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\title[Generalised Surfaces in ${\Bbb{R}}^3$]
   {Generalised Surfaces in ${\Bbb{R}}^3$}
\author{Brendan Guilfoyle}
\address{Brendan Guilfoyle\\
          Department of Mathematics and Computing \\
          Institute of Technology, Tralee \\
          Clash \\
          Tralee  \\
          Co. Kerry \\
          Ireland.}
\email{brendan.guilfoyle@@ittralee.ie}
\thanks{The first author was supported by the Isabel Holgate Fellowship from  Grey College, Durham and the Royal Irish Academy Travel Grant Scheme}
\author{Wilhelm Klingenberg}
\address{Wilhelm Klingenberg\\
 Department of Mathematical Sciences\\
 University of Durham\\
 Durham DH1 3LE\\
 United Kingdom.}
\email{wilhelm.klingenberg@@durham.ac.uk }

\keywords{congruence, real surface}
\subjclass{Primary: 53C28; Secondary:57R25}
\date{February 18, 2004}

\begin{abstract}
The correspondence between 2-parameter families of oriented lines in ${\Bbb{R}}^3$ and surfaces in $T{\Bbb{P}}^1$ is studied, and the
geometric properties of the lines are related to the complex geometry of the surface. Congruences
generated by global sections of $T{\Bbb{P}}^1$ are investigated and a number
of theorems are proven that generalise results for closed convex
surfaces in ${\Bbb{R}}^3$.
\end{abstract}

\maketitle

\section{Introduction}

In this paper we study a generalisation of the concept of a
surface in Euclidean ${\Bbb{R}}^3$ through the twistor
construction. The fundamental aspect of this generalisation is that,
through its normal, an oriented surface in ${\Bbb{R}}^3$  determines a surface
in the space of all oriented lines: ${\Bbb{T}}=T{\Bbb{P}}^1$. This
4-manifold inherits a natural complex structure from the Euclidean
metric in ${\Bbb{R}}^3$ and has been useful in
connection with monopoles \cite{hitch1}.

A key feature of the construction is that
a point in ${\Bbb{R}}^3$ corresponds to a holomorphic sphere in
${\Bbb{T}}$. In this paper we consider (not necessarily holomorphic)
2-sphere's in ${\Bbb{T}}$ and the {\it line congruences} they
generate in ${\Bbb{R}}^3$ \cite{pottwall}. Our main aim is to show that, in this
setting, curvature and umbilics can all be sensibly defined
and used to obtain results generalising those of closed convex surfaces.

The sense in which we are generalising is given by Frobenius' theorem:
we are allowing consideration of {\it twisting} line congruences. This
twist is encoded in the anti-symmetric part of a suitably defined second
fundamental form of the line congruence. The curvature $K$ of a congruence can then be defined as
the determinant of the second fundamental form.

This has the following bundle interpretation:

\noindent{\bf Theorem 1.}
{\it
A line congruence $\Sigma$ is locally the graph of a section of the bundle
$\pi:{\Bbb{T}}\rightarrow {\Bbb{P}}^1$ if and only if the curvature is
non-zero.
}

We say a line congruence is {\it globally
convex} if it is the graph of a global section of the bundle
$\pi:{\Bbb{T}}\rightarrow {\Bbb{P}}^1$. Thus, a globally convex congruence $\Sigma\subset {\Bbb{T}}$ is a topological
sphere and the Gauss map $\pi|\Sigma$ yields natural global coordinates ($\xi$,$\bar{\xi}$) on $\Sigma$. We show that a generalised Gauss-Bonnet theorem
holds for globally convex congruences:

\noindent{\bf Theorem 2.}
{\it
Let $\Sigma$ be a globally convex congruence with curvature
$K$. Then
\[
\int_\Sigma Kd\mu=4\pi
\]
where $d\mu$ at $\gamma\in\Sigma$ is the pull-back via $\pi$ of the volume form induced on the
plane orthogonal to $\gamma$ by the Euclidean metric on ${\Bbb{R}}^3$.
}

In general these globally convex congruences, aside from twisting, will be
non-holomorphic. However, we can always perturb a congruence so that
it has isolated complex points. The total number of complex points,
counted with index, is a topological invariant of the congruence. We
show that this index is the number of shear-free points on a globally
convex congruence and that:

\noindent{\bf Theorem 3.}
{\it
The total number of shear-free lines (counted
with index) on a globally convex congruence with only isolated
shear-free lines is 4.
}

This generalises the well-known result that the number of isolated
umbilics (counted with index) on a closed convex surface is 4.

This paper is organised as follows: in the next section we recall the
relevant twistor construction, for further details see \cite{hitch1}
\cite{hitch2}.  To prove our main results we use a canonical
co-ordinate system, as in \cite{gak1} and null frame adapted to the congruence under
consideration. After describing these in section 3, we turn to the
first order description of congruences. Here the method of
spin-coefficients \cite{nap} \cite{perjes}, applied to ${\Bbb{R}}^3$,
yields a compact description of the geometric data. Finally in section
5 we prove the results regarding globally convex congruences.

\section{The Minitwistor Construction}

We begin by recalling the minitwistor construction of straight lines in ${\Bbb{R}}^3$
(see Hitchin \cite{hitch1} for further details).  Given a choice of origin in ${\Bbb{R}}^3$,
a straight line can be uniquely described by two vectors: the (oriented)
direction of the line
$\vec{W}$ and its perpendicular displacement from the origin $\vec{U}$.
A straight line $\gamma$ is given by
\[
\gamma=\{\vec{U}+r\vec{W}\in {\Bbb{R}}^3\;|\; <\vec{U},\vec{W}>=0 \quad |\vec{W}|=1\quad r\in {\Bbb{R}}\},
\]
where $<,>$ is the Euclidean inner product and $|.|$ the associated norm.

The space of all oriented straight lines in ${\Bbb{R}}^3$ is the minitwistor space
\[
{\Bbb{T}}=\{(\vec{U},\vec{W})\in {\Bbb{R}}^3\times {\Bbb{R}}^3\;| \;<\vec{U},\vec{W}>=0 \quad |\vec{W}|=1 \}
\cong TS^2.
\]
Throughout this paper we utilise this bijection to identify an oriented line $\gamma\subset{\Bbb{R}}^3$ with the point $\gamma\in{\Bbb{T}}$. This 4-manifold has a natural almost complex structure ${\Bbb{J}}$ defined by
rotation in ${\Bbb{R}}^3$
through 90$^o$ about the direction of the line.  In fact, the almost complex
structure ${\Bbb{J}}$ is integrable and so $\Bbb{T}$ is a complex surface.  Alternatively, ${\Bbb{J}}=j\oplus j$, where the splitting $T{\Bbb{T}}=TS^2\oplus TS^2$ and the complex structure $j$ on $S^2$ are induced by the Euclidean metric on ${\Bbb{R}}^3$.  In addition,
there is an anti-holomorphic involution $\tau: \Bbb{T} \rightarrow \Bbb{T}$
given by reversing the orientation of the line.

A point $p$ in ${\Bbb{R}}^3$ is uniquely determined by the 2-sphere
$S^2_p$ of oriented  lines
passing through it.  This sphere $S^2_p\subset \Bbb{T}$ has the following
properties:
\begin{enumerate}
 \item $S^2_p$ is a complex line in $ \Bbb{T}$ i.e. the complex
 structure on $ \Bbb{T}$  leaves invariant the tangent space of $S^2_p$
 \item $S^2_p\cap S^2_q$ consists of two points in $\Bbb{T}$ - the two oriented lines in ${\Bbb{R}}^3$ passing through $p$ and $q$
 \item $S^2_p$ is invariant under the involution $\tau$
\end{enumerate}

From the above it follows that $S^2_p$ can be given, in terms of a
holomorphic coordinate $\xi$ on ${\Bbb{P}}^1$, as the graph of a section of $\pi:{\Bbb{T}}
\rightarrow {\Bbb{P}}^1$
\begin{equation}\label{e:cp1}
s(\xi)= \frac{1}{2}\left((x^1+ix^2)-2x^3\xi-(x^1-ix^2)\xi^2\right)\frac{\partial}{\partial \xi},
\end{equation}
where the Euclidean coordinates of $p$ are ($x^1,x^2,x^3$).  This
follows from the fact that global holomorphic sections can be at most
quadratic in $\xi$ (self-intersection 2) and then the invariance of
the section under the antipodal map $\tau(\xi)= -\overline{\xi}^{-1}$
restricts the coefficients to be of the above form.

In this paper we will investigate {\it line congruences}, that is,
surfaces $\Sigma\subset \Bbb{T}$ or two parameter families of lines in
${\Bbb{R}}^3$. Every surface $S\subset{\Bbb{R}}^3$ gives rise to a
congruence $\Sigma\subset \Bbb{T}$ by way of its normal, but not every
congruence arises in this way. It is in this sense that we are working
with generalised surfaces.

In the next section we introduce new local coordinates and a null frame on
${\Bbb{R}}^3$ which fit nicely with the description of a congruence as a
surface in $T{\Bbb{P}}^1$.

\section{Coordinates and an Adapted Frame on ${\Bbb{R}}^3$}
Let $(x^1,x^2,x^3)$ be the standard
coordinates on ${\Bbb{R}}^3$ and and set $z=x^1+ix^2$, $\overline{z}=x^1-ix^2$ and $x^3=t$.

\begin{Def} Consider
the following transformation $ \Phi:(u,v,r)\rightarrow
(z,\overline{z},t)$ on an open subset of ${\Bbb{R}}^3$ given by
\begin{equation}\label{e:coord1}
z=\frac{2(F-\overline{F}\xi^2)+2\xi(1+\xi\overline{\xi})r}{(1+\xi\overline{\xi})^2}
\end{equation}
\begin{equation}\label{e:coord3}
t=\frac{-2(F\overline{\xi}+\overline{F}\xi)+(1-\xi^2\overline{\xi}^2)r}{(1+\xi\overline{\xi})^2},
\end{equation}
where $F(u,v)$ and $\xi(u,v)$ are smooth complex-valued functions of
two real parameters $u$ and $v$.  We call the coordinates ($u,v,r$) the {\it congruence  coordinates}.
\end{Def}

For each ($u,v$), $\Phi(r)$ is a straight line in ${\Bbb{R}}^3$.
Moreover, these lines are parameterised by arclength $r$ and
$\frac{\partial}{\partial r}$ is the unit tangent to the lines. The
shortest distance from the origin to each line is given by the point
$r=0$.

These coordinates come from the twistor construction in the following way:  consider one of these
lines with direction $\vec{W}$ and perpendicular displacement from the origin $\vec{U}$.  Translate
$\vec{W}$ along $\vec{U}$ to the origin and then translate $\vec{U}$
along $\vec{W}$.  This vector tangent to $S^2$ gives us the
line as a point in $\Bbb{T}$.  Moreover, $\xi$ is the standard holomorphic coordinate on ${\Bbb{P}}^1$ induced from
stereographic projection from the South pole and in these coordinates the point in $\Bbb{T}$ is given by
\[
\left(\xi(u,v),F(u,v)\frac{\partial}{\partial
    \xi}\right)\in T_{\xi}{\Bbb{P}}^1.
\]
Thus $F(u,v)$ determines the perpendicular distance of the line from
the origin. This can be viewed as parametric equations for a line congruence in terms of coordinates ($\xi$,$F$) on ${\Bbb{T}}$, which are holomorphic with respect to ${\Bbb{J}}$. For further details see \cite{gak1}.

A change of origin leads to a quadratic holomorphic
translation of the function $F$.  In particular, if the origin is translated
$(0,0,0) \rightarrow (x^1_0,x^2_0,x^3_0)$ then
\begin{equation}\label{e:trans1}
F\rightarrow F+\frac{1}{2}\left(\alpha_0-2t_0\xi-\overline{\alpha}_0\xi^2\right),
\end{equation}
where $\alpha_0=x^1_0+ix^2_0$ and $t_0=x_0^3$. This can be seen from
equation (\ref{e:cp1}) since the lines through the origin (the zero
section of the bundle) will change to the section
\[
s(\xi)= \frac{1}{2}\left(\alpha_0-2t_0\xi-\overline{\alpha_0}\xi^2\right)\frac{\partial}{\partial \xi}.
\]
 Our coordinates will then transform by:
\begin{equation}\label{e:trans3}
(u,v,r)\rightarrow \left(u,v,r+\frac{\overline{\alpha}_0\xi+\alpha_0\overline{\xi}+t_0(1-\xi\overline{\xi})}
                         {1+\xi\overline{\xi}}\right)
\end{equation}
\[
(z,t)\rightarrow(z+\alpha_0,t+t_0).
\]
The change in $r$ is just $<\vec{T},\vec{W}>$, where $\vec{T}$ is
the translation vector determined by $\alpha_0$ and $t_0$.

The quantities that have geometric significance are invariant under
this translation.  In particular, we have the following translation
invariant derivatives of the twistor function $F$:

\begin{Prop}\label{p:traninv}
Let $\nu=u+iv$,$\overline{\nu}=u-iv$, and $\partial =\frac{\partial
}{\partial\nu}$ and $\overline{\partial}=\frac{\partial
}{\partial\overline{\nu}}$. Then
\[
\partial^+F\equiv\partial F+r\partial\xi-\frac{2F\overline{\xi}\partial \xi}{1+\xi\overline{\xi}}
\]
\[
\partial^-F\equiv\overline{\partial} F+r\overline{\partial}\xi-\frac{2F\overline{\xi}\;\overline{\partial} \xi}{1+\xi\overline{\xi}},
\]
are invariant under the translations (\ref{e:trans1}) and (\ref{e:trans3}).

\end{Prop}
\begin{pf}
This is a straight-forward calculation.

\end{pf}

\begin{Note}
The Jacobian of the tranformation $\Phi$ is
\[
\Delta=\frac{4}{(1+\xi\overline{\xi})^2}\left(\partial^+F\overline{\partial^+F}-\partial^-F\overline{\partial^-F}\right)
\]
Thus the transformation is a diffeomorphism wherever $\Delta\neq0$.
\end{Note}

A {\it null frame} in ${\Bbb{R}}^3$ is a trio $\{e_0,e_+,e_-\}$ of
complex vector fields in ${\Bbb{C}}\otimes T\:{\Bbb{R}}^3$, where $e_0$ is real,
$e_+$ is the complex conjugate of $e_-$ and they satisfy the following orthogonality properties:
\[
<e_0, e_0>=1 \qquad <e_0, e_+>=0 \qquad <e_+, e_+>=0 \qquad <e_+,
e_->=1,
\]
where we have extended the Euclidean inner product of ${\Bbb{R}}^3$ bilinearly over
${\Bbb{C}}$. Orthonormal frames $\{e_0,e_1,e_2\}$ on $T\:{\Bbb{R}}^3$ and null frames are
related by
\begin{equation}\label{e:ortho}
e_+=\frac{1}{\sqrt{2}}(e_1-ie_2) \qquad
e_-=\frac{1}{\sqrt{2}}(e_1+ie_2).
\end{equation}

\begin{Def}
A {\it congruence null frame} for $\Sigma\subset{\Bbb{T}}$ is a null frame
$\{e_0,e_+,e_-\}$ if, for each $\gamma\in\Sigma$, we have $e_0$ tangent
to $\gamma$ in ${\Bbb{R}}^3$, and the orientation of $\{e_0,e_1,e_2\}$
is the standard orientation on ${\Bbb{R}}^3$.
\end{Def}

\begin{Prop}\label{p:nullframe}
Let $\Sigma$ be a line congruence and consider an open set $U\subset\Sigma$ with $\Delta\neq 0$. Suppose that $F$ is the twistor function describing
$\Sigma$ on $U$.
A null frame is a congruence null frame to $\Sigma$ if and only if it
has the following expression in terms of canonical coordinates
\[
e_0=\frac{\partial}{\partial r}
\qquad\qquad
e_+= \left(\alpha\frac{\partial}{\partial \nu}
     +\beta\frac{\partial}{\partial \overline{\nu}}
     +\Omega\frac{\partial}{\partial r}\right)e^{i\phi}
\qquad\qquad
e_-=\overline{e_+},
\]
where
\[
\Omega=\frac{\sqrt{2}\left[\overline{\partial^-F}(F\overline{\partial}\;\overline{\xi}+\overline{F}\;\overline{\partial}\xi)-\overline{\partial^+F}(F\partial\overline{\xi}+\overline{F}\partial\xi)\right]}
{(1+\overline{\xi}\xi) (\partial^-F\overline{\partial^-F}-\partial^+F\overline{\partial^+F}) }
\]
\begin{equation}\label{e:coef}
\alpha=\frac{\overline{\partial^+F}(1+\overline{\xi}\xi)}
{\sqrt{2}(\partial^-F\overline{\partial^-F}-\partial^+F\overline{\partial^+F}) }
\qquad
\beta=-\frac{\overline{\partial^-F}(1+\overline{\xi}\xi)}
{\sqrt{2}(\partial^-F\overline{\partial^-F}-\partial^+F\overline{\partial^+F}) },
\end{equation}
and $\phi$ is a function of $\nu$, $\overline{\nu}$ and $r$.
\end{Prop}
\begin{pf}
We can prove this by a change of coordinates from
 $(\nu,\overline{\nu},r)$ to $(z,\overline{z},t)$ using equations
 (\ref{e:coord1}) to (\ref{e:coord3}). This gives the following mixed
 form for the frame:
\[
e_0= \frac{2\xi}{1+\xi\overline{\xi}}\frac{\partial}{\partial z}
     +\frac{2\overline{\xi}}{1+\xi\overline{\xi}}\frac{\partial}{\partial \overline{z}}
  + \frac{1-\xi\overline{\xi}}{1+\xi\overline{\xi}}\frac{\partial}{\partial t}
\]
\[
e_+= \frac{\sqrt{2}}{1+\xi\overline{\xi}}\frac{\partial}{\partial z}
     -\frac{\sqrt{2}\;\overline{\xi}^2}{1+\xi\overline{\xi}}\frac{\partial}{\partial \overline{z}}
  - \frac{\sqrt{2}\;\overline{\xi}}{1+\xi\overline{\xi}}\frac{\partial}{\partial t}.
\]
Now, since the Euclidean inner product on ${\Bbb{R}}^3$ in the coordinates
$z,\overline{z},t$ is simply
\[
g_{ij}=\left[\begin{matrix}
                 0 & \frac{1}{2} & 0\\
                 \frac{1}{2} & 0 & 0\\
                 0 & 0 & 1
              \end{matrix}\right],
\]
we can check that these
vectors form a null frame.

Alternatively, we can work backwards as follows. Set
\[
e_0=\frac{\partial}{\partial r}
\qquad\qquad
e_+= \alpha\frac{\partial}{\partial \nu}
     +\beta\frac{\partial}{\partial \overline{\nu}}
     +\Omega\frac{\partial}{\partial r},
\]
for some functions $\alpha$, $\beta$ and $\Omega$ to be determined. Now the first condition on these functions is $<e_0, e_+>=0$ which gives us that
\[
\Omega =-\alpha\;\left<e_0,\frac{\partial}{\partial\nu}\right>- \beta\left<e_0,\frac{\partial}{\partial\overline{\nu}}\right>.
\]
Thus
\begin{equation}
e_+=\alpha\left(\frac{\partial}{\partial\nu}-\left<e_0,\frac{\partial}{\partial\nu}\right>\frac{\partial}{\partial r}\right)
 +\beta\left(\frac{\partial}{\partial\overline{\nu}}-\left<e_0,\frac{\partial}{\partial\overline{\nu}}\right>\frac{\partial}{\partial r}\right)
\quad=\alpha Z_++\beta Z_-\nonumber.
\end{equation}
The second condition $<e_+, e_+>=0$ now becomes
\begin{equation}\label{e:quad1}
<Z_+, Z_+>\alpha^2+2<Z_+, Z_->\alpha\beta\;+<Z_-, Z_->\beta^2=0.
\end{equation}
A lenghty calculation, involving a change to Euclidean coordinates via
$\Phi$, shows that
\begin{equation}\label{e:inprod}
 <Z_+, Z_+>=\frac{4\partial^+F\overline{\partial^-F}}{(1+\xi\overline{\xi})}
\qquad
<Z_+,Z_->=\frac{2(\partial^-F\overline{\partial^-F}+\partial^+F\overline{\partial^+F})}
 {(1+\xi\overline{\xi})}.
\end{equation}
With the aid of this, we can solve equation (\ref{e:quad1}) for the
ratio of $\alpha$ and $\beta$:
\[
\frac{\alpha}{\beta}=-\frac{\overline{\partial^+F}}{\overline{\partial^-F}}
 \qquad or \qquad
\frac{\alpha}{\beta}=-\frac{\partial^-F}{\partial^+F}.
\]
The fact that we get two answers simply represents the choice of
orientation for the null frame. We choose the first in order to have
our orientation agree with the standard orientation of
${\Bbb{R}}^3$. This orientation coincides with the graph orientation
$\Sigma$ in ${\Bbb{T}}\rightarrow{\Bbb{P}}^1$.

The final equation we have to solve is $<e_+, e_->=1$ and this gives
us $\beta\overline{\beta}$, that is $\phi$.  This represents rotation
of the frame  about $e_0$, which we can set to agree with the argument
of $-\overline{\partial^-F}$ by parallel translation of the frame.

\end{pf}

\section{The Local Geometry of Congruences}

We now use the twistor coordinates and null frame of the last section
to investigate the local geometry of congruences in ${\Bbb{R}}^3$.
To start we find expressions for the first order geometric properties:

\begin{Prop}\label{p:opscal}
The Lie derivative of $e_+$ in the $e_0$ direction is given by
\[
L_{e_0}e_+=\overline{\rho}e_++\sigma e_-,
\]
with
\begin{equation}\label{e:rho}
\rho=\frac{ \partial^+F\overline{\partial}\;\overline{\xi} -\partial^-F\partial\overline{\xi}}
{\partial^-F\overline{\partial^-F}-\partial^+F\overline{\partial^+F}}
\end{equation}
\begin{equation}\label{e:shear}
\sigma=\frac{\overline{\partial^+F}\partial\overline{\xi} -\overline{\partial^-F}\;\overline{\partial}\;\overline{\xi}}
{\partial^-F\overline{\partial^-F}-\partial^+F\overline{\partial^+F}}.
\end{equation}
\end{Prop}

\begin{pf}
This follows by using the computations in the previous Proposition.  For the twist and divergence,
\begin{align}\nonumber
\overline{\rho} &= \left<L_{e_0}e_+, e_-\right> \\\nonumber
  &= \left<\frac{\partial \alpha}{\partial r}Z_++\frac{\partial \beta}{\partial r}Z_-\;,\;\overline{\alpha}Z_-+\overline{\beta}Z_+\right>\nonumber.
\end{align}
Now use the expressions for $\alpha$ and $\beta$ and equation (\ref{e:inprod}) to get the result.

Similarly
\begin{align}\nonumber
\sigma &= \left<L_{e_0}e_+, e_+\right> \\\nonumber
  &= \left<\frac{\partial \alpha}{\partial r}Z_++\frac{\partial \beta}{\partial r}Z_-\;,\;\alpha Z_++\beta Z_-\right>\nonumber.
\end{align}

\end{pf}

The complex scalar functions $\rho$ and $\sigma$ describe the first order geometric behaviour of the congruence
of lines. In particular, the real part of $\rho$ is the {\it divergence}, the imaginary part is the
{\it twist} and $\sigma$ is the {\it shear} of the congruence (see
\cite{guil} for details). By Proposition \ref{p:traninv} these are invariant under translations of
the origin.  For line congruences in ${\Bbb{R}}^3$ the
evolution of these quantities along the line can be more directly
derived.  In particular, they satisfy the {\it Sachs equations} \cite{par}
\begin{equation}\label{e:sachs}
\frac{\partial \rho}{\partial r}=\rho^2+\sigma\overline{\sigma}
\qquad
\frac{\partial \sigma}{\partial r}=(\rho+\overline{\rho})\sigma.
\end{equation}
Note that if the shear or the twist vanish at some point on the line, they
vanish at every point on the line.

Following Penrose and Rindler \cite{par} these can be integrated in
terms of the initial values of $\rho$ and $\sigma$ at $r=0$:

\[
\rho=\frac{\rho_0-(\rho_0\overline{\rho}_0-\sigma_0\overline{\sigma}_0)r}
  {1-(\rho_0+\overline{\rho}_0)r+(\rho_0\overline{\rho}_0
                              -\sigma_0\overline{\sigma}_0)r^2}
\qquad
\sigma=\frac{\sigma_0}
  {1-(\rho_0+\overline{\rho}_0)r+(\rho_0\overline{\rho}_0-\sigma_0\overline{\sigma}_0)r^2}.
\]

The shear has the following interpretation. Consider a circle in the plane orthogonal to the line $\gamma_0$ at some
point. Lie-propogation along the line can alter this circle in a number of
ways: if the shear is zero, it will remain a circle and
if the shear is non-zero it will become an ellipse. In particular,
$|\sigma|$ measures the eccentricity of the ellipse, while
$\phi=Arg(\sigma)$ measures the inclination of the semi-major and
semi-minor axes (details can be found in \cite{guil}).

The twist can be understood as follows:
\begin{Def}
A line congruence is {\it integrable} iff locally there exists an embedded surface $S$ in ${\Bbb{R}}^3$ such that $S$ is orthogonal to the lines of the congruence.
\end{Def}

\begin{Prop}\cite{guil}
A line congruence is integrable iff $\rho$ is real (the twist vanishes).
\end{Prop}

\begin{Prop}
The orthogonal surface $S$ to an integrable line congruence is given by $r=r(\nu,\bar{\nu})$, where
\[
\bar{\partial} r=\frac{2F\bar{\partial}\bar{\xi}+2\bar{F}\bar{\partial}\xi}{(1+\xi\bar{\xi})^2}.
\]
\end{Prop}

\begin{pf}
It is not hard to show that the parametric surface in ${\Bbb{R}}^3$,
obtained by inserting $r=r(\nu,\bar{\nu})$ in (\ref{e:coord1}) and (\ref{e:coord3}),  is orthogonal to the
congruence iff the above condition holds.
\end{pf}

\begin{Def}
The {\it curvature} of a congruence is defined to be $K=\rho\overline{\rho}-\sigma\overline{\sigma}$.
\end{Def}

This definition reflects the fact that, in the
twist-free case, the curvature of a congruence is the curvature of the one
parameter family of  surfaces in ${\Bbb{R}}^3$
orthogonal to the congruence. More generally,

\begin{Thm}\label{t:section}
A line congruence is locally the graph of a section of the bundle
$\pi:{\Bbb{T}}\rightarrow {\Bbb{P}}^1$ if and only if the curvature is
non-zero.
\end{Thm}
\begin{pf}
From equations (\ref{e:rho}) and (\ref{e:shear}) we find that
\[
K=\frac{\partial\xi\overline{\partial}\;\overline{\xi}-\partial\overline{\xi}\;\overline{\partial}\xi}
{\partial^-F\overline{\partial^-F}-\partial^+F\overline{\partial^+F}}.
\]
Thus, the left hand side will vanish at a point $\gamma\in\Sigma$ if
and only if the Jacobian relating $\xi$ and $\nu$ is zero at $\gamma$,
that is, the tangent space to $\Sigma$ at $\gamma$ has a
vertical component.
\end{pf}

\section{Globally Convex Congruences}

We say a congruence is {\it globally convex} if it is the graph of a global
section of the canonical bundle $\pi:{\Bbb{T}}\rightarrow
{\Bbb{P}}^1$. A globally convex congruence is a toplogical 2-sphere which generalises the concept of an closed convex surface.

\vspace{0.1in}
\noindent{\bf Example}: The derivative of the action of $PSL(2,{\Bbb{C}})$ on ${\Bbb{P}}^1$
generates a holomorphic vector field on ${\Bbb{P}}^1$ which is a
global section of the canonical bumdle. This 6-parameter family of line congruences splits into twisting and twist-free congruences. The former
contains the standard overtwisted contact structure on ${\Bbb{R}}^3$, see
\cite{elia}, while the latter consists of the holomorphic ${\Bbb{P}}^1$
generated by lines through a point in ${\Bbb{R}}^3$.

\vspace{0.1in}

Globally convex congruences have nice properties. For example,

\begin{Prop}
Suppose that $\Sigma\subset{\Bbb{T}}$ is a globally convex congruence,
then every
point in ${\Bbb{R}}^3$, is contained in some line of the congruence,
i.e. the congruence is spacefilling.
\end{Prop}
\begin{pf}
Consider any point $p\in {\Bbb{R}}^3$.  By a
translation we can make $p$ the origin.  Now the resulting surface
$\Sigma$ is a vector field on $S^2$ and therefore must have a zero.  In terms of the lines on ${\Bbb{R}}^3$, a
zero represents a line with zero perpendicular distance from the origin, that is a line passing through the origin.
Thus every point $p$ must have a line passing through it.

\end{pf}

\begin{Thm}
Let $\Sigma$ be a globally convex congruence with curvature
$K$. Then:
\[
\int_\Sigma Kd\mu=4\pi,
\]
where $d\mu$ at $\gamma\in\Sigma$ is the pull-back via $\pi$ of the volume form induced on the
plane orthogonal to $\gamma$ by the Euclidean metric on ${\Bbb{R}}^3$.
\end{Thm}

\begin{pf}
Suppose we choose coordinates $(\xi,\overline{\xi})$ and twistor
function $F$ to describe an open subset of $\Sigma$.

The basis of 1-forms $\{\theta^0,\theta^+,\theta^-\}$ dual to the vector
basis in Proposition \ref{p:nullframe} has coordinates
\[
\theta^+=\frac{\sqrt{2}}{K(1+\xi\overline{\xi})}\left(\rho
  d\xi-\overline{\sigma} d\overline{\xi}\right).
\]
Now, $d\mu=i\pi^*(\theta^+\wedge\theta^-)$ and so we get that
\begin{equation}
\int_\Sigma K d\mu=\int_\Sigma K\;i\pi^*(\theta^+\wedge\theta^-)\quad
              =2\int_{{\Bbb{P}}^1}\frac{d\xi
              d\overline{\xi}}{(1+\xi\overline{\xi})^2}\quad
              =4\pi \nonumber.
\end{equation}

\end{pf}

Consider a line congruence in ${\Bbb{R}}^3$ given by an oriented surface
$\Sigma\subset{\Bbb{T}}$.

\begin{Def} A point $\gamma\in\Sigma$ is {\it complex} if the complex
structure acting on ${\Bbb{T}}$ preserves $T_\gamma\Sigma$, the
tangent space to $\Sigma$ at $\gamma$. A complex point $\gamma\in\Sigma$ is
called {\it positive} or {\it negative} if the complex structure
induced on $T_\gamma\Sigma$ by ${\Bbb{J}}$ agrees or disagrees with
the given orientation on $\Sigma$, respectively.
\end{Def}

In the general case a line congruence $\Sigma\subset{\Bbb{T}}$ will
have places where it is complex and others where it fails to
be. We can perturb such a surface so that the complex
points are isolated. Moreover, there is a well-defined index for each
isolated complex point and we identify these as follows:

\begin{Prop}\label{p:minikerr}
A point on $\Sigma$ is complex iff the shear vanishes along the
line. The index $I$ of an isolated complex point $\gamma_0$ is minus the
winding number of the semi-major axes of shear as we go around $\gamma_0$.
\end{Prop}

\begin{pf}
Let $f:\Sigma\rightarrow {\Bbb{T}}$ be a $C^2$ smooth immersion with
differential $df:T\Sigma\rightarrow T{\Bbb{T}}$.  Let $j$ be a
conformal structure on $\Sigma$ compatible with $\nu$. We define the
sections $\delta^+f\in\Omega^{10}(\Sigma)\otimes T^{10}{\Bbb{T}}$ and
$\delta^-f\in\Omega^{01}(\Sigma)\otimes T^{10}{\Bbb{T}}$ by
\[
\delta^\pm f=\frac{1}{2}\left(df \mp {\Bbb{J}}\circ df \circ j\right).
\]
Then
$\delta^+f\wedge\delta^-f\in\Omega^2(\Sigma)\otimes\mbox{det}\;T^{10}{\Bbb{T}}$,
works out to be
\[
\delta^+f\wedge\delta^-f=(\partial\xi\bar{\partial}\eta-\bar{\partial}\xi\partial\eta)d\nu\wedge d\bar{\nu}
\]

A point is complex iff this 2-form vanishes, see \cite{kling}. By
simplifying the numerator of $\sigma$ in (\ref{e:shear}), this is
equivalent to
the vanishing of the shear.

The index of a complex point is equal to the intersection index of
$\delta^+f\wedge\delta^-f$ with the zero section of
$\Omega^2(\Sigma)\otimes\mbox{det}\;T^{10}{\Bbb{T}}$. This is just the
winding number of the complex function
$\partial\xi\bar{\partial}\eta-\bar{\partial}\xi\partial\eta$,
i.e. the winding number of $\bar{\sigma}$ about its isolated zero.
\end{pf}

In the integrable case the argument of the shear is the angle between
the real axis of $\nu$ with a principal curvature direction of the
orthogonal surface in
${\Bbb{R}}^3$. We have the following generalisation of the fact that the total number
of isolated umbilics (counted with index) on an closed convex surface is 4.

\begin{Thm}
The total number of shear-free lines (counted
with index) on a globally convex congruence with only isolated
shear-free lines is 4.
\end{Thm}
\begin{pf}

Let $d_+$ be the sum of the indices over all positive complex points
of $\Sigma$. Define $d_-$ similarily for negative complex points. Then
by \cite{lai},
\[
d_++d_-=\chi(T\Sigma)+\chi(N\Sigma)
\]
\[
d_+-d_-=c_1(TM)\left[\Sigma\right],
\]
where $N\Sigma$ is the normal bundle and $T\Sigma$ is
the tangent bundle of $\Sigma$,
$\chi$ is the Euler number of the appropriate bundle, and $c_1(TM)$ is
the first Chern class of the tangent bundle to $M$.

In the case of a globally convex congruence with the graph orientation, we
have $\chi(T\Sigma)=2$, $\chi(N\Sigma)=2$ and
$c_1(T{\Bbb{T}})\left[\Sigma\right]=4$. The theorem then follows.

\end{pf}


text:
\begin{document}

\title{The \texttt{huge} Package for High-dimensional Undirected Graph Estimation in \texttt{R}}

\author{\name Tuo Zhao \email tourzhao@jhu.edu\\
       \name Han Liu \email hanliu@cs.jhu.edu \\
       \addr Johns Hopkins University\\
       Baltimore, MD 21218, USA
       \AND
       \name Kathryn Roeder \email roeder@stat.cmu.edu\\
       \name John Lafferty \email lafferty@cs.cmu.edu\\
       \name Larry Wasserman \email larry@stat.cmu.edu\\
       \addr Carnegie Mellon University\\
       Pittsburgh,PA, 15213}

\editor{Mikio Braun}

\maketitle

\begin{abstract}
We describe an \texttt{R} package called \texttt{huge} (ver 1.1.2),
that provides easy-to-use functions for estimating high dimensional
undirected graphs from data.  This package implements recent results
in the literature, including
\citet{Friedman07a}, \citet{Liu09} and \citet{Liu10}.  Compared with \texttt{glasso}, the \texttt{huge} package provides
several extra features: (i) instead of using \texttt{Fortran}, it is
written in \texttt{C}, which makes the code more portable and easier
to modify; (ii) besides fitting Gaussian graphical models, it also
provides functions for fitting high dimensional  semiparametric
Gaussian copula models, data-dependent model selection, data
generation and graph visualization; and  (iii) to achieve better
scalability, it incorporates correlation screening into graph
estimation. In particular, the package allows the user to apply both
lossless and lossy screening rules to scale up for high-dimensional problems,
making a tradeoff between computational and statistical efficiency.
\end{abstract}

\vspace{0.1in}
\begin{keywords}
high-dimensional undirected graph estimation, glasso,  huge, semiparametric graph estimation, data-dependent model selection, lossless screening, lossy screening.
\end{keywords}

\section{Overview}

Significant progress has been made recently on designing efficient
algorithms to learn undirected graphical models from high-dimensional
observational datasets. Existing packages include \texttt{glasso}
\citep{Friedman07a}, \texttt{Covpath} and
\texttt{CLIME}.  In particular, the \texttt{glasso}
package has been widely adopted by statisticians and computer
scientists due to its friendly user-inference and efficiency.  In this
paper, we describe a newly developed \texttt{R} package named
\texttt{huge}\footnote{The development was partially supported by google summer of code project} (High-dimensional Undirected Graph Estimation).
Compared with \texttt{glasso}, the core engine of \texttt{huge} is
coded in \texttt{C}, making modifications of the package more accessible to researchers
from the computer science and signal processing communities. The
package includes a wide range of functional modules, including data
generation, data preprocessing, graph estimation, model selection, and
visualization. Many recent methods have been implemented, including
the nonparanormal \citep{Liu09} for estimating a high dimensional
Gaussian copula graph, the StARS \citep{Liu10} approach for
stability-based graphical model selection, and correlation screening for high dimensional graph estimation. The package supports two modes of
screening, lossless \citep{Witten11} and lossy screening.  When using lossy screening, the user
can select the desired screening level to scale up for high-dimensional problems, but this introduces some estimation bias.

\section{Design and Implementation}

The package \texttt{huge} aims to provide a general framework for high-dimensional undirected graph estimation. The package includes six functional modules (M1-M6), see (Figure \ref{flowchart}).
\vspace{-0.3in}
\begin{figure}[!htb]
\centering
\includegraphics[width=5.5in]{flowchart.pdf}
\vspace{-0.2in}
\caption{\small The graph estimation pipeline.}
\label{flowchart}
\end{figure}

{\bf M1. Data Generator}: The function \texttt{huge.generator()} can simulate multivariate Gaussian data with different undirected graph structures, including hub, cluster, band, scale-free, and Erd{\"{o}s-R{\'{e}}nyi random graphs. The sparsity level of the obtained graph and signal-to-noise ratio can also be set up by users.

{\bf M2. Semiparametric Transformation}: The function \texttt{huge.npn()} implements the nonparanormal method \citep{Liu09} for estimating a semiparametric Gaussian copula model. Motivated by  additive models, the nonparanormal family extends the Gaussian distribution by marginally transforming the variables using smooth functions. Computationally, the  nonparanormal transformation only requires one pass through the data matrix.

{\bf M3. Graph Screening}: The \texttt{scr} argument in the main
function \texttt{huge()} controls the use of large-scale correlation
screening before graph estimation. The function supports two types of
screening rules, lossless screening and lossy screening. The lossless
screening method is from \citet{Witten11}.  Such screening procedures can greatly
reduce the computational cost and achieve equal or even better
estimation by reducing the variance at the expense of increased bias.

{\bf M4. Graph Estimation}: Similar to the \texttt{glasso} package,
the \texttt{method} argument in the  \texttt{huge()} function supports
two estimation methods: (i) the Meinshausen-B{\"{u}}hlmann  covariance
selection algorithm \citep{Meinshausen06} and (ii) the graphical lasso
algorithm \citep{Friedman07a}. In our implementation, we
exploit many suggested tricks and practices from \citet{Friedman10a}. For example, we solve each individual
lasso problem using coordinate descent combined with active set and
covariance update tricks. One difference between \texttt{huge} and
\texttt{glasso} is that we implement all the core components using
\texttt{C} instead of \texttt{Fortran}.   The code is also
memory-optimized  using the sparse matrix data structure so that it can
handle larger datasets  when estimating and storing full
regularization paths.  We also provide an additional graph estimation
method based on thresholding the sample correlation matrix.  Such an approach
is computationally efficient and has been widely applied in biomedical research.

{\bf M5. Model Selection}:  The function \texttt{huge.select()}
provides three regularization parameter selection methods: the
stability approach for regularization selection (StARS)
\citep{Liu10}; a modified rotation information criterion (RIC); and the extended Bayesian information criterion. The latter approach is a likelihood-based model
selection criterion that is only applicable for the graphical lasso
method. StARS conducts many subsampling steps to calculate
U-statistics, which is computationally intensive but can be trivially
parallelized.  RIC is closely related to the permutation approach for model selection and scales to large datasets.

{\bf M6. Graph Visualization}:  The plotting functions
\texttt{huge.plot()} and \texttt{plot()} provide visualizations of the
simulated data sets, estimated graphs and paths. The implementation is
based on the \texttt{igraph} package.  Due to the limits of
\texttt{igraph}, sparse graphs with only up to 2,000 nodes can
be visualized.

\section{User Interface by Example}

We illustrate the user interface by analyzing a stock
market data which we contribute to the \texttt{huge} package\footnote{We thank Mladen Kolar for providing the python code to craw the data from the web.}. We
acquired closing prices from all stocks in the S\&P 500 for all the
days that the market was open between Jan 1, 2003 and Jan 1,
2008. This gave us 1258 samples for the 452 stocks that remained in the S\&P 500 during the entire time period.
\begin{verbatim}
> library(huge)
> data(stockdata)                                           # Load the data
> x = log(stockdata$data[2:1258,]/stockdata$data[1:1257,])  # Preprocessing
> x.npn = huge.npn(x, npn.func="truncation")                # Nonparanormal
> out.npn = huge(x.npn,method = "glasso", nlambda=40,lambda.min.ratio = 0.4)
\end{verbatim}
Here the data have been transformed by calculating the log-ratio of the
price at time $t$ to price at time $t-1$.
The nonparanormal transformation is applied to the data, and a
graph is estimated using the graphical lasso (the default is the Meinshausen-B\"{u}hlmann estimator). The program automatically sets up a sequence of 40 regularization parameters and estimates the  graph path. The lossless screening method is applied by default.

\section{Performance Benchmark}
We adopt similar experimental settings as in \citet{Friedman10b} to compare \texttt{huge} with \texttt{glasso} (ver 1.4).  We consider four scenarios with varying sample sizes $n$ and dimensionality $d$, as shown in Table 1. We simulate the data from a normal distribution $N(0,I_d)$. Timings (in seconds) are computed over 10 values of the corresponding regularization parameter, and the range of regularization parameters is chosen so that each method produced approximately the same number of non-zero estimates (the sparsity level is from 0 to about 0.03). The convergence threshold of  both \texttt{glasso} and \texttt{huge} is chosen to be $10^{-4}$. All experiments were carried out on a PC with Intel Core i5 3.3Hz processor. We also tried \texttt{CLIME} (ver 1.0) and \texttt{Covpath} (ver 0.2), but were unable to obtain timing results due to numerical issues.
\begin{table}[!htb]
\footnotesize
  \centering
  \caption{Experimental Results}
  \vspace{0.05in}
  \begin{tabular}[htb!]{l | l | c | r | r | r | r}
   \Xhline{1pt}

    \multicolumn{3}{c|}{\multirow{2}{*}{Method}} & $d=1000$ &$d=2000$ &$d=3000$ &$d=4000$\\[-2pt]
    \multicolumn{3}{l|}{} &$n=100$ & $n=150$ &$n=200$ &$n=300$\\
    \Xhline{1 pt}
    \multicolumn{3}{l|}{\texttt{huge}-Meinshausen-B{\"{u}}hlmann (lossy)} &0.938(0.054) &3.562(0.581) &8.238(0.822) &19.06(1.845)\\
    \multicolumn{3}{l|}{\texttt{huge}-Meinshausen-B{\"{u}}hlmann} &1.247(0.060) &11.88(2.136) &38.86(3.882) &104.0(5.574)\\
    \multicolumn{3}{l|}{\texttt{glasso}-Meinshausen-B{\"{u}}hlmann} &27.91(0.286) &216.9(2.557) &717.0(2.870) &1688(9.991)\\
     \multicolumn{3}{l|}{\texttt{huge}-graphical lasso (lossy)} &23.03(2.261) &222.4(23.92) &709.3(37.68) &1552(86.14)\\
    \multicolumn{3}{l|}{\texttt{huge}-graphical lasso (lossless)} &24.24(2.951) &267.5(39.48) &819.5(47.39) &1750(55.78)\\
    \multicolumn{3}{l|}{\texttt{glasso}-graphical lasso} &79.68(2.338) &742.5(9.831) &2348(12.08) &5455(17.25)\\
    \Xhline{1pt}
  \end{tabular}
\end{table}

For Meinshausen-B{\"{u}}hlmann graph estimation, we can see that
\texttt{huge} achieves the best performance. In particular, when the
lossy screening rule is applied,  \texttt{huge} automatically reduces
each individual lasso problem from the original dimension $d$ to the
sample size $n$, therefore a better efficiency can be achieved in
settings when $d \gg n$.  Based on our experiments, the speed up due
to the lossy screening rule can be up to $500\%$.

Unlike the Meinshausen-B{\"{u}}hlmann graph approach, the graphical
lasso estimates the inverse covariance matrix.
The lossless screening rule \citep{Witten11} greatly reduces the
computation required by the graphical lasso algorithm, especially when
the estimator is highly sparse.  The lossy screening rule can further
speed up the algorithm and provides an extra performance boost.

\section{Summary}

We developed a new package named \texttt{huge} for high dimensional
undirected graph estimation. The package is complementary to the existing \texttt{glasso} package by providing extra features and functional modules. We plan to maintain and support this package in the future.

\bibliography{HUGE}
\end{document}

config: arxiv


TITLE: Topology, Proving Interior and Exterior in $Q^n$
QUESTION [0 upvotes]: We have the subsets $\mathbb Q$ (Rationals), $I=\mathbb R -\  \mathbb Q$ (Irrationals), and I want to prove that $(\mathbb Q)°=(I)°=Ø $and also find $Ext \mathbb Q $.
In order to prove that both subsets’ interior are empty, I proposed $x∈\mathbb Q$ such that for an $r > 0, B(x,r) ⊂ \mathbb Q,$ but $I=\mathbb R-\mathbb Q$ are dense, such as $\mathbb Q,$ thus $B(x,r) ⊄ \mathbb Q $ and $x $ is never an interior point of $Q,$ we can’t find a complete ball inside $Q,$ which means that $(\mathbb Q)°=(I)°=Ø $(following the same steps for I).
Then, to find $Ext Q$, I defined $E=\mathbb R-\mathbb Q$ and proposed y∈E and tried to fit a ball inside it so it could prove the existence of E by finding its interior. So, for an $r > 0, B(y,r) ⊂ E$, but $E=I$ and we already know that $(I)°=Ø$, thus $Ext \mathbb Q = Ø $(same for $Ext I$).
But then I’m asked to find $(\mathbb Q^n)°$ and$ Ext (\mathbb Q^n)$, if $\mathbb Q^n = \{x∈R^n | x=(x_1,x_2,...,x_n), x_i ∈ \mathbb Q, 0\leq x \leq 1\},$ could I use the same method? Is my proof for the first two paragraphs correct and sufficient? Thanks

REPLY [1 votes]: Fact of life: every open ball contains rational and irrational points.
This implies all of the following, straight from the definitions:

$\mathbb{Q}^\circ = \emptyset$.
$\mathbb{P}^\circ  =\emptyset$.
$\overline{\mathbb{Q}} = \mathbb{R}$.
$\overline{\mathbb{P}} = \mathbb{R}$.

So $\operatorname{Ext}(\mathbb{Q}) = \mathbb{R} \setminus \overline{\mathbb{Q}} = \emptyset$ and similarly for the irrationals $\mathbb{P}$.


TITLE: Probability of two integers' square sum divisible by $10$
QUESTION [5 upvotes]: Given two random integers $a$, $b$ calculate probability of $a^2$ + $b^2$ is divisible by $10$.
I've tried to simulate this process and got a result about $0.18$ (maybe incorrect), but have no idea why.

REPLY [3 votes]: It is natural to assume that both $a$ and $b$ are independently and uniformly distributed mod $2$ as well as mod $5$.
Then $a^2+b^2$ is even iff $a\equiv b$ mod $2$, which happens with probability ${1\over2}$.
The $5$ elements of ${\mathbb Z}_5$ have squares $0,1,-1,-1,1$ mod $5$. The probability that two such squares add up to $0$ mod $5$ therefore is $={1\over25}+{4\over5}\cdot{2\over5}={9\over25}$.
The probability that $a^2+b^2\equiv0$ mod $10$  is then given by ${1\over2}\cdot{9\over25}={9\over50}$.


TITLE: Let $u(x,t)$ be the solution of $u_t(x,t)-u_{xx}(x,t) = \cos(x)$ on the interval $(x,t) \in [0.\pi] \times (0, \infty)$ with boundary conditions
QUESTION [1 upvotes]: Let $u(x,t)$ be the solution of $u_t(x,t)-u_{xx}(x,t) = \cos(x)$ on
   the interval $(x,t) \in [0,\pi] \times (0, \infty)$ with boundary
   conditions $u(0,t) = 0$ and $u(\pi,t) = \sin t$ together with initial
   conditions $u(x,0) = \sin(x)$. Expand the solution $u(x,t)$ by the
   Fourier sine series: $$u(x,t) = \sum U_n(t)\sin(nx)$$
Find the ODE for each $U_n(t)$.

I am having some problem intepretating this question. Do I need to solve the inhomogenous diffusion problem first? Then i express the solution as a fourier sine series? Anyone can give me a push in the right direction would be appreciated. As I am reading Strauss PDE book, i see that the question although looks like a diffusion question, however the interval is neither half interval nor full interval... So i am really confused.

REPLY [1 votes]: Check out this guide for solving inhomogeneous equations.
Since there are two different inhomogeneties, we can split up the solution to two parts $$u(x,t) = v(x,t) + w(x,t)$$
where $w(x,t)$ is a function that only satisfies the inhomogeneous boundary condition. For convenience, we choose a function such that $w_{xx} = 0$, therefore $w(x,t) = \dfrac{x}{\pi}\sin t$
This leaves
$$ v_t - v_{xx} = u_t - u_{xx} - w_t = \cos x - \frac{x}{\pi}\cos t $$
with boundary conditions
$$v(0,t)=v(\pi,t)=0$$
$$v(x,0) = u(x,0) - w(x,0) = \sin x $$
We assume an ansatz of the form
$$ v(x,t) = \sum_{n=1}^{\infty} T_n(t)\sin(nx) $$
This result came from performing separation of variables on the homogeneous equation, and that any function can be expressed as a linear combination of
$$ f(x,t) = \sum f_n(t) X_n(t) $$
where $X_n(x)=\sin(nx)$ are eigenfunctions of the homogeneous problem
Plugging into the equation and the remaining B.C,
$$ v_t - v_{xx} = \cos x - \frac{x}{\pi}\cos t = \sum_{n=1}^{\infty}\big[{T_n}'(t)+n^2T_n(t)\big]\sin(nx) \tag{1} $$
$$ v(x,0) = \sin x = \sum_{n=1}^{\infty} T_n(0)\sin(nx) \tag{2} $$
The Fourier coefficients give a family of ODEs
$$ {T_n}'(t)+n^2T_n(t) = \frac{2}{\pi} \int_0^\pi \left(\cos x - \frac{\cos t}{\pi} x\right)\sin(nx)\ dx $$
$$ T_n(0) = \frac{2}{\pi}\int_0^\pi \sin x\sin(nx) \ dx $$
Solving the integrals, we obtain
$$ {T_1}' + T_1 = -\frac{2}{\pi}\cos t, \qquad T_1(0) = 1 $$
$$ {T_n}' + n^2T_n = \frac{2}{\pi}\left[\big(1+(-1)^n\big)\frac{n}{n^2-1}+\frac{(-1)^n}{n}\cos t \right], \qquad \ T_{n\ge 2}(0) = 0 $$
These are easily solved using undetermined coefficients
$$ T_n = c_n e^{-n^2t} + A_n\sin t + B_n\cos t + D_n $$


TITLE: Prove that $f(x,y) = 10x^2+ 10y^2 + 12xy + 2x + 6y + 1 \geq 0$
QUESTION [0 upvotes]: I need to prove that
$f(x,y) = 10x^2+ 10y^2 + 12xy + 2x + 6y + 1$ has a global minimum at the point $(1/8, -3/8)$.
Since $f(1/8, -3/8) = 0$, that would be proving that $f(x,y) \geq 0$ $\forall (x,y)\in \mathbb{R}^2 $
I tried completing squares but wasn't successful. I always ended up with at least term varying in sign.
Is there an alternative to the method of completing squares in case like this?
Thanks a lot in advanced.

REPLY [0 votes]: Completing the square works:
$(x+3y+1)^2=x^2+9y^2+1+2x+6y+6xy$, hence
$$f(x,y)= (x+3y+1)^2 + 9x^2+y^2+6xy=(x+3y+1)^2+(3x+y)^2.$$


text:
Going back to the heuristic discussion of \ac{MXL0} in the previous section, we see that the one-shot estimator $\basicvopfunction$ is bounded as $\abs{\basicvopfunction} \leq \sup\modulus{\genericfunction}/\delta = \magnitude{1/\delta}$.
This unveils a significant trade-off between the $\magnitude{\delta}$ bias of the estimator and its $\magnitude{1/\delta}$ deviation from the true derivative:
the more accurate $\basicvopfunction$ becomes (smaller bias), the less precise it will be (higher variance).
In the context of iterative optimization algorithms, this \emph{bias\textendash variance dilemma} induces strict restrictions on the design of the query-radius and step-size policies, with deleterious effects on the algorithm's convergence rate (\cf \cref{section:MXL0,section:experiments}).
Motivated by this drawback of the \ac{SPSA} approach,
we proceed in the sequel to design a gradient estimator which requires a single function query per iteration, whilst at the same time enjoying a uniform bound on the norms of the estimates.

\subsection{\ac{SPSA} with callbacks}
\label{section:SPSAplus}

To proceed with our construction, let $z$ take the value $-1$ or~$+1$ with equal probability, and consider the estimator
\begin{equation}
\label{basicvopO}
\basicvopO{x}{\delta}
 	=\frac{\generic{x + \delta z}-\offset}{\delta} z.
\end{equation}

The offset value $\offset$ is decided \emph{a priori}, independently of the random variable $z$, so that $\expectation{\offset z} = \offset \expectation{z} = 0$.
In turn, this implies that $\expectation{\basicvopO{x}{\delta}} = \basicvtp{x}{\delta}$, and hence:
\begin{equation}
\modulus{\expectation{\basicvopO{x}{\delta} - \dgeneric{x}}}
	\leq \gradlipschitz \delta /2
\end{equation}
\ie the accuracy (bias) of $\basicvopO{x}{}$ is again $\magnitude{\delta}$.

The novelty of \eqref{basicvopO} is as follows:
if we take $\offset = \generic{x}$, then $\modulus{\basicvopO{x}{\delta}} = (1/\delta)\, \modulus{\generic{x + \delta z}-\generic{x}} \leq \Lipschitz $ where~$\Lipschitz$ denotes the Lipschitz constant of~$\genericfunction$,
so the choice $\offset = \generic{x}$ would be ideally suited for our purposes;
however, taking $\offset = \generic{x}$ would also involve an additional function query.
To circumvent this, we will instead approximate $\generic{x}$ with the closest available surrogate, namely the function value observed at the previous iteration of the process.

To make this precise in our \ac{MIMO} context, we will consider the \emph{enhanced} \ac{SPSA} estimator
\acused{SPSA+}

\begin{equation}
\label{estgradratekMO}
\tag{\SPSAplus}
\estgradkn{\nbk}{\itr}
	= \frac{\dimensionk{\nbk}}{\deltan{\itr}} \big[ \rate{ \testmatn{\itr}} - \rate{ \testmatn{\itr-1}} \big] \, \zmatkn{\nbk}{\itr},
\end{equation}
where:
\begin{enumerate}[wide,labelwidth=!,labelindent=5pt]
\item
$\deltan{\itr}$ is the given query radius at time $\itr$.
\item
$\zmatkn{\nbk}{\itr}$ is drawn randomly from the sphere $\trmspherei{\dimensionk{\nbk}-1}$
\item
$\testmatn{\itr}$
is the transmit covariance matrix defined along \eqref{linearperturbator}.

\end{enumerate}

\begin{algorithmenv}[t]
\caption{\Acf{MXL+}\label{alg:trMXLMO}}
\removelatexerror
\RestyleAlgo{tworuled}
\centering
\vspace{-2.5pt}
\begin{algorithm}[H]

\Parameters{

$\stepsizen{\itr}$, $\deltan{\itr}$}
 \KwInit{$\itr\leftarrow 1$, $\dualmat\leftarrow 0$;\\

 \hfill
 $\forall\nbk\colon \matk{\nbk} \leftarrow (\powerk{\nbk}/\mmk{\nbk})\, \identityk{\nbk}$, $\offsetk{\nbk}\leftarrow\rate{\mat}$}

\mainnl \Repeat{}{
\mainnl \inlineAteachDIP{$\nbk\in\{1,\dots,\kk\}$}{$\DAplusk{\nbk}{\stepsizen{\itr},\deltan{\itr}}$ }\;
\mainnl $\itr\leftarrow \itr+1$ \;
 }

\smallskip
\rule{0.96\linewidth}{.4pt} \\

\setcounter{AlgoLine}{0}
\Fctn{$\DAplusk{\nbk}{\stepsize,\delta}$}{

\routinenl \Sampleuniformly{$\trzmatk{\nbk}$}{$\trmspherei{\dimensionk{\nbk}-1}$} \;

\routinenl \Play $ \testmatk{\nbk} \leftarrow \matk{\nbk} + \frac{\delta}{\radiusk{\nbk}} (\cmatk{\nbk} - \matk{\nbk}) + \delta \trzmatk{\nbk} $ \;

\routinenl \Get $ \ratevalue \leftarrow \rate{ \testmat}$ \;

\routinenl \Set $\estgradk{\nbk} \leftarrow \frac{\dimensionk{\nbk}}{\delta} (\ratevalue-\offsetk{\nbk}) \, \zmatk{\nbk} $ \;
\routinenl \Set $\offsetk{\nbk} \leftarrow \ratevalue $ \;
\routinenl \Set $\dualmatk{\nbk} \leftarrow \dualmatk{\nbk} + \stepsize \estgradk{\nbk}$ \;
\routinenl \Set $\matk{\nbk} \leftarrow \normalize{\powerk{\nbk}\,}{} \matsolk{\nbk}{\dualmatk{\nbk}}$ \;

}

\end{algorithm}
\end{algorithmenv}

Then, integrating \eqref{estgradratekMO} in the chassis of \ac{MXL}, we obtain a similarly enhanced version of \ac{MXL0}, which we call \acdef{MXL+}.
For concreteness, we present a pseudocode implementation of the resulting method in \cref{alg:trMXLMO}.

In terms of parameter values, \ac{MXL+} supports a broad class of policies satisfying the so-called Robbins\textendash Monro conditions:
\noeqref{nsss}
\begin{equation}
\hyplabel{nsss}
\begin{array}{lcr}\nocolsep
\textup{(a)}\ \,
\sum\nolimits_{\itr} \stepsizen{\itr} = \infty
,
&\quad&
\textup{(b)}\ \,
\sum\nolimits_{\itr} \stepsizen{\itr}^2 <\infty
.
\end{array}
\end{equation}
In addition, \ac{MXL+} also requires the following precautions regarding the allowable step-size and query-radius sequences:
\begin{align}
\hyplabel{trconditionbias}
\textstyle
\sum\nolimits_{\itr} \stepsizen{\itr} \deltan{\itr}
	&< \infty,
	\\[\smallskipamount]
\hyplabel{generalfirstcondition}
\sup\nolimits_{\itr} \stepsizen{\itr}/\deltan{\itr+1}
	&< 2\strongly / (\dimension\Lipschitz\kk),
	\\[\smallskipamount]
\hyplabel{generalsecondcondition}
\sup\nolimits_{\itr} \deltan{\itr}/\deltan{\itr+1}
	&< \infty,
\end{align}
Of the above conditions, \eqref{generalfirstcondition}\textendash \eqref{generalsecondcondition} guarantee the uniform boundedness of the gradient estimator, while \eqref{trconditionbias} is an additional condition needed for convergence of the algorithm.

In practice, these conditions are easy to verify when $\stepsizen{\itr} = \stepsize / \itr^{\alpha}$ and $\deltan{\itr} = \delta/\itr^{\beta}$ for some $\alpha,\beta > 0$.
In this case, the conditions \eqref{deltacondition}\textendash\eqref{generalsecondcondition} reduce to:
\begin{subequations}
\refstepcounter{hypothesis}
\label{fullhypothesis}
\begin{align}

\label{firstcondition}
\tag{Ha}
\begin{array}{l}
\nocolsep
(\dimension\Lipschitz \kk/2) \, \stepsize
	< \strongly \, \delta
	< \strongly 1/\sqrt{\maxmm(\maxmm-1)},
\end{array}
\\
\label{secondcondition}
\tag{Hb}
\begin{array}{l}
\nocolsep
0 \leq \beta \leq \alpha \leq 1
\quad\text{and}\quad
\alpha + \beta >1,
\end{array}
\end{align}
\end{subequations}

With all this in hand, we are finally in a position to state our main convergence results for the \ac{MXL+} algorithm.
We begin by establishing the algorithm's almost sure convergence:

\begin{theorem}
[Convergence of \ac{MXL+}]
\label{thm:lasttrMXLMO}
Suppose that \ac{MXL+} \textpar{\cref{alg:trMXLMO}} is run with step-size and query-radius policies satisfying \eqref{deltacondition}\textendash\eqref{generalsecondcondition}.
Then, with probability $1$, the sequence of the users' transmit covariance matrices $\testmatn{\itr}$ converges to the solution set of \eqref{eq:trproblem}.
\end{theorem}

As in the case of \cref{thm:lasttrMXL}, \cref{thm:lasttrMXLMO} provides a strong asymptotic convergence result, but it leaves open the crucial question of the algorithm's convergence speed.
Our next result justifies the introduction of \eqref{estgradratekMO} and shows that \cref{alg:trMXLMO} achieves the best of both worlds:
one-shot throughput measurements with an $\tilde{\mathcal{O}}(1/\sqrt{T})$ convergence rate.

\begin{theorem}[Convergence rate of \ac{MXL+}]
\label{timecomplexity:trMXLMO} \
Suppose that \ac{MXL+} \textpar{\cref{alg:trMXLMO}} is run for $T$ iterations.
We then have:

\begin{enumerate}

\item
\label{thm:trMXLMOi}
If $\stepsizen{\itr} = \stepsize/\sqrt{\itr}$ and $\deltan{\itr} = \delta/\sqrt{\itr}$ with $\stepsize$ and $\delta$ satisfying \eqref{firstcondition}:
\begin{equation}
\label{asymptoticrate}
\expectation{\optrate - \rate{ \meanmatn{T}}}
	= \Magnitude{\frac{\log T}{\sqrt{T}}}.
\end{equation}

\item
\label{thm:trMXLMOiinew}
If $\stepsizen{\itr} = \stepsize/\sqrt{T}$ and $\deltan{\itr} = \delta/\sqrt{T}$ with $\stepsize$ and $\delta$ satisfying \eqref{firstcondition}:
\begin{enumerate}

\addtolength{\itemsep}{.5ex}
\item
\label{thm:trMXLMOiifirst}
In expectation,
\begin{equation}
\label{troptrateboundmeanexpectation}
\expectation{\optrate - \rate{ \meanmatn{T}} }
	\leq \frac{\Bcoefficient{\stepsize}{\delta}}{\sqrt{T}}
	=\Magnitude{\frac{1}{\sqrt{T}}},
\end{equation}
where
\(
\Bcoefficient{\stepsize}{\delta}
	= (\kk/\stepsize) \log\maxmm
		+ 4\kk^2\meanlipschitz \delta
		+ \frac
			{\twosquaredtwitchFlipschitz \kk \dimension \squaredstrongly \stepsize}
			{[2/(\dimension\Lipschitz\kk) - \stepsize/\delta]^{2}}
\).

\item
\label{thm:trMXLMOiisecond}
In probability, for any small enough tolerance $\zmargin>0$,
\begin{equation}\label{troptrateboundmeaninterval}
\Proba{\optrate -\rate{ \meanmatn{T}} \geq \frac{\Bcoefficient{\stepsize}{\delta}}{\sqrt{T}} + \zmargin}
	\leq \exp\left(- \frac{\zmargin^2 T}{\Ccoefficient{\stepsize}{\delta}} \right),
\end{equation}
where
\(
\Ccoefficient{\stepsize}{\delta}
	= \twopowersevensquaredtwitchFlipschitz \dimension \kk^{2} \squaredstrongly
		\big/ [2/(\dimension\Lipschitz\kk) - \stepsize/\delta]^{2}

\).
\end{enumerate}
\end{enumerate}
\end{theorem}

Importantly, \cref{timecomplexity:trMXLMO} shows that \ac{MXL+} recovers the $\magnitude{1/\sqrt{T}}$ convergence rate of \ac{MXL} with \emph{full} gradient information, even though the network's users are no longer assumed to have \emph{any} access to a gradient oracle.
In fact, the guarantees of \cref{timecomplexity:trMXLMO} can be optimized further by finetuning the choice of $\stepsize$ and $\delta$;
doing just that (and referring to \cref{appendix:MXLzeroplus} for the details), we have:

\begin{table}[!t]
\heavyrulewidth=0.8pt
\lightrulewidth=0.4pt

\renewcommand{\arraystretch}{1.3}
\setlength{\tabcolsep}{2pt}

\caption{Parameters of \ac{MXL+} for \cref{convergenceratres:trMXLMO}}
\centering

\begin{tabular}{rl}
\toprule
\ref{cr:trMXLMOi})
&
\(
\displaystyle
\stepsizecoefficient{\errorproba}
	= \frac{\sqrt{\log\maxmm \big/ (\dimension\Lipschitz\kk^{2})}}{\sqrt{\meanlipschitz} + \sqrt{2\dimension\Lipschitz\kk}}

;
\quad
\deltacoefficient{\errorproba}
{\,=\,}

\frac{\sqrt{(\dimension\Lipschitz/\meanlipschitz)\log\maxmm}}{2}
;
\quad

T\geq \frac{\Lipschitz\maxmm^4\log\maxmm}{4\meanlipschitz}
\)
\\
\midrule
\ref{cr:trMXLMOii})
&
$
\stepsizecoefficient{\errorproba}
=

\frac{\renrmlzdcsteps{\errorproba}}{ \sqrt{\Lipschitz}}   \bigg[ {\scriptstyle \sqrt{ \twitchFlipschitz   \Lipschitz }\dimension\, \renrmlzdcsteps{\errorproba} } +  \sqrt{\sqrt{{2\halftwitchFlipschitz\strongly}/{\logarithm{1/\errorproba}}} \, \strongly \meanlipschitz }  \bigg]^{-1} \frac{ \sqrt{ \log\maxmm} }{\kk \sqrt{\dimension}}
$ ;

\\
&
$
\deltacoefficient{\errorproba}
=

 \frac{\renrmlzdcsteps{\errorproba}}{2}  \sqrt{{   \Lipschitz}/{\strongly \meanlipschitz}  \sqrt{{ \logarithm{1/\errorproba}  }/{ \twitchFlipschitz  \strongly}  }  \, \dimension    \log\maxmm  }
 $ ;
\\
&
$

T
=

{ 4  \repownrmlzdcsteps{\errorproba}{4}  \Lipschitz^2 } {\bigg[ \forcesqrthalftwitchFlipschitzoneoverstrongly {+}         \frac{1}{\renrmlzdcsteps{\errorproba} \dimension }\sqrt{  {2\meanlipschitz}/{\strongly\Lipschitz}\sqrt{{2}/{ \logarithm{1/\errorproba}} }} \bigg]^2}  \Big(\frac{\logarithm{1/\errorproba}\kk^4 \dimension^{3}  \log\maxmm }{\errortolerance^2} \Big)
$ ;

\hspace{6mm}
\\ &
\hfill
with
$
\renrmlzdcsteps{\errorproba}
=

\Big[     {1}/{\sqrt{\logarithm{1/\errorproba}}} + {4\sqrtstrongly}/{\sqrt{ \logarithm{\maxmm}}} \Big]^{1/ 2}
$
\\
\bottomrule
\end{tabular}
\label{table:convergenceratres:trMXLMO}
\end{table}

\begin{corollary}
\label{convergenceratres:trMXLMO}
Suppose that  \ac{MXL+} is run with $\stepsizen{\itr} = \stepsize/\sqrt{T}$, $\deltan{\itr} = \delta/\sqrt{T}$, and $T$, $\stepsize$, $\delta$ as in \cref{table:convergenceratres:trMXLMO}.
Then:

\begin{enumerate}
[a),ref=\alph*,wide,labelwidth=!,labelindent=5pt]
\item
\label{cr:trMXLMOi}
In expectation, we have:
\begin{equation}
\expectation{\optrate - \rate{ \meanmatn{T}}}
	\leq
		2\Lipschitz\left( \forcesqrthalftwitchFlipschitzoneoverstrongly +

        \frac{2^{3/4}\sqrt{ { \meanlipschitz}/{  \strongly\Lipschitz}  }}{\maxdimension}  \right)
		\sqrt{\frac{\kk^4\maxmm^6 \log\maxmm  }{T}}
.
\end{equation}

\item
\label{cr:trMXLMOii}

In probability, given a small enough tolerance $\errortolerance>0$ and a confidence level $1- \errorproba\in(0,1)$, we have:
\begin{equation}
\label{standardtroptrateboundmeaninterval}
\proba{\optrate - \rate{\meanmatn{\ceil{T}}} \leq \errortolerance}
	\geq 1 - \errorproba
.
\end{equation}
\obsolete{with $
T
\equiv
T(\kk,\maxmm,\errortolerance,\errorproba)
=

\Bigtheta{{\logarithm{1/\errorproba}\kk^4 \maxmm^6  \log\maxmm}/{\errortolerance^{2} }}
$.}
\end{enumerate}
\end{corollary}

An important feature of the convergence rate guarantee \eqref{standardtroptrateboundmeaninterval} is that it does not depend on the number of antennas $\nn$ at the receiver.
As such, \cref{alg:trMXLMO} exhibits a \emph{scale-free} behavior relative to $\nn$, which makes it particularly appealing for distributed massive-\ac{MIMO} systems.
In the next section, we further relax the requirement that all users update their transmit covariance matrices in a synchronous manner, and we derive a fully distributed version of the \ac{MXL+} algorithm.


TITLE: Integral $\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))dx$
QUESTION [4 upvotes]: Evaluate
  $$\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx$$

I tried the following ideas:
$u=a+b\tan(x)$, $du=b\sec^2(x)dx$
$$\frac{1}{b}\int \cos^4(x)\cos(u)du$$
using $1+\tan^2(x)=\sec^2(x)$
$$b^3\int \frac{\cos(u)du}{(u-a)^4}$$
I am stuck at this point. Any help?

REPLY [1 votes]: Setting $I_{a,b} $ equal to the original integral and assuming $b > 0$, we get, with your substitution $b\tan(x) = u-a$,
$$\begin{split}
I_{a,b} &= \frac 1b \int_{-\infty}^{+\infty} \cos^4(x) \cos(u) \mathop{}\!du = \frac 1b \int_{-\infty}^{+\infty} \frac{\cos(u) \mathop{}\! du}{\sec^4(x)} = \frac1 b\int_{-\infty}^{+\infty} \frac{\cos(u)\mathop{}\! du}{(1+\tan^2(x))^2} \\
&= \frac 1 b \int_{-\infty}^{+\infty} \frac{b^4\cos(u)\mathop{}\!du}{(b^2 + (u-a)^2)^2} = b^3 \int_{-\infty}^{+\infty} \frac{\cos(s+a) \mathop{}\!ds}{(b^2 + s^2)^2} = \int_{-\infty}^{+\infty} \frac{\cos(bt + a)}{(1+t^2)^2}dt.
\end{split}$$
The last integral can be solved in two ways:
Direct contour integration. Consider the complex integral
$$J_{a,b} = \lim_{r\to\infty} e^{ia}\int_{\Gamma_r} \frac{e^{ibz}}{(1 + z^2)^2}\mathop{}\!dz \tag 1$$
where the closed contour $\Gamma_r = [-r,r] \cup \gamma_{0,r,\curvearrowleft}$ indicates the concatenation of a segment of length $2r$ lying on the real line and a semicircle lying in the upper half-plane. The integral can be split into two components,
$$\int_{\Gamma_r} \frac{e^{ibz}}{(1 + z^2)^2}\mathop{}\!dz = \int_{[-r,r]} \frac{e^{ibz}}{(1 + z^2)^2}\mathop{}\!dz + \int_{\gamma_{0,r,\curvearrowleft}} \frac{e^{ibz}}{(1 + z^2)^2}\mathop{}\!dz,  $$
and since the second integral vanishes thanks to the Jordan lemma, we are led to
$$J_{a,b} = \lim_{r\to\infty} \int_{[-r,r]} \frac{e^{i(bz+a)}}{(1+z^2)^2} \mathop{}\!dz = \lim_{r\to\infty} \int_{-r}^r \frac{e^{i(bt+a)}}{(1+t^2)^2} \mathop{}\!dt = \int_{-\infty}^{+\infty} \frac{e^{i(bt+a)}}{(1+t^2)^2} \mathop{}\!dt, $$
whence it is clear that $$\operatorname{Re}(J_{a,b}) = I_{a,b}. \tag2$$
Applying the residue theorem to the integral in $(1)$ entails
$$\begin{split}
J_{a,b} &= e^{ia} \lim_{r\to\infty} \left(2\pi i \operatorname*{Res}\limits_{z=i} \frac{e^{ibz}}{(1 + z^2)^2}  \right) \\
&= 2\pi i\ e^{ia} \left( -\frac{b+1}{4e^b}i \right) = \frac \pi 2 e^{ia} \frac{b+1}{e^b},
\end{split}$$
and by $(2)$ we obtain
$$I_{a,b} = \boxed{ \frac \pi 2 \cos(a) \frac{b+1}{e^b}} $$
The Feynman trick, then contour integration. Let us introduce the following parameter-dependent integral:
$$ K_{a,b}(\lambda) = \int_{-\infty}^{+\infty} \frac{\cos(bt+a)}{\lambda^2 + t^2} \mathop{}\!dt. $$
Differentiating w.r.t. $\lambda$ yields
$$K_{a,b}'(\lambda)=\frac{dK_{a,b}}{d\lambda}(\lambda) = \int_{-\infty}^{+\infty} \frac{\partial}{\partial\lambda} \left(\frac{\cos(bt+a)}{\lambda^2 + t^2}\right) \mathop{}\!dt = -2\lambda \int_{-\infty}^{+\infty} \frac{\cos(bt+a)}{(\lambda^2 + t^2)^2} \mathop{}\!dt.  $$
By a procedure which is almost identical to the one seen above, we may calculate
$$K_{a,b}(\lambda) = \operatorname{Re} \left( 2\pi i\ e^{ia} \operatorname*{Res}\limits_{z=\lambda i} \frac{e^{ibz}}{\lambda^2 + z^2} \right) = \operatorname{Re}\left(2\pi i\ e^{ia} \frac{-i}{2\lambda e^{b \lambda}} \right) = \pi \frac{\cos(a)}{\lambda e^{b\lambda}} ,$$
so that
$$K_{a,b}'(\lambda) = - \pi\cos(a) \frac{e^{-b\lambda} (b\lambda +1)}{\lambda^2}.  $$
We conclude by observing
$$ I_{a,b} = -\frac 1 2 K'_{a,b}(1) = \boxed{ \frac \pi 2 \cos(a) \frac{b+1}{e^{b}}} $$


TITLE: Proving a Simple Inequality involving Fourier Transforms
QUESTION [2 upvotes]: This is a minor technical matter that I know to be true, I just don't know how to prove it. Help would be much appreciated.
Let:$$\psi\left(x\right)=\sum_{n=0}^{\infty}e^{-2^{n}v\left|x\right|}$$
where $v$ is a positive constant. $\psi$ is in both $L^{1}\left(\mathbb{R}\right)$ and $L^{2}\left(\mathbb{R}\right)$, and its Fourier transform is:
$$\Psi\left(\xi\right)=\int_{-\infty}^{\infty}\psi\left(x\right)e^{-2\pi i\xi x}dx=\frac{2}{v}\sum_{n=0}^{\infty}\frac{2^{-n}}{\left(\frac{2\pi\xi}{2^{n}v}\right)^{2}+1}$$
I am trying to prove that $2\Psi\left(2\xi\right)>\Psi\left(\xi\right)$ for all $\xi\in\mathbb{R}$. Any ideas, people?
Thanks in advance!

REPLY [3 votes]: $$2\Psi\left(2\xi\right)=2\frac{2}{v}\sum_{n=0}^{\infty}\frac{2^{-n}}{\left(\frac{2\pi(2\xi)}{2^{n}v}\right)^{2}+1} = \frac{2}{v}\sum_{n=0}^{\infty}\frac{2^{-n+1}}{\left(\frac{2\pi\xi}{2^{n-1}v}\right)^{2}+1} = \frac{2}{v}\sum_{n=-1}^{\infty}\frac{2^{-n}}{\left(\frac{2\pi\xi}{2^{n}v}\right)^{2}+1} >\Psi\left(\xi\right)$$
since the first term of the series is positive and sum of the rest of the terms is equal to $\Psi\left(\xi\right)$.


TITLE: Analysis: Is $A$ dense in $[0,1]$?
QUESTION [1 upvotes]: Let $f_n:[0,1]\to\mathbb R$ define by $f_n(x)=\cos(nx)$. Let $A_n=\{x\mid f_n(x)=0\}$ and $A=\bigcup_{n\in\mathbb N}A_n$. I have shown that $|A|=+\infty$. Do you think that $A$ is dense in $[0,1]$ ? I think that it is, and my proof would go like this:
Suppose by contradiction that it's not. Then there is an $\varepsilon>0$ and a $x\in [0,1]\backslash A$ such that $]x-\varepsilon,x+\varepsilon[\cap A=\emptyset.$ Then $f_n(y)\neq 0$ for all $n\in\mathbb N$ and for all $y\in]x-\varepsilon,x+\varepsilon[$. With loss of generality, suppose that $\cos(ny)>0$ for all $n\in\mathbb N$ and all $y\in]x-\varepsilon,x+\varepsilon[.$ Let $n$ enough big for that $\frac{1}{n}<\varepsilon$. Then for all $k>n$, $\cos(kx)>0$ and thus $\cos(ky)>0$ for all $y\in \left[x-\frac{1}{n},x+\frac{1}{n}\right]+\frac{2k \pi}{n}$ and so on $$[0,1]\cap \bigcup_{k\in\mathbb N}\left(\left[x-\frac{1}{n},x+\frac{1}{n}\right]+\frac{2k \pi}{n}\right)$$ which is actually a finite union what is a contradiction with the fact that $|A|=\infty $.
Do you think that it work ?

REPLY [0 votes]: Why approach by contradiction when you can do it directly?
You know explicitly that $\cos(nx) = 0$ when $x = \frac{1}{n} (k+ \frac12) \pi$ for any $k\in \mathbb{Z}$. This means that any $y\in [0,1]$ is within $\pi/n$ distance of a zero of $f_n(x)$. This is enough to show that $y$ is arbitrarily close to elements of $A$.


TITLE: Is there a space filling "tube"?
QUESTION [2 upvotes]: I am trying to construct a curve that fills the space as much as possible while maintaining certain minimum distance between two segments of the curve. Intuitively, this should be some sort of space filling "tube".
I could wrap a boundary around a space filling curve, e.g. Hilbert curve. However this certainly is not an optimal solution as the allowed maximum width of such "tube" is not continuous when the order of the curve increases.
How can I construct such a tube which fills the space as much as possible?

REPLY [1 votes]: You should be able to realize the density of the optimal planar disk packing in the limit.
Denote the diameter of the rope by $\rho$. The basic module is a rectangular $a\times b\times L$ box filled with parallel strands  of length $L-2\rho$. The strands are arranged such that a normal cross section of the box shows  a densest  packing of circles of diameter $\rho$. The strands  are spliced together in two $a\times b$ layers of thickness $\rho$ at the ends of the box, in order to get a single long piece of rope within the box.
Now fill space with such boxes in a clever way (making sure that the rope stays connected), such that the boxes get ever larger as you go out. In this way you can make the "overhead" disappear in the limit.


TITLE: Determining if a map is linear from vector mapping
QUESTION [1 upvotes]: I know we can say a map is linear if $f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)$ is true.
The issue I'm having is converting the maps to the functions so that I can apply that definition.
$\mathbb{R}^2\ni \begin{bmatrix}x \\ y \end{bmatrix} \rightarrow \begin{bmatrix}x^2 \\ xy \\y^2 \end{bmatrix} \in \mathbb{R}^3$
$\mathbb{R}^3\ni \begin{bmatrix}x \\ y \\z \end{bmatrix} \rightarrow \begin{bmatrix}1+x \\ 1+y \\z \end{bmatrix} \in \mathbb{R}^3$
How do I test if these maps are linear?
EDIT: More specifically, how do I write these mappings as functions so that that I can apply the definition of linearity?
EDIT2: How do I give the matrix expression for the map?

REPLY [1 votes]: If you want to write them as functions, you may write $f\left(\begin{bmatrix}x \\ y \end{bmatrix}\right)=\begin{bmatrix}x^2 \\ xy \\y^2 \end{bmatrix}$ and similar for the other one.
One thing we can say about linear functions into vector spaces is that they will also project to linear functions in each component. Since $f(x,y)=x^2$ is not linear, the first component function of the first function is not linear, so the function is not linear.


text:
\chapter{Future work}
There are several areas to explore, and we've listed a few below.

\begin{enumerate}
  \item An interesting challenge is to remove the symmetry constraint in the algorithm. Since the idea behind the algorithm is so simple, it's reasonable to believe that it can be generalized.
  \item Another related question is to find a rich enough subclass of problems (Hamiltonians) where correctors exist. Can the algorithm be tuned to produce correctors for these problems? Questions about regularity and strict convexity appear more accessible if the existence of correctors can be guaranteed.
  \item What does the existence of correctors for the cell-problem tell us about the percolation problem?
  \item As stated in~\Secref{sec:generalization}, there are several possible generalizations of our work. It's probably quite easy to remove the bounds on $\tau(\cdot,\cdot,\cdot)$ in~\Eqref{eq:basic-assumption-on-edge-weights} and replace it with a moment condition. Other types of lattices and other control problems can also be explored. For example, the directed versions of first-passage percolation results has a monotone Hamiltonian, and these appear to be easier to work with.
  \item It will be interesting to explore the behavior of the so-called integrable models under the variational formula. This has already been begun in the context of last-passage percolation and polymer models by~\citet{georgiou_variational_2013}.
\end{enumerate}


text:
\paragraph{Proof of \propref{p:sandwich}}
We first show that the lower bound holds. If for all $d$, $1/R_d > \mytd/DB$, then clearly $\sum_{d=1}^D 1/R_d > \sum_{d=1}^D \mytd/DB = \mytd/B$. Therefore
\begin{align}
\begin{split}
\prob{\sum_{d=1}^D \frac{1}{R_d} > \frac{\mytd}{B}}
&\geq
\prob{\bigcap_{d=1}^D \left\{\frac{1}{R_d} > \frac{\mytd}{DB}\right\}} \\

&= \prod_{d=1}^D \prob{R_d < \frac{DB}{\mytd}},
\end{split}
\end{align}
where the last equality follows because $\set{R_d}$ are independent random variables. In a similar way, we show that the upper bound holds. If $\sum_{d=1}^D 1/R_d > \mytd/B$, then $\exists d$ s.t. $R_d < DB/\mytd$. To see that this is true, assume otherwise. If $R_d \geq DB/\mytd$ for all $d$, then $\sum_{d=1}^D 1/R_d \leq \mytd/B$, a contradiction. Therefore
\begin{align}
\begin{split}
\prob{\sum_{d=1}^D \frac{1}{R_d} > \frac{\mytd}{B}}
&\leq
\prob{\bigcup_{d=1}^D \left\{R_d < \frac{DB}{\mytd} \right\}} \\

&=
1-\prod_{d=1}^D \prob{\bigcap_{d=1}^D \left\{R_d > \frac{DB}{\mytd} \right\}}.
\end{split}
\end{align}


TITLE: Expected value of Brownian motion at a time decided by a rate one Poisson process.
QUESTION [1 upvotes]: Situation:

We have that $\{W_{t},t\geq 0\}$ is a Brownian motion and
  $\{N_{t},t\geq 0\}$ is a Poisson process such that $N_{t}$ follows a
  Poisson distribution with parameter $t$. This process is independent
  from our Brownian motion.

Question: a) Show that $\mathbb{E}[W^{2}_{N_{t}}]=t$
b) For $a\geq 0$ we define the stopping time: $\tau=\inf\{t\geq 0:|W_{t}|>a\}$. Show that this stopping time is finite.
My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:
$\mathbb{E}[W^{2}_{N_{t}}]=\mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0\leq s<N_{t}$.
Using the Brownian motion properties a few more times, this reduces to $\mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.
b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.
Any help is appreciated.

REPLY [2 votes]: Note that $N_t$ is a random variable and therefore the statement "where $0 \leq s < N_t$" doesn't make sense.

$N_t$ takes values in $\mathbb{N}_0$ and because of the independence from $(W_t)_{t \geq 0}$ we have
$$\mathbb{E}(W_{N_t}^2) = \mathbb{E} \left( \sum_{k \geq 0} W_k^2 1_{\{N_t = k\}} \right) = \sum_{k \geq 0} \underbrace{\mathbb{E}(W_k^2)}_{=k} \underbrace{\mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} \sum_{k \geq 1} \frac{t^{k-1}}{(k-1)!} =t.$$
Alternatively, we can use the tower property of the conditional expectation to show that
$$\mathbb{E}(W_{N_t}^2) = \mathbb{E} \big[ \mathbb{E}(W_{N_t}^2 \mid N_t) \big] = \mathbb{E} \big[ \underbrace{\mathbb{E}(W_k^2)}_{=k} \big|_{k=N_t} \big] = \mathbb{E}(N_t)=t.$$

To prove that the stopping time $\tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$\mathbb{E}(W_{t \wedge \tau}^2) = \mathbb{E}(t \wedge \tau).$$ Since $|W_{t \wedge \tau}| \leq a$ this gives $$\mathbb{E}(t \wedge \tau) \leq a^2,$$ and now the monotone convergence theorem yields $$\mathbb{E}(\tau) \leq a^2$$ which shows, in particular, that $\tau<\infty$ almost surely.


TITLE: The Hodge $*$-operator and the wedge product
QUESTION [12 upvotes]: On every Riemannian manifold $M$, we can consider the Hodge $*$-operator, which is characterised by the following formula:
$$a \wedge *b = (a,b)\nu.$$
Here $a$ and $b$ are smooth forms on $M$, $(\ ,\ )$ is a metric on $\wedge T^*\!M$ and $\nu$ is the volume form with respect to the Riemannian metric.
My question: Is a formula of $*(a \wedge b)$ known?
I suspect that we can have a formula like "$*(a \wedge b)=(*a)\wedge(*b)$" or "$*(a \wedge b) = *a \wedge b \pm a \wedge *b$". Of course these formulae never hold. (Look at the degree.)

REPLY [3 votes]: A toy example. Given an orthonormal basis $e_1,\dots,e_n$ for the vector space $V$, then $\nu=e_1\wedge\dots\wedge e_n$ and
$$*(e_1\wedge e_2)=e_3\wedge\dots\wedge e_n\in  \bigwedge^{n-2}V, $$
with  $*e_1=e_2\wedge\dots\wedge e_n$ and $*e_2=-e_1\wedge e_3\wedge\dots\wedge e_n$ both in $\bigwedge^{n-1}V$. Introducing the insertion operator $i_\bullet: \bigwedge^{k}V\rightarrow \bigwedge^{k-1}V$ we arrive at
$$*(e_1\wedge e_2)=\frac{1}{2}\left( \underbrace{i_{e_2}(\underbrace{*e_1}_{\in \bigwedge^{n-1}V}}_{\in \bigwedge^{n-2}V})-i_{e_1}(*e_2)\right). $$
I do not if this is what you were searching for, but I hope it helps (a bit).


TITLE: What does the following set symbol notation mean
QUESTION [2 upvotes]: [6] x [6] -> Z
I know it's the cartesian product of [6], but I don't quite understand what [6] means?
Does it mean all numbers until 6, or is another way to write {6}?

REPLY [2 votes]: $[n]$ is a common notation for the set $\{1,2,\dots,n\}$.


TITLE: Implications of continuity on normed spaces
QUESTION [1 upvotes]: My goal for this question is to understand the implications of continuity its relationship to normed spaces. This question is based off of the notes for 18.155 available at ocw.mit.edu.
Let $u$ be a linear functional $u:V\rightarrow R$ on a normed space $V$. Let's assume we know that $u$ is continuous at $0$. This implies that $u^{-1}(-1, 1)$ is a neighborhood of $0 \in V$.
The text continues to say that this implies that $\exists \epsilon>0$ such that $u(\{f \in V: ||f|| < \epsilon\}) \subset (-1,1)$. Simply put, I don't understand why this is implied. So two questions: Why is this implied? Why does the value of the norm have anything to do with a neighborhood based on the distance function?
Any help at all would be appreciated. Thanks much!

REPLY [1 votes]: $\newcommand{\eps}{\epsilon}$A set $U$ in a metric space $(X,d)$ is open if for every $x \in U$ there exists an $\epsilon>0$ such that $B(x,\epsilon) \subset U$. A normed vector space is a metric space with metric given by $d(v,w)=||v-w||$. In your question we have that $U=u^{-1}((-1,1))$ is an open set containing $0$ thereby there is an $\epsilon>0$ such that $B(0,\eps) \subset U$ but
$$B(0,\eps)=\{f \in V : ||f-0|| < \eps\},$$
in particular $||f-0||=||f||$. Finally we see that since $B(0,\eps) \subset U$ so when we push forward under $u$ we must have that $u(B(0,\eps)) \subset u(U)=(-1,1)$.


TITLE: Simple-looking nonlinear ODE with fractional power
QUESTION [4 upvotes]: I am trying to solve the following nonlinear ODE for a function $P(x)$:
$$\left(1-x^2\right)P''(x)+k(k+1)P(x)=cP(x)^\frac{k-1}{k+1}.$$
Here, $k$ and $c$ are arbitrary parameters. By rescaling $P(x)$, one can without loss of generality set $c=1$, so really this equation is only parameterized by $k$.
I wonder if there exists a closed form solution for this equation, at least for integer $k$? Although it looks very simple, my efforts to solve it have so far been unsuccessful. Nonetheless, I have been able to find some very simple solutions: for instance,

when $k=1$, the equation can be integrated, resulting in

$$P(x)=\frac{1}{2}+a\left(1-x^2\right)+b\left[x+\left(1-x^2\right)\mathrm{arctanh}\,x\right],$$

when $k=-2$, by inspection I found

$$P(x)=\sqrt{2}\frac{4x}{3+x^2},$$

when $k=-3$, by inspection I found

$$P(x)=120\frac{1-x^2}{\left(5-x^2\right)^2}.$$
These solutions look simple enough that I'm hopeful that a general solution exists. Perhaps there even exists some standard transformation to linearize such equations?

REPLY [5 votes]: Using contact transformations, Lie showed that second order equations of this type can be transformed to the trivial equation $\frac{d^2 P^\prime}{dx^{\prime 2}} = 0$.
Edit: an algorithm for this is described in Kumei and Bluman When Nonlinear Differential Equations are Equivalent to Linear Differential Equations, SIAM J. Math. 42(5) (1982) 1157-1173.
Alternatively, a more general technique is described in Bluman and Dridi New Solutions for ordinary differential equations, J. Symb. Comp. 47 (2012) pp. 76-88. It uses a combination of invertible and non-invertible mappings. They treat several examples in detail.
Edit(2): The paper by N. Euler, Transformation Properties of $\ddot{x}+f_1(t)\dot{x}+f_2(t)x+f_3(t)x^n=0$ J. Nonlin. Math. Phys. 4(3-4) (1997) pp. 310-337 provides the conditions for an equation of this type to be integrable, as well as the conditions for it to have Lie symmetries.
He also considers non-point transformations $X=F(x,t)$, $dT = G(x,t)dt$ to transform the equation to $\frac{d^2X}{dT^2}+k_1\frac{dX}{dT}+k_2X^p=0$ for constants $k_1$, $k_2$ and $p\in \mathscr{Q}$. I've tried substituting the corresponding values for your equation into his conditions, but the results underline your point about the lack of symmetries.

REPLY [3 votes]: For $k=0$, according to Maple the general solution is
$$ \text{arctanh}(x) \pm \int^{P(x)/\sqrt{x^2-1}} \dfrac{du}{\sqrt{u^2 - 2 c \ln(u) + b}} = a $$
Another family of solutions (for all $k$) is
$$ P(x) = \left( {\frac {k \left( k+1 \right) }{c}} \right) ^{-(k+1)/2}{x}^{k+1
}
$$
For $k = -3$, there are some other closed-form solutions, e.g.
$$ P(x) =  \frac{8}{c(1-x^2)} $$
and
$$ P(x) = \frac{24 (21 x^2-5)}{c (3x^2+5)^2}$$


TITLE: How to prove the matrix inequality $\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}+\det{(B)}|$
QUESTION [3 upvotes]: Let $A,B\in M_{2}(R)$ be two square matrices so that
$AB=BA$, prove the inequality
$$\sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}
+\det{(B)}|$$
I only know prove this
$$\sqrt{a^2-ab+b^2}+\sqrt{a^2+ab+b^2}\ge |a+b|,a,b\in R$$
because
$$\sqrt{x}+\sqrt{y}\ge\sqrt{x+y}$$
so
$$LHS\ge\sqrt{2(a^2+b^2)}\ge (a+b)$$
But for matrices I can't prove it. Thank you

REPLY [2 votes]: As user 1551 points out, it suffices to prove the inequality for $AB = BA$ where, $A,B$ are invertible. Since, commuting matrices preserve each others eigenspaces, they are simultaneous diagonalisable, $\exists \, U \in U_2(\mathbb{R})$, such that $U^{-1}AU = D[x_1,y_1] = D_1$ and $U^{-1}BU = D[x_2,y_2] = D_2$.
Thus, $\displaystyle \sqrt{\det{(A^2-AB+B^2)}}+\sqrt{\det{(A^2+AB+B^2)}}\ge |\det{(A)}
+\det{(B)}| \iff \sqrt{\det{(D_1^2-D_1D_2+D_2^2)}}+\sqrt{\det{(D_1^2+D_1D_2+D_2^2)}}\ge |\det{(D_1)}
+\det{(D_2)}|$
(since, $|\det(U)| = 1$)
That is,
$\displaystyle \sqrt{(x_1^2-x_1x_2+x_2^2)(y_1^2-y_1y_2+y_2^2)}+\sqrt{(x_1^2+x_1x_2+x_2^2)(y_1^2+y_1y_2+y_2^2)} \ge |x_1y_1+x_2y_2| \iff \left(\frac{3}{4}(x_1-x_2)^2+\frac{1}{4}(x_1+x_2)^2\right)^{1/2}\left(\frac{3}{4}(y_1-y_2)^2+\frac{1}{4}(y_1+y_2)^2\right)^{1/2} + \left(\frac{1}{4}(x_1-x_2)^2+\frac{3}{4}(x_1+x_2)^2\right)^{1/2}\left(\frac{1}{4}(y_1-y_2)^2+\frac{3}{4}(y_1+y_2)^2\right)^{1/2} \ge_{C.S.} \frac{3}{4}|x_1-x_2||y_1-y_2|+\frac{1}{4}|x_1+x_2||y_1+y_2|+\frac{1}{4}|x_1-x_2||y_1-y_2|+\frac{3}{4}|x_1+x_2||y_1+y_2| \ge |x_1y_1+x_2y_2|$ (by triangle inequality).


TITLE: Particle size that can be driven by electrical field or magnetic field
QUESTION [0 upvotes]: Does anyone know what size-range of metal particles can be driven by either electrical field or magnetic field? For example, should the size be nano-meter range or micrometer?

REPLY [2 votes]: There is almost no limit to the size of a "particle" that can be moved by a magnetic field. See for example this picture (source)

which shows an entire hospital gurney being sucked into an MRI scanner. In this case the patient was injured but survived. Not all are that lucky.
Bottom line: with a sufficiently strong (and diverging) field, a ferromagnetic "particle" can be very large, and yet be moved.


TITLE: how to get $dx\; dy=r\;dr\;d\theta$
QUESTION [3 upvotes]: In polar coordinate how we can get $dx\;dy=r\;dr\;d\theta$?
with these parameters:
$r=\sqrt{x^2+y^2}$
$x=r\cos\theta$
$y=r\sin\theta$
Tanks.

REPLY [4 votes]: In general, under the change of coordinates $u=u(x,y)$ and $v=v(x,y)$, the area element changes according to the formula: (See here)
$$du \;dv=\left|\frac{\partial (u,v)}{\partial(x,y)}\right|dx\; dy,$$
where the Jacobaian is the determinant given by
$$\left|\frac{\partial (u,v)}{\partial(x,y)}\right|=\left|
  \begin{array}{cc}
    \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
    \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\
  \end{array}
\right|.$$
Therefore, for polar coordinates $(x,y)=(r\cos\theta,r\sin\theta)$, we have
$$\left|\frac{\partial (x,y)}{\partial(r,\theta)}\right|=\left|
  \begin{array}{cc}
    \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\
    \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \\
  \end{array}
\right|=\left|
  \begin{array}{cc}
    \cos\theta & -r\sin\theta \\
    \sin\theta & r\cos\theta \\
  \end{array}
\right|=r,$$
which implies that
$$dx\; dy=r\;dr\; d\theta.$$


TITLE: Multidimensional function as manifold
QUESTION [0 upvotes]: Given $g(x,y)=2x^2+y^2-1$. I know that $g^{-1}(0)$ is a manifold. But why?
Could you please give me a step by step answer so i can try to understand how its done and try it on my own next time.
Thank you

REPLY [1 votes]: More generally, this can be seen as a consequence of the following result :

Let $f : V \mapsto \mathbb{R}$ be a smooth function over an open subset $V$ of $\mathbb{R}^3$, and let $S=f^{-1}(\{ 0 \})$. If, for all $x \in S$, $Df(x) \neq 0$, then S is a manifold.

The proof of this uses the implicit fonctions theorem to show that $S$ is locally the graph of some  smooth function $g$.


TITLE: Which of these properties hold for sequences satisfying $x_n \leq y_n \leq x_{n+2}$?
QUESTION [0 upvotes]: $\{ x_n \}$ and $\{ y_n\}$ are two sequences of real numbers such that $x_n\leq y_n \leq x_{n+2}, n=1,2,3...$ then
(a) $\{ y_n\}$ is a bounded sequence.
(b) $\{ x_n \}$ is an increasing sequence.
(c) $\{ x_n \}$ and $\{ y_n\}$ converge together.
(d) $\{ y_n\}$  is an increasing sequence.
I found this question in a book and I am not really able to reach any conclusion regarding the options.
I have concluded that since $x_n\leq  x_{n+2}$, $n=1$ gives the subsequence $\{ x_{2n-1}\}$ is decreasing and similarly, $\{ x_{2n}\}$ will be decreasing. But I am not able to proceed further.
Also, I tried ruling out the options using certain examples. I think option (c) should be right.

REPLY [0 votes]: Let's consider two examples:

Let $x_n=n$ and let $y_n=n+1$ for all $n\in \mathbb{N}$, then $x_n=n\leq y_n=n+1\leq x_{n+2}=n+2$ but $(y_n)_n$ is not bounded. In general we can't conclude (a).
Let $$x_n=y_n=\begin{cases}0& \mbox{ if $n$ is even,}\\ 1 & \mbox{ if $n$ is odd.}\end{cases}$$  Clearly $x_n=y_n=x_{n+2}$ for all $n\in \mathbb{N}$ and thus $x_n\leq y_n\leq x_{n+2}$ holds for all $n\in \mathbb{N}$. Clearly, both sequences are not convergent and not increasing. Hence we can't conclude  (b), (c) or (d) either.

You can't conclude any of those properties.
EDIT: I misread the meaning of (c).
Let's show the following lemma.

If $a_n\to a$ and $\lim_{n\to \infty}|b_n-a_n|=0$, then $b_n\to a$.

Proof: Choose $\varepsilon>0$. By assumption, there exists an $n_0\in \mathbb{N}$ such that for all $n\geq n_0$, we have that $|a_n-a|<\frac{\varepsilon}{2}$. Similarly, there exists an $n_1\in \mathbb{N}$ such that for all $n\geq n_1$ we have that $|b_n-a_n|<\frac{\varepsilon}{2}$. Let $n\geq \max\{n_0,n_1\}$, then \begin{eqnarray}
|b_n-&a|&\leq &|b_n-a_n|+|a_n-a|\\
&<& \frac{\varepsilon}{2}+\frac{\varepsilon}{2}\\
&=& \varepsilon.
\end{eqnarray}
This concludes the proof of the lemma.
We now show that if $(x_n)_n$ converges to some $x\in \mathbb{R}$, then so does $(y_n)_n$. Indeed, note that
\begin{eqnarray}
|y_n-x_n| &= & (y_n-x_n)\\
&\leq & (x_{n+2}-x_n).
\end{eqnarray}
Here we used that $x_n\leq y_n\leq x_{n+2}$. But as $(x_n)_n$ is convergent to $x$, the right-hand side goes to zero and thus $y_n\to x$ by the above lemma.
Conversely, you can easily show that if $y_n\to y$, then so does $x_n$. Indeed, note that
\begin{eqnarray}
|y_n-x_n| &= & (y_n-x_n)\\
&\leq & (y_{n+2}-y_{n-2}).
\end{eqnarray}
Again, as $(y_n)_n$ is convergent, the right-hand side goes to zero (as $(y_n)_n$ is a Cauchy sequence) and thus $x_n\to y$ by the above lemma.
This shows that under these conditions, $(x_n)_n$ is convergent with limit $x\in \mathbb{R}$ if and only if $(y_n)_n$ is convergent with limit $x\in \mathbb{R}$. I'll leave it to you to figure out what happens when these limits are infinite or do not exist.


TITLE: Is there any identity for this series?
QUESTION [0 upvotes]: While solving inequality and finite series problem I often come across this series-
$$(n+1)(n+2)(n+3)...(n+n)$$.
Is there a general solution to this form of a series?
Thanks for any help!!

REPLY [3 votes]: $\prod_{k = n + 1}^{2n} k = \frac{\prod_{k = n + 1}^{2n} k \prod_{k = 1}^{n} k}{\prod_{k = 1}^{n} k} = \frac{\prod_{k = 1}^{2n} k }{\prod_{k = 1}^{n} k} = \frac{(2n)!}{n!}$


TITLE: An application of the Grauert's upper semi-continuity theorem
QUESTION [1 upvotes]: Let $X$ be a smooth projective variety, $A$ a complete discrete valuation ring, $Y=\mbox{Spec} A$ and $f:X \to Y$ a smooth, projective, surjective morphism. Denote by $y$ the closed point of $Y$. Let $\mathcal{L}$ and $\mathcal{M}$ be two line bundles on $X$ such that its restrictions to the fiber $X_y$ over $y$ are isomorphic, i.e.,  $\mathcal{L}\otimes \mathcal{O}_{X_y} \cong \mathcal{M} \otimes \mathcal{O}_{X_y}$. Does this imply that there exist a line bundle $\mathcal{N}$ on $Y$ such that $\mathcal{L} \cong \mathcal{M} \otimes f^*\mathcal{N}$?
Note that this is in some sense a generalization of Ex. III.12.4 of Hartshorne, "Algebraic Geometry".

REPLY [5 votes]: No. First of all note that your line bundle $\mathcal{N}$ on $Y$ is trivial, so your assertion is $\mathcal{L}\cong\mathcal{M}$.
Take a smooth projective curve $C$ of genus $\geq 1$ (say over $\mathbb{C}$), with a closed point $p$. Consider the first projection $f:C\times C\rightarrow C$, and take for $\mathcal{L}$, resp. $\mathcal{M}$, the line bundle on $C\times C$ associated to the diagonal, resp. to $C\times \{p\} $. Then $\mathcal{L}$ and $\mathcal{M}$ have isomorphic restrictions to $f^{-1}(p)$, but not to the other fibers. Now take $A=\mathcal{O}_{C,p}$, and restrict to $Y\times C\rightarrow Y$.


text:
\newpage
\onecolumn
\appendix
\section{Proofs for Determining Stationary Points}

\subsection{Proof of Theorem~\ref{thm:stationary}}\label{pf:thm:stationary}
\begin{proof}
	 Remind that the eigendecomposition of $\tilde{A}$ is $(\Lambda^B)^{-\frac{1}{2}} O^{B\top}A O^B (\Lambda^B)^{-\frac{1}{2}}= O^{\tilde{A}} \Lambda^{\tilde{A}} (O^{\tilde{A}})^{\top}$. Given the eigendecomposition of $B$ is $B = O^{B} \Lambda^{B} (O^{B})^{\top}$, we can write $B^{-1}$ as
	\begin{align*}
		B^{-1} = O^B (\Lambda^B)^{-1} (O^B)^{\top}.
	\end{align*}

We denote $\tilde{X}$ as $\tilde{X} = O^{\tilde{A}}_{:,\cI}$ for some $\cI \subseteq [d]$ with $|\cI| = r$. For $X = (B^{-1/2} O^{\tilde{A}}_{:,\cI})\cdot \Psi ,$ where $\Psi\in \cG$. It is easy to see that $\nabla_Y \cL(X,Y)=0$. Ignore the constant 2 in the gradient $\nabla_X \cL(X,Y)$ for convenience, we have,
	\begin{align*}
		\nabla_X \cL(X,Y) &= -(I_d - BXX^\top) A X = -(I_d - B B^{-1/2} O^{\tilde{A}}_{:,\cI} (O^{\tilde{A}}_{:,\cI})^{\top} B^{-1/2} ) A B^{-1/2} O^{\tilde{A}}_{:,\cI} \\
		& = -A B^{-1/2} O^{\tilde{A}}_{:,\cI} + B^{1/2} O^{\tilde{A}}_{:,\cI} (O^{\tilde{A}}_{:,\cI})^{\top} O^{\tilde{A}} \Lambda^{\tilde{A}} (O^{\tilde{A}})^{\top} O^{\tilde{A}}_{:,\cI} \\
		& = -B^{1/2} O^{\tilde{A}} \Lambda^{\tilde{A}} (O^{\tilde{A}})^{\top} O^{\tilde{A}}_{:,\cI} + B^{1/2} O^{\tilde{A}}_{:,\cI} \Lambda^{\tilde{A}}_{\cI,\cI} \\
		&= -B^{1/2} O^{\tilde{A}} \Lambda^{\tilde{A}}_{:\cI} + B^{1/2}O^{\tilde{A}}_{:,\cI} \Lambda^{\tilde{A}}_{\cI,\cI} = 0.
	\end{align*}

	Next we show that if $X$ is not as specified, then $\nabla_X \cL(X,Y) \neq 0$. We only need to show that if $\tilde{X} = [O^{\tilde{A}}_{:,\cS}, ~\phi] \Psi$, where $\cS \subseteq [d]$ with $|\cS| = r-1$ and $\phi = c_1 O^{\tilde{A}}_{:,i} + c_2 O^{\tilde{A}}_{:,j}$ with $i,j \not\in \cS$, $i \neq j$, $c_1^2 + c_2^2 = 1$, and $c_1,c_2 \neq 0$, then we have $\nabla_X \cL(X,Y) \neq 0$. The general scenario can be induced from this basic setting. It is easy to see that such an $X = B^{-1/2} \tilde{X}$ satisfies the constraint,
	\begin{align*}
		X^{\top} B X = \Psi^{\top} [O^{\tilde{A}}_{:,\cS}, ~\phi]^{\top} B^{-1/2} B B^{-1/2}[O^{\tilde{A}}_{:,\cS}, ~\phi] \Psi = \Psi^{\top} \left[ \begin{array}{cc}
			I_{r-1} & 0_{(r-1) \times 1} \\
			0_{1 \times (r-1)} & \phi^{\top} \phi
		\end{array} \right] \Psi = I_r,
	\end{align*}
	where the last equality follow from $\phi^{\top} \phi = c_1^2 + c_2^2 = 1$.

	Plugging such an $X$ into the gradient, we have
	\begin{align*}
		\nabla_X \cL(X,Y) &= -(I_d - B X X^\top) AX = -(I_d - B B^{-1/2} [O^{\tilde{A}}_{:,\cS}, ~\phi] [O^{\tilde{A}}_{:,\cS}, ~\phi]^{\top} B^{-1/2}) A B^{-1/2} [O^{\tilde{A}}_{:,\cS}, ~\phi] \Psi \\
		&= -B^{1/2} (O^{\tilde{A}}_{:,\cS^{\perp}} (O^{\tilde{A}}_{:,\cS^{\perp}})^{\top} - \phi \phi^{\top}) O^{\tilde{A}} \Lambda^{\tilde{A}} [(I_d)_{\cS}, ~c_1 e_i + c_2 e_j] \Psi \\
		&= -B^{1/2} [0_{d \times (r-1)}, ~O^{\tilde{A}}_{:,\cS^{\perp}} \Lambda^{\tilde{A}}_{\cS^{\perp},:}(c_1 e_i + c_2 e_j)]\Psi + [0_{d \times (r-1)},~\phi (c_1^2 \lambda^{\tilde{A}}_{i}+c_2^2 \lambda^{\tilde{A}}_{j})]\Psi \\
		&= -B^{1/2} [0_{d \times (r-1)}, ~c_1 c_2^2(\lambda^{\tilde{A}}_{i} + \lambda^{\tilde{A}}_{j}) O^{\tilde{A}}_{:,i} + c_2 c_1^2(\lambda^{\tilde{A}}_{j} - \lambda^{\tilde{A}}_{i}) O^{\tilde{A}}_{j,j}] \Psi \neq 0,
	\end{align*}
	where the last $\neq$ is from $c_1,c_2 \neq 0$, $c_1^2+ c_2^2 = 1$, $\lambda^{\tilde{A}}_{j} \neq \lambda^{\tilde{A}}_{j}$ for $i \neq j$.
\end{proof}

\subsection{Proof of Theorem~\ref{lem:invertB_saddle}} \label{pf:lem:invertB_saddle}

\begin{proof}
	We have the Hessian of $\cL(X,Y)$ on $X$ with $Y=\cD(X)$ as
	\begin{align}
		H_X &= 2\sym\big( I_r \otimes (( BXX^\top - I_d ) A) + (X^\top A X) \otimes B + (AX) \boxtimes (BX) \big) \label{eqn:hessian}

	\end{align}
	where $\sym(M) = M + M^{\top}$, $\otimes$ is the Kronecker product, and for $U \in \RR^{d \times r}$ and $V \in \RR^{m \times k}$, $U \boxtimes V \in \RR^{dk \times mr}$ is defined as
	\begin{align*}
		U \boxtimes V = \left[ \begin{array}{cccc}
			U_{:,1} V_{:,1}^\top & U_{:,2} V_{:,1}^\top & \cdots & U_{:,r} V_{:,1}^\top \\
			U_{:,1} V_{:,2}^\top & U_{:,2} V_{:,2}^\top & \cdots & U_{:,r} V_{:,2}^\top \\
			\vdots & \vdots & \ddots & \vdots \\
			U_{:,1} V_{:,k}^\top & U_{:,2} V_{:,k}^\top & \cdots & U_{:,r} V_{:,k}^\top
		\end{array} \right].
	\end{align*}

	To determine whether a stationary point is an unstable stationary or a minimax global optimum, we consider its Hessian. We start with checking that $\cS = [r]$ corresponds to the global optimum, $X = B^{-1/2} O^{\tilde{A}}_{:,[r]} \Psi$. Without loss of generality, we set $\Psi = I_r$. We only need to check that for any vector $v = [v_1^{\top},\ldots,v_r^{\top}]^{\top} \in \RR^{nr}$ with $v_i \in \RR^n$ denoting the $i$-th block of $v$, which satisfies
	\begin{align*}
		v_i = c_{j_i} B^{-1/2} O^{\tilde{A}}_{:,{j_i}} \text{ for any $j_i \in [d]$ and a real constant $c_j$}
	\end{align*}
	such that $\norm{v}_2 = 1$, then we have $v^\top H_X v \geq 0$. The general case is only a linear combination of such $v$'s. Specifically, for $X = O^{\tilde{A}}_{:,[r]}$, we have
	\begin{align*}
		v^\top H_X v &= -v^\top \sym\left( I_r \otimes ((I_d - B XX^\top) A) - (X^\top A X) \otimes B - (AX) \boxtimes (BX) \right) v \\
		&= -v^\top \sym\Big( I_r \otimes ((I_d - B^{1/2} O^{\tilde{A}}_{:,[r]} O^{\tilde{A} \top}_{:,[r]} B^{-1/2}) A) - (O^{\tilde{A} \top}_{:,[r]} B^{-1/2} A B^{-1/2} O^{\tilde{A}}_{:,[r]} ) \otimes B\\
		&\hspace{2.0in} - (A B^{-1/2} O^{\tilde{A}}_{:,[r]}) \boxtimes (B^{1/2} O^{\tilde{A}}_{:,[r]}) \Big) v \\
		&= -v^\top \sym\Big( I_r \otimes ( B^{1/2} O^{\tilde{A}}_{:,[d]\backslash [r]} O^{\tilde{A} \top}_{:,[d]\backslash [r]} B^{-1/2} A ) - \Lambda^{\tilde{A}}_{:, [r]} \otimes B  - (B^{1/2} O^{\tilde{A}} \Lambda^{\tilde{A}}_{:, [r]}) \boxtimes (B^{1/2} O^{\tilde{A}}_{:,[r]}) \Big) v \\
		&= -2 \sum_{i=1}^r c_{j_i}^2 O^{\tilde{A} \top}_{:,j_i} O^{\tilde{A}}_{:,[d]\backslash [r]} \Lambda^{\tilde{A}}_{[d]\backslash [r],:} O^{\tilde{A} \top} O^{\tilde{A}}_{:,j_i} + 2 \sum_{i=1}^r c_{j_i}^2 \lambda^{\tilde{A}}_{i} + 2 \sum_{i=1}^{r} \sum_{k=1}^{r} c_{j_i} c_{j_k} e_{j_i}^{\top} \Lambda^{\tilde{A}}_{:, k} O^{\tilde{A} \top}_{:,i} O^{\tilde{A}}_{j_k} \\
		&\geq 0 + 2 \sum_{i=1}^r c_{j_i}^2 \lambda^{\tilde{A}}_{i} + 2 \sum_{i=1}^{r} \sum_{k=1}^{r} c_{j_i} c_{j_k} \lambda^{\tilde{A}}_{j_i} = 0,
	\end{align*}
	where the last inequality is obtained by taking $j_k \in [r]$, $i=j_k$, and $k=j_i$ in the last term, and the last equality is obtained by setting $c_{j_k} = -c_{j_i}$ when $j_i=k$, which implies that the restricted strongly convex property at $X$ holds.

	For any other $\cI \neq [r]$, we only need to show that the largest eigenvalue of $\nabla^2 \cL$ is positive and the smallest eigenvalue of $\nabla^2 \cL$ is negative, which implies that such a stationary point is unstable. Using the same construction as above, we have
	\begin{align*}
		&\lambda_{\min} (H_X) \leq -v^\top \sym \left( I_r \otimes ((I_d - B XX^\top) A) - (X^\top A X) \otimes B - (AX) \boxtimes (BX) \right) v\\
		&= -v^\top \sym \Big( I_r \otimes ( B^{1/2} O^{\tilde{A}}_{:,\cI^{\perp}} O^{\tilde{A} \top}_{:,\cI^{\perp}} B^{-1/2} A ) - \Lambda^{\tilde{A}}_{:, \cI} \otimes B - (B^{1/2} O^{\tilde{A}} \Lambda^{\tilde{A}}_{:, \cI}) \boxtimes (B^{1/2} O^{\tilde{A}}_{:,\cI}) \Big) v \\
		&= -2 \sum_{i \in \cI} c_{j_i}^2 O^{\tilde{A} \top}_{:,j_i} O^{\tilde{A}}_{:,\cI^{\perp}} \Lambda^{\tilde{A}}_{\cI^{\perp},:} O^{\tilde{A} \top} O^{\tilde{A}}_{:,j_i} + 2 \sum_{i \in \cI} c_{j_i}^2 \lambda^{\tilde{A}}_{i} + 2 \sum_{i \in \cI} \sum_{k \in \cI} c_{j_i} c_{j_k} e_{j_i}^{\top} \Lambda^{\tilde{A}}_{:, k} O^{\tilde{A} \top}_{:,i} O^{\tilde{A}}_{j_k} \\
		&\overset{(i)}{=}  2 c_{j_r}^2 ( \lambda^{\tilde{A}}_{\max \cI} - \lambda^{\tilde{A}}_{\min \cI^{\perp}}),
	\end{align*}
	where $(i)$ is from setting $c_{j_i}=0$ for all $j_i \in \cI^{\perp}$ except $j_r$, and $c_{j_r}= 1/\| B^{-1/2} O^{\tilde{A}}_{:,\min \cI^{\perp}}\|_2$.

	On the other hand, we have
	\begin{align*}
		&\lambda_{\max} (H_X) \geq v^\top H_X v \\
		&= -2 \sum_{i \in \cI} c_{j_i}^2 O^{\tilde{A} \top}_{:,j_i} O^{\tilde{A}}_{:,\cI^{\perp}} \Lambda^{\tilde{A}}_{\cI^{\perp},:} O^{\tilde{A} \top} O^{\tilde{A}}_{:,j_i} + 2 \sum_{i \in \cI} c_{j_i}^2 \lambda^{\tilde{A}}_{i} + 2 \sum_{i \in \cI} \sum_{k \in \cI} c_{j_i} c_{j_k} e_{j_i}^{\top} \Lambda^{\tilde{A}}_{:, k} O^{\tilde{A} \top}_{:,i} O^{\tilde{A}}_{j_k} \\
		&\overset{(i)}{=} 2 c_{j_1}^2\lambda^{\tilde{A}}_{\min \cI} + c_{j_1}^2\lambda^{\tilde{A}}_{\min \cI} = 4 c_{j_1}^2 \lambda^{\tilde{A}}_{\min \cI},
	\end{align*}
	where $(i)$ is from setting $c_{j_i}=0$ for all $j_i \in \cI$ except $j_1$, and $c_{j_1} = 1/\| B^{-1/2} O^{\tilde{A}}_{:,\min \cI}\|_2$.

\end{proof}


TITLE: Partial orders - minimal elements
QUESTION [0 upvotes]: The question I have problem with is
a) How many partial orders for the set {x,y} has x as a minimal element
b) How many partial orders for the set {x,y,z} has x as a minimal element
I don't really get how I should think. I believe I understand how it works when you use a diagram, but now I can't get a grip on how the diagram would look for a problem like this. Or can't you use that way of thinking in this problem?

REPLY [1 votes]: a.  There are three orders for {x,y}, x < y, y < x and
{x,y} is an antichain (x and y cannot be compared, x||y).
Which of those orders have x as a minimal order?
b.  What are the possible orders for {x,y,z}?
If you must have diagrams, then draw a diagram for each order.