TITLE: Probability Density Function of $Y=X^3$
QUESTION [0 upvotes]: $X \sim \text{Uniform}(-\frac{1}{2},\frac{1}{2})$, $Y=X^3$. To find the PDF of $Y$, I did the following:
$$F_Y(y) = P(Y\leq y) = P(X^3 \leq y) = P(-y^{1/3} \leq x \leq y^{1/3}) $$
Then $$ = \int_{-y^{1/3}}^{y^{1/3}} f_x(x) dx = 2y^{1/3}$$
Thus I get $f_y(y) = 2/3y^{-2/3}$. However the correct answer is $1/3y^{-2/3}$. It seems my bounds should be from $0$ to $y^{1/3}$ instead of $-y^{1/3}$ to $y^{1/3}$.  Why is this the case when $x \in  (-\frac{1}{2}, \frac{1}{2})$?

REPLY [1 votes]: $ P(X^3 \leq y) = P(-y^{1/3} \leq X \leq y^{1/3})$

This is incorrect. $x\mapsto x^3$ is monotone. So:

$ P(X^3 \leq y) = P(X \leq y^{1/3})$


TITLE: Find the first two non vanishing maclaurin terms
QUESTION [4 upvotes]: Find the first two nonvanishing terms in the Maclaurin series of $\sin(x + x^3)$.
Suggestion: use the Maclaurin series of $\sin(y)$ and write $y = x + x^3$
Using
this result, find
$\lim\limits_{x\to 0}\frac{\sin(x + x^3)−x}{x^3}$
$\sin(y)= y-\frac{y^3}{3!}+\frac{y^5}{5!}+\frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-\frac{(x+x^3)^3}{3!}+\frac{(x+x^3)^5}{5!}-\frac{(x+x^3)^7}{7!}$
Now what do I do?

REPLY [1 votes]: The limit in question is more easily solved by a little bit of algebra. Note that $$\frac{\sin(x+x^3)-x}{x^3}=\frac{\sin (x+x^3)-(x+x^3)}{(x+x^3)^3}\cdot (1+x^2)^3+1$$ Putting $t=x+x^3$ we can see that $t\to 0$ and hence the first factor $(\sin t - t) /t^3$ in first term tends to $-1/6$ and the desired limit is thus $(-1/6)+1=5/6$.
There is no need to find the Maclaurin series for $\sin(x+x^3)$, just the well known series for $\sin x $ is sufficient to tackle the problem.


text:
\begin{document}

\title{Path counting and rank gaps in differential posets}
\author{Christian Gaetz}
\address{Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA.}
\email{\href{mailto:gaetz@mit.edu}{gaetz@mit.edu}}
\thanks{C.G. was supported by the National Science Foundation Graduate Research Fellowship under Grant No. 1122374.}

\author{Praveen Venkataramana}
\address{Division of Physics, Mathematics, and Astronomy, California Institute of Technology, Pasadena, CA.}
\email{\href{mailto:pvenkata@caltech.edu}{pvenkata@caltech.edu}}
\date{\today}

\begin{abstract}
We study the gaps $\Delta p_n$ between consecutive rank sizes in $r$-differential posets by introducing a projection operator whose matrix entries can be expressed in terms of the number of certain paths in the Hasse diagram.  We strengthen Miller's result that $\Delta p_n \geq 1$, which resolved a longstanding conjecture of Stanley, by showing that $\Delta p_n \geq 2r$.  We also obtain stronger bounds in the case that the poset has many substructures called \textit{threads}.
\end{abstract}

\keywords{Differential poset; rank growth; thread element; Hasse diagram; path counting}

\maketitle

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

Differential posets are a class of partially ordered sets, originally defined by Stanley \cite{Stanley1988}, which generalize many of the enumerative and combinatorial properties of Young's lattice $Y$, the lattice of integer partitions.  We refer the reader to \cite{Stanley2012} for definitions and basic facts about posets.

If $P$ is a graded poset and $S \subseteq P$, let $\Q S$ denote the $\Q$-vector space with basis $S$.  Let $P_n$ denote the $n$-th rank of $P$ and define the \textit{up and down operators} $U_n: \Q P_n \to \Q P_{n+1}$ and $D_n: \Q P_n \to \Q P_{n-1}$ by
\begin{align*}
U_n x &= \sum_{x \lessdot y} y \\
D_n x &= \sum_{z \lessdot x} z
\end{align*}
where $x \in P_n$ and $\lessdot$ denotes the covering relation in $P$.  We often omit the subscripts and write $U,D$ when no confusion can result.  For $r \in \Z_{>0}$ a locally-finite $\N$-graded poset $P$ with $\hat{0}$ is an \textit{r-differential poset} if, for all $n \geq 0$ we have
\begin{align}
D_{n+1}U_n - U_{n-1}D_n = r \cdot I
\label{eq:DU-UD}
\end{align}
as linear operators.

\begin{ex} \label{ex:Y-and-Z}
Young's lattice $Y$ is a 1-differential poset (see Figure \ref{fig:Y}).  More generally the product poset $Y^r$ is an $r$-differential poset.

Another family of examples are the Fibonacci lattices $Z(r)$ (see Figure \ref{fig:Z2}).  They are produced by the following iterative procedure:
\begin{enumerate}
\item Having defined a differential poset $P$ up to rank $n$, add an element $\hat{x}$ to $P_{n+1}$ for each element $x \in P_{n-1}$ so that $\hat{x}$ covers exactly those $y \in P_n$ which cover $x$.
\item For each $y \in P_n$, add an additional $r$ singletons covering $y$.
\end{enumerate}
The $r$-differential poset $Z(r)$ is obtained by iterating this procedure beginning with a single point \cite{Stanley1988}.
\end{ex}

\begin{center}
\begin{figure}
\begin{tikzpicture} [every node/.style={draw,shape=circle,fill=black,scale=0.4}, scale=1.4]
\node(a0) at (0,0) {};
\node[fill=white](b0) at (-0.7,1) {};
\node[fill=white](b1) at (0.7,1) {};
\node(c0) at (0,2) {};
\node[fill=white](c1) at (-1.4,2) {};
\node[fill=white](c2) at (-.7,2) {};
\node[fill=white](c3) at (.7,2) {};
\node[fill=white](c4) at (1.4,2) {};
\node(d0) at (-0.7,3) {};
\node(d1) at (0.7,3) {};

\node[fill=white](d2) at (-1.4,3) {};
\node[fill=white](d3) at (-2.1,3) {};
\node[fill=white](d4) at (-1,3) {};
\node[fill=white](d5) at (-.4,3) {};
\node(d6) at (-.2,3) {};
\node(d7) at (.2,3) {};
\node[fill=white](d8) at (1,3) {};
\node[fill=white](d9) at (.4,3) {};
\node[fill=white](d10) at (1.4,3) {};
\node[fill=white](d11) at (2.1,3) {};

\draw (a0)--(b0);
\draw (a0)--(b1);
\draw (b0)--(c0);
\draw (b1)--(c0);
\draw (b0)--(c1);
\draw (b0)--(c2);
\draw (b1)--(c3);
\draw (b1)--(c4);
\draw (c0)--(d0);
\draw (c1)--(d0);
\draw (c2)--(d0);
\draw (c0)--(d1);
\draw (c3)--(d1);
\draw (c4)--(d1);

\draw (c1)--(d2);
\draw (c1)--(d3);

\draw (c2)--(d4);
\draw (c2)--(d5);

\draw (c3)--(d8);
\draw (c3)--(d9);

\draw (c4)--(d10);
\draw (c4)--(d11);

\draw (c0)--(d6);
\draw (c0)--(d7);
\end{tikzpicture}
\caption{The Hasse diagram of the Fibonacci lattice $Z(2)$, a 2-differential poset, up to rank 3.  The thread elements (see Section \ref{sec:threads}) are unfilled.}
\label{fig:Z2}
\end{figure}
\end{center}

\begin{center}
\begin{figure}
\begin{tikzpicture} [every node/.style={draw,shape=circle,fill=black,scale=0.4}, scale=1.4]
\node(zero) at (0,-1) {};
\node[fill=white](a0) at (0,0) {};
\node[fill=white](b0) at (-0.7,1) {};
\node[fill=white](b1) at (0.7,1) {};
\node(c0) at (0,2) {};
\node[fill=white](c1) at (-1.4,2) {};
\node[fill=white](c4) at (1.4,2) {};
\node(d0) at (-0.7,3) {};
\node(d1) at (0.7,3) {};

\node(d2) at (-.7,3) {};
\node[fill=white](d3) at (-2.1,3) {};
\node(d4) at (0,3) {};
\node(d10) at (.7,3) {};
\node[fill=white](d11) at (2.1,3) {};

\node[fill=white](e0) at (-2.8,4) {};
\node(e1) at (-1.4,4) {};
\node(e2) at (-.7,4) {};
\node(e3) at (0,4) {};
\node(e4) at (.7,4) {};
\node(e5) at (1.4,4) {};
\node[fill=white](e6) at (2.8,4) {};

\draw (d0)--(e2);
\draw (d0)--(e3);
\draw (d0)--(e1);
\draw (d4)--(e2);
\draw (d4)--(e4);
\draw (d1)--(e3);
\draw (d1)--(e4);
\draw (d1)--(e5);
\draw (d3)--(e0);
\draw (d3)--(e1);
\draw (d11)--(e5);
\draw (d11)--(e6);

\draw (zero)--(a0);
\draw (a0)--(b0);
\draw (a0)--(b1);
\draw (b0)--(c0);
\draw (b1)--(c0);
\draw (b0)--(c1);
\draw (b1)--(c4);
\draw (c0)--(d0);
\draw (c1)--(d0);
\draw (c0)--(d1);
\draw (c0)--(d4);
\draw (c4)--(d1);
\draw (c1)--(d2);
\draw (c1)--(d3);
\draw (c4)--(d10);
\draw (c4)--(d11);
\end{tikzpicture}
\caption{The Hasse diagram of Young's lattice $Y$, a 1-differential poset, up to rank 5.  The thread elements (see Section \ref{sec:threads}) are unfilled.}
\label{fig:Y}
\end{figure}
\end{center}

One of the most basic properties of differential posets that one might hope to study are the possible \textit{rank functions} $p_n=|P_n|$, not least because the rank functions for the examples $Y$ and $Z(1)$ from Example \ref{ex:Y-and-Z} are the integer partition numbers and Fibonacci numbers respectively, two sequences of integers which have received an immense amount of study.  The first general result obtained about $p_n$ was the following:

\begin{prop}[Stanley \cite{Stanley1988}] \label{prop:weak-growth}
For any differential poset $P$ the rank sizes $p_0,p_1,p_2,...$ weakly increase.
\end{prop}

Two main lines of research about the rank function have developed, both motivated by the intuition that $Y^r$ should be the ``smallest" $r$-differential poset, and $Z(r)$ the ``largest".

The first concerns the asymptotic behavior of $p_n$.  Stanley and Zanello \cite{Stanley2012a} have shown that in an $r$-differential poset,
\[
p_n \gg n^a \cdot e^{\sqrt{2rn}}
\]
a bound which is fairly close to the well-known asymptotics for $p_r(n)=|(Y^r)_n|$:
\[
p_r(n) \gg c \cdot n^b \cdot e^{\pi \sqrt{2rn/3}}
\]
where $a,b,c$ are constants.  On the other hand, Byrnes (\cite{Byrnes2012}, Theorem 1.2) has shown that for any $r$-differential poset
\begin{equation*}
p_n \leq |Z(r)_n|.
\end{equation*}
From the recursive definition of $Z(r)$ in Example \ref{ex:Y-and-Z}
it is immediate that $|Z(r)|_n=r \cdot |Z(r)|_{n-1}+|Z(r)|_{n-2}$.  Standard techniques allow us to solve this recurrence and observe that
\begin{align}
|Z(r)|_n &= \frac{r+\sqrt{r^2+4}}{2\sqrt{r^2+4}}\cdot \ph_r^n+\frac{\sqrt{r^2+4}-r}{2\sqrt{r^2+4}}\cdot \psi_r^n \nonumber \\
&\leq 2 \cdot \ph_r^n \label{eq:pn-upper-bound}
\end{align}
where $\ph_r=(r+\sqrt{r^2+4})/2$ and $\psi_r=(r-\sqrt{r^2+4})/2$.

The second, more difficult, line of research has sought explicit bounds on the \textit{rank gaps} $\Delta p_n = p_n-p_{n-1}$, which, by Proposition \ref{prop:weak-growth} are nonnegative. These rank gaps are the eigenvalue multiplicities of $DU_n$ (see \cite{Stanley1988}) and also appear in the study of the finer algebraic structure of differential posets (see \cite{Agarwal2017, Miller2009}).  It was conjectured in 1988 in Stanley's original paper \cite{Stanley1988} that, except when $r=n=1$, the rank gaps $\Delta p_n$ are strictly positive.  This conjecture was proven by Miller only as recently as 2013, who showed:

\begin{theorem}[Miller \cite{Miller2013}] \label{thm:strict-growth}
For any $r$-differential poset, $\Delta p_n \geq 1$, unless $r=n=1$.
\end{theorem}

Despite the fact that Stanley's conjecture was open for so long, it seems that much stronger results may be true.  For example, Stanley and Zanello \cite{Stanley2012a} have asked whether \emph{all} iterated partial differences $\Delta^t p_n$ are eventually positive for large $n$, and whether this is already true for $n=2$ in the case $t=r$.  One implication of a positive answer to this question would be that
\begin{equation}
\lim_{r \to \infty} \min_{\substack{P \text{ $r$-differential} \\ n\geq 2}} \Delta p_n = \infty \label{eq:r-limit}
\end{equation}
and that
\begin{equation}
\lim_{n \to \infty} \Delta p_n = \infty \label{eq:n-limit}
\end{equation}
for any differential poset $P$.

Our main theorem is the first improvement on Miller's bound for general differential posets, and resolves (\ref{eq:r-limit}).  In addition, it seems plausible that the new methods we use to prove this result might lead to a resolution of (\ref{eq:n-limit}), see Remark \ref{rem:strategy}.

\begin{theorem} \label{thm:main-theorem}
For any $r$-differential poset, $\Delta p_n \geq 2r$ for $n \geq 4$ if $r=1$, $n \geq 3$ if $r=2$ and for $n \geq 2$ if $r>2$.
\end{theorem}

In Theorem \ref{thm:gaps-from-threads} we also prove a stronger result in the case where $P$ has many \textit{threads} (see Section \ref{sec:threads}).

\section{Projection matrices and path counting}

Let $\langle, \rangle$ denote the inner product on $\Q P$ which makes the elements of $P$ into an orthonormal basis.  It is easy to see that the operators $U,D$ are adjoint with respect to this inner product:
\[
\langle Ux,y \rangle = \langle x, Dy \rangle.
\]
Formulas for counting paths in the Hasse diagram of $P$ (viewed as a graph) can often be expressed in terms of the inner product.  For example $e(x):=\langle U^n \hat{0}, x \rangle$ for $x \in P_n$ is easily seen to be the number of paths in the Hasse diagram of $P$ from $\hat{0}$ to $x$ using only upward steps.  We will make use of the following enumerative identity:

\begin{prop}[Stanley \cite{Stanley1988}] \label{prop:sum-of-squares}
Let $P$ be an $r$-differential poset, then
\[
\sum_{x \in P_n} e(x)^2 = r^n n!.
\]
\end{prop}

The following linear algebraic facts will also be used:

\begin{prop}[Stanley \cite{Stanley1988}]
\label{prop:U-injective}
Let $P$ be a differential poset and $n \geq 0$, then:
\begin{itemize}
\item[(a)] $U_n: \Q P_n \to \Q P_{n+1}$ is injective, and
\item[(b)] $DU_n$ is invertible.
\end{itemize}
\end{prop}

Our main tool will be to study the projection operator onto $\im(U)$, whose entries can be expressed in terms of path counting in the Hasse diagram of $P$.

\begin{prop} \label{prop:M-is-projection}
Let $M_n=U(DU)^{-1}D_n$, then:
\begin{itemize}
\item[(a)] $M$ is the orthogonal projection $\Q P_n \sur \im(U_{n-1})$.
\item[(b)] In the standard basis $P_n$ for $\Q P_n$, the entries for $M_n=(m_{xy})_{x,y \in P_n}$ are given by
\[
m_{xy}=\sum_{k=1}^n (-1)^{k-1} \frac{\langle D^k x, D^k y \rangle}{r^k k!}.
\]
Futhermore $\langle D^k x, D^k y \rangle$ counts the number of pairs of paths of length $k$ beginning at $x$ and $y$ and ending at a common element of $P_{n-k}$.
\end{itemize}
\end{prop}
\begin{proof} We have
\[
M^2=U(DU)^{-1}DU(DU)^{-1}D=M,
\]
so $M$ is a projection.  Since $D$ and $(DU)^{-1}$ are surjective by Proposition \ref{prop:U-injective} ($D$ is the transpose of an injective map), $\im(M_n)=\im(U_{n-1})$, so $M$ is a projection onto $\im(U_{n-1})$.  Orthogonality follows easily from the adjointness of $U,D$, so (a) is proven.  For (b), we first show that
\begin{align*}
(DU_{n-1})^{-1} = \sum_{k=0}^{n-1} (-1)^k \frac{U^kD^k}{r^{k+1}(k+1)!}.
\end{align*}
Indeed, let $R$ denote the right-hand side, then:
\begin{align*}
DU_{n-1} \cdot R &= \sum_{k=0}^{n-1} (-1)^k \frac{DUU^kD^k}{r^{k+1}(k+1)!} \\
&= \sum_{k=0}^{n-1} (-1)^k \frac{U^{k+1}D^{k+1}+r(k+1)U^kD^k}{r^{k+1}(k+1)!} \\
&= I + (-1)^{n-1} \frac{U^nD^n}{r^nn!} \\
&= I.
\end{align*}
where the second equality follows from applying (\ref{eq:DU-UD}), the third from collapsing the telescoping sum, and the last from observing that $D^n=0$ on $\Q P_{n-1}$.  Now, by definition we have
\[
M_n = \sum_{k=0}^{n-1} (-1)^k \frac{U^{k+1}D^{k+1}}{r^{k+1}(k+1)!}.
\]
Reindexing and taking matrix entries by applying adjointness, we obtain the desired result.
\end{proof}

\begin{remark}
The formula for $(DU_{n+1})^{-1}$, which plays a key role in the proof of Proposition 2.3, was suggested to us in personal communication by Alexander Miller, who more generally gave a formula for $(UD_{n+1}+kI)^{-1}$ (our formula is the case where $k=r$).  The expression for $m_{xy}$ is also similar to an expression obtained by Miller in \cite{Miller2013} for some of the entries in $(DU+kI)^{-1}$.  Motivated by an earlier representation-theoretic proof of strict rank growth in the case of \textit{differential towers of groups} \cite{Miller2009} (see also \cite{Gaetz2018}), Miller studied the denominators in $(DU+kI)^{-1}$, rather than the eigenvalues of submatrices of $M$, as we do.
\end{remark}

We will always think of $M_n$ as a matrix in the standard basis for $\Q P_n$.  Given $S \subseteq P_n$, let $M_S$ denote the principal submatrix of $M_n$ with rows and columns indexed by the elements of $S$.

\begin{lemma} \label{lem:submatrix}
Let $S \subseteq P_n$ and suppose $M_S$ has no eigenvalue equal to 1, then $\Delta p_n \geq |S|$.
\end{lemma}
\begin{proof}
If $M_S$ has no eigenvalue equal to 1, then $\Q S$ must intersect $\im(U_{n-1})$ trivially.  Since $U_{n-1}$ is injective, this means that
\[
\dim(\Q P_n) \geq \dim(\Q P_{n-1}) + \dim(\Q S).
\]
And so $\Delta p_n \geq |S|$.
\end{proof}

\section{Thread elements and rank gaps}
\label{sec:threads}

It seems very difficult to estimate the entries $m_{xy}$ for arbitrary $x,y \in P_n$ using the formula in Proposition \ref{prop:M-is-projection}, since the sum is alternating and individual terms can be much larger than the sum (it follows from the fact that $M^2=M$ that $|m_{xy}| \leq 1$, however individual terms in the sum can be much larger than 1).  We therefore define special elements for which $m_{xy}$ can be more easily analyzed.

An element $x \in P$ is called a \textit{singleton} if $x= \hat{0}$ or $x$ covers a unique element of $P$.  A \textit{thread element} is a singleton $x \in P$ which covers another singleton.  A key property of thread elements is the following:

\begin{prop}[Miller and Reiner \cite{Miller2009}]
\label{prop:threads-extend}
Let $t_0 \in P$ be a thread element.  Then there exist thread elements
\[
t_0 \lessdot t_1 \lessdot t_2 \lessdot \cdots
\]
We call this infinite sequence a thread.
\end{prop}

Given a differential poset $P$, let $\tau_n$ denote the number of thread elements in $P_n$.  Proposition \ref{prop:threads-extend} implies that $\tau$ is a nondecreasing function of $n$.

\begin{ex}
If $P=Y^r$, then the thread elements in ranks $n \geq 2$ are the elements $(\emptyset,...,\emptyset, \lambda, \emptyset,...,\emptyset)$ where $\lambda=(n)$ or $(1^n)$.  Thus $\tau_n=2r$ for $n \geq 2$.

If $P=Z(r)$, we get $r$ thread elements in rank $n$ for each singleton in $P_{n-1}$, and there are $r$ of these singletons for each element of $P_{n-2}$.  Thus $\tau_n=r^2 p_{n-2}$; in particular, $\tau$ grows exponentially in $n$ by (\ref{eq:pn-upper-bound}).
\end{ex}

\begin{theorem} \label{thm:gaps-from-threads}
Let $P$ be an $r$-differential poset.  Then for $N \geq 4n$:
\[
\Delta p_N \geq \tau_n.
\]
\end{theorem}
\begin{proof}
Assume $n \geq 1$, since $\tau_0=0$.  Let $T \subset P_n$ be the set of thread elements in $P_n$, so $|T|=\tau_n$.  Suppose $N \geq 4n$ and for each element $t \in T$, extend $t$ to a thread via Proposition \ref{prop:threads-extend}, and let $\hat{t}$ be the element of this thread at rank $N$.  It is clear from the definition of threads that $\hat{s} \neq \hat{t}$ for $s \neq t$.  Let $\hat{T}=\{\hat{t}\:| \: t \in T\}$.  We now want to show that $M_{\hat{T}}$ has no eigenvalue equal to 1, and apply Lemma \ref{lem:submatrix}.

Let $s,t \in T$.  We first bound the tail of the sum formula for $m_{\hat{s} \hat{t}}$ given in Proposition \ref{prop:M-is-projection}.  We have
\begin{align}
\left| \sum_{k=N-n+1}^N (-1)^{k-1} \frac{\langle D^k \hat{s}, D^k \hat{t} \rangle}{r^k k!} \right| &= \left| \sum_{k=N-n+1}^N (-1)^{k-1} \frac{\langle D^{k-N+n} s, D^{k-N+n} t \rangle}{r^k k!} \right| \label{eq:using-thread} \\
& \leq \sum_{k=N-n+1}^N \frac{r^n n!}{r^k k!} \label{eq:using-e} \\
& \leq n \cdot \frac{r^n n!}{r^{3n} (3n)!}. \label{eq:a(r,n)}
\end{align}
The equality (\ref{eq:using-thread}) follows because $\hat{s}, \hat{t}$ are thread elements, so $D^{N-n} \hat{s}=s$ and $D^{N-n}\hat{t}=t$.  The inequality (\ref{eq:using-e}) uses Proposition \ref{prop:sum-of-squares}: the number of paths down from $s$ to $\hat{0}$ is at most $\sqrt{r^nn!}$, and thus the number of paths down from $s$ and ending in any particular rank is at most this number; as this also holds for $t$, the bound follows.  Letting $a(r,n)$ denote the expression on the right-hand side of (\ref{eq:a(r,n)}), all off-diagonal entries of $M_{\hat{T}}$ satisfy
\[
|m_{\hat{s} \hat{t}}| \leq a(r,n),
\]
since $\hat{s}$ and $\hat{t}$ lie on separate threads and have no common lower bounds above rank $n$.  The diagonal entries satisfy
\begin{align*}
m_{\hat{t} \hat{t}} &\leq \left( \frac{1}{r}-\frac{1}{2r^2}+ \cdots \pm \frac{1}{r^{N-n}(N-n)!} \right) + a(r,n) \\
&< \left(\frac{1}{r}-\frac{1}{2r^2}+\frac{1}{6r^3} \right) + a(r,n) \\
&\leq \frac{2}{3} + a(r,n).
\end{align*}
Let $\phi_r=\frac{1}{2}(r+\sqrt{r^2+4})$.  By Byrnes' result (\ref{eq:pn-upper-bound}), $p_n \leq 2 \phi_r^n$, and so $\hat{T}$ has at most this size.  By Gershgorin's Circle Theorem, in order to see that $M_{\hat{T}}$ has no eigenvalue equal to 1, it suffices to show that
\[
\frac{2}{3} + 2 \phi_r^n a(r,n) \leq 1.
\]
If $r>1$, then $r^{2n} > \phi_r^n$ and $2nn!/(3n)! \leq 1/3$.  If $r=1$, we have $\tau_1=1$, so Theorem \ref{thm:strict-growth} gives the result when $n=1$; for $n\geq 2$ it is easy to see that $2 \phi_1^n a(1,n)<1/3$.
\end{proof}

We now give the proof of Theorem \ref{thm:main-theorem}.

\begin{prop}
\label{prop:2r-threads}
Let $P$ be an $r$-differential poset.  Then at least two thread elements cover each $x \in P_1$.
\end{prop}
\begin{proof}
In any $r$-differential poset $P$, Equation (\ref{eq:DU-UD}) implies that there are $r$ elements in $P_1$, and all of these are singletons, so they must each be covered by $r+1$ elements.  Given $x \in P_1$, for each $y$ in $P_1$ different from $x$, there is at most one element covering both $x,y$, as can easily be seen from the defining relation (\ref{eq:DU-UD}).  Thus there are at most $r-1$ elements covering $x$ which are not singletons, and so at least two thread elements cover $x$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:main-theorem}]
This proof is similar to the proof of Theorem \ref{thm:gaps-from-threads}, except that we can be more precise, since the threads involved begin in ranks 0 or 1.  We prove the case $r \geq 3$ here, the other cases are nearly identical, except that the different values of $r$ in the denominators imply that larger ranks $n$ must be considered in order for the Gershgorin bound to hold.

Let $r \geq 3, n \geq 2$ and let $T=\{t_1,...,t_r,s_1,...,s_r\} \subset P_2$ be a set of $2r$ thread elements, with $t_i,s_i$ covering each element $x_i$ of $P_1$; such elements exist by Proposition \ref{prop:2r-threads}.  Extend each $t \in T$ to an infinite thread, and let $\hat{T}=\{\hat{t} \: | \: t \in T\}$ be the elements of rank $n$ in these threads.  Then $M_{\hat{T}}$ has diagonal elements:
\[
m_{\hat{t} \hat{t}} = \sum_{k=0}^{n-1} (-1)^k \frac{\langle D^{k+1}\hat{t},D^{k+1}\hat{t}\rangle}{r^{k+1}(k+1)!} = \frac{1}{r} - \frac{1}{2!r^2} + \cdots + (-1)^{n-1}\frac{1}{n!r^n}  < \frac{1}{r} \leq \frac{1}{3}
\]

for some $k$. For off-diagonal entries we have
\[
|m_{\hat{t_i}, \hat{s_i}}| = \left|\frac{1}{r^{n-1}(n-1)!} - \frac{1}{r^n n!}\right| \leq \frac{1}{r^{n-1}(n-1)!} \leq \frac{1}{3},
\]
and all other off-diagonal entries satisfy
\[
|m_{\hat{s} \hat{t}}|=\left| \frac{1}{r^n n!} \right| \leq \frac{1}{2r^2}.
\]
The sum of the absolute values of the matrix entries of $M_{\hat{T}}$ across a row is thus at most $1/3+1/3+(2r-2)/2r^2 < 1$.  Therefore by the Gershgorin Circle Theorem, $M_{\hat{T}}$ has no eigenvalue equal to 1.  Applying Lemma \ref{lem:submatrix} completes the proof.

\end{proof}

\begin{remark} \label{rem:strategy}
One can obtain stronger bounds in an ad-hoc manner using Lemma \ref{lem:submatrix}.  For example, it is possible to show that when $r=1$ we have $\Delta p_n \geq 3$ for $n \geq 6$ by letting $s_n, t_n$ be the elements of rank $n$ in the two threads which begin in rank 2, and letting $u_n$ be a certain sequence of elements which are ``close" to $s_n$ so that the entries of $M_{\{s_n,t_n,u_n\}}$ can be well approximated by exhaustively considering all possible paths.  What is needed to progress further on this problem is identify families $S_n \subset P_n$ whose size grows with $n$ for which the entries of $M_{S_n}$ can be estimated.
\end{remark}

\begin{conj}
In any differential poset $\lim_{n \to \infty} \Delta p_n = \infty$.
\end{conj}

\section*{Acknowledgements}
The authors wish to thank Fabrizio Zanello for helping to initiate this joint project and Richard Stanley for his helpful conversations.  We are also grateful to Patrick Byrnes for making his computer code available and Alexander Miller for providing useful references.

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\end{document}

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text:
\section{Bounds on $\ell_2$-error for truncated linear regression}\label{sec:ell2}

In this section, we give bounds on the $\ell_2$-error for the truncated sparse linear regression problem. The proofs are straightforward given the techniques we have used elsewhere in this paper.

We use the same notation as Section \ref{sec:linreg}, with:
\begin{equation}\label{eqn:minhat}
\hat{\vec{\Omega}} \in \arg \min_{\vec{\Omega} \in \R^d} \{l_n(\vec{\Omega})+ \lambda \|\vec{\Omega}\|_1\}
\end{equation}
where $l_n$ is the negative log-likelihood function:
\[
l_n(\vec{\Omega}) = \frac1n \sum_{i=1}^n \left(\frac{1}{2} {y^{(i)}}^2 - y^{(i)} \vec{\Omega}^\top \vec{x}^{(i)}  + \log \int \exp\left(-\frac{1}{2} z^2 +z \vec{\Omega}^\top \vec{x}^{(i)}\right) S(z) dz\right),
\]
which is convex everywhere. Recall that each $y^{(i)}$ is distributed as $\mathcal{N}(\vec{x}^{(i)}\vec{\Omega}^*, 1, S)$; we denote $K$ as the support of $\vec{\Omega}^*$ and let $k = |K|$. The matrix $\vec{X}$ is defined to have $n$ rows $\vec{x}^{(1)}, \dots, \vec{x}^{(n)}$.

For the theorem in this section, we need the survival probability assumption (Assumption \ref{ass:sur}) and the normalization assumption (Assumption \ref{ass:norm}), but we replace the minimum eigenvalue and mutual incoherence assumptions by the following.

\begin{assumption}[Restricted Eigenvalue]\label{ass:reseig}
There exists $\beta > 0$ such that the matrix $\vec{X}$ satisfies:
\[\frac{1}{n} \|\vec{X} \vec{\Delta}\|_2^2 \geq \beta \|\vec{\Delta}\|_2^2
\quad\quad \text{for all } \vec{\Delta} \text{ such that } \|\vec{\Delta}_{\overline{K}}\|_1 \leq 3 \|\vec{\Delta}_K\|_1.
\]
\end{assumption}
The restricted eigenvalue assumption is a very common one in the study of Lasso-type algorithms, and it is known that
many families of random design matrices satisfy it (see Chapter 7 of \cite{Wbook}). We can now state the main result of this section:

\begin{theorem}
Suppose that Assumptions \ref{ass:sur}, \ref{ass:norm} and \ref{ass:reseig} hold. Then, for any $\eps > 0$, if $\lambda = O\left(\frac{\beta \alpha^4 \eps}{Ck}\right)$ and $n = \tilde{O}\left(\frac{C^4k^2 \log d}{\beta^2 \alpha^8\eps^2}\right)$, then $\|\hat{\vec{\Omega}} - \vec{\Omega}^*\|_2 \leq \eps$.
\end{theorem}
\begin{proof}
We begin with a lemma showing ``restricted strong convexity'' of $l_n$ in a neighborhood of $\vec{\Omega}^*$.
\begin{lemma}\label{lem:locresconv}
There exists $\kappa \geq \Omega(\alpha^4\beta)$ for which the following holds. For any $\vec{\Omega}$ such that $\vec{\Delta} = \vec{\Omega}-\vec{\Omega}^*$ satisfies: (i) $\|\vec{\Delta} \|_2 \leq \frac{1}{C\sqrt{k}}$ and (ii) $\|\vec{\Delta}_{\overline{K}}\|_1 \leq 3 \|\vec{\Delta}_K\|_1$, it holds that:
 \[\vec{\Delta}^\top \left(\nabla^2 l_n(\vec{\Omega})\right)\vec{\Delta} \geq \kappa \|\vec{\Delta}\|_2^2.\]
\end{lemma}
\begin{proof}
Using the same steps as in the proof of Lemma \ref{lem:locstrong}, we get:
	\begin{align*}
\vec{\Delta}^\top \left(\nabla^2 l_n(\vec{\Omega})\right)\vec{\Delta}
		\geq \frac{1}{12e^2n} \sum_{i=1}^n ({\vec{x}^{(i)}}\vec{\Delta})^2 (\alpha(\vec{\Omega}^*, \vec{x}^{(i)}))^4 \cdot \exp\left(-2|\vec{x}^{(i)}(\vec{\Omega}-{\vec{\Omega}^*})|^2-4\right)

	\end{align*}
	Now, $|\vec{x}^{(i)}\vec{\Delta}|\leq \|\vec{x}^{(i)}\|_\infty \|\vec{\Delta}\|_1$. From Assumption \ref{ass:norm}, $\|\vec{x}^{(i)}\|_\infty \leq C$, while from the lemma's conditions, $\|\vec{\Delta}\|_1 = \|\vec{\Delta}_K\|_1 + \|\vec{\Delta}_{\overline{K}}\|_1 \leq 4 \|\vec{\Delta}_K\|_1 \leq 4 \sqrt{k}\|\vec{\Delta}\|_2 \leq 4/C$. Therefore:
	\begin{align*}
	\vec{\Delta}^\top \left(\nabla^2 l_n(\vec{\Omega})\right)\vec{\Delta}
		\geq \frac{\alpha^4}{12e^{38}n} \|\vec{X}\vec{\Delta}\|_2^2 \geq \frac{\alpha^4\beta}{12e^{38}}\|\vec{\Delta}\|_2^2,
		\end{align*}
		where the last inequality uses Assumption \ref{ass:reseig}.
\end{proof}

In the following, let $\vec{\Delta} = \hat{\vec{\Omega}} - \vec{\Omega}^*$ where $\hat{\vec{\Omega}}$ is as defined in (\ref{eqn:minhat}). The next lemma is well-known and uses only the convexity of $l_n$.
\begin{lemma}[e.g., Proposition 9.13 of \cite{Wbook}]\label{lem:wainwright913}
If $\lambda \geq 2 \|\nabla l_n(\vec{\Omega}^*)\|_\infty$, then $\|\vec{\Delta}_{\overline{K}}\|_1 \leq 3 \|\vec{\Delta}_K\|_1$.
\end{lemma}
Combining the above, we can now follow the proof of Lemma \ref{lem:mainbig} to get a similar statement.
\begin{lemma}
Suppose $\lambda \geq 2 \|\nabla l_n(\vec{\Omega}^*\|_\infty$. For any $\eps > 0$, if $\lambda = O\left(\frac{\beta \alpha^4 \eps}{Ck}\right)$, then $\|\Delta\|_2 \leq \eps$.
\end{lemma}
\begin{proof}
We first argue that if $\|\vec{\Delta}\|_2 > \frac{1}{C\sqrt{k}}$:
\[
|\vec{\Delta}^\top (\nabla l_n(\hat{\vec{\Omega}}) - \nabla l_n(\vec{\Omega}^*))| \geq \Omega\left(\frac{\beta\alpha^4}{C\sqrt{k}}\right) \|\vec{\Delta}\|_2.
\]
This follows by combining Lemma\footnote{We actually need an extension of Lemma \ref{lem:wainwright910} where $\vec{\Delta}$ is restricted to the cone $\mathbb{C} = \{\vec{z}: \|\vec{z}_{\overline{K}}\|_1 \leq 3 \|\vec{z}_K\|_1$\}. The proof goes through unchanged.} \ref{lem:wainwright910}, Lemma \ref{lem:locresconv} and Lemma \ref{lem:wainwright913}. We also have the upper bound as in the proof of Lemma \ref{lem:mainbig}:
\[
|\vec{\Delta}^\top (\nabla l_n(\hat{\vec{\Omega}}) - \nabla l_n(\vec{\Omega}^*))| \leq
\|\vec{\Delta}\|_1 \left(\lambda + \|\nabla l_n(\vec{\Omega}^*)\|_\infty\right) \leq \frac{3 \lambda}{2} \|\vec{\Delta}\|_1 \leq 6\sqrt{k}\lambda \cdot \|\vec{\Delta}\|_2
\]
where we used the lower bound on $\lambda$ and the fact $\|\vec{\Delta}\|_1 \leq 4\sqrt{k}\|\vec{\Delta}\|_2$ due to Lemma \ref{lem:wainwright913}. The above two inequalities contradict if $\lambda =O\left(\frac{\beta \alpha^4}{Ck}\right)$ and hence, $\|\vec{\Delta}\|_2 \leq \frac{1}{C\sqrt{k}}$.

Continuing along the lines of Lemma \ref{lem:mainbig}, we see that:
\[
|\vec{\Delta}^\top (\nabla l_n(\hat{\vec{\Omega}}) - \nabla l_n(\vec{\Omega}^*))| \geq  \Omega(\beta \alpha^4) \|\vec{\Delta}\|_2^2
\]
Hence, we again get a contradiction to $\|\vec{\Delta}\|_2 > \eps$ if $\lambda = O\left(\frac{\beta \alpha^4 \eps}{Ck}\right)$.
\end{proof}
Finally, we need an upper bound to $\|\nabla l_n(\vec{\Omega}^*\|_\infty$. We already carried out essentially this calculation in the proof of Lemma \ref{lem:lambda}. It follows from there that $\|\nabla l_n(\vec{\Omega}^*\|_\infty < t$ with high probability if $n \geq \Omega(C^2 \log d \log(n/\alpha)/t^2)$. Setting $t = \lambda/3 = O\left(\frac{\beta \alpha^4\eps}{Ck}\right)$, we get that it's sufficient to set $n \geq \tilde{\Omega}\left(\frac{C^4k^2 \log d}{\beta^2 \alpha^8\eps^2}\right)$.

\end{proof}


TITLE: If $0 \le A \le B$ then $\sqrt{A} \le \sqrt{B}$
QUESTION [0 upvotes]: If $0 \le A \le B$ then $\sqrt{A} \le \sqrt{B}$, where $A, B$ are linear transformations.
Hint: compute $(\sqrt{B} + \sqrt{A} + \epsilon)(\sqrt{B} - \sqrt{A} + \epsilon)$.

I used a hint from the book and proved that expression $C = (\sqrt{B} - \sqrt{A} + \epsilon)$ is invertible for $\epsilon > 0$.
Now I don't know how to show that $C$ is positive?
Once $0 \le C$, we can write $\sqrt{A} \le \sqrt{B} + \epsilon$, for all $\epsilon > 0$, and conclude that $\sqrt{A} \le \sqrt{B}$. Also for me it is not clear if we can apply the "techniques with limits" to the linear transformations.
Update:
Proving that $C = (\sqrt{B} - \sqrt{A} + \epsilon)$ is invertible. Let
$$(\sqrt{B} + \sqrt{A} + \epsilon)(\sqrt{B} - \sqrt{A} + \epsilon) = X + i\cdot Y$$
where:
$$X = B - A + 2\epsilon\sqrt{B} + \epsilon^2$$
$$Y = -i(\sqrt{A}\sqrt{B} - \sqrt{B}\sqrt{A})$$
Now $X$ is strictly positive for $\epsilon > 0$ and $Y$ is self-adjoint/Hermitian and $X + iY$ is invertible, which follow from previous exercise in the book.

REPLY [2 votes]: I assume that $A$ and $B$ are n x n Hermitian (equivalently real symmetric).  Your post says you've already proven $C := (\sqrt{B} - \sqrt{A} + \epsilon I)$  is invertible so skip to 3

A better hint is to try to maintain (conjugate) symmetry, so consider proving
$(\sqrt{B} + \sqrt{A} + \epsilon I)(\sqrt{B} - \sqrt{A} + \epsilon I) + (\sqrt{B} - \sqrt{A} + \epsilon I)(\sqrt{B} + \sqrt{A} + \epsilon I) \succeq 2 \epsilon^2 I \succ 0$
We know $(\sqrt{B} - \sqrt{A} + \epsilon I) = C$ has a trivial nullspace -- i.e. if there was some $\mathbf x \neq \mathbf 0$ such that $C\mathbf x = \mathbf 0$ consider the quadratic form implications which imply a contradiction with (1).
Since you've proven C invertible for any $\epsilon \gt 0$ that is equivalent to proving
$\det\Big(\big(\sqrt{B} - \sqrt{A}\big)-\lambda I  \Big) = (-1)^n \cdot \det\Big(\lambda I - \big(\sqrt{B} - \sqrt{A}\big) \Big) \neq 0$
since rescalaing by plus or minus 1 doesn't change whether a number is zero, we can just say  $p\Big(\lambda \Big) = p\Big(-\epsilon \Big) \neq 0$
for any $\lambda \lt 0$  (where we assign $\lambda:= -\epsilon$)
and $p$ is the characteristic polynomial of $\big(\sqrt{B} - \sqrt{A}\big)$, which is evidently Hermitian and hence all eigenvalues are real but none can be negative, which proves  $\big(\sqrt{B} - \sqrt{A}\big)\succeq \mathbf 0$
You mention concerns about 'techniques with limits' -- as shown in 3 that isn't really what's going on here.  The positive $\epsilon$ in effect acts as a negative number to evaluate under the image of the characteristic polynomial of $\big(\sqrt{B} - \sqrt{A}\big)$ and we show this negative number can never be a root (eigenvalue).  Techniques involving continuity can be used, however, though some care is needed.  On the other hand continuity related techniques can often dramatically simplify proofs.


TITLE: If $x$ is an approximate number, is $2x=x+x$ true?
QUESTION [0 upvotes]: Suppose I have a mass and I measured it. I found that it weighs $1.001 kg$. Since there may be error in my measurement process, the mass could be a few milligrams heavier or lighter. In grams, its weight is $1001 g$. I have no knowledge beyond the given 4 digits. It could weigh $1001.000 g$ i.e., the error in my process could be less than a milligram. But I don't know that!
Now suppose I have $1000$ measurements of $1.001 kg$. If I add them, I will end up with $1001.000 kg$.
So in one context, $1000×1.001=1001$ and in the other $1000×1.001=1001.000$.
Now, if someone asks me "what's $1000×1.001$, where $1.001$ is an approximate number" what should I say?

REPLY [1 votes]: It is better to add some rigor to the notion of approximate number. Let's say that an approximate number is just an interval $[a, b]$, where $a, b \in \mathbb R$. We can express it as $[a-\varepsilon,a+\varepsilon]$ - that is, $a$ with possible error $\varepsilon$.
Now, all we have to do is to define operations on intervals. If $x \in [a,b]$ and $y \in [c,d]$, then $x+y \in [a+c, b+d]$, which defines a reasonable way to add intervals.
So, if in your case the approximate value is, say, $[1.000, 1.002]$, which is the same as $1.001 \pm 0.001$, then adding $1000$ copies of it would yield $[1000, 1002]$, or $1001 \pm 1$ - notice that the error is multiplied by $1000$, too.


TITLE: Can the Multiplicative Fragment of Linear Logic be shown to be non-truth-functional?
QUESTION [1 upvotes]: The short form of my question is: Can we find two formulae (in the multiplicative fragment of linear logic (MLL), that is, without additives or exponentials) A and B such that (1st) A is provable and B is provable, (2nd) A ⅋ B (A par B) is provable?
The context of the question is this: If we can find two such formulae, we can easily show that the multiplicative fragment of linear logic is not truth functional.
By truth-functionality, we should have:

ϕ(A) = ϕ(B) ==> ϕ(~A) = ϕ(~B)
ϕ(A1) = ϕ(A2) AND ϕ(B1) = ϕ(B2) ==> ϕ(A1 ⊗ B1) = ϕ(A2 ⊗ B2)
ϕ(A1) = ϕ(A2) AND ϕ(B1) = ϕ(B2) ==> ϕ(A1 ⅋ B1) = ϕ(A2 ⅋ B2)

Let us assume, for reductio, that there exists a function ϕ, from the set of linear formulae to the elements of an arbitrary set E, that satisfies (1)-(3), and which is such that ϕ(A) = e if and only if A is provable in linear logic.
(Remark: We know that it is possible, in fact, to find a truth-functional value-assignment that gives every provable formula the same value---i.e. that assigns "true" to a formula whenever it is provable. This won't concern us here. What we want to show is that any such function always assigns truth to too many formulae, that "truth", on any assignment, may be construed as a necessary condition for provability, but by no means a sufficient condition for provability.)
What we want to do now is find formulae {A1, A2, B1, B2} which are each, on their own, provable, and so which will be such that:
ϕ(A1) = ϕ(A2) = e
 ϕ(B1) = ϕ(B2) = e
But which are such that (A1 ⅋ B1) is not provable, while (A2 ⅋ B2) is provable. If we find such formulae, then we will have shown that our assumption leads to a contradiction: by (I), ϕ(A1 ⅋ B1) ≠ ϕ(A2 ⅋ B2), but by (II), ϕ(A1 ⅋ B1) = ϕ(A2 ⅋ B2).
This is where I hit the wall. Finding multiplicative A1, B1 is simple enough: Let them each be the constant 1. 1 ⅋ 1 is easily shown to be unprovable (which by itself tells us a lot about the peculiarities of linear logic---1 ⅋ 1 looks like the linear counterpart of True v True). If we allow ourselves to dip into the additives, we can let A2 = 1 and let B2 = (1⊕⊥), and the resulting A2 ⅋ B2 will be provable. This is enough to show that LL, as a whole, is not truth-functional, because it leads to the destructive dilemma:
Either ϕ(1 ⅋ 1) = ϕ(1 ⅋ (1 ⊕ ⊥)) = e, in which case ϕ(x) = e is not equivalent to “x is provable”; since ϕ and e are arbitrary, this shows that no truth-functional value assignment captures linear provability.
Or else ϕ(1 ⅋ 1) ≠ ϕ(1 ⅋ (1 ⊕ ⊥)), which contradicts truth-functionality requirement (3). To the extent that it violates (3), ϕ is not truth-functional.
This is all well and good, but can we get a similar result for the multiplicative fragment, that is, can we find two multiplicative, MLL-provable formulae whose multiplicative disjunction is also provable?
(PS: I'm using Girard's notation, not Troelstra's, so 1 and ⊥ should be read as the multiplicative constants, the neutrals for tensor and par, respectively.)

REPLY [3 votes]: I realise I am several years late, but there is a lovely simple principle that answers your question definitively, and I can’t resist explaining it. I’ll assume we’re working in multiplicative linear logic, with units.
Consider a sequent $\Gamma$. Suppose $\Gamma$ has $m$ occurrences of ⅋ – counting commas as occurrences of ⅋ – and that it has $n$ negative occurrences of atoms (e.g. $p^\perp$ or $\perp$).
Lemma. If $\Gamma$ is provable, then $m=n$.
Proof. Suppose $\Gamma$ is provable, thus has a proof net. Ignoring axiom and $\perp$ links, each ⅋-switching of the proof net has $m+1$ connected components. To make it connected and acyclic, there must be $m$ axiom and $\bot$ links. But the number of axiom and $\bot$ links is just $n$, hence $m=n$ as required. ■
So at most two of $\vdash A$, $\vdash B$ and $\vdash A⅋B$ can be provable.
PS. Maybe it’s worth adding that you don’t need proof nets to prove this lemma. It can also be done as an easy induction over proofs.


TITLE: Does $\sum \log(n)^n/n^{n/2}$ converge?
QUESTION [1 upvotes]: Does the infinite series
$$\sum_{n=1}^\infty \frac{(\log n)^n}{n^{n/2}}$$
converge?
I'm not even sure where to go with this. I know there are a bunch of tests you can apply to find if a series converges or diverges, but I'm not sure where to start. Is there a trick to recognizing what test you should use?

REPLY [2 votes]: Hint:
For large $n$ we have
$$
\log n < \frac1{2} n^{1/2}
$$
Therefore
$$
\sum_{n=N}^\infty \frac{(\log n)^n}{n^{n/2}} < \sum_{n=N}^\infty \frac{1}{2^n} < 1
$$


TITLE: How does a Zener Diode work as a voltage stabilizer?
QUESTION [1 upvotes]: I mean I do have this diagram though and all the stuff about how it works, but I read it and not all of it makes sense. In my notes it says that when the input voltage increases beyond a certain limit the breakdown occurs, I get that line. But I can't understand the next line which says ""
How does the breakdown increase the potential drop across the resistor ? Shouldn't the potential drop be decreased as the voltage across the zener diode is also high and that due to the DC supply is also high ? And how can the voltage across the Zener diode be constant ? Didn't the supplied voltage just increase beyond the breakdown voltage, that's how the breakdown happens right ? When the voltage across the diode reaches a value higher than the zener diode the covalent bonds are broken and the electrons and holes flow to the opposite sides, so the current increases ? This is very confusing, any books that I can refer to study about semi-conductors

REPLY [0 votes]: [quote] when the input voltage increases beyond a certain limit the breakdown occurs, I get that line. But I can't understand the next line which says "https://i.stack.imgur.com/EaRVY.png"  [/quote]
Note that the voltage "across" the zener diode must increase above the breakdown voltage. If there was no dropping resistor, then that would be your " input voltage". But since the dropping resistor and the load resistor is fixed, the input voltage would have to be higher for the voltage across the zener (or load) to cause the breakdown.
At the zener breakdown voltage, the zener starts conducting ( it was acting like an open circuit all this while). But the voltage drop across the dropping resistor doesn't drop or rise magically. There will be 'sufficient increase in the voltage drop' only when you increase the input voltage further, because the rest will be maintained constant across the load.
Hence why I think that quoted statement is not talking about the instant the breakdown happens but after it.
Rest of your questions are nicely answered by @Vamsi Krishna.
Hope it helps.


TITLE: How to calculate the argument and its limit for the sequence $z_n=-2+i\frac{(-1)^n}{n^2}$
QUESTION [0 upvotes]: I am trying to show that the limit of the sequence
$$z_n=-2+i\frac{(-1)^n}{n^2}$$
exists, using the polar representation. Note that $\lim_{n\rightarrow \infty }z_n=-2$. $$$$I am finding difficulty in arriving at the final answer (specifically in finding the limit of the argument - I am getting a wrong result)

In polar form:
$$ r=\sqrt{4+\frac{1}{n^4}}$$
$$\theta_0 =tan^{-1}(\frac{(-1)^n}{-2n^2})=-tan^{-1}(\frac{(-1)^n}{2n^2})$$
where $\theta_0$ is the principal argument defined on $-\pi <\theta _0\le \pi $.
Now taking the limit of the modulus, we get
$$\lim_{n\rightarrow \infty }r=\sqrt{4+0}=2$$
and for the argument, we consider even and odd integers:

If n is even:

If n is odd:

But in the text book, the answers are

What did I do wrong? Isn't the inverse of an angle of $0$ is just $0$, which is is also in the range of the principal argument as in the figure below?

Moreover, in another example in the textbook (same sequence but with a positive real part)
$$z_n=2+i\frac{(-1)^n}{n^2}$$
the limits of the argument were both $0$ and hence it exists. What is the difference here? Why both $0$ and not, say, $\pi$, as above?

REPLY [1 votes]: Using $\theta_0$ as the principal argument of $z_n$ with $-\pi < \theta_0 \leq \pi$, it seems to me that when $n$ is even, we have
$$\tan\theta_0 = \frac{\mathrm{Im }\ z}{\mathrm{Re }\ z} = \frac{-1}{2n^2}$$ so that $$\theta_0 = \pi - \tan^{-1}\frac{-1}{n^2},$$
while for odd $n$, we have
$$\theta_0 = \tan^{-1}\left(\frac{1}{n^2}\right) - \pi.$$
If you really want, I can provide the crude drawings that lead me to believe this, but it hinges on the fact that $\tan^{-1}\theta \in (-\pi/2, \pi/2)$, while we need the principal argument $\theta_0$ to be between $-\pi$ and $\pi$.
If you look at what your angles are converging to, you'll see $0$. But that certainly shouldn't be the case, since $z_n$ converges to a value on the negative $x$-axis, where the angle is $\pi$. That should send up a red flag.


TITLE: Locally constant sheaves over an irreducible space is constant.
QUESTION [11 upvotes]: In Hartshorne's  Algebraic Geometry Chapter II Proposition 6.15: If $X$ is an integral scheme, the homomorphism $CaCl X \rightarrow Pic X$ is an isomorphism.
In the proof he wants to prove that $\mathcal L \otimes \mathcal K= \mathcal K$. It is clear that on an open cover $\{U_i\}$ $(\mathcal L \otimes \mathcal K)|_{U_i}\cong \mathcal K$.
From this he concludes that $\mathcal L \otimes \mathcal K \cong \mathcal K$, which follows from a general fact that if "$X$ is irreducible, a sheaf whose restriction to each open set of a covering of $X$ is constant, (*) to  is in fact a constant sheaf".
Can someone please give a  proof of the above fact that locally constant sheaves over an irreducible space is actually constant .

I guess Harthsorne wants to say that it is isomorphic to a constant sheaf.

REPLY [1 votes]: It suffices to show that for any such sheaf $\mathcal{F}$ and any global section $s\in\Gamma (X,\mathcal{F} )$, there exists an open cover $\{U_{\alpha}\}$ of $X$ and a set $A$ so that $s\vert_{U_{\alpha}}\in A$ for each $\alpha$. This is equivalent to saying that $\mathcal{F}$ is the sheaf associated to the constant presheaf $\underline{A}$.
Now since we have had an open cover $\{U_{\alpha}\}$ so that $\mathcal{F}\vert_{U_{\alpha}}$ is constant, there exists an open cover $\{V_{\gamma}\}$ of $X$ which is a refinement of $\{U_{\alpha}\}$ so that the restriction of $s$ to $V_{\gamma}$ lies in $A_{\alpha}$ if $V_{\gamma}\subseteq U_{\alpha}$. Since $X$ is irreducible, any two open subsets has non-empty intersection, and since $\mathcal{F}\vert_{U_{\alpha}}$ is constant for all $\alpha$, we have for $x\in U_{\alpha\beta }=U_{\alpha }\cap U_{\beta}$, $\mathcal{F}_x =A_{\alpha} =A_{\beta}$, hence the sets are all the same, and we have constructed such an open cover and find such a set, which implies that $\mathcal{F}$ is constant.


TITLE: Does the series $\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}$ converge?
QUESTION [1 upvotes]: Consider the series $$\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{n}$$
Does this series converge or diverge?
I tried all the tests like ratio test, Raabe's test, logarithmic test, comparison test but in vain. In ratio test I got the limit as $1$, so can't do anything with that. Can anybody give some hints?
Thank you!

REPLY [3 votes]: Hint: Write $$\frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{1}{n(\sqrt{n+1}+\sqrt{n})}$$


TITLE: Are these spaces differential manifolds?
QUESTION [0 upvotes]: I'm thinking about this spaces.

My doubt arise since the intersection point of the triangles.
Regards!

REPLY [3 votes]: The main point here is that these are not even topological manifold: there are points (in fact 1 point) which does not have any neighbourhood homeomorphic to an open set of $R^n$. As you guessed the problem is the "intersection point of the triangles" (vague enough but it should be clear what it means.
But even if you removed say the lower triangle there would be still problems: a smooth manifolds does not have "corners". So the upper triangle in the first picture would be a topological manifold but not a smooth one (thinking of the topology and smooth structure as a submanifold of $R^2$). In the second one is not clear whether the boundary is to be included or not. If yes the triangle is not a topological manifold but a topological manifold with boundary; it is not a smooth manifold with boundary as it has corners. The open triangle would be a smooth manifold.


TITLE: An open cover that is not locally finite
QUESTION [4 upvotes]: I could not understand why example $13.4$ is not locally finite.
Can you give me an explanation please.

REPLY [5 votes]: An open cover $\mathscr{U}$ of a space $X$ is point-finite if $\{U\in\mathscr{U}:x\in U\}$ is finite for each $x\in X$: each point of $X$ belongs to only finitely many members of $\mathscr{U}$. Every locally-finite cover of a space is automatically point-finite as well. And this open cover of $\Bbb R$ is not even point-finite. In fact, it fails in the worst way to be point-finite: each $x\in\Bbb R$ belongs to infinitely many members of the cover.
To see this, let $x\in\Bbb R$. For each $n\in\Bbb Z^+$ there is a rational number $q_n$ such that $\left|x-q_n\right|<\frac1n$, and we may further assume that $q_n\notin\{q_k:k<n\}$, since there are infinitely many rational numbers within $\frac1n$ of $x$. But then $x\in U_{q_n,n}$ for each $n\in\Bbb Z^+$, and the sets $U_{q_n,n}$ are all distinct, so $x$ belongs to infinitely many members of the cover.
(This is a real distinction, by the way: there are point-finite covers that are not locally finite.)


TITLE: Charge distribution that generates Electrostatic Field with constant magnitude
QUESTION [0 upvotes]: I have founde the following problem. I am given a sphere of radius $R$, with charge density $\rho$ and the fact that the electrostatic field inside the sphere has constant magnitude $E_0$. The question is to find the charge distribution.
Gauss' Law states that $\nabla\cdot \vec{E} = \frac{\rho}{\epsilon_0}$.  Thus, if I know a mathematical expression for the electrostatic field, I can calculate the charge density.
Obviously, the uniform field $\vec{E}=E_0 \cdot \hat{n} $ satisfies the constant magnitude constraint, but can there be another solution other than that trivial case? How can I prove that this is the only solution?

REPLY [0 votes]: From Gauss' law, using a Gaussian surface of radius $R$ with $R$ inside your charged sphere:
$$
\epsilon\int \vec E\cdot d\vec S=\oint \rho dV
$$
Assuming $\rho$ is spherically symmetric we obtain
$$
\epsilon \vert \vec E\vert\, 4\pi R^2 = 4\pi \int_0^R r^2 \rho dr
$$
Thus,  if $R_0$ is the radius of your charged sphere, any charge density $\rho$ of the form
$$
\rho=\rho_0 R_0/r
$$
for $\rho_0$ constant will produce
$$
4\pi \int_0^R r^2 \rho dr = 4\pi \int_0^R r \rho_0 R_0 = 2\pi \rho_0 R_0 R^2
$$
and thus
$$
\vert \vec E \vert = \frac{2\pi \rho_0 R_0 R^2 }{4\pi \epsilon R^2}= \frac{\rho_0 R_0}{2\epsilon}
$$
will be constant.


TITLE: Discrete Random Variables (Coin Flipping)
QUESTION [0 upvotes]: Consider flipping a coin sequentially. Let the probability of seeing a head be p, 0 < p < 1.
Let N denote the number of times that the coin is flipped until the first time that two
consecutive heads or two consecutive tails appear.
How do we derive the probability mass function of N?
So I assume N should be at least 2 which means P(N = 2) = 2/4. [{H,H},{H,T},{T,H},{T,T}]
P(N=3) = 6/8. [{H,H,H}.{H,H,T},{H,T,H},{H,T,T},{T,H,H},{T,H,T},{T,T,H},{T,T,T}]
Is this correct? I'm not sure

REPLY [1 votes]: Let $E_n$ denote the event that the first time you get two heads or two tails is in n throws. Then $\mathbb{P}[N=i]=\mathbb{P}[E_i]\quad\forall i\in[n]$. Note that $\{E_2,E_3,\dots\}$ are disjoint.
As you pointed out, $\mathbb{P}[N=2]=\mathbb{P}[E_2]=\frac{1}{2}$. Now let us look at the other cases. We will look at $E_3$ in detail and then generalize. We hit $E_3$ only when $\neg E_2$. The first two flips for such an event are $\{HT, TH\}$.
Specifically, the last flips are $\{T, H\}$ respectively. If the last flip is $T$, then with probability $\frac{1}{2}$, we can flip the third coin as $T$ for $E_3$, and in case last flip is $H$, then too with probability $\frac{1}{2}$, we can flip the third coin as $H$ for $E_3$. In case this doesn't happen, we have $\neg E_2\wedge\neg E_3$, and we can try for $E_4$ and so on.
Since nothing we mentioned in the last case is specific to the third throw, this generalizes. In other words,
\begin{align*}
\mathbb{P}[E_i|\neg E_2\wedge\neg E_3\dots\neg E_{i-1}] &= \frac{1}{2}\\
\text{and }\mathbb{P}[\neg E_i|\neg E_2\wedge\neg E_3\dots\neg E_{i-1}] &= \frac{1}{2}\text{$\qquad\forall i\geq3$}\\
\end{align*}
Then,
\begin{align*}
\mathbb{P}[E_2] &= \frac{1}{2}\\
\mathbb{P}[E_3|\neg E_2] &= \frac{1}{2}\\
\mathbb{P}[E_3] &= \mathbb{P}[E_3|\neg E_2] \times \mathbb{P}[\neg E_3]
= \frac{1}{2} \times \frac{1}{2} =\frac{1}{4}\\
\mathbb{P}[E_4] &= \mathbb{P}[E_4|\neg E_2\wedge\neg E_3] \times \mathbb{P}[\neg E_2\wedge\neg E_3]
= \frac{1}{2} \times \frac{1}{4} =\frac{1}{8}\\
\end{align*}
In general, the required distribution is: $$\mathbb{P}[N=i] =\mathbb{P}[E_i] = \frac{1}{2^{i-1}}\quad \forall i \geq 2$$


TITLE: Convert expression using only basic logical symbols - have trouble making the two sides apart
QUESTION [4 upvotes]: I have this expression:
$F \subseteq P(A)$
Where $F$ is some family of sets and $P(A)$ is the power set of $A$. I need to express this with using only:
$\in, \notin, =, \neq, \wedge, \vee, \rightarrow, \leftrightarrow, \forall, \exists$
I first started out at writing $F$ as:
$\forall x(\exists A(\forall y \in A(y\in F)))...$
However I think the notation is totally wrong. Am I supposed to somehow use indices to describe F with only these symbols?

REPLY [1 votes]: Recall that the elements in both $F$ and $P(A)$ are sets.
Let the universe of discourse (universal set) be "sets".
$$F\subseteq P(A) \iff \forall x(x \in F \rightarrow x \in P(A))$$
Essentially, what we are saying is that for any set $x$, if $x\in F$, then $x \in P(A).$


text:
\section{Direct Product Norm is Norm}
Tags: Examples of Norms

\begin{theorem}
Let $\struct {X, \norm {\, \cdot \,}}$ and $\struct {Y, \norm {\, \cdot \,}}$ be [[Definition:Normed Vector Space|normed vector spaces]].
Let $V = X \times Y$ be a [[Definition:Direct Product of Vector Spaces/Finite Case|direct product of vector spaces]] $X$ and $Y$ together with [[Definition:Operation Induced by Direct Product|induced component-wise operations]].
Let $\norm {\tuple {x, y} }$ be the [[Definition:Direct Product Norm|direct product norm]].
Then $\norm {\tuple {x, y} }$ is a [[Definition:Norm on Vector Space|norm]] on $V$.
\end{theorem}

\begin{proof}
=== Positive Definiteness ===
Let $\tuple {x , y} \in V$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {\tuple {x, y} }
      | r = \map \max {\norm x, \norm y}
      | c = {{defof|Direct Product Norm}}
}}
{{eqn | o = \ge
      | r =  0
      | c = [[Definition:Norm on Vector Space|Norm Axiom]] $N1$: Positive Definiteness
}}
{{end-eqn}}
Suppose $\norm {\tuple {x, y} } = 0$.
Then:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = \norm x
      | c = [[Definition:Norm on Vector Space|Norm Axiom]] $N1$: Positive Definiteness
}}
{{eqn | o = \le
      | r = \map \max {\norm x, \norm y}
}}
{{eqn | r = \norm {\tuple {x, y} }
      | c = {{defof|Direct Product Norm}}
}}
{{eqn | r = 0
}}
{{end-eqn}}
Hence, $\norm x = 0$
{{qed|lemma}}
=== Positive Homogeneity ===
{{begin-eqn}}
{{eqn | l = \norm {\alpha \tuple {x, y} }
      | r = \norm {\tuple {\alpha x, \alpha y} }
      | c = [[Definition:Operation Induced by Direct Product|Induced component-wise operations]]
}}
{{eqn | r = \map \max {\norm {\alpha x}, \norm {\alpha y} }
      | c = {{defof|Direct Product Norm}}
}}
{{eqn | r = \map \max {\size \alpha \norm x, \size \alpha \norm y}
      | c = [[Definition:Norm on Vector Space|Norm Axiom]] $N2$ : Positive Homogeneity
}}
{{eqn | r = \size \alpha \max \set {\norm x, \norm y}
}}
{{eqn | r = \size \alpha \norm {\tuple {x ,y} }
      | c = {{defof|Direct Product Norm}}
}}
{{end-eqn}}
{{qed|lemma}}
=== Triangle Inequality ===
{{begin-eqn}}
{{eqn | l = \norm {x_1 + x_2}
      | o = \le
      | r = \norm {x_1} + \norm {x_2}
}}
{{eqn | o = \le
      | r = \map \max {\norm {x_1}, \norm {y_1} } + \map \max {\norm {x_2}, \norm {y_2} }
}}
{{eqn | r = \norm {\tuple {x_1, y_1} }+ \norm {\tuple {x_2, y_2} }
      | c = {{defof|Direct Product Norm}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \norm {y_1 + y_2}
      | o = \le
      | r = \norm {y_1} + \norm {y_2}
}}
{{eqn | o = \le
      | r = \map \max {\norm {x_1}, \norm {y_1} } + \map \max {\norm {x_2}, \norm {y_2} }
}}
{{eqn | r = \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }
      | c = {{defof|Direct Product Norm}}
}}
{{end-eqn}}
Together:
:$\map \max {\norm {x_1 + x_2}, \norm {y_1 + y_2}} \le \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }$
Then:
{{begin-eqn}}
{{eqn | l = \norm {\tuple {x_1, y_1} + \tuple {x_2, y_2} }
      | r = \norm {\tuple {x_1 + x_2, y_1 + y_2} }
      | c = [[Definition:Operation Induced by Direct Product|Induced component-wise operations]]
}}
{{eqn | r = \map \max {\norm {x_1 + x_2}, \norm {y_1 + y_2} }
      | c = {{defof|Direct Product Norm}}
}}
{{eqn | o = \le
      | r = \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }
}}
{{end-eqn}}
{{qed}}
\end{proof}

config: wiki


TITLE: Maximum size of a cube in which a cube is mapped by linear transformation in Euclidean space
QUESTION [1 upvotes]: Suppose $L:E \rightarrow \mathbb{R}^d$ where $L$ is a linear transformation and $E \subset \mathbb{R}^d$. Then it is automatic that
$$
\|L(x)-L(x')\| \leq M \| x-x'\|
$$
for $x, x' \in E$ and some $M>0$. A text that I am reading (Elias and Shakarchi) says that a cube of side $l$ will mapped into a cube of side at most $2 \sqrt{d} M l$. I understand all factors (take the diagonal of the original cube) but do not understand where the 2 is coming from. Kindly point out what I am missing. This assertion is given in Exercise 8 of Chapter 1 in Stein and Shakarchi's Real Analysis (Book III). In my view, the bound should be $\sqrt{d/2} M l$.

REPLY [0 votes]: Choosing an orthonormal basis we get $$L(x)-L(y)=L(t_1e_1+…+t_ne_n)-L(s_1e_1+…+s_ne_n)=$$$$(t_1-s_1)L(e_1)+…+(t_n-s_n)L(e_n)=$$$$<x-y,e_1L(e_1)+…e_nL(e_n)>.$$ Then use cauchy-schwarz. Then check where the unit cube gets mapped. Every cube can then be streched from the unit cube.


TITLE: Write a statement in propositional logic that says $x \in A Δ B$. Solution should use at most two connectives.
QUESTION [0 upvotes]: We have two sets $A$ and $B$ and some object $x$. Let’s introduce two propositional variables:

$a$, which states that $x \in A$, and
$b$, which states that $x \in B$.

I get as far as $(a \lor b) \land \lnot(a \land b)$. But the questions asks for a solution that the total number of connectives used is just two. I have used 4: $\lor$, $\land$, $\lnot$ and another $\land$. How can I reduce even further?
Connected to this question:
What does the connection between 𝐴△𝐵△C and A <=> B <=> C?

REPLY [0 votes]: You can do $x \in A \oplus x \in B$
Here, the $\oplus$ is the symbol for the XOR


TITLE: Self-intersection number of a complex curve in complex projective space
QUESTION [6 upvotes]: I'm currently trying to get a grip on actually calculating some differential-geometric definitions. I'm looking at the following map from $\mathbb{CP}^{1}$ to $\mathbb{CP}^2$ :
$f([z_0,z_1])=[z_0^3,z_0 z_1^2,z_1^3]$
What I don't understand is how one would go about calculating the self-intersection number of this. It's not an immersion, so do I need to take an immersion in the same homology class first? And how do I actually calculate it - by perturbing the map a little and then counting transversal intersections with sign (this turned out to be very messy with my choices), or by actually finding the Poincaré dual (how would one go about that?) and integrating it?
In case anyone is wondering, this problem arose while trying to understand the adjunction inequality for J-holomorphic curves.
Thanks already for any help you can give!

REPLY [4 votes]: Well, the cohomology/intersection theory of $\mathbb{CP}^2$ is rather simple. Its generated by the class of the hyperplane. So every curve is $dH$ where $H$ is the class of the hyperplane and $d$ is an integer.
Furthermore, $\mathbb P^2$ has no curves of class $dH$ for $d<0$. You've described a cubic: so its class is $3H$, and self intersection is $9H^2$.
So $H^2 = pt$. Either by your favourite argument from topology for the cup product or by observing that two (generic) lines in $\mathbb P^2$ intersect at a point transversally.
In general, actually finding the Poincare dual to some class and integrating can be a bit of a pain. $\mathbb P^2$ is actually not too bad though. I've essentially described it above in very vague terms. There's only one real class to think about which is the class of the hyperplane.


TITLE: Let $b_{n}$ denote the number of compositions of $n$ into $k$ parts, where each part is one or two. Find the generating series for $b_{n}$
QUESTION [5 upvotes]: I am stuck with this combinatorics problems -
Let $n$ be a positive integer and let $b_{n}$ denote the number of compositions of $n$ into $k$ parts, where each part is one or two. For example, $(1, 2, 1, 2, 1)$ and $(2, 2, 1, 1, 1)$ are two compositions of $n = 7$ into $k = 5$ parts.
Firstly, we need to find the generating series for $b_{n}$
Secondly, Prove that $b_{n} = {k \choose n-k}$ for $k\le n \le2k$ and $b_{n} = 0$ otherwise.

REPLY [2 votes]: There's a nice combinatorial proof for the second question:
The number of compositions of $n$ into $k$ parts each equal to $1$ or $2$ is equal to the number of compositions of $n-k$ into $k$ parts each equal to $0$ or $1$. Clearly there must be $n-k$ $1$s in such a composition, and there are $\binom k{n-k}$ ways of arranging those $1$s among the $k$ parts.


TITLE: Quotient of arbitrary free involution on $S^n$
QUESTION [9 upvotes]: If we consider arbitrary free involution on $S^n$, then the quotient need not be diffeomorphic to $\Bbb RP^n$ if $n\geq 5$ and a reference for this is "some curious involutions of spheres"
by Morris W. Hirsch and John Milnor. In that same paper, it was mentioned as an unsolved problem that whether the quotient space is homeomorphic to $\Bbb RP^n$?. I would like to know if the answer of the above question is known by now and  references for that will be very helpful.
Thank you so much!

REPLY [11 votes]: In every dimension $\ge 4$ there is a fake real projective space, i.e., a manifold that is homotopy equivalent but not homeomorphic to $RP^n$.
Here you can find a computation for the topological surgery structure set of $RP^n$. It is stated for $n>4$ but I think it extends to $n=4$ because $\mathbb Z_2$ is good in the sense of Freedman. To get a homeomorphism classification of homotopy $RP^n$'s we also need to know the group of homotopy self equivalences of (any manifold homotopy equivalent to) $RP^n$: it is $\mathbb Z_2$ if $n$ is odd and trivial if $n$ is even (see e.g. p.61 of Rutter's survey ``Spaces of homotopy self-equivalences''). Moreover, if $n$ is odd, then $RP^n$ admits a self-map of degree $-1$, namely, the map induced by reflection in an equator of $S^n$. Thus any homotopy self-equivalence of $RP^n$ is homotopic to a diffeomorphism.
Let $f_i: M_i\to RP^n$ be homotopy equivalences representing different elements in the structure set. If $d_{ij}: M_i\to M_j$ is a homeomorphism, then $f_i^{-1}f_j d_{ij}$ is a homotopy self-equivalence of $M_i$. By the previous paragraph $f_i^{-1}f_j d_{ij}$ is homotopic to a homeomorphism, so $f_i$, $f_j$ represent the same element in the structure set. Thus in every dimension there is a fake $RP^n$.
A word of caution: the above argument does not show that the fake $RP^n$ is smoothable. For example, the only fake $RP^4$ is not smoothable as I learned from ``Invariant knots of free involutions of $S^4$'' by Ruberman, see here. There do exist many smoothable fake $RP^n$'s, see e.g.,  Smooth free involutions on homotopy $4k$-spheres by
Fintushel-Stern.


TITLE: Problem regarding mathematical skill of a math lover
QUESTION [10 upvotes]: I don’t know whether it would be a good idea to seek solution for some personal problem here. But as there’s no harm in asking I’m putting it under the soft tag.
It’s about forming a career in mathematics for an individual like me for whom morning didn’t actually show the day. It not before 14 that I actually got interested in Mathematics when I choose (don’t know why) ‘Mechanics’ as my optional subject. Earlier to it I had a consistent mediocrity in the subject. But everything started to change ever since Mechanics came into my way. I started to realize the scope of the subject & the harder problems started to drive me crazy. I fell in love with Mathematics & my performance started to increase at a consistent level. Ever since I’m with it and today I’m very happy with my decision with a dream of being a mathematician. Right now I’ve completed my master in Mathematics with a high grade with a dream of having a PhD from some reputed institution.
But now as I’m preparing for the competition exams like NET (sample question), NBHM (sample question) or ISI (sample question) for pursuing PhD, I began to feel that I’m slower than I should be. Many a time, the solution of a problem occurred to me much later than others (preparing for the same). Even when I found the trick for a solution I didn’t get why I took so much time to crack it. But I think how fast someone crack problems determine his career in the field. All this things have provoked me to present my questions in the forum:

Does all this happen because of a lack of talent or because of a Mathematicians are designed from the childhood in some other way?
Can my mathematical skill be improved to a Mathematician's level?

REPLY [12 votes]: I think mathematics is an artificial language that makes possible very profound thinking. As with any other human technique, there will always be people who seem to do everything effortlessly and are always brilliant. These are geniuses and very rare and special human specimens; they are to admire.... but mathematics is not about discovering geniuses, but about a particular way of deep thinking which's only purpose is to enlighten the thinker himself.
But society and people will take every possible excuse to show off how marvelous they are. People will use whatever technique to prove themselves better than everyone else.
I am convinced, that the pleasure a genius feels when he understands some particular mathematical concept and plays with it in his mind, is the same one that you feel when you do it. The pleasure that one can
feel by being the best at something has little to do with mathematics. It is the same pleasure a guy can feel by being the one who jumps higher or the one who made some pointless Guiness record. It has nothing to do with the pleasure and satisfaction one can find in mathematics' deep thought.
So keep on studying, love what you study and know that doing what you love is the best thing you can do in life.
To the first question: Some seem to have a special ability to do exams, that's all. In my opinion exams are merely in the way and must be passed so that one can do what is really important, in this case, mathematics.
To the second question: Absolutely, you can train yourself to "mathematician's level" and you will love it.


TITLE: Why is my tupperware sealed after heating it up in the microwave?
QUESTION [6 upvotes]: I had a vegetable soup in a crystal tupperware and I put it in the microwave to heat it up.
The funny thing is that when I took it out, it was impossible to open it up. The lid is bent towards the inside, as if the air has escaped the tupperware somehow.
Can anyone tell me why this might have happened?
Thanks

REPLY [1 votes]: When heated, air expands. Now when you take the tupperware out of the microwave and it cools down, the air inside cools down again and contracts, and 'sucks' the lid inwards.
Actually, that's not quite right - it's more precise to say the air outside presses the lid shut. The bodies of air outside and inside seek to equalize their pressure, and normally the tupperware would just shrink / contract until it's inside pressure is equal to the room pressure. But since the container is somewhat rigid, it can't shrink enough and there remains a pressure difference. This causes a force acting on the lid, keeping it shut.
So:
cooling down => volume shrinks
cooling down, but forcing the volume to be constant => pressure goes down
inequal pressure => force towards the lower pressure


TITLE: Bounded linear functional on a $C^*$ algebra
QUESTION [1 upvotes]: I want to prove the following statement:

Let $\mathcal{A}$ be a unital $C^*$ algebra with unit $e_\mathcal{A}$ and let $\varphi \in \mathcal{A}^*$ be a bounded linear functional in $\mathcal{A}$ such that
  $$||\varphi||_{\mathcal{A}^*} = \varphi(e_\mathcal{A})=1.$$
  Show that $\varphi$ is a state.

So all I have to show is that $\varphi$ is positiv, that means $\varphi(A^*A)\geq 0$ for all $A\in\mathcal{A}.$
I got the hint to show first that $\varphi(A)\in\mathbb{R}$ for all self-adjoint $A=A^*\in \mathcal{A}$ by considering $\varphi(A+i\gamma e_\mathcal{A})$ with $\gamma\in\mathbb{R},$ but I don't know how to show that and how can I apply this for the statement above?

REPLY [1 votes]: Suppose $A$ is positive, then
$${\big\|}\,\|A\|e_\mathcal{A}-A\,\big\|≤\|A\|,$$
which you can see by considering the spectrum of $A$. Specifically $\sigma(A)$ consists only of numbers $≥0$ and we have $\|A\|=\sup_{\lambda\in\sigma(A)}|\lambda|$ for normal elements.
If you assume $\varphi(A)<0$ then $\varphi(\|A\|e_\mathcal{A}-A)=\|A\|-\varphi(A)>\|A\|$, contradicting $\|\varphi\|=1$.
So you see that for any positive $A$, $\varphi(A)$ must also be positive or admit a non-vanishing imaginary component. But positive elements are self-adjoint and here your hint tells you that $\varphi(A)$ can never have non-vanishing imaginary component if $A$ is positive.
To see how the hint can be proven, suppose $A$ is self-adjoint with $\varphi(A)= x+ iy$ with $y\neq0$. Note that for real $\lambda$
$$\|A+i\lambda\, e_{\mathcal A}\|= \sqrt{\|A\|^2+\lambda^2}$$
and that
$$|\varphi(A+i\lambda\, e_\mathcal{A})|=\sqrt{x^2+(\lambda+y)^2}.$$
It follows that
$$\frac{\varphi(A+i\lambda\,e_\mathcal{A})}{\|A+i\lambda\,e_\mathcal{A}\|}=\sqrt{\frac{x^2+(\lambda+y)^2}{\|A\|^2+\lambda^2}}.$$
If you do some analysis you will find that for very large $\lambda$ this is roughly $\left|\mathrm{sign}(\lambda)+\frac y\lambda\right|$, once again contradicting $\|\varphi\|=1$.