TITLE: Let X be a topological space such that A $ \in $ X and A' $ \subset $ P(X) and A' $\neq$ $\emptyset$. Prove or disprove the statements below.
QUESTION [0 upvotes]: Let X be a topological space such that A $ \subset $ X and A' $ \subset $ P(X). Prove or disprove the statements below.
int($\bigcup_{A \in A' }$ A)   $\supset$ $\bigcup_{A \in A'}$int(A)
int($\bigcap_{A \in A'}$A)    = $\bigcap_{A \in A'}$int(A)
$\overline {\bigcup_{A \in A' }A}$   =  $\bigcup_{A \in A'} \bar A$
$\overline {\bigcap_{A \in A' }A}$ $\subset$ $\bigcap_{A \in A'} \bar A$
====================================================================
As we know that the following statements are true, then all of the above statements are also true? Right?
int(A $\cup$ B) $\supset$ int(A) $\cup$ int(B)
int(A $\cap$ B) = int(A) $\cap$ int(B)
$\overline{A \cup B}$ = $\bar A$ $\cup$ $\bar B$
$\overline{A \cap B}$ $\subset$ $\bar A$ $\cap$ $\bar B$

REPLY [1 votes]: The 3rd one is wrong because it implies that the union of a collection of closed sets is closed. Suppose every $A\in A'$ is closed. Let $B=\cup_{A\in A'}A.$ Since $\bar A=A$ for every $A\in A',$ we have $\cup_{A\in A'}\bar A=\cup_{A\in A'}A=B$ on the RHS, but the LHS is $\bar B.$
Example. Let $X=\Bbb R$ with the usual topology. Let $B$ be $any$ subset of $\Bbb R.$ Let $A'=\{\,\{r\}: r\in B\}.$
Inferences about finite collections rarely apply to infinite collections.


TITLE: Question about orthogonal projection
QUESTION [0 upvotes]: Let $\mathcal{H}$ be a Hilbert space. I am trying to show that every self-adjoint idempotent continuous linear transformation is the orthogonal projection onto some closed subspace of $\mathcal{H}$. If $P$ is such an operator, the obvious thing is to consider $S=\{Px:x\in\mathcal{H}\}$. However, I'm having trouble showing that S is in fact closed even though I'm sure this should be almost trivial. I tried to show that if $x_n\to x$ and $x_n\in S$ then $x\in S$ but somehow I just can't quite do it...

REPLY [2 votes]: A convenient way to check for closure of subspaces is to try to write the subspace as the kernel of some continuous operator.
Note that $(I-P)x = 0$ iff $Px=x$.
Note that $x \in S$ iff $x = Py$ for some $y$ iff $Px = x$, and so
$S = \ker (I-P)$. Hence $S$ is closed.


TITLE: Diffraction of light
QUESTION [0 upvotes]: If $l_1$ and $l_2$ are parallel how can they both can hit same spot?
Why their difference in path length is $l_2 - l_1 = d \sin \theta$ instead of 0?

REPLY [0 votes]: First of all, the hint is in the words "compressed scale" i.e $L>>d$ . So the lines
intersect at a point. They subtend certain angles with the horizontal.
When $L$ and $d$ are comparable, the picture is like below:

As for your second question, $$l_2 - l_1 = d \tan \theta \,\ (\text{from diagram})$$
$$ =d \sin \theta \,\ (\text{since $\theta$ is very small, we approximate $\sin \theta = \tan \theta$})$$


TITLE: How to rewrite the intersection of one set with the the union of some arbitrary number of other sets.
QUESTION [0 upvotes]: This question stems from this statement contained in an answer to a different question:

Observe that $P(A\setminus B)=P(A)-P(A\cap B)$. Therefore
$$ \begin{aligned}
 P\left(\bigcup_{i=1}^{n+1}A_i\right)&=P\left(\bigcup_{i=1}^n
 A_i\right)+P\left(A_{n+1}\setminus\bigcup_{i=1}^n A_i\right)\\
 &=P\left(\bigcup_{i=1}^n A_i\right)+P(A_{n+1})-P\left(\bigcup_{i=1}^n
 (A_i\cap A_{n+1})\right). \end{aligned} $$

Ok, so what I don't understand is how the last doesn't read
$$=P\left(\bigcup_{i=1}^n A_i\right)+P(A_{n+1})-P\left(A_{n+1}\cap(\bigcup_{i=1}^n
 A_i) \right)$$
based on the "observe that" statement in the preface. If these are equivalent statements, my question is how do I understand and show that? It seems like the author of the answer skipped a pretty big step (at least for someone like me).

REPLY [1 votes]: i apologise if this isn't helpful but i think this wouldnt be counted as a comment either.
but your first line
$$ \begin{aligned}
 P\left(\bigcup_{i=1}^{n+1}A_i\right)&=P\left(\bigcup_{i=1}^n
 A_i\right)+P\left(A_{n+1}\setminus\bigcup_{i=1}^n A_i\right)\\
 \end{aligned} $$
so lets think of it in a different way. let $E = A_{n+1}$ and $F=\bigcup_{i=1}^{n}A_{i}$
then what we have is
$$P(F \cup E)= P(F)+P(E \setminus F)$$
which by youre own statement you accept.
applying the arguement: $P(A\setminus B)=P(A)-P(A\cap B)$
gives
$$P(F \cup E)= P(F)+P(E)-P(E \cap F)$$
which again makes sense as this is just the inclusion-exclusion principle. we know that intersections and unions are distributive so
$$P(\bigcup_{i=1}^{n}A_{i}) \cup A_{n+1})= P(\bigcup_{i=1}^{n+1}A_{i})= P(\bigcup_{i=1}^{n}A_{i})+P(A_{n+1})-P(A_{n+1} \cap (A_{1}\cup A_{2}\cup ...\cup A_{n}))$$
$$= P(\bigcup_{i=1}^{n}A_{i})+P(A_{n+1})-P((A_{n+1} \cap A_{1})\cup (A_{n+1} \cap A_{2})\cup ...\cup A_{n+1} \cap A_{n}))$$
$$=P(\bigcup_{i=1}^{n}A_{i})+P(A_{n+1})-P(\bigcup_{i=1}^{n}(A_{n+1}\cap A_{i}))$$


TITLE: limsup proof for "Lebesuge integration on Euclidean Space" Problem 2.a
QUESTION [0 upvotes]: I picked up this book called "Lebesgue Integration on Euclidean Space" but I cannot find a solutions manual for it anywhere. Problem 2 talks about limsup and liminf.
It goes like this: for a sequence of sets $A_1,A_2,...$, prove that
$$\cap_{j=1}^{\infty} \left( \cup_{k=j}^{\infty} A_k \right) = \{ x| x \in A_k \text{for infinitely many} \ \ k \}$$
My attempt at it:
If we look at the union of sets inside the brackets then we will have:
$$A_j \cup A_{j+1} \cup ...$$
and so we can rewrite LHS:
$$\cap_{j=1}^{\infty} \left(  A_j \cup A_{j+1} \cup ... \right) = (A_1 \cup A_{2} \cup ...) \cap (A_2 \cup A_3 \cup ...) \cap ...$$
And so naturally due to intersection we will only have $x \in A_k$ for infinitely large $k$ in our final set, however, that is not what the problem asks to prove. It asks to prove that the final set will be composed of $x \in A_k$ for infinitely many $k$. So I do not think I have proved anything.

REPLY [0 votes]: Suppose $x \in A_i$ for infinitely many $i$. Then $x \in \bigcup_{k = j}^\infty A_k$ since we find an $i \geq k$ because there are infinitely many ones. So especially $x \in \bigcap_{j = 1}^\infty \bigcup_{k = j}^\infty A_k$. Conversly suppose $x \in \bigcap_{j = 1}^\infty \bigcup_{k = j}^\infty A_k$. Then $x \in \bigcup_{k = j}^\infty A_k$ for any $j \in \mathbb{N}$. Which means that $x \in A_k$ for some $k \leq j$ for all $j \in \mathbb{N}$. Hence $x \in A_k$ for infinitely many $k$.


TITLE: If light and gravity travels in the same speed then why light cant escape the gravitational pull of black holes?
QUESTION [0 upvotes]: Speed of gravity is infinite?
If fastest thing is light then how universe is spreading faster than the speed of light?

REPLY [0 votes]: A black hole does not necessarily emit gravitational waves, and if one did that is not the reason light cannot escape. What causes things to fall for any massive object is Einstein's description of gravity with General Relativity. What he tells us is space and time curve around massive objects, hence any object moving in a gravitational field will be influenced to curve with spacetime. We have observed light curve around other massive objects too, light just isn't trapped into any of those objects. Blackholes create a large enough weld in spacetime that when light passes through it curves just as it curves around any other object however it just curves hard enough that it directs in word.


TITLE: Completely additive functions
QUESTION [5 upvotes]: Other then the logarithm, does anybody know of any other important completely additive functions, by completely additive I mean $f(ab)=f(a)+f(b)$, for any integers $a,b$ ?

REPLY [1 votes]: Consider the fact that any positive rational number $q$ can be rewritten as $2^{e_1}3^{e_2}5^{e_3}7^{e_4}… p_n^{e_n}$.  Then, taking the exponents $e_i$ of each prime, we can construct a vector as follows:
$\vec{v}_q=<C_1e_1,C_2e_2,…,C_ne_n>$
(for arbitrary constants $C_i$)
Our general function is then $f(q)=\sum_iC_ie_i$ (the sum of the vector components).  You can think of this as a varation of the L1-norm where sign matters.  It becomes equivalent to the logarithm function when $C_i =\ln p_i$.


TITLE: How many different arrangements are possible with 3 digits numbers from 1-9 if we have to arrange them in ascending order?
QUESTION [0 upvotes]: if there are numbers 1,2,3,4,5,6,7,8,9 how many different three digit values we can make such that the digits are in ascending order?
since there are numbers from 1-9, I understand that the first digit should be from 1-7
So like that the different arrangements I found are,
123
234
345
456
567
678
789
even though I have arranged these I am confused with finding all the combinations here. they have given the answer as 84.

REPLY [1 votes]: Pick any $3$ numbers out of these $9$ numbers and arrange them in ascending order. So there are $9\choose 3$ possibilities.


TITLE: Generating false positives to the Rabin-Miller primality test
QUESTION [0 upvotes]: For what composite numbers $x$ will $a^{x-1} \equiv 1 \pmod x$ for
$a \in [2,n]$?
Can we generate $x$s that give false positives to the Rabin-Miller test for the first, say 10, consecutive integers $a > 1$?

REPLY [1 votes]: $230004476882957071 = 234271*546631*1796071$ is a base ${2,3,5,7}$ strong pseudoprime (thus all bases $1-10$).
There is one other counterexample here, but they are not easy to construct (as DanaJ pointed out). Let me explain how I constructed the one above:
For any odd prime $p$, by Fermat's Little Theorem, we have
$a^{p-1}=1\pmod p$.
By the law of quadratic reciprocity,
$a^{(p-1)/2}= $$\left( \frac{a}{p}\right)$$ \pmod p$ where $\left( \frac{a}{p}\right)$ is the Legendre Symbol.
If $p=3\pmod 4$, then the Miller-Rabin test is equivalent to a Euler-probable prime test (the identity above) since $(p-1)/2$ is odd.
In general, if $n=3\pmod 4$ and $a^{(n-1)/2}= $$\left( \frac{a}{n}\right)$$ \pmod n$, then for each prime $p | n$, $$\left( \frac{a}{p}\right) = \left( \frac{a}{n}\right)$$
A Miller Rabin counterexample would require that many of these congruences hold, in addition to $n$ being a Carmichael number or pseudoprime to many bases.
Lastly, I would like to point out that counterexamples  exist with $n=1\pmod 4$, but they are much harder to construct. Suppose $n=5\pmod 8$. There are two cases to consider. Like with $n=3 \pmod 4$:
$a^{(n-1)/2}= $$\left( \frac{a}{n}\right)$$ \pmod n$
If $\left( \frac{a}{n}\right) = -1$, then for each prime $p | n$, $p=1\pmod 4$, $\left( \frac{a}{p}\right) = -1$ if $p=5\pmod 8$ and $\left( \frac{a}{p}\right) = 1$ if $p=1\pmod 8$.
If $\left( \frac{a}{n}\right) = 1$, then $a^{(n-1)/4}= ±1 \pmod n$. One can check if it is $1$ or $-1$ using quartic reciprocity, and the same conditions as if $n=3\pmod 4$ also hold here.
More complex properties hold for $n=9\pmod {16}$, and when $n-1$ has more factors of $2$, we eventually can prove $n$ is prime (this explains why counterexamples are easier to construct if $n=3\pmod 4$). If $n=2^sd+1$ with $d < 2^s$, and $a^{(n-1)/2}= -1 \pmod n$, then $n$ is prime (Proth's Theorem).


text:
\textcolor{Internal}{With the pursuit of increasingly innovative and complex space missions, the focus of the space industry has been turning towards low-thrust technologies. Electric propulsion systems provide large savings in propellant mass, which can be decisive for the mission's feasibility. Since the first spacecraft using low-thrust was successfully flown in 1998 \cite{DS1}, this technology has allowed the planning of a range of missions that would otherwise be infeasible, including visits to the outer planets, comets and asteroids \cite{noton}.}

Designing a low-thrust trajectory is a more complex task than doing so for a high-thrust one. For the latter case, the few short thrust phases can be approximated by singular instantaneous $\Delta v$'s. On the contrary, low-thrust missions require the propulsion system to operate for a significant part of the transfer. Consequently, the thrust vector is a continuous function of time and the trajectory optimisation problem has to find the optimal control law \cite{dachwald}. This is an extremely complex problem that has no closed-form solutions, except for some very specific cases \cite{dachwald_cite}. Thus, the optimal control problem has to be carefully conceived in order to achieve convergence, which includes an attentive definition of bounds and constraints for the trajectory.

Following the process described in Section \ref{sec:opt_traj_design}, one of the main steps in the formulation of the optimal control problem is choosing the model of motion in which the trajectory is developed. Certain design applications, like missions to near-Earth asteroids (NEAs) require models of motion of higher complexity than the classical two-body problem (2BP), since third-body perturbations have a non-negligible effect. This is due to the fact that NEAs \textcolor{Internal}{usually} move in \textit{low-energy regimes}. These are here defined as regimes of motion in which ballistic capture is theoretically possible \cite{ballistic}, which may occur for objects whose orbital energy does not differ much from that of the third-body perturbation, e.g. nearly co-orbital bodies \cite{TCO} or, in the case of the Sun-Earth system, spacecraft departing from or arriving to Earth with a low excess velocity ($v_{\infty}$). \textcolor{External}{However, the utilisation of an alternative higher-fidelity method (e.g. the circular restricted three-body problem (CR3BP)) may bring difficulties related to the definition of the optimal control problem. For instance, boundary conditions are not trivial to set, since the coordinates are presented in the synodic Cartesian reference frame.}

\textcolor{External}{As an alternative, the Gauss' variational equations third-body (GVE-3B) framework can describe third-body motion in terms of the Keplerian elements that define the osculating orbit of the spacecraft, in a barycentric coordinate system. This is advantageous when devising a control strategy near global optima, since it provides a better intuitive understanding of the trajectory. Given that the solution to an optimal control problem typically relies on the setting of parameters by an expert in astrodynamics, outside of the actual optimisation process \cite{dachwald}, an intuitive understanding of the trajectory's evolution until a solution is reached is crucial. Furthermore, using Keplerian elements, the bounds and boundary values of the optimal control problem can be more easily assessed, facilitating the convergence of the non-linear process to generate the control law.}

Thus, this chapter presents the usage of the GVE-3B framework, as presented in Section \ref{sec:GVE}, in the low-thrust trajectory design of missions to NEAs. The latter are now considered the easiest celestial bodies to reach from the Earth \textcolor{Internal}{and may represent a potential impact threat to our planet \cite{RN1}.}

A test example of the usability of the GVE-3B framework is presented with the design of two trajectories: an asteroid capture and a rendezvous mission. The first case entails having a spacecraft attaching itself to an asteroid and moving it---from its nominal orbit to the stable invariant manifold of the $L_2$ point of the Sun-Earth system. Mirroring this mission, the rendezvous scenario consists on a spacecraft departing \textcolor{Internal}{$L_2$} through its unstable manifold orbit and matching its motion to the one of another NEA.

\textcolor{External}{Some choices were made to simplify the trajectory design and maximise the chances of having an easy convergence for the optimal control problem. This is in line with the main focus of this chapter: to demonstrate the application of the GVE-3B framework with fully fledged trajectories that meet the initial requirements, which may not be the overall best solutions. The next sections analyse and explain the design process in detail.}

\section{Asteroid Mission Trajectory Design}
\label{sec:traj_design}

This section presents the results from each phase of the trajectory design approach for the asteroid capture and rendezvous missions. The benefits of using the GVE-3B framework as the model of motion for low-thrust design are highlighted, and \textcolor{External}{fully} optimised transfers are shown for asteroids 2018 AV2 and 2017 SV19.

\subsection{Mission Summary}

The trajectory design for two asteroid missions is here presented: one for capture, another for rendezvous. Asteroid missions are invariably linked to the use of low-thrust propulsion systems, since the trajectory will benefit from their high exhaust velocity \cite{frontiers}. Furthermore, both missions consider a spacecraft departing from (the rendezvous case) or arriving to (the capture case) \textcolor{External}{one of the libration points of the Sun-Earth system}. \textcolor{External}{The close-range proximity operations that precede or conclude a capture or rendezvous mission are not considered in this work.}

\textcolor{External}{The asteroids were chosen using the Accessible NEAs NASA database\footnote{cneos.jpl.nasa.gov/nhats, Accessed 01-11-2018}, which was searched to find objects whose estimated mission costs do not surpass 5 km$\cdot$s$^{-1}$ (defined in the database as the total $\Delta v$ for departing a notional 400 km altitude circular Earth parking orbit, matching the NEA's velocity at arrival, departing the NEA and controlling the atmospheric entry speed at Earth return).}

\textcolor{Both}{The choice fell on asteroids 2018 AV2 and 2017 SV19 as the capture and rendezvous targets. At the time of search, they were the most recently discovered ones with a clearly defined optical opportunity, i.e. a set calendar date in which the asteroid will next be observable from the Earth. Since both asteroids have semi-major axes greater than the one of the Earth, the endgame was set for the $L_2$ point. This location is connected to several past missions (e.g. Herschel and Planck in 2009 \cite{herschel, planck}) and benefits from} the existence of invariant manifold orbits that can be travelled without spending any fuel, both arriving to and departing from it. It is important to denote that the mission $\Delta v$ costs may be further minimized by picking out optimal libration point orbits (LPOs) \textcolor{Internal}{and} connected manifold orbits (i.e. closer to the asteroids' energy), instead of the generic $L_2$ point. Since this component would bring an additional choice to the design process, it was not included in order to keep the focus on showcasing the methodology in a straightforward manner.

\subsubsection{Asteroid Capture}

Asteroid 2018 AV2 is an Apollo asteroid discovered in January, 2018. Currently, there is little data regarding its composition and nature: it is known that its diameter is on the \textcolor{Internal}{small} range (3.2 - 14 m) and, assuming an average material composition, its mass is of about 318 tonnes \cite{mass, mass2}. It is postulated that this object may likely be artificial \cite{mini-moon}; still, its dimensions make it a feasible notional body to study a capture mission.

The mission design entails the following: the spacecraft meets the asteroid at a certain point of the asteroid's nominal orbit and grabs it by any chosen means \cite{ARRM, artemis}. Then, the coupled system uses its propulsive capabilities to insert itself into a stable invariant manifold trajectory of the $L_2$ point, the target destination. This happens around the asteroid's next close approach with the Earth, starting in 2036.

\subsubsection{Asteroid Rendezvous}

Asteroid 2017 SV19 is an Amor asteroid discovered in September, 2017. Its diameter is in the 17-78 m range, with an estimated mass of 52,850 tonnes. This makes the body very heavy for capture, but adequate for a rendezvous mission with a spacecraft.

While the capture trajectory takes the spacecraft-asteroid system to $L_2$, the rendezvous one takes the spacecraft away from this point. The remaining mission mirrors the previously described one: after the spacecraft departs $L_2$ into an unstable invariant manifold orbit, it changes its trajectory to meet the nominal motion of the asteroid. The rendezvous happens during the asteroid's next close approach with the Earth, in 2040.

As previously stated, both bodies are moving in a very similar energy regime to the one of the Earth, making them great candidates to test the use of the GVE-3B framework. In order to better visualize and compare both mission scenarios, Figure \ref{fig:caperend} can be observed. The orbital transfer that will be designed in this chapter concerns the blue dashed segment from $\mathrm{t_1}$ to $\mathrm{t_2}$. The required steps in order to determine its initial and final points (A and B, respectively) will be explained in the following sections.

\begin{figure}[h]
	\centering
	\includegraphics[width=\linewidth]{Images/capandrend.pdf}
	\caption{\label{fig:caperend}Trajectory diagram for the asteroid capture case (left) and the asteroid rendezvous mission (right)}
\end{figure}

\subsection{Low-Thrust Trajectory Design}
\label{sub:ltdesign}

The problem of designing a spacecraft's trajectory can be simply stated as the determination of a transfer that satisfies some initial and final conditions, while minimizing a chosen parameter \cite{conway}. Its mathematical formulation is given in Chapter \ref{chap:intro}, together with a detailed overview of the particular problem of low-thrust modelling.

Several methods exist to solve optimal control problems, as surveyed by several authors \cite{ltreview, betts}. Currently, direct methods are the ones most employed for mission design \cite{conway}. While robust, these still require the generation of a reasonable initial guess of the solution parameters, i.e. a set of state and control time-histories that yield a sub-optimal trajectory to be improved by the NLP problem solver.

\textcolor{Internal}{In short, feasible initial guesses that can be posteriorly transcribed and solved using an optimal control software (e.g GPOPS-II \cite{gpops}) have to be generated.} This computation is achieved by implementing the following sequential approach, which will be detailed in the following sections:

\begin{description}
	\item{\textbf{\textcolor{External}{Step 1: Transfer Optimisation.}}} The optimal initial and final dates for the orbital transfer manoeuvre ($\mathrm{t_1}$ and $\mathrm{t_2}$) are determined by computing several possible Lambert arcs with different boundary conditions and choosing the one with the lowest $\Delta v$.
	\item{\textbf{\textcolor{External}{Step 2: Sims-Flanagan Approach.}}} The set transfer dates are used to compute an initial guess trajectory in the style of a Sims-Flanagan approach \cite{sims}. The result will be a transfer divided into equal-time segments, with an impulsive $\Delta v$ applied in each of them.
	\item{\textbf{\textcolor{External}{Step 3: Continuous Low-Thrust Transfer.}}} The multiple-impulse transfer is transcribed to a continuous one, by converting the $\Delta v$'s and segment times determined in the previous step into continuous accelerations and uniting the trajectory with a multiple shooting method.
\end{description}

\subsubsection{Step 1: Transfer Optimisation}
\label{sub:LA}

Finding the best possible trajectory for both asteroid missions requires defining several potential starting and ending points and calculating the transfer costs between them. Looking at Figure \ref{fig:caperend}, one can see that this means choosing the best Lambert arc out of different values of $\mathrm{t_1}$ and $\mathrm{t_2}$ and related asteroid ephemerides. In this case, the best arc is chosen as the one with the lowest $\Delta v$: the corresponding values of $\mathrm{t_1}$ and $\mathrm{t_2}$ are the initial and final transfer times and the related asteroid ephemerides are fixed to the state of points A and B. \textcolor{External}{These ephemerides remain fixed in time in the trajectory design, which may be sub-optimal for the final solution, as low-thrust trajectories are generally slower than impulsive ones.}

More concretely, looking first to the capture case: the first step is to obtain the real ephemerides of the asteroid 2018 AV2 in a far-away position from the Earth. Referring back to Figure \ref{fig:caperend}, this position corresponds to point $\mathrm{A_0}$. Then, this state is propagated forward in time to $\mathrm{t_1}$ (point A). In a similar fashion, the invariant manifold of the $L_2$ point is backpropagated to $\mathrm{t_2}$ (point B). Posteriorly, the states corresponding to $\mathrm{t_1}$ and $\mathrm{t_2}$ are connected via a Lambert arc, obtaining the impulsive transfer cost. This is done for many different values of $\mathrm{t_1}$ and $\mathrm{t_2}$, yielding a plot of transfer $\Delta v$'s as a function of the initial and final dates: a \textit{porkchop plot}.

\textcolor{External}{An analogous methodology is implemented for the rendezvous case: the ephemerides of asteroid 2017 SV19 are retrieved in a far-away position from the Earth (at $\mathrm{t_3}$, point $\mathrm{B_0}$). These are then backpropagated to $\mathrm{t_2}$ (point B). Simultaneously, the spacecraft's position is propagated forward from the $L_2$ point (at $\mathrm{t_0}$) to $\mathrm{t_1}$. The ephemerides corresponding to $\mathrm{t_1}$ and $\mathrm{t_2}$ are connected with a Lambert arc.}

\textcolor{External}{In short, the initial asteroid positions were retrieved from the Horizons JPL database\footnote{https://ssd.jpl.nasa.gov/?horizons, Accessed 01-12-2018}. In this step of the process, the Lambert arcs connecting A and B are computed in the 2BP. The propagations from $\mathrm{A_0}$ to A and $\mathrm{B_0}$ to B are done either in the GVE-3B model or in the 2BP, so as to further highlight their different results. The contour plots indicating the transfer $\Delta v$ cost as a function of the initial and final manoeuvre dates can be found on Figure \ref{fig:PC}: images a) and b) depict the capture and rendezvous cases in which the propagations from $\mathrm{A_0}$ to A and $\mathrm{B_0}$ to B are done in the GVE-3B framework, while c) and d) show the computation using the 2BP.}

\begin{figure}[hbt!]
	\centering
	\begin{minipage}[t]{0.49\linewidth}
		\includegraphics[width=\textwidth]{Images/fig9a.pdf}
		\caption*{a) Capture of Asteroid 2018 AV2 in the GVE-3B model}
	\end{minipage}
	\hfill
	\begin{minipage}[t]{0.49\linewidth}
		\includegraphics[width=\textwidth]{Images/fig9b.pdf}
		\caption*{b) Rendezvous with Asteroid 2017 SV19 in the GVE-3B model}
	\end{minipage}
	\centering
	\begin{minipage}[t]{0.49\linewidth}
		\includegraphics[width=\textwidth]{Images/fig9c.pdf}
		\caption*{c) Capture of Asteroid 2018 AV2 in the 2BP model}
	\end{minipage}
	\hfill
	\begin{minipage}[t]{0.49\linewidth}
		\includegraphics[width=\textwidth]{Images/fig9d.pdf}
		\caption*{d) Rendezvous with Asteroid 2017 SV19 in the 2BP model}
	\end{minipage}
	\caption{Contour plots of the $\Delta v$ cost as a function of departure and arrival transfer dates\label{fig:PC}}
\end{figure}

The initial conditions for asteroids 2018 AV2 and 2017 SV19 were taken to be April, 2036 ($\mathrm{t_0}$ of Figures \ref{fig:caperend} a) and c)) and July, 2041 ($\mathrm{t_3}$ of Figures \ref{fig:caperend} b) and d)). \textcolor{External}{The minimum and maximum transfer durations are set to zero and three years, respectively (the former condition being obviously infeasible, but being employed to establish the limit). The departing times range from the initial condition dates up to one year after these.} In total, 10,000 Lambert arcs were computed, with 100 departure and 100 arrival dates. The optimal $\Delta v$ manoeuvres are marked with a red circle; in the case of the GVE-3B framework propagation, the capture trajectory yields a cost of about 255 m$\cdot$s$^{-1}$, while the rendezvous requires 680 m$\cdot$s$^{-1}$. The optimal values obtained with the 2BP propagation are slightly different, \textcolor{Internal}{while} the optimal dates obtained in the two models are quite distinct. For instance, the optimal departure and arrival dates of the GVE-3B rendezvous porkchop yield, in the 2BP plot, a $\Delta v$ of 1502 m$\cdot$s$^{-1}$. This demonstrates that ignoring the Earth's influence in these trajectories could clearly yield sub-optimal transfers.

\subsubsection{Step 2: Sims-Flanagan Approach}

The Sims-Flanagan method has been extensively used to further refine impulsive thrust first guesses into low-thrust trajectories \cite{sims}. This approach consists on adding several short impulses along the path, making it more similar to a low-thrust trajectory and, thus, more likely for an optimal control solver to converge easily \cite{hargraves}. The trajectory is divided into segments \textcolor{External}{with pre-defined durations}, as it can be seen on the adapted Sims-Flanagan schematic of Figure \ref{fig:SF}. In the middle of each segment, a small impulsive manoeuvre is implemented. Each transfer (called a leg, defined from point A to B) is propagated forward and backward to a match point, located usually halfway through the leg. \textcolor{External}{This process can be further optimised by not fixing each segments' duration, at the cost of a more complex optimisation problem.}

In other applications of the method, the problem is inserted into an optimiser with a constraint on the match point distance \cite{sims2}. In this work, the match points are connected with a Lambert arc, and a genetic algorithm is used to minimise the sum of each segments' $\Delta v$. As such, the method is merely inspired on the Sims-Flanagan approach, with the named fundamental differences.

\begin{figure}[h]
	\centering
	\includegraphics[width=0.8\linewidth]{Images/fig7.pdf}
	\caption{\label{fig:SF}Sims-Flanagan trajectory scheme}
\end{figure}

The application of the Sims-Flanagan method can be further detailed in the following manner: first, points A and B are obtained from the optimal impulsive trajectory found via the porkchop plot. \textcolor{External}{One thing to look out for when defining these transfer conditions is not to choose points A and B with very close ephemerides---the connecting Lambert arc would become very short, leading to an increased sensitivity of the $\Delta v$ parameter in the optimisation problem.} Then, a genetic algorithm is employed: \textcolor{External}{it is part of the MATLAB code suite \cite{matlab_ga} and was implemented using an initial population of 1000 parameters, with the same number of maximum generations to convergence.} Following the characterisation of a typical optimisation problem, the design variable is the array of $\Delta v$'s that represent the mid-segment impulses (purple arrows of Figure \ref{fig:SF}). The objective function propagates the motion forward from A and backwards from B using the GVE-3B model, implementing the impulsive $\Delta v$'s in each segment. This yields match points A' and B', which are then connected using a Lambert arc. Finally, the cost function can be computed as the sum of this connecting Lambert arc $\Delta v$ with the $\Delta v$'s provided by the design variable array.

\subsubsection{Step 3: Continuous Low-Thrust Transfer}

After obtaining the Sims-Flanagan solution, a low-thrust trajectory can finally be determined. Since the guiding motivation is to showcase the capabilities of the GVE-3B framework, the optimal control problem was simplified in a way that only the acceleration is optimised, while assuming a constant mass. Thus, the controlled trajectory design is computed in the following manner:

\begin{enumerate}
	\item Each impulsive $\Delta v$ from the Sims-Flanagan approach is converted into continuous accelerations using the formula $\bm{a} = \frac{\bm{v}_1 - \bm{v}_0}{\Delta t}$, \textcolor{External}{in which $\bm{v}_0$ is the velocity at the beginning of the segment, $\bm{v}_1$ is the velocity at the end of the segment and $\Delta t$ is its time duration.}
	\item A is propagated forwards and B backwards, using the respective accelerations for each segment. The final points of the trajectory will be the new match points A'' and B'', \textcolor{Internal}{different from the previous A' and B' since they are achieved with continuous accelerations}.
	\item A'' and B'' are linked using the multiple shooting method \textcolor{External}{shown in Section \ref{sec:MSM} using the GVE-3B model, which} changes the thrusting accelerations to match the initial and final states, achieving a fully connected trajectory.
\end{enumerate}

A multiple shooting method \textcolor{External}{is} used to connect two ephemerides by computing several intermediate points in between them and, finally, linking them with individual segments. \textcolor{External}{This multiple shooting method was first described in Section \ref{sec:MSM}:} the solution to this scheme involves iterating through Eq. \eqref{eq:lsm} until the error ($\text{\textbf{G}}(\bm{X}^{i+1})$) is smaller than a convergence tolerance ($\epsilon$).

The convergence tolerance was chosen to be $\epsilon = 10^{-8}$. In this particular scenario of a multiple shooting scheme with pre-determined initial and final points and free accelerations, the beginning and ending position and velocities are fixed. No acceleration constraints are imposed for any of the patch points.

The particular matrices required for this multiple shooting scheme are described by Eqs. \eqref{eq:J1} to \eqref{eq:J2}, taking into consideration the Jacobian of the GVE-3B framework in Eq. \ref{eq:jac_gve}. This is here represented as taking into account the system's accelerations, yielding a $9\times9$ \textcolor{Internal}{matrix}. If these are not considered, the matrix may be reduced to a $6\times6$ one.
\begin{align}\label{eq:jac_gve}
\text{\textbf{D}}f(\bm{s})_{3B} &= \begin{bmatrix}
& & & & & & & & \\
& \text{\textbf{0}}_{3\text{x}3} & & & \text{\textbf{I}}_{3\text{x}3} & & & \text{\textbf{0}}_{3\text{x}3} &\\
& & & & & & & & \\
& & & & & & & & \\
& \text{\textbf{R}}_I^{\Earth^{-1}} \text{\textbf{A}}_{3B} & & & \text{\textbf{0}}_{3\text{x}3} & & &\text{\textbf{I}}_{3\text{x}3} &\\
& & & & & & & & \\
& & & & & & & & \\
& \text{\textbf{0}}_{3\text{x}3} & & & \text{\textbf{0}}_{3\text{x}3} & & & \text{\textbf{0}}_{3\text{x}3} &\\
& & & & & & & & \\
\end{bmatrix}\text{, where }
\text{\textbf{A}}_{3B} = \begin{bmatrix}
\textcolor{External}{\partial_x a_x} & \textcolor{External}{\partial_y a_x} & \textcolor{External}{\partial_z a_x}\\
\textcolor{External}{\partial_x a_y} & \textcolor{External}{\partial_y a_y} & \textcolor{External}{\partial_z a_y}\\
\textcolor{External}{\partial_x a_z} & \textcolor{External}{\partial_y a_z} & \textcolor{External}{\partial_z a_z}\\
\end{bmatrix}
\end{align}
\textcolor{External}{where $\partial_\beta a_{\alpha}$ ($\alpha \in \{x, y, z\} \text{ and }\beta \in \{x, y, z\}$) are the first partial derivatives of the accelerations $\{a_x, a_y, a_z\}$ presented in Eq. \eqref{eq:acc}}. These are expanded into the following:
\begin{align}\nonumber
\partial_x a_{x} &= \mu\bigg(\frac{3(x - 1)^2}{r_3^5} - \frac{1}{r_3^3} - \frac{15 x^3}{r^7} + \frac{9x - 3x^2}{r^5} + \frac{1}{r^3}\bigg)\\
\partial_y a_{x} &= - 3 y\mu \bigg(\frac{1 - x}{r_3^5} + \frac{5 x^2}{r^7} - \frac{1- x}{r^5}\bigg)\nonumber\\
\partial_z a_{x} &= -3 z \mu \bigg(\frac{1 - x}{r_3^5} + \frac{5 x^2}{r^7} - \frac{1- x}{r^5}\bigg)\nonumber\\
\partial_x a_{y} &= y\mu \bigg(\frac{3(x - 1)}{r_3^5} + \frac{3 + 2x}{r^5} - \frac{5x(3x + r^2)}{r^7}\bigg)\nonumber\\
\partial_y a_{y} &= \mu\bigg(\frac{3x + r^2}{r^5} -\frac{1}{r_3^3} + 3 y^2 \Big(\frac{1}{r_3^5} - \frac{5x + r^2}{r^7}\Big)\bigg)\nonumber\\
\partial_z a_{y} &= 3 y z \mu \bigg(\frac{1}{r_3^5} - \frac{5x + r^2)}{r^7}\bigg)\nonumber\\
\partial_x a_{z} &= z\mu \bigg(\frac{3(x - 1)}{r_3^5} + \frac{3 + 2x}{r^5} - \frac{5x(3x + r^2)}{r^7}\bigg)\nonumber\\
\partial_y a_{z} &= 3 y z \mu \bigg(\frac{1}{r_3^5} - \frac{5x + r^2)}{r^7}\bigg)\nonumber\\
\partial_z a_{z} &= \mu\bigg(\frac{3x + r^2}{r^5} - \frac{1}{r_3^3}  + 3 z^2 \Big(\frac{1}{r_3^5} - \frac{5x + r^2}{r^7}\Big)\bigg)
\end{align}

The number of patch points for the multiple shooting method was chosen to be 3. This can clearly be seen on the Jacobian of the constraint vector in Eq. \eqref{eq:J1}, where the first 3 conditions represent the patching requirements and the remaining ones establish that \textcolor{Internal}{A'' and B''} are fixed. This number was chosen in order to simulate the conditions of an on-off control, which will be later validated.

\begin{align}\label{eq:J1}
\text{\textbf{DG}}(\bm{X}) &=
\begin{bmatrix}
\bm{\Phi}_{1, 6\text{x}9} & -\text{\textbf{IO}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9}\\
\text{\textbf{0}}_{6\text{x}9} & \bm{\Phi}_{2, 6\text{x}9} & -\text{\textbf{IO}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9}\\
\text{\textbf{0}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9} & \bm{\Phi}_{3, 6\text{x}9} & -\text{\textbf{IO}}_{6\text{x}9}\\
\text{\textbf{IO}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9}\\
\text{\textbf{0}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9} & \text{\textbf{0}}_{6\text{x}9} & \text{\textbf{IO}}_{6\text{x}9}\\
\end{bmatrix}, \text{  }
\text{\textbf{G}}(\bm{X}) =
\begin{bmatrix}
\bm{s}_1(t_1 + T_1) - \bm{s}_2(t_2) \\
\bm{s}_2(t_2 + T_2) - \bm{s}_3(t_3)\\
\bm{s}_3(t_3 + T_3) - \bm{s}_4(t_4)\\
\bm{s}_1(t_1) - \bm{A''}\\
\bm{s}_{4}(t_4) - \bm{B''}\\
\end{bmatrix}\\\nonumber\\
\label{eq:J2}
\text{\textbf{IO}} &= \begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{bmatrix}
\end{align}

\section{Results and Discussion}

\textcolor{External}{After determining the ephemerides of points A and B for both mission scenarios, the Sims-Flanagan approach was implemented using 4 trajectory segments,} having the extra Lambert arc to link points A' and B'. The total $\Delta v$ for the optimal solutions was 380 m$\cdot$s$^{-1}$ and 875 m$\cdot$s$^{-1}$, respectively for the capture and rendezvous cases. The obtained trajectory can be found in Figure \ref{fig:SFRes}. Figure \ref{fig:SFRes} a) depicts the capture transfer: starting from point A in the nominal motion of 2018 AV2 and finishing at point B of the invariant manifold orbit, the purple arrows indicate the small manoeuvres of the Sims-Flanagan approach, with the red dashed line outlining the Lambert arc that connects the patch points A' and B'. Figure \ref{fig:SFRes} b) contains the rendezvous trajectory from point A in the invariant manifold orbit to point B of the nominal motion of 2017 SV19, using the same notation.

The low-thrust design is computed as detailed in Section \ref{sub:ltdesign}. Two control laws (sequence of accelerations and times to implement them) were devised, using the Sims-Flanagan approach and the multiple shooting method. One is computed entirely using the GVE-3B framework. The other is modelled with the same multiple shooting method, but using the 2BP model. Figure \ref{control_law} shows the propagation of both control laws using the CR3BP model.

By comparing the propagations, done with the CR3BP, of the GVE-3B and 2BP control laws, one can see that the two trajectories finish at distinct end-points. It is noteworthy that the desired target position is not reached, which is particularly evident in the asteroid rendezvous case. Considering that this scenario demands a much higher amount of $\Delta v$ in a considerably shorter mission time, the translation from impulsive to low-thrust trajectory is more prone to accumulating errors. \textcolor{External}{In the capture case, the minimum distance to target for the trajectory computed with the GVE-3B control law is of about 20,000 km, while the one calculated in the 2BP is about four times that length. Similarly, the minimum distance in the rendezvous case computed with the 2BP control law exceeds the one of the GVE-3B by over 10,000 km. While it is clear that the GVE-3B control law performs much better with respect to the target, there is likely a timing problem in the trajectory propagation.}

\begin{figure}[p!b]
	\centering
	\begin{minipage}[t]{0.424\linewidth}
		\includegraphics[width=\textwidth]{Images/fig101.pdf}
		\caption*{a) Capture of Asteroid 2018 AV2}
	\end{minipage}
	\hfill
	\begin{minipage}[t]{0.42\linewidth}
		\includegraphics[width=\textwidth]{Images/fig102.pdf}
		\caption*{b) Rendezvous with Asteroid 2017 SV19}
	\end{minipage}
	\caption{Trajectory segments in the Sims-Flanagan approach\label{fig:SFRes}}
	\vspace{4mm}
	\begin{minipage}[t]{0.42\linewidth}
		\includegraphics[width=\textwidth]{Images/fig111.pdf}
		\caption*{a) Capture of Asteroid 2018 AV2}
	\end{minipage}
	\hfill
	\begin{minipage}[t]{0.422\linewidth}
		\includegraphics[width=\textwidth]{Images/fig112.pdf}
		\caption*{b) Rendezvous with Asteroid 2017 SV19}
	\end{minipage}
	\caption{Final \textcolor{External}{trajectories with control laws determined with the GVE-3B framework (in red) and with the 2BP (in black), propagated with the CR3BP}\label{control_law}}
\end{figure}

A different analysis involves considering \textcolor{External}{solely trajectories propagated using the GVE-3B control law.} The corresponding control history is detailed in Figure \ref{fig:cl_time}, where the different trajectory phases can be distinguished. The solution is close to a on-off control pattern, but there is a clear disparity in between phases. Future implementations could have the multiple shooter be used throughout the entire motion, so that a smoother and more consistent on-off control can be attained.

\begin{figure}[ht!]
	\centering
	\begin{minipage}[t]{0.48\linewidth}
		\includegraphics[width=\textwidth]{Images/fig121.pdf}
		\caption*{a) Capture of Asteroid 2018 AV2}
	\end{minipage}
	\hfill
	\begin{minipage}[t]{0.48\linewidth}
		\includegraphics[width=\textwidth]{Images/fig122.pdf}
		\caption*{b) Rendezvous with Asteroid 2017 SV19}
	\end{minipage}
	\caption{Control history for the GVE-3B model, as depicted in Figure \ref{control_law}\label{fig:cl_time}}
\end{figure}

\textcolor{External}{Considering only this GVE-3B control law, each mission's trajectory can be propagated using the GVE-3B, the 2BP and the CR3BP models. The absolute errors of the propagations with the GVE-3B and the 2BP are computed, using the CR3BP model as the reference solution.} The results can be found on Figure \ref{fig:cl_compare}: the error yielded by the 2BP propagation is much higher for both mission scenarios, reaching around $10^6$ and $10^4$ km for capture and rendezvous, respectively. Plus, the 2BP model error grows very fast in the capture case, where the motion gets very close to the Earth. The rendezvous trajectory also proves to have a smaller error when using the GVE-3B, despite the fact that the motion starts at a distance of 30 times the Earth's Hill radius. This further demonstrates that the planet's impact on the mission design is noteworthy, even when quite far from its sphere of influence. Therefore, it can be inferred that the GVE-3B framework is more accurate \textcolor{External}{than the 2BP when compared} to the CR3BP.

\begin{figure}[ht!]
	\centering
	\begin{minipage}[t]{0.48\linewidth}
		\includegraphics[width=\textwidth]{Images/fig131.pdf}
		\caption*{a) Capture of Asteroid 2018 AV2}
	\end{minipage}
	\hfill
	\begin{minipage}[t]{0.48\linewidth}
		\includegraphics[width=\textwidth]{Images/fig132.pdf}
		\caption*{b) Rendezvous with Asteroid 2017 SV19}
	\end{minipage}
	\caption{\textcolor{External}{Absolute error, with respect to the CR3BP baseline, for the propagation of the 2BP and the GVE-3B. Error corresponding to the trajectory of Figure \ref{control_law} using the GVE-3B control law}\label{fig:cl_compare}}
\end{figure}

\textcolor{Both}{It is now established that, in terms of accuracy, the GVE-3B far surpasses the 2BP. In terms of convergence of the optimal control problem, the situation is similar. A preliminary study using the optimal control solver GPOPS-II was performed for the capture of asteroid 2018 AV2. Under the initial and final time and ephemerides conditions and using the CR3BP model, the solver takes about 50\% more iterations to converge with the 2BP control law than under the GVE control law. The fact that the initial guess is already a fully fledged trajectory also contributes to the easiness in establishing bound and boundary constraints---however, there is clearly still much room for improvement in the search of optimal solutions.}

\section{Summary}
\label{sec:GVEconclusions}

This chapter \textcolor{Internal}{demonstrates the application of a novel formulation of the third-body perturbation, using a GVE framework, to the propagation of trajectories within low-energy regions and subject to low-thrust propulsive accelerations.}

The nature of the GVE-3B equations makes them very advantageous for low-thrust trajectory design. Particularly, the intuitive observation of the orbital elements evolution and the easy definition of boundaries and constraints for the optimal control problem make the GVE-3B framework straightforward to set up and solve. \textcolor{External}{In summary, its usage facilitates the set up and convergence of the optimal control problem solver.} Thus, its utilisation in preliminary mission design can be extremely valuable, providing important insights into the time evolution of the orbital elements.

The presented framework can be utilised in many different \textcolor{External}{applications for space mission design}. One such concept entails Jovian or Saturnian moon tours---as shown by Alessi and S\'anchez \cite{alessi_semi}, both of these planetary systems have small enough gravitational parameters that the developed equations of motion remain accurate (as opposed to, for example, the Earth-Moon system). However, computing trajectories for moon tours would demand a higher complexity in the mission design, since this would require the integration of the third-body framework with multiple planetary systems.

config: arxiv


TITLE: Probability and Statistics: Over-booking flights
QUESTION [2 upvotes]: I am finding difficulty trying to answer the second part of this question. The answer is one this website but I do now get how they worked it out. https://nrich.maths.org/4932
I also know that binomial and normal distribution plays a big part in this.
An airline flies a plane with 400 seats. Each passenger who buys a ticket arrives for the flight (that is, does not miss the flight) with probability 0.95. If the airline sells 400 tickets what is the expected number of empty seats? (I found the mean to be 380 and the Sd to be 20)
The airline regularly books more than 400 passengers for its flights. How many tickets can the airline sell if it wants to have to refuse passengers who arrive for the flight with tickets in no more than about two per cent of the flights? (I do not know how to reach the answer of 411. Can someone explain with step by step working?)

REPLY [3 votes]: Let the number of tickets sold be $n$. The number of arriving passengers is distributed as $\operatorname{binom}(n,0.95)$, which we approximate as $\mathcal N(0.95n,0.0475n)$ because we do not know what $n$ is yet.
For a standard normal variable $Z$, $P(Z>x)=0.02$ gives $x=2.055$. Thus, for the airline to have to refuse passengers less than 2% of the time, 400 has to be greater than 2.055 standard deviations above the mean:
$$0.95n+2.055\sqrt{0.0475n}<400$$
This is a quadratic in $\sqrt n$ and we solve this as $\sqrt n<20.285$, i.e. $n<411.489$. Thus the largest number of tickets the airline can sell is 411.


TITLE: Evaluating stability using Jacobian in dimension 1
QUESTION [1 upvotes]: Generally, when we are considering a system of differential equations, to evaluate the stability of fixed points, we find the eigenvalues of the Jacobian to evaluate stability. I was wondering how this applies to 1 dimensional systems.
For example, when we have $\frac{dx}{dt}= -3x + \frac{5}{7+x}$, we can evaluate the stability of the fixed points by simply analyzing how it grows around the point (by plugging in points nearby).
How would I use the Jacobian in such a system?
In 2D systems, where $\frac{dx}{dt} = -3x + \frac{5}{7+y}$ and $\frac{dy}{dt} = -3y + \frac{5}{7+x}$, I would have to derive the respective equations with respect to $x$ and $y$ to form my Jacobian. Would I follow the same method but only with respect to $x$ for the 1D system? In that case, is the eigenvalue whatever value the derivative of $\frac{dx}{dt}$ with respect to $x$ is at a point?

REPLY [1 votes]: The Jacobian for scalar systems is itself a scalar.
We have $\dot{x}=f(x)\implies J = \left.\dfrac{df}{dx}\right|_{x=x_\text{eq}}$.
The equilibrium points of your system are given by  $$0=-3x_\text{eq}+\dfrac{5}{7+x_\text{eq}}\implies 3x_\text{eq}(7+x_\text{eq})=5\implies 3x^2_\text{eq}+21x_\text{eq}-5=0.$$
Solve this equation. Then determine the derivative of $f(x)=-3x+\dfrac{5}{7+x}$ and evaluate it at the equilibrium points. Your eigenvalue is given by the equation
$$\det\left[J-\lambda I \right]=0\implies \det[J-\lambda]=0\implies J-\lambda=0 \implies \lambda = J.$$
If the real part of your eigenvalue is positive the equilibrium point is unstable if the real part of your eigenvalue is negative the equilibrium is asymptotically stable. If the equilibrium point has a real part of $0$ we cannot conclude anything about the nonlinear system by Lyapunov's linearization theorem.

Just a remark: The differential equation is separable. Hence, you can determine the solution by integrating
$$\int_{\bar{x}=x_0}^{x}\dfrac{d\bar{x}}{-3\bar{x}+\dfrac{5}{7+\bar{x}}}=\int_{\tau=t_0}^td\tau$$


TITLE: Closed sets in product topology
QUESTION [1 upvotes]: I have an assignment, I have to proof that arbitrary product of close sets is closed in the product topology, I think I have to use complements and treat with opens, what do you think?

REPLY [4 votes]: You can also use that
$$
\prod_{i \in I} A_i = \bigcap_{i \in I} \pi_i^{-1}(A_i),
$$
where $\pi_i : \prod_j X_j \to X_i, (x_j)_j \mapsto x_i$ is continuous by definition of the product topology.
Why does that help you?

REPLY [1 votes]: Let $X=\prod_i X_i$ and suppose that $F_i$ is closed in $X_i$. Observe that
$$
(\prod_{i\in I}F_i)^c=\cup_{i\in I}(\prod_{j\in I}G_{i,j}). \tag{*}
$$
Here $G_{i,j}=X_j$ if $j\neq i$ and $G_{i,j}=F_i^c$ if $j=i$. Because $F_i^c$ is open in $X_i$, the definition the product topology implies $\prod_{j\in I}G_{i,j}$ is open in $X$. Since a union of open sets is open, the RHS of (*) is open in $X$. This means $\prod_i F_i$ is closed in $X$.


TITLE: What is the highest probability
QUESTION [1 upvotes]: At a faculty the probability of seeing a person greater than 190cm is 12%. From a class of $24$ people, is it more reliable to meet $2$ people above $190cm$ or $4$ people above $190cm$?
I dont undersand. Consider that the solution may be related to any distribution (Binomial or Poisson). Help me please to solve. Thanky very much for my hard?

REPLY [0 votes]: Here is a binomial distribution for $p=0,12$ and $n=24$:


TITLE: Prove that sequence defined by composition of a function is bounded given that function's derivative is bounded
QUESTION [5 upvotes]: Let $f : \mathbb{R} → \mathbb{R}$ be a differentiable function such that its derivative $f'(x)$.
is a continuous function. Moreover, assume that for all $x \in \mathbb{R}$, $$0 \leq |f'(x)| \leq \frac{1}{2}.$$ Define a sequence of real numbers $\{a_n\}_{n\in\mathbb{N}}$ such that $$a_1=1$$ $$a_{n+1} = f(a_n) $$for all $n \in \mathbb{N}$. Prove that there exists a positive real number $M$ such that for all $n \in \mathbb{N}$, $$|a_n| \leq M.$$
Source

My attempt:
We use the Mean Value Theorem. $$\frac{f(a_n)-f(a_{n-1})}{a_n-a_{n-1}} = f'(x)$$ for some $x \in [a_n,a_{n-1}]$. We know that $-\frac{1}{2} \leq f'(x)\leq \frac{1}{2}$, hence $$-\frac{1}{2} \leq\frac{f(a_n)-f(a_{n-1})}{a_n-a_{n-1}} \leq \frac{1}{2}.$$ Simplifying, we find that $|a_{n+1} - a_n| \leq \frac{|a_n-a_{n-1}|}{2}$.
Here we define a sequence $x_n = |a_{n+1} -a_n|.$ It is easy to see that $x_n \leq \frac{x_{n-1}}{2}$ which implies that $$ \lim_{n \rightarrow \infty} x_n = 0$$ Which means that $|a_{n-1} - a_n|$ is arbitrarily small and grows slower than $\{X_n : X_n = \Sigma \frac{1}{2^n} \}$. Therefore the sequence must converge to a finite limit, say $L$. We choose $M = L+1$.
QED.

Is this proof correct? I have a few qualms about it, but a few of my peers looked over it and weren't able to find any problem with it. Please help me with a nudge in the right direction if there turns out to be an error.

REPLY [1 votes]: Hints : use the fact, for $m\gt n $, $|a_{m}-a_{n}|\le |a_{m}-a_{m-1}|+|a_{m-1}-a_{m-2}|+\cdot\cdot\cdot+|a_{n+1}-a_{n}|\le \frac{1}{2^{m-1}}+\frac{1}{2^{m-2}}+\cdot\cdot\cdot+\frac{1}{2^{n}} $
Right hand side series is a G.P. series with ratio $\frac{1}{2}$.

REPLY [1 votes]: You've essentially proved that $\displaystyle \sum_{k=1}^\infty |a_{k+1} - a_k|$ converges. This means $\displaystyle \sum_{k=1}^\infty (a_{k+1} - a_k)$ also converges, and if you define
$$L = a_1 + \sum_{k=1}^\infty (a_{k+1} - a_k)$$
you get
$$L - a_n = \sum_{k=n}^\infty (a_{k+1} - a_k)$$ for all $n$, and in particular
$$|L - a_n| \le \sum_{k=n}^\infty |a_{k+1} - a_k| \to 0$$ as $n \to \infty$.


TITLE: Where is the mistake in this limit calculation?
QUESTION [2 upvotes]: I got this limit:
$$\lim_{x\to1}\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}*\frac{1}{\sqrt[4]{x}-1}*\frac{\sqrt[4]{x}+1}{\sqrt[4]{x}+1} \implies \lim_{x\to1}\frac{\sqrt[4]{x}+1}{\sqrt[3]{x²}+\sqrt[3]{x}+1} \xrightarrow[x\to 1]{}\frac{2}{3}.$$
But in the book the answer is $$ \frac{4}{3}.$$
I cannot find the mistake in my calculation.

REPLY [0 votes]: Your calculation mistake is already pointed out by TonyK.
I don't really understand what prevents you from using the standard limit: $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ This is in fact the least used of all standard limits (most used one is probably $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$). We have to only notice that here $a = 1$ and then $$\lim_{x \to 1}\frac{\sqrt[3]{x} - 1}{\sqrt[4]{x} - 1} = \lim_{x \to 1}\dfrac{\dfrac{x^{1/3} - 1}{x - 1}}{\dfrac{x^{1/4} - 1}{x - 1}} = \frac{1/3}{1/4} = \frac{4}{3}$$


text:
\section{The equation}
\label{sec:equation}
In this section we prove theorem~\ref{thm:main} via the implicit
function theorem.  We consider a setup similar to Ye \cite{Ye:1991}. Let $(M,g)$ be given with injectivity radius
$\rho>0$. Fix a base point $p\in M$ and an orthonormal frame
$\{e_j\}_{j=1}^3$ for $T_p(M)$. Consider the map:
\begin{equation*}
  c : \bR^3 \supset B_\rho(0) \to M  : \tau \mapsto \exp_p(\tau),
\end{equation*}
where $\exp_p:T_p M \to M$ denotes the exponential map of $M$ at $p$.
Let $e_j^\tau$ be the parallel transports of the $e_j$ to $c(\tau)$
along the geodesic $t\mapsto c(t\tau)|_{t\in[0,1]}$. Define the map
\begin{equation*}
  F_\tau : \bR^3 \supset B_{\rho}(0) \to M : x \mapsto
  \exp_{c(\tau)} (x^i e_i^\tau).
\end{equation*}
Let
$\Omega_1 := \{ \varphi \in C^{4,\frac12}(S_1) \mid
\| \varphi \|_{C^{4,\frac12}(S_1)}< 1\}$
and for $\varphi\in \Omega_1$ let
$S_\varphi:=\{ (1+\varphi(x))x \mid x\in S_1\}$. For
$\tau\in B_\rho\subset\bR^3$ and $r\in(0,\rho/2)$ let
$S(r,\tau,\varphi)=F_\tau(\Psi_r (S_\varphi))$ where $\Psi_r$ denotes
scaling by $r$ as in section~\ref{sec:expansion}.  Define
\begin{equation*}
  \tilde \Phi : (0,\rho/2)\times B_\rho(0) \times \Omega_1 \times \bR \to
  C^{\frac12}(S_1) : (r,\tau,\varphi,\lambda) \mapsto
  \tilde\Phi(r,\tau,\varphi,\lambda)
\end{equation*}
where $\tilde\Phi(r,\tau,\varphi,\lambda)$ is the function
\begin{equation}
  \label{eq:3}
  \Delta H + H|\Acirc|^2+H \Ric(\nu,\nu)+\lambda H
\end{equation}
evaluated on $S(r,\tau,\varphi)$ with respect to the metric $g$ and
pulled back to $S_1$ via the parameterization
$x \mapsto F_\tau(\Psi_r((1+\varphi(x))x))$.

Our goal is to find $r_0\in(0,\rho/2)$ and a map
\begin{equation*}
 (0,r_0) \to B_\rho \times \Omega_1 \times \R : r \mapsto (\tilde\tau(r),\tilde\varphi(r),\tilde\lambda(r))
\end{equation*}
so that
\[
\tilde\Phi(r,\tilde\tau(r),\tilde\varphi(r),\tilde\lambda(r))=0.
\]
Then for all $r\in(0,r_0)$ the surfaces $\Sigma_r:=S(r,\tilde\tau(r),\tilde\varphi(r))$
solve the equation
\begin{equation*}
  \Delta H +H|\Acirc|^2+H \Ric(\nu,\nu)+\tilde\lambda(r) H = 0
\end{equation*}
as claimed. Up to reparameterization, the family
$(\Sigma_r)_{r\in(0,r_0)}$ is the family of solutions as in
Theorem~\ref{thm:main}, see Corollary~\ref{thm:reparametrize} for
details.

An equivalent way to define $\tilde\Phi(r,\tau,\varphi,\lambda)$ is to
evaluate the operator~\eqref{eq:3} on $S_\varphi$ with respect to the
metric $\tilde g^{r,\tau} := (\phi_\tau\circ \Psi_r)^* g$. To get a
uniform scale in $r$, we consider instead the rescaled metric
$g^{r,\tau} := r^{-2} \tilde g^{r,\tau}$ and define the rescaled
function $\Phi(r,\tau,\varphi,\lambda)$ to be the operator
\begin{equation}
  \label{eq:7}
  \Delta_{r,\tau} H_{r,\tau} + H_{r,\tau}|\Acirc_{r,\tau}|^2+H_{r,\tau} \Ric_{r,\tau}(\nu,\nu)+r^2 \lambda H_{r,\tau}
\end{equation}
evaluated on $S_\varphi$ with respect to $g^{r,\tau}$ and pulled back to
$S_1$ via the parameterization $x\mapsto (1+\varphi(x))x$ of
$S_\varphi$. From the scaling of the geometric quantities, we get
\begin{equation*}
  \Phi(r,\tau,\varphi,\lambda) = r^3 \tilde\Phi(r,\tau,\varphi,\lambda).
\end{equation*}
By definition
\begin{equation}
  \label{eq:6}
  \Phi(r,\tau,0,\lambda)=\mathcal{W}_{r,\lambda}(S_1),
\end{equation}
where $\mathcal{W}_{r,\lambda}(S_1)$ is from
Proposition~\ref{prop:expansion-Willmore} and the geometric quantities
in the expression for $\mathcal{W}_{r,\lambda}(S_1)$ are evaluated at $c(\tau)$.
Note that after shifting by $\tau$, the metric $g$
in Proposition~\ref{prop:expansion-Willmore} corresponds to
the metric $g^{r,\tau}$ here.

The linearization of the Willmore operator $\tilde\Phi$ is denoted by
$W_\lambda$. It was calculated in \cite[Section~3]{LMS:2011}. For a
variation of an arbitrary surface $\Sigma$ with normal speed $f$ it is
given by
\begin{equation}
  \label{eq:9}
  W_\lambda f=LLf+\frac12 \nabla^\star (H^2 \nabla f)-2 \nabla^\star(H\Acirc(\nabla f,\cdot))+\lambda Lf+fQ,
\end{equation}
where $\nabla^\star =- \text{div}$, $L=-\Delta -|A|^2-\Ric(\nu,\nu)$, and
\begin{equation}
  \label{eq:10}
  \begin{aligned}
    Q&= |\nabla H|^2 +2 \omega(\nabla H)+H\Delta H+2\langle \nabla^2 H ,\Acirc \rangle +2 H^2|\Acirc|^2+2H\langle \Acirc,T\rangle\\
    &\quad- H \nabla\Ric(\nu,\nu,\nu)-\frac12 H^2 |A|^2 + \frac12 H^2 \Ric(\nu,\nu).
  \end{aligned}
\end{equation}
Here $\omega = \Ric(\nu,\cdot)^T$ is the tangential projection of the $1$-form $\Ric(\nu,\cdot)$ to $\Sigma$ and $T = R(\cdot,\nu,\nu,\cdot)$.
All the geometric quantities in $W_\lambda$ are evaluated on $\Sigma$
with respect to the corresponding ambient geometry. For given
$f\in C^{4}(S_1)$ the family $t\mapsto S(r,\tau,t f)$ is a normal
variation of $S(r,\tau,0)$ with normal speed $rf$, so that
\begin{equation}
  \label{eq:4}
  \tilde\Phi_\varphi(r,\tau,0,\lambda) f = r W_\lambda f.
\end{equation}
Here we evaluate $W_\lambda$ with respect to the metric $g$ in
$M$. Rescaling to the $g^{r,\tau}$ metric, we find that
\begin{equation}
  \label{eq:5}
  \Phi_\varphi(r,\tau,0,\lambda) f = r^4 W_\lambda f =
  W_{r,\tau,\lambda} f,
\end{equation}
where $W_{r,\tau,\lambda}$ is the linearized Willmore operator with
respect to $g^{r,\tau}$:
\begin{equation}
  \label{eq:8}
  \begin{aligned}
    W_{r,\tau,\lambda} f
    &= L_{r,\tau}L_{r,\tau} f +\frac12
    \nabla^\star_{r,\tau} (H_{r,\tau}^2\nabla_{r,\tau} f)
    \\
    &\quad
    -2\nabla^\star_{r,\tau}(H_{r,\tau}\Acirc_{r,\tau}(\nabla_{r,\tau}f,\cdot)+r^2\lambda L_{r,\tau}f+Q_{r,\tau}f.
  \end{aligned}
\end{equation}
Here we use the subscript $_{r,\tau}$ to denote quantities evaluated
with respect to the metric $g^{r,\tau}$.

In the limit $r\to 0$ the metric $g^{r,\tau}$ converges to the
Euclidean metric so that in the limit we have
\begin{equation*}
  W_{0,\tau,\lambda} f = L_0(L_0+2)f=(-\Delta) (-\Delta-2)f.
\end{equation*}
The kernel of this operator is given by
\begin{equation*}
  K:=\operatorname{ker} W_{0,\tau,\lambda} = \operatorname{Span} \{ 1,x^1,x^2,x^3\},
\end{equation*}
where the $x^i$ are the standard coordinate functions on $S_1$. We split this kernel into two parts:
\begin{equation}
  \label{eq:1}
    K_0 := \operatorname{Span} \{1\}\quad\text{and}\quad
    K_1 := \operatorname{Span} \{x_1,x_2,x_3\}.
\end{equation}
As in \cite{Ye:1991}, the function space $C^{4,\frac12}(S_1)$ splits
as a direct sum into $K$ and its $L^2$-orthogonal complement
$K^\perp$. It is standard to verify that we have the direct sum
decomposition of the target with respect to the $L^2$-scalar product:
\[
C^{0,\frac12}=  K + W_{0,\tau,\lambda}(K^\perp).
\]
Define the $L^2$-orthogonal projection maps
\begin{equation*}
  P_0:C^{0,\frac12}(S_1)\to K_0
  \quad\text{and}\quad
  P_1:C^{0,\frac12}(S_1)\to K_1.
\end{equation*}
The maps $T_0:K_0\to \bR$ and $T_1:K_1\to \bR^3$ identify $K_0$ and
$K_1$ with $\bR$ and $\bR^3$ according to the basis given in
equation~\eqref{eq:1}. Moreover, for $i\in \{0,1\}$ let
$\tilde P_i=T_i\circ P_i$. Denote by $\{e_1,e_2,e_3\}$
the standard basis of $\bR^3$.
\begin{lemma}
  \label{thm:projection-to-kernel}
  We have
  \begin{align*}
    \tilde P_0 (\Phi(r,\tau,\varphi,\lambda))
    &=  8\pi r^2\left(\lambda+\frac13\Scal(c(\tau))\right)+O(r^4)
      +\tilde P_0 \left(\int_0^1 \Phi_{\varphi}(r,\tau,t\varphi,\lambda)\varphi\, dt\right)
    \intertext{and}
    \tilde P_1 (\Phi(r,\tau,\varphi,\lambda))&=
                                               \frac{4\pi}{3}r^3 \nabla_{e_i}\Scal(c(\tau)) e_i +O(r^5)
                                               +\tilde P_1 \left(\int_0^1 \Phi_{\varphi}(r,\tau,t\varphi,\lambda)\varphi\, dt\right).
  \end{align*}
\end{lemma}
\begin{proof}
  Start by writing
  \begin{align*}
    \Phi(r,\tau,\varphi,\lambda)&= \Phi(r,\tau,0,\lambda)+\int_0^1 \Phi_{\varphi}(r,\tau,t\varphi,\lambda)\varphi\, dt.
  \end{align*}
  By~\eqref{eq:6}, for $i\in \{0,1\}$
  \[
  \tilde P_i( \Phi(r,\tau,0,\lambda))=\tilde P_i (\mathcal{W}_{r,\lambda}(S_1)),
  \]
  Where $\mathcal{W}_{r,\lambda}(S_1)$ is evaluated at the base point
  $c(\tau)$. The right hand side can be calculated term by term from
  the expansion of $\mathcal{W}_{r,\lambda}(S_1)$ given in
  Proposition~\ref{prop:expansion-Willmore}:
  \begin{align*}
    \tilde P_0(\mathcal{W}_{r,\lambda}(S_1))&= 8\pi r^2\left(\lambda+\frac13 \Scal(c(\tau))\right)+O(r^4)\quad\text{and} \\
    \tilde P_1(\mathcal{W}_{r,\lambda}(S_1))&= \frac{4\pi}{3}r^3 \nabla_{e_i}\Scal(c(\tau))e_i+O(r^5).
  \end{align*}
  Note that all terms that contain an odd number of $x^i$-factors
  integrate to zero. For the other terms we used that
  $\int_{S_1} x^i x^p =\frac{4\pi}{3} \delta_{ip}$ and a similar
  expression for integrals involving four factors of components of
  $x$.
\end{proof}
\begin{lemma}
  \label{thm:derivativePhi1}
  For every $\tau\in\bR^3$ and every $\lambda\in\bR$ we have that
  \begin{equation*}
    \Phi_{\varphi r}(0,\tau,0,\lambda) = \left.\frac{\partial}{\partial r}\right|_{r=0}
    W_{r,\tau,\lambda} = 0.
  \end{equation*}
\end{lemma}
\begin{proof}
  For the proof, we have to calculate
  $\left.\frac{\partial}{\partial r}\right|_{r=0} W_{r,\tau,\lambda}$
  from its expression~\eqref{eq:8} taking into account its
  definition~\eqref{eq:9} and~\eqref{eq:10}. Since we compute
  $\frac{\partial}{\partial r} W_{r,\tau,\lambda}$ at $r=0$ we see
  that all terms that are product of at least two quantities that
  vanish at $(r,\varphi)=(0,0)$ do not contribute to the
  derivative. In particular
  \begin{equation}
    \label{eq:11}
    \begin{aligned}
      \left.\tfrac{\partial}{\partial r}\right|_{r=0}
      \big(|\nabla H_{r,\tau} |^2 + 2\omega_{r,\tau}(\nabla H_{r,\tau}) + 2\langle
      \nabla^2 H_{r,\tau} ,\Acirc_{r,\tau} \rangle &
      \\ \qquad
      + 2 H_{r,\tau}^2|\Acirc_{r,\tau}|^2+2H_{r,\tau}\langle
      \Acirc_{r,\tau},T_{r,\tau}\rangle + r^2 \lambda L_{r,\tau} \big) & = 0.
    \end{aligned}
  \end{equation}
  From the proof of \cite[Lemma 1.3]{Ye:1991}, we quote equation
  (1.15)
  $\left.\tfrac{\partial}{\partial r}\right|_{r=0} g^{r,\tau} = 0$,
  its consequence
  $\left.\tfrac{\partial}{\partial r}\right|_{r=0} \Delta_{r,\tau} =
  0$,
  equation (1.17)
  $\left.\tfrac{\partial}{\partial r}\right|_{r=0} A_{r,\tau} = 0$,
  $\left.\tfrac{\partial}{\partial r}\right|_{r=0} \Ric_{r,\tau} = 0$,
  and assertion (1), that is
  $\left.\tfrac{\partial}{\partial r}\right|_{r=0} L_{r,\tau} = 0$.
  These identities also imply that
  $\left.\tfrac{\partial}{\partial r}\right|_{r=0} H_{r,\tau} = 0$
  and
  $\left.\tfrac{\partial}{\partial r}\right|_{r=0} \Acirc_{r,\tau} =
  0$.  From these formulas we find that
  \begin{equation*}
    \left.\tfrac{\partial}{\partial r}\right|_{r=0}
      \big( L_{r,\tau} L_{r,\tau}  +
      2H_{r,\tau}\Delta_{r,\tau} H_{r,\tau} - \tfrac12
      H_{r,\tau}^2|A_{r,\tau}|^2 \big) = 0.
  \end{equation*}
  Since $\Ric_{r,\tau}=O(r^2)$ as in~\eqref{eq:12} and
  $\nabla_{r,\tau}\Ric_{r,\tau}(\nu_{r,\tau},\nu_{r,\tau},\nu_{r,\tau}) = O(r^3)$ by a similar
  argument, also
  \begin{equation*}
    \left.\tfrac{\partial}{\partial r}\right|_{r=0} \big(
    - H_{r,\tau} \nabla_{r,\tau}\Ric_{r,\tau}(\nu_{r,\tau},\nu_{r,\tau},\nu_{r,\tau}) +
    \tfrac12H_{r,\tau}^2 \Ric_{r,\tau}(\nu,\nu) \big) = 0.
  \end{equation*}
  To treat the final remaining terms in $W_{r,\tau,\lambda}$ rewrite it to
  \begin{equation}
    \label{eq:23}
    \begin{split}
      \tfrac12 \nabla_{r,\tau}^*(&H_{r,\tau}^2 \nabla_{r,\tau} f) - 2 \nabla^*_{r,\tau}(H_{r,\tau}
      \Acirc_{r,\tau} (\nabla_{r,\tau} f, \cdot))
      \\
      &
      =
      -H_{r,\tau}\langle \nabla H_{r,\tau}, \nabla_{r,\tau} f \rangle
      -\tfrac12 H_{r,\tau}^2 \Delta_{r,\tau} f
      + 2 \Acirc_{r,\tau}(\nabla H_{r,\tau},\nabla_{r,\tau} f) \\
      &\quad
      + 2 H_{r,\tau}\langle \nabla_{r,\tau}^* \Acirc_{r,\tau}, \nabla_{r,\tau}
      f\rangle + 2 H_{r,\tau} \langle \Acirc_{r,\tau},\nabla_{r,\tau}^2 f\rangle\\
      &
      =
      -\tfrac12 H_{r,\tau}^2 \Delta_{r,\tau} f + 2 \Acirc_{r,\tau}(\nabla
      H_{r,\tau},\nabla_{r,\tau} f) + 2 H_{r,\tau} \langle \Acirc_{r,\tau},\nabla_{r,\tau}^2 f\rangle\\
      &\ \ \  +
      2H_{r,\tau}\omega_{r,\tau}(\nabla_{r,\tau} f).
    \end{split}
  \end{equation}
  In the last equality we used the Codazzi equation in the form
  $-\nabla^* \Acirc = \tfrac12 \nabla H + \omega$. By inspection we
  see that each term in this expression has vanishing derivative in
  $r$-direction. Hence
  \begin{equation*}
    \left.\frac{\partial}{\partial r}\right|_{r=0} W_{r,\tau,\lambda}f
    = 0
  \end{equation*}
  as claimed.
\end{proof}
Let $\varphi_0$ be the unique solution of the PDE
\begin{equation}
  \label{eq:13}
  W_{0,\tau,\lambda} \varphi_0=\left.\left(-\frac43 \Scal+4   \Ric_{pq}x^px^q\right)\right|_{r=0}.
\end{equation}
Note that the right hand side of this equation is an element of $K^\perp$ and hence $\varphi_0\in K^\perp$ is indeed uniquely defined.
\begin{lemma}
  \label{thm:solution-lambda-tau}
  Let $(M,g)$ be a three dimensional manifold and $p\in M$ such that
  $\nabla\Scal(p) = 0$ and $\nabla^2 \Scal(p)$ is non-degenerate. For
  this base point there exists $r_0\in(0,\infty)$, an open
  neighborhood $U\subset C^{4,\frac12}(S_1)$ of $\varphi_0$, and
  functions
  \begin{equation*}
    \begin{aligned}
      \lambda : [0,r_0) \times U \to \bR : (r,\varphi) \mapsto \lambda(r,\varphi) \quad\text{and}\quad
      \tau : [0,r_0) \times U \to \bR^3  : (r,\varphi) \mapsto \tau(r,\varphi)
    \end{aligned}
  \end{equation*}
  so that
  \begin{align}
    \label{projection0}
    &\tilde P_i
    \big(\Phi(r,\tau(r,\varphi),r^2\varphi,\lambda(r,\varphi))\big)=0
    \quad\text{for}\quad i\in\{0,1\},
    \\
    & \tau(0,\varphi_0) = 0, \qquad\text{and}\qquad
      \lambda(0,\varphi_0) = -\frac13 \Scal(p).
  \end{align}
 \end{lemma}
\begin{proof}
  Calculate:
  \begin{align*}
    &\Phi_{\varphi}(r,\tau,tr^2\varphi,\lambda)
    \\
    &\quad=
      \Phi_\varphi(0,\tau,0,\lambda)
      +
      r\int_0^1  \Phi_{\varphi r}(sr,\tau,str^2\varphi,\lambda)\, ds
      + tr^2 \int_0^1\Phi_{\varphi \varphi}(sr,\tau,str^2\varphi,\lambda)\varphi \, ds.
  \end{align*}
  Moreover, we have
  \begin{align*}
    r\int_0^1  \Phi_{\varphi r}(sr,\tau,str^2\varphi,\lambda)\, ds
    &=
      r^2 \int_0^1 \int_0^1 s\Phi_{\varphi r r}(usr,\tau,ustr^2 \varphi,\lambda)\, du \, ds\\
    &\quad
      + r^3\int_0^1 \int_0^1 st \Phi_{\varphi \varphi r} (usr,\tau,ustr^2\varphi,\lambda) \varphi \, du \, ds\\
    &\quad
      +r\Phi_{\varphi r}(0,\tau,0,\lambda).
  \end{align*}
  Hence
  \begin{align*}
    r^2 \Phi_{\varphi}(r,\tau,tr^2\varphi,\lambda)
    &=
      r^2 \Phi_\varphi(0,\tau,0,\lambda)+r^3 \Phi_{\varphi r}(0,\tau,0,\lambda)\\
    &\quad
      +r^4  \int_0^1 t \Phi_{\varphi \varphi}(sr,\tau,str^2\varphi,\lambda)\varphi \, ds \\
    &\quad
      +r^4 \int_0^1 \int_0^1 s\Phi_{\varphi r r}(usr,\tau,ustr^2 \varphi,\lambda)\, du \, ds +O(r^5).
  \end{align*}
  By equation~\eqref{eq:5} we have
  $\Phi_\varphi(0,\tau,0,\lambda)= W_{0,\tau,\lambda}$ and
  from  Lemma~\ref{thm:derivativePhi1} we get
  $\Phi_{\varphi r}(0,\tau,0,\lambda)=0$. Therefore
  \begin{equation*}
    \tilde P_i\big(r^2
    \Phi_{\varphi}(r,\tau,tr^2\varphi,\lambda)\big)=O(r^4)
    \quad\text{for}\quad i\in \{0,1\}.
  \end{equation*}
 It follows from
  Lemma~\ref{thm:projection-to-kernel} that the
  system~\eqref{projection0} is equivalent to
  \begin{align*}
    8\pi\left(\lambda+\frac13 \Scal\right)
    &= -r^{-2} \tilde P_0 \left(\int_0^1 \Phi_{\varphi}(r,\tau,tr^2\varphi,\lambda)\varphi\, dt\right)+O(r^2) =O(r^2)\ \ \ \text{and}\\
    \frac{4\pi}{3} \nabla_{e_i}\Scal e_i
    &= r \tilde P_1 \left(\int_0^1 \int_0^1 t \Phi_{\varphi \varphi}(sr,\tau,str^2\varphi,\lambda)\varphi \varphi \, ds \, dt\right)\\
    &\quad +r \tilde P_1 \left(\int_0^1 \int_0^1 \int_0^1
      s\Phi_{\varphi r r}(usr,\tau,ustr^2 \varphi,\lambda)\varphi \,
      du \, ds\, dt\right) +O(r^2)
    \\
    &=O(r).
  \end{align*}
  By assumption $\nabla \Scal(p)=0$. Hence, at $r=0$,
  this system is satisfied for an arbitrary $\varphi_0\in K^\perp$, if
  $\lambda|_{r=0}=-\frac13 \Scal(p)$ and $\tau|_{r=0} = 0$.

  The derivative with respect to $\lambda$ and $\tau$ at $r=0$ of the
  left hand side of this system is given by the matrix
  \[
  \begin{pmatrix}
    8\pi & \frac{8\pi}{3} \nabla \Scal|_{r=0} \\
    0 & \frac{4\pi}{3} \nabla^2 \Scal|_{r=0}
  \end{pmatrix}
  =
  \begin{pmatrix} 8\pi & 0 \\ 0 & \frac{4\pi}{3} \nabla^2 \Scal|_{r=0} \end{pmatrix}.
  \]
  By assumption $\nabla^2 \Scal|_{r=0}$ is non-degenerate. Hence, it
  follows from the implicit function theorem that there exist
  functions $\lambda=\lambda(r,\varphi)$ and $\tau=\tau(r,\varphi)$ as
  claimed at least for $(r,\varphi)$ in a neighborhood of
  $(0,\varphi_0)\in \bR\times C^{\frac12}(S_1)$.
\end{proof}

\begin{lemma}
  \label{thm:solution-final}
  Assume that $(M,g)$, $p$, $\phi_0$, $r_0$, $U$, $\lambda$ and $\tau$ are as in
  Lemma~\ref{thm:solution-lambda-tau}.

  Then there exists $r_1\in(0,r_0]$ and a function
  \begin{equation*}
    \varphi : [0,r_1) \to U : r\mapsto \varphi(r)
  \end{equation*}
  such that
  \begin{equation*}
    \begin{aligned}
      &\Phi(r,\tau(r,\varphi(r)),r^2\varphi(r), \lambda(r,\varphi(r)))
      = 0 \quad\text{and}
    \end{aligned}
    \varphi(0) = \varphi_0.
  \end{equation*}
\end{lemma}
In particular, for small enough $r$, we have constructed a surface of
Willmore type with Lagrange multiplier $\lambda(r,\varphi(r))$.
\begin{proof}
  Consider the expansion
  \begin{align*}
    \Phi(r,\tau,r^2\varphi,\lambda)
    &= \Phi(r,\tau,0,\lambda)+r^2 \int_0^1 \Phi_\varphi(r,\tau,tr^2 \varphi,\lambda)\, dt\\
    &= r^2\big(2\lambda -\frac{2}{3}\Scal+ 4\Ric_{pq}x^px^q\big)+O(r^3)+r^2 \Phi_\varphi(0,\tau,0,\lambda)\varphi\\
    &\quad
      +r^4 \int_0^1 \int_0^1 t\Phi_{\varphi \varphi} (sr,\tau,str^2\varphi,\lambda)\varphi \varphi \, dt\, ds\\
    &\quad
      +r^4\int_0^1 \int_0^1 \int_0^1 s \Phi_{\varphi r r}(usr,\tau,ustr^2 \varphi,\lambda)\varphi \, du\, ds\, dt\\
    &\quad
      +r^5\int_0^1 \int_0^1 \int_0^1 st \Phi_{\varphi \varphi r}(usr,\tau,ustr^2 \varphi,\lambda)\varphi \varphi \, du\, ds\, dt,
  \end{align*}
  where we used the fact that $\Phi_{\varphi r}(0,\tau,0,\lambda)=0$
  from Lemma~\ref{thm:derivativePhi1}.

  Since $\Phi_\varphi(0,\tau,0,\lambda)=W_{0,\tau,\lambda}$ as in equation~\eqref{eq:5},
  $\lambda(0,\varphi_0)=-\frac13 \Scal(p)$ and
  \begin{equation*}
    W_{0,\tau, \lambda} \varphi_0 =(-\frac43 \Scal+4
    \Ric_{pq}x^px^q)|_{r=0},
  \end{equation*}
  we conclude with the help of the implicit function theorem that,
  after dividing the above equation by $r^2$, there exists
  $r_1\in(0,r_0]$ and solution $\varphi:[0,r_1)\to U$ as claimed.
\end{proof}


TITLE: Question about a double-index recursion sequence
QUESTION [1 upvotes]: I have a question regarding a 2D sequence matrix equation defined as:

$v_{i,j} = 2(i-j)(2(i-j)-1)v_{i,j-1} - 4(i-j-1)^2 v_{i-1,j-1}$
$v_{0,0} = -1$
and $\forall k \neq 0, v_{0,k} = 0$
$v_{i,j} = 0$ if $i < 0$ or $j<0$

I solved this numerically a couple of times using algorithms, and I tried to guess a form.
I noticed $v_{i,j}/ \Gamma(2(i-j))$ seems to be a polynomial (at a fixed row) in terms of i but really high order.
What I want to do is to find $v_{i,j}$ analytically as a closed formula.
So far, I have reached the following conclusions using recursion,
$v_{i,0} = −(2i + 1)!$ and
$v_{i,1} =  \dfrac{(2i + 1)!}{3} (2i(i+1)(2i+1))$ and $v_{i,2}$ is a really high-order polynomial.
Matrix $M$:

Scheme of the recursion:

Also the diagonal is $-(-4)^k$.
It looks a bit like Pascal's triangle, it might be involved?
Mind helping me, please?

REPLY [1 votes]: Lemma: If $\{c_i\}$ are distinct constants and $\{u(i, j)\}$ satisfy\begin{gather*}
u(i, 0) = 1\ (i \geqslant 0),\ u(-1, j) = 0\ (j \geqslant 0),\\
u(i, j) = u(i - 1, j) + c_i u(i, j - 1)\ (i \geqslant 0,\ j \geqslant 1),
\end{gather*}
then\begin{gather*}
u(i, j) = \sum_{k = 0}^i \frac{c_k^{i + j}}{\prod_{\substack{0 \le l \le i\\l ≠ k}} (c_k - c_l)}. \quad \forall i, j \geqslant 0 \tag{1}
\end{gather*}
Proof: It can be calculated by the boundary condition that $u(0, j) = c_0^j$ ($j \geqslant 0$). For $i \geqslant 0$, using Lagrange polynomials to interpolate the function $y = x^i$ yields$$
\sum_{k = 0}^i c_k^i \cdot \frac{\prod_{\substack{0 \le l \le i\\l ≠ k}} (x - c_l)}{\prod_{\substack{0 \le l \le i\\l ≠ k}} (c_k - c_l)} \equiv x^i,
$$
and comparing the coefficient of $x^i$ yields$$
\sum_{k = 0}^i \frac{c_k^i}{\prod_{\substack{0 \le l \le i\\l ≠ k}} (c_k - c_l)} = 1 = u(i, 0).
$$
Then (1) can be proved by induction on $i$ and $j$.
Now back to the question. The recurrence relation to be solved is\begin{gather*}
v(0, 0) = -1,\ v(i, 0) = 0\ (i \geqslant 1),\ v(-1, j) = 0\ (j \geqslant 0),\\
v(i, j) = 2(i - j)(2(i - j) - 1) v(i, j - 1) - 4(i - j - 1)^2 v(i - 1, j - 1)\ (i \geqslant 0,\ j \geqslant 1).
\end{gather*}
(This is the most likely what you mean based on your expressions for “$v(i, 0)$” and “$v(i, 1)$” since there seems to be a few typos in the indices.) It is easy to prove that $v(i, j) = 0$ for $i > j$. Define\begin{gather*}
u(-1, j) = 0\ (j \geqslant 0),\\
u(i, j) = \frac{(-1)^{j + 1}}{4^j \cdot (2i + 1)!} \cdot v(j, i + j)\ (i, j \geqslant 0),
\end{gather*}
then it is not hard to prove that $\{u(i, j)\}$ satisfy the conditions in the lemma with $c_i = (i + 1)^2$. By the lemma,$$
u(i, j) = \sum_{k = 0}^i \frac{(k + 1)^{2(i + j)}}{\prod_{\substack{0 \le l \le i\\l ≠ k}} ((k + 1)^2 - (l + 1)^2)}, \quad \forall i, j \geqslant 0
$$
thus for $i \leqslant j$,\begin{align*}
v(i, j) &= (-1)^{i + 1} 4^i \cdot (2(j - i) + 1)! \cdot u(j - i, i)\\
&= (-1)^{i + 1} 4^i \cdot (2(j - i) + 1)! \sum_{k = 0}^{j - i} \frac{(k + 1)^{2j}}{\prod_{\substack{0 \le l \le j - i\\l ≠ k}} ((k + 1)^2 - (l + 1)^2)}\\
&= (-1)^{i + 1} 4^i \cdot (2(j - i) + 1)! \sum_{k = 0}^{j - i} \frac{(k + 1)^{2j}}{\dfrac{(-1)^{j - i - k}}{2(k + 1)^2} \cdot (j - i - k)!\, (j - i + k + 2)!}\\
&= \frac{4^i}{j - i + 1} \sum_{k = 0}^{j - i} (-1)^{j - k + 1} (k + 1)^{2j + 2} C(2(j - i) + 2, j - i - k).
\end{align*}