TITLE: Solving $-2y'=xy^3+y$ with an initial value
QUESTION [0 upvotes]: I'm having a bit of trouble with this differential equation and it's initial value:
The equation itself is $-2y'=xy^3+y$, and the initial value $y(1)=0$
I solved it using Bernoulli's equation:
Let $z=y^{-2}$ then $z'=-2y^{-3}y'$ so, $y'=\frac{z'}{-2y^{-3}}$
Using that change of variables, I get $z'=x+z$, and solving that diff eq and undoing the change of variable:
$y(x)=\frac{1}{\sqrt{ke^x-x-1}}$
When trying to solve for $k$, with the given initial value I can't get past this:
$y(1)=\frac{1}{\sqrt{ke-2}}=0$, and of course there's no $k$ that satisfy this.

REPLY [1 votes]: Since $y(1) =0$, the substitution $z = y^{-2}$ does not work.
Indeed, the constant function $y(x)=0$ is one of the solutions. By uniqueness, it's THE solution.


TITLE: Nonnegative solution for wave equation.
QUESTION [0 upvotes]: Consider the following boundary problems for the wave equation:
$$\begin{cases} u_{tt}=u_{xx},\; (x,t)\in (0,\pi)\times (0,\infty) \\ u(0,t)=u(\pi,t)=0,\; t\geq 0 \\ u(x,0)=g(x),\; x\in [0,\pi] \\ u_t(x,0)=0,\; x\in [0,\pi] \end{cases} (*) \quad \quad  \begin{cases} u_{tt}=u_{xx},\; (x,t)\in (0,\pi)\times (0,\infty) \\ u_x(0,t)=u_x(\pi,t)=0,\; t\geq 0 \\ u(x,0)=g(x),\; x\in [0,\pi] \\ u_t(x,0)=0,\; x\in [0,\pi] \end{cases}(**) $$
where $g\in C^2([0,\pi])$.
I was asked to solve these two questions:
a) In $(*)$, if $g(0)=g(\pi)=0$ and $g\geq0$, then $u\geq0$.
b) In $(**)$, if $g'(0)=g'(\pi)=0$ and $g\geq0$, then $u\geq0$.
For a), I proved that it is not true. Solving by separation of variables, the only possible solution has the form $$u(x,t)=\sum_{k=0}^{\infty} a_k\cos(kt)\sin(kx),\quad a_k=\frac{2}{\pi}\int_0^{\pi} g(x)\sin(kx)\,dx.$$ Then, if $g(x)=\sin x$, we have $g\geq0$ on $[0,\pi]$ and $u(x,t)=\sin x\cos t$, so for example $u(\pi/4,3\pi/4)=(1/\sqrt{2})(-1/\sqrt{2})<0$.
For b), using separation of variables again, the only possible solution has the form $$u(x,t)=\sum_{k=0}^{\infty} a_k\cos(kt)\cos(kx),\quad a_k=\frac{2}{\pi}\int_0^{\pi} g(x)\cos(kx)\,dx.$$ Looking at some functions $g$, I was not able to find $u<0$ at some point, so I think b) should be true. However, I cannot prove this. For instance, looking at the explicit solution, since $a_k$ and $\cos$ may take any sign, I cannot proceed. I also tried by contradiction assuming that there is a minimum of $u$ smaller than $0$, and then use that $\nabla u$ at that minimum is $0$ and $\Delta u$ at that point is $\geq0$, but nothing.
Could you provide any hint to solve b)?

REPLY [1 votes]: Solve (**) in a different way.
Extend the initial value to $[\pi,2\,\pi]$ by parity ($g(x)=g(x-\pi)$,) and then to all of $\Bbb R$ by periodicity. The initial value for $u_t(x,0)$ is extended to be equal to $0$ on $\Bbb R$. Then the solution of the problem is given by $u(x,t)=g(x+t)+g(x-t)$. Since $g\ge0$, so is $u$.


text:
\section{Proof of Theorem~\ref{theSingleStream}}\label{proofSingleStreamPenalty}

Toward proving Theorem~\ref{theSingleStream}, we adapt the approach of \cite{wanOptimalityITW2016}, to lower bound the delay for the case where, out of the $K$ users, only $K_1$ users have a cache. The bound will then also prove tight for all
\begin{align}
	L\geq \frac{K_2}{T_{K_{1}}}-1 = \frac{K_2(1+K_1\gamma)}{K_1(1-\gamma)}-1
\end{align}
as we will see in Appendix~\ref{appendix2}.

The proof (for $L=1$) tracks closely steps\footnote{We note in advance that a naive adaptation of the approach in \cite{wanOptimalityITW2016}, where we would simply account for a reduced sum-cache constraint $K_1 M$ corresponding to a redundancy $t=\frac{K_1 M}{N}$, would yield a loose bound; for example when $L=1$, this naive bound would be $T\geq\frac{K-t}{t+1}$ which would then translate to $T\geq \frac{K_1(1-\gamma)}{1+K_1\gamma} + \frac{K_2}{1+K_1\gamma}$ which is loose as the optimal delay will turn out to be $T = \frac{K_1(1-\gamma)}{1+K_1\gamma} + K_2$.} from~\cite{wanOptimalityITW2016} which --- for the case of $K_1 = K$ (where all users have caches) --- employed index coding to bound the performance of coded caching.
Some of these steps are sketched here for the sake of completeness. Particular care is taken here to properly construct the bound's counting arguments in a way that accounts for the fact that specific symmetries that are essential to the approach in \cite{wanOptimalityITW2016}, do not directly hold here, simply because the set $\mathcal{K}_{1} = [K_1]$ of users that enjoy side information is only a subset of the users that request files.

We will begin with lower bounding, first for the case of $L=1$, the delay $T(\dv,\chi)$ for any generic caching-delivery strategy $\chi$ and any demand vector $\dv \in \mathcal{D}_{wc} \defeq \{\dv:  \ d_i \neq d_j, ~i,j\in[K],~i\neq j\}$ whose $K$ entries are all different.  In the following, we use $\Zc_i$ to denote the cache of each user $i$, where naturally $\Zc_i = \emptyset$ for $i\in \mathcal{K}_2 \defeq [K] \setminus \mathcal{K}_1$.

\paragraph*{Distinct caching problems and their corresponding index coding equivalents \label{sec:CachingProblem}}

We first follow closely the approach in \cite{wanOptimalityITW2016} to describe the association between index coding and our specific caching scenario here. As in \cite{wanOptimalityITW2016}, each caching problem (defined by a demand vector $\dv \in \mathcal{D}_{wc}$) is converted into an index coding problem, by having each requested file $W^{d_i}$ split into $2^{K_1}$ disjoint subfiles $W^{d_i}_\Tau,\Tau\in 2^{[K_1]}$, where $\Tau\subset[K_1]$ indicates the set of users that have $W^{d_i}_\Tau$ cached.
Since no subfile of the form $W^{d_i}_\Tau, \; ~\Tau \ni i$ is requested, the index coding problem here is defined by $$K_1 2^{K_1-1} + K_2 2^{K_1}$$ requested subfiles, which form the nodes of the side-information graph $\mathcal{G}=(\mathcal{V}_{\mathcal{G}},\mathcal{E}_{\mathcal{G}})$, where $\mathcal{V}_{\mathcal{G}}$ is the set of vertices (each vertex/node representing a different demanded subfile $W^{d_i}_\Tau, \Tau\not\ni i$) and $\mathcal{E}_{\mathcal{G}}$ is the set of direct edges of the graph. We recall that an edge from node $W^{d_i}_\Tau$ to $W^{d_{i'}}_{\Tau'}$ exists if and only if $i'\in\Tau$.

As in \cite{wanOptimalityITW2016}, this allows us to lower bound $T(\dv,\chi)$ by using the index-coding converse from~\cite{li2017cooperative} which says that for a given $\dv,\chi$ --- with corresponding side information graph $\mathcal{G}_{\dv}=(\mathcal{V}_{\mathcal{G}},\mathcal{E}_{\mathcal{G}})$ with $\mathcal{V}_{\mathcal{G}}$ vertices/nodes and $\mathcal{E}_{\mathcal{G}}$ edges --- the delay is bounded as
\begin{equation}\label{eq:indexbound}T\geq \sum_{\smallV \in \mathcal{V_{J}}}|\smallV|
\end{equation}
for every acyclic induced subgraph $\mathcal{J}$ of $\mathcal{G}_{\dv}$, where $\mathcal{V}_{\mathcal{J}}$ denotes the set of nodes of the subgraph $\mathcal{J}$, and where $|\smallV|$ is the size of the message/subfile/node $\smallV$.

The following describes the acyclic graphs, and also directly shows that these remain acyclic after they are enlarged to account for the content requested by the cache-less users.
In the following we will consider permutations $\sigma\in S_{K_1}$ from the symmetric group $S_{K_1}$, and, for a given demand vector $\dv$, we will use $\mathcal{A}_{\dv} \defeq \cup_{i \in [K] \setminus [K_1]} W^{d_i}$ to denote the union of all content in $\dv$ that is requested by the users in $[K] \setminus [K_1]$.

\begin{lemma}\label{lem:cons_acyclic}
For any $\dv$ and any $\sigma\in S_{K_1}$, and for an acyclic subgraph $\mathcal{J}_{\dv,\sigma}$ of $\mathcal{G}_{\dv}$, is designed here to consist of all subfiles $\{W^{d_{\sigma(i)}}_{\Tau},~\forall i\in[K_1], \forall \Tau\subseteq [K_1]\setminus \{\sigma(1),\sigma(2),\dots,\sigma(i)\} \}$, then the enlarged graph $\mathcal{J}_{\dv,\sigma} \cup \mathcal{A}_{\dv} $ is also acyclic.
\end{lemma}
\vspace{3pt}\emph{Proof.} The proof that the subgraph $\mathcal{J}_{\dv,\sigma}$ is acyclic is direct from \cite[Lemma 1]{wanOptimalityITW2016}. The proof that $\mathcal{J}_{\dv,\sigma} \cup \mathcal{A}_{\dv} $ is also an acyclic graph, i.e., that the addition (on the original $\mathcal{J}_{\dv,\sigma}$) of all the nodes corresponding to $\mathcal{A}_{\dv}$ does not induce any cycles, follows by first recalling that a directed edge from node $W^{d_i}_\Tau$ to $W^{d_{i'}}_{\Tau'}$ exists if and only if $i'\in\Tau$, which thus tells us that an edge cannot be drawn from any node representing content from~$\mathcal{A}_{\dv}$, because any cache-less user $i \in {K} \setminus [K_1]$ cannot belong to any such $\Tau$ simply because $\Tau \subset [K_1]$.\vspace{3pt} \qed

Given the acyclic subgraph $\mathcal{J}_{\dv,\sigma} \cup \mathcal{A}_{\dv} $, we combine Lemma~\ref{lem:cons_acyclic} with \eqref{eq:indexbound} to get
\begin{equation}
T(\dv,\chi)\geq T^{LB}(\sigma,\dv,\chi)
\end{equation} where
\begin{align}
&T^{LB}(\sigma,\dv,\chi) \defeq \sum_{\smallV \in \mathcal{V_{\mathcal{J}_{\dv,\sigma}}}\cup \mathcal{A}_{\dv}}|\smallV| \nonumber\\
& =  \!\!\!\!\! \sum_{\Tau\subseteq [K_1]\setminus \{\sigma(1)\}}|W^{\boldsymbol{d_{\sigma(1)}}}_{\Tau}| + \!\!\!\! \sum_{\Tau\subseteq [K_1]\setminus \{\sigma(1),\sigma(2)\}}   \!\!\!\! \!\!\!\! |W^{\boldsymbol{d_{\sigma(2)}}}_{\Tau}|+\dots \nonumber \\
&+ \sum_{\Tau\subseteq [K_1]\setminus \{\sigma(1),\dots,\sigma(K_1)\}}|W^{\boldsymbol{d_{\sigma(K_1)}}}_{\Tau}|
+|\mathcal{A}_{\dv}| . \label{eq:TLB}
\end{align}

Then, as in \cite{wanOptimalityITW2016}, we average over worst-case demands to get
\begin{align}
T^*&\defeq \min_{\chi} \max_{\dv \in \mathcal{D}_{W_c}} T(\dv,\chi)\nonumber \\
&\geq \min_{\chi} \max_{\dv \in \mathcal{D}_{W_c}} \max_{\sigma \in S_{K_1}} T^{LB}(\sigma,\dv,\chi)\nonumber \\
&\geq \min_{\chi} \frac{1}{|\mathcal{D}_{W_c}|}\frac{1}{|S_{K_1}|}  \sum_{\sigma \in S_{K_1}} \sum_{\dv \in \mathcal{D}_{W_c}} T^{LB}(\sigma,\dv,\chi)\nonumber \\
&\geq \min_{\chi} \frac{1}{ P(N,K) K_1! }  \sum_{\sigma \in S_{K_1}} \sum_{\dv \in \mathcal{D}_{W_c}} T^{LB}(\sigma,\dv,\chi)\label{eq:alternativedefinitionofT}
\end{align}
where in the above we use $P(N,K) \defeq \frac{N!}{(N-K)!}$.

Rewriting the summation in \eqref{eq:alternativedefinitionofT}, we get
\begin{align}\label{eq:longinequality}
&\sum_{\sigma \in S_{K_1}} \sum_{\dv \in \mathcal{D}_{W_c}} T^{LB}(\sigma,\dv,\chi) =  \\
& \sum_{i=0}^{K_1}\sum_{n\in[N]}\sum_{\Tau\subseteq[K_1]:|\Tau|=i} \!\!\!\! |W^n_{\Tau}| \cdot \!\!\!\! \underbrace{\sum_{\sigma \in S_{K_1}} \sum_{\dv \in \mathcal{D}_{W_c}} \!\!\!\! \mathds{1}_{\mathcal{V}_{\mathcal{J}_{\dv,\sigma}}}(W^n_{\Tau})}_{\defeq Q_{i}(W^n_\Tau)} + |\mathcal{A}_{\dv}| \nonumber
\end{align}
where $\mathcal{V}_{\mathcal{J}_{\dv,\sigma}}$ is the set of vertices in the acyclic component subgraph $\mathcal{J}_{\dv,\sigma}$ for a given $\dv,\sigma$ pair, and where $\mathds{1}_{\mathcal{V}_{\mathcal{J}_{\dv,\sigma}}}(W^n_{\Tau})$ denotes the indicator function which takes the value of 1 only if $W^n_{\Tau} \subset \mathcal{V}_{\mathcal{J}_{\dv,\sigma}}$, else it is set to zero.

\paragraph*{Counting arguments accounting for cache-less users \label{sec:countingArguments}}
Our aim is to count the number of times, $Q_{i}(W^n_\Tau)$, that any specific subfile $W^n_\Tau$ appears in the summation in~\eqref{eq:longinequality}. To do this, we draw from the counting arguments in \cite[Section VII-C]{parrinello2018coded} which derives $Q_{i}(W^n_\Tau)$ for the case where $K$ users share $\Lambda \leq K$ caches, where each cache $r\subset [\Lambda]$ serves $\Lambda_r$ users. Adapting these steps\footnote{The following expression could not have been derived, had we simply substituted $K$ for $K_1$, in the corresponding $Q_i$ expression in \cite{wanOptimalityITW2016}. Such a naive approach would have essentially corresponded to treating the cache-less and cache-aided cases separately, and would not have allowed us to guarantee, among other things, that both cache-less and cache-aided users request different files.} in \cite[Section VII-C]{parrinello2018coded} gives that
\begin{align}\label{eq:Qi}
Q_{i} &= Q_{i}(W^n_\Tau)\dfn \sum_{\sigma \in S_{K_1}} \sum_{\dv \in \mathcal{D}_{W_c}} \mathds{1}_{\mathcal{V}_{\mathcal{J}_{\dv,\sigma}}}(W^n_{\Tau}) \nonumber\\
=&{N-1 \choose K-1}\sum_{r=1}^{K_1}P(K_1-i-1,r-1)(K_1-r)! \nonumber \\ & \cdot (K-1)!(K_1-1)! (K_1-i).
\end{align}

Setting $x_i\dfn\sum_{n\in[N]}\sum_{\Tau\subseteq[K_1]:|\Tau|=i}|W^n_{\Tau}|$ and recalling that
\begin{equation}\label{eq:sumfiles}
N=\sum_{i=0}^{K_1}x_i=\sum_{i=0}^{K_1}\sum_{n\in[N]}\sum_{\Tau\subseteq[K_1]:|\Tau|=i}|W^n_{\Tau}|
\end{equation}
we combine \eqref{eq:alternativedefinitionofT}, \eqref{eq:longinequality} and \eqref{eq:Qi},
to get
\begin{equation}\label{eq:compacteq}
T \geq \sum_{i=0}^{K_1}\frac{Q_{i}}{P(N,K)K_1!}x_{i}.
\end{equation}

We now resume counting to calculate $\frac{Q_{i}}{\Lambda!P(N,K)}$ for each $i = 0,1,\dots,K_1$.
Applying~\eqref{eq:Qi}, we see that
\begin{align} \label{eq:finanumber}
&\frac{Q_{i}}{K_{1} !P(N,K)} =\frac{(N-1)!(N-K)!}{(K-1)!(N-K)!K_{1}!N!} \nonumber \\
&\cdot \sum_{r=1}^{K_{1}}(K-1)!(K_{1}-i)P(K_{1}-i-1,r-1)(K_{1}-r)!\nonumber\\
&=\frac{1}{K_{1}!N}\sum_{r=1}^{K_{1}}(K_{1} -i)P(K_{1} -i-1,r-1)(K_{1}-r)!\nonumber\\
&=\frac{1}{K_{1} !N}\sum_{r=1}^{K_{1}} \frac{(K_{1} -i)(K_{1} -i-1)!(K_{1} -r)!}{(K_{1} -i-r)!}\nonumber\\
&=\frac{1}{K_{1} !N}\sum_{r=1}^{K_{1}}\frac{(K_{1} -i)!(K_{1} -r)!}{(K_{1} -i-r)!}\nonumber\\
&=\frac{1}{N}\sum_{r=1}^{K_{1}}\frac{(K_{1} -i)!(K_{1} -r)!i!}{K_{1} !(K_{1} -i-r)!i!}\nonumber\\
&=\frac{1}{N}\sum_{r=1}^{K_{1}}\frac{{K_{1} -r\choose i}}{{K_1 \choose i}}=\frac{{K_{1}\choose i+1}}{{K_1 \choose i}N}=\frac{K_1-i}{(i+1)N}.
\end{align}

Now substituting \eqref{eq:finanumber} into \eqref{eq:compacteq}, we get that
\begin{align}
T(\chi)&\geq \sum_{i=0}^{K_1}\frac{K_1-i}{(i+1)N} x_{i} + \frac{K_{1} !P(N,K)}{K_{1} !P(N,K)} \underbrace{|\mathcal{A}_{\dv}|}_{K_2} \label{eq:LBwithxi_2}
\end{align}
where the use of the fraction $\frac{K_{1} !P(N,K)}{K_{1} !P(N,K)} = 1$ is meant to remind us the number of times acyclic graphs corresponding to $\mathcal{A}_{\dv}$ were invoked in the summation in~\eqref{eq:longinequality}, and where we also note that the expression above follows from the fact that all $\dv\in \mathcal{D}_{wc}$ force $|\mathcal{A}_{\dv}|  = K_2 $.

\paragraph*{Optimization\label{sec:optimization}}
At this point we observe that the crucial constant $\frac{K_1-i}{(i+1)N}$ derived for the part of the subgraph corresponding to cache-aided users, matches exactly the number $\frac{K_1-i}{(i+1)N}$ derived in \cite{wanOptimalityITW2016} for the $K=K_1$ case where all users can have a cache. Consequently, under the same file-size constraint given in \eqref{eq:sumfiles}, and given the current cache-size constraint $\sum_{i=0}^{K_1}i \cdot x_{i}\leq  K_1 M $, the expression in~\eqref{eq:LBwithxi_2} serves as a lower bound on the delay of scheme $\chi$ whose cache placement implies the set $\{x_i\}$.

Then, following the exact minimization steps in \cite[Proof of Lemma 2]{yuFactorOf2TransIT2019}, we get
\begin{equation}\label{eq:optimization1}
T(\chi)\geq \frac{K_1(1-\gamma)}{1+K_1\gamma} + K_2
\end{equation}
for integer $K_1\gamma$, whereas for all other values of $K_1\gamma$, this is extended to its convex lower envelop.

This concludes lower bounding $\max_{\dv} \in \mathcal{D}_{wc} T(\dv,\chi)$, and thus --- given that the right hand side of \eqref{eq:optimization1} is independent of $\chi$ --- lower bounds the performance for any scheme $\chi$, hence concluding the proof of the converse for Theorem~\ref{theSingleStream} for the case of $L = 1$.

config: arxiv


TITLE: About proof of fundamental lemma of calculus of variation
QUESTION [2 upvotes]: I have found several proofs of fundamental lemma of calculus of variation for $C^1$ functions, but could someone help me to show this lemma with the following hypothesis?:
Let be $g:[x,y]\rightarrow{\mathbb{R}}$ a continuous function such that $\int_x^yg(t)h(t)=0$  $\forall{h\in{C^\infty_c(x,y)}}$, show that $g=0$.
Thanks.

REPLY [1 votes]: Preparation:
Consider the function
$$
h(x)=\begin{cases} exp\left(\frac1{x^2-1}\right) & |x|<1\\0 & otherwise\end{cases}
$$
Then $h\in C_c^\infty(\mathbb R)$, $h(x)>0$ for $|x|<1$ and $supp(h)=[-1,1]$.
For $x_0\in\mathbb R$ and $\varepsilon>0$ you can consider $h_{x_0,\varepsilon}(x):=h\left(\frac{x-x_0}\varepsilon\right)$.
Then we get $h_{x_0,\varepsilon}\in C_c^\infty(\mathbb R)$, $h_{x_0,\varepsilon}(x)>0$ for $|x-x_0|<\varepsilon$ and $supp(h_{x_0,\varepsilon})=[x_0-\varepsilon,x_0+\varepsilon]$.
Proof of the lemma:
Assume $g\not\equiv 0$. Then there exists $x_0\in(x,y)$ such that $g(x_0)\neq 0$. Wlog $g(x_0)>0$. By continuity of $g$, there exists $\varepsilon>0$ such that $g(\xi)>0$ for all $\xi\in[x_0-\varepsilon,x_0+\varepsilon]\subset (x,y)$.
Now, we choose $h_{x_0,\varepsilon}\in C_c^\infty(x,y)$.
We deduce $gh_{x_0,\varepsilon}$ is continuous with $(gh_{x_0,\varepsilon})(\xi)> 0$ for $|\xi-x_0|<\varepsilon$ and $supp(gh_{x_0,\varepsilon})=[x_0-\varepsilon,x_0+\varepsilon]$.
Finally, we conclude
$$
\int_x^y gh_{x_0,\varepsilon}=\int_{x_0-\varepsilon}^{x_0+\varepsilon}\underbrace{gh_{x_0,\varepsilon}}_{\geq 0}\geq \int_{x_0-\frac12\varepsilon}^{x_0+\frac12\varepsilon}\underbrace{gh_{x_0,\varepsilon}}_{> 0}>0.
$$
But this is a contradiction. Hence $g\equiv 0$.


TITLE: $M \leq X$. Is it true that $M^*$ can be isometrically embedded in $X^*$ where $*$ denotes the dual space?
QUESTION [1 upvotes]: Let $X$ be a Banach space and $M$ be a closed subspace i.e. $M \leq X$.
Is it true that $M^*$ can be isometrically embedded as a subspace of $X^*$ where $*$ denotes the dual space?
What I have tried:
I know by Hahn Banach theorem that every $f \in M^*$ has an extension $\hat f \in X^*$ such that $\hat f |_M =f$ and the operator norm doesn't change i.e. $\|\hat f\| =\|f\|$.
I found that such an extension can be chosen to be linear iff $X$ is a Hilbert space here, hence my claim might not hold in the general case.
Any help is appreciated.

REPLY [1 votes]: I have an assumption that I have not proved but which implies your conjecture.
Notation: For  $S\subset  X$ let $[ S]$ be the closed linear subspace of $X$ generated by  S. Let $\Bbb M$ be the set of closed linear subspaces  of $M.$
Let $I$ be the set of all $(Y,f)$ such that $Y\in \Bbb M $ and such that  $f:Y^*\to X^*$ is an isometric embedding satisfying  $\forall g\in Y^*\,(\,f(g)|_Y=g).$
For $(M_1,f_1),(M_2,f_2)\in I\, $  let $(M_1,f_1)\le (M_2,f_2)$ iff $M_1\subset M_2$ and $$\forall z\in (M_2)^*\,(\,f_2(z)|_{M_1}=f_1(z|_{M_1})|_{M_1}\,).$$
Unproven assumption: If $Y\ne M$ and $(Y,f)\in I$  there exists $x\in M$ \ $Y$ and $([Y\cup \{x\}],g)\in I$ with $(Y,f)\le ([Y\cup \{x\}],g).$
Suppose $C=\{(M_j,f_j):j\in J\}\,$ is any non-empty $\le$-chain in $I.$  For $z\in [\cup_{j\in J}M_j]^*$ and $k\in J\,$   let $$f(z)|_{M_k} =f_k(z|_{M_k})|_{M_k}.$$ Then each $f(z)$ is a bounded linear function on $\cup_{j\in J}M_j$ so it extends uniquely to $g(z)\in [\cup_{j\in J}M_j]^*.\, $ And by  the Hahn-Banach Theorem $g(z)$ extends to $h(z)\in X^*$ with $\|g(z)\|=\|h(z)\|.$  Then $([\cup_{j\in J} M_j],h)$ is a $\le$-upper bound for $C.$
So every non-empty $\le$-chain has a $\le$-upper bound.
Now $\le$ is transitive and reflexive. And $I$ is not empty, for if $Y=\{0\}$ and $f(0)=0\in X^*$ then $(Y,f)\in I. $
So by Zorn's Lemma, $I$ has a $\le$-maximal member $(Y',f').$ My unproven assumption implies that $Y'=M.$


TITLE: Does the property $\int_{a}^b[ f(x) + g(x)] dx = \int_{a}^b f(x) dx + \int_{a}^b g(x) dx$ hold for improper integrals?
QUESTION [5 upvotes]: I'm trying to show the convergence or divergence of the following integral:
$$ \int_{0}^1 \frac{\sec^2(x)}{x\sqrt{x}} dx$$
I begin by writing it as:
$$ \int_{0}^1 \frac{1+\tan^2(x)}{x^{3/2}} dx$$
And this is the step I'm not entirely sure is correct:

$$ \int_{0}^1\frac{1}{x^{3/2}}dx +\int_{0}^1\frac{\tan^2(x)}{x^{3/2}}dx$$

Assuming it is correct, the integral would be:
$\lim_{t\to 0^{+}}\int_{t}^1\frac{1}{x^{3/2}}dx +\int_{t}^1\frac{\tan^2(x)}{x^{3/2}}dx$
And by just evaluating $\lim_{t\to 0^{+}}\int_{t}^1\frac{1}{x^{3/2}}dx$ the divergence could be shown.
$\lim_{t\to 0^{+}}\int_{t}^1\frac{1}{x^{3/2}}dx = \lim_{t\to 0^{+}}\Bigg[\frac{-2}{\sqrt{x}} \Bigg]_{t}^{1}$
$=\lim_{t\to 0^{+}}\frac{-2}{\sqrt{1}} - (\frac{-2}{\sqrt{t}}) $
$=\lim_{t\to 0^{+}}-2 + \frac{2}{\sqrt{t}} $
This limit is $\infty$ so the integral does not converge.
Is this correct? Can I split the integral like it is done with a proper definite integral?

REPLY [2 votes]: Since the improper integral is a limit, you can split it into a sum, provided both summands have a limit: just apply
$$
\lim_{t\to0}\int_t^1 (f(x)+g(x))\,dx=
\lim_{t\to0}\int_t^1 f(x)\,dx+
\lim_{t\to0}\int_t^1 g(x)\,dx
$$
which you can do provided both limits on the right-hand side exist, with the usual rules for $\infty$.
If one of the limits is $\infty$ and the other is either finite or $\infty$, then you can conclude for the divergence of the left-hand side. Similarly for both $-\infty$ or one $-\infty$ and the other one finite.
But if one is $\infty$ and the other one is $-\infty$ you can't conclude.


TITLE: Characterizing weak and weak$^\ast$ topologies in normed spaces
QUESTION [2 upvotes]: I have a general question about how can we characterize the weak and weak$^\ast$ topologies and continuous maps between then in normed spaces:
Let's start with a normed space $X$. The weak topology on $X$ is the smallest topology such that every $x^\ast\in X^\ast$ is a continuous linear functional. Here $X^\ast$ is the topological dual of $X$, therefore previously equipped with the operator norm topology. Similarly, the weak$^\ast$ topology is the weakest topology on $X^\ast$ such that every linear functional $J_X(x):X^\ast\rightarrow\mathbb{C}$ is continuous, where $J_X:X\rightarrow X^{\ast\ast}$ is the canonical embedding. Both these definitions are a bit unpractical when dealing with claims about continuity, so can we characterize continuity by sequential continuity, i.e., a map $T:X\rightarrow Y$ is weakly continuous (respectively, weak$^\ast$ continuous) iff $Tx_j\xrightarrow{w} Tx$ (respectively, $Tx_j\xrightarrow{w^\ast}Tx$) for every weakly convergent (respectively, weak$^\ast$ convergent) sequence $x_j\xrightarrow{w}x$? Of course this would be true for nets, but usually sequences are easier to work with.
I know generally sequences are not enough to characterize a topology, however I've seen this notion of continuity being used around and it got me wondering if that's the notion of weak and weak$^\ast$ topology that is being referred to in this context.

REPLY [1 votes]: Sequences are not enough to characterize the weak topology or weak continuity.
Indeed, recall that the normed space $\ell^1$ has this peculiar property that weak convergence of sequences is equivalent to strong convergence of sequences. However, weak topology is not equal to the topology of the norm (in infinite dimension it never is since the open unit ball is not weakly open).
Similarly, the function $\|\cdot\|_1 : \ell^1 \to \Bbb{R}$ is strongly continuous, weakly sequentially continuous but not weakly continuous. Indeed, otherwise for any weakly convergent net $x_j \xrightarrow{w} x$ we would have $$\|x_j - x\| \to 0 \implies x_j \xrightarrow{s} x$$
which would imply that the topologies are equal, but they aren't.


TITLE: Concerning the countability of the set of primitive recursive functions
QUESTION [2 upvotes]: If we generally would define the smallest set, that has to have some properties, as the set obtained by intersecting all the sets that have those properties (if the intersection is non-empty) and the tiniest set, that has to have some properties, as the set with the least cardinality, that has to have those properties (if there is only one set with the least cardinality), would then the set $F$ of primitive recursive function be also the tiniest set (besides being the smallest one) $F \subseteq \cup_{k\in \mathbb{N}} \left\{f:\mathbb{N}^k \rightarrow  \mathbb{N} \right\}$, that contains the base functions and is closed under composition and primitive recursion ?
(Or if I may rephrase the question: Is $F$ the only countable set, that contains the base functions and is closed under composition and primitive recursion ?)
Side question: I know $F$ is countable set, since to every function $f \in F$ there corresponds a very basic program containing only bounded loops and such; and a program is just a finite (but arbitrary long)  string over a finite (and fixed) alphabet; thus $F$ can be seen as a reunion of all finite strings (that make synthactically sense; i.e. are programs) over a finite fixed alphabet - and therefore has to be countable.
But how can I prove that $F$ is countable without falling back to interpret a function $f\in F$ as a program (just by staying "inside mathematics") ?

REPLY [5 votes]: You can prove $F$ is countable without appealing to its characterization as the set of all programs by the following argument.  $F$ is constructed as follows:  Let $F_0$ be the collection of base functions which is easily seen to be countable.  Let $F_{n+1}$ be $F_n$ together with every composition of finitely many elements of $F_n$ and all applications of primitive recursion to elements of $F_n$.  It's easy to prove by induction that each $F_n$ is countable, and hence the union of the $F_n$ is countable.  It's clear that this union satisfies the desired closure properties, hence this union is in fact $F$.
This can be generalized: Closing an infinite set $X$ under $\leq |X|$ many finitary operations results in a set of the same size $|X|$.  This can be generalized further still, but I'll leave at this for now.


TITLE: how to prove $\mathbb Z_n$ has divisors of zero if and only if $n$ is not prime
QUESTION [4 upvotes]: A non-zero number $a \in \mathbb Z_n$ is called a divisor of zero if there is a non-zero number $b \in \mathbb Z_n$ such that $ab\equiv 0\pmod n.$
How can I prove $\mathbb Z_n$ has divisors of zero if and only if $n$ is not prime.
I filled out the addition and multiplication tables for modulo 6 and 7 and tried to find out the relation, and it's definitely true. I know I need to prove it in both directions since it is an 'iff' question. But I still don't get it.

REPLY [0 votes]: If $n$ is not prime, then, we can factorize in the form $ab=n$, then, $ab\equiv n\equiv 0 (\text{mod} \ n)$. The another implication follows from this reasoning (and the definition of a congruence).


TITLE: Limit aproaching infinity $f(x) = -\infty$ ?!?
QUESTION [0 upvotes]: On one of my homework problems it's asking for me to sketch a graph of a function which fulfills five requirements. All of the requirements make sense to me other than the following:

$$\lim_{x\rightarrow \ \infty}  f(x) = {-\infty}$$

How could there be a limit approaching infinity at negative infinity? Doesn't one of the number need to be finite for me to know where to place the infinity limit? This makes no sense to me and is really throwing me off.

$$f'(x) \ge 2 \ \text{when} -2 \lt x \lt 0 \ \text{and} \ \text{when} \ 2 \lt x \lt 4 $$

I don't even know how to approach this one. How do you apply information about the derivative/tangent line to the graph of the original equation?
Other requirements (purely for reference, I understand these):

$$\lim_{x\rightarrow \ 0+}  f(x) = -2$$
  f(x) is an even function
  f(x) is differentiable everywhere but two x-values, and is continuous for all x

Thank you.

REPLY [2 votes]: Saying that $$\lim\limits_{x\to\infty}f(x)=-\infty$$ just means that for each real number $M$ there is an $x_M$ such that $f(x)<M$ whenever $x\ge x_M$. Taking $M=-100$, for instance, there is an $x_{-100}$ such that $f(x)<-100$ whenever $x\ge x_{-100}$. A very simple example of such a function is the function $f(x)=-x$: for this function you can take $x_{-100}$ to be $101$ for instance, since $f(x)=-x<-100$ whenever $x\ge 101$.
For the other requirement, remember that $f\,'(a)$ is the slope of the tangent line to the curve $y=f(x)$ at $x=a$. The requirement that $f\,'(x)\ge 2$ when $2<x<4$ just means that the graph has a slope of at least $2$ over the entire interval $(2,4)$. Many functions have this property: $f(x)=ax$ with $a\ge 2$ (since $f\,'(x)=a$), $f(x)=x^2$ (since $f\,'(x)=2x$, which is certainly greater than $2$ when $2<x<4$, and many, many more.
Remember, you don’t have to write down a single expression for $f$: you can define it piecewise, with different definitions on different intervals.


TITLE: Number of spanning trees by dividing graph into subgraphs
QUESTION [3 upvotes]: I'm trying to figure out the number of spanning trees that are in the graph depicted below,

I am currently working on the idea of splitting up the graphs into subgraphs, because then I'm able to apply certain formulas to calculate the number of spanning trees.
My problem is, though, how do I add them together in the end?
For the beforementioned graph, I've divided them into 1 complete bipartite graph, 1 complete graph and 2 simple graphs.

The complete graph can be calculated using Cayley's formula,
$\tau(k_n) = n^{n-2}$ giving me $\tau(k_4) = 4^{4-2} = 16$ spanning trees
The complete bipartite graph can be calculated using Scoin's formula,
$\tau(k_{n,m}) = n^{m-1} m^{n-1}$ giving me $\tau(k_{3,3}) = 3^{3-1} 3^{3-1} = 81$ spanning trees
The two simple graphs each have 8 spanning trees (found from lookup)

I've found nowhere though how this can be added together (and even if it can).
So my question is: how, if at all possible, do you sum up the number of spanning trees from several subgraphs?
First time on math.stackexchange, so any feedback is appreaciated.

REPLY [2 votes]: On each subgraph, a spanning tree must visit each vertex, so a choice of a spanning tree on each is equivalent to a choice of a spanning tree on the whole graph.
How many choices are there?  Just multiply.  The general formula (assuming the subgraphs meet their neighbors in a single vertex without introducing any new cycles) looks like:
$$
\tau(G_1 \cup \cdots \cup G_k) = \tau(G_1) \cdots \tau(G_k).
$$
Can you finish from there?


TITLE: Is this contradiction proof correct?
QUESTION [1 upvotes]: Statement : suppose $a,b$ belongs to $\mathbb{Z}$ (integers). If $4/ (a^2 + b^2)$ then $a$ and $b$ are not both odd.
Proof by contradiction:
Assume that if $4/(a^2 + b^2)$ then a and b are both odd.
If $a^2$ and $b^2$ is odd then by definition a and b must be odd too.
It follows that $a^2$ (or $b^2$) $= (4k+1)^2$  -----  (unsure about this step)
Then a (or b) $= 4k+1$  ---- (unsure about this step)
So if $a$ and $b$ are both odd then this is a contradiction hence the supposition is false and the statement is true.

REPLY [1 votes]: The proof by contradiction should start with

assume that $a$ and $b$ are both odd and $4\mid (a^2+b^2)$

You can continue by saying

write $a=2h+1$ and $b=2k+1$

and

so that $a^2+b^2=4h^2+4h+1+4k^2+4k+1=\dots$

Now you can add

and we have a contradiction, because this would imply $\dots$


TITLE: Can the product of an increasing monotonic function and a strictly increasing monotonic function have turning points
QUESTION [0 upvotes]: Suppose $f:\Re \rightarrow [0,1]$ is an increasing monotonic function and $g:\Re \rightarrow [0,1]$ is a strictly increasing monotonic function.
For the product function $h(x) = f(x)g(x)$, will there exists a local maximum? If the gradient of $f$ and $g$ are different, I think there exists a turning point for $h$?
I am thinking if it is possible that $h$ can become a concave function, with a local maximum?
Previously, I misunderstood turning point to be a local maximum.

REPLY [2 votes]: I guess turning point (German: Wendepunkt?) is point of inflection and you mean $h(x)=f(x)g(x)$. Then the answer is yes: Take
$$f(x) \equiv 1, \quad g(x) = \frac{1}{2} + \frac{\arctan x}{\pi}$$
then both functions are $\mathbb{R} \rightarrow [0,1]$, $f$ is increasing monotonic,
$g$ is strictly increasing monotonic, $h''(0) = 0$ and $h'''(0) \ne 0$.


TITLE: Does the inverse-cube law for dipoles collapse into the inverse-square law for a 2-dimensional system?
QUESTION [0 upvotes]: Such as a piece of sheet-metal, or the mammalian cortex?

REPLY [1 votes]: That's not really how two-dimensional electrostatic systems work - even if you restrict charges to a surface, they still 'emit' electric field flux outside it. Instead, two-dimensional electrostatics actually describes three-dimensional systems with translational symmetry along one axis, so e.g. a point charge in 2D electrostatics corresponds in 3D to an infinitely long line charge; you can check easily that they have the same $\propto 1/r$ electric field.
And yes, within that formalism, electric dipoles (now represented by to infinitely strong line charges, at infinitely small separation) do have a $\propto 1/r^2$ dependence.


TITLE: Create 2 set from 1 to 10
QUESTION [0 upvotes]: In school I learnt to create a set of number or how many sets of number we can create from given numbers. But now, I forgot. Please help me here.
I want to know how many sets can be possible to create and I want create 2 number in 1 set from 1 to 10
Example :
Number are 1 to 10
Sets should be like (1,1), (1,2), (1,3), ....................(5,1), (5,2), ....................(10,8), (10,9), (10,10).
Thank you for your help.
[EDITED]
As I remember it can be possible like aª. And a=10 in my case. So the answer will be aª=10×10=100.

REPLY [0 votes]: What you want to know is that if we are given two lists of numbers $$1, \dots, m$$ and $$1, \dots, n$$
Then how many ways are there to pick two numbers $(a, b)$ where $a$ is in the first list and $b$ is in the second list.
Note that if we fix the number $1$ from the first list. Then we can pair that number up with $n$ different numbers. More specifically, we would have the pairs $$(1, 1), \dots, (1, n)$$
This tells us that if fix a number $k$ from the first list, then we get $n$ different pairs $$(k, 1),\dots, (k, n)$$
Because there are $m$ numbers in the first list. And each number will give us $n$ pairs. The total number of pairs is $$\underbrace{n+\dots+n}_\text{$m$ times}=m\times n$$
In your specific example, the two lists are $$1, 2, 3, \dots, 10$$ and $$1, 2, 3, \dots, 10$$
Thus, the total number of ways to pick a pair $(a, b)$ where $a$ comes from the first list and $b$ comes from the second list is $10\times 10=100$.
This result holds in general. This is known as the principle of multiplication or the rule of product. Thus, if you have $n$ lists $$1, \dots, m_1\\ \vdots \\ 1,\dots ,m_n$$
Then the total number of ways to pick $n$ numbers where each number comes from each list is $$m_1\times \cdots \times m_n$$
Hope this helps!


TITLE: Does matter follow the expansion of space?
QUESTION [4 upvotes]: Non-physics person here. I am trying to understand the expansion of space. I've read about the redshift of light and the balloon model (2 points on expanding balloon move away from each other).
The detection of the expanding universe depends on light. But what about matter? Does matter follow the expansion of space? Or does it "slide" so that space expands and whizzes by the matter, but the matter remains in the same distance and configuration relative to other matter?
I'm thinking of a very smooth disc of ice and hockey pucks on top. Suppose the disc somehow expands. Due to it being very smooth, the expansion doesn't move the pucks. They remain in the same relative positions but the disc underneath it expands and points on the ice whiz by the pucks.
Can someone explain how matter fits into space expansion and its detection?

REPLY [0 votes]: To start with, there was the original Big bang model ,which postulated an expansion of the universe in the sense of an explosion. As in an explosion the fragments follow their trajectories, it was posited that the particles, after being created, followed their trajectories,  and then the forces between particles, much stronger than the impulse of separation, took over, creating clusters of matter which evolved into the observable universe.
The four forces of nature are strong enough to attract and keep matter together, even when an accelerating expansion of the universe was detected.

The currentBig Bang.model with accelerated expansion,seen in the changing opening angle.

This accelerated  expansion of the universe also does not expand bound states, as atoms, molecules , cups and tables, planetary systems, galaxies or even cluster of galaxies,bound systems.
The expansion of space does not take place on bound systems , because ,as a force, accelerated expansion is very very very weak. Even weaker than gravity.
After all, we discovered  expansion and accelerated expansion because clusters of galaxies and galaxies  were moving away from each other  . If all matter were expanding uniformly, we would never know about an expansion.


TITLE: Why absolute-value function is not a continuous function
QUESTION [1 upvotes]: I am doing a pre-calculus course from:  https://www.edx.org/course/pre-university-calculus
It tells me that a continuous function has continuous graphs while a piece-wise function has kind of step-based graph. Then it tells that absolute value function |x| is a piece-wise function. I noticed the |x| has a continuous graph without any breaks.  How come it  is defined piece-wise then ?
Wikipedia says it is both continuous and piece-wise linear function. I am confused:
https://en.wikipedia.org/wiki/Absolute_value#Absolute_value_function

REPLY [1 votes]: Imagine you have a piecewise function
$$f(x)=\left\{\begin{matrix}g(x) & x\leq a \\ h(x) & x>a\end{matrix}\right.$$
$f(x)$ is continuous if $g(x)$ is continuous when $x\leq a$, $h(x)$ is continuous when $x>a$, and $g(a)=h(a)$ (which is indeed the case for the absolute value function).


TITLE: Characteristic functions of a non-ohmic material
QUESTION [1 upvotes]: We know that Ohm's laws are experimental laws. Ohm's first law for ohmic materials follows a linear approximate progression of the kind:
$$\frac{\Delta V}{I}=R$$

What are the functions of type $$\Delta V(I)=f(I)$$ that are characteristics of non-ohmic materials?
Are there any known functions for non-ohmic materials and how are they written?

In other words, for tungsten is there an analytical expression that you can study or you need to make an experiment in the laboratory and to plot the characteristic curve that best approximates the experimental points: polynomial, exponential, logarithmic $\ldots$?
For example I add this image:

where the word "pendenza=slope". It is easy that I have a function $\Delta V(I)=RI$, or straight lines that across the origin of the axes. In good approximation, ohmic materials are straight lines across the origin of the axes.
If I have this picture

semiconductor device where the curve is not linear, how do I find the equation associated with it?

REPLY [1 votes]: Typically, in a non-ohmic system, the current-voltage relationship doesn't just depend on the material, but also on the geometry.
For example, an incandescent light filament is non-ohmic because as it heats itself up, its resistance increases, limiting the current. But the change in temperature as a function of current depends not only on the material of the filament, but also its diameter, its length, the thermal masses of the objects it's mounted to on each end, the density of the gas around it, etc.
As another example, the PN-junction semiconductor diode is non-ohmic because it has (to simplify things greatly) very high conductance in one direction but almost no conductance in the other direction. But the details of the I-V curve (for example, how much leakage current is seen in the reverse direction, or exactly how much forward voltage is required to reach the region of high conductance) depend on the doping concentrations of the two sides (and how those concentrations vary with position near the junction), the cross-sectional area of the junction, the operating temperature, etc.


TITLE: Writing this specific sequence in another form
QUESTION [0 upvotes]: How would one write this sequence: 0, 1, 3, 12, 44, 165, 615 ...
https://oeis.org/A109437/internal
Without using a recursive formula, and generating a formula that does not  refer back to the previous term to work out the next term.
Kind regards
Joshua

REPLY [0 votes]: Are you familiar with the theory of linear homogeneous recurrences? They can be solved exactly:
$$a(n) = \frac{(3+\sqrt{3}) (-1)^{n+1}+(3+2 \sqrt{3}) (2+\sqrt{3})^n -\sqrt{3} (2-\sqrt{3})^n}{6 (3+\sqrt{3})}$$
If you would like to learn how to solve any linear homogeneous recurrence relation, see this.


TITLE: Is there any continuous function having local minima at c but is neither decreasing in (c-h,c) nor increasing in (c,c+h)?
QUESTION [2 upvotes]: I can find a discontinuous function having the above conditions.....
f(x) =x for x lying betwen (0,1) and (1,2)
and
-10 at x=1
But can anyone help me in finding a continuous function satisfying the conditions....

REPLY [3 votes]: I think you can consider the function
$$y=\bigg\{ \begin{array}{ccc} x^2 \sin^2\frac{1}{x}& &x\ne0
\\0 && x=0\end{array}$$
It is continuous and $(0,0)$ is a local minimum, but does not satisfy your conditions.


TITLE: Is $\operatorname{Hom} _A(M,N)$ a finitely generated $A$-module?
QUESTION [0 upvotes]: If $M,N$ are finitely generated $A$-modules, is $\operatorname{Hom} _A(M,N)$ a finitely generated $A$-module?

REPLY [3 votes]: If $N$ is Noetherian, then there is an injective embedding $\operatorname{Hom}_A(M,N) \rightarrow N^G$ for some finite $G \subset M$, and $N^G$ is Noetherian, so $\operatorname{Hom}_A(M,N)$ is Noetherian.
On the other hand, weird things can happen in the non-Noetherian case.
Consider $N=A$, and define, for all ideals $I \subset A$, $I^{\perp}=\{x \in A,\, \forall y \in I,\, xy=0\}$.
Now, there are isomorphisms $\operatorname{Hom}_A(A/I,A)=\{f(x)=ax,\,a \in A, aI=0\}=I^{\perp}$.
If for some $I \subset A$, $I^{\perp}$ is not finitely generated, $M=A/I$, $N=A$ is a counterexample.
Now consider the ring $A=\mathbb{Z}[(X_i)_i]/\langle (X_i^2),(X_iX_jX_k)_{i,j,k}\rangle$, and $I=\langle x_1 \rangle$, $I^{\perp}=Ax_1 \oplus \bigoplus_{j > i > 1}(Ax_ix_j)$ is not finitely generated.


TITLE: Understanding Spin, (from Griffiths), a Paradox?
QUESTION [3 upvotes]: In Griffiths', Introduction to Quantum Mechanics, 3rd ed, Sec. 4.4.1 Spin 1/2 the author represents the spin state by a spinor
$$
\chi = \binom{a}{b} = a\chi_+ + b\chi_-, \qquad \chi_+ = \binom{1}{0},\quad \chi_- = \binom{0}{1},\quad |a|^2 + |b|^2 = 1
$$
where the latter two represent the spin up and the spin down along the $z$-axis, i.e., they are the eigenspinors of the $S_z$ for the eigenvalues $+\hbar/2$ and $-\hbar/2$, respectively. After he constructs the Pauli spin matrices he finds the eigenspinors of the $S_x$ to be
$$
\chi_+^{(x)} = \binom{1/\sqrt{2}}{1/\sqrt{2}},\quad \chi_-^{(x)} = \binom{1/\sqrt{2}}{-1/\sqrt{2}}
$$
for the eigenvalues $+\hbar/2$ and $-\hbar/2$, respectively. And therewith
$$
\chi = \frac{a+b}{\sqrt{2}}\ \chi_+^{(x)} + \frac{a-b}{\sqrt{2}}\ \chi_-^{(x)} \tag{*}
$$
Question: If I choose $a = \cos \theta, b = \sin \theta$ then $|a|^2 + |b|^2 = 1$ is fullfiled. But in the case $\theta = \frac{\pi}{4}$ (which means equal probabilities for $\chi_+$ and $\chi_-$) we get from ($*$) that $\chi = \chi_+^{(x)}$ which is a determinate (certain) state of $S_x$, and it means spin up along the $x$-axis, i.e., in the positive direction. But how does it know which is the positive direction ? why isn't it the other direction, since we could choose the opposite direction along $x$ as the positive ?
Stated another way: If the particle is in a state such that it is equally likely to get by measurement $+\hbar/2$ or $-\hbar/2$ along some axis ($z$-axis, say), How does measuring the spin at this state along some perpendicular line gives  the positive direction along that line, though one may have not yet chosen which direction is the positive and which is the negative ?

REPLY [3 votes]: I'm summing up, What @Cosmas Zachos said,
The first direction in space is unique but not it's representation. The one I'm calling positive $x$-axis, you can call $-x$-axis or vice-versa. We know that Spinors are generators of SU(2) group which correspond to rotation in $3D$ (naively) without changing right hand to left hand system.
For instance:
In our case :
$$\chi=\frac{a+b}{\sqrt{2}}\chi^x_++\frac{a-b}{\sqrt{2}}\chi^x_-$$
Now apply a rotation of $\pi$ about $z$-axis so that
$$R(\pi\hat{z})=\begin{pmatrix}
\cos(\pi) & \sin(\pi)\\
-\sin(\pi) & \cos(\pi)
\end{pmatrix}=\begin{pmatrix}
-1 & 0  \\
0 & -1
\end{pmatrix}$$
$$R(\pi\hat{z})\chi=-\frac{a+b}{\sqrt{2}}\chi^x_+-\frac{a-b}{\sqrt{2}}\chi^x_-$$
If you work out the whole algebra for the case $a=b$ lead you to
$$\chi'=-\chi^x_-$$
That's the negative direction which is equivalent to other.

REPLY [2 votes]: though one may have not yet chosen which direction is the positive and which is the negative ?

You did choose the direction when you chose the spinor basis. You have to prepare a system in the state $\binom{1/\sqrt{2}}{1/\sqrt{2}}$ before you do your measurement, and whatever method you use to do that will make use somehow of the fact that that state points in the +x direction in your basis. There's no way to create that state through measurements along the z axis, or a magnetic field directed along the z axis, or anything else that doesn't break the x/y symmetry.
The complex-vector representation of spin obscures the geometry. It's somewhat clearer to imagine a Pauli spinor as a quaternion representing a rotation from a canonical spinor orientation to the actual orientation. If you take the canonical spinor to be pointing in the +z direction, and $\binom10$ to represent the quaternion $1$, then $\binom01$ has to be a unit quaternion perpendicular to $1$, i.e. a pure imaginary quaternion. If you pick $j$ – i.e., a 180° rotation in the xz plane – then $\binom{1/\sqrt{2}}{1/\sqrt{2}}$ is $(1{+}j)/\sqrt2$, or a 90° rotation in the xz plane, and this leaves the spinor pointing in the +x direction (or -x; I wish there weren't so many sign conventions). The quaternionic representation makes it clearer that $\binom10$ and $\binom01$ aren't (can't be) just "+z" and "-z"; they have another spatial direction hidden in them.


TITLE: Let the column vectors of a $3 × 3$ matrix $A$ form an orthonormal basis. Explain why $A^T = A^{−1}$ .
QUESTION [3 upvotes]: Let the column vectors of a $3 × 3$ matrix $A$ form an orthonormal basis. Explain why $A^T = A^{−1}$
.
My Attempt:  $AA^T=I$ if and only if $A^TA=I$. So $A$ is orthogonal if and only if $A^T$ is orthogonal.
Is this a correct explanation/ can it be put in a more better way?

REPLY [1 votes]: A more detailed version of $AA^T = I \iff A^T A = I$:
\begin{align}
A A^T &= I \iff \\
A^{-1} (A A^T) &= A^{-1} I \iff \\
(A^{-1} A) A^T &= A^{-1} \iff \\
A^T &= A^{-1} \iff \\
A^T A &= A^{-1} A = I
\end{align}
What is missing in your agrument is a reason, why $A A^T = I$ holds here.


TITLE: Cotangent Fields: Exactness vs. Conservation
QUESTION [5 upvotes]: Given a smooth manifold.
Then a cotangent field is exact iff conservative:
$$\alpha\in\mathcal{X}^*(M):\quad\alpha=\mathrm{d}h\iff\oint\alpha=0$$
How to prove this properly?

REPLY [0 votes]: This proof is taken from Lee's Smooth Manifolds.
Consider up to a constant:
$$h(x)=\int_z^x\alpha$$
Regard a smooth coordinate path:
$$\gamma:[0,\varepsilon)\to M:\quad\hat{\gamma}(t):=(0,\ldots,0,t,0,\ldots,0)\quad(\gamma(0)=z)$$
So one obtains as partial derivatives:
$$\frac{\partial h}{\partial x^i}(z)=\gamma'(0)h=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}(h\circ\gamma)=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\int_z^{\gamma(t)}\alpha\\=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0}\int_0^t\alpha(\gamma'(s))\mathrm{d}s=\alpha(\gamma'(0))=\alpha^i(z)$$
(This especially shows smoothness.)


text:
\section{Preliminaries}
\subsection{Assumptions}
We conduct our analysis on \fedac in two settings with two sets of assumptions.
The following \cref{asm1} consists of a set of standard assumptions: convexity, smoothness and bounded variance.
Comparable assumptions are assumed in existing studies on \fedavg \citep{Haddadpour.Kamani.ea-NeurIPS19,Stich.Karimireddy-arXiv19,Khaled.Mishchenko.ea-AISTATS20,Woodworth.Patel.ea-ICML20}.\footnote{In fact, \citet{Woodworth.Patel.ea-ICML20} imposes the same assumption in \cref{asm1}; \citet{Khaled.Mishchenko.ea-AISTATS20} assumes $f(w; \xi)$ are convex and smooth for all $\xi$, which is more restricted; \citet{Stich.Karimireddy-arXiv19} assumes quasi-convexity instead of convexity; \citet{Haddadpour.Kamani.ea-NeurIPS19} assumes P-\L\ condition instead of strong convexity. In this work we focus on standard (general or strong) convexity to simplify the analysis.}
\begin{assumption}[$\mu$-strong convexity, $L$-smoothness and $\sigma^2$-uniformly bounded gradient variance] \label{asm1}
    \;
    \begin{enumerate}
        \item[(a)] $F$ is $\mu$-strongly convex, i.e., $F(u) \geq F(w) + \langle \nabla F(w), u - w  \rangle + \frac{1}{2} \mu \|u - w\|^2$ for any $u, w \in \reals^d$. In addition, assume $F$ attains a finite optimum $w^* \in \reals^d$.
        (We will study both the strongly convex case $(\mu > 0)$ and the general convex case $(\mu = 0)$, which will be clarified in the context.)
        \item [(b)]  $F$ is $L$-smooth, i.e., $F(u) \leq F(w) + \langle \nabla F(w), u - w  \rangle + \frac{1}{2} L \|u-w\|^2$ for any $u, w \in \reals^d$.
        \item [(c)]  $\nabla f(w; \xi)$ has $\sigma^2$-bounded variance, i.e., $\sup_{w \in \reals^d}  \expt_{\xi \in \mathcal{D}} \| \nabla f(w; \xi) - \nabla F(w)\|^2 \leq \sigma^2$.
    \end{enumerate}
\end{assumption}
The following \cref{asm2} consists of an additional set of assumptions: \nth{3} order smoothness and bounded $\nth{4}$ central moment.
A similar version
of \cref{asm2}
was studied in \citep{Dieuleveut.Patel-NeurIPS19}.\footnote{In fact, \citep{Dieuleveut.Patel-NeurIPS19} assumes bounded \nth{4} central moment at optimum only. We adopt the uniformly bounded \nth{4} central moment for consistency with \cref{asm1}.}
\begin{assumption}\label{asm2}     In addition to \cref{asm1}, assume that
    \begin{enumerate}
        \item[(a)] $F$ is $Q$-\nth{3}-order-smooth, i.e., $F(u) \leq F(w) + \langle \nabla F(w), u-w \rangle +  \frac{1}{2} \langle \nabla^2 F(w) (u-w), (u-w) \rangle + \frac{1}{6} Q \|u-w\|^3$ for any $u, w \in \reals^d$.
        \item[(b)] $\nabla f(w; \xi)$ has $\sigma^4$-bounded \nth{4} central moment, i.e, $\sup_{w \in \reals^d} \expt_{\xi \in \mathcal{D}} \| \nabla f(w; \xi) - \nabla F(w)\|^4 \leq \sigma^4$.
    \end{enumerate}
\end{assumption}

\subsection{Notations}
We use $\|\cdot\|$  to denote the operator norm of a matrix or the $\ell_2$-norm of a vector, $[n]$ to denote the set $\{1, 2, \ldots, n\}$.
Let $w^{*}$ be the optimum of $F$ and denote $F^*:=F(w^*)$. Let $D_0 := \|w_0 - w^*\|$.
be the Euclidean distance of the initial guess $w_0$ and the optimum $w^*$.
For both \fedac and \fedavg, we use $M$ to denote the number of parallel workers, $R$ to denote synchronization rounds, $K$ to denote the synchronization interval (i.e., the number of local steps per synchronization round), and $T = KR$ to denote the parallel runtime.
We use the subscript to denote timestep, italicized superscript to denote the index of worker and unitalicized superscript ``md'' or ``ag'' to denote modifier of iterates in \fedac (see definition in \cref{alg:fedac}).
We use overline to denote averaging over all workers, \eg, $\overline{w_t^{\mathrm{ag}}} := \frac{1}{M} \sum_{m=1}^M w_t^{\mathrm{ag}, m}$. We use $\tildeo, \tilde{\Theta}$ to hide multiplicative $\polylog$ factors, which will be clarified in the formal context.


TITLE: Video lectures of mathematics courses available online for free
QUESTION [408 upvotes]: It can be difficult to learn mathematics on your own from textbooks, and I often wish universities videotaped their mathematics courses and distributed them for free online. Fortunately, some universities do that (albeit to a very limited extent), and I hope we can compile here a list of all the mathematics courses one can view in their entirety online.
Please only post videos of entire courses; that is, a speaker giving one lecture introducing a subject to the audience should be off-limits, but a sequence of, say, 30 hour-long videos, each of which is a lecture delivered in a class would be very much on-topic.

REPLY [2 votes]: I recently found the YouTube channel of the university of Uppsala (Uppsala Algebra : https://www.youtube.com/channel/UCPWnhR29VHTAk7rZUEDQdDQ/playlists)
It contains mostly courses by Walter Mazorchuk on representation theory of finite groups, Lie algebras and the category O (and on linear algebra which if of less interest for me) and a course on commutative algebra and algebraic geometry by Seidon Alsaody.
I mostly watched the courses by W. Mazorchuk and they are very good.


TITLE: Mathematical idea behind "Cut a hole in a piece of paper large enough you can walk through"?
QUESTION [1 upvotes]: "Cut a hole in a piece of paper large enough you can walk through?". I have seen the solution.
What is mathematical idea behind the solution?I am looking for a mathematical explanation.Is it related to topology?

REPLY [0 votes]: The area of the strip you cut is limited to whatever the piece of paper you started with had (and can actually neither increase nor decrease, assuming an ideal pair of scissors, and that you dont' cut anything away). That means that the narrower you cut it, the longer it necessarily has to be (half as thick means twice as long).


TITLE: If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$
QUESTION [3 upvotes]: If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex.
Can anyone help me with a simpler proof?

REPLY [3 votes]: Notice that
$$  \left( \sqrt[3]{a+b}+\sqrt[3]{a-b} \right)^3= 2a + 3\cdot \sqrt[3]{a^2-b^2}\left(\sqrt[3]{a-b}+\sqrt[3]{a+b} \right)$$
Now with your question
Consider
$$x= \sqrt[3]{a+\sqrt {a^2+b^3}}  +\sqrt[3]{a-\sqrt {a^2+b^3}} $$
Let $\sqrt{a^2+b^3} = c$
$$ x= \sqrt[3]{a+c}  +\sqrt[3]{a-c} $$
$$ \Leftrightarrow x^3 = 2a + 3 \cdot \sqrt[3]{a^2-c^2}\left(\sqrt[3]{a-c}+\sqrt[3]{a+c} \right)  $$
$$ x^3 = 2a + 3\cdot \sqrt[3]{a^2-c^2}(x)$$
I think you can take it from here.


TITLE: Parametric curve of a spiral with straight and semicircular parts
QUESTION [2 upvotes]: I need to design a fully parameterized spiral like this

where I can choose/change the central part length and the radius of the innermost curve.
I understand that it will be somehow related to the Archimedean spiral, whose equations are $x=r\cos(\theta)$ and $y=r\sin(\theta)$, but I can't pull this out.
Can anyone help me with this? Thanks in advance!

REPLY [1 votes]: Our aim is to find two curves $\mathbf l(t, k)$ and $\mathbf s(t, k)$ that depend on $t$ and an integer $k \geq 1$ such that $\mathbf l(t, k)$ gives the $k$th line segment in the curve and $\mathbf s(t, k)$ gives the $k$th semicircle in the curve. Henceforth, I will assume without loss of generality that this curve is symmetric about the $x$-axis.
Observe that we can parametrize each of the semicircles by $\mathbf s(t, k) = \langle C(k) + R(k) (-1)^{k + 1} \sin t, R(k) \cos t \rangle$ for $0 \leq t \leq \pi,$ where $C(k)$ is the $x$-coordinate of the center of the $k$th semicircle, and $R(k)$ is the radius of the $k$th semicircle. Each of these curves unravels clockwise from top to bottom for $k$ odd or bottom to top for $k$ even, as desired. Considering the diagram, we have that $\{C(k)\}_{k \geq 1} = \{5, -5, 5, -5, \dots \}$ so that $C(k) = (-1)^{k + 1} 5.$ Likewise, we have that $\{R(k)\}_{k \geq 1} = \{\frac 2 2, \frac 3 2, \frac 4 2, \frac 5 2, \dots\}$ so that $R(k) = \frac{k + 1}{2}.$ We conclude therefore that the parametrization of the $k$th semicircle of the curve is given by $$\mathbf s(t, k) = \biggl \langle (-1)^{k + 1} \biggl( 5 + \frac{k + 1}{2} \sin t \biggr), \frac{k + 1}{2} \cos t \biggr \rangle \text{ for } 0 \leq t \leq \pi.$$
Observe that the sequence of $y$-coordinates for the line segments is $\{1, -1, 2, -2, 3, -3, \dots\}.$ Consequently, the $y$-coordinate for the line segments is controlled by the equation $y(k) = (-1)^{k + 1} \lfloor \frac{k + 1}{2} \rfloor,$ where $\lfloor \cdot \rfloor$ denotes the usual floor function. Each of the line segments has length 10 (except for the first one) and has endpoints $(-5, y(k))$ and $(5, y(k)),$ so we have that $x(k) = (-1)^{k + 1} \bigl(\frac{10}{\pi} t - 5 \bigr).$ (We need to divide by $\pi$ because we assume in the parametrization of the semicircles that $0 \leq t \leq \pi.$) Consequently, we obtain $$\mathbf l(t, k) = \begin{cases} \bigl \langle (-1)^{k + 1} \bigl(\frac{10}{\pi} t - 5 \bigr), (-1)^{k + 1} \bigl \lfloor \frac{k + 1}{2} \bigr \rfloor \bigr \rangle \text{ for } \frac{\pi}{2} \leq t \leq \pi & \text{if } k = 1 \text{ and} \\ \\ \bigl \langle (-1)^{k + 1} \bigl(\frac{10}{\pi} t - 5 \bigr), (-1)^{k + 1} \bigl \lfloor \frac{k + 1}{2} \bigr \rfloor \bigr \rangle \text{ for } 0 \leq t \leq \pi & \text{if } k \geq 2. \end{cases}$$


TITLE: Using differentials to approximate the change in f(x,y)
QUESTION [0 upvotes]: I have been
"Use differentials to approximate the change in $f(x, y)$  as $(x, y)$ changes
from $(−2, 3)$ to $(−2.02, 3.01)$."
$$f(x,y)=x^2-3x^3y^2+4x-2y^3+6$$
I have obtained:
$$\partial_xf(x,y)=2x-9x^2y^2+4,\qquad \partial_y f(x,y)=-6x^3-6y^2$$
Where do I go from here?

REPLY [0 votes]: The total differential is the sum of the partials.  So, $df = \partial_x f + \partial_y f$.  Since $\partial_x f$ literally means the amount of changes in $f$ due to changes in $x$, then stick in the values for $x$, $y$, and $dx$ (i.e., the amount of change in $x$ from your starting point to your next point) into the formula for $\partial_x f$.  Then, do the same thing for $\partial_y f$.  Then add then together.


TITLE: Simple problem on congruence.
QUESTION [0 upvotes]: On a highway around the outskirts of a city are a bridge, a tunnel and a dam in clockwise order, and only clockwise traffic is allowed. There is a toll charge of 5 to drive over the bridge, 13 to drive through the tunnel and 8 to drive on the dam. Miss Dos lives and works along the highway, driving to work in the morning and driving home in the evening.
One day at the office, she calculates that her total toll charge since she started working is 2101.
Counting in clockwise order, where are her home and office, between the bridge and the tunnel, between the tunnel and the dam, or between the dam and the bridge?

-->---B,5--->--T,13--->---D,8--->---
As calculation is done at office, so there is an extra trip to office. Let there be $n$ days to office, hence $(n+1)*x + n*y= 2101$.
An easy way can be to subtract $x$ from $2101$, and find out by substituting all values of $x$, which can be the fit value.
$x \in \{8, 21, 26,0\}$, with the corresponding value of $y = \{18, 5, 0,26\}$.
1. $x=8\implies (x+y=)26 \nmid 2093$,
2. $x=21\implies 26 \mid 2080$,
3. $x=26\implies 26 \nmid 2075$,
4. $x=0\implies 26 \nmid 2101$
So, home is between Bridge & Tunnel, while office is between Dam and Bridge.
Want to know an alternate way, more compact and may be advanced too.

REPLY [1 votes]: A more compact way is to observe that the remainder of $2101$ on division by $26$ is $21,$ so she paid $21$ to get to the office today.  This is often written $2101 \equiv 21 \pmod{26}.$


TITLE: finding the area of a trapezoid using only 2 of the 4 triangles that makes up its interior
QUESTION [0 upvotes]: Given a trapezoid $ABCD$ with its diagonals drawn and $E$ is the point where the diagonals intersect. Then the trapezoid is divided up into 4 triangles. Theres a well known theorem that if $k_{1},k_{2},k_{3},k_{4}$ are the areas of each of the four triangles then, $k_{1} \cdot k_{3} = k_{2} \cdot k_{4}$ Is it possible to find the area of the trapezoid if you only knew the area of two of the four triangles that make it up using the theorem above?

REPLY [1 votes]: I have the figure drawn according to the given.

 $k_1 … k_4$ has to be arranged as such according to (1) the given equality and (2) ratio of the areas is proportional to bases if the altitude is the same.
Also, by (2) again, $k_2 = k_4$ ….. (*)
If 2 of the areas of triangles are given, depending which two, we have to separate the study into the following cases. For simplicity, I will use {i} to represent $k_i$ for i = 1, … 4
Case-1 If {1} and {2} are known, then {4} is also known because of (*) and then {3} can be found from the given relation.
Case-2 If {1} and {4} are known, this same as case-1.
Case-3 If {2} and {4} are known, then $k_1 \cdot k_3 = (k_2)^2$. No further result can be developed.
Case-4 If {3} and {4} are known, similar to case-1.
Case-5 if {2} and {3} are known, similar to case-1.
Case-6 If {1} and {3} are known, then {2}= {4} $= \sqrt (k_1 \cdot k_3)$.


TITLE: Expected value, given the sum of the random variable of interest, and another RV?
QUESTION [0 upvotes]: Suppose we have two random variables, X and Y, which have the same distribution (Gaussian) but different parameters. We define a third RV, Z=X+Y. Of course, the mean of Z is the sum of the means of X, Y, and the S.D. of Z is the sum of the S.D.s of X, Y.
Then we observe z $\in$ Z. How can we figure out the expected component x, given z? I guess I'm wondering where in the X distribution the value that contributes to z is most likely to fall. (As a function of means and S.D.s of X, Y) I'm stumped. Thanks

REPLY [1 votes]: HINT
It sounds like what you are looking for is
$$
\mathbb{P}[X \le x|X+Y=z] = F_{X|X+Y}(x|z)
$$
and then you can compute $\mathbb{E}[X|X+Y=z]$.