
=&\left(\frac{e_6^5e_{36}e_{54}^2}{e_{12}^2e_{18}e_{108}}\right)
+3\left(\frac{e_{12} e_{18}^7}{e_{6}e_{36}^3}\right)
-2\left(\frac{e_3^2e_{12}^2e_{18}^2e_{27}e_{108}}
{e_6e_9e_{36}e_{54}}\right)
-6\left(\frac{e_6^2e_9^3e_{36}^3}{e_3e_{12}e_{18}^2}\right).
\end{split}
\end{equation}
Identities between modular forms, such as \eqref{F(1,3) equivalent
eta function identity}, are usually established by checking that
both sides of a formula have the same McLauren series expansion for
sufficiently many terms.  If we use the inversion formula for the
eta function, then \eqref{F(1,3) equivalent eta function identity}
can also be viewed as an example of a \textit{mixed modular
equation}.  Of course, the main difficulty of extending this type of
approach, is to actually find the necessary modular equations. Since
the first version of this paper appeared, it has become clear that
this method also applies to $F(2,9)$, $F(5,9)$ and $F(4,7,7,28)$
\cite{RY}.

\section{New Mahler measures, and a twisted $L$-series}\label{section: more mahlers}\init

In this section we will use values of class invariants to deduce
some explicit formulas for Mahler measures.  Recall that if
$q=e^{-\pi\sqrt{m}}$, then the class invariants are defined by
\begin{align*}
g_m:=&2^{-1/4}q^{-1/24}\left(q;q^2\right)_{\infty},&
G_m:=&2^{-1/4}q^{-1/24}\left(-q;q^2\right)_{\infty}.
\end{align*}
It is a classical fact that $G_m$ and $g_m$ are algebraic numbers
whenever $m\in\mathbb{Q}$, and that they satisfy the following
algebraic relation:
\begin{equation}\label{Gm and gm algebraic relation}
\left(g_m G_m\right)^8\left(G_m^8-g_m^8\right)=\frac{1}{4}.
\end{equation}
Since most tables only contain values of $g_m$ when $m$ is even, and
$G_m$ when $m$ is odd, our calculations will require \eqref{Gm and
gm algebraic relation}.

\begin{theorem}\label{Theorem: addition F(1,2)(m) cases} Suppose that $m\in\mathbb{N}$, then
\begin{equation}
m\left(8 i g_m^8
G_m^4\right)=\frac{16\sqrt{m}}{\pi^2}\sum_{n=1}^{\infty}\frac{b_n}{n^2},
\end{equation}
where
\begin{equation*}
\sum_{n=1}^{\infty}b_n q^n=\frac{e_8^3 e_{4m}^2}{e_{8m}}.
\end{equation*}
The following table gives evaluations of $8g_m^8 G_m^4$, and states
whether or not $b_n$ is multiplicative:
\begin{equation*}

    \begin{tabular}{|c|c|c|p{6 in}|}
        \hline
        $m$ &  $8g_m^8 G_m^4$ &   $b_n$ multiplicative?\\
        \hline
        $1$ & $4$ &   Yes\\
        $2$ & $4\sqrt{2+2\sqrt{2}}$ &   No\\
        $3$ & $4\left(2+\sqrt{3}\right)$ &   No\\
        $7$ & $4\left(8+3\sqrt{7}\right)$ & No\\
        $9$ & $4\left(7+4\sqrt[4]{12}+2\sqrt[4]{12^2}+\sqrt[4]{12^{3}}\right)$ &  No\\
        $15$ & $4\left(28+16\sqrt{3}+12\sqrt{5}+7\sqrt{15}\right)$ &  No\\
        \hline
    \end{tabular}

\end{equation*}
\end{theorem}
\begin{proof}
If $m\in\mathbb{N}$, then we can use the definition of
$F_{(1,2)}(x)$ to show that
\begin{equation*}
F_{(1,2)}(\sqrt{m})=\sum_{n=1}^{\infty}\frac{b_n}{n^2},
\end{equation*}
where $b_n$ has the stated generating function. Furthermore,
equation \eqref{F(1,2)(x) as a Mahler measure} reduces to
\begin{equation}
F_{(1,2)}\left(\sqrt{m}\right)=\frac{\pi^2}{16\sqrt{m}}m\left(8 i
g_m^8 G_m^4\right).
\end{equation}
Therefore, we can obtain Mahler measure formulas by appealing to
tables of class invariants \cite{Be5}. In order to check the values
of $8g_m^8 G_m^4$, we can solve \eqref{Gm and gm algebraic relation}
to show that
\begin{align*}
8g_m^8 G_m^4&=4\left(G_m^{12}+\sqrt{G_m^{24}-1}\right)\\
 &=4\sqrt{2}
g_m^{6}\sqrt{g_m^{12}+\sqrt{g_m^{24}+1}}.
\end{align*}
For example, since $G_1=1$, it follows that $8g_1^8 G_1^4=4$.  While
this type of argument frequently leads to nested radicals, many of
those identities simplify with sufficient effort. $\blacksquare$
\end{proof}

A cursory inspection of Theorem \ref{Theorem: addition F(1,2)(m)
cases}, reveals only one instance where $b_n$ is multiplicative. In
particular, when $m=1$ the generating function for $b_n$ reduces to
$e_4^2 e_8^2$.  This cusp form is associated to a conductor $36$
elliptic curve with complex multiplication \cite{Ono}.  Thus, we
have given a new proof of the formula for $m(4i)$ \cite{RV}.  When
$m=2$, a brief computation reveals that $b_3=b_{11}=0$, but
$b_{33}=-8$, and therefore $e_8^5/e_{16}$ is not multiplicative.
Fortunately, Somos has pointed out that $e_8^5/e_{16}$ can be added
to another eta product to obtain a multiplicative cusp form in
$\mathbb{Q}(i)$:
\begin{equation*}
\begin{split}
g(q)&=\frac{e_8^5}{e_{16}}+2i\frac{e_{32}^2 e_8^3}{e_{16}}\\
&=q+2i q^3-q^9-6iq^{11}\dots
\end{split}
\end{equation*}
Mellit observed that $g(q)$ is attached to a CM elliptic curve of
conductor $256$, twisted by an imaginary character. Letting
$x\rightarrow 1/\sqrt{2}$ in \eqref{F(1,4)(x) hypergeometric
integral} produces a Mahler measure for $L(e_{32}^2 e_8^3
e_{16}^{-1},2)$, and therefore
\begin{equation}\label{L(g,2)}
\frac{16\sqrt{2}}{\pi^2}L(g,2)=m\left(4i\sqrt{2+2\sqrt{2}}\right)+im\left(4(\sqrt{2}-1)\right).
\end{equation}
By elementary $q$-series identities, it is also possible to prove
that $g(q)=q\varphi(i q^2)f^3(-q^8)$.  This final observation shows
that \eqref{L(g,2)} arises from the dissection of a $q$-series
\cite{So}.

\section{Higher polylogarithms and conclusion}
\label{section: conclusion} \init

We will conclude the paper, by showing that the method from Section
\ref{section:hypergeometric reductions} can be used to produce
identities for elliptic polylogarithms. Let us briefly reexamine
Proposition \ref{q-series proposition}.  If we had used involutions
for weight $1/2$ theta functions, rather than involutions for weight
$3/2$ theta functions, we would have obtained formulas including
\begin{equation*}
\frac{F(1,1,1,x)}{(3+x)^2}=\frac{\pi}{36\sqrt{x}}\sum_{n=1}^{\infty}\chi(n)
D\left(i e^{-\frac{\pi n}{\sqrt{12x}}}\right),
\end{equation*}
where $D(z)=\Im\left(\Li_2(z)+\log|z|\log(1-z)\right)$ is the
Bloch-Wigner dilogarithm, and $\chi(n)$ is a character modulo $12$,
with $\chi(5)=\chi(7)=-1$, and $\chi(1)=\chi(11)=1$.  It is also
interesting to note that the following conjecture of
Rodriguez-Villegas \cite{Fi}:
\begin{equation*}
\m\left(1+x_1+x_2+x_3+x_4\right)\stackrel{?}{=}\frac{675\sqrt{15}}{16\pi^5}L(f,4),
\end{equation*}
where $f(q)=e_3^3e_5^3+e_1^3e_{15}^3$, can be reformulated using
such an argument.  By the method of Proposition \ref{q-series
proposition}, we have
 \begin{equation*}
 L(f,4)=-\frac{128\pi}{15^3}\sum_{n=1}^{\infty}\chi_{-4}(n)\left(9R\left(i
 q^{n}\right)+R\left(i
 q^{3n}\right)\right),
 \end{equation*}
 where
 $R(z)=\Im\left(\log|z|\Li_4(z)-\log^2|z|\Li_3(z)+\frac{\log^3|z|}{3}\Li_2(z)\right)$,
 and $q=e^{-\pi\sqrt{15}/6}$.

\bigskip

\begin{acknowledgements}

The author would like to thank David Boyd, Wadim Zudilin, Larry
Glasser, Bruce Berndt, Michael Somos, and Anton Mellit for their
useful comments and encouragement.  Additionally, the author thanks
the Max Planck Institute of Mathematics for their hospitality.
\end{acknowledgements}

config: arxiv


TITLE: Dual Basis, Polynomials
QUESTION [0 upvotes]: Consider the ordered basis $B=\{1,x-1,x^2+1\}$ of polynomials of degree $2$ or less, find the dual basis. Am I supposed to write the basis vectors in their canonical form first i.e.. $[1,0,0],[-1,1,0],[1,0,1]$?
Any hints or help would be much appreciated.

REPLY [2 votes]: The matrix having as columns the coefficients of the polynomials in $B$ relative to the canonical basis of $\Bbb R_2[x]$ is:
$$P = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
If $Q$ denotes the matrix having as columns the coefficients of the vectors in the dual basis of $B$, relative to the canonical basis of the dual space $\Bbb R_2^*[x]$ of $\Bbb R_2 [x]$, then:
$$Q= (P^{-1})^T$$
So, you just need to invert and transpose.
Added:
You should get:
$$Q = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$$
This means that the dual basis of $B$ is: $B^* = \{f_1, f_2,f_3\}$, where:
$$\begin{cases} f_1 = e_1^* + e_2^* - e_3^* \\ f_2 = e_2^* \\ f_3 = e_3^* \end{cases}$$
Where: $e_i^*: \Bbb R_2[x] \to \Bbb R$, $(a_1 \times 1 + a_2 \times x + a_3 \times x^2) \mapsto a_i$


TITLE: Prove that the Jaccard Similarity of bags is always less than or equal to 1/2
QUESTION [0 upvotes]: Prove that the Jaccard Similarity of bags is always less than or equal to 1/2
I know that the Jaccard Similarity of bags is $= \dfrac{|B\cap C|}{|B\cup C|}$ and if the bags are the same then the similarity $=1$
A bag is a set of elements with order unimportant and repetition allowed.
For example if we have bags $B=\{A,B,B,C,C,C,C,D,D\}$ and $C=\{A,A,B,C,C,D,D,D\}$ then the Jaccard Similarity is $=\dfrac{1+1+2+2}{9+8}=\dfrac{6}{17}$
So we count an element n times in the intersection if n is the minimum number of times the element appears in B and C and the size of the union is the sum of the size of the bags.
But how would I prove it is always $\leq \dfrac{1}{2}$

REPLY [1 votes]: Without loss of generality we can consider bags with only one kind of element. Let $b=|B|, c=|C|$. Then the formula gives $\min(b,c)/(b+c)$ but then $2\min(b,c)\leq(b+c)$. Note that the formula gives $1/2$ if the bags are equal.


TITLE: Hanging a Hammock
QUESTION [1 upvotes]: I have a hammock that I want to hang using a tree on one end and a wooden post set in a bucket of cement on the other end.
How would I determine how heavy the cement needs to be so that it would support the weight of a 250lbs person in the hammock?
Any ideas would be appreciated.

REPLY [2 votes]: ^       |\                   /
    |       |α\                 /
    |       |  \_______________/
    |       |
 H  |       |
    |  _____|_____
    |  |         |
    |  |    M    |
    V  |_________|

            <---->
               b

The figure should explain the notions (except for $T$, but never mind on it). $m$ is your mass, $M$ is the mass of the block.
$$T   \cos \alpha =  \frac{1}{2} m g$$
$$T H \sin \alpha =  M g b$$
Thus the answer for the minimum mass required for the concrete block is
$$M = \frac{1}{2} m \frac{H}{b} \tan(\alpha)$$
But this won't do since it the low boundary, so multiply it by 5-10 and bury the bucket.

To community: Maybe we have a policy on such questions?


TITLE: Diffeomorphism from level set onto $S^2$
QUESTION [5 upvotes]: I'm given a map $\phi : R^4 \rightarrow R^2$ defined by $\phi(x,y,s,t) = (x^2 + y, x^2+y^2+s^2+t^2+y)$.
It's easy to show that the level set $C = \phi^{-1}(0,1)$ is a smooth submanifold of $R^4$ with dimension $2$ by showing that $(0,1)$ is a regular value.
I'm trying to show that this level set is diffeomorphic to $S^2$, the unit sphere in $R^3$. The only thing I can think of is to construct a smooth embedding from $S^2$ into $R^4$ whose image is $C$, since smooth embeddings are diffeomorphisms onto their images.
It's easy enough to get that $C$ is diffeomorphic to $\{(x,y,z) \mid x^4 + y^2 + z^2 = 1\}$, but from there I'm stuck. The map $(x, y, z) \rightarrow (\text{sign}(x) x^2, y, z)$ is a differentiable homeomorphism from $S^2$ onto $C$, but it's not even $C^2$ let alone smooth.

REPLY [6 votes]: Once you're working with the surface $C=\{ (x,y,z) : x^4+y^2+z^2=1 \}$, how about the map
$$(x,y,z) \mapsto \frac{(x,y,z)}{ \sqrt{x^2+y^2+z^2}} ?$$
$C$ is a ball-type surface centered at the origin, so why not just project to the sphere in this simple way?


TITLE: How to calculate inertia tensor of composite shape?
QUESTION [0 upvotes]: How can I calculate inertia tensor of composite shape. I have done math of moment of inertia, but not inertia tensor.
Can anyone help me with the calculation of this shape :

REPLY [2 votes]: you can use the parallel axes theory :
$$I=I_1+I_2+I_3-m_1 \,\tilde{r}_{01}\,\tilde{r}_{01}-m_3\,\,\tilde{r}_{03}\,\tilde{r}_{03}$$
where :
$$\tilde{r}_{01}=\begin{bmatrix}
   0 & -z & 0 \\
   z & 0 & 0 \\
   0 & 0 & 0 \\
 \end{bmatrix}$$
and
$$\tilde{r}_{03}=\begin{bmatrix}
   0 & +z & 0 \\
   -z & 0 & 0 \\
   0 & 0 & 0 \\
 \end{bmatrix}$$


TITLE: True or false( invertibility of $f$)
QUESTION [0 upvotes]: Let $f$ : $(1,10)$ $\to$ $[2,11]$
be a continuous function, then , $f$ cannot be an invertible function.
True/false
Both domain and codomain has same cardinality so we can find a bijection between both sets.
But, the statement is true. What property is used in this problem?

REPLY [2 votes]: If it is a bijection then the inverse would be continuous too and range of the inverse, which is $(0,10)$ would be compact because continuous image of compact sets is compact, This is a contradiction. [A continuous bijection is strictly monotonic and this makes the inverse also monotonic and continuous].


TITLE: Coordinates for Teichmuller space for compact conformal surfaces
QUESTION [2 upvotes]: Fenchel-Nielsen coordinates give a coordinatization of Teichmuller space for compact conformal surfaces admitting a pants decomposition. But not all compact conformal surfaces (possibly with boundary, which we assume totally geodesic) of genus less than 2 admit such a decomposition, such as the cylinder. Is there an extension of Fenchel-Nielsen coordinates to these cases?

REPLY [2 votes]: As Lee points out, the only nontrivial examples are the cylinder and the torus. These admit Euclidean metric, which can be normalized to have area one, and for a cylinder we can assume that the boundary components are the same length and parallel. (in other words, metrically the cylinder has the form $a S^1 \times [0, 1/a]$). Then the Fenchel-Nielsen coordinate (there is only one) is $a,$ the length of the boundary curve. A torus can be made from a cylinder as above by gluing the two boundary components with a twist, so now we have both the length and the twist coordinate. It is pretty clear that this last construction is what gave rise to the hyperbolic version of Fenchel-Nielsen coordinates (I am not sure if this is actually written down somewhere).


TITLE: Sheaf of rings with vanishing stalk?
QUESTION [2 upvotes]: How common is that a sheaf of rings has a vanishing stalk? To define the rank of a locally free sheaf of $\mathscr{O}$-modules, for instance, $\mathscr{O}_x=0$ may cause some problem, since the rank of a free $A$-module is not well-defined if $A=0$. It would make life easier if $\mathscr{O}_x\neq0\ \forall x\in X$, but is this condition somehow incorporated in the definition of sheaf of rings?

REPLY [6 votes]: I will presume that by "a sheaf of rings" you mean "a sheaf of rings with $1$" (since this is what is usually so meant).  If the stalk of $\mathcal O_X$ vanishes at $x$, then this means that $1 = 0$ in the stalk, and hence in $\mathcal O_X(U)$ for some
neighbourhood $U$ of $x$.  Thus the stalk of $\mathcal O_X$ vanishes at $x$ if and only if the sheaf $\mathcal O_X$ restricts to the zero sheaf in some n.h of $x$.
If e.g. $X$ is not only ringed, but locally ringed, then the stalks of $\mathcal O_X$ (which are then local rings by definition) never vanish at a point (since local rings are non-zero, again by definition).
Added: If $U$ is an open subset of $X$ with complement $Z$, and $i: Z \to X$ is the
inclusion, then for any sheaf of rings $\mathcal O_Z$ on $Z$, the pushforward $i_* \mathcal O_Z$ will be a sheaf of rings on $X$ whose stalks vanish on $U$.  Thus we can always find examples realizing the discussion of the first paragraph.  More generally, if we let $Z$ be the support of any sheaf of rings $\mathcal O_X$ on $X$, this will coincide with the support of the identity section $1 \in \mathcal O_X(X)$, and hence will be a closed subset of $X$, and we will have that $\mathcal O_X = i_* i^{-1} \mathcal O_X,$ with $i^{-1}\mathcal O_X$ a sheaf of rings on $Z$, none of whose stalks vanish.  Consequently, any sheaf of rings on $X$ can be obtained by a sheaf of rings with non-vanishing stalks on a closed subspace by pushing forward.


TITLE: Subsequence and diagonal process
QUESTION [0 upvotes]: We consider a sequence of functions défined on $\mathbb R^n$ by $f_m(x)=f(\frac{x}{m}),\   \forall m\in \mathbb{N}$ such that  :
1) $f=1 $ in $B(0,1)$
2) $\mathrm{supp\,} f\subset B(0,2)$
3) $f \in C^{\infty}(\mathbb R^n)$
We  consider also a  sequence $(u_n)$ such that :
1) $u_n$ converges weakly to $0$ in $L^2(\mathbb R^n)$
2) $||u_n||_{L^2(R^n)}=1$
I want to prove by a diagonal process that we can  extract a  subsequence of $f_mu_n$ which converges strongly in $L^2(\mathbb R^n)$ to $0$ as $m$ goes to infinity.
Can someone help me?
Thanks.

REPLY [1 votes]: The result is wrong. You cannot even get pointwise convergence on a sequence $(m_k,n_k)\to (\infty,\infty)$ if, e.g., the $u_n$ are a typical highly oscillatory sequence of functions with a joint compact support. In that case you just have $f_m u_n = u_n$ for $k$ large enough and the $u_n$ do not converge pointwise (on any subsequence).


TITLE: Does the Renormalization of QFT Contradict Canonical Quantization?
QUESTION [5 upvotes]: Does the renormalization of QFT contradict canonical quantization?
In canonical quantization, you take the classical fields and canonical momenta and turn them into operators, and you require that the commutators ( or anti-commutators) equal the Poisson brackets of the original fields. Also, you need to quantize the classical Hamiltonian (of course, since classical fields commute but quantum operators don't, you need to quantize the Hamiltonian according to a well defined prescription, such as normal-ordering). The form of the quantum Hamiltonian will resemble the form of the classical one, but it will be an operator rather than a number. In particular, if a physical constant appears in the classical Hamiltonian, it will also appear in the Quantum one, and have the same value.
Now, in renormalization the constants that appear in the (bare) Lagrangian are not the physical ones, and so are different from the constants that appear in the "classical" Lagrangian (and therefore Hamiltonian). So the "bare" quantum Hamiltonian will have a similar form to the classical Hamiltonian, but the constants will be totally (and perhaps infinitely) different!
So does the quantization scheme need to be reformulated?

REPLY [3 votes]: (After thinking about the above issue, I came up with this explanation which I think is correct. I'm aware that the question itself hasn't been 100% well stated).
There is no contradiction.
Although "pedagogically" it is helpful to think about the procedure of quantization, in which the quantum theory is obtained from the classical one, it is more "natural" to think the other way around - that the classical theory emerges from the quantum one via the process of "classicization" (This is because the quantum theory is a more fundamental description of reality).
Now, quantum field theory and renormalization group flow tell us that the effective Lagrangian of a theory depends on the momentum scale of the interacting fields. In other words, coupling constants and mass depend on the momentum scale. When we do classical physics we usually use Lagrangians that correspond to low momentum scales. These Lagrangians should be thought of as obtained from the quantum theory in the process of "classicization", and the effective Lagrangian that was used in the process corresponded to the low momentum scale. If we would work in a higher momentum scale, we would "flow" to a different Lagrangian in the quantum theory, and then when "classicizating" the theory we would get a different Lagrangian.
(Of course, there is no difference between the classical and quantum Lagrangians; but the quantum physics and the classical physics that relate to these identical Lagrangians are different).
So it's true that if you use only a specific Lagrangian of a classical theory and you quantize it, then you get an Hamiltonian which has a different dependence on the physical constants than the classical one. This is because the quantum Hamiltonian has additional counterterms that "absorb" the fact that the theory behaves differently at different momentum scales. This different behaviour was forgotten in the classical theory since it was "classicized" from a specific effective Lagrangian at a specific momentum scale.


TITLE: Does any finite subset of a curve fit into an open affine?
QUESTION [0 upvotes]: Let $C$ be an irreducible and non-singular algebraic curve over an algebraically closed field $k$, and let $S \subset C$ be any finite set of points. Is it true that there is always an open affine $U \subset C$, which contains all of $S$?
If $C$ is projective, this seems to be true, see this answer. This also settles the case for quasi-projective curves.
Some context: I think this is implicitly used in Serre's Algebraic Groups and Class Fields. He wants to construct another curve $C'$, such that $C \to C'$ is the normalization of $C'$, and $S$ is the preimage of the singular points. He claims that it suffices to construct $C'$ for the affine case, and glue the affines together. However I think this only works if we find an affine which contains $S$.

REPLY [2 votes]: Every curve is quasi-projective: see Stacks 0A25, for instance. One strategy for the proof is to find an ample line bundle, take a high-enough tensor power so that it becomes very ample, and get an immersion in to $\Bbb P^r_k$. In fact, $\mathcal{O}(p)$ for $p$ a smooth point will do for our ample line bundle: $\mathcal{O}(np)$ for $n\gg 0$ will be very ample by Riemann-Roch. (I'm sure that this is recounted somewhere here on MSE or MO, but I can't find it at the moment - if you are reading this and can find the link, please put it in the comments!)


TITLE: Prove that $\Bbb N × \Bbb N$ is countable.
QUESTION [3 upvotes]: I am given this problem:
Prove that $\Bbb N × \Bbb N$ is countable by using the function $f(m, n) = 2^m3^n$ and Theorem that says any subset of a countable set is countable.
I'm not exactly sure how the function can be used with the theorem. Would I be able to use the Fundamental Theorem of Arithmetic to show that f is injective? If so, how?
Then, could I say that $f(\Bbb N × \Bbb N)$ is an infinite set of the countable set $\Bbb N$, so then it is therefore countably infinite. Thus, there exists a bijection $g$ (after showing there is an injection) from $\Bbb N × \Bbb N$ to $\Bbb N$, therefore $\Bbb N × \Bbb N$ is countably infinite?

REPLY [3 votes]: Would I be able to use the Fundamental Theorem of Arithmetic to show that $f$ is injective?

This proves itself, right? If $2^a3^b=2^c3^d$, what does the FTA say?

Then, could I say that $f(\Bbb N × \Bbb N)$ is an infinite set of the countable set $\Bbb N$, so then it is therefore countably infinite. Thus, there exists a bijection $g$ (after showing there is an injection) from $\Bbb N × \Bbb N$ to $\Bbb N$, therefore $\Bbb N × \Bbb N$ is countably infinite?

It would be clearer to say "$f(\Bbb N × \Bbb N)$ is a subset of the countable set $\Bbb N$..." , but yes, that would be the conclusion. You don't care about is infinity that much, but rather that it's a subset.


TITLE: Give a recurrence relations & base cases for the number of $n$ digit decimal strings containing an even number of $0$ digits.
QUESTION [0 upvotes]: Give a recurrence relations & base cases for the number of $n$ digit decimal strings containing an even number of $0$ digits.

My solution is:
Let $a_n$ denote the the number of $n$ digit numbers containing an even number of $0$ digits.
$a_n = (9a_{n-1}) + (10^{n-1}-a_{n-1}) = 8a_{n-1}+10^{n-1}$.
Is the solution correct? What is the initial condition?

REPLY [1 votes]: If you mean decimal strings, where leading zeros are permitted, your recurrence is good.  It would be good to explain the logic behind it.  The $9a_{n-1}$ term takes a string of length $n-1$ and appends a non-zero digit to it.  The $10^{n-1}-a_{n-1}$ term takes an unacceptable string of length $n-1$ (which must have an odd number of zeros) and appends a zero to get an acceptable string of length $n$.
The bases case is just a place to start that you compute some other way.  If you imagine computing $a_3$, you need $a_2$ to do that.  To compute $a_2,$ you need $a_1$.  It is easy to count $a_1$ by hand-there are $9$ strings of length $1$ that have an even number of zeros.  That can be your base case.  It also works to take $a_0$ as your base case.  There is one string of length $0$ that has an even number of zeros-the empty string.  It works fine, as it gives $a_1=8(1)+10^0=9$


TITLE: Calculate Wronksian of Second Order Differential Equation
QUESTION [3 upvotes]: Use variation of parameters to find a particular solution to:
$\frac{d^{2}y}{dx^{x}} + 2 \frac{dy}{dx} + y = \frac{1}{x^{4}e^{4}}.$
There are no solutions given so finding a wronskian that way is nil.
But since it is still in the order $p(x)y'' + q(x)y' + r(x)y = g(x)$ I think there is still a way to calculate a Wronskian. I have not worked with second order differential equations before and some hints/tips/help would be appreciated.

REPLY [0 votes]: Since the discriminant of the differential equation $y'' + 2y' + y = 0$ is $2^{2} - 4 = 0,$ it follows that $$u_{1} := e^{-x},\ u_{2} := xe^{-x}$$ are the basis solutions. If $x \mapsto w$ is the Wronskain of $u_{1}$ and $u_{2}$, then
$$w = u_{1}u_{2}' - u_{2}u_{1}' = e^{-2x}.$$
Let $R(x) := 1/x^{4}e^{4},$ let $t_{1} := -D^{-1}u_{2}R(x)/w,$ and let $t_{2} := D^{-1}u_{1}R(x)/w,$ where $D^{-1}$ means the primitive "operator". Then
$$t_{1} = -D^{-1}x^{-3}e^{x-4},\ t_{2} = D^{-1}x^{-4}e^{x-4}.$$ Then the particular solution $y_{1}$ is simply
$$y_{1} = t_{1}u_{1} + t_{2}u_{2}.$$
As to the underlying theorems, please simply check any book on ordinary differential equations.


TITLE: How to prove the Mantel's theorem of graph theory 's bound is best possible?
QUESTION [5 upvotes]: The theorem state that every graph of order $n$ and size greater than floor function $\lfloor \frac{n^2}{4} \rfloor$ contain a triangle.
I already know a proof of the number of the edge of graph $\leq\frac{n^2}{4}$ if it does not contains a triangle, but how to prove this bound is best possible if n is bigger than 100, in another word, how to prove there are infinite example satisfy the best bound?

REPLY [5 votes]: We can't use a smaller bound because we can show that for each natural number $n$, there exists a graph of order $n$ with exactly $\lfloor \frac{n^2}{4} \rfloor$ edges and no triangles. In response to your edit, this gives us an infinite number of examples.
So let $n$ be a natural number, and we will find a graph of order $n$ with no triangles and $\lfloor \frac{n^2}{4} \rfloor$ edges.
If $n$ is even, then $\lfloor \frac{n^2}{4} \rfloor = \frac{n^2}{4}$, and the graph $K_{n/2,n/2}$ has $\left( \frac{n}{2} \right) \left( \frac{n}{2} \right) = \frac{n^2}{4}$ edges and no triangles.
If $n$ is odd, then $n = 2m + 1$ for some integer $m$, so
\begin{align*}
    \lfloor \frac{n^2}{4} \rfloor &= \lfloor \frac{4m^2 + 4m + 1}{4} \rfloor \\
                                  &= \lfloor m^2 + m + \frac{1}{4} \rfloor \\
                                  &= m(m + 1).
\end{align*}
Then the graph $K_{m,m+1}$ has order $m + m + 1 = n$, no triangles, and has $m(m + 1) = \lfloor \frac{n^2}{4} \rfloor$ edges.


TITLE: By Kuratowski's theorem this graph seems to be planar so why is it 5 colourable considering the Four Colour theorem?
QUESTION [0 upvotes]: o
    |
 o--o--o
    |
    o

(the dashed lines represent arcs and o represents nodes)
I can't see any way this graph contains a subgraph homeomorphic to K5 or K3,3 so by Kuratowski's theorem this suggests the graph is planar.
Four Colour theorem says that the vertices of every planar graph can be coloured with at most four colours so that no two adjacent vertices receive the same colour.
So why is the graph above 5-colourable?

REPLY [3 votes]: I'm not by any means an expert in graph theory (in fact, I have little to no knowledge on the subject), but still I'll point out that you can indeed color this graph with only two colours. That's because no arc connects to any of the sorrounding ones except for the middle one, so your "leaf" vertices do not "touch" if you were to draw a map resembling this diagram.
You could also have all the vertices as drawn areas touching each others, making this diagram:

   o
 / | \
o--o--o
 \ | /
   o

and you could still colour this one with only three colours: two vertices opposing the  centre one don't actually connect.
For example, you could have green for the middle one, yellow for the top and the bottom ones, and blue for the left and right ones.
Anyway, keep in mind that the Four Colour Theorem says that you can colour it with at most four colors - if you were to be extremely conservative about the number of colours to use. But in facts, you may as well colour the map with as many colours as the number of vertices in your graph.