TITLE: A System of differential Equations
QUESTION [1 upvotes]: How can I analyze the phase diagram for this system of differential eqs?

This field is not my area of my expertise, so please be generous with the answers. I appreciate quick references as well. Thank you.

REPLY [4 votes]: Well, you start by finding the zeroes of the vector field
$X(x, y) = \begin{pmatrix} X_x \\ X_y \end{pmatrix} = \begin{pmatrix} x(7 - x - 2y) \\ y(5 - y - x) \end{pmatrix}, \tag{1}$
which corresponds to the given differential equation since we may write
$\mathbf r = \begin{pmatrix} x \\ y \end{pmatrix} \tag{2}$
so that the ODE becomes
$\dot{\mathbf r} = X(\mathbf r). \tag{3}$
It's pretty easy to see that the zeroes of $X$ occur at $(0, 0)$, $(0, 5)$, $(7, 0)$, and $(3, 2)$.  Next you can work out the Jacobean matrix $J_X$ of $X$ in general terms, that is
$J_X = \begin{bmatrix} \dfrac{\partial X_x}{\partial x} &  \dfrac{\partial X_x}{\partial y} \\  \dfrac{\partial X_y}{\partial x} &  \dfrac{\partial X_y}{\partial y} \end{bmatrix}. \tag{4}$
which, using the specific form of $X(x, y)$ from (1) yields
$J_X = \begin{bmatrix} 7 - 2x - 2y & -2x \\ -y & 5 - 2y - x \end{bmatrix}. \tag{5}$
Then we plug in the numbers at each zero of $X$, and find the eigenvalues of $J_X$ at each zero.  The eigenvalues are easily found using the characteristic polynomial of $J_X$ at each point, a simple quadratic equation.  I tabulate the results below:
$J_X(0, 0) = \begin{bmatrix} 7 & 0 \\ 0 & 5 \end{bmatrix}; \; p_J(\lambda) = \lambda^2 - 12\lambda + 35; \; \lambda = 5, 7. \tag{7}$
$J_X(0, 5) = \begin{bmatrix} -3 & 0 \\ -5 & -5 \end{bmatrix}; \; p_J(\lambda) = \lambda^2 + 8\lambda + 15; \; \lambda = -3, -5. \tag{8}$
$J_X(7, 0) = \begin{bmatrix} -7 & -14 \\ 0 & -2 \end{bmatrix}; \; p_J(\lambda) = \lambda^2 + 9\lambda + 14; \; \lambda = -7, -2. \tag{9}$
$J_X(3, 2) = \begin{bmatrix} -3 & -6 \\ -2 & -2 \end{bmatrix}; \; p_J(\lambda) = \lambda^2 + 5\lambda -6; \; \lambda = -6, 1. \tag{10}$
Finally, using (7)-(10) we can describe the types of the zeroes of $X$ and the flow surrounding them.  $(0, 0)$ is a repelling point; all trajectories starting near $(0, 0)$ move away from it as $t \to \infty$; $(0, 5)$ and $(7, 0)$ are attractors, points sufficiently near each of them converge toward the one they start near under the action of the flow, as $t \to \infty$; the remaining zero of $X$, at $(3, 2)$, is a saddle and there are, locally (that is, sufficiently near), exactly two trajectories which approach $(3, 2)$ as $t \to \infty$ and two which approach $(3, 2)$ as $t \to -\infty$.  Near $(3, 2)$ the rest of the integral curves of $X$ have a vaguely hyperbolic appearance, heading towards $(3, 2)$ for awhile and then abruptly turning away.  Taking the trouble to calculate (and sketch) all the eigenvectors for the matrices in (7)-(10) strongly enhances the purely verbal imagery I exploit as I give this somewhat crude description of the phase portrait of $X$, but I type these words as I concurrently work on several other projects, and I am not really in a mood for more arithmetic at the moment.  Perhaps I'll leave my old buddy Amzoti an note in the comments and he'll come on down here and get his programs to make us a nice phase portrait; I don't have the graphics SW to do this at the moment.  But I think anyone who works out the eigenvectors of these four $J_X$ will quickly get my meaning.  It should be observed that there is one special trajectory $\gamma(t)$ such that $\gamma(t) \to (0, 0) \; \text{as} \; t \to -\infty$ and $\gamma(t) \to (3, 2) \; \text{as} \; t \to \infty$; this is the seperatrix which effectively serves as a "switch point" between integral curves emanating from $(0, 0)$ ultimately headed for $(0, 5)$ and those which are destined for $(7, 0)$.  Well, I hope this verbage was at least a tad on the clear side; I hope Amzoti will show up with a phase portrait.
A good book for this kind of stuff is Differential Equations, Dynamical Systems and an Introduction to Chaos by Hirsch, Smale, and Devaney, Academic Press, 2004.  A more advanced treatment of many of these topics may be found in Ordinary Differential Equations by Jack Hale, available from Dover for a song.  Finally,the theorem which puts teeth into the derivation of these phase portraits is called The Stable Manifold Theorem; it can be found in wikipedia here.  There is also a somewhat cursory discussion of it in my answers to this question.
Well, even if my answer is not the $\Omega$ of this topic, it is at least an $A$!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
Thanks to Amzoti for the fine graphics below!
Update
Here is the systems phase portrait using two different tools.


TITLE: Jordan normal form (Basis)
QUESTION [0 upvotes]: Define $A = \begin{pmatrix}  -7 & -32 & -32 & -35 \\ 1 & 5 & 4 & 4 \\ 1 & 4 & 5 & 5 \\ 0 & 0 & 0 & 1  \end{pmatrix} \in \mathbb{C^{4x4}}$
I computed the Jordan normal form of this matrix:
$J=\begin{pmatrix}  1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1  \end{pmatrix}$.
But how to find a Jordan basis for this matrix?
Since $ker(A-I)^3=0$ I can't choose a $v \in ker(A-I)^3$ \ $ker(A-I)^2$
Also, I tried to use $v \mapsto Av$
$Ab_1=b_1$,
$Ab_2=b_2$,
$Ab_3=b_2+b_3$,
$Ab_4=b_3+b_4$
But I don't know how to continue.

REPLY [0 votes]: From here we can solve as follow

$Ab_1=b_1\implies (A-I)b_1=0$ and find $b_1$
$Ab_2=b_2\implies (A-I)b_2=0$ and find $b_2$
$Ab_3=b_2+b_3 \implies (A-I)b_3=b_2$ and find $b_3$
$Ab_4=b_3+b_4 \implies (A-I)b_4=b_3$ and find $b_4$


TITLE: $D^n \times D^m \cong D^{n+m}$
QUESTION [2 upvotes]: I am trying to prove that the cartesian product of the closed n dimensional disk with the closed m dimensional disk is homeomorphic to the n+m dimensional disk via a function that maps the boundary to the boundary homeomorphically.
Symbolically I am looking for a continuous function $f:D^n \times D^m \rightarrow D^{n+m}$ such that $f$ restricted to $S^{n-1} \times D^m \cup D^n \times S^{m-1} \stackrel{f}{\longrightarrow} S^{n+m-1}$ is also a homeomorphism. Here is what I have cooked up so far
First, define $k:(D^n \times D^m - \{0,0\}) \rightarrow (0,\infty)$ to be $k(x,y) = \frac{1}{min\{c_1,c_2\}}$ where $c_1|x| = 1$ and $c_2|y| = 1$ and if no such number exists set $c_i =\infty$. The purpose of this function is to generate the appropriate stretching factor for a point in $D^n \times D^m$ relative to the boundary. Now we can define $f(x,y) = (k(x,y)\frac{x}{|x,y|} , k(x,y)\frac{y}{|x,y|})$ for $(x,y) \ne (0,0)$ and $f(0,0)=(0,0)$.
I believe that this should work. However proving continuity is another issue, and there may be a more streamlined way to go about this. By the way, this problem comes from Strom's Modern Classical Homotopy Theory

REPLY [4 votes]: I presume $D^n$ is the closed unit disc.
Define the closed cube by
$$C^n=[-1,1]^n.$$
For each $n$ there is a homeomorphism $\Phi_n:C^n\to D^n$ (preserving
boundaries). As $C^m\times C^n$ is $C^{m+n}$, then you can take
your homeomorphism to be $\Phi^{m+n}\circ(\Phi_m^{-1}\times \Phi_n^{-1})$.
To construct $\Phi_n$, consider a nonzero point $x=(x_1,\ldots,x_n)\in C^n$. The set $D^n$ is $\{x\in\Bbb R^n:\|x\|_2\le1\}$, and
$C^n$ is $\{x\in\Bbb R^n:\|x\|_\infty\le1\}$. Here $\|x\|_2=\sqrt{x_1^2
+\cdots+x_n^2}$ and $\|x\|_\infty=\max(|x_1|,\ldots,|x_n|)$. Define
$\Phi_n(x)=(\|x\|_2/\|x\|_\infty)x$. Also define $\Phi_n(0)=0$.
The continuity at $0$ is not entirely obvious: it follows from the boundedness of $\|x\|_2/\|x\|_\infty$ on the rest of the cube.
Anyway, $\Phi_n$ is a continuous bijection between compact Hausdorff
spaces, and so a homeomorophism.


TITLE: In rainy days John is late $30\%$ of the time. In non rainy, $10\%$. There is a $70\%$ chance it will rain.
QUESTION [2 upvotes]: In rainy days, John is late for work with a probability of $0.3$. In non
  rainy days, he is late with a probability of $0.1$ There is a $0.7$
  probability that it will rain tomorrow. Determine the probability that
  John won't be late for work tomorrow.

What I need is:
$$P(\lnot L|R) = \frac{P(\lnot L\cap R)}{P(R)}$$
I know $P(R) = 0.7$.
I also have
$$P(L|R)= 0.3\\
P(L|\lnot R)=0.1 \\
$$
From this I can find
$$P(L|R)=\frac{P(L\cap R)}{P(R)} \Leftrightarrow 0.3=\frac{P(L\cap R)}{0.7}\Leftrightarrow P(L\cap R) = .21$$
$$P(L|\lnot R) = \frac{P(L \cap \lnot R)}{P(\lnot R)}\Leftrightarrow0.1 = \frac{P(L \cap \lnot R)}{0.3}\Leftrightarrow P(L \cap \lnot R) = 0.03$$
So the probability of him being late should be $.21+.03=.24$. Then $P(\lnot L) = 0.76$.
What is $P(\lnot L \cap R)$? I feel like I made a mistake somewhere.

REPLY [0 votes]: Not so terribly confusing if you try thinking rather than just applying formulas!
Imagine this happening 10000 days.  70% of the time, 7000 days, it rains.  The other 3000 days it does not rain.  Of the 7000 days when it rains, John is late 70% of the time, 4900 days, on time the other 2100 days.  Of the 3000 days when it does not rain, John is late 0.1% of the time, 3 days, not late the other 2997 days.
So of the entire 10000 days John is late 4903 days or 49.3% of the time.


text:
\section{Problem Description}
\label{sec:model}
We aim to solve the swimming problem, i.e., to recover the swimming motion from a given history of shape changes. To do so, we define the mathematical and numerical modeling required to properly study micro-organisms swimming in a fluid medium. In particular, the swimming model is composed by a kinematic part describing the possible motions for the swimmer and a dynamic part defining the conservation laws for the fluid domain. The final resolution strategy couples these two models and is solved numerically either in a monolithic approach~\eqref{monolithic_system_complete_1}
 or in a split approach.

\subsection{Swimming model}
Following \cite{Maso2011,DalMaso2015} the swimmer is represented as a time-dependent bounded open set $B_t \in \mathbb{R}^3$. The map $\chi:  \bar{B}_0 \subset \mathbb{R}^3 \times
[0,T] \rightarrow \mathbb{R}^3$ defines the position $x$ at time $t$ of a material point $X$ of the swimmer, namely
\begin{equation}
x(X,t) = \chi(X,t) = q(t) + R(t)s(X,t),
\label{GenericDispl}
\end{equation}
where $q(t)$, $R(t)$, $s(X,t)$ are the position of a point of the swimmer, the rotation of the system local frame and the shape at time $t$. The time derivative of \eqref{GenericDispl} defines the velocity of any material point on the swimmer,
\begin{equation}
\begin{aligned}
u_{s} = &\dot{x} = \dfrac{\partial \chi(X,t)}{\partial t} \\=&  \dfrac{d
  q}{d t} + R(t)\dfrac{\partial s(X,t)}{\partial t} + \dfrac{d
  R(t)}{d t} s(X,t)=\\
=&\dot{q}(t) + R(t)\dot{s}(X,t) + \omega(t) \wedge (R(t) s(X,t)).\\
\end{aligned}
\label{u_swimmer}
\end{equation}
We assume $s(X,t)$ to be known, we define ``shape velocity" the quantity $R(t)\dot{s}(X,t)$, therefore the unknowns are the rigid movements $q(t)$, $R(t)$, which define the linear and angular velocities $\dot{q}(t), \omega(t)$.
In the case of self-propelled swimmers, only viscous drag is acting on the swimmer so the usual momentum balance laws read
\begin{subequations}
\begin{equation}
\int_{\Gamma} f(x) d\gamma(x)  =0,
\label{LinearMomentumBalanceDummy}
\end{equation}
\begin{equation}
\int_{\Gamma} f(x) \wedge (x-x_0) d\gamma(x) =0.
\label{AngularMomentumBalanceDummy}
\end{equation}
\label{DummyMomentumSystem}
\end{subequations}
The viscous tractions $f$ are given by the action of the Cauchy stress tensor $\sigma$, see~\cite{Gurtin}, namely
\begin{equation}
f = \sigma(u,p) n,
\label{ForceContinuumMechanics}
\end{equation}
where $n$ is the outer unit normal vector to the surface, and $u,p$ represent the velocity and the pressure
in the fluid.

\subsection{Fluid model}
We study micro-swimmers immersed in a Newtonian incompressible fluid,
by writing the non-dimensional formulation of the incompressible Navier-Stokes equation in a generic domain $\Omega$, see~\cite{Gurtin}, as
\begin{subequations}
\begin{equation}
\nabla \cdot u = 0 \quad  \text{ in } \Omega  \
\label{MassConservationNStokesintro}
\end{equation}
\begin{equation}
Re \left( \bar{\sigma}\frac{\partial u}{\partial t} + (\nabla u) u \right)= \Delta u -\nabla p \quad \text{ in } \Omega,
\label{LinearMomentumBalanceNStokesintro}
\end{equation}
\label{NavierStokesSystemintro}
\end{subequations}
where $u, p$ are non dimensional velocity and pressure and $Re$ is the Reynolds number defined as
\begin{equation}
Re = \frac{\rho U L}{\mu},
\label{Reynolds}
\end{equation}
while
\begin{equation}
\bar{\sigma} = \frac{\omega L}{U}
\label{Womersley}
\end{equation}
is the Womersley number.
We consider the typical length, velocity and frequency for a micro-swimmer as $L = 10^{-5} m$, $U = 10^{-5} m/sec$ and $\omega = 10^{2} Hz$, we obtain $Re=10^{-4}$ and $\bar{\sigma}Re = 10^{-2}$. Therefore the system of equations governing the flow is the Stokes system,
\begin{subequations}
\begin{equation}
\nabla \cdot u = 0 \quad \text{ in } \Omega  \
\label{MassConservationStokesintro}
\end{equation}
\begin{equation}
\Delta u -\nabla p = 0 \quad \text{ in } \Omega.
\label{LinearMomentumBalanceStokesintro}
\end{equation}
\label{StokesSystemintro}
\end{subequations}
We identify the boundary of the fluid domain with the boundary of the swimmer $\partial \Omega = \Gamma$, we write Dirichlet's boundary conditions for \eqref{StokesSystemintro} as
\begin{equation}
u = u_{s},
\end{equation}
where $u_s$ denotes the velocity of the swimmer on the boundary defined in \eqref{u_swimmer}.
The system~\eqref{StokesSystemintro} with Dirichlet boundary conditions is well posed and admits a unique solution $(u,p) \in ((H^1(\Omega))^3, L_2(\Omega))$.

Following \cite{Pozrikidis1992} we rewrite \eqref{StokesSystemintro} using the fundamental solution to get the representation formula for the Stokes system
\begin{equation}

u_i(x) - \int_{\Gamma} W_{ijk}(x,y) n_k(y) u_j(y) d\gamma_y = \int_{\Gamma}
G_{ij}(x,y) f_j(y)d\gamma_y \ \forall x \in \mathbb{R}^d
  \setminus \Gamma,

\label{StokesRepGeneral}
\end{equation}
where $G, W$ are the first two fundamental solution for the Stokes system.
We take the trace of \eqref{StokesRepGeneral} to obtain the Boundary Integral Equation (BIE) of the Stokes system as
\begin{equation}
\alpha(x)u_i(x) - \int_{\Gamma}^{PV} T_{ijk}(x,y) n_k(y) u_j(y) d\gamma_y = \int_{\Gamma}
G_{ij}(x,y) f_j(y)d\gamma_y \ \forall x \in \Gamma,
\label{StokesBIEGeneral}
\end{equation}
where the integral on the left is computed in the Cauchy principal value
sense, and $\alpha$ represents its Cauchy principal value.
We define the single and double layer operator $V, H$ as
\begin{subequations}
\begin{equation}
H: (H^{\frac{1}{2}}(\Gamma))^3 \rightarrow (H^{\frac{1}{2}}(\Gamma))^3,
\label{doublelayerfunctional}
\end{equation}
\begin{equation}
V: (H^{-\frac{1}{2}}(\Gamma))^3 \rightarrow (H^{\frac{1}{2}}(\Gamma))^3,
\label{singlelayerfunctional}
\end{equation}
\label{boundaryoperatorsfunctional}
\end{subequations}
where $ (H^{\frac{1}{2}}(\Gamma))^3$ represents the space of the traces of the functional $(H^1(\Omega))^3$ and $ (H^{-\frac{1}{2}}(\Gamma))^3$ is its dual space.
We rewrite \eqref{StokesBIEGeneral} using the operators~\eqref{doublelayerfunctional} and~\eqref{singlelayerfunctional} as
\begin{equation}
\left[ \alpha I - H \right] u = -K  u=-  V  f.
\label{StokesOperator}
\end{equation}
We remark that we use the complete traction $f$ as given by~\eqref{ForceContinuumMechanics}. For this reason, differently from standard Finite Element simulations of~\eqref{NavierStokesSystemintro}, we don't need to solve for the pressure $p$ and the problem is not a saddle point problem. The usage of the fundamental solutions $G, T$ guarantees that the velocity in the fluid domain is automatically divergence free, see~\cite{Steinbach2008} for greater details. For these reasons operator $V$ is coercive, similarly to what happens to the classic Stokes system~\eqref{NavierStokesSystemintro}  if we consider a velocity domain which is divergence free.

\subsection{Numerical resolution of the swimming problem}
\label{sec:fom_resolution}
To solve the swimming problem means to recover the translation $q(t)$ and rotation $R(t)$ given the history of shape changes.
For this purpose we use the numerical methodology presented and validated in~\cite{Giuliani2018SoRo} to discretize \eqref{StokesBIEGeneral} into a Boundary Element Method (BEM) \footnote{In particular we use the open source software \bemstokes which is freely available under LGPL license v.2.1 on github~\cite{BEMStokesrepo}}. We discretize both the geometry of the swimmer and the unknowns $f, u$ using  Lagrangian finite element and we apply a collocation method to derive the BEM linear system
\begin{equation}
\hat{K}  \hat{u}= \hat{V} \hat{f},
\label{StokesOperator2}
\end{equation}
where $\hat{\dots}$ represents the finite dimensional counterpart of the operators and functionals represented in~\eqref{StokesOperator}. Following~\cite{Steinbach2008} we express the tractions as a function of the velocity introducing the discretized Dirichlet to Neumann map
\begin{equation}
\hat{T} = \left[ \hat{V}^{-1} \hat{K} \right],
\label{DNMap}
\end{equation}
so that $\hat{f} = \hat{T}\hat{u}$. We will drop the $\hat{\dots}$ in the rest of the paper for the sake of simplicity.
Following~\cite{Giuliani2018SoRo} we rewrite \eqref{u_swimmer} using a suitable set of basis functions for the rigid velocity $P(X,t)$ and introducing the shape velocity in the swimmer body frame $v(X,t) = R(t) \dot{s}(X,t)$, as
\begin{equation}
\begin{aligned}
u_{s} &= \dot{q}(t) + \omega(t) \wedge R(t) s(X,t) + v(X,t) \\
&=\sum_{i=1}^{N_{r}} p_i(X,t) \dot{p}_i(t)+ v(X,t) \\
&=P(X,t) \dot{p}(t) + v(X,t).
\end{aligned}
\label{u_swimmer2}
\end{equation}
We assume $\dot s(X,t)$ to be known, so at time $t$, given $R(t)$, we compute $v(X,t)$ and we write
\begin{subequations}
\begin{equation}
f_{shape} = T v,
\label{shapeforce}
\end{equation}
together with
\begin{equation}
F_{rigid} = T P.
\label{rigidforce}
\end{equation}
\label{DNforces}
\end{subequations}
We obtain the rigid velocity coefficients $\dot{p} (t)$ using~\eqref{DummyMomentumSystem} as
\begin{equation}
P^T M F_{rigid} \dot{p} + P^T M f_{shape} = 0,
\label{numericalmomentumbalance}
\end{equation}
where $M$ is the mass matrix taking care of the surface integrals of~\eqref{DummyMomentumSystem}.
We then solve ~\eqref{numericalmomentumbalance} to get
\begin{equation}
\dot{p} = - \left[ P^T M F_{rigid}\right]^{-1} P^T M f_{shape}.
\label{solvingequation}
\end{equation}
We remark that $ \left[ P^T M F_{rigid}\right]$ is a symmetric negative definite matrix called grand resistive matrix, so its inversion in~\eqref{solvingequation} is legit. We call this resolution strategy, where we compute the tractions associated to any possible velocity and then combine them, the \emph{split approach}.

Another possible resolution strategy is the so-called \emph{monolithic approach}~\cite{Pimponi2016,Giuliani2018SoRo}, which is based on a single resolution to obtain the overall traction $f$ together with the rigid velocity coefficients. We write the discretized BIE together with the constraint~\eqref{numericalmomentumbalance} as
\begin{equation}

\begin{bmatrix}
[V] & -[K]P \\
P^T [M]  & 0
\end{bmatrix}
\begin{bmatrix}
f\\
 \dot{p}
\end{bmatrix}
=
 \begin{bmatrix}
[K] v\\
 0
\end{bmatrix},
\label{monolithic_system_complete_1}
\end{equation}
the solution of~\eqref{monolithic_system_complete_1} is equivalent to the split approach. The differences lie in the fact that the monolithic approach only needs one resolution while the split needs seven. Moreover even the spectral properties of the matrices change if we impose the constraint during the BEM resolution. We now apply ROM to both full order approaches to highlight pros and cons of the two resolution strategies.

Standard numerical integration of $\dot{p}(t)$ provides the rigid displacements and rotations.
The \bemstokes library efficiently combines several algorithms and ideas in a flexible, modular, and extendible way.
It exploits distributed memory parallelism (MPI) using an automatic splitting of the workload at algebraic level using on the graph partitioning tool METIS~\cite{GeorgeKarypis1998},
with the high performance computing libraries Trilinos~\cite{Heroux2005}, and \deal~\cite{dealII90}, used to  tackle distributed linear algebra.
A similar combination has been successfully applied to achieve high computational efficiency in fluid dynamics, as demonstrated in ASPECT~\cite{Kronbichler2012} and $\pi$-\texttt{BEM}~\cite{Giuliani2018}.

config: arxiv


TITLE: Let $A$ be a convex and closed subset of $\mathbb R²$ which not contains $0$. Show that there is a vectorial line separated from A.
QUESTION [1 upvotes]: I have not studied Hahn-Banach theorem yet, so I have to do without.
I made a drawing, it's quite clear on it :)
Thank you for your help.

REPLY [0 votes]: Let $r = \inf \{d(0,a) \mid a \in A\}$. By assumption $r > 0$. By definition $A \subset B(0,r)^c$.
Moreover there is a unique point $p \in A \cap \partial B(0,r)$. For the unicity, if there is two points on the boundary, then by convexity the segment between them is inside the circle, contradicting definition of $r$. The existence is because $A$ is closed.
As you said, take $L$ perpendicular to the segment $\overline{ Op}$. Assume there is a point $q \in L \cap A$. Then, by convexity the segment $\overline{qp} \subset A$. Now the segment $\overline{qp}$ intersect $\partial B(0,r)$ in some point $p' \neq p$ which is a contradiction by unicity of the point of the boundary.
Remark : if you are not convinced that $\overline{qp}$ intersect $\partial B(0,r)$ in another point that $p$, here is an algebraic proof. By rotating we can assume $p = (1,0)$ and $q = (0,y)$. The line between $p$ and $q$ is parametrized by $(1+t, - yt)$ (for example). Now we want to solve the equation $f(t) = 0$ where $f(t) = (1+t)^2 + (yt)^2 - 1$. A little calculation gives $t = \frac{-2}{y^2 + 1}$ is the desired point.


text:
\begin{document}

\title{Visualize Geometric Series}

\author{H\`ung Vi\d{\^e}t Chu}
\email{\textcolor{blue}{\href{mailto:hungchu2@illinois.edu}{hungchu2@illinois.edu}}}
\address{Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL 61820, USA}

\subjclass[2010]{40A05}

\keywords{}

\maketitle

\begin{abstract}
    We review some ``proofs without words'' for the formula for geometric series and find a common theme lurking behind them. We also answer negatively the question raised by Edgar on the existence of other proofs similar to Mabry's and his.
\end{abstract}

\section{Introduction}
The well known formula for geometric series
\begin{align}\label{k00}1+x+x^2+x^3\cdots \ =\ \frac{1}{1-x},\end{align}
for any $0<x<1$, is very useful due to its ability to convert an infinite sum into a fraction. A common proof of this formula is as follows:
\begin{align*}
&(1-x)(1+x+x^2+x^3+\cdots+x^n)\\
\ =\ &(1+x+x^2+x^3+\cdots+x^n) - (x+x^2+x^3+x^4+\cdots+x^{n+1})\ =\ &1 - x^{n+1}.
\end{align*}
Hence, we have $(1-x)(1+x+x^2+x^3+\cdots+x^n) = 1-x^{n+1}$. Dividing both sides by $1-x$, we arrive at
$$1+x+x^2+x^3+\cdots + x^n \ =\ \frac{1-x^{n+1}}{1-x}.$$
Letting $n\rightarrow \infty$ and noting that $\lim_{n\rightarrow \infty} x^{n+1} = 0$ because $0<x<1$, we have the desired formula. Though the proof is elegant, it does not give a clear intuition behind the formula. For this reason, a number of ``proofs without words'' have been proposed. The goal of this article is to review proofs without words by various authors, find the common theme in these proofs, and answer an open question by one of these authors about the existence of similar proofs.

In 2001, the Viewpoints 2000 Group published Figure \ref{fig: Viewpoints 2000 Group} to prove \eqref{k00} without words. In 1999, Mabry \cite{Ma} published Figure \ref{fig:Mabry's proof} as a visual proof of $\frac{1}{4} + (\frac{1}{4})^2+ (\frac{1}{4})^3+\cdots   = \frac{1}{3}$. Though Mabry's proof is only for a case of \eqref{k00}, the proof provides nice intuition behind the formula. For convenience, we assume that the area of the outmost triangle is $1$. The idea is to find the colored area in two ways. The first way is to find the ratio of the colored area in each layer, which is $1/3$. The second way is to add up the areas of all colored triangles, which gives $1/4+(1/4)^2+(1/4)^3+\cdots$. The two ways must give the same answer, so we have the desired identity. Continuing the work, in 2016, Edgar \cite{Ed} proved that $\frac{4}{9}+(\frac{4}{9})^2+(\frac{4}{9})^3 + \cdots = \frac{4}{5}$ using Mabry's technique (see Figure \ref{fig: Edgar's}). At the end of his proof, Edgar asked: ``Is it possible to determine which other series allow ananalogous proof without words?''. We will show that the answer is negative in Section 2.

\begin{figure}[h!]
\begin{center}
\includegraphics[scale=.7]{VP.eps}
\end{center}
\caption{Viewpoints 2000 Group's proof} \label{fig: Viewpoints 2000 Group}
\end{figure}

\begin{center}
\begin{figure}[h!]
\begin{tikzpicture}
[acteur/.style={circle, fill=black,thick, inner sep=0.3pt, minimum size=0.01cm}]
\draw (-2,0)--(2,0);
\draw (-2,0)--(0,3.4641);
\draw (2,0)--(0,3.4641);
\filldraw[draw=black, fill=cyan] (-1,1.732) -- (1, 1.732) -- (0,0) -- cycle;
\filldraw[draw=black, fill=cyan] (-0.5,2.598) -- (0, 1.732) -- (0.5,2.598) -- cycle;
\filldraw[draw=black, fill=cyan] (0.25,3.031) -- (0, 2.598) -- (-0.25,3.031) -- cycle;
\draw (2,0)--(1, 1.732) node[pos=0.5,right] {layer 1};
\draw (1,1.732)--(0.5, 2.598) node[pos=0.5, right] {layer 2};
\draw (0.5,2.598)--(0,3.464) node[pos=0.5, sloped, above] {$\cdots$};
\node (a1) at (0,3.15) [acteur][label = below: ]{};
\node (a2) at (0,3.25) [acteur][label = below: ]{};
\node (a3) at (0,3.35) [acteur][label = below: ]{};
\end{tikzpicture}
\caption{Mabry's proof.} \label{fig:Mabry's proof}
\end{figure}
\end{center}
\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.5]{Edgar.eps}
\end{center}
\caption{Edgar's proof} \label{fig: Edgar's}
\end{figure}

\section{Demystify Previous Visual Proofs}
In this section, we analyze the technique used by Mabry and Edgar thoroughly. Understanding the technique is the key to answering the question posed by Edgar and allows us to connect it with the proof by the Viewpoints 2000 Group to find a common theme.
\subsection{Backbones of the beautiful proofs by Mabry and Edgar}
\begin{center}
\begin{figure}[h!]
\begin{tikzpicture}
[acteur/.style={circle, fill=black,thick, inner sep=0.3pt, minimum size=0.01cm}]
\draw (-2,0)--(2,0);
\draw (-2,0)--(0,3.4641);
\draw (2,0)--(0,3.4641);
\draw (1.6, 0.693)--(-1.6,0.693);
\draw (1.28, 1.247)--(-1.28, 1.247);
\draw (1.024, 1.69)--(-1.024, 1.69);
\draw (0.819, 2.045)--(-0.819, 2.045);
\draw (0.655, 2.329)--(-0.655, 2.329);
\node (a1) at (0,2.613) [acteur][label = below: ]{};
\node (a2) at (0,2.755) [acteur][label = below: ]{};
\node (a3) at (0,2.897) [acteur][label = below: ]{};
\node (a4) at (-2,0) [acteur][label = below: $C$]{};
\node (a5) at (2,0) [acteur][label = below: $B$]{};
\node (a6) at (0,3.4641) [acteur][label = above: $A$]{};
\node (a7) at (1.6, 0.693) [acteur][label = right: $D$]{};
\draw (a5)--(a7) node[pos=0.5, right] {layer 1};
\draw (1.28, 1.247)--(a7) node[pos=0.5, right] {layer 2};
\draw (1.28, 1.247)--(1.024,1.69) node[pos=0.5, right] {layer 3};
\node (a8) at (-1.6, 0.693) [acteur][label = left: $E$]{};
\filldraw[draw=black, fill=cyan] (1.2,0) -- (1.6, 0.693) -- (2,0) -- cycle;
\node (a9) at (0.25,0.35) [acteur][label = below: ]{};
\node (a10) at (0,0.35) [acteur][label = below: ]{};
\node (a11) at (-0.25,0.35) [acteur][label = below: ]{};
\filldraw[draw=black, fill=cyan] (0.96,0.6928) -- (1.28, 1.247) -- (1.6, 0.693) -- cycle;
\node (a12) at (0.25,0.95) [acteur][label = below: ]{};
\node (a13) at (0,0.95) [acteur][label = below: ]{};
\node (a14) at (-0.25,0.95) [acteur][label = below: ]{};
\filldraw[draw=black, fill=cyan] (0.768, 1.247) -- (1.024, 1.69) -- (1.28, 1.247) -- cycle;
\filldraw[draw=black, fill=cyan] (0.614, 1.69) -- (0.819, 2.045) -- (1.024, 1.69) -- cycle;
\filldraw[draw=black, fill=cyan] (0.491, 2.045) -- (0.655, 2.329) -- (0.819, 2.045) -- cycle;
\node (a15) at (0.25,1.45) [acteur][label = below: ]{};
\node (a16) at (0,1.45) [acteur][label = below: ]{};
\node (a17) at (-0.25,1.45) [acteur][label = below: ]{};
\node (a18) at (0.25,1.85) [acteur][label = below: ]{};
\node (a19) at (0,1.85) [acteur][label = below: ]{};
\node (a20) at (-0.25,1.85) [acteur][label = below: ]{};

\node (a21) at (0.25,2.2) [acteur][label = below: ]{};
\node (a22) at (0,2.2) [acteur][label = below: ]{};
\node (a23) at (-0.25,2.2) [acteur][label = below: ]{};
\end{tikzpicture}
\caption{Mabry's and Edgar's proof in a nutshell} \label{fig: backbones}
\end{figure}
\end{center}

Let $n$ be the total number of triangles in each layer and $a$ be the number of colored triangles in each layer. Note that $1\le a<n$. Also, let $r$ be the ratio of the height of each layer over the distance from $A$ to the base of that layer. The below analysis uses Figure \ref{fig: backbones}, which is a sketch of Figures \ref{fig:Mabry's proof} and \ref{fig: Edgar's}. We assume that $\triangle{ABC}$ has area $1$.

We will compute the colored area, denoted by $T$, in two ways. Because in each layer, there are $n$ triangles, $a$ of which are colored, we know that $T = a/n$. Another way to compute $T$ is to find the total areas of all colored triangles as follows.

The area of layer 1 is clearly $1-(1-r)^2$. Hence, each triangle in layer 1 has area
$$\frac{1-(1-r)^2}{n},$$
which is also the ratio of the area of each triangle in layer 1 over the area of $\triangle{ABC}$ (because the area of $\triangle{ABC}$ is $1$). Due to similarity, the area of each triangle in layer $2$ over the area of $\triangle{ADE}$ is also
$$\frac{1-(1-r)^2}{n}.$$
Since the area of $\triangle{ADE}$ is $(1-r)^2$, each triangle in layer $2$ has area
$$\frac{1-(1-r)^2}{n}\cdot (1-r)^2.$$
In general, each triangle in layer $n$ has area
$$\frac{1-(1-r)^2}{n}\cdot (1-r)^{2(n-1)}.$$
Therefore, the total colored area is (recall that in each layer, $a$ triangles are colored)
\begin{align}\label{k0} a\cdot \frac{1-(1-r)^2}{n}+a\cdot \frac{1-(1-r)^2}{n}(1-r)^2+a\cdot \frac{1-(1-r)^2}{n}(1-r)^4+\cdots.\end{align}
As the two ways of computing $T$ must produce the same result, we have
\begin{align}\label{k1}
    a\cdot \frac{1-(1-r)^2}{n}+a\cdot \frac{1-(1-r)^2}{n}(1-r)^2+a\cdot \frac{1-(1-r)^2}{n}(1-r)^4+\cdots \ =\ \frac{a}{n}.
\end{align}

We now apply \eqref{k1} to have the two identities proven by Edgar \cite{Ed} and Mabry \cite{Ma}. In Mabry's proof, we have $(n,a,r) = (3, 1, 1/2)$. Plugging these numbers into \eqref{k1} gives
$$\frac{1}{4} + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{4}\right)^3 + \cdots \ =\ \frac{1}{3}.$$
In Edgar's proof, we have $(n, a, r) = (5,4,1/3)$. Plugging these numbers into \eqref{k1} gives
$$\frac{4}{9} + \left(\frac{4}{9}\right)^2 + \left(\frac{4}{9}\right)^3 + \cdots \ =\ \frac{4}{5}.$$

\subsection{More beautiful pictures?}

A natural question is whether we can produce more beautiful pictures like Mabry's and Edgar's for other geometric series of the forms
\begin{align}\label{k2}v + v^2 + v^3 + \cdots.\end{align}
The answer is, unfortunately, negative. To show this, we prove two restrictions on $r$, which are simultaneously satisfied only when $r$ is $1/2$ or $1/3$.

Comparing \eqref{k0} and \eqref{k2} shows that the second term of the geometric series is the square of the first term; hence, we have
$$a\cdot \frac{1-(1-r)^2}{n}(1-r)^2 \ =\ \left(a\cdot \frac{1-(1-r)^2}{n}\right)^2,$$
which is equivalent to
\begin{align}\label{k5}\frac{(1-r)^2}{1-(1-r)^2}\ =\ \frac{a}{n}.\end{align}
Because $a/n < 1$, we know that \begin{align}\label{k3}0.29\ <\ 1-\frac{1}{\sqrt{2}}\ <\ r\ <\ 1.\end{align}
This is the first restriction on $r$.

Furthermore, each triangle in layer 1 has area $r^2$, while the area of layer 1 is $1-(1-r)^2$. Therefore, to have pictures like Mabry's and Edgar's, we need
$$\frac{1-(1-r)^2}{r^2} \in \mathbb{N},$$
which implies that
\begin{align}\label{l4}\frac{2}{r}\in \mathbb{N}.\end{align}
Then the number of triangles in each layer is $2/r-1$, or $n = 2/r-1$.
Plugging this back to \eqref{k5}, we have
\begin{align*}
    \frac{(1-r)^2}{1-(1-r)^2} \ =\ \frac{a}{2/r-1},
\end{align*}
which is simplifed to
\begin{align*}
    a \ =\ \frac{1}{r^2} + 1 - \frac{2}{r}.
\end{align*}
Due to \eqref{l4} and $a\in \mathbb{N}$, it follows that $r = 1/m$ for some $m\in \mathbb{N}$. This is the second restriction on $r$. Combining this restriction with \eqref{k3}, we know that $m\in \{2,3\}$. Therefore, we answer negatively the question raised by Edgar \cite{Ed} on whether there are more similar ``proofs without words'' for other series.

\section{The common theme}
At the first glance, the proof by the Viewpoints 2000 Group and the proof by Mabry are little related. However, there is a common theme lurking behind them. Figure \ref{fig: reposition} is obtained by repositioning the Viewpoints 2000 Group's. For convenience, we substitute $a = 1$ and replace $r$ by $\sqrt{r}$.
\begin{center}
\begin{figure}[h!]
\begin{tikzpicture}
[acteur/.style={circle, fill=black,thick, inner sep=0.3pt, minimum size=0.01cm}]
\draw (0,6)--(6,0);
\draw (0,6)--(3.6,0);
\draw (0,0)--(7.2,0);
\draw (0,0)--(0,7.2);
\filldraw[draw=black, fill=cyan] (3.6, 0) -- (6, 0) -- (3.6, 2.4) -- cycle;
\filldraw[draw=black, fill=cyan] (3.6, 2.4) -- (2.16, 3.84) -- (2.16, 2.4) -- cycle;
\filldraw[draw=black, fill=cyan] (1.296,3.84) -- (2.16, 3.84) -- (1.296, 4.704) -- cycle;
\filldraw[draw=black, fill=cyan] (1.296,4.704) -- (0.7776, 4.704) -- (0.7776, 5.2224) -- cycle;
\node (a1) at (0.66,5.1804) [acteur][label = below: ]{};
\node (a2) at (0.54,5.34) [acteur][label = below: ]{};
\node (a3) at (0.42,5.484) [acteur][label = below: ]{};

\node (a4) at (6,0) [acteur][label = below: $\frac{1}{1-\sqrt{r}}$]{};
\node (a5) at (0,6) [acteur][label = left: $\frac{1}{1-\sqrt{r}}$]{};
\draw (3.6,0)--(6,0) node[pos=0.5, below] {$1$};
\draw (3.6,0)--(3.6,2.4) node[pos=0.5, left] {$1$};
\draw (3.6,2.4)--(2.16,2.4) node[pos=0.5, below] {$r^{\frac{1}{2}}$};

\draw (2.16, 3.84)--(1.296,3.84) node[pos=0.5, below] {$r$};

\node (a6) at (0,6) [acteur][label = right: $A$]{};
\node (a7) at (6,0) [acteur][label = above: $B$]{};
\node (a8) at (3.6,0) [acteur][label = below: $C$]{};

\draw (3.6,2.4)--(6, 0) node[pos=0.5,right] {layer 1};
\draw (3.6,2.4)--(2.16,3.84) node[pos=0.5,right] {layer 2};
\draw (1.296,4.704)--(2.16,3.84) node[pos=0.5,right] {layer 3};
\end{tikzpicture}
\caption{Repositioning the Viewpoints 2000 Group's picture} \label{fig: reposition}
\end{figure}
\end{center}
We can now apply the Mabry's technique to prove the formula for geometric series.
The colored area is clearly $\frac{1}{2}+\frac{1}{2}r+\frac{1}{2}r^2 + \frac{1}{2}r^3 + \cdots$. On the other hand, in each layer, the colored area is $\frac{1}{1+r^{\frac{1}{2}}}$ the area of the layer. Hence, the colored area in total is $\frac{1}{1+r^{\frac{1}{2}}}$ the area of $\triangle{ABC}$, which is $\frac{1}{2(1-r^{\frac{1}{2}})}$. Therefore, we have
$$\frac{1}{2}+\frac{1}{2}r+\frac{1}{2}r^2 + \frac{1}{2}r^3 + \cdots\ =\ \frac{1}{1+r^{\frac{1}{2}}}\cdot \frac{1}{2(1-r^{\frac{1}{2}})},$$
which is simplified to
$$1+r+r^2+r^3+\cdots \ =\ \frac{1}{1-r}.$$


TITLE: What is the furthest object from which fermion rays were detected?
QUESTION [3 upvotes]: What is the furthest object from which non-electromagnetic cosmic rays were detected?

REPLY [2 votes]: For example, neutrinos from a Supernova was detected in 1987 (and it seems that was the only observation of this kind). Cosmic rays from outer space are also observed, but I don't know if their source can be identified precisely.


TITLE: Multivariable limit $\lim\limits_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2}$
QUESTION [2 upvotes]: The problem is:
$$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2}.$$
The tutor guessed it didn't exist, and he was correct. However, I'd like to understand why it doesn't exist.
I think I have to turn it into spherical coordinates and then see if the end result depends on an angle, like I've done for two variables with polar coordinates. I don't know how though.
I know $\rho = \sqrt{x^2+y^2+z^2}$ and $\theta = \arctan \left(\frac{y}{x} \right)$ and $\phi = \arccos \left( \frac{z}{\rho} \right)$, but how on earth do I break this thing up?

REPLY [1 votes]: The limit for problems like these do not exist since the limit depends on the direction you approach. For the problem you have mentioned, say you approach $(0,0,0)$ along the direction $y = m_y x$ and $z = m_z x$, where $m_y$, $m_z$ are some constants, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \lim_{x \rightarrow 0} \frac{m_y x^2+2m_y m_z x^2 + 3m_z x^2}{x^2+4m_y^2 x^2+9 m_z^2 x^2}$$
Hence,
$$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \lim_{x \rightarrow 0} \frac{m_y + 2m_y m_z + 3m_z}{1 + 4 m_y^2 + 9 m_z^2} = \frac{m_y + 2m_y m_z + 3m_z}{1 + 4 m_y^2 + 9 m_z^2}$$
You can key in different values for $m_y$ and $m_z$ and you will get different values.
Below are some instances of the different limits you get by approaching in different directions.
If you tend to zero along the direction $(1,0,0)$ i.e. $m_y = m_z = 0$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = 0.$$
If you tend to zero along the direction $(1,1,0)$ i.e. $m_y = 1$ and $m_z = 0$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac15.$$
If you tend to zero along the direction $(1,0,1)$ i.e. $m_y = 0$ and $m_z = 1$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac3{10}.$$
If you tend to zero along the direction $(1,1,1)$ i.e. $m_y = 1$ and $m_z = 1$, then we get $$\displaystyle \lim_{(x,y,z) \rightarrow (0,0,0)} \frac{xy+2yz+3xz}{x^2+4y^2+9z^2} = \frac37.$$


TITLE: Divergence of vector in spherical coordinates
QUESTION [0 upvotes]: How should I calculate the divergence for $$\vec{V}=\frac {\vec{r}}{r^2}$$
Is it possible to convert it from spherical coordinates to cartesian?

REPLY [2 votes]: Suppose its $n$ dimension, note that $\overrightarrow{r}=(x_1,x_2,\cdots,x_n)$ and $\frac{\partial r}{\partial x_i}=\frac{x_i}{r}$, so
$$\frac{\partial}{\partial x_i}(\frac{x_i}{r^2})=\frac{1}{r^2}-\frac{2x_i^2}{r^4}$$
$$\text{div}\overrightarrow{V}=\sum_{i=1}^n\frac{\partial}{\partial x_i}(\frac{x_i}{r^2})=\frac{n}{r^2}-\frac{2\sum_{i=1}^nx_i^2}{r^4}=\frac{n-2}{r^2}$$

REPLY [1 votes]: You certainly can convert $\bf V$ to Cartesian coordinates, it's just
${\bf V} = \frac{1}{x^2 + y^2 + z^2} \langle x, y, z \rangle,$
but computing the divergence this way is slightly messy.
Alternatively, you can use the formula for the divergence itself in spherical coordinates. If we write the (spherical) components of $\bf V$ as
$${\bf V} = \langle V^r, V^{\theta}, V^{\phi} \rangle,$$
then
$$\text{div } {\bf V} = \frac{1}{r^2} \partial_r (r^2 V^r) + \frac{1}{r \sin \theta} \partial_{\theta} (V^{\theta} \sin \theta) + \frac{1}{r \sin \theta} \partial_{\phi} V^{\phi}.$$
In our case, the components of $\bf V$ are $$V^r = \frac{1}{r}, V^{\theta} = 0, V^{\phi} = 0,$$ so its divergence is
$$\text{div } {\bf V} = \frac{1}{r^2} \partial_r \left(r^2 \cdot \frac{1}{r}\right) = \frac{1}{r^2} \partial_r r = \frac{1}{r^2}.$$


TITLE: Angle and distance between subspaces.
QUESTION [2 upvotes]: Let $\text{Sin}(\Theta)$ be the diagonal matrix containing the $Sine$'s of principal angles between two subspaces $\mathcal{A}$ and $\mathcal{B}$, and let $\Pi_A$ and $\Pi_B$ be their respective projectors. How do I show that $\|\text{Sin}(\Theta)\| = \frac{1}{\sqrt{2}}\cdot\|\Pi_A - \Pi_B\|$ (where $\|\cdot\|$ is the spectral norm)?
Edit: the $\frac{1}{\sqrt{2}}$ factor should not be there as per the answer below.

REPLY [2 votes]: I seem to obtain a slightly different result. In particular, I think that the factor of $\frac 1{\sqrt{2}}$ should not be there.
I will focus on the case that both subspaces are $\ell$-dimensional subspaces of $\Bbb C^n$ with $2 \ell \leq n$. Bhatia's Matrix Analysis proves this result (rather, sets up an exercise in which one may prove this result) by using the following consequence of the existence of the CS-decomposition.

Theorem VII.1.8: Let $X,Y$ be $n \times \ell$ matrices with orthonormal columns. Then there exist $\ell \times \ell$ unitary matrices $U,V$ and an $n \times n$ unitary matrix $Q$ with the following properties.
If $2\ell \leq n,$ then
$$
QX U = \pmatrix{I_\ell \\ 0\\0}, \quad QY V = \pmatrix{C \\ S\\ 0}
$$
where $C,S$ are diagonal matrices with diagonal entries $0 \leq c_1 \leq \cdots \leq c_\ell \leq 1$ and $1 \geq s_1 \geq \cdots \geq s_\ell \geq 0$ (respectively) and $C^2 + S^2 = I$.

There is also a version of this theorem for the case that $2\ell > n$ which I am ignoring for the purposes of my answer. Now, suppose that $X$ and $Y$ have columns that form orthonormal bases for $\mathcal A, \mathcal B$ respectively. Let $Q,U,V$ be as in the above theorem. Let $\mathcal A',\mathcal B'$ denote the images of $\mathcal A,\mathcal B$ respectively under $Q$. That is, $\mathcal A' = \{Qx : x \in \mathcal A\}$, and $\mathcal B'$ is defined similarly.
Note that the columns of $QXU$ form an orthonormal basis for $\mathcal A'$, and the columns of $QYV$ form an orthonormal basis for $\mathcal B'$. Moreover, the principal angles between $\mathcal A'$ and $\mathcal B'$ must be equal to those between $\mathcal A$ and $\mathcal B$. With that, we can see that $S = \sin(\Theta)$, as you have defined it.
Now, we note that $\Pi_{\mathcal A} = XX^*$, so that
\begin{align}
\left\|\Pi_{\mathcal A} - \Pi_{\mathcal B} \right\| & =
\left\|XX^* - YY^* \right\|
\\ & =
\left\|Q(XX^* - YY^*)Q^* \right\|
\\ & =
\left\|(QXU)(QXU)^* - (QYV)(QYV)^* \right\|
\\ & =
\left\|\pmatrix{I_\ell & 0 & 0\\0 & 0 & 0\\0 & 0 & 0} - \pmatrix{C^2 & CS& 0\\CS & S^2 & 0\\0 & 0 & 0} \right\|
\\ & =
\left\|\pmatrix{I - C^2 & -CS & 0\\ -CS & -S^2 & 0\\0 & 0 & 0}\right\|
= \left\|\pmatrix{S^2 & -CS & 0\\ -CS & -S^2 & 0\\0 & 0 & 0}\right\|.
\end{align}
So, we now need to compute the spectral norm of the matrix
$$
\pmatrix{S^2 & -CS\\ -CS & -S^2}.
$$
We see that there is a permutation-similarity between this matrix and the block-diagonal matrix
$$
\pmatrix{M_1 \\ & \ddots \\ && M_\ell}, \quad M_k = s_k\pmatrix{s_k & -c_k\\ -c_k & -s_k}, \quad k = 1,\dots,\ell.
$$
We see that $M_k$ is $s_k$ multiplied by an orthogonal matrix, which means that we have $\|M_k\| = s_k$. Thus, we conclude that $\|\Pi_{\mathcal A} - \Pi_{\mathcal B}\|$ is equal to $\max_{k = 1,\dots,\ell} s_k$, which coincides with $\|S\| = \|\sin(\Theta)\|,$ which was what we wanted.


TITLE: $W\subset U \subset V, u=proj_U(v), w=proj_W(u). \ \ w=proj_W(v)?$
QUESTION [1 upvotes]: if $V$ is an inner product vector space, $W,U$ are subspaces S.T.  $W\subset U, v\in V.\ $ $u$ is the orthogonal projection of $v$ onto $U$, $w$ is the orthogonal projection of $u$ onto $W$. The question is whether the orthogonal projection of $v$ onto $W$ is equal to $w$. I am almost certain this is not the case but failed to find a counter example. Any thoughts?

REPLY [0 votes]: I will restrict this answer to the finite dimensional case.
Choose an orthonormal basis $\{w_1,\dots,w_r\}$ of $W$.
Extend to an orthonormal basis $\{w_1,\dots,w_r,u_1,\dots,u_m\}$ of $U$, the $u_i$'s coming from $W^{\perp}$.
Finally, extend to an orthonormal basis $\{w_1,\dots,w_r,u_1,\dots,u_m,v_1,\dots,v_l\}$ of $V$, the $v_i$'s coming from $U^{\perp}$.
Note that we have $U^{\perp}\subset W^{\perp}$.
Fix $v\in V$. We have
$$\text{proj}_U(v)=\sum_{i=1}^{r}\langle v,w_i\rangle w_i+\sum_{j=1}^{m}\langle v,u_j\rangle u_j=u$$
$$\text{proj}_W(u)=\sum_{i=1}^{r}\langle v,w_i\rangle w_i=w\tag{$u_j$'s are in $W^{\perp}$}$$
$$\text{proj}_W(v)=\sum_{i=1}^{r}\langle v,w_i\rangle w_i=w\tag{$u_j$'s and $v_k$'s are in $W^{\perp}$}$$


TITLE: Find the minimal polynomial of $\sqrt[3]{3} + \sqrt{5}$ over $\mathbb{Q}$.
QUESTION [1 upvotes]: Let $x = \sqrt[3]{3} + \sqrt{5}$
Notice that $(x+y)^2 = x^2 + 2xy + y^2$
Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$
Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$
Or notice that $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$
Then, $x^3 = (\sqrt[3]{3} + \sqrt{5})^3 = 3 + 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Then, $x^3 - 3 = 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Which I think the degree of the minial polynomial of $x$ over $\mathbb{Q}$ should have a degree of $6$.
Which means the minimal polynomial can be written as $x^6 + a_5 x^5 + \cdots + a_0$ for some $a_5, ..., a_0 \in \mathbb{Q}$.
But I just don't how get it from both the ways I provide above.
Or that $\sqrt[3]{3} + \sqrt{5}$ is not algebraic. So, I can not write such minimal polynomial out algebraic ?

REPLY [3 votes]: If $x=\sqrt[3]3+\sqrt5$, then $\left(x-\sqrt5\right)^3=3$. In other words $x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0$. But\begin{align}x^3-3 \sqrt{5} x^2+15 x-5 \sqrt{5}-3=0&\iff x^3+15x-3=(3x^2-5)\sqrt5\\&\implies(x^3+15x-3)^2=5(3x^2-5)^2\\&\iff x^6-15 x^4-6 x^3+375 x^2-90 x-116=0.\end{align}

REPLY [2 votes]: Set of algebraic numbers form a field, so the sum of two algebraic numbers $\sqrt[3]{3} + \sqrt{5}$ is again an algebraic number.
To find the minimal polynomial, you want to continue in either of the two ways that you have started.
If it is the first way,
you have to take the suitable powers of $x^2-5$ and continue until getting a rational polynomial.


TITLE: What set do generic types define?
QUESTION [2 upvotes]: Let $G$ be an infinite $\omega$-stable group and $p \in S(G)$ a generic type. What set does $p$ define? Or do we know anything about it? Like where it lies.
I am struggling with generic types/planes/etc in model theory. It would be also great if someone could suggest me some sources about it to read.

REPLY [1 votes]: In any $\omega$-stable theory $T$, the formula $\top$ or $x = x$ in the context of a single free variable $x$ has a Morley rank $\alpha$ (an ordinal) and a Morley degree $n$ (a natural number). The degree $n$ is the maximal number of pairwise disjoint definable sets of Morley rank $\alpha$. Working in a sufficiently saturated model $M\models T$, $M$ is partitioned into $n$ definable pieces $X_1,\dots,X_n$, each of Morley rank $\alpha$, and this partition is unique up to adjustments by definable sets of rank $<\alpha$. We call the pieces of the partition irreducible components of $M$.
Now a type $p\in S_1(M)$ is generic if it has Morley rank $\alpha$. Since the $X_i$ form a partition, $p$ must live in in one of the sets $X_i$. But for every definable subset $Y\subseteq X_i$, either $Y$ or $X_i\setminus Y$ has rank $<\alpha$ (otherwise we could split $X_i$ into two pieces of full rank and get a bigger partition). So since $p$ does not live in any definable set of rank $<\alpha$, $p$ is uniquely determined by which irreducible component $X_i$ it lives in. So $S_1(M)$ contains exactly $n$ generic types.
To your question "where does a generic type lie", the answer is...well, almost everywhere!  (In its irreducible component.) A realization of a generic type $p\in S_1(M)$ (in an elementary extension of $M$) lives in the irreducible component $X$ of $p$, but not in any "small" (in the sense of rank) $M$-definable subset of $X$. So $p$ defines the set of elements of $X$ which do not lie in any lower-rank $M$-definable subset of $X$.
If you know something about algebraic geometry, a good analogy is that the generic types in an $\omega$-stable theory are like the generic points of irreducible components of an algebraic variety. The Zariski closure of a generic point is its entire irreducible component, because it does not live in any proper Zariski closed subset.
All of the above was in the general context of $\omega$-stable theories. Specializing to $\omega$-stable groups, we get some additional information about generic types. Let $G$ be a sufficiently saturated $\omega$-stable group.

$G$ has a "connected component of the identity" $G^0$, which is the minimal definable subgroup of finite index in $G$. The Morley degree is the index of this subgroup and the irreducible components are its cosets. So there is exactly one generic type living in each coset of $G^0$.
A definable subset $X\subseteq G$ has full rank $\alpha$ if and only if finitely many translates of $X$ cover $G$, i.e. $a_1\cdot X \cup \dots \cup a_n\cdot X = {G}$ for some $a_1,\dots,a_n\in {G}$. So a generic type only contains formulas defining "wide" sets in the sense that finitely many translates cover the group. This is a precise sense in which  "where a generic type lives" can't be pinned down very tightly.
Any element of $G$ is a product of two generic elements over $G$, i.e. for all $g\in G$ there exist $a,b\in \mathbb{G}$, a sufficiently saturated elementary extension of $G$, such that $a\cdot b = g$ and $\text{tp}(a/G)$ and $\text{tp}(b/G)$ are generic.

For your reference request: If you are particularly interested in $\omega$-stable theories and $\omega$-stable groups, Marker's book Model Theory: An Introduction gives a good overview (Chapters 6 and 7 - all the facts I wrote above are proven there). $\omega$-stability is a special case of the more general theory of stability. For more information on stable groups in general, the standard reference is Poizat's book Stable Groups. For stable theories in general, there are many references. Which one is best for you depends on your background and interests.


TITLE: Angle of Revolution for a truck
QUESTION [0 upvotes]: I am doing a problem out of my textbook, which I don't understand.  In my Alg2/Trig class, we are learning about angular speed, and linear speed in terms of angular speed.  I can't figure out how to solve this using those.  Please note, a full explanation is not required.  Even just a hint can help.  Lastly, I appreciate any help, but helping me using math that is way beyond me doesn't help me.
Problem:

A tite on a truck has a diameter of 31.125 in.  Through what angle (radians) does the tire turn when traveling 1 mile

REPLY [1 votes]: Notice, in general, the angle $\theta$ turned by the wheel $$\theta=2\pi \times \frac{\text{distance traveled by the wheel}\ (D)}{\text{circumference of wheel}\ (\pi d)}=\frac{2 D}d$$
setting the corresponding values of distance, $D=1\ \mathrm{mile}=1609.344\ m$ & diameter, $d=31.125\ in.=31.125\times 0.0254\ m$, the angle turned by the wheel
$$\theta=\frac{2\times 1609.344\ m}{31.125\times 0.0254\ m}=\color{red}{4071.325302\ \mathrm{radians}}$$


TITLE: Is work done defined for a state?
QUESTION [0 upvotes]: The following excerpt is from Energy principles and variational methods in applied mechanics, 2nd edition, by J. N. Reddy:

4.1 CONCEPTS OF WORK AND ENERGY
Consider a material particle, moving from point $A$ to point $B$ along some path in space under the influence of a force $\mathbf{F}$, which can be time-dependent. The position of the particle is measured from a fixed origin by position vector $\mathbf{r}$. Then the work $dW$ performed by the force $\mathbf{F}$ in moving the particle by an infinitesimal distance (or displacement) $d\mathbf{r}= d\mathbf{u}$ along the path over an interval of time $dt$ is defined as
  $$dW= \mathbf{F} . d \mathbf{u}=F_1du_1+F_2du_2+F_3du_3$$
  In other words, work done is the product of the displacement and force in the direction of the displacement. The total work done, $W$, by the force $\mathbf{F}$ in moving the particle from point $A$ to point $B$ is given by
  $$W=\int_A^B \mathbf{F} . d \mathbf{u}$$
  By definition, work done is a scalar quantity, and it is positive whenever both displacement and force have the same direction and negative if they hare in the opposite directions. Since $\mathbf{u}$ depends on the chosen reference frame, $W$ also depends on the choice of the reference frame. Thus, work is a relative quantity. However, work done does not depend on the path but only the end points, $W=W_B-W_A$. If the reference frame is chosen such that $W_A=0$, then $W=W_B$.

Link to the page
What is he trying to say? Is he saying that work in general is a state variable? And has exact differential? I got confused! So far, all of books I have read, also all of my professors, and many other sources and expert people claim that work in general is path dependent and only in some special cases can be determined by means of initial and final states. Is there something here that I don’t consider? Is there any meaning for $W_B$? Even for those special cases we have for example:
$W=U_B-U_A$ (not $W=W_B-W_A$)
where $U$ is some kind of energy.

REPLY [0 votes]: Your mention of "state function" implies you are coming at this from a thermodynamic background. Indeed, in thermodynamics, work is not a state function because a "state" in thermodynamics is not a position in space and the "work" is not done against a conservative force field (or any fixed force field, really). What forces are involved in the work being done depends on the thermodynamic process happening, and is rarely known or indeed useful to know.
However, this is not thermodynamics, you're reading a text about classical mechanics, where a "state" is merely a collection of positions and momenta of some objects, mostly point particles, and where the work is being done by moving the particle from one position to another against (or with) a force field. If the force is conservative, then the work done will be independent from the path taken, and indeed work with respect to that force will be a "state function" on the states of classical mechanics, but this terminology is not commonly used. The necessary and sufficient condition for a force being conservative is that there is some potential scalar function $U$ such that $F=-\nabla U$, where the negative sign is pure convention.


TITLE: Assume that Pr(A ∪ B) = 5/6 and Pr(A`|B`) = 1/3. Determine Pr(B).
QUESTION [0 upvotes]: Assume that Pr(A ∪ B) = 5/6 and Pr(A'|B') = 1/3. Determine Pr(B).
Where A' is A complement and B' is B complement.
Not sure how to approach this one. Appreciate any input

REPLY [3 votes]: hint: $P(A'\mid B')\cdot P(B') = P(A'\cap B')= 1 - P(A\cup B)$. You can find $P(B')$, and then $P(B) = 1- P(B')$.


TITLE: Finding continuous f where f is not bounded .
QUESTION [0 upvotes]: I am finding a function f which is continuous on a closed interval but not bounded on the interval.

REPLY [4 votes]: Isn't $[0, \infty)$ a closed interval? Then just take $x$ in that case.


TITLE: The number of paths on a graph of a fixed length w/o repeatings
QUESTION [5 upvotes]: Sorry for bad English.
Consider a graph $G$ with the adjacency matrix $A$. I know that the number of paths of the length $n$ is the sum of elements $A^n$.
But what if we can't walk through a vertex more than one times?

REPLY [3 votes]: There doesn't seem to be a simple expression involving the adjacency matrix for the number of simple (i.e. self-avoiding) paths in an arbitrary graph, but there is an algorithm.
Googling "adjacency matrix number of paths" turned up this SE question which points out (with a nice example) that powers of the adjacency matrix count the number of walks of length $n$, not paths.  That fact also turned up in this wikipedia entry.  These walks allow a vertex to be visited more than once, so it's unlikely they can help you to count simple paths.
The paper Self-Avoiding Paths and the Adjacency Matrix of a Graph - J. Ponstein contains an algorithm on pages 9-10 which counts all simple paths in an arbitrary graph.
Edit: The good news is explicit formulae for the paths of length $n$ exist, the bad news is the number of terms in the formulae grow exponentially with $n$!
The paper On the determination of redundancies in sociometric chains - Ross, Ian C.; Harary, Frank counts the number of 'redundant paths' (i.e. non-simple paths, walks of length $n$ where a vertex is visited more than once).
The English language powerpoint slideshow of the Russian language paper "The number of fixed length cycles in an undirected graph. Explicit formulae in case of small lengths" by S. N. Perepechko & A. N. Voropaev contains formula for path lengths 2-5 using the Ross & Harary method.  I reproduce, for your delectation or horror, 2 of the 17476 terms from the path length 10 formula ($\times$ is element-wise matrix multiplication, $\cdot$ is ordinary matrix multiplication):