text:
\begin{document}

\begin{frontmatter}

\title{A Posteriori Subcell Limiting of the Discontinuous Galerkin Finite Element Method
for Hyperbolic Conservation Laws}

\author[trento]{Michael Dumbser$^{*}$}
\ead{michael.dumbser@unitn.it}
\cortext[cor1]{Corresponding author}

\author[trento]{Olindo Zanotti}
\ead{olindo.zanotti@unitn.it}

\author[imt]{Rapha{\"e}l Loub{\`e}re}
\ead{raphael.loubere@math.univ-toulouse.fr}

\author[lanl]{Steven Diot}
\ead{diot@lanl.gov}

\address[trento]{Department of Civil, Environmental and Mechanical Engineering, University of Trento, Via Mesiano, 77 - 38123 Trento, Italy.}
\address[imt]{CNRS and Institut de Math\'{e}matiques de Toulouse (IMT)
 Universit{\'e} Paul-Sabatier, Toulouse, France}
\address[lanl]{Fluid Dynamics and Solid Mechanics (T-3), Los Alamos National Laboratory, NM 87545 U.S.A.}

\begin{abstract}
The purpose of this work is to propose a novel \textit{a posteriori} finite volume subcell limiter
technique for the Discontinuous Galerkin finite element method for nonlinear systems of hyperbolic
conservation laws in multiple space dimensions that works well for \textit{arbitrary} high order
of accuracy in space \textit{and time} and that does \textit{not} destroy the natural \textit{subcell resolution}
properties of the DG method. High order time discretization is achieved via a one-step ADER approach that uses
a local space-time discontinuous Galerkin predictor method to evolve the data locally in time within each cell.

Our new limiting strategy is based on the so-called MOOD paradigm, which \aposteriori verifies
the validity of a discrete candidate solution against physical and numerical detection criteria
after each time step. Here, we employ a relaxed discrete maximum principle in the sense of
piecewise polynomials and the positivity of the numerical solution as detection criteria. Within the
DG scheme on the main grid, the discrete solution is represented by piecewise polynomials of
degree $N$.
For those troubled cells that need limiting, our new limiter approach recomputes the discrete
solution by scattering the DG polynomials at the previous time step onto a set of $N_s=2N+1$ finite
volume subcells per space dimension.
A robust but accurate ADER-WENO finite volume scheme then updates the subcell averages
of the conservative variables within the detected troubled cells. The recomputed subcell averages
are subsequently gathered back into high order cell-centered DG polynomials on the main grid via a subgrid
reconstruction operator. The choice of $N_s=2N+1$ subcells is optimal since it allows to match the
maximum admissible time step of the finite volume scheme on the subgrid with the maximum admissible
time step of the DG scheme on the main grid, minimizing at the same time the local truncation error
of the subcell finite volume scheme. It furthermore provides an excellent subcell resolution of
discontinuities.

Our new approach is therefore radically different from classical DG limiters, where the limiter is
using TVB or (H)WENO \textit{reconstruction} based on the discrete solution of the DG
scheme on the \textit{main grid} at the new time level. In our case, the discrete solution is \textit{recomputed}
within the troubled cells using a \textit{different and more robust} numerical scheme on a \textit{subgrid level}.

We illustrate the performance of the new \aposteriori subcell ADER-WENO finite volume limiter approach for
very high order DG methods via the simulation of numerous test cases run on Cartesian grids in two and
three space dimensions, using DG schemes of up to \textit{tenth} order of accuracy in space and time
($N=9$). The method is also able to run on massively parallel large scale supercomputing infrastructure,
which is shown via one 3D test problem that uses 10 billion space-time degrees of freedom per time step.
\end{abstract}

\begin{keyword}
 Arbitrary high-order Discontinuous Galerkin schemes \sep
 a posteriori subcell finite volume limiter \sep
 MOOD paradigm \sep
 ADER-DG \sep
 ADER-WENO \sep
 high performance computing (HPC) \sep
 hyperbolic conservation laws

\end{keyword}

\end{frontmatter}

\input{introduction}

\input{framework}

\input{limiter}

\input{numerics}

\input{conclusion}

\section*{Acknowledgments}
M.D. and O.Z. have been financed by the European Research Council (ERC) under the
European Union's Seventh Framework Programme (FP7/2007-2013) with the
research project \textit{STiMulUs}, ERC Grant agreement no. 278267.
R.L. has been partially funded by the ANR under the JCJC project
``ALE INC(ubator) 3D''.
This work has been authorized for publication under the reference LA-UR-14-24778.

The authors would like to acknowledge PRACE for awarding access to the SuperMUC
supercomputer based in Munich, Germany at the Leibniz Rechenzentrum (LRZ).

Parts of the material contained in this work have been elaborated, gathered
and tested during the SHARK-FV conference (\textit{Sharing Higher-order Advanced
Know-how on Finite Volume} held in Ofir, Portugal in 2014,
\texttt{www.math.univ-toulouse.fr/SHARK-FV/}.

\bibliographystyle{plain}
\bibliography{DG_MOOD}

\end{document}


TITLE: Bouncing a point across a rectangle
QUESTION [1 upvotes]: Let's say you have a rectangle $W$ by $H$, $W > H$. Take a point in one of the smaller sides, and consider an angle theta relative to the base (in degrees, $0 < \theta < 90$). Consider that this point is "thrown" to the other side with this initial theta angle, and that every time it hits a "wall", it is "bounced" the same way as the light (see image below). Express $y$ in function of $x, \theta, W, H$. Notice that the point can bounce from 0 to $\infty$ times (when theta tends to $90^\circ$).

Note:
1) I hope the question was understandable
2) I am sorry for the quality of the image.

REPLY [0 votes]: Reflecting the rectangle repeatedly on its long side gives a straight line:

The vertical distance travelled after the first reflection is $W\tan\theta-x$. Let $q$ and $r$ be the quotient and remainder after dividing by $H$.
If $q$ is odd, as in the above picture, the top rectangle has the same orientation as the bottom one, giving $y=H-r$. If $q$ is even, the top rectangle has opposite orientation and $y=r$.


TITLE: Proof that $ \textbf{Set}\ $ is not isomorphic to its dual
QUESTION [6 upvotes]: I am trying to understand in an intuitive and straightforward way why the category $ \textbf{Set}\ $ is not isomorphic to its dual. An answer that I find appealing is [here] (Why is every category not isomorphic to its opposite?) where one respondent states:
"Perhaps the easiest way to see that the category of sets isn't isomorphic to its dual is to observe that there is an object, namely the empty set, such that all morphisms into it are isomorphisms, but there is no object such that all morphisms out of it are isomorphisms."
However, I fail to see the reasoning here.  To prove or disprove isomorphism between categories we need to look at functors between the two categories.  And any functor is of course just a pair of functions that links objects in one category to objects in the other, and morphisms in one category to morphisms in the other.
How, then, do morphisms and isomorphism within a category figure into the problem. Obviously, I am missing a link somewhere.

REPLY [7 votes]: This is a way of thinking that you have to get used too: if two categories are equivalent and one has a categorical property then so has the other. All the goal is to understand what is a categorical property. (Of course we could say that they are the properties perserved by equivalence, but hum...)

An example of such a categorical property: there is an terminal object.
A counterexample: there is a unique object.

The property you describe is categorical. A precise proof follows.

Suppose $F : \mathbf{Set} \to \mathbf{Set}^{\rm op}$ is an equivalence with pseudo inverse $G:\mathbf{Set}^{\rm op} \to \mathbf{Set}$.
Remark that $\emptyset \in \mathbf{Set}$ has the curious property that any map $f: X \to \emptyset$ forces $X$ to be isomorphic to $\emptyset$ and $f$ to be such an isomorphism$^{\text{1}}$. Let us then prove that this property should also be satisfied by $F(\emptyset)$ in $\mathbf{Set}^{\rm op}$:

If $g : Y \to F(\emptyset)$ in $\mathbf{Set}^{\rm op}$, then $G(g)$ is a map $G(Y) \to GF(\emptyset)$ in $\mathbf{Set}$. But there is an natural isomorphism $\alpha: GF \Rightarrow \mathrm{Id}_\mathbf{Set}$ (by definition of a pseudo inverse), so that $\alpha_\emptyset \circ G(g)$ is a map $GF(X) \to \emptyset$. The curious property says that this is an isomorphism. But as $\alpha_\emptyset$ already is an iso, it means that $G(g)$ is an isomorphism. Now you should know that an equivalence (in fact a fully faithful functor is enough) reflects isomorphisms (if you do not, prove it as an exercise!): that means that $G(g)$ is iso if and only if $g$ is. So we get that $g$ is an isomorphism.

Now you just have to prove that such an object cannot exist in $\mathbf{Set}^{\rm op}$. But it the same as proving that in $\mathbf{Set}$ we can not have an object $A$ such that any morphism $A \to X$ is a isomorphism. Which is obvious: one always have an injection $A \to A + 1$ which is never a surjection ($A+1$ is the set $A$ in which we add a new element distinct from any other and the map $A \to A+1$ is just the inclusion of $A$ in that new set).

depending on your fondations, every set $X$ isomorphic to $\emptyset$ could be equal to $\emptyset$, but this is irrelevant here and reasoning up to isomorphism is usually better in category theory.


TITLE: What is the derivative of the inverse of the unit quaternion?
QUESTION [1 upvotes]: the unit quaternion is a good way to represent the rotation. but does anyone know the following formula is right or not?I am trying to prove it by myself but I am stuck on it.
$\dot{(q^{-1})}=-q^{-1}*\dot{q}*q^{-1}$ where q is a unit quaternion,* is the quaternion multiplication.

REPLY [1 votes]: I assume you are assuming $q(t)$ is a quaternion-valued function of a real variable $t$.
One thing you can check is that the product rule works for quaternion-valued functions. Indeed, the usual proof works, you just have to be careful not to use the commutative proprety.
Then, defining $p(t):=q(t)^{-1}$, we may differentiate $p(t)q(t)=1$ to obtain
$$ \frac{\mathrm{d}p}{\mathrm{d}t}q+p\frac{\mathrm{d}q}{\mathrm{d}t} = 0. $$
Can you finish?


TITLE: dense set on $[0,1]$ and more...
QUESTION [1 upvotes]: I want to prove that $A=\{\{n\alpha\} \mid \text{where }n\text{ is an integer number}\}$ is dense in $[0,1]$ and  I do not know how to continue.
$\alpha$ is irrational.
I obtained that:
for a given $\varepsilon>0$ there are $i$, $j$ such that: $|\{i\alpha\}-\{j\alpha\}| <\varepsilon$.
How should I continue. I would really appreciate an answer in the easiest way possible and one that is complete as well.

REPLY [0 votes]: A relatively simple proof hinges on the so called Dirichlet lemma:
If $\alpha$ is irrational then for any $n$ one can find $p$ and $q$ such that
$|\alpha -\frac{p}{q}| < \frac{1}{nq}$ with $q<n$. (p,q,n - integers)
So, in particular, one can find $p_n->\infty$ and $q_n->\infty$ such that
$$|\alpha -\frac{p_n}{q_n}| < \frac{1}{q_n^2} $$
From this one easily deducts that ${\{\alpha n\}}$  has a cluster point either at $0$ or at $1$.
Say, it has a cluster point at $0$. Then, if ${\{\alpha n\}}$ is "small" then
$${\{\alpha n\cdot k\}} = k{\{\alpha n\}}$$ which means density.
 Lastly, using similar trick one shows that if $1$ is a cluster point of ${\{\alpha n\}}$ then so is $0$.


TITLE: Geometry problem, find lengths
QUESTION [0 upvotes]: Here's my problem:

I know the angles $\alpha$ and $\beta$, and the line length $a + b$. I need to find the lengths $a$ and $b$ and the height of the triangle. I came up with the identity $$\frac{a}{b} = {\tan\alpha \over \tan\beta}$$, but I don't know what to do next. Can you help?

REPLY [0 votes]: Law of Sines:
$$\frac{|\overline{CB}|}{\sin A} = \frac{|\overline{CA}|}{\sin B} = \frac{|\overline{AB}|}{\sin C} \quad \left(= \frac{a+b}{\sin(A+B)}\;\right)$$
Then
$$\begin{align}
a &= |\overline{CA}| \cos A = \frac{(a+b)\cos A \sin B}{\sin(A+B)} \\[6pt]
b &= |\overline{CB}| \cos B = \frac{(a+b)\sin A \cos B}{\sin(A+B)}
\end{align}$$


TITLE: Series calculation
QUESTION [0 upvotes]: I have some random integers separated by space read from a text file like
99 20 30 1 100 400 5 10

I have to find a sum from any combination of these integers that is closest to a specifiq number like
183

what is the fastest and accurate way of doing this?

REPLY [1 votes]: You can also have a look at http://comeoncodeon.wordpress.com/2009/07/21/counting-problem/
Problem 1 describes different approaches to see whether a sum K is poosible. The DP approach for the Problem 1 should give you all the numbers that are possible sum of a subset. Then you can simply do search for the nearest.


TITLE: Find the derivative of $G(x) = \tan^{-1}(x^2 + 1)$
QUESTION [0 upvotes]: Find the derivative of $G(x) = \tan^{-1}(x^2 + 1)$

I’m having trouble with derivatives. Could someone please help me with this? I’m not sure exactly how to go on in solving this problem.

REPLY [1 votes]: HINT
Let use the chain rule
$$g(x)=f(h(x))\implies g'(x)=f'(h(x))\cdot h'(x)$$
that is
$$(\arctan f(x))'=\frac{f'(x)}{1+f^2(x)}$$
with $$f(x)=x^2+1\implies f'(x)=2x$$

REPLY [0 votes]: $$\frac{d}{dx}[ \tan^{-1} (x^2 + 1)] = \frac{1}{1+(x^2+1)^2}\times \frac{d}{dx}[x^2 + 1] = \frac{2x}{x^4 + 2x^2 + 2}$$


text:
\begin{document}
\maketitle

\begin{abstract}
All codes with minimum distance $8$ and codimension up to $14$ and
all codes with minimum distance $10$ and codimension up to $18$
are classified. Nonexistence of codes with parameters [33,18,8]
and [33,14,10] is proved. This leads to 8 new exact bounds for
binary linear codes. Primarily two algorithms considering the dual
codes are used, namely extension of dual codes with a proper
coordinate, and a fast algorithm for finding a maximum clique in a
graph, which is modified to find a maximum set of vectors with the
right dependency structure.
\end{abstract}

\section{Introduction}

Let $\F_2^n$ denote the $n$-dimensional vector space over the
field $\F_2$ and let the inner product
 $\left< \cdot ,\cdot\right>: \F_2^n \times \F_2^n \rightarrow \F_2$ be defined in the natural way as
 $\left<u,v\right> = \sum_{i=1}^n u_i v_i$, where addition is in $\F_2$. The Hamming
distance between two vectors of $\F_2^n$ is defined as the number
of coordinates in which they differ, and the weight $\wt(v)$ of a
vector $v \in \F_2^n$ is the number of the nonzero coordinates of
$v$. A
linear binary $[n,k,d]$ code $C$ is a $k$-dimensional subspace of
$\F_2^n$ with minimum distance $d = \min\{\wt(c) :
 c \in C, c\neq 0 \}$. A generator matrix $G$ for an
$[n,k]=[n,k,d \geq 1]$ code is any matrix whose rows form a basis
for the code. The orthogonal complement $C^\perp$ of $C$ in
$\F_2^n$ is called the dual code of $C$ and is an
$[n,n-k,d^\perp]$ code, where $d^\perp$ is called the dual
distance of $C$ and $n-k$ the codimension. A generator matrix $H$
for the dual code is called a parity check matrix of the code $C$.
In this paper we will say that a code $C$ is an
$[n,k,d]^{d^\perp}$ code if it is an $[n,k,d]$ code with dual
distance $d^\perp$. Further, two binary linear codes $C_1$ and
$C_2$ are said to be equivalent if there is a permutation of
coordinates which sends $C_1$ to $C_2$.
 Throughout this paper all codes are assumed to be binary.

A central problem in coding theory is that of optimizing one of
the parameters $n$, $k$ and $d$ for given values of the other two.
Usually this optimization is related to the following functions:
$n_2(k,d)$ - the minimum length of linear codes for given minimum
distance $d$ and dimension $k$ and $d_2(n,k)$  the largest value
of $d$ for which a binary [$n,k,d$] code exists.  Codes with
parameters $[n_2(k,d),k,d]$ and $[n,k,d_2(n,k)]$ are called
\textit{optimal}. There are many reasons to study optimal codes.
These codes are interesting not only for detection and correction
of errors. Some of them have rich algebraic and combinatorial
structure. The problems of optimality are strongly connected and
can be considered as packing problem in statistics and in finite
projective spaces \cite{HirSt}. Unfortunately, all these problems
are, as many others in coding theory, computationally difficult
\cite{Barg} and the exact values of $n_2(k,d)$ for all $d$ are
known only for $k\le 8$ \cite{BJV_IEEE00}. Tables with bounds and
exact values for $d_2(n,k)$  are given in \cite{Brou98} and
\cite{new-Tables}.

Another application of optimal codes is directly related to the
design method of cryptographic Boolean functions suggested by
Kurosawa and Satoh \cite{Kurosawa}. In this case the optimal
linear codes have to be with largest possible dual distance. More
precisely one have to study  the function $N(d,d^\perp)$ as the
minimal $n$ such that there exists a linear binary code of length
$n$ with minimum distance $d$ and dual distance $d^\perp$. The
investigation of $N(d,d^\perp)$ seems to be much harder than the
investigation of $n_2(k,d)$. There are some general bounds (see
\cite{Matsumoto}) but these bounds can be reached only for a few
values of $d$ and $d^\perp$. In a previous work, we studied
$N(d,d^\perp)$ for $d^\perp\leq d \le 12$ by computer using the
package \textsc{Q-Extension} \cite{Bo_SJC07}. With this package we
attempted to construct generator matrices and classify codes with
fixed parameters. Practically, we had no success in the cases
where the dual distance was more than 6. In the case of codes with
fixed minimum distance larger than 2, it is quite natural to look
at the duals of the codes with needed  properties and to the
parity check matrices. In other words, extending an $[n-1,k-1,d]$
code $C_1$ to an $[n,k,d]$ code $C_2$ can be considered as
extending an $[n-1,n-k]^{d}$ to an $[n,n-k]^{d}$ code with one
coordinate. This approach helps us to develop two different
methods, which are much more effective when we study
$N(d,d^\perp)$, and also $n_2(k,d)$, for $d= 8$ and $d=10$. For
small dimensions, it is convenient to use a brute force algorithm
that takes generator matrices of all inequivalent $[n,k,d]^{\dd}$
codes as input, extends them in all possible ways and then checks
the constructed codes for minimum and dual distance and for
equivalence. As a result we get all inequivalent $[n+1,k,\ge
d]^{\dd}$ codes. This method is of course impossible to use for
larger dimensions $k$, because of the large number of possible
extensions. To avoid this problem we use a second method for
larger dimensions adopting a strategy for bounding of the search
space similar to the strategy for finding a maximum clique in a
graph suggested in \cite{Ostergard-clique}.

 In this paper we present two algorithms which can be used for
 constructing of linear codes with fixed dual distance. We give
 classification results for  all codes with minimum
 distance 8 and codimension up to 14 and minimum
 distance 10 and codimension up to 18.
We would like to refer to \cite{Jaff97} and \cite{Jaffe} for a detailed bibliography of works which study linear codes with minimum
distance 8 and 10.

\section{Preliminaries}

In this Section we give some properties of linear codes and the
relations with their dual codes which help us to design the
construction algorithms.

The first proposition considers the even linear codes, namely the
linear binary codes which consist only of even weight vectors.

\begin{prop}
If $d\geq 2$ is even and a linear $[n,k,d]$ code exists, then
there exists an even $[n,k,d]$ code.
\end{prop}

\begin{proof}
If $C$ is a linear $[n,k,d]$ code, the code produced by puncturing $C$ in one coordinate has parameters $[n-1,k,d$ or $d-1]$. Adding a
parity check bit to all codewords, we obtain an even $[n,k,d]$
code.
\end{proof}

Practically, we use the following corollary:

\begin{cor}
If $d\geq 2$ is even and even linear $[n,k,d]$ codes do not exist,
then no $[n,k,d]$ code exists.
\end{cor}

Later on we give the definition and some properties of residual
codes.

\begin{deff}\label{res_def}
The residual code $Res(C,c)$ with respect to a codeword $c \in C$
is the restriction of $C$ to the zero coordinates of $c$.
\end{deff}

A lower bound on the minimum distance of the residual code is
given by
\begin{thm}(\cite{Hand},Lemma 3.9) \label{res_d}
Suppose $C$ is a binary $[n,k,d]$ code and suppose $c \in C$ has
weight $w$, where $d > w/2$. Then $Res(C,c)$ is an $[n-w,k-1,d']$
code with $d' \ge d - w + \lceil{w/2}\rceil$,
\end{thm}
and on the dual distance by
\begin{prop}\label{res_dd} Suppose $C$ is a
binary $[n,k,d]$ code with dual distance $d^{\perp}$, $c \in C$,
and the dimension of $Res(C,c)$ is $k-1$. Then the dual distance
of $Res(C,c)$ is at least $d^{\perp}$.
\end{prop}

There is also a well known elementary relationship between the
minimum distance of a linear code and the parity check matrix.
\begin{prop}\label{pcm}
A linear code has minimum distance $d$ if and only if its parity
check matrix has $d$ linearly dependent columns but no set of
$d-1$ linearly dependent columns.
\end{prop}

The next proposition gives a connection between weights of the
rows of generator matrices and columns of parity check matrices.
\begin{prop}\label{even}
Any even linear code $C$ has a parity check matrix whose columns
have odd weights.
\end{prop}
\begin{proof}
Let $G = \left[ I_k P \right]$ be a generator matrix for $C$ in
standard form. Then every row of $P$ has odd weight. And
accordingly the parity check matrix $H= \left[ P^T I_{n-k}
\right]$ has only odd weight columns. The sum of all rows of $H$
gives the all-ones vector.
\end{proof}

\begin{prop}\label{weight1}
If a linear code $C$ has $t$ codewords of weight $1$, then its
parity check matrix has $t$ zero-columns.
\end{prop}
\begin{proof}
Without loss of generality, let $u=(1,0, \hdots ,0) \in C$. Then
$\forall v \in C^\perp$, $0 = \left< u,v \right> = \sum_{i=1}^n
u_i v_i = v_1$.
\end{proof}
This means that if we have found all inequivalent codes with dual
distance $8$ and dimension $14$, then we have also
found all codes with minimum distance $8$ and codimension up to $14$.\\
Another well known fact is that for $d$ odd,
\begin{align*}
|\left\{ \text{inequivalent } [n,k,d]  \text{ codes} \right\}| \\
 \ge |\left\{\text{inequivalent even }[n+1,k,d+1] \text{ codes} \right\}|
\end{align*}
with equality only if the automorphism groups are transitive.

We define for convenience the function $L(k,d^\perp)$.
\begin{deff}
Let $d^\perp \ge 3$. Then $L(k,d^\perp)$ is  the maximum length
$n$  such that a binary $[n,k]^{\dd}$ code exists.
\end{deff}

The next theorem holds.
\begin{thm}\label{b_th}
Let $C$ be a binary linear $[n,k,d]^{\dd}$ code. Then $d \ge n -
L(k-1,d^\perp)$.
\end{thm}
\begin{proof}
Let $C$ be a code with parameters $[n,k,d]^{\dd}$ and $c_d$ be a
codeword in $C$ of weight $d$. If $d < n - L(k-1,d^\perp)$ then
the residual code $Res(C,c_d)$ has dual distance at least
$d^\perp$, length $n-d > L(k-1,d^\perp)$ and dimension $k-1$. This
is impossible because $L(k,d_1^\perp) \ge L(k,d_2^\perp)$ for
$d_1^\perp \ge d_2^\perp$.
\end{proof}

In fact, the function $L(k,d^\perp)$ has already been investigated
in another setting for dual distance greater than or equal to $4$
in connection with studies on $\kappa$-caps in projective
geometries since it is known that a $\kappa$-cap in $PG(k-1,q)$ is
equivalent to a projective $q$-ary $[n=\kappa,k]^{d^\perp}$ code
with $d^\perp \ge 4$. And $L(k,d^\perp)$ have thus been considered
in connection with the function
\begin{align*}
\mu_\nu(N,q) = & \text{ the maximum value of } \kappa \text{ such that there exist} \\
&\text{ a $\kappa$-cap in } PG(N,q),
\end{align*}
where $q$ is the order of the underlying Galois field, in our case $2$.
 More information on this can be found in the survey by Hirschfeld and Storme \cite{HirSt}.

\section{Computational  Algorithms}

We use mainly two algorithms for the extension of codes.
If the dimension of the considered codes with given dual distance
is small we can find the next column of the generator matrix of
the new code relatively easy because we can represent any
generator matrix $G$ in the packing form as a vector $G_b$ of $n$ computer words.
This algorithm, named \textsc{Bruteforce}, can be described with the following steps:\\

Algorithm \textsc{Bruteforce}:\\

 \noindent
    INPUT: $ C_{inp}$ - Set of all inequivalent $[n,k]^{d^\perp}$ codes represented by their generator matrices in packing form.\\
    OUTPUT: $C_{out}$ - Set of all inequivalent $[n,k+1]^{d^\perp}$
    codes.\\
    var a:array$[1..2^k-1]$ of  integer;\\

    {In the beginning, $C_{out}$ is the empty set.
    For any code $C_r$ in $C_{inp}$ with generator matrix in
    packing form $G_r$ do the following:}
\begin{enumerate}

\begin{item}{    Set a[i]:=1 for any i.}
\end{item}
\begin{item}{
    Find all linear combinations $b$ of up to $d^\perp-2$ column vectors of $G_r$ and set a[b]:=0.}
\end{item}
\begin{item}{
    For all $j$ such that $a[j]=1$ extend $G_r$ with one coordinate, equal to $j$, to $G_r'$. If
    there are no codes in $C_{out}$ equivalent to $C_r'$ (generated by $G_r'$) do $C_{out}:=C_{out}\cup C_r'$}
\end{item}
    \end{enumerate}

\noindent  The big advantage of this algorithm is given by Step 2.
In that step, all possible solutions for the $(n+1)$th column of
the generator matrices are determined with approximately with
$\sum_{i=1}^{d-2}{n \choose i}$ operations. Actually, to find all
vector solutions for the $(n+1)$th column, we take all
$k$-dimensional vectors and delete those which are not
solutions. We find all sums  of less than $\dd-2$ columns of the
known part of the generator matrix. Each sum gives us one vector
which is not a solution and have to be deleted. All remaining
vectors are solutions.

 In Step 3, we use canonical representation
of the objects. The main priority of the canonical representation
is that the equivalence (isomorphism) test is reduced to check of
coincidence of the canonical representations of the structures. In
the case of many inequivalent codes,  the computational time for
comparing is growing fast. A technique for surmounting  this problem
is worked out. We split the set of inequivalent codes into a big
amount of cells according to a proper invariant.

To explain the next algorithm, we need the following definition.

 \begin{deff}\label{p-proper}Let $E$ be a set of $k$-dimensional vectors.
    \begin{enumerate}
    \item We call $E$ \textit{p-proper} if all subsets of  $p$ vectors of $E$ are linearly independent.
    \item Let $M$ be a $k \times n$ matrix. The set $E$ is called $p$-proper with respect to $M$ if $E \cup \{\text{columns in } M \}$ is a $p$-$proper$ set.
    \end{enumerate}
\end{deff}
\noindent Observe that, by Proposition \ref{pcm}, the columns of a parity check matrix for an $[n,k,d]$ code form a $(d-1)$-proper set.\\

 We consider the following problem: How to find a set of  $t=d-1$ binary vectors which have a certain property, i.e.,
     $(\dd-1)$-proper  subset  of the set of all possible binary vectors with respect to a fixed generator matrix.
    To attempt to solve this problem in reasonable time, we adopt an idea suggested by \"Osterg{\aa}rd in
    \cite{Ostergard-clique}
     for finding a maximum clique in a graph in the algorithm \textsc{Extend}.

   Let $C$ be an $[n,k,d]^{\dd}$ code with a generator matrix $G$ in the form
    \begin{equation} G =
        \left[
        \begin{array}{cc|c}
        0 0 \hdots 0 & 1 & 1 1 \hdots 1 \\
         & 0 &   \\
        Res_d(C)  &  \vdots &  X  \\
          &  0      &
        \end{array}
        \right]
        =
        \left[
        \begin{array}{c|c}
        A  &  \hat{X}
        \end{array}
        \right]
        \end{equation}
where $Res_d(C)$ is a generator matrix of the residual
$[n-d,k-1,\geq \frac{d}{2}]^{\ge\dd}$ code. Given that we know all
such inequivalent generator matrices the problem is reduced to
finding all $(\dd-1)$-proper sets $\hat{X}$ with respect to $A$ of
$d-1$ binary vectors on the form $(1,x_2,\hdots,x_k)^T$ .

Let
\begin{equation*} V^* = \left\{ (1,x_2,x_3,\hdots,x_k): x_i \in \F_2 \right\}
\end{equation*}

(\textit{Remark}: If the dual code is even we may, by Proposition \ref{even}, reduce the search space to the set of odd-weight binary vectors).
Delete from $V^*$ all linear combinations of $\dd-2$, or less, vectors from $A$. The remaining set
\begin{equation}
V = \left\{ v_1, v_2, \hdots, v_N \right\}
\end{equation}
 is the search space for our search strategy.
Now, for each integer $1 \leq i \leq N$, let

\begin{equation}
V_i = \left\{ v_i, v_{i+1}, \hdots, v_N \right\}
\end{equation}
and let $\textbf{r}$ be the $N$-tuple, defined by $\textbf{r}[i]=
\min\{s,t \}$, $1\leq i \leq N$, where $s$ is the size of the
largest $(\dd-1)$-proper subset of $V_i$. First we consider
$(\dd-1)$-proper subsets with respect to $A$ of $V_N$ that contain
the vector $v_N$, this obviously is $\left\{ v_N \right\}$, and we
record the size of the largest proper subset found up to now in
the tuple ${\bf r}$, so $\textbf{r}[N]=1$.

In the $i$-th step we consider $(\dd-1)$-proper subsets with
respect to $A$ in $V_i$ containing $v_i$ and record the minimum
between  size of the largest proper subset found up to now and $t$
in ${\bf r}[i]$ (row *** in the algorithm).

The tuple ${\bf r}$ for the already calculated steps enables the
pruning strategy for the search. Since we are looking for a proper
subset of size $t = d-1$, and if the vector $v_i$ is to be the
$(size)$st vector
 in the subset and $size+{\bf r}[i] < t$, then we can prune the search (row * in the algorithm).
 When the search terminates, the size $s$ of the largest $(\dd-1)$-proper
 subset  with respect to $A$ of $V$ or $t$ (if $t<s$) will then be recorded in ${\bf r}[1]$.

In Step $size$, $size>1$, we choose all $k$ dimensional vectors
with first coordinate 1 which are not linearly dependent with
$\dd-2$ column vectors of the constructed until now part of the
generator matrix. Our idea is with one pass to find all proper
vectors (all elements of $U_{size}$) using all column vectors from
 the generator matrix obtained until this step using $U_{size-1}$ (all proper
 vectors  from previous step). To find all proper
 vectors $U_{size}$ we take  $U_{size-1}$
 and delete those which are not proper, with respect to the
already constructed part (row ** in the algorithm). We find all
sums  of less than $\dd-2$ columns. Each sum gives us one vector
which is not $(\dd-1)$-proper to the current step and have to be
deleted. All remaining vectors are proper. To improve the speed of
the algorithm, we pack each column in a computer word and use the
bit operation XOR for computer words. The presented algorithm is
much faster than the algorithm in \cite{BV_IEEE05}.

\section{Results}

In this section, we present obtained result for codes with dual
distance 8 and 10 using the above algorithms. With algorithm
\textsc{Bruteforce} we construct  all codes with length $n\leq
28$, dimension $k\le 14$ and dual distance at least 8, and all
codes with dimension up to 18 and dual distance at least 10. The
summarized results for the number of inequivalent codes for given
parameters are presented in the tables below. The stars in some
cells mean that for the corresponding parameters $n$ and $k$ there
are codes with dual distance greater than the considered one. In
the remaining cases,  the number of inequivalent codes of length
$n$ and dimension $k$, given in the table,  coincide with the
number of optimal codes with minimum distance 8 (respectively 10)
which have dimension $n-k$ and length not larger than $n$. We can
use the numbers in the tables to determine the exact number of
inequivalent optimal codes with length $n'=n$ and dimension
$k'=n-k$ in some of the cases. For the cells without $*$, the
number of inequivalent optimal $[n'=n,k'=n-k,8]$ codes is equal to
the number of inequivalent $[n,k]^d$ codes minus the number of
inequivalent $[n-1,k-1]^d$ codes. The calculations took about 72
hours in contemporary PC. The number of all inequivalent codes
with dimension 15 and dual distance 8, and dimension 19 and dual
distance 9, grows exponentially, so we could not calculate all
cases. That is why we consider the problems for existence of codes
with parameters $[33,18,8]$ and $[33,15,10]$.

The existence of a $[33,18,8]$ code leads to the existence of a
$[33,18,8]$ even code and $[32,17,8]$ even code (from the
properties of shortened codes and Lemma \ref{even}) and its dual
code $C_{32}$  with parameters $[32,15,d]^8$. We know that
$L(14,8)=28$ (see Table 1) and $d \le 8$ from the tables for
bounds of linear codes \cite{new-Tables}. Theorem \ref{b_th} gives
us that the minimum distance $d$ of $C_{32}$ has to be $4 \le d
\le 8$. Using the algorithm \textsc{Extend} and already
constructed even codes with parameters $[28,14,8]^8$,
$[27,14,7]^8, \dots, [24,14,d \ge 4]^8$, we obtain that there are
exactly two inequivalent even codes $C_{32}^1$ and $C_{32}^2$ with
generator matrices $G_{32}^1$ and $G_{32}^2$ and weight
enumerators:

\noindent
$1+124z^{8}+1152z^{10}+3584z^{12}+6016z^{14}+11014z^{16}+6016z^{18}+3584z^{20}+1152z^{22}+124z^{24}+z^{32}$, and\\
$1+116z^{8}+1216z^{10}+3360z^{12}+6464z^{14}+10454z^{16}+6464z^{18}+3360z^{20}+1216z^{22}+116z^{24}+z^{32}.$\\

\begin{table}[htb]
 {\small \[
 G_{32}^1=\left( \begin{array}{c}
10000000000000000000000001111111\\
01000000000000001010101101100011\\
00100000000000000011100001010101\\
00010000000000000011010001011010\\
00001000000000011000101011110100\\
00000100000000011011101011111011\\
00000010000000010111111110111111\\
00000001000000010111110010110000\\
00000000100000010111110100001100\\
00000000010000011010010100101110\\
00000000001000001101100010101101\\
00000000000100011110101101010000\\
00000000000010000110011010011101\\
00000000000001001000110110010111\\
00000000000000100110101100111001\\
  \end{array}
\right)
\]}
\end{table}

\begin{table}[htb]

 {\small \[ G_{32}^2=
\left( \begin{array}{c}
10000000000000000000000001111111\\
01000000000000001010101101110100\\
00100000000000000011100001010101\\
00010000000000011011100111001110\\
00001000000000010111101011111000\\
00000100000000001100011101100011\\
00000010000000011000111110100100\\
00000001000000011000110010101011\\
00000000100000011000110100010111\\
00000000010000010101010100101110\\
00000000001000001101100010110110\\
00000000000100010001101101100011\\
00000000000010000111110110110000\\
00000000000001000110011010001001\\
00000000000000100110101100101110\\
  \end{array}
\right)
\]}
\end{table}

None of these two codes can be extended to a code with parameters
$[33,15]^8$. This leads to

\begin{thm}
Codes with parameters $[33,18,8]$ do not exist and $n_2(18,8)=34$.
\end{thm}

It follows that codes with parameters $[32,18,7]$ and $[33,19,7]$
do not exist and the minimum distance of a putative
$[33,19,d]^{10}$ code has to have $d<7$. From Theorem \ref{b_th} and
value of $L(18,10)=28$ we have that $d$ has to be 5 or 6.

 There are exactly 30481 codes with parameters $[27,18]^{10}$ and
 11 codes with parameters $[26,18]^{10}$ and the dual codes of all  these codes are even.
But none can be extended to a code with parameters
$[33,19,d]^{10}$.
 This leads us to the conclusion

\begin{thm}
Codes with parameters $[33,14,10]$ do not exist and $n_2(14,10)=34$.
\end{thm}

The calculations  for nonexistence of codes with parameters
$[33,18,8]$ and $[33,14,10]$ took about 4 weeks in a contemporary
PC.

From Theorem 9 and Theorem 10  and tables for bounds of codes
\cite{new-Tables}, we have:

\begin{cor}
$n_2(19,8)=35$, $n_2(20,8)=36$, $n_2(21,8)=37$, $n_2(22,8)=38$,
$n_2(15,10)=35$ and  $n_2(16,10)=36$.
\end{cor}

\begin{table}
\begin{center}
{\bf TABLE 1 - Classification results for $[n,k]^{d^\perp \ge 8}$ codes} \\

\bigskip

\begin{tabular}{| c | c | c | c | c | c |}\hline
$\bf{n} \setminus \bf{k}$ & \bf{14} & \bf{13} & \bf{12} & \bf{11} & \bf{10} \\
\hline \hline
10 & 0 & 0 & 0 & 0 & 1* \\
\hline
11 & 0 & 0 & 0 & 1* & 4* \\
\hline
12 & 0 & 0 & 1* & 5* & 1 \\
\hline
13 & 0 & 1* & 6* & 3 & 0 \\
\hline
14 & 1* & 7* & 7* & 1 & 0 \\
\hline
15 & 8* & 14* & 4 & 1 & 0 \\
\hline
16 & 24* & 16 & 5 & 1 & 0 \\
\hline
17 & 50* & 23  & 5 & 0 & 0 \\
\hline
18 & 131 & 39 & 2 & 0 & 0 \\
\hline
19 & 450 & 30 & 1 & 0 & 0 \\
\hline
20 & 1863 & 27 & 1 & 0 & 0 \\
\hline
21 & 11497 & 13 & 1 & 0 & 0 \\
\hline
22 & 46701 & 10 & 1 & 0 & 0 \\
\hline
23 & 40289 &  9& 1 & 0 & 0 \\
\hline
24 & 5177 & 10 & 1 & 0 & 0 \\
\hline
25 & 536 & 8 & 0 & 0 & 0 \\
\hline
26 & 274 & 0 & 0 & 0 & 0 \\
\hline
27 & 1 & 0 & 0 & 0 & 0 \\
\hline
28 & 1 & 0 & 0 & 0 & 0 \\
\hline
29 & 0 & 0 & 0 & 0 & 0 \\

\hline
\end{tabular}

\end{center}
\end{table}

\begin{table}
\begin{center}

{\bf TABLE 2 - Classification results for $[n,k]^{d^\perp \ge 10}$ codes} \\

\bigskip

\begin{tabular}{| c | c | c | c | c | c |}\hline
$\bf{n} \setminus \bf{k}$  & \bf{18} & \bf{17} & \bf{16} & \bf{15} & \bf{14} \\
\hline \hline
15 & 0 & 0 & 0 & 0 & 1* \\
\hline
16 &  0 & 0 & 0 & 1* &6* \\
\hline
16 &  0 & 0 & 1* & 7* &3 \\
\hline
17 &  0 & 1* & 8* & 7& 0\\
\hline
18 &  1* & 9* & 14 & 1 &0\\
\hline
19 &  10* & 24* & 7 & 0 &0\\
\hline
20 &  38* & 29* & 3 &0 & 0\\
\hline
21 &  90* & 30& 2 &0 & 0 \\
\hline
22 &  237* &39 & 0 &0 &0\\
\hline
23 &  1031* & 29& 0 &0 &0\\
\hline
24 &  11114 &6 &0 &0 &0 \\
\hline
25 & 188572  & 0 & 0 &0 &0\\
\hline
26 &  563960  &0 & 0 &0 &0\\
\hline
27 &  30481 & 0 & 0 &0 &0\\
\hline
28 &  11 & 0   & 0 & 0  &0\\
\hline
29 &  0 &  0 &0  &  0  &0\\

\hline
\end{tabular}

\end{center}
\end{table}

\begin{table}

{\small
Algorithm \textsc{Extend} (Input: $V$, $t$):\\

\noindent

$|$~~~ \textbf{var} U: array of sets; \  max,i:integer;\\
$|$~~~~~~~~~~  found:boolean;  $\hat{X}$:set;
  r:array of integer;\\
\textbf{Procedure}  ext($\hat{X}$:set; size:integer);\\
$|$~ \textbf{var} i:integer;\\
$|$~ \textbf{ \{}\\
$|$~~~~~\textbf{if} size= $t$ \textbf{then}   \textbf{ \{} print($\hat{X}$); \textbf{exit}; \textbf{ \}};\\
$|$~~~~~\textbf{if}  $|U{[size-1]}|=0$ \textbf{then}\\
$|$~~~~~~ \textbf{ \{}\\
$|$~~~~~~~~~~~\textbf{if} ((size$>$max) and (size$<t$)) \textbf{then}\\
$|$~~~~~~~~~~~~\textbf{ \{}\\
$|$~~~~~~~~~~~~~~~~~~~~max:=size;\\
$|$~~~~~~~~~~~~~~~~~~~~\textbf{if} $max<t$ \textbf{then} found:=true;\\
$|$~~~~~~~~~~~~\textbf{ \}};\\
$|$~~~~~~~~~~~~~exit;\\
$|$~~~~~~~\textbf{ \}};\\
$|$~~~~~~~\textbf{while} $|U{[size-1]}| <> 0$ \textbf{do}\\
$|$~~~~~~~\textbf{ \{}\\
$|$~~~~~~~~~~\textbf{if} $(((size + |U{[size-1]}|<=max)$ and $(max<t))$ or\\
$|$~~~~~~~~~~~~$((size + |U{[size-1]}|)<t)$ and $(max=t)))$ \textbf{then} exit;\\
$|$~~~~~~~~~~$i:=\min\{j: v_j \in U{[size-1]}\};$\\
$|$~~~~~~~~~~$\textbf{if} \ (((size + r[i])<=max)$ and $(max<t))$ or\\
$|$*~~~~~~~~~~~~~~~$((size + r[i])<t)$ and $(max=t)))$  \textbf{then}   exit;\\
$|$~~~~~~~~~~$U{[size-1]}= U{[size-1]} \backslash \{v_i \};$\\
$|$**~~~~~~~~$U{[size]}:=\{v_j: v_j \in U{[size-1]}$ and $\{\hat{X}\cup v_i \cup v_j\} \in P \}$;\\
$|$~~~~~~~~~~\textbf{if} $size< t$ \textbf{then} ext($\{\hat{X}\cup v_i \}, size + 1$);\\
$|$~~~~~~~~~~\textbf{if} $found=true$ \textbf{then} exit;\\
$|$~~~~~~~\textbf{ \}};\\
$|$~\textbf{ \}};\\

\noindent
\textbf{Procedure} Main;\\
$|$ \textbf{\{}\\
$|$~~~max:=0;\\
$|$~~~\textbf{for} $i:=|V|$ \textbf{downto} 1 \textbf{do}\\
$|$~~~\textbf{\{}\\
$|$~~~~~~~~found:=false;\\
$|$~~~~~~~~$\hat{X}:=\{v_i \}$;\\
$|$~~~~~~~~$U{[0]}:=\{v_j: v_j \in V \,  \ j>i$ and  $\{\hat{X} \cup v_j \}\in P$ \};    \\
$|$~~~~~~~~ext($\hat{X}$, 1);\\
$|$***~~~~~r[i]:=max;\\
$|$~~~\textbf{\}};\\
$|$~\textbf{\}};\\

}

\end{table}

 \addcontentsline {toc}{section}{References}

config: arxiv


TITLE: Gcd between Ideals
QUESTION [1 upvotes]: "If two ideals I and J of a ring R are relatively prime then I+J=R "
How can I determinate if I and J are relatively prime when I and J are not principal?
If they are principal I have to take generators and check if gcd is 1.
I have to proof that if I and J are relatively prime, I^n and J^n are too.
It is easy when I and J are principal, but when they aren't?
I also have to proof that if IJ=K^m (K ideal and m>=1) I and J are X^m and Y^m for some ideals X and Y of R, and I totally don't know how to do.
Sorry for my english

REPLY [1 votes]: First of all, to prove two ideals are equal you may show they are contained one into the other and viceversa. Of course you have $I+J\subset R$, to prove the opposite you just have to show that $1\in I+J$, so as ideals are closed to scalar multiplication you have $r\cdot 1\in I+J$ for every $r\in R$. Notice that each element of $I+J$ is linear combination of element of $I$ and $J$. for the second question, just try to figure it out trying to show that $1$ belongs to that sum as. For the last point, are you sure it is true for every ring? if you take just $R=\mathbb{Z}$ and $I=J=2\mathbb{Z}$, we have $IJ=I^2$ but we can't prove your claim (for $m=2$).


TITLE: Is the endomorphism algebra of a dualizable bimodule necessarily finite dimensional?
QUESTION [18 upvotes]: Let $k$ be  field. Let $A$, $B$ be $k$-algebras, and let ${}_AM_B$ be a dualizable bimodule.
Pre-Question (too naive): Is the algebra of $A$-$B$-bilinear endomorphisms of $M$ necessarily finite dimensional?
Answer: No. Take $A$ some infinite dimensional commutative algebra, and $M={}_AA_A$. Then $End({}_AA_A)=A$ is not finite dimensional.

Question: Assume that $A$ and $B$ have finite dimensional centers.
  Is it then true that the algebra of $A$-$B$-bilinear endomorphisms of $M$ has to be finite dimensional?

Special case for which I know the answer to be positive:
If $k=\mathbb C$ or $\mathbb R$, and if we're in a C*-algebra context, then I know how to prove that $End({}_AM_B)$ is finite dimensional. But my proof relies on certain inequalities, and it does not generalize.

Definitions:
A bimodule ${}_AM_B$ is called left dualizable if there is an other bimodule ${}_BN_A$ (the left dual) and maps $r:{}_AA_A\to {}_AM\otimes_BN_A$ and $s:{}_BN\otimes_AM_B\to {}_BB_B$ such that $(1\otimes s)\circ(r\otimes 1) = 1_M$ and $(s\otimes 1)\circ(1\otimes r) = 1_N$.
There's a similar definition of right dualizability. I'm guessing that right dualizability is not equivalent to left dualizability, but I don't know a concrete example that illustrates the difference between these two notions.
In my question above, I've just used the term "dualizable", which you
should interpret as

"both left and right dualizable".
There's also the notion of fully dualizable, which means that the left dual has its own left dual, which should in turn have its own left dual etc., and similarly for right duals. Once again, I'm a bit vague as to whether all these infinitely many conditions are really needed, or whether they are implied by finitely many of them.
PS: I hope that I didn't mix my left and my right.

REPLY [4 votes]: This is more or less orthogonal to Kevin's answer; he puts a stronger restriction on $A$ and $B$, while I put a stronger restriction (perhaps too strong!) on ${_A}M_B$. I likewise apologize for the lack of pictures.
If we assume that

${_A}M_B$ has a 2-sided dual ${_B}N_A$, and
the duality pairings exhibit ${_B}N \otimes_A M_B$ as a retract of ${_B}B_B$,

then $\operatorname{End}({_A}M_B)$ is dualizable as a $Z(A)$-module, and hence, if $Z(A)$ is finite dimensional, it will follow that $\operatorname{End}({_A}M_B)$ is finite dimensional.
To prove this, we exhibit a duality pairing between ${_{Z(A)}}\operatorname{End}({_A}M_B)$ and $\operatorname{End}({_B}N_A)_{Z(A)}$.
The evaluation map $\epsilon: {_{Z(A)}}\operatorname{End}({_A}M_B) \otimes_k \operatorname{End}({_B}N_A)_{Z(A)} \to {_{Z(A)}}Z(A)_{Z(A)}$ is given by first mapping to $\operatorname{End}({_A}M \otimes_B N_A)$ and then pre- and post-composing by the appropriate unit and counit, respectively, from the duality pairings between $M$ and $N$ to get an endomorphism of ${_A}A_A$, i.e., an element of $Z(A)$.
The coevaluation map $\eta: k \to \operatorname{End}({_B}N_A) \otimes_{Z(A)} \operatorname{End}({_A}M_B)$ is given by identifying the codomain with $\operatorname{End}({_B}N \otimes_A M_B)$, which follows from condition 2, whence $\eta$ is just the unit of this algebra.
That $\epsilon$ and $\eta$ give a duality pairing follows from condition 2. (The nontrivial morphism in one of the triangle diagrams sends a map $f: {_A}M_B \to {_A}M_B$ to the identity on ${_A}M_B$ tensored with the "trace" of $f$ via the duality pairings. Condition 2 tells us that the identity string can absorb the trace bubble, and similarly for the other triangle diagram.)
Unfortunately, condition 2 (or its "mirror," which would show dualizability as a $Z(B)$-module) seems very strong, but it is crucial for the above proof.


text:
\begin{document}
\title{The equation $(x-d)^5+x^5+(x+d)^5=y^n$}

\author{Michael A. Bennett}
\address{Department of Mathematics, University of British Columbia, Vancouver, BC Canada V6T 1Z2}
\email{bennett@math.ubc.ca}
\thanks{Supported in part by a grant from NSERC}

\author{Angelos Koutsianas}
\address{Department of Mathematics, University of British Columbia, Vancouver, BC Canada V6T 1Z2}
\email{akoutsianas@math.ubc.ca}

\date{\today}
\keywords{}

\begin{abstract}
In this paper, we solve the equation of the title under the assumption that $\gcd(x,d)=1$ and $n\geq 2$. This generalizes  earlier work of the first author, Patel and Siksek \cite{BennettPatelSiksek16}. Our main tools include Frey-Hellegouarch curves and associated modular forms, and an assortment of Chabauty-type techniques for determining rational  points on curves of small positive genus.
\end{abstract}

\maketitle

\section{Introduction}

 If $f(x)$ is a polynomial in $\mathbb{Z}[x]$ of degree at least three with, say, distinct complex roots, then the {\it superelliptic equation}
 $$
 f(x)=y^n
 $$
 has at most finitely many solutions in integers $x, y$ and $n$ with $|y|, n \geq 2$. This result, due to Schinzel and Tijdeman \cite{SchinzelTijdeman76},  is a consequence of bounds for linear form in logarithms and, as such, is effective -- one can readily deduce explicit bounds upon $y$ and $n$, depending only upon $f$. Actually classifying these solutions for a given polynomial $f(x)$, however, is, in most cases, still beyond current technology. By way of example, while we strongly suspect  that the equation
 $$
 x^2-2=y^n
 $$
 has only the obvious solutions with $y=-1$, this has not yet been proven.

 If we replace the polynomial $f(x)$ with a binary form $f(x,d)$ of degree $k \geq 3$, we would expect that analogous finiteness statements hold for the {\it generalized superelliptic equation}
\begin{equation} \label{GSE}
 f(x,d)=y^n, \; \; \gcd(x,d)=1,
\end{equation}
at least provided $n$ is suitably large relative to $k$. That such a result might be inherently difficult to prove  in any generality is suggested by the fact that the case $f(x,d)=xd(x+d)$ is essentially equivalent to Fermat's Last Theorem \cite{Wiles95}. Some statements for forms of degree $k \in \{ 3, 4, 6, 12 \}$ may be found in a paper of the first author and Dahmen \cite{BennettDahmen13}, though they are applicable to a vanishingly small proportion of such forms unless $k=3$.

One accessible source of  $f(x,d)$ for which equation (\ref{GSE}) has proven occasionally tractable is provided by forms derived from sums of powers of consecutive terms in arithmetic progression.  Defining
$$
S_j(x,d,k)= \sum_{i=0}^{j-1} (x+id)^k,
$$
equation (\ref{GSE}) with $f(x,d)=S_j(x,d,k)$ is a long-studied problem
(see e.g. \cite{Schaffer56,VoorhoeveGyoryTijdeman79,GyoryTijdemanVoorhoeve80,Brinza84,Urbanowicz88,Pinter97,	BennettGyoryKalmanPinter04,Pinter07,Hajdu15,PatelSiksek17,Argaez19}). The case of three terms, i.e. $j=3$,
has attracted particular attention over the years, dating back to work of Cassels and Uchiyama independently \cite{Cassels85,Uchiyama79}, who studied the case $d=1$, $n=2$ and $k=3$. More recently,
all solutions of the equation
\begin{equation} \label{three}
S_3(x,d,k)=y^n
\end{equation}
have been determined for $2 \leq k \leq 7$, and various values of $d$ (where the prime divisors of $d$ are small and fixed). The interested reader may consult \cite{Zhang14,BennettPatelSiksek16,KoutsianasPatel18,Koutsianas19,Zhang17,Argaez19, ArgaezPatel19, ArgaezPatel20, DasKoutsianas20}.

Closely related to the focus of the paper at hand is recent work of van Langen \cite{Langen19}, where equation (\ref{GSE}) is completely resolved in the case $f(x,d)=S_3(x,d,4)$, through careful consideration of two Frey-Hellegouarch $\mathbb{Q}$-curves.
We will deduce an analogous result for
$$
f(x,d)=S_3(x,d,5),
$$
though our argument is essentially rather more straightforward, utilizing elliptic curves defined over $\mathbb{Q}$.
We prove the following theorem.

\begin{theorem}\label{thm:main}
The only solutions to the equation
$$
(x-d)^5+x^5+(x+d)^5 = y^n,~x,y,d,n\in\ZZ,~n\geq 2,
$$
with $\gcd(x,d)=1$ satisfy $x=0, |d|=1$ or  $|x|=1, |d|=2$.
\end{theorem}

We note that this result generalizes an earlier theorem of the first author, Patel and Siksek \cite{BennettPatelSiksek16}, who treated the same equation under the restriction that $d=1$. The main new ingredients we employ, in comparison to \cite{BennettPatelSiksek16}, are a variety of Chabauty-type techniques to handle the small exponent cases $n \in \{ 2, 3, 5 \}$, which reduce to analyzing rational points on curves of positive genus, and an elementary observation about local behaviour at $p=3$.

For each of the four cases $n=2, 3, 5$ and $n \geq 7$, we will make use of the computational algebra package Magma \cite{Magma}. The relevant code for all the computations in this paper can be found at
\begin{center}
\url{https://github.com/akoutsianas/ap_5th_powers}.

\end{center}

\section{Preliminaries}

Here and henceforth, let us suppose that we have
\begin{equation}\label{eq:main}
(x-d)^5+x^5+(x+d)^5 = y^n,
\end{equation}
with $x$ and $d$ coprime nonzero integers, and $n\geq 2$ an integer. Without loss of generality, we may suppose that $n$ is prime.
Then
$$
x ( 3x^4+20x^2d^2+10d^4)=y^n
$$
and so, writing $\kappa = \gcd (x,10)$, there necessarily exist integers $a$ and $b$ such that
\begin{equation}\label{eq:x_power}
x = \kappa^{n-1} a^n
\end{equation}
and
\begin{equation} \label{fish}
 3x^4+20x^2d^2+10d^4 = \kappa b^n.
\end{equation}
Here, $\kappa a$ and $b$ are coprime. If $xy=0$, then from (\ref{eq:main}), $x=y=0$. We will thus suppose henceforth that $a \neq 0$ and rewrite the last equation as
\begin{equation}\label{eq:pp2_a}
10(d^2 + x^2)^2 - 7x^4 = \kappa b^n.
\end{equation}
From  \eqref{eq:x_power} and \eqref{eq:pp2_a}, we are led to the Diophantine equation
\begin{equation} \label{fowl}
7 \kappa^{4n-5} a^{4n} +  b^n  = \left( \frac{10}{\kappa} \right) T^2,
\end{equation}
where $T = d^2 + x^2$. A simple but very helpful observation is as follows. Since $\gcd(x,d)=1$ and $n \geq 2$, equation (\ref{fish}) implies that $3 \nmid d$.
It follows from (\ref{fish}), then, that
$$
\kappa b^n \equiv 1-x^2 \mod{3}.
$$
Thus either $3 \mid x$ (in which case, from (\ref{eq:x_power}), $3 \mid a$), or $3 \nmid x$, so that $x^2 \equiv 1 \mod{3}$, whence $3 \mid b$.
We thus may conclude, in all cases, that $3 \mid ab$ (i.e. that $3 \mid y$ in equation (\ref{eq:main})).

\section{The case $n \geq 7$}

We begin by treating the case of larger prime exponents, i.e. those with $n \geq 7$.
Following \cite{BennettSkinner04}, we construct signature $(n,n,2)$ Frey curves $E_\kappa$ for each value of $\kappa$, namely
\begin{equation} \label{F1}
E_1 \; \; : \; \; Y^2 = X^3 + 20TX^2 + 10b^nX,
\end{equation}
\begin{equation} \label{F2}
E_2 \; \; : \; \; Y^2 + XY = X^3 + \frac{5T - 1}{4}X^2 + 35\cdot 2^{4n-11}a^{4n}X,
\end{equation}
\begin{equation} \label{F3}
E_5 \; \; : \; \; Y^2 = X^3 + 4TX^2 + 2b^nX
\end{equation}
and
\begin{equation} \label{F4}
E_{10} \; \; : \; \; Y^2 + XY = X^3 + \frac{T-1}{4}X^2 +  7\cdot 10^{4n - 11}a^{4n}X.
\end{equation}

We denote by $\brhoEk$ the Galois representation acting on the $n$--torsion subgroup of $\Ek$. For prime $n\geq 7$, from modularity \cite{Wiles95,TaylorWiles95,BreuilConradDiamondTaylor01}, irreducibility of $\brhoEk$ \cite[Corollary 3.1]{BennettSkinner04}, the fact that $3 \mid ab$ (so that, in particular, $ab \neq \pm 1$), and Ribet's level lowering theorem \cite[Theorem 1.1]{Ribet90}, we may conclude that  there exists a  weight $2$ cuspidal newform $f$ of level $N_{E\kappa}$,
with $q$-expansion
\begin{equation} \label{frog}
f = q + \sum_{i\geq 2}a_i(f)q^i
\end{equation}
and eigenvalue field $K_f$, such that  $\Ek \sim_n f$.
Here,
$$
N_{E_\kappa} = \left\{
\begin{array}{cc}
2^8 \cdot 5^2 \cdot 7 & \mbox{ if } \kappa = 1,\\
2\cdot 5^2 \cdot 7 & \mbox{ if } \kappa = 2,\\
2^8 \cdot 5 \cdot 7 & \mbox{ if } \kappa = 5,\\
2 \cdot 5 \cdot 7 & \mbox{ if } \kappa = 10,\\
\end{array}
\right.
$$
and, by $\Ek \sim_n f$, we mean that there exists a prime ideal $\fn \mid n$ of $K_f$ such that, for almost all primes $p$,
$$
a_p(E_\kappa)\equiv a_p(f)\mod{\fn}.
$$

In order to bound $n$ we appeal to the following by-now  standard result.

\begin{proposition} \label{goblet}
Suppose $E/\QQ$ is an elliptic curve of conductor $N_E$ and $f$ is a newform of weight $2$, level $N_f\mid N_E$, and $q$-expansion as in (\ref{frog}).
Suppose $E\sim_n f$ for some prime $n$. Then there exists a prime ideal $\fn\mid n$ of $K_f$ such that, for all primes $p$:
\begin{itemize}
\item if $p\nmid nN_EN_f$ then $a_p(E)\equiv a_p(f)\mod{\fn}$,
\item if $p\nmid nN_f$ and $p \ \| \ N_E$ then $\pm(p+1)\equiv a_p(f)\mod{\fn}$.
\end{itemize}
\end{proposition}

It is important to note here that, {\it a priori}, this result does not automatically provide an upper bound upon $n$, as we might have that $a_p(E)=a_p(f)$ for all $p$. In our case, however, where we apply Proposition \ref{goblet} with $E=E_\kappa$,  the fact that $3 \mid ab$ implies that $3 \nmid nN_f$ and $3 \mid N_{E_\kappa}$, so that
\begin{equation} \label{goody}
a_3(f) \equiv \pm 4 \mod{\mathfrak{n}}.
\end{equation}
This congruence alone eliminates many possible forms $f$ from consideration and, in all cases, provides an upper bound upon $n$. Explicitly, we find after computing representatives for each Galois conjugacy class of weight $2$ newforms of levels $70, 350, 8960$ and $44800$ (there are $1$, $8$, $64$ and $196$ such classes, respectively), that $n \leq 16547$ (with the value $n=16547$ corresponding to a pair of classes of forms of level $44800$).
For the remaining pairs $(f,n)$, we may then appeal to Proposition \ref{goblet} with $E=E_\kappa$ and use the fact that $E_\kappa$ has nontrivial rational $2$-torsion (so that $a_p(E_\kappa) \equiv 0 \mod{2}$, for each prime $p$ of good reduction). For each prime $p \geq 11$,  $p \neq n$, Proposition \ref{goblet} thus implies a congruence of the shape
\begin{equation} \label{conger}
a_p(f)\equiv 2 m \mod{\fn} \; \mbox{ or } \; a_p(f)\equiv \pm (p+1) \mod{\fn},
\end{equation}
where $\fn \mid n$ and, by the Weil bounds, $m$ is an integer with $|m| \leq \sqrt{p}$. Applying these congruences to each remaining pair $(f,n)$ for each prime $11 \leq p \leq 97$, $p \neq n$, we eliminate all but a number of pairs with $n=7$ ($2$ forms at level $350$, $4$ forms at level $8960$ and $30$ forms of level $44800$), $n=11$ ($12$ forms of level $44800$), and $n=13$ ($4$ forms at level $8960$ and $8$ forms of level $44800$).

To show that these remaining $60$ pairs $(f,n)$ fail to give rise to solutions to equation (\ref{fowl}),
we will appeal to Proposition \ref{goblet} with more carefully chosen primes $p$.
This argument has its genesis in work of  Kraus \cite{Kraus98}. The key observation is that if $p \equiv 1 \mod{n}$ is not too large, relative to $n$, then $n$-th powers take on relatively few values modulo $p$ and hence, from (\ref{fowl}) and the models for the $E_\kappa$, the choices for $m$ in (\ref{conger}) are greatly restricted. An extreme example for this is the case $n=11$ and $\kappa=1$, where, choosing $p=23$, we find that (\ref{fowl}) has no solutions modulo $23$, unless $23 \mid ab$.
There are thus no values of $m$ to consider in (\ref{conger}) and hence
\begin{equation} \label{bad}
a_{23}(f)\equiv \pm 24 \equiv \pm 2 \mod{\fn},
\end{equation}
for a prime $\fn$ in $K_f$ with $\fn \mid 11$. This eliminates $8$ of the remaining  $12$ pairs $(11,f)$.
For the other $4$ pairs, (\ref{bad}) cannot be ruled out and we must choose a different prime $p$. With, for example, $p=89$, we find that for a given integer $x$, either $89 \mid x$ or
$$
x^{11} \equiv \pm 1, \pm 12, \pm 34, \pm 37 \mod{89}.
$$
From (\ref{fowl}), it follows that either $89 \mid ab$, or that the pair $(b^{11},T) $ is congruent to one of
$$
(1, \pm 28), (-1, \pm 27), (12, \pm 32), (-12, \pm 20), (37, \pm 29), (-37, \pm 7),
$$
modulo $89$. A short calculation with the model (\ref{F1}) reveals that either $89 \mid ab$ or
$$
a_{89} (E_1) \in \{ -4, 12, 14 \},
$$
i.e. that $m$ in (\ref{conger}) is restricted to the set  $\{ -2, 6, 7 \}$.
We thus have that
$$
a_{89} (f) \equiv -24, -4, 12, 14, 24 \equiv 1, \pm 2, 3, -4   \mod{\fn},
$$
for a prime $\fn$ in $K_f$ with $\fn \mid 11$. We arrive at the desired contradiction upon calculating that, for the $4$ forms in question, $a_{89} (f) \equiv 0  \mod{\fn}$.

More formally, we can state this argument as follows. Let $n\geq 7$ and $p=nt+1$ be prime, and define
$$
\mu_n(\mathbb{F}_p) = \{x_0\in\mathbb{F}_p^*~:~x_0^t=1\}.
$$
For a pair $(a,b)\in\mu_n(\mathbb{F}_p)$, let $T_{a,b}\in\mathbb{F}_p$ be such that
$$
\frac{10}{\kappa}T^2_{a,b} = 7\kappa^{4n-5}a^4 + b.
$$
For a triple $(a,b,T_{a,b})$ we denote by $E_{\kappa,a,b,T_{a,b}}$ the elliptic curve over $\mathbb{F}_p$ derived from $E_\kappa$ by replacing $(a^n,b^n,T)$ with $(a,b,T_{a,b})$.

\begin{proposition}\label{prop:Kraus}
With the above notation, suppose that $f$ is a newform with $n\nmid \mbox{Norm}_{K_f/\QQ}(4 - a_p^2(f))$. If, for all triples $(a,b,T_{a,b})$, we have that
$$
n\nmid \mbox{Norm}_{K_f/\QQ}(a_p(f) - a_p(E_{\kappa,a,b,T_{a,b}})),
$$
then no solutions to equation \eqref{fowl} arise from $f$.
\end{proposition}

\begin{proof}
The proof is similar to \cite[Proposition 8.1]{BugeaudMignotteSiksek06}.
\end{proof}

Applying Proposition \ref{prop:Kraus} with $p \in \{ 29, 43 \}$ (if $n=7$), $p \in \{ 23, 89 \}$  (if $n=11$) and $p \in \{ 53, 79, 157 \}$ (if $n=13$), completes the proof of Theorem \ref{thm:main} in case $n \geq 7$ is prime. The above computations can be found at
\begin{center}
\url{https://github.com/akoutsianas/ap_5th_powers}
\end{center}
 in the file \textit{elimination\_step.m}. We note that there are two methods one can use to compute data about classes of weight $2$ modular forms in Magma, either the classical approach, or by viewing such forms as Hilbert modular forms over $\mathbb{Q}$. In our situation, they are both reasonable computations (taking less than $24$ hours on a 2019 Macbook Pro); the Hilbert modular form package is substantially faster in the current Magma implementation. It is perhaps worth observing that the labels of forms generated by these two methods do not match.

\section{The case $n=2$}

In case $n=2$, we find immediately that  equation (\ref{fish}) has no integer solutions with  $\kappa = 1$. From (\ref{eq:pp2_a}), working modulo $8$, it follows that necessarily
 $b,d$ are odd, $x$ is even and $\kappa \in \{ 2, 10 \}$.
Since then
$$
3x^4+20x^2d^2+10d^4 \equiv 10 \mod{16},
$$
we thus have $\kappa=10$, i.e.
\begin{equation} \label{fix}
b^2 = d^4 + 200 d^2 a^4 + 3000a^8.
\end{equation}
Writing
$$
y= \frac{4d(b+d^2+100a^4)}{a^6} \; \mbox{ and } \;
x=\frac{2(b+d^2+100a^4)}{a^4},
$$
we find that
$$
y^2 = x^3 -400x^2+28000x.
$$
The latter equation defines a model for an elliptic curve, given as
134400ed1 in Cremona's  database, of rank $0$ over $\mathbb{Q}$ (with torsion subgroup of order $2$). It follows that necessarily either $ad=0$, or that $b+d^2+100a^4=0$. The first cases are excluded by assumption. If $b+d^2+100a^4=0$, then, from (\ref{fix}), again we have that $a=0$, a contradiction.

\section{The case $n=3$}

If we next assume that $n=3$, equations (\ref{eq:x_power}) and (\ref{fish}) imply that
\begin{equation}\label{eq:Picard_curve_n_3}
b^3=\left( \frac{10}{\kappa} \right) d^4 + 20 \kappa^3 d^2 a^6 + 3 \kappa^{7} a^{12}
\end{equation}
and so
\begin{equation} \label{mule}
b^3 = \left( \frac{10}{\kappa} \right) \left( d^2 + \kappa^4 a^6 \right)^2 - 7 \kappa^{7} a^{12},
\end{equation}
i.e.
\begin{equation} \label{Ellie}
Y^2= X^3 + 7 \kappa  \left( \frac{10}{\kappa} \right)^3,
\end{equation}
where
$$
Y= \frac{100 (d^2+\kappa^4a^6)}{\kappa^5 a^6} \; \mbox{ and } \; X= \frac{10b}{\kappa^3a^4}.
$$
In case $\kappa \in \{ 1, 10 \}$, the elliptic curve corresponding to (\ref{Ellie}) has Mordell-Weil rank $0$ over $\mathbb{Q}$ and trivial torsion. Since we have assumed that $a \neq 0$, we may thus suppose that $\kappa \in \{ 2, 5 \}$. In these cases, the curves corresponding to (\ref{Ellie}) have Mordell-Weil rank $1$ over $\mathbb{Q}$; we will tackle these values of $\kappa$ with two different approaches.

\vspace{0.5cm}
\paragraph{\textbf{The case $\kappa=2$.}} For this choice of $\kappa$,  from equation \eqref{mule} we have
\begin{equation}\label{eq:k_2_param}
5Z^2 = X^3 + 14Y^3,
\end{equation}
where $Z = d^2 + 2^4a^6$, $X = b$ and $Y = 2^2a^4$. It follows that $\gcd (X,Y)=1$. Suppose $K = \QQ(\rmt)$; the class number of $K$ is $3$, $\OK=\ZZ[\rmt]$ and a generator of the free part of the unit group of $\OK$ is given by
$$
r = -1 + 2\rmt + (\rmt)^2.
$$
We factorize \eqref{eq:k_2_param} over $K$ to obtain
\begin{equation}\label{eq:k_2_factorization}
5Z^2 = (X - \rmt Y) (X^2 + \rmt XY + (\rmt)^2 Y^2).
\end{equation}
There are unique prime ideals of $K$ above $2$, $3$ and $7$ which we denote by $\fp_2$, $\fp_3$ and $\fp_7$, respectively. If $\fp$ is an prime ideal of $K$ that divides $\gcd(X - \rmt Y,X^2 + \rmt XY + (\rmt)^2 Y^2)$, then we can readily conclude that $\fp\mid (3\rmt X,3\rmt Y)$. Because $\gcd (X,Y)=1$, it follows that $\fp\mid 3\rmt$, whence $\fp=\fp_2$, $\fp_3$ or $\fp_7$. There are two prime ideals of $K$ above $5$ with norm $5$ and $5^2$ which we denote by $\fp_5$ and $\fq_5$, respectively; we have  $\langle 5\rangle = \fp_5\fq_5$. From the above discussion, we may write
\begin{equation}\label{eq:k_2_fac_ideal}
\langle X - \rmt Y\rangle = \fp_2^{a_2}\fp_3^{a_3}\fp_7^{a_7}\fp_5^{a_5}\fq_5^{a^\prime_5}\fI^2,
\end{equation}
where $0\leq a_i < 2$ for $i=2,3,7$, $(a_5,a^\prime_5) = (1,0)$ or $(0,1)$ and $\fI$ is an integral ideal of $\OK$. Because $N(\fp_2)=2$, $N(\fp_3)=3$, $N(\fp_7)=7$ and $N(\fq_5)=5^2$, taking norms on both sides of equation \eqref{eq:k_2_fac_ideal} and appealing to \eqref{eq:k_2_param}, we necessarily have that $a_2=a_3=a_7=0$ and $(a_5,a^\prime_5)=(1,0)$. In conclusion,
$$
\langle X - \rmt Y\rangle = \fp_5\fI^2.
$$
The ideal $\fp_5$ is not principal but $\fp_5^3 = \langle 5 - \rmt + (\rmt)^2\rangle$. Therefore, we have
\begin{equation} \label{fox}
(X - \rmt Y)^3 = \pm r^i(5 - \rmt + (\rmt)^2)\gamma^2,
\end{equation}
where $\gamma\in\OK$ and $i=0,1$. Replacing $(X,Y)$ be $(-X,-Y)$ we can only consider the positive sign in the last equality. Suppose that
$$
\gamma = s + t\rmt + u(\rmt)^2
$$
 with $s,t,u\in\ZZ$. After expanding the left and right hand sides of (\ref{fox}), we find that either
$$
\begin{cases}
X^3 + 14Y^3 = 5s^2 - 28st + 14t^2 + 28su - 140tu + 196u^2\\
-3X^2Y =-s^2 + 10st - 14t^2 - 28su + 28tu - 70u^2,\\
 3XY^2 =  s^2 - 2st + 5t^2 + 10su - 28tu + 14u^2,
\end{cases}
$$
or
$$
\begin{cases}
X^3 + 14Y^3 = -19s^2 - 56st + 42t^2 + 84su + 532tu + 392u^2,\\
-3X^2Y = -3s^2 - 38st - 28t^2 - 56su + 84tu + 266u^2,\\
 3XY^2 = 2s^2 - 6st - 19t^2 - 38su - 56tu + 42u^2,
\end{cases}
$$
if $i=0$ or $i=1$, respectively. Since $Y = 2^2a^4$ is even, in either case we have that $s$ is necessarily even and hence so is $X^3+14Y^3$, contradicting the fact that $\gcd (X,Y)=1$.

\vspace*{0.5cm}
\paragraph{\textbf{The case $\kappa=5$.}} Finally, let us suppose that $\kappa = 5$. A solution $(a,b)$ of \eqref{eq:Picard_curve_n_3}  corresponds to a rational point on the Picard curve
$$
C:~X_1^3 = Y_1^4 + 200\cdot 5^2 Y_1^2 + 3000 \cdot 5^4,
$$
where $X_1 = \frac{2b}{a^4}$ and $Y_1 = \frac{2d}{a^3}$. Let $J = Jac(C)$ be the Jacobian of $C$. Using Magma's  \textit{PhiSelmerGroup} routine, we can prove that $\rank(J)\leq 1$, while \textit{ZetaFunction} shows that $\# J(\QQ)_{\tor}=1$. On the other hand, we have the non-trivial point
$$
P:=[(-150,\rmpic) + (-150,-\rmpic) - 2\infty]\in J(\QQ).
$$
Therefore, it follows that $J(\QQ)\cong \ZZ$ and that $P$ generates an infinite subgroup of $J(\QQ)$. Applying the code\footnote{The code can be found at \url{https://github.com/travismo/Coleman}.} from \cite{HashimotoMorrison20} with $p=19$ leads to the conclusion that $C(\QQ) = \{\infty\}$ which completes the proof for $n=3$.

\section{The case $n=5$}

From (\ref{fowl}), we are concerned, in all cases, with the genus $2$ hyperelliptic curve
$$
C:~ Y^2 = X^5 + 7 \cdot 10^5,
$$
where $Y=\frac{10^3(d^2 + x^2)}{\kappa^8a^{10}}$ and $X = \frac{10b}{\kappa^3a^4}$. Unfortunately, the rank of the Jacobian of $C$ is $3$ and Chabauty methods are not immediately applicable. Instead, we note that  the determination of $C(\QQ)$ can be reduced to the problem of computing all the points on a family of genus one curves over $\QQ(\sqrt[5]{7})$ with rational $X$-coordinates.

Let $K = \QQ(\theta)$ where $\theta^5 + 7=0$ and $\alpha = 10 \theta$. Then we can write
$$
X^5 + 7 \cdot 10^5 = (X - \alpha) \left( X^4 +\alpha X^3+ \alpha^2 X^2 + \alpha^3 X + \alpha^4 \right),
$$
from which it follows that
$$
u^2 = \delta_j (X - \alpha) \; \; \mbox{ and } \; \;  v^2 = \delta_j \left( X^4 +\alpha X^3+ \alpha^2 X^2 + \alpha^3 X + \alpha^4 \right),
$$
where $u,v\in K^*$ and $\delta_j\in K^*/K^{*2}$. Since the polynomial $x^5 + 7\cdot 10^5$ is irreducible over $\QQ$, from \cite[Section 2]{BruinStoll09} we have that $\delta_j$ is an element of $\Sel^{(2)}(C/\QQ)\subset K^*/K^{*2}$. Using Magma's function \textit{TwoSelmerGroup} we can prove that $\Sel^{(2)}(C/\QQ)=\langle a_1,a_2,a_3\rangle$, where
\begin{align*}
a_1 & = -\theta^4 - \theta^3 + \theta^2 + \theta + 1,\\
a_2 & = \theta^4 - \theta^2 - 4\theta - 5,\\
a_3 & = -5\theta^4 - 7\theta^3 - 7\theta^2 - \theta + 15.
\end{align*}
We can therefore choose $\delta_j$ to be an $\mathbb{F}_2$-linear combination of $a_1,a_2$ and $a_3$, so that in particular, we may suppose that
\begin{align*}
\delta_1 & = 1,\\
\delta_2 & = -5\theta^4 - 7\theta^3 - 7\theta^2 - \theta + 15,\\
\delta_3 & = \theta^4 - \theta^2 - 4\theta - 5,\\
\delta_4 & = 75\theta^4 + 99\theta^3 + 73\theta^2 - 41\theta - 257,\\
\delta_5 & = -\theta^4 - \theta^3 + \theta^2 + \theta + 1,\\
\delta_6 & = -33\theta^4 - 65\theta^3 - 77\theta^2 - 49\theta + 43,\\
\delta_7 & = 9\theta^4 + 7\theta^3 - 3\theta^2 - 23\theta - 47,\\
\delta_8 & = 545\theta^4 + 913\theta^3 + 993\theta^2 + 381\theta - 1251.
\end{align*}
For each $\delta_j$, we have a corresponding genus one curve
\begin{equation}
C_j:~v^2 = \delta_j\left( X^4 +\alpha X^3+ \alpha^2 X^2 + \alpha^3 X + \alpha^4 \right).
\end{equation}
It is enough to determine the points $(X,v)$ of $C_j$ such that $X\in\QQ$. The Jacobian $J_j$ of $C_j$ is an elliptic curve over $K$ and using Magma we are able to compute a subgroup of $J_j(K)$ of full rank. The rank of $J_j(K)$ for each $j$ is less than $5$ (see Table \ref{table:rank_Jj}), hence $C_j$ satisfies the Elliptic Curve Chabauty condition over $K$ for every $j$, $1 \leq j \leq 8$ \cite{Bruin03}. If $j\neq 2$,  we are able to determine a $K$-point on $C_j$; in case  $j=2$, we are unable to find an element of  $C_2(K)$ and must argue somewhat differently.

\begin{center}
\begin{table}
\begin{tabular}{| c | c || c | c |}
\hline
$j$ & $\rank(J_j(K))$ & $j$ & $\rank(J_j(K))$ \\
\hline
$1$ & $2$ & $5$ & $3$ \\
\hline
$2$ & $1$ & $6$ & $2$ \\
\hline
$3$ & $2$ & $7$ & $1$ \\
\hline
$4$ & $4$ & $8$ & $3$ \\
\hline
\end{tabular}
\caption{The rank of $J_j(K)$.}\label{table:rank_Jj}
\end{table}
\end{center}

\subsection{$C_j(K)\neq\emptyset$:} For the cases $j\neq 2$ we have computed a point $P_j\in C_j(K)$. Then it is known that $C_j$ is birationally equivalent to an elliptic curve $E_j$ over $K$ and the map $\phi_j: E_j\mapsto C_j$ may be found  explicitly in \cite[Theorem 1]{Connell92}. We also know that $E_j\simeq_K J_j$. Let $\pr:C_j\mapsto \bar{K}$ with $\pr(X,v) = X$ be the projection map and $\psi_j := \pr\circ\phi_j$. We observe that $\psi_j$ is defined over $K$.

Let $(X,Y)$ be a rational point of $C$ that corresponds to a point $(X,v)$ on $C_j$. From the above discussion,  it is enough to determine $H_j = E_j(K)\cap\psi_j^{-1}(\QQ)$ because then $X\in\psi(H_j)$. As we have already mentioned, the Elliptic Curve Chabauty condition is satisfied for all cases and hence we can apply the method in \cite{Bruin03} together with Mordell-Weil sieve \cite{BruinStoll10}, as implemented in Magma, to determine $H_j$ (we also recommend \cite[Section 4]{BremnerTzanakis04} for a nice exposition of the method). For $j \in \{ 1, 3, 5, 6, 7, 8 \}$ it is enough to apply a Chabauty argument using the primes $\fp\mid p$ of good reduction of $E_j$ with $p\leq 47$, but for $j=4$ we are forced to consider $p\leq 199$. After carrying this out, we conclude, for $j\neq 2$, that  $X=-6$ or $30$.

\subsection{$C_j(K) = \emptyset$:} For the case $j=2$, we are not able to find a point $P\in C_2(K)$. However, we know that the curve $C_2$ is a $2$-covering of the elliptic curve $J_2$ which means that if $C_2(K)\neq \emptyset$ then $C_2$ corresponds to a non-trivial point of the $2$-Selmer group of $J_2$ over $K$ that lies on the image of $J_2(K)$.

Since $J_2(K)=\ZZ/2\ZZ\times\ZZ$, the image of $J_2(K)/2J_2(K)$ corresponds to three non-trivial elements of $\Sel^{(2)}(J_2/K)$ which in turn correspond to three $2$-covering curves of $J_2$ that have a $K$-point. Using Magma's function \textit{TwoDescent} we can compute all $2$-coverings of $J_2$ that are elements of $\Sel^{(2)}(J_2/K)$. We identify the three of them that have a $K$-point (using the function \textit{Points}) and note that none of them is isomorphic to $C_2$; we may thus  conclude that $C_2(K)=\emptyset$ and so no rational points of $C$ arise from $C_2$.

The data for the various $C_j$, $P_j$ and $E_j$, together with the corresponding Magma code can be found at
\begin{center}
\url{https://github.com/akoutsianas/ap_5th_powers}
\end{center}
 in the files \textit{ap\_5th\_powers\_data.m} and \textit{ap\_5th\_powers.m}.

To sum up, we have proved that
$$
C(\QQ) = \{(30,\pm 5000),(-6,\pm 832),\infty\}.
$$
From the fact that $X = \frac{10b}{\kappa^3a^4}$ and $a \neq 0$, we thus have $b=3$ and $a=\kappa=1$, corresponding to the solutions to equation (\ref{eq:main}) with
$$
(|x|,|d|,|y|,n)=  (1, 2, 3, 5).
$$
This completes the proof of Theorem \ref{thm:main}.

\section{Concluding remarks}

In this paper, we have studied  equation (\ref{GSE}) in the case $f(x,d)=S_3(x,d,5)$.
Important to our argument is that we are able to assert that $3 \mid y$, while, at the same time, the associated ternary equation (\ref{fowl}) has the property that its coefficients are coprime to $3$. A nearly identical approach enables one to treat (\ref{GSE}) if $f(x,d)=S_j(x,d,5)$ for every fixed $j \equiv \pm 3 \mod{18}$. Indeed, writing
$$
x=z-\frac{(j-1)}{2} d,
$$
we have that
$$
S_j(x,d,5) = \frac{jz}{3}  \left( 3z^4 +5 \left( \frac{j^2-1}{2} \right) z^2d^2 + \left( \frac{(j^2-1)(3j^2-7)}{16} \right)  d^4 \right).
$$
Once again equation (\ref{GSE}) and the fact that $n \geq 2$ implies that $3 \nmid d$ and hence
$$
y^n \equiv \pm z \left( 1 - z^2 \right)  \equiv 0 \mod{3}.
$$

\section{Acknowledgment}

The second author is grateful to Sachi Hashimoto, Travis Morrison and Prof. Michael Stoll for useful discussions and to Prof. John Cremona for providing access to the servers of the Number Theory group of Warwick Mathematics Institute where all the computations took place.

\bibliographystyle{alpha}
\bibliography{my_bibliography}

\end{document}


TITLE: Which algebraic variety can become a algebraic group?
QUESTION [5 upvotes]: First, I know the algebraic group must be non-singular and the index of the identity component must be finite.
Now given a algebraic variety (especially for a algebraic curve or a algebraic surface whose picture is beautiful) with these conditions, how to judge whether we can give it a group structure and make it as a algebraic group?

REPLY [7 votes]: An algebraic group must be smooth (as you mention) and also homogeneous in the sense that given two points on it there is an automorphism of the variety sending the first to the second.
 This homogeneity condition already prevents complete smooth curves of genus $\geq 2$ from being algebraic groups (because they have finite groups of automorphisms).
Over $\mathbb C$ the complete connected algebraic groups have been  classified: they are exactly the tori $\mathbb C^g/\Lambda$, where $\Lambda$ is a lattice satisfying the Riemann bilinear conditions: see Theorem (4.2.1) page 73 of Birkenhake-Lange's Complex Abelian Varieties.


TITLE: Fourier transformation of Yukawa potential
QUESTION [1 upvotes]: I'm trying to understand the first step of the fourier transformation of the Yukawa potential.
The Yukawa potential is a spherically symmetric potential in 3 dimensions defined as $V_Y(r) = e^{-\alpha*r}/r$ with $r=|\vec{r}|$ and $\alpha >0$
We obtain the Fourier transformation by using polar coordinates (the z-axis points along the direction of k⃗):
\begin{align*}
F(V_Y(r))&=\int_{\mathbb{R}^n}^{}\frac{e^{-\alpha*r}}{r}e^{-ik\vec{r}}d^3r
\\
&=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\frac{e^{-\alpha*r}}{r}e^{-ikr\cos(\theta)}r^2\sin(\theta)dr d\theta d\phi
\end{align*}
The rest is just super painful integration.
I have a very basic understanding of how Fourier works and what it's expressing but I don't have too much of an idea how the Fourier space looks like. In this first step, I don't get how the integral becomes a spherical integral but at the same time we use polar coordinates to express $e^{-ik\vec{r}}$ as $e^{-ikr\cos(\theta)}$ and how it holds?
Thank you in advance

REPLY [1 votes]: "The z axis points along the direction of k"
so k has a direction and is a vector... $\vec k$, and the angle between it and $\vec r$ is the polar angle $\theta$. The exponent will be
 $$
exp(-i\vec k \cdot\vec r)
$$
i.e a scalar product which is naturally $k \,r \cos \theta$.


TITLE: Is Kirchoff's Voltage law valid in inductive circuits?
QUESTION [0 upvotes]: Consider a circuit with a generator and an inductor as shown -

Now, the e.m.f. that both the generator and the inductor create are due to changing magnetic fields. So, the electric field in the wires is non-conservative.
So, is Kirchoff's Voltage Law or Loop rule valid here? What is the meaning of "potential difference" across A and B or $v_\mathrm{L}$? Is it the work done to move from A to B through the wire? Is $\mathcal{E} = v_\mathrm{L}$ true or not?
For the difference between e.m.f. and electric potential, see this.
EDIT-
(From comments) I'm trying to understand the $\nabla$ operator but an explanation not involving it would be highly appreciated. There are many students at high-school level(like me) who do not understand $\nabla$ as it is not taught at that level but Kirchoff's law and inductance are taught. It would be of great help if someone gives an explanation at a lower mathematics level(which I believe can be done). The answer can include integration.

REPLY [1 votes]: This is just an answer to the second question.
In Situations like these, Potential-difference always means the terminal voltage between two points of the circuit, that is, the voltage you can measure if you take some multimeter, and connect it's ends to the two points.
As you may know, the electric field $\vec{E}$ (a vector field) can always be written as a sum of a conservative field $\vec{E}_c = - \nabla \Phi$ (which doesn't have rotation) and a source-free field  $\vec{E}_i$, with $\nabla \vec{E}_i = 0$. This is the helmholtz decomposition. We can easily identify those components as the ones generated by a potential, and the ones induced by changing magnetic fields:
In the following answer I will make the following assumption: We have a circuit whose wires are 100% conducting, their resistance is 0. This results in no electric field components in the wires, as charges can counterbalance any field infinitely fast (I'm ignoring the mass that electrons are having at this point, If you want, you can use it as the additional assumption that electrons don't have mass).
Additionally we assume, that there are no changing fields inside the multimeter, and inside the wires connecting the multimeter to two abitrary points $A$ and $B$ of your circuit.
The measurement of voltage inside your multimeter (voltage in the sense of a length integral of the electromotive force) is then exactly the difference $\Phi(\vec{r}_A) - \Phi(\vec{r}_B)$:
$\int_{\text{closed path}} d\vec{l} \vec{E}_c = 0 = \int_{\text{path from A to B}} d\vec{l} \vec{E}_c + \int_{\text{path from B over Multimeter to A}} d\vec{l} \vec{E}_c  \\ = \Phi(\vec{r}_B) - \Phi(\vec{r}_A) + \int_{\text{path over Multimeter}} d\vec{l} \vec{E} \\ = \Phi(\vec{r}_B) - \Phi(\vec{r}_A) - U_{Multimeter} = 0 $
This makes use of the the fact that inside the Multimeter $E = E_c$ because of the absence of changing fields, and the fact that there is no field in the wire connecting the multimeter. In the multimeter, the measured voltage $U_{\text{Multimeter}}$ is the electric force integrated over a certain distance, and that is why $U_{\text{Multimeter}} = \int_{\text{path over Multimeter}}$.
To be very exactly about this, most multimeters (if not all) measure the steady state current that flows through a defined ohmic resistance. This is equal to measuring a field strength over the length of the ohmic resistance, and this is equal to measuring $\int_{\text{path over Multimeter}}$.
Long story short: The Potential difference from $A$ to $B$ is the quantity you can measure when plugging in a multimeter at the two points (That's how you measure it). Furthermore, you have a field called "Potential": $\Phi(\vec{x})$ whose derivatives $ - \nabla \Phi(\vec{x})$ will give you the conservative part of the electric field. The Potential difference is by definition the difference between the values $\Phi(\vec{r}_A)$ and $\Phi(\vec{r}_B$. The Potential difference is not the work done by the electric field along a path from $A$ to $B$, but instead just the work done by the conservative Part of the electric field.


TITLE: Is $ax+b=0$ still a linear equation if $a$ and $b$ are both $2$?
QUESTION [0 upvotes]: Is $ax+b=0$ still a linear equation if $a$ and $b$ are both $2$, like below?
$$2x+2=0$$

REPLY [0 votes]: Like the first answer said, yes it is.
The equation $2x+2=0$ can also be expressed as $x=-1$, which is a vertical line intersecting the $x$ axis at $(0, -1)$.


TITLE: Confusion about a Random variable proof for CDF and Probability
QUESTION [0 upvotes]: I am confused about the following problem in my probability textbook (solution provided below it).  We're given that the random variable $X$ is less than the random variable $Y$ at the given value.  But I do not understand how they can say that $\{y\leq w\} \subset \{x \leq w\}$.
Where does that fact come from?

Show that if $X(\zeta) \leq Y(\zeta)$ for every $\zeta \in \mathcal{I}$, then $F_X(w) \geq F_Y(w)$ for every $w$.

The proof is given as:

If $Y(\zeta_i) \leq w$, then $X(\zeta_i) \leq w$ because $X(\zeta_i) \leq Y(\zeta_i)$.
Hence, $$\{Y \leq w\} \subset \{X \leq w\} \qquad \qquad \mathbb{P}(Y \leq w) \leq \mathbb{P}(X \leq w).$$
Therefore, $$F_Y(w) \leq F_X(w).$$

REPLY [2 votes]: We are told that for every outcome, $\xi$, in the sample space, then $x(\xi)<y(\xi)$.
That means for any such outcome where $y(\xi)\leq w$, then it is also true that $x(\xi)\leq w$.   (However, the converse is not necessarily so.)
Therefore every outcome in the event $\{\xi:y(\xi)\leq w\}$ is an outcome in the event $\{\xi:x(\xi)\leq w\}$.
Thus by definition of subset, $\{\xi:y(\xi)\leq w\}\subseteq\{\xi:x(\xi)\leq w\}$ .
We quite usually abbreviate this as $\{y\leq w\}\subseteq\{x\leq w\}$.
Then the properties required of a probability measure mean that: $\mathsf P(y\leq w)\leq \mathsf P(x\leq w)$


TITLE: Borel sigma-algebra for compacts
QUESTION [0 upvotes]: How can we show that the borel sigma-algebra on the interval (0, 1] is the smallest sigma-algebra that contains compact sets ?

REPLY [0 votes]: The $\sigma$-algebra generated by the compacts contains every open subinterval, so it contains every open subset.  Therefore, it contains the Borel sets.

REPLY [0 votes]: Since $(0,1]$ is a subset of $\Bbb R$, we know compactness is equivalent to being closed and bounded.  Since every subset of $(0,1]$ is bounded, this implies subsets of $(0,1]$ are compact if and only if they are closed.
Let $\mathcal{C}$ be the collection of closed (i.e., compact) subsets of $(0,1]$.  Suppose by contradiction that there is a smaller $\sigma$-algebra $\mathcal{B}'$ than the Borel $\sigma$-algebra which contains $\mathcal{C}$.
Since $\mathcal{B}'$ contains $\mathcal{C}$ and it is also closed under set complements, $\mathcal{B}'$ contains all open sets in $(0,1]$.  But then this is a $\sigma$-algebra that's smaller than the Borel $\sigma$-algebra which contains the open sets, and this is a clear contradiction.  So $\mathcal{B}'$ doesn't exist, and thus the Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing the closed subsets of $(0,1]$, and thus the compact subsets of $(0,1]$.


TITLE: Does an infinite conducting plate entirely shield off the magnetic field?
QUESTION [3 upvotes]: Let's say we have an infinite current-carrying wire parallel to an infinite conducting plate. Is the magnetic field behind the plate zero? I would say yes, because the field in the plate is zero. Also is the field in front of the plate, so the region containing the wire, given by the field produced by the wire only or is the field altered by the presence of the plate?

REPLY [2 votes]: If the plate has a finite thickness, then there will be a magnetic field on the other side. This is because the magnetic field inside a superconductor does not vanish discontinuously, but instead decays exponentially over a distance the order of the London penetration depth $\lambda_\mathrm{L}$:
$$B(x) = B_0 \exp\left(- \frac{x}{\lambda_\mathrm{L}}\right),$$
where $B_0$ is the field at the surface. $\lambda_\mathrm{L}$ can be determined from the properties of the conductor. For a plate of thickness $d$ the field on the other side would therefore be
$$B(d) = B_0 \exp\left(- \frac{d}{\lambda_\mathrm{L}}\right).$$
If $d \gg \lambda_\mathrm{L}$ the field on the other side of the plate would be negligibly small. But for $d \sim l $ there will be an appreciable magnetic field making it through to the other side.


TITLE: How Many Countable Models of the successor function
QUESTION [3 upvotes]: Consider the successor function (s(x)=x+1), $T_{S}$ to be the set of axioms given by;
S1: ∀xy[s(x)=s(y)→x=y] (injective)
S2: [s(x)≠0] (never 0)
S3: ∀x[x≠0→∃y[s(y)=x]] (everything bar 0 is in image)
S4n : ∀x[sn(x)≠x] (no cycles)
clearly $\langle N;s,0 \rangle$ is a model of TS and the upward Löwenheim–Skolem theorem theorem tells us there are models of TS of every infinite cardinality
I claim that there are $\aleph _{0}$ many countable models, how would i prove this claim?

REPLY [2 votes]: If $\langle X,s,0\rangle$ is a model, we can define an equvalence relaton on $X$ via $x\sim y\iff \exists z\in X, n,m\in \mathbb N_0\colon x=s^n(z), y=s^m(z)$. The equivalnece classes are all isomorphic to $\mathbb Z$ (with standard successor function) or $\mathbb N_0$ (for the unique class that contains $0$). Conclude that $X\cong \mathbb N_0\sqcup \mathbb Z^I$ for some set $I$ and that $X$ is countable iff $I$ is countable.


TITLE: Random walk with 3 cases.
QUESTION [8 upvotes]: Let the walking start be at $x=0$. With probability $p_1$ new $x=x+1$, with probability $p_2$: $x=x-1$ and with probability $1-p_1-p_2 \geq 0$ walking ends.
The question is what is the probability of ending on point $n$.
I've computed it numerically and getting something close to normal distribution with peak at zero, but with different dispersion on sides of $x=0$. How to get it analytically?
The provided answer seems ok, but I hope I can find some smooth function for it. Maybe with another type of solution.
Upd: my idea of solution.
$$P_{n+1} (x) = p_1 P_{n} (x-\delta x) + p_2 P_{n} (x + \delta x)$$
$$P_{n+1} (x) - P_{n} (x) = p_1 [P_n(x-\delta x) - P_n(x)] + p_2 [P_n(x+\delta x) - P_n(x)] - q\cdot P_{n}(x)$$
Dividing by $\delta n$ and $\delta x$ getting something like:
$$\frac{\partial P(n,x)}{\partial n}= \frac{\delta x}{\delta n}\left[ (p_2 + p_1) \frac{\partial P(n,x)}{\partial x} - (1 - p_2 - p_1)\cdot P(n,x)\right]$$
Solving it leads to something like this:
$$P(n,x) = Ae^{\frac{-q x}{p_1 + p_2}}(n + \frac{x}{p_1 + p_2})$$
I'm not sure yet  if it means something.

REPLY [6 votes]: We derive a closed formula and show the following is valid for $n\geq 0$
  \begin{align*}
\sum_{k=0}^\infty\binom{n+2k}{k}x^k&=
\frac{1}{\sqrt{1-4x}}
\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n\tag{1}\\
\end{align*}

Note that
\begin{align*}
c(x)=\frac{1-\sqrt{1-4x}}{2x}=1+x+2x^2+5x^3+14x^4+\cdots
\end{align*}
is the generating function of the ubiquitous Catalan numbers.
We use the coefficient of operator $[w^k]$ to denote the coefficient of $w^k$ in a series. This way we can write e.g.
\begin{align*}
[w^k](1+w)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{\infty}\binom{n+2k}{k}x^k}
&=\sum_{k=0}^\infty[w^{k}](1+w)^{n+2k}x^k\tag{2}\\
&=[w^0](1+w)^n\sum_{k=0}^\infty\left(\frac{x(1+w)^2}{w}\right)^k\tag{3}\\
&=[w^0]\frac{(1+w)^n}{1-\frac{x(1+w)^2}{w}}\tag{4}\\
&=-[w^{-1}]\frac{(1+w)^n}{x+(2x-1)w+xw^2}\\
&=-[w^{-1}]\frac{\left(1+w\right)^n}{x(w-w_0(x))(w-w_1(x))}\tag{5}\\
&=\frac{\left(1+w_0(x)\right)^n}{x(w_1(x)-w_0(x))}\tag{6}\\
&\color{blue}{=\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n}\tag{7}
\end{align*}
  and the claim (1) follows.

Comment:

In (2) we apply the coefficient of operator.
In (3) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[w^{p+q}]A(w)=[w^p]w^{-q}A(w)
\end{align*}
In (4) we apply the geometric series expansion.
In (5) we represent the denominator with poles $w_0(x)$ and $w_1(x)$.
\begin{align*}
w_0(x)&=\frac{1-\sqrt{1-4x}}{2x}-1\tag{8}\\
w_1(x)&=\frac{1+\sqrt{1-4x}}{2x}-1\\
\end{align*}
In (6) we calculate the residual of the simple pole at $w=w_0(x)$ using
\begin{align*}
\operatorname{res}_{w=a}f(w)=\lim_{w\to a}(w-a)f(w)
\end{align*}
In (7) we use the representation from (8).


TITLE: "M is reflexive" implies "M is maximal Cohen-Macaulay". Is the converse true?
QUESTION [4 upvotes]: Let $R$ be a local integrally closed domain of dimension $2$. Let $M$ be a nonzero finitely generated $R$-module. We know that "$M$ is reflexive" implies "$M$ is maximal Cohen-Macaulay". Is the converse true?

REPLY [1 votes]: If $R$ is a local normal domain with $\dim R=2$, then every MCM is reflexive.

First prove that $M$ is torsion-free. This shows that $M_{\mathfrak p}$ is free over $R_{\mathfrak p}$ for any prime $\mathfrak p$ of height $\le 1$.
Next, if $\mathfrak p$ is a prime of height $2$ it's obvious that $M_{\mathfrak p}$ satisfies Serre's condition $(S_2)$.
In the end, use Proposition 1.4.1(b) from Bruns and Herzog.
(Maybe this can help you.)


TITLE: Is Top_4 (normal spaces) a reflective subcategory of Top_3 (regular spaces)?
QUESTION [13 upvotes]: I’m studying some category theory by reading Mac Lane linearly and solving exercises.
In question 5.9.4 of the second edition, the reader is asked to construct left adjoints for each of the inclusion functors $\mathbf{Top}_{n+1} $ in $\mathbf{Top}_n$, for $n=0, 1, 2, 3$, where $\mathbf{Top}_n$ is the full subcategory of all $T_n$-spaces in Top, with $T_4$=Normal, $T_3$=Regular, etc.
For $n=0, 1, 2$, it seems to me that I can use the AFT, with the solution set constructed similarly to the one constructed for proving that Haus (=$\mathbf{Top}_2$) is a reflective subcategory of Top (Proposition 5.9.2, p. 135 of Mac Lane).
But I can’t figure out what should I do with the case of $n=3$, that is, with the inclusion functor $\mathbf{Top}_4$ in $\mathbf{Top}_3$:  $\mathbf{Top}_4$ doesn’t even have products, so it seems that I cannot use the AFT.
Is there some direct construction of this left adjoint (by universal arrows, perhaps)?  Answers including a reference would be especially helpful.

REPLY [15 votes]: I think that MacLane made a mistake.  I think that he just forgot that the category of $T_4$ spaces lacks closure properties.
Claim: If $\mathcal{A} \subseteq \mathcal{C}$ is a (full) reflective subcategory and $\mathcal{C}$ has finite products, then $\mathcal{A}$ is closed under $\mathcal{C}$'s finite products, up to isomorphism.
If $\mathcal{A}$ is reflective, it means that an object $X \in \mathcal{C}$ has an "$\mathcal{A}$-ification" $X'$, e.g., an abelianization in the case where $\mathcal{A}$ is abelian groups and $\mathcal{C}$ is groups.  If $A, B \in \mathcal{A}$ are objects, then they have a product $A \times B$ in $\mathcal{C}$, and then that has an $\mathcal{A}$-ification $(A \times B)'$.  There is an $\mathcal{A}$-ification morphism $A \times B \to (A \times B)'$, and there are also projection morphisms $A \times B \to A, B$.  Since $A, B \in \mathcal{A}$, the projection morphisms factor through $(A \times B)'$, and then the universal property gives you a morphism $(A \times B)' \to A \times B$.  So you get canonical morphisms in both directions between $A \times B$ and $(A \times B)'$, and I think that some routine bookkeeping shows that they are inverses.
So the Sorgenfrey plane (which is the non-normal Cartesian square of the Sorgenfrey line) does not have a "$T_4$-ification".
I got onto this track after I found the paper, Reflective subcategories and generalized covering spaces, by Kennison.  Kennison lists closure under products as a necessary condition for reflectiveness.