text:
\subsection{Main case: \texorpdfstring{$c_l\neq 0$ or $c_g\neq 0$}{at least one out of the two values cl and cg is not equal 0}}
\label{sec:maincase}
If both parameters $c_l$ and $c_g$ would be zero then PSO performs a deterministic movement independent of the objective function. As this is not profitable we exclude this case for our main analysis and move the respective discussions to Section~\ref{sec:c0}.

Similarly as in \cite{Ozcan99} we split the positions of the particles
into an angle and a magnitude.

Here we first describe how it is possible to define the current state by the
current position and some current angle instead of the current solution. We
present an iterative procedure to evaluate the distribution of this angle,
which is even independent of the current position. The distribution on this
angle enables us to evaluate the expected change in the convergence indicator
$\E[\Phi_{t+1}-\Phi_t]$.

If no atypical initialization is used then $x_t$ almost
surely~(in the mathematical sense, see~\cite{durrett2010probability})
has a value which is not \emph{exactly} equal to $0$, i.\,e., $\Pr[x_t \neq 0]=1$.
Excluding the event that $x_t=0$ for any $t$, we can specify $v_t$ by $x_t$
and an unique additional angle
$\alpha_t\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
such that $v_t=x_t\cdot\tan(\alpha_t)$.
Then $v_{t+1}=x_t\cdot(\chi\tan(\alpha_t)-H_{t+1})$ and
$x_{t+1}=v_{t+1}+x_t=x_t\cdot(\chi\tan(\alpha_t)+1-H_{t+1})$, where
$H_{t+1}=c_l r_{t+1}+c_g s_{t+1}$ and $s_{t+1},r_{t+1}$ are independent and uniformly
distributed over $[0,1]$ as already specified.
Therefore
\begin{equation}
  \tan(\alpha_{t+1})
=\frac{v_{t+1}}{x_{t+1}}
=\frac{\chi \tan(\alpha_t)-H_{t+1}}{\chi\tan(\alpha_t)+1-H_{t+1}}
=1-\frac{1}{1+\chi \tan(\alpha_t)-H_{t+1}}=f^+(\tan(\alpha_{t}),H_{t+1}),
  \label{eq:fplusA}
\end{equation}
where
\begin{equation}
f^+(m,h):=1-\frac{1}{1+\chi m-h}
  \label{eq:fplusB}
\end{equation}

This means that the new angle is only dependent on the old angle and the random
numbers of the current iteration but it is not dependent on the old position or
velocity. $\chi$ does not appear as parameter for $f^+$ as it is a constant.

For any $H_t=c_l\cdot r_t+ c_g\cdot s_t$ we can explicitly specify the probability density function.
Let $l_H:=\min(0,c_l)+\min(0,c_g)$ which is the \textbf{l}ower bound on the values which could be observed while evaluating $H_t$
and $u_H:=\max(0,c_l)+\max(0,c_g)$ which is the respective \textbf{u}pper bound.
Furthermore, let $c_{\min}:=\min(\vert c_l\vert, \vert c_g\vert)$ and
$c_{\max}:=\max(\vert c_l\vert, \vert c_g\vert)$.
Then the probability density function of $H_t$, where each $H_t$ is the
sum of two independent and uniformly distributed random variables, is
\begin{equation}
f_H(h):=\begin{cases}
  \frac{h-l_H}{\vert c_l\cdot c_g\vert} & \text{if }h\in (l_H,l_H+c_{\min})\\
  \frac{1}{c_{\max}} & \text{if }h\in [l_H+c_{\min},u_H-c_{\min}]\\
  \frac{u_H-h}{\vert c_l\cdot c_g\vert} & \text{if }h\in (u_H-c_{\min},u_H)\\
  0&\text{otherwise.}
\end{cases}
  \label{eq:densityH}
\end{equation}
Please note that this probability density function is also correct if one out of the two values $c_l$ and $c_g$ is zero.

Let $F_{\triangle,t}$ be the cumulative distribution function of
$\alpha_t$.
$F_{\triangle,t}$ is a continuous function for any $t>0$ as
\begin{align}
  &\lim_{\delta\rightarrow 0}\left(F_{\triangle,t+1}(\beta+\delta)-F_{\triangle,t+1}(\beta-\delta)\right)
  =\Pr[\alpha_{t+1}=\beta]
  \\&\overset{m=\tan(\beta)}{=}\int_{l_H}^{u_H}\Pr[f^+(\tan(\alpha_t),h)= m]f_H(h)\di h=0.
  \label{eq:probabilityexactvalue}
\end{align}
This integral has the value zero, because $f^+(m,h)$ is injective in respect to
$h$ and therefore the value $\Pr[f^+(\tan(\alpha_t),h)= m]$ can be positive (the value
can be at most one) only for a countably finite set of values for $h$, which
has no influence on the value of the integral.

For $t=0$ the cumulative distribution function $F_{\triangle,t}$ might not be a
continuous function as there exists for example the standard idea to use zero
as initial value for the velocity resulting in $\alpha_0=0$ deterministically.
This would result in $F_{\triangle,t}(\beta)=1$ if $\beta\geq 0$ and
$F_{\triangle,t}(\beta)=0$ otherwise, which is not continuous.

$F_{\triangle,t}$ is by the definition of cumulative distribution functions a monotonic function.
In combination with the property that $F_{\triangle,t}$ is continuous (for
$t>0$) we obtain by a theorem of Lebesgue (a proof can be found, e.\,g.,
in~\cite{B2003}) that $F_{\triangle,t}$ is
differentiable almost everywhere. We assume to have no pathological example
similar to the Cantor (ternary) function~\cite{cantor1884}.
Therefore it is assumed that this differentiability almost everywhere can be
extended to a derivative of $F_{\triangle,t}$, which will be denoted as
$f_{\triangle,t}$.
The function $f_{\triangle,t}$ is then the probability density function for $\alpha_t$ and any $t>0$.
In practice (see Section~\ref{sec:numeric}) this probability density function
exists and it is also a continuous function.

$F_{\triangle,t+1}$ can be calculated using $f^+$ from
Equation~\eqref{eq:fplusB} and in a second step $F_{\triangle,t}$ by the
following formula
\begin{equation}
  F_{\triangle,t+1}(\beta)=\Pr[\alpha_{t+1}\leq\beta]\overset{m=\tan(\beta)}{=}\int_{l_H}^{u_H}\Pr[f^+(\tan(\alpha_t),h)\leq m]f_H(h)\di h,
  \label{eq:alphadistributioniteration}
\end{equation}
where
\begin{align}&
  \Pr[f^+(\tan(\alpha_t),h)\leq m]=\nonumber\\&=\begin{cases}
	\begin{array}{@{}l@{} l l@{}}
	  1&=1&\text{if }\chi=0\wedge1-\frac{1}{1-h}\leq m\\
	  0&=0&\text{if }\chi=0\wedge1-\frac{1}{1-h}> m\\
	  \Pr[\alpha_t >  \gamma_1]&=1-F_{\triangle,t}(\gamma_1)&\text{if }m=1\wedge \chi>0\\
	  \Pr[\alpha_t <  \gamma_1]&=F_{\triangle,t}(\gamma_1)&\text{if }m=1\wedge \chi<0\\
	  \Pr[\alpha_t\not\in[\gamma_{\min},\gamma_{\max}]]&=1-F_{\triangle,t}(\gamma_{\max})+F_{\triangle,t}(\gamma_{\min})&\text{if }m>1\\
	  \Pr[\alpha_t\in[\gamma_{\min},\gamma_{\max}]]&=F_{\triangle,t}(\gamma_{\max})-F_{\triangle,t}(\gamma_{\min})&\text{if }m<1\\
  \end{array}
  \end{cases}
  \label{eq:probabilitypropagation}
\end{align}
and
\begin{align}&
  \gamma_1=\arctan\left(\frac{h-1}{\chi}\right)
  ,\quad \gamma_2=\arctan\left( \frac{h-1}{\chi}+\frac{1}{\chi\cdot(1-m)} \right)
  ,\\& \gamma_{\min}=\min(\gamma_1,\gamma_2)
  ,\quad \gamma_{\max}=\max(\gamma_1,\gamma_2).
  \label{eq:gammadefinitions}
\end{align}

This result can be obtained by the following rearrangements (* Please note that
for the final simplifications $\leq$ and $<$ are arbitrarily replaced as well
as $\geq$ and $>$, because the event that a single value is exactly achieved is
either zero in our setting (if $t>0$) or using a different value has no effect
as argued after Equation~\eqref{eq:probabilityexactvalue}.
Also we use that $\gamma_1\leq\gamma_2\Leftrightarrow \chi\cdot(1-m)>0$ in the final step):
\begin{align}
  &\Pr[f^+(\tan(\alpha_t),h)\leq m]\\&=\Pr\left[ 1-\frac{1}{1+\chi\tan(\alpha_t)-h}\leq m\right]\\
  &=\Pr\left[
	\begin{array}{l}
	  \left(\chi=0\wedge1-\frac{1}{1-h}\leq m\right)\\
	  \lor(1-m=0\wedge 0\leq \frac{1}{1+\chi\tan(\alpha_t)-h})\\
	  \lor\left(1-m>0\wedge 0\leq \frac{1}{1+\chi\tan(\alpha_t)-h} \wedge 1-m\leq\frac{1}{1+\chi\tan(\alpha_t)-h}\right)\\
	  \lor\left(1-m<0\wedge \left(0\leq \frac{1}{1+\chi\tan(\alpha_t)-h} \lor 1-m\leq\frac{1}{1+\chi\tan(\alpha_t)-h}\right)\right)
	\end{array}
  \right]\\
  &=\Pr\left[
	\begin{array}{l}
	  \left(\chi=0\wedge1-\frac{1}{1-h}\leq m\right)\\
	  \lor(1-m=0\wedge 0< {1+\chi\tan(\alpha_t)-h})\\
	  \lor\left(1-m>0\wedge 0< {1+\chi\tan(\alpha_t)-h} \wedge 1-m\leq\frac{1}{1+\chi\tan(\alpha_t)-h}\right)\\
	  \lor\left(1-m<0\wedge \left(0< {1+\chi\tan(\alpha_t)-h} \lor 1-m\leq\frac{1}{1+\chi\tan(\alpha_t)-h}\right)\right)
	\end{array}
  \right]\\
  &=\Pr\left[
	\begin{array}{l}
	  \left(\chi=0\wedge1-\frac{1}{1-h}\leq m\right)\\
	  \lor(1-m=0\wedge\chi>0\wedge \alpha_t>\gamma_1)\\
	  \lor(1-m=0\wedge\chi<0\wedge \alpha_t<\gamma_1)\\
	  \lor\left(1-m>0\wedge \chi>0\wedge \alpha_t>\gamma_1 \wedge \alpha_t\leq\gamma_2\right)\\
	  \lor\left(1-m>0\wedge \chi<0\wedge \alpha_t<\gamma_1 \wedge \alpha_t\geq\gamma_2\right)\\
      \lor\left(1-m<0\wedge \chi>0\wedge \left(\alpha_t>\gamma_1 \lor   \alpha_t\leq\gamma_2\right))\right)\\
      \lor\left(1-m<0\wedge \chi<0\wedge \left(\alpha_t<\gamma_1 \lor   \alpha_t\geq\gamma_2\right))\right)
	\end{array}
  \right]\\
  &\overset{*}{=}\Pr\left[
	\begin{array}{l}
	  \left(\chi=0\wedge1-\frac{1}{1-h}\leq m\right)\\
	  \lor(1-m=0\wedge\chi>0\wedge \alpha_t>\gamma_1)\\
	  \lor(1-m=0\wedge\chi<0\wedge \alpha_t<\gamma_1)\\
	  \lor\left(1-m>0\wedge \gamma_{\min}<\alpha_t<\gamma_{\max}\right)\\
	  \lor\left(1-m<0\wedge \left(\alpha_t<\gamma_{\min}\lor \gamma_{\max}<\alpha_t\right)\right)
	\end{array}
  \right]
  \label{eq:rearrangements}
\end{align}

Finally, we return to our initial aim to determine whether the convergence indicator converges to $-\infty$.
The differences of subsequent convergence indicators are only
dependent on the current angle, which we can see in the following series of
equations
\begin{align}
  &\E[\Phi_{t+1}-\Phi_t]=\E\left[\ln\left(x_{t+1}^2+v_{t+1}^2\right)-\ln\left({x_t^2+v_t^2}\right)\right]
=\E\left[\ln\left(\frac{x_{t+1}^2+v_{t+1}^2}{x_t^2+v_t^2}\right)\right]
\\
=&\E\left[\ln\left(\frac{ (\chi\tan(\alpha_t)-H_{t+1})^2 + (\chi\tan(\alpha_t)-H_{t+1}+1)^2}{1+\tan(\alpha_t)^2}\right)\right]
\\
=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\E\left[\ln\left(\frac{ (\chi\tan(\alpha)-H_{t+1})^2+(\chi\tan(\alpha)-H_{t+1}+1)^2}{1+\tan(\alpha)^2}\right)\right]f_{\triangle,t}(\alpha)\di\alpha,\label{eq:expectedxvdifference}
\end{align}
where $f_{\triangle,t}$ is the probability density function for $\alpha_t$.

With the helping functions
\begin{equation}
  g(\alpha,h):=\ln\left(\frac{(\chi\tan(\alpha)-h)^2+(\chi\tan(\alpha)-h+1)^2}{1+\tan(\alpha)^2}\right)
  \quad
 f^{\emph{\scriptsize{in}}}(\alpha)=\int_{l_H}^{u_H}g(\alpha,h)\cdot f_H(h)\di h
  \label{eq:helpingfunction}
\end{equation}
we can shorten Equation~\eqref{eq:expectedxvdifference}. If we additionally
replace the expected value by the respective integral we obtain
\begin{equation}
  \eqref{eq:expectedxvdifference}=
  \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\int_{l_H}^{u_H}g(\alpha,h)\cdot f_H(h)\di h \right)f_{\triangle,t}(\alpha)\di \alpha
  =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f^{\emph{\scriptsize{in}}}(\alpha)\cdot f_{\triangle,t}(\alpha)\di \alpha .
  \label{eq:shorterexpectedxvdifference}
\end{equation}
In this final equation only $f_{\triangle,t}$ is changing for varying values of $t$.
The corresponding cumulative distribution function $F_{\triangle,t}$ on the angles $\alpha_t$ converge to a
unique stationary cumulative distribution function $F_{\triangle,\infty}$.
The reason for this cumulative distribution function to be unique is that for
the underlying process and for any starting angle $\beta$ the set of reachable
angles
$A_{\alpha_0=\beta,t}:=\lbrace \alpha\in(-\pi/2,\pi/2)\mid
\forall\varepsilon>0:
\Pr[\alpha_t\in(\alpha-\varepsilon,\alpha+\varepsilon)\mid \alpha_0=\beta]>0\rbrace$
converges to exactly the same set if $t$ tends to infinity.
If, e.\,g., $(0<\chi<1\wedge \max(c_l,c_g)>1)$, which is true for
many common parameter sets, this limit set contains all possible angles in
$(-\pi/2,\pi/2)$ as already $A_{\alpha_0=\beta,6}=(-\pi/2,\pi/2)$ for any $\beta\in(-\pi/2,\pi/2)$. In three
steps one can move from any starting angle to any open set containing angle
zero with positive probability.  In three further steps one can move with
positive probability from there to any other open set of angles.
To make this visible we define the function
$$f^-(m,h):=\frac{1}{\chi\cdot(1-m)}+\frac{h-1}{\chi}\enspace.$$
This function is the inverse function of $f^+$ in respect to $m$, i.\,e., $f^-(f^+(m,h),h)=m$ and also $f^-(\tan(\alpha_{t+1}),H_{t+1})=\tan(\alpha_t)$.
Then we can observe that the preimage of the angle zero is at least the set $[0,\pi/4]$ as $[0,1]\subset
f^-([0,0]\times(l_H,u_H))$ and $\arctan(1)=\pi/4$,
the preimage of the angles in $[0,\pi/4]$ is at least the set $[0,\pi/2)$ as
$[0,+\infty)\subset f^-([0,1]\times(l_H,u_H))$ and the preimage of the angles $[0,\pi/2)$ is the set of all possible angles in $(-\pi/2,\pi/2)$ as
$\R= f^-([0,+\infty)\times(l_H,u_H))$.
As $H_t$ has a probability density function which is positive in the interval
$(l_H,u_H)$, this justifies the claim that one can reach any open set
containing the angle zero with positive probability within exactly three
steps.
Additionally, from angle zero one can reach at least the angles in $(-\pi/2,0)$
as $(-\infty,0)\subset f^+([0,0]\times(l_H,u_H))$, from the angles $(-\pi/2,0)$
one can reach at least the angles in
$\left( (-\pi/2,0)\cup(\pi/4,\pi/2)\right)$ as
$\left((-\infty,0)\cup(1,+\infty)\right)\subset f^+( (-\infty,0)\times(l_H,u_H))$ and
from the angles in $\left( (-\pi/2,0)\cup(\pi/4,\pi/2)\right)$ one can reach all possible angles except exactly $\pi/4=\arctan(1)$ as
$(\R\setminus\lbrace 1\rbrace)= f^+(\left((-\infty,0)\cup(1,+\infty)\right) \times(l_H,u_H))$.
This justifies that one can reach any open set of angles from any open set containing angle zero with positive probability within exactly three steps.

Using
the associated stationary probability density function $f_{\triangle,\infty}$ in
Equation~\eqref{eq:shorterexpectedxvdifference} answers the question whether the particles converge.
\begin{theorem}
  \label{thm:classification}
  Let $\omega$ be the respective result of Equation~\eqref{eq:shorterexpectedxvdifference} using $f_{\triangle,\infty}$ as density function:
  \begin{equation*}
    \omega=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f^{\emph{\scriptsize{in}}}(\alpha)\cdot f_{\triangle,\infty}(\alpha)\di \alpha\enspace.
  \end{equation*}
  \begin{itemize}
    \item If $\omega<0$ then the particles position and velocity converge to the constant attractors.
    \item If $\omega>0$ then the particles position and velocity do not converge to the constant attractors.
  \end{itemize}
\end{theorem}
\begin{proof}
For any $\varepsilon>0$ there exists an iteration $t_0$ such that
$\vert\E[\Phi_{t+1}-\Phi_t]-\omega\vert<\varepsilon$ for all $t\geq t_0$ which
implies that
$\E[\Phi_t]
=\E[\Phi_0]+\sum_{i=0}^{t-1}\E[\Phi_{i+1}-\Phi_i]
\sim\omega\cdot t$.
Furthermore, the distribution of $(\Phi_t-\E[\Phi_t])/\sqrt{\Var[\Phi_t]}$
converges to a Gaussian/normal distribution as it can be written as sum of
consecutive convergence indicators and the covariance between
$\E[\Phi_{t_1+1}-\Phi_{t_1}]$ and $\E[\Phi_{t_2+1}-\Phi_{t_2}]$ tends to zero
if $\vert t_2-t_1\vert$ tends to $\infty$ (similar arguments apply as for the
reasoning on a unique limit distribution on $\alpha_t$).
The vanishing covariance also implies that the variance of $\Phi_t$ grows only linearly.
Therefore
$\E[\Phi_t]/\sqrt{\Var[\Phi_t]}$ grows or decreases in the order of $\sqrt{t}$
if $\omega>0$ or $\omega<0$ respectively.

These properties imply that if $\omega$ is negative then the probability
$\lim_{t\rightarrow\infty}\Pr[\Phi_t >\delta]=0$ for any value $\delta\in\R$.
Consequently the position and the velocity needs to converge to zero.

Analogously, if
$\omega$ is positive the particles positions and velocities will finally diverge.
\end{proof}
If the result $\omega$ evaluates to zero exactly then it has to be decided by different
means whether the particles do converge, do not converge or stay in some range
as also the initialization can have
an effect on this question in this case. Parameters, where the result is
exactly zero, are not recommended as also in cases where convergence still
appears, the time until sufficient optimization is accomplished would be quite
large as the drift of $\Phi_{t}$ to $-\infty$ is very slow and therefore also
the convergence of the position to the surroundings of the attractors.

Please note that by Theorem~\ref{thm:classification} also
Theorem~\ref{thm:figureClassification} is proven as $\omega$ is strictly less
than zero strictly between the red curve and the $\chi$-axis and $\omega$ is
strictly greater than zero strictly outside that region.

Theoretically at least the inner integral of
Equation~\eqref{eq:shorterexpectedxvdifference} can be solved analytically but
already displaying the respective solution, which can be obtained by maple or
other computer algebra systems, would require more than a page full of
formulas.
The question how this formula can be evaluated in practice is answered in Section~\ref{sec:numeric}.


TITLE: are these complex roots correct?
QUESTION [0 upvotes]: I have a doubt about this exercise, it's about finding complex roots of a a complex equation (raised to a power greater than 1). I don't know if this method is correct, even if the solutions seem right.
I need to solve this exercise $$z^2 = -6$$
my attempt:

I used an external variable $w$, $w = z^2$ $$ w = -6 $$
now, I found a, b, rho, and theta. (rho is the norm of the complex number, and theta is the angle between the hypotenuse and the x-axis). $$a = -6\space (a\space is\space the\space real\space part)$$ and $$b = 0 \space(the\space imaginary\space part) $$
and now rho, and theta $$rho = ||-6|| = \sqrt(-6^2) = 6$$ and $$theta = arctan(b/a) = 0$$

okay, now I rewrote it in trigonometric form, and raising to 1/2 (because 2 is the power).
$$w = \sqrt(6)*(cos(1/2 * 0+2k\pi)+sin(1/2 * 0+2k\pi)$$
and now I found easily complex roots, with k = 0, and with k = 1.
$z = \sqrt(6) \space \space \space \space z = -\sqrt(6)$.

REPLY [1 votes]: That is wrong, since the square of each of those two numbers is $6$, rather than $-6$.
You have $-6=6(\cos(\pi)+\sin(\pi)i)$, and therefore a square root of $-6$ is$$\sqrt6\left(\cos\left(\frac \pi2\right)+\sin\left(\frac\pi2\right)i\right)=\sqrt6i,$$and another square root is$$\sqrt6\left(\cos\left(\frac{3\pi}2\right)+\sin\left(\frac{3\pi}2\right)i\right)=-\sqrt6i.$$
Of course, you can also say that, since the square roots of $-1$ are $\pm i$, and since $-6=\sqrt6^2\times(-1)$, the square roots of $-$ are $\pm\sqrt6i$.


TITLE: If the Higgs Boson and Higgs Field could be manipulated, could that potentially make $m=0$ in all physics equations?
QUESTION [0 upvotes]: The Higgs Field and Higgs Boson are theorized to be responsible for mass, or $m$. If these could be manipulated, altered, or controlled for, if the Higgs Field could somehow be parted, would this result in $m=0$ in all physics equations?
Since the units of $c$ are just time and distance, would $c$ be affected?

This is NOT asking about redefining and applying the standard model universally, this is about a specific localized instance. The standard model with non-zero Higgs VEV still applies universally, so discussion of global effects is not relevant. This is a 'bubble', if you will, where there is a 'wave' (for a real groaner, a 'disturbance') in the Higgs field, such that the Higgs boson inside does not interact with the Higgs field outside. The field is manipulated, NOT made zero.
The Alcubierre drive has been accepted as theoretically plausible, this is related. Would the Higgs field be enclosed in this space-time bubble, or would it remain outside?
Yes, this would seem to break Lorentz symmetry, but the same rules would apply inside as outside, it's just that the outside is not the inside. That is, the Higgs field inside appears to move through the Higgs filed outside. Thus, m inside the bubble would appear to be zero from outside.

REPLY [1 votes]: I am replying at the level of a highschool student.
I will use your last comment:

The fact that c is the speed of SOMETHING is derived from working through the basic units of the equation e=mc^2, until you are left with c being distance over time, once all the other units are cancelled out.

This is a complicated way of looking at a dated equation, which is no longer used in particle physics since it is not a fundamental property but result of the mathematics. Fundamental are the four vectors in the pseudoeuclidean space of special relativity, space and time and energy and momentum. The inertial form can be derived from these simple assumptions

That is speed, but does it HAVE to be the speed of light? That it is the same value as the speed of light in a vacuum could be just an artifact.

The speed calculated from E=mc^2 is for an effective inertial mass of a massive macroscopic body. It is useful to show how a massive body behaves under large accelerations, but irrelevant for basic physics. The speed of light in vacuum is an axiom of special relativity, and special relativity has been validated innumerable times in nuclear and particle physics.

Does the Higgs field have a 'speed', or is it absolutely stationary?

Fields are mathematical constructs that fill up all space, all the elementary particles have a field over all space. A photon field, a neutrino field etc are hypothesized mathematically over all space. It is the creation and annihilation operators acting on these fields that have measurable meaning. Higgs bosons appear with the creation operators on the Higgs field. Fields are stationary by construction, like extra coordinate systems.

If there were no Higgs boson, EVERYTHING would travel at the speed of c.

If there were no Higgs boson a new theory would be needed. The fact that the Higgs boson has been found at LHC is a validation of the present standard model of particle physics.
At the universe at present the Higgs field gives mass only to the elementary particles of the table. The atoms and molecules we observe would still have an invariant mass due to the strong force in the nucleus and the rules of special relativity. Whether bound states would form is a moot question.
The only laboratory we have is the cosmological history of the universe, where due to the recovery of the symmetry, which is broken at the low energies we live in, the elementary particles have zero mass up to  ~10^-12 seconds after the big bang. The particles bouncing around there still have invariant masses when in pairs and systems.
To answer your question, no the Higgs field cannot be manipulated in our present universe, and it certainly has nothing to do with the speed of light. In this sense the symbol "m" is miss used in E=mc^2, it is a variable, it is just a mathematical result of the special relativity equations, which do assign an invariant mass to each four vector and thus to each particle.


text:
\section{Introduction}
A recent mathematical framework that ensures recovery of sparse vectors from incomplete information is \emph{Compressed sensing}. In this context, incomplete information  refers to the fact that linear systems of the form
$$
Ax_0=y
$$
can only be solved uniquely for general $x_0\in \R^N$ if $A\in \R^{M,N}$ is quadratic and invertible, i.e., if $M<N$ the information is incomplete. By imposing an a-priori structure on $x_0$, however, the ill-posed problem for $M<N$ can be turned into a well-posed one. An important structure of $x_0$ is that of \emph{sparsity}, which means that only a few entries of $x_0$ are different from zero. Another relevant assumption on $x_0$ is that its entries stem from a finite alphabet. In this paper we will study sparse signals whose entries stem from a binary alphabet, e.g., $x_0\in \{0,1\}^N$. Such signals appear for example in wireless communications, where the transmitted signals are sequences of bits. Moreover, in certain types of communication networks it is appropriate to assume that only a few transmitters are active at a certain instance, which naturally induces sparsity.

Note that binary signals are in particular symmetric in the sense, that if $x_0$ is a dense signal, $\mathds{1}-x_0$ is sparse. In the recent publication \cite{FlKei}, we have proven that using shifted random matrices such signals can be recovered from a particularly small number of measurements and that this number reflects the mentioned symmetry of $x_0$. This means that we need the same number of measurements to recover an $s$-sparse signal as we need to recover an $(N-s)$-sparse signal.

The considered measurement matrices, however, are of somewhat limited use in applications. The reasons are diverse. Often the design of the measurement matrix is given by the applications with little or even
no freedom to design it. Moreover, unstructured matrices, such as random or particularly Gaussian and Rademacher matrices, do not allow for a fast matrix multiplication, which may speed up recovery algorithms significantly. Beyond that storing a large unstructured matrix might be difficult. Hence, from a computational and application-technological point of view it would be desirable to use structured random matrices.

Up until now there are only few good recovery conditions for completely deterministic measurement matrices available. One, therefore, should allow for some randomness to come into play. In this work, we will consider biased partial random circulant matrices for the measurement process. A precise definition of such matrices will be given in Subsection \ref{sec:IntMainRes} (Equation \eqref{eqn:BiasedCirc}). A main difference is that those matrices depend on only one random vector and accordingly $N$ random variables, whereas (sub-) Gaussians either depend on $MN$ random variables (following the definition in \cite{FR}) or on $M$ random row vectors (following the definition in \cite{Tropp}).

The concern of this work is to proof recovery guarantees for binary, sparse signals from biased random partial circulant and Toeplitz measurements.

Partial random circulant matrices (centered and not biased) have already been successfully applied to the classical compressed problem (see for example \cite{StrctRau}) and in one-bit compressed sensing (\cite{DJR,DS}).
In \cite{DJR}, \cite{DS} and \cite{FKS}, the reconstruction of sparse signals from binary Gaussian circulant measurements was also considered. The main difference is that in those papers the measurements $y=\sign(Ax_0)$ are assumed to be binary whereas we assume that the signal $x_0$ itself is binary. Besides, the proof techniques are very different.

\subsection{Preliminaries}
To put our results in a precise setting we first aim to introduce some notation. We define $[N]:=\left\{1,\dots, N\right\}$ and denote the standard unit vectors with $e_i$ for $i\in [N]$, i.e., the vector which is zero everywhere except at the $i$-th entry it is equal to one. Further, we denote with $\one$ the matrix or vector, respectively, which is equal to one in each entry. For a subset $S\subset [N]$ and a vector $x=(x_1,\dots,x_N)=(x(1),\dots, x(N))\in \R^N$ the notation $x_S$ refers to the following vector
$$x_S(i)=\begin{cases}
x_i, & \text{if}\quad i\in S\\ 0, & \text{else}.
\end{cases}$$
Further $\|\cdot\|_0$ denotes the $\ell_0$-norm and $\|\cdot\|_p$ the $\ell_p$-norm for $p>0$, i.e., for $x=[x_1,\dots,x_N]\in \R^N$
$$\|x\|_0:=\abs{\left\{i: x_i\neq 0\right\}} \qquad \text{and} \qquad \|x\|_p^p:=\sum_{i=1}^N \abs{x_i}^p.$$

For a matrix $A\in \R^{M,N}$ the adjoint matrix is denoted by $A^*$. The notation $\supp(x)$ refers to the support of a vector $x\in\R^N$, i.e., to the set of non-zero entries of $x$.

For two real numbers $a,b\in \R$ we write $a\gtrsim b$ if there exists a constant $c>0$, which is independent from $a$ and $b$, such that $a\ge c b$.

Let $\Theta\subset [N]$ then we define $$i+\Theta:=\{i+k\imod{N}: k\in \Theta\}.$$
Moreover, for a linear operator $L:\R^N\to \R^M$ with matrix representation $L=[L_{i,j}]_{i,j=1}^{MN}$ the Hilbert-Schmidt norm of $L$ is denoted by $$\norm{L}_{HS}:=\sqrt{\tr(L^*L)}=\sqrt{\sum_{i=1}^M\sum_{j=1}^NL_{i,j}^2},$$ and $$\norm{L}:= \sup_{\norm{w}_{2} \leq 1} \norm{Lw}_{2}$$ is the operator norm of $L$. Note that the operator norm of $L:\R^N\to \R^N$ corresponds to the largest eigenvalue in absolute of $L$. Further note that for $A,B\in \R^{N,N}$ with $B=[b_1,\dots,b_N]$ it holds true that
\begin{align}\label{eqn:EstHS}
\|AB\|_{\text{HS}}^2=\sum_{i=1}^N\|Ab_i\|_2^2\le\sum_{i=1}^N \|A\|^2\|b_i\|_2^2=\|A\|\|B\|^2_{\text{HS}},
\end{align}
where we used the consistency of the operator norm with the Euclidean norm $\|\cdot\|_2$.

For $A\in \R^{M,N}$ and $S\subset[N]$ we let $A_S$ be either the matrix which consists of the rows of $A$ corresponding to the indices in $S$, i.e. $A_S\in \R^{s,N}$, or the matrix whose rows corresponding to the indices in $S$ equal those of $A$ and all other columns are zero, i.e., $A_S\in \R^{M,N}$. We further let $A^S\in \R^{M,s}$ or $A^S\in \R^{M,N}$ be the matrix whose columns corresponding to $S$ are deleted or substituted with zero-columns.

Finally, we aim to recall Gershgorin circle theorem, which will be an important tool for our proofs.

\begin{theorem}[Gershgorin circle theorem \cite{Gersch}]\label{thm:Gers}
Let $A=[a_{i,j}]_{i,j=1}^N\in \C^{N,N}$ and for $i\in[N]$ let $R_{i}=\sum _{j\neq {i}}\left|a_{ij}\right|$ be the sum of the absolute values of the non-diagonal entries in the $i$-th row.  Further define $D(a_{ii},R_{i})\subseteq \mathbb {C}$ be a closed disc centered at $a_{ii}$ with radius $R_{i}$. Every eigenvalue of $A$ lies then within at least one of the \emph{Gershgorin discs} $D(a_{ii},R_{i})$.
\end{theorem}

\subsection{Reconstruction of Binary Signals}
There are several compressed sensing approaches for the reconstruction of nonnegative sparse signals from random measurements \cite{Stojnic+,DT,JK}. As binary vectors are particularly nonnegative, those approaches can readily be applied to binary vectors. Let us therefore shortly review one of the approaches for the recovery of nonnegative signals.

It has become evident that basis pursuit restricted to the positive orthant $$\R^N_+:=\left\{x=(x_i)_{i=1}^N\in \R^N: x_i \ge 0, i\in [N]\right\}$$ has a strong performance at recovering nonnegative-valued sparse signals $x_0$ from the measurements $y=Ax_0$. This is the following program:
\begin{align}\label{P+}
\min\|x\|_1 \quad \text{subject to} \quad Ax=y \quad \text{and} \quad x\in \R^N_+,
\end{align}

Even so the mentioned approach is applicable to binary signals, it is already known that there are methods which yield even stronger recovery guarantees for binary signals. The canonical approach is to use the following adaptation of basis pursuit, to which one typically refers to as \emph{box-constrained basis pursuit}:
\begin{align}\label{Pbin}
\min\|x\|_1 \quad \text{subject to} \quad Ax=y \quad \text{and} \quad x\in [0,1]^N.
\end{align}
In \cite{Stojnic_Binary, KKLP}. the following equivalent condition for the success of \eqref{Pbin} has been shown. The vector $\one_S$, $S\subset [N]$, is the unique solution of \eqref{Pbin} if and only if
\begin{align}
\ker(A)\cap N^+\cap H_S=\{0\},
\end{align}
where $\ker(A)$ denote the nullspace of $A$,
\begin{align}
N^+:=\left\{w\in \R^N: \sum_{i=1}^N w_i\le 0\right\} \quad \text{and} \quad H_S:=\left\{w\in \R^N: w_i\le 0\text{ for } i \in S \text{ and } w_i\ge 0 \text{ for } i \notin S\right\}.
\end{align}

Least-squares on the other hand is an algorithm which is usually not well-applicable to sparse vectors since the $\ell_2$-norm does not promote sparsity. In \cite{FlKei}, however, it has been shown that least-squares with box-constraints works comparably well for binary-valued sparse signals, when reconstructing from biased sub-Gaussian measurements, even, if the measurements are contaminated with noise. \emph{Least-squares with box-constraints} is the following program
\begin{align}\label{LSBin}
\min\|Ax-y\|_2 \quad \text{subject to} \quad x\in [0,1]^N.
\end{align}
Note, that this fact has some practical impact. On the one hand least-squares is less complex and on the other hand it ensures a priori robustness in case of noisy measurements.

\subsection{Biased Random Matrices}
In compressed sensing the measurement matrix is often assumed to be (sub-) Gaussian, meaning that each entry of the measurement matrix $A\in\R^{M,N}$ is an independent copy of some (sub-) Gaussian. More precisely, we call  $A\in\R^{M,N}$ \emph{Gaussian}, if its entries are independently drawn from a renormalized normal distribution, i.e.,
\begin{align*}
A=\frac{1}{\sqrt{M}}\left[a_{i,j}\right]_{i,j=1}^{M,N} \qquad \text{with} \qquad a_{i,j}\sim \mathcal{N}(0,1).
\end{align*}
A more general type of measurement matrices are sub-Gaussians, whose entries follow a sub-Gaussian distribution:
\begin{definition}
Let $(\Omega, \Sigma, \mathbb{P})$ be a probability space. Further let $X:\Omega\to \R$ be a random variable. The \emph{sub-Gaussian} or \emph{Orclitz-2}-norm of $X$ is given by
\begin{align}
\|X\|_{\Psi_2} = \sup_{p\ge 1} p^{-\frac{1}{2}}\E\left(|X|^p\right)^{\frac{1}{p}}.
\end{align}
We call $X$ \emph{sub-Gaussian} if $\|X\|_{\Psi_2}<\infty$.
\end{definition}
A particular example for a sub-Gaussian matrix is a Rademacher matrix.  $A\in\R^{M,N}$ is called \emph{Rademacher matrix}, if its entries follow a Rademacher distribution, i.e., are chosen to be $-1$ or $1$ with equal probability:
\begin{align}
A=\frac{1}{\sqrt{M}}\left[a_{i,j}\right]_{i,j=1}^{M,N} \qquad \text{with} \qquad \mathbb{P}(a_{i,j}=1)=\mathbb{P}(a_{i,j}=-1)=\frac{1}{2}.
\end{align}
Rademacher variables $X$ are indeed sub-Gaussians with norm $\|X\|_{\Psi_2}=1$, because $|X|=1$. Further, note that sub-Gaussians are sometimes also defined by assuming that the rows are independent random vectors, which fulfill some specific properties such as sub-Gaussian marginals (cf. \cite{Tropp}). This basically is some generalization of the definition used in the underlying paper.

Sub-Gaussian matrices are often considered to model the measurement process. They have the specific property to be centered, which means that the expected value of each entry is $0$. However, it was shown in a recent work \cite{FlKei} that non-centered matrices have some advantage for the recovery of binary signals. Moreover, in \cite{JK}, a similar phenomenon was observed for the recovery of non-negative signals by \eqref{P+}. To be  more precise the following \emph{biased random matrices} have been considered in \cite{FlKei}:

\begin{align}\label{eqn:biased}
A=\mu\one+D,
\end{align}
where  $\mu\ge 0$ is a freely chosen parameter that controls the expected value of the entries, $\one \in \R^{M,N}$ is the matrix having only entries equal to one, and $D\in \R^{M,N}$ is assumed to be centered and having sub-Gaussian entries whose expected value is $0$.

Roughly speaking the following was proven for the recovery of binary, sparse signals from biased random measurements:

\begin{theorem}[Simplified Version of Theorem III.2 and III.8 of \cite{FlKei}]\label{thm:BiasedRand}
	Let $x_0\in \{0,1\}^N$ be a binary vector, and $A\in \R^{M,N}$ be a biased random matrix of the form \eqref{eqn:biased} with $\mu >0$, and $x_0\in \{0,1\}^N$ a $s$-sparse binary vector.
		\renewcommand{\labelenumi}{\roman{enumi})}
	\begin{enumerate}
		\item If $M$ is slightly larger than $N/2$, $x_0$ will be the unique solution of \eqref{Pbin} with high probability.
		\item Under the assumption
		\begin{align}
		M\gtrsim\max\left(\frac{R^2}{\mu^2},\min(s,N-s)\right)\log(N),
		\end{align}

		 the solution $x_*$ of \eqref{LSBin} for $y=Ax_0+n$ with $n\in \R^M$ and $\|n\|_2\le \eta$ obeys with high probability
		\begin{align}
		\|x_0- x_*\|_2\le \sqrt{\frac{\left(\frac{\sigma^2}{\mu^2}+32\min(s,N-s)\right)}{m\sigma^2}}\cdot \eta,
		\end{align}
		where $\sigma$ is the variance of the entries of $A$. Particularly, in the case of noiseless measurements, i.e., $\eta=0$, $x_0$ is the unique solution of \eqref{Pbin} and of \eqref{LSBin} with high probability.
	\end{enumerate}
\end{theorem}

\subsection{Main Result}\label{sec:IntMainRes}
As above-mentioned there are several applications where we do not have full freedom to design the measurement matrix. It is therefore of some importance to study structured random matrices. In applications such as radar or wireless communications the measurement process can be represented using partial random circulant matrices and partial random Toeplitz matrices  (cf. \cite{HBRN},\cite{JR}). In those applications binary sparse signals appear also in a natural way. The main goal of this work is therefore to prove comparable results as in \cite{FlKei} (see Theorem \ref{thm:BiasedRand}), for such matrices. Before stating our main results, let us introduce the considered matrices.

For $b=(b_0,b_1,\dots,b_{N-1})\in \R^N$ we define the associated \emph{circulant matrix} $\Phi=\Phi(b)\in \R^{N,N}$ by setting
\begin{align}
\Phi_{k,j}=b_{j-k\imod N } \qquad k,j\in [N].
\end{align}
Similarly, for a vector $c=(c_{-N+1},c_{-N+2},\dots, c_{N-1})\in \R^{2N-1}$ the associated \emph{Toeplitz matrix} $T=T(c)\in \R^{N,N}$ has entries
\begin{align}
T_{k,j}=c_{j-k} \qquad k,j\in [N].
\end{align}
For an arbitrary subset $\Theta\subset[N]$ of  cardinality $M<N$, we let the \emph{partial circulant matrix} $\Phi_{\Theta}=\Phi_{\Theta}(b)\in \R^{M,N}$, and the \emph{partial Toeplitz matrix} $T_{\Theta}$, respectively, be the submatrix of $\Phi$, and $T$ respectively, consisting of the rows indexed by $\Theta$.
In \cite{StrctRau} one can find a comprehensive overview of compressed sensing with structured random matrices. It is particularly shown that partial circulant matrices with Rademacher input vector $b$ work comparable well for the classical compressed sensing task as completely random sub-Gaussian matrices.

For our purpose we choose the vectors $b$ and $c$ to be sub-Gaussian sequences. Hence, the matrices $\Phi_{\Theta}$ and $T_{\Theta}$ are centered. Similarly to the results in \cite{FlKei}, we consider biased partial random matrices given by

\begin{align}\label{eqn:BiasedCirc}
A=A(b)=\mu \mathds{1}+\Phi_{\Theta}(b),
\end{align}
or
\begin{align}\label{eqn:BiasedToep}
A=A(c)=\mu \mathds{1}+T_{\Theta}(c).
\end{align}
Here, the parameter $\mu\ge0$ controls the expected value of the entries of $A$ and $\one\in \R^{M,N}$ is the matrix having all entries equal to one.

The main purpose of this paper is to prove the symmetric phase transition observed in \cite{FlKei}, for biased partial random matrices given by Equation \eqref{eqn:BiasedCirc}, meaning that we need as many measurements to recover sparse signals as we need to recover dense signals. The main result of this paper is the following theorem:

\begin{theorem}\label{thm:phaseTrans} Let $\mu>0$ and fix some tolerance $\varepsilon>0$. Let $A\in \R^{M,N}$ be a biased measurement matrix
		\renewcommand{\labelenumi}{\roman{enumi})}
		\begin{enumerate}
		\item  of the form \eqref{eqn:BiasedCirc}, where $b=[b_i]_{i=1}^N\in \R^N$ is a sub-Gaussian vector with $\erw{b_{i}}=0$, $\erw{b_{i}^2}=\sigma^2$, for $i\in [N]$, and sub-Gaussian norm $R$. Or
		\item  of the form \eqref{eqn:BiasedToep}, where $c=[c_i]_{i=-N+1}^{N-1}\in \R^{2N-1}$ is a sub-Gaussian vector with $\erw{c_{i}}=0$, $\erw{c_{i}^2}=\sigma^2$, for $i\in \{-N+1,\dots,N-1\}$, and sub-Gaussian norm $R$.
	\end{enumerate}
    A binary signal $x_0\in\{0,1\}^N$ with $\|x_0\|_0=s$  is the unique solution of \eqref{Pbin} and \eqref{LSBin} for $y=Ax_0$ with  probability larger than $1-\varepsilon$, provided
	\begin{align}\label{eqn1:main}
	M \gtrsim \max\left(\frac{R^2}{\mu^2}, \abs{\min(s, N-s)} \frac{2R^4}{\sigma^4}\right)\log \left(\frac{N}{\varepsilon}\right)
	\end{align}
	with a constant depending only on $\sigma$ and $\mu^{-1}$.

	\begin{enumerate}
		\item[iii)] 	Under the additional assumption $M \gtrsim \left(\frac{R}{\sigma}\right)^{4/3}\log(\varepsilon^{-1})$ the solution $x_*$ of \eqref{LSBin} for $y= Ax_0 + n$ with $\norm{n}_2 \leq \eta$ for some $\eta >0$ obeys
		\begin{align}\label{eqn:noisyMeas}
		\norm{x_0-x_*}_2 \leq \sqrt{\frac{9\left(\frac{16\sigma^2}{\mu^2}+ \min(s, N-s)\right)}{M \sigma^2}}\cdot \eta.
		\end{align}
	\end{enumerate}
\end{theorem}

config: arxiv


TITLE: Notation for ends of a string
QUESTION [2 upvotes]: I work now a lot with strings of characters and other finite sequences and found that I need many times a good notation for "cutting the end" a string. If $a$ is a finite sequence and $a'$ is its right end, then there is a sequence $x$ such that $a = x a'$ (where the product is defined by concatenation). What I would like to have is a suggestive notation for $x$: $x$ might be $a$ "minus" $a'$, or $a$ "divided by $a'$, if we view the set of all strings as a semigroup. I have therefore thought of writing $x$ as $a / a'$ but am not completely happy with it. Obvoiusly I need also a notation for "cutting at the left end"; by continuing with the previous idea the notation $a' \setminus a$ could stand for the solution of $a = a' x$, but it collides a bit with the notation for set difference.
So far I have thought, and my question is: Is there a good notation for these concepts already in use in some areas of mathematics or computer science, or has someone else already defined a suggestive notation?
P.S. And I also would like to have a notation for the "overlapping concatenation" of strings: If $a = a'x$ and $b = xb'$, what is $a'xb'$?

Addition, 2014:
After reading the answers so far, and not being satisfied with them, I came up with the following scheme:

Right ends: If $a = a' x$, then I write $a \mathrel{//} x$ to express that $x$ appears at the right end of $a$. I also write $a' = a / x$ for the result of cutting $x$ from the right end of $a$.
Left ends: If $a = x a'$, then I write $x \mathrel{\backslash\backslash} a$ to express that $x$ appears at the left end of $a$. I also write $a' = x \setminus a$ for the result of cutting $x$ from the left end of $a$.
Overlapping product: If $a = a'x$ and $b = x b'$ (or $a \mathrel{//} x \mathrel{\backslash\backslash} b$ in the new notation), then $a' x b' = a \mathbin{\langle x\rangle} b$.

The result is a workable notation and aesthetically pleasing (at least for me).

REPLY [6 votes]: A good reference is "G. Rozenberg, A. Salomaa Eds., Handbook of Formal Languages - Vol 1 Word Language Grammar,  Springer, 1997" in Chapter 6: Combinatorics of Words:
Let $\sum$ be a finite alphabet.
If $u$ is a word then $u=a_{1}\ldots a_{n}$, with
$a_{i}\in\sum$.
For a pair $\left(u,v\right)$ of words we define four relations:
$u$ is a prefix of $v$, if there exists a word $z$ such that $v=uz$;
$u$ is a suffix of $v$, if there exists a word $z$ such that $v=zu$;
$u$ is a factor of $v$, if there exists words $z$ and $z'$ such that $v=zuz'$;
$u$ is a subword of $v$, if $v$ as a sequence of letters contains
$u$ as a subsequence, i.e., there exist words $z_{1},\ldots,z_{t}$
and $y_{0},\ldots,y_{t}$ such that $u=z_{1}\ldots z_{t}$ and $v=y_{0}z_{1}y_{1}\ldots z_{t}y_{t}$.
Sometimes factors are called subwords, and then subwords are called
sparse subwords.
If $v=uz$ we write $u=vz^{-1}$ or $z=u^{-1}v$, and say that $u$
is the right quocient of $v$ by $z$, and that $z$ is the left quotient
of $v$ by $u$.
For the "overlapping concatenation" of
strings: $b'=x^{-1}b$ and $a'=ax^{-1}$. Then $a'xb'=ax^{-1}b$.


text:
\section{Archimedean Principle}
Tags: Real Analysis, Number Theory

\begin{theorem}
Let $x$ be a [[Definition:Real Number|real number]].
Then there exists a [[Definition:Natural Numbers|natural number]] greater than $x$.
:$\forall x \in \R: \exists n \in \N: n > x$
That is, the [[Definition:Natural Numbers|set of natural numbers]] is [[Definition:Bounded Above Set|unbounded above]].
\end{theorem}

\begin{proof}
Let $x \in \R$.
Let $S$ be the set of all [[Definition:Natural Numbers|natural numbers]] less than or equal to $x$:
:$S = \set {a \in \N: a \le x}$
It is possible that $S = \O$.
Suppose $0 \le x$.
Then by definition, $0 \in S$.
But $S = \O$, so this is a [[Proof by Contradiction|contradiction]].
From the [[Trichotomy Law for Real Numbers]] it follows that $0 > x$.
Thus we have the element $0 \in \N$ such that $0 > x$.
Now suppose $S \ne \O$.
Then $S$ is [[Definition:Bounded Above Set|bounded above]] (by $x$, for example).
Thus by the [[Continuum Property]] of $\R$, $S$ has a [[Definition:Supremum of Set|supremum]] in $\R$.
Let $s = \map \sup S$.
Now consider the number $s - 1$.
Since $s$ is the [[Definition:Supremum of Set|supremum]] of $S$, $s - 1$ cannot be an [[Definition:Upper Bound of Set|upper bound]] of $S$ by definition.
So $\exists m \in S: m > s - 1 \implies m + 1 > s$.
But as $m \in \N$, it follows that $m + 1 \in \N$.
Because $m + 1 > s$, it follows that $m + 1 \notin S$ and so $m + 1 > x$.
\end{proof}

config: wiki


TITLE: Distribution of collision times for 2-D diffusion in a area of radius $R$ with concentric disc of radius $r<R$ and absorbing boundaries
QUESTION [1 upvotes]: I'm trying to derive the time probability distribution function of collision times $p_R(t)$ and $p_r(t)$ for a time-zero uniform random concentration of $N$ diffusing particles with diffusion coefficient $D$ for a circular area of radius $R$ and concentric disc $r<R$. Once a particle hits a surface at $r$ or $R$ it is annihilated/absorbed.
$p_\rho(t)dt$ is the probability of a particle hitting and being absorbed by the circle of radius $\rho$ (where $\rho$ is outer $R$ or inner $r$) between time $t$ and $t+dt$. I've written a Monte Carlo engine to simulate but would like a closed-form solution.

REPLY [1 votes]: You want $\dot\rho = D\nabla^2 \rho$, where $\rho$ is the density; given the radial symmetry, $\nabla^2 \rho =\partial_r^2 \rho + \frac{1}{r}\partial_r \rho$.  Decaying eigenfunctions (such that $\dot\rho = -D k^2\rho$) will be solutions to $\partial_r^2 \rho + \frac{1}{r}\partial_r \rho + k^2 \rho = 0$, which are the ordinary Bessel functions of the first and second kinds, $J_0(kr)$ and $Y_0(kr)$.  (The Bessel functions of the second kind are singular at the origin, so we can only include them because we're working in $0 < R_{\text{min}} \le r \le R_{\text{max}}$.)  You need linear combinations that vanish at $R_{\text{min}}$ and $R_{\text{max}}$: $$a_k J_0(k R_{\text{min}})+b_k Y_0(k R_{\text{min}}) = a_k J_0(k R_{\text{max}})+b_k Y_0(k R_{\text{max}}) = 0.$$
Only for (infinitely many) special values of $k$ will the necessary condition be met for a nontrivial solution: $$J_0(kR_{\text{min}}) Y_0(kR_{\text{max}}) - Y_0(kR_{\text{min}}) J_0(kR_{\text{max}})=0.$$  The functions $a_k J_0(kr) + b_k Y_0(kr)$ at these special values will form an orthogonal basis; in particular, you'll find that
$$
\int_{R_{\text{min}}}^{R_{\text{max}}}(a_k J_0(kr) + b_k Y_0(kr))(a_l J_0(lr) + b_l Y_0(lr))\cdot r dr \propto \delta_{k,l}.
$$
You can use this to write your initial condition as $\rho(r, t=0)=\sum C_k (a_k J_0(kr)+b_k Y_0(kr))$ for some coefficients $C_k$, and then $\rho(r,t)=\sum C_k(a_k J_0(kr) + b_k Y_0(kr))\exp(-Dk^2t)$ for all later times.


TITLE: Showing that mean value theorem does not hold for $c(t)=(\cos t,\sin t, t)$
QUESTION [4 upvotes]: I am trying to show that the mean value theorem does not hold for the continuously differentiable function
$\qquad\qquad\qquad\qquad\qquad\qquad c:\mathbb{R}\to\mathbb{R}^3, \;\;c(t)=(\cos t,\sin t, t)$
This means that there exist $a,b\in \mathbb{R}$ for which there is no $\tau\in (a,b)$, such that
$\qquad\qquad\qquad\qquad\qquad\qquad c(b)-c(a)=(b-a)*c'(\tau)$
As there is $\cos$ and $\sin$ involved I chose $a=0$ and $b= \pi$. Hence:
$\qquad\qquad\qquad\qquad\qquad c(b)-c(a)=(-1, 0, \pi)-(1, 0, 0)=(-2, 0, \pi)$
$\qquad\qquad\qquad\qquad (b-a)*c'(\tau)=\pi * (-\sin\tau, \cos\tau, 1)=(-\pi\sin\tau, \pi\cos\tau, \pi)$
For $\pi\cos\tau$ to be equal to $0$, $\tau$ must be equal to a multiple of $\pi/2$, which means that $\pi\sin\tau = -\pi$, because $\sin \pi/2=1$.
$(a,b)=(0,\pi)\implies c(b)-c(a)=(-2, 0, \pi)\neq (-\pi,0,\pi)=(b-a)*c'(\tau)$
Therefore the mean value theorem does not hold for $c:\mathbb{R}\to\mathbb{R}^3, \;\;c(t)=(\cos t,\sin t, t)$
I'd appreciate if you could point out any mistakes in my working

REPLY [2 votes]: You need $t\in(a,b)$ such that
$$
\frac{1}{b-a}(c(b)-c(a))=c'(t)
$$
that is,
$$
\left(\frac{\cos b-\cos a}{b-a},\frac{\sin b-\sin a}{b-a},1\right)=(-\sin t,\cos t,1)
$$
With $a=0$ and $b=\pi$, this becomes
$$
\left(-\frac{2}{\pi},0,1\right)=(-\sin t,\cos t,1)
$$
which clearly fails, because $\cos t=0$ implies $\sin t=1$ in the specified interval.
More generally, the equality implies that
$$
(\cos b-\cos a)^2+(\sin b-\sin a)^2=(b-a)^2
$$
that simplifies to
$$
2-2\cos(b-a)=(b-a)^2
$$
If $k=(b-a)/2$, we get
$$
\sin^2k=k^2
$$
Note that, for $k>0$, $\sin k<k$. Therefore the mean value theorem fails on every interval.


TITLE: Find area of triangle ABC with $∠CAB = α $ and $∠ABC = β$
QUESTION [1 upvotes]: From the point M lying inside the triangle ABC, perpendiculars are drawn to the sides BC, AC, AB, the lengths of which are k, l and m, respectively. Find the area of a triangle ABC if ∠CAB = α and ∠ABC = β. α = π / 4, β = 2π / 3, k = 5, l = 3, m = 4. If the answer is not an integer, round it to the nearest integer.
I built circles on the segments MA, MB and MC as on diameters ... they will pass through the bases of the perpendiculars lowered to the sides ...
it turns out the task is to find the area of an inscribed quadrilateral along two sides and the angle between them, if the other two angles adjacent to these sides are straight lines ...
maybe there is some ready-made formula for the solution, but I don't know it ... but there is enough data

REPLY [1 votes]: Let [] denote areas. Consider the quadrilateral AYMZ
$$[AYMZ] = \frac12 m^2 \cot a_1 + \frac12 l^2 \cot a_2$$
$$l\sin a_1 = m \sin a_1, \>\>\>\>\> a_1+a_2 = \alpha$$
Eliminate $a_1$ and $a_2$ to get
$$[AYMZ]=\frac12(m^2+l^2)\cot \alpha + ml\csc \alpha$$
Similarly,
\begin{align}
[BZMX] &=\frac12(k^2+m^2)\cot \beta+ km\csc \beta\\
[CXMY] &=\frac12(l^2+k^2)\cot \gamma+ lk\csc \gamma\\
\end{align}
where $\gamma = \pi - (\alpha+\beta)$. Thus, the area of the triangle ABC is
\begin{align}
[ABC] &= [AYMZ] + [BZMX] + [CXMY] \\
& = \frac12[ (m^2+l^2)\cot \alpha+(k^2+m^2)\cot \beta+ (l^2+k^2)\cot \gamma]\\
&\hspace{1cm}+ ml\csc \alpha+ km\csc \beta + lk\csc \gamma
\end{align}


TITLE: Bacteria branching process
QUESTION [2 upvotes]: The population starts with one individual. Per one period, the individual can either split into two, survive, or die, each with probability one third. What is the expected population size after n periods?

REPLY [3 votes]: Hint: Define $Y_n$ the amount of individuals at step $n$, and notice that if $Y_n=m$ then you have $m$ R.V-s $X_{i}$, where each $X_i$ satisfies
$$ \mathbb{P}(X_i=1),\mathbb{P}(X_i=2),\mathbb{P}(X_i=-1)=\frac{1}{3}. $$
Note that $\sum_{i=1}^{Y_n} X_i=Y_{n+1}$ and that this gives you a simple identity for $\mathbb{E}[Y_{n+1}\vert Y_n=m].$ Can you carry on from this point?


TITLE: Using rgb triplets in "scalar" math
QUESTION [1 upvotes]: So I'm reading a computer graphics research paper and I'm confused as to how to interpret certain formulas that are being used.
In the paper, a value σ is defined as an RGB triplet. Later on, that same triplet is used in the calculation of scalar values. For instance, a floating point value S is defined using the following equation:
float S = -log(ξ)/σ, where ξ is a random variable between [0,1].

I see this kind of thing come up a lot in computer graphics papers, not just this one, and I don't really know how to interpret it. Do I take the average of the 3 RGB values? Do I use the luminance value? Do I add them all up?
If anyone here has experience with this, I appreciate your input.
Edit: Here is the link to the paper - http://www.graphics.stanford.edu/papers/bssrdf/bssrdf.pdf
On page 2, σt is defined as σa + σs.
On page 5, you can see that σa and σs values are defined in terms of RGB. This means that σt is also an RGB triplet.
Then on page 6, they say this: "We pick a random distance, s'o = log(ξ) / σt.
Distance is a scalar quantity, so how does this equation work?

REPLY [2 votes]: Up to my understanding, $\sigma$ are not triplets. Instead, they are functions of the frequency $\omega$ of light over the visible spectrum. These $\sigma(\omega)$ are really infinite dimensional instead of 3 dimensional objects.
For ordinary material, if you shoot a monochromatic laser beam with a fixed frequency $\omega$ to it, it will reflect/absorb/scatter light back with exactly the same frequency $\omega$. The corresponding physical color won't change. This means one can analysis the problem for each frequency channel independently.
Within each frequency channel, all $\sigma(\omega)$ behave like scalars.
Of course, human eyes has only 3 types of cells to detect colors. So biologically, we will
approximate all the physical colors by a 3-dimensional RGB triplets.
For application in the computer graphics, one usually just take the lowest order approximation. You treat these $\sigma$ as a 3-dimensional vector of
coefficients. All formula will be interpreted as one operating on the three color channels independently. i.e
$$S = -\frac{\log\xi}{\sigma} \quad\iff\quad
\begin{cases}
S_R &= -\frac{\log\xi_R}{\sigma_R}\\
S_G &= -\frac{\log\xi_G}{\sigma_G}\\
S_B &= -\frac{\log\xi_B}{\sigma_B}
\end{cases}$$


TITLE: Does function composition with the set of all functions from $A$ to $B$ form a monoid?
QUESTION [0 upvotes]: Let $A = \{a, b, c\}$ and $B = \{a, b\}$. Denote the set of all functions $f : A \to B$ as $F$ and denote function composition in the typical way, i.e., $f \circ g = f(g(x))$.
Is this a monoid? From my perspective, I don't think so. If $A = B$ then we can easily find an identity element but I am having difficulty finding one here where $A \neq B$.
Edit: ignore this part In particular, I can find a left inverse $e$ but not a right inverse. I'm not sure if I'm just missing how to find the right inverse. If there is truly no right inverse, how do I prove that?

REPLY [0 votes]: Inverses aren't relevant for monoids, but since you wrote about inverses, consider what I initially wrote below.

According to Wikipedia, an function $f$ has a left inverse (or is the empty function) if and only if it is injective. A function $f$ has a right inverse if and only if it is surjective.
Based on these, you can find right inverses for some elements of $F$, but no element of $F$ will have a left inverse.

There are only $2^3 = 8$ functions from $A$ to $B$, so you could just check them all and show none of them is an identity element.


TITLE: How to solve the following expression
QUESTION [1 upvotes]: I have the following expression: $2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}(n-k+1)^2$ and I have no idea how to solve it. I tried plugging it in WolframAlpha to get an idea but it gives me $-n^2 - 4 n + 7 \cdot 2^{n - 1} - 6$ and I have even less of an idea how I would get there.
I don't need the complete solution done. I just would like pointers on how to solve it.

REPLY [0 votes]: You'll just to simplify the expression
$f(n) = 2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}*(n-k+1)^2$
Now let's expand
$f(n) = 2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}*(n^2-2*k*n+2*n+k^2-2*k+1)$
Let's reduce it
$f(n) = 2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}*(n^2+2*n+1) +\sum_{k=1}^{n-3} 2^{k-1}*(-2*k*n-2*k) + \sum_{k=1}^{n-3} 2^{k-1}*(k^2)$
Simplify more
$f(n) = 2^{n-3}+ (n^2+2*n+1)*\sum_{k=1}^{n-3} 2^{k-1} -2*(n+1)*\sum_{k=1}^{n-3} k*2^{k-1}+\sum_{k=1}^{n-3} (k^2)*2^{k-1}$
Remember that
$\sum_{k=1}^{n-3} 2^{k-1} = 1+2+2^2+2^3+...........+2^{n-3-1} = 2^{n-3}-1$
It's becomes difficult here
$\sum_{k=1}^{n-3} k*2^{k-1} = h(x) = k*2^{k-1}+(k-1)*2^{k-2}+(k-2)*2^{k-3}+........+3*2^2+2*2^1+1*2^0$
But It's not so hard to see that
$h(x)*2-k*2^k = 2*h(x-1)$
You'll need to solve some recurring equation now
$h(x) = 2^x*(x-1)+1$
$\sum_{k=1}^{n-3} (k^2)*2^{k-1} = g(x) = (k^2)*2^{k-1}+(k-1)^2*2^{k-2}+(k-2)^2*2^{k-3}+........+ 3^2*2^2+2^2*2+1$
Same process involved we can also see that
$g(x)*2 -x^2*2^x = 2*g(x-1)$
So that $g(x) = 2^x*(x^2-2*x+3)-3$
Don't forget that the plugin value is $h(n-3)$ and $g(n-3)$
Therefore
$f(n) = 2^{n-3}+(n^2+2*n+1)*(2^{n-3}-1)-2*(n+1)*(2^{n-3}*(n-3-1)+1)+2^{n-3}*((n-3)^2-2*(n-3)+3)-3$
Expand and simplify this to get our solution
$f(n) = 7*2^{n-1}-n^2-4*n-6$
So you see that the trick is recurring equation, which Wolfram automatically solves


TITLE: How to find $\int_0^{\frac{\sqrt2}{2}} \frac{\log(1-x^2)}{2(1-x)^{\frac{3}{2}}} dx$ and the relationship with the second integral?
QUESTION [0 upvotes]: I´m trying to enhance my integral skills, as I´m planning to tutor some students next semester,  but I´m stuck with this integrals. I`m searching for a method or an idea that would work for both of them. I tried integrating by parts but that failed, the usual substitutions to get rid of the square roots did not look very promising either. So can someone please help me to solve the following 2 integrals, and not just solve, but providing a method that works for both of them and, if possible, explaining why they are equal and what was the idea behind it?
$$\int_0^{\frac{\sqrt2}{2}} \frac{\log(1-x^2)}{2(1-x)^{\frac{3}{2}}}dx$$ and
$$\int_1^{\frac{\sqrt2}{2}} \frac{\log (\frac{1}{x})}{x^2\sqrt{1-x^2}}dx$$
Thanks!

REPLY [1 votes]: First of all, many thanks for the answers, especially the answer of Aritro Pathak gave me some ideas to solve it by myself.
For the first one, using the substitution $x=sin(t)$ gave me
$$\int_0^{\frac{\pi}{4}} \frac{\log(cos(t)^2)}{cos(t)^2}dt$$
which is solvable by using integration by parts and gave me
$\frac{1}{4}(4-\pi-\log(4)) $
The second one can be solved in the same spirit, but starting with $x=cos(t)$ and using the hint of amWhy, $\log(\frac{1}{x})=-\log(x)$.
Thanks to all, it`s always more fun if you can inspire me to solve it by myself!


TITLE: Difference between Hellinger Distance and Wasserstein Distance between Two Distributions
QUESTION [1 upvotes]: I really want to understand the difference between Hellinger Distance and Wasserstein Distance. I am from a Physics background. I am expecting an intuitive explanation for the difference. Is Wasserstein distance give more pieces of information than Hellinger distance? In what way, Hellinger distance is different from Wasserstein distance? Is there any relation between Hellinger distance and Wasserstein distance between two distributions?

REPLY [1 votes]: Since you asked for an intuitive explanation this is going to be somewhat imprecise, but hopefully helpful.
The Hellinger distance is a bounded metric where you're kind of looking at the cumulative difference in density (of two probability measures), over all points in a probability space.
So lets say we have two probability measures with densities $f$ and $g$, where $f$ has support on $[0,1]$ and g has support on $[x, x+1]$ with $x\in\mathbb{R}$.
For $x=0$ the Hellinger and any Wasserstein distance between our distributions is zero.
As we increase $x$, the Hellinger distance increases until $x>1$, then it just stays $1$. The Wasserstein distance on the other hand keeps on increasing as $x$ increases.
The Wasserstein distance can be intuitively seen as mass times the distance you displace the mass.
If this explanation is too handwavy, I can be more specific!


TITLE: Why can the $W^{+/-}$ particles change quarks of different generations, while the $Z^0$ only changes leptons of one generation?
QUESTION [5 upvotes]: A $W^{+/-}$ particle can change a quark from one generation into a quark of a different generation, as long as these quarks (obviously) differ 1 or -1 in electric charge. So an up quark ($+\frac 2 3$) can be changed in a strange quark ($-\frac 1 3$) by means of a $W^-$ vector boson.
But why can't (for example) a $Z^0$ change a muon in an electron neutrino. I think conservation of lepton number offers no answer because is a rule which follows from experiment. Conservation of baryon number is also a conservation rule, but nevertheless, the decay of the proton is believed to be possible. So maybe it will also be possible to change a muon in an electron neutrino (or a muon in an electron).

REPLY [6 votes]: The standard model summarizes the fact you are stating as

At tree level, there are no flavor-changing neutral currents.

The reason is trivial, but slightly technical. The fermion flavor linear combinations the gauge bosons couple to are not the mass (yukawa flavor) eigenstates. One diagonalizes through Unitary mixing matrices the up-like and down-like quark (and lepton) triplets separately, and inserts these rotated triplets into all gauge boson vertices.
The respective unitary matrix for the W couplings, consisting of the product of the up (hermitian transposed) and down unitary matrices, yield an off-diagonal unitary matrix, the CKM one (so, then, generation changing).
The analog matrices at the Z vertex, however, are the products of the up-diagonalizing matrix with its own hermitian conjugate, so inverse; and, respectively that of the down-diagonalizing matrix with its inverse. They are thus both the identity: diagonal. They are gone. The neutral currents couple to mass/flavor eigenstates.
So neutral couplings do not change flavor.  Vanishing charge flow  blesses fermions  with diagonal couplings.
One should leave lepton numbers out of it.
Loop corrections are another matter, but they are subdominant.


TITLE: Compositum of abelian Galois extensions is also abelian?
QUESTION [7 upvotes]: Suppose I have a field $k$ and two extensions $K/k$ and $L/k$ which are both abelian Galois extensions of $k$. Then (assuming $K$ and $L$ are both contained in some bigger field) is the compositum $KL$ an abelian Galois extension of $k$?

REPLY [0 votes]: Yes, find the Fixed fields in $\mathrm{Gal}(KL/k)$. You can proceed

If $K$ is splitting field of separable polynomials $\{f_i\}$ in $F$, is $KL$ also a splitting field of those polynomials over $L$?
What is the image of the homomorphism $\mathrm{Gal}(KL/L)$ to $\mathrm{Gal}(K/F)$ by fixing automorphisms of $K$.


TITLE: Special Gram's inequality
QUESTION [5 upvotes]: For $1 \le s < k$ and $v_1$, $v_2,\dots,v_k$ vectors in $\mathbb{R}^n$, show that
$$\det G(v_1, v_2,\dots,v_k) \le \det G(v_1,v_2,\dots,v_s)\det G(v_{s+1}, v_{s+2},\dots,v_k).$$
Here, $G(v_1, v_2,\dots,v_k)$ is a Gram matrix of vectors $v_1, v_2,\dots,v_k$ with the standard inner product.
This is my homework problem, but I have no idea how to approach it. I found that this is one of the Gram's inequalities, but I can't figure out how it could be derived either from this:
$$\det G(v_1, v_2,\dots,v_k) \ge 0,$$
or Hadamard's inequality:
$$\det G(v_1, v_2,\dots,v_k) \le \prod_{i=1}^{k}\|x_i\|^2.$$
So any help would be much appreciated.

REPLY [1 votes]: Here are some hints. This approach uses a lemma which is interesting by itself.
Let $A:=G(v_1,\ldots,v_k)$, $A_1:=G(v_1,\ldots,v_s)$, $A_2:=G(v_{s+1},\ldots,v_k)$. Note that these are positive semidefinite matrices.

Note that $A=\pmatrix{A_1&B\\B^*&A_2}$ and we need to prove $\det A\leq \det A_1\cdot \det A_2$. Without loss of generality, we can assume that $\det A>0$, i.e. $\{v_1,\ldots,v_k\}$ are linearly independent, for otherwise the inequality is trivial.
Lemma   For every linearly independent set $\{u_1,\ldots,u_m\}$ and every vector $x$ in $\mathbb{R}^n$, the Euclidean distance between $x$ and $V$ the span of $\{u_1,\ldots,u_m\}$ is given by
$$d(x,V)=\sqrt{\frac{\det G(u_1,\ldots,u_m,x)}{\det G(u_1,\ldots,u_m)}}$$
Sketch  Note that $d=d(x,V)=\|x-y\|$ where $y$ is the orthogonal projection onto $V$, which satisfies  $\|x\|^2=\|y\|^2+d^2$ and $\langle x,u_j\rangle =\langle y,u_j\rangle$ for every $j$. Then use the multilinearity of the determinant when inspecting the last column of $G(u_1,\ldots,u_m,x)$. $\Box$

Use the lemma a couple of times to get
$$
\det A=d(v_1,*)^2\det G(v_2,\dots,v_k)=\ldots=*****\det A_2
$$
and
$$
\det A_1=d(v_1,*)^2\det G(v_2,\ldots,v_s)=*****
$$
Then compare the $*$'s by observing the appropriate subspace inclusions.


TITLE: $Show: a\in \mathbb{Z} \Rightarrow 6\mid a^3 -a$
QUESTION [0 upvotes]: $Show: a\in \mathbb{Z} \Rightarrow 6\mid a^3 -a$
My attempt:
NTS: $6\mid a^3 -a$, so $(2\mid a^3 -a) \land (3\mid a^3 -a)$
Assume that $a\in \mathbb{Z}$, therefore I have 2 cases:

a is even $\Rightarrow a=2k, k\in \mathbb{Z}$
a is odd $\Rightarrow a=2k+1, k\in \mathbb{Z}$

Therefore in case 1: $a^3-a=(2k)^3-2k=8k^3-2k$
and in case 2: $a^3-a=(2k+1)^3-(2k+1)=8k^3+2k$
And it led me to a blind alley. I can only factor 2 out of the expressions but there's no way to factor 3 out, so $k\in \mathbb{Z}$
Is there a way to show that using direct proof and the assumption?
I also tried to factor the $a^3-a=(a-1)(a)(a+1)$ but it isn't telling me much either.

REPLY [1 votes]: Let $a$ $\in\mathbb {Z}$.
Given:
$a^{3}-a=a(a-1)(a+1)$
For $a$ even, then 2 | $a$. If $a$ is odd, clearly we got that $a + 1$ or $a - 1$ is even.
So for a $\in\mathbb {Z}$, 2 | $a^{3}-a$.
Since gdc(2,3)=1, we need to show  $3$| $a^{3}-a$.
For a $\in\mathbb {Z}$, $a$ can be written in one of the following forms $3k_{0}$,   $3k_{1}+1$,  $3k_{2}+2$, with $  k_{0}$,$k_{1}$,$k_{2}$ $\in$ $\mathbb {Z}$.
Using this forms for $a$ and the above factorization the result follows.


TITLE: Approximation of smooth diffeomorphisms by polynomial diffeomorphisms?
QUESTION [7 upvotes]: Is it possible to (locally) approximate an arbitrary smooth diffeomorphism by a polynomial diffeomorphism?
More precisely: Let $f:\mathbb{R}^d\rightarrow\mathbb{R}^d$ be a smooth diffeomorphism. For $U\subset\mathbb{R}^d$ bounded and open and $\varepsilon>0$, is there a diffeomorphism $p=(p_1, \cdots, p_d) : U\rightarrow\mathbb{R}^d$ (with inverse $q:=p^{-1} : p(U)\rightarrow U$) such that both

$\|f - p\|_{\infty;\,U}:=\sup_{x\in U}|f(x) - p(x)| < \varepsilon$, $\ \textbf{and}$
each component of $p$ and of $q=(q_1,\cdots,q_d)$is a polynomial, i.e. $p_i, q_i\in\mathbb{R}[x_1, \ldots, x_d]$ for each $i=1, \ldots, d$?

Clearly, by Stone-Weierstrass there is a polynomial map $p : \mathbb{R}^d\rightarrow\mathbb{R}^d$ with $\|f - p\|_{\infty;\,U} < \varepsilon$ and such that $q:=(\left.p\right|_U)^{-1}$ exists; in general, however, this $q$ will not be a polynomial map.
Do you have any ideas/references under which conditions on $f$ an approximation of the above kind can be guaranteed nonetheless?

REPLY [1 votes]: As pointed out by Robert Bryant over at
https://mathoverflow.net/questions/364099/approximation-of-smooth-diffeomorphisms-by-polynomial-diffeomorphisms ,
the answer to this question is 'no'.


TITLE: Question Concerning Positivity of the G2 Three-Form
QUESTION [0 upvotes]: It is often said that the condition of being a positive threeform (in the context of G2 Holonomy Manifolds) is an open condition. My understanding of this statement is that if $\phi$ is a positive threeform and $\phi'$ is a threeform $C_{0}$ close to $\phi$ then $\phi'$ is also a positive threeform. Mathematically this can be expressed via
$\mid \phi-\phi'\mid_{C0} \leq \epsilon$
Where here the norm is given by that of the metric associated to the original g2 threeform. My question is what does $\epsilon$ depend on? I would figure that it does not depend on $\phi'$ but it could depend on $\phi$ or the manifold itself. However, positivity is basically a pointwise condition so I could see how it would be possible for $\epsilon$ to be computed from a computation in $\mathbb{R}^{7}$. Assuming that $\epsilon$ does depend on $\phi$ is it possible to calculate/estimate? I am paricularly interested in the case where the new three-form is of the form $\phi'=\phi+d\beta$ for some two-form $\beta$ in which the argument of the norm is simply d$\beta$

REPLY [2 votes]: Remember a stable 3-form $\phi$ actually determines a Riemannian metric $g$ since we have the conformal class by
$$
i_v\phi\wedge i_w\phi\wedge\phi=g(v,w)\,\mathrm{dvol}_g.
$$
So the norm $\lvert\cdot\rvert_{C^0}$ is actually taken with respect to the metric determined by $\phi$.  Hence it reduces to pointwise $\mathbb{R}^7$, with $\epsilon>0$ since stable 3-forms has dimension $\dim GL^+(7,\mathbb{R})-\dim G_2=35$, the same dimension as $\bigwedge^3\mathbb{R}^7$.  If you want to compute an explicit $\epsilon$, you proceed to bound everything you need in inverse function theorem (i.e., $\lVert f(x)-f(y)\rVert$, $Df$, etc.).  However, in practice we just need existence not the numerical value of an $\epsilon$.


TITLE: radius of nth circle inside a circle of given radius
QUESTION [1 upvotes]: A circle of radius 100 cm is given. The goal is to place 7 circles of unequal radius inside the initial circle, so that each 3 of circles don't overlap each other and all of them stay inside the initial circle.
The first smaller circle is of radius 50 cm and so on.
I thought it has to do something with Ford circles, but answer r = 14.36 cm for the 7th circle is incorrect.

REPLY [0 votes]: Hint:

One way is to consider
this construction as a part of the (infinite)
Steiner’s chain of circles defined by two circles,
$\bigcirc O((0,0), R)$ and $\bigcirc o((-r,0),r)$,
with the distance between their centers $d=r$,
where $R=100,\ r=50$.
The other $6$  circles $\bigcirc O_0,\dots,\bigcirc O_5$
are sandwiched between the two,
starting with $\bigcirc O_0((r,0),r)$.
All tangent points of the kissing circles in the chain
are located on the circle (shown green in the picture) centered at $X$ with radius
$r_x=|XO_i|$, $i=0,\dots,5$,
\begin{align}
r_x&=
\sqrt{rR-\frac{rRd^2}{(r+R)^2}}
=\frac{200}3\approx 66.7
,
\end{align}
while all the centers $O_i$, $i=0,\dots,5$,
belong to the ellipse (shown yellow in the picture)
focused at $o,\,O$.

Edit:
Using Descartes' theorem,
for the curvatures
$k_1,k_2,k_3,k_k$
of four mutually tangent circles,
\begin{align}
k_4&=k_1+k_2+k_3\pm 2\sqrt{k_1 k_2+k_2 k_3+k_3 k_1}
\tag{1}\label{1}
.
\end{align}
In this particular arrangement of circles,
\begin{align}
k_1&=-\tfrac1R
,\quad k_2=\tfrac2R
,\quad k_3=k_2=\tfrac2R
\end{align}
since each next circle in the chain is smaller,
the sign in \eqref{1} would be always positive
and we have
\begin{align}
k_4&=\tfrac3R
,\\
k_5&=\tfrac{6}R
,\\
k_6&=\tfrac{11}R
,\\
k_7&=\tfrac{18}R
,\\
k_8&=\tfrac{27}R
,
\end{align}
so the radii of
$\bigcirc O_i$, $i=1,\dots,5$
can be found as
\begin{align}
r_i&=\frac R{i^2+2}
.
\end{align}
For example, the seventh inner circle, the smallest one,
must have the radius
\begin{align}
r_5&=\frac R{27}
\approx 3.7037
.
\end{align}


TITLE: How to Write Coordinates in $C^2$
QUESTION [1 upvotes]: I have in my textbook the following quote
As you can verify, $\frac{(i,1)}{\sqrt2}$, $\frac{(-i,1)}{\sqrt2}$ is an orthonormal basis of $C^2$ consisting of eigenvectors of $T$.
I don't know how to read this, to be a basis we need that the list is four vectors long. But if we take the four vectors then $\frac{1}{\sqrt2}$ is repeated twice. So these have to be two vectors each in $C^2$ with dimension 2? This notation has never been introduced in the book and I am pretty confused.

REPLY [1 votes]: This is really two vectors in two dimensional complex space,
$$
\left( \frac{i}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \text{ and }
\left( -\frac{i}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \ .
$$
I think the author has chosen poor notation.


TITLE: Physical interpretation of statistical mechanics
QUESTION [1 upvotes]: What is the physical interpretation of
$$ \langle p_\theta^2\rangle = \dfrac{8\pi^2 I^2}{\beta^2} \tag{1}$$
and
$$\langle p_\varphi^2 \rangle = \dfrac{16\pi^2 I^2}{3\beta^2} \tag{2}$$
where the Hamiltonian (for $1$ particle) that describe the rotation of a diatomic molecule is
$$ H_1 (\theta, \varphi, p_\theta, p_\varphi)=\dfrac{1}{2I} \left(p_\theta^2 +\dfrac{p_\varphi^2}{\sin^2 \theta} \right)
\tag3 $$
where $I$ is the inertia momentum of the molecule ($I=mr_0^2$). Are the units correct? What do the numerical values mean?
Note that
$\beta =1/k_BT$
as usual in statistical mechanics.
For the definition of the expected value in the canonical ensemble, I've used
$$
\langle p_\theta^2 \rangle =\dfrac{1}{h^2}\dfrac{1}{Z(\beta,\alpha,1)}\int_0^{2\pi} \text{d}\varphi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\varphi \cdot p_\theta^2 \text{ e}^{-\beta H_1},$$
where
$$ Z(\beta,\alpha,1)= \int_0^{2\pi} \text{d}\varphi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\varphi  \text{ e}^{-\beta H_1}.$$
I don't know if this is right or not.
For example:
[1] Statistical Mechanics, Ryogo Kubo, p. 200, nº 3.

REPLY [0 votes]: By definition we have
$$ \langle a^2 \rangle = \dfrac{\int_0^{2\pi} \text{d}\phi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\phi \cdot a^2 \text{ e}^{-\beta H}}{\int_0^{2\pi} \text{d}\phi \int_0^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\theta \int_{-\infty}^\infty \text{d}p_\phi  \text{ e}^{-\beta H}} \tag{S.1}$$
but, for $a=p_\theta$, this expression can be simplified
$$\langle p_\theta^2 \rangle = \dfrac{\int_{-\infty}^\infty p_\theta^2 \text{ e}^{-\beta p_\theta^2/(2I)} \text{ d}p_\theta}{\int_{-\infty}^\infty \text{e}^{-\beta p_\theta^2/(2I)}\text{ d}p_\theta}. \tag{S.2}$$
For the Hamiltonian $(3)$ we obtain
$$ Z(\beta,\alpha,1)=8\pi^2 \dfrac{I}{\beta} \tag{S.3}$$
and using this expression
$$\int_\mathbb{R} \text{d}x \text{ e}^{-ax^2} = 2a\int_\mathbb{R} \text{d}x \text{ } x^2 \text{ e}^{-ax^2}=\sqrt{\dfrac{\pi}{a}}\tag{S.4}$$
into $(S.2)$ we have
$$ \boxed{\langle p_\theta^2 \rangle =\dfrac{I}{\beta}= k_B TI}. \tag{S.5}$$
and, for $a=p_\phi$, the expression $(S.1)$ turns into
$$\langle p_\phi^2 \rangle = \dfrac{\int_{0}^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\phi \text{ } p_\phi^2 \text{ e}^{-\beta p_\phi^2/(2I\sin^2 \theta)}}{\int_{0}^\pi \text{d}\theta \int_{-\infty}^\infty \text{d}p_\phi \text{ }\text{e}^{-\beta p_\phi^2/(2I\sin^2 \theta)}}. \tag{S.6}$$
We obtain
$$ \boxed{\langle p_\phi^2\rangle = \dfrac{2}{3}\dfrac{I}{\beta}=\dfrac{2}{3} k_B TI}. \tag{S.7}$$
Equations $(S.5)-(S.7)$ are $\textbf{the dispersion of each (conjugate) angular momentum}$ because
$$ \langle p_\theta \rangle = \langle p_\phi \rangle =0.$$
Therefore,
$$ \sigma_{p_\theta}^2 = \langle p_\theta^2 \rangle= k_B TI > \sigma_{p_\phi}^2 = \langle p_\phi^2 \rangle = \dfrac{2}{3}k_B TI = \dfrac{2}{3}\langle p_\theta^2 \rangle \geq 0.$$
The physical interpretation is that the $\textbf{dispersion}$ of $\theta$-$\textbf{angular momentum}$  $p_\theta$ is bigger than the $\textbf{dispersion}$ of $\phi$-$\textbf{angular momentum} $ $p_\phi$ in a factor of $3/2$:
$$ \boxed{\sigma_{p_\theta}^2 = \dfrac{3}{2}\sigma_{p_\phi}^2}. \tag{S.8}$$
According to @rob the energy of one atom is
$$ E_1 = \langle H_1 \rangle = \dfrac{1}{2I} \left( k_B TI + k_B TI\right)=k_B T \tag{S.9}$$ and for the molecule
$$ E_{1+1}=\langle 2H_1 \rangle = 2k_B T.\tag{S.10}$$


text:
\begin{definition}[Definition:Subjective Unknown]
An '''subjective unknown''' is an [[Definition:Event|event]] which does not have a well-defined objective [[Definition:Probability|probability]] of occurring.
An '''subjective unknown''' is appropriately modelled by means of a [[Definition:State-Variable Model|state-variable model]].
=== [[Definition:Subjective Unknown/Examples|Examples]] ===
{{:Definition:Subjective Unknown/Examples}}
\end{definition}