text:
\chapter{Sexual Activities Hidden in Social Organization: A Preferential Attachment Mechanism for Human Sexual Network Formation in  Social Network Context}\label{ap2}
The current appendix adds to our previous appendix on generating  heterosexual  networks with prescribed joint-degree distribution  considering impact of people's social behavior (i.e., non-sexual partners) on their sexual partner selection. We generate a sexual network which has two properties: first property is that our sexual network is a subgraph of social network with a structure defined in \cite{eubank2010detail}, second it follows a joint-degree (degree-degree) distribution  originated from sexual activity data.

\section{Degree and joint-degree distribution and $BJD$ matrix for a sexual network}
A conventional heterosexual network  $\textbf{G}$, which is from bipartite family networks, is a sexual network of men and women, in which men only have partnership with women and vise versa, there is no partnership between two men or two women. Every edge $\textbf{ij}$ in this network means that two persons \textbf{i} and \textbf{j} are sexual partners.
The \textit{degree}  of a person \textbf{i}, is defined as the number of his/her sexual partners.
The \textit{degree distribution} $d_k$ defines the number of people with  degree $k$.
The \textit {joint-degree distribution}  $(k,j)$ is the number of men with degree $j$ who are connected to women with  degree $k$.
This distribution  can be represented by the \textit{Bipartite Joint Degree} or $BJD$ matrix:

\[BJD_G = \parenMatrixstack{
    e_{11} & e_{12} & e_{13} & \dots  & e_{1m} \\
    e_{21} & e_{22} & e_{23} & \dots  & e_{2m} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    e_{w1} & e_{w2} & e_{w3} & \dots  & e_{wm} },\]

where, $w$ is the maximum degree in women nodes, and $m$ is the maximum degree in men nodes, each element $e_{ij}$ is the number of edges between women with $i$ partners and men with $j$ partners.
The degree distribution of the number of women nodes, $d^w_k$, and men nodes, $d^m_k$, with  $k$ partners can be obtained from $BJD_G$:
\begin{eqnarray}
d^w_k=\frac{\sum_{j=1}^m e_{kj}}{k}~,\text{~~and~~~}~~~d^m_k=\frac{\sum_{i=1}^w e_{ik}}{k}~.
\end{eqnarray}

\section{Social network embedding sexual network}
The  Social Network  called  \so is a graph whose nodes are synthetic people, labeled by their demographics,
and whose edges represent contacts determined in which each synthetic person is deemed to have made contact with a subset of other synthetic people through some \textit{Activity} types.
We have five different activity locations: Home($H$), Work($W$), School($SC$), Shopping($SH$), and Others($O$).
Each edge is labeled with one of these activity locations and is weighted by  the time spent on these contact per day. For example edge \textbf{i}\textbf{j} labeled by $W$ and weighted by  $T$   means two persons \textbf{i} and \textbf{j} have contact for $T$ fraction of their total time  spent at work  \cite{eubank2010detail}.

\section{Generating Bipartite Sexual Network }\label{method}
Our goal is to introduce an algorithm for generating a heterosexual network called \se that is  a partial subgraph of social network \so and meets a particular joint-degree distribution represented by matrix $BJD$.
\subsection{The algorithm }
Now, we  provide an algorithm that uses the sexual activity data (BJD matrix) and social network to generate the sexual network of individual  called \se. This  graph is partially embedded in \so-depends on what percentage of sexual partners of a typical  person are selected from his/her social friends- and has a joint-degree distribution that meets BJD matrix.
The algorithm includes three phases.
\subsubsection{Phase 1: extension and revision of SocNet }

This Phase  is a procedure to extend \so and  then to find its subgraph  in a way that  only consist the potensial edges for sexual activities:
\begin{enumerate}
\item The first step  is to condense \so by making friendship between friends of an index case. For two persons \textbf{i} and \textbf{j} who are not currently connected, suppose the probability of their meet through a common social friend like \textbf{k} within an activity $A$ is $p^A_{ij}$. If they have $k(i,j)$ common social friends through a particular activity $A$, therefore, with probability of $1-(1-p^A_{ij})^{k(i,j)}$ they meet each other and make friendship, that is, with probability of $1-(1-p^A_{ij})^{k(i,j)}$ we make a new social edges between \textbf{i} and \textbf{j}. We define $p^A_{ij}=p^A_ip^A_j$ where $p^A_k$ for a person \textbf{k} is the average fraction of time spent  per social friend for social friends within  activity $A$, that is, if $N_A(k)$ is set of all  social friends for person \textbf{k} through an activity location $A$, then
\begin{eqnarray}
p^A_k=\frac{\sum_{l \in N_A(k) } T_{k,l}}{|N_A(k)|}.
\end{eqnarray}

 The Figure (\ref{frined_of_friend}) represents an schematic of this approach: for the persons \textbf{i} and \textbf{j} who are not currently  social friends but have three different common friends \textbf{k}$_0$,\textbf{k}$_1$, and \textbf{k}$_2$ though different activities. Suppose $A_{ik_0}=A_{jk_0}=A_{ik_1}=A_{jk_1}=A\neq A_{ik_2}\neq A_{ik_2}$, that is, \textbf{k}$_0$ meets \textbf{i} and  \textbf{j} at the same location, similarly \textbf{k}$_1$ meets \textbf{i} and  \textbf{j} at the same location to \textbf{k}$_0$'s, however, \textbf{k}$_2$ meets them in different places. To compute  $p^A_i$ and $p^A_j$ we only count the friends who meet them at the same location A, therefore, $p^A_i=\frac{T_{ik_0}+T_{ik_1}}{2}$, and $p^A_j=\frac{T_{jk_0}+T_{jk_1}}{2}$.
\begin{figure}[H]
\centering
  \begin{tikzpicture}[
      mycircle/.style={
         circle,
         draw=black,
         fill=gray,
         fill opacity = 0.3,
         text opacity=1,
         inner sep=0pt,
         minimum size=20pt,
         font=\small},
      myarrow/.style={-},
      node distance=1.5cm and 1.9cm
      ]
      \node[mycircle] (i) {$\textbf{i}$};
      \node[mycircle, right=of i] (k1) {\textbf{k}$_1$};
      \node[mycircle,above =of k1] (k0) {\textbf{k}$_0$};
       \node[mycircle,below =of k1] (k2) {\textbf{k}$_2$};
       \node[mycircle,right =of k1] (j) {\textbf{j}};

     \draw [myarrow] (i) -- node[sloped,above] {$A_{ik_0}$,~$T_{ik_0}$} (k0);
     \draw [myarrow] (i) -- node[sloped,above] {$A_{ik_1}$,~$T_{ik_1}$} (k1);
     \draw [myarrow] (i) -- node[sloped,above] {$A_{ik_2}$,~$T_{ik_2}$} (k2);
     \draw [myarrow] (j) -- node[sloped,above] {$A_{jk_0}$,~$T_{jk_0}$} (k0);
     \draw [myarrow] (j) -- node[sloped,above] {$A_{jk_1}$,~$T_{jk_1}$} (k1);
     \draw [myarrow] (j) -- node[sloped,above] {$A_{jk_2}$,~$T_{jk_2}$} (k2);
\iffalse
    \foreach \i/\j/\txt/\p in {
      i/c2/8/below,
      i/c4/11/above,
      c2/c3/11/below,
      c3/c6/4/below,
      c4/c5/12/above,
      c5/c6/15/above,
      c5/c2/4/below,
      c3/c5/7/below,
      c2.70/c4.290/1/below}
       \draw [myarrow] (\i) -- node[sloped,font=\small,\p] {\txt} (\j);

     \draw [myarrow] (c4.250) -- node[sloped,font=\small,above,rotate=180] {10} (c2.110);
     \fi
    \end{tikzpicture}
    \caption[\textbf{Schematic of step 1}]{An schematic of step 1, the only case that \textbf{i} and \textbf{j} have a chance to meet is when at least one of their common friends meet them within the same activity location.}
    \label{frined_of_friend}
\end{figure}

\item If sexually active population under study has an age range $\alpha=[\alpha_1,\alpha_2]$, therefore people in  \so but out of this age range cannot play a role in sexual network, for example a child with age six years old cannot be part of sexual network. Thus, the second step is to remove all people with ages $\notin\alpha$.

\item The sexual network we aim to extract from \so is heterosexual. That is, we do not have homosexual individuals in \textbf{SexNet}, therefore,  the edges between two men or two women in \so can not be a potensial sexual edge, so we remove this edge from \so.
\item Our last assumption is that sexual partners are not living together, therefore household edges in \so (edges with label H) cannot be the proper edge ending up with sexual activity, so we remove household edges.
\end{enumerate}

\subsubsection{Phase 2: main algorithm }

The main algorithm takes three inputs \textbf{SocNet}, $BJD$ matrix corresponding to  joint-degree distribution for \textbf{SexNet}, and $p\in[0,1]$ the embedding percentage of \textbf{SocNet}, in fact the value p tells what percentage of edges in \se is in \textbf{SocNet}. The output would be heterosexual network \se  which is a partially subgraph of \so and matrix $BJD$ represents its joint-degree distribution.
Starting with an empty set of nodes for \textbf{SexNet}, we select nodes in \so to add to \textbf{SexNet}: we sort nodes in \so based on their degree and as the first node in \se add the man node in \so with maximum degree to set of nodes in \textbf{SexNet} and assign its desired degree in \se equal to column size of $BJD$, and then select his first sexual partner with probability $p$ from his social neighbors and with probability $1-p$ from closest nodes in \so but not a social friend.

At any stage, we first find the partners for nodes in \se from their social friend, if the degree of all nodes in \se meet their desired degree, we add a new node to \textbf{SexNet}. We repeat adding edges until all edges are placed.
Algorithms \ref{main_alg} and \ref{help_alg}
explain this procedure, and table  \ref{notation} is the table of definition for symbols in the algorithms.

\begin{table}[htbp]
\begin{center}

\begin{tabular}{r c p{10cm} }
\toprule
\footnotesize
$G.n$ & $\defeq$ &set of nodes  in  $G$\\
$G.e$ & $\defeq$ &set of edges  in network $G$\\
$d_G(\textbf{i})$ & $\defeq$ &degree of node \textbf{i} in network $G$\\
$d_G$ & $\defeq$ &degree frequency list for network $G$\\
$G.N(\textbf{i})$ & $\defeq$ & set of neighbors of node \textbf{i} in Network G\\
$dis(G,\textbf{u},\textbf{v})$ & $\defeq$ & distance between two nodes \textbf{u} and \textbf{v} in Network G\\
$M.col(M.row)$ & $\defeq$ & column size (row size) of a matrix M\\
$M(i,:) (M(:,i))$ & $\defeq$ & $i^{th}$ row (column) of matrix M\\
$V( i)$ & $\defeq$ & $i^{th}$ element of vector V\\
$|S|$ & $\defeq$ & size (the number of elements) of  a set S\\
$S.remove(m)$ &$\defeq$ & remove member m from   a set S\\
$S.sample(P)$ &$\defeq$ & randomly select an element with property P (if P$=1$ there is no property) from set S \\
$urn$ &$\defeq$ & uniform random number in $[0,1]$\\
\iffalse
\multicolumn{3}{c}{}\\
\multicolumn{3}{c}{\underline{Decision Variables}}\\
\multicolumn{3}{c}{}\\
$y_f$ & $=$ & \(\left\{\begin{array}{rl}
1,  & \text{if blah blah blah } \\
0,  & \text{otherwise} \end{array} \right.\)\\
\fi
\bottomrule
\end{tabular}
\end{center}
\caption[\textbf{Notations table}]{Table of notation for a conventional network $G$ in algorithms.}
\label{notation}
\end{table}
To keep or remove an edge  we have to calculate the degree of nodes attached to it for each possible edge in the \textbf{SocNet}, thus,  the full set of experiments run in O($|E|P^mP^w$) time, where $|E|$ is the number of edges in \textbf{SocNet}, $P^m$ number of its men nodes and $P^w$ number of its women nodes. This method is feasible if average degree ($\frac{2|E|}{P^m+P^w}$) of the network is not high.

\newpage

\begin{algorithm}[H]
 \SetAlgoLined
 $\textbf{SexNet}.n=\emptyset,~\textbf{SexNet}.n \leftarrow \textbf{u}={max_{d_\so(\textbf{k})}}\{\textbf{k} \in \textbf{SocNet}.n \}$\;
  $d_{\se}(\textbf{u}) \triangleq BJD.col,~stub(\textbf{u}) \triangleq d_\se(\textbf{u})$\;
   $E\triangleq \sum_i\sum_jBJD(i,j)$\;
  \While{$|\textbf{SexNet}.e|\leq E$}{
 $NF\triangleq\{\textbf{k} \in \textbf{SexNet}.n~if~stub(\textbf{k})>0\}$\;
  \While {$|NF|\geq 1$}{
  $\textbf{u}={max_{stub(\textbf{k})}}\{\textbf{k} \in NF \}$\;
  $(d',\textbf{v})=\Call{FP}{\textbf{u}, \textbf{SocNet},            \textbf{SexNet}, BJD, NF, p}$\;
  \eIf{$(d',\textbf{v})\neq False$}{
    $NF.remove$(\textbf{u})\;
   }{
    Make edge (\textbf{u},\textbf{v}) in \textbf{SexNet}$,stub(\textbf{u})\leftarrow stub(\textbf{u})-1,$ $stub(\textbf{v})\leftarrow stub(\textbf{v})-1$\;
    \If{stub(\textbf{u})=0 [stub(\textbf{v})=0]}
    {$d_\se.remove(d_\se(\textbf{u}))$ [$d_\se.remove(d_\se(\textbf{v}))$]\;}
    \eIf{\textbf{u} is woman}{$BJD(d,d')-1$}{$BJD(d',d)-1$}
  }}
   \textbf{SexNet}$.n \leftarrow \textbf{u}={max_{d_\so(k)}}\{k \in \textbf{SocNet}.n-\textbf{SexNet}.n\}$\;
   $d_\se(\textbf{u})\triangleq max\{d_\se\},~stub(\textbf{u})\triangleq d_\se(\textbf{u}).$
  }
   \caption{ Extracting sexual network from social network (\textbf{Soc2sex})}\label{main_alg}
\end{algorithm}

\begin{minipage}{\linewidth}
\begin{algorithm}[H]
 \SetAlgoLined
   $d=d_{\se}(\textbf{u})$\;
      \lIf{\textbf{u} is woman}
       {$R\triangleq BJD(d,:)$, \textbf{else} $R\triangleq BJD(:,d)$}
\For{\texttt{iter} in range($|R|$)}
{\eIf{$R\neq0$}
    {$R.sample(d': R(d')\neq 0)$\;
    \eIf{$urn\leq$p}
          { $K1\triangleq\{\textbf{k}\in NF: \textbf{k}\in \textbf{SocNet}.N(\textbf{u})-\textbf{SexNet}.N(\textbf{u}), d_\se                      (k)=d'\}$\;
         $K2\triangleq\{\textbf{k}\in \textbf{SocNet}.N(\textbf{u})-\textbf{SexNet}.n:  d_\so(k)\geq d'\}$\;}
         {$K1\triangleq \{\textbf{k}\in NF: \textbf{k}\notin \textbf{SocNet}.N(\textbf{u})\cup\textbf{SexNet}.N(\textbf{u}), d_\se (k)=d'\}$\;
        $K2\triangleq \{\textbf{k}\in \textbf{SocNet}.n-\textbf{SexNet}.n:  d_\so(k)\geq d'\}$\;}
          \uIf{$K1\neq \emptyset$}
           {\footnotesize{\textbf{v}$=K1.sample(\textbf{w}: dis(\textbf{SocNet},\textbf{u},\textbf{w})=min\{dis(\textbf{SocNet},\textbf{u},\textbf{k})~for~k\in K1\})$}\;
              Break\; }
              \uElseIf{$K2\neq \emptyset$}
              { \footnotesize{\textbf{v}$=K2.sample(\textbf{w}: dis(\textbf{SocNet},\textbf{u},\textbf{w})=min\{dis(\textbf{SocNet},\textbf{u},\textbf{k})~for~k\in K2\})$\;
       $d_\se(\textbf{v})\triangleq d'$, $stub(\textbf{v})\triangleq d'$}\;
          Break\;}
              \Else{$R(d')\triangleq 0$\;}
         }
         {$(d',\textbf{v})=False$\;
         break\; }
         }
\textbf{Return} $(d',\textbf{v})$\;

   \caption{  Finding partner with proper degree for a given node (\textbf{FP})}\label{help_alg}
\end{algorithm}
\end{minipage}
\subsubsection{Phase 3: correcting $BJD$ }
We intent to have  a generated sexual network  such that $BJD$ matrix represents its  joint-degree distribution. But, because in the main algorithm, we may not find a proper partner for some people, therefore, we may end up with a different joint-degree distribution for the network. The goal of third Phase is to correct joint-degree distribution of generated \se through some edge rewiring.  We define $\widetilde{BJD}$ as matrix representing joint-degree distribution of \se and $\mathcal{E}=BJD-\widetilde{BJD}$ as error matrix. Matrix  $\mathcal{E}$ may have nonzero elements:
\begin{enumerate}
\item If element $(i,j)$ of $\mathcal{E}$ for $i, j>1$ is a positive value $k$, it means that \se  needs $k$ more edges between women with $i$ partners and men with $j$ partners. To make these edges  we do the following process $k$ times: In \se we find a woman with $i$ partners like \textbf{w}$_i$ which has a degree-1 partner like \textbf{m}$_1$, then we find a man with $j$ partners  like \textbf{m}$_j$ which is social friend of \textbf{w}$_i$  but not her sexual partner and also has a degree-1 partner like \textbf{w}$_1$. Then we do a rewiring: remove edges (\textbf{w}$_i$,\textbf{m}$_1$) and (\textbf{m}$_j$,\textbf{w}$_i$) and add edge (\textbf{w}$_i$, \textbf{m}$_j$).
\item If element $(i,j)$ of $\mathcal{E}$ for $i, j>1$ is a negative value $k'$, it means that we have  extra $k'$  edges between women with $i$ partners and men with $j$ partners. To remove these extra edges  we do the following  process $k'$ times: we find a woman with $i$ partners like \textbf{w}$_i$ which has a degree-j partner like \textbf{m}$_j$, and  that at least one of their social friends are not in \se. Then we find a man  like \textbf{m}$_1$ which is social friend of \textbf{w}$_i$ that  not in \textbf{SexNet}$.n$, and also  we find a woman  like \textbf{w}$_1$ which is social friend of \textbf{m}$_j$ that is not in \textbf{SexNet}$.n$. Then we do a rewiring: remove edge (\textbf{w}$_i$,\textbf{m}$_j$) and add edges (\textbf{w}$_i$, \textbf{m}$_1$),(\textbf{w}$_1$,\textbf{m}$_j$).
\item In the previous steps, we pushed  back nonzero elements in $\mathcal{E}$ to its first row and column, which this causes new nonzero elements in the first row and column.
To remove these nonzero values, we have to add or remove small components. For example if  element (i,1) of $\mathcal{E}$ is a positive value $k$ it means  that we need a small component of a woman with $i$ partners whose all partners  are degree-1 men, therefore, we simply make this component from the people who are not currently in \textbf{SexNet}$.n$. If  element $(1,j)$ of $\mathcal{E}$ is a negative value $k$ it means  we have to remove a small component of a man with $j$ partners whose all partners  are degree-1 women, therefore, we simply look for such a component and remove it from \textbf{SexNet}.
\end{enumerate}
This Phase 3 may not make matrix $\mathcal{E}$ exactly equal to zero, because in its approach the proper nodes may not exist. However, it improves joint-degree distribution of \textbf{SexNet}. In the Result Section of Chapter (\ref{abm}), we observe that in practice Phase 3 would not be needed.


TITLE: On quotient manifold theorem
QUESTION [0 upvotes]: I want to solve the following exercise from Lee's book on smooth manifolds. If a Lie group $G$ acts smoothly and freely on a smooth manifold $M$, and the orbit space $M/G$ has a smooth manifold structure, such that the quotient map $\pi:M\to M/G$ is a smooth submersion, then $G$ acts properly.
This question was asked before somewhere but the answer is not satisfactory. Can someone give some hint on how to solve this problem?

REPLY [1 votes]: We show this by using the characterization of proper actions via sequences:
Let $(g_{i})\subseteq G$ and $(p_{i})\subseteq M$ be sequences such that $p_{i}\to p$ and $g_{i}\cdot p_{i}\to q$. We need to find a convergent subsequence  of $(g_{i})$.
Observe that $\pi\colon M\to M/G$ is a smooth submersion. This means that there is an open set $V\subseteq M/G$ and a smooth section of $\pi$ defined on $V$ (that is, a map $\chi\colon V\to M$ such that $\pi\circ \chi=\operatorname{Id}_{V}$ and $p\in \chi(V)$). Define a map $\Phi\colon G\times V\to \pi^{-1}(V)$ by letting $\Phi(g,z)=g\cdot\chi(z)$. This map is bijective and $G$-equivariant, so it has constant rank along the orbits $G\times \{s\}$ for all $s\in V$. Therefore, if we can prove that the rank of $\Phi$ is globally constant, we get that $\Phi$ is a diffeomorphism. This is because, by taking a point of the form $(e,s)$, we get that $\Phi_{*(e,s)}(T_{(e,s)}(G\times V))=T_{\chi(s)}(G\cdot \chi(s)))+\chi_{*s}(T_{s}(M/G))$. Since $\chi$ is an immersion and those two summands have trivial intersection, we get that $\Phi$ has constant rank in $\{e\}\times V$, so $\Phi$ has constant rank everywhere and is therefore a diffeomorphism.
Now, since $p_{i}\to p$, we can suppose that $\pi(p_{i})\in W$ for all $i$, where $W\subseteq V$ is an open subset containing $p$ and having compact closure in $V$. Therefore, we can write $\Phi^{-1}(p_{i})=(h_{i},s_{i})$ for $h_{i}\in G$, $s_{i}\in V$. Notice that $\pi(p_{i})=\pi(\Phi(h_{i},s_{i}))=\pi(h_{i}\cdot \chi(s_{i}))=\pi(\chi(s_{i}))=s_{i}$, so that $p_{i}=h_{i}\cdot \chi(\pi(p_{i}))$. Also, $p=\chi(\pi(p))=e\cdot \chi(\pi(p))$. Since $p_{i}\to p$, $h_{i}\to e$, because $\Phi$ is a diffeomorphism.
Furthermore, $g_{i}\cdot p_{i}\to q$, so that $\pi(p_{i})\to \pi(q)\in \overline{W}\subseteq V$. We can therefore write $q=h\cdot \chi(\pi(q))$ for some $h\in G$. Also, $g_{i}\cdot p_{i}=(g_{i}h_{i})\cdot \chi(\pi(p_{i}))\to q=h\cdot \chi(\pi(q))$. Again, since $\Phi$ is a diffeomorphism, this implies that $g_{i}h_{i}\to h$, so $g_{i}=g_{i}h_{i}h_{i}^{-1}\to h$. We get that $(g_{i})$ is a convergent subsequence.
Hope this helps!


TITLE: Is it true that $\langle [x]\rangle +\langle [y]\rangle =\mathbb Z_n?$
QUESTION [2 upvotes]: Consider the ring $\mathbb Z_n$ where $n=p^a q^b$.
  Let $[x]\in \langle [p]\rangle \setminus \langle [pq]\rangle$ and $[y]\in \langle [q]\rangle \setminus \langle [pq]\rangle$ where $\langle [p]\rangle $ denotes the ideal generated by $[p]\in \mathbb Z_n$.
  Here $p<q$ are primes.
Is it true that $\langle [x]\rangle +\langle [y]\rangle =\mathbb Z_n?$

My try::
$[x]\in \langle [p]\rangle \setminus \langle [pq]\rangle\implies [x]=s[p]$ and $[y]\in \langle [q]\rangle \setminus \langle [pq]\rangle\implies [y]=t[q]$.
Also $[x]\notin \langle [pq]\rangle\implies \gcd(s,q)=1$. Also $\gcd(t,p)=1$.
Is it true that  $\langle [x]\rangle +\langle [y]\rangle =\mathbb Z_n?$
Can someone please help?

REPLY [0 votes]: Hint $ $ By Euclid and $\, (p,q) = (t,p) = (s,q)  = 1\ $ we have in $\Bbb Z$
$$\begin{align}
(sp+tq,\,p)\ \  &=\, (tq,p)\, = (t,p)=1\\[.2em]
 (sp+tq,\,q)\ \  &=\, (sp,q) = (s,q)=1\\[.2em]
\Rightarrow\ (sp+tq,\,p^aq^b)&=\,1\end{align}\qquad$$


TITLE: intersect of subspaces
QUESTION [0 upvotes]: How can I prove that if X and Y are n-dimensional subspaces of a vector space Z and X intersect Y is also dimension n, that X = Y?
It makes sense logically, but I'm not sure how I would prove it - going by properties of subspaces hasn't led me anywhere

REPLY [1 votes]: A way to do that consists in remarking that :
$Y \cap X \subset Y$.
Since both spaces have same dimension, we must have that $Y = Y \cap X$.
And by same argument with $X$ you obtain that $X = Y$.


TITLE: Confusion about pointwise vs. uniform convergence
QUESTION [1 upvotes]: I'm confused about expressions like $$L=\sum_{k=1}^{\infty} R_k, \quad \quad (1)$$where $L:H\rightarrow H$ and $R_k:H \rightarrow H$ are continuous linear operators in a Hilbert space $H$. We assume $\sum_{k=1}^{\infty} R_k x$ is convergent every $x\in H$.
The thing I'm confused about, is: If we know that for  every $x\in H$ we have $$Lx=\sum_{k=1}^{\infty} R_k x,$$can we that automatically conclude that $(1)$ holds ?
I would conjecture "yes" since two maps are identical if they have the same domain and range (trivial here) and are identical for every argument - which they are, as the above shows.
But if the answer is indeed "yes", why is it in sum texts shown, that the $R_k$ additionally converge in the operator norm to $L$, i.e. $$\left\|L- \sum_{k=1}^{n} R_k\right\| \rightarrow 0\quad (n\rightarrow \infty),$$ before they conclude that $(1)$ is true ? Was that really necessary ?

REPLY [0 votes]: One of the issues with convergence is that it need not preserve continuity.
The simplest example (although not too relevant here) is $x^n$ on $[0,1]$ converging to a discontinuous function.
Another example is convergence for a series of rationals in $\mathbb Q$. If the limit is irrational, then it does not exist (although it does exist in a larger space). This case is similar, if the limit operator is not continuous then it does not exist in our space -- and the sequence is not convergent.
Pointwise convergence is indeed enough to show that $L$ is well-defined, but it is not enough to show that it is continuous. If we wish to talk about convergence in a space of continuous linear operators then we must also verify that the convergence is strong enough for that, namely that the limit operator is itself continuous.


TITLE: Trigonometric equation 3
QUESTION [2 upvotes]: $$2 \arcsin x = \arccos 2x$$
I tried applying $\sin$ to the equation, so it results
$$\sin (2 \arcsin x) = \sin(\arccos 2x)$$
$$2 \sin(\arcsin x) \cdot \cos(\arcsin x) = \sin (\arccos 2x)$$
$$2x \cdot \sqrt{1 - x^2} = \sqrt{1- 4x^2}$$
But this doesn't result in the correct answer $\frac{\sqrt{3} -1}{2}$
What am I missing ?

REPLY [3 votes]: Hint:
First of all, $0\le2\arcsin x\le\pi\implies x\ge0$
Let $\arcsin x=y\implies x=\sin y$
$\implies2\sin y =\cos2y=1-2\sin^2y$
Can u take it from here?


TITLE: Determining if a differential equation has unique solution
QUESTION [13 upvotes]: Our teacher asked a seemingly simple question:

What is the value of $f(2)~$ if $~f( f(x) ) = 16x-15~$?

I found a solution assuming $f(x)$ in the form of $ax+b$, but how do I show that $f(x)$ must be in that form?
I thought of taking derivative of both sides and reached
$$f'(f(x)) = \frac{16}{ f'(x) }$$
How can I show that this function has a unique solution or it doesn't have a unique solution?

REPLY [12 votes]: Apply $f$ from the inside and outside, set $x=f(w)$ then
$$
f(f(f(w)))=16f(w)-15
$$
and also
$$
f(f(f(x)))=f(16x-15)
$$
so that $f(16x-15)=16f(x)-15$, from which one concludes directly $f(1)=1$. Set $g(x)=f(1+x)-1$, then $16g(x)=f(1+16x)-1=g(16x)$, $g(0)=0$ or equivalently
$$
g(x)=16\,g\biggl(\frac{x}{16}\biggr)=16^n\,g\biggl(\frac{x}{16^n}\biggr)~~\forall n\in\Bbb N
$$
Assuming differentiability of $f$ and thus of $g$ one concludes that $g(x)=ax$, $f(x)=ax+1-a$, validating your first guess as the only solution,
$$
f(f(x))=a(ax+1-a)+1-a=a^2x+1-a^2\implies a=\pm4, ~~f(2)=a+1\in\{-3,5\}.
$$


TITLE: A cubic nonlinear Euler sum
QUESTION [6 upvotes]: Any idea how to solve the following Euler sum

$$\sum_{n=1}^\infty \left( \frac{H_n}{n+1}\right)^3 =
 -\frac{33}{16}\zeta(6)+2\zeta(3)^2$$

I think It can be solved it using contour integration but I am interested in solutions using real methods.

REPLY [2 votes]: \begin{align}
S&=\sum_{n=1}^\infty\left(\frac{H_n}{n+1}\right)^3=\sum_{n=1}^\infty\left(\frac{H_{n-1}}{n}\right)^3\\
&=\sum_{n=1}^\infty\frac{H_n^3}{n^3}-3\sum_{n=1}^\infty\frac{H_n^2}{n^4}+3\sum_{n=1}^\infty
\frac{H_n}{n^5}-\sum_{n=1}^\infty\frac{1}{n^6}
\end{align}
Substituting the following results:
$$\sum_{n=1}^\infty\frac{H_n^3}{n^3}=\frac{93}{16}\zeta(6)-\frac52\zeta^2(3)$$
$$\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$
We get $$\boxed{S=2\zeta^2(3)-\frac{33}{16}\zeta(6)}$$
Note that the first and second sum are proved here and here respectively. As for the third sum, can be obtained using the Euler identity.


TITLE: A driver approaches a traffic light
QUESTION [0 upvotes]: A driver approaches a traffic light, which is green with speed $ v_0 $ when it turns yellow.
a) If the driver's reaction occurs within temp $ T $, during which he decides to stop and apply the brake foot, and if the maximum brake deceleration is $ a $, what is the minimum distance $ S_ {min} $ before hitting Does the intersection, the moment the light turns yellow, can it make the car stop without crossing it?
b) If the yellow light stays on for a time $ t $ before turning red, what is the distance $ S_ {\max} $, before the intersection, the instant the yellow light comes on so that it can cross the intersection with speed $ v_0 $ without the red light coming on?
c) Show that in case the initial velocity is greater than$$ v_ {0\max} = 2a (t-T) $$there will be a range of distances before the rally so that the driver does not stop in time and cannot cross it without the red light coming on.
Solution:
''a) The car travels a distance $d_1=v_0T$ while the driver is reacting. It then travels a further distance $d_2=\frac{v_0^2}{2a}$ while decelerating at a rate $a$ before it comes to rest. So the total distance travelled is:
$S_{min}=d_1+d_2 = v_0T + \frac{v_0^2}{2a}$
b) This is simpler than you think. The car is travelling at a constant velocity $v_0$, so it will reach the intersection before the red light comes on as long as its distance from the intersection when the yellow light comes on (t seconds earlier)  is less than
$S_{max} = v_0t$
c) The driver can stop before the intersection if his initial distance is greater than $S_{min}$. And he can cross the intersection before the red light comes on if his initial distance is less than $S_{max}$. If $S_{min} \le S_{max}$ then for any initial distance he can either stop before the intersection or cross the intersection before the red light comes on (and for distances between $S_{min}$ and $S_{max}$ he can choose from both actions).
For what value of $v_0$ is $S_{min}$ equal to $S_{max}$ ? If $v_0$ is greater than this value then $S_{min} > S_{max}$. What happens now if the driver's initial distance is between $S_{min}$ and $S_{max}$ ?''
I still don't know how to prove it. I couldn't think of a distance left or missing to play in the equations

REPLY [1 votes]: Note that, if $v_0=2a(t-T)$, the result from a) $S_{min}= v_0T + \frac{v_0^2}{2a}$ leads to,
$$S_{min}= 2a(t-T)T + \frac1{2a}[2a(t-T)]^2 = 2a(t-T)t = S_{max}$$
Therefore,  $v_0=2a(t-T)$ is the critical speed with which the driver can either break to stop before the intersection or cruise cross the intersection before the red light.
In the case where $v_0 > 2a(t-T)$, we have
$$S_{min}>2a(t-T)t=S_{max}$$
That is, if the drive is closer than $S_{min}$ and further than $S_{max}$ from the
the intersection, there is no way over this range $(S_{max}, S_{min})$ for the drive to stop in time or cruise cross the intersection without the red light coming on.


text:
\section{Preliminaries}\label{sec:preliminaries}
\subsection{Notation}
For a quick reference for oriented matroids, polytopes and Delaunay triangulations,
we refer to the chapters \cite{RichterGebertZiegler1997}, \cite{HenkRichterGebertZiegler1997}
and \cite{Fortune1997} of the handbook~\cite{HanbookDCG}, respectively. Our notation
coincides mostly with theirs.

Let $V$ be a configuration of $n$ vectors in $\RR^r$, which are labeled by
elements in $[n]=\{1,\dots,n\}$. Consider the map $\chi^V:[n]^{r}\mapsto
\{0,1,-1\}$ that for each tuple $(i_1,\dots,i_{r})$ assigns the sign
\[\chi^V(i_1,\dots,i_{r})=\signe\det(v_{i_1},\dots,v_{i_{r}}).\]
The map $\chi^V$ is called the \defn{chirotope} of $V$ and determines its
\defn{oriented matroid}, which has \defn{rank}~$r$
(see~\cite{OrientedMatroids1993} for a comprehensive introduction to oriented
matroids).

Now, let $A$ be a configuration of $n$ points in $\RR^d$. The \defn{homogenization} of
$A$ is a vector configuration $\homog{A}=\{\bar a_1,\dots,\bar
a_n\}\subset\RR^{d+1}$ obtained by appending $1$ as the last coordinate of the
points of $A$: $\bar a_i=(a_i,1)$.
The \defn{oriented matroid} of $A$ is defined to be the oriented matroid of its
homogenization~$\homog{A}$.

The oriented matroid of a point configuration is always \defn{acyclic}.
A point configuration $A$ is in \defn{general position} if no $d+1$ points of
$A$ lie in a common hyperplane, and then it defines a \defn{uniform} matroid.

The convex hull of $A$ is a \defn{polytope} $P=\conv(A)\subset\RR^d$ and the
intersection of $P$ with a supporting hyperplane is a \defn{face} of $P$. Faces
of dimensions $0$ and $d-1$ are called \defn{vertices} and \defn{facets},
respectively. A point configuration $A$ is in \defn{convex position} if it
coincides with $\verts(P)$, the set of vertices of $P=\conv(A)$. If $A$ is in
convex position, each face $F$ of $P$ can be identified with the set of labels
$\set{i\in[n]}{a_i\in F}$. The \defn{face lattice} of $P$ is then a poset of
subsets of $[n]$. In this context, two vertex-labeled polytopes are
\defn{combinatorially equivalent}, denoted $P\simeq Q$, if their face lattices
coincide. We call $P$ \defn{inscribed} if all its vertices lie in the unit
sphere~$\Sp{d-1}$, and \defn{inscribable} if it is combinatorially equivalent to
an inscribed polytope.

The face lattice of a polytope $P$ coincides with that of the \defn{convex cone} obtained
as the positive hull of its homogenization
$\cone(\hom(P)):=\set{\sum \lambda_i x_i}{\lambda_i \geq 0, \, x_i\in \hom(P)}$.

The oriented matroid of a polytope $P$ is \defn{rigid} if its face lattice of
determines the oriented matroid of its set of vertices (see \cite[Section
6.6]{Ziegler1995}). In the language of the next section, a polytope $P$ is rigid
if and only if $\rsom(\verts({P}))= \rspol(P)$.

A \defn{subdivision} of a point configuration $A$ is a collection $\cT$ of
polytopes with vertices in $A$, which we call \defn{cells}, that cover the
convex hull of $A$ and such that any pair of polytopes of $\cT$ intersect in a
common face. A \defn{triangulation} of $A$ is a subdivision where all the cells
are simplices. Again, a subdivision of $A$ can be identified with a poset of
subsets of~$[n]$. Two subdivisions $\cT$ and $\cT'$ of two labeled
configurations $A$ and $A'$ are \defn{combinatorially equivalent}, denoted by
$\cT\simeq\cT'$, if their respective posets coincide.

The \defn{Delaunay subdivision} $\Del(A)$ of a point configuration
$A\subset\RR^d$ is the subdivision that consists of all cells defined by the
\defn{empty circumsphere condition}: $S \in \Del(A)$ if and only if there exists
a $(d-1)$-sphere that passes through all the vertices of $S$ and all other
points of $A$ lie outside this sphere. If $A$ is in general position and no
$d+2$ points of $A$ lie on a common sphere, then the empty circumsphere
condition always defines a simplex of~$A$, and hence the Delaunay subdivision is
a triangulation,  the \defn{Delaunay triangulation} of $A$.

We denote \defn{homeomorphic} sets $S$ and $T$ by $S\cong T$ and \defn{homotopic}
sets by $S\htpy T$ (see \cite[Section~58]{Munkres} for
definitions).  We also recall that a continuous map $f : S\to T$ is a  \defn{retraction}
of $S$ onto $T$ if there
is a continuous map $g:T\to S$ such that $f\circ g=\id$.
If a retraction exists, then $T$ is a \defn{retract} of~$S$. If moreover $g \circ f$ is homotopic
to the identity, then $T$ is a \defn{deformation retract} of $S$, and $T\htpy S$.

\subsection{Realization spaces}\label{sec:realizationspaces}

We will work with the following \defn{realization spaces}.
Observe that for oriented matroids and polytopes, we work with the
acyclic vector configurations arising from homogenization. (This approach is
convenient for technical reasons, and used often, for example in~\cite{OrientedMatroids1993}.)

We also identify a $d$-dimensional configuration of $n$ points or vectors with the
corresponding tuple in $\RR^{d\times n}$ containing the
coordinates, ordered according to their labels.

\begin{itemize}[$\bullet$]
\item The \defn{realization space of an oriented matroid} $M$ (of rank $d+1$ with
$n$ elements), that we denote $\rsom(M)\subset \RR^{(d+1)\times n}$, is the set of vector
configurations that realize $M$, up to linear transformation:
\[\rsom(M)=\bigslant{\set{V\in \RR^{(d+1)\times n}}{V \text{ realizes }M}}{\GL{\RR^{d+1}}}.\]
\item The \defn{realization space of a polytope} $P$
(with $n$ vertices in~$\RR^d$), that we denote
$\rspol(P)\subset \RR^{(d+1)\times n}$, is the set of acyclic configurations whose positive span
is combinatorially equivalent to the cone over $P$, up to linear transformation:
\[\rspol(P)=\bigslant{\set{V\in \RR^{(d+1)\times n}}{\cone(V)\simeq P}}{\GL{\RR^{d+1}}}.\]
\item The \defn{realization space of an inscribed polytope} $P$
(with $n$ vertices in~$\RR^d$), that we denote
$\rsins(P)\subset (\Sp{d-1})^n \subset \RR^{d\times n}$, is the set of inscribed
point configurations whose convex hull
is combinatorially equivalent to $P$, up to M\"obius transformation:
\[\rsins(P)=\bigslant{\set{A\in (\Sp{d-1})^n}{\conv(A)\simeq P}}{\Mob{\Sp{d-1}}}.\]
\item The \defn{realization space of a Delaunay subdivision} $T$ (of $n$ points
in $\RR^d$), that we
denote $\rsdel(T)\subset \RR^{d\times n}$, is the set of point configurations whose
Delaunay triangulation is combinatorially equivalent to $T$,
up to similarity:
\[\rsdel(T)=\bigslant{\big\{ A\in \RR^{dn} \,\big|\, \Del(A)\simeq T\}}{\Sim{\RR^d}}.
\]\end{itemize}

For a given point configuration $A$, we abuse notation and use $\rsom(A)$,
$\rspol(A)$, $\rsins(A)$,
and $\rsdel(A)$
to denote the realization spaces $\rsom(\chi^A)$, $\rspol(\conv(A))$,
$\rsins(\conv(A))$ and $\rsdel(\Del(A))$, respectively.

\begin{rmk}
A number of alternative definitions are possible. For example,
by factoring different transformation groups or by considering non-homogenized configurations.
Most of these definitions are actually homotopy-equivalent, as we discuss below, and hence our results hold anyway.
These definitions are natural because the groups preserve spheres.

We also leave out the combinatorics of the boundary for Delaunay subdivisions,
which amounts to take also into account the empty spheres that go through the ``point at infinity''.
Although these two definitions are not necessarily homotopy-equivalent, again,
our results hold for both kinds of definition, see also Remark~\ref{rmk:fixedboundary}.
\end{rmk}

It is sometimes useful to commute between different realization spaces; we state the straightforward lemmata, without detailed proof, here:
\begin{lem}[Realization spaces of matroids]\label{lem:homogenization}
Let $M$ be an acyclic oriented matroid.  Then the following three
spaces are homotopy equivalent:
\begin{compactitem}[$\circ$]
\item The realization space of homogeneous configurations, modulo linear transformations:
\begin{equation}\label{eq:lin-relspace}
	\rsom(M)
	=\bigslant{\set{V\in \RR^{(d+1)\times n}}{V \text{ realizes }M}}{\GL{\RR^{d+1}}}
\end{equation}
\item  The realization space of affine configurations, modulo admissible projective transformations:
\begin{equation}\label{eq:proj-relspace}
	\rsom^{\textsf{proj}}(M)=\bigslant{\set{V\in \RR^{d\times n}}{V \text{ realizes }M}}{\PGL{\RR^{d}}}
\end{equation}
\item The realization space of affine configurations, modulo affine transformations:
\begin{equation}\label{eq:aff-relspace}
	\rsom^{\textsf{aff}}(M)=\bigslant{\set{A\in \RR^{d\times n}}{A \text{ realizes }M}}{\GA{\RR^{d}}}
\end{equation}

\end{compactitem}
\end{lem}

\begin{proof}[Proof (sketch)]
To see \eqref{eq:lin-relspace}$\htpy$\eqref{eq:proj-relspace}, consider the
map $\rsom^{\textsf{lin}}(M)\to\rsom^{\textsf{proj}}(M)$
that sends a vector configuration $V$ to the intersection of its positive span
with a hyperplane that intersects every positive ray spanned by $V$.
The map is well defined because two point configurations arising from different hyperplanes
are related by an admissible projective transformation,
and all linear transformations of $V$ also induce admissible projective transformations of $A$.
The homogenization map provides a section, and the fibers are easily seen to be contractible.

For \eqref{eq:proj-relspace}$\htpy$\eqref{eq:aff-relspace}, observe that each fiber of the quotient map
$\rsom^{\textsf{aff}}(M)\to \rsom^{\textsf{proj}}(M)$ is homeomorphic to the
set of admissible projective transformations up to affine transformation. That is,
the set of ``hyperplanes at infinity'' that do not cut $\conv(A)$. This, in turn, is homeomorphic to a polytope, the polar polytope
of $\conv(A)$, which depends continuously on $A$. Hence, a continuous section can be defined by selecting its barycenter.

\end{proof}

Similarly, we have the following lemma for inscribed poltopes:

\begin{lem}[Realization spaces of inscribed polyopes]\label{lem:eqiv_relins}
Let $P$ be an inscribed polytope in $\RR^d$.  Then the following three spaces are homotopy equivalent:
\begin{compactitem}[$\circ$]
\item The realization space of all inscribed polytopes combinatorially equivalent to it, modulo M\"obius transformations:
\begin{equation}
	\rsins(P)
	=\bigslant{\set{A\in (\Sp{d-1})^n}{\conv(A)\simeq P}}{\Mob{\Sp{d-1}}}.
\end{equation}
\item  The realization space of all inscribed polytopes combinatorially equivalent to it, modulo orthogonal transformations:
\begin{equation}
	\rsins^{\textsf{ort}}(P)=\bigslant{\set{A\in (\Sp{d-1})^n}{\conv(A)\simeq P}}{\Ort{\RR^{d}}}.
\end{equation}
\end{compactitem}
\end{lem}

\subsection{Mn\"ev's universality theorem}\label{sec:universality}

A \defn{primary basic semi-algebraic} set is a subset of $\RR^d$ defined by integer polynomial equations and inequalities \[S=\set{\mathbf{x}\in\RR^d}{f_1(\mathbf{x})=0,\dots,f_k(\mathbf{x})=0, f_{k+1}(\mathbf{x})>0,\dots,f_r(\mathbf{x})>0},\text{  where }f_i\in\ZZ[\mathbf{x}].\]
Realization spaces of polytopes and oriented matroids are examples of primary basic semi-algebraic sets. Mn\"ev's Universality Theorem \cite{Mnev1988} is a reciprocal statement:
every primary basic semi-algebraic set appears as the realization space of some oriented matroid/polytope up to \defn{stable equivalence}, which implies homotopy equivalence
(see~\cite{RichterGebert1998}). We refer to \cite{Mnev1988}\cite{RichterGebert95}\cite{RichterGebert1998} for its proof.

\begin{thm}[{Universality Theorem~\cite{Mnev1988}}]\label{thm:universality}
For every primary basic semi-algebraic set $S$ defined over $\ZZ$ there is a
rank~$3$ oriented matroid whose realization space is stably equivalent to~$S$.
If moreover $S$ is open, then the oriented matroid may be chosen to be uniform.

Given any presentation of $S$, such an oriented matroid of size linear in the
size of the presentation can be found in polynomial time.
In particular, there is such a matroid whose size is linear in the sum of the arithmetic
complexities of the polynomials.

\end{thm}

The \defn{arithmetic complexity} of a polynomial $f\in\ZZ[x]$ is, roughly speaking, the minimal number of operations $+$ and $\times$ needed to compute it from $x$ and $1$, when we are allowed to reuse computations. For example, $(x + 1)^2=(x + 1)(x + 1)$ can be computed with one addition and one multiplication. See \cite{BuergisserClausenShokollahi1997}\cite{Valiant1979} for details.

The following statements are among the consequences of the Universality Theorem:
\begin{cor}\label{cor:omhard}
The {realizability problem for oriented matroids of rank~$3$} is polynomially equivalent to the \emph{existential theory of the reals (ETR)}.
\end{cor}
\begin{cor}\label{cor:omalgebraic}
For every finite field extension $F/\QQ$ of the rationals, there exists an
oriented matroid of rank~$3$ that cannot be realized with coordinates in $F$.
\end{cor}
In other words, that ``all algebraic numbers'' are needed to coordinatize oriented matroids.

config: arxiv


TITLE: Prove using De Moivre's formula,that $\sum\limits_{k=0}^{n}\sin(kx)=\frac{1}{2}\cot(x/2)-\frac{\cos(nx+(x/2))}{2\sin(x/2)}$
QUESTION [2 upvotes]: I've been asked to prove that:
$$
\sum\limits_{k=0}^{n}\sin(kx)=\frac{1}{2}\cot(x/2)-\frac{\cos(nx+(x/2))}{2\sin(x/2)}
$$
When $0<x<2\pi$.
I know there are many similar posts on this site, but using $\cos(kx)$ instead, that's why I created this post, I can't get to the $1/2\cot$ in this one.
Thanks!
$$\begin{align}\sum\limits_{k=0}^{n}\sin(kx) &= \Im\left(\frac{e^{i(n+1)x}-1}{e^{ix}-1}\right)\\ &= \Im\left(\frac{e^{\frac{i(n+1)x}2}}{e^{\frac{ix}2}}\frac{\left(e^{\frac{i(n+1)x}2}-e^{-\frac{i(n+1)x}2}\right)}{\left( e^{\frac{ix}2}-e^{-\frac{ix}2}\right)}\right)\\&= \Im\left(e^{\frac{inx}2}\frac{2i\sin\frac{(n+1)x}2}{2i\sin{\frac{x}2}}\right)\end{align}$$ Applying $e^{iy}-e^{-iy}=2i\sin y,$
$$\begin{align}\sum\limits_{k=0}^{n}\sin(kx) &= \Im\left(\left(\cos\frac{nx}2+i\sin\frac{nx}2\right)\frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}}\right)\\ &= \sin\frac{nx}2 \frac{\sin\frac{(n+1)x}2}{\sin{\frac{x}2}},\end{align}$$ then I don't know what to do.

REPLY [0 votes]: Here is an elementary way to see it. We know that
$$
\sin[x]\sin[y]=\frac{1}{2}(\cos[x-y]-\cos[x+y])
$$
Therefore the sum $A=\sin[x]+\sin[2x]+\sin[3x]$ be be treated by
$$
2\sin[x/2]A=\cos[x-x/2]-\cos[x+x/2]+\cos[2x-x/2]-\cos[2x+x/2]+\cos[3x-x/2]-\cos[3x+x/2]
$$
After cancel out intermediate terms we get
$$
A=\frac{\cos[0.5x]-\cos[3.5x]}{\sin[x/2]}
$$
And your sum can be treated in obvious similar ways.


TITLE: Resolving forces in the $x'$ and $y'$ directions
QUESTION [0 upvotes]: I am trying to get my head around some introductory mechanics, however I need some help.
For this question:
Number 4
It asks me to resolve the $200kN$ force shown acting on the pin into components in the (a) $x$ and $y$ directions, (b) $x’$ and $y$ directions and (c) $x$ and $y’$ directions.
Now, I know for part (a) $R_{x}=200\cos{40}=153.21$ (2 d.p.) and $R_{y}=200\sin{40}=128.56$ (2 d.p.).
However, for parts (b) and (c), isn't it true that $R_{x'}=200\cos(40+30)$ and $R_{y'}=200\sin(40+30)$?
Also, I don't understand why $R_{x}$ and $R_{y}$ change when you are resolving them with respect to $R_{x'}$ and $R_{y'}$ and not each other.

REPLY [1 votes]: No. The problem does not ask you to write $R$ in terms of $R_{x'}$ and $R_{y'}$. Forget everything you did in part a. What the problem asks in part b is to find two forces, one along $x'$ the other along $y$, such as the sum of them is $R$. You can write $R_{x'}$ as made up of a component along $x$: $R_{xx'}=R_{x'}\cos(30)$ and along $y$: $R_{yx'}=-R_{x'}\sin(30)$. Now write $R$ along $x$ and $y$ and you get $$R\cos(40)= R_{x'}\cos(30)$$ and $$R\sin(40)=R_y-R_{x'}\sin(30)$$
Equivalently, you can project the required components along $R$ and the direction perpendicular to $R$