TITLE: How To Simplify Logic Equation
QUESTION [0 upvotes]: Consider the logic equation below:
$$(\lnot A \land B \land \lnot C) \lor (A \land \lnot B \land C)$$
It is already in Disjunctive Normal Form. I'm wondering if there is any way to simplify the equation further. Since $A,B,C$ are all the negations of themselves in each of the individual brackets I don't think there are any simpler ways of writing this, but I'm not sure.
If there is a more simplified version, can someone please let me know and how to reach such a simplified form.
Thanks in advance!

REPLY [2 votes]: It's already in its simplest form.
Generally if you write it the "arithmetic way" i.e. $$A'BC'+AB'C$$ you see that you cannot really find something nicer since you cannot factor anything.
While if was written as product of sums $$(A+B+C)(A+B+C')(A+B'+C')(A'+B+C)(A'+B'+C)(A'+B'+C')$$
you might think with reason that expanding it would lead to simplifications.
Note: you can play with the software Logic Friday to evaluate these kind of expressions.


text:
\begin{document}

\note[Semi-symmetric spacetimes]{Investigation of all Ricci
semi-symmetric and all conformally semi-symmetric spacetimes}

\author{Jan E. {\AA}man}

\address{Department of Physics, Stockholm University, SE-106 91 \ Stockholm,
Sweden}
\eads{\mailto{ja@fysik.su.se}}

\begin{abstract}

We find all Ricci semi-symmetric as well as all conformally
semi-symmetric spacetimes. Neither of these properties implies the
other. We verify that only conformally flat spacetimes can be Ricci
semi-symmetric without being conformally semi-symmetric and show that
only vacuum spacetimes and spacetimes with just a $\Lambda$-term can
be Ricci semi-symmetric without being conformally semi-symmetric.

\end{abstract}

\pacs{04.20.Cv, 02.40.Ky, 04.20.Jb}

\section{Introduction}

Semi-symmetric spaces where introduced by Cartan \cite{Cartan1946} and
are characterized by the curvature condition
\begin{equation} \label{SemiSymmetric}
    \nabla_{[a}\nabla_{b]}R_{cdef} = 0,
\end{equation}
where $R_{cdef}$ denotes the Riemann tensor and round and square
brackets enclosing indices indicate symmetrization and
antisymmetrization, respectively.

A semi-Riemannian manifold is said to be {\em conformally
semi-symmetric} if the Weyl tensor $C_{abcd}$ satisfies
\begin{equation} \label{ConformallySemiSymmetric}
    \nabla_{[a}\nabla_{b]}C_{cdef} = 0;
\end{equation}
and {\em Ricci semi-symmetric} if the Ricci tensor satisfies
\begin{equation} \label{RicciSemiSymmetric}
    \nabla_{[a}\nabla_{b]}R_{cd} = 0.
\end{equation}

The geometrical properties of semi-symmetric spaces where discussed
in \cite{Sz,HV,Senovilla2006,Senovilla2008,DDVV}.

In this paper we find all semi-symmetric, Ricci semi-symmetric and all
conformally semi-symmetric spacetimes. Eriksson and Senovilla
\cite{Eriksson2010} found all such non-conformally flat spacetimes.

It is an advantage to instead of tensors rather to use the Weyl spinor
$\Psi_{ABCD}$, the curvature scalar spinor $\Lambda$, and the spinor
$\Phi_{ABA'B'}$ for the tracefree part of the Ricci tensor.  We also
use the spinor ${\rm X}_{ABCD} = \Psi_{ABCD} +\Lambda \left(
\varepsilon_{AC}\varepsilon_{BD} +\varepsilon_{AD}\varepsilon_{BC}
\right)$. The spinor commutator $\Box_{AB}$ operating on spinors with
one index is $\Box_{AB}\, \kappa_C = -{\rm X}_{ABC}{}^E\kappa_E$ and
$\Box_{AB}\, \tau_{C'} = -\Phi_{ABC'}{}^{E'}\tau_{E'}$ \cite{Penrose1984}.

In spinors (\ref{ConformallySemiSymmetric}) is equivalent to
$\Box_{AB} \Psi_{CDEF} = 0$ and $\Box_{A'B'} \Psi_{CDEF} = 0$, or
\begin{eqnarray}
     {\rm X}_{AB(C}{}^G \Psi_{DEF)G} &= 0, \label{WeylCondition1} \\
      \Phi_{A'B'(C}{}^G \Psi_{DEF)G} &= 0. \label{WeylCondition2}
\end{eqnarray}

\section{Conditions for semi-symmetry}

Calculation of the components of (\ref{WeylCondition1}) in terms of
$\Psi_{ABCD}$ and $\Lambda$ shows show that it not will have 15
independent components but rather just 5 components and
(\ref{WeylCondition1}) can be replaced by its contraction over $BC$ or
\begin{equation}
\Psi^{GH}{}_{(AD}\Psi_{EF)GH} -2\Lambda \Psi_{ADEF} = 0.
\label{PsiCondition1}
\end{equation}

We observe that the spinor commutator $\Box_{AB}$ yields 0 if
operating on a scalar as the curvature scalar $\Lambda$, therefore
only its effect on $\Phi_{ABA'B'}$, i.e. on the tracefree part of the
Ricci tensor has to be considered.  The second condition for Ricci
semi-symmetry (\ref{WeylCondition2}) corresponds to spinor equations
\begin{eqnarray}
   \Box_{AB} \Phi_{CDC'D'} &=
        -2 {\rm X}_{AB(C}{}^E \Phi_{D)EC'D'}
        -2 \Phi_{AB(C'}{}^{E'} \Phi_{D')E'CD}, \label{RicciCondition1} \\
   \Box_{A'B'} \Phi_{CDC'D'} &=
        -2 \overline{\rm X}_{A'B'(C'}{}^{E'} \Phi_{D')E'CD}
        -2 \Phi_{A'B'(C}{}^E \Phi_{D)EC'D'}. \label{RicciCondition2}
\end{eqnarray}
The condition (\ref{RicciCondition2}) is however just the complex
conjugate of (\ref{RicciCondition1}) and will not give any additional
conditions for Ricci semi-symmetry.

Equation (\ref{RicciCondition1}) has three groups of two symmetric
indices but will in fact not have 27 independent components as
$\Box^{AB} \Phi_{ABC'D'} = 0$, it can be replaced by the fully
symmetric spinors
\begin{eqnarray}
 \Box_{(AB} \Phi_{CD)C'D'} = - 2 \Psi_{(ABC}{}^E\Phi_{D)EC'D'}
  \label{PhiCondition1} \\
 \Box_{(A}{}^F \Phi_{C)FC'D'} = 4 \Lambda\Phi_{ACC'D'}
   -\Psi^{EF}{}_{AC}\Phi_{EFC'D'}
   -2 \Phi^E{}_A{}^{F'}{}_{(C'}\Phi_{D')F'CE} \label{PhiCondition2}
\end{eqnarray}
with 15 and 9 components respectively.

The various spacetimes are for simplicity studied in a frame where
first $\Psi_{ABCD}$ has been brought to a standard form depending on
its Petrov type. Thereafter $\Phi_{ABA'B'}$ is brought to standard
form depending on its Segre type \cite{Quartic,Exact}.

All of the above formulas have been implemented in CLASSI
\cite{Classi}.

\section{Main Results}

The equations for Ricci semi-symmetric spacetimes
(\ref{PhiCondition1}) and (\ref{PhiCondition2}) and Petrov Types
\textbf{I}, \textbf{II} or \textbf{III} in standard frame requires
that $\Phi_{ABA'B'}=0$, while spacetimes with a $\Lambda$-term only or
vacuum ($\Lambda=0$) are Ricci semi-symmetric.

For Petrov type \textbf{D} spacetimes (with only $\Psi_2$ nonzero) the
conditions both for conformal and for Ricci semi-symmetry require all
components of $\Phi_{ABA'B'}$ to be zero except for $\Phi_{11'}$. The
conditions for conformal semi-symmetry reduces to $2 \Lambda +\Psi_2 =
0$ and for Ricci semi-symmetry to $\Phi_{11'} (2 \Lambda +\Psi_2) =
0$. Spacetimes of Segre type A1[(11)(1,1)] are therefore both Ricci
and conformally semi-symmetric if they conform to
$\Lambda=-\frac{1}{2}\Psi_2$ with arbitrary $\Phi_{11'}$.  Spacetimes
with $\Lambda$-term only are semi-symmetric if once again
$\Lambda=-\frac{1}{2}\Psi_2$ but only Ricci semi-symmetric for other
relations between $\Psi_2$ and $\Lambda$.

For Petrov type \textbf{N} spacetimes (with only $\Psi_4$ nonzero) the
conditions for Ricci semi-symmetry as well as for conformal
sem-symmetry require all components of $\Phi_{ABA'B'}$ to be zero
except for $\Phi_{22'}$. The conditions here reduces to $\Lambda
\Psi_4 = 0$ for conformal semi-symmetry and $\Lambda \Phi_{11'} = 0$
for Ricci semi-symmetry. They are therefore both Ricci and conformally
semi-symmetric for arbitrary $\Psi_4$ and $\Phi_{22'}$ as long as
$\Lambda=0$. Once again all spacetimes with $\Lambda$-term only are
Ricci semi-symmetric.

\subsection{Conformally flat spacetimes}

This is the case not treated by Eriksson and Senovilla \cite{Eriksson2010}.

All conformally flat (Petrov type \textbf{0}) spacetimes are
conformally semi-symmetric.

Calculations with CLASSI shows that all conformally flat spacetimes of
Segre type A1[(11)(1,1)] (only $\Phi_{11'}$ nonzero) with $\Lambda=0$
are also Ricci semi-symmetric.

For conformally flat perfect fluids (Segre type A1[(111),1)],
$\frac{1}{2}\Phi_{00'}=\Phi_{11'}=\frac{1}{2}\Phi_{22'}$) as well as for
tachyons (Segre type A1[1(11,1)],
$-\frac{1}{2}\Phi_{00'}=\Phi_{11'}=-\frac{1}{2}\Phi_{22'}$) the
condition for Ricci semi-symmetry reduces to $\Lambda = \Phi_{11'}$.
So perfect fluids with
$\Lambda=\frac{1}{2}\Phi_{00'}=\Phi_{11'}=\frac{1}{2}\Phi_{22'}$ as
well as tachyons with
$\Lambda=-\frac{1}{2}\Phi_{00'}=\Phi_{11'}=-\frac{1}{2}\Phi_{22'}$ are
also Ricci semi-symmetric.

All spacetimes of Segre type A3[(11,2)] with $\Lambda=0$ (only
$\Phi_{22'}$ nonzero) are also Ricci semi-symmetric.

All conformally flat spacetimes with $\Lambda$-term only, including
flat spacetimes, are Ricci semi-symmetric.

For all other Segre types the conditions for Ricci semi-symmetry are
not satisfied.

\section{Summary}

All conformally flat spacetimes are conformally semi-symmetric, all
spacetimes with $\Lambda$-term only are Ricci semi-symmetric.

Semi-symmetric spacetimes are flat spacetimes, $\Lambda$-term and
Segre A1[(11)(1,1)] spacetimes of Petrov type \textbf{0} and of type
\textbf{D} with $\Lambda=-\frac{1}{2}\Psi_2$, Segre type A3[(11,2)]
with $\Lambda=0$ of Petrov types \textbf{N} and \textbf{0}, as well as
conformally flat perfect fluids and tachyons with
$\Lambda=\Phi_{11'}$. For a table see appendix A.

\section*{Acknowledgments}

I thank Brian Edgar (1945-2010) for pointing out ref
\cite{Eriksson2010} and introducing me to semi-symmetric spacetimes. I
thank Ingemar Bengtsson and Jos\'e Senovilla for comments on the
manuscript.

\newpage

\section*{Appendix A}
\begin{table}[h]

\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
Segre type \ \ \ \ \ \ \ \ \ Petrov type & \textbf{I} & \textbf{II} &
  \textbf{III} & \textbf{D} & \textbf{N} & \textbf{0} \\
\hline
$\Lambda$-term A1[(111,1)] or vacuum & Ric s-s & Ric s-s & Ric s-s &
  Ric s-s & Ric s-s & semi-sym \\
$\Lambda$-term, $\Lambda=-\frac{1}{2}\Psi_2$ & Ric s-s & Ric s-s &
  $\nexists$ & semi-sym & $\nexists$ & $\nexists$ \\
A1[(11)(1,1)], $\Lambda=-\frac{1}{2}\Psi_2$ & - & - & $\nexists$ & semi-sym &
  $\nexists$ & $\nexists$ \\
A1[(11)(1,1)] & - & - & - & see above & - & semi-sym \\
A3[(11,2)], $\Lambda=0$ & - & - & - & - & semi-sym & semi-sym \\
A1[(111),1] perfect fluid, &  &  &  &  &  & \\
  \ \ $\Lambda=\frac{1}{2}\Phi_{00'}=\Phi_{11'}=\frac{1}{2}\Phi_{22'}$
  & - & - & - & - & - & semi-sym \\
A1[1(11,1)] tachyon fluid, &  &  &  &  &  & \\
  \ \ $\Lambda=-\frac{1}{2}\Phi_{00'}=\Phi_{11'}=-\frac{1}{2}\Phi_{22'}$ &
  - & - & - & - & - & semi-sym \\
All other Ricci tensors & - & - & - & - & - & conf s-s \\
\hline
\end{tabular}

\caption{Relations between Petrov type, Segre type and conformal
semi-symmetry (conf s-s), Ricci semi-symmetry (Ric s-s) and
semi-symmetry. A hyphen (-) indicates neither conformal nor Ricci
semi-symmetry.}

\end{table}

\section*{References}

config: arxiv


TITLE: Different interpretations of the power in a resistor law?
QUESTION [1 upvotes]: On this KhanAcademy article, under the Power in a resistor heading, it states three equations, all of which can be derived from $V=iR$ and $P=iV.$
They are:

$P=Vi$
$P=i^2R$
$P=\frac{V^2}{R}$

Then if you click on [How much has the power dissipated by the resistor been changed? ] at the bottom of the same section, it reveals this puzzling tract:

Power in a resistor is proportional to $v^2,$ so lowering voltage by a factor of $2$ cuts the power by a factor of $4.$
Power in a resistor is also porportional* to $i^2,$ so lowering the current by a factor of two also cuts the power by a factor of $4.$
Overall, Aaron and Beth have reduced the power by a factor of $4\times4=16.$

*they meant proportional
However, if you these changes through equation $1$ in our list above, you find that power has only decreased by a factor of $4.$
Is KhanAcademy right? And if so, why are we taking both equation $2$ and $3$ into account at the same time? Why not take into account all three equations and make power decreased by a factor of $2^3\times2^3=64?$ (One $2^3$ for $i$ and one for $V$.) Is this just another time in science where we discover something and randomly mash equations together to try to model something, and if it works we just don't question it, or am I missing something really big?

REPLY [1 votes]: This is my understanding:
You already have $V = IR$, and take note that $R$ is constant because you are not adding any more resistor.
When you lower the voltage by a factor of 2, you could lower the $I$ and keep the $R$ constant, or you can lower $R$ and keep $I$ constant. The latter is not possible, so the only way for us to decrease the voltage is to also decrease the current.
If we use eqn 1, it relates $P$ to both $I$ and $V$ but we only know that V is decreased, so we also have to take into account the decrease in current when the voltage is decreased:
since $V = IR$;  $\frac{1}{2} V = (\frac{1}{2} I)R$; which means that current is also decreased by 1/2.
So using eqn 1:$P = IV$, so $P_1 = \frac{1}{2} I\frac{1}{2} V = \frac{1}{4} IV = \frac{1}{4}P$ thus the power is decreased by a factor of 4
Using eqn 2:$P = I^2R$ so $P_1 = (\frac{1}{2}I)^2R = \frac{1}{4}I^2R = \frac{1}{4}P$ thus the power is also decreased by a factor of 4
Using eqn 3:$P = V^2/R$ so $P_1 = (\frac{1}{2}V)^2/R$ = $\frac{1}{4}V^2/R = \frac{1}{4}P$ thus the power is also decreased by a factor of 4
So by reducing V by 1/2, we are reducing $P$ by a factor of 4, using either 3 equations.
Then, in step two, we reduce the current by 1/2. Again, we know that by reducing the current, we also reduce the voltage, while keeping $R$ constant: $V = IR$ so $\frac{1}{2}V = (\frac{1}{2}I)R$.
since $V = IR$;  $\frac{1}{2} V = (\frac{1}{2} I)R$; which means that current is also decreased by 1/2.
So using eqn 1:$P_1 = I_1V_1$, so $P_2 = \frac{1}{2} I_1\frac{1}{2} V_1 = \frac{1}{4} I_1V_1 = \frac{1}{4} P_1 = \frac{1}{4}\frac{1}{4}P = \frac{1}{16}P$ thus the total power is decreased by a factor of 16
Using eqn 2:$P_1 = I_1^2R$ so $P_2 = (\frac{1}{2}I_1)^2R = \frac{1}{4}I_1^2R = \frac{1}{4} P_1 = \frac{1}{4}\frac{1}{4}P = \frac{1}{16}P$ thus the total power is also decreased by a factor of 16
Using eqn 3:$P_1 = V_1^2/R$ so $P_2 = (\frac{1}{2}V_1)^2/R$ = $\frac{1}{4}V_1^2/R = \frac{1}{4} P_1 = \frac{1}{4}\frac{1}{4}P = \frac{1}{16}P$ thus the total power is also decreased by a factor of 16
So by reducing $I$ in step two by 1/2, we are reducing the power by a factor of 4 using either 3 equations, and the initial power is decreased by a factor of $4\times4 = 16$.
In the article, they used eqn 3 for step 1 because by using it, you don't need to take account of the decease in current involved, but only the voltage. Also for step 2, they used eqn 2 so that you don't need to take into account the reduction of voltage due to the reduction of current.


TITLE: Is pendulum only chaotic when its force is a single sinusoidal frequency?
QUESTION [1 upvotes]: Chaotic systems will be chaotic under certain parameters, but when the forced, damped pendulum is discussed, the initial conditions of its position and velocity are only mentioned. Is it assumed that the force will always be a single sinusoidal frequency?
I did a quick Matlab simulation of the pendulum's EOMs where the input was a superposition of sine waves at different frequencies and the Poincaré plot yielded random behavior (no indication of periodic points or a pattern, just points scattered). I then created individual Poincaré plots for each of the frequencies used in the superposition; each outputted random behavior. Is this a reason why the force needs to be a single sinusoidal frequency?

REPLY [3 votes]: We always try to find the simplest model possible that manages to reflect the relevant characteristics of the real system. So whenever a sinusoidal forcing will do, it makes sense to adopt it.
Even if it looks random, the behavior of a system with complicated forcing is still deterministic by construction — but a messy response from a system subjected to a messy forcing is not surprising, much more interesting is how chaotic behavior emerges in simple systems.


TITLE: Expansions of iterated, or nested, derivatives, or vectors--conjectured matrix computation
QUESTION [3 upvotes]: The entry OEIS A139605 (also related OEIS A145271) has a matrix computation for the partition polynomials that represent the expansions of iterated derivatives, or vectors in differential geometry,
$$(g(x)D_x)^n.$$
The formula section of A139605 contains the matrix formula. Multiply the $n$-th diagonal (with $n=0$  the main diagonal) of the lower triangular Pascal matrix A007318 by $g_n = D_x^n g(x)$  to obtain the matrix $VP$  with $VP_{n,k} = \binom{n}{k}g_{n-k} $. Then
$$(g(x)D)^n = (1, 0, 0,..) [VP \dot \; S]^n (1, D, D^2, ..)^T,$$
where S is the shift matrix A129185, representing differentiation in the divided powers basis $x^n/n!$.
Example:
$$(g(x)D_x)^3$$
$$= (1, 0, 0, 0) [VP \dot \; S]^3 (1, D, D^2, D^3)^T$$
$$= \begin{pmatrix}
1 &  0 &  0 & 0
\end{pmatrix} \begin{pmatrix}
0 & g_0 & 0 & 0 \\
0 & g_1 & g_0 & 0\\
0 & g_2 & 2g_1 & g_0 \\
0 & g_3 & 3g_2 & 3g_1
\end{pmatrix}^3 \begin{pmatrix}
1 \\
D \\
D^2 \\
D^3 \end{pmatrix} $$
$$ = [g_0g_1^2  + g_0^2 g_2] D + 3 g_0^2g_1 D^2 + g_0^3D^3 $$
And, the pdf Mathemagical Forests gives a diagrammatic method for creating forests of trees through "natural growth" that represent the partition polynomials.
I have either lost a proof of the validity of this formula or got sidetracked before I developed one.
Question:
Can someone prove this conjecture?
Some background:
The refined Eulerian numbers (RENs) of A145271 are related analytically to the compositional inversion of functions and formal generating series and to flow fields generated by tangent vectors. The $n$-th row of RENs are the numerical coefficients of the expansion of $(g(x)\frac{d}{dx})^ng(x)$ in terms of the monomials in the derivatives of $g(x)$, i.e.,
$$g_k=\frac{d^k}{dx^k}g(x).$$
For example,
$$(g(x)\frac{d}{dx})^3g(x) = 1 g_0^1 g_1^3 + 4 g_0^2 g_1^1 g_2^1 + 1 g_0^3 g_3^1.$$
With $(\omega,x) = (f(x),f^{(-1)}(\omega))$ and $g(x) = 1/f^{'}(x)$,
$$\exp[t g(x)d/dx]x = \exp[td/d\omega]f^{(-1)}(\omega) = f^{(-1)}(t+\omega)=f^{(-1)}(t+f(x)).$$
Evaluated at the origin of $x$, this gives the compositional inverse
$$\exp[tg(x)d/dx] x |_{x=0}=f^{(-1)}(t).$$
See also
1) MO-Q Guises of the refined Eulerian numbers generated by tangent vectors
2) MO-Q Important formulas in combinatorics
3) MO-Q Why is there a connection between enumerative geometry and nonlinear waves?

REPLY [0 votes]: Here is probably something close to my original thinking on the identity.
Taylor series analysis gives us, with $D_x^k g(x) = \frac{d^k}{dx^k} g(x)= g_k$,
$$g(t+x) = e^{tD_x}g(x)= \sum_{k \geq
 0} g_k\frac{t^k}{k!}.$$
Then
$$g(x+t)g'(x+t)= g(x+t)D_xe^{tD_x}g(x)$$
$$= \sum_{k \geq
 0} g_k\frac{t^k}{k!}D_x\sum_{k \geq
 0} g_k\frac{t^k}{k!}$$
can be written as a binomial convolution (umbrally, $e^{a.t}e^{b.t}=e^{(a.+b.)t}$) in (singly infinite) matrix form in the divided powers basis  $\frac{t^k}{k!}$ as
$$ \begin{pmatrix}
g_0 & 0 & 0 & 0 \\
g_1 & g_0 & 0 & 0\\
g_2 & 2g_1 & g_0 & 0\\
g_3 & 3g_2 & 3g_1 & g_0
\end{pmatrix}
\begin{pmatrix}
g_1\\
g_2 \\
g_3 \\
g_4 \end{pmatrix} $$
$$ = \begin{pmatrix}
g_0 & 0 & 0 & 0 \\
g_1 & g_0 & 0 & 0\\
g_2 & 2g_1 & g_0 & 0\\
g_3 & 3g_2 & 3g_1 & g_0
\end{pmatrix} \begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
1 \\
D \\
D^2 \\
D^3 \end{pmatrix} g(x)$$
$$= \begin{pmatrix}
0 & g_0 & 0 & 0 \\
0 & g_1 & g_0 & 0\\
0 & g_2 & 2g_1 & g_0 \\
0 & g_3 & 3g_2 & 3g_1
\end{pmatrix} \begin{pmatrix}
1 \\
D \\
D^2 \\
D^3 \end{pmatrix} g(x)$$
Repeating operation $n$  times by $g(x+t)D_x$ and ultimately setting $t=0$, or, equivalently, extracting the first term in the resulting column vector of operators acting on $g(x)$, gives us
$$[g(x)D_x]^n .$$
The full repeated (infinite) matrix operator that acts on $g(x)$ is a rep in the basis $\frac{t^k}{k!}$ of
$$[g(x+t)D_x]^n e^{tD_x}.$$
Letting
$\omega = f(x)$, $x = f^{-1}(\omega)$, and $g(x) = 1/f'(x)$, for $h(x)$ an arbitrary analytic function,
$$  \exp[ug(x+t)D_x]e^{tD_x}h(x)= \exp[u\frac{d}{d(\omega(x+t))}]h(x+t)$$
$$= \exp[u\frac{d}{d(\omega(x+t))}]h[f^{-1}[\omega(x+t)]]= h[f^{-1}[\omega(x+t)+u]] $$
$$ = h[f^{-1}[f(x+t)+u]]  = e^{tD_x} h[f^{-1}[f(x)+u]] $$
$$ = e^{tD_x} \exp[ug(x)D_x] h(x),$$
so (as we might surmise initially), equating coefficients of $u^n$,
$$ [g(x+t)D_x]^n e^{tD_x} = e^{tD_x} [g(x)D_x]^n,$$
implying the $m-$th component of the operator matrix rep, the operator coefficient of $\frac{t^m}{m!}$ is
$$ D_x^m [g(x)D_x]^n,$$
in agreement with Grinberg's general result.
More simply, the proof follows through a double translation--a similarity transformation;
$$  e^{-tD_x}  [g(x+t)D_x]^n e^{tD_x} h(x) =[g(x)D_x]^n h(x)$$
since $D_x = \frac{d}{dx}= \frac{d}{d(x+t)} = D_{x+t}$.


TITLE: Probability theory problem
QUESTION [0 upvotes]: A bag contains $5$ white and $5$ black balls. A draws $5$ balls retain any that are white and returns any black ones to the bag. B then draws $5$ balls, retains any that are white and returns any black ones to the bag. Now there are $6$ balls in the bag. Find the probability A drew exactly 2 white balls.

REPLY [0 votes]: OK, I answer the question assuming it said 'A drew exactly 5 black balls'. In this case before B samples from the basket, it hasthe same same composition as before A did his sampling.
Then the question reduces to 'what is the probability B samples 4 white and 1 black balls, since this is the only way it is possible to observe 6 balls after 1 sampling.
This is called Negative Binomial distribution, and the probability of this event is
$$
P(S) = \frac{\binom{5}{4} \cdot \binom{5}{1}}{\binom{10}{5}}
$$


TITLE: Given any two $n \times n$ matrices $A$ and $B$ can we always find $c \neq 0$ and $Y \neq 0$ such that $AY = cBY$ is true?
QUESTION [0 upvotes]: Given any two $n \times n$ matrices $A$ and $B$ can we always find a scalar $c \neq 0$ and $n \times 1$ vector $Y \neq 0$ such that $AY = cBY$ is true ?

REPLY [1 votes]: $AY=cBY$ if and only if $(A-cB)Y=0$. So, you are essentially asking whether there always exists a scalar $c$ such that $A-cB$ is singular.
When $B$ is nonsingular and the underlying field is algebraically closed, the answer is yes, because $\det(A-cB)=\det(AB^{-1}-cI)\det(B)$ is zero whenever $c$ is an eigenvalue of $AB^{-1}$.
However, the answer to your question is no in general. In particular, when $B=0$ and $A$ is nonsingular, the $c$ and $Y$ you require obviously do not exist.


text:
\begin{document}
\begin{titlepage}
\title{\bf The global well-posedness of strong solutions to 2D MHD equations in Lei-Lin space}
\author{ Baoquan Yuan\thanks{Corresponding Author: B. Yuan}\ and Yamin Xiao
       \\ School of Mathematics and Information Science,
       \\ Henan Polytechnic University,  Henan,  454000,  China.\\

        (bqyuan@hpu.edu.cn, ymxiao106@163.com)
          }

\date{}
\end{titlepage}
\maketitle
\begin{abstract}
In this paper, we study the Cauchy problem of the 2D incompressible magnetohydrodynamic equations in Lei-Lin space. The global well-posedness of a strong solution in the Lei-Lin space $\chi^{-1}(\mr^2)$ with any initial data in $\chi^{-1}(\mr^2)\cap L^2(\mr^2)$ is established. Furthermore, the uniqueness of the strong solution in $\chi^{-1}(\mr^2)$ and the Leray-Hopf weak solution in $L^2(\mr^2)$ is proved.
\end{abstract}

\vspace{.2in} {\bf Key words:} 2D MHD equations, strong solutions, Lei-Lin space, weak-strong uniqueness.

\vspace{.2in} {\bf MSC(2010):} 76W05, 35D35, 74H25.

\section{Introduction}
\setcounter{equation}{0}
\vskip .1in
In this paper, we consider the following incompressible magnetohydrodynamic equations
\be\label{1.1}
\begin{cases}
{ \begin{array}{ll}
\partial_{t}u-\mu\Delta u+(u\cdot\nabla) u+\nabla p=(b\cdot\nabla) b, \\
\partial_{t}b-\nu\Delta u+(u\cdot\nabla) b=(b\cdot\nabla) u, \\
\nabla\cdot u=\nabla\cdot b=0, \\
u(0,x)=u_{0}(x), \ b(0,x)=b_{0}(x).
 \end{array} }
\end{cases}
\ee
for $t\geq0$, $x\in\mr^2$. We denote $u=u(t,x)$, $b=b(t,x)$ and $p=p(t,x)$ the velocity field,
magnetic field and scalar pressure of the fluid, respectively. The positive constants $\mu$ and $\nu$ are the viscosity and the resistivity coefficients, and $u_0(x)$ and $b_0(x)$ are the initial velocity and initial magnetic field satisfying $\mbox{div}u_0=\mbox{div}b_0=0$, respectively.

When $b=0$, it reduces to the classical Navier-Stokes equation, which has been investigated extensively with many interesting results. Leray \cite{L} and Hopf \cite{H} established the global existence of weak solutions. Kato \cite{K}, Fujita and Kato \cite{F-K} obtained the local well-posedness for any initial data and the global well-posedness for small initial data in $L^n(\mr^n)$ and $\dot{H}^{s}(\mr^n)$ with $s\geq\frac{n}{2}-1$, respectively. Recently, Lei and Lin \cite{L-L} constructed the global mild solution with small initial data in the critical space $\chi^{-1}(\mr^3)$, and Zhang and Yin \cite{Z-Y} studied the local well-posedness with large initial data and the  global well-posedness with small initial data by the semi-group method. \\
{\bf Theorem A} (\cite{L-L} and \cite{Z-Y}) Let $u_0$ be in $\chi^{-1}(\mr^3)$. There exists a positive time $T$ such that the Navier-Stokes equation has a unique solution $u$ in $L^2([0,T];\chi^0(\mr^3))$ which also belongs to
$$\mathcal{C}([0,T];\chi^{-1}(\mr^3))\cap L^1([0,T];\chi^1(\mr^3))\cap L^\infty([0,T];\chi^{-1}(\mr^3)).$$
Let $T_{u_0}$ denote the maximal time of existence of such a solution. Then:\\
-There exists a constant $C$ such that, if $\|u_0\|_{\chi^{-1}}\leq C\mu$, then $$T_{u_0}=\infty.$$
-If $T_{u_0}$ is finite, then $$\int_0^{T_{u_0}}\|u\|_{\chi^{0}}^2\md t=\infty.$$

There are several important global well-posedness and decay results for the 3D Navier-Stokes equation and MHD equations (\ref{1.1}) in Lei-Lin space $\chi^{-1}(\mr^3)$ (see, e.g., \cite{B1,B2,B-B,W,W-W,X-Y-Z,Y}). In particular, Ye and Zhao \cite{Y-Z} proved the global well-posedness for the $n$ dimensional generalized MHD equations with small initial data. As we know that, in 2D case, the global Leray-Hopf weak solution to the MHD equations (\ref{1.1}) is a unique one in $L^2(\mr^2)$ space. We wonder if there is a global strong solution to the 2D MHD equations (\ref{1.1}) in the Lei-Lin space $\chi^{-1}(\mr^2)$ without the smallness condition, and if it is a unique solution with the Leray-Hopf weak solution with a same initial data in the space $L^2(\mr^2)\cap\chi^{-1}(\mr^2)$.

In this paper, we will prove the global well-posedness of a strong solution to (\ref{1.1}) in $\chi^{-1}(\mr^2)\cap L^2(\mr^2)$ for large initial data. Our main results are presented as follows.
\begin{thm}\label{thm1.1}
Let $(u_0,b_0)$ be in $\chi^{-1}(\mr^2)\cap L^2(\mr^2)$. A unique solution $(u,b)$ then exists in the space $L^2(\mr^+;\chi^0(\mr^2))$ which also belongs to
$$\mathcal{C}(\mr^+;\chi^{-1}(\mr^2))\cap L^1(\mr^+;\chi^1(\mr^2))\cap L^\infty(\mr^+;L^2(\mr^2))\cap L^2(\mr^+;\dot{H}^1(\mr^2)),$$
and satisfies the energy equality
\begin{align}\label{1.2}
\|(u,b)\|_{L^2}^2+2\mu\int_0^t\|\nabla u\|_{L^2}^2\md \tau+2\nu\int_0^t\|\nabla b\|_{L^2}^2\md \tau=\|(u_0,b_0)\|_{L^2}^2.
\end{align}
$(u,b)$ also satisfies the global a priori estimate
\begin{align}\label{1.3}
\|(u,b)\|_{\widetilde{L}^\infty(\chi^{-1})}+\frac{1}{2}\min\{\mu,\nu\}\int_0^t\|(u,b)\|_{\chi^1}\md\tau
\leq\|(u_0,b_0)\|_{\chi^{-1}}+\frac{C}{2\min\{\mu\nu\}}\|(u_0,b_0)\|_{L^2}^4.
\end{align}
\end{thm}
The weak-strong uniqueness theorem is in order.
\begin{thm}\label{thm1.2}
Let $(u_0,b_0)$ and $(v_0,h_0)$ be the divergence-free vector field in $L^2(\mr^2)$ and $(u_0,b_0)$ also be in $\chi^{-1}(\mr^2)$. Let $(v,h)\in L^\infty(\mr^+;L^2(\mr^2))\cap L^2(\mr^+;\dot{H}^1(\mr^2))$ be a Leray-Hopf weak solution associated with $(v_0,h_0)$, and let $(u,b)$ be a strong solution constructed in Theorem \ref{thm1.1} associated with $(u_0,b_0)$ and $$(u,b)\in L^2(\mr^+;\chi^0(\mr^2))\cap L^\infty(\mr^+;L^2(\mr^2))\cap L^2(\mr^+;\dot{H}^1(\mr^2)).$$
Write $(w,g)\triangleq (u-v,b-h)$, then we have for $0\leq t<\infty$
\begin{align}\label{1.4}
\nonumber&\|(w,g)\|_{L^2}^2+\frac12\min\{\mu,\nu\}\left(\int_0^t\|\nabla w\|_{L^2}^2\md \tau+\int_0^t\|\nabla g\|_{L^2}^2\md \tau\right)\\
\leq&\|(u_0,b_0)-(v_0,h_0)\|_{L^2}\times\mbox{exp}\left(C\int_0^t\|(u,b)\|_{\chi^0}^2\md \tau\right).
\end{align}
\end{thm}
\begin{rem}
If two solutions $(u,b)$ and $(v,h)$ constructed in Theorem \ref{thm1.1} have a same initial data $(u_0,b_0)=(v_0,h_0)\in L^2(\mr^2)\cap \chi^{-1}(\mr^2)$, then the estimate inequality (\ref{1.4}) implies the uniqueness of solutions $(u,b)\equiv (v,h)$.
\end{rem}
\begin{rem}
The definitions and notations used in this paper are presented as follows:\\
(1) For $s\in \mr$, the functional space $\chi^s(\mr^n)$ is defined by
$$\chi^{s}:=\left\{f\in \mathcal{D}'(\mr^n)\ \bigg| \ \int_{\mr^n}|\xi|^s|\hat f(\xi)|\mathrm{d} \xi<\infty\right\}.$$
(2) Let $s\in\mr$ and $p\in [1,\infty]$. $f(t,x)\in L^p([0,T];\chi^s(\mr^n))$ if and only if
$$\|f\|_{L^p(\chi^s)}\triangleq\Big( \int^T_0\|f(t,\cdot)\|^p_{\chi^s}\md t\Big)^{\frac1{p}}<\infty;$$
$f(t,x)\in \tilde L^p([0,T];\chi^s(\mr^n))$ if and only if
$$\|f\|_{\tilde L^p(\chi^s)}\triangleq\Big\|\Big(\int^T_0\left(|\xi|^s|\hat{f}(t,\xi)|\right)^p\md t\Big)^{\frac1{p}}\Big\|_{L^1}<\infty.$$
(3) we will use $\|\cdot\|_{X}$ to denote $\|\cdot\|_{X(\mr^n)}$, $\|\cdot\|_{\tilde{L}^q(\chi^s)}$ to denote $\|\cdot\|_{\tilde{L}^q([0,T];\chi^s(\mr^n))}$ and $\|(u,b)\|^p_{X}$ to denote $\|u\|^p_{X}+\|b\|^p_{X}$ for conciseness. And throughout the paper, $C$ stands for a generic positive constant, which may be different from line to line.
\end{rem}
\begin{rem}
The space $L^2(\mr^2)\cap\chi^{-1}(\mr^2)$ is not empty. Indeed, let $f=\mathcal{F}^{-1}\left(\frac{1}{|\xi|^2}\mathbf{1}_{\{|\xi|>2\}}\right)$, then we have
$$\|f\|_{\chi^{-1}}=2\pi\int_2^{+\infty}\frac{1}{r^2}\md r<\infty$$ and $$\|f\|_{L^2}=2\pi\int_2^{+\infty}\frac{1}{r^3}\md r<\infty.$$
Therefore, $f\in L^2(\mr^2)\cap\chi^{-1}(\mr^2)$.
\end{rem}

\section{Preliminaries}
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\vskip .1in
In this Preliminary section, we present some elementary lemmas which will be used in our proofs.
\begin{lem}\label{lem2.1}
Let $s_1<s_0<s_2$. If $f\in \chi^{s_1}(\mr^n)\cap\chi^{s_2}(\mr^n)$, then $f\in\chi^{s_0}(\mr^n)$, and
$$\|f\|_{\chi^{s_0}}\leq\|f\|_{\chi^{s_1}}^{\frac{s_2-s_0}{s_2-s_1}}\|f\|_{\chi^{s_2}}^{\frac{s_0-s_1}{s_2-s_1}}.$$
\begin{proof}
\begin{align}\label{2.1}
\nonumber\|f\|_{\chi^{s_0}}=&\int_{\mr^n}|\xi|^{s_0}|\hat{f}(\xi)|\md \xi\\
\nonumber=&\int_{|\xi|\leq\la}|\xi|^{s_0-s_1}|\xi|^{s_1}|\hat{f}(\xi)|\md \xi+\int_{|\xi|>\la}|\xi|^{s_0-s_2}|\xi|^{s_2}|\hat{f}(\xi)|\md \xi\\
\leq&\la^{s_0-s_1}\|f\|_{\chi^{s_1}}+\la^{s_0-s_2}\|f\|_{\chi^{s_2}}.
\end{align}
Let $\la=\left(\frac{\|f\|_{\chi^{s_2}}}{\|f\|_{\chi^{s_1}}}\right)^{\frac{1}{s_2-s_1}}$, then the Lemma \ref{lem2.1} follows from (\ref{2.1}).
\end{proof}
\end{lem}
\begin{lem}\label{lem2.2}
Let $f\in L^2(\mr^2)\cap\dot{H}^1(\mr^2)$, then $f\in\chi^{-\frac12}(\mr^2)$, and $\|f\|_{\chi^{-\frac12}}\leq C\|f\|_{L^2}^{\frac12}\|f\|_{\dot{H}^1}^{\frac12}$.
\end{lem}
\begin{proof}
We write
\begin{align}\label{2.2}
\|f\|_{\chi^{-\frac12}}=\int_{\mr^2}|\xi|^{-\frac12}|\hat{f}(\xi)|\md \xi:=I_1(\lambda)+I_2(\lambda),
\end{align}
where
$$I_1(\lambda)=\int_{|\xi|\leq\lambda}|\xi|^{-\frac12}|\hat{f}(\xi)|\md \xi\leq C\lambda^{\frac12}\|f\|_{L^2}$$
and
$$I_2(\lambda)=\int_{|\xi|>\lambda}|\xi|^{-\frac12-1}|\xi||\hat{f}(\xi)|\md \xi\leq C\lambda^{-\frac12}\|f\|_{\dot{H}^{1}}$$
by the H\"{o}lder's inequality. Choosing $\lambda=\frac{\|f\|_{\dot{H}^{1}}}{\|f\|_{L^2}}$ completes the proof of Lemma \ref{lem2.2}.
\end{proof}
\begin{lem}\label{lem2.3}
Let $f,g\in L^2([0,T];\chi^0(\mr^n))$, then $fg\in L^1([0,T];\chi^0(\mr^n))$ and
$$\|fg\|_{L^1(\chi^{0})}\leq \|f\|_{L^2(\chi^{0})}\|g\|_{L^2(\chi^{0})}.$$
In particular, $\|f^2\|_{L^1(\chi^{0})}\leq \|f\|_{L^2(\chi^{0})}^2$.
\end{lem}
\begin{proof}
\begin{align}\label{2.3}
\nonumber\|fg\|_{L^1(\chi^{0})}=&\int_0^T\int_{\mr^n}|\hat{f}(\tau,\xi)|\ast_\xi|\hat{g}(\tau,\xi)|\md \xi\md \tau\\
\nonumber\leq &\int_0^T\|f\|_{\chi^0}\|g\|_{\chi^0}\md \tau\\
\leq &\|f\|_{L^2(\chi^0)}\|g\|_{L^2(\chi^0)}.
\end{align}
\end{proof}
\begin{lem}\label{lem2.4}
There exists a constant $C$ such that
$$\|B(u,b)\|_{L^2(\chi^0)}\leq \frac{C}{\min\{\mu,\nu\}^\frac12}\|(u,b)\|_{L^2(\chi^0)}^2,$$
where $B(u,b)\triangleq\int_0^t[\mathrm{e}^{\mu(t-\tau)\triangle}\mathbb{P}\nabla\cdot(u\otimes u+b\otimes b)+\mathrm{e}^{\nu(t-\tau)\triangle}\nabla\cdot(u\otimes b+b\otimes u)]\md \tau$.
\end{lem}
\begin{proof}
\begin{align}\label{2.4}
\nonumber&\|B(u,b)\|_{L^2(\chi^0)}\\
\nonumber\leq&\left\|\int_{\mr^2}\int_0^{T}\mathrm{e}^{-\min\{\mu,\nu\}(t-\tau)|\xi|^{2}}|\xi|\left(|\hat{u}|\ast|\hat{u}
|+|\hat{b}|\ast|\hat{b}|+|\hat{u}|\ast|\hat{b}
|+|\hat{b}|\ast|\hat{u}|\right)\md \tau\md\xi\right\|_{L^2}\\
\nonumber\leq&\int_{\mr^2}\left(\int_0^{T}\mathrm{e}^{-2\min\{\mu,\nu\}\tau|\xi|^{2}}|\xi|^{2}\md \tau\right)^{\frac12}\int_0^T\left(|\hat{u}|\ast|\hat{u}|+|\hat{b}|\ast|\hat{b}|+|\hat{u}|\ast|\hat{b}
|+|\hat{b}|\ast|\hat{u}|\right)\md \tau\md\xi\\
\nonumber\leq&\frac{C}{\min\{\mu,\nu\}^\frac12}\int_0^T\left(\|u\|_{\chi^0}^2+\|b\|^2_{\chi^0}+2\|u\|_{\chi^0}\|b\|_{\chi^0}\right)\md\tau\\
\leq&\frac{C}{\min\{\mu,\nu\}^\frac12}\|(u,b)\|_{L^2(\chi^0)}^2.
\end{align}
\end{proof}
We also need the following Banach contraction mapping principle (see \cite[Lemma 5.5]{B-C-D}).
\begin{lem}\label{lem2.5}
Let $E$ be a Banach space, $\mathcal{B}$ a continuous bilinear map from $E\times E$ to $E$, and $\alpha$ a
positive real number such that
\bes
\alpha<\frac{1}{4\|\mathcal{B}\|} \ with \ \|\mathcal{B}\|=\sup_{\|f\|,\|g\|\leq1}\|\mathcal{B}(f,g)\|.
\ees
For any $a$ in the ball $B(0,\alpha)$ (i.e., with center $0$ and radius $\alpha$) in $E$, a unique $x$ then
exists in $B(0,2\alpha)$ such that
\bes
x=a+\mathcal{B}(x,x).
\ees
\end{lem}
Finally, we show some estimates of the heat equations in the space $\mathcal{C}([0,T];\chi^{s}(\mr^n))\cap L^1((0,T];\chi^{s+2}(\mr^n))$.
\begin{lem}\label{lem2.6}
For $s\in\mr$, let $v$ be a solution of the Cauchy problem
\be\label{2.5}
\begin{cases}
{ \begin{array}{ll}
\partial_{t}v-\kappa\Delta v=f, \\
v(x,0)=v_0(x),
 \end{array} }
\end{cases}
\ee
with $f\in L^1([0,T];\chi^{s}(\mr^n))$ and $v_0\in\chi^s(\mr^n)$. Then $v$ belongs to $$\widetilde{L}^\infty([0,T];\chi^{s}(\mr^n))\cap L^1((0,T];\chi^{s+2}(\mr^n))\cap\mathcal{C}([0,T];\chi^{s}(\mr^n)),$$ and satisfies the following estimate
\begin{align}\label{2.6}
\|v\|_{\widetilde{L}^\infty(\chi^{s})}+\kappa\|v\|_{L^1(\chi^{s+2})}\leq C\left(\|v_0\|_{\chi^s}+\|f\|_{L^1(\chi^s)}\right).
\end{align}
\end{lem}
\begin{proof}
By the Duhamel's formula in Fourier space, the equations (\ref{2.5}) can be written as
\begin{align}\label{2.7}
|\hat{v}(t,\xi)|\leq\mathrm{e}^{-\kappa t|\xi|^{2}}|\hat{v}_0(\xi)|+\int_0^t\mathrm{e}^{-\kappa(t-\tau)|\xi|^{2}}|\hat{f}(\tau,\xi)|\md \tau.
\end{align}
Multiplying (\ref{2.7}) by $|\xi|^s$ and taking the $L^\infty$ norm in time, one has
\begin{align}\label{2.8}
\sup_{0\leq t\leq T}|\xi|^{s}|\hat{v}(t,\xi)|\leq|\xi|^{s}|\hat{v}_0(\xi)|+\int_0^t|\xi|^s|\hat{f}(\tau,\xi)|\md \tau.
\end{align}
Taking the $L^1$ norm in $\xi$ to (\ref{2.8}), we get
\begin{align}\label{2.9}
\nonumber\|v\|_{\widetilde{L}^\infty(\chi^{s})}\leq&\int_{\mr^n}|\xi|^{s}|\hat{v}_0(\xi)|\md\xi
+\int_{\mr^n}\int_0^t|\xi|^s|\hat{f}(\tau,\xi)|\md \tau\md \xi\\
\leq&\|v_0\|_{\chi^s}+\|f\|_{L^1(\chi^s)}.
\end{align}
Because, for almost all fixed $\xi\in\mr^n$, the map $t\mapsto\hat{v}(t,\xi)$ is continuous over $[0,T]$, thus by the Lebesgue dominated convergence theorem, it implies that $v\in\mathcal{C}([0,T];\chi^s(\mr^n))$. Next we estimate $\|v\|_{L^1(\chi^{s+2})}$. Multiplying (\ref{2.7}) by $|\xi|^{s+2}$ and taking the $L^1$-norm in time, applying the Young's inequality in time and noting $\int_0^T |\xi|^{2}\mathrm{e}^{-\kappa t|\xi|^{2}}\md t$
$=\frac{1}{\kappa}$, we deduce that
\begin{align}\label{2.10}
\nonumber\int_0^T|\xi|^{s+2}|\hat{v}(t,\xi)|\md t\leq&\int_0^T\mathrm{e}^{-\kappa t|\xi|^{2}}|\xi|^{s+2}|\hat{v}_0(\xi)|\md t+\int_0^T\int_0^t\mathrm{e}^{-\kappa(t-\tau)|\xi|^{2}}|\xi|^{s+2}|\hat{f}(\tau,\xi)|\md \tau\md t\\
\leq&\frac1\kappa\left(|\xi|^{s}|\hat{v}_0(\xi)|+\int_0^T|\xi|^s|\hat{f}(t,\xi)|\md t\right),
\end{align}
then taking $L^1$ norm in $\xi$ to (\ref{2.10}), we obtain
\begin{align}\label{2.11}
\nonumber\|v\|_{L^1(\chi^{s+2})}\leq&\frac1\kappa\left(\int_{\mr^n}|\xi|^{s}|\hat{v}_0(\xi)|\md\xi
+\int_{\mr^n}\int_0^T|\xi|^s|\hat{f}(t,\xi)|\md t\md \xi\right)\\
\leq&\frac1\kappa\left(\|v_0\|_{\chi^s}+\|f\|_{L^1(\chi^s)}\right).
\end{align}
Combining (\ref{2.9}) and (\ref{2.11}), we get (\ref{2.6}).
\end{proof}

\section{Proof of Theorems \ref{thm1.1} and \ref{thm1.2}}
In this section, we focus on the global well-posedness of a mild solution to (\ref{1.1}) in $L^2(\mr^+;\chi^{0}(\mr^2))$. First, we give the following lemma on the well-posedness and blow-up criterion, which plays a key role in proving our theorem.
\setcounter{equation}{0}
\begin{lem}\label{lem3.1}
Assume $(u_0,b_0)$ be in $\chi^{-1}(\mr^2)$. Then there exists a positive time $T$ such that the equations (\ref{1.1}) have a unique local solution $(u,b)$ in the space $L^2([0,T];\chi^0(\mr^2))$, which also belongs to $$\tilde{L}^\infty([0,T];\chi^{-1}(\mr^2))\cap L^1([0,T];\chi^1(\mr^2))\cap \mathcal{C}([0,T];\chi^{-1}(\mr^2)).$$
Moreover, there exists a constant $C(\mu,\nu)$ such that if $\|(u_0,b_0)\|_{\chi^{-1}}\leq C(\mu,\nu)$, then the solution is a global one; if $T^*<\infty$ is the maximal time of existence, then
\begin{align}\label{3.1}
\lim_{T\rightarrow T^*}\int_0^{T}\|(u,b)\|_{\chi^{0}}^2\md t=\infty.
\end{align}
\end{lem}
\begin{proof}
It is easy to see that the equations (\ref{1.1}) can be rewritten as the integral form
\be\label{3.2}
\begin{cases}
{ \begin{array}{ll}
u(t,x)=\mathrm{e}^{\mu t\Delta}u_0-\int_0^t\mathrm{e}^{\mu(t-\tau)\Delta}[\mathbb{P}\nabla\cdot
(u\otimes u)-\mathbb{P}\nabla\cdot(b\otimes b)]\md \tau,\\
\\
b(t,x)=\mathrm{e}^{\nu t\Delta}b_0-\int_0^t\mathrm{e}^{\nu(t-\tau)\Delta}[\nabla\cdot
(u\otimes b)-\nabla\cdot(b\otimes u)]\md \tau.
 \end{array} }
\end{cases}
\ee
By applying the Banach contraction mapping principle Lemma \ref{lem2.5}, we will carry out the proof of global or local well-posedness of the Cauchy problem (\ref{3.2}) in the Lei-Lin space $\chi^{-1}(\mr^2)$.
First, one has
\begin{align}\label{3.3}
\nonumber&\|(\mathrm{e}^{\mu t\triangle}u_0,\mathrm{e}^{\nu t\triangle}b_0)\|^2_{L^2(\chi^{0})}\\
\nonumber=&\bigg\|\int_{\mr^2}\mathrm{e}^{-\mu t|\xi|^2}|\hat{u}_0(\xi)|\md \xi\bigg\|^2_{L^2}+\bigg\|\int_{\mr^2}\mathrm{e}^{-\nu t|\xi|^2}|\hat{b}_0(\xi)|\md \xi\bigg\|^2_{L^2}\\
\nonumber\leq& \bigg(\int_{\mr^2}\left(\int_0^t\mathrm{e}^{-2\nu \tau|\xi|^2}|\hat{u}_0(\xi)|^2 \md \tau\right)^{\frac{1}{2}}\md\xi\bigg)^{2}+\bigg(\int_{\mr^2} \left(\int_0^t\mathrm{e}^{-2\nu \tau|\xi|^2}|\hat{b}_0(\xi)|^2 \md \tau\right)^{\frac{1}{2}}\md\xi\bigg)^{2}\\
\leq&\frac1{2\min\{\mu,\nu\}}\|(u_0,b_0)\|^2_{\chi^{-1}}.
\end{align}
Thus, Combining the estimate (\ref{3.3}) with Lemma \ref{lem2.4}, we can obtain that if $\|(u_0,b_0)\|_{\chi^{-1}}\leq\frac{\min\{\mu,\nu\}}{2^\frac32C_0}$, then
\begin{align}\label{3.4}
\|(\mathrm{e}^{\mu t\triangle}u_0,\mathrm{e}^{\nu t\triangle}b_0)\|_{L^2(\chi^{0})}\leq \frac{\min\{\mu,\nu\}^\frac12}{4C_0}<\frac{\min\{\mu,\nu\}^\frac12}{4C}
\end{align}
with $C_0>C$, and we obtain the global solution.

Now, we consider the case of a large initial data $(u_0,b_0)$ in $\chi^{-1}(\mr^2)$. Setting
$$(u_0,b_0)=(u_0^\ell,b_0^\ell)+(u_0^\hbar,b_0^\hbar),$$
where
$$(u_0^\ell,b_0^\ell)\triangleq\mathcal{F}^{-1}\left(\mathbf{1}_{\{|\xi|\leq\rho_{u_0,b_0}\}}(\hat{u}_0,\hat{b}_0)\right) \quad \mbox{and} \quad (u_0^\hbar,b_0^\hbar)\triangleq\mathcal{F}^{-1}\left(\mathbf{1}_{\{|\xi|>\rho_{u_0,b_0}\}}(\hat{u}_0,\hat{b}_0)\right).$$
One fixes some positive real number $\rho_{u_0,b_0}$ such that
\begin{align}\label{3.5}
\|(u_0^\hbar,b_0^\hbar)\|_{\chi^{-1}}=\int_{|\xi|>\rho_{u_0,b_0}}|\xi|^{-1}\left(|\hat{u}_0|+|\hat{b}_0|\right)\md\xi\leq\frac{\min\{\mu,\nu\}}{2^\frac52C_0}.
\end{align}
By (\ref{3.3}) we derived that
\begin{align}\label{3.6}
\|(\mathrm{e}^{\mu t\triangle}u_0,\mathrm{e}^{\nu t\triangle}b_0)\|_{L^2(\chi^{0})}\leq\frac{\min\{\mu,\nu\}^\frac12}{8C_0}+\|(\mathrm{e}^{\mu t\triangle}u_0^\ell,\mathrm{e}^{\nu t\triangle}b_0^\ell)\|_{L^2(\chi^{0})},
\end{align}
and
\begin{align}\label{3.7}
\nonumber&\|(\mathrm{e}^{\mu t\triangle}u_0^\ell,\mathrm{e}^{\nu t\triangle}b_0^\ell)\|_{L^2(\chi^{0})}\\
\nonumber=&\big\|\int_{|\xi|\leq\rho_{u_0,b_0}}|\xi||\xi|^{-1}\left(\mathrm{e}^{-\mu t|\xi|^2}|\hat{u}_0(\xi)|+\mathrm{e}^{-\nu t|\xi|^2}\hat{b}_0(\xi)|\right)\md \xi\big\|_{L^2}\\
\leq&\rho_{u_0,b_0}T^{\frac12}\|(u_0,b_0)\|_{\chi^{-1}}.
\end{align}
Hence, if we choose
\begin{align}\label{3.8}
T\leq\left(\frac{\min\{\mu,\nu\}^\frac12}{8\rho_{u_0,b_0}C_0\|(u_0,b_0)\|_{\chi^{-1}}}\right)^{2},
\end{align}
then we have a unique solution $(u,b)$ in the ball $B(0,\frac{\min\{\mu,\nu\}^\frac12}{2C_0})$ of the space $L^2([0,T];\chi^0(\mr^2))$.

Next, we prove the persistence that if $(u,b)$ is a solution to (\ref{1.1}) in $L^2([0,T];\chi^0(\mr^2))$ with initial data $(u_0,b_0)\in \chi^{-1}(\mr^2)$, then $(u,b)$ also belongs to $$\mathcal{C}([0,T];\chi^{-1}(\mr^2))\cap L^1([0,T];\chi^1(\mr^2)).$$ In fact, let $s=-1$ and $n=2$ in Lemma \ref{2.6} and by Lemma \ref{2.3} the result is followed.

Finally, we prove the blow-up criterion (\ref{3.1}). Assume that we have a solution to the equations (\ref{1.1}) on a time interval $[0,T)$ such that
$$\int_0^{T}\|(u,b)\|_{\chi^{0}}^2\md t<\infty.$$
We claim that the lifespan $T^*$ of $(u,b)$ is larger than $T$. Indeed, due to the estimate (\ref{2.9}) in Lemma \ref{2.6} and  by Lemma \ref{lem2.3}, we have
\begin{align}\label{3.9}
\int_{\mr^n} \sup_{0\leq t\leq T}|\xi|^{-1}(|\hat{u}|+|\hat{b}|)(t,\xi)\md \xi\leq \|(u_0,b_0)\|_{\chi^{-1}}+C\|(u,b)\|_{L^2(\chi^{0})}^2<\infty.
\end{align}
Thus, a positive number $\rho$ exists such that
\begin{align}\label{3.10}
\forall t\in [0,T), \quad \int_{|\xi|>\rho}|\xi|^{-1}(|\hat{u}|+|\hat{b}|)(t,\xi)\md \xi\leq\frac{\min\{\mu,\nu\}}{2^\frac52C_0}.
\end{align}
The condition (\ref{3.8}) now implies that for any $t\in[0,T)$, the lifespan for a solution to (\ref{1.1}) with initial data $\left(u(t),b(t)\right)$ is bounded from below by a positive real number $C$ which is independent of $t$. Thus the lifespan $T^*>T$, and the whole proof of Lemma \ref{lem3.1} is finished.
\end{proof}
\subsection {Proof of Theorem \ref{thm1.1}} Taking the $L^2$ inner products of the equations $(\ref{1.1})_{1,2}$ with $u$ and $b$, respectively, adding the results and integrating by parts, we obtain, for any $t\in (0,+\infty)$, the energy equality
\begin{align}\label{3.11}
\|(u,b)\|_{L^{2}}^{2}+2\mu\int_0^t\|\nabla u\|_{L^{2}}^{2}\md\tau+2\nu\int_0^t\|\nabla b\|_{L^{2}}^{2}\md\tau=\|(u_0,b_0)\|_{L^{2}}^{2},
\end{align}
for details refer to \cite[Theorem 5.14]{B-C-D}. And by Lemma \ref{lem2.2}, it yields that, for any $t\in (0, +\infty)$
\begin{align}\label{3.12}
\nonumber\int_0^t\|(u,b)\|_{\chi^{-\frac12}}^4 \md \tau
\nonumber\leq&\|(u,b)\|_{L^2}^2\int_0^t\|(u,b)\|_{\dot{H}^1}^2 \md \tau\\
\leq&\frac{1}{2\min\{\mu\nu\}}\|(u_0,b_0)\|_{L^2}^4.
\end{align}
According to Lemmas \ref{lem2.6}, \ref{lem2.3}  and \ref{lem2.1}, by the Young's inequality, we deduce that
\begin{align}\label{3.13}
\nonumber&\|(u,b)\|_{\widetilde{L}^\infty(\chi^{-1})}+\mu\int_0^t\|u\|_{\chi^{1}}\md \tau+\nu\int_0^t\|b\|_{\chi^{1}}\md\tau\\
\nonumber\leq&\|(u_0,b_0)\|_{\chi^{-1}}+C\int_0^t\|(u,b)\|_{\chi^{0}}^2\md\tau\\
\nonumber\leq&\|(u_0,b_0)\|_{\chi^{-1}}+C\int_0^t\|(u,b)\|_{\chi^{-\frac12}}^{\frac43}\|(u,b)\|_{\chi^1}^{\frac23}\md \tau\\
\leq&\|(u_0,b_0)\|_{\chi^{-1}}+C\int_0^t\|(u,b)\|_{\chi^{-\frac12}}^4\md\tau+\frac{1}{2}\min\{\mu,\nu\}\int_0^t\|(u,b)\|_{\chi^1}\md\tau.
\end{align}
Inserting the estimate (\ref{3.12}) into (\ref{3.13}) leads to the result
\begin{align}\label{3.14}
\|(u,b)\|_{\widetilde{L}^\infty(\chi^{-1})}+\frac{1}{2}\min\{\mu,\nu\}\int_0^t\|(u,b)\|_{\chi^1}\md\tau
\leq\|(u_0,b_0)\|_{\chi^{-1}}+\frac{C}{2\min\{\mu\nu\}}\|(u_0,b_0)\|_{L^2}^4.
\end{align}
By Lemma \ref{2.1}, we arrive at
\begin{align}\label{3.15}
\int_0^{t}\|(u,b)\|_{\chi^{0}}^2\md \tau\leq\int_0^{t}\|(u,b)\|_{\chi^{-1}}\|(u,b)\|_{\chi^{1}}\md \tau\leq C\left(\mu,\nu,\|(u_0,b_0)\|_{\chi^{-1}},\|(u_0,b_0)\|_{L^2}\right),
\end{align}
for any $0\leq t\leq T$. Thus, the blow-up criterion (\ref{3.1}) implies that the strong solution $(u,b)$ is a global one in $L^2([0,T];\chi^{0}(\mr^2))$ for any $T<\infty$. And by Lemma \ref{2.6} the solution $(u,b)\in\mathcal{C}(\mr^+;\chi^{-1}(\mr^2))\cap L^1(\mr^+;\chi^1(\mr^2))$, which completes the proof of Theorem \ref{thm1.1}.
\subsection {Proof of Theorem \ref{thm1.2}} Subtracting the two equations satisfied by $(u,b)$ and $(v,h)$, respectively, we get
\be\label{4.1}
\begin{cases}
{ \begin{array}{ll}
\partial_{t}w-\mu\Delta w+(w\cdot\nabla) u+(v\cdot\nabla) w+\nabla (p_1-p_2)=(g\cdot\nabla) b+(h\cdot\nabla)g,\\
\partial_{t}g-\nu\Delta g+(w\cdot\nabla) b+(v\cdot\nabla) g=(g\cdot\nabla) u+(h\cdot\nabla) w, \\
 \end{array} }
\end{cases}
\ee
where $(w, g)\triangleq (u-v, b-h)$. Taking the $L^2$ inner products of the equations $(\ref{4.1})_{1,2}$ with $w$ and $g$, respectively, adding the results and by H\"{o}lder's and Young's inequalities and the fact $\|f\|_{L^\infty}\leq\|\hat{f}\|_{L^1}=\|f\|_{\chi^0}$, and
$$\int_{\mr^{2}}(v\cdot\nabla)w\cdot w \md x=0, \ \int_{\mr^{2}}(v\cdot\nabla)g\cdot g\md x=0$$
and $$\int_{\mr^{2}}(h\cdot\nabla)g\cdot w\md x+\int_{\mr^{2}}(h\cdot\nabla)w\cdot g\md x=0,$$
it follows that
\begin{align}\label{4.2}
\nonumber&\frac12\frac{\md}{\md t}\|(w,g)\|_{L^2}^2+\mu\|\nabla w\|_{L^2}^2+\nu\|\nabla g\|_{L^2}^2\\
\nonumber=&-\int_{\mr^{2}}(w\cdot\nabla) u\cdot w\md x+\int_{\mr^{2}}(g\cdot\nabla) b\cdot w\md x-\int_{\mr^{2}}(w\cdot\nabla) b\cdot g\md x+\int_{\mr^{2}}(g\cdot\nabla) u\cdot g\md x\\
\nonumber\leq&C\left(\|\nabla w\|_{L^2}\|w\|_{L^2}\|u\|_{L^\infty}+\|\nabla w\|_{L^2}\|g\|_{L^2}\|b\|_{L^\infty}+\|\nabla g\|_{L^2}\|w\|_{L^2}\|b\|_{L^\infty}+\|\nabla g\|_{L^2}\|g\|_{L^2}\|u\|_{L^\infty}\right)\\
\leq&\frac12\min\{\mu,\nu\}\|(\nabla w,\nabla g)\|_{L^2}^2+C\|(w,g)\|_{L^2}^2\|(u,b)\|_{\chi^0}^2.
\end{align}
Thanks to the Gr\"{o}nwall's inequality, we conclude the proof of (\ref{1.4}). And we thus complete the proof of Theorem \ref{thm1.2}.

\textbf{Acknowledgements} The research of B Yuan
was partially supported by the National Natural Science Foundation
of China (No. 11471103).


TITLE: Help in understanding demonstration: every Cauchy sequence is bounded
QUESTION [1 upvotes]: I found that demonstration. I know one different. I need some explanation.
The demonstration:
$(a_{n})$ is a Cauchy sequence.
$|a_{k}-a_{j}|< \epsilon$ $\forall k,j>n_{\epsilon}$
We can consider $\epsilon=1$.
$|a_{k}-a_{j}|< 1$
Then $a_{j}-1< a_{k}<a_{j}+1$
Now we define $m=min(a_{1},a_{2},...,a_{j}-1,a_{k},a_{j}+1)$ and $M=max(a_{1},a_{2},...,a_{j}-1,a_{k},a_{j}+1)$
$a_{k}$ with $k>j$ belongs to the interval (m,M) because $m\leqslant a_{j}-1< a_{k}<a_{j}+1\leqslant M$.
In particular $m<a_{k}<M$ which ends our proof.
My doubts:

Why $k>j$ ? It should be true for any $k,j>n_{1}$

Why does it consider the minimum and
maximum?

What does it happen for $k<j$?

REPLY [0 votes]: $j$ is fixed (to define $m$ and $M$) and chosen $>n_\epsilon=n_1$ (e.g. $j=n_1+1$), and in that part of the proof (but see point 3), $k>j$, for $a_j-1< a_k<a_j+1$ to hold (but $k>n_1$ would be ok as well).

If we take only $m=a_j-1$ and $M=a_j+1$ (instead of min and max), the inequalities $m\le a_k\le M$ will  only hold for all $k>n_\varepsilon$. We want them for all $k\in\mathbb N$.

for $k\le j$, we still have $m\le a_k$ because $m=\min(a_1,\dots,a_k,\dots,a_{j-1}, a_j-1)$. Similarily, we still have $a_k\le M$.


TITLE: Condition for $8p+1$ divides $(2^p+1)/3$?
QUESTION [5 upvotes]: Here is what I observed :
Let $8p+1 = 256a^2+(2b-1)^2$ with $a$ and $b$ be a positive integers, $p$ and $8p+1$ both prime numbers.
Then $8p+1$ divides $(2^p+1)/3$ only if you can write $8p+1$ as $256a^2+(2b-1)^2$.
For example :

$1049 = 256 \cdot 2^2+(2 \cdot 3 - 1)^2$ and $1049 = 131 \cdot 8+1$ so $1049$ divides $(2^{131}+1)/3$
$12569 = 256 \cdot 7^2+(2 \cdot 3 - 1)^2$ and $12569 = 1571 \cdot 8+1$ so $12569$ divides $(2^{1571}+1)/3$
$2120057 = 256 \cdot 91^2+(2 \cdot 6 - 1)^2$ and $2120057 = 265007 \cdot 8+1$ so $2120057$ divides $(2^{265007}+1)/3$
$137 = (8 \cdot 17 + 1)$ but you can't write $137$ as $256a^2+(2b-1)^2$ so $137$ does not divide $(2^{17}+1)/3$

For the moment, I didn't find a counterexample with this condition.
I need help for proving it but I don't know how to start.
I thought about Wagstaff numbers and Sophie Germain primes that say if $p \equiv 1 \pmod{4}$ and $2p+1$ is prime then $2p+1$ divides $(2^p+1)/3$ but for $8p+1$ it doesn't work.
This question looks like the one I asked on Math stack exchange (for $2^p-1$ and the divisibility by $8p+1$) but I didn't get an answer for the question.
If you found a counterexample please tell me.

REPLY [8 votes]: This follows easily from the octic reciprocity law for $2$. More precisely, it states that if $q$ is a prime of the form $q=8n+1$ and $n$ is odd, then $q=a^2+256b^2$ for some integers $a,b$ (note that $a$ is necessarily odd) if and only if $2$ is a biquadratic residue but not an octic residue, i.e. there is $x \in \mathbb F_q$ with $x^4=2$ but no $y\in \mathbb F_q$ with $y^8=2$.
This condition is equivalent to $2^{2n}\equiv 1\pmod q$ and $2^{n}\not\equiv 1\pmod q$, which is the same as $q\mid \frac{2^n+1}{3}$ since $2^{2n}-1=3(2^n-1)\frac{2^n+1}{3}$. For instance, your observation works for non-prime (but odd) values of $p$.
For the reference on octic reciprocity, see, for example, the first paragraph of Whiteman AL. The Sixteenth Power Residue Character of 2


TITLE: Defects of smart glass (electrochromic glass)?
QUESTION [0 upvotes]: Smart glass is a type of glass that uses electrochromic mechanism to change it's light transmission properties in response to voltage, and it's made by sandwiching certain materials between two panes of glass.
My question is:
Is the electrochromic glass at it's transparent state more transparent than a normal glass? Why?

REPLY [1 votes]: No, it cannot be. Electrochromic glass is made by coating electrochromically active films onto an electrochromically inert glass substrate. As such, changes in the electrochromic state only affect the transparency of these added layers, and the base transmission of the underlying glass in unchanged.


TITLE: Is there something like acceleration constraint in a rigid body?
QUESTION [0 upvotes]: Suppose we have two points on a body which having acceleration a1 and a2 direction we know and magnitude as well , is there some sort of relation between them we can establish , like for velocities along the line joining velocities must be same sort of ? For example a rod which has two points of acceleration value known and directions too suppose angle theta1 for a1 and theta 2 for a2 from the rod , is there a constraint which can be established on that ? As we know velocity along the line joining the rod should be same everywhere , otherwise it fails to consider rigid body . Hope now clarity is given

[is there some relation between them which we can deduce thats what i am asking ] body is just intially has such acc vectors , so intiallly angulsr velocity is zero so not shown in figure

REPLY [2 votes]: Of course, there is. All velocities/accelerations on the extended rigid body are a result of the same motion of the rigid body, which we refer to as the rotating frame.
Just as the velocities of two points on a rigid body are related by $$ \boldsymbol{v}_A + \boldsymbol{r}_A \times \boldsymbol{\omega} = \boldsymbol{v}_B + \boldsymbol{r}_B \times \boldsymbol{\omega}$$
where $\boldsymbol{r}_A$ is the position vector of point A, $\boldsymbol{r}_B$ the position vector of point B, $\boldsymbol{v}_A$ and $\boldsymbol{v}_B$ the velocity vectors of the points, and $\boldsymbol{\omega}
$ the rotationa velocity of the rigid body.
Take the time derivative of the above using the chain rule to get
$$ \boldsymbol{a}_A + \boldsymbol{r}_A \times \boldsymbol{\dot \omega} + \boldsymbol{v}_A \times \boldsymbol{\omega} = \boldsymbol{a}_B + \boldsymbol{r}_B \times \boldsymbol{\dot \omega} + \boldsymbol{v}_A \times \boldsymbol{\omega}$$
Which is the kinematic constraint between two points. The above is commonly rewritten as follows, using the relative position vector $\boldsymbol{d} = \boldsymbol{r}_A - \boldsymbol{r}_B$
$$\boldsymbol{v}_A = \boldsymbol{v}_B + \boldsymbol{\omega} \times \boldsymbol{d}$$
$$\boldsymbol{a}_A = \boldsymbol{a}_B + \boldsymbol{\dot \omega} \times \boldsymbol{d} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \boldsymbol{d}) $$


TITLE: Evaluating $\int_0^1\frac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$
QUESTION [13 upvotes]: $$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$$

Attempt:
If we write: $f(x)= x^4 + x^3+ x^2$, we get:
$$I = \displaystyle\int_0^1 \dfrac{3f(x)+x^3}{(f'(x)+1)^2}\, dx$$
I have no idea how to proceed. Integration by parts/ substitution can't help.
I don't need the full solution. Just a guiding hint would suffice.

REPLY [11 votes]: The hint of @achillehui to substitute $x\to y=\frac{1}{x}$  is valuable and should be an answer by its own.

We obtain
  \begin{align*}
\color{blue}{\int_0^1\frac{3x^4+4x^3+3x^2}{\left(4x^3+3x^2+2x+1\right)^2}\,dx}
&=\int_1^\infty\frac{3y^{-4}+4y^{-3}+3y^{-2}}{\left(4y^{-3}+3y^{-2}+2y^{-1}+1\right)^2}y^{-2}\,dy\tag{1}\\ %fixed denominator here
&=\int_1^\infty\frac{3y^2+4y+3}{\left(y^3+2y^2+3y+4\right)^2}\,dy\tag{2}\\
&=-\left.\frac{1}{y^3+2y^2+3y+4}\right|_1^\infty\tag{3}\\
&\,\,\color{blue}{=\frac{1}{10}}
\end{align*}

Comment:

In (1) we substitute $y=\frac{1}{x},\quad dy=-\frac{1}{x^2}dx$.
In (2) we expand with $y^6$.
In (3) we integrate by noting that $\frac{d}{dy}\left(y^3+2y^2+3y+4\right)=3y^2+4y+3$.


TITLE: explicit description of the cosimplicial simplicial set $Q^{\bullet}$
QUESTION [3 upvotes]: I'm struggling to understand the explicit description of the cosimplicial simplicial set $Q^{\bullet}$ on page 76 (section 2.2.2) of Lurie's book Higher Topos Theory, and would be grateful if someone could elaborate on that a bit.
More concretely, we have that $$Q^n = \operatorname{Map}_{\mathfrak{C}\left[J^n\right]}(x,y),$$ where $$J^n := (\Delta^n \star \{y\}) \coprod_{\Delta^n} \{x\},$$ and it is claimed that $Q^n$ can be obtained as a certain quotient of the simplicial set $K_{\left[n\right]} := N(P_{\left[n\right]})$, where $P_{\left[n\right]}$ is the partially ordered set of nonempty subsets of $\left[n\right]$.

REPLY [4 votes]: Here is how I think about this; don't know if it will help.
Start with the straightening construction: this takes a map $f\colon X\to S$ of simplicial sets to a simplicial functor $\def\St{\mathrm{St}}\def\op{\mathrm{op}}\St(f)\colon \mathfrak{C}(S)^{\op}\to s\mathrm{Set}$.
The key example is when $f=\mathrm{id}\colon \Delta^n\to \Delta^n$.  This is a simplicial functor $\St(\mathrm{id})\colon\mathfrak{C}(\Delta^n)^\op\to s\mathrm{Set}$, which I'll call $\mathfrak{D}^n$. Recall that $\mathfrak{C}(\Delta^n)$ has mapping spaces $$\mathrm{map}(x,y) =N\bigl\{ \{x,y\}\subseteq S\subseteq [x,y] \bigr\},$$ i.e., nerves of posets of subsets of the interval $[x,y]$ containing the endpoints.  The functor $\mathfrak{D}^n$ is defined similarly:
$$\mathfrak{D}^n(x) = N\bigl\{\{x\}\subseteq S\subseteq [x,n]\bigr\}.$$
The action $\mathfrak{D}^n(y)\times \mathrm{map}(x,y)\to \mathfrak{D}^n(x)$ is defined by union.
In general, we have
$$ \St(\Delta^n\xrightarrow{f} S) \approx \St(\Delta^n\xrightarrow{\mathrm{id}}\Delta^n)\otimes_{\mathfrak{C}(\Delta^n)} \mathfrak{C}(S),$$
where the tensor product really means left Kan extension along the (simplicial) functor $\mathfrak{C}(f)\colon \mathfrak{C}(\Delta^n)\to \mathfrak{C}(S)$ (or, more precisely, along its opposite).
The simplicial set
$$Q^n= \St(\Delta^n\to \Delta^0) \approx \mathfrak{D}^n\otimes_{\mathfrak{C}(\Delta^n)} 1,$$
where $1=\mathfrak{C}(\Delta^0)$ is the terminal simplicial category. In other words, the simplicial set $Q^n$ is the colimit of the (simplicial) functor $\mathfrak{D}^n\colon \mathfrak{C}(\Delta^n)^\op\to s\mathrm{Set}$.  This is just something you have to compute: it's a coequalizer of
$$\small
\coprod_{x=0,\dots,n} N\bigl\{ \{x\}\subseteq S\subseteq [x,n] \bigr\} \leftleftarrows \coprod_{0\leq x\leq y\leq n} N\bigl\{ \{y\}\subseteq S\subseteq [y,n]\bigr\} \times N\bigl\{ \{x,y\}\subseteq T \subseteq [x,y] \bigr\}.
$$
This turns out to be isomorphic to Lurie's quotient of $K_{[n]}$.  Explicitly, a $k$-simplex in Lurie's quotient is a sequence $\varnothing \subsetneq S_0\subseteq \cdots\subseteq S_k\subseteq [n]$, subject to the relation that, for each $x\in S_0$, we identify it with the $k$-simplex $\varnothing \subsetneq S_0\cap[x,n]\subseteq \cdots \subseteq S_k\cap [x,n]\subseteq [n]$.


text:
\begin{document}

\maketitle

\begin{abstract}
The paper presents a deep learning-aided iterative detection algorithm for massive overloaded MIMO channels.
The proposed algorithm is based on the iterative soft thresholding algorithm for sparse signal recovery.
The notable feature of the proposed scheme is that the detector has a reasonably low computational cost and contains trainable parameters
 which can be optimized with standard deep learning techniques.
The number of trainable parameters is constant to the channel size, which promotes fast and stable training processes for the detector.
The numerical simulations show that the proposed detector achieves a comparable detection performance
 to the state-of-the-art IW-SOAV detector for massive overloaded MIMO channels.
\end{abstract}

\begin{keywords}
massive MIMO, overloaded MIMO, detection algorithm,  deep learning
\end{keywords}

\section{Introduction}\label{sec_intro}

Multiple input multiple output (MIMO) systems have attracted great interests because they potentially achieve
high spectral efficiency.
Recently, as a consequence of high growth of mobile data traffic,
\emph{massive} MIMO is regarded as a key technology in the 5th generation (5G) wireless communication systems~\cite{Yang}.
In massive MIMO systems, tens or hundreds of antennas are used in the transmitter and the receiver.
This fact complicates the detection problem for MIMO channels because the computational complexity of a MIMO detector, in general,
increases as the numbers of antennas grows.  A practical massive MIMO
detection algorithm should possess both low energy consumption and low computational complexity in addition to
reasonable bit error rate (BER) performance.

In a down-link massive MIMO channel with mobile terminals, a transmitter in a base station can have many
antennas but a mobile terminal cannot have such a number of receive antennas because of the restrictions on
cost, space limitation, and power consumption.
This scenario is known as the \emph{overloaded} (or \emph{underdetermined}) scenario.
Development of an overloaded MIMO detector with computational efficiency and considerable BER performance
is a highly challenging problem
because conventional simple MIMO decoders such as
the minimum mean square error (MMSE) detector~\cite{Shnidman} and
the ZF detector ~\cite{Etten} exhibits poor BER performance over overloaded MIMO channels,
and an optimal detection based on the exhaustive search is evidently computationally intractable.

Several search-based detection algorithms such as slab-sphere decoding~\cite{SSD}
and enhanced reactive tabu search~\cite{ERTS} have been proposed for overloaded MIMO channels.
Though these schemes show reasonable detection performance,
they are computationally demanding,
and it prevents us from implementing them into a practical massive overloaded MIMO system.
As a computationally efficient approach based on the $\ell_1$-regularized minimization
for underdetermined systems,
 Fadlallah et. al proposed a detector using a convex optimization solver~\cite{Fad1,Fad2}.
Recently, Hayakawa and Hayashi~\cite{IW-SOAV2} proposed an iterative detection algorithm with practical computational complexity
called  \emph{iterative weighted sum-of-absolute value} (IW-SOAV) optimization (see also ~\cite{IW-SOAV1}).
The algorithm is based on the SOAV optimization~\cite{SOAV} for sparse discrete signal recovery.
In addition, the algorithm includes a re-weighting process
based on the log-likelihood ratio, which improves the detection performance.
The IW-SOAV yields the state-of-the-art BER performance among overloaded MIMO detection algorithms
with comparable computational complexity.

The use of deep neural networks has spread to numerous fields such as image recognition~\cite{Image1,Image2}
and speech recognition~\cite{Speech1} with the progress of computational resources.
It also gives a great impact on design of algorithms for communications and signal processing~\cite{Com1,Com2}.
In~\cite{Com2}, for instance, it is shown that the belief propagation decoder with trainable parameters
provides excellent BER performances for several short codes such as BCH codes.
Gregor and LeCun first proposed the learned iterative soft thresholding algorithm (LISTA)~\cite{LISTA}, which
exhibits better recovery performance than that of the original ISTA~\cite{ISTA2} for sparse signal recovery problems.
Recently, the authors proposed the \emph{trainable ISTA} (TISTA)~\cite{TISTA},
which shows significantly faster convergence than those of known algorithms such as AMP.

The goal of this paper is
to propose a novel trainable iterative detection algorithm for massive overloaded MIMO systems,
which is called the \emph{Trainable Iterative-Detector} (TI-detector).
Although deep learning architectures for massive MIMO systems were recently proposed
as deep MIMO detectors (DMDs) in~\cite{DMD1,DMD2},
no deep learning-aided iterative detectors
for massive \emph{overloaded} MIMO channels have been proposed as far as
the authors are aware of.
The architecture of the proposed algorithm
is essentially based on TISTA~\cite{TISTA} because the success of TISTA would imply
that the TISTA approach appears promising not only for sparse signal recovery
but also for various types of detection problems and inverse problems for underdetermined systems.

\section{Problem Setting}\label{sec_set}

The section describes the channel model and introduces several definitions and notation.
The number of transmit and receive antennas is denoted by $n$ and $m$, respectively.
We only consider the overloaded MIMO scenario in this paper where $m <n$ holds.
It is also assumed that the transmitter does not use precoding and that the receiver perfectly knows
the channel state information, i.e., the channel matrix.

Let $\t{x} \triangleq [\t{x}_1,\t{x}_2,\dots,\t{x}_n]^{T} \in \t{{\mathbb S}} ^{n}$ be
a vector which consists of  a transmitted symbol $\t{x}_j$ ($j=1,\dots,n$) from the $j$th antenna.
The symbol $\t{\bb{S}} \subset \bb{C}$ represents a symbol alphabet, i.e., a signal constellation.
Similarly, $\t{y} \triangleq [\t{y}_1,\t{y}_2,\dots,\t{y}_m]^{T} \in \bb{C}^{m}$ denotes
a vector composed of a received symbol $\t{y}_i$ $(i=1,\dots,m)$ by the $i$th antenna.
A flat Rayleigh fading channel is assumed here and the received symbols $\t{y}$ then reads
$
  \t{y} = \t{H}\t{x} + \t{w},
$
where $\t{w}\in \bb{C}^m$ consists of complex Gaussian random variables
with zero mean and covariance $\sigma_{w}^2I$.
The matrix $\t{H} \in \bb{C}^{m \times n}$ is a channel matrix whose $(i,j)$ entry $\t{h}_{i,j}$ represents
a path gain from the $j$th transmit antenna to the $i$th receive antenna.
Each entry of $\t{H}$ independently follows the complex Gaussian distribution with zero mean and unit variance.
For the following discussion, it is convenient  to derive an equivalent channel model defined over $\mathbb{R}$,
i.e.,  $y = Hx + w$, where
    \begin{eqnarray} \nonumber
      y &\triangleq& \begin{bmatrix}
        \r{Re}(\t{y}) \\
        \r{Im}(\t{y})
        \end{bmatrix} \in \bb{R}^{M},\
      H \triangleq \begin{bmatrix} \label{eq:H_real}
        \r{Re}(\t{H}) & - \r{Im}(\t{H})\\
        \r{Im}(\t{H}) &  \r{Re}(\t{H})\\
      \end{bmatrix}, \\ \nonumber
      {x} &\triangleq& \begin{bmatrix}
        \r{Re}(\t{{x}}) \\
        \r{Im}(\t{{x}})
        \end{bmatrix} \in \bb{S}^{N}, \
      {w} \triangleq \begin{bmatrix}
        \r{ Re}(\t{{w}}) \\
        \r{ Im}(\t{{w}})
        \end{bmatrix}\in \bb{R}^{M},
    \end{eqnarray}
 and $(N,M) \triangleq (2n,2m)$.
The signal set $\bb{S}$ is the real counter part of $\t{\bb{S}}$.
The matrix $H \in \bb{R}^{M \times N}$ is converted from $\t{H}$.
 Similarly, the noise vector $w$ consists of i.i.d. random variables following the Gaussian distribution
 with zero mean and variance $\sigma^2_w/2$.
Signal-to-noise ratio (SNR) per receive antenna is then represented by
  $
  \mathrm{SNR}\triangleq E_s/N_0
  =  {N}/{\sigma_{w}^{2}}
  $,
  where $E_s\triangleq \bb{E}[||\t{H}\t{x}||^{2}_{2}]/m$ stands for the signal power per receive antenna and
  $N_0 \triangleq \sigma_w^2$ stands for the noise power per receive antenna.
Throughout the paper, we assume the QPSK modulation format, i.e., $\t{\bb{S}} \triangleq \{1+j,-1+j,-1-j,1-j\}$,
which is equivalent to the BPSK modulation $\bb{S} \triangleq \{-1,1\}$.

\section{Trainable Iterative-Detector}

\subsection{Brief review of ISTA}
The iterative soft thresholding algorithm (ISTA) is the origin of our proposed algorithm described later.
ISTA~\cite{ISTA2} is known as a powerful iterative algorithm for sparse signal recovery from linear observations.
The underlying model is described as $u = A v + z$
where {$u$} and {$v$} are the linear observation and the sparse signal, respectively.
The matrix $A$ is called a sensing matrix.
ISTA is an iterative algorithm defined by the following recursion:
\begin{eqnarray} \label{l_eq}
r_t &=& s_t + \beta A^T({u} - As_t)\\ \label{n_eq}
s_{t + 1} &=& \xi(r_t; \tau).
\end{eqnarray}
Eq.(\ref{l_eq}) represents a linear estimator and
Eq.(\ref{n_eq}) represents a nonlinear estimator based on the soft thresholding function $\xi(r_t; \tau)$.
The soft thresholding function matches to the Laplacian prior
for the source sparse signal.

\subsection{Details of TI-detector}

Our proposal, TI-detector, is described by the following recursive formulas that are based on those of ISTA:
\begin{eqnarray}
  \label{eq:hoge_1}
  r_t &=& s_t + \gamma_t W(y - H s_t),\\
  \label{eq:hoge_2}
  s_{t+1} &=& \tanh\left(\frac{r_t}{|\theta_t|}\right),
\end{eqnarray}
where $\tanh(\cdot)$ is an element-wise function operator.
The index variable $t$ represents the index of an iterative step (or \emph{layer}) and we set $s_1=0$ as the initial value.
This algorithm estimates a transmitted signal $x$ from the received signal $y$ and outputs
the estimate $\hat x = s_{T+1}$ after $T$ iterative steps.
The matrix $W$ in the linear estimator (\ref{eq:hoge_1})
is the Moore-Penrose
pseudo inverse matrix of $H$, i.e., $W \triangleq H^{T}(HH^{T})^{-1}$.
We use the nonlinear estimator (\ref{eq:hoge_2}) based on the hyperbolic tangent function
derived from the BPSK prior of
the transmitted signals. Note that this type of nonlinear estimator has been commonly used
in several iterative multiuser detection algorithms such as the soft parallel interference canceller~\cite{PIC}.

The trainable parameters of TI-detector   are $2T$
real scalar  variables $\{\gamma_t\}_{t=1}^T$ and  $\{\theta_t\}_{t=1}^T$
in (\ref{eq:hoge_1}) and (\ref{eq:hoge_2}), respectively.
Note that the number of trainable parameters of TI-detector is constant to $N$ and $M$
though a DMD contains $O(N^2T)$ parameters in $T$ layers~\cite{DMD1}.
The trainable parameters are optimized in a training process described later.
The parameters $\{\gamma_t\}_{t=1}^T$,  which {control} the strength of the linear estimator,
are introduced according to the structure of TISTA~\cite{TISTA}, i.e.,
TISTA uses the same linear estimator in (\ref{eq:hoge_1}).
The parameters $\{\theta_t\}_{t=1}^T$ {control} the nonlinear estimator to perform appropriately.
For simplicity, we use the notation
$\Theta_t \triangleq  \{\{\gamma_1,\theta_1\}, \{\gamma_2,\theta_2\} \dots ,\{\gamma_t,\theta_t\} \}$
as a set of the parameters up to the $t$th layer.

The computational complexity of TI-detector  per iteration
is $O(MN)$ because one needs to calculate vector-matrix products $Hs_{t}$ and $W(y-Hs_t)$
that take $O(MN)$ computational steps.
We need to calculate
the pseudo inverse matrix $W$ taking $O(M^3)$ computational steps
only when $H$ changes.

\subsection{Training process of TI-detector}
\begin{figure}[bt]
        \begin{center}
          \includegraphics[clip, width=7.8cm]{./TI-detector.pdf}
          \caption{The $t$-th layer of the TI-detector. The trainable parameters are $\gamma_t$ and $\theta_t$.}
          \label{fig:TI-detector}
        \end{center}
\end{figure}

In order to retrieve reasonable detection performance from TI-detector,
it is critical to optimize the trainable parameters $\Theta_T$.
We here describe how to train the trainable parameters
based on standard deep learning techniques.

A training data $d^{i} \triangleq (x^i, y^i)$ is stochastically generated as follows.
A transmitted signal $x^i \in \{+1, -1 \}^N$ is generated uniformly at random and
the corresponding received signal $y^i$ is then generated
according to the probabilistic channel model $y^i = Hx^i + w$
with a given $H$ and a fixed variance $\sigma_w^2$ corresponding to a given SNR.
A mini-batch consists of $D$ training data $\mathcal{D}\triangleq \{d^{1}, d^{2},\dots, d^{D}\}$
and it is fed into the TI-detector.
In the training process, we use an \emph{incremental training} as TISTA~\cite{TISTA}, i.e.,
the parameters are sequentially trained from the initial layer ($t = 1$) to the last layer ($t = T$)
in an incremental manner.
In the $t$th round of the incremental training, we optimize the trainable parameters $\Theta_t$
to minimize the squared loss function
\[
  L(\Theta_t) \triangleq \frac 1 D \sum_{d^i \in \mathcal{D}} \| x^i -  \hat x (y^i) \|^{2}_{2}.
\]
Note that
$\hat x (y^i)$ represents the output of TI-detector  after $t$th layer with the input $y^{i}$,
i.e., $\hat x (y^i) \triangleq  s_{t+1}$.
Since the loss function $L(\Theta_t)$ is a differentiable function, we can exploit
back propagation to compute the gradient of $\Theta_t$ efficiently and the gradient can be
used in a stochastic gradient optimizer.
For each round of the incremental training,  $K$ mini-batches are processed.

As described in Sec.~\ref{sec_set}, we assume a practical situation in which a channel matrix $H$ is a random variable.
The problem setting is called  a \emph{varying channel} (VC) scenario in~\cite{DMD1,DMD2}.
Following the VC scenario, a matrix $H$ is randomly generated for each mini-batch in a
training process of TI-detector.

\section{Numerical Results}
In this section, we show several numerical results of TI-detector and compare them to those of the IW-SOAV,
which is known as an efficient iterative algorithm for massive overloaded MIMO systems.

\subsection{Experimental setup}
We evaluate the detection performance of TI-detector
under situations
with $(N,M)=(200,128)$ and $(300,192)$, where the ratio $M/N=0.64$ is fixed.
A transmitted signal is generated as i.i.d. random variables with a distribution $p(x=1)=p(x=-1)=1/2$.
The BER is then evaluated for a given SNR.
Following the VC scenario, we use randomly generated channel matrices for BER estimation,
where a channel matrix is kept constant for consecutive $D$ trials.

TI-detector was implemented by PyTorch 0.4.0 \cite{PyTorch}.
The following numerical experiments were carried out on a PC with GPU NVIDIA GerForce GTX 1080 and Intel Core i7-6700K CPU 4.0GHz $\times$ 8.
In this paper, a training process is executed with $T=50$ rounds using the Adam optimizer~\cite{Adam} with learning rate $\epsilon$.
A training process takes within 25 minutes under our environment.
The hyperparameters of TI-detector  are $(D,K,\epsilon) = (1250,2000,0.025)$ for $(N,M)=(200,128)$
 and $(1000,1500,0.025)$ for $(300,192)$.
To calculate the BER of TI-detector, a sign function $\mathrm{sgn}(z)$
which takes $-1$ if $z\le 0$ and $1$ otherwise is applied to the final estimate $s_{T+1}$.

As the baselines of detection performance, we use
the IW-SOAV  \cite{IW-SOAV2}  and the standard MMSE detector.
The IW-SOAV is a double loop algorithm
whose inner loop is the W-SOAV optimization recovering a signal using a proximal operator.
Each round of the W-SOAV takes $O(MN)$ computational steps, which is comparable to that of TI-detector.
After finishing an execution of the inner loop with $K_{\mathrm{itr}}$ iterations,
several parameters are then updated in a re-weighting process based on a tentative recovered signal.
This procedure is repeated $L$ times in the outer loop.
The total number of steps of the IW-SOAV is thus $K_{\mathrm{itr}} L$.
In the following, we use the simulation results in~\cite{IW-SOAV2} with $K_{\mathrm{itr}}=50$.

\subsection{Main results}

We first present the BER performance of each detector as a function of SNR for $(N,M)=(200,128)$
in Fig.~\ref{fig:BER_N200}.
The results show that the MMSE detector fails to detect transmitted signals reliably (BER $\simeq 10^{-1}$)
because the system is underdetermined.
TI-detector exhibits the BER performance superior to that of the IW-SOAV ($L=1$),
i.e., TI-detector achieves approximately $5$ dB gain at $\mathrm{BER}=10^{-4}$ over the IW-SOAV ($L=1$).
Note that the computational cost for executing TI-detector with $T=50$ is almost comparable
to that of the IW-SOAV ($L=1$).
More interestingly, the BER performance of TI-detector
is fairly close to that of IW-SOAV ($L\ge 2$).
For example, with $\mathrm{SNR}=20$ dB, the BER estimate of TI-detector
is $4.3 \times 10^{-5}$ whereas that of the IW-SOAV ($L=2$)
is $6.8 \times 10^{-5}$.  It should be noted that the total number iterations of the IW-SOAV ($L=2$) is 100.

Figure \ref{fig:BER_N300} shows the BER performance for $(N,M)=(300,196)$.
TI-detector successfully recovers transmitted signals with lower BER than that of the IW-SOAV($L=1$).
It again achieves about $5$ dB gain against the IW-SOAV($L=1$) at $\mathrm{BER}=10^{-5}$.
Although the IW-SOAV ($L \ge 2$) shows considerable performance improvements in this case,
the gaps between the curves of TI-detector and the IW-SOAV ($L \ge 2$)  are within 2 dB at $\mathrm{BER}=10^{-5}$.

Figure~\ref{fig:parameter} displays the learned parameters {$\{\gamma_t,|\theta_t|\}$} of TI-detector
after a training process as a function of a layer index $t(=1,\dots,T)$.
We find that they exhibit a zigzag shape with damping amplitude similar to that observed in TISTA~\cite{TISTA}.
The parameter $\gamma_t$, the step size of a linear estimator, is expected to accelerate the convergence of the signal recovery.
Theoretical treatments for providing reasonable interpretation on
these characteristic shapes of the learned parameters
are interesting open problems.

\begin{figure}[tb]
  \centering
  \includegraphics[width=8.5cm]{./200_128_SNR.pdf}
  \caption{BER performance for $(N,M) = (200,128)$.}
  \label{fig:BER_N200}
\end{figure}
\begin{figure}[tb]
  \centering
  \includegraphics[width=8.5cm]{./300_192_SNR.pdf}
  \caption{BER performance for $(N,M) = (300,192)$. }
  \label{fig:BER_N300}
\end{figure}
\begin{figure}[tb]
  \centering
  \includegraphics[width=8.5cm]{./Parameter_N300_SNR20.pdf}
  \caption{Sequences of learned parameters $\gamma_t$ (upper) and $|\theta_t|$ (lower);
  $(N,M)=(300,192)$, $\mathrm{SNR}=20$[dB], $1\le t\le T=50$.}
  \label{fig:parameter}
\end{figure}

\section{Concluding remarks}

In this paper, we proposed TI-detector, a deep learning-aided iterative decoder for massive overloaded MIMO channels.
TI-detector contains two trainable parameters for each layer: $\gamma_t$ controlling a step size of the linear estimator and
 $\theta_t$ dominating strength of the nonlinear estimator.
The total number of the trainable parameters in $T$ layers
is thus $2T$, which is significantly smaller than that used in the previous studies such as \cite{DMD1,DMD2}.
This fact promotes fast and stable training processes for TI-detector.
The numerical simulations show that TI-detector outperforms the state-of-the-art IW-SOAV ($L=1$) by a large margin and
achieves a comparable detection performance to the IW-SOAV ($L\ge 2$).
TI-detector therefore can be seen as a promising iterative detector for overloaded MIMO channels
providing an excellent balance between a low computational cost and a reasonable detection performance.
There are several open problems regarding this study.
Fast, adding a re-weighting process similar to the one used in the IW-SOAV ($L\ge 2$)
to TI-detector seems an interesting direction to improve the detection performance.
Secondly, enhancing TI-detector toward a large constellation such as QAM is a practically important problem.

\section*{Acknowledgement}
The authors are very grateful to Mr.~Ryo Hayakawa and Prof.~Kazunori Hayashi
for providing their numerical results and simulation programs~\cite{IW-SOAV2}.

\vfill\pagebreak


TITLE: Differentiability in $R^2$
QUESTION [2 upvotes]: Let $U=\{(x,y) \text{ in } \mathbb{R}^2 : x_2 + y_2 < 4\}$, and let $f(x,y)= \sqrt{4-x_2-y_2}$.
  Prove that $f$ is differentiable, and find its derivative.

I do know how to prove it is differentiable at a specific point in $\mathbb{R}^2$, but I could not generalize it to prove it differentiable on $\mathbb{R}^2$. Any hint?

REPLY [0 votes]: A function from $\mathbb{R^{2}}\to\mathbb{R}$ is said to be differentiable at a point $(a,b)$ iff there exist $\phi(h,k)$ and $\psi(h,k)$ from $\mathbb{R}^{2}\to\mathbb{R}$ such that
$f(a+h,b+k)=h\frac{\partial f}{\partial x}\vert_{(a,b)}+k\frac{\partial f}{\partial y}\vert_{(a,b)}+h\phi(h,k)+k\psi(h,k)$ such that  $\phi(h,k)\to 0\quad \text{and}\quad\psi(h,k)\to 0$ as $(h,k)\to(0,0)$.
The existence and continuity of the partial derivatives give only a sufficient condition for differentiability.
For example $f(x,y)=x^{2}\sin(\frac{1}{x})+y^{2}\sin(\frac{1}{y})$ is differentiable . But the partial derivatives are not continuous. To see that it is differentiable it is easy to note that the partial derivatives at $(0,0)$ is $0$. And take $\phi(h,k)=h\sin(\frac{1}{h})$ and $\psi(h,k)=k\sin(\frac{1}{k})$.


TITLE: When is $X^n-a$ is irreducible over F?
QUESTION [24 upvotes]: Let $F$ be a field, let $\omega$ be a primitive $n$th root of unity in an algebraic closure of $F$. If $a$ in $F$ is not an $m$th power in $F(\omega)$ for any $m\gt 1$ that divides $n$, how to show that $x^n -a$ is irreducible over $F$?

REPLY [20 votes]: Below is a classical result:
Theorem $\ $ Suppose $\,c\in F\,$ a field,  and $\,0 < n\in\mathbb Z$.
$\ \ \ x^n - c\ $ is irreducible over $F\! \iff c \not\in F^p\,$ for all primes $\,p\mid n,\,$ and $\ c\not\in -4F^4$ when $\, 4\mid n$
A proof is in many Field Theory textbooks, e.g. Karpilovsky, Topics in Field Theory, Theorem 8.1.6.


TITLE: Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions?
QUESTION [42 upvotes]: The title pretty much says it all.  I am revisiting complex analysis for the first time since I "learned" some as an undergraduate.  I am trying to wrap my head around why it should be the case that a function which is differentiable once should be differentiable twice.  I know a proof (Use Cauchy integral formula and differentiate under the integral sign), but that proof doesn't do a whole lot to explain the magic.
Say you had never heard of complex numbers before, and someone told you that you had a function $f: \mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x,y) = (u(x,y), v(x,y))$, which locally looks like a rotation/expansion (i.e. it satisfies the Cauchy-Riemann equations everywhere).  Then why on Earth should this function be $C^\infty$?  I would love to see a picture, or just a proof that makes this feel less like a magic trick.  I hoped that "Visual Complex Analysis" would help me out here, but this seems to be the one theorem in the book which is not given a geometric motivation.

REPLY [2 votes]: This appears to be the earliest "topological" proof:
A TOPOLOGICAL  PROOF OF  THE CONTINUITY OF  THE
DERIVATIVE OF A FUNCTION OF A COMPLEX
VARIABLE
BY ROBERT L. PLUNKETT
1958
http://www.ams.org/journals/bull/1959-65-01/S0002-9904-1959-10251-2/S0002-9904-1959-10251-2.pdf
G.T. Whyburn himself (whose research was foundational to the result) gives his own proof, allegedly inspired by Connell and Porcelli:
Developments in topological analysis, 1961
http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50125.pdf


TITLE: Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$
QUESTION [1 upvotes]: This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147).  I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \frac{Dx+E}{(x^2+x+1)^2}$$
then:
$$3x^2+2x+1 = A(x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
and I can solve for A by setting $x=-2$, which yields $A=1$.
I know the final answer is this:
$$\frac{1}{x+2} - \frac{x-1}{x^2+x+1} - \frac{1}{(x^2+x+1)^2}$$
But I've worked this problem many ways and cannot make progress on the numerators for the other fractions.

REPLY [2 votes]: I'll expand all the things from the comment.
Method 1:
By expanding we get
$$
\begin{cases}
A+B=0&&\hbox{ for }x^4\\\
2A+3B+C=0&&\hbox{ for }x^3\\\
3 A + 3 B + 3 C + D - 3=0&&\hbox{ for }x^2\\\
2 A + 2 B + 3 C + 2 D + E - 2=0&&\hbox{ for }x\\\
A + 2 C + 2 E - 1=0&&\hbox{ for }1
\end{cases}
$$
which boils down to $A = 1, B = -1, C = 1, D = 0, E = -1$.
Method 2 "substitution":
with $x=-\frac12\pm\frac{\sqrt{3}}{2}i$ we get rid of $(x^2+x+1)$, can have equations for $D,E$ and after finding $D,E$ and bringing them to the LHS we can cancel by $(x+2)$ which boils down to Mohammad Riazi-Kermani's answer


TITLE: How to find the point on a line closest to a set of n points in 2 space?
QUESTION [0 upvotes]: There are n points distributed in 2 space, with known coordinates. A straight line is drawn through the space.
How do I find the point on the line which minimizes the sum of the distances between itself and all other points?

REPLY [2 votes]: WLOG, the line is the axis $x$ (otherwise, translate/rotate).
You want to minimize
$$D(x):=\sum_{k=1}^n\sqrt{(x-x_k)^2+y_k^2}.$$
For this you need to solve
$$\frac{dD}{dx}=\sum_{k=1}^n\frac{x-x_k}{\sqrt{(x-x_k)^2+y_k^2}}=0,$$
that has no closed-form solution (except for $n=2$). The terms in this sum are sigmoids, all growing from $-1$ to $1$, so that there is a single root.
You can try with Newton's iterations, from the initial approximation $x=\overline{x}_k$, which is in fact the value that minimizes the sum of the squared distances (by $\dfrac{dD}{dx}=\displaystyle\sum_{k=1}^n(x-x_k)=0$).

In the case of $n=1$, the solution is obviously $x=x_0$, i.e. you project the point orthogonally onto the line.
In the case of $n=2$,
$$\frac{x-x_0}{\sqrt{(x-x_0)^2+y_0}}+\frac{x-x_1}{\sqrt{(x-x_1)^2+y_1}}=0$$
can be transformed to the quadratic equation
$$(x-x_0)((x-x_1)^2+y_1^2)-(x-x_1)((x-x_0)^2+y_0^2)
\\=(x_0-x_1)x^2+(y_1^2-y_0^2+x_1^2-x_0^2)x+x_1y_0^2+x_1x_0^2-x_0y_1^2-x_0x_1^2=0.$$
But an alien root is introduced.
Geometrically, the solution is immediate: if the points are on opposite sides, intersect the line and the segment between the two points; otherwise reflect one of the points.


TITLE: Can the powersets of two different cardinals be of the same cardinality?
QUESTION [1 upvotes]: Is the following statement consistent with ZF (obviously, this is not true in ZFC)?

There are cardinals $\kappa, \lambda$ such that $\kappa \neq \lambda$ and $2^\kappa = 2^\lambda$.

REPLY [6 votes]: Yes, this is possible, even with ZFC. In Cohen's model, where the continuum hypothesis fails, he gets ZFC plus $2^\omega=2^{\omega_1}$. This statement is known as Luzin's hypothesis, an alternative to CH.
More generally, the situation for regular cardinals is described by Easton's theorem, which says that if ZFC plus the GCH holds, then for any function $E:\kappa\mapsto E(\kappa)$ on the regular cardinals satisfying the properties that $\kappa\leq\lambda\to E(\kappa)\leq E(\lambda)$ and $\kappa<\operatorname{cof}(E(\kappa))$, then there is a forcing extension in which $2^\kappa=E(\kappa)$.
In particular, one can easily make functions $E$ that will thereby ensure ZFC plus many instances of $2^\kappa=2^\lambda$.


TITLE: Integral Convergence test for Sum
QUESTION [2 upvotes]: I'm trying to implement the integral test for sum convergence on the following series:
$$
\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\cdot e^{\sqrt{n}}}
$$
I've proved that the function $f:[1,\infty]\rightarrow \mathbb{R} | f(x)=\dfrac{1}{\sqrt{x}\cdot e^{\sqrt{x}}}$ is monotonically descending and positive for all $x\ge 1$,
and also that the improper integral
$\int_{1}^{\infty}\dfrac{1}{\sqrt{x}\cdot e^{\sqrt{x}}}dx$ converges,  therefor the sum  stated above is convergent.
However, according to an online calculator, the sum $\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\cdot e^{\sqrt{n}}}$ is divergent, which makes no sense to me.
(said calculator only provides further information for money, as a student, I cannot afford such luxuries.)
Currently I am uncertain in my solution.
Any assistance/guidance is happily accepted!

REPLY [2 votes]: For any $ X>1$,
$$F(X)=\int_1^X\frac{dx}{2\sqrt{x}e^{\sqrt{x}}}$$
$$=\Big[-e^{-\sqrt{x}}\Big]_1^X$$
$$=e^{-1}-e^{-\sqrt{X}}$$
$$\lim_{X\to\infty}F(X)=e^{-1}$$
the integrale is surely convergent.

REPLY [0 votes]: Tell me I'm not blind, please.


TITLE: Find functions $f,g$ such that $f(x+y)=g(x·y)$ for all $x,y$?
QUESTION [2 upvotes]: My approach towards this question was that first I putted x=0 and then y=0 which yields f(x)=f(y)and f(x)=g(0), and again for x=1 & y=1 it gives g (x)=f(1+x), and so on. My query is that what should the compact & rigorous form of proof or solution looks like because I found myself stuck between results and not comprehending results very well. Thanks.

REPLY [7 votes]: $$f(x) = f(x + 0) = g(x \cdot 0) = g(0)$$ for all $x$.  Thus $f(x)$ is constant and always equal to $g(0)$.  Moreover, $$g(0) = f(1 + y) = g(1 \cdot y) = g(y)$$ for all $y$.  Thus $g$ is also constant and equal to $g(0)$.  Thus $f, g$ are the same constant function.  All constant functions $f = g = c$ will satisfy the relation.  This is all you need.


TITLE: PDF of an exponential distribution with varying paramter, lambda
QUESTION [1 upvotes]: Suppose that the lifetime of a device is exponential with rate λ, but suppose
also that the value of λ is not fixed but is itself a random variable that is
uniform in the range [a, b) with 0 < a.
Can the pdf of the lifetime of the device be written as:
$ \lambda = \begin{cases} 0, x < a\\ \dfrac {1} {b-a} , a\leq x < b\\ 0,  x\geq b \end{cases} $

REPLY [1 votes]: Let $T|\lambda \sim \text{Exp}(\lambda)$ be the lifetime of the device and let $\lambda \sim \text{U}[a,b)$ be the parameter.  Then for all $t \geqslant 0$ you have:
$$\begin{equation} \begin{aligned}
f_T(t)
&= \int \limits_a^b p(T=t|\lambda) \pi(\lambda) d\lambda \\[6pt]
&= \frac{1}{a-b} \int \limits_a^b \exp(-\lambda t) d\lambda \\[6pt]
&= \frac{1}{a-b} \Bigg[ - \frac{1}{t} \cdot \exp(-\lambda t) \Bigg]_{\lambda=a}^{\lambda=b} \\[6pt]
&= \frac{1}{a-b} \Bigg[ \frac{\exp(-at) - \exp(-bt)}{t} \Bigg] \\[6pt]
&= \frac{e^{-at} - e^{-bt}}{a-b} \cdot  \frac{1}{t}. \\[6pt]
\end{aligned} \end{equation}$$