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\begin{document}
  \begin{center}
        {\fontsize{18}{22}\selectfont
       \bf Choreography solutions of the $n$-body problem on $S^2$}
       \end{center}

       \vspace{4mm}

        \begin{center}
        {\bf Juan Manuel S\'anchez-Cerritos$^1$ and Shiqing Zhang$^1$}\\
\bigskip
$^1$Department of Mathematics\\
Sichuan University, Chengdu, People's Republic of China\\
\bigskip
sanchezj01@gmail.com, zhangshiqing@msn.com
       \end{center}

\abstract{We try to prove the existence of choreography solutions for the $n-$body problem on $S^2$}. For the three-body problem, we show the existence of the 8-shape orbit on $S^2$.\\

\noindent {\bf Key words:} celestial mechanics, curved n-body problem, periodic solutions, choreographies.\\

1991 \textit{Mathematics Subject Classification} Primary 70F10, Secondary 70H12

\section{Introduction}

The curved $n-$body problem is a generalization of the Newtonian gravitational problem. It has been studied for several authors, for example in \cite{Diacu4,Diacu5,Diacu6,Diacu3,Diacu,Diacu2,paper1,naranjo,paper,Zhu}. Particularly the interesting history of this problem can be found on \cite{Diacu4, Diacu5}. Here we consider the positive curvature case, i.e. particles moving on the unit sphere, $S^2$.

The motion of the $n$ particles with masses $m_i>0$ and positions $q_i \in S^2$, $i=1\dots,n$,  is described by the following system

\begin{equation}
 m_i\ddot{q}_i=\frac{\partial U}{\partial q_i}-m_i(\dot{q}_i\cdot \dot{q_i})q_i, \ \ \ i=1\dots,n, \label{eq}
\end{equation}
where $U$ is the force function which generalizes the Newtonian one, and it is given by

\begin{equation}
U=\sum_{i<j}m_i m_j\cot(d(q_i,q_j)).
\end{equation}

On classical n-body problems, Chenciner and Montgomery proved the existence of the eight-shape choreography for the three body problem in 2000 \cite{chenciner}, which was described numerically by Moore in 1993 \cite{moore} and Sim\'o in 2000 \cite{Simo}. In the recent years Montenelli and Gushterov computed numerically the analogue solutions in the positive curved space \cite{montanelli}.

The goal of this work is, based on the work of Zhang and co-authors \cite{Zhang1, Zhang2, Zhang3}, to prove the existence of periodic solutions for the $n-$body problem on $S^2$.

Motivated by Sim\'o \cite{Simo} for planar $N-$body problems, in this paper we seek for periodic solutions of (\ref{eq}) moving on the same orbit, i.e., setting the period as $T=1$, we look for solutions such that

\begin{equation}
q_i(t)=Q\left(t+k_i \right), \ \ \ i=1,\dots,n,
\end{equation}
with $\ 0=k_1<\dots<k_n<1$, and for some periodic function $Q:[0,1]\rightarrow S^2$.

Since our problem is on the curved space, the solution is much more complicated than Euclidean space.

We define the following set

\begin{equation}
\begin{split}
D=&\left\lbrace q=(q_1,\dots,q_n) \in (S^2)^n \ | \ q_i  \ \text{is absolutely continuous and} \ q_i(t)\neq q_j(t), \right. \\
 &\left. \text{for} \ \ 1 \leq i\neq j \leq n  \right\rbrace.
\end{split}
\end{equation}

The Lagrangian action associated to system (\ref{eq}) on $D$ is

\begin{equation}
f(q)=\int_0^1\left( \dfrac{1}{2}\sum_1^nm_i|\dot{q}_i(t)|^2+U(q(t)) \right) dt. \label{functional}
\end{equation}

We are interested in showing the existence of new choreography solutions of (\ref{eq}). In other words, we will not only show that the Lagrangian action functional reaches its minimum in $D$, but in a subset where the $n$ particles follow the same orbit.

There are some works where circular choreography solutions have been found, see for instance \cite{Diacu,paper1}. In order to find new families of choreographies we will introduce the following sets

\begin{equation}
\begin{split}
E_1=&\left\lbrace q=(q_1,\dots,q_n) \in D \ | \ q_1(t)=q_n(t+1/n), \  \ q_i(t)=q_{i-1}(t+1/n),\right.\\
&\left. i=2,\dots, n \ \right\rbrace, \nonumber\\
E_2=&\left\lbrace q=(q_1,\dots,q_n) \in D \ | \ q_1(t+1/2)=diag\{1,-1,1 \} q_1(t)\  \right\rbrace, \nonumber\\
E_3=&\left\lbrace q=(q_1,\dots,q_n) \in D \ | \ q_1(-t)=diag\{ -1,-1,1 \} q_1(t)\ \right\rbrace.\nonumber
\end{split}
\end{equation}

It is not difficult to see that $q_1(0)=(0,0,1)=q_1(1/2)$ for $q \in E_2 \cap E_3$ . Hence circular orbits mentioned above do not belong to $E_2 \cap E_3$. The set of choreographies are orbits on

\[  H=\left\lbrace q=(q_1,\dots,q_n) \in D \ | q_1 \in E_1 \cap E_2 \cap E_3\ \right\rbrace.  \]

Let $B=diag\{1,-1,1 \}$ and $C=diag\{-1,-1,1 \}$. We now define the following actions $\Phi_1$, $\Phi_2$ and $\Phi_3$ on $D$

 \[  \Phi_1(q(t))=(q_n(t+1/n),q_1(t+1/n),\cdots, q_{n-1}(t+1/n)), \]
 \[  \Phi_2(q(t))=(Bq_1(t+1/2),Bq_2(t+1/2),Bq_3(t+1/2)), \]
 \[ \Phi_3(q(t))=(Cq_1(-t),Cq_2(-t),Cq_3(-t)) . \]

Then the fixed point of $\Phi_i$ is $E_i$, $i=1,2,3$.

 We refer to Palais' principle of symmetric criticality \cite{palais}, in order to conclude that the critical points of $f$ restricted to $H$ are critical points of $f$ on $D$ as well.

 We state our main theorem as follows

\begin{theorem}\label{theorem1}
Consider the $n-$body problem on $S^2$. The Lagrange action functional (\ref{functional}) reaches its minimum on $H$. This minimum is a periodic non-collision solution of the equations of motion (\ref{eq}).
\end{theorem}

We first show that any critical point of (\ref{functional}) on $D$ satisfies (\ref{eq}).

\begin{prop}
A critical point of the Lagrange action functional on $D$ is a solution of the equations of motion.
\end{prop}

\begin{proof}
Let $q_0=({q_0}_1,\dots,{q_0}_n)$ be a critical point of the Lagrange action functional on $D$.

For a given $q$, a displacement $\delta f$ is given by (the Gateaux derivative)

\begin{equation}
 \delta f=\dfrac{d}{d \varepsilon}\int_0^1\left( \dfrac{1}{2}\sum_1^nm_i|\dot{q}_i(t)+\varepsilon \dot{p}_i|^2+U(q(t)+\varepsilon p(t)) \right) dt\Big|_{\varepsilon=0} , \label{gateaux}
\end{equation}
restricted to any $p=(p_1,\dots,p_n)$ such that $|q_i(t)+\varepsilon p_i(t)|^2=1$, for every $\varepsilon \rightarrow 0$, and $i=1,\dots,n$.
Let $g_i$ be the function defined as $g_i(q_i)=|q_i(t)|^2-1$ (the constraint $g_i(q_i)=0$ maintains the particle $q_i$ on the sphere $S^2$). At a given time, for displacements  of the constraint equation, the following should be held

\begin{equation}
\delta g_i=\dfrac{d}{d \varepsilon}\left( |q_i(t)+\varepsilon p_i(t)|^2-1\right)\Big|_{\varepsilon=0}=0, \ \ i=1,\cdots,n.  \nonumber
\end{equation}

Integrating both sides with respect time we have

\begin{equation}
\delta h_i=\int_0^1 \dfrac{d}{d \varepsilon}\left( |q_i(t)+\varepsilon p_i(t)|^2-1\right)\Big|_{\varepsilon=0}dt=0,  \ \ i=1,\cdots,n.  \nonumber
\end{equation}

From Hamilton principle we have

\begin{equation}
\begin{split}
0=&\delta f+\sum_{i=1}^n\lambda_i \delta h_i\\

=&\dfrac{d}{d \varepsilon}\int_0^1\left( \dfrac{1}{2}\sum_{i=1}^nm_i|\dot{q}_i(t)+\varepsilon \dot{p}_i|^2+U(q(t)+\varepsilon p(t)) +\sum_{i=1}^n \lambda_i g_i \Big|_{q_i=q_{0_i}} \right) dt\Big|_{\varepsilon=0}\\
\end{split}  \nonumber
\end{equation}
where each $\lambda_i$ is the Lagrange multiplier corresponding to the body $i$, it will be computed later in the proof	.

Then we have
\begin{equation}
\begin{split}

0=&\dfrac{d}{d \varepsilon}\int_0^1\left( \dfrac{1}{2}\sum_1^nm_i|\dot{q}_i(t)+\varepsilon \dot{p}_i|^2+\dfrac{1}{2}\sum_{i=1}^n\sum_{j=1,j\neq i}^nm_im_j\cot(d(q_i+\varepsilon p_i,q_j+\varepsilon p_j))\right.\\

\end{split}  \nonumber
\end{equation}
\begin{equation}
\begin{split}

&\left. \left. +\lambda_i( |q_i(t)+\varepsilon p_i(t)|^2-1) \Big|_{q_i=q_{0_i}} \right) \Big|_{\varepsilon=0} dt \right]\\
=&\sum_{i=1}^n \left[ \int_0^1  \dfrac{d}{d \varepsilon}\left( \dfrac{1}{2}m_i |\dot{q}_i(t)+\varepsilon \dot{p}_i|^2 + \right. \right.\\
& \left. \left. \dfrac{1}{2}\sum_{j=1,j\neq i}^n\dfrac{m_im_j\dfrac{(q_i+\varepsilon p_i)\cdot (q_j+\varepsilon p_j)}{\sqrt{(q_i+\varepsilon p_i)\cdot (q_i+\varepsilon p_i)}\sqrt{(q_j+\varepsilon p_j)\cdot (q_j+\varepsilon p_j)}}}{\left(1-\left(\dfrac{(q_i+\varepsilon p_i)\cdot (q_j+\varepsilon p_j)}{\sqrt{(q_i+\varepsilon p_i)\cdot (q_i+\varepsilon p_i)}\sqrt{(q_j+\varepsilon p_j)\cdot (q_j+\varepsilon p_j)}}\right)^2\right)^{1/2}}  \right.  \right.\\
&\left. \left. +\lambda_i( |q_i(t)+\varepsilon p_i(t)|^2-1)  \Big|_{q_i=q_{0_i}}\right) \Big|_{\varepsilon=0} dt \right].\\
\end{split}   \nonumber
\end{equation}

After considering $\varepsilon \rightarrow 0$, and $q_i \cdot q_i=1$ we have
\begin{equation}
\begin{split}
0=&\sum_{i=1}^n \left[ \int_0^1 \left(m_i \dot{q}_i\cdot \dot{p}_i + \dfrac{1}{2}\sum_{j=1,j \neq i}^nm_im_j \dfrac{[q_i \cdot p_j+q_j \cdot p_i]-(q_i\cdot q_j)[q_j \cdot p_j+q_i\cdot p_i]}{(1-(q_i \cdot q_j)^2)^{3/2}} \right. \right.\\
&\left. \left. +2\lambda_i (q_i \cdot p_i) \Big|_{q_i=q_{0_i}} \right) dt \right]\\
=&\sum_{i=1}^n \left[ \int_0^1 \left( m_i\dot{q}_i\cdot \dot{p}_i +\sum_{j=1,j \neq i}^nm_im_j \dfrac{[q_j \cdot p_i]-(q_i\cdot q_j)[ q_i \cdot p_i]}{(1-(q_i \cdot q_j)^2)^{3/2}} +2\lambda_i (q_i \cdot p_i) \Big|_{q_i=q_{0_i}} \right) dt \right]\\
=&\sum_{i=1}^n \left[ \int_0^1 \left( m_i\dot{q}_i\cdot \dot{p}_i +\sum_{j=1,j \neq i}^nm_im_j \left(\dfrac{q_j -(q_i\cdot q_j) q_i}{(1-(q_i \cdot q_j)^2)^{3/2}} \right) \cdot p_i +2\lambda_i (q_i \cdot p_i) \Big|_{q_i=q_{0_i}} \right) dt \right].
\end{split}  \nonumber
\end{equation}

Integrating the first term and considering that the variations vanish at the the end points
\begin{equation}
\begin{split}
0=&\sum_{i=1}^n \left[m_i \dot{q}_i  \cdot p_i \Big|_0^1+ \int_0^1 \left( -m_i\ddot{q}_i\cdot p_i +\sum_{j=1,j \neq i}^nm_im_j \left(\dfrac{q_j -(q_i\cdot q_j) q_i}{(1-(q_i \cdot q_j)^2)^{3/2}} \right) \cdot p_i \right. \right. \\
&\left. \left.+2\lambda_i (q_i \cdot p_i) \Big|_{q_i=q_{0_i}} \right) dt \right]\\
=&\sum_{i=1}^n \left[ \int_0^1 \left( -m_i\ddot{q}_i\cdot p_i +\sum_{j=1,j \neq i}^nm_im_j \left(\dfrac{q_j -(q_i\cdot q_j) q_i}{(1-(q_i \cdot q_j)^2)^{3/2}} \right) \cdot p_i \right. \right. \\
&\left. \left.+2\lambda_i (q_i \cdot p_i) \Big|_{q_i=q_{0_i}} \right) dt \right]\\
=&\sum_{i=1}^n \left[ \int_0^1 \left( -m_i\ddot{q}_i +\sum_{j=1,j \neq i}^n m_im_j\left(\dfrac{q_j -(q_i\cdot q_j) q_i}{(1-(q_i \cdot q_j)^2)^{3/2}} \right) +\lambda_i q_i \right) \cdot p_i \ \Big|_{q_i=q_{0_i}} dt \right]\\
=&\sum_{i=1}^n \left[ \int_0^1 \left( -m_i\ddot{q}_i +\dfrac{\partial U}{\partial q_i} +2\lambda_i q_i \right) \cdot p_i \ \Big|_{q_i=q_{0_i}} dt \right].\\

\end{split}\nonumber
\end{equation}

Since this must hold for any $p=(p_1,\dots, p_n)$ in the interval $(0,1)$, it follows that the critical point  should satisfy

\begin{equation}
-m_i\ddot{q}_i +\dfrac{\partial U}{\partial q_i} -2\lambda_i q_i \Big|_{q_i=q_{0_i}}=0, \ \ \ i=1,\cdots,n, \label{eq4}
\end{equation}
where the multiplier $\lambda_i$ can be computed multiplying the last expression by $q_i$

\begin{equation}
-m_i\ddot{q}_i \cdot q_i +\dfrac{\partial U}{\partial q_i}\cdot q_i -2\lambda_i q_i \cdot q_i \Big|_{q_i=q_{0_i}}=0, \ \ \ i=1,\cdots,n. \nonumber
\end{equation}

Using the fact that the potential is a homogeneous function of degree zero, and that the expression $\ddot{q}_i\cdot q_i=-\dot{q}_i\cdot \dot{q}_i$ holds we have

\begin{equation}
\lambda_i=\dfrac{m_i\dot{q}_i\cdot \dot{q}_i}{2}. \nonumber
\end{equation}

Substituting this expression into (\ref{eq4}), we have

\begin{equation}
-m_i\ddot{q}_i  +\dfrac{\partial U}{\partial q_i} -m_i(\dot{q}_i\cdot \dot{q_i}) q_i \Big|_{q_i=q_{0_i}}=0, \ \ \ i=1,\cdots,n. \nonumber
\end{equation}

Hence, any critical point $q_0$ of the Lagrangian action satisfies the equation of motion.

\end{proof}

\section{Proof of Theorem \ref{theorem1}}
Now we prove that the action functional reaches its minimum on $D$. The proof of the theorem will be a consequence of the following result,

\begin{prop} \cite{Ramm}
A weakly lower semicontinuous from below functional $F(u)$, in a reflexive  Banach space $U$ is bounded from below on any  bounded  weakly  closed  set $M \subset DomF$ and attains its minimum  on $M$ at a point of $M$.
\end{prop}

Our task now is to prove that the functional ($\ref{functional}$) is weakly lower semicontinuous from below and that $D \cup \partial D$ is weakly closed.

\begin{prop}
$f(q)$ is weakly lower semicontinuous from below on $D \cup  \partial D$
\end{prop}

\begin{proof}
Recall that $f$ is called  weakly lower semicontinuous from below if for any  $q^n \in D \cup  \partial D$ such that $q^n \rightarrow q$ weakly, the following inequality holds

\[ \liminf_{n \rightarrow \infty} f(q^n)\geq f(q). \]

If $q \in D$, then there exists $N$ such that for $n>N$, $q^n \in D$. The functions $q^n_i$ are continuous and converges to $q_i$ uniformly.

This implies that $U(q^n_i)\rightarrow U(q_i)$ for $t \in [0,1]$.

By Fatou's lemma we have

\[ \liminf_{n \rightarrow \infty} f(q^n) \geq \int_0^1 \dfrac{1}{2}\sum_1^3|\dot{q}_i(t)|^2+ \int_0^1 \liminf_{n\rightarrow \infty}\left(\sum_{i<j} \cot d(q_i^n,q_j^n)\right)dt=f(q).\]

Now let us suppose that $q_i^n \in \partial D$ and $q_i^n \rightarrow q_i$ weakly.

There exist $t_0 \in [0,1)$ such that $q_{i_0}^n(t_0)=q_{j_0}^n(t_0)$ for $i_0\neq j_0$. Consider the set $C=\{ t \in [0,1) \ | \text{  there exist} \ i_0\neq j_0 \ \text{with} \ q_{i_0}(t)=q_{j_0}(t)  \}$.

Consider the Lebesgue measure, $\mu(C)$, of $C$. Firstly, let us suppose that $\mu(C)=0$. Since $q^n$ converges to $q$ uniformly, then the following holds almost everywhere,

$$\cot(d(q_i^n(t),q_j^n(t)))\rightarrow \cot(d(q_i(t),q_n(t))).$$

This implies, by Fatou's lemma
\begin{equation}
\begin{split}
\int_0^1 \cot(d(q_i(t_0),q_n(t_0)))&=\int_0^1\liminf_{n}\cot(d(q_i^n(t_0),q_j^n(t_0)))\\
&\leq \liminf_{n}  \int_0^1 \cot(d(q_i^n(t_0),q_j^n(t_0))).
\end{split}
\end{equation}

Hence $f(q)\leq \liminf_n f(q^ n)$. Secondly, if $\mu(C)>0$, then

$$\int_0^1 \cot(d(q_i(t),q_j(t)))= +\infty.$$

Additionally we have,

$$\cot(d(q^n_i(t),q^n_j(t)))\rightarrow \cot(d(q_i(t),q_j(t))),$$
uniformly. This implies that

$$\int_0^1 \cot(d(q^n_i(t),q^n_j(t)))\rightarrow +\infty.$$

It follows that $$f(q)\leq \liminf_n f(q^n).$$

\end{proof}

\begin{prop}
$D \cup  \partial D$ is a weakly closed subset of $(W^{1,2}(\mathbb{R}/\mathbb{Z},S^2))^3:=\{(q_1,q_2,q_3) \in (S^2)^3 | \ q_i \in L^2, \ \dot{q}_i \in L^2, \ q_i(t+1)=q_i(t), \ i=1,2,3   \}$
\end{prop}

\begin{proof}

Since $q^n \rightarrow q$ weakly, then $q^n \rightarrow q$ uniformly, then $q \in D \cup \partial D$. Hence $D \cup  \partial D$ is a weakly closed subset of $(W^{1,2}(\mathbb{R}/\mathbb{Z},S^2))^3$.

\end{proof}

\section{Choreography solution for the three-problem on $S^2$}

In order to show a choreography solution for the three-body problem on $S^2$, we will firstly estimate the lower bound of the Lagrangian action for a binary collision generalized solution. We will consider masses equal to 1.

\begin{prop}\label{p}
Consider three bodies on $S^2$. Let $q \in T^*(S^2)^3$ be a periodic binary collision generalized solution, then the Lagrangian action satisfies $f(q)\geq \frac{3}{2}(12\pi)^{2/3}-3$.
\end{prop}

The following lemma will be useful to proof Proposition \ref{p}.

\begin{lemma}\label{cot}
Consider $q_i$ and $q_j$ on $S^2$ satisfying equations of motion (\ref{eq}), then

$$\dfrac{1}{r_{ij}}-1<\cot(d(q_i,q_j))<\dfrac{1}{r_{ij}},$$
where $r_{ij}$ is the Euclidean distance  between $q_i$ and $q_j$.
\end{lemma}

\begin{proof}
For this proof we will consider the origin of the system at the north pole  of the unit sphere, i.e., at $R=(0,0,1)$. The equations of motion takes the form

\begin{equation}
\ddot{q}_i=\sum_{i=1,j\neq i}^{n}\dfrac{q_j-\left(1-\frac{r_{ij}^2}{2}\right)q_i+\frac{r_{ij}^2R}{2}}{r_{ij}^2\left(1-\frac{r_{ij}^2}{4}\right)^{3/2}}-(\dot{q}_i\cdot \dot{q}_i)(q_i+R).
\end{equation}

The potential energy in $S^2$ is given by

\begin{equation}
\begin{split}
U=\sum_{i<j}\cot(d(q_i,q_j))&=\sum_{i<j}\dfrac{1-\frac{r_{ij}^2}{2}}{r_{ij}\left(1-\frac{r_{ij}^2}{4} \right)^{1/2}},
\end{split}
\end{equation}
for more details about the equations of motion and potential energy written in this coordinates, please see \cite{Diacu5}.

Consider $n=2$, then

\begin{equation}
\begin{split}
\cot(d(q_i,q_j))&=\dfrac{1-\frac{r_{ij}^2}{2}}{r_{ij}\left(1-\frac{r_{ij}^2}{4} \right)^{1/2}}> \dfrac{1-\frac{r_{ij}^2}{2}}{r_{ij}}=\left(\dfrac{1}{r_{ij}}-\dfrac{r_{ij}}{2}\right)> \dfrac{1}{r_{ij}}-1.
\end{split}
\end{equation}

On the other hand, we have

\begin{equation}
\begin{split}
\cot(d(q_i,q_j))&=\dfrac{1-\frac{r_{ij}^2}{2}}{r_{ij}\left(1-\frac{r_{ij}^2}{4} \right)^{1/2}}< \dfrac{1-\frac{r_{ij}^2}{4}}{r_{ij}\left(1-\frac{r_{ij}^2}{4} \right)^{1/2}}=\dfrac{\left(1-\frac{r_{ij}^2}{4} \right)^{1/2}}{r_{ij}}< \dfrac{1}{r_{ij}}.
\end{split}
\end{equation}

Hence we conclude the proof of the lemma.
\end{proof}

Now we can proceed with the proof of Proposition \ref{p}.

\begin{proof}

Consider three point particles $q_1, q_2, q_3 \in S^2$ with masses $m_1=m_2=m_3=1$ satisfying the equations of motion (\ref{eq}), and suppose that the particles $q_1$ and $q_2$ collide, without loss of generality, at the north pole.

The Lagrangian action is given by

$$f(q)=\int_0^1\left(\dfrac{1}{2}\sum_{i=1}^3 |\dot{q}_i^2|+\sum_{1\leq i<j\leq 3}\cot(d(q_i,q_j)) \right)dt,$$
where the constrains $|q_i|^2=1$ and $q_i\cdot \dot{q}_i=0$, $i=1,2,3$, hold.

Notice that \cite{Zhang1,Zhang2} $$\sum_{1\leq i<j\leq 3}|\dot{q}_i-\dot{q}_j|^2+\Bigl| \sum_{i=1}^3\dot{q}_i \Bigr|^2=3\sum_{i=1}^3|\dot{q}_i|^2.$$
We have

\begin{equation}
\begin{split}
f(q)=&\int_0^1\left(\sum_{k=1}^3\dfrac{1}{2}|\dot{q}_k|^2+\sum_{1\leq i<j\leq 3}\cot(d(q_i,q_j)) \right)dt\\
\geq&\int_0^1\left(\sum_{1\leq i<j\leq 3}\dfrac{1}{6}|\dot{q}_i-\dot{q}_j|^2+\sum_{1\leq i<j\leq 3}\cot(d(q_i,q_j)) \right)dt \\
\geq&\int_0^1\left(\sum_{1\leq i<j\leq 3}\dfrac{1}{6}|\dot{q}_i-\dot{q}_j|^2+\sum_{1\leq i<j\leq 3}\dfrac{1}{r_{ij}}-3 \right)dt \ \text{(by Lemma \ref{cot})}\\
=&\dfrac{1}{3}\int_0^1\left(\sum_{1\leq i<j\leq 3}\dfrac{1}{2}|\dot{q}_i-\dot{q}_j|^2+\sum_{1\leq i<j\leq 3}\dfrac{3}{r_{ij}} \right)dt-3.\\
\end{split}
\end{equation}

If $q_1(t_0)=q_2(t_0)$, then $q_1(t_0+1/2)=q_2(t_0+1/2)$. Then using some estimates of \cite{Zhang1,Zhang2} we have

\begin{equation}
\begin{split}
\dfrac{1}{3}\int_0^1  \left( \dfrac{1}{2}|\dot{q}_1-\dot{q}_2|^2+\dfrac{3}{r_{12}} \right)dt&=\dfrac{2}{3}\int_0^{1/2}  \left( \dfrac{1}{2}|\dot{q}_1-\dot{q}_2|^2+\dfrac{3}{r_{12}} \right)dt\\
&=\dfrac{(12\pi)^{2/3}}{2}. \nonumber
\end{split}
\end{equation}

Since $q_1(t)=q_3(t+1/3)=q_2(t+2/3)$, then

\begin{equation}
f(q)\geq \dfrac{3(12\pi)^{2/3}}{2}-3.
\end{equation}

\end{proof}

\begin{prop}
$f^{-1}((0,\frac{3}{2}(12 \pi )^{2/3}-3)) \neq \emptyset$
\end{prop}

\begin{proof}
Consider the test loop

\begin{equation}
q_1(t)=(x(t),y(t),z(t)), \ \ q_2(t)=q_1(t+1/3), \ \ q_3=q_1(t+2/3),\label{eq3}
\end{equation}
 where

\begin{equation}
\begin{split}
x(t)=&0.15 \sin(4 \pi t),\\
y(t)=&0.2275 \sin(2 \pi t),\\
z(t)=&\sqrt{1-x^2(t)-y^2(t)}. \nonumber
\end{split}
\end{equation}

In \cite{Zhang1} the authors show that if $\sin(2 \pi t)=\sin(2 \pi (t+\frac{i-1}{3}))$, then $\sin(4 \pi t)\neq \sin(4 \pi (t+\frac{i-1}{3}))$, for $t \in (0,1)$, $i=2,3$. Hence $q_i(t)\neq q_j(t)$, $i \neq j$.

With the expressions (\ref{eq3}), we have  $f(q)\approx 13.76572<\frac{3}{2}(12 \pi )^{2/3}-3\approx 13.8647$.

\end{proof}

\subsection*{Acknowledgements} The first author has been partially supported by {\it The 2017's Plan of Foreign Cultural and Educational Experts Recruitment for the Universities Under the Direct Supervision of Ministry of Education of China} (Grant WQ2017SCDX045).

config: arxiv


TITLE: Brownian motion property
QUESTION [0 upvotes]: I was reading a book and I came across this part which says that a Brownian motion $B(t)$ has this property:
 $P\{B(1) >c \}> 0$ for any real number $c$.
I know that $B(0) = 0$ but I don't understand why the above statement is true? how we are sure about that ?
Thanks for your help.

REPLY [0 votes]: We know that $B(1) \sim N(0,1)$. The support of the normal density is all $\mathbb R$ so there is a positive probability of $B(1)$ passing any threshold


text:
\begin{document}

\maketitle

\begin{abstract}

Topology based analysis of time-series data from dynamical systems is
powerful: it potentially allows for computer-based proofs of the
existence of various classes of regular and chaotic invariant sets for
high-dimensional dynamics.  Standard methods are based on a cubical
discretization of the dynamics and use the time series to construct
an outer approximation of the underlying dynamical system. The
resulting multivalued map can be used to compute the Conley index of
isolated invariant sets of cubes. In this paper we introduce a
discretization that uses instead a simplicial complex
constructed from a witness-landmark relationship. The goal is to
obtain a natural discretization that is more tightly connected with
the invariant density of the time series itself. The time-ordering of
the data also directly leads to a map on this simplicial complex that
we call the witness map. We obtain conditions under which this witness
map gives an outer approximation of the dynamics, and thus can be used
to compute the Conley index of isolated invariant sets. The method is
illustrated by a simple example using data from the classical H\'enon
map.

\end{abstract}

\bigskip
\input{intro}
\input{multivalued-maps}
\input{witness-map}
\input{example}
\input{conc}

\section*{Acknowledgment}
This material is based upon work supported by the National Science
Foundation under Grants \#CMMI-1245947, \#CNS-0720692, and
\#DMS-1211350.  Any opinions, findings, and conclusions or
recommendations expressed in this material are those of the authors
and do not necessarily reflect the views of the National Science
Foundation.

The authors would like to thank the referees for extensive comments
that improved the paper. They also thank Robert Ghrist for pointing
out the Dowker reference and Konstantin Mischaikow and Robert Easton
for useful conversations.

\input{appendices}

\bibliographystyle{alpha}
\bibliography{Witnessbib}

\end{document}


TITLE: If function has positve integral on domain then there exist interval and positve number such that function is bounded below by that number on Interval
QUESTION [1 upvotes]: Suppose $\int_a^b f$ exist and positive Prove that there exist interval [c,d]  and m>0 such that $f(x)\geq m$.

I was thinking to prove by contradiction . Suppose there is no interval with above property
If we have partition then in any partion our function take negative value.
I donot know how to argue then
Please can anyone suggest me how to prove?

REPLY [2 votes]: If the Riemann integral $\int_a^b f(x)\; dx  > 0$, there is a partition $P: a = x_0 < x_1 < \ldots < x_n = b$ so that the lower Darboux sum
$\sum_{i=1}^n \inf_{t \in [x_{i-1},x_i]} f(t) (x_i - x_{i-1}) > 0$.
For this to be positive, at least one of the summands must be positive, and thus we have an interval $[x_{i-1}, x_i]$ such that $\inf_{t \in [x_{i-1},x_i]} f(t) > 0$.  Take
$m = \inf_{t \in [x_{i-1},x_i]} f(t)$.


TITLE: What is the meaning of ∀x∃x?
QUESTION [2 upvotes]: Say we are in first-order logic, $x$ is a variable and its values are from a set of values, doesn't matter what.
$∀x∃x$ means for every $x$ (from the set of values) there exist at least one x (from the set of values). Since we took all $x$s from the set, of course each one itself does exist for itself because $x$ is $x$. Hence this should actually shorten to $∀x$ am I right?
The same reasoning for $∃x∀x$: there exist at least one $x*$ such that every $x$ from those $*$ is $x$. Well of course, since $x$ is $x$. So it shortens to $∃x$. Am I correct?
Conclusion
Thank you for all the answers. I was wrong, it is the other way around. I think the gist of it is that the innermost quantifier matters only when quantifying the same variable for the same formula, and the outer quantifiers (of the same variable & formula) become null quantifiers since there is nothing to quantify anymore. This is, analog to computer programs, similar to variable shadowing, which is when you over-declare/define a variable, here we are over-defining quantifiers in an expression like the above.

REPLY [0 votes]: A more succinct explanation: it's obvious that $\exists x P(x)$ is equivalent to $\exists y P(y)$. You can't change the truth value of a statement merely by renaming some entities.
So $\forall x [\exists x P(x)]$ should be equivalent to $\forall x [\exists y P(y)]$ - if you're queasy about this, step through it explicitly by setting $\phi := \exists x P(x)$.
But that is manifestly equivalent to $\exists y P(y)$ (assuming as usual that the domain of discourse is not empty), which is just $\exists x P(x)$.


TITLE: How to compare dispersion of data?
QUESTION [0 upvotes]: From my statistic book, I learn that standard deviation is used to estimate how much the data spread around the mean value. If I have two or more sets of data, for each one, they normalized to the same number (I mean all data add up to the same value). Is it a good way to compare how much the data spread-out around the center by calculating the standard deviation? For example, I have the following two data set
A = [0 0 0 0 0 15 35 15 0 0 0 0 0];
B = [6 5 8 6 5 4 3 4 5 8 5 6];

The plot of them look like

Now I estimate the standard deviation separately on those two data, I got
STD(A) = 10.6066
STD(B) = 1.5050

But that just against what I though that A should has less STD for it is not widely spread as B does. So my question is if STD only works for normal distribution or it could be used for any data? Secondly, if it works for any samples, why A will have STD higher than B?
p.s. for second question, I understand why it gives higher value for A from the definition of STD but I wonder why it should go like that because it is pretty localized.

REPLY [0 votes]: Bayes Theorem can help you out of this dilemma, but we do not even have to go that far to understand what is going on.
You can calculate the standard deviation for any two datasets and compare them. But it depends on what you want to know and what your assumptions are about the data. For example if you want to know what you can measure more exactly - the size of a human or the size of a human cell, you would copare the relative standard deviation. If you want to know what measures lengths more exactly, a microscope or a measuring stick, you compare the absolute standard deviation.
In a related matter it depends on how you interpret your dataset, the way you ask your second question and the way you drew the datasets suggests you think of them as histograms. Then you have to estimate the standard deviation and the mean in a different way and you get the answer you were expecting. If you interpret the data as a list of points then obviously dataset A has the higher std as 15 is far away from 35 and 5, 6, 8 etc. are not that far away from each other.


TITLE: Applied PDE's book suggestion
QUESTION [1 upvotes]: I am searching for a book on applied PDEs which do not focus on how to solve them, but instead different areas where they appear and what can we do with them. Maybe a book in which each chapter shows applications of pdes in areas like economics, finance, physics, biology, social sciences, etc... .I also would be nice if it contains some numerical ways to use them. Can you suggest one book like that to me?

REPLY [0 votes]: It depends on which level you are aiming for. For undergraduate level, the book "Partial Differential Equations An Introduction" by Walter A. Strauss is decent (but it does not contain too much material on numerical PDE methods). For graduate level PDE, the classical book "Partial Equation Equations" by L.C.Evans seems quite popular.


TITLE: Which is positive? which is negative? about laws thermodynamics...from problem solve
QUESTION [0 upvotes]: 1) A hot block of iron does 50 kJ of work on a cold floor. The block of iron also transfers 20 kJ of heat energy to the air. Calculate the change in energy (in kJ) of the iron block.
2) A chef vigorously stirs a pot of cold water and does 150 J of work on the water. The water also gains 75 J of thermal energy from the surroundings. Calculate the change in energy of the water.
3) A scientist very carefully does mechanical work on a container of liquid sodium. The liquid sodium loses 300 J of energy to its surroundings but gains 250 J of energy overall. How much work did the scientist do?

Hi, I am so confused about this problem to solve, because of which one for negative and which one for positive of the Heat and Work. Usually, this problem is about internal energy, $\Delta U$. like why the heat is negative, why the heat is positive? and also the 'work' has negative or positive. Please, can you explain how it's work? and... can you please give me some example or some pictures, so that I can easy to understand. :)
formula of internal energy, $\Delta U$ = $Q$ - $W$
well I did saw the answers from questions, but I didn't know how it works and why is negative or positive? here the answer:
1) $ΔU = Q − W  = −20 − (+50)  = −70 kJ$
2) $ΔU = Q − W  = 75 − (−150)  = 225 J$
3) $ΔU = Q − W 250 = −300 − W, W = −550 J$ The scientist does 550 J of work on the sodium.

REPLY [1 votes]: The first law of thermodynamics is an energy balance, which is a restatement of the law of conservation of energy.  Given that the symbol for energy is "E", and using the standard symbol for internal energy of "U", all energy entering a system adds to internal energy and all energy leaving a system subtracts from internal energy.  In equation form, this is:
$\Delta U = E_{in} - E_{out}$
Note that work and heat are both forms of energy, and under normal conditions, work that is added to a process ends up as heat through dissipative processes (e.g., friction).  Because work and heat are equivalent from an energy standpoint, the first law of thermodynamics can easily be written for each individual process, as detailed below:
1) For a process where shaft work is being used to pump a fluid, work is entering the system.  Assume that for some reason, the fluid also has to be heated, so heat is also entering the system.  For this process, the equation that applies would be
$\Delta U = W_{in} + Q_{in}$
2) For a refrigeration system, shaft work is used to compress a refrigerant, heat is rejected to the environment to condense the refrigerant, the refrigerant pressure is lowered via an expansion valve, and low temperature heat is absorbed by the refrigerant in order to keep something cool.  This means that there are two heat terms in the first law for this system, resulting in the following equation:
$\Delta U = W_{in} + Q_{in} - Q_{out}$
Since this is a cyclic process, $\Delta U = 0$, so the amount of heat leaving this system must be greater than the amount of heat absorbed by the refrigerant because the work done by the compressor turned into heat that must be rejected to the environment.
The above examples illustrate one way to construct the first law of thermodynamics that minimizes the confusion that can result from deciding whether work is positive or negative.  In effect, since work and heat are equivalent from a thermodynamic standpoint, all energy terms are arbitrarily assigned positive values.  Thereafter, any energy entering a system is added to internal energy and any energy leaving a system is subtracted from internal energy.
Using this method for problem 3 in the posted question, the container of liquid sodium is the system, and that container transferred 300 J of heat to the environment.  Even with this heat transfer to the environment, the system gained 250 J of internal energy.  This means that the problem works as follows:
$\Delta U = E_{in} - E_{out}$
It is obvious that heat left the system, but it is not obvious whether work entered or left the system.  Assume that work entered the system.  When the first law equation is solved for work, if the work term is positive, this assumption was correct.  If the work term is negative, then the assumption was incorrect and work left the system.  Thus, the following solution applies:
$\Delta U = W_{in} - Q_{out}$
$\Delta U + Q_{out} = W_{in}$
250 J + 300 J = 550 J = $W_{in}$
Thus, the work put into the system was 550 J.


TITLE: Prove that $C_{A_7}((567))\cong H \times A_4$
QUESTION [2 upvotes]: Let $H=(<567>)\subset{A_7}$. And let $C=C_{A_7}((567))$ denote the centralizer of $(567)$ in $A_7$.
Prove that $C_{A_7}((567))\cong H \times A_4$
I'm fairly certain that I can use the First Isomorphic Theorem (FIT) here if I can create/find a map between the two groups. I know there are 3 elements in $H$ and there are 24 elements in $A_4$. I know that all the elements in $A4$ are going to be disjoined with (567), and so will commute with it.
I'm trying to picture what the centralizer looks like. I know that 3-cycles exist in $A_n$. In this case we'd need elements that commute with (567). So if I'm thinking clearly, we could create a 3-cycle with the elements $\{1,2,3,4\}$. So there are 4 choices for the first spot in the 3-cycle, 3 choices for spot 2 and 2 choices for spot 3. So there should be 24 possible choices for the 3-cycles in this centralizer. Similarity, we can consider double transpositions, $(ab)(cd)$. Using the same line of thought, 4 choices for a, 3 choices for b, etc. We conclude that there would be 24 double transpositions. I know I am missing the identity as well, but I cannot think of what the other elements look like.
Meanwhile, considering $H \times A_4$, we can clearly see 36 elements. Namely
$$(\epsilon ,A_4) \\
((567) ,A_4)\\
((567)^{-1} ,A_4)
$$
If I can create a map where it sends types to a certain set, ie 3-cycles of the centralizer could be mapped to $((567) ,A_4)$, I might be able to define an isomorphism with a trivial kernel to use the FIT.
Edit: Much of my counting here is just wrong.

REPLY [2 votes]: OP's counting argument is not correct. You can count the elements of a given cycle structure as Jyrki has or like this:
In Jyrki's answer you want to form a $3$-cycle $(a\ b \ c)$ from $\{1,2,3,4,5,6,7\}$. Start like you did in the question: you have $7$ choices for $a$, $6$ for $b$ and $5$ for $c$. But remember that $(a \ b \ c) = (b \ c \ a) = (c \ a \ b)$, and hence you must divide the number by $3$. Thus there are $\frac{7 \cdot 6 \cdot 5}{3} = 70$ $3$-cycles.


TITLE: Particle current operator in general vs Particle current operator for tight binding Hamiltonian
QUESTION [4 upvotes]: I am referring Mahan Many-Particle Physics. There are 2 particle current operators -one in general and one for the tight binding Hamiltonian. How do we go from the general current operator (1.195 in Mahan)
$ j_{i}(\mathbf{r})=\frac{1}{2mi}\left(\psi^{\dagger}(\mathbf{r})\nabla\psi(\mathbf{r})-\psi(\mathbf{r})\nabla\psi^{\dagger}(\mathbf{r})\right)
$
[whose Fourier transform is
$\int d^{3}r\,\,\exp(-i\mathbf{q}.\mathbf{r})\,\, j_{i}(\mathbf{r}) = \frac{1}{2mi}\sum_{\sigma\beta}c_{\sigma}^{\dagger}c_{\beta}\int d^{3}r\,\,\exp(-i\mathbf{q}.\mathbf{r})\,\left(\phi_{\alpha}^{\star}(\mathbf{r})\nabla\phi_{\beta}(\mathbf{r})-\phi_{\beta}(\mathbf{r})\nabla\phi_{\alpha}^{\star}(\mathbf{r})\right)$
giving                                                                  $j_{l}(q) = \frac{1}{m}\sum_{k\alpha}\left(k+\frac{q}{2}\right)c_{k+q,\sigma}^{\dagger}c_{k,\sigma}
 $]
to (1.204) current operator $
 j=-iw\sum_{j\delta\sigma}\delta c_{j+\delta,\sigma}^{\dagger}c_{j\sigma}$
 for the Tight Binding Hamiltonian $H=w\sum_{j\delta\sigma}c_{j+\delta,\sigma}^{\dagger}c_{j,\sigma}+\frac{1}{2}\sum_{ij,ss^{\prime}}n_{is}n_{js^{\prime}}V_{ij}$?  (here $\phi(r)$
  are plane waves.)  I mean we use an alternative definition of current in terms of polarization to get to the form of current operator in real space for the tight binding Hamiltonian but on fourier transforming this operator, it should return us a form similar to the current operator generally defined in Fourier space-- I don't see how we will get the mass m of $j_{l}(q)$ on taking the fourier transform of $ j=-iw\sum_{j\delta\sigma}\delta c_{j+\delta,\sigma}^{\dagger}c_{j\sigma}$

REPLY [1 votes]: The fermions in the tight-binding model are not the same ones as in the "general" model, and therefore there should not be an equality between the two current operators.
The tight-binding model is an approximation that allows one to take into account much of the effect of the lattice ions. In general there are overlaps between the atomic wavefunctions from which the tight binding fermions are constructed.
The current operator given by Mahan is consistent with the tight-binding Hamiltonian, and should be used whenever this approximation is used. The "bare" current operator will not satisfy particle conservation within this model.
This kind of thing often repeats itself in physics: you pick the relevant, low-energy effective model that describes your system (in this case, the tight-binding Hamiltonian and the operators' commutation relations) and derive other quantities from there.


TITLE: Magnetic quantum numbers - axes correspondence
QUESTION [0 upvotes]: We know that the magnetic quantum number describes the space orientation of an orbital within an atom. For the $p$-orbital, the magnetic quantum numbers can be -1,0,1 (one for every axis). We have three axes $x$, $y$, $z$ and three quantum numbers (for the $p$-orbital) -1,0,1. How are these two groups related to each other? Does -1 correspond with the $x$-axis, 0 with the $y$-axis and 1 with the $z$-axis?
What about more complicated orbitals having more possible space orientations? How are the axes and magnetic quantum numbers related?

REPLY [0 votes]: The $p$ orbitals that you refer to are actually certain linear combinations of the ones we usually use with the magnetic quantum number, i.e., the wave=functions are of the form
$$\begin{eqnarray}p_z &=& p_0\\
p_x &=& \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right)\\
p_y &=& \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} \right)\\\end{eqnarray}$$
where the subscripts correspond to $m$. The picture is not so nice for higher values of $\ell$, but it depends on the symmetries of the spherical harmonic functions $Y_{\ell m}(\theta,\varphi)$, which form representations of the rotation group $SO(3)$. We can always pick a different set of axes, and find a different set of spherical harmonic functions as linear combinations of the old ones.
I would like to try to answer your second question. In general, an atom is in a superposition (linear combination) of these orbital states, so one doesn't normally picture an atom with it's orbitals pointing in a certain direction, even though they are often drawn that way. However, we can try to measure the angular momentum by placing the atom in a magnetic field. Then the different orbitals correspond to different energies, and we can put the atom in a particular state depending on the orientation of the magnetic field (e.g. in a Stern-Gerlach type experiment).


TITLE: derivatives and constants
QUESTION [2 upvotes]: I am confused on finding the derivative when constants are involved:
with $3\ln(x^4 + \sec x)$
the derivative is
$(3(4x^3 + \sec x \tan x))/(x^4+\sec x)$
notice the three stayed, but with
$x^2+2x+7$
the derivative of seven turns to zero in the answer
$2(x+1)$
my question is what is the difference.

REPLY [0 votes]: As 3 is constant with in multiplication with function therefore 3 can taken outside of differentiation by the rule d/dx(kf(x)) = kd/dx(f(x)) while in second example 7 is constant therefore its differentiation is zero due to rule d/dx(C) = 0 where c = constant


TITLE: On trying to achieve control and speed of a mathematical topic
QUESTION [1 upvotes]: I've spent most of my studies focused on learning proofs of theorems, as that was what took the most time to learn well enough in order to do well at exams. After seeing how some mathematicians I know are able to solve some problems with ease, I've started trying to improve my problem solving ability and general control of a topic.
The first goal I set for myself was to go through Isaac's Finite Group Theory, and do all the exercises. But I'm not sure it's working well enough for my goal - after doing the exercises in a few of the sub-chapters, I still feel like if I were to given a similar type of problems as the ones I did, I would probably not be much faster than I was when I did the first set.
My aim isn't to just somewhat understand the topic, and feel like I can do the exercises - I want it to become something relatively easy, that doesn't take effort, that's almost 'intuitive' in some sense, and that I can do it without much focus.
But I don't know whether it's a good idea to aim for this so directly - maybe it will just (hopefully) come with time as I improve generally and get a slight understanding of some topic... but I have considered that maybe I should just do tons of exercises of a given type, until it becomes easy and fast, and then move on, rather than moving on after I do the exercises at the end of a chapter.
Also, I do ask my own questions concerning the topics, the exercises, and play around with it a little bit, but only as much as I feel like, as I enjoy it more to just move to the next exercise - which might not be enough.
I would appreciate opinions on whether trying something like this (very large amount of exercises of 'similar' difficulty and topic, and possibly even repetitive practice of a certain type exercises) is a good idea - and whether even directly trying to get control of a certain topic is a good goal, etc.

REPLY [2 votes]: Different people learn differently, so take the responses to your question, even this one, in that light.

Ramanujan learned math by reading a book that was just a list of Theorems, which he set about proving, and then went on to prove his own.
Many people think that solving problems is the best way to learn mathematics.
Whether you should try to do every problem in a text book depends on the number and quality of the problems. If there a lot or they seem repetitious, you could leave half or more for future review. Even if there are only a few, the author might sneak in a very hard one that you could skip. Another problem with text book exercises is that they typically focus on what you have just learned, so you are focused on that material.
More important are the problems you set for yourself as you are reading and trying to understand what the author is saying and its implications. Making up your own examples is a good exercise too. Some of the best math books are Do It Yourself books; they come with no problems at all.
I recommend reading the 10 page essay "How to Learn Math" which is chapter 7 of Ian Stewart's book "letters to a young mathematician."
I strongly recommend against spending time doing problems over again, esp. text book exercises. What happens is that you tend to learn the problem, not the math.
But the best advice to improve your proficiency, intuition and speed, is as the first commenter said: PRACTICE, PRACTICE, PRACTICE, not just solving problems but talking about the subject, reading other peoples solutions, reading published papers, and books.


TITLE: Use Euler’s formula to write an exponential function in the form $a + ib$
QUESTION [0 upvotes]: Euler's formula is $e^{it} = \cos(t) + i \sin(t)$.

$$2^{1-i} = 2\left(2^{-i}\right)=2\cos(-1) + 2\ i\sin(-1)$$
  $$\pi^{-1+2i} = \pi^{-1}\left(\pi^{2i}\right)=\pi^{-1}\left(\cos 2+i\sin 2\right)=\frac{\cos 2}{\pi}+i\ \frac \sin 2 \pi$$

Is the above the correct way to write $2^{1-i}$ and $\pi^{-1+2i}$ in the form $a + ib$? I wasn't sure if I could change Euler's formula to $n^{it} = \cos(t) + i \sin(t)$, where $n$ is a constant.

REPLY [1 votes]: No. It is not true that $n^{it} = \cos(t) + i \sin(t)$. Notice that
$$
1^{it} = 1 \neq \cos(t) + i \sin(t),
$$
for example consider this at $t=0$.
As a hint for figuring this out, notice that
$$
x^{a+ib} = e^{\ln(x^{a+ib})}
$$
then recall your rules for logarithms to get this to the form $e^{(a+ib)\ln(x)}$.


TITLE: Maximum Velocity of Parabolic Balloon
QUESTION [0 upvotes]: We had this exercise in class, which went something like:

A balloon's height is described by the function $$ h(t)=\frac{1}{5} t
  \cdot (t-5)^2 \space (0 \leq t \leq 5) $$ $t$ denotes the passed time in
  hours, $h(t)$ the height in kilometers.

At what time is the balloon's rising velocity at maximum? What is its value?
When does the balloon drop the fastest? At what height does this happen?

We all know that $v = \frac{s}{t}$. $s$ is the traveled distance, so $s = h(t)$, at least until the balloon hits its maximum height. $v = \frac{s}{t}$ cannot describe velocity continuously, but if we let $s = h(t)$ and plug it into $v$'s equation, we got $\frac{h(t)}{t}$, from where we can get to $\lim_{\epsilon \to 0}\frac{h(t + \epsilon) - h(t)}{t}$, which is the very (or at least one) definition of differentiation. Hence, $h(t)$'s derivative $h'(t)$ should give me the the velocity $v$ at any time, so $v(t) = h'(t)$.
Long story short, I tried it but failed. I got
$$
h'(t) = \frac{3}{5}t^2 - 4t + 5 = v(t)
$$
and to get the maximum velocity required in task 1, I need to set the derivative of $v(t)$ to $0$:
$$
v'(t) = \frac{6}{5}t - 4 = 0 \iff t = \frac{10}{3}
$$
Plugging this into $v(t)$ now gives me:
$$
v(\frac{10}{3}) = \frac{3}{5}(\frac{10}{3})^2 - 4(\frac{10}{3}) + 5 = \frac{25}{3}
$$
which is wrong. The solutions says that the highest rising velocity is at $t = 0$. (It should actually be solved graphically, but that seemed so unmathematical and vague to me I didn't think of it in the first place.)
I think it doesn't work because $v''(t) = \frac{6}{5} > 0$, so there is only a minimum and we can answer task 2, but not task 1.
However, why is this? I mean, it is right conceptually, isn't it? Why don't I get a maximum at all, there must be one, right?

REPLY [1 votes]: When we set the derivative of a differentiable function equal to $0$ we are finding a so called "critical point". An absolute maximum for a differentiable function occurs either at a critical point (which you found by setting the derivative equal to $0$), OR it can occur at one of the boundary points, in this case $t=0$ or $t=5$. Therefore, by finding the critical points and the boundary points, plugging those into the velocity function, and finding whichever is largest, we can determine the maximum velocity. This is because the maximum MUST occur at one of these points. In your case, try the critical point you found ($t=\frac{10}{3}$), $t=0$, and $t=5$. Whichever gives the highest velocity is a maximum.
By the way, using graphs is not "unmathematical" at all, but I understand that you want to find it "analytically".


TITLE: Does the series $\sum_{k=1}^{\infty}(\arctan(\frac{1}{k}))^{2}$ converge or diverge?
QUESTION [0 upvotes]: I'm trying to figure out whether the following series converges or diverges:
$$\sum_{k=1}^{\infty}(\arctan(\frac{1}{k}))^{2}$$
My first thought was to compare $\sum_{k=1}^{\infty}(\arctan(\frac{1}{k}))^{2}$ with $\sum_{k=1}^{\infty}\arctan(\frac{1}{k})$ and see if the later converges, which would implicate that $\sum_{k=1}^{\infty}(\arctan(\frac{1}{k}))^{2}$ converges. This because $(\arctan(\frac{1}{k}))^{2}\leq\arctan(\frac{1}{k})$ for $k\geq1$. That was not the case.
I've also tried the quotient test and the root test but didn't see any success.
Can I just use that $\arctan(\frac{1}{k})\approx\frac{1}{k}$ as $k\rightarrow\infty$ and then prove that it converges by proving that $\sum_{k=1}^{\infty}(\frac{1}{k})^{2}$ converges? Is that valid? I guess it is, because we really just care about what happens at infinity, right?

REPLY [4 votes]: Yes, you can compare $\left(\arctan^2\left(\frac1k\right)\right)_{k\in\Bbb N}$ with $\left(\frac1{k^2}\right)_{k\in\Bbb N}$. And you can do it using the fact that\begin{align}\lim_{k\to\infty}\frac{\arctan^2\left(\frac1k\right)}{\frac1{k^2}}&=\lim_{k\to\infty}\left(\frac{\arctan\left(\frac1k\right)}{\frac1k}\right)^2\\&=\left(\lim_{k\to\infty}\frac{\arctan\left(\frac1k\right)}{\frac1k}\right)^2\\&=1.\end{align}So, by the comparison test, and since $\sum_{k=1}^\infty\frac1{k^2}$ converges and both series are series of numbers greater than $0$, the series$$\sum_{k=1}^\infty\arctan^2\left(\frac1k\right)$$converges.


TITLE: planar curve has infinitely many Bertrand mates
QUESTION [0 upvotes]: Show that a planar curve has infinitely many Bertrand mates.
Can anyone give hints or direction on how to prove this? Thanks

REPLY [0 votes]: A parallel Bertrand curve is obtained by erecting a normal of constant length at each point on a curve C. Any smooth continuous curve C is a set of infinite connected points, so correspondingly there are infinitely many curved parallels.