text:
\begin{document}

\title[Remarks on the Liechti-Strenner's examples having small dilatations]{Remarks on the Liechti-Strenner's examples having small dilatations}

\author{Ji-Young Ham and Joongul Lee}
\address{Department of Science, Hongik  University,
94 Wausan-ro, Mapo-gu, Seoul,
04066,\\
  Korea}
\email{jyham@hongik.ac.kr}

\address{Department of Mathematics Education, Hongik University,
94 Wausan-ro, Mapo-gu, Seoul,
04066\\
   Korea}
\email{jglee@hongik.ac.kr}

\subjclass[2010]{37E30, 37B40, 57M60}

\keywords{Minimal dilatation, Nonorientable surface, Liechti-Strenner, pseudo-Anosov stretch factors}

\begin{abstract}
We show that the Liechti-Strenner's example for the closed nonorientable surface in~\cite{LiechtiStrenner18} minimizes the dilatation within the class of pseudo-Anosov homeomorphisms with an orientable invariant foliation and all but the first coefficient of the characteristic polynomial of the action induced on the first cohomology nonpositive. We also show that the Liechti-Strenner's example of orientation-reversing homeomorphism for the closed orientable surface in~\cite{LiechtiStrenner18} minimizes the dilatation within the class of pseudo-Anosov homeomorphisms with an orientable invariant foliation and all but the first coefficient of the characteristic polynomial $p(x)$ of the action induced on the first cohomology nonpositive or all but the first coefficient of $p(x) (x \pm 1)^2$, $p(x) (x^2 \pm 1)$, or $p(x) (x^2 \pm x + 1)$ nonpositive.
\end{abstract}

\maketitle

\section{Introduction}

Let $\Sigma_g$ be a closed surface of finite genus.
A homeomorphism $h$ of $\Sigma_g$ is called \emph{pseudo-Anosov} if there is a pair of transversely measured foliations
$\mathcal{F}^u$ and $\mathcal{F}^s$ in $\Sigma$ and a real number $\lambda >1$ such that $h(\mathcal{F}^u)=\lambda \mathcal{F}^u$ and  $h(\mathcal{F}^s)=1/\lambda \mathcal{F}^s$~\cite{Thurston88,CassonBleiler88}.
The number $\lambda$ is called the \emph{dilatation} of $h$ and the logarithm of $\lambda$ is called the \emph{topological entropy}. The set of dilatations of pseudo-Anosov homeomorphisms of the group of isotopy classes $\Sigma_g$ is discrete~\cite{ArnouxYoccoz81, Ivanov88}. In particular, there exists the minimal dilatation.

The dilatation of a pseudo-Anosov homeomorphism of $\Sigma_g$ measures its dynamical complexity.
Furthermore, the collection of topological entropies has a geometric interpretation as the collection of Teichm\"uller distances
between Riemann surfaces of the same topological type as $\Sigma_g$~\cite{Abikoff80} if $\Sigma_g$ is orientable (as a subcollection of Teichmu\"ller distances between those of the same topological type as the orientable double cover of $\Sigma_g$ if $\Sigma_g$ is nonorientable).
In particular, the logarithm of the minimal dilatation of a genus $g$ surface gives the length of the systole for the genus $g$ moduli space if $\Sigma_g$ is orientable.

For an orientable surface $S_g$, several results have been known on the bounds of the minimal dilatation, $\delta_g$, for all pseudo-Anosov homeomorphisms of
$S_g$.
Penner gave upper and lower bounds for the dilatations of $S_g$, and
proved that as $g$ tends to infinity, the minimal dilatation tends to one (the logarithm of the minimal dilatation tends to zero on the order of
$1/g$)~\cite{Penner91}.
The upper bound was improved by Bauer for closed surfaces of genus $g\geq3$~\cite{Bauer92}.

However, the exact value of the minimal dilatation $\delta_g$ of $S_g$ has been
found only when the genus $g$ is two~\cite{ChoHam08}.

More is known for the minimal dilatation of orientation-preserving pseudo-Anosov homeomorphisms on $S_g$ with orientable invariant foliations. Denote the minimal dilatation of orientation-preserving pseudo-Anosov homeomorphisms on $S_g$ with orientable invariant foliations by $\delta^{+}(S_g)$. The following Table~\ref{table:deltaplus} shows the known values.

\begin{table}
\begin{tabular}{|c||c||l|} \hline
$g$ &  $\delta^{+}(S_g) \approx$ & Minimal polynomial of $\delta^{+}(S_g)$\\\hline
1 & $2.61803$ & $x^2-3 x+1$ \\
2 & $1.72208$ & $x^4 - x^3 - x^2 - x + 1$ \\
3 &$1.40127$ & $x^6-x^4-x^3-x^2+1$ \\
4 &$1.28064$ & $x^8-x^5-x^4-x^3+1$ \\
5 &$1.17628$ & $x^{10}+x^9-x^7-x^6-x^5-x^4-x^3+x+1=\frac{x^{12}-x^7-x^6-x^5+1}{x^2- x+1}$ \\
7 & $1.11548$ & $x^{14}+x^{13}-x^9-x^8-x^7-x^6-x^5+x+1$ \\
8 & $1.12876$ & $x^{16}-x^9-x^8-x^7+1$ \\
\hline
\end{tabular}
\caption{The known values of $\delta^{+}(S_g)$}
\label{table:deltaplus}
\end{table}

The pseudo-Anosov homeomorphisms realizing $\delta^{+}(S_g)$ in Table~\ref{table:deltaplus} were constructed by
Zhirov~\cite{Zhirov95} for $g=2$,
Lanneau and Thiffeault~\cite{LanneauThiffeault11b} for $g=3$ and $4$,
Leiniger~\cite{Leininger04} for $G=5$,
Kin and Takasawa~\cite{KinTakasawa13} and Aaber and Dunfield~\cite{AaberDunfield10} for $g=7$, and
Hironaka~\cite{Hironaka10} for $g=8$. Hironaka~\cite{Hironaka10} then showed that all of the examples above except the g = 7 example arise from the fibration of the link complement of $6_2^2$.
 From genus $6$ to genus $8$, each example identified the lower bound calculated by Lanneau and Thiffeault~\cite{LanneauThiffeault11b} as the minimal dilatation.

Recently, Liechti and Strenner~\cite{LiechtiStrenner18} determined the minimal dilatation of pseudo-Anosov homeomorphisms with orientable invariant foliations on the closed nonorientable surfaces of genus 4, 5, 6, 7, 8, 10, 12, 14, 16, 18 and 20 and the minimal dilatation of orientation-reversing pseudo-Anosov homeomorphisms with orientable invariant foliations on the closed orientable surfaces of genus 1, 3, 5, 7, 9, and 11.
 Denote by $N_g$ the closed nonorientable surface of genus $g$ and by
$\delta^{+} (N_g)$ the minimal dilatation among pseudo-Anosov homeomorphisms of $N_g$ with an orientable invariant foliation.
Denote the minimal dilatation among orientation-reversing pseudo-Anosov homeomorphisms on $S_g$ with orientable invariant foliations by $\delta^{+}_{rev}(S_g)$.  The values determined by Liechti and Strenner~\cite{LiechtiStrenner18} are in  Table~\ref{table:nonorientable} and Table~\ref{table:reversing}.

\begin{table}
\begin{tabular}{|c||c||l|} \hline
$g$ &  $\delta^{+}(N_g) \approx$ & Minimal polynomial of $\delta^{+}(N_g)$\\\hline
4 & $1.83929$ & $x^3-x^2- x-1$ \\
5 & $1.51288$ & $x^4 - x^3 - x^2 + x - 1$ \\
6 &$1.42911$ & $x^5-x^3-x^2-1$ \\
7 &$1.42198$ & $x^6-x^5-x^4-x^3+x-1$ \\
8 &$1.28845$ & $x^7-x^4-x^3-1$ \\
10 & $1.21728$ & $x^{11}-x^6-x^5-1$ \\
12 & $1.17429$ & $x^{13}-x^7-x^6-1$ \\
14 & $1.14551$ & $x^{15}-x^8-x^7-1$ \\
16 & $1.12488$ & $x^{16}-x^9-x^8-x^7+1$ \\
18 & $1.10938$ & $x^{17}-x^9-x^8-1$ \\
20 & $1.09730$ & $x^{19}-x^{10}-x^9-1$ \\
\hline
\end{tabular}
\caption{The known values of $\delta^{+}(N_g)$}
\label{table:nonorientable}
\end{table}

\begin{table}
\begin{tabular}{|c||c||l|} \hline
$g$ &  $\delta^{+}_{rev}(S_g) \approx$ & Minimal polynomial of $\delta^{+}_{rev}(S_g)$\\\hline
1 & $1.61803$ & $x^2- x-1$ \\
3 & $1.25207$ & $\frac{x^8 - x^5 - x^3- 1}{x^2+1}$ \\
5 &$1.15973$ & $\frac{x^{12} - x^7 - x^5- 1}{x^2+1}$ \\
7 &$1.11707$ & $\frac{x^{16} - x^9 - x^7- 1}{x^2+1}$ \\
9 &$1.09244$ & $\frac{x^{20} - x^{11} - x^9- 1}{x^2+1}$ \\
11 & $1.07638$ & $x^{24}-x^{13}-x^{11}-1$ \\
\hline
\end{tabular}
\caption{The known values of $\delta^{+}_{rev}(S_g)$}
\label{table:reversing}
\end{table}

The main purpose of the paper is to show that the Liechti-Strenner's example for the closed nonorientable surface in~\cite{LiechtiStrenner18} minimizes the dilatation within the class of pseudo-Anosov homeomorphisms with an orientable invariant foliation and all but the first coefficient of the characteristic polynomial of the action induced on the first cohomology nonpositive. We also show that the Liechti-Strenner's example of orientation-reversing homeomorphism for the closed orientable surface in~\cite{LiechtiStrenner18} minimizes the dilatation within the class of pseudo-Anosov homeomorphisms with an orientable invariant foliation and all but the first coefficient of the characteristic polynomial $p(x)$ of the action induced on the first cohomology nonpositive or all but the first coefficient of $p(x) (x \pm 1)^2$, $p(x) (x^2 \pm 1)$, or $p(x) (x^2 \pm x + 1)$ nonpositive.
\begin{theorem} \label{thm:N}
Denote by $N_{2k}$ the closed nonorientable surface of genus $2k$.
For all $k \geq 2$, the largest positive root of
$$ x^{2k-1}-x^k-x^{k-1}-1$$ minimizes the dilatation within the class of pseudo-Anosov homeomorphisms with an orientable invariant foliation and all but the first coefficient of the characteristic polynomial of the action induced on the first cohomology nonpositive.
\end{theorem}

\begin{theorem} \label{thm:S}
 Denote by $S_{2k-1}$ the closed orientable surface of genus $2k-1$.
For all $k \geq 2$, the largest positive root of
$$ x^{4k}-x^{2k+1}-x^{2k-1}-1$$ minimizes the dilatation within the class of pseudo-Anosov homeomorphisms with an orientable invariant foliation and all but the first coefficient of the characteristic polynomial $p(x)$ of the action induced on the first cohomology nonpositive or all but the first coefficient of $p(x) (x \pm 1)^2$, $p(x) (x^2 \pm 1)$, or $p(x) (x^2 \pm x + 1)$ nonpositive.
\end{theorem}

\section{Perron-Frobenius matrix} \label{section:perron}

\begin{definition}
Let $M=\begin{pmatrix}m_{ij}\end{pmatrix}$ and $N= \begin{pmatrix}n_{ij}\end{pmatrix}$ be two nonnegative $d \times d$ matrices. We say $M > N$
if  $m_{ij} \geq n_{ij}$ for all $i,\, j \in \{1,2,\ldots,d\}$ and strict inequality holds for at least one entry.

\end{definition}

\begin{lemma} [Perron-Frobenius] \label{lemma:Perron}
Let $T$ and $L$ be two Perron-Frobenius matrices such that $T > L$, then the spectral radius $\lambda_T$ of $T$ is strictly bigger than the spectral radius
$\lambda_L$ of $L$.
\end{lemma}

\begin{proof}
Let $t > 0$ be a left eigenvector of $T$ corresponding to $\lambda_T$ and let $l > 0$ be a right eigenvector of $L$ corresponding to $\lambda_L$. Then
$$Tl > Ll = \lambda_Ll$$
and
$$\lambda_Ttl=tTl>\lambda_Ltl.$$
Therfore $\lambda_T>\lambda_L$.
\end{proof}

\begin{proposition}
The transpose of the Frobenius companion matrix of $$x^{2k-1}-a_{2k-2}x^{2k-2} \cdots -a_{1}x-1$$
 \noindent is
$$\begin{bmatrix}
0 & 1 & 0 & \hdots & 0 \\
0 & 0 & 1 & \hdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0& 0 & 0 & \hdots 1 \\
1 & a_1 & a_2 & \hdots & a_{2k-2}
\end{bmatrix},$$
and the transpose of the Frobenius companion matrix of $$x^{4k}-a_{4k-1}x^{4k-1} \cdots -a_{1}x-1$$
\noindent is
$$\begin{bmatrix}
0 & 1 & 0 & \hdots & 0 \\
0 & 0 & 1 & \hdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0& 0 & 0 & \hdots 1 \\
1 & a_1 & a_2 & \hdots & a_{4k-1}
\end{bmatrix}.$$
\end{proposition}

\begin{proposition} \cite[Proposition 4.1]{LiechtiStrenner18} \label{pro:LS1}
Let $\psi : N_g \rightarrow N_g$ be a pseudo-Anosov map with a transversely orientable invariant foliation on the closed nonorientable surface $N_g$ of genus $g$. Then its dilatation $\lambda$ is a root of a (not necessarily irreducible) polynomial $p(x) \in \Z[x]$ with the following properties:
\begin{enumerate}
\item $\textnormal{deg}(p)=g-1$
\item $p(x)$ is monic and its constant coefficient is $\pm1$
\item The absolute values of the roots of $p(x)$ other than $\lambda$ lie in the open interval
$(\lambda^{-1},\lambda)$. In particular, $p(x)$ is not reciprocal or anti-reciprocal.
\item $p(x)$ is reciprocal mod $2$.
\end{enumerate}
\end{proposition}

\begin{proposition} \cite[Proposition 4.3]{LiechtiStrenner18} \label{pro:LS2}
Let $\psi : S_g \rightarrow S_g$ be an orientation-reversing pseudo-Anosov map with  transversely orientable invariant foliations. Then its dilatation
$\lambda$ is a root of a (not necessarily irreducible) polynomial $p(x) \in \Z[x]$ with the following properties:
\begin{enumerate}
\item $\textnormal{deg}(p)=2g$
\item $p(x)$ is monic and its constant coefficient is $(-1)^g$
\item $p(x)=(-1)^g x^{2g} p(-x^{-1})$
\item The absolute values of the roots of $p(x)$ other than $\lambda$ and $-\lambda^{-1}$ lie in the open interval $(\lambda^{-1},\lambda)$.
\end{enumerate}
\end{proposition}

\begin{lemma} \label{lemma:DM}
Let $A$ be the adjacency matrix for a graph $G$ with $m$ vertices.
Then $A$ is primitive if and only if G is strongly connected and the gcd of the lengths of the loops in $G(A)$ is one.
\end{lemma}

\begin{proof}
See~\cite[Chapter 6, Section 1, Remarks, 6, 8]{DulmageMendelsohn67}.
\end{proof}

\begin{lemma} \label{lemma:mod2}
\begin{enumerate}
\item Let $\psi : N_{2k} \rightarrow N_{2k}$ be a pseudo-Anosov map with a transversely orientable invariant foliation on the closed nonorientable surface $N_{2k}$ of genus $2k$. Let $x^{2k-1}-a_{2k-2}x^{2k-2} \cdots -a_{1}x-1$ be the characteristic polynomial of the action of $\psi$ induced on first cohomology of $N_{2k}$,
then $a_{i} \equiv a_{2k-1-i}$ mod $2$ and at least for one $a_{i}$ with $gcd(2k-1,2k-1-i)=1$, $a_{i} \neq 0$ for $1 \leq i \leq k-1$.
\item Let $\psi : S_{2k-1} \rightarrow S_{2k-1}$ be an orientation-reversing pseudo-Anosov map with  transversely orientable invariant foliations.  If $p(x)=x^{4k-2}-a_{4k-3}x^{4k-3} \cdots -a_{1}x-1$ be the characteristic polynomial of the action of $\psi$ induced on first cohomology of $S_{2k-1}$,
then $a_{i} \equiv a_{4 k-2-i}$ mod $2$ and at least for one $a_{i}$ with $gcd(4k-2,4k-2-i)=1$, $a_{i} \neq 0$ for $1 \leq i \leq 2 k-1$.
\item Let $\psi : S_{2k-1} \rightarrow S_{2k-1}$ be an orientation-reversing pseudo-Anosov map with  transversely orientable invariant foliations. Let $p(x)$ be the characteristic polynomial of the action of $\psi$ induced on first cohomology of $S_{2k-1}$. If $ x^{4k}-a_{4k-1}x^{4k-1} \cdots -a_{1}x-1$ is one of $p(x) (x \pm 1)^2$, $p(x) (x^2 \pm 1)$, or $p(x) (x^2 \pm x + 1)$, then $a_{i} \equiv a_{4 k-i}$ mod $2$ and at least for one $a_{i}$ with $gcd(4k,4k-i)=1$, $a_{i} \neq 0$ for $1 \leq i \leq 2 k-1$.
\end{enumerate}
\end{lemma}

\begin{proof}
\begin{enumerate}
\item[]
\item[] (1) can be obtained from Proposition~\ref{pro:LS1} (1) and (4) and Lemma~\ref{lemma:DM}.

\item[] (2) can be obtained from Proposition~\ref{pro:LS2} (1) and (3) and Lemma~\ref{lemma:DM}.

\item[] Note that $p(x) (x \pm 1)^2$, $p(x) (x^2 \pm 1)$, or $p(x) (x^2 \pm x + 1)$ is either $p(x) (x^2 + 1)$, or $p(x) (x^2 + x + 1)$ mod $2$. Hence,
\item[] (3) can be obtained from (2) and Lemma~\ref{lemma:DM}.
\end{enumerate}
\end{proof}

By a spectral radius of a polynomial, we mean the spectral radius of the Frobenius companion matrix of it.

\begin{lemma} \label{lemma:minispec}
Let $n \geq 2$.
$$ x^{2n}-x^{n+1}-x^{n-1}-1$$ gives the minimum spectral radius among the polynomials $x^{2n}-a_{2n-1}x^{2n-1} \cdots -a_{1}x-1$ with $a_{i} \equiv a_{2 n-i}=1$
for only one $i$ ($1 \leq i \leq  n-1$) and $a_{j}=0$ for all $j$ with $j \neq i$ and $j \neq 2 n-i$.
\end{lemma}

\begin{proof}
Let $g(x)$ be $g(x)=x^{2n}-x^{2n-i}-x^{i}-1$. Then $g(1)=-2 <0$ and $g(2)=2^{2n}-2^{2n-i}-2^{i}-1=(2^{2n-i}-1)(2^{i}-1)-2>0$.
$g^{\prime}(x)=(2 n) x^{2n-1}-(2n-i) x^{2n-i-1}-i x^{i-1}$. $g^{\prime}(1)=2n-(2n-i) -i=0$ and
$g^{\prime}(x)=x^{2 n-1} \left((2n) -(2n-i-1) x^{-i}-i x^{-2n+i}\right) \geq 0$ for $x \geq1$.
Hence the largest root of $g(x)$ lies between $1$ and $2$ and it is the only root bigger than $1$.
We can regard $g$ as a two variable polynomial $G(x,i)$.
$$\frac{\partial G}{\partial i}=x^i \ln{x} \left(x^{2 n-2 i}-1\right) \geq 0$$ if $x \geq 1$.
Hence, $$x^{2n}-x^{2n+1}-x^{2n-1}-1$$ gives the minimum spectral  cosidering Lemma~\ref{lemma:Perron}.
\end{proof}

\section{Proof of Theorem~\ref{thm:N} and Theorem~\ref{thm:S}}
Given a pseudo-Anosov homeomorphism $\psi$, let $\psi^{*}$ be the induced action of $\psi$ on the first cohomology and
$p_{\psi}(x)$ be the characteristic polynomial of $\psi^{*}$.

\subsection{Proof of Theorem~\ref{thm:N}}

Let $\Psi$ be the following set:

$$\begin{array}{ll}
 \Psi = &
\! \! \! \begin{Bmatrix}
 \psi \, : &\! \! \! \text{$\psi$  is a pseudo-Anosov homeomorphism of $N_{2k}$ with an orientable invariant }   \! \! \\
 &\! \! \!  \text{foliation and $p_{\psi}(x)$ has nonpositive coefficients except the first one \qquad \qquad    } \! \! \! \\
\end{Bmatrix}.
\end{array}$$

Let $\Lambda$ be the following set:
$$\Lambda=\{\lambda \, : \, \lambda \text{ is the dilatation of a pseudo-Anosov homeomorphism $\psi$, } \psi \in \Psi \}.$$

Let $\Gc$ be the following set:

$$\begin{array}{ll}
 \Gc = &
 \begin{Bmatrix}
  p_{\psi}(x) \, : \, \psi \in \Psi\\
 \end{Bmatrix}.
\end{array}$$

Note that $$x^{2k-1}-x^k-x^{k-1}-1$$ is in $\Gc$ by~\cite[Proposition 2.6]{LiechtiStrenner18}.
 Since any polynomial in $\Gc$ can be considered as the characteristic polynomial of nonnegative Frobenius companion matrix which is also primitive,
the largest root of $$x^{2k-1}-x^k-x^{k-1}-1$$ is the minimal dilation among $\lambda$'s in $\Lambda$
by Lemma~\ref{lemma:mod2}  (1), Lemma~\ref{lemma:minispec}, and Lemma~\ref{lemma:Perron}.

\subsection{Proof of Theorem~\ref{thm:S}}
Let $\Psi$ be the following set:

$$\begin{array}{ll}
 \Psi = &
 \! \! \! \begin{Bmatrix}
  &\! \! \! \text{$\psi$  is a pseudo-Anosov homeomorphism of $S_{2k-1}$, which is orientation} \\
\psi \, : &\! \! \!  \text{ reversing with orientable  invariant foliations and either $p_{\psi}(x)$ or one}\\
  &\! \! \!  \text{ of $p(x) (x \pm 1)^2$, $p(x) (x^2 \pm 1)$, or $p(x) (x^2 \pm x + 1)$ has nonpositive \quad \ }\\
  &\! \! \!  \text{coefficients except the first one  \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad }\\
 \end{Bmatrix}.
\end{array}$$

Let $\Lambda$ be the following set:
$$\Lambda=\{\lambda \, : \, \lambda \text{ is the dilatation of a pseudo-Anosov homeomorphism $\psi$, } \psi \in \Psi \}.$$

Let $\Gc$ be the following set:

$$\begin{array}{lll}
 \Gc = &
\! \! \!  \begin{Bmatrix}
& \! \! \!  \text{g(x) has nonpositive coefficients except the first one and } g(x) \text{ is }\\
 g(x) \, : \,  &\! \! \!  \text{ one of } p_{\psi}(x),\ p_{\psi}(x) (x \pm 1)^2,\ p_{\psi}(x) (x^2 \pm 1), \text{ or } p_{\psi}(x) (x^2 \pm x + 1), \\
&\! \! \!  \psi \in \Psi  \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \quad\\
 \end{Bmatrix}.
\end{array}$$

Note that $$x^{4k}-x^{2k+1}-x^{2k-1}-1$$ is in $\Gc$ by~\cite[Proposition 3.3]{LiechtiStrenner18}. Observe that
the largest positive root of $$x^{4k}-x^{2k+1}-x^{2k-1}-1$$ is less than the largest positive root of $$x^{4k-2}-x^{2k}-x^{2k-2}-1.$$
Since any polynomial in $\Gc$ can be considered as the characteristic polynomial of nonnegative Frobenius companion matrix which is also primitive, the largest positive root of
$$x^{4k}-x^{2k+1}-x^{2k-1}-1$$ is the minimal dilation among $\lambda$'s in $\Lambda$ by Lemma~\ref{lemma:mod2} (2) \& (3), Lemma~\ref{lemma:minispec}, and Lemma~\ref{lemma:Perron}.

\section{acknowledgement}
We thank Erwan Lanneau, Livio Liechti, Julien Marché, Alan Reid, Darren Long and anonymous referees. This work was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. NRF-2018R1A2B6005847). The second author was supported by 2018 Hongik University Research Fund.

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TITLE: Proving prime p exists within $n+1\le p \le n!+1$
QUESTION [1 upvotes]: The title is another homework problem I got from graduate number theory. My friend gave me this hint:
Since any two consecutive integers are relatively prime and n! and n!+1 are integers, n! and n!+1 are relatively prime.  Therefore, there is at least one prime that divides n!, so there is a prime p where p < n!.  Moreover, if p ≤ n, p would divide n!, which is a contradiction.  Therefore, n < p ≤ n! or n+1 ≤ p ≤ n!+1.
But I am still lost. Any help would be appreciated and thanks in advance for all your time.

REPLY [1 votes]: The argument is a special case of the idea used in Euclid's classical proof that there are infinitely many primes. Namely, note that $\,m = 1+a_1\cdots a_n\,$ is coprime to $\,k = a_1\cdots a_n\,$ since any common factor divides their difference $\,m-k= 1.\,$ Therefore $\,m\,$ and $\,k\,$ have no common prime factors. In particular, if $\,a_i = i,\,$ so $\, k = n!,\,$ the prime factors of $\,1+n!\,$ are $> n,\,$ by prime $\,p \le n\,\Rightarrow\,p\mid n!.$


TITLE: Cubic modular equation
QUESTION [2 upvotes]: If I found the sol of (a), I could find sol of (b) and (c) by using Hensel's lemma.
I want to know the way of solving (a) without testing all the cases (ex testing 1,2,3,4,...10)
I think that there will be other way not testing all residues. It takes such a long time...
Or is there short cut for testing..?

REPLY [1 votes]: Some ideas to solve this:

Use the Chinese Remainder Theorem. Seems pretty applicable. https://en.wikipedia.org/wiki/Chinese_remainder_theorem
Notice that $11^3=1331$. So, if you can find a solution to the $x^3+8x^2-x-1\equiv 0\pmod{1331}$, then we know $1331\mid x^3+8x^2-x-1$ and since $11\mid 121$ and $121\mid 1331$, then $11\mid x^3+8x^2-x-1$ and $121\mid x^3+8x^2-x-1$ also. So, solutions to the third equation will also be solutions to the first two.


TITLE: Decomposing a "nonnegative" polynomial into a canonical form
QUESTION [0 upvotes]: EDIT: This question is related to my research, but seemed too elementary to pose at MathOverflow, so I posted here.
If $p \in \mathbb{R}[x]$, then $p(x) \ge 0$, $\forall x \ge 0$ if and only if $p(x) = \left([f_1(x)]^2 +[f_2(x)]^2\right) + x\left([g_1(x)]^2 + [g_2(x)]^2\right)$ (this result is known as the Pólya–Szegö theorem).
Consider the polynomial $p(x) = 4x^3 -2x+1$, whose graph is given by:

Clearly, $p(x) \ge 0$, $\forall x \ge 0$. Of course, this can be proven rigorously by showing that $p$ has a local minimum at $x = \sqrt{6}/6$. It seems that the only way to write $p$ in the canonical form above is $p(x) = x(4x^2 -2) + 1 = x\left((2x)^2 + (\sqrt{2}i)^2\right) + (1^2 + 0^2)$.

Question: Is there a way to write $p$ in the canonical form above without complex numbers?

REPLY [1 votes]: Your misconception seems to arise from thinking that the split of $ p(x) = (f_1 ^2 + f_2^2 ) + x (g_1 ^2 + g_2^2 ) $ is a split of even/odd degree, which is why you wrote
$$p (x) = x ( 4x^2 - 2 ) + 1 = \ldots $$
However, that need not be the case. For example, we could have
$$ p(x) = x ( 4x^2 - x) + (x^2 + 2x + 1) = \ldots $$
Here's a way to write it using only polynomials with real coefficients (though you really should give it a try by yourself first):

 $p(x) = x [ (2x-1)^2  + 1^2 ] + [(2x-1)^2 + 0^2 ] $


TITLE: Sum and difference equal to squares
QUESTION [0 upvotes]: Can anyone help me to find the integer solutions of these equations?
\begin{align}
(w^2-r^2)+(s^2-k^2)&=t_1^2\\
(w^2-r^2)-(s^2-k^2)&=t_2^2
\end{align}
It can be written as
\begin{align}
a+b&=t_1^2 \\
a-b&=t_2^2
\end{align}

REPLY [0 votes]: I am assuming that you are looking for some kind of a parametric solution. I can get you half-way there and give you parametric solution for $t_1$, $t_2$, $w$ and $r$.  Imposing the condition for $s$ and $k$ is harder and I have to think a bit. Hope that getting you this far helps.
Clearly
$$
t_1^2-t_2^2 = w^2 - r^2
$$
Factoring and rearranging (standard trick to convert quadratic to linear)
$$
\frac{t_1-t_2}{w-r} = \frac{t_1+t_2}{w+r}
$$
Call the common ratio $u/v$. This gives 4 linear equations in 4 unknowns
$$
\begin{align}
t_1-t_2 &= \alpha u &
w-r &= \alpha v\\
w+r &= \beta u & t_1+t_2 &= \beta v
\end{align}$$
You can solve for $t_1$, etc. and impose parity conditions to ensure that when dividing by $2$, the numerator is even.
$$\begin{align}
t_1 &= \frac{\alpha u + \beta v}{2} & t_2 &= \frac{\beta v - \alpha u}{2} \\
w &= \frac{\alpha v + \beta u}{2} & t_2 &= \frac{\beta u - \alpha v}{2} \\
\end{align}$$
Substitute in the original to get a condition for $s^2-k^2$ and proceed as
$$
s^2-k^2 =\left(\frac{\beta v - \alpha u}{4}\right)^2$$
Rest of the calculation is same, i.e. parametrize
$$
4 \frac{s-k}{\beta v - \alpha u} = \frac{\beta v - \alpha u}{(s+k)} $$
Only tricky part is you will have two equations for $\beta v$ and $\alpha u$ and you have to reconcile those. I will leave it up to you.


TITLE: Let $a$ be a three digit integer number with digits $x; y; z $(in that order). Prove that $a$ is divisible by 9...
QUESTION [0 upvotes]: Let $a$ be a three digit integer number with digits $x; y; z $(in that order). Prove that $a$ is divisible by 9 if and only if $x + y + z$ is divisible by 9.
Following a proof of this:
Let $a; b; d; k$ be integers such that $a = dk + b$. Prove that $a$ is divisible by $d$ if and
only if $b$ is divisible by $d$.

REPLY [0 votes]: Notice that the number $xyz$ equals $$100x+10y+z = 9(11x+y)+(x+y+z),$$
which is clearly divisible by $9$ if and only if $x+y+z$ is.


TITLE: Exact value of the limit of the Mandelbrot iteration for c=1/4?
QUESTION [2 upvotes]: I know and understand the proofs, that c $\to c^2+c \to(c^2+c)^2+c \to ...$ converges, if $0<=c<=\frac{1}{4}$. But I wonder: is the exact value (a closed expression) for the limit known for the case $c=\frac{1}{4}$?
Edit, clarification :
  I thought of a squeeze-like argument. Or of the usage of techniques as used to show the divergence of the harmonic series. So, that one (in the special case of $c=\frac{1}{4}$) sees, that the limit is both at least and at most $\frac{1}{2}$ (without refering to fixpoints).

REPLY [1 votes]: I find the value c = .25 instructive. z converges to .5 but only comparatively slowly. After ten iterations of $ f(z) = z^2 + c $ from z = 0 we reach z = .4305. For some c I find it impossible to determine whether z converges to a point or not because of  slowness. "Moving slowly" defined by 0 < modulus (z - f(z)) < 0.01 is easy to calculate. I find the interesting observation that no z with c on the boundary of the Mandelbrot set are moving slowly after an arbitrary number of iterations, apart from c=.25  and c = -2 where z stops at +2.
If all of the z moving slowly are converging to a point, this implies that if z is cycling it never moves slowly, which does indeed seem to be the case. Moving slowly is an efficient way of testing whether z is converging to a point.


TITLE: Proof by induction: $(a+b)^n=a^n+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+...+nab^{n-1}+b^n$
QUESTION [1 upvotes]: I'm studying in preparation for a Mathematical Analysis I examination and I'm solving past exam exercises.
If it's any indicator of difficulty, the exercise is Exercise 1 of 4, part $a$ and graded for 8%, so it's supposed to be pretty easy.

Use induction to prove that $(a+b)^n=a^n+na^{n-1}b+\dfrac{n(n-1)}{2!}a^{n-2}b^2+\dfrac{n(n-1)(n-2)}{3!}a^{n-3}b^3+...+nab^{n-1}+b^n$ for any $a,b\in\mathbb{R}$ and $n\in\mathbb{N}$

I've got the three steps in mind, but I'm failing at the third.
It's obviously true for n=1: $(a+b)^1=a^1+1a^0b$
Let it be true for n=k: $(a+b)^k=a^k+ka^{n
k-1}b+\dfrac{k(k-1)}{2!}a^{k-2}b^2+\dfrac{k(k-1)(k-2)}{3!}a^{k-3}b^3+...+kab^{k-1}+b^k$
The third step is obviously the problematic one for me. By expanding the formula for n=k+1, I fail to produce something recognisable that will lead me to something valid like $(a+b)(a+b)^k$
Is there something I'm missing?

REPLY [2 votes]: Hint: In the third step, try multiplying the RHS of what you got in the second step with $(a+b)$.
That will create two distinct series of sums, say $S_1$ and $S_2$, where
$$S_1=a(a^k+ka^{k-1}b+\cdots+kab^{k-1}+b^k)=a^{k+1}+ka^kb+\cdots+ka^2b^{k-1}+ab^k$$
$$S_2=b(a^k+ka^{k-1}b+\cdots+kab^{k-1}+b^k)=a^kb+ka^{k-1}b^2+\cdots+kab^k+b^{k+1}$$
If you leave out the first term in $S_1$ and last term in $S_2$, then its just a matter of using
$${x \choose y}+{x \choose y-1}={x+1 \choose y}$$
which can be easily proved.


TITLE: Are the sums equal to each other?
QUESTION [0 upvotes]: They are $2$ different results for the integral
$$\int xe^{2x}\sin\left(\frac x3\right)\,dx$$

$\displaystyle\frac{-3}{1369}e^{2x}\left(3(35-74x)\sin\left(\frac x3\right)+(37x-36)\cos\left(\frac x3\right) \right)$
$\displaystyle\frac{3}{37}e^{2x}\left(x\left(6\sin\left(\frac x3\right)-\cos\left(\frac x3\right) \right) -\frac{3}{37}\left(35\sin\left(\frac x3\right)-13\cos\left(\frac x3\right)  \right)\right) $

Are the sums equal to each other?

REPLY [2 votes]: @Samir;
Set them equal to each other and divide both sides by e^(2x).
$ -\frac{3 \left((-36+37 x) \cos \left(\frac{x}{3}\right)+3 (35-74 x) \sin \left(\frac{x}{3}\right)\right)}{1369}=\frac{3}{37} \left(x \left(-\cos \left(\frac{x}{3}\right)+6 \sin \left(\frac{x}{3}\right)\right)-\frac{3}{37} \left(-12 \cos \left(\frac{x}{3}\right)+35 \sin \left(\frac{x}{3}\right)\right)\right) $
Multiply both sides by 1369.
$-3 \left((-36+37 x) \cos \left(\frac{x}{3}\right)+3 (35-74 x) \sin \left(\frac{x}{3}\right)\right)=111 \left(x \left(-\cos \left(\frac{x}{3}\right)+6 \sin \left(\frac{x}{3}\right)\right)-\frac{3}{37} \left(-12 \cos \left(\frac{x}{3}\right)+35 \sin \left(\frac{x}{3}\right)\right)\right) $
Divide both sides by 3.
$ -(-36+37 x) \cos \left(\frac{x}{3}\right)-3 (35-74 x) \sin \left(\frac{x}{3}\right)=37 \left(x \left(-\cos \left(\frac{x}{3}\right)+6 \sin \left(\frac{x}{3}\right)\right)-\frac{3}{37} \left(-12 \cos \left(\frac{x}{3}\right)+35 \sin \left(\frac{x}{3}\right)\right)\right) $
Start cleaning up.
$ (36-37 x) \cos \left(\frac{x}{3}\right)-(105-222 x) \sin \left(\frac{x}{3}\right)=37 \left(x \left(-\cos \left(\frac{x}{3}\right)+6 \sin \left(\frac{x}{3}\right)\right)-\frac{3}{37} \left(-12 \cos \left(\frac{x}{3}\right)+35 \sin \left(\frac{x}{3}\right)\right)\right) $
$ (36-37 x) \cos \left(\frac{x}{3}\right)-(105-222 x) \sin \left(\frac{x}{3}\right)=37 \left(-x \cos \left(\frac{x}{3}\right)+6 x \sin \left(\frac{x}{3}\right)-\frac{3}{37} \left(-12 \cos \left(\frac{x}{3}\right)+35 \sin \left(\frac{x}{3}\right)\right)\right) $
$ (36-37 x) \cos \left(\frac{x}{3}\right)-(105-222 x) \sin \left(\frac{x}{3}\right)=-37 x \cos \left(\frac{x}{3}\right)+222 x \sin \left(\frac{x}{3}\right)-3 \left(-12 \cos \left(\frac{x}{3}\right)+35 \sin \left(\frac{x}{3}\right)\right) $
$ (36-37 x) \cos \left(\frac{x}{3}\right)-(105-222 x) \sin \left(\frac{x}{3}\right)=36 \cos \left(\frac{x}{3}\right)-37 x \cos \left(\frac{x}{3}\right)-105 \sin \left(\frac{x}{3}\right)+222 x \sin \left(\frac{x}{3}\right) $
$ 36 \cos \left(\frac{x}{3}\right)-37 x \cos \left(\frac{x}{3}\right)-(105-222 x) \sin \left(\frac{x}{3}\right)=36 \cos \left(\frac{x}{3}\right)-37 x \cos \left(\frac{x}{3}\right)-105 \sin \left(\frac{x}{3}\right)+222 x \sin \left(\frac{x}{3}\right) $
Easy to see now that they are equal.