TITLE: Xi Function on Critical Strip - Mellin Transform
QUESTION [4 upvotes]: Story
I'm trying to prove following identity
$$\int_0^\infty \frac{\Xi(t)}{t^2 + \frac{1}{4}} \cos(xt) dt = \frac{1}{2} \pi (e^{\frac{1}{2}x} - 2e^{-\frac{1}{2}x} \psi(e^{-2x}))$$
where
$$\psi(x)=\sum_{n = 1}^{\infty} e^{-n^2 \pi x}$$
and
$$\Xi(t) = \xi(\frac{1}{2} + it)$$
is the xi function on the critical line.
Problem
I only don't understand the following equality
$$-\frac{1}{4 i \sqrt{y}} \int_{\frac{1}{2} - i\infty}^{\frac{1}{2} + i\infty} \Gamma(\frac{1}{2}s) \pi^{-\frac{1}{2}s} \zeta(s) y^s ds = -\frac{\pi}{\sqrt{y}} \psi(\frac{1}{y^2}) + \frac{1}{2} \pi \sqrt{y}.$$
Where does latter summand $\frac{1}{2} \pi \sqrt{y}$ come from? When I write $(\frac{1}{y})^{-s}$ and substitute with $s = 2w$ I only get the first summand. We know that by the Mellin transform theorem $\psi(x)$ can be recovered by the Mellin transform integral over $\Gamma(s) \pi^{-s} \zeta(2s)$.
Source
This is from Titchmarsh's book "The Theory of the Riemann Zeta-function" p. 35−36

REPLY [3 votes]: Let $f(x) = 2\sum_{n = 1}^\infty e^{- \pi n^2 x^2}$ and $E(s) = \pi^{-s/2} \Gamma(s/2)\zeta(s)$.
For $Re(s) >0$ : $2\int_0^\infty x^{s-1} e^{-\pi n^2 x^2} dx = n^{-s} \pi^{-s/2}\Gamma(s/2) $ so we have for $Re(s) > 1$ :
$$E(s) = \Gamma(s/2) \pi^{-s/2} \zeta(s) = 2\sum_{n=1}^\infty \int_0^\infty x^{s-1} e^{-\pi n^2 x^2} dx = \int_0^\infty f(x) x^{s-1} dx \quad (1)$$
By inverse  Mellin transform :
$$f(x) = \frac{1}{2 i \pi}\int_{\sigma-i\infty}^{\sigma+i\infty} E(s) x^{-s} ds$$
but this is only true for $\sigma  > 1$, since $E(s)$ has a pole at $s=1$.
Note that for $Re(s) > 1$ :  $\displaystyle\quad\frac{1}{s-1} = \int_0^1 x^{s-2} dx = \int_0^\infty \frac{1_{x < 1}}{x} x^{s-1} dx $,
and for $Re(s) < 1$ : $\displaystyle\quad-\frac{1}{s-1} = \int_1^\infty x^{s-2} dx = \int_0^\infty \frac{1_{x > 1}}{x} x^{s-1} dx $
Hence, at least for $Re(s) > 1$ :
$$E(s) - \frac{1}{s-1} = \int_0^\infty \left( f(x)- \frac{1_{x < 1}}{x}\right) x^{s-1} dx \qquad (2) $$
indeed, this is true also for $Re(s) > 0$  (see below) and we get, for $Re(s) \in (0,1)$ :
$$E(s) = \int_0^\infty \left( f(x) - \frac{1}{x}\right) x^{s-1} dx$$
Finally, by inverse Mellin transform, for $\sigma \in (0,1)$ :
$$f(x) -  \frac{1}{x} = \frac{1}{2 i\pi}\int_{\sigma -i \infty}^{\sigma +i \infty} E(s) x^{-s}dx$$
and $\displaystyle f(1/y) -  y  = \frac{1}{2 i\pi} \int_{\sigma -i \infty}^{\sigma +i \infty} E(s) y^{s}dy$ as expected.

You can show $(2)$ converges for $Re(s) > 0$ by proving $\displaystyle\theta(x)  = 1+ f(x) = \sum_{n =- \infty}^\infty e^{-\pi n^2 x^2}$ fulfills the functional equation $$\theta(1/x) = x\,\theta(x)$$ using the Poisson summation formula (see this proof).
So that $\displaystyle f(1/x) = \theta(1/x)-1= x \,\theta(x)-1 = x(f(x)+1)-1$ and : $$\int_0^1 \left(f(x)- \frac{1}{x}\right) x^{s-1} dx = \int_1^\infty (f(1/y) - y) y^{-s+1} \frac{dy}{y^2} $$ $$= \int_1^\infty (y (f(y)+1)-1-y) y^{-s-1} dy = \int_1^\infty (y f(y)-1) y^{-s-1} dy$$
whence $$E(s) - \frac{1}{s-1} = \int_0^1+\int_1^\infty \left(f(x)-\frac{1_{x<1}}{x}\right) x^{s-1}dx $$ $$=  \int_1^\infty \left( f(x) (x^{s-1} +  x^{-s}) - x^{-s-1}\right) dx$$
converges for $Re(s) > 0$.
Riemann obtained the functional equation $E(s) = E(1-s)$ from $$E(s) - \frac{1}{s-1} + \frac{1}{s} = \int_1^\infty  f(x) (x^{s-1} +  x^{-s})  dx$$
which is entire.


TITLE: Forces between wheels and road for a vehicle
QUESTION [1 upvotes]: Let’s assume we have a front-wheel drive (FWD) car, and we apply a torque (Cf) to the front wheels. At this point, assuming no slip condition, a friction force (Tf) will occur due to friction between the rotating wheel and the road. This force will be from left to right (according to the picture).
Now what’s about the rear wheels? They are pushed forward by the car that is now moving thanks to the Tf and they can only begin rolling if the force between them and the road is backward. Is it correct or am I missing something?
What happens if I have no torque (when I release the throttle) and when I brake?
Thank you

REPLY [0 votes]: There is a force to the left on the axle of the back wheels $F$ which is accelerating the rear wheels to the right.
The friction force between the tyre and the ground to the left $F_{\rm friction}$ applies a torque about the axle of the wheel to cause it to increase its rotational speed.
The frictional force is try reduce the relative movement between the tyre and the ground at the point of contact or maintain no relative movement.
With no torque being applied and the wheels not rotating there would be relative movement between the tyre and the ground - a skid has occurred with the friction force on the tyre due to the ground in the backward direction.
That friction force is trying to slow the centre of mass of the vehicle down whilst at the same time trying to get the wheels to rotate ie attempting to reduce the relative movement between the tyre and the ground.


TITLE: In a two-body problem, is the argument of periapse the same for both bodies in orbit?
QUESTION [2 upvotes]: I'm trying to find the argument of periapse $\omega$ for a certain planet which has been studied with imaging, spectroscopy and radial velocity. In the literature I've only been able to find the arguement of periapse of the star it's orbiting. Are these equivalent?
Or rather, in general, is the argument of periapse the same for both bodies in orbit?

REPLY [1 votes]: No, the Arguments of Perapses are not the same.
The Argument of Periapsis is the angle measured, in the plane of the orbit,  in the direction of travel around the orbit, from the ascending node, through the primary focus, to the perapsis
In the two-body keplerian-newtonian simulation, both bodies orbit the center of mass opposite one another,  in orbits that share the same line of apsides,  and the same eccentricity, in the same plane. Both bodies reach periapsis and apoapsis from their center of mass at the same time. Both orbits will have the same value for Longitude of the Ascending node.

Two objects in a Keplerian two-body mutual orbits around their barycenter

Shown a pair of objects orbiting one another, perpendicular to the orbital plane. Star 2 is 60% of the mass of Star 1, and the orbital eccentricity of both orbits is 0.4. The barycenter (labeled CoM) is in both the orbital plane and the reference plane. The line of nodes is in purple, passing through both orbits. Both stars orbit the barycenter counterclockwise in this image. $AN_1$ and $AN_2$ are the ascending nodes of both orbits. $q_1$ and $q_2$ are the periapses of the orbits, and $\omega_1$ and $\omega_2$ are the respective arguments of periapsis.
Since their Periapses are on opposite sides of the star, if the orbits aren't in the reference plane, one object will have its periapsis "above" the reference plane, and the other "below" it. So one will have to go through a larger angle from its Ascending Node, passing through its Descending node, to reach the periapsis, and the other will not.
As such, the values of their respective Arguments of Periapsis will differ by 180°.


TITLE: A long horizontal rod has a bead which can slide along its length and is initially placed at a distance L from one end A of the rod
QUESTION [1 upvotes]: Question: A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end $A$ of the rod.The rod starts from rest in angular motion about $A$ with a constant angular acceleration $α$.
If the coefficient of friction between the rod and the bead is $μ$ and gravity is neglected then what is the time after which the bead starts slipping?
Online Solution:

The bead will start sliding when centripetal force is equal to limiting friction.
$$mω²L = μN\quad\dots(1)$$
We know that here normal reaction is provided by tangential force. $$mLα = N$$
Substituting value of $N$ in eq.$(1)$

Now, at time $t, \,\omega = 0 + αt \implies w = αt$.
Now,
$\displaystyle mω²L = μmLα\\
 m(αt)²L = μmLα\\
 (αt)² = μ\\
\therefore t = √(μ/α)$
My questions:
My issue with this solution is that why is the centripetal force equated with the frictional force? Aren't both forces directed radially inward? Intuitively the ball would be moving outwards but why exactly?

REPLY [1 votes]: Your confusion is thinking of the frictional force and centripetal force as two different forces.  In this case the frictional force is the centripetal force.  The centripetal force is the force necessary to keep the ball in circular motion at a constant distance $L$ from the center of motion.  That is what the frictional force is doing.
It is analogous to a ball held in circular motion by a string.  In that case the centripetal force is provided by the tension in the string.  There is not a centripetal force and a tensile force.  They are one in the same.
They are one in the same in your case also.


text:
\begin{document}

\begin{center}
{\LARGE\bf  Leibniz, Randomness \& the Halting Probability\vspace{1mm}}
\\
{\scriptsize\bf
Gregory Chaitin, IBM Watson Research Center, Yorktown Heights\vspace{2mm}}
\\
\emph{\textbf{Dedicated to Alan Turing on the 50th Anniversary of his Death}}
\end{center}

\bigskip

Turing's remarkable 1936 paper \emph{On computable numbers, with an application
to the Entscheidungsproblem}
marks a dramatic turning point in modern mathematics. On the one hand, the
computer enters center stage as a major mathematical concept.  On the other
hand, Turing establishes a link between mathematics and physics by talking about what
\emph{a machine} can accomplish.
It is amazing how far these ideas have come in a comparatively short amount of time;
a small stream has turned into a major river.

I have recently completed a small book about some of these developments, \emph{Meta Math!}
It is currently available as an e-book on my personal website, and also from arxiv.org,
and is scheduled to be published next year.
Here I will merely give a few highlights.

My story begins with Leibniz in 1686, the year before Newton published his \emph{Principia.}
Due to a snow storm, Leibniz is forced to take a break in his attempts to improve the
water pumps for some important German silver mines, and writes down an outline of some
of his ideas, now known to us as the \emph{Discours de m\'etaphysique.}
Leibniz then sends a summary of the major points through a mutual friend to the
famous fugitive
French \emph{philosophe} Arnauld, who is so horrified at what he reads that Leibniz never sends him nor
anyone else the entire manuscript.
It languishes among Leibniz's voluminous personal papers and is only discovered and published
many years
after Leibniz's death.

In sections V and VI of the \emph{Discours de m\'etaphysique,}
Leibniz discusses the crucial question of how we can distinguish a world in which science
applies from one in which it does not. Imagine, he says, that someone has splattered
a piece of paper with ink spots determining in this manner a finite set of points on the page.
Nevertheless, Leibniz observes, there will always be a mathematical
equation that passes through this finite set of points.  Indeed, many good ways to
do this are now known. For example, what is called Lagrangian interpolation will do.

So the existence of a mathematical curve passing through a set of points cannot enable us to
distinguish
between points that are chosen at random and those that obey some kind of a scientific law.
How then can we tell the difference? Well, says Leibniz,
if the equation must be extremely complex (``fort compos\'ee'') that is not a valid scientific law
and the points are random (``irr\'egulier'').

Leibniz had a million other interests and earned a living as a consultant to princes, and as
far as I know after having this idea he never returned to this subject.  Indeed, he was
always tossing out good ideas, but rarely, with the notable exception of the infinitesimal
calculus, had the time to develop them in depth.

The next person to take up this subject, as far as I know, is Hermann Weyl in his 1932
book \emph{The Open World,} consisting of three lectures on metaphysics that Weyl gave at Yale
University.
In fact, I discovered Leibniz's work on complexity and randomness by reading this little book
by Weyl.
And Weyl points out that Leibniz's way of distinguishing between points that are random
and those that follow a law by invoking the complexity of a mathematical formula is unfortunately
not too well defined, since it depends on what
primitive functions
you are allowed to use in writing that formula and
therefore varies as a function of time.

Well, the field that I invented in 1965 and which I call algorithmic information theory provides
a possible solution for the problem noticed by Hermann Weyl.
This theory defines a string of bits to be random, irreducible, structureless, if it is
algorithmically incompressible, that is to say, if the size of the smallest computer
program that produces that particular finite string of bits as output is about the same
size as the output it produces.

So we have added two ideas to Leibniz's 1686 proposal.  First, we measure complexity in terms
of bits of information, i.e.,  0s and 1s.   Second, instead of mathematical equations,
we use binary computer programs.
Crucially, this enables us to compare the complexity of a scientific theory (the computer program)
with the complexity of the data that it explains (the output of the computer program).

As Leibniz observed, for any data there is always a complicated theory, which is a computer program
that is the same size as the data. But that doesn't count.  It is only a real theory
if there is compression, if the program is much smaller than its output, both measured in 0/1 bits.
And if there can be no proper theory, then the bit string is algorithmically random or irreducible.
That's how you define a random string in algorithmic information theory.

I should point out that Leibniz had the two key ideas that you need to get this modern definition
of
randomness, he just never made the connection.  For Leibniz produced one of the first calculating
machines, which he displayed at the Royal Society in London, and he was also one of the first
people to appreciate base-two binary arithmetic and the fact that everything can be represented
using only 0s and 1s.  So,  as Martin Davis argues in his book \emph{The Universal Computer: The Road
from Leibniz
to Turing,} Leibniz was the first computer scientist, and he was also the first information
theorist.  I am sure that Leibniz would have instantly understood and appreciated the modern definition of
randomness.

I should mention that A. N. Kolmogorov also proposed this definition of randomness.
He and I did this independently in 1965.
Kolmogorov was at the end of his career, and I was a teenager at the
beginning of my own career as a mathematician.
As far as I know, neither of us was aware of the Leibniz \emph{Discours.}
Let me compare Kolmogorov's work in this area with my own.
I think that there are two key points to note.

Firstly,
Kolmogorov never realised as I did that our original definition of randomness was
incorrect. It was a good initial idea but it was technically flawed.
Nine years after he and I independently proposed this definition, I realised that
it was important for the computer programs that are used in the theory to be what I call
``self-delimiting''; without this it is not even possible to define my $\Omega$ number that I'll
discuss below. And there are other important changes that I had to make in the original definitions
that Kolmogorov never realised were necessary.

Secondly, Kolmogorov thought that the key application of these ideas was to be to obtain a new, algorithmic
version of probability theory.
It's true, that can be done, but it's not very interesting, it's too systematic a re-reading
of standard probability theory.  In fact, every statement that is true with probability one, merely
becomes a statement that must necessarily be true, for sure, for what are defined to be the random
infinite sequences of bits.
Kolmogorov never realised as I did that the really important application of these ideas was the new
light that they shed on G\"odel's incompleteness theorem and on Turing's halting problem.

So let me tell you about that now, and I'm sure that Turing would have loved these ideas if his
premature death had not prevented him from learning about them.
I'll tell you how my $\Omega$ number, which is defined to be the halting probability of a binary
program whose bits are generated using independent tosses of a fair coin, shows that in a sense
there is randomness in pure mathematics.

Instead of looking at individual instances of Turing's famous halting problem, you just
put all possible computer programs into a bag, shake it well, pick out a program, and ask
what is the probability that it will eventually halt.  That's how you define the halting probability
$\Omega,$ and for this to work it's important that the programs have to be self-delimiting.
Otherwise the halting probability diverges to infinity instead of being a real number between
zero and one like all probabilities have to be. You'll have to take my word for this; I can't
explain this in detail here.

Anyway, once you do things properly you can define a halting probability $\Omega$ between zero and one.
The particular value of $\Omega$ that you get depends on your choice of computer programming language,
but its surprising properties don't depend on that choice.

And what is $\Omega$'s most surprising property? It's the fact that if you write $\Omega$ in binary, the
bits in its base-two expansion, the bits after the binary decimal point, seem to have absolutely
no mathematical structure. Even though $\Omega$ has a simple mathematical definition, its individual
bits seem completely patternless. In fact, they are maximally unknowable, they have, so to speak,
maximum entropy.  Even though they are precisely defined once you specify the programming language,
the individual bits are maximally unknowable, maximally irreducible. They seem to be mathematical facts that
are true for no reason.

Why? Well, it is impossible to compress $N$ bits of $\Omega$ into a computer
program that is substantially smaller than $N$ bits in size (so that $\Omega$ satisfies the
definition of randomness of algorithmic information theory).  But not only does computation fail
to compress $\Omega,$
reason fails as well.
No formal mathematical theory whose axioms have less than $N$ bits of complexity can
enable us to determine $N$ bits of $\Omega.$  In other words, essentially the only way to be able to
prove what the values of $N$ bits of $\Omega$ are, is to assume what you want to prove as an axiom, which of
course is cheating and doesn't really count, because you are not using reasoning at all.
However, in the case of $\Omega,$ that is the best that you can ever do!

So this is an area in which mathematical truth has absolutely no structure, no structure that
we will ever be able to appreciate in detail, only statistically.  The best way of thinking about
the bits of $\Omega$ is to say that each bit has probability 1/2 of being zero and probability 1/2 of
being one, even though each bit is mathematically determined.

So that's where Turing's halting problem has led us, to the discovery of pure randomness
in a part of mathematics.
I think that Turing and Leibniz would be delighted at this remarkable turn of events.

Now I'd like to make a few comments about what I see as the philosophical
implications of all of this.
These are just my views, and they are quite controversial.
For example, even though
a recent critical review of two of my books in the \emph{Notices of the American Mathematical Society}
does not claim that there are any technical mistakes in my work, the reviewer strongly
disagrees with my philosophical conclusions, and in fact he claims that my work has no
philosophical implications whatsoever.
So these are just my views, they are certainly not a community  consensus, not at all.

My view is that $\Omega$ is a much more disagreeable instance of mathematical incompleteness
than the one found by G\"odel in 1931, and that it therefore forces our hand philosophically.
In what way? Well, in my opinion, in a quasi-empirical direction, which is a phrase coined
by Imre Lakatos when he was doing philosophy in England after leaving Hungary in 1956.
In my opinion, $\Omega$ suggests that even though math and physics are different, perhaps they
are not as different as most people think.

What do I mean by this?
(And whether Lakatos would agree or not, I cannot say.)
I think that physics enables us to compress our experimental data, and math enables us to
compress the results of our computations,
into scientific or mathematical theories as the case may be.
And I think that neither math nor science gives absolute certainty; that is an asymptotic
limit unobtainable by mortal beings.
And in this connection I should mention the book (actually two books) just published by
Borwein and Bailey on experimental math.

To put it bluntly, if the incompleteness phenomenon
discovered by G\"odel in 1931 is really serious---and I believe that Turing's work and my
own work suggest that incompleteness is much more serious than people think---then perhaps mathematics
should be pursued somewhat more in the spirit of experimental science rather than always
demanding proofs for everything. In fact, that is what theoretical computer scientists
are currently doing.  Although they may not want to admit it, and refer to
\emph{\textbf{P}} $\neq$ \emph{\textbf{NP}}
as an unproved hypothesis, that community is in fact behaving as if this were a new axiom,
the way that physicists would.

At any rate, that's the way things seem to me. Perhaps by the time we reach the centenary
of Turing's death this quasi-empirical view will have made some headway,
or perhaps instead these foreign ideas will be utterly
rejected by the immune system of the math community. For now they certainly are rejected.
But the past fifty years have brought us many surprises,
and I expect that the next fifty years will too,
a great many indeed.

\section*{References}

\begin{enumerate}
\item
Chapter on Leibniz in E. T. Bell, \emph{Men of Mathematics,} Simon \& Schuster, 1986.
\item
R. C. Sleigh, Jr., \emph{Leibniz and Arnauld: A Commentary on Their Correspondence,}
Yale University Press, 1990.
\item
H. Weyl, \emph{The Open World: Three Lectures on the Metaphysical Implications of Science,}
Yale University Press, 1932,
Ox Bow Press, 1994.
\item
Article by Lakatos in T. Tymoczko, \emph{New Directions in the Philosophy of Mathematics,}
Princeton University Press, 1998.
\item
M. Davis, \emph{The Universal Computer: The Road from Leibniz to Turing,} Norton, 2000.
\item
Play on Newton vs.\ Leibniz in C. Djerassi, D. Pinner, \emph{Newton's Darkness: Two Dramatic Views,}
Imperial College Press, 2003.
\item
J. Borwein, D. Bailey, \emph{Mathematics by Experiment: Plausible Reasoning in the 21st Century,}
A. K. Peters, 2004.
\\
J. Borwein, D. Bailey, R. Girgensohn, \emph{Experimentation in Mathematics: Computational Paths
to Discovery,} A. K. Peters, 2004.
\item
G. Chaitin, \emph{Meta Math! The Quest for Omega,}
\\
http://cs.umaine.edu/\verb!~!chaitin/omega.html,
\\
http://arxiv.org/abs/math.HO/0404335.
\\
To be published by Pantheon Books in 2005.
\end{enumerate}

\end{document}

config: arxiv


text:
\begin{document}

\title{Capacity Scaling Law by Multiuser Diversity in Cognitive Radio Systems}

\author{\IEEEauthorblockN{Jun-pyo Hong and Wan Choi}
\IEEEauthorblockA{Dept. of Electrical Engineering, Korea Advanced Institute of Science and Technology (KAIST)\\

Email: \{jp\_hong, w.choi\}@kaist.ac.kr}} \maketitle

\begin{abstract}
This paper analyzes the multiuser diversity gain in a cognitive radio (CR) system where secondary transmitters opportunistically utilize the spectrum licensed to primary users only when it is not occupied by the primary users. To protect the primary users from the interference caused by the missed detection of primary transmissions in the secondary network, minimum average throughput of the primary network is guaranteed by transmit power control at the secondary transmitters. The traffic dynamics of a primary network are also considered in our analysis. We derive the average achievable capacity of the secondary network and analyze its asymptotic behaviors to characterize the multiuser diversity gains in the CR system.
\end{abstract}

\section{Introduction}
The demand for wireless spectrum is constantly increasing as various
wireless communication services have appeared. Correspondingly the
available spectrum has become scarce. Current spectrum allocation
policies aggravate spectrum scarcity since a particular spectrum is
dedicated to only specific users as Federal Communications
Commission (FCC) reported \cite{FCC_Report}. As a solution to
improve the spectrum efficiency, \emph{cognitive radio} (CR) has
emerged, where secondary (unlicensed) users try to access the
spectrum without interfering with communications of primary users.
The term CR is classified into several techniques depending on the way to access the spectrum licensed to the primary users:
\emph{overlaid}, \emph{underlaid}, and \emph{interweaved} CR systems
\cite{Mag_Jafar}.

Our work focus on interweaved CR system where the secondary users are allowed to access the
spectrum licensed to primary users only when primary users do not
occupy the spectrum. This approach comes from the idea of
\emph{opportunistic communication} \cite{Mitola}. Secondary users
monitor whether the spectrum is occupied by primary users in order
to opportunistically communicate over vacant spectrum without
interfering with primary users.

Stable throughput of an
interweaved CR system was analyzed by reflecting random packet
arrival of a primary user using queueing process in \cite{Osvaldo}
and \cite{Osvaldo_1}.

Srinivasa and Jafar \cite{CR_Jafar} studied the optimum number of
secondary users that maximizes the total throughput in a
decentralized CR system.

Secondary user scheduling and medium access control play key roles
in CR systems but they have not been well studied yet. It has been
well known that opportunistic user selection for transmission
provides a multiuser diversity gain due to fluctuations of fading
channels \cite{Viswanth2002}. For non-CR systems, there have been
many studies on characterizing the multiuser diversity gains
\cite{MUD_1, MUD_2, MUD_3, Choi2008}. These studies showed that the
multiuser diversity gain in terms of average capacity grows like
$\log_2(\ln N)$ and $\sqrt{2\log_2 N} $ in Rayleigh fading channels
and lognormal shadowing channels, respectively, where $N$ is the
number of users. The multiuser diversity gain of a secondary network
in an underlaid CR system was recently discussed in \cite{MUD_SS,
MUD_SS_Liang}. It was shown that the average capacity of the
secondary network scales like $\log_2(\ln N)$ and $\log_2 N$ under
the finite and the infinite peak transmit power constraints at the
secondary transmitters, respectively. Moreover, the scaling laws of
underlaid CR system in a cognitive ad hoc network was studied in
\cite{Scalinglaw_Tarokh} although it did not focus on the multiuser
diversity gain coming from opportunistic user selection. Despite the
fact that the term CR typically refers to interweaved CR systems,
however, the multiuser diversity gain in interweaved CR systems has
not been identified.

In this paper, we investigate characteristics of multiuser diversity
gains in interweaved CR systems by deriving average achievable
capacity of a secondary network. Multiuser scheduling and transmit
power control are employed in a secondary network to maximize the
achievable capacity of a secondary user network and to satisfy a
quality of service (Qos) constraint of a primary user network.
Unlike underlaid CR systems, secondary users are not allowed to
access the spectrum licensed to primary users when a secondary
network detects primary users' transmission. So our analysis
reflects both traffic dynamics of a primary user network and sensing
reliability in a secondary network. Furthermore, this paper takes
into account the interference from a primary transmitter contrary to
\cite{MUD_SS}. Our asymptotic analysis and numerical results
insightfully capture the key characteristics of multiuser diversity
gains in interweaved CR systems. It is shown that multiuser
diversity gains in interweaved CR systems are quite different from
those in non-CR based systems due to the QoS constraint on a primary
user network. Moreover, even if the secondary users of interweaved
and underlaid CR systems have different ways to access the spectrum
licensed to primary users, the multiuser diversity gains of both CR
systems show very similar asymptotic characteristics.

\vspace{0.05in}
\section{System Description}

\subsection{System and Channel Model}

\begin{figure}[!t]
    \centerline{\psfig{figure=./figure1.eps,width=0.9\columnwidth} }
    \caption{System model}
    \label{Fig:System_Model}
\end{figure}

The system model considered in this paper is illustrated in Fig.
\ref{Fig:System_Model}, which consists of a primary receiver
(primary base station), a primary transmitter, a secondary receiver
(secondary base station), and $N$ secondary transmitters which try
to access the spectrum licensed to a primary transmitter. The
additive thermal noise is assumed to be zero mean mutually
independent, circularly symmetric, complex Gaussian random variable
with unit variance. The channel gains from the $i$th secondary
transmitter to the primary receiver and to the secondary receiver
are denoted by $\alpha_{s,i}$ and $\beta_{s,i}$, respectively.
Similarly, $\alpha_{p}$ and $\beta_{p}$ are the channel gains from
the primary transmitter to the primary receiver and to the secondary
receiver, respectively. All channel gains are assumed to be
independent and identically distributed (i.i.d.) exponential random
variable with unit mean in Rayleigh flat fading channels. It is also
assumed that the $i$th secondary transmitter knows the instantaneous
channel state information of $\alpha_{s,i}$ and $\beta_{s,i}$ and
knows the statistical channel information of $\alpha_{p}$ and
$\beta_{p}$. The secondary transmitters are able to obtain the
instantaneous channel state information of $\alpha_{s,i}$ and
$\beta_{s,i}$ through a periodic sensing of pilot signal from the
primary and secondary receivers, according to the channel
reciprocity.

In our system model, all the secondary transmitters which have
packets to transmit compute the maximum allowable transmit power
$P_{s,i}$ based on $\alpha_{s,i}$ to maximize their throughput while
ensuring the QoS of primary users. Then, the secondary transmitters
report their 1-bit sensing results with calculated transmit power
$P_{s,i}$. The secondary receiver fuses received sensing results
from the secondary transmitters and makes a global decision.
The reliability of cooperative sensing is captured by detection probability $p_d$ and false alarm probability $p_f$.
However, detailed discussion about how to fuse the received sensing
results and how to reflect the performance of cooperative sensing
are beyond the scope of this paper. If the global decision declares
the absence of primary users' traffic in the spectrum, the secondary
receiver selects the secondary transmitter which has the largest
signal-to-noise power ratio (SNR) $\beta_{s,i}P_{s,i}$ to maximize
secondary network throughput. Otherwise, all secondary transmitters
are silent during a unit time slot. On the other hand, a primary
transmitter transmits its packets with its maximum power, $P_{p}$.

\subsection{Traffic Model}
We consider a \emph{packet-by-packet} access network where the spectrum licensed to the primary users is accessed in a time slotted manner similarly in \cite{Osvaldo}.
In other words, all primary and secondary transmitters can transmit one packet per each time slot.

The traffic at the primary transmitter is modeled by a
Bernoulli process with average packet arrival rate,
$0\le\lambda\le1$ [packets/slot], and departure rate, $0\le\mu\le1$
[packets/slot]. Note that the departure rate can be interpreted as
the probability of successful packet transmission during a time
slot; moreover, it can directly be translated into average
throughput of a primary user. In contrast to arrival rate $\lambda$,
departure rate $\mu$ is significantly affected by the behavior of
the secondary network.

So the primary user sets a minimum tolerable
departure rate $\mu_{\min}$ to guarantee a certain service rate for
the case of missed detection in the secondary network by
\begin{eqnarray}\label{min_departure}
\mu \ge \mu_{\min}.
\end{eqnarray}

It should be noted that the departure rate without any interference
is greater than the arrival rate for an essentially stable primary
user's queue.

\vspace{0.05in}
\section{QoS Constraint and Secondary User Scheduling}
\subsection{Departure Rate at a Primary Transmitter}

The outage probabilities when there is no interference and when
there is interference from the secondary network are given,
respectively, by
\begin{align}
P_{out} &= \Pr\left[ \log_2\left(1+\alpha_pP_p\right)<R\right]\label{P_outage1} \\
P_{out}^{inf} &= \Pr\left[\log_2\left(1+\frac{\alpha_pP_p}{1+\alpha_{s,i}P_{s,i}}\right)<R\right],
\label{P_outage2}
\end{align}
where $R$ is the required rate at the primary receiver. Then, the
departure rate of the primary transmitter is given by
\begin{eqnarray}\label{departure}
\mu = p_d(1-P_{out})+(1-p_d)(1-P_{out}^{inf}).
\end{eqnarray}
From a perspective of a primary user, a minimum departure rate,
$\mu_{\min}$, needs to be set as a OoS parameter and is notified to
the secondary network so that interference from a secondary network
is regulated to guarantee the minimum departure rate.

\subsection{Power Adaptation and Secondary User Scheduling}

Based on the channel information including the instantaneous value
of $\alpha_{s,i}$ and the statistical properties of $\alpha_p$, each
secondary transmitter can estimate the corresponding outage
probabilities of the primary user, when it is scheduled.

The outage probabilities, (\ref{P_outage1}) and (\ref{P_outage2}),
can be rewritten by
\begin{align}
P_{out} &= \Pr\left[\alpha_p<\frac{2^R-1}{P_p}\right] = 1-e^{-R_p}
\label{out_noinf}
\end{align}
\begin{align}
P_{out}^{inf} &= \Pr\left[\alpha_p<\left(2^R-1\right)\frac{1+\alpha_{s,i}P_{s,i}}{P_p}\right]\nonumber\\
&= 1-e^{-R_p(1+\alpha_{s,i}P_{s,i})}
\label{out_inf}
\end{align}
where $R_p=(2^R-1)/P_p$.

If the $i$th secondary transmitter is scheduled for transmission,
the departure rate of the primary user is obtained by plugging
(\ref{out_noinf}) and (\ref{out_inf}) into (\ref{departure}),
\begin{eqnarray}\label{mu_func}
\mu(\alpha_{s,i}, P_{s,i}) = p_de^{-R_p} + (1-p_d)e^{-R_p(1+\alpha_{s,i}P_{s,i})}.
\end{eqnarray}
The departure rate of the primary user is affected by transmit power
of the secondary transmitter $P_{s,i}$ and instantaneous
interference channel gain $\alpha_{s,i}$.

Hereinafter, we use $\mu(\alpha_{s,i}, P_{s,i})$ instead of $\mu$ to
represent the departure rate as a function of $\alpha_{s,i}$ and
$P_{s,i}$.

The transmit power of the $i$th secondary transmitter is determined
by the value making $\mu(\alpha_{s,i}, P_{s,i})$ equal to
$\mu_{\min}$ since $\mu(\alpha_{s,i}, P_{s,i})$ is inversely
proportional to $P_{s,i}$. The power $P_{\mu,i}$ corresponding to
$\mu_{\min}$ is given by
\begin{eqnarray}\label{P_mu}
P_{\mu,i}=\frac{1}{\alpha_{s,i}}\left[\frac{1}{R_p}\ln\left(\frac{1-p_d}{\mu_{\min}-p_de^{-R_p}}\right)-1\right].
\end{eqnarray}

Since the transmit power of the secondary user is limited by a
maximum power constraint, i.e., $P_{s,i}\le P_{s,\max}$, the
transmit power of the $i$th secondary transmitter is given by
\begin{align} \label{P_si}
P_{s,i}=\left\{ \begin{array}{ll}
0 ~,& P_{\mu,i}<0 \\
P_{\mu,i} ~,& 0\le P_{\mu,i}<P_{s,\max}\\
P_{s,\max} ~,& P_{\mu,i} \geq P_{s,\max}
\end{array} \right.
.
\end{align}
The first case $P_{\mu,i}<0$ corresponds to the situation where the
primary user cannot achieve $\mu_{\min}$ even if a secondary network
do not interfere with the primary user. In this case, all the
secondary transmitters can not transmit anything, regardless of the
sensing result, so that there is no secondary network throughput. In
the other cases, a secondary transmitter can transmit a packet with
adaptively adjusted power based on the value of $\alpha_{s,i}$. Our
analysis focuses on the last two cases where the spectrum licensed
to primary user is opportunistically accessed.

Hence, from (\ref{P_mu}), it is assumed that
\begin{align}
K = \frac{1}{R_p}\ln\left(\frac{1-p_d}{\mu_{\min}-p_de^{-R_p}}\right)-1>0
\end{align}
where a positive real constant $K$ is pre-determined by the system
parameters.

Based on the computed $P_{s,i}$ in (\ref{P_si}), all the secondary
transmitters report their 1-bit sensing results with its transmit
power, $P_{s,i}$. Using the collected sensing results, the secondary
receiver checks the availability of the spectrum licensed to the
primary network. If the spectrum is decided to be vacant, the
secondary receiver selects the $i^*$th secondary transmitter whose
SNR is the largest among $N$ secondary transmitters such as
\begin{align}
i^* = \arg\max_i \gamma_{s,i}~
\end{align}where the received SNR $\gamma_{s,i}$ from the
$i$th secondary transmitter is given by
\begin{align}\label{class_SU_SNR}
\gamma_{s,i}  = \left\{ \begin{array} {ll}
P_{\mu,i}\beta_{s,i}=\frac{K\beta_{s,i}}{\alpha_{s,i}} ~,& P_{\mu,i}<P_{s,\max}\\
P_{s,\max}\beta_{s,i} ~,& P_{\mu,i}\ge P_{s,\max}
\end{array} \right.
.
\end{align}

\vspace{0.05in}
\section{Characteristics of Multiuser Diversity in a Cognitive Radio System}
Unfortunately, analyzing the average capacity of secondary network is not easy to handle because extracting the PDFs of $\alpha_{s,i^*}$ and $\beta_{s,i^*}$from that of $\gamma_{s,i^*}$ is intractable. Instead, to see the exact scaling law of multiuser diversity gain, this section provides lower and upper bounds on the asymptotic average achievable capacity of the secondary network and derives the
scaling law of the average achievable capacity of the secondary network from the two bounds.

\subsection{A Lower Bound}
To give the independency between $\alpha_{s,i}$ and $\beta_{s,i}$ in
the selection of a secondary transmitter, scheduling procedure is
divided into two stages. In the first stage, the secondary receiver
determines a set $\mathcal{S}$ consisting of the candidates for
transmission which have low enough $\alpha_{s,i}$ so that the
calculated $\underset{i}{\max} P_{\mu,i} >P_{s,\max}$, and hence the transmit
power is saturated with $P_{s,\max}$. In the second stage, if the
set $\mathcal{S}$ is not empty, the $i^{\dagger}$ secondary
transmitter which has the largest $\beta_{s,i}$ among the elements
of $\mathcal{S}$ is finally scheduled with its maximum allowable
transmit power $P_{s,\max}$. Otherwise, the $i^{\dagger}$th
secondary transmitter which has the maximum $\gamma_{s,i}$ among the
$N$ secondary transmitters is scheduled with the transmit power of
$P_{\mu,i^{\dagger}}$. Hence, the index of the scheduled secondary
transmitter is given by
\begin{align}
i^{\dagger} = \left\{ \begin{array}{ll}
\arg\underset{i}{\max} \gamma_{s,i} ,& \underset{i}{\max}P_{\mu,i}\le P_{s,\max}\\
\arg\underset{i\in\mathcal{S}}{\max} \beta_{s,i} ,&
\underset{i}{\max}P_{\mu,i}> P_{s,\max}
\end{array}\right..
\end{align}
This two-stage scheduling certainly provides a lower bound on the
average achievable capacity of the secondary network, which is given
in the following theorem. \vspace{0.05in}
\begin{theorem}
For finite $P_{s,\max}$, a lower bound on the average achievable
capacity of the secondary network converges to
\begin{align}\label{Cap_low_complete}
&\mathbb{E}[C_{s,low}] = e^{\frac{K}{P_{s,\max}\ln(1\!-\!\frac{1}{N})}}\Bigg[\frac{\lambda}{\mu_{\min}}(1-p_d)\Bigg(\! \log_2(1\!+\!b_{N,low1})\nonumber\\
&-\frac{e^{\frac{1}{P_p}}\textrm{E}_1\!\!\left(\frac{1}{P_p}\right)}{\ln 2} \Bigg) + \left(\! 1\!-\!\frac{\lambda}{\mu_{\min}} \!\right)(1-p_f)\log_2(1+b_{N,low1}) \Bigg] \nonumber\\
&+ \left(1-e^{\frac{K}{P_{s,\max}\ln(1-\frac{1}{N})}} \right)\Bigg[\frac{\lambda}{M_{avg,l}}(1-p_d)\nonumber\\
&\times\Bigg( \log_2 \left( 1+P_{s,\max}\ln\left( N\left( 1-e^{-\frac{K}{P_{s,\max}}} \right) \right) \right)\nonumber\\
&-\frac{e^{\frac{1}{P_p}}\textrm{E}_1\left(\frac{1}{P_p}\right)}{\ln 2} \Bigg)+\left(1-\frac{\lambda}{M_{avg,l}}\right)(1-p_f)\nonumber\\
&\times\log_2 \left( 1+P_{s,\max}\ln\left( N\left( 1-e^{-\frac{K}{P_{s,\max}}} \right) \right) \right)\Bigg]
\end{align}
as $N$ goes to infinity, where
\begin{align}
b_{N,low1}=P_{s,\max}\mathcal{W}\left(\frac{KN}{P_{s,\max}}e^{\frac{K}{P_{s,\max}}}\right)-K,
\end{align}
\begin{align}
M_{avg,l} = p_de^{-R_p}\!+\!(1\!-\!p_d)\frac{\left(\!
e^{\frac{K(R_pP_{s,\max}+1)}{P_{s,\max}}}-1
\!\right)e^{-R_p(1\!+\!K)}}{\left(\!
e^{\frac{K}{P_{s,\max}}}\!-\!1\! \right)\left( R_pP_{s,\max}\!+\!1
\right)}
\end{align}\label{theorem3}
\end{theorem}where $\mathcal{W}(\cdot)$ denotes a Lambert W function.
\begin{proof}
The proof is given in Appendix.
\end{proof}\vspace{0.05in}

As shown in (\ref{Cap_low_complete}), the lower bound can be divided
into the capacities when the transmit power of the scheduled user is
not bounded and when the transmit power is bounded by $P_{s,\max}$.
The following corollary further simplifies the lower bound.
\vspace{0.05in}
\begin{corollary}\label{C_low_infty}
If we assume that $N$ goes to infinity, the result of Theorem
\ref{theorem3} is approximated as
\begin{align}
&\mathbb{E}[C_{s,low}]\approx \frac{\lambda}{M_{avg,l}}(1-p_d)\log_2(\ln N)\nonumber\\
&~~~\qquad\qquad+ \left( 1-\frac{\lambda}{M_{avg,l}} \right)(1-p_f)\log_2(\ln N)\nonumber\\
&=\left[ \frac{\lambda}{M_{avg,l}}(1-p_d) + \left( 1-\frac{\lambda}{M_{avg,l}} \right)(1-p_f) \right]\log_2(\ln N).
\end{align}
\end{corollary}
\begin{proof}
In (\ref{Cap_low_complete}), the probability of a secondary
transmission with unbounded power,
$e^{\frac{K}{P_{s,\max}\ln(1-\frac{1}{N})}}$, decreases with $N$.
\end{proof}\vspace{0.05in}

Corollary \ref{C_low_infty} indicates that the scaling law of the
lower bound is $k_{l}\log_2(\ln N)$, where $0\le k_{l}\le 1$ is
constant determined by system parameters.

\subsection{An Upper Bound}
An upper bound can be obtained from the case that
$\underset{i}{\max}\beta_{s,i}$ and $\underset{i}{\min}\alpha_{s,j}$ are the effective
forwarding and interference channel gains of the scheduled secondary
transmitter, respectively. This scenario certainly constructs an
upper bound on the achievable capacity of the secondary network and
the effective SNR of the scheduled secondary transmitter in this
case is given by
\begin{align}\label{snr_type}
\gamma_{s,up} = \left\{ \begin{array}{ll}
\frac{K\underset{i}{\max}\beta_{s,i}}{\underset{j}{\min}\alpha_{s,j}} ,& \underset{i}{\max}P_{\mu,i}\le P_{s,\max}\\
P_{s,\max}\underset{i}{\max}\beta_{s,i} ,& \underset{i}{\max}P_{\mu,i}> P_{s,\max}
\end{array}\right..
\end{align}

The following theorem shows an asymptotic behavior of the upper
bound on the average achieved capacity. \vspace{0.05in}
\begin{theorem}
For finite $P_{s,\max}$, the upper bound of the secondary average
achievable capacity converges to
\begin{align}\label{Cap_up_complete}
&\mathbb{E}[C_{s,up}]= e^{\frac{K}{P_{s,\max}\ln(1\!-\!\frac{1}{N})}}\Bigg[ \frac{\lambda}{\mu_{\min}}(1-p_d)\Bigg(\!\log_2 (P_{s,\max}\ln N)\nonumber\\
&-\!\frac{\textrm{E}_1\!\!\left(\frac{1}{P_p}\right)e^{\frac{1}{P_p}}}{\ln 2} \Bigg)+\left( \!1\!-\!\frac{\lambda}{\mu_{\min}} \right)(1-p_f)\log_2(P_{s,\max}\ln N) \Bigg]\nonumber\\
&+\left( 1-e^{\frac{K}{P_{s,\max}\ln(1-\frac{1}{N})}} \right)\Bigg[ \frac{\lambda}{\textrm{M}_{avg,u}(N)}(1-p_d)\nonumber\\
&\times\left( \log_2(1+P_{s,\max}\ln N)-\frac{\textrm{E}_1\left(\frac{1}{P_p}\right)e^{\frac{1}{P_p}}}{\ln 2} \right)\nonumber\\
&+\left(\!1- \frac{\lambda}{\textrm{M}_{avg,u}(N)}
\!\right)(1-p_f)\log_2(1+P_{s,\max}\ln N) \Bigg]
\end{align}
as $N$ grows to infinity, where
\begin{align}
&\textrm{M}_{avg,u}(N)\nonumber\\
&= p_de^{-R_p}+(1-p_d)\frac{ Ne^{-R_p(1+K)}\left( e^{\frac{K(R_pP_{s,\max}+N)}{P_{s,\max}}}-1 \right) }{ (R_pP_{s,\max}+N)\left( e^{\frac{KN}{P_{s,\max}}}-1 \right) }.
\end{align}
\label{theorem4}
\end{theorem}
\begin{proof}
The proof is similar to that of theorem 1.
\end{proof}\vspace{0.05in}

The asymptotic upper bound on the average achievable capacity can be
divided into the two parts according to the type of received SNR
described in (\ref{snr_type}). The following corollary further
simplifies the lower bound. \vspace{0.05in}
\begin{corollary}\label{corollary3}
If we assume that $N$ goes to infinity, the asymptotic upper bound
on the average achievable capacity in (\ref{Cap_up_complete}) is
given by
\begin{align}
&\mathbb{E}[C_{s,up}]\approx \frac{\lambda}{e^{-R_p}}(1-p_d)\log_2(\ln N)\nonumber\\
&~~~\qquad\qquad+ \left( 1-\frac{\lambda}{e^{-R_p}} \right)(1-p_f)\log_2(\ln N)\nonumber\\
&=\left[ \frac{\lambda}{e^{-R_p}}(1-p_d) + \left(1-\frac{\lambda}{e^{-R_p}} \right)(1-p_f) \right]\log_2(\ln N).
\end{align}
\end{corollary}
\begin{proof}
In (\ref{Cap_up_complete}), the capacity via bounded secondary transmit power,
$\gamma_{s,up}=P_{s,\max}\underset{i}{\max}\beta_{s,i}$, becomes
more dominant than the other as $N$ grows.
\end{proof}\vspace{0.05in}

Corollary \ref{corollary3} indicates that the upper bound grows like
$k_u\log_2(\ln N)$, where $0\le k_u\le 1$ is a constant determined
by system parameters. Consequently, we can characterize the
multiuser diversity gain of the secondary network under finite
$P_{s,\max}$ from the scaling law of the lower and upper bounds. The
asymptotic capacity of the secondary network is laid between the
asymptotic lower and upper bounds as
\begin{align}
k_l\log_2(\ln N)\le \mathbb{E}[C_s] \le k_u\log_2(\ln N).
\end{align}

This result implies that the scaling law of secondary network is $k\log_2(\ln N)$, where $k_l\le k \le k_u$.
This scaling law is similar to that in underlaid CR system, $\log_2(\ln N)$, except a constant scaling term $k$ determined by system parameters \cite{MUD_SS}.

\begin{figure}[!t]
    \centerline{\psfig{figure=./figure4.eps,width=0.9\columnwidth} }

    \caption{Average capacities of the secondary network ($p_d=0.8, p_f=0.3, \lambda=0.5, P_p=10\textrm{dB}, P_{s,\max}=P_p, R = 0.5, \mu_{\min}=0.95$)}
    \label{Fig:feasible_power}
\end{figure}

Fig. \ref{Fig:feasible_power} shows the average achievable capacity
of the secondary network, the lower bound, and the upper bound
versus $N$ for $P_{s,\max}=P_p$. It is verified that the lower and
upper bounds on the secondary network capacity are quite well
characterized by the asymptotic approximations even if the number of
secondary transmitter is small. In addition, this figure confirms
that the scaling laws of lower and upper bounds are the same as that
of the exact capacity.

\section{Conclusion}
This paper has investigated the multiuser diversity gain and its
capacity scaling law in an interweaved CR system. We have analyzed the capacity of a secondary network by taking into account
the traffic dynamics of a primary user and reliability of spectrum
sensing in the secondary network. Then, by applying extreme value
theory, we have derived asymptotic capacity of the secondary network
and characterized the multiuser diversity gain. The asymptotic capacity has
been shown to grow like $k\log_2(\ln N)$, where $0\le k\le 1$.
This scaling law of the capacity is similarly observed in underlaid CR systems even
though the mechanism of spectrum access is totally different. Our
numerical results have also verified that the asymptotic capacity
agrees well with the exact capacity even for small number of
secondary transmitters, $N$.

\appendix[Proof of Theorem 1]

The average achievable capacity by the suboptimal two stage
scheduling is given by
\begin{align}\label{cap_sub_sec}
\mathbb{E}[C_{s,low}]=&\Pr\left[ \max_{i}P_{\mu,i}\le P_{s,\max}
\right]\mathbb{E}[C_{s,low}^{(1)}]\nonumber\\
& + \Pr\left[ \max_{i}P_{\mu,i}>
P_{s,\max} \right]\mathbb{E}[C_{s,low}^{(2)}]
\end{align}
where $\mathbb{E}[C_{s,low}^{(1)}]$ and
$\mathbb{E}[C_{s,low}^{(2)}]$ are the average achievable capacities
when there is no element in $\mathcal{S}$ and when there exists at
least one element in $\mathcal{S}$ and given, respectively, by
\begin{align}\label{cap_sub_sec_1st}
\mathbb{E}&[C_{s,low}^{(1)}]=\frac{\lambda}{\mu_{\min}}(1-p_d)\nonumber\\
&~\times\mathbb{E}\left[ \log_2 \left( 1+ \frac{\underset{i}{\max}\frac{K\beta_{s,i}}{\alpha_{s,i}}}{1+\beta_pP_p} \right)\bigg|  \max_i \frac{K}{\alpha_{s,i}}\le P_{s,\max} \right]\nonumber\\
&~+\left( 1-\frac{\lambda}{\mu_{\min}}\right)(1-p_f)\nonumber\\
&~\times\mathbb{E}\left[ \log_2
\left(1+\max_{i}\frac{K\beta_{s,i}}{\alpha_{s,i}} \right)\bigg|
\max_i\frac{K}{\alpha_{s,i}}\le P_{s,\max} \right],
\end{align}
\begin{align}\label{cap_sub_sec_2nd}
\mathbb{E}&[C_{s,low}^{(2)}]=\frac{\lambda}{\mathbb{E}\left[ \mu(\alpha_{s,i^{\dagger}}, P_{s,\max})\Big|\underset{i}{\max}P_{\mu,i}>P_{s,\max} \right]}(1-p_d)\nonumber\\
&~~\times\mathbb{E}\left[ \log_2\left( 1+\frac{\underset{i\in\mathcal{S}}{\max}\beta_{s,i}P_{s,\max} }{ 1+\beta_pP_p } \right) \right]\nonumber\\
&~~+\left(1-\frac{\lambda}{\mathbb{E}\left[
\mu(\alpha_{s,i^{\dagger}},X
P_{s,\max})\Big|\underset{i}{\max}P_{\mu,i}>P_{s,\max}
\right]}\right)\nonumber\\
&~~\times(1-p_f)\mathbb{E}\left[ \log_2 \left( 1+
\underset{i\in\mathcal{S}}{\max}\beta_{s,i}P_{s,\max}\right)
\right].
\end{align}
As $N$ grows, the distribution of maximum random variables in
(\ref{cap_sub_sec_1st}) and (\ref{cap_sub_sec_2nd}) are converges to
the Gumbel distribution or Frech\'et distribution
\cite{Order_statistic}. Using these approximated distributions, we
can asymptotically approximate the expectations as shown in
(\ref{Cap_low_complete}).


TITLE: Prove that $\left(\sum_{k=1}^{n}k\right)^2=\sum_{k=1}^{n}k^3$ holds true for $n ≥ 1$
QUESTION [2 upvotes]: I've been trying to figure out this proof for way too long now, I'm just not sure where to begin for the inductive step. Any guidance would be greatly appreciated.

REPLY [0 votes]: Hint: We know that $\sum_{k=1}^nk={n(n+1)\over 2}$.


TITLE: On the matrix logarithm
QUESTION [5 upvotes]: Let $A\in M_{n\times n}(\mathbb{C})$. We define

$\displaystyle\ln A=\sum_{k=1}^\infty(-1)^{k+1}\frac{(A-I)^k}{k}$

This series is convergent in particular when $A-I$ is nilpotent. Is it true (and how do you prove it) that if $A-I$ is nilpotent, then $e^{\ln A}=A$? (Also what about $\ln e^A=A$?)

REPLY [2 votes]: For convenience, let $N = I-A$ and suppose $N^M=0$.  Consider
$$\ln(I-tN) = - \sum_{k=1}^{M-1} (tN)^k
/k$$
Note that $$ \dfrac{d}{dt} \ln(I-tN) = - \sum_{k=1}^{M+1} (tN)^{k-1} = -(I-tN)^{-1}$$
Let $g(t) = \exp(\ln(I-tN))$.  Then $g(0) = \exp({\bf 0}) = I$ and
$$ g'(t) = g(t) \dfrac{d}{dt} \ln(I-tN) = - (I-tN)^{-1} N g(t)$$
The unique solution of this initial value problem is
$$g(t) = I-tN$$
In particular, for $t=1$ we have $\exp(\ln(A)) = A$.
The other direction $\ln(\exp(A))$ doesn't quite make sense yet: you've defined $\ln$ for a matrix whose only eigenvalue is $1$; it would work if all eigenvalues are in the open disk $\{z \in \mathbb C: |z-1|<1\}$, but the eigenvalue of $\exp(A)$ is $e$, and the series
$$ \sum_{k=1}^\infty (-1)^k (\exp(A)-I)^k/k $$ will diverge.
It is possible to define a branch of logarithm
so that $\ln(\exp(A)) = A$.  However, in general one has
to be careful, because the logarithm is multivalued.  Even for scalars, it's impossible to define
a single logarithm function so that $\log(\exp(z)) = z$ for all $z$, because $\exp(2\pi i) = \exp(0) = 1$.


TITLE: Isomorphism of irreducible R-modules
QUESTION [1 upvotes]: Let $R$ be a $k-$algebra and $M,N$ two irreducible $R-$modules, isomorphic as vector spaces. If we know that for every $r\in R$ we have the same eigenvalues on $M$ and $N$ (with multiplicities) is it true that $M\simeq N$?
I have two thoughts:

Denoting $St(m)=\{ rm=m \mid r\in R\}$, if we have $St(m)=St(n)$ for some $m\in M, n\in N$, we could find an isomorphism that sends $m$ to $n$ and is extended by $f(rm)=rf(m)$ (every element of $M$ is $rm$ for some $r$ by irreducibility). I cannot prove the existence of such a pair though.
If $R$ was commutative, we could simultaneously diagonalize all the actions according to some basis and take the isomorphism of vector spaces that sends the one basis to the other. This will then be an isomorphism of modules. I cannot extend that to the general case.

REPLY [1 votes]: Consider the commutative algebra $R=k[x,y]/(x^2,y^2)$ and the modules $R/(x)$ and $R/(y)$.


TITLE: Check if a $f \in U$
QUESTION [1 upvotes]: Let $U$ be a subspace where $U = \{ f \in Abb(\mathbb{R},\mathbb{R}) |\; \forall\; x \in \mathbb{R}: f(x) = -f(-x)\}$ and $f_0: x \mapsto \frac{1}{1+x^2}$.
Check if $b: x \mapsto |x|$ is in $f_0 + U$.

When I write out $f_0 + U = \frac{1}{1+x^2} + f(x) = \frac{1}{1+x^2} -f(-x)$ I can't see a way to bring the function $b$ into the game. I know I can write $b = x$, $x \geq 0$ and $b = -x$, $x<0$ or when I plot $f(x) = x$ which is in U, I can see that it matches the graph for $x>0$ but not when $f_0$ comes into play. I am not even sure how to tackle this problem.
EDIT: $Abb(\mathbb{R},\mathbb{R})$ means all maps from $\mathbb{R} \to \mathbb{R}$

REPLY [1 votes]: If $b \in f_0+U,$ then $b-f_0 \in U.$ That is $(b-f_0)(x)=(f_0-b)(-x).$ Now $$(b-f_0)(x)=|x|-\frac{1}{1+x^2}$$ and $$(f_0-b)(-x)=\frac{1}{1+x^2}-|-x|=\frac{1}{1+x^2}-|x|,$$ therefore $$|x|-\frac{1}{1+x^2}=\frac{1}{1+x^2}-|x|$$ which implies $$|x|=\frac{1}{1+x^2}$$but this is only true for exactly $2$ values of $x \in \mathbb{R}.$


TITLE: Degree of a variety vs degree of its blow-up
QUESTION [4 upvotes]: Let $V\subset \mathbf{P}^n$ be a variety. Let $f:\mathbb{P}^n\times \mathbb{P}^{n-1}\to \mathbb{P}^n$ be a blow-up of a point $P$ on $V$. Write $\widetilde{V}$ for the strict transform of $V$ under $f$, i.e., the Zariski closure of $f^{-1}(V\setminus \{P\})$.
It is tempting to guess that there ought to be an ample divisor $D$ on $\mathbb{P}^n\times \mathbb{P}^{n-1}$ such that, for $\deg_D$ the associated degree, $$\deg_D \widetilde{V} = \deg V.$$
However, as is shown in a comment, this need not be the case.
What is the minimal $c_n$ such that, for every $V$,
$$\deg_D \widetilde{V} \leq c_n \deg V$$
for some divisor $D$?

REPLY [3 votes]: First, the ample divisors on $\mathbf P^n \times \mathbf P^{n-1}$ are of the form $aH_1+bH_2$ for $a,b>0$, which means that every ample divisor is of the form $H_1+H_2+N$ for some nef divisor $N$. This shows that $\operatorname{deg}_{H_1+H_2}\widetilde{V} \leq \operatorname{deg}_D\widetilde{V}$ for every other ample divisor $D$.
Next, the restriction of $H_1+H_2$ to the blowup $\operatorname{Bl}_p\mathbf P^n \subset \mathbf P^n \times \mathbf P^{n-1}$ is represented by the strict transform $\widetilde{Q}$ of any quadric $Q$ smooth at  $p$.
For a variety $V \subset \mathbf P^n$ of dimension $d$, we can find $d$ such quadrics $Q_1,\ldots,Q_d$ such that the projectivised tangent cones of $V$ and $Q_1 \cap \cdots \cap Q_d$ at $p$ are disjoint, and away from $p$ the intersection $V \cap Q_1 \cap \cdots \cap Q_d$ is a finite set.
Hence
$$\operatorname{deg}_D \widetilde{V} = \widetilde{V} \cdot \widetilde{Q_1} \cdots \widetilde{Q_d}= V \cdot Q_1 \cdots Q_d -\operatorname{mult}_pV =2^d \operatorname{deg} V -\operatorname{mult}_p V.$$
In particular taking $V=\mathbf P^n$ as in Remy van Dobben de Bruyn's comment, we get
$$\operatorname{deg}_D \widetilde{V} = (2^n-1)\operatorname{deg} V$$
while for any other $V$ we have
$$\operatorname{deg}_D \widetilde{V} \leq 2^d \operatorname{deg} V<(2^n-1) \operatorname{deg} V.$$
So $c_n=2^n-1$.


TITLE: need help on limit question
QUESTION [0 upvotes]: need to find limit of $f(x)=((x^2-x-6)^2)/(x+2)^2$ as $x$ tends to $-2$.
so $f(x)$ can be cancelled down to $(x-3)^2$.
I need to be really formal when taking the limit of $(x-3)^2$. this is what I did:
$$\lim_\limits{x\to-2} f(x)=\lim_\limits{x\to-2} (x-3)^2=\left(\lim_\limits{x\to-2} x-3\right)^2$$
Now since we know the polynomial function $x-3$ is continuous and in particular continuous at $x=-2$, by the algebra of continuous functions, we have:
$$\left(\lim_\limits{x\to-2} x-3 \right)^2=((-2)-3))^2=(-5)^2=25.$$
Is this OK?

REPLY [0 votes]: Yes, it is correct.
By noting that $x^2-x-6=(x-3)(x+2)$, the function
$$
f(x)=\frac{(x^2-x-6)^2}{(x+2)^2}
$$
can be rewritten as
$$
f(x)=\left(\frac{(x-3)(x+2)}{x+2}\right)^{\!2}=(x-3)^2
$$
in its whole domain (that is, for $x\ne-2$). Thus
$$
\lim_{x\to-2}f(x)=\lim_{x\to-2}(x-3)^2=25
$$
because $g(x)=(x-3)^2$ is everywhere continuous.


TITLE: $f$ continuous, moderate decrease and $\int_{-\infty}^{\infty}f(y)e^{-y^{2}}e^{2xy}dy=0$ implies $f=0$.
QUESTION [3 upvotes]: Let $f$ a continuous function, moderate decrease and satisfying
\begin{equation}
\int_{-\infty}^{\infty}f(y)e^{-y^{2}}e^{2xy}dy=0
\end{equation}
for all $x\in\mathbb{R}$ I need to prove that $f=0$.
The hint is to consider $f*e^{-x^2}$. I know that both $f$ and $e^{-x^{2}}$ are moderate decrease, so the convolution is too, i.e., $\exists A>0$ such that $|(f*e^{-x^2})(x)|\leq\frac{A}{1+x^{2}}$ for all $x\in\mathbb{R}.$ So,
$$|(f*e^{-x^2})(x)|=\left|\int_{-\infty}^{\infty}f(x-y)e^{-y^{2}}dy\right|\leq\frac{A}{1+x^{2}} $$
How can I use the $\int_{-\infty}^{\infty}f(y)e^{-y^{2}}e^{2xy}dy=0$ condition?

REPLY [1 votes]: Define $g=f * e^{-x^{2}}$.
\begin{align*}
    g(x)&=\int_{-\infty}^{\infty} f(y) e^{-(x-y)^{2}} d y\\
    &=e^{-x^{2}} \int_{-\infty}^{\infty} f(y) e^{-y^{2}+2 x y} d y\\
    &=0.
\end{align*}
Therefore
\begin{align*}
  0&=\hat{g}(\xi)\\
  &=\hat f(\xi) \cdot \hat{e^{-x^{2}}}(\xi)\\
  &=\hat{f(\xi)} \cdot \sqrt{\pi} e^{-\pi^{2} \xi^{2}}
\end{align*}
Therefore $$
\hat{f}(\xi)=0.
$$
By Fourier inversion or Plancheral identity  $$f\equiv 0.$$


TITLE: Can someone please explain this vector space definition to me?
QUESTION [1 upvotes]: Snippet taken from "Linear Algebra" by Jim Hefferon (4th edition):

1.2 Definition Suppose that $V$ and $W$ are vector spaces of dimensions $n$ and $m$
with bases $\mathrm{B}$ and $\mathrm{D},$ and that $\mathrm{h}: \mathrm{V} \rightarrow \mathrm{W}$ is a linear map. If
$$
\operatorname{Rep}_{\mathrm{D}}\left(\mathrm{h}\left(\vec{\beta}_{1}\right)\right)=\left(\begin{array}{c}
\mathrm{h}_{1,1} \\
\mathrm{~h}_{2,1} \\
\vdots \\
\mathrm{h}_{\mathrm{m}, 1}
\end{array}\right) \quad \ldots \quad \operatorname{Rep}_{\mathrm{D}}\left(\mathrm{h}\left(\vec{\beta}_{\mathrm{n}}\right)\right)=\left(\begin{array}{c}
\mathrm{h}_{1, \mathrm{n}} \\
\mathrm{h}_{2, \mathrm{n}} \\
\vdots \\
\mathrm{h}_{\mathrm{m}, \mathrm{n}}
\end{array}\right)_{\mathrm{D}}
$$
then
$$
\operatorname{Rep}_{\mathrm{B}, \mathrm{D}}(\mathrm{h})=\left(\begin{array}{cccc}
\mathrm{h}_{1,1} & \mathrm{~h}_{1,2} & \ldots & \mathrm{h}_{1, \mathrm{n}} \\
\mathrm{h}_{2,1} & \mathrm{~h}_{2,2} & \ldots & \mathrm{h}_{2, \mathrm{n}} \\
& \vdots \\
\mathrm{h}_{\mathrm{m}, 1} & \mathrm{~h}_{\mathrm{m}, 2} & \ldots & \mathrm{h}_{\mathrm{m}, \mathrm{n}}
\end{array}\right)_{\mathrm{B}, \mathrm{D}}
$$
is the matrix representation of h with respect to B, D.

Here $h$ is homomorphoism.
What I am not able to understand that if there is a linear mapping, $f:V\to W$, then according to a previous theorem in the same book, the bases of $V$ maps to bases of $W$.
Therefore, if $B$ is the bases of $V$ containing $n$ vectors $\langle B_1, B_2, \dots, B_n\rangle$, then
$\langle h(B_1), h(B_2), \dots, h(B_n)\rangle$ is the bases for $W$.
Now shouldn't in the definition of the attached screenshot $h_ij = 1$; if $i=j$ and $h_ij = 0$; if $i\neq j$? So that the final matrix have to be an identity matrix? What's going on?
(The book is available from https://hefferon.net/linearalgebra/ and this definition is on page 214. The notation $\mathrm{Rep}_B(\vec v)$ is defined on page 124.)

REPLY [1 votes]: You've not told us how this notation is defined in the book.
I take it that $B=(\beta_1,\dots, \beta_n)$ is an ordered basis of $V$ and $D=(\delta_1,\dots,\delta_m)$ is an ordered basis of $W$.
I think that if you look at the definition of
$\text{Rep}_D (v)$
you will see that the hypothesis is that for each $i$ the vector $h(\beta_i)$ can be written
$$
h(\beta_i)=h_{1,i}\delta_1+\dots +h_{m,i}\delta_m . (*)
$$
If you now look at the definition of $\text{Rep}_{B,D}(f)$ you will see that $(*)$ are exactly the equations you need to make $\text{Rep}_{B,D}(h)$ exactly what the book says.


TITLE: What is the codomain of the exponential function?
QUESTION [0 upvotes]: Do we have

$\exp:\mathbb R \rightarrow \mathbb R$; or
$\exp:\mathbb R \rightarrow \mathbb R^+$?

Option 1 seems to be common but seems to have the problem that $\exp$ wouldn't be the inverse of $\ln:\mathbb R^+ \rightarrow \mathbb R $ ?

REPLY [1 votes]: The codomain may be any set that contains the positive real numbers. The range of course is the set of all positive reals, provided the domain is restricted to only the real numbers.