text:
\begin{document}
\title[Blow up for 2D Euler Equation]
{Blow up for the 2D Euler Equation \\ on Some Bounded Domains}

\author{Alexander Kiselev}
\address{\noindent Department of Mathematics \\ Rice University \\
Houston, TX 77005, USA \newline Email: kiselev@rice.edu}

\author{Andrej Zlato\v s}
\address{\noindent Department of Mathematics \\ University of
Wisconsin \\ Madison, WI 53706, USA \newline Email:
zlatos@math.wisc.edu}

\begin{abstract}
We find a smooth solution of the 2D Euler equation on a bounded domain which exists and is unique in a natural class locally in time, but blows up in finite time in the sense of its vorticity losing continuity. The domain's boundary is smooth except at two points, which are interior cusps.
\end{abstract}

\maketitle

\section{Introduction} \lb{S1}

Consider the 2D Euler equation on a bounded domain $D\subseteq\bbR^2$, in vorticity formulation:
\beq\lb{1.1}
\omega_t+u\cdot\nabla\omega = 0, \qquad \omega(0,\cdot)=\omega_0.
\eeq
The incompressible velocity $u$ satisfies the no-flow condition $u\cdot n=0$ on $\partial D$ (with $n$ the unit outer normal to $\partial D$), and is found from the vorticity $\omega$ via
the Biot-Savart law
\beq\lb{1.2}
u(t,x)= \int_D K_D(x,y) \omega(t,y)dy,
\eeq
where the Kernel $K_D$ is given by $K_D(x,y):=\nabla^\perp G_D(x,y)$, with $\nabla^\perp=(-\partial_{x_2},  \partial_{x_1})$ and $G_D\ge 0$ the Green's function for $D$.

If the domain $D$ and the initial data $\omega_0$ are smooth, results on existence, uniqueness, and global regularity of solutions of the
2D Euler equation go back to the 1930s works of Wolibner \cite{Wol} and H\"older \cite{Hold}. Even though the solutions in this case

are globally regular, infinite-time growth of their derivatives and small scale creation appear to be ubiquitous in two dimensional fluids.

The best known upper bound on the growth of the gradient of the vorticity is double-exponential in time (see \cite{YudDE}, even though the result is also implicit in earlier works).
The question of how fast it can actually grow has received much attention in the literature. First results of this type were due
to Yudovich \cite{Yud1,Yud2}, who constructed examples with unbounded growth of the vorticity gradient on the boundary of the domain. Recently, such growth at
the boundary was shown to be fairly generic in \cite{MSY} and \cite{Koch}. Better quantitative estimates were obtained by Nadirashvili \cite{Nad1}, who built
examples with at least  linear growth, and Denisov \cite{Den1}, whose construction gives a super-linear lower bound for the vorticity gradient. Denisov also constructed
examples with a double-exponential growth for an arbitrary but finite time \cite{Den2}. In these examples, growth happens
in the bulk of the fluid. Very recently, Kiselev and \v Sver\' ak \cite{KS} constructed examples with a double-exponential in time growth of the vorticity gradient
on the boundary, which is of course  the fastest possible rate of growth. Whether such growth can happen in the bulk remains open. The best result here so far is due to Zlato\v s \cite{Zlatos}, who built examples with at least exponential in time growth.

The 2D Euler equation in less regular domains and with less regular initial data has also been  studied extensively.
Yudovich theory \cite{Yudth} (see also \cite{MB,MP}) guarantees existence and uniqueness
of a bounded weak solution $\omega$ if the domain is $C^\infty$ and the initial data $\omega_0$ is $L^\infty.$
Uniqueness comes from the fact that the fluid velocity corresponding to a bounded vorticity $\omega$ is log-Lipshitz.
This makes bounded vorticities a very natural class  to consider (although incremental generalizations are possible).
The smoothness assumptions on the boundary of the domain can also be weakened, and Yudovich theory can be extended to domains which are only $C^{1,1}$ \cite{GL}.
Moreover, if the initial data and domain have a higher regularity, the solution inherits it as well \cite{MP}.

The 2D Euler equation has also been studied on certain types of domains with singular points, including domains with corners.
We refer to recent works \cite{GL,Lacave,LMW} for some results on
existence, uniqueness, and regularity of solutions, as well as for further references.

Our main purpose in this  note is to provide an example showing that if the boundary of the domain is not {\it everywhere} sufficiently regular,
then much more dramatic effects than just a fast growth of the derivatives of solutions can take place.
We show that there are solutions of the 2D Euler equation on domains with two cusps (and smooth elsewhere)
which are smooth locally in time but blow up in finite time, in the sense of the vorticity losing continuity. We note that our construction will
rely only on the above existence and uniqueness results for smooth domains and initial data, due to a special structure of our example.

The example involves a ``double stadium" domain. Let
\[

\tilde{D}_r^\pm:= \left[ (-r,r)\times(\pm1-r,\pm1+r) \right] \cup B_r(-r,\pm1) \cup B_r(r,\pm1)

\]
be the ``stadium'' domain in $\bbR^2$ with width $r>0$ and centered at $(0,\pm 1)$.
While $\tilde{D}_r^\pm$ are defined explicitly, they are only $C^{1,1}.$
Let us therefore define domains $D_r^\pm$ by making the connections between the circular arcs and the horizontal intervals (which form $\partial \tilde D_r^\pm$) infinitely smooth. This can be done by adjusting the circular arcs
near the points of contact with the intervals, and such an adjustment can be made as small as we want in $C^1.$
Let us do this so that  again all the $D_r^\pm$ are rescaled copies of each other, centered at $(0,\pm 1)$.
Finally, let
\[

D':=D_1^+\cup D_1^- \cup \left[ (-1,1)\times\{0\} \right]

\]
and denote $D:=D_1^+$ the upper half of $D'$.

If $\omega_0\in L^\infty(D')$ is odd in $x_2$, then the Yudovich theory shows that there is a unique odd-in-$x_2$ weak solution $\omega$ to \eqref{1.1} on $D'\times\bbR$, namely the one whose restriction to $D\times\bbR$ solves \eqref{1.1} there. Moreover, if $\omega_0 \in C(D'),$ then  it follows from Yudovich theory that $\omega(t,\cdot)\in C(D)$

 for all $t>0.$
The following theorem provides a necessary and sufficient condition for $\omega$ to remain continuous on the whole domain $D'.$

\begin{theorem}\lb{T.1.1}
Let $\omega_0\in C(\bar D')$ be non-negative on $D$ and odd in $x_2$.  Then the unique odd-in-$x_2$ solution $\omega$ to \eqref{1.1} on $D'\times\bbR$ is continuous for all $t>0$ if and only if $\omega_0(\partial D)=0$.
\end{theorem}

\it Remark. \rm If in addition we have $\omega_0 \in C^\infty (\bar D),$ then  classical regularity theory gives us $\omega(t,\cdot)\in C^\infty (D)$ for all $t>0$.
If furthermore $\omega_0$ is supported away from $[-1,1]\times\{0\} $, then finite speed of propagation (i.e., bounded $u$) implies $\omega(t,\cdot)\in C^\infty(D')$ for all small enough $t>0$. By Theorem~\ref{T.1.1}, such a solution will become discontinuous in finite time unless $\omega_0$  vanishes on all of $\partial D.$

\section{Proof of Theorem \ref{T.1.1}} \lb{S2}

If $\omega_0(\partial D)=0$, then the result follows from Yudovich theory because both $D$ and $\partial D$ are invariant under the flow map generated by $u$.

Let us now assume that $\omega_0(x_0)>0$ for some $x_0\in\partial D$.

We will show that if $X'(t;x)=u(t,X(t;x))$ and $X(0;x)=x$ for $x\in\bar D$, then there is $T>0$ such that $X(T;x_0)=0$.  This, the continuity of $\omega_0$ and of $X(t;\cdot)$  on $\bar D$, and oddness of $\omega$ in $x_2$ then show that $\omega(T,\cdot)$ is discontinuous at the origin.

Let $\delta>0$ be such that  $|\omega_0^{-1}([\delta,\infty))|\ge 30\delta\, (> (4+2\pi)2\delta+\delta)$.  Because $u$ is incompressible and $\omega(t,\cdot)\ge 0$ on $D$, it follows that for each $t\ge 0$,
\beq\lb{2.1}
|\{ y\in D_{1-2\delta}^+ \,|\, \omega(t,y)\ge\delta \}| \ge \delta.
\eeq

Next let
\[
\kappa:=\inf_{x\in \partial D_{1-\delta}^+ \,\&\, y\in \bar D_{1-2\delta}^+} G_D(x,y).
\]
Continuity of $G_D$ away from the diagonal $x=y$, positivity of $G_D$ on $D$, and compactness of $\bar D_{r}^+$ show that $\kappa>0$.  Let $v>0$ solve $\Delta v=0$ on $D\setminus \bar D_{1-\delta}^+$ with $v=0$ on $\partial D$ and $v=\kappa$ on $\partial D_{1-\delta}^+$.  By the Hopf Lemma it follows that $\eps:=\inf_{x\in\partial D} |\nabla v(x)|>0$.

 We then have $G_D(x,y)\ge v(x)$ for $x\in D\setminus \bar D_{1-\delta}^+$ and $y\in  \bar D_{1-2\delta}^+$, because the inequality holds for $x\in \partial(D\setminus \bar D_{1-\delta}^+)$ (and both functions are harmonic in $x$).  So we have $|K_D(x,y)|\ge\eps$ for $x\in\partial D$ and $y\in \bar D_{1-2\delta}^+$.  This and \eqref{2.1} show that for $x\in \partial D$,  the right-hand side of \eqref{1.2} equals $a(t,x)(n_2(x),-n_1(x))$, with  $a(t,x)\ge \eps\delta^2$.  Thus $X(t;x_0)$ travels clockwise along $\partial D$ with velocity $\ge \eps\delta^2$, which means that $X(T;x_0)=0$ for some $T>0$.  The proof is finished.

\medskip

{\bf Acknowledgements.}
AK acknowledges partial support by NSF  grants DMS-1104415 and DMS-1159133, and by a Guggenheim Fellowship.
AZ acknowledges partial support  by NSF grants DMS-1056327  and DMS-1159133.

config: arxiv


TITLE: 2D Laplace Equation with Dirichlet Condition
QUESTION [0 upvotes]: I'm having trouble following the solution for this question. Firstly (underlined in orange) why is the general solution of this form. I would have thought it would be the more general $A(w)e^{wy}+B(w)e^{-wy}$
And for the green part I don't really follow it all.

REPLY [1 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$$
\Phi\pars{x,y}=\int\tilde{\Phi}\pars{\vec{k}}\expo{\ic\vec{k}\cdot\vec{r}}\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}\quad\imp\quad
\int\tilde{\Phi}\pars{\vec{k}}\pars{-k_{x}^{2} -k_{y}^{2}}
\expo{\ic\vec{k}\cdot\vec{r}}\,{\dd^{2}\vec{k} \over \pars{2\pi}^{2}}
=0
$$
$\ds{\imp\quad k_{y} = \pm\verts{k_{x}}\ic}$

$$
\Phi\pars{x,y}=\int_{-\infty}^{\infty}{\rm B}\pars{k_{x}}
\exp\pars{\ic k_{x}x \color{#c00000}{\Large -} \verts{k_{x}}y}\,{\dd k_{x} \over 2\pi}
$$

$$
{\rm p}\pars{x}=\int_{-\infty}^{\infty}{\rm B}\pars{k_{x}}
\exp\pars{\ic k_{x}x}\,{\dd k_{x} \over 2\pi}\quad\imp\quad
{\rm B}\pars{k_{x}} = \int_{-\infty}^{\infty}{\rm p}\pars{\xi}\expo{-\ic k_{x}\xi}\,\dd\xi
$$

\begin{align}
\color{#00f}{\large\Phi\pars{x,y}}&=\int_{-\infty}^{\infty}\bracks{%
\int_{-\infty}^{\infty}{\rm p}\pars{\xi}\expo{-\ic k_{x}\xi}\,\dd\xi}
\expo{\ic k_{x}x - \verts{k_{x}}y}\,{\dd k_{x} \over 2\pi}
\\[3mm]&={1 \over 2\pi}\int_{-\infty}^{\infty}{\rm p}\pars{\xi}
\int_{-\infty}^{\infty}\expo{\ic k_{x}\pars{x - \xi} - \verts{k_{x}}y}\,\dd k_{x}
\,\dd\xi
\\[3mm]&={1 \over \pi}\Re\int_{-\infty}^{\infty}{\rm p}\pars{\xi}
\int_{0}^{\infty}\!\!\!\!\!\expo{k_{x}\bracks{\ic\pars{x - \xi} - y}}\,\dd k_{x}
\,\dd\xi
\\[3mm]&={1 \over \pi}\Re\int_{-\infty}^{\infty}{\rm p}\pars{\xi}
{-1 \over \ic\pars{x - \xi} - y}\,\dd\xi
={1 \over \pi}\Re\int_{-\infty}^{\infty}{\rm p}\pars{\xi}
{\ic\pars{x - \xi} + y \over \pars{x - \xi}^{2} + y^{2}}\,\dd\xi
\\[3mm]&=\color{#00f}{\large{y \over \pi}\int_{-\infty}^{\infty}
{{\rm p}\pars{\xi} \over \pars{x - \xi}^{2} + y^{2}}\,\dd\xi}
\end{align}


TITLE: Derivative and divison [SOLVED]
QUESTION [1 upvotes]: [SOLUTION: Like Tom-Himler said: numerator is wrong because chain-rule was not followed]
I am reading, Calculus Made Easy and there is this example on page 73:
$$\omega = \frac{\sqrt{1-\theta^2}}{1+\theta}$$ (notice square-root is only in numerator). The author made the calculation easier by cancelling out,  $$\omega = \sqrt\frac{1-\theta}{1+\theta}$$ (notice square-root is in both numerator and denominator) and then taking the derivative $$\frac{d\omega}{d\theta} = \frac{-1}{(1+\theta)\sqrt{1-\theta^2}}$$
But when I do the derivative  without cancelling out, it does not come out to be correct.
Quotient rules states that the derivative of a function that is the ratio of two different functions is  $$\frac{d}{dx}\frac{u(x)}{v(x)} = \frac{(v \frac{du}{dx}) -(u\frac{dv}{dx})}{v^2}$$
Hence $u = (1 - \theta^2), v = (1+\theta)$
$$\frac{d\omega}{d\theta}$$ evaluates to

Numerator   $$\frac{(1+\theta)}{2\cdot \sqrt{1-\theta^2}}  - \sqrt
    {1 - \theta^2}\cdot 1$$
Denominator  $$(1+\theta)^2 = (1+\theta)(1+\theta)$$
Simplifying numerator: $$\frac{(1+\theta) - 2(1-\theta^2)}{2\sqrt{1
        - \theta^2}}$$
Putting back numerator and denominator  $$\frac{(1+\theta)
        -2(1+\theta)(1-\theta)}{2(1+\theta)(1+\theta)\sqrt{1-\theta^2}}\\
     \frac{1-2(1-\theta)}{2(1+\theta)\sqrt{1-\theta^2}}\\
     \frac{1-2+2\theta}{2(1+\theta)\sqrt{1-\theta^2}}\\
     \frac{-1+2\theta}{2(1+\theta)\sqrt{1-\theta^2}}$$

This does not match the answer the book gave.
Where did I go wrong ?

REPLY [0 votes]: Don't forget to use the chain rule. Your numerator is wrong. -Tom Himler (from comments).


text:
\begin{document}

\hbox{Preprint, {\tt arXiv:2006.16089}}
\medskip

\title
[{A new extension of the Sun-Zagier result}]
{A new extension of the Sun-Zagier result involving Bell numbers and derangement numbers}

\author
[Zhi-Wei Sun] {Zhi-Wei Sun}

\address{Department of Mathematics, Nanjing
University, Nanjing 210093, People's Republic of China}
\email{zwsun@nju.edu.cn}

\subjclass[2020]{Primary 11B73; Secondary 05A15, 05A18, 11A07.}
\keywords{Bell number, Bell polynomial, congruence, derangement number.
\newline \indent Supported by the Natural Science Foundation of China (grant no. 11971222).}

\begin{abstract}  Let $p$ be any prime and let $a$ and $n$ be positive integers with $p\nmid n$. We show that $$\sum_{k=1}^{p^a-1}\frac{B_k}{(-n)^k}\equiv a(-1)^{n-1}D_{n-1}\pmod {p},$$
where $B_0,B_1,\ldots$ are the Bell numbers and $D_0,D_1,\ldots$ are the derangement numbers.
This extends a result of Sun and Zagier published in 2011. Furthermore, we prove that
$$(-x)^n\sum_{k=1}^{p^a-1}\frac{B_k(x)}{(-n)^k}\equiv -\sum_{r=1}^ax^{p^r}\sum_{k=0}^{n-1}\frac{(n-1)!}{k!}(-x)^k\pmod{p\mathbb Z_p[x]},$$
where $B_k(x)=\sum_{l=0}^kS(k,l)x^l$ is the Bell polynomial of degree $k$ with $S(k,l)\ (0\ls l\ls k)$ the Stirling numbers of the second kind,
and $\mathbb Z_p$ is the ring of all $p$-adic integers.
\end{abstract}
\maketitle

\section{Introduction}
\setcounter{lemma}{0}
\setcounter{theorem}{0}
\setcounter{corollary}{0}
\setcounter{remark}{0}
\setcounter{equation}{0}

Let $B_0=1$. For each $n\in\Z^+=\{1,2,3,\ldots\}$ let $B_n$ denote the number of
  partitions of a set of cardinality $n$.  For example, $B_3=5$ since there are totally 5 partitions of $\{1,2,3\}$:
  $$\{\{1\},\{2\},\{3\}\},\{\{1,2\},\{3\}\},\ \{\{1,3\},\{2\}\},\ \{\{2,3\},\{1\}\},\{\{1,2,3\}\}.$$
The Bell numbers $B_0,B_1,\ldots$, named after E. T. Bell who studied them in the 1930s,
play important roles in combinatorics.
Here are values of $B_1,\ldots,B_{7}$:
$$B_1=1, \ B_2=2,\ B_3=5,\ B_4=15,\ B_5=52,\ B_6=203,\ B_7=877.$$
It is known that
$$\sum_{n=0}^\infty B_n\f{x^n}{n!}=e^{e^x-1}\  \t{and}\ \ B_{n+1}=\sum_{k=0}^n\bi nk B_k\ (n=0,1,2,\ldots).$$
The author's conjecture (cf. \cite[Conjecture 3.2]{S11}) that the sequence $(\root{n+1}\of{B_{n+1}}/\root n\of{B_n})_{n\gs1}$
is strictly decreasing (with limit $1$), is still open.
For any prime $p$ and  $m,n\in\N=\{0,1,2,\ldots\}$, we have the classical Touchard congruence (cf. \cite{T})
$$B_{p^m+n}\eq mB_n+B_{n+1}\pmod p.$$

Let $D_0=1$, and define $D_n$ $(n\in\Z^+)$ by
$$D_n=|\{\pi\in S_n:\ \pi(k)\not=k\ \t{for all}\ k=1,\ldots,n\}|.$$
Those $D_0,D_1,D_2,\ldots$ are called the
derangement numbers, and they were first introduced by Euler. It is well known that
$$D_n=n!\sum_{k=0}^n\f{(-1)^k}{k!}\ \ \t{for all}\ n\in\N.$$

In 2011, the author and D. Zagier \cite{SZ} showed that for any prime $p$ and $n\in\Z^+$ with $p\nmid n$ we have
\begin{equation}\label{SZ}\sum_{k=1}^{p-1}\f{B_k}{(-n)^k}\equiv (-1)^{n-1}D_{n-1}\pmod {p},
\end{equation}
which relates the Bell numbers to the derangement numbers.
The surprising congruence \eqref{SZ} was called the Sun-Zagier congruence
by Y. Sun, X. Wu and J. Zhuang \cite{SWZ} who used the umbral calculus to give a generalization,
by I. Mez\H o and T. L. Ramirez \cite{MR} in 2017 who extended it to the so0called $r$-Bell numbers,
and by Q. Mu \cite{Mu} in 2018 who re-proved via an identity of R.J. Clarke and M. Sved \cite{CS} relating the Bell numbers to the derangement numbers.

In this paper we extend the fundamental Sun-Zagier result in a new way.

\begin{theorem}\label{Th1.1}
 Let $p$ be any prime and let $a$ be a positive integer. For any $n\in\Z^+$ with $p\nmid n$, we have
\begin{equation}\label{BD}\sum_{k=1}^{p^a-1}\f{B_k}{(-n)^k}\equiv a(-1)^{n-1}D_{n-1}\pmod {p}.
\end{equation}
\end{theorem}
\begin{remark}\label{Rem1.1} Note that \eqref{BD} in the case $a=1$ gives \eqref{SZ}.
\end{remark}

For $n\in\Z^+$ and $k\in\{0,\ldots,n\}$, the Stirling number $S(n,k)$ of the second kind
denotes the number of ways to partition the set $\{1,\ldots,n\}$ into $k$ disjoint nonempty parts.
In addition, we adopt the usual convention $S(0,0)=1$. For $n\gs k\gs0$, it is well known that
\begin{equation}\label{Stirling}k!S(n,k)=\sum_{j=0}^k\bi kj(-1)^{k-j}j^n.
\end{equation}

For any $n\in\N$, the Bell polynomial (or the Touchard polynomial) of degree $n$ is given by
\begin{equation}\label{BellP}B_n(x)=\sum_{k=0}^n S(n,k)x^k.
\end{equation}
Clearly, $B_n(1)=B_n$ for all $n\in\N$, and $B_n(x)=x\sum_{k=1}^nS(n,k)x^{k-1}$ for all $n\in\Z^+$.
Theorem \ref{Th1.1} actually follows from our following theorem concerning the Bell polynomials.

\begin{theorem}\label{Th1.2} Let $a$ be any positive integer.
For any $n\in\Z^+$ and prime $p\nmid n$, we have
\begin{equation}\label{BDx}(-x)^n\sum_{k=1}^{p^a-1}\f{B_k(x)}{(-n)^k}\eq -\sum_{r=1}^ax^{p^r}\sum_{k=0}^{n-1}\f{(n-1)!}{k!}(-x)^k\pmod{p\Z_p[x]},
\end{equation}
where $\Z_p$ denotes the ring of all $p$-adic integers.
\end{theorem}
\begin{remark}\label{Th1.2} The congruence \eqref{BDx} in the case $a=1$
was deduced by Sun and Zagier \cite{SZ} via the usual explicit formula \eqref{Stirling} for Stirling numbers of the second kind. Our Theorem 1.2 can be further extended in the spirit of \cite{SWZ,MR}, we omit the details.
\end{remark}

We will show Theorem 1.2 in the next section.

\section{Proof of Theorem 1.2}
\setcounter{lemma}{0}
\setcounter{theorem}{0}
\setcounter{corollary}{0}
\setcounter{remark}{0}
\setcounter{equation}{0}

\begin{lemma}\label{Lem2.1} Let $p$ be a prime and let $a\in\Z^+$.

{\rm (i)} For any $j,k\in\N$ with $j+k\ls p^a-1$, we have
$$\bi{p^a-1-k}j\bg/\bi{-1-k}j\eq1\pmod p.$$
In particular,
$$\bi{p^a-1}j\eq(-1)^j\pmod p\ \ \ \t{for all}\ j=0,\ldots,p^a-1.$$

{\rm (ii)} We have
$$B_{p^a}(x)\eq \sum_{r=0}^a x^{p^r}\pmod {p\Z_p[x]}.$$
\end{lemma}
\Proof. (i) Since $j+k<p^a$ we have
$$\f{\bi{p^a-1-k}j}{\bi{-1-k}j}=\f{\prod_{0<i\ls j}\f{p^a-i-k}i}{\prod_{0<i\ls j}\f{-i-k}i}
=\prod_{0<i\ls j}\l(1-\f{p^a}{i+k}\r)\eq1\pmod p.$$
When $k=0$, this yields
$$\bi{p^a-1}j\eq\bi{-1}j=(-1)^j\pmod p.$$

(ii) By A. Gertsch and A. M. Robert's extension \cite{GR} of Touchard's  congruence, for any $n\in\N$ we have
$$B_{p^a+n}(x)\eq B_{n+1}(x)+B_n(x)\sum_{r=1}^ax^{p^r}\pmod {p\Z_p[x]}.$$
In particular,
$$B_{p^a}(x)\eq B_1(x)+B_0(x)\sum_{r=1}^ax^{p^r}=x+\sum_{r=1}^ax^{p^r}=\sum_{r=0}^a x^{p^r}\pmod {p\Z_p[x]}.$$

In view of the above, we have completed the proof of Lemma \ref{Lem2.1}. \qed

\medskip
\noindent{\it Proof of Theorem \ref{Th1.2}}. It is known that
\begin{equation}\label{recur}B_{m+1}(x)=x\sum_{k=0}^m\bi mkB_k(x)\ \ \t{for all}\ m\in\N.
\end{equation}
In light of this and Lemma 2.1, for any prime $p$ we have
\begin{align*}\sum_{k=1}^{p^a-1}(-1)^kB_k(x)\eq&\sum_{k=1}^{p^a-1}\bi{p^a-1}kB_k(x)=\f{B_{p^a}(x)}x-B_0(x)
\\\eq& \sum_{r=1}^a x^{p^r-1}\pmod{p\Z_p[x]}.
\end{align*}
So the desired result holds when $n=1$.

Now we fix $n\in\Z^+$ and assume that \eqref{BDx} holds
 for every prime $p\nmid n$.

  Let $p$ be any prime not dividing $n+1$.
If $p\mid n$, then $n!/k!\eq0\pmod p$ for all $k=0,\ldots,n-1$, and hence
\begin{align*}(-x)^{n+1}\sum_{k=1}^{p^a-1}\f{B_k(x)}{(-n-1)^k}\eq&(-x)^{n+1}\sum_{k=1}^{p^a-1}\f{B_k(x)}{(-1)^k}
\eq (-x)^{n+1}\sum_{r=1}^ax^{p^r-1}
\\\eq& -\sum_{r=1}^ax^{p^r}\sum_{k=0}^n\f{n!}{k!}(-x)^k\pmod{p\Z_p[x]}.
\end{align*}

Now we suppose that $p\nmid n$. In view of \eqref{recur} and Lemma \ref{Lem2.1}, we have
\begin{align*} \sum_{k=1}^{p^a-1}\f{B_k(x)/x}{(-n)^k}=&\sum_{k=1}^{p^a-1}\f{\sum_{l=0}^{k-1}\bi{k-1}lB_l(x)}{(-n)^k}
=\sum_{l=0}^{p^a-2}\f{B_l(x)}{(-n)^l}\sum_{k=l+1}^{p^a-1}\f{\bi{k-1}l}{(-n)^{k-l}}
\\=&\sum_{l=0}^{p^a-2}\f{B_l(x)}{(-n)^{l+1}}\sum_{r=1}^{p^a-1-l}\f{\bi{l+r-1}{r-1}}{(-n)^{r-1}}
\\=&\sum_{l=0}^{p^a-2}\f{B_l(x)}{(-n)^{l+1}}\sum_{r=1}^{p^a-1-l}\f{\bi{-l-1}{r-1}}{n^{r-1}}
\\\eq&\sum_{l=0}^{p^a-2}\f{B_l(x)}{(-n)^{l+1}}\sum_{r=1}^{p^a-1-l}\bi{p^a-1-l}{r-1}n^{-(r-1)}
\\\eq&\sum_{l=0}^{p^a-1}\f{B_l(x)}{(-n)^{l+1}}\(\l(1+\f1n\r)^{p^a-1-l}-\f1{n^{p^a-1-l}}\)
\\\eq&\sum_{l=0}^{p^a-1}\f{n^lB_l(x)}{(-n)^{l+1}}\l(\f1{(n+1)^l}-1\r)\ (\mo\ p)
\end{align*}
with the aid of Fermat's little theorem.
Therefore
$$-n\sum_{k=1}^{p^a-1}\f{B_{k}(x)/x}{(-n)^{k}}\eq\sum_{k=1}^{p^a-1}\f{B_k(x)}{(-n-1)^{k}}
-\sum_{l=1}^{p^a-1}\f{B_l(x)}{(-1)^l}
\pmod {p\Z_p[x]}$$
and hence
$$\sum_{k=1}^{p^a-1}\f{B_k(x)}{(-n-1)^k}\eq-n\sum_{k=1}^{p^a-1}\f{B_k(x)/x}{(-n)^k}
+ \sum_{r=1}^ax^{p^r-1}\pmod{p\Z_p[x]}.$$
Combining this with \eqref{BDx}, we obtain
\begin{align*}(-x)^{n+1}\sum_{k=1}^{p^a-1}\f{B_k(x)}{(-n-1)^k}
\eq&-n\sum_{r=1}^ax^{p^r}\sum_{k=0}^{n-1}\f{(n-1)!}{k!}(-x)^k
+(-x)^{n+1}\sum_{r=1}^ax^{p^r-1}
\\=&-\sum_{r=1}^ax^{p^r}\sum_{k=0}^n\f{n!}{k!}(-x)^k\pmod{p\Z_p[x]}.
\end{align*}
This concludes the induction step.

By the above, the proof of Theorem 1.1 is now complete. \qed


TITLE: If $P$ is a polynomial of degree $n>0$, then there exists circle $C$ of radius $R$ such that $\int_{C} \frac{P^{\prime}(z)}{P(z)} dz=2n\pi i $
QUESTION [0 upvotes]: Let $P$ be a polynomial of degree $n>0.$ Could anyone advise me on how to show there exists $R>0$ such that if $C$ is the circle $|z|=R$ anticlockwise oriented, then $\begin{align}\int_{C} \dfrac{P^{\prime}(z)}{P(z)} dz=2n\pi i\end{align} \  ?$
My attempt: By Argument principle, $\begin{align}\int_{C} \dfrac{P^{\prime}(z)}{P(z)} dz \end{align} \leq 2n\pi i.$ Since $P$ has at most countable number of zeroes, there exists a circle of radius $R$, sufficiently large, so it encloses all the zeros of $P \ ?$
Thank you.

REPLY [1 votes]: WLOG we may assume $P$ is monic. Let $P(z)=(z-a_1)\cdots(z-a_n)$ then $\frac{P'(z)}{P(z)}=\frac{1}{z-a_1}+\cdots\frac{1}{z-a_n}.$($\frac{f'}{f}$this is sometimes called logarithmic derivative of $f.$) Then choose a circle big enough that contains all $a_k$. So $\int_C \frac{1}{z-a_k}=2\pi i$ for $k=1\cdots n.$ Thus $\int_C \frac{P'(z)}{P(z)}dz=2n\pi i$


text:
\begin{document}
\title{Regular elements determined by generalized inverses}
\author{Adel Alahmadi, S. K. Jain, and Andr\'e Leroy}
\address{Adel Alahmedi\\
Department of Mathematics\\
King abdulaziz University\\
Jeddah, SA\\
Email:adelnife2@yahoo.com;\\
S. K. Jain\\
Department of Mathematics\\
King Andulaziz University Jeddah, SA,and\\
Ohio University, USA\\
Email:jain@ohio.edu\\
Andre Leroy\\
Facult\'e Jean Perrin\\
Universit\'e d'Artois\\
Lens, France\\
Email:andre.leroy@univ-artois.fr}

\begin{abstract}
In a semiprime ring, von Neumann regular elements are determined by their
inner inverses. In particular, for elements $a,b$ of a von Neumann regular ring $R$, $a=b$ if
and only if $I(a)=I(b)$, where $I(x)$ denotes the set of inner
inverses of $x\in R$. We also prove that, in a semiprime ring, the same is
true for reflexive inverses.
\end{abstract}

\maketitle

\section{Introduction and preliminaries}

\bigskip

In this short note, $R$ will stand for an associative ring with unity. An
element $a\in R$ is (von Neumann) regular if there exists $x\in R$ such that
$a=axa$. Such an element $x$ is called an inner inverse (also called von
Neumann inverse or generalized inverse) of $a$. The set of regular elements
of a ring $R$ is denoted by $Reg(R)$. A ring $R$ is regular if $Reg(R)=R$.
Note that a regular ring is semiprime. In general, a regular element may
have more than one inner inverse. We denote the set of inner inverses of $a$
by $I(a)$. An element $x\in R$ is called an outer inverse of $a$ if $xax=x$.
Note that if $x\in I(a)$ then $xax$ is both an inner and an outer inverse of
$a$. An element $x\in R$ is called a reflexive inverse of $a$ if it is both
an inner and an outer inverse of $a$. Denote the set of reflexive inverses
of $a$ by $Ref(a)$. We first obtain a necessary and sufficient condition for
$I(a)\subseteq I(b)$ (Lemma \ref{Comparison of quasi inverse set}) and use
this to prove that in a semiprime ring, for $a,b\in Reg(R)$, $I(a)=I(b)$ if
and only if $a=b$ if and only if $Ref(a)=Ref(b)$. (Theorem \ref{I(a)=I(b)}
and Theorem \ref{ref(a)=ref(b)} ).

We begin with a few key lemmas.

The following is well-known (cf. \cite{BIG} Corollary 1, Chapter 2. p. 40.)

\begin{lemma}
For $a\in R$ and $a_{0}\in I(a),$ we have $I(a)=\{a_{0}+t-a_{0}ataa_{0}\mid
t\in R\}.$
\end{lemma}

As usual, $l(a)$ and $r(a)$ denote respectively the left and right
annihilator of an element $a\in R$. We define the inner annihilator of an
element $a\in R$, as $\{x\in R\mid axa=0\}$ and denote it by $Iann(a)$.

The next Proposition gives a link between $I(a)$ and $Ref(a)$.

\begin{proposition}
\label{I(a) and ref(a)} For $a\in Reg(R)$, let $\varphi_a:I(a)
\longrightarrow Ref(a)$ be such that $\varphi_a(x)=xax$. Then

\begin{enumerate}
\item The map $\varphi_a$ is onto.

\item $Ref(a)=I(a)aI(a)$.

\item If $x,y\in I(a)$ are such that $\varphi_a(x)=\varphi_a(y)$ then $
x-y\in l(a)\cap r(a)$.

\item Let $x\in Ref(a)$, then $\varphi_a(x)=x$.
\end{enumerate}
\end{proposition}

\begin{proof}
This is clear.
\end{proof}

The next lemma is straightforward.

\begin{lemma}
Let $a\in Reg(R)$ and $a_{0}\in I(a).$ \ Write $e=aa_{0},$ $f=a_{0}a$ and $
e^{\prime }=1-e,$ $f^{\prime }=1-f$. Then

\begin{enumerate}
\item[(i)] $Iann(a)=l(a)+r(a)=Re^{\prime}+f^{\prime}R$.

\item[(ii)] $I(a)= a_0+Iann(a)=a_0 + Re^{\prime}+f^{\prime}R$.

\item[(iii)] If $a_{0}\in Ref(a),$ then $Ref(a)=a_{0}+fRe^{\prime
}+f^{\prime }Re+f^{\prime }RaRe^{\prime }$.
\end{enumerate}
\end{lemma}

\begin{proof}
We simply mention that the statement (iii) can be proved by using statement
(2) of the above proposition \ref{I(a) and ref(a)}.
\end{proof}

\section{Characterization of $I(a)$ and $Ref(a)$}

We can now state our first result in the following proposition.

\begin{proposition}
\label{Invariance} Let $R$ be a semiprime ring. If $a\in Reg(R)$, then for
any $b\in R,$ $bI(a)b$ is a singleton set if and only if $b\in Ra\cap aR$.
\end{proposition}

\begin{proof}
Firstly, suppose that there exist $x,y\in R$ such that $b=xa=ay$ and let $
a_{0}\in I(a)$. We then have that, for any $t\in R$, $
b(a_{0}+t-a_{0}ataa_{0})b=(xaa_{0}+xat-xataa_{0})ay=xay+xatay-xatay=xay$.
This shows that indeed $bI(a)b$ is a singleton set.

\noindent Conversely, suppose that $bI(a)b=\{ba_{0}b\}$. We then have $
b(a_{0}+t-a_{0}ataa_{0})b=ba_{0}b$, for any $t\in R$. This implies that, for
any $t\in R$, we have
\begin{equation*}
b(t-a_{0}ataa_{0})b=0.
\end{equation*}
Substituting $(1-a_{0}a)t$ for $t$ in this equality leads to $
b(1-a_{0}a)tb=0 $, for any $t\in R$. The semiprimeness of $R$ then implies
that $b(1-a_{0}a)=0$, i.e. $b=ba_{0}a\in Ra$. Similarly, substituting $t$ by
$t(1-aa_{0})$ in the above equality gives $b=aa_{0}b$. In particular, $b\in
aR$.
\end{proof}

\vspace{3mm}

We recall the following result obtained by S.K. Jain and M. Prasad (\cite{JP}
).

\begin{lemma}
\label{Jain-Prasad} Let $R$ be a ring and let $b,d\in R$ such that $b+d$ is
a Von Neumann regular element. Then the following are equivalent:

\begin{enumerate}
\item $bR\oplus dR=(b+d)R$.

\item $Rb\oplus Rd = R(b+d)$.

\item $bR \cap dR = \{0\}$ and $Rb\cap Rd=\{0\}$.
\end{enumerate}
\end{lemma}

The next proposition provides necessary and sufficient conditions as to when
$I(a)\subseteq I(b),$ where $a,b\in Reg(R)$ and $R$ is semiprime.

\begin{proposition}
\label{Comparison of quasi inverse set} Let $R$ be a semiprime ring and let $
a,b\in Reg(R)$. Then $I(a)\subseteq I(b)$ if and only if $bR\cap dR=0$ and $
Rb\cap Rd=0$ where $a=d+b$.
\end{proposition}

\begin{proof}
Since $I(a)\subseteq I(b)$, we have $bxb=b$ for every $x\in I(a)$. By
Proposition \ref{Invariance}, $b\in Ra\cap aR$. Write $b=\alpha a=a\beta $
for some $\alpha,\beta \in R$. Then $bI(a)a=b.$ Next, $
bI(a)d=bI(a)a-bI(a)b=b-bI(a)b=0.$

Consider now $dI(a)b=aI(a)b-bI(a)b=a\beta -bI(a)b=b-b=0.$ We thus have
\begin{equation*}
bI(a)d=0 \quad \mathrm{and} \quad dI(a)b=0 \quad \quad \quad (1)
\end{equation*}
Then, for any $x\in I(a)$, we have $b+d=a=axa=(b+d)x(b+d)=bxa+dxb+
dxd=b+0+dxd.$ This yields,
\begin{equation*}
dI(a)d=d \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)
\end{equation*}
Now, we proceed to show $dR\cap bR=0.$ Let $bx=dy\in bR\cap dR.$ Multiplying
both sides of the equality (2) by $y$ on the right and using $bx=dy$ we
obtain $dI(a)bx=dy.$ As proved above, we have $dI(a)b=0.$ and so $dy=0$.
This proves our assertion.

\noindent Similarly, we show that $Rb\cap Rd=0$. Let $xb=yd\in Rb\cap Rd.$
Now, multiplying both sides of the equality (2) on the left by $y$, we get $
ydI(a)d=yd$. This proves $xbI(a)d=yd.$ Since $bI(a)d=0$, we obtain $yd=0,$
proving $Rb\cap Rd=0$.

\noindent The converse is easy using the above lemma \ref{Jain-Prasad}.
\end{proof}

Next, we show, in particular, that the regular elements of a semiprime ring
are equal if their sets of inner inverses are the same.

\begin{theorem}
\label{I(a)=I(b)} Let $R$ be a semiprime ring and $a,b\in Reg(R)$. Then $
I(a)=I(b)$ if and only if $a=b$.
\end{theorem}

\begin{proof}
We only need to prove the sufficency. So assume that $I(a)=I(b)$.
Proposotion \ref{Comparison of quasi inverse set} implies that we can write $
a=b+d$ with $bR\cap dR=0,\; Rd\cap Rb=0$.

Lemma \ref{Jain-Prasad} then gives that $(b+d)R=bR\oplus dR$. Since $
I(a)=I(b)$ we also have $aI(b)a=\{a\}$ and $bI(a)b=\{b\}$ and Proposition
\ref{Invariance} implies that $Ra=Rb$ and $aR=bR$. This leads to $
aR=(b+d)R=bR\oplus dR=aR\oplus dR$. This forces $d$ to be zero and hence $
a=b $, as desired.

Alternatively we may invoke Hartwig's result (cf. \cite{H}) in place of
Lemma \ref{Jain-Prasad}.

This was pointed out to us by T.Y. Lam. Indeed, by our Proposition \ref
{Invariance} we have $aR=bR$ and $Ra=Rb$, and thus by invoking Hartwig's
result, there exist units $u,v\in R$ such that $b=au=va$. If $x\in I(a)=I(b)$
, then $axa=a$ and $bxb=b$. The last equality implies that $vaxau=au$ and
hence $va=a$. Thus $b=a$.
\end{proof}

\vspace{3mm}

\begin{corollary}
Let $R$ be a regular ring. Then $I(a)=I(b)$ if and only if $a=b$.
\end{corollary}

\vspace{2mm}

\vspace{2mm}

\begin{remark}
{\rm Pace Nielsen remarked that, in the above theorem, the semiprime
hypothesis can be replaced by the assumption that $a-b$ is regular. So
assume $I(a)=I(b)$, and $a-b\in Reg(R)$. As in our Proposition \ref
{Invariance} we obtain
\begin{equation*}
bt(1-aa_0)b=0,\quad\quad\quad (1)
\end{equation*}
for any $t\in R $. If $b_1$ is a reflexive inverse of $b$, we obtain, for
any $t$ in $R$, $ataa_0b=(ab_1a)taa_0b=a(b_1bb_1)ataa_0b= ab_1(bb_1ataa_0b)$
. Replace $t$ by $b_1at$ in (1) and obtain
\begin{equation*}
ataa_0b=atb. \quad\quad\quad (2)
\end{equation*}
For $z\in I(a-b)$ we have $bzb=bza + azb-aza+a-b$. Using this equality we
compute
\begin{equation*}
bzb=bzba_0b=(bza+azb-aza +a-b)a_0b=bzaa_0b+azb-azaa_0b+aa_0b-b.
\end{equation*}
Using formulae (1) and (2) we get $aa_0b=b$ so that $bR\subseteq aR$.
Symmetric arguments leads to $aR=bR$ and $Ra=Rb$ and Hartwig's theorem
finishes the proof.
}
\end{remark}

\bigskip

\section{Reflexive inverses for semiprime rings}

We conclude by characterizing the equality of $Ref(a)=Ref(b),$ and obtain the analogue
of Theorem \ref{I(a)=I(b)} for reflexive inverses of semiprime rings.

\begin{theorem}
\label{ref(a)=ref(b)} Let $R$ be a semiprime ring such that $a,b\in Reg(R)$.
Then $Ref(a)=Ref(b)$ if and only if $a=b$.
\end{theorem}

\begin{proof}
Let $a_0\in Ref(a)=Ref(b)$. Since $a=0$ if and only if $Ref(a)=0$, we may assume
that $a$ and $b$ are not zero. Since $bRef(a)b=bRef(b)b=b$ and $
Ref(a)=I(a)aI(a) $, we have that, for any $t$ in $R$,
\begin{equation*}
b(a_0 + t - a_0ataa_0) a (a_0 + t - a_0ataa_0)b=b \quad \quad \quad (1)
\end{equation*}
Replacing $t$ by $(1-a_0a)t$ and noting that $a(1-a_0a)=0$, we obtain
successively $b(a_0a + (1-a_0a)ta)(a_0 +(1-a_0a)t)b=b$ and $b(a_0a +
(1-a_0a)ta)(a_0)b=b$ and so $ba_0b + b(1-a_0a)taa_0b=b$. Since $ba_0b=b$
this gives $b(1-a_0a)taa_0b=0$ for all $t\in R$. This leads to
\begin{equation*}
aa_0b(1-a_0a)taa_0b(1-a_0a)=0 \quad \forall t\in R.
\end{equation*}
The semiprimeness of $R$ implies that $aa_0b(1-a_0a)=0$. Left multiplying by
$a_0\in ref(a)$, we get that $a_0b(1-a_0a)=0$ and hence since $a_0\in I(b)$
we conclude that $b(1-a_0a)=0$.

Therefore we obtain that $Rb\subseteq Ra$ and by symmetry $Ra\subseteq Rb$
and hence $Ra=Rb$. In the same way replacing $t$ by $t(1-aa_0)$ in (1), we
obtain $aR=bR$. The Hartwig's Theorem then gives us that there exist
invertible elements $u,v\in R$ such that $a=bu$ and $b=av$. The argument at
the end of the proof of the semiprime case (cf. Theorem \ref{I(a)=I(b)})
proves the theorem.
\end{proof}

We now give an example of a ring,  showing that without the
semipriness hypothesis both of the above theorems are false.

\begin{example}
{\rm
Consider the $\mathbb{F}_{2}$-algebra
\begin{equation*}
R=\mathbb{F}_{2}\langle a,b,x\mid
axa=a,bxb=b,xax=x,xbx=x,a^{2}=b^{2}=ab=x^{2}=0\rangle
\end{equation*}
This ring is finite and $\{a,b,x,ax,bx,xa,xb,axb,bxa\}$ is a basis of $R$ as
an $\mathbb{F}_{2}$-vector space. It is easy to determine that $r(a)=r(b)\langle
a,b,ax,bx,axb,bxa\rangle $, $l(a)=l(b)=\langle a,b,xa,xb,axb,bxa\rangle $ $
I(a)=x+R$, $I(b)=x+R$ , $ref(a)=\{x\}=ref(b)$. Of course, $(RaR)^{2}=0$
, showing that $R$ is not semiprime. }
\end{example}

\smallskip
The next corollary is a direct consequence of
Theorems \ref{I(a)=I(b)} and \ref{ref(a)=ref(b)}.
\begin{corollary}
Let $a,b$ be elements of a semiprime ring $R$.   Then the following are equivalent:
\begin{enumerate}
\item $I(a)=I(b)$,
\item $a=b$,
\item $ref(a)=Ref(b)$.
\end{enumerate}
\end{corollary}

\begin{remark}
{\rm
We close with the following comment. The question of the equality of
two elements in a regular ring that have the same set of inner inverses
arose while the authors have been working on the question: if, for a regular
self-injective ring $R$, $I(c)=I(a)+I(b)$, $a,b,c\in R$, is it true that $c$
is unique? If not, obtain a complete solution for $\mathrm{c.}$ We will
discuss that in another paper which is in progress.
}
\end{remark}

\bigskip

\centerline{ {\bf Acknowledgment} } \smallskip

We thank the referee for his/her valuable comments including the remark that
our proof of Theorem 10 for prime rings can be carried to semiprime rings as
well.

\bigskip


TITLE: How can I find the constant of integration?
QUESTION [1 upvotes]: I'm attempting a novel approach to some tough integration problems.  I'm using the idea of series expansions to help integrate.  In other words, I will attempt to approximate integration by integrating the series expansion of an integrand, rather than direct integration or standard numerical methods.
I believe I can approximate integration of a series very easily, compared to the other methods.  However, there's a catch.  I will use at least two different series expansions.  One for the lower limit of integration, and one for the upper limit.  Now, when I attempt to integrate these expansions, the constant of integration comes into play, and it's not obvious what it is.  Since I am using at least two different series expansions, the constant of integration may differ for each expansion.  So I'm wondering if there is an easy way to get the constants of integration without much more work.  Any help, ideas, or suggestions are welcome.
EDIT
A few additional notes... I know ahead of time that the series will converge.  I consider that I could integrate in sections, like quadrature, while still using the series to aid in integration.  However, I am considering the idea of only using only the endpoints, with two different series.  So the constants of integration would be different for each series.  If I could somehow find them or find how they differ relative to one another, that would save me the trouble of breaking the integral into sections and using something akin to conventional numerical methods.

REPLY [1 votes]: If you are talking about looking for the anti-derivative (though you also mentioned upper and lower limits), then you just write some constant $C$ after all your calculations. E.g. if
$$
f(x) = \sum\limits_{k=0}^\infty a_k x^k
$$
then
$$
F(x) = \int f(t)\,dt = \sum\limits_{k=0}^\infty\frac{a_k}{k+1}x^{k+1} +C.
$$
The idea of an anti-derivative is the following:
$$
\int\limits_{a}^bf(t)\,dt = F(b)-F(a)
$$
for any anti-derivative $F$ of the function $f$. Since any anti-derivative is determined up to a constant, you just should pick one anti-derivative before calculating $F(b)-F(a)$. In our example, you can put $C=1$ or $C=2$ and the result will stay the same.
Finally, to apply such a technique you should be careful since you cannot always integrate series in such a simple way - there are sufficient conditions to integrate series by parts.


TITLE: Approximating an arbitrary set by a Borel set
QUESTION [1 upvotes]: Q) Show that for each (not necessarily measurable) subset $A\subset \mathbb{R}$, there exists a Borel subset $B\subset \mathbb{R}$ such that $A\subset B$ and $\lambda^*(A) = \lambda(B)$ where $\lambda^*$ is the outer measure.
I know that for a Lebesgue measurable $A\subset \mathbb{R}$, there exists a Borel set $B\subset \mathbb{R}$ s.t. $B\supset A$ and $\lambda(B-A)=0$. How can I go from there?

REPLY [2 votes]: For each $n$ there exist open interval $U_k$ with $A \subset \bigcup_k U_k$ and $\lambda (B_n) \leq \lambda^{*} (A)+\frac  1 n$ where $B_n=\bigcup_k U_k$. Take $B=\bigcap B_n$. Then $A \subseteq B$, $B$ is a Borel set and $\lambda (B) \leq \lambda^{*} (A)$. The reverse inequality is obvious.


text:
\section{Introduction}\label{sec:intro}

The study of taut foliations has led to a number of important advances in knot theory and three-manifold topology.  Consequently, any connection between taut foliations and other invariants in low-dimensional topology can be quite useful.  One example of this is that taut foliations guarantee the non-triviality of certain analytically-defined invariants associated to three-manifolds, namely various types of Floer homology.  This non-triviality plays a key role in the proofs of Property P \cite{KM} and the Dehn surgery characterization of the unknot \cite{KMOS}.  One such non-triviality statement is that if a rational homology sphere $Y$ admits a co-orientable taut foliation, then $Y$ is not an {\em L-space} (i.e. $\rank \widehat{HF}(Y) > |H_1(Y;\mathbb{Z})|$, where $\widehat{HF}$ denotes the hat-flavor of Heegaard Floer homology) \cite[Theorem 1.4]{OSGenus}.  Consequently, L-spaces form an interesting family of manifolds that have been intensely studied.  Some useful examples of L-spaces to keep in mind for this paper are manifolds with finite fundamental group \cite[Proposition 2.3]{OSLens}.

Taut foliations are also related to the fundamental group of the underlying three-manifold.  For instance, any loop transverse to the foliation must represent a non-trivial element in $\pi_1$.  More structure can often be found.  Thurston's universal circle construction shows that for a co-orientable taut foliation with hyperbolic leaves, there is an orientation-preserving action of the fundamental group on a circle; for more details see \cite{CD}.  Furthermore, in many cases, one can actually find an orientation-preserving action on the real line, such as when the manifold is an integer homology sphere \cite[Lemma 0.4]{BB} or the foliation is $\mathbb{R}$-covered \cite[Corollary 7.11]{CD}.  Since $\operatorname{Homeo}_+(\mathbb{R})$ is {\em left-orderable} (i.e. admits a left-invariant strict total order), \cite[Theorem 1.1]{BRW} shows that admitting such an action implies that $\pi_1(Y)$ is left-orderable.

Thus, there is a strong relationship between left-orderability, taut foliations, and L-spaces.  In fact, there are at present no counterexamples known to the possibility that, for an irreducible rational homology sphere $Y$, the following three conditions are equivalent:
\begin{flalign}\label{equiv:lo} &\text{$\pi_1(Y)$ is left-orderable,}&& \end{flalign} $ $ \\ \vspace{-.6in} \\
\begin{flalign}\label{equiv:ctf} &\text{$Y$ admits a co-orientable taut foliation,}&& \end{flalign} $ $ \\ \vspace{-.6in} \\
\begin{flalign}\label{equiv:lspace} &\text{$Y$ is not an L-space.} &&  \end{flalign}
The equivalence of \eqref{equiv:lo} and \eqref{equiv:lspace} was explicitly conjectured in \cite{BGW}.  The equivalence of \eqref{equiv:lo}, \eqref{equiv:ctf}, and \eqref{equiv:lspace}, has been established for many families of three-manifolds, including Seifert manifolds \cite{BGW,BRW, LS}, Sol manifolds \cite{BGW}, and graph manifold homology spheres \cite{BB}.  Also, it is shown in \cite{BC} that \eqref{equiv:lo} and \eqref{equiv:ctf} are equivalent for graph manifolds.  However, the only relation between \eqref{equiv:lo}, \eqref{equiv:ctf}, and \eqref{equiv:lspace} that is unconditionally known is the implication mentioned above: if $Y$ is an L-space, then $Y$ does not admit a co-orientable taut foliation.  The original proof of this required that the foliation be at least $C^2$. However, Kazez and Roberts have recently shown that a weaker condition suffices \cite{KR}. All the foliations in the present paper satisfy this weaker condition; see \cite{KR} for details.

One reason why the possible equivalence of \eqref{equiv:lo}, \eqref{equiv:ctf}, and \eqref{equiv:lspace} is desirable is that they exhibit different types of behavior.  For example, left-orderability is well-understood under non-zero degree maps, while Floer homology is well-understood under Dehn surgery.  The equivalence of \eqref{equiv:lo}, \eqref{equiv:ctf}, and \eqref{equiv:lspace} would allow one to use these properties in tandem.  We mention that it would follow from the equivalence of \eqref{equiv:lo} and \eqref{equiv:lspace} that a toroidal integer homology sphere can never be an L-space (see \cite[Section 4]{CLW}).

\begin{definition}
A closed, connected, orientable three-manifold $Y$ is {\em excellent} if $Y$ admits a co-orientable taut foliation and $\pi_1(Y)$ is left-orderable.  Consequently, $Y$ is prime and is not an L-space.  On the other hand, $Y$ is called a {\em total L-space} if all three conditions \eqref{equiv:lo}, \eqref{equiv:ctf}, and \eqref{equiv:lspace} fail.  In other words, a total L-space is an L-space whose fundamental group is not left-orderable.
\end{definition}

We remark that if $b_1(Y) > 0$ and $Y$ is prime, then $Y$ always admits a co-orientable taut foliation \cite{Gabai1} and has left-orderable fundamental group \cite[Corollary 3.4]{BRW}, and thus is excellent.  Many examples of total L-spaces have come from branched covers of knots, including all two-fold branched covers of non-split alternating links \cite{BGW, OSBranched} and some higher-order branched covers of two-bridge knots \cite{DPT2005,Peters}.  In this paper, we study when cyclic branched covers of certain other families of knots give excellent manifolds or total L-spaces.  We will denote the $n$-fold cyclic branched cover of a knot $K$ by $\Sigma_n(K)$ and will always assume $n \geq 2$.  Note that by the Smith conjecture \cite{BM}, $\Sigma_n(K)$ is simply-connected only if $K$ is trivial.

A key input throughout the paper is that, as mentioned above, the equivalence of \eqref{equiv:lo}, \eqref{equiv:ctf}, and \eqref{equiv:lspace} is known for Seifert fibered manifolds.  This enables us to give a complete analysis of the case of torus knots.  Throughout, we let $T_{p,q}$ denote the $(p,q)$-torus link.

\begin{theorem}\label{thm:torus} Let $n, p, q \geq 2$ and suppose $p,q$ are relatively prime.  Then, $\Sigma_n(T_{p,q})$ is excellent if and only if its fundamental group is infinite.  Thus, $\Sigma_n(T_{p,q})$ is excellent except in the cases:
\begin{enumerate}[(i)]
\item $\{p,q\} = \{2,3\}$, $2 \leq n \leq 5$,
\item $\{p,q\} = \{2,5\}$, $2 \leq n \leq 3$,
\item $\{p,q\} = \{2,r\}$, $r \geq 7$, $n = 2$,
\item $\{p,q\} = \{3,4\}$, $n = 2$,
\item $\{p,q\} = \{3,5\}$, $n = 2$,
\end{enumerate}
in which case $\Sigma_n(T_{p,q})$ is a total L-space.
\end{theorem}

Both left-orderability and taut foliations have been studied in the context of gluing manifolds with toral boundary (see for instance \cite{BC, CLW}).  For this reason, we will also study certain families of satellite knots.  While bordered Floer homology is suitable machinery for studying the Heegaard Floer homology of manifolds glued along surfaces \cite{LOT}, we will not focus on its use here.  Let $K'$ be a satellite knot with non-trivial companion $K$; then the complement of $K'$ contains a copy of the exterior $X$ of $K$.  In $\Sigma_n(K')$, the preimage of $X$ consists of copies of some (possibly trivial) cyclic cover of $X$, and the proof of \cite[Theorem 1]{GL} shows that if $K'$ is prime, then the boundary components of this preimage are incompressible in $\Sigma_n(K')$.

Work of Li and Roberts \cite{LR} provides abundant examples of taut foliations on the exteriors of non-trivial knots in $S^3$ (see Section~\ref{sec:background} for more precise statements).  We will utilize these foliations heavily.  In the case of cables, the remaining piece of $\Sigma_n(K')$ is Seifert fibered, and so we are able to get a nearly complete result in this case.  Let $C_{p,q}(K)$ denote the $(p,q)$-cable of a knot $K$, where $q$ denotes the longitudinal winding.  Assume $q > 1$.

\begin{theorem}\label{thm:cables}
Let $K$ be a non-trivial knot in $S^3$.  Then, $\Sigma_n(C_{p,q}(K))$ is excellent, unless $n = q = 2$.
\end{theorem}

It turns out that without additional information about the companion $K$, Theorem~\ref{thm:cables} is sharp.

\begin{theorem}\label{thm:cablefail}
Let $K$ be the right-handed trefoil.  Then, $\Sigma_2(C_{p,2}(K))$ is excellent if and only if $p \leq 1$.
\end{theorem}

\begin{remark}\label{rmk:adam}
In the proof of Theorem~\ref{thm:cablefail}, we will show that for $p \geq 3$, $\Sigma_2(C_{p,2}(T_{2,3}))$ does not have left-orderable fundamental group and does not admit a co-orientable taut foliation.  Adam Levine has computed, using bordered Floer homology, that these manifolds are also L-spaces \cite{Levine}.
\end{remark}

L-spaces seem to be rare among integer homology spheres: it is conjectured that the only irreducible integer homology sphere L-spaces are $S^3$ and the Poincar\'e homology sphere.  This, and the possibility that \eqref{equiv:lspace} implies \eqref{equiv:lo}, suggests that most integer homology spheres should have left-orderable fundamental group.  Now, if a knot has trivial Alexander polynomial, then all its cyclic branched coverings are integer homology spheres.  A well-known class of knots with trivial Alexander polynomial are the Whitehead doubles.  For the cyclic branched coverings of these knots, we show the following.

\begin{theorem}\label{thm:whitehead}
Let $K$ be a non-trivial knot and let $Wh(K)$ denote the positive, untwisted Whitehead double of $K$.  Then, $\Sigma_2(Wh(K))$ is an excellent manifold and $\pi_1(\Sigma_n(Wh(K)))$ is left-orderable for all even $n$.  If $\pi_1(S^3_\alpha(K))$ is left-orderable for $\alpha = 1, \frac{1}{2},$ or $\frac{1}{3}$, then $\pi_1(\Sigma_n(Wh(K)))$ is left-orderable for all $n \geq 2$.
\end{theorem}

In light of Theorems~\ref{thm:cables} and ~\ref{thm:whitehead}, we make the following conjecture.

\begin{conjecture}\label{conj:asymptotics}
Let $K$ be a prime, satellite knot.  For all $n \gg 0$, $\Sigma_n(K)$ is excellent.
\end{conjecture}

The condition that $K$ be prime is necessary, as $\Sigma_n(K)$ is prime (in fact irreducible) if and only if $K$ is prime; this follows from the equivariant sphere theorem \cite{MSY} and the Smith conjecture \cite{BM}.  If $K$ has a summand which is the figure-eight knot, then no cyclic branched cover of $K$ will have left-orderable fundamental group \cite[Theorem 2(c)]{DPT2005}.

As mentioned, there have been a number of results determining when cyclic branched covers of two-bridge knots have non-left-orderable fundamental group and/or are L-spaces \cite{DPT2005,Hu2013,Peters, Tran2013}.  Let $K_{[p_1,\ldots,p_m]}$ denote the two-bridge knot of the type $\frac{a}{b}$, where $[p_1,\ldots,p_m]$ is the continued fraction expansion for $\frac{a}{b}$.  We follow the convention that $\frac{a}{b} = p_1 + \frac{1}{p_2 + \frac{1}{\ddots + \frac{1}{p_m}}}$.  We give the first examples of cyclic branched covers of hyperbolic two-bridge knots which are excellent manifolds.

\begin{theorem}\label{thm:twobridgesurgery}
Suppose $k,\ell \geq 1$ and $n$ divides $(2k+1)$ for $n > 1$.  Then $\Sigma_n (K_{[2(2k+1),2\ell+1]})$ is excellent.
\end{theorem}

Our next result gives a family of cyclic branched covers of two-bridge knots which are total L-spaces.  If $a_i \geq 1$, $1 \leq i \leq m$, then for any $n \geq 2$, the $n$-fold cyclic branched cover of the two-bridge knot $K_{[2a_1,2a_2,\ldots,2a_m]}$ is the two-fold branched cover of an alternating knot or link \cite{MV} and is therefore a total L-space; see \cite{T14}, also \cite{DPT2005,Peters}.  The simplest case where this positivity condition does not hold is the family of two-bridge knots $K_{[2\ell,-2k]}$, $\ell, k \geq 1$.  For these, it is known that $\Sigma_3(K_{[2\ell,-2k]})$ is an L-space \cite{Peters} and has non-left-orderable fundamental group \cite{DPT2005} and is therefore a total L-space.  Recently, Teragaito has shown that $\Sigma_4(K_{[2\ell,-2k]})$ is an L-space \cite{T14}.  We show that $\pi_1(\Sigma_4(K_{[2\ell,-2k]}))$ is not left-orderable, so we have the following.

\begin{theorem}\label{thm:twobridge}
For $k, \ell \geq 1$, $\Sigma_4(K_{[2\ell,-2k]})$ is a total L-space.
\end{theorem}

Finally, Teragaito \cite{Ter} has also shown that the three-fold cyclic branched cover of a three-strand pretzel knot of the form $P (2k+1, 2\ell+1, 2m+1)$, where $k,\ell,m \geq 1$, is an $L$-space.  We show that its fundamental group is not left-orderable.  We thus establish the following.

\begin{theorem}\label{thm:not left}
For $k, \ell, m \geq 1$, $\Sigma_3(P(2k+1,2\ell+1,2m+1))$ is a total L-space.
\end{theorem}

With this, we can improve \cite[Corollary 1.2]{Ter} to the following.
\begin{corollary}
Let $K$ be a genus one, alternating knot.  Then $\Sigma_3(K)$ is a total L-space.
\end{corollary}
\begin{proof}
We repeat the argument of \cite[Corollary 1.2]{Ter}, where it is shown that $\Sigma_3(K)$ is an L-space.  We are interested in showing $\pi_1(\Sigma_3(K))$ is not left-orderable.  If $K$ is a genus one, alternating knot, then $K$ is either a genus one, two-bridge bridge knot or, up to mirroring, a pretzel knot $P(2k+1,2\ell+1,2m+1)$ with $k,\ell,m \geq 1$ \cite[Lemma 3.1]{BZ} (independently \cite{Patton}).  The former case follows from \cite{DPT2005}, while the latter is Theorem~\ref{thm:not left}.
\end{proof}

\noindent {\bf Organization:} In Section~\ref{sec:background}, we review the relevant definitions and results that we will invoke throughout the paper.  In Section~\ref{sec:torus} we prove Theorem~\ref{thm:torus}.  In Section~\ref{sec:cables} we prove Theorems~\ref{thm:cables} and ~\ref{thm:cablefail}.  In Sections~\ref{sec:whitehead}, ~\ref{sec:twobridgesurgery}, ~\ref{sec:twobridge}, and ~\ref{sec:pretzels} we prove Theorems~\ref{thm:whitehead}, ~\ref{thm:twobridgesurgery}, ~\ref{thm:twobridge}, and ~\ref{thm:not left} respectively.  Finally, we briefly discuss two other families of cyclic branched covers in Section~\ref{sec:misc}. \\

\noindent {\bf Acknowledgments:}  We would like to thank Steve Boyer, Nathan Dunfield, and Rachel Roberts for helpful discussions.  We also thank Adam Levine for the computation needed for Remark~\ref{rmk:adam}.  We are particularly grateful to Nathan Dunfield for providing the information described in Remark~\ref{rmk:nathan}\eqref{item:nathan}.  The first author was partially supported by NSF Grant DMS-1309021.  The second author was partially supported by NSF Grant DMS-0636643.


TITLE: Proving second derivatives
QUESTION [1 upvotes]: I'm asked to prove a theorem (if that is the right word) about double derivatives. I'm still struggling with understanding Leibniz notation and I could use a push in the right direction. It's easy enough for me to differentiate the function when I write it down as $f(g(x))$ but not so much with Leibniz notation.
The problem is as follows: If $y = f(u)$ and $u = g(x)$, where $f$ and $g$ are twice differentiable functions, show that:
$$\frac{d^2y}{dx^2} = \frac{d^2y}{du^2}\frac{du}{dx}^2 + \frac{dy}{du}\frac{d^2u}{dx^2}$$
Could someone fill me in on the details of each part of this equation? For example, why is the second derivative of $\frac{dy}{dx}$ written as $\frac{d^2y}{dx^2}$ instead of $\frac{d^2y}{d^2x}$?
Thanks!

REPLY [0 votes]: Let $y=f\circ g$ and $y(x)=f(g(x))$, with $u:=g(x)$. Then
$$\frac{dy}{dx}:=\frac{d(f\circ g)}{dx}=\frac{df}{du}\frac{du}{dx};$$
whenever we write $\frac{df}{du}$ we mean $\frac{df(u)}{du}$, i.e. $\frac{dy}{du}$. Introducing the function $h(u):=\frac{df(u)}{du}$ we arrive at
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(h(u)\frac{dg}{dx}\right)=
\frac{dh}{dx}\frac{dg}{dx}+
h\frac{d^2g}{dx^2}=\\
\frac{dh}{du}\frac{du}{dx}\frac{dg}{dx}+
\frac{df}{du}\frac{d^2g}{dx^2}=\frac{d^2f}{du^2}\left(\frac{dg}{dx}\right)^2+
\frac{df}{du}\frac{d^2g}{dx^2}=\frac{d^2y}{du^2}\left(\frac{du}{dx}\right)^2+
\frac{dy}{du}\frac{d^2u}{dx^2}.$$


TITLE: A complex number $a+bi$ as a continued fraction.
QUESTION [4 upvotes]: I tried to learn about continued fractions of rational and irational numbers, but when it comes to complex numbers i cannot find anything resonable about that. Is there a way to write complex numbers as a finite or infinite continued fraction and what is the algorithm for this ?

REPLY [1 votes]: Let $z=\frac {p}{q+\frac {p}{q+...}}$
$z=\frac p{q+z}$
$zq+z^2=p$
$z^2+qz-p=0$
If a complex number $z=x+yi$ is a root of a quadratic equation with real coefficients, then another root is $z^*=x-yi$.
The equation is $(z-x-yi)(z-x+yi)=0$
$z^2-2zx+x^2+y^2=0$
Comparing with $z^2+qz-p=0$
gives $p=-(x^2+y^2)$, $q=-2x$


TITLE: Massless Kerr black hole
QUESTION [14 upvotes]: Kerr metric has the following form:
$$
ds^2 = -\left(1 - \frac{2GMr}{r^2+a^2\cos^2(\theta)}\right) dt^2 +
       \left(\frac{r^2+a^2\cos^2(\theta)}{r^2-2GMr+a^2}\right) dr^2 +
    \left(r^2+a^2\cos(\theta)\right) d\theta^2
  + \left(r^2+a^2+\frac{2GMra^2}{r^2+a^2\cos^2(\theta)}\right)\sin^2(\theta) d\phi^2 -
    \left(\frac{4GMra\sin^2(\theta)}{r^2+a^2\cos^2(\theta)}\right) d\phi\, dt
$$
This metric describes a rotating black hole.
If one considers $M=0$:
$$
ds^2 = - dt^2 +
       \left(\frac{r^2+a^2\cos^2(\theta)}{r^2+a^2}\right) dr^2 +
    \left(r^2+a^2\cos(\theta)\right) d\theta^2
  +
    \left(r^2+a^2\right)\sin^2(\theta) d\phi^2
$$
This metric is a solution of the Einstein equations in vacuum.
What is the  physical interpretation of such a solution?

REPLY [3 votes]: A reference which answers this is Visser (2008). It discusses the limits of vanishing mass $M \rightarrow 0$, and rotation parameter $a \rightarrow 0$. Your example is in $\S5$. Visser comments "This is flat Minkowski space in so-called “oblate spheroidal” coordinates...", as described in a different answer here.


TITLE: How to take a limit of a diagram with more than one category?
QUESTION [2 upvotes]: The German Wikipedia describes how one can define the quotient field of a ring over a universal property:

A quotient field $(\mathrm{Quot}(R), i)$ of a ring $R$ is a field $\mathrm{Quot}(R)$ together with an injective ring homomorphism $i:R\to\mathrm{Quot}(R)$ with the universal property that for each field $K$ and each injective ring homomorphism $f:R\to K$ exists a unique field homomorphism $g:\mathrm{Quot}(R)\to K$ such that $f=g\circ i$.

Intuitively, this makes sense: I imagine the quotient field to be some kind of "smallest field that includes the ring $R$".
However, I don't understand how to define this in a formal way: The formal definition of a universal property deals with limits: one defines a diagram as a functor $\mathcal{F}:D\to C$ from a finite category $D$ (with the diagram structure) to the category $C$ at hand, then one defines a cone to this functor, and the limit is a terminal object in the category of cones.
The limit we're interested in is not a ring, it is a field, so I would expect the diagram to be in the category of fields and field homomorphisms. But $R$ is not a field, so this cannot be true. But if the diagram is in the category of rings, then the limit is a ring instead of a field.
How does one treat limits of diagrams where the image of the diagram is in more than one category?

REPLY [6 votes]: The term "universal property" does not refer just to limits.  More generally, it refers to any structure that can be described as a terminal object in some category (probably not your original category).  For instance, the limit of a diagram is just a terminal object in the category of cones under the diagram.
In this case, the category in question is the category whose objects are fields $K$ together with a ring-homomorphism $R\to K$, and the morphisms are homomorphisms between the fields that make the diagram commute.  A quotient field of $R$ is then just an initial object in this category (or if you prefer, a terminal object in the opposite category).
As you observe, this definition is secretly talking about two different categories at once: the category of rings and the category of fields.  To make this more explicit, you can mention you are using the forgetful functor $F$ from the category of fields to the category of rings to identify a field $K$ as also giving a ring $F(K)$.  You can then say an object of your category is then a pair $(f,K)$ where $K$ is an object in the category of fields and $f$ is a morphism in the category of rings from $R$ to $F(K)$.  A morphism from $(f,K)$ to $(g,L)$ is then a morphism $h:K\to L$ in the category of fields such that $g=F(h)\circ f$.