text:
\begin{definition}[Definition:Weird Number/Sequence]
The [[Definition:Integer Sequence|sequence]] of [[Definition:Weird Number|weird numbers]] begins:
:$70, 836, 4030, 5830, 7192, 7912, 9272, \ldots$
\end{definition}

config: wiki


TITLE: Allen Hatcher algebraic topology proof of theorem 4.5 Whitehead
QUESTION [1 upvotes]: It was stated as the last answer in this post:
Idea behind the proof of Whitehead's Theorem and Compression Lemma
I copy here what is asked there:
I am reading as well the proof of Whitehead's theorem in AHAT and trying to grasp the idea of what happens in the general case when $f$ is not cellular without using cellular approximation.
The final line of the argument ''$( X \times I \sqcup Y, X \times \partial I \sqcup Y) \rightarrow (M_f, X \cup Y) \rightarrow (M_f, X)$ gives a deformation retraction of $M_f$ onto $X$'' is, if I understand right, quotient by $(x,1) \sim f(x)$ in the first arrow and the homotopy $g$ explained in the book in the second. Clearly, this composition would give the deformation retraction we seek if taking the quotient in the first arrow would give a homotopy equivalence, which I am not so sure why is so.. this may be an easy algebraic fact but I'm not on algebra so if it is a silly questions I excuse myself from now lol, anyway,  Thank you very much in advance for any help on this matter.

REPLY [1 votes]: Applying the compression lemma to the inclusion $(X \cup Y,X) \to (M_f,X)$, we obtain a homotopy rel $X$ from $X \cup Y \to M_f$ to a map $X\cup Y \to X$. Since $(M_f,X \cup Y)$ has the homotopy extension property, this homotopy extends to a homotopy rel $X$ from $\text{id}_{M_f}$ to a map $g:M_f \to M_f$ such that $g(X\cup Y)\subset X$. Now apply the compression lemma again to the composition
$$ g \circ q: (X\times I \amalg Y, X\times \partial I \amalg Y) \to (M_f, X\cup Y) \to (M_f, X)$$
where $q: X\times I \amalg Y \to M_f$ is the standard quotient map. Then we obtain a homotopy $F: (X\times I \amalg Y)\times I \to M_f $ rel $X\times \partial I \amalg Y$ from $g \circ q$ to a map $X\times I \amalg Y \to X$. Since the homotopy $F$ is "rel $X\times \partial I \amalg Y$", it passes to the quotient and induces a homotopy $\bar{F}:M_f \times I \to M_f$  rel $X\cup Y$ from $g$ to a map $h:M_f \to X$. Therefore, we have $\text{id}_{M_f} \simeq g \simeq h$ rel $X$, and hence $X$ is a deformation retract of $M_f$, as desired.


TITLE: Since $2^{100} \equiv 1 \pmod {125}$ and $2^{100}$ is divisible by $8$, deduce that $2^{100} \equiv 376 \pmod {1000}$.
QUESTION [1 upvotes]: In this book, "A Course In Algebra" the author Vinberg makes this deduction without any explanation and I was wondering if anyone could help me understand it:
Since $2^{100} \equiv 1 \pmod {125}$ and $2^{100}$ is divisible by $8$, we deduce that $2^{100} \equiv 376 \pmod {1000}$ i.e., that the decimal representation of $2^{100}$ ends with $376$.
Thanks.

REPLY [1 votes]: Using the Easy CRT formula for the Chinese Remainder solution we have
$$\begin{align}&x\equiv \color{#0a0}0\!\!\!\pmod{\!8}\\ &x\equiv \color{#c00}1\!\!\!\pmod{\!125}\\[.2em]
\iff\ &x\equiv \color{#c00}1\!+\!125\left[\dfrac{\color{#0a0}0\!-\!\color{#c00}1}{\!125}\!\bmod{8}\right]\equiv\, 1+125\,[\color{#90f}3]\equiv 376\end{align}\qquad\quad$$
using the fraction calculation  $\!\bmod 8\!:\ \dfrac{-1}{125}\equiv \dfrac{-1}{-3}\equiv \dfrac{1}{3}\equiv\dfrac{9}3\equiv\color{#90f}3$
Beware $ $ Modular fractions $\,a/b := ab^{-1}$ are well-defined only for fractions whose denominator $\,b\,$ is coprime to the modulus. See here for further discussion, and see here and here for many ways to calculate modular fractions and inverses.


TITLE: About Gorenstein ring
QUESTION [4 upvotes]: Is it true that in a (non-local) Gorenstein ring, every maximal ideal has the same height? It seems a little strange, but I don't see any reason why it shoudn't.

REPLY [5 votes]: I think the answer is no. Take the product of two local Gorenstein rings of different dimensions; or take any Gorenstein ring $A$ and two prime ideals $P, Q$ of different heights, and no inclusion relation between them and localize at $A\setminus (P\cap Q)$ (you get a semi-local ring whose maximal ideals are given by $P$ and $Q$).

REPLY [5 votes]: No, it is not true.
Consider a discrete valuation ring $R$ with uniformizing parameter $\pi$ and residue field $k$.
Take for your Gorenstein ring  the polynomial ring $A=R[X] $:  since $R$ is regular, so is $A$ and thus $A$ is a fortiori Gorenstein.
Now consider the  ideals $\mathfrak p=(\pi, X)$ and $\mathfrak q=(\pi X-1)$.
We have $A/\mathfrak p=k$, a field,  and $A/\mathfrak q=Frac(R)$, a field too.  So both $\mathfrak p$ and $\mathfrak q$ are maximal ideals. However their heights are different: $ht(\mathfrak p)=2$  (use $dim A=2$) and $ht(\mathfrak q)=1$ (use Krull's principal ideal theorem)


TITLE: Partition of N with ratios close to $2$
QUESTION [2 upvotes]: a) Prove that given constants $a,b$ with $1<a<2<b$, there is no partition of the set of positive integers into two subsets $A_0$ and $A_1$ such that: if $j \in \{0,1\}$ and $m,n$ are in $A_j$, then either $\frac{n}{m} <a$ or $\frac{n}{m}>b$.
(b) Find all pairs of real numbers $(a,b)$ with $1<a<2<b$ for which the following property holds: there exists a partition of the set of positive integers into three subsets $A_0, A_1, A_2$ such that if $j \in \{0,1,2\}$ and $m,n$ are in $A_j$, then either $\frac{n}{m} <a$ or $\frac{n}{m}>b$.
I thought a partition of a set is to write such a set as a union of disjoint subsets two by two

REPLY [3 votes]: First question:
Suppose there is such a partition. Let $M = \frac{a + 1}{b - a}$. Then for any integer $m > M$, we have $bm - a(m + 1) = (b - a)m - a > 1$.
This means that there exists at least one integer $n$ such that $n \geq a(m + 1)$ and $n \leq bm$.
Since $am < a(m + 1) \leq n \leq bm$, we have $a \leq \frac{n}{m} \leq b$. Hence $n$ and $m$ are not in the same subset.
Since $a(m + 1)\leq n \leq bm < b(m + 1)$, we have $a \leq \frac{n}{m + 1} \leq b$. Hence $n$ and $m + 1$ are not in the same subset.
But there are only two subsets, so $m$ and $m + 1$ must be in the same subset.
This being true for all $m > M$ leads to a contradiction.
Note that we only used $b > a$, instead of the full inequality $1 < a < 2 < b$.

Second question:
I only have obtained a sufficient condition: $\sqrt2 < a < 2 < b \leq a^2$.
Under this condition, let us define three subsets of $\mathbb R$: for $j = 0, 1, 2$, define $B_j$ as the union $\bigcup_{k \geq 0} [a^{3k + j}, a^{3k + j + 1})$.
Now define $A_j$ as the set of positive integers in $B_j$. Let us verify that this partition indeed satisfies our condition.
Suppose $n > m$ are in one subset $A_j$. Let $k$ (resp. $t$) be the integer such that $n$ (resp. $m$) is in the interval $[a^{3k + j}, a^{3k + j + 1})$ (resp. $[a^{3t + j}, a^{3t + j + 1})$).
If $k = t$, then we have $\frac{n}{m} < \frac{a^{3k + j + 1}}{a^{3k + j}} = a$.
If $k > t$, then $k \geq t + 1$ and we have $\frac{n}{m} > \frac{a^{3(t + 1) + j}}{a^{3t + j + 1}} = a^2 \geq b$.

I don't know whether this condition is also necessary. This seems to be the key point of the question, the rest being somewhat easy.


TITLE: Proving that the maximum of a continuous function is well-separated?
QUESTION [0 upvotes]: Suppose that we have that $f(x)$ is a continuous function over the support of $x$. Assume also that $f(x)$ has a unique maximizer, $x^*$. I would like to show that maximizer is well-separated. In other words:
For every $\epsilon >0$, there exists a $k > 0$ such that:
$$
\sup_{x \ : \ d(x,x^*) \geq \epsilon} f(x) < f(x^*)-k
$$
My idea is to construct sequences using a proof by contradiction, but am running into problems there. Is there a more straightforward way to prove this? Thanks!

REPLY [0 votes]: Consider $f:[0,\infty) \to \mathbb R$ defined by
$$ f(x) = \begin{cases}
1 - x, & x \le 1 \\
1 - \frac{1}{x}, & x > 1.
\end{cases}
$$
Then, for every $\epsilon > 0$ we have
$$ \sup_{x \ge \epsilon} f(x) = 1 = f(0). $$
Notes:
It is true, if you assume a compact domain, say $X$.
As $X_{\epsilon} = \{ x\in X \mid d(x,x^*) \ge \epsilon \}$ is closed and thus compact, there exists some $x_\epsilon \in X_{\epsilon}$ which maximizes $f$ on $X_\epsilon$ and by uniqueness we have
$$ \sup_{x\in X_\epsilon} f(x) = f(x_\epsilon) < f(x^*). $$


TITLE: How many combinations are there of the word "clementine" with a maximum with 3 repeated consonants?"
QUESTION [0 upvotes]: I'm having an issue with a math question:
"How many combinations are there of the word "clementine" with a maximum with 3 repeated consonants?"
The issue I have is understanding how to account for the combinations of the consonants.
As there are 4 vowels and 6 consonants but at most only 3 consonants should be next to each other, so...
(1 C)
(1 L)
(2 N)
(1 M)
(1 T) <- Consonants
(3 E) (1 I) <- Vowels

So far I believe I'm on the correct path however I am confused on calculating how many spaces the 3 letters can fit into:
4!/(2!1) X 6!/(3!1)X (the number of spaces that 3 consonants can group in?)
Any help in understanding the process for this question better would be appreciated. I'm aware there are many questions on the subject on stackoverflow and have looked at a number including :
How many permutations of a word do not contain consecutive vowels?
However, I'm uncertain how to handle the specific instance of having multiple consonants together in the space.

REPLY [2 votes]: We will count the ways we can assign positions to the consonants, then count the number of ways the consonants and vowels can be assigned to their designated positions.
An arrangement of the word CLEMENTINE in which no more than three consonants are consecutive has the form
$$\square v \square v \square v \square v \square$$
where $v$ represents the position of a vowel and the squares represent the possible positions of the consonants.  Let $x_k$ represent the number of consonants placed in the $k$th square from the left.  Since there are six consonants and no more than three consonants can be placed in each position represented by a square,
$$x_1 + x_2 + x_3 + x_4 + x_5 = 6 \tag{1}$$
where $x_k \leq 3, 1 \leq k \leq 5$.  Equation 1 is an equation in the non-negative integers.  A particular solution corresponds to the placement of four addition signs in a row of six ones.  For instance,
$$1 + 1 + + 1 1 1 + 1$$
represents the solution $x_1 = x_2 = 1$, $x_3 = 0$, $x_4 = 3$, and $x_5 = 1$, while
$$1 + 1 + 1 + 1 + 1 1$$
represents the solution $x_1 = x_2 = x_3 = x_4 = 1$, and $x_5 = 2$. Therefore, the number of solutions of equation 1 in the nonnegative integers is the number of ways four addition signs can be placed in a row of six ones, which is
$$\binom{6 + 4}{4} = \binom{10}{4}$$
since we must choose which four of the ten positions (six addition signs and four addition signs) will be filled with addition signs.
However, we have counted solutions of equation 1 in which one or more of the variables is larger than $3$.  We must exclude these.  Note that at most one variable can exceed $3$ since $2 \cdot 4 = 8 > 6$.
Suppose $x_1 > 3$.  Since $x_1$ is an integer, $x_1 \geq 4 \implies y_1 = x_1 - 4$ is a nonnegative integer.  Substituting $y_1 + 4$ for $x_1$ in equation 1 yields
\begin{align*}
y_1 + 4 + x_2 + x_3 + x_4 + x_5 & = 6\\
y_1 + x_2 + x_3 + x_4 + x_5 & = 2 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers.  The number of solutions of equation 2 is the number of ways four addition signs can be placed in a row of two ones, which is
$$\binom{2 + 4}{4} = \binom{6}{4}$$
By symmetry, this is the number of solutions in which $x_k > 3$, $1 \leq k \leq 5$.  Hence, the number of solutions of equation 1 in which none of the variables exceeds $3$ is
$$\binom{10}{4} - \binom{5}{1}\binom{6}{4}$$
Thus far, we have counted the number of ways we can designate positions for the consonants.  Once we have chosen the positions where the consonants will be located (which determines the positions where the vowels will be located), we must arrange the consonants and vowels in those positions.  There are six consonants, of which there are $2$ N's, $1$ C, $1$ L, $1$ M, and $1$ T.  We choose two of the six positions for the $N$'s, then can place the remaining four consonants in the four remaining positions designated for the consonants.  This can be done in
$$\binom{6}{2}4!$$
ways.  That leaves the four positions for the four vowels, of which there are $3$ E's and $1$ I.  Choosing which of the four positions will be assigned an I completely determines how the vowels will be arranged in their designated positions.  Hence, the number of possible arrangements of the letters of the word CLEMENTINE in which no more than three vowels are consecutive is
$$\left[\binom{10}{4} - \binom{5}{1}\binom{6}{4}\right] \cdot \binom{6}{2}4! \cdot 4$$


TITLE: Is del defined in non-cartesian coordinates
QUESTION [3 upvotes]: I am trying to understand a list of differential operator identities in different coordinate systems. I already understand how they work in Cartesian coordinates. I will try my best to ask the question systematically:

In cylindrical coordinates, the gradient of a scalar function $f$ is given by $\nabla f=\frac{\partial f}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial f}{\partial\theta}\hat{\theta}+\frac{\partial f}{\partial z}\hat{z}$
The above point would seem to imply that I can define an operator called 'del' which is given by $\nabla=\frac{\partial}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial}{\partial\theta}\hat{\theta}+\frac{\partial}{\partial z}\hat{z}$
If I then want to calculate the divergence of a vector function $\mathbf{F}$, I take the dot product and obtain $\nabla\cdot\mathbf{F}=\frac{\partial F_r}{\partial r}+\frac{1}{r}\frac{\partial F_{\theta}}{\partial\theta}+\frac{\partial F_z}{\partial z}$
This final point is incorrect, according to the books I have checked. Apparently the divergence is instead given by $\nabla\cdot\mathbf{F}=\frac{1}{r}\frac{\partial (r F_r)}{\partial r}+\frac{1}{r}\frac{\partial F_{\theta}}{\partial\theta}+\frac{\partial F_z}{\partial z}$

My question is: why does this discrepancy arise? If somebody said "the form you gave in your fourth point is indeed the divergence, but in non-Cartesian coordinates it cannot be written in the form of a dot product between del and the vector function", then that would make sense to me, but actually I have seen this dot-product written in a textbook, where $div(\mathbf{F})$ and $\nabla\cdot\mathbf{F}$ are used interchangeably, regardless of the coordinate system. It is also written in this way on wikipedia.
Pre-empting the obvious objection: I am not using the above points to "derive del from the gradient". I am just trying to show that this form of del doesn't seem to give the correct result for the gradient and the divergence simultaneously.
As a final note I am a physics student not a mathematician, so it would be really appreciated if the answer was posed at the difficulty level of the question (or below). Thanks.

REPLY [4 votes]: Consider the expansion of the vector in cylindrical coordinates
$${\mathbf F} = F_r\,{\mathbf e}_r +F_\theta\,{\mathbf e}_\theta +F_z\,{\mathbf e}_z$$
When you apply a differential operator (grad, div, curl) you must differentiate the entire expression, the basis vectors as well as the coordinate functions.
For example $$\frac{\partial {\mathbf e}_r}{\partial \theta}={\mathbf e}_\theta$$ This is different from the situation for the Cartesian base vectors, all of whose derivatives are zero.
After accounting for the changing basis vectors, you will obtain valid results.


text:
\section{Polynomial functors and polynomial monads}
\label{sec:polynomial-functors-and-monads}

We survey elements of the theory of polynomial functors, trees, and monads. For
more comprehensive references, see \cite{Kock2011, Gambino2013}.

\subsection{Polynomial functors}

\begin{definition}
    [Polynomial functor]
    \label{def:polynomial-functor}
    A \emph{polynomial (endo)functor $P$ over $I$}\index{polynomial
    endofunctor}\footnote{The denomination ``functor'' comes from the fact that
    such a diagram induces a functor $\Set / I \xrightarrow{s^*} \Set / E
    \xrightarrow{\Pi_p} \Set / B \xrightarrow{\Sigma_t} \Set / I$ by
    composition of the pullback along $s$, dependent product along $p$, and
    dependent sum along $t$, respectively.} is a diagram in $\Set$ of the form
    \begin{equation}
        \label{eq:polynomial-functor}
        \polynomialfunctor{I}{E}{B}{I.}{s}{p}{t}
    \end{equation}
    $P$ is said to be \emph{finitary}\index{finitary polynomial functor} if the
    fibres of $p:E \longrightarrow B$ are finite sets. We will always assume
    polynomial functors to be finitary.

    We use the following terminology for a polynomial functor $P$ as in
    \cref{eq:polynomial-functor}, which is motivated by the intuition that a
    polynomial functor encodes a multi-sorted signature of function symbols.
    The elements of $B$ are called the \emph{nodes}\index{nodes} or
    \emph{operations}\index{operations} of $P$, and for every node $b$, the
    elements of the fibre $E (b) \eqdef p^{-1} (b)$ are called the
    \emph{inputs}\index{inputs} of $b$. The elements of $I$ are called the
    \emph{colours}\index{colours} or \emph{sorts}\index{sorts} of $P$. For
    every input $e$ of a node $b$, we denote its colour by $s_e (b) \eqdef s
    (e)$.
    \[
        \tikzinput{polynomial-functors}{operation}
    \]
\end{definition}

\begin{definition}
    [Morphism of polynomial functor]
    \label{def:polynomial-endofunctor-morphism}
    A morphism from a polynomial functor $P$ over $I$ (as in
    \cref{eq:polynomial-functor}) to a polynomial functor $P'$ over $I'$ (on
    the second row) is a commutative diagram of the form
    \[
        \begin{tikzcd}
            I \ar[d, "f_0"'] &
            E \pullbackcorner \ar[r, "p"] \ar[d, "f_2" left] \ar[l, "s" above] &
            B \ar[r, "t"] \ar[d, "f_1" left] &
            I \ar[d, "f_0" left] \\
            I' &
            E' \ar[r, "p'"] \ar[l, "s'" above] &
            B' \ar[r, "t"] &
            I'
        \end{tikzcd}
    \]
    where the middle square is cartesian (i.e. is a pullback square). If $P$
    and $P'$ are both polynomial functors over $I$, then a morphism from $P$ to
    $P'$ \emph{over $I$} is a commutative diagram as above, but where $f_0$ is
    required to be the identity. Let $\PolyEnd$\index{$\PolyEnd$} denote the
    category of polynomial functors and morphisms of polynomial functors, and
    $\PolyEnd (I)$\index{$\PolyEnd (I)$} the category of polynomial functors
    over $I$ and morphisms of polynomial functors over $I$.
\end{definition}

\subsection{Trees}
\label{sec:trees}

The combinatorial notion of a tree fits nicely in the framework of polynomial
functors. We now state the definition of a polynomial tree, and refer the
reader to \cite[section 1.0.3]{Kock2011} for more details about the intuition
behind it.

\begin{definition}
    [Polynomial tree]
    \label{def:polynomial-tree}
    A polynomial functor $T$ given by
    \[
        \polynomialfunctor{T_0}{T_2}{T_1}{T_0}{s}{p}{t}
    \]
    is a \emph{(polynomial) tree}\index{polynomial tree} (or just \emph{tree})
    \cite[section 1.0.3]{Kock2011} if
    \begin{enumerate}
        \item the sets $T_0$, $T_1$ and $T_2$ are finite (in particular, each
        node has finitely many inputs);
        \item the map $t$ is injective;
        \item the map $s$ is injective, and the complement of its image $T_0 -
        \im s$ has a single element, called the \emph{root}\index{root};
        \item let $T_0 = T_2 + \{ r \}$, with $r$ the root, and define the
        \emph{walk-to-root}\index{walk-to-root function} function $\sigma$ by
        $\sigma (r) = r$, and otherwise $\sigma (e) = t p (e)$; then we ask
        that for all $x \in T_0$, there exists $k \in \bbNN$ such that
        $\sigma^k (x) = r$.
    \end{enumerate}
    We call the colours of a tree its \emph{edges}\index{edge} and the inputs
    of a node the \emph{input edges}\index{input edge} of that node.

    Let $\Tree$\index{$\Tree$} be the full subcategory of $\PolyEnd$ whose
    objects are trees. Note that it is the category of \emph{symmetric} or
    \emph{non-planar} trees (the automorphism group of a tree is in general
    non-trivial) and that its morphisms correspond to inclusions of non-planar
    subtrees. An \emph{elementary tree}\index{elementary polynomial tree} is a
    tree with at most one node. Let $\elTree$\index{$\elTree$} be the full
    subcategory of $\Tree$ spanned by elementary trees.
\end{definition}

\begin{definition}
    [$P$-tree]
    \label{def:p-tree}
    For $P \in \PolyEnd$, the category $\tree P$\index{$\tree P$} of
    \emph{$P$-trees}\index{$P$-tree} is the slice $\Tree / P$. The fundamental
    difference between $\Tree$ and $\tree P$ is that the latter is always rigid
    i.e. it has no non-trivial automorphisms \cite[proposition
    1.2.3]{Kock2011}. In particular, this implies that $\PolyEnd$ does not have
    a terminal object.
\end{definition}

\begin{notation}
    \label{not:p-tree}
    Every $P$-tree $T \in \tree P$ corresponds to a morphism from a tree (which
    we shall denote by $\underlyingtree{T}$\index{$\underlyingtree{T}$}) to
    $P$, so that $T : \underlyingtree{T} \longrightarrow P$. We point out that
    $\underlyingtree{T}_1$ is the set of nodes of $\underlyingtree{T}$, while
    $T_1 : \underlyingtree{T}_1 \longrightarrow P_1$ is a decoration of the
    nodes of $\underlyingtree{T}$ by nodes of $P$, and likewise for edges.
\end{notation}

\begin{definition}
    [Address]
    \label{def:address}
    Let $T \in \Tree$ be a polynomial tree and $\sigma$ be its walk-to-root
    function (\cref{def:polynomial-tree}). We define the
    \emph{address}\index{address} function $\addr$\index{$\addr$|see {address}}
    on edges inductively as follows:
    \begin{enumerate}
        \item if $r$ is the root edge, let $\addr r \eqdef []$,
        \item if $e \in T_0 - \{ r \}$ and if $\addr \sigma (e) = [x]$, define
        $\addr e \eqdef [x e]$.
    \end{enumerate}
    The address of a node $b \in T_1$ is defined as $\addr b \eqdef \addr t
    (b)$. Note that this function is injective since $t$ is. Let
    $T^\nodesymbol$\index{$T^\nodesymbol$|see {node address}} denote its image,
    the set of \emph{node addresses}\index{node address} of $T$, and let
    $T^\leafsymbol$\index{$T^\leafsymbol$|see {leaf address}} be the set of
    addresses of leaf edges\index{leaf address}, i.e. those not in the image of
    $t$.

    Assume now that $T : \underlyingtree{T} \longrightarrow P$ is a $P$-tree.
    If $b \in \underlyingtree{T}_1$ has address $\addr b = [p]$, write
    $\src_{[p]} T \eqdef T_1 (b)$. For convenience, we let $T^\nodesymbol
    \eqdef \underlyingtree{T}^\nodesymbol$, and $T^\leafsymbol \eqdef
    \underlyingtree{T}^\leafsymbol$.
\end{definition}

\begin{remark}
    The formalism of addresses is a useful bookkeeping syntax for the
    operations of grafting and substitution on trees. The syntax of addresses
    will extend to the category of opetopes and will allow us to give a precise
    description of the composition of morphisms in the category of opetopes
    (see \cref{def:o}) as well as certain constructions on opetopic
    sets.
\end{remark}

\begin{notation}
    \label{not:marked-tree}
    We denote by $\treewithleaf P$\index{$\treewithleaf P$} the set of
    $P$-trees with a marked leaf, i.e. endowed with the address of one of its
    leaves. Similarly, we denote by $\treewithnode P$\index{$\treewithnode P$}
    the set of $P$-trees with a marked node.
\end{notation}

\begin{definition}
    [Elementary $P$-trees]
    \label{def:elementary-p-trees}
    Let $P$ be a polynomial endofunctor as in equation
    \cref{eq:polynomial-functor}. For $i \in I$, define $\itree{i} \in \tr
    P$\index{$\itree{i}$} as having underlying tree
    \begin{equation}
        \label{eq:polynomial-functor:itree}
        \polynomialfunctor
            {\{i\}}{\emptyset}{\emptyset}{\{i \},}
            {}{}{}
    \end{equation}
    along with the obvious morphism to $P$, that which maps $i$ to $i \in I$.
    This corresponds to a tree with no nodes and a unique edge decorated by
    $i$. Define $\ytree{b} \in \tr P$\index{$\ytree{b}$}, the
    \emph{corolla}\index{corolla} at $b$, as having underlying tree
    \begin{equation}
        \label{eq:polynomial-functor:ytree}
        \polynomialfunctor
            {s(E(b)) + \{*\}}{E (b)}{\{b \}}{s(E(b)) + \{*\},}
            {s}{}{}
    \end{equation}
    where the right map sends $b$ to $*$, and where the morphism $\ytree{b}
    \longrightarrow P$ is the identity on $s (E (b)) \subseteq I$, maps $*$ to
    $t (b) \in I$, is the identity on $E (b) \subseteq E$, and maps $b$ to $b
    \in B$. This corresponds to a $P$-tree with a unique node, decorated by
    $b$. Observe that for $T \in \tr P$, giving a morphism $\itree{i}
    \longrightarrow T$ is equivalent to specifying the address $[p]$ of an edge
    of $T$ decorated by $i$. Likewise, morphisms of the form $\ytree{b}
    \longrightarrow T$ are in bijection with addresses of nodes of $T$
    decorated by $b$.
\end{definition}

\begin{remark}
    \label{rem:elementary-p-trees:addresses}
    Let $P$ be a polynomial endofunctor as in equation
    \cref{eq:polynomial-functor}.
    \begin{enumerate}
        \item Let $i \in I$ be a color of $P$. Since $\itree{i}$ does not have
        any nodes, the set of its node addresses $\itree{i}^\nodesymbol =
        \emptyset$. On the other hand, the set of its leaf addresses is
        $\itree{i}^\leafsymbol = \left\{ [] \right\}$, since the unique leaf is
        the root edge.
        \item Let $b \in B$ be an operation of $P$. Then $\ytree{b}^\nodesymbol
        = \left\{ [] \right\}$ since the only node is that above the rood edge.
        For leaves, we have $\ytree{b}^\leafsymbol = \left\{ [e] \mid e \in E
        (b) \right\}$.
    \end{enumerate}
\end{remark}

\begin{definition}
    [Grafting]
    \label{def:grafting}
    For $S, T \in \tr P$, $[l] \in S^\leafsymbol$ such that the leaf of $S$ at
    $[l]$ and the root edge of $T$ are decorated by the same $i \in I$, define
    the \emph{grafting}\index{grafting} $S \graft_{[l]} T$\index{$S
    \graft_{[l]} T$|see {grafting}} of $S$ and $T$ on $[l]$ by the following
    pushout (in $\tree P$):
    \begin{equation}
        \label{eq:polynomial-functor:grafting}
        \pushoutdiagram
            {\itree{i}}{T}{S}{S \graft_{[l]} T .}
            {[]}{[l]}{}{}
    \end{equation}
    Note that if $S$ (resp. $T$) is a trivial tree, then $S \graft_{[l]} T = T$
    (resp. $S$). We assume, by convention, that the grafting operator $\circ$
    associates to the right.
\end{definition}

\begin{lemma}
    \label{lemma:grafting-addresses}
    For $S, T \in \tr P$, $[l] \in S^\leafsymbol$ such that the grafting $S
    \graft_{[l]} T$ is defined, we have
    \[
        (S \graft_{[l]} T)^\nodesymbol = S^\nodesymbol +
            \left\{ [lp] \mid [p] \in T^\nodesymbol \right\} ,
        \qqquad
        (S \graft_{[l]} T)^\leafsymbol = S^\leafsymbol - \left\{ [l] \right\} +
            \left\{ [lp] \mid [p] \in T^\leafsymbol \right\} .
    \]
\end{lemma}

\begin{notation}
    [Total grafting]
    \label{not:total-grafting}
    Take $T, U_1, \ldots, U_k \in \tr P$, where $T^\leafsymbol = \left\{ [l_1],
    \ldots, [l_k] \right\}$, and assume the grafting $T \graft_{[l_i]} U_i$ is
    defined for all $i$. Then the \emph{total grafting} will be denoted
    concisely by
    \begin{equation}
        \label{eq:big-grafting}
        T \biggraft_{[l_i]} U_i
		= ( \cdots
			(T \graft_{[l_1]} U_1) \graft_{[l_2]} U_2
		\cdots ) \graft_{[l_k]} U_k .
    \end{equation}\index{$T \biggraft_{[l_i]} U_i$}
    It is easy to see that the result does not depend on the order in which the
    graftings are performed.
\end{notation}

\begin{proposition}
    [{\cite[proposition 1.1.21]{Kock2011}}]
    \label{prop:polynomial-trees-are-graftings}
    Every $P$-tree is either of the form $\itree{i}$, for some $i \in I$, or
    obtained by iterated graftings of corollas (i.e. $P$-trees of the form
    $\ytree{b}$ for $b \in B$).
\end{proposition}

\begin{remark}
    \label{rem:tree-contexts}
    As a consequence of \cite[proposition 1.1.3]{Kock2011}, a morphism $T
    \longrightarrow S$ of $P$-trees exhibits $T$ as a subtree of $S$ as in
    \[
        S = U \graft_{[p]} T \biggraft_{[l]} V_{[l]}
    \]
    where $U$ is spanned by all the edges of $S$ that are either descendant of
    the root edge of $T$, or incomparable to it \cite[paragraphs 1.0.7 and
    1.1.11]{Kock2011}, and where $[l]$ ranges over $T^\leafsymbol$. Conversely,
    any such decomposition of $S$ induces a morphism $T \longrightarrow S$.
\end{remark}

\subsection{Polynomial monads}
\label{sec:polynomial:monads}

\begin{definition}
    [Polynomial monad]
    \label{def:polynomial-monad}
    A \emph{polynomial monad over $I$}\index{polynomial monad} is a monoid in
    $\PolyEnd(I)$. Note that a polynomial monad over $I$ is thus necessarily a
    cartesian monad on $\Set/I$.\footnote{We recall that a monad is
    \emph{cartesian}\index{cartesian monad} if its endofunctor preserves
    pullbacks and its unit and multiplication are cartesian natural
    transformations.} Let $\PolyMnd(I)$\index{$\PolyMnd(I)$} be the category of
    monoids in $\PolyEnd(I)$. That is, $\PolyMnd(I)$ is the category of
    polynomial monads over $I$ and morphisms of polynomial functors over $I$
    that are also monad morphisms.
\end{definition}

\begin{definition}
    [$(-)^\star$ construction]
    \label{def:star-construction}
    Given a polynomial endofunctor $P$ as in equation
    \cref{eq:polynomial-functor}, we define a new polynomial endofunctor
    $P^\star$\index{$P^\star$} as
    \begin{equation}
        \label{eq:free-monad}
        \polynomialfunctor
            {I}{\treewithleaf P}{\tree P}{I}
            {s}{p}{r}
    \end{equation}
    where $s$ maps a $P$-tree with a marked leaf to the decoration of that
    leaf, $p$ forgets the marking, and $r$ maps a tree to the decoration of its
    root. Remark that for $T \in \tree P$ we have $p^{-1} T = T^\leafsymbol$.
\end{definition}

\begin{theorem}
    [{\cite[section 1.2.7]{Kock2011}, \cite[sections 2.7 to 2.9]{Kock2010}}]
    \label{th:polynomial-monads-star-algebras}
    The polynomial functor $P^\star$ has a canonical structure of a polynomial
    monad. Furthermore, the functor $(-)^\star$ is left adjoint to the
    forgetful functor $\PolyMnd (I) \longrightarrow \PolyEnd (I)$, and the
    adjunction is monadic.
\end{theorem}

\begin{definition}
    [Readdressing function]
    \label{def:readdressing}
    We abuse notation slightly by letting $(-)^\star$ denote the associated
    monad on $\PolyEnd (I)$. Let $M$ be a polynomial monad as on the left
    below. Buy \cref{th:polynomial-monads-star-algebras}, $M$ is a
    $(-)^\star$-algebra, and we will write its structure map $M^\star
    \longrightarrow M$ as on the right:
    \begin{equation}
        \label{eq:polynomial-monad-structure-map}
        \polynomialfunctor{I}{E}{B}{I,}{s}{p}{t}
        \qqquad
        \begin{tikzcd}
            I
                \ar[d, equal] &
            \treewithleaf M
                \pullbackcorner
                \ar[l] \ar[d, "\readdress"] \ar[r] &
            \tree M
                \ar[r] \ar[d, "\tgt"] &
            I
                \ar[d, equal]
            \\
            I &
            E
                \ar[l] \ar[r] &
            B \ar[r] &
            I.
        \end{tikzcd}
    \end{equation}
    For $T \in \tree M$, we call $\tgt T \in B$ the \emph{target}\index{target}
    of $T$. We call $\readdress_T : T^\leafsymbol
    \stackrel{\cong}{\longrightarrow} E (r T)$\index{$\readdress$|see
    {readdressing}} the \emph{readdressing}\index{readdressing} function of
    $T$, and $\tgt T \in B$ is called the \emph{target} of $T$. If we think of
    an element $b \in B$ as the corolla $\ytree{b}$, then the target map $\tgt$
    ``contracts'' a tree to a corolla, and since the middle square is a
    pullback, the number of leaves is preserved. The map $\readdress_T$
    establishes a coherent correspondence between the set $T^\leafsymbol$ of
    leaf addresses of a tree $T$ and the elements of $E(\tgt T)$.
\end{definition}

\begin{definition}
    [Baez--Dolan $(-)^+$ construction]
    \label{def:baez-dolan-construction}
    Let $M$ be a polynomial monad as in \cref{eq:polynomial-functor}, and
    define its \emph{Baez--Dolan construction}\index{Baez--Dolan construction}
    $M^+$\index{$M^+$|see {Baez--Dolan construction}} to be
    \begin{equation}
        \label{eq:polynomial-functor:+}
        \polynomialfunctor{B}{\treewithnode M}{\tree M}{B}{\src}{p}{\tgt}
    \end{equation}
    where $\src$ maps an $M$-tree with a marked node to the label of that node,
    $p$ forgets the marking, and $\tgt$ is the target map of
    \cref{def:readdressing}. If $T \in \tree M$, remark that $p^{-1} T =
    T^\nodesymbol$ is the set of node addresses of $T$. If $[p] \in
    T^\nodesymbol$, then $\src [p] \eqdef \src_{[p]} T$.
\end{definition}

\begin{theorem}
	[{\cite[section 3.2]{Kock2010}}]
    \label{th:polynomial-functor:+:is-monad}
    If $M$ a polynomial monad, then $M^+$ has a canonical structure of a
    polynomial monad.
\end{theorem}

\begin{remark}
    [Nested addresses]
    \label{rem:nested-addresses}
    Let $M$ be a polynomial monad, and $T \in \tree M^+$. Then the nodes of $T$
    are decorated in $M$-trees, and its edge by operations of $M$. Assume that
    $U \in \tree M$ decorates some node of $T$, say $U = \src_{[p]} T$ for some
    node address $[p] \in T^\nodesymbol$.
    \begin{enumerate}
        \item The input edges of that node are in bijection with
        $U^\nodesymbol$. In particular, the address of those input edges are of
        the form $[p[q]]$, where $[q]$ ranges over $U^\nodesymbol$. This really
        motivates enclosing addresses in brackets.
        \item On the other hand, the output edge of that node is decorated by
        $\tgt U$ (where $\tgt$ is defined in \cref{def:readdressing}).
    \end{enumerate}
\end{remark}

\begin{notation}
    \label{not:edg}
    Let $M$ be a polynomial monad, and $T \in \tree M^+$. For $[a]$ the address
    of an edge of $T$, let $\edg_{[a]} T$\index{$\edg_{[l]}$} be the operation
    of $M$ decorating that edge. Explicitely, if $[a] = []$, then $\edg_{[]} T
    \eqdef \tgt \src_{[]} T$. Otherwise, $[a] = [p[q]]$ for some $[p] \in
    T^\nodesymbol$ (the node below the edge) and $[q] \in (\src_{[p]}
    T)^\nodesymbol$, and let $\edg_{[p[q]]} T \eqdef \src_{[q]} \src_{[p]} T$.
\end{notation}

config: arxiv


TITLE: Prove by the definition that $\ln:[\varepsilon, 1]\to\mathbb{R}$ is Cauchy-integrable and compute its integral
QUESTION [3 upvotes]: I'll first define the Cauchy integral (which is merely a relaxed version of the Riemann integral).
(Partitions and their norms are defined as usual, just like in the definition of the Riemann integral.)
Definition. Given a function $f:[a, b]\to\mathbb{R}$ ($ a < b$) and a partition $P = \{a = x_0 < x_1 < \dots < x_N = b\}$ of $[a, b]$, we define the Cauchy sum for $f$ corresponding to $P$ to be $C(f, P) = \sum_{n = 1}^Nf(x_{n-1})(x_n - x_{n-1})$.
Definition. A function $f:[a, b]\to\mathbb{R}$ ($ a < b$) is Cauchy-integrable iff there exists $l\in\mathbb{R}$ such that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that if $P$ is a partition of $[a, b]$ with $|P| < \delta$, then $|C(f, P) - l| < \varepsilon$. We call $l$ the Cauchy integral of $f$. (One can prove $l$ is unique, if it exists.)
For more on the Cauchy integral, see http://emp.byui.edu/BrownD/Mathematics/Calculus-R-R/Calc-R-R-Advanced/Cauchy-integral-intro.pdf
Now here's the exercise I'm having trouble with:
Exercise. Suppose $0 < \varepsilon < 1$ and consider the function $\ln:[\varepsilon, 1]\to\mathbb{R}$. Prove by the definition that such function is Cauchy-integrable and that its Cauchy integral is $-1 + \varepsilon - \varepsilon \ln(\varepsilon)$.
(This exercise comes from a course in Measure and Integration I'm taking right now. Our professor has defined the Cauchy integral and others for historical and pedagogical reasons.)
Of course, I could argue thus: since $\ln:[\varepsilon, 1]\to\mathbb{R}$ is continuous, it must be Riemann-integrable (and, therefore, Cauchy-integrable) and its integral can be computed using the Fundamental Theorem of Calculus. However, the exercise requires me to do it with only the definition of the Cauchy integral. It's not clear to me how to proceed here.
Thoughts?

REPLY [1 votes]: I would do it as follows (writing "$\log$" instead of "$\rm ln$").
First, note that for any function $f:[a,b]\to\mathbb R$ and any partition $P=(x_0,\dots ,x_N)$, we may write
\begin{eqnarray}C(f,P)&=&\sum_{n=1}^N f(x_{n-1})(x_n-x_{n-1})\\
&=&\sum_{n=1}^{N-1} x_n(f(x_{n-1})-f(x_n))+x_Nf(x_{N-1})-x_0f(x_0)\\
&=&\sum_{n=1}^{N-1} x_n(f(x_{n-1})-f(x_n))+bf(x_{N-1})-af(a)\, .
\end{eqnarray}
When specialized to $f={\log}$ on $[a,b]=[\varepsilon,1]$, this becomes
$$C({\log},P)=\sum_{n=1}^{N-1} x_n(\log(x_{n-1})-\log(x_n))+\log(x_{N-1})-\varepsilon\log(\varepsilon)\,. $$
By the mean value theorem, for each $n\in\{ 1,\dots ,N-1\}$ one can find $c_n\in (x_{n-1},x_n)$ such that $\log(x_{n-1})-\log(x_n)=\frac{1}{c_n} (x_{n-1}-x_n)$. This gives $$C(\log, P)=\sum_{n=1}^{N-1} \frac{x_n}{c_n} (x_{n-1}-x_n)+\log(x_{N-1})-\varepsilon\log(\varepsilon)\,. $$
Now, the idea is that if $\vert P\vert$ is small, then all $x_n/c_n$ are close to $1$ (because $c_n\in (x_{n-1},x_n)$) and $x_{N-1}$ is close to $x_N=1$ (so that $\log(x_{N-1})$ is close to $0$). Hence, we get
\begin{eqnarray}C(\log, P)&\sim& \sum_{n=1}^{N-1} (x_{n-1}-x_n)-\varepsilon\log(\varepsilon)\\
&=&x_0-x_{N-1}-\varepsilon\log(\varepsilon)\\
&\sim&\varepsilon-1-\varepsilon\log(\varepsilon)\, . \end{eqnarray}
I'm pretty sure that you will be able to write down the missing details.


TITLE: Inequality Weight of Cans
QUESTION [0 upvotes]: A grocery scale is designed to be accurate to within 0.25oz. If two identical cans of soup placed on the scale have combined weight of 35.15oz, what are the largest and smallest possible weights of one of the cans?

REPLY [0 votes]: The scales are accurate to 0.25 oz, and display a weight of 33.15 oz. Therefore the "actual" weight of the two cans is anything between 32.9 oz and 33.4 oz.
The largest possible weight of a single can (given they are identical) is therefore: $\frac{33.4}{2} = 16.7\mbox{ oz}$. The smallest possible weight is $\frac{32.9}{2} = 16.45\mbox{ oz}$.


TITLE: Mean Q of all points q on a cone whose origin vectors 0q are perpendicular to a given point P.
QUESTION [0 upvotes]: Consider a point $P~(P_x,P_y,0)$ which lies somewhere in the Cartesian region $x > 0. $
Consider a simple 3D cone surface with half angle $\gamma$, originating at point $\mathcal{O}$ $(0,0,0)$ where the cone axis is the positive x-axis $(x,0,0)$.
The cone surface can be described by a bundle of an infinite number of straight line segments originating at origin point $\mathcal{O}$ and each making an angle $\gamma$ with the x-axis.
Now there is an infinite set of points $q(qx,qy,qz)$ - one point per line segment - with origin vectors $\textbf{q}$ which are perpendicular to the corresponding vectors $\textbf{Pq}$ which run from point $P$ to the various $q$ points.
I wish to find a formula for the coordinates ($Qx$, $Qy$, $0$) of the "Mean Point" $Q$ defined by the average coordinates of all the $q$ points.
=====
Ideally I would like to follow the pattern used in ElThor's answer to my previous question here (i.e. using the parametrized vector $\textbf{a}=(\cos\gamma, \sin\alpha\,\sin\gamma, \cos\alpha\,\sin\gamma)$.  In that question ElThor has provided the formula for the reflected point.  Clearly this current question requires the general formula for a perpendicular point, but I haven't been able to figure it out (or find it) so far.
====
Edit: Based on ElThor's answer to the older question and taking his formula for the reflected point, adding $P$ and dividing by 2 (these operations prompted by drawing a schematic of the vector geometry) gives the following formula for the vector from the origin to the perpendicular point $\textbf{q}$:-
$$\textbf{q}=\frac{\textbf{p}\cdot\textbf{a}}{|\textbf{a}|^2}\textbf{a}  ,$$
where $\textbf{p}$ and $\textbf{q}$ are just position vectors of points $P$ and $Q$ respectively.
expanding for each xyz coordinate:-
$$\textbf{q}=\begin{bmatrix}
(P_x\cos\gamma + P_y\sin\gamma\,\sin\alpha)\cos\gamma\\
(P_x\cos\gamma + P_y\sin\gamma\,\sin\alpha)\sin\gamma\,\sin\alpha\\
(P_x\cos\gamma + P_y\sin\gamma\,\sin\alpha)\sin\gamma\,\cos\alpha\
\end{bmatrix},$$
expanding...
$$\textbf{q}=\begin{bmatrix}
(P_x\cos\gamma.\cos\gamma\ + P_y\sin\gamma.\sin\alpha.\cos\gamma)\\
(P_x\cos\gamma.\sin\gamma.\sin\alpha + P_y\sin\gamma.\sin\alpha.\sin\gamma.\sin\alpha)\\
(P_x\cos\gamma.\sin\gamma.\cos\alpha + P_y\sin\gamma.\sin\alpha.\sin\gamma.\cos\alpha)\
\end{bmatrix},$$
simplifying, and using $\sin^2 \alpha = (1/2) + (1/2)\cos2\alpha$...
$$\textbf{q}=\begin{bmatrix}
P_x\cos^2\gamma. + P_y\sin\gamma.\cos\gamma.\sin\alpha\\
P_x\cos\gamma.\sin\gamma.\sin\alpha + P_y\sin^2\gamma.((1/2) + (1/2)\cos2\alpha\\
P_x\cos\gamma.\sin\gamma.\cos\alpha + P_y\sin^2\gamma.\sin\alpha.\cos\alpha\
\end{bmatrix},$$
We obtain the mean point $Q$ by noticing that when we integrate with respect to $\alpha$ any terms which are functions of $\sin\alpha, \cos\alpha, \sin2\alpha$ or $\cos 2\alpha$ integrate to zero
$$\textbf{Q}=\begin{bmatrix}
P_x\cos^2\gamma\\
(1/2)P_y\sin^2\gamma\\
0\
\end{bmatrix}.$$
Note 1: I still lack a formal source or derivation of the formula for $q$.
Note 2: Aretino's answer provides a derivation of the formula for $q$.

REPLY [1 votes]: A formula for $\mathbf{q}$ is easy to derive. Notice first of all that $\mathbf{q}=k\mathbf{a}$, for some real number $k$. To find $k$ we can exploit perpendicularity between $\mathbf{a}$ and $\mathbf{q}-\mathbf{p}$:
$$
\mathbf{a}\cdot(\mathbf{q}-\mathbf{p})=0,
\quad\text{that is:}\quad
\mathbf{a}\cdot(k\mathbf{a}-\mathbf{p})=0,
\quad\text{whence:}\quad
k={\mathbf{p}\cdot\mathbf{a}\over|\mathbf{a}|^2}.
$$


TITLE: $\pi$-system and monotone class
QUESTION [0 upvotes]: Assume $P$ is a $\pi$-system (that is, $P$ is closed under finite intersections) and $M$ is monotone class (that is $M$ is a non-empty collection of subsets of $\Omega$ closed under monotone limits).
Show that $P\subset M$ does not imply $\sigma(P)\subset M$.

REPLY [1 votes]: Let $\Omega$ be any infinite set, and let $\preceq$ be a linear order on $\Omega$. Say that $\Delta\subseteq\Omega$ is downward closed if $\alpha\preceq \beta\in\Delta$ implies that $\alpha\in\Delta$, and let $P=\{\Delta\subseteq\Omega:\Delta\text{ is downward closed}\}$. (Note that $\varnothing$ is vacuously downward closed, so $\varnothing\in P$.) You should have no trouble showing that $P\,$ is itself a monotone class that is not closed under complementation. (In fact this is true as long as $\Omega$ has at least two elements.)


text:
\begin{document}

\sloppy

\maketitle
\pagenumbering{arabic}

\begin{abstract}
The classification of finite simple strongly real groups is complete. It is
easy to see that strong reality for
every nonabelian
finite simple group is equivalent to the fact that each element can be written
as a product of two involutions. We thus obtain a solution
to Problem 14.82 from the Kourovka notebook from  the classification
of finite simple strongly real groups.
\end{abstract}

\section*{Introduction}
In this article we solve Problem 14.82 from the Kourovka notebook [1].

\begin{probl}
{\rm [1, 14.82]} Find all finite simple groups whose every element
is a product of two involutions.
\end{probl}

Since each involution of a nonabelian finite simple group lies in an elementary
abelian subgroup of order $4$; i.e., for every
involution $t$ there exist an involution $s\not=t$ commuting with $t$, Problem
14.82 is equivalent to that of the classification of
finite simple strongly real groups. Recall that an element $x$ of $G$ is called
{\em real}
({\em strongly real}), if $x$ and
$x^{-1}$ are conjugate in $G$ (respectively are conjugate by an involution in
$G$). A group $G$ is called {\em
real} ({\em strongly real}), if all elements of $G$ are real
(strongly real). Thus, if the order of $x$ is not
equal to  $1$ or $2$, then $x$ can be written as a product of two
involutions $s$ and $t$ if and only if
$x$ is strongly real. Indeed, if $t$ is an involution inverting $x$ and $\vert
x\vert>2$ then $t$ and $tx$ are involutions and $x=t\cdot tx$.
Conversely, if there exists involutions $s$ and $t$ with $x=st$, then
$x^t=ts=x^{-1}$;
i.e., $x$ is strongly real.
The fact that in finite simple groups each element of order at most $2$ can be
written as a product of two involutions follows from the
Feit-Thompson Odd Order Theorem and the note  above.

The problem of reality and strong reality of finite simple groups and the groups
in some sense close to simple was studied by many
authors, see [2--13]. In particular, the classification of finite simple real
groups is obtained in [2]. Thus it suffices to check what
finite simple real groups are strongly real in order to solve Problem 14.82. All
strongly real alternating and sporadic groups are found respectively in
[3,\,4]. It is proven in  [5--7] that the symplectic groups
$\operatorname{PSp}_{2n}(q)$ are strongly real if and only if
$q\not\equiv3\pmod4$. The strong reality of $\Omega_{4n}^{\varepsilon}(q)$ for
$q$ even is proven in [8]. The strong reality of
${\mathrm P}\Omega_{4n}^{-}(q)$ for $q$ odd is proven in [9]. Moreover, [10,
Theorem~8.5] implies  that if  $q$ is odd, then  ${\mathrm
P}\Omega_{4n}^{+}(q)$ and $\Omega_{2n+1}(q)$, together with ${\mathrm
P}\Omega_{4n}^{-}(q)$, are strongly real if $q\equiv1\pmod4$, while
$\Omega_9(q)$ and ${\mathrm P}\Omega_8^+(q)$ are also strongly real
if $q\equiv3\pmod4$. In the
present paper the following is proven.

\begin{theorem} {\em (Main Theorem)}
$G={}^3D_4(q)$ is strongly real.
\end{theorem}

\noindent The theorem together with
[2--10]
imply the following theorems.

\begin{theorem}
Each finite simple real group is strongly real.
\end{theorem}

\begin{theorem}
Every element of a finite simple group $G$ can be written as a product of two
involutions if and only if $G$ is isomorphic to one of the
following groups:
\begin{itemize}
 \item [{\rm (1)}] $\operatorname{PSp}_{2n}(q)$ for $q\not\equiv3\pmod4$,
$n\ge1$;
\item[{\rm (2)}] $\Omega_{2n+1}(q)$ for $q\equiv1\pmod4$, $n\ge3$;
\item[{\rm (3)}] $\Omega_9(q)$ for  $q\equiv3\pmod4$;
\item[{\rm (4)}] ${\mathrm P}\Omega_{4n}^-(q)$ for $n\ge 2$;
\item[{\rm (5)}] ${\mathrm P}\Omega_{4n}^+(q)$ for  $q\not\equiv3\pmod4$,
$n\ge3$;
\item[{\rm (6)}] ${\mathrm P}\Omega_8^+(q)$;
\item[{\rm (7)}] ${}^3D_4(q)$;
\item[{\rm (8)}] $A_{10}$, $A_{14}$, $J_1$, $J_2$.
\end{itemize}
\end{theorem}

Theorem 3 gives a complete solution to Problem 14.82 from the Kourovka
notebook.

\section{Preliminary results}

Our notation for finite groups agrees with that of [14]. The notation and basic
facts for finite groups of Lie type and linear
algebraic groups can be found in [15]. A finite group $G$ is said to be a {\it
central product} of subgroups
$A$ and $B$ (which is denoted by $A\circ  B$) if $G=AB$ and the derived
subgroup $[A,B]$ is trivial. The order of a group $G$ and of
an element $g\in G$ we denote by  $| G|$ and $| g|$. If $X$ is a
subset of $G$ and
$H$ is a subgroup of $G$ then the centralizer of $X$ in $G$ and the normalizer
of $H$ in $G$ are denoted by $C_G(X)$ and $N_G(H)$,
respectively. Given a subset $X$ of $G$ by $\langle X\rangle$  we denote the
subgroup
generated by $X$. A finite field of order $q$ we
denote
by ${\mathbb F}_q$, while $p$ always denotes its characteristic; i.e.,
$q=p^\alpha$
for some positive integer~$\alpha$.
By $e$ we denote the identity element of a group, while $1$ stands for  the unit
of a field.

Let $\overline{G}$ be a simple connected algebraic group over the algebraic
closure $\overline{{\mathbb F}}_p$ of a finite field
${\mathbb F}_p$. A surjective endomorphism $\sigma$ of $\overline{G}$ is called
a
{\it  Steinberg endomorphism} (see [15,
Definition~1.15.1]), if the set of $\sigma$-stable points $\overline{G}_\sigma$
is finite.  $O^{p'}(\overline{G}_\sigma)$ is known
to be a finite group of Lie type, and each finite group of Lie type can be
obtained in this way (notice that given a finite group of Lie type a
corresponding algebraic group and a Steinberg map are not uniquely determined in
general).
More detailed definitions and related results can be found in [15,
Sections~1.5,~2.2]. If $\overline{G}$ is simply connected then
$\overline{G}_\sigma=O^{p'}(\overline{G}_\sigma)$ by
[15, Theorem~2.2.6(f)].
Moreover,
[16, Proposition~2.10] implies that the centralizer of every semisimple element
is a connected reductive subgroup of maximal rank
in~$\overline{G}$.

If $G$ is isomorphic to ${}^3D_4(q)$, then a corresponding algebraic group
$\overline{G}$ can be chosen simply connected. We always
assume that $\overline{G}$ is simply connected in this case, i.e., for every
$G={}^3D_4(q)$ a simply connected connected simple linear
algebraic group
$\overline{G}=D_4(\overline{{\mathbb F}}_q)$, where $\overline{{\mathbb F}}_q$
is the
algebraic closure of
${\mathbb F}_q$, and a Steinberg endomorphism $\sigma$ are chosen so that
$G=\overline{G}_\sigma$. In particular, the
centralizer of every semisimple element in $\overline{G}$ is connected. If
$\overline{T}$ is a
$\sigma$-stable maximal torus of $\overline{G}$ then  $T=\overline{T}\cap G$ is
called a  {\it  maximal torus} of a finite group of Lie type
$G$. If $\overline{R}\leq \overline{S}$ are $\sigma$-stable subgroups of
$\overline{G}$, $R=\overline{R}\cap G$, and  $S=\overline{S}\cap
G$; then  $N_{\overline{S}}(\overline{R})\cap G$ is denoted by $N(S,R)$. Notice
that $N(S,R)\leq N_S(R)$, but the equality is not true in
general. For every $x\in G$ there exist unique elements $s,u\in G$ such that
$x=su=us$, $s$ is semisimple, and  $u$ is unipotent. Furthermore, $s$ is the
$p'$-part of $x$, while  $u$ is the  $p$-part of
$x$. This is called the {\it Jordan decomposition} of~$x$.

By [9, Lemma~10], all semisimple elements of ${}^3D_4(q)$ are strongly real.
Moreover, the conjugating involution found in the proof
of [9, Lemma~10] satisfies to the following property.

\begin{lem} For every maximal torus $T$ of ${}^3D_4(q)$ there exists an
involution
$x \in N(G,T)$ such that $t^x=t^{-1}$ for every ~${t \in T}$. In particular, for
every $t\in T$, both $xt$ and $tx$ are involutions
inverting every element of~$T$.
\end{lem}

All statements from the next lemma are immediate from the structure of
projective linear groups of degree~$2$.

\begin{lem}
The following hold:
  \begin{itemize}
 \item[{\em (1)}] $\operatorname{PSL}_2(q)$ is strongly real if and only if
$q\not\equiv3\pmod4$.
\item[{\em (2)}] $\operatorname{PGL}_2(q)$ is strongly real.
\item[{\em (3)}] If $q$ is odd, $u$ is a nonidentity unipotent element of
$\operatorname{PGL}_2(q)$, and $t\in\operatorname{PGL}_2(q)$ is chosen so that
$u^t=u^k$ for some $k\in{\mathbb N}$; then $t$ lies in a Cartan
subgroup {\rm (}which is cyclic of order $q-1)$ of $\operatorname{PGL}_2(q)$
normalizing a unique maximal unipotent subgroup
of~$\operatorname{PGL}_2(q)$ containing $u$.
 \end{itemize}
\end{lem}

\section{Proof of the main theorem}

Let $G={}^3D_4(q)$ and $g\in G$. If $g$ is semisimple then by [9, Lemma~10] it
is strongly real. If
$g$ is unipotent and  $q$ is even then [12, Theorem~1] implies that $g$ is
strongly real. Assume that $g$ is unipotent, $q$ is odd and
$C_G(g)$ does not contain nonidentical semisimple elements; i.e., $C_G(g)$ is a
$p$-group. By [2, Lemma~5.9] there exists
$x \in G$ such that $g^x=g^{-1}$.  Clearly we may assume that  $|x|=2^k$ for
some $k \in {\mathbb N}$. Then $x^2 \in C_G(g)$ and
$| x^2|$ is a power of~$2$. Therefore, $x^2$ is semisimple, whence $x^2=e$. If
$g$ is unipotent, $q$ is odd
and $C_G(g)$ contains a nonidentical semisimple element $s$; then we consider
$g_1=sg$. Decomposition $sg$ is the Jordan decomposition of
$g_1$. If we show the existence of an involution $x$ inverting $g_1$ then the
uniqueness of the Jordan decomposition implies
$s^x=s^{-1}$ and $g^x=g^{-1}$. Thus we may assume that $g$ has a ``mixed
order''; i.e., in the Jordan decomposition
$g=su$ both $s$ and $u$ are nonidentical.

Assume that $C=C_G(s)$. Then $u \in C$. Moreover,
$\overline{C}=C_{\overline{G}}(s)$ is a connected reductive subgroup of maximal
rank in
$\overline{G}$ and $C=\overline{C}_\sigma$. Clearly, every maximal torus $T$ of
$G$, which contains $s$, is included in $C$.
The structure of the centralizers of semisimple elements is given in [17,
Proposition~2.2]. Tables~2.2a and~2.2b from [17] are
the main technical instrument in the forthcoming arguments. If $q$ is even then
up to conjugation in $G$ there exist $8$
centralizers of order divisible by $p$ of nonidentical semisimple elements. If
$q$ is odd then there exist $9$ centralizers of this sort. We
consider each centralizer separately. Note that
$\overline{C}=\overline{M}\circ\overline{S}$, where
$\overline{M}=[\overline{C},\overline{C}]$ is connected and semisimple, while
$\overline{S}=Z(\overline{C})^0$ is a torus. Furthermore, $C$ possesses a
normal subgroup
$M\circ S$, where $S=\overline{S}_\sigma\leq Z(C)$ and
$M=\overline{M}_\sigma=O^{p'}(C)$, and the structure of $M$ ($=M_{\sigma}$ in
the
notation from [17]) and $S$ ($=S_\sigma$ in the notation from [17]) is given
in [17, Tables~2.2a,~2.2b], where the structure or the order
of~${C/(M\circ S)}$ is also given. The indeces of elements below are chosen as
in [17, Tables~2.2a, ~2.2b]. Moreover, the subgroups and factor
groups of $C$ are isomorphic to classical groups in a natural way, and we
identify the subgroups and the factor groups of $C$ with the
corresponding classical groups.

Let $s$ be such that its centralizer is conjugate to the centralizer of $s_2$
(hence $s$ is an involution and this case can occur only if
$q$ is odd). Then $M\simeq{\operatorname{SL}_2(q^3)}\circ
\operatorname{SL}_2(q)$, $|Z(M)|=2$ and $S=\{e\}$. Moreover, $|C:M|=2$ and, by
[18, Theorem~2],
$C/\operatorname{SL}_2(q)\simeq \operatorname{PGL}_2(q^3)$ and
$C/\operatorname{SL}_2(q^3)\simeq
\operatorname{PGL}_2(q)$. We write $u$ as $u_1\cdot u_2$, where $u_1 \in
\operatorname{SL}_2(q^3)$, $u_2 \in \operatorname{SL}_2(q)$, and let $v_1, v_2$
be the images of
$u_1, u_2$ in $C/\operatorname{SL}_2(q)$ and
$C/\operatorname{SL}_2(q^3)$, respectively. Assume first that $q\equiv 1\pmod4$.
Then $\operatorname{PSL}_2(q)$ and
$\operatorname{PSL}_2(q^3)$ are strongly real. Therefore,  there exist
involutions
$t_1 \in \operatorname{PSL}_2(q^3)$, $t_2 \in
\operatorname{PSL}_2(q)$ such that $v_1^{t_1}=v_1^{-1}$ and
$v_2^{t_2}=v_2^{-1}$. Let $z_1$, $z_2$ belong to the preimages of
$t_1$, $t_2$ in $\operatorname{SL}_2(q^3)$ and
$\operatorname{SL}_2(q)$, respectively. Then $| z_1|=4=| z_2|$ and $z_1^2\in
Z(\operatorname{SL}_2(q^3))$, $z_2^2\in Z(\operatorname{SL}_2(q))$. It follows
that $z_1^2=z_2^2$ in $M$, whence $(z_1z_2)^2=e$. Thus $z_1z_2$ is an inverting
involution for $u$.
Assume now that  $q\equiv 3\pmod4$. In this case there exist involutions $t_1\in
\operatorname{PGL}_2(q^3)\setminus \operatorname{PSL}_2(q^3)$, $t_2\in
\operatorname{PGL}_2(q)\setminus
\operatorname{PSL}_2(q)$ such that $v_1^{t_1}=v_1^{-1}$ and
$v_2^{t_2}=v_2^{-1}$. Furthermore, $t_1$ lies in a Cartan subgroup of
$\operatorname{PGL}_2(q^3)$, i.e., in a maximal torus of
$\operatorname{PGL}_2(q^3)$ of order $q^3-1$, while $t_2$ lies in a Cartan
subgroup of $\operatorname{PGL}_2(q)$, i.e., in a maximal torus of
$\operatorname{PGL}_2(q)$ of order $q-1$. Let $T$ be a maximal torus of $C$ such
that its images under the natural homomorphisms
$C\rightarrow C/\operatorname{SL}_2(q)$ and $C/\operatorname{SL}_2(q^3)$
contain elements $t_1$ and $t_2$, respectively.
Then  $| T|=(q^3-1)(q-1)$ and, by [17, Table~1.1], $T\simeq {\mathbb
Z}_{q^3-1}\times {\mathbb Z}_{q-1}$.
In particular,  $T$ does not contain elements of order $4$. Let $z$ be a
preimage of $t_1$ in $T$. We may assume that $z$ is a
$2$-element; hence $z^2=e$. Moreover, since $t_1$ does not lie in
$\operatorname{PSL}_2(q^3)$,
we see that $z$ does not lie in $M$. Consider a natural homomorphism
$\widetilde{\phantom{G}}:C\rightarrow C/\operatorname{SL}_2(q^3).$
Since $z\not\in M$, we obtain $\tilde{z}\not\in \operatorname{PSL}_2(q)$, and so
$t_2
\operatorname{PSL}_2(q)=\tilde{z} \operatorname{PSL}_2(q)$. Moreover,
$t_2,\tilde{z}\in
\widetilde{T}\simeq {\mathbb Z}_{q-1}$; hence $t_2=\tilde{z}$. Thus $z$ lies in
the
preimage of $t_2$ as well. We obtain that $z$ is an
inverting involution for~$u$.

Let $s$ be such that its centralizer is conjugate either to the centralizer
of $s_5$ or to the centralizer of $s_{10}$. Then
$| C: (M\circ S)|=(2,q-1)$,  $M\simeq
\operatorname{SL}_2(q)$, and $S\simeq {\mathbb Z}_{q^3-\varepsilon}$, where
$\varepsilon=1$ if $C_G(s)$ is conjugate to $C_G(s_5)$ and $\varepsilon=-1$
if
$C_G(s)$ is conjugate to $C_G(s_{10})$. Moreover,
$C/S\simeq \operatorname{PGL}_2(q)$. We choose a maximal torus  $T$ of $C$ so
that $T\cap M$ is a Cartan subgroup of $M$. Since $M\simeq
\operatorname{SL}_2(q)$, we use matrices from $\operatorname{SL}_2(q)$ to write
elements of $M$, assuming that $T\cap M$ is the group of diagonal matrices.  By
Lemma~1, there exists an involution $x \in N(G,T)$ inverting each $t\in T$. In
particular,  $x$ inverts $s$ so $x$ normalizes $C_{\overline{G}}(s)$, i.e.,
$x\in N(G,C)$. Therefore, $x$ normalizes $\overline{S}$, and so it normalizes
$S$.
Put $C_0=\langle C, x\rangle$ and let $\widetilde{\phantom{C}}: C_0\rightarrow
C_0/S$ be the natural homomorphism. Since $M=O^{p'}(C)$ is characteristic in
$C$, $x$ induces an automorphism of $M$ of order  2. By [19, Lemma~2.3],
$N(G,C)$ does not induce field automorphisms on $M$. Moreover, $x\not\in
\widehat{M}$, where $\widehat{M}$ is a group of inner-diagonal automorphisms of
${M}$, since $C/S\simeq
\operatorname{PGL}_2(q)\simeq \widehat{M}$. So
$C_0/S=\widetilde{C}_0\simeq \operatorname{PGL}_2(q)\times {\mathbb Z}_2.$ The
elements of $\widetilde{C}$ are written below as projective images of matrices
from
$\operatorname{GL}_2(q)$. Up to conjugation in $\widetilde{C}_0$ we may assume
that
$$\tilde{u}=\left[\begin{array}{rr} 1 & \alpha \\
0 & 1 \end{array}\right]$$
for some $\alpha \in {\mathbb F}_q$. Let
$\overline{T}\in\overline{G}$  be such that $T=\overline{T}_\sigma$.
Then $x$ normalizes $\overline{T}$. So, acting  by conjugation,
$x$ leaves invariant the set of maximal unipotent subgroups of
$\overline{M}$, that are normalized by
$\overline{T}\cap \overline{M}$. Furthermore, since $x$ is stable under
$\sigma$; therefore,  $x$ normalizes the subgroups of $\sigma$-stable points of
these unipotent subgroups. Since
$\overline{M}=[\overline{C},\overline{C}]\simeq
\operatorname{SL}_2(\overline{{\mathbb F}}_q)$, there exists exactly two maximal
unipotent subgroups of $\overline{M}$ that are normalized by
$\overline{T}$:
one of them consists of upper-triangular matrices, another consists of
lower-triangular matrices. So
$x$ either leaves these subgroups invariant or interchanges these subgroups.
Thus, either $\tilde{u}^{\tilde{x}}=\left[\begin{array}{rr} 1 &
\beta \\
0 & 1 \end{array}\right]$,  or
$\tilde{u}^{\tilde{x}}=\left[\begin{array}{rr} 1 & 0 \\
\beta & 1 \end{array}\right],$  for some $\beta \in {\mathbb F}_q$
Going back to elements $u$ and $x$ in $C_0$ and using the fact that $p$ is
coprime to
$|S|$ we derive that  $
u=\left(\begin{array}{rr} 1 & \alpha \\
0 & 1 \end{array}\right)$  and either
$u^x=\left(\begin{array}{rr} 1 & \beta \\
0 & 1 \end{array}\right),$ or $\left(\begin{array}{rr} 1 & 0 \\
\beta & 1 \end{array}\right).
$

Let $u^x=\left(\begin{array}{rr} 1 & \beta \\
0 & 1 \end{array}\right)$. Then there exists $\tilde{t}\in
\operatorname{PGL}_2(q)\cap
\widetilde{T}$ such that $${\left[\begin{array}{rr} 1 & \beta \\
0 & 1 \end{array}\right]}^{\tilde{t}}=\left[\begin{array}{rr} 1 & -\alpha \\
0 & 1 \end{array}\right].$$ Therefore,
$u^{xt}=\left(\begin{array}{rr} 1 & -\alpha \\
0 & 1 \end{array}\right)=u^{-1}$ and $(xt)^2=t^xt=t^{-1}t=e$.

Assume that $u^x=\left(\begin{array}{rr} 1 & 0 \\
\beta & 1 \end{array}\right)$. Then there exists $\tilde{t}\in
\operatorname{PGL}_2(q)\cap \widetilde{T}$ such that
$${\left[\begin{array}{rr} 1 & 0 \\
\beta & 1 \end{array}\right]}^{\tilde{t}}=\left[\begin{array}{rr} 1 & 0 \\
\alpha & 1 \end{array}\right].
$$
Therefore $u^{xt}=\left(\begin{array}{rr} 1 & 0 \\
\alpha & 1 \end{array}\right)=(u)^T$, where $^T$ denotes the transposition of a
matrix and $(xt)^2=t^xt=t^{-1}t=e$. Replacing $x$ by $xt$, we may assume
that $u^x=\left(\begin{array}{rr} 1 & 0 \\
\alpha & 1 \end{array}\right)=u^T$. Since  $| x|=2$, we also derive that
$(u^T)^x=u$. Set $z=\left(\begin{array}{rr} 0 & 1 \\
-1 & 0 \end{array}\right) \in \operatorname{SL}_2(q)\cap N(C_0, T)$. Then
$$u^{xz}=u^{-1}=u^{zx}.$$ Since
$| N(C_0, T)/T|=4$, $N(C_0, T)/T$ is abelian. Moreover,
$x$ and $z$ lie in $N(C_0, T)$, and their images in $N(C_0, T)/T$ are
involutions;
hence  $N(C_0,T)/T\simeq
{\mathbb Z}_2\times{\mathbb Z}_2$ and $x$ normalizes $z (T\cap M)$, i.e.,
$z^{x}=zt$
for some $t\in T\cap M$. Therefore $xzt=zx$. As we noted above, both
$xz$ and $zx$ invert $u$, and so $t\in Z(M)$.
If $q$ is even then $Z(M)=\{e\}$, whence $x$
centralizes $\langle
z\rangle$. Therefore  $| xz|=2$, so $xz$ is an inverting involution. Assume
that  $q$ is odd. Then $|z|=4$,
$| Z(M)|=2$, and for $t\in Z(M)\setminus\{e\}$ the identity
$zt=z^{-1}$ holds. Thus either $z^x=z$, or
$z^x=z^{-1}$. We show that $z^x=z^{-1}$, whence
$(xz)^2=z^xz=z^{-1}z=e$ and $xz$ is an inverting involution. Let $Q$ be a
maximal torus of $C$ that contains $z$. Note that
$\tilde{x}$, $\widetilde{Q}$ lie in $C_{\widetilde{C}_0}(\tilde{z})$
and $C_{\widetilde{C}_0}(\tilde{z})\leq N(\widetilde{C}_0, \widetilde{Q})$.
Moreover, $\widetilde{C}_0=\operatorname{PGL}_2(q)\times \langle
\tilde{y}\rangle$ and
$\tilde{y} \in N({\widetilde{C}_0}, \widetilde{Q})$. Consider the cosets
$\tilde{y}\widetilde{Q}$ and $\tilde{x}\widetilde{Q}$.
Suppose that these cosets coincide. Then
$\tilde{x} \in \tilde{y}\widetilde{Q}$. Since $\widetilde{Q}$ is cyclic, it
contains a unique involution $\tilde{z}$. Therefore,
$\tilde{y}\widetilde{Q}$ contains the two involutions
$\tilde{y}$, $\tilde{y}\tilde{z}$; hence $\tilde{x}=\tilde{y}$
or $\tilde{x}=\tilde{y}\tilde{z}$. The first equality is impossible, since
$\tilde{y}$ centralizes $\operatorname{PGL}_2(q)$; and if
$\tilde{x}=\tilde{y}\tilde{z}$, then
$\tilde{u}^{-1}=\tilde{u}^{\tilde{x}\tilde{z}}=\tilde{u}^{\tilde{y}\tilde{z}^2}
=\tilde{u}^{\tilde{z}^2}=\tilde{u}$, which is impossible.
Therefore $\tilde{y}\widetilde{Q}\neq\tilde{x}\widetilde{Q}$. By Lemma 1, there
exists an involution  $x' \in N(G, Q)$, inverting each element of
$Q$. We have $s \in Q$, and so
$x' \in C_0$ and $x, x' \in N(G,Q)\cap C_0$.
Since $\tilde{y}\widetilde{Q}\neq\tilde{x}\widetilde{Q}$,
$N(C_0,Q)/Q\simeq{\mathbb Z}_2\times{\mathbb Z}_2$ and $x,x'\not\in C$,
we see that
$\tilde{x}'\widetilde{Q}=\tilde{x}\widetilde{Q}$  and $xQ=x'Q$.
Therefore $x$ inverts each element of $Q$; in particular,~${z^x=z^{-1}}$.

Let  $s$ be such that its centralizer either is conjugate to the centralizer
of $s_3$, or is conjugate to the centralizer of  $s_{7}$. Then
$| C: (M\circ S)|=(2,q-1)$,  $M\simeq
\operatorname{SL}_2(q^3)$,  and $S\simeq {\mathbb Z}_{q-\varepsilon}$, where
$\varepsilon=1$ if  $C_G(s)$ is conjugate to  $C_G(s_3)$ and
$\varepsilon=-1$ if
$C_G(s)$ is conjugate to $C_G(s_{7})$. Moreover,
$C/S\simeq \operatorname{PGL}_2(q)$.
This case can settled by using exactly the same arguments as in the previous
case.

Assume that $C_G(s)$ is conjugate to $C_G(s_4)$. Then $M\simeq
\operatorname{SL}_3(q)$,
$S\simeq
{\mathbb Z}_{q^2+q+1}$. Moreover, if  $3$ divides $q-1$ then
$|C:M\circ S|=3$ and $C/S\simeq \operatorname{PGL}_3(q)$;
and if $3$ does not divide $q-1$ then  $C=
M\times S$ and $\operatorname{SL}_3(q)\simeq \operatorname{PGL}_3(q)$.
In both cases the proof is the same. Choose a maximal torus
$T$ of $C$ so that $T\cap M$ is a Cartan subgroup of $M$. We identify elements
of $M$ with matrices of $\operatorname{SL}_3(q)$ and we assume that  $T\cap M$
is a subgroup of diagonal matrices under this identification. By Lemma~1,
there exists $x \in N(G,T)$ such that $x^2=e$ and $t^x=t^{-1}$ for every
$t \in T$. Consider $C_0=\langle C,x\rangle$. By [19, Lemma~2.3],
$N(G,C)$ does not induce field automorphisms on $M$. Since $x$ inverts each
element of a Cartan subgroup $T\cap M$ of $M$, we see that $x$ induces a
graph automorphism on $M$. Let
$\iota$ be a graph automorphism of
$\operatorname{SL}_3(q)$ acting by
$y\mapsto(y^{-1})^T$, where  ${}^T$ denotes the transposition of a
matrix. Then  $\iota$ normalizes $T\cap M$ and inverts each element from
$T\cap M$. Hence, multiplying  $x$ by a suitable element of  $T\cap M$, we may
assume that  $x$ acts on $M$ in the same way as $\iota$. The element
$u$ is conjugate to its Jordan form in $C$, and so we may assume that
$u=\left(\begin{array}{rrr} 1 & 1 & 0
\\ 0 & 1 & \alpha \\ 0 & 0 &
1\end{array}\right)$, where $\alpha \in
\{0, 1\}$. Let $u=\left(\begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0
& 1\end{array}\right)$, and set
$z=\left(\begin{array}{rrr} 0 & 0 & -1 \\
0 & -1 & 0 \\ -1 & 0 &
0\end{array}\right) \in \operatorname{SL}_3(q)$. We have
$$
u^{xz}=((u^{-1})^T)^z
=\left(\left(\begin{array}{rrr} 1 & -1 & 1 \\ 0 & 1 & -1 \\
0 & 0 & 1\end{array}\right)^T\right)^z=\left(\begin{array}{rrr} 1 & 0 & 0 \\ -1
& 1 & 0 \\
1 & -1 & 1\end{array}\right)^z=\left(\begin{array}{rrr} 1 & -1 & 1 \\ 0 & 1 & -1
\\
0 & 0 &
1\end{array}\right)=u^{-1}.
$$
Therefore $xz$ is a sought involution, since
$(xz)^2=z^xz=(z^{-1})^Tz=e$. Let
$u=\left(\begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &
1\end{array}\right)$,  and set
$z=\left(\begin{array}{rrr} 0 & 1 & 0 \\
1 & 0 & 0 \\ 0 & 0 &
-1\end{array}\right) \in \operatorname{SL}_3(q)$. We have
$$u^{xz}=((u^{-1})^T)^z=\left(\left(\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 1 & 0
\\
0 & 0 & 1\end{array}\right)^T\right)^z=\left(\begin{array}{rrr} 1 & 0 & 0 \\ -1
& 1 & 0 \\
0 & 0 & 1\end{array}\right)^z=\left(\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 1 & 0
\\
0 & 0 & 1\end{array}\right)=u^{-1}.$$ Therefore $xz$ is a sought involution,
since  $(xz)^2=z^xz=(z^{-1})^Tz=e$.

Assume that  $C_G(s)$ is conjugate to $C_G(s_9)$. Then
$M\simeq \operatorname{SU}_3(q)$, $S\simeq
{\mathbb Z}_{q^2-q+1}$. Moreover, if $3$ divides $(q+1)$ then
$|C:M\circ S|=3$ and $C/S\simeq \operatorname{PGU}_3(q)$;
and  if  $3$ does not divide  $(q+1)$ then  $C=
M\times S$ and $\operatorname{SU}_3(q)\simeq \operatorname{PGU}_3(q)$.
In both cases the proof is the same. Choose a maximal torus
$T$ of $C$ so that $T\cap M$ is a Cartan subgroup of $M$. By Lemma~1,
there exists $x \in N(G,T)$ such that $x^2=e$ and $t^x=t^{-1}$ for every
$t \in T$. Again, by [19, Lemma~2.3],
$N(G,C)$ does not induce field automorphisms on $M$. Let
$A=\left(\begin{array}{rrr} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 &
0\end{array}\right)$ and let $\iota$ be an automorphism of
$\operatorname{SL}_3(q^2)$, acting by  $y\mapsto A(y^{-1})^TA$, where $^T$
denotes the transposition of a
matrix. Denote the automorphism of $\operatorname{SL}_3(q^2)$ that maps each
element of a matrix from  $\operatorname{SL}_3(q^2)$ into the power $q$ by  $f$.
In view of  [20, p.~268--270] we may assume that  $\operatorname{SU}_3(q)$
coincides with the set of  $\iota\circ f$-stable points.
We identify elements of $M$ with the set of $\iota\circ f$-stable points of
$\operatorname{SL}_3(q^2)$ and we assume that  $T\cap M$ is a group of diagonal
matrices under this identification. The restriction of $\iota$ on
$\operatorname{SU}_3(q)$ we denote by the same symbol $\iota$. Then
$\iota$ normalizes  $T\cap M$. So, multiplying $x$ by a suitable element of
$T$, we may assume that $x$ acts on $M$ in the same way as $\iota$.
Up to conjugation in  $C$, each unipotent element $u$ of $M$ has the form
$u=\left(\begin{array}{rrr} 1 & \alpha & \beta
\\ 0 & 1 & \alpha^q \\ 0 & 0 &
1\end{array}\right)$, where $\alpha, \beta \in
F_{q^2}$ and $\beta + \beta^q=\alpha^{q+1}$. If $\alpha\neq 0$, there exists
an element $t \in T$ such that
$u^t=\left(\begin{array}{rrr} 1 & 1 & \gamma \\ 0 & 1 & 1 \\
0 & 0 & 1\end{array}\right)$ for some $\gamma \in F_{q^2}$. So we may assume
that $u=\left(\begin{array}{rrr} 1 & 1 & \gamma
\\ 0 & 1 & 1 \\ 0 & 0 &
1\end{array}\right)$ and
$u^{-1}=\left(\begin{array}{rrr} 1 & -1 &
\gamma' \\ 0 & 1 & -1 \\ 0 & 0 &
1\end{array}\right)$ for some
$\gamma' \in F_{q^2}$. Put $z=\left(\begin{array}{rrr} -1 & 0 & 0 \\ 0 &
1 & 0 \\ 0 & 0 &
-1\end{array}\right) \in \operatorname{SU}_3(q)$. Then
$$u^{xz}=(A(u^{-1})^TA)^z=(A(u^{-1})^TA)^z=\left(\begin{array}{rrr} 1 & 1 &
\gamma' \\ 0 & 1 & 1 \\
0 & 0 & 1\end{array}\right)^z=\left(\begin{array}{rrr} 1 & -1 & \gamma' \\ 0 & 1
& -1 \\
0 & 0 & 1\end{array}\right)=u^{-1}.$$ For $\alpha=0$ the identity
$u^{xz}=u^{-1}$ is also true. Therefore, $xz$ is  a sought involution, since
$(xz)^2=z^xz=(z^{-1})^Tz=e$.

Theorem 1 and so Teorems 2 and 3 are proven.


TITLE: Projectivity and semisimple in an abelian category
QUESTION [0 upvotes]: Let $C$ be a locally finite $k$-linear abelian category. Here locally finite means that the home sets are finite dimensional vector space and every object of $C$ has finite length.
Suppose that each object of $C$ is projective. I want to prove that $C$ is semisimple.
Here us what I have thought:
Let $X$ be an object of $C$. Since it has finite length, there is a simple object $Y$ and monomorphism $i: Y \to X$.
Then we have an exact sequence $Y \to X \to X/Y \to 0$.
Since $X/Y$ is projective, we have a unique morphism from $X/Y to X$ so that the sequence splits.
I don't know how to conclude that $X$ is a direct sum of $Y$ and $X/Y$ from here. I want to construct a morphism from $X$ to $Y$ as well.
Also, I have no idea how to prove that there are only finitely many simple objects. Could you give me a proof?

REPLY [1 votes]: Suppose that $\mathbb{A}$ is an abelian category and that $0\to K \to A\to B\to 0$ is an exact sequence which splits. Let us write $\kappa: K\to A$, $\alpha:A\to B$  and $\beta : B\to A$ for the morphisms involved. By exactness we have that $\kappa$ is the kernel of $\alpha$, $\alpha$ is the cokernel of $\kappa$, and since the sequence splits $\alpha\beta =1_B$. Since $\kappa$ is the kernel of $\alpha$ and $\alpha(1_A - \beta\alpha) = \alpha - \alpha\beta\alpha = 0 $ it follows that there exists a unique morphism $\gamma : A\to K$ such that $\kappa\gamma = 1_A -\beta\alpha$. Note that since $\kappa$ is a monomorphism and $\kappa\gamma\kappa = (1_A-\beta\alpha)\kappa=\kappa=\kappa1_K$ it follows that $\gamma\kappa = 1_K$. Using that $\kappa$ is a monomorphism again and $\kappa \gamma \beta = (1_A -\beta\alpha) \beta= \beta - \beta\alpha\beta = \beta - \beta = 0 = \kappa 0$ we find that $\gamma \beta = 0$.
We now have what Mac Lane calls a biproduct diagram (which is more often now called a direct sum diagram) i.e. we have objects $K$, $A$ and $B$, morphisms $\alpha:A\to B$, $\gamma:A\to K$, $\beta: B\to A$ and $\kappa: K\to A$ satisfying $\alpha \beta=1_B$, $\gamma\kappa=1_K$, $\alpha \kappa = 0$, $\gamma \beta =0$ and $\beta \alpha + \kappa\gamma = 1_A$. It is a relatively easy exercise to show that for such a diagram the diagram $X \leftarrow A \rightarrow B$ is a product and the diagram $K \rightarrow A \leftarrow B$ is a coproduct.


TITLE: Number of Dyck paths with maximal odd sequence of $(1,-1)$ ending on the $x$-axis
QUESTION [2 upvotes]: A Dyck path from $(0,0)$ to $(2n+2,0)$ is a lattice path with steps
  $(1,1)$ and $(1,-1)$, never falling below the $x$-axis.
Find the number of Dyck paths from $(0,0)$ to $(2n+2,0)$ such that any
  maximal sequence of consecutive steps $(1,-1)$ ending on the $x$-axis
  has odd length.

The above question is from Richard Stanley's Enumerative Combinatorics. In fact, a solution is given by Stanley, but I am not sure why we should proceed like that.

Let $A(x)=x+x^3+2x^4+6x^5+\cdots$ (respectively,
  $B(x)=x^2+x^3+3x^4+8x^5+\cdots$) be the generating function for Dyck
  paths from $(0,0)$ to $(2n, 0)$ such that the path only touches the
  $x$-axis at the beginning and end, and the number of steps $(1,-1)$ at
  the end is odd (respectively, even).

How do we obtain these two generating functions? Why does it suffice to consider the paths that touch the $x$-axis at the beginning and the end?

Let $C(x)=1+x+2x^2+5x^3+\cdots$ be the generating function for all
  Dyck paths from $(0,0)$ to $(2n,0)$. Note that the coefficients are
  Catalan numbers. It is easy to see that $A=x(1+CB)$ and $B=xCA$.
  Solving for $A$ gives $$A=\frac{x}{1-x^2C^2}.$$

Although I may not understand how to come up with $A$ and $B$, this part is simply computation.

The generating function we want is $$\frac{1}{1-A}.$$

I again do not understand this part. Why do we have such a generating function?

The above simplifies to $1+xC$, which means that the required number
  follows the Catalan numbers.

Any help for the above question is appreciated.

REPLY [1 votes]: Approach: The idea is to analyse specific types of Dyck paths which can be used as building blocks and to find generating functions for them. These generating functions will be used to specify the generating function we are actually looking for.

We put the focus on Dyck paths of length $2n$, starting at $(0,0)$ and ending at $(2n,0)$ which do not touch the x-axis between the start and end point. The number $n$ specifies the number of up steps which is equal to the number of down steps.
It is convenient to split these Dyck paths in paths which end with an odd number of down steps and those which end with an even number of down steps.

Generating functions A(x) and B(x)
Let's denote the generating function with an odd number of terminating down steps $A(x)$. In order to get a first impression of this function we look for the number of Dyck paths counted by $A(x)$ for small lenghts  $2n$.
In the following we encode up steps with u and down steps with d. So, all paths counted by $A(x)$ start with an $u$ and end with an odd sequence of $d$'s.

$n=2$: We get one path $\color{blue}{ud}$
$n=4$: We get no path, since $udud$ touches the $x$-axis between start and end point and $uudd$ does not end with an odd sequence.
$n=6$: We consider the pattern $\color{blue}{u(..)ddd}$ and obtain one path $\color{blue}{uuuddd}$.
$n=8$: We consider the pattern $\color{blue}{uu(..)uddd}$. Since we have to start with two $u$'s in order to not touch the $x$-axis in between and we have due to the same reason to terminate with three $d$'s, which implies that there is an up step $u$ before them. We find the paths $\color{blue}{uu(ud)uddd}$ and $\color{blue}{uu(du)uddd}$.
$n=10$: We consider the path $\color{blue}{uuuuuddddd}$  and the pattern $\color{blue}{uu(....)uddd}$  giving $1+\left(\binom{4}{2}-1\right)=6$ Dyck paths. Note, that we have to exclude $uu(dduu)uddd$ to not touch the $x$-axis in between.

We conclude
\begin{align*}
   A(x)=x+x^3+2x^4+6x^5+\cdots
   \end{align*}
With similar reasoning we can find the first coefficients of the generating function $B(x)$ counting all Dyck words which do not touch the $x$-axes in between and which terminate with an even number of down steps.
\begin{align*}
   B(x)=x^2+x^3+3x^4+8x^5+\cdots
   \end{align*}

$$ $$

Relationship $A(x),B(x)$ with $C(x)$:
We denote with $C(x)$ the generating function of the Catalan numbers $C_n=\frac{1}{n+1}\binom{2n}{n}$
  \begin{align*}
  C(x)=\sum_{n=0}^{\infty}C_nx^n=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)
  \end{align*}
  The coefficients $C_n$ give the number of all Dyck paths of length $2n$.
Let's look at the expression
\begin{align*}
  C(x)A(x)&=\left(\sum_{n=0}^{\infty}C_nx^n\right)\left(\sum_{n=0}^{\infty}A_nx^n\right)\\
  &=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}C_kA_{n-k}\right)x^n
  \end{align*}

If we consider e.g. the coefficient of $x^4$ we see
\begin{align*}
  C_0A_4+C_1A_3+C_2A_2+C_3A_1+C_4A_0
  \end{align*}
So, the coefficient of $x^4$ specifies the number of all Dyck paths $C_k$ of length $k\, (0\leq k \leq 4)$ concatenated with all specific Dyck paths $A_{4-k}$.

We conclude: The multiplication $C(x)A(x)$ specifies the number of all Dyck paths which terminate with an odd sequence of down steps. Note, here are also Dyck paths which touch the $x$-axis thanks to the left multiplication with $C(x)$ and the termination by an odd number of down steps is guaranteed by the multiplication with $A(x)$.
It follows, that
  \begin{align*}
  B(x)=xC(x)A(x)
  \end{align*}
  since the multiplication with $x$ represents a starting up step $u$ and an ending down step $d$. This guarantees that we end with an even number of down steps and it is also assured, that we do not touch the $x$-axis in between.
With a similar reasoning we see that
  \begin{align*}
  A(x)=x+xC(x)B(x)
  \end{align*}
  whereby the summand $x$ represents the Dyck path $\color{blue}{ud}$.

$$ $$

Concatenation of building blocks:
If we consider the geometric series
\begin{align*}
  G(x)=\frac{1}{1-A(x)}=1+A(x)+A^2(x)+A^3(x)+\cdots
  \end{align*}
we can think of $A^k(x)$ as counting the concatenation of $k$ Dyck paths each terminating with an odd number of down steps.
We observe that each $A^k(x)$ counts Dyck paths, which touch the $x$-axis between start and end point, $k-1$ times. So,  the geometric series $G(x)$ provides us with all Dyck paths of the wanted shape.


TITLE: Confusion in understanding the derivation of wave speed from Newton's Second Law
QUESTION [0 upvotes]: In the book 'Principles of Physics' by Resnick,Halliday and Jearl Walker,the opening para of derivation of wave speed from Newton's Second Law is such:

Let us consider a symmetrical pulse moving from left to right along a string of speed $v$. For convenience,we choose a reference frame in which the pulse remains stationary;i.e,we run along the pulse,keeping it constantly in view.In this frame,the string appears to move past us,from right to left with speed $v$

My confusion  is :

If I am moving with the pulse,I should moving left to right with velocity $v$ wrt the reference for which pulse is moving left to right with velocity $v$.
So I should have zero speed wrt pulse and string both.
How can the string be moving right to left with speed $v$ if I am stationary to the pulse?

Are not the velocity of the pulse and velocity of the string are in the same direction and of same magnitude?

REPLY [0 votes]: If the pulse is moving to the right relative to the string (and you), then string (and you) are moving to the left relative to the pulse.  It's like when you are in a car; if you are moving forward relative to the landscape, the landscape is moving backward relative to you.  The next time you are in a car traveling at a constant speed along a straight road, imagine that the car (and you) are stationary and the landscape is moving backward. It's a weird feeling.


TITLE: $T_1$ spaces where the closure of a compact set is not compact
QUESTION [4 upvotes]: I have many problems with this exercise:

Give an example of $T_{1}$ Topological Space $(X,\tau)$ and a subset $Y\subset X$ compact such that $\bar{Y}$ is not compact.

Now, honestly, I know this example:

$X= A\sqcup B$ where $A$ and $B$ are infinite sets, and $\tau_{X} = \lbrace\emptyset\rbrace\cup\lbrace \mathcal{U}\in X\mid A\setminus\mathcal{U}$ is finite$\rbrace$

But I'd like to find other examples... the problem is that I don't know much Topologies $T_{1}$ but not $T_{2}$; I know Cofinite Topology and Zariski Topology but these examples don't work, I know that the Alexandroff extension of $\mathbb{Q}$ is $T_{1}$ but not $T_{2}$, but I don't think that it's work.
Can someone help me?

REPLY [2 votes]: Let $X=\Bbb N \cup A$ where $A$ is some non-empty set disjoint from $\Bbb N$, with at least two elements. A basic neighbourhood of $n \in \Bbb N$ is $\{n\}$ (so is an isolated point), while a basic neighbourhood of $a \in A$ is $\{a\} \cup (\Bbb N \setminus F)$ where $F \subseteq \Bbb N$ is finite.
Then it's easily checked that $X$ is $T_1$, that any two neighbourhoods of $a \neq a'$ in $A$ intersect (so $X$ is not $T_2$) and that $N_a:=\Bbb N \cup \{a\}$ is compact for every $a \in A$ and for all $a$, $\overline{N_a}=X$ and $X$ is not compact if $A$ is infinite (as $A$ is a closed discrete subspace of $X$). So this gives an example.


TITLE: Fixed point iteration method converging to infinity
QUESTION [1 upvotes]: So I am trying to use Fixed Point Iteration Method for this function:
$ f(x)=x^5 -x^4-x-3 $ .
By using bisection method I found out that the root is around $1.64056$.
So I started by doing:
$f(x)=0$
then found out that:
$x=x^5-x^4-3$ .
So here I used Microsoft Excel and it generated very big numbers, which are converging to Inf almost instantly, even if I select my $x_0$ really close to my actual root I need to find. I am adding an Excel screenshot of what I am getting. Is it even possible to use Iteration method for this kind a function? If so,  where can I go from here?

REPLY [2 votes]: In a polynomial with only one sign change in the coefficient sequence one can always find an equivalent form that is monotonous and convex or concave on the positive half-axis. With such a function it is known that the Newton iteration will converge monotonically towards the root if started on the right side of the root. This follows just from the geometry of the tangent as supporting line of the graph.
Here this can be achieved by dividing by $x^4$ or $x^5$. The first one gives
$f(x)=x-1-x^{-3}-3x^{-4}$, which is monotonically increasing and concave over the positive half-axis. Any tangent there is thus above the graph, tangent roots thus smaller than the function roots, but larger than the base point of the tangent. Then starting, for example, from $x_1=1$ the Newton iteration will be growing towards the root of the function.
The Newton method of $f$ to be used as fixed-point iteration is
$$N(x)=x-\frac{x-1-x^{-3}-3x^{-4}}{1+3x^{-4}+12x^{-5}}.$$

$k$
$x_k$

1
1.0000000000000000

2
1.2500000000000000

3
1.4919752765802732

4
1.6203796983619547

5
1.6402369415053781

6
1.6405855189039007

7
1.6405856214263030

8
1.6405856214263119

9
1.6405856214263119


TITLE: Finding the initial conditions of a conservation law hyperbolic system?
QUESTION [1 upvotes]: Consider $u(\boldsymbol{x},t)$ to be a function that is known at $t=0 $ by the initial conditions
$$u(\boldsymbol{x},0)= f(\boldsymbol{x}),\quad u_t(\boldsymbol{x},0)= g(\boldsymbol{x})$$
Where $\boldsymbol{x}\in \mathbb{R}^3$
Let $q, r, s$ be functions such that
\begin{align*}
    &q_t= cu_x  \\
    &r_t= cu_y  \\
    &s_t= cu_z  \\
    &u_t =cq_x+ cr_y+ cs_z
\end{align*}
Then, apparently, up to a constant, we can determine $q(\boldsymbol{x},0),\ r(\boldsymbol{x},0)$ and $s(\boldsymbol{x},0)$ at the initial time, in term of $f$ and $g$ .
But I stuck at how to find that.
The question is : Can we determine these initial values ??

\ tentative :
We can notice that the system of equations can be looked at as two simple equations if we consider $v= (q,r,s)$ :
\begin{align*}
    &v_t= c\ \nabla u \\
    &u_t =c\ \nabla \cdot v
\end{align*}
Notice also, that for the one dimensional case:
\begin{align*}
    &q_t= cu_x \\
    &u_t =cq_x
\end{align*}
We can easily derive that
\begin{align*}
    &c\ q(x,t)= \int_{x_0}^x u_t(\xi,t)d\xi \\
\text{So that }\quad  & q(x,0)= \frac{1}{c}\int_{x_0}^x u_t(\xi,0)d\xi= \frac{1}{c}\int_{x_0}^x g(\xi)d\xi
\end{align*}
And we see that indeed in the one dimensional case $q(x,0)$ is known in term of $g$ up to a constant. So now the question is to do the same for the 3 dimensional case.

REPLY [1 votes]: If you take the spatial and temporal derivatives of $q,\,r,\,s$, you'll quickly find,
\begin{align}
  q_{tx} &= cu_{xx} \\
  r_{ty} &= cu_{yy} \\
  s_{tz} &= cu_{zz}
\end{align}
And if you then take the second temporal derivative of $u$, you can see pretty easily,
$$ u_{tt}=c\left(q_{x}+r_{y}+s_{z}\right)_t=c\left(q_{tx}+r_{ty}+s_{tz}\right)=c^2\left(u_{xx}+u_{yy}+u_{zz}\right)$$
which is to say that this is really just the wave equation. In one dimension, you will find that this is solved using d'Alembert's formula,
$$ u\left(x,t\right)=\frac{1}{2}\left(f(x+ct)+f(x-ct)\right)+\frac{1}{2c}\int_{x-ct}^{x+ct}g(\xi)\,\mathrm d\xi.\tag{1}$$
In higher dimensions, this becomes known as Kirchhoff's formula (PDF link), which is a bit more complicated to derive, but more or less follows the same idea as the 1D case. You want to convert to spherical coordinate system and consider spherical means, defining $U$, $F$ and $G$ such that,
\begin{align}
  U(x,r,t)&=\frac{1}{\vert\partial B(x,r)\vert}\int_{\partial B(x,r)}u(\xi,t)\mathrm dS_\xi \\
  F(x,r,t)&=\frac{1}{\vert\partial B(x,r)\vert}\int_{\partial B(x,r)}f(\xi,t)\mathrm dS_\xi \\
  G(x,r,t)&=\frac{1}{\vert\partial B(x,r)\vert}\int_{\partial B(x,r)}g(\xi,t)\mathrm dS_\xi
\end{align}
where $B$ is the ball in $\mathbb{R}^d$ and $\partial B$ the surface. To recover $u$ from this, we simply take the limit of $r\to0$,
$$u(x,t)=\lim_{r\to0}U(x,r,t)$$
If we then let $\bar{U}=rU$ and likewise for $F$ and $G$, then $\bar{U}$ satisfies the wave equation and has a solution similar to Eq (1),
$$\bar{U}(x,r,t)=\frac{1}{2}\left[\bar{F}(t+r)+\bar{F}(t-r)\right]+\frac{1}{2}\int_{t-r}^{t+r}\bar{G}(y)\mathrm dy $$
Taking the limit gives,
$$ u(x,t)=\lim_{r\to0}\frac{\bar{U}}{r}=\frac{\partial}{\partial t}\left(\frac{t}{\vert\partial B\vert}\int_{\partial B}f(\xi)\mathrm dS_\xi\right)+\frac{t}{\vert\partial B\vert} \int_{\partial B}g(\xi)\mathrm dS_\xi.$$
With quite some more work, this can be reduced to what is called Kirchhoff's formula,
$$ u(x,t)=\frac{1}{\vert\partial B\vert}\int_{\partial B}\left[tg(\xi)+f(\xi)+\nabla f\cdot\left(\xi-x\right)\right]\,\mathrm dS_\xi$$

When solving the wave equation $\psi_{tt}=c^2\nabla^2\psi$, a common trick is to define variables $p$, $q$, $r$ and $s$ such that,
$$ p=c\psi_x,\quad q=c\psi_y,\quad r=c\psi_z,\quad s=\psi_t$$
and combine them into $u=\left[p,\,q,\,r,\,s\right]^T$ to form the evolution equation,
$$ \frac{\partial u}{\partial t}=-\nabla F\tag{A.1}$$
where $F$ is flux in terms of the variables $p,\,q,\,r,\,s$. For instance,
\begin{align}
  p_t &= c\psi_{xt}=c\psi_{tx}=cs_x \\
  q_t &= c\psi_{ty}=cs_y \\
  r_t &= c\psi_{tz}=cs_z \\
  s_t &= cp_x+cq_y+cr_z
\end{align}
implies,
$$ F_x=\left(\begin{array}{cccc} 0 & 0 & 0 & -c \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -c & 0 & 0 & 0\end{array}\right)\cdot\left(\begin{array}{c} p \\ q \\ r \\ s \end{array}\right) $$
I believe that what you are supposed to be doing is that, given the initial conditions $\psi(x,0)=f(x)$ and $\psi_t(x,0)=g(x)$, you apply these directly to the variables:
$$ p(x,0)=c\frac{\partial f}{\partial x},\;q(x,0)=c\frac{\partial f}{\partial y},\;r(x,0)=c\frac{\partial f}{\partial z},\;s(x,0)=g(x) $$
And integrate according to Eq (A.1) using Lax-Wendroff, rather than trying to apply Kirchhoff's formula to the equations.