TITLE: Global symmetries in type IIB string theory vs type IIB supergravity
QUESTION [3 upvotes]: In the AdS/CFT correspondence I know that the mapping of global symmetries involves also the S duality that in the field theory side is $SL(2,Z)$. In Type IIB supegravity this duality is $SL(2,R)$. I don't understand how to matching these symmetries.

REPLY [6 votes]: The AdS/CFT correspondence (or this particular, most famous example of it) is the equivalence between the non-gravitational ${\mathcal N}=4$ supersymmetric Yang-Mills theory on one side ("the boundary CFT"); and the type IIB superstring theory in $AdS_5\times S^5$ ("the bulk theory") on the other side. The equivalence is exact and the gauge theory is well-defined, so the dual bulk AdS description has to be consistent, too. String/M-theory is the only consistent theory of quantum gravity so the bulk theory has to be a string theory and not just "supergravity", too.
Supergravity is just a (nonrenormalizable, and for this reason and others, inconsistent) low-energy approximation (effective field theory) of the full theory, string theory. This particular one has the noncompact symmetry $SL(2,{\mathbb R})$. However, quantum effects break it to the discrete subgroup $SL(2,{\mathbb Z})$, the same symmetry that we see on the boundary CFT side. It's not hard to see how it happens.
For example, the supergravity contains black 1-branes carrying one of the two string-like charges under the 2-form B-fields. One of them is the "fundamental (F1) string" charge and the other one is the "D1-brane" charge. These two charges transform as a doublet under the $SL(2)$ group.
In supergravity used as a classical description, these 1-branes are seen as black-hole-like extended solutions – the black 1-branes – but the amount of charge these BPS objects carry is arbitrary, a continuous, real number. The same holds for the two types of 5-branes, those with the NS5-brane charge and the D5-brane charge, another doublet.
However, in the quantum theory, the charges cannot be continuous. One way to prove this assertion is to appreciate the Dirac quantization rule (which results from the requirement of single-valuedness of the wave function of one object when it orbits around the electromagnetically dual object's Dirac string). F1-strings are electromagnetically dual to the NS5-branes and D1-branes are electromagnetically dual to D5-branes. For the charges of mutually dual objects (just like for the electric charges and magnetic monopoles in 4D), the Dirac quantization rule has to hold:
$$ Q_E Q_M \in 2\pi {\mathbb Z} $$
It follows that the allowed F1,NS5; D1, D5-charges are not continuous. They actually belong to a lattice that is "self-dual" in the sense of the Dirac quantization condition above.
Such a lattice isn't quite unique. When the allowed minimum charge $Q_E$ is increased $k$ times and the minimum $Q_M$ is decreased $k$ times, the Dirac quantization rule continues to hold. Also, one may mix the electric and magnetic charges (in a way that fully matches the QCD-like theta-angle in the gauge theory).
When the conditions are taken into account, one sees that there is a 2-real-dimensional family of possible charge lattices and it may be written as the usual moduli space
$$ SL(2,{\mathbb Z}) \backslash SL(2,{\mathbb R}) / SO(2, {\mathbb R}) $$
It may be represented as the "fundamental domain" of the modular group. It's the same as the space parameterized by $g_{YM}$ and $\theta_{QCD}$ in the gauge theory, of course. The moduli space is a quotient on both sides. The left-multiplication identifies the points with respect to the actual "U-duality group" $SL(2,{\mathbb Z})$ because elements of this group act in such a way that they map the lattice of allowed charges onto the same lattice.
The approximation of the moduli space by "one point" is basically equivalent to assuming that the charges of all the brane-like objects are continuous, not quantized, and ignoring to the Dirac quantization rules and all similar quantum effects. This approximation is good enough at "low energies". When we consider the low-energy limit, the only objects that carry such charges are "large branes" that look like macroscopically large extended black holes. And those carry so huge charges, relatively to the minimum units, that these charges may be considered continuous.
I mentioned the quantization of the brane charges as a reason why the dual theory is the full string theory and not just the supergravity. There exist other differences between string theory and supergravity. All of these discriminating features may be checked in AdS/CFT and in all of them, one may see that the gauge theory is dual to the full string theory and not just "supergravity". For example, one may see all the excited string states in terms of traces of long products in the gauge theory (complex composite operators), verify that they interact just like in string theory, see all the wrapped branes, check that the string theory cures all the problems of SUGRA with its nonrenormalizability, and many other things.
On the gauge theory side, the supergravity limit is obtained by focusing on the planar limit of simple enough single-trace operators etc. But if one sees beyond this subset of observables and beyond the approximations with which it is being computed to compare the gauge theory with SUGRA, one may see all of string theory, too.


TITLE: Rationalizing complex number
QUESTION [0 upvotes]: I have this equation from electronic task:
\$I_{1} = \frac{u}{1200-j4416}\$.
How to remove the complex from the denominator?
I'm getting this, is this right? -48(92j-25)

REPLY [1 votes]: You have\begin{align}\frac u{1200-4416j}&=\frac{u(1200+4416j)}{1\,200^2+4\,416^2}\\&=\left(\frac{25}{436\,272}+\frac{23}{109\,068}j\right)u\\&=\frac{25+92j}{436\,272}u.\end{align}

REPLY [0 votes]: Well, assuming that $i^2=\text{j}^2=-1$ and $\text{u}\in\mathbb{C}$, we get:
$$\frac{\Re\left(\text{u}\right)+\Im\left(\text{u}\right)i}{1200-4416i}=\frac{\Re\left(\text{u}\right)+\Im\left(\text{u}\right)i}{1200-4416i}\cdot\frac{1200+4416i}{1200+4416i}=$$
$$\frac{\left(\Re\left(\text{u}\right)+\Im\left(\text{u}\right)i\right)\left(1200+4416i\right)}{1200^2+4416^2}\tag1$$


TITLE: Conditional probability and a game of tennis
QUESTION [1 upvotes]: In a best of 5 tennis match, Alex and Bob are equally likely to be the better players. If Alex (Bob) is the better player, he wins a set with probability 0.75 (0.75) independently of the outcome of the other sets. Three sets are played, and Alex leads with two sets to one. What is the probability that Alex will go on to win the match?
Here is an attempt and I will try  show you all the steps:
Let $A = \{\text{Alex is the better player}\}$, $B = \{\text{Bob is the better player}\}$, $T = \{\text{Alex leads two sets to one}\}$, $W = \{\text{Alex wins the match}\}$.
From the problem statement, $P(A) = P(B) = 1/2$, and $$P(T) = P(T|A)P(A)+P(T|B)P(B) = (0.75\times 0.75 \times0.25) \times 0.5 + (0.25\times 0.25 \times 0.75) \times 0.5 = 3/32$$
We want to calculate:
$$P(W|T) = P(W|T,A)P(A) + P(W|T,B)P(B)$$
Now to calculate $P(W|T,A)$, this is just the event that Alex goes on to win the match given that he is the better player and he leads two sets to one. The possible outcomes here is that he wins the next set, or loses a set and wins a set, and so $P(W|T,A) = 0.75 + 0.25\times0.75 = 15/16$. Similarly, we have $P(W|T,B) = 0.25 + 0.75\times0.25 = 7/16$.
The final answer, therefore, is $P(W|T) = 11/16$.
Any feedback is appreciated

REPLY [1 votes]: Outline: The fact that Alex is leading is evidence he is the better player. We need to take that into account by doing a conditional probability calculation.
You averaged the two numbers $7/16$ and $15/16$, giving equal weight to each.
But for example the weight in front of the $15/16$ should be the probability that Alex is the better player, given that he is leading. This is greater than $1/2$.
The required probability $\Pr(B|T)$ can be calculated  using $\Pr(B|T)=\frac{\Pr(B\cap T)}{\Pr(T)}$.
For the weight in front of the $7/16$, one can do a similar calculation, but it is easier to note that it is $1-\Pr(B|T)$.


text:
\begin{document}
\newcounter{MYtempeqncnt}
\pagestyle{empty}
\thispagestyle{empty}

\author{\authorblockA{Himal A. Suraweera~\IEEEauthorrefmark{1}, Jason Gao~\IEEEauthorrefmark{1}, Peter J. Smith~\IEEEauthorrefmark{2}, Mansoor Shafi~\IEEEauthorrefmark{3} and Michael Faulkner~\IEEEauthorrefmark{1}\\
\IEEEauthorrefmark{1}~School of Electrical Engineering, Victoria University, Melbourne, Australia\\
}
\authorblockA{\IEEEauthorrefmark{2}Department of Electrical and Computer Engineering, University of Canterbury, Christchurch, New Zealand}
\authorblockA{\IEEEauthorrefmark{3}~Telecom New Zealand, PO Box 293, Wellington, New Zealand\\ Email: himal.suraweera@vu.edu.au,~p.smith@elec.canterbury.ac.nz,~mansoor.shafi@telecom.co.nz}}

\maketitle
\thispagestyle{empty}
\pagestyle{empty}

\begin{abstract}
Cognitive radio technology is an innovative radio design concept which aims to increase spectrum utilization by exploiting unused spectrum in dynamically changing environments. By extending previous results, we investigate the capacity gains achievable with this dynamic spectrum approach in asymmetric fading channels. More specifically, we allow the secondary-to-primary and secondary-to-secondary user channels to undergo Rayleigh or Rician fading, with arbitrary link power. In order to compute the capacity, we derive the distributions of ratios of Rayleigh and Rician variables. Compared to the symmetric fading scenario, our results indicate several interesting features of the capacity behaviour under both average and peak received power constraints. Finally, the impact of multiple primary users on the capacity under asymmetric fading has also been studied.
\end{abstract}

\section{Introduction}
Conservative spectrum policies employed by regulatory authorities have resulted in spectrum underutilization of the overall available spectrum for wireless communications. Measurements performed by agencies such as the Federal Communications Commission \cite{fcc} in the United States and Ofcom \cite{ofcom} in the United Kingdom have revealed that at any given time, large portions of spectrum are sparsely occupied. Findings of such campaigns on spectrum usage have challenged the traditional spectrum management approaches.

The concept of cognitive radio (CR) \cite{mitola} refers to a smart radio which can sense the external electromagnetic environment and adapt its transmission parameters according to the current state of the environment. CRs can access parts of the spectrum for their information transmission, provided that they cause minimal interference to the primary users in that band~\cite{jondral,akyildiz}. Therefore, spectrum sharing among the primary licensee and the secondary CR must be carried out in a controlled fashion. In the technical literature, the \emph{interference temperature} introduced by Kolodzy \cite{kolodzy,haykin} indicates the interference level at the primary licensee's receiver. From the licensees' point of view, the secondary access can be controlled in two ways. The total interference power can be required to remain below a certain threshold (an interference temperature constraint) or the signal-to-noise-and-interference (SINR) can be constrained.

The capacity of wireless systems has been extensively studied under fixed spectrum access. For CR, this work is less mature and many information/communication theoretic problems and implementation issues \cite{cabric} remain to be solved. However, several interesting results on the capacity, outage probability and throughput of CR systems have recently emerged. See for example,~\cite{ghasemi_b,jafar,gastpar_1}. In \cite{gastpar_1}, Gastpar derived the capacity of different non-fading additive-white-Gaussian-noise (AWGN) channels with the average received-power at a primary receiver being constrained. In \cite{ghasemi_b}, Ghasemi and Sousa showed that with the same limit on the received-power level, channel capacity for a range of fading models (e.g., Rayleigh, Nakagami-$m$ and log-normal fading) exceeds that of the non-fading AWGN channel. In some scenarios, primary user spectral activity in the vicinity of the cognitive transmitter may differ from that in the vicinity of the cognitive receiver. Considering this, in \cite{jafar}, the capacity of opportunistic spectrum acquisition in the presence of distributed spectral activity has been investigated.

We extend the work in \cite{ghasemi_b} which assumed that fading conditions for the interference path (CR transmitter-primary receiver) and the desired path (CR transmitter-CR receiver) are the same. In practice, these two links could experience different fading conditions (types) and different link powers (due to path length or shadowing). This is referred as \emph{asymmetric} fading in this paper.

In this paper we consider Rayleigh and Rician fading. Hence, we are able to quantify the effects of propagation paths consisting of both line-of-sight (LoS) and scattered components. Building on the work
in \cite{ghasemi_b}, this paper makes the following main contributions:

\begin{enumerate}
\item
The secondary capacity under average-received power and peak-power constraints is studied for asymmetric conditions including different fading types (Rayleigh and Rician) and different link powers. Here we show that under low interference to the primary receiver, the secondary capacity is sensitive to the fading type on the desired and interference paths.
\item

The impact of multiple primary licensee receivers in Rayleigh and Rician fading is studied for
peak power constraints.
\item
Closed-form expressions for the cumulative distribution function
(CDF) and the probability density function (PDF) of a random
variable (RV), $g_1/g_0$ is derived, for the cases where
$(\sqrt{g_1},\sqrt{g_0})$ experience (Rayleigh, Rician) and (Rician,
Rayleigh) fading. This is needed to derive the above mentioned
capacities.

\end{enumerate}
This paper is organized as follows. The system and channel model is
described in Section~II. In Section~III we derive the exact PDFs for
the ratio of Rayleigh and Rician RVs. In Section~IV, these results
are used to study the capacity gains under average/peak
received-power constraints respectively. Extensions of these results
to multiple primary users are presented in Section~V. Finally, some
conclusions are drawn in Section~VI. Throughout the paper, the
reference to average/peak received-power refers to as the
average/peak interference power at the primary receiver. The CR link
is also referred to as a secondary link.

\begin{figure}[t]
\centering
\includegraphics[width=0.85\columnwidth]{icc_08p2}
\caption{Shared spectrum usage between primary and secondary users.}
\label{fig_1}
\vspace{-3mm}
\end{figure}

\section{System and Channel Model}
In this section, the system and channel model considered in the paper are briefly outlined (cf. Fig. 1). The system model is borrowed from \cite{ghasemi_b}, however we have considered asymmetric fading scenarios. A point-to-point flat fading channel with perfect channel side information available to both the receiver and the
transmitter is assumed. Let $g_0$ and $g_1$ denote the instantaneous channel gains from the secondary transmitter to the primary and secondary receivers respectively. Furthermore, we denote the respective PDFs by $p_{g_0}(g_0)$ and $p_{g_1}(g_1)$. For a unit power channel gain, the Rayleigh PDF is given by
\begin{equation}
p_{\sqrt{g}}(x)=2xe^{-x^2}
\end{equation}
for $x \ge 0$. For a Rician distribution the PDF is given by
\begin{align}
p_{\sqrt{g}}(x)=2x(1+K)e^{-K-(1+K)x^2}I_{0}\left(2x\sqrt{K+K^2}\right)
\end{align}
for $x \ge 0$, where $K$ is the Rician $K$-factor defined as the ratio of signal power in dominant component over the scattered power and $I_0(\cdot)$ is the zeroth-order modified Bessel function of the first kind. For $K=0$, Rayleigh fading is experienced and $K=\infty$ gives the AWGN (no fading) situation. Values of the $K$-factor in indoor/outdoor land mobile applications normally range from $0-12$ dB \cite{parsons}. In a practical environment the CR transmitter to CR receiver link may not be of the same length as the CR interference path to the primary receiver. When the link powers, $E\{g_0\}$ and $E\{g_1\}$, differ, it can be shown that capacity only depends on the power ratio. Hence, we define the relative power parameter, $c$, by $c=E\{g_1\}/E\{g_0\}$. Note that $E\{\cdot\}$ denotes the expectation operator.

Two important items of notation should be stressed at this point. Since the main results of the paper depend on the ratio $g_1/g_0$, we use the shorthand notation Rayleigh/Rician to indicate that $\sqrt{g_1}$ is Rayleigh and $\sqrt{g_0}$ is Rician. Similarly, Rician/Rayleigh indicates that $\sqrt{g_1}$ is Rician and $\sqrt{g_0}$ is Rayleigh. The second issue is that the secondary transmitter must constrain its power so that the interference at the primary is acceptable. Hence the power constraints in this scenario are really interference constraints. This is different to many other problems where the constraints are for transmit power.

\section{CDF and PDF of $g_1/g_0$}
Here, we anticipate the results of Section~IV, where it is shown that capacity depends on the ratio, $g_1/g_0$. Hence, in this section the CDF and the PDF for a Rayleigh/Rician RV and a Rician/Rayleigh RV are derived.

Consider the distribution of a Rayleigh/Rician RV, $X=g_1/g_0$. Mathematically, $P(X<x)$, i.e., the CDF of $X$, is given by
\begin{equation}
\label{ae1}
P(X<x)=\int^{\infty}_{0}P\left(\frac{g_1}{g_0}<x|g_0\right)p_{g_0}(g_0)dg_0
\end{equation}
Equation \eqref{ae1} can be simplified as
\begin{align}
\label{re1}
P(X<x)& =(K+1)e^{-K} \int^{\infty}_{0}(1-e^{-xg_0})\\\nonumber
& \cdot e^{-(K+1)g_0}I_{0}(2\sqrt{K(K+1)g_0})dg_0
\end{align}
The integral in \eqref{re1} can be solved using \cite[eq. 2.15.5.4]{prud} and we obtain the CDF of $X$ as
\begin{align}
\label{ty1}
F_{X}(x) = 1-\frac{K+1}{x+K+1}e^{-K+\frac{K^2+K}{x+K+1}}
\end{align}
for $x \ge 0$. The PDF of $X$ can be found by taking the derivative of \eqref{ty1} with respect to $x$, yielding
\begin{equation}
\label{rayrician}
p_{X}(x)=(K+1)\frac{x+(K+1)^2}{(x+K+1)^3}e^{-K+\frac{K^2+K}{x+K+1}}
\end{equation}
for $x \ge 0$. As expected, for $K=0$ the PDF $p_{X}(x)$ is given by $p_{X}(x)=1/(x+1)^2$ \cite[eq. 11]{ghasemi_b}.

Now consider the distribution of $Y=g_1/g_0$ when $\sqrt{g_1}$ is Rician and $\sqrt{g_0}$ is Rayleigh. Using the same approach, $P(Y<y)$ is given by
\begin{align}
\label{iy1}
P(Y<y)=1-\int^{\infty}_{0}Q_1\left(\sqrt{2K},\sqrt{2(1+K)yg_0}\right)e^{-g_0}dg_0
\end{align}
where $Q_{1}(a,b)=\int^{\infty}_{b}xe^{-\frac{a^2+b^2}{2}}I_{0}(a x)dx$ is the first-order Marcum $Q$-function which satisfies the following identity \cite[eq. 5]{nuttall}
\begin{align}
\label{h1}
Q_1(a,b)+Q_1(b,a)=1+e^{-\frac{a^2+b^2}{2}}I_{0}(ab)
\end{align}
Using \eqref{h1}, we can express \eqref{iy1} as shown in \eqref{inr4}.
\begin{figure*}[!t]
\normalsize
\setcounter{MYtempeqncnt}{\value{equation}}
\setcounter{equation}{8}
\begin{align}
\label{inr4}
P(Y<y)=-\int^{\infty}_{0}e^{-(y+Ky+1)g_0}I_{0}(2\sqrt{(Ky+K^2y)g_0})dg_0+\int^{\infty}_{0}Q_{1}(\sqrt{2(y+Ky)g_0},\sqrt{2K})e^{-g_0}dg_0
\end{align}
\setcounter{equation}{\value{MYtempeqncnt}}
\hrulefill
\end{figure*}
\addtocounter{equation}{1}
The first integral in \eqref{inr4} can be evaluated in closed-form using the result of \cite[eq. 2.15.5.4]{prud} and is
\begin{align}
I_1=\frac {e^{-K+\frac{Ky+K^2y}{y+Ky+1}}}{{y+Ky+1}}
\end{align}
The second integral in \eqref{inr4} can be evaluated using the result of \cite[eq. 25]{nuttall_2} as $I_2=e^{-\frac{K}{y+Ky+1}}$. Hence, $P(Y<y)$ is given by
\begin{align}
\label{yr1}
F_{Y}(y)=e^{-\frac{K}{y+Ky+1}}-\frac {e^{-K+\frac{Ky+K^2y}{y+Ky+1}}}{{y+Ky+1}}
\end{align}
for $y \ge 0$. After taking the derivative of \eqref{yr1} with respect to $y$, we obtain the PDF of $Y$ as
\begin{align}
\label{ricianray}
p_{Y}(y)& = \frac{K(1+K)}{(y+Ky+1)^2}e^{-\frac{K}{y+Ky+1}}\\\nonumber
& +\frac{(1+K)^2(1-K+y)}{(y+Ky+1)^3}e^{-K+\frac{Ky+K^2y}{y+Ky+1}}
\end{align}
for $y \ge 0$. For Rayleigh/Rayleigh fading, the PDF $p_{Y}(y)$ simplifies to $p_{Y}(y)=1/(y+1)^2$ \cite[eq. 11]{ghasemi_b}. To confirm the derivations, the PDFs of \eqref{rayrician} and \eqref{ricianray} were validated by Monte Carlo simulations, and a perfect match was obtained.

\section{Capacity Gains of Spectrum Sharing}

\subsection{Capacity Under an Average Received-Power Constraint}
In this section, we investigate the capacity gains achievable by the secondary user under an average received-power constraint. In \cite{ghasemi_b}, the channel capacity was expressed as
\begin{align}
\label{first}
C=\int\int B\log_2\left(1+\frac{g_1 P(g_0,g_1)}{N_0 B}\right)p_{g_0}(g_0)p_{g_1}(g_1)dg_0dg_1
\end{align}
such that
\begin{align}
\int_{0}^{\infty}\int_{0}^{\infty}g_0P(g_0,g_1)p_{g_0}(g_0)p_{g_1}(g_1)dg_0dg_1 \leq Q
\end{align}
where $Q$ is the maximum average interference power tolerated by the primary receiver~\footnotemark[2]\footnotetext[2]{The quality of transmission at the primary receiver can also be measured using the SINR. This requires a knowledge of the primary transmitter to primary receiver channel.}, $B$ is the available bandwidth, $N_0$ is the noise power at the secondary receiver and $P(g_0,g_1)$ denotes the optimal power allocation. Using the Lagrangian technique, \cite{ghasemi_b} has found $P(g_0,g_1)$ to be
\begin{equation}
P(g_0,g_1)=\left(\frac{1}{\lambda_0g_0}-\frac{N_0 B}{g_1}\right)^+
\end{equation}
where $(\cdot)^{+}$ denotes $\max\{\cdot,0\}$. Note that $\lambda_0$ is determined such that the average receive power is equal to $Q$. That is mathematically
\begin{equation}
\label{u3}
\int_{g_0}\int_{g_1}\left(\frac{1}{\lambda_0}-N_{0}B\frac{g_0}{g_1}\right)^{+}p_{g_1}(g_1)p_{g_0}(g_0)dg_1dg_0=Q
\end{equation}
Hence, the channel capacity can be calculated from
\begin{equation}
\label{u1}
C=B\int^{\infty}_{\frac{1}{\gamma_0}}\log_2(\gamma_0g_{10})p_{\frac{g_1}{g_0}}(g_{10})dg_{10}
\end{equation}
where $\gamma_0=1/(\lambda_0N_{0} B)$ and $p_{\frac{g_1}{g_0}}(\cdot)$ denotes the PDF of $g_1/g_0$. To the best of the authors' knowledge, there are no closed-form solutions for the integral in \eqref{u3} for the two fading scenarios considered in this paper. Rewriting    \eqref{u3} gives
\begin{align}
\label{fd1}
\int^{\gamma_0}_{0}(\gamma_0-x)p(x)dx=\frac{Q}{N_0B}=\alpha
\end{align}
where $p(x)$ in \eqref{fd1} denotes the PDF of $g_0/g_1$. Therefore, $\alpha$ is the allowable interference-to-noise power ratio at the primary receiver. Using integration by parts, \eqref{fd1} can be further simplified as
\begin{align}
\label{fd2}
\int^{\gamma_0}_{0}F(x)dx=\alpha
\end{align}
where $F(x)$ denotes the CDF of $g_0/g_1$. Hence, using \eqref{fd2} we have calculated $\gamma_0$ numerically. Note that the calculations in \eqref{first}-\eqref{fd2} are general and apply to both the equal power ($c=1$) and the unequal power ($c \ne 1$) case. In Section~III the required PDFs and CDFs were derived for the equal power case. When $c \ne 1$, it is a simple process to repeat the steps in  \eqref{first}-\eqref{fd2} and to show that using $c\alpha$ instead of $\alpha$ in \eqref{fd2} with the equal power results from Section~III gives the correct results. Hence, we only require the PDF and CDF of $g_1/g_0,g_0/g_1$ for the equal power case. To obtain results for the unequal power case, we simply use $c\alpha$ rather than $\alpha$ in \eqref{fd2}. This is equivalent to replacing $N_0$ by $N_0/c$, which makes intuitive sense since a power ratio of $c$ implies that the secondary receiver receives a signal $c$ times stronger than the primary. Hence, relative to the equal power case the SNR is $c$ times bigger and the equivalent noise level is $N_0/c$.

Figs. \ref{fig_2} and \ref{fig_3} show the secondary capacity versus $\alpha$ and under an average received interference power constraint. In all plots, AWGN refers to the scenario where $g_0$ and $g_1$ are equal to unity all the time \cite{ghasemi_b}. We make the following noteworthy observations:

\begin{figure}[t]
\centering
\includegraphics[width=0.95\columnwidth]{ave_ray_ri_1.eps}
\caption{Capacity under an average received-power constraint against $\alpha$ in Rayleigh/Rician fading. $c=0$ dB.}
\label{fig_2}
\end{figure}

\begin{figure}[t]
\centering
\includegraphics[width=0.95\columnwidth]{c2}

\caption{Capacity under an average received-power constraint against $\alpha$ in Rician/Rayleigh fading. $c=0$ dB.}
\label{fig_3}
\end{figure}

\begin{figure}[t]
\centering
\includegraphics[width=0.95\columnwidth]{icc08_himal.eps}
\caption{Capacity under a peak received-power constraint against $\alpha$ in Rayleigh/Rician fading. $c=0$ dB.}
\label{fig_4}
\end{figure}

\begin{figure}[t]
\centering
\includegraphics[width=0.95\columnwidth]{icc08_him2.eps}
\caption{Capacity under a peak received-power constraint against $\alpha$ in Rician/Rayleigh fading. $c=0$ dB.}
\label{fig_5}
\end{figure}

\begin{enumerate}

\item The secondary capacity increases if the primary receiver can tolerate more interference. This is because the secondary transmitter is able to transmit with higher power (probabilistically).

\item The case of interest in engineering practice is for a low value of $\alpha$, i.e., when the acceptable CR interference is correspondingly low. Here we see that the capacity can be sensitive to the type of fading and indeed the symmetric fading, i.e, the Rayleigh/Rayleigh case significantly overestimates the capacity compared to
the Rayleigh/Rician case in the low $\alpha$ regime. This observation is central to our contribution in
this paper. The difference in capacity reduces to almost zero when the acceptable interference at
the primary is large.

\item
The capacity of Rician/Rayleigh fading (cf. Fig. \ref{fig_3}) is not so sensitive to the $K$-factor $(0-15)$ dB. For a given $\alpha$, the Rayleigh fading on the primary link determines the transmit power of the secondary user. Once this is determined, the resulting secondary user capacity is less sensitive to the $K$-factor within
the considered range of $0-15$ dB. This is in contrast with Rayleigh/Rician fading, (cf. Fig. \ref{fig_4}), where we see that the $K$-factor induces an appreciable capacity difference especially in the low $\alpha$ regime. As $K$-factor decreases and in the low $\alpha$ regime, more opportunities for the secondary user to
transmit with relatively high power are created. However, for large $\alpha$, the impact of changing $K$-factor on the secondary user transmit power is reduced.

\item Under fading, the secondary capacity is higher than the AWGN case. This observation is consistent with the findings of \cite{ghasemi_b}. In a fading environment, the secondary user can transmit with high power, when its signal received by the primary user is subject to deep fades.

\end{enumerate}

\subsection{Capacity Under a Peak Received-Power Constraint}
As discussed in \cite{ghasemi_b}, although \emph{average} received-power is reasonable for delay insensitive applications, in other cases it is desirable to impose a peak received-power constraint. Under the peak received-power constraint \cite{ghasemi_b}
\begin{equation}
g_{0}P(g_0,g_1)\leq Q
\end{equation}
and the channel capacity was given in \cite{ghasemi_b} as
\begin{equation}
\label{peakeq}
C=B\int_{0}^{\infty}\log_2\left(1+\alpha x\right)p_{\frac{g_1}{g_0}}(x)dx
\end{equation}
Therefore, under Rayleigh/Rician fading the channel capacity is obtained by substituting the PDF in \eqref{rayrician} into \eqref{peakeq}. This gives
\begin{align}
C = (K+1)B\int^{\infty}_{0} & \log_2\left(1+\alpha x\right) \frac{x+(K+1)^2}{(x+K+1)^3}\\\nonumber &
\times e^{-K+\frac{K^2+K}{x+K+1}} dx
\end{align}
\begin{figure}[t]
\centering
\includegraphics[width=0.935\columnwidth]{e2.eps}

\caption{Capacity under average and peak received-power constraints against $\alpha$ in Rician/Rayleigh and Rayleigh/Rician fading for two different values of $c$. $K=6$ dB.}
\label{fig_6}
\vspace{-3mm}
\end{figure}
\begin{figure*}[!t]
\normalsize
\setcounter{MYtempeqncnt}{\value{equation}}
\setcounter{equation}{22}
\begin{align}
\label{uy1}
C& =K(1+K)B \int^{\infty}_{0} \log_2\left(1+\alpha x\right) \frac{e^{-\frac{K}{x+Kx+1}}}{(x+Kx+1)^2}dx
 +(1+K)^{2}B\int^{\infty}_{0}\log_2\left(1+\alpha x\right)\frac{(1-K+x)}{(x+Kx+1)^3}e^{-K+\frac{Kx+K^2x}{x+Kx+1}}dx
\end{align}
\setcounter{equation}{\value{MYtempeqncnt}}
\hrulefill
\end{figure*}

\noindent Similarly, under Rician/Rayleigh fading the channel capacity is given by \eqref{uy1} on the next page. The case where the shadowing on the two links is different can be derived using the same arguments as above. Hence, numerical results are obtained assuming $g_1$ and $g_0$ have equal power but $\alpha$ is replaced by $c\alpha$.

Figs. \ref{fig_4} and \ref{fig_5} show the secondary capacity versus $\alpha$ and under a peak received interference power constraint. We make the following noteworthy observations:
\begin{enumerate}

\item  Like the average interference power case, the capacities increase if the primary can tolerate more interference.

\item The secondary capacity is sensitive to the type of fading on the two links and depending on the fading type on either link one could overestimate the capacity especially for low values of $\alpha$ and Rayleigh/Rician fading.

\item
From \cite[Fig. 4]{ghasemi_b} in symmetric fading conditions, the capacity under a peak received power constraint is always higher than the AWGN case. However in Rayleigh/Rician fading, we see that the capacity is higher/lower than the AWGN case depending on the $\alpha$.
\end{enumerate}

\begin{figure}[t]
\centering
\includegraphics[width=0.93\columnwidth]{d2.eps}

\caption{Capacity under a peak received-power constraint and Rician/Rayleigh fading for different numbers of primary receivers. $K=6$ dB.}
\label{fig_7}
\end{figure}

\begin{figure}[t]
\centering
\includegraphics[width=0.93\columnwidth]{icc08_msim2.eps}
\caption{Capacity under a peak received-power constraint and Rayleigh/Rician fading for different numbers of primary receivers. $K=6$ dB.}
\label{fig_8}
\vspace{-3mm}
\end{figure}

Fig. \ref{fig_6} shows the impact of signal power differences on CR capacity. Such differences usually arise from shadowing and  path length differences. We assume  two values for the power ratio between the links, $c=10$ dB and $c=-10$ dB. The effect on CR capacity is a simple scaling by the $c$ parameter. Hence, we have a simple and efficient approach to investigating such asymmetric links.

\begin{figure*}[!t]
\normalsize
\setcounter{MYtempeqncnt}{\value{equation}}
\setcounter{equation}{25}
\begin{align}
\label{uys1}
C=nB\sum^{n-1}_{k=0}(-1)^{k}\binom{n-1}{k}\int^{\infty}_{0}\log_{2}\left(1+\alpha x \right) \frac{1}{(1+k+(1+K)x)^2} e^{-\frac{(1+k)K}{1+k+(1+K)x}}\left(1+K+\frac{K(K+1)^2x}{1+k+(K+1)x}\right)dx
\end{align}
\setcounter{equation}{\value{MYtempeqncnt}}
\hrulefill
\end{figure*}
\addtocounter{equation}{1}

\section{Effect of Multiple Primary Users}
When $n>1$ primary users are present, the transmit/receive powers of the secondary user would be subject to additional constraints. This leads to a capacity reduction \cite{ghasemi_b}. Let $g_{0i}$ denote the channel gain of the secondary transmitter to the $i$-th primary receiver. In this case, the peak received-power constraint is reformulated by the following constraint
\begin{align}
P(g_{01},g_{02},\ldots,g_{0n},g_{1}) \leq \min_{i} \frac{Q}{g_{0i}},\ \ i=1,\ldots,n
\end{align}
The channel capacity is given by
\begin{align}
\label{capz}
C=B\int_{0}^{\infty}\log_{2}\left(1+\alpha z\right)p_{Z}(z)dz
\end{align}
where $Z=g_1/\max_{i} g_{0i}$. In Appendix A we have derived the PDF for $Z$ when $\sqrt{g_{0i}}$, $i,=1,\dots,n$ are independent and identically distributed (i.i.d) Rayleigh RVs and $\sqrt{g_1}$ is a Rician RV. Substituting this PDF into \eqref{capz} results in the capacity given by \eqref{uys1}. Such a result can be extended to the unequal power case by considering the maximum of independent Rayleigh variables with differing means. This is possible using standard order statistic results, but is beyond the scope of the paper.

Unfortunately, the PDF for the case when $\sqrt{g_{0i}}$, $i=1,\dots,n$ are i.i.d Rician RVs and $\sqrt{g_1}$ is a Rayleigh RV could not be found in closed-form. Instead we have resorted to \emph{time consuming} Monte-Carlo simulations to obtain the capacity. The average received-power case appears to be rather complex and is not considered here. Figs. \ref{fig_7} and \ref{fig_8} illustrate the CR capacity for $n=1,2,3$ and Rayleigh/Rician and Rician/Rayleigh fading respectively. In all cases, the capacity reduces compared to the AWGN case as $n$ gets larger.

\section{Conclusions}
In this paper we have investigated the impact of asymmetric fading on the secondary user
capacity under average and peak interference power constraints. Compared to symmetric fading conditions assumed in the previous literature, our analysis have added several new insights, especially for a low value of $\alpha$, i.e., the regime that most CRs would expect to operate in practice. The results show that under Rayleigh/Rician fading and low $\alpha$, the capacity is significantly lower than that in a symmetric Rayleigh/Rayleigh fading scenario, and as $\alpha$ increases, the impact of $K$-factor on the capacity is reduced. Under Rician/Rayleigh fading, the capacity results change only slightly with different $K$-factors within considered range of $0-15$ dB. The capacity results were also extended to include the effects of different power gain and multiple primary users.

\appendices

\addtocounter{equation}{1}

\section{Derivations with multiple primary users}
Let $\sqrt{g_{0i}}$, for $i=1,\ldots, n$, be i.i.d Rayleigh RVs and let $\sqrt{g_1}$, which is independent of all $g_{0i}$, have a Rician distribution. Define $g_0=\max_{i} g_{0i}$ for $i=1,\ldots,n$ and $U=g_1/g_0$. Then the CDF of $U$ is given by
\begin{align}
\label{s1}
P(U<u)=\int^{\infty}_{0}P\left(g_1<g_0u|g_0\right)p_{g_0}(g_0)dg_0
\end{align}
The PDF of $g_0$, $p_{0}(g_0)$ is given by \cite[eq. 9.326]{simon} as
\begin{align}
\label{s2}
p_{g_0}(g_0)=n\sum^{n-1}_{k=0}(-1)^{k}\binom{n-1}{k}e^{-(1+k)g_{0}}
\end{align}
Substituting \eqref{s2} into \eqref{s1} we obtain
\begin{align}
\label{tr1}
P(U<u)& =1-n\sum^{n-1}_{k=0}(-1)^{k}\binom{n-1}{k}\\\nonumber
& \times \int^{\infty}_{0} Q_1\left(\sqrt{2K},\sqrt{2(1+K)ut}\right)e^{-(1+k)t}dt
\end{align}
After solving the integral in \eqref{tr1}, we express $P(U<u)$ in closed-form as
\begin{align}
F_{U}(u)&=1-n\sum^{n-1}_{k=0}\frac{(-1)^{k}}{1+k}\binom{n-1}{k}\\\nonumber
& \times \left(1-\frac{(1+K)u}{1+k+(1+K)u}e^{-\frac{(1+k)K}{1+k+(1+K)u}}\right)
\end{align}
Finally, differentiation of $P(U<u)$ with respect to $u$, yields the PDF of $U$. Therefore, the PDF of $U$ is given by
\begin{align}
p_{U}(u)&=n\sum^{n-1}_{k=0}(-1)^{k}\binom{n-1}{k}\frac{1}{(1+k+(1+K)u)^2} \\\nonumber & \times  e^{-\frac{(1+k)K}{1+k+(1+K)u}}\left(1+K+\frac{K(K+1)^2u}{1+k+(K+1)u}\right)
\end{align}

\section*{Acknowledgement}
This research is supported under the Australian Research Council's Discovery funding scheme (DP0774689).

config: arxiv


TITLE: Physical Significance of non-normalized state
QUESTION [0 upvotes]: What does the coefficient physically mean for an operator that isn't an observable. For an observable the coefficient is the eigenvalue and is the value that will be measured, but for operators that aren't observables they can't be measured. An example is the Ladder Operator where $a\vert n \rangle = \sqrt{n+1}\vert n+1 \rangle$. Here the ladder operator maps the state to a new one, but now there is a coefficient in front of $\sqrt{n+1}$ which means the new state isn't normalized, what is the physical significance of it?

REPLY [1 votes]: Normalisation has no effect on physical states. We say that states are rays in Hilbert space, where a ray is an equivalence class of all nonzero vectors which are identical up to a complex scalar factor. Any vector in the ray chosen as a state vector representing the the state. We usually pick a normalised vector, but then phase is still not determinate. Multiplicative factors only become important in superpositions.
This is so because all physical predictions of quantum mechanics are probabilities given by the Born rule. If the state at time $t$ is known to be $|f\rangle$, then the probability that a measurement at time $t$ will give the result $|g\rangle$ is given by
$$P(g|f) = \frac{ |\langle g |f\rangle |^2} {\langle g |g\rangle\langle f |f\rangle} $$
In the case of operators, like ladder operators, which do not correspond to observable quantities, normalisation is chosen for convenience. It has no physical meaning.


TITLE: If $x \in R $ and $\frac{(x-a)(x-c)}{(x-b)}$ can assume any real value, then which of the following may be correct
QUESTION [3 upvotes]: If $x \in R $ and $\frac{(x-a)(x-c)}{(x-b)}$ can assume any real value, then which of the following may be correct

$a>b>c$
$a>c>b$
$b>c>a$
$b>a>c$

I have tried simplifying this equation like -
$$\frac{(x-a)(x-c)}{(x-b)} = y$$
$$x^2 -xc -ax + ac = xy - by$$
$$x^2 -x(c+a-y) + (ac+by) = 0$$
Since this is always real, $D\ge 0$
$$(c+a-y)^2 - 4(ac+by) \ge  0$$
$$c^2 + a^2 + y^2 + 2(ac -cy-ay) - 4ac - 4by \ge 0$$
However, I have no idea how to continue from here. Can someone possibly advice on how to continue?

REPLY [5 votes]: You have started in the right way but you have made a small slip with the algebra. You should have $$y^2+y(2c+2a-4b)+(a-c)^2\geq0$$
For this quadratic to be non-negative, it must have a discriminant which is negative or zero so
$$(2c+2a-4b)^2-4(a-c)^2\leq0$$
This simplifies to $$(b-a)(b-c)\leq0$$
So we can conclude, ignoring the equal signs, that $b$ must lie between $a$ and $c$ So the first option could be correct.


TITLE: Get polynom from polynom, roots of second one are multiplication of first one.
QUESTION [0 upvotes]: I have a polynomial $P$, with unknown roots $r_1,r_2, ... ,r_n$.
My goal is to find a polynomial $X$ with roots $s_1,s_2, ... ,s_n$, where each $s_i = 2r_i$
I shall get $X$ with no need to find the roots of $P$.
I tried to find some relation between these polynomials by using Horner's method, but I seem to be completely lost.
Thanks for help.

REPLY [2 votes]: If you substitute $x=ay$ (with $a$ some constant) into a polynomial $P$ in $x$, then you get a polynomial $Q$ in $y$. Moreover if $x_0$ is a root of $P$, that is setting $x=x_0$ in $P$ makes the result evaluate to $0$, then substituting $y=x_0/a$ into $Q$ will have the same effect; therefore $x_0/a$ will be a root of $Q$. Therefore what you want, doubling all the roots of $P$ is achieved by taking $a=\frac12$, so substituting $x=y/2$. Afterwards you can replace $y$ by $x$, if you want a polynomial in $x$. This operation then replaces any polynomial
$$
  c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0 \quad\text{by}\quad
  \frac{c_n}{2^n}x^n + \frac{c_{n-1}}{2^{n-1}}x^{n-1}\cdots+\frac{c_1}2x+c_0.
$$
You don't need to know the roots explicitly for this.


TITLE: Is there a standard name for a category all of whose contravariant hom functors are sheaves?
QUESTION [5 upvotes]: What prompted this question is the definition of a
pseudogroup in nlab:
Given a X a topological space. Then a pseudogroup is a subgroupoid of the groupoid of transitions between open sets in X, contains the groupoid of identity transitions, and satisfies a sheaf condition.
(Pseudogroups of continuous/smooth transitions are used to define the atlases for manifolds of the respective kind).
It seems to me a pseudogroup is morally a groupoid G that satisfies the sheaf condition for each presheaf G[-,V] for V an object of G.

REPLY [4 votes]: The largest (weakest) Grothendieck topology where all contravariant hom functors are sheaves corresponds to the canonical topology. See:
http://ncatlab.org/nlab/show/canonical+topology


TITLE: Show that a polynomial family is a base to a subspace
QUESTION [1 upvotes]: QUESTION
Taking $P4 = ax^4+bx^3+cx^2+dx+e$ the vector space of polynomials of degree $4$ or lower and $H$ the subspace of $P_4$ such that its elements satisfy the conditions $p(1)= 0$, $p(-1)=0$
Show that $B = \{q_1,q_2,q_3\}$ is a base of $H$ if the polynomials are the following:
$q_1(x) = x^2-1$
$q_2(x)=x^3-x$
$q_3(x) = (x^2-1)^2$
ANSWER ?
So I know that to be a base you need to be linearly independent and you have to generate $P_4$ (in this specific case).
To prove so, I tried to calculate the matrix of the canonical coordinates of $B$, which gave me
$$\begin{pmatrix}
0&0&1\\
0&1&0\\
1&0&-2\\
0&-1&0\\
-1&0&1
\end{pmatrix}$$
Then I used Gaussian elimination and got to the following point (I had to transform $(x^2-1)^2$ into $x^4-2x^2+1$:
$$\begin{pmatrix}
-1&0&1\\
0&1&0\\
0&0&1\\
0&0&0\\
0&0&0
\end{pmatrix}$$
But I am stuck because how can I show that $B$ is a base of $H$ if $\dim(B) = 3$.
It should be $5$? I recall reading that to be a base of something, you need to have the same dimension.

REPLY [1 votes]: You are correct in that the number of elements in the basis must equal the dimension of the space. You are asked to show that $\{q_1,q_2,q_3\}$ is a basis for $H$, not for $P_4$, so you should verify that

The space $H$ is $3$-dimensional.
The basis elements $q_1$, $q_2$ and $q_3$ are contained in $H$.
The basis elements $q_1$, $q_2$ and $q_3$ span $H$.

For the latter, it suffices to show that $q_1$, $q_2$ and $q_3$ are linearly independent.

The space $P_4$ is $5$-dimensional, and $H\subset P_4$ is defined as
the subspace of polynomials $p\in P_4$ satisfying $p(1)=p(-1)=0$, or
equivalently the $ax^4+bx^3+cx^2+dx+e\in P_4$ satisfying
$$a+b+c+d+e=0\qquad\text{ and}\qquad a-b+c-d+e=0.$$ These are two
(inequivalent) linear equations, so $H$ is indeed $3$ dimensional,
not $5$-dimensional!
It is easily verified that $q_1,q_2,q_3\in H$ by plugging in $1$ and
$-1$.
Your Gaussian elimination yields a matrix of rank $3$, which means that its column vectors $q_1$, $q_2$ and $q_3$ span a $3$-dimensional space. So they span all of $H$.

Alternatively, you could argue that $q_1$, $q_2$ and $q_3$ are linearly independent along the following lines: Because $q_1$ and $q_2$ have different degrees they are certainly not scalar multiples of eachother, so they are linearly independent. Any linear combination of $q_1$ and $q_2$ has degree at most $3$, whereas $q_3$ has degree $4$, so $q_3$ is linearly independent of $q_1$ and $q_2$, and we are done.


TITLE: Problem with Integral attempt
QUESTION [4 upvotes]: Problem: Evaluate:

$$\displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1}(\pi x) - \tan^{-1}x\right)dx.$$

Attempt: $$\displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1}(\pi x) - \tan^{-1}x\right)dx.$$
$$=\displaystyle\int_{0}^{\infty} \dfrac{tan^-1 (\pi x)}{x} dx -\int_0^\infty  \dfrac{\tan^{-1}x}{x}dx.$$
Consider $$J(b) = \int_0^\infty \dfrac{\tan^-1(bx)}{x} dx$$
Differentiating $J(b)$ w.r.t (b)
$$J'(b)=\int_0^\infty \dfrac{dx}{1+(bx)^2}=\dfrac{\pi}{2b}$$
$$\Longrightarrow J(b) = \dfrac{\pi}{2}\ln b + C$$
Now how do we proceed further to find C? $J(0) = 0$, but $\ln(0)$ is not defined. $$$$

REPLY [0 votes]: Recall Frullani's Integral
$$\int_0^{\infty}\frac{f(\alpha t)-f(\beta t)}{t}dt=\left(f(0)-f(\infty)\right)\log(\beta/\alpha) \tag 1$$
where $f$ is continuous, the integral converges, and $f(\infty)=\lim_{t\to \infty}f(t)$.
For the integral of interest we have $f=\arctan t$, $\alpha =\pi$, and $\beta =1$.  We therefore have immediately from $(1)$
$$\int_0^{\infty}\frac{\arctan(\pi t)-\arctan( t)}{t}dt=\frac{\pi}{2}\log \pi$$

NOTE:
To answer the question at the end of the original post, we ought not evaluate $J$ since that integral diverges.  This is so since the arctangent function is bounded as approaches $\pi /2$ as $x \to \infty$, while the integral of $1/x$ diverges.
We could form, however, a new function $K(a)=\int_0^{\infty}\frac{\arctan(ax)-\arctan(x)}{x}dx$.  Differentiating with respect to $a$ gives the result that
$$K'(a)=\int_0^{\infty}\frac{1}{1+a^2x^2}dx=\frac{\pi}{2a} \tag2$$
Integrating $(2)$ yields $K(a)=\frac{\pi}{2}\log a +C$.
Now, we see that $K(1)=0 \implies C=0$ and thus we have
$$K(\pi)=\frac{\pi}{2}\log \pi$$
as expected!


TITLE: Is $d$ a metric for $\mathbb{R}^2?$
QUESTION [0 upvotes]: In $\mathbb{R}^2 $  define $d(x,y)$ = smallest integer greater or equal to usual
distance between $x$ and $y$. Is $d$ a metric for $\mathbb{R}^2?$
My attempt : I think No, take $x=[0.1] , y=[0.2]$ then $d(x,y)=0$ but $x\neq  y$
Is its correct ?

REPLY [3 votes]: We have that $d(x,y)=\lceil d_2(x,y)\rceil$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ is a metric. Indeed, $d$ is a metric if we change $d_2$ for any metric on $\mathbb{R}^2.$

$d(x,x)\ge 0?$

Yes, since $d(x,y)=\lceil d_2(x,x) \rceil \ge \lceil 0 \rceil \ge 0.$

$d(x,x)=0 \iff x=y?$

Yes: $x=y\iff d_2(x,y)=0 \iff \lceil d_2(x,y)\rceil=0\iff d(x,y)=0.$

$d(x,y)=d(y,x)?$

Yes, because $d(x,y)=\lceil d_2(x,y) \rceil=\lceil d_2(y,x) \rceil=d(y,x).$

$d(x,y)\le d(x,z)+d(y,z)?$

Yes: $d(x,y)=\lceil d_2(x,y) \rceil\le \lceil d_2(x,z)+d_2(y,z) \rceil \le \lceil d_2(x,z)\rceil + \lceil d_2(y,z) \rceil=d(x,z)+d(y,z).$


TITLE: Am I obtaining the Incorrect Equation for this Joint Probability density Function?
QUESTION [0 upvotes]: I am confused over how I am messing the equation up to solve this problem listed below

I was asked in Part C to find P(X > Y).
Therefore I made a double integral as such:

The answer to the problem is 15/16, but why I am I getting -1 by doing it this way?
Am I calculating the equation incorrectly?

REPLY [1 votes]: Your integration domain is just wrong, you don't consider the restriction $X < 1$, e.g. $y = \frac{3}{2}$ is part of your domain and for this the inner integral becomes negative.
Nevertheless $y = \frac{3}{2}$ is not valid for the domain $\{X > Y\}$
So your double integral should look like $$\int_0^1 \int_y^1 \ldots$$ because $X>Y$ implies $y < 1$.


TITLE: What are the branch points of the function w(z)=arctan(z)
QUESTION [0 upvotes]: I have the next question:
What are the branch points of the function $w(z)=\arctan(z)$.
I don't have any idea of how to start to solve this question.

Can someone please guide me on how to approach this question? Thank you very much in advance.

REPLY [0 votes]: They did just algebra:
$$iz=\frac{e^{2iw}-1}{e^{2iw}+1}\stackrel{\text{common denominator}}\implies(iz-1)e^{2iw}=-iz-1\implies e^{2iw}=\frac{1+iz}{1-iz}$$


TITLE: Is Huygens's Wave Theory still correct?
QUESTION [3 upvotes]: We have to study on details about Huygens's Wave Theory though we have Electromagnetic theory, quantum theory today. Is it still correct or not?

REPLY [1 votes]: Huygen's principle has certain limitations.
However, you cannot say that it is incorrect. Like dalton theory, Thomson's theory, Rutherford's theory, Huygen's theory, they were stepping stones to the study of optics. Even we can consider light as wave, for example - when we light a candle in dark room , it would spread light around all sides thus we can consider light as a wave here.


TITLE: Construct the smallest graph homeomorphic to a given graph by smoothing
QUESTION [5 upvotes]: The homeomorphism class $ \mathcal{H}(G) $ of a graph $G$ is the set of isomorphism classes of graphs that are topologically homeomorphic to $G$. It is natural to ask: Is there a "smallest" representative in the homeomorphism class? If yes, how to find it? Unfortunately, I found no useful result on this problem after a quick google search.
Nevertheless, guided by intuition, I have the following hypothesis:

The smallest graph homeomorphic to a graph is obtained by smoothing
every maximal ear.

In this post I attempt to sketch a proof, but there is a gap in the proof, so I don't know whether my hypothesis is actually correct. I would appreciate anyone for pointing out my mistakes and filling out the gap.
Warning: this would be a long post
First, let's clear up some terminology. The term "ear" means different things in different graph theory textbooks. In this post, we adopt the following definition:

Definition 1
An ear in a graph is either:

a cycle whose all except possibly one vertices are of degree $2$, or
a path whose all internal vertices are of degree $2$.

A maximal ear is an ear that is not a proper subgraph of a larger ear.
Equivalently, a maximal ear is one of the following:

a cycle that is a whole connected component on its own
a cycle in which exactly one vertex is of degree $ \geq 3 $, while all other vertices are of degree $2$
a path in which all internal vertices are of degree $2$, while both end vertices are of degree $ \neq 2 $

Two common operations that preserve topology on graphs are subdividing and smoothing:

Definition 2
Subdividing an edge means replacing it by an ear. More precisely, let
$e = uv$ be an edge.
If $u = v$, then subdividing the self-loop $e$ means replacing it by a
cycle $C$, and $u = v$ becomes a vertex on $C$, which may or may not
have degree $2$, depending on whether $e$ is isolated.
On the other hand, if $u \neq v$, then subdividing $e$ means replacing
it by a path $P$, and $u, v$ become the end vertices of $P$.
Subdividing a graph means preforming a sequence of subdividing on
edges.

Definition 3
Smoothing an ear means replacing it by a single edge. More precisely,
let $C$ be an ear.
If $C$ is a cycle, then smoothing $C$ means replacing it by a
self-loop $e$, and the vertex of degree $ \neq 2 $ on $C$ becomes the
only vertex incident on $e$ (if all vertices on $C$ are of degree $2$,
just choose any vertex).
On the other hand, If $C$ is actually a path $P$, then smoothing $P$
means replacing it by a loopless edge $e$, and the end vertices of $P$
become the end vertices of $e$.
Smoothing a graph means preforming a sequence of smoothing on ears.

Next, we have the following classic result on topology of graphs:

Theorem 1
Two graphs are homeomorphic if and only if one of them can be obtained
from a sequence of subdividing and smoothing operations on the another.

Proof: See this post.

Theorem 2
Let $G$ and $H$ be two homeomorphic graphs. Then $ |V(G)| = |V(H)| $
if and only if $ |E(G)| = |E(H)| $.

Sketch of proof: Subdividing (resp. smoothing) always increases (resp. decreases) the number of vertices and edges by the same number. $\square$
In light of Theorem 2, we can define an ordering on the homeomorphism class of a graph:

Definition 4
Let $ \mathcal{H}(G) $ be the homeomorphism class of a graph $G$.
Define the ordering $\preceq$ on $ \mathcal{H}(G) $ by:
$$ G_1 \preceq G_2 \iff |V(G_1)| \leq |V(G_2)| $$
for any $ G_1, G_2 \in \mathcal{H}(G) $.
If $ G_1 \preceq G_2 $ and $ G_2 \preceq G_1 $, then we denote
$ G_1 \sim G_2 $.

The ordering $\preceq$ is a total preorder, meaning it is transitive and any two homeomorphic graphs are comparable. Unfortunately it is not a total order, since $ G_1 \sim G_2 $ does not imply $ G_1, G_2 $ are isomorphic, even through Theorem 2 implies $ |E(G_1)| = |E(G_2)| $.

Theorem 3
Any graph without isolated vertex can be uniquely decomposed into an
edge-disjoint union of maximal ears.

Sketch of proof:
Let $G$ be a graph without isolated vertex. Define the relation $R$ on $E(G)$ by:
$$ eRf \iff \exists \text{ ear } C \subseteq{G} \text{ s.t. } e, f \in E(C) $$
for any $ e, f \in E(G) $.
Then $R$ is an equivalence relation on $E(G)$, in which each equivalence class contains the edges of exactly one maximal ear. Thus, $R$ induces a decomposition of $G$ into an edge-disjoint union of maximal ears. Conversely, any such decomposition must be induced by $R$, so the decomposition is unique. $\square$
Based on the above decomposition, we can define the following:

Definition 5
A graph without isolated vertex is called smooth if every maximal ear
is of length $1$. For a graph $G$ without isolated vertex, the smooth
graph obtained from smoothing every maximal ear in $G$ is denoted as
$ \text{Smooth} (G) $.

The term "smooth graph" is not standard, but I could not find any existing term for such a graph, so I just make it up on my own.

Theorem 4
Let $G$ be a smooth graph without isolated vertex and
$ H \in \mathcal{H}(G) $, then $ G \preceq H $. Moreover,
$ G \sim H $ if and only if $H$ is a smooth graph.

Sketch of proof:
By Theorem 1, $H$ can obtained from a sequence of subdividing and smoothing operations on $G$. Each step of the operations can only change one of the maximal ear to another maximal ear of possibly different length.
On the other hand, in a smooth graph all the maximal ears already have the shortest possible length (namely, $1$), so any sequence of subdividing and smoothing can never further decrease the number of vertices. Thus, $ |V(G)| \leq |V(H)| $ and the equality holds if and only if $H$ is smooth. $\square$
The following claim is based on intuition, but I don't know how to prove it. It is where the gap of my whole proof lies.

Claim 0
Let $G$ and $H$ be two smooth graphs without isolated vertex. If
they are homeomorphic, then they are isomorphic.

Finally, assuming the above claim, we can prove the main hypothesis:

Main Hypothesis
Assume Claim 0 is correct and let $G$ be a graph without isolated
vertex. Then $ \text{Smooth} (G) $ is the unique smallest graph in
$ \mathcal{H} (G) $ with respect to the ordering $ \preceq $.

Proof:
The fact that $ \text{Smooth} (G) \preceq H $ for all $ H \in \mathcal{H} (G) $ follows from Theorem 4.
To prove uniqueness, let $ H \in \mathcal{H} (G) $ be such that $ \text{Smooth} (G) \sim H $. Since $ \text{Smooth} (G) $ is smooth and $ H \in \mathcal{H} (\text{Smooth} (G)) $, by Theorem 4, $H$ is smooth as well. Claim 0 then implies $H$ is isomorphic to $ \text{Smooth} (G) $. $\square$
Questions:

Is Claim 0 correct? How to prove it?
Even if Claim 0 is wrong, is my main hypothesis still correct?
Are there any other mistakes in my proof?
Is there a better term for graphs whose every maximal ear is of length $1$, other than "smooth graphs"?

REPLY [2 votes]: Your proof appears correct to me. I don't see why you assume that the graph has no isolated vertices – does it make a difference anywhere? Also, "smooth graph" seems to be a fancy name for a graph with no vertices of degree two. (More precisely, the only vertices of degree two are isolated vertices with a loop.)
I'll give a proof of your claim. We may assume that the graphs in question are connected and that they have at least one edge. To any graph $G$, associate a vertex-colored graph $Ear(G)$ in the following way:

The vertices of $Ear(G)$ correspond to the ears in the unique decomposition of $G$ into maximal ears. They are colored blue and red according to whether the ear is a path or a cycle.
Two vertices are adjacent if the corresponding ears have a common vertex; if they have two common vertices, then we draw two parallel edges. (This can only happen if the corresponding ears are paths.)

There are two observations to be made which are more or less implicit in your proof of Theorem 4:

If $G$ and $H$ are homeomorphic, then $Ear(G)$ and $Ear(H)$ are isomorphic, with the isomorphism preserving the vertex colors. This follows from Theorem 1 after checking that both smoothing and subdividing preserve $Ear(G)$.
If $G$ is smooth, then (disregarding the coloring) $Ear(G)$ is just the line graph of $G$, defined appropriately for graphs with loops and multiple edges.

Conveniently, a theorem of Whitney states that if the line graphs of two connected simple graphs isomorphic, then the graphs themselves are isomorphic, except if the graphs are the triangle $K_3$ and the claw $K_{1,3}$. Note that the triangle is not smooth. In the case of graphs with loops and parallel edges, the situation is more complicated (though not terribly so, according to this article* to which I could only find a paywalled link; funnily enough, the name of Whitney is misspelled in the title), but in our case the vertex-coloring and Theorem 4 give us enough information to uniquely reconstruct the original graph. You could probably sort this out yourself, but I'll give the details for completeness.
So suppose that $G$ and $H$ are smooth and that $Ear(G)$ and $Ear(H)$ are isomorphic. First, we deal with loops: these correspond precisely to the red vertices of $Ear(G)$ and $Ear(H)$. It follows that if we denote by $G'$ and $H'$ the graphs obtained by deleting the loops in $G$ and $H$, then $Ear(G')$ and $Ear(H')$ are obtained by deleting the red vertices from $Ear(G)$ and $Ear(H)$; in particular, they are isomorphic. Now it is enough to show that $G'$ and $H'$ are isomorphic, since then the positions of the loops are uniquely determined by $Ear(G)$: a vertex in $G'$ has a loop if and only if there is an edge incident to it which is adjacent to a red vertex in $Ear(G)$, or if $G'$ consists of this single vertex (since we assumed that our graphs have at least one edge).
Thus, we may assume that $G$ and $H$ contain no loops. Now we just have to take care of parallel edges. If two edges are parallel in $G$, then by our construction the corresponding vertices in $Ear(G)$ are connected by two parallel edges. More generally, two or more parallel edges in $G$ correspond to a clique in $Ear(G)$ in which every edge is doubled. Every vertex in $Ear(G)$ is contained in a unique maximal such "double clique" (potentially of size one), and by contracting these cliques and replacing newly formed parallel edges by single ones, we obtain the line graph of the simple graph underlying $G$. Since this works backwards as well (i.e. from the simple graph and $Ear(G)$ we can recover $G$), we may assume that $G$ and $H$ are simple.
So we are done by Whitney's theorem, right? Well, not so fast. It could happen that after leaving loops and parallel edges from $G$ and $H$, we are left with a triangle and $K_{1,3}$: after all, a triangle with doubled edges is smooth. But this is ruled out by Theorem 4: the original $G$ and $H$ had the same number of vertices, and leaving edges doesn't change that. So $G$ and $H$ are indeed isomorphic.
*Note that, as far as I can tell, the notion of line graph used in the article is different from the one used here, in that the vertices corresponding to parallel edges are still connected with only one edge.


TITLE: Find $\mathcal{F}$ s.t. for each family $\mathcal{G}$ of $n$ sets each of size $n/2$, there's a $F\in \mathcal{F}$ intersecting each $G\in\mathcal{G}$
QUESTION [2 upvotes]: Show that given $N=\{1,\ldots, n\}$, there exists a family $\mathcal{F}$ such that $|\mathcal{F}| = n^{10}$, for each $F\in \mathcal{F}$ we have $|F| \leq 10\log n$ and such that for every family $\mathcal{G}$ of $n$ subsets of $N$, each of cardinality $n/2$, there is an $F\in \mathcal{F}$ such that $F$ intersects all members of $\mathcal{G}$. Then find a deterministic polynomial time algorithm that determines such an $\mathcal{F}$.
This is a question from Alon and Spencer from a chapter on derandomization. I am having a hard time even getting started. In general, the chapter is about how we can deterministically find structures we are interested in by first showing that a suitable random structure is expected to have the desired properties. Then we make a sequence of choices that fix part of our structure in such a way that the expectation conditioned on the first choices moves in the right direction. Once all choices are made, the expectation then just counts something, which is then bounded by the bound in the random structure.
So I guess maybe we should let $\mathcal{F}$ consist of random sets of size $10\log n$. Given $F$ of size $10\log n$ selected randomly and $G$ of size $n/2$, we have
\begin{equation}
\mathbb{P}(F\cap G = \emptyset) = \frac{ \binom{n/2}{10\log n} }{\binom{n}{10\log n}}\sim \frac{ 1 }{2^{10\log n} }.
\end{equation}
Then, given a family $\mathcal{G}$ as described, we have
\begin{equation}
\mathbb{P}(\emptyset \notin F\cap \mathcal{G}) \geq 1-\frac{ n }{2^{10\log n} }.
\end{equation}
This is where I am stuck... I do not see where to get an expectation of something that I can use throughout my construction. Moreover, we cannot take $\mathcal{G}$ to be random, we need to build an $\mathcal{F}$ that works simultaneously for all $\mathcal{G}$. Also, the number of subsets of $N$ of size $n$ with each element of size $n/2$ is exponential in $n$, so I think we should be able to find some alternative sufficient condition on the $\mathcal{F}$ that we can use...
Thanks in advance for any help!
EDIT: So as per Yuval Filmus' answer, indeed the existence part is quite easy. However, it does not really help in the construction of a deterministic poly time algorithm. I think we probably need some estimates on the conditional probabilities that are computable in polytime.

REPLY [3 votes]: Did you take a look at the hint provided by Alon and Spencer? I didn't work it out too formally, but I think the following roughly works: using, say, the Lubotzky or Margulis explicit construction of expanders, one may check that the preconditions for Corollary 9.2.8 in their book applies, with parameters $d=30$, $\lambda\leq 2\sqrt{30-1}=2\sqrt{29}$, $C$ an arbitrary subset of size $n/2$, and $c=\vert C\vert/n=1/2$. What that result tells you is that the probability a randomly chosen walk of length of length $2.5\log n$ avoids $C$ is at most $2^{-2.5\log n/2}=1/n^{1.25}$. In particular, for any $n$ subsets $C_1,\ldots,C_n$ all with $\vert C_i\vert=n/2$, it follows that there exists some walk of length $2.5\log n$ that intersects all of them simultaneously. The number of such walks is $30^{2.5\log n}n\leq n^{9.6}$. So the algorithm is just to use the explicit expander construction and simply list all such walks (where of course, each vertex is labeled with a unique element of $[n]$), which can evidently be done in polynomial time.


TITLE: Phase transition on magnetic materials
QUESTION [1 upvotes]: Is ferromagnetic to paramagnetic phase transition a reversible process?
If I start with a ferromagnetic material with a spontaneous magnetization below the Curie temperature, and then I start to heat it, it will become paramagnetic above the critical temperature. If I then start to drop the temperature slowly to below the Curie temperature then will I achieve the ferromagnetic behaviour with same spontaneous magnetization as before?

REPLY [0 votes]: Going above and below the Curie point, back and forth, in a cyclical manner was done by Giauque in his use of cryogenic cooling of a paramagnetic salt, see magneto caloric effect.
Similar adiabatic cooling effect can also be observed in Gadolinium metal and also in its alloys at room or near room temperature. The effect is used in developing commercial refrigerators. Gschneidner and Pecharsky in MAGNETOCALORIC MATERIALS claim that the transition between ferromagnetic and paramagnetic is a reversible process:

In the case of a ferromagnet near its magnetic ordering temperature,
the adiabatic application of a magnetic field reduces the magnetic
entropy of a solid and, in turn, it is heated via the increase of its
lattice entropy to maintain the entropy of a closed system at a
constant value. In a reversible process, a ferromagnet is cooled as
the magnetic entropy increases, and the lattice entropy decreases upon
adiabatic removal of the magnetic field.


TITLE: finding integral of $ \int \frac{1}{\sin x + \sqrt{3} \cos x}\ dx $
QUESTION [2 upvotes]: In $ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx $, If I multiply and divide by $1/2$ I get $$ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx ,$$
then I can write $ 1/2=\sin (\frac{\pi}{6}) $ and $ \sqrt{3}/2=\cos (\frac{\pi}{6}) $ and apply $\cos(a-b)=\cos a \cos b + \sin a \sin b$,
I'll get $ \int\frac{1}{\cos(\frac{\pi}{6} - x)}\ dx $, I get $ \int \sec(\frac{\pi}{6} - x)\ dx $, now we know $ \int \sec(x)\ dx = \log|\tan(\frac{\pi}{4}+\frac{x}{2})| + c $ ,
so I finally get $$ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx = \frac{1}{2}\log|\tan(\frac{\pi}{4}+\frac{\pi}{6}-\frac{x}{2})| + c $$
However if at this step $ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx $
I take $ 1/2=\cos (\frac{\pi}{3}) $ and $ \sqrt{3}/2=\sin (\frac{\pi}{3})$, then what I get is $ \int\frac{1}{\sin(\frac{\pi}{3} + x)}\ dx $, which is equal to $ \int \csc(\frac{\pi}{3} + x)\ dx $, now we know $ \int \csc(x)\ dx = \log|\tan(\frac{x}{2})| + c $
so I finally get $$ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx = \frac{1}{2}\log|\tan(\frac{\pi}{6}+\frac{x}{2})| + c $$
why answer is different, I have rechecked and could not find any mistake

REPLY [1 votes]: Both of your methods have a mistake in them. In your first method, you should have
$$\frac{\sin\left(\frac{\pi}{6}\right)}{\sin\left(\frac{\pi}{6}\right)\sin\left(x\right)+\cos\left(\frac{\pi}{6}\right)\cos\left(x\right)} = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}-x\right)}.$$
Similarly, your second method should have
$$\frac{\cos\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)\sin\left(x\right)+\sin\left(\frac{\pi}{3}\right)\cos\left(x\right)} = \frac{\cos\left(\frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{3}+x\right)}.$$
I'm guessing you forgot about the numerator, right? I think you can figure out the rest from there. Integrating both of them should result in the same answer, probably differing by a constant (I didn't try it out).


TITLE: Back-projecting Pixel to 3D Rays in World Coordinates using PseudoInverse Method
QUESTION [5 upvotes]: For perspective projection with given camera matrices and rotation and translation we can compute the 2D pixel coordinate of a 3D point.

using the projection matrix,
$$
P = K [R | t]
$$
where $K$ is intrinsic camera matrix, $R$ is rotation $t$ is translation. The projection is simple matrix multiplication $x = P X $.
Zisserman's book, pg. 161 suggests using $3 \times 4$ projection matrix and taking pseudoinverse. Then one would compute $X$ which defined up to scale which can then be interpreted as the ray starting from camera center going to infinity. I quickly coded this up, I took $Z$ as depth, so I translated the camera in $Y$ direction (up 1 meter), and after retrieving $X$ flipped $Y,Z$ for plotting (most projective geom. math seems to be built to make $Z$ depth),
K = [[ 282.363047,      0.,          166.21515189],
     [   0.,          280.10715905,  108.05494375],
     [   0.,            0.,            1.        ]]
K = np.array(K)
R = np.eye(3)
t = np.array([[0],[1],[0]])
P = K.dot(np.hstack((R,t)))

import scipy.linalg as lin

x = np.array([300,300,1])
X = np.dot(lin.pinv(P),x)
X = X / X[3]
from mpl_toolkits.mplot3d import Axes3D
w = 20
f = plt.figure()
XX  = X[:]; XX[1] = X[2]; XX[2] = X[1]
ax = f.gca(projection='3d')
ax.quiver(0, 0, 1., XX[:3][0], XX[:3][1], XX[:3][2],color='red')
ax.set_xlim(0,10);ax.set_ylim(0,10);ax.set_zlim(0,10)
ax.quiver(0., 0., 1., 0, 5., 0.,color='blue')
ax.set_xlabel("X")
ax.set_ylabel("Y")
ax.set_zlabel("Z")
ax.set_title(str(x[0])+","+str(x[1]))
ax.set_xlim(-w,w);ax.set_ylim(-w,w);ax.set_zlim(-w,w)

ax.view_init(elev=29, azim=-30)
fout = 'test_%s_01.png' % (str(x[0])+str(x[1]))
plt.savefig(fout)
ax.view_init(elev=29, azim=-60)
fout = 'test_%s_02.png' % (str(x[0])+str(x[1]))
plt.savefig(fout)

These images below are the result (blue arrow shows the normal vector perpendicular to the image plane, the images demonstrate all x=10,300 y=10,300 combinations):

I give the camera/ray plot for each pixel from two different angles.
Do these results look sensible? 10,10 and 200,200 looked odd, I played around with signs a little bit, if I translate up using negative -1, and using -Z after X calc., things improve somewhat?
t = np.array([[0],[-1],[0]])
..
XX  = X[:]; XX[1] = X[2]; XX[2] = -X[1]

I do not know why that is.

REPLY [0 votes]: Since this is the first and only Google result that actually helps, and it took quite a while to create, here it is in Numpy
import numpy as np

def convert_nx2_to_homo_3xn(points_nx2: np.array):
    return np.hstack([points_nx2, np.ones((points_nx2.shape[0], 1))]).T

def convert_nx3_to_homo_4xn(points_nx3: np.array):
    return np.hstack([points_nx3, np.ones((points_nx3.shape[0], 1))]).T

def cast_2d_points_as_3d_rays(sub_pixels_nx2: np.array, proj_3x4: np.array):
    """
    see Harley & Zisserman pg 162, section 6.2.2, figure 6.14
    see https://math.stackexchange.com/a/597489/541203

    cast rays from camera center through the sub_pixels_nx2. Return camera center and ray directions.
    """
    m_3x3 = proj_3x4[:, :3]
    p4_3x1 = proj_3x4[:, 3]
    m_inv_3x3 = np.linalg.inv(m_3x3)

    # projection matrix to camera center
    camera_center_3x1 = np.expand_dims(-m_inv_3x3 @ p4_3x1, 1)
    camera_center_homo_4x1 = np.vstack([camera_center_3x1, 1])

    # projection matrix + pixel locations to ray directions
    sub_pixels_homo_3xn = convert_nx2_to_homo_3xn(sub_pixels_nx2)
    ray_directions_3xn = m_inv_3x3 @ sub_pixels_homo_3xn
    ray_directions_homo_4xn = convert_nx3_to_homo_4xn(ray_directions_3xn.T)
    ray_directions_homo_4xn[3, :] = 0  # this is a direction

    return ray_directions_homo_4xn, camera_center_homo_4x1

```


TITLE: Tangent half angle formula
QUESTION [3 upvotes]: So we start with the following tangent half angle formula:
$$ \tan\left(\frac \theta2\right) = \pm\sqrt{\frac {1 - \cos \theta}{1 + \cos \theta}} $$
If I do some algebraic manipulation I end up with the following below:
$$ \tan \left(\frac \theta2\right)= \pm\frac {1 - \cos \theta} {\sin \theta}$$
Now according to Michael Corral's Trigonometry the minus sign is not possible so I only end up with:
$$ \tan \left(\frac \theta2\right)= \frac {1 - \cos \theta} {\sin \theta} $$
Can you please explain why that is true?

REPLY [3 votes]: The fussiness over $\pm$ arises because $\tan \frac\theta2$ can be either positive or negative, whereas the square root is always considered positive. Your algebraic manipulation of $\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$ is OK except in the final step: the result should be $\left|\frac{1-\cos\theta}{\sin\theta}\right|$, as in @haqnatural's derivation, which leads to
$$
\tan\frac\theta2 = \pm\left|\frac{1-\cos\theta}{\sin\theta}\right|.\tag1
$$
Equation (1) is true, but doesn't get us any closer to resolving the question of which sign to choose! One way to decide this is to consider different ranges of $\theta$, as in @imranfat's answer.
If this is not satisfying, you can try an alternative derivation that doesn't fuss with plus and minus signs:
$$
\tan\frac\theta2=\frac{\sin\frac\theta2}{\cos\frac\theta2}=\frac{2\sin\frac\theta2\sin\frac\theta2}{2\sin\frac\theta2\cos\frac\theta2}=\frac{1-\cos\theta}{\sin\theta},
$$
using the double-angle identities
$$
\cos2t=1-2\sin^2t
$$
and
$$\sin2t = 2\sin t\cos t.$$


TITLE: find the area of the region lying inside the circle $r=6$ and inside the cardioid $r=4-3\sin \theta$.
QUESTION [1 upvotes]: Well, I drew a graph to visualise it and I found the interceptions $\theta=\arcsin \left(-\frac{2}{3}\right)$. From the graph, by symmetry, I found that the area of region from $\theta$ to $\pi/2$ and from $\pi/2$ to $\pi-\theta$.
So How would I apply the formula now?

REPLY [0 votes]: You want to compute the area integral
$$\int_A {1} dA$$
where $A$ is the intersection of the interior of the circle and the cardioid i.e.
\begin{align*}
A &= \{(x, y) : r = \sqrt{x^2 + y^2} \leq 6\} \cap \{(x, y) : r = \sqrt{x^2 + y^2} \leq 4 - 3 \sin \theta \}\\
&= \{(x, y) : r \leq \min(6, 4 - 3 \sin \theta) \}
\end{align*}
So let
$$f(\theta) = \min(6, 4 - 3 \sin \theta)$$
and you can rewrite
$$A = \{(x, y) : r \leq f(\theta)\}$$
and the problem is equivalent to computing area of the region bounded by the closed curve $r = f(\theta)$ and so is the same as computing
$$A = \int_0^{2\pi} \frac{1}{2} f(\theta)^2 d\theta$$
[See this link for details.]
Now you have to split the integration range to the region where $f$ takes value 6 and where $f$ take value $4 - 3 \sin \theta$. It is easily solved: for $\theta \in [0,2\pi)$
$$6 \leq 4 - 3 \sin \theta \iff 2 \leq -3 \sin \theta \iff -\frac{2}{3} \geq \sin\theta \iff \theta \in [\theta_1, \theta_2]$$
where
$$\theta_1 = \pi - \arcsin\left(-\frac{2}{3}\right); \theta_2 = 2 \pi + \arcsin\left(-\frac{2}{3}\right); 0 < \theta_1 < \theta_2 < 2\pi.$$
This means
$$f(\theta) = \begin{cases}
6 &\text{if }\theta_1 \leq \theta \leq \theta_2\\
4 - 3 \sin \theta &\text{otherwise}
\end{cases}$$
Going back to the integral
\begin{align*}
A &= \int_0^{2\pi} \frac{1}{2} f(\theta)^2 d\theta\\
&= \int_0^{\theta_1} \frac{1}{2} (4 - 3 \sin)^2 d\theta + \int_{\theta_1}^{\theta_2} \frac{1}{2} 6^2 d\theta + \int_{\theta_2}^{2\pi} \frac{1}{2} (4 - 3 \sin)^2 d\theta
\end{align*}
(You should be able to complete the computation.)


TITLE: Expectation value of position in infinite square well
QUESTION [0 upvotes]: I'm looking for some help to a question.
I'm working in the infinite square well, and I have the wavefunction:
$$\psi(x,t=0)=A\left( i\sqrt{2}\phi_{1}+\sqrt{3}\phi_{2} \right).$$
For every time t, the wavefunction is:
$$\psi(x,t)=A\left( i\sqrt{2}\phi_{1}e^{-iE_{1}t/\hbar}+\sqrt{3}\phi_{2}e^{-iE_{2}t/\hbar} \right).$$
Now, I'm asked to calculate the expectation value of the particles position $\left\langle x\right\rangle (t)$.
My guess was to just do it like this:
$$\left\langle x \right \rangle(t) = \int_{-\infty}^{\infty}x|\Psi(x,t)|^{2}dx,$$
and then use last of my two wavefunctions, and use the Kronecker Delta argument to remove the terms that got $i \neq j$.
But, that doesn't seem to work. The answer should be:
$$\langle x \rangle (t) =\frac{32\sqrt{6}a}{45\pi^{2}}\sin((E_{1}-E_{2})t/\hbar).$$
But if I do what I said, I will get something like 1 I think.
So I'm guessing I'm going all wrong about, so I was hoping someone could give me a hint :)

REPLY [2 votes]: Since you're only asking for a hint I will point out what I think you're missing:
The integral for the expectation value will yield 4 terms
$$\langle x \rangle (t) = \int_{-\infty}^\infty \psi(x,t)^* x \psi(x,t) dx $$
$$= |A|^2 \int_{\infty}^{\infty} \left( -i\sqrt{2}\phi_1^* e^{+iE_1t/\hbar} + \sqrt{3} \phi_2^* e^{+iE_2t/\hbar}\right)x\left(i\sqrt{2}\phi_1e^{-iE_1t/\hbar} + \sqrt{3}\phi_2 e^{-iE_2t/\hbar}\right)dx$$
$$ = |A|^2 \int_{-\infty}^{\infty}\left(2\phi_1^*x\phi_1 + 3\phi_2^*x\phi_2 + i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$
The expressions $\phi_1^*\phi_1$ and $\phi_2^*\phi_2$ are even functions in position space, when standing alone. Remember that if you integrate over an even function multiplied by an odd function (e.g. $x$) it evaluates to zero:
$$\int_{-\infty}^{\infty} \phi_1^* \phi_1 dx = 1$$
However
$$\int_{-\infty}^{\infty} \phi_1^* x \phi_1 dx = 0$$
Hence, you actually lose the first two terms (not the 'cross' terms like you mentioned).
On the other hand, for your cross terms, you have
$$\int_{-\infty}^{\infty} \phi_1^* \phi_2 dx = 0$$
However, this is your mistake (which Will pointed out in his comment)
$$\int_{-\infty}^{\infty} \phi_1^* x \phi_2 dx \neq 0$$
After you remove the appropriate terms you are left with the following expression to evaluate.
$$ = |A|^2 \int_{-\infty}^{\infty}\left(i\sqrt{6}\left( \phi_2x\phi_1^*e^{i(E_2-E_1)t/\hbar} - \phi_1^* x \phi_2 e^{-i(E_2-E_1)t/\hbar}\right)\right)dx$$
And remember that
$$ \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$


TITLE: Completely positive maps-equivalent definition
QUESTION [12 upvotes]: The most usual definition of the completely positive map (c.p.) between two C*-algebras (say, unital) is the following: $\sigma: A \to B$ should satisfy $\sigma(1)=1$ and for each $n \in \mathbb{N}$ the map $\sigma_n: M_n(A) \to M_n(B)$ defined ``cordinatewise''should be positive.
However I met also the following definition: $\sigma(1)=1$ and for each $n \in \mathbb{N}, a_1,...,a_n \in A, b_1,...,b_n \in B$ we have $\sum_{i,j}b_i^*\sigma(a_i^*a_j)b_j \geq 0$. I was able to show that the first definition implies the second: how to prove that the converse is also true (if it is so)?

REPLY [13 votes]: The converse follows from two observations:
1) If $a=[a_{ij}]$ is a positive element in $M_n(A)$, then $a=x^*x$ for some $x=M_n(A)$, and if we write this out in matrix entries we have
$$
a_{ij} = \sum_k x_{ki}^*x_{kj}.
$$
2) The matrix $[a_{ij}]$ is positive in $M_n(A)$ if and only if it is positive in every GNS representation $\pi$ of $A$; and this is implied by the condition
\begin{equation}
\sum_{ij}c_i^*a_{ij}c_j\geq 0
\end{equation}
for all $c_1,\dots c_n\in A$. To see this fix a state $\rho$ on $A$, let $H$ denote the GNS space associated to $\rho$ and $\pi:A\to B(H)$ the GNS representation. $H$ is spanned by equivalence classes $[c]$ with $c\in A$.  Note that since $A$ is unital, by the GNS construction we have $[c]=\pi(c)[1]$. To check that $[a_{ij}]$ is positive in the representation $\pi$, fix vectors $[c_1], \dots [c_n]$ in $H$ and complute:
\begin{align}
\sum_{ij}\langle \pi([a_{ij}])[c_j],[c_i]\rangle_H &= \sum_{ij}\langle \pi(c_i^*)\pi(a_{ij})\pi(c_j)[1],[1]\rangle_H \\
&=\sum_{ij}\langle \pi(c_i^*a_{ij}c_j)[1].[1]\rangle_H\\
&= \rho \left(\sum_{i,j} c_i^*a_{ij}c_j\right)\\
&\geq 0
\end{align}
by assumption.
Now, for the proof: suppose $\sum_{ij} b_i^*\sigma(a_i^*a_j)b_j\geq 0$ for all $a's$ and $b's$.  Fix a positive element $[a_{ij}]\in M_n(A)$, factor it as in the first observation. To check the positivity of $\sigma([a_{ij}])$, we check the condition of the second observation; so fix $b_1, \dots b_n$; we have
\begin{equation}
\sum_{ij}b_i^*\sigma(a_{ij})b_j = \sum_k\sum_{ij}b_i^*\sigma(x_{ki}^*x_{kj})b_j\geq 0,
\end{equation}
where the last inequality follows by hypothesis.


TITLE: Adjoint action as a natural transformation
QUESTION [1 upvotes]: Let $\mathbf{G}$ be a group functor and $R$ an algebra with inclusion map $i_R : R \to R[\epsilon],$ where $R[\epsilon]$ is the algebra of dual numbers. The adjoint action of $\mathbf{G}(R)$ on $\mathbf{Lie}(\mathbf{G})(R)$ is defined as
$$ \mathrm{Ad}(g)(x) = \mathbf{G}(i_R)(g) \cdot x \cdot \mathbf{G}(i_R)(g^{-1}),$$
for $g \in \mathbf{G}(R)$ and $x \in \mathbf{Lie}(\mathbf{G})(R).$
The adjoint action can be extended to a natural transformation $\mathrm{Ad} : \mathbf{G} \times \mathbf{Lie}(\mathbf{G}) \to \mathbf{Lie}(\mathbf{G}).$ To prove this, we need to show that for an algebra morphism $f : R \to S,$
$$ \mathrm{Ad}(\mathbf{G}(f)(g))(\mathbf{Lie}(\mathbf{G})(f)(x)) = \mathbf{Lie}(\mathbf{G})(f) (\mathrm{Ad}(g)(x)) $$
In Zhihau Chang's notes on Lie algebras of affine group schemes, Section 2.3, the proof of this is as follows, where $\tilde f : r + \epsilon r' \mapsto f(r) + \epsilon f(r')$ is the extension of $f$ to $R[\epsilon] \to S[\epsilon],$
\begin{align}
\mathrm{Ad}(\mathbf{G}(f)(g))(\mathbf{Lie}(\mathbf{G})(f)(x))
&= \mathbf{G}(i_S \circ f)(g) \cdot \mathbf{G}(\tilde f)(x) \cdot \mathbf{G}(i_S \circ f)(g^{-1})\\
&= \mathbf{G}(\tilde f \circ i_R)(g) \cdot \mathbf{G}(\tilde f)(x) \cdot \mathbf{G}(\tilde f \circ i_R)(g^{-1})\\
&= \mathbf{G}(\tilde f) \left( \mathbf{G}(i_R)(g) \cdot x \cdot \mathbf{G}(i_R)(g^{-1}) \right) \\
&= \mathbf{G}(\tilde f)(\mathrm{Ad}(g)(x)) \\
&= \mathbf{Lie}(\mathbf{G})(f) (\mathrm{Ad}(g)(x))
\end{align}
I am struggling to see how the third equality holds.
My understanding is that $\cdot$ is composition in $\mathbf{Lie}(\mathbf{G})(R),$ so distributivity does not make sense.
Another reference is J.S. Milne's Basic Theory of Affine Group Schemes, Chapter XI, Section 7.

REPLY [1 votes]: $\cdot$ here is multiplication in $\mathbf{Lie(G)}(R)$, which is the same as multiplication in $\mathbf{G}(R(\tau))$, in particular, $\mathbf{G}$ being a $k$-group functor, $\mathbf{G}(\tilde{f})$ respects that multiplication.
You have $\newcommand{\G}{\mathbf{G}} \G(\tilde{f}\circ i_R)(g)\cdot \G(\tilde{f})(x)\cdot G(\tilde{f}\circ i_R)(g^{-1}) = \G(\tilde{f})(\G(i_R)(g))\cdot \G(\tilde f)(x)\cdot \G(\tilde f)(\G(i_R)(g^{-1}) $ and so $\G(\tilde{f})$ factors out.


TITLE: Why is $ r \propto v$ in the equation $r=\frac{mv}{qB}$?
QUESTION [0 upvotes]: For a charged particle in a magnetic field,
$qvB=m\frac{v^2}{r}$ and hence $r=\frac{mv}{qB}$.
In confirming that this result is consistent with Newton's laws, my notes say

A fast moving charged particle would travel further (in the direction of motion) than a slow moving charge, for the same perpendicular acceleration, leading to the same result as above, a larger radius of curvature.

Hence explaining why $r \propto v$.
My question is, if $v$ increases, doesn't the centripetal force increase as $F = qvB$? So their perpendicular acceleration increases. I do not understand what my notes mean by same perpendicular acceleration.

REPLY [1 votes]: I think the deciding factor is that the force $qvb$ is "linear" with the velocity $v$.
Let us say that the charge is moving with some velocity $v$ in uniform magnetic field and perpendicular to the direction of the magnetic field, then the force is $qvb$ and we know that the charge moves on a circle which can't be completed without a centripetal force. So the magnetic force serves as the center seeking force here.
Now if you double the velocity then in order to move on the same circle the center seeking force must rise to four times it's initial value but as I said earlier the force $qvb$ is linear with $v$, thus you can only get twice the value of the initial force.
So indeed increasing the velocity increases the force but not much to restrict the particle on the same trajectory. And thus the radius increases.


TITLE: A sinusoidal trigonometric function
QUESTION [0 upvotes]: I need to plot a graph of the following function for values time, $0\le t\le2\pi$ (radians).
$$y=5\sin\left(2t + \frac\pi6\right)$$
I need to describe and define the amplitude, periodic time and frequency.
Can the formula be put into Excel somehow to plot the graph?
Thanks

REPLY [0 votes]: Yep.
Enter the initial value of $t$ in a cell. In the right neighbor, enter the formula =5*SIN(2*?+PI()/6) where the question mark replaces a click in the left cell.
In the cell below the initial value, enter the initial value plus the desired step. Select both cells and extend downward to the desired final value.
Select the cell with the formula and extend downward. You should then have two columns with pairs $(t, f(t))$.
Select the two columns and insert a scatter plot.


TITLE: Why is water not used as a dielectric in the condenser?
QUESTION [6 upvotes]: The dielectric constant of water is very high. Then why is it not used as a dielectric in the condenser?

REPLY [0 votes]: Pure water is a very effective dielectric at high frequencies, though to keep it pure normally involves pumping it round an ion-exchange resin to remove the ions dissolving into it from the enclosure. It also has a very high breakdown voltage compared to air (50 million volts per meter or more).
By comparing the resistivity (18 megohms cm) to the capacitance of a unit cube (8pF or so for the same 1cm cube) we obtain a self-discharge time constant of about 150usec . For any event fast compared to this it is better to think of the water as capacitance, while on timescales slower than this it is better to think of it as resistance.


TITLE: Is there an intuitive way to understand why a frequency cannot be writen as a sum of other frequencies?
QUESTION [4 upvotes]: Let's say we have a sequence of functions $(f_n)$ and for $n$, $f_n$ is a cosine at frequency $k$,  i.e. $f_n(t)=\cos(2\pi kt)$ for some $k$ which depends of $n$.
Let $n_1,\dots n_k$ be natural numbers, why can't $f_{n_1}$ be expressed as a sum of $f_{n_2},\dots,f_{n_k}$?
I know the "official" proof that involves a well chosen scalar product, but is there a more intuitive way, that even a high school student who knows circular functions could understand?
Edit: Sorry, I meant a linear combination and not a simple sum.

REPLY [1 votes]: You could use $2\cos(2\pi k t)=\exp(2\pi i kt)+\exp(-2\pi i kt) = z^k+z^{-k}$ with $z=\exp(2\pi i t)$ to reduce the problem to the linear independence of polynomials. Is the question why $z^{n_1}+z^{-n_1}$ can't be expressed as a sum of $z^{n_2}+z^{-n_2}, \ldots,z^{n_k}+z^{-n_k}$ easier to solve? Not necessarily, but at least its solution doesn't involve a well chosen scalar product.


TITLE: A question on triangulation of compact surfaces
QUESTION [0 upvotes]: We consider the quotient space of the hexagon. Its edges are identified according to the string $abcab^{-1}c$. I have no idea what surface does it describe? And I also wonder does there exist some general ways to do such questions?

REPLY [0 votes]: Augmenting Nicolas's answer: you can avoid triangulating the surface to compute Euler characteristic by thinking of it as a CW complex.  Here's an algorithm that even a computer can do.
Step 1. Determine the number of vertices in the boundary of the polygon after identification: label an unlabeled vertex with a fresh name, then propagate that label to other vertices using the identification word; count the number of labels once everything is labeled.
Step 2. Let $n$ be the number of vertices after identification and $e$ the number of edge labels.  Then $\chi=n-e+1$. (The $1$ is for the one polygonal face.)
Step 3. Determine whether the surface is orientable: if every letter $a$ in the word also has an $a^{-1}$ in the word, then the surface is orientable, otherwise it is unorientable.
Step 4. If orientable, $g=1-\frac{1}{2}\chi$ is the number of $T^2$'s.  If unorientable, $k=2-\chi$ is the number of $\mathbb R\mathrm P^2$'s.
Example. The vertices on either end of $a$ end up being the only distinct vertices in the identification, so $n=2$, hence $\chi=2-3+1=0$.  The word has $a$ twice, so it is unorientable.  Hence $k=2-0=2$ is the number of $\mathbb R\mathrm P^2$'s.


TITLE: Derivative of $f(\mathbf{x})$ wrt $\mathbf{x}$ where $\mathbf{x} \in \mathcal{V}$
QUESTION [2 upvotes]: I would like to know how to find the derivative of $f(\mathbf{x})$ w.r.t. $\mathbf{x}$ where $\mathbf{x} \in \mathcal{V}$ and $\mathcal{V} \subseteq \mathbb{R}^n$.
In particular, what is the derivative of: $$f(\mathbf{x})=\frac{\mathbf{x}^{\top}\mathbf{A}\mathbf{x}}{\lVert\mathbf{x}\rVert^2}$$
where $\mathbf{x} \in \mathcal{V}^{\perp}$ [Edited: as pointed out H.H. Rugh] and $\mathcal{V}$ is the eigenspace corresponding to the top $d(<n)$ eigenvalues of $\mathbf{A}$?
I heard this is called the Gateaux derivative. Any suggestion in this regard will be highly appreciated.

REPLY [2 votes]: When taking the derivative of $g(x)=x^T B x$ (B symmetric) in the direction of $h$ you calculate as follows:
 $$ g(x+h) = x^T B x + h^T B x + x^T B h + o(h)= g(x) + g'(x).h + o(h),$$
whence $g'(x).h = 2 h^T B x $. I suspect in your case $A$ is symmetric (or else replace it by $1/2(A+A^T)$. Apply the previous to that ratio and after reduction you should arrive at:
 $$ f'(x).h = 2\frac{h^T ( A x - f(x) x)}{\|x\|^2}, $$
from which conclusion about critical points may be reached.
To look at the set-complement of $V$ is a bit funny. I think it would be more interesting to look at the orthogonal complement (minus the origin).


TITLE: Absolutely Convergent Series Converges
QUESTION [0 upvotes]: Let $\sum_{n=1}^\infty a_n$ an converges absolutely, $(a_n)$ being a series.  Prove for every $\epsilon >0, \exists N \in \mathbb{N}$ s.t. $\sum_{k=N+1}^\infty |a_n| < \epsilon$.

So if it converges absolutely, then $\sum_{n=1}^\infty |a_n|=L$ and $\sum_{k=N+1}^\infty |a_n| = L- \sum_{n=1}^N |a_n| < \epsilon$.
Find $n$ such that $L- \sum_{n=1}^N |a_n|<\epsilon'$ since this is a Cauchy sequence if the series of partial sums converges. AM i on the right track thank!

REPLY [0 votes]: $S_n =\sum_{k=1}^{n}|a_k|,$ i.e. is Cauchy.
For $\epsilon >0$ there exists a $n_0$ s t. for $m \ge n \ge n_0$.
$|S_m-S_n| \lt \epsilon$,
$\sum_{k=n+1}^{m}|a_k| <\epsilon.$
Set $n=n_0$, and take the limit $m \rightarrow \infty$:
$\sum_{n_0+1}^{\infty}|a_k| \le \epsilon$.
Note the $\le$ sign in above expression . If you want a < sign,
what do you change?


TITLE: Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$
QUESTION [13 upvotes]: Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$

REPLY [1 votes]: Here is yet another proof. Let $n = u-\frac{1}{2}$. Then we have:
$$\lim_{n \to \infty} \sqrt{n(n+1)} - n$$
$$=\lim_{u \to \infty} \sqrt{\left(u-\frac{1}{2}\right) \left(u+\frac{1}{2}\right)} - u + \frac{1}{2}$$
$$=\lim_{u \to \infty} \sqrt{u^2-\frac{1}{4}} - u + \frac{1}{2}$$
$$=\lim_{u \to \infty} u - u + \frac{1}{2} = \frac{1}{2}$$