TITLE: Label x axis in radians WolframAlpha
QUESTION [2 upvotes]: I am trying to graph a trig function in WolframAlpha. I want the x axis to show radians. It will do this for a basic function, such as $y=\sin x$, but for many functions it will just label the x axis from -10 to 10. I have googled this and this is the best I can come up with:
plot sin (x/2 -pi/16) ,Ticks ->{{-pi, -pi/2, 0, pi/2, pi, 3pi/2, 2pi, 5pi/2},{-1,1}}
It doesn't work, WolframAlpha just plots the function from -10 to 10 on the x axis.

REPLY [0 votes]: I know this is a little tedious, but in the free version of Wolfram Alpha, you can still have it give you results at any point even if you can't do the ticks.
In particular, before the graph it will still nicely list all the global maxima and minima with the max{} and min{} functions. For example: max { sin(x) } from x=0 to 8 pi will list out where the global maxima is. The min function works identically. Here, the result is given in radians, even though the axis unit ticks are not.


TITLE: Optimization, prove the following equation...
QUESTION [0 upvotes]: Proving the equation in part (a) is difficult, it may require using (t=d/s).
However in part (b) my answer i got was 50 gallons per hour. Can someone please provide help with part(a).

REPLY [0 votes]: The traveling time for the $800$ miles is $(800/w)$ hr, and the fuel rate is $(1 + 4\times 10^{-6}w^3)$ gal/mi, so the total amount $A$ of fuel used in the $800$ mile journey is $$\left(\frac{800}{w} \operatorname{hr}\right)\left(1 + 4\times 10^{-6}w^3\frac{\operatorname{gal}}{\operatorname{hr}}\right) = \frac{800}{w} + 32 \times 10^{-4}w^2\operatorname{gallons}$$ or $$\frac{800}{w} + 0.0032 w^2 \operatorname{gallons}.$$


TITLE: Finding the value of $6b-3a$
QUESTION [2 upvotes]: Consider $f(x)= [x]^2 - [x+6]$ and $g(x)= 3kx^2+ 2x + 4(1-3k)$ where
  $[x]$ denotes the floor function.
Let $A= \{x ~|~ f(x)= 0\}$ and $k \in [a,b]$ for which every element
  of set A satisfies the inequality $g(x)\ge 0$.
1) The set $A$ is equal to ?
2) The value of $6b- 3a =$?

The first part is easy. It's just the solution of the quadratic; $[x]^2- [x]- 6 = 0 $
$\therefore \mathbb A = [-2, -1) \cup [3,4)$ which is the right answer.
But I am stuck on the second part of the question.  I am unable to use the constraint on the values of $x$ to find k.
using the condition I have got:
$k(12- 3x^2) \ge -{2(x+2)} $
For $x=2$ this is true,
 For $x\ne 2$:
$\implies3k(2-x) \ge -2 $
What do I do next?
The answer is :

 $2$

REPLY [1 votes]: You need for $k$ to satify:
$$k (3x^2-12) =3k(x-2)(x+2)\geq -2(x+2)$$ for all $x \in \Bbb A= [-2, -1) \cup [3,4)$.
You can then consider the three cases:

$x=-2$ then the inequality is verified.
For all $x \in (-2, -1)$, as $x-2 <0$ and $x+2>0$ the inequality is equivalent to:
$$k \leq \frac{2}{3(2-x)}$$
and as it must be true for any $x$ and $\inf_{x \in(-2,-1)} \frac{2}{3(2-x)}=\frac{2}{3(2-(-2))}=\frac{1}{6}$ it is the same as:
$$k \leq \frac{1}{6}$$
For all $x \in (3, 4)$, as $x-2 >0$ and $2+x>0$ the inequality is equivalent to:
$$k \geq -\frac{2}{3(x-2)}$$
and once again as $\sup_{x \in(3,4)} \frac{-2}{3(x-2)}= \frac{-2}{3(4-2)}=\frac{-1}{3}$ it is true iff:
$$k \geq -\frac{1}{3}$$

Thus $g(x) \geq 0$ for all $x \in \{-2\} \cup (-2, -1) \cup [3,4)$ iff:
$$-\frac{1}{3} \leq k \leq \frac{1}{6}$$
i.e:
$$a=-\frac{1}{3}$$
$$b=\frac{1}{6}$$
and:
$$6b-3a=1-(-1)=2$$


TITLE: I want to learn more mathematics
QUESTION [1 upvotes]: I'm a computer science student, but I'm not satisfied with my math skills.
I need suggestions on how to improve my math skills, including books for beginners and the like.

REPLY [2 votes]: I am also learning mathematics but I have few suggestions to you as a fellow learner.

Solve as much problems you can. Mathematics is not a spectator sport. You need to get your hands dirty, scribble things on paper etc.
Read "How to Solve it" by G. Polya and "What is Mathematics" by Courant . These books will help you in developing some interests which is most important thing. You may also like to read "Letters to young mathematician " by Ian Stewart.


TITLE: Extracting useful energy inline to a pressure relief valve?
QUESTION [0 upvotes]: So imagine if we have a pressure vessel being pressured with a pump.
Let's say there is a pressure relief valve that's necessary in the design.
Would you actually be able to extract useful work from putting an expander in line to the pressure relief valve?
I want to say no because any additional constrictions would just be more work the pump would have to do.
However on the other hand, when there is no requirements on flow and you have all the time you want, the pressure will equalize in the system's volume, and restrictions don't matter as there is a limited volume of the system, and a specific volume of gas in that system.
So it seems like you could put an expander over a pressure relief outlet and recycle energy.
Is this correct? if so, couldn't you just put expanders/generators in air conditions instead of expansion valves?

REPLY [2 votes]: Pressure relief valves are safety devices that prevent rupture of equipment and occasionally prevent explosions.  Their design typically accounts for any downstream back-pressure, and that back-pressure should be minimized, precisely because it is a safety device.  It may even be illegal to install downstream equipment that would impede process flow from the device.  I recommend that you consult OSHA 1910 (management of change) to determine if this issue is explicitly addressed.
In the event that your suggestion is legal, there is an important consideration to keep in mind.  When a given process is running normally, the pressure relief valve is not open, and indeed, pressure relief valves should not open at all if process conditions remain within their designed range.  This means that you want to install an expensive piece of rotating equipment at a spot in the process where it either will not be used at all, or will be used VERY infrequently.  No manager wants to waste money on a given process, so such an idea is very impractical.


TITLE: Continuous limsup, Discrete limsup relation
QUESTION [2 upvotes]: Is it true that for a real function f:
$$
\limsup_{t \to \infty} f(t) = \sup_{(t_n)_{n \in \mathbb{N}}, (t_n) \to \infty}\limsup_{t_n \to \infty} f(t_n)
$$
?
Where the first limsup is seen in the continuous sense and the second in a discrete sense.
I'm interested in this kind of properties as I'm working on some Brownian Motion proofs where I need easy ways for manipulating both kinds of limsup. I'm not directly interested in the proof of the result but rather on it's truthfulness. Still any intuition or sketch of proof is welcome...
What useful characterizations allow one to work and move between both kinds of limsup without much trouble?

REPLY [0 votes]: not generally, the discrete version loses too much information to work with $\lim\sup$. take for instance the function $f=\sin x$, sequence $t_n=2 \pi n$. it might work better with Brownian motion-type functions, since my example comes from the periodicity.


TITLE: Divisor Function Symmetry Neighbor Divisors
QUESTION [0 upvotes]: Error symmetry wave divisor function.
The divisor function can be written as a summation of waves (see link below previous questions Stacks Exchange). The error in the wave divisor function is mainly determined by it's neighbor divisors. The error is proportional to:
$$\varepsilon (x) \propto \sum_{\mathbb{X}\vert (x-1)}^{} \cos(k \mathbb{X}) + \sum_{\mathbb{X}\vert (x+1)}^{} \cos(k \mathbb{X})$$
Here $\mathbb{X} \vert (x-1)$ means: $\mathbb{X}$ divides $(x-1)$. Basically the divisors are added from the neighbors of $x$:
$$\varepsilon (9) = \cos(k1)+\cos(k2)+\cos(k4)+\cos(k8)+\cos(k1)+\cos(k2)+\cos(k5)+\cos(k10)$$
The total error then is a (cosine)summation of errors like Brownian motion. Where k is a constant and determines the pulse width of each divisor wave, see link below for more information.
$$k=-\frac{2 \log(L)}{\pi \Delta x^{2}}$$
We can simulate the error for a number $x$ by keeping $L=0.5$ and vary $\Delta x$ between: 0.15 and 0.2 in 10000 steps. For every $k$ the error can be calculated. See simulation below.
It is observed that for $x=odd$ the error $\varepsilon (x)$ tends to nonsymmetrical/skewed distribution. For $x=even$ the error $\varepsilon (x)$ tends to a symmetrical distribution.
Normally I exclude 1 as an divisor, but the symmetrical and skewed distribution are always present (with and without 1 as divisor).
Typical example Symmetrical even $x$:

Typical Example Non Symmetrical odd $x$:

Question:
Why does the error for odd and even numbers $x$ tend to behave symmetric and asymmetric/skewed?
More information.
More information stackexchange:
Error in Divisor Function Modelled With Waves
Interactive Simulation Github/Mybinder:
Mybinder Jupyternotebook

REPLY [2 votes]: Partial Answer Skew Error Distribution.
This is the divisor counting of neighbors left and right of x. So sum of divisors at (x-1) and (x+1).
Assumption 1:
Divisors of odd numbers will always be odd:
Do odd numbers have only odd divisors?
Case 1 (odd divisors only):
Function analysis show that the following function is symmetrical (positive and negative spikes occur).
$$\varepsilon(k) = \sum_{\mathbb{X}}^{30} \cos(k (2\mathbb{X}-1))$$
Case 2 (even divisors only):
Function analysis show that the following function only has positive spikes.
$$\varepsilon(k) = \sum_{\mathbb{X}}^{30} \cos(k 2 \mathbb{X})$$
So for even divisors the distribution will look skewed asymmetrical. Even divisors have a mix of odd and even divisors.
Thus the error for even numbers $x$ is symmetric and the error for odd numbers $x$ is skewed.
Though no proof is supplied for both formula above. Maybe the derivative (and determine max and min) supply the proof.


text:
\section{How universal is universality?}
Random matrix theory traditionally studies certain random matrices of interest to
physicists and statisticians.
The central question of classical random matrix theory is to prove that the eigenvalues
of random matrices' show \emph{universal} behavior as the size of the random matrices grow.
Universality is not a precise concept. The classical central limit theorem gives
an example of it:  with only light conditions
on a collection of random variables (being i.i.d.\ with finite variances), their centered and
normalized sums converge in law to Gaussian.

The most basic symmetric random matrix model
is the \emph{Gaussian Orthogonal Ensemble}, abbreviated GOE.
Let $G$ be an $n\times n$ matrix whose entries are independent and distributed as $N(0,2)$.
Define $X$ as $(G+G^T)/2$, a random symmetric matrix with independent
entries on and above the diagonal. The random matrix $X$ has centered Gaussian entries with
variance~$1$ above the diagonal and variance~$2$ on the diagonal, and it
is said to be drawn from the GOE.
Any $n\times n$ random matrix with centered independent entries on and above the diagonal and variance~$1$
entries above the diagonal is called a \emph{Wigner matrix}.
(The word ``ensemble'' does not have any precise meaning,
but it is usually refers to a collection of probability distributions on $n\times n$ matrices,
as $n$ ranges from $1$ to infinity. Each distribution typically obeys some sort of invariance.
For instance, if $O$ is an arbitrary orthogonal matrix and $X$ is drawn from the GOE, then
$OX$ has the same distribution as $X$.)

An example of universality for random matrices is that the eigenvalues of $n\times n$ Wigner matrices
show the same limiting behavior as those of matrices from the GOE as $n\to\infty$.
Most results along these lines were confirmed only recently, in a series
    of papers including \cite{TV1,TV2,ESY1,ESY2,ESY3,ESY4,ESY5}.

  The adjacency matrix of a random regular graph is similar to a Wigner matrix, but its entries
  are uncentered and lightly dependent. How does this affect the adjacency matrix's spectral properties?
   To put it another way, how universal is universality
  of random matrices? This is our main motivation for investigating properties of eigenvalues
  of random regular graphs from the perspective of random matrix theory.
\section{Stein's method applied to random regular graphs}
Graph eigenvalues have a close connection to the graph's structural properties
(see \cite{Chung,Spielman}). We exploit this by determining spectral properties
of random regular graphs by looking at the distribution of their cycle counts.
The main novelty of our approach is the use of \emph{Stein's method}, which to our knowledge
had never been applied to random regular graphs before.
Stein's method is a collection of techniques for distributional approximation. Stein's method
naturally gives not just asymptotic results but also quantitative error bounds on the approximations.
This was essential for the eigenvalue fluctuation results described in this thesis.

Stein's method  was originally developed by Charles
Stein for normal approximation; its first published
use is \cite{St1}. Louis Chen adapted the method
for Poisson approximation \cite{Chen1}.
Because of this, Stein's method is sometimes called the
Stein-Chen or Chen-Stein method when applied to Poisson approximation.
Now that Stein's method is understood in a more general and applied
to a wide range of distributions, it is more typical to see it called
just Stein's method, regardless of the type of approximation.
The survey
paper \cite{Ross} gives a broad introduction to Stein's method,
and \cite{CDM} and
\cite{BHJ} focus specifically on using it for Poisson approximation, as we do in
this thesis.

The classical scenario for Poisson approximation is for sums of increasingly many, increasingly
unlikely independent indicators: in other words, the convergence of $\Bin(n, \lambda/n)$
to $\Poi(\lambda)$ as $n\to\infty$.
There are several approaches to Stein's method for Poisson approximation, each allowing
this approximation to hold in the presence of some dependence.
The most straightforward is the local approach: each indicator is independent of all others
but a small ``neighborhood''. This was the original approach in \cite{Chen1}, and it is generalized
and put in a very usable form in \cite{AGG}. This approach does not seem to work in the context
of random regular graphs, where nearly everything is lightly dependent on everything else.
Another approach is size-bias coupling. This theory  is developed at length for Poisson approximation
in \cite{BHJ}, though it is not viewed through the lens of size-biasing there.
See\cite{Ross} and \cite{AGK} for how it fits into this framework. We use this method on the
\emph{permutation model} of random regular graph (see Section~\ref{sec:models} for its definition).
Another technique is the method of exchangeable pairs; see \cite{CDM} and \cite{Ross} for good
expositions. This technique is perhaps the most flexible and the most finicky of the three.
We use it for Poisson approximation in the \emph{uniform model} of random regular graph, defined
in Section~\ref{sec:models}. This technique has some clear similarities to a combinatorial
technique called  \emph{the method of switchings}, and we make some rigorous connections between the two.
\section{The results of this thesis}
  Consider an $n\times n$ Wigner random matrix $X_n$ (a symmetric matrix with independent, mean zero,
  variance one entries above the diagonal). Choose an interval in the real line,
  and let $N_n$ denote the
  number of eigenvalues of $n^{-1/2}X_n$ lying in this interval. A fundamental result in random matrix
  theory is that $N_n /n$ converges in probability to a deterministic value as $n$ tends to infinity. This
  value is the measure of the interval
  under Wigner's semicircle law, the measure on $[-2,2]$ given by the density $\frac{1}{2\pi}
  \sqrt{4-x^2}\,dx$. This measure is a universal limit, in the sense
  that it does not depend on the distributions of the individual matrix entries, besides their
  means and variances.

  The analogue of this result for random regular graphs appears in \cite{McK}:
  Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of a random $d$-regular graph on $n$ vertices.
  If $f\colon \RR\to\RR$ is an indicator
    on an interval or is bounded and continuous, then as $n\to\infty$,
    \begin{align*}
      \frac{1}{n}\sum_{i=1}^n f(\lambda_i) \toPr \int_{-2\sqrt{d-1}}^{2\sqrt{d-1}} f(x)p_d(x)\,dx.
    \end{align*}
  The limiting measure $p_d(x)\,dx$ is not the semicircle law, but a different
  measure known now as the
  Kesten--McKay law. Its density is given on $\abs{x}\leq 2\sqrt{d-1}$ by
  \begin{align}
    p_d(x)=\frac{d\sqrt{4(d-1)-x^2}}{2\pi(d^2-x^2)}.\label{eq:kmlaw}
  \end{align}
  The expression $\sum f(\lambda_i)$ is called a \emph{linear eigenvalue statistic}.

  The topic of this thesis is the second-order behavior of these linear statistics.
  We will show that when the degree of the random graphs is held fixed,
  their fluctuations converge to compound
  Poisson distributions, in contrast to the Gaussian limit known for Wigner matrices.
  If the degree grows with the size of the graph, however, the limit of the fluctuations is
  Gaussian, in line with the universal behavior.
  We show that this holds in two models of random regular graphs,
  defined in Section~\ref{sec:models}.

  The path to these results is through an analysis of the distribution of cycle counts in these
  models by Stein's method. These results are interesting in their own right, and they make up
  Chapter~\ref{chap:poisson}. In Chapter~\ref{chap:fluctuations}, we apply them to prove
  the eigenvalue fluctuation results.

  In Chapter~\ref{chap:GFF}, we consider a process of growing random regular graphs.
  The eigenvalue fluctuations are then a stochastic process whose marginals are given
  by the results of Chapter~\ref{chap:fluctuations}. This is analogous to a \emph{corners
  process} in random matrix theory; see \cite{BG} for a good introduction. The idea
  is to think of a sequence of random matrices as the principal minors of an infinite random matrix.
  One can then consider not just the marginal distribution of the eigenvalues of each random matrix,
  but the joint distribution of eigenvalues of a matrix and its minors.
  The limiting fluctuations of some of these processes can be expressed in terms
  of the Gaussian free field \cite{Bor1,Bor2,BG}. We show that the same holds for the
  eigenvalues of the growing random regular graphs.

Most of this thesis is joint work.

Chapters~\ref{chap:poisson} and~\ref{chap:fluctuations} are a synthesis of \cite{DJPP},
\cite{JP},
and \cite{Joh}. The results on the permutation model are from \cite{DJPP}, which
is joint with Ioana Dumitriu, Elliot Paquette, and Soumik Pal, and from \cite{JP}, which
is joint with Pal. The results on the uniform
model are from \cite{Joh}. (See Section~\ref{sec:models} for the definitions of these two
models of random regular graphs).
\thref{thm:bestpermutationpoiapprox}, a version
of \cite[Corollary~24i]{JP} with an improved rate, appears only in this thesis.

Chapter~\ref{chap:GFF} is mostly taken from \cite{Joh}, which is joint work with Pal.
Section~\ref{sec:GFFconvergence} is new to this thesis and was also done jointly
with Pal. (The exception is Section~\ref{sec:GFFbackground2}, an extended introduction
to the Gaussian free field. It and any errors contained in it are mine alone.)
The main result here is \thref{thm:GFFconvergence}, which shows
the convergence of eigenvalue fluctuations to the Gaussian free field
in a more explicit form than in \cite{JP}.

\section{Models of random regular graphs}
\label{sec:models}
In Chapters~\ref{chap:poisson} and~\ref{chap:fluctuations}, we
will present results on two models of random regular graphs, the \emph{permutation mode} and
the \emph{uniform model}.
Traditionally, combinatorialists were most concerned with the uniform model of random regular
  graphs. The permutation model is typically easier to work with, however, and it is the setting
  for many spectral results on random regular graphs (for example,
  \cite{BrSh, Fri91, Fri}). There seems to have been a sense that that the two models
  had basically the same properties, besides the permutation model having loops
  and multiple edges. The contiguity result in \cite{GJKW} justifies this somewhat.

We now review the definitions of these two models and of our sequence of growing
graphs.
\subsection{The uniform model}
A random $d$-regular graph on $n$ vertices drawn from the uniform model
is just a graph chosen uniformly from the set of all $d$-regular graphs
  (i.e., graphs where every vertex has degree exactly $d$)
  on $n$ vertices
  without loops or multiple edges. Such graphs only exist when $nd$ is even.
\subsection{The permutation model}
\label{sec:permutationmodel}
  The permutation model is given by choosing
  $d/2$ independent, uniformly random permutations on $n$ vertices, making a graph from
  the cycle structure of each permutation, and overlaying them. It exists only
  for even values of $d$. For a more formal definition, let $\pi_1,\ldots,\pi_{d/2}$ be
  independent, uniformly random permutations on $n$ vertices. Define a graph on vertices
  $\{1,\ldots,n\}$ by making an edge between vertices $x$ and $y$ for every $k$
  such that $\pi_k(x)=y$. This model allows loops and multiple edges.
  We consider a loop at vertex~$x$ as counting as two edges when computing the degree of $x$,
  so that the graph really is $d$-regular. We also count a loop at vertex~$i$ as increasing the
  graph's adjacency matrix by $2$ at position $(i,i)$. The adjacency matrix of a graph from
  this model is then a sum of independent permutation matrices.

\subsection{Growing random regular graphs}
\label{sec:growingrrg}
A \emph{tower of random permutations} is a sequence of random permutations $(\pi^{(n)}, n\in \NN)$ such that
\begin{enumerate}[(i)]
\item $\pi^{(n)}$ is a uniformly distributed random permutation of $\{1,\ldots,n\}$, and
\item for each $n$, if $\pi^{(n)}$ is written as a product of cycles then $\pi^{(n-1)}$ is derived from $\pi^{(n)}$ by deletion of the element $n$ from its cycle.
\end{enumerate}
The stochastic process that grows $\pi^{(n)}$ from $\pi^{(n-1)}$ by sequentially inserting an element $n$ randomly is called the Chinese Restaurant Process.
 We give a further review of it in Section~\ref{sec:CRPprimer}.

Now suppose we construct towers of random permutations $(\pi_d^{(n)},\,n\geq 1)$, independent for each $d$.
For any $n$ and $d$, we can define a random $2d$-regular graph $G(n, 2d)$
from $\{ \pi^{(n)}_j,\, 1\le j \le d \}$ as in Section~\ref{sec:permutationmodel}.
Marginally, $G(n,2d)$ is then a random graph from the permutation model.
We will often keep $d$ fixed and consider $n$ as a growing parameter, referring to $G(n,2d)$
as $G_n$. Here and later, $G_0$ will represent the empty graph.

We construct a continuous-time version of this by inserting new vertices
into $G_n$ with rate $n+1$.  Formally,
define independent \label{page:poissonization}
times $T_i\eqd\Exp(i)$, and let
\begin{align*}
  M_t=\max\Big\{m\colon\ \sum_{i=1}^mT_i\leq t\Big\},
\end{align*}
and define the continuous-time Markov chain
$G(t)=G_{M_t}$.\label{page:chaindef}
When we vary $d$ as well as $n$, we will also refer to this as $G(t,2d)$.

config: arxiv


TITLE: How many subatomic particles can absorb/emit photons?
QUESTION [6 upvotes]: Is the electron the only subatomic particle that can absorb and emit a photon?

REPLY [8 votes]: The photon couples to all particles with electric charge or magnetic moment. This includes all of the quarks, the charged leptons $e,\mu,\tau$, and their antiparticles. It also includes particles composed of quarks and charged leptons: the proton and neutron (though the neutron only magnetically), the charged mesons, etc. Many electrically neutral mesons, like the $\pi^0$, decay to states of only photons.
I'm not sure whether there is a good limit on the magnetic moment of the neutrino; I think it's connected to the question of whether the neutrino-antineutrino relationship is as described by Dirac, like the charged leptons and quarks, or as suggested by Majorana, with the neutrino and antineutrino the same particle.

REPLY [0 votes]: If you are interested [let me know], I can explain to you exactly why this occurs (via quark annihilation) by using Feynman diagrams. Sub atomic particles that decay into photons include:
As you already know, the leptons (which includes the electron and various neutrinos) are mediated by photons.
Mesons: The neutral pion (which can decay into 2 photons), the neutral eta particle (can decay into 2 photons)
Baryons: The neutral sigma (which is like an excited state of the neutral lambda particle (decays into a a neutral lambda and a photon).
There are many more, but I cannot think of more off the top of my head (the truly intriguing aspect is which decays do NOT occur, even though they theoretically could occur). [I posted them at the bottom after an edit]
In the center of momentum of the 2 photon decays, each photon goes off in opposite directions with half of the total energy of the original particle. In the center of momentum of the 1 photon decays, the photon and the new particle go off in opposite directions, and the photon carries the difference of the rest masses between the two particles (note here that I say rest energy, because you do not have to add any energy to an accelerator for a decay to occur, nor can you ever add enough energy to make a decay occur.
My sincere apologies for misinterpreting your question in the comments. Hopefully this will give you what you need to research more photon particle decay modes (which are very intriguing).
Here are links to the wikipedia pages where you will have access to all of the particle decay modes:
Leptons : http://en.wikipedia.org/wiki/Lepton
Mesons: http://en.wikipedia.org/wiki/List_of_mesons
Baryons: http://en.wikipedia.org/wiki/List_of_baryons


TITLE: Area of a circle as it changes to an ellipse when viewing at different angles
QUESTION [4 upvotes]: Sorry for the long paragraph, I'm not too good at describing things:
Firstly, to explain the situation, I need to know how the area of a circle can change as you view it from different angles. For example, if you were viewing a circle at a 0 degree angle, it would look like a circle and it's area would be $\pi r^2$. However, as the angle of observation increases or decreases, one side of the circle will "collapse" inwards, and become smaller. It would create an ellipse, where the variable 'a' (longest distance from centre to side) stays the same, and where variable 'b' (the smallest distance from centre to side) decreases. I want to know the best way to find the area, based on the angle theta. For clarification, theta refers to the angle from a line perpendicular to the circle. The best way to approach the issue seemed to be the diagram in the link.

In the diagram, the 2 cm long line, is the circle. Therefore, the radius is 1 cm. The best approach I could find, was by using a cos ratio in the triangle highlighted in red as whenever I try any trig rules, it always gets too complicated with multiple variables. The angle closest to the centre of the circle would have to be 90-theta and, if the triangle is right angle (which it was made to be this way), the opposite angle must be theta. Hence by using $\cos \theta = a/h$, where h is the 1cm line, 'a' (which refers to adjacent not the other longer 'a') must be cos theta. This only works if the entire line is 2a. But if it's not and either side is different, then I have a problem. So, I wanted to know, am I right with the above working, or am I wrong. And if I'm wrong, how should I approach this problem differently. Thanks.

REPLY [0 votes]: I'm responding to amd's response without commenting for the same reason as with fleablood, I'm going to attach a photo:
Thanks for the answer! In the diagram, I've labelled a bunch of points for easier questioning. I believe I have assigned all the variables discussed to the image so if any are in the wrong place please tell me. For the line which you state is perpendicular to OV, and is written as "y=r−xcotθ", are you referring to AC or the line from A to the 'd' line? Not sure if my uncertainty about that is the reason, but how did you obtain the equation "y=r−xcotθ"? Why is there a z direction when the diagram is only dealing with changes on a 2d plane? Also, on a diagram what does this 'l' value represent?
Also, I hadn't learnt what homogeneous meant so I did some research and found this website: http://www.tomdalling.com/blog/modern-opengl/explaining-homogenous-coordinates-and-projective-geometry/
It talks about letting w=1 which is where I'm assuming you got the z coordinate for the next part of your answer. However I have a similar problem with 'm' as I did with 'l' where I don't understand how/why you chose the value. If we're talking vectors, m would give a vector coming out/into the screen at a slight angle (to account for the +1 in the z) perpendicular to the vector OV and OB.
For the next part of your explanation I understand how you reduced the z direction back down to 1, then turned it into a 2d vector/point. However, I haven't had time to go through the "algebraic manipulation", which I'm assuming at the moment is the difference between this vector, and viewpoint V vector. Finally, for the equation d/d+rsinθ I can't get this far as I haven't understood l and m.
Sorry for so many questions, I'm in year 12, Math C and B so my knowledge only goes so far. If you're wondering what I mean by C and B, go to the following link which briefly showcases what kind of information is learnt in each class:
https://en.wikipedia.org/wiki/Mathematics_education_in_Australia#Mathematics_C


TITLE: need to equate MATLAB and MATHEMATICA solutions
QUESTION [0 upvotes]: Am trying to utilize MATLAB and MATHEMATICA to solve $$y''(x)+ω^2y(x)=0.$$
Mathematica gives the expected answer of  $C_1\cos(xω) + C_2\sin(xω)$ while Matlab gives $C_1e^{xωi}+C_2e^{−xωi}$.
I'm guessing (!) that Euler's theorem might provide a way of equating these two answers, but i cannot find it.
Help requested.

REPLY [1 votes]: The Mathematica's solution  $$   C_1\cos(xω) + C_2\sin(xω) $$   is a real solution and that is what we really deal with in real life problems.
The Matlab's solution $$ C_1e^{xωi}+C_2e^{−xωi}$$ is complex valued solution.
We may derive the real valued solution from the Matlab's solution by considering the real parts and the imaginary parts of the complex valued solution.
They are solutions because they are indeed a linear combination of the complex valued solutions.
Thus the Mathmica's solution is a linear combination of Matlab's solution.
Note that we are really looking for real valued solutions in real life problems so the Mathematica wins in this case.


TITLE: Solve Laplace's equation in spherical coordinates, $\nabla^2 u(r,\theta,\phi)=0$, in the general case.
QUESTION [2 upvotes]: To find the equilibrium temperature distribution of a sphere I am required to solve the following PDE BVP (Laplace's equation in spherical coordinates).

$$\nabla^2 u(r,\theta,\phi)=0, 0<r<a,0<\theta<2\pi,0<\phi<\pi$$
  $$u(a,\theta,\phi)=f(\theta,\phi), 0<\phi<\pi$$

Note that I am required to solve this problem for the general case.
Work thus far
Note that in spherical coordinates $\nabla^2 u(r,\theta,\phi)=u_{rr}+\frac{2}{r}u_r+\frac{1}{r^2}[u_{\phi \phi}+\cot(\phi)u_{\phi}+\csc^2(\phi)u_{\theta \theta}]=0$
We use separation of variables letting $u(r,\theta,\phi)=w(r)v(\theta,\phi)$.
Substituting in
\begin{align}
0&=u_{rr}+\frac{2}{r}u_r+\frac{1}{r^2}[u_{\phi \phi}+\cot(\phi)u_{\phi}+\csc^2(\phi)u_{\theta \theta}] \\
0&=w''v+\frac{2}{r}w'v+\frac{1}{r^2}[wv_{\phi \phi}+\cot(\phi)wv_{\phi}+\csc^2(\phi)wv_{\theta \theta}] \\
\frac{r^2}{w}(-w''-\frac{2}{r}w')&=\frac{1}{v}[v_{\phi \phi}+\cot(\phi)v_{\phi}+\csc^2(\phi)v_{\theta \theta}]=-\lambda \\
\end{align}
to obtain the following ODEs
\begin{align}
0&=r^2w''+2rw'-\lambda w\\
0&=v_{\phi \phi}+\cot(\phi)v_{\phi}+\csc^2(\phi)v_{\theta \theta}+\lambda v\\
\end{align}
We now do another separation of variables $v(\theta,\phi)=g(\theta)h(\phi)$. We get the following separation
$$\frac{h''+\cot(\phi)h'+\lambda h}{\csc^2(\phi)h}=\frac{-g''}{g}=\mu $$
and obtain the following ODEs
\begin{align}
0&=g''+\mu g\\
0&=h'' + \cot(\phi)h'+\lambda h-\mu \csc^2(\phi)h\\
\end{align}
Solving the eigenvalue problem $g''+\mu g=0$ with periodic boundary conditions $g(0)=g(2\pi)$ and $g'(0)=g'(2\pi)$, we obtain the eigenvalues $\mu_n=n^2$ and the eigenvectors $g_n(\theta)=\cos(n\theta)$ or $\sin(n\theta)$ for $n=0,1,2,3,...$.
What I am confused about
What order do I solve the eigenvalue problems to find the eigenvalues $\mu,\lambda$?
How do I transform the ODE $0=h'' + \cot(\phi)h'+\lambda h-\mu \csc^2(\phi)h$ into a easier ODE to solve.
Note that following posts are not relevant as

Radial symmetry is used
Derive the formula for $\nabla^2 u(r,\theta,\phi)$

REPLY [1 votes]: Assume
$$
  x=r\sin\theta\cos\phi,\;\;\; y=r\sin\theta\sin\phi,\;\;\; z=r\cos\theta, \\
   0 \le r < \infty, \;\;\; 0 \le \theta \le \phi,\;\;\; 0 \le \phi \le 2\pi.
$$
Because the spherical coordinate system is an orthogonal coordinate system (meaning that the coordinate curves where only one variable varies are mutually orthogonal where they meet,) the Laplacian is determined by the metric scale factors
$$
          m_r = 1,\;\; m_{\theta}=r,\;\; m_{\phi} = r\sin\theta.
$$
The Laplacian is
\begin{eqnarray*}
 & & \nabla^{2}F(r,\phi,\theta) \\
   & = & \frac{1}{m_{r}m_{\phi}m_{\theta}}\left[
      \frac{\partial}{\partial r}\left(\frac{m_{\phi}m_{\theta}}{m_{r}}
            \frac{\partial F}{\partial r}\right)
     +\frac{\partial}{\partial\phi}\left(\frac{m_{r}m_{\theta}}{m_{\phi}}
            \frac{\partial F}{\partial\phi}\right)
     +\frac{\partial}{\partial\theta}\left(\frac{m_{r}m_{\phi}}{m_{\theta}}
            \frac{\partial F}{\partial\theta}\right)
       \right]
       \\
%   & = & \frac{1}{r^{2}\sin\theta}\left[
%      \frac{\partial}{\partial r}\left(r^{2}\sin\theta
%            \frac{\partial F}{\partial r}\right)
%     +\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\theta}
%            \frac{\partial F}{\partial \phi}\right)
%     +\frac{\partial}{\partial\theta}\left(\sin\theta
%            \frac{\partial F}{\partial\theta}\right)
%       \right]
%       \\
   & = & \frac{1}{r^{2}}\frac{\partial}{\partial r}
        \left(r^{2}\frac{\partial F}{\partial r}\right)
     + \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}F}{\partial\phi^{2}}
     + \frac{1}{r^{2}\sin\theta}
        \frac{\partial}{\partial\theta}
        \left(\sin\theta\frac{\partial F}{\partial\theta}\right).
\end{eqnarray*}
Standard separation of variables equations are obtained by setting $F(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)$. The separated equations involve a separation parameter $m=0,\pm 1,\pm 2,\cdots$ imposed by periodicity in $\phi$, and a separation parameter $l$ determined by the ODE in $\theta$. The separated equations are:
\begin{eqnarray*}
       -\frac{d^2}{d\phi^2}\Phi & = & m^2\Phi, \\
   -\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)
       +\frac{m^2}{\sin\theta}\Theta & = & l\sin\theta \Theta, \\
   -\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+l R & = & 0.
\end{eqnarray*}
The equation in $\theta$ is the Associated Legendre equation, and this equation determines the possible values of $l$. The radial equation is determined by the requirement that $R$ be bounded near $r=0$. The equation in $\theta$ is transformed to standard Legendre form by setting $x=-\cos\theta$ so that $x$ varies between $-1$ and $1$ as $\theta$ varies between $0$ and $\pi$. Then
$X$ defined by $X(-\cos\theta)=\Theta(\theta)$ satisfies the Associated Legendre Equation:
$$
       -\frac{d}{dx}(1-x^2)\frac{dX}{dx}+\frac{m^2}{1-x^2}X=l X.
$$
For a given $m$, the values $l$ for which there are bounded solutions $X$ are those given by $l=n(n+1)$ where $n=m,m+1,m+2,\cdots$. The resulting bounded solutions are scalar multiples of
$$
           e^{im\phi}r^{n}\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^{n}.
$$


TITLE: Ant going to the vertex
QUESTION [0 upvotes]: An ant has a $\frac{1}{2}$ chance of moving to the vertex to its left and to its right in a hexagon. What is the probability for the ant to move from a vertex to the opposite vertex
I found this resolution:
I assume you wanted to find the probability that the ant goes to the other vertex, where the ant can go indefinitely.
Solution
Let the vertex where the ant is on be $P_1$, and let the two vertices adjacent to it be $P_2$, the two vertices adjacent to these be $P_3$, and let the last vertex be $P_4$.
Using States, we obtain the following equations:
$$\begin{cases}P_1=P_2\\
P_2=\frac{P_1+P_3}2\\
P_3=\frac{P_2+P_4}2\\
P_4=1\end{cases}$$By solving these systems of equations, we get:
$$(P_1,P_2,P_3,P_4)=(1,1,1,1)$$Therefore, the probability that the ant would get to the opposite vertex is $\boxed1$.
She is right? Or is the answer $ \frac{1}{10}$?

REPLY [1 votes]: If you want the probability that the ant ever reaches the opposite vertex, that is indeed $1$.  All you need to conclude this is that this is a finite irreducible Markov chain.  That is:

There are a finite number of vertices.
At each step, the probability of the ant moving to a given vertex depends only on where it is now.
It is possible for the ant to get from any vertex to any other vertex.


TITLE: Would heating and cooling a ferromagnetic material allow it to more easily take on a magnetic field
QUESTION [0 upvotes]: I'm trying to magnetize various objects, such that they take on a strong magnetic field.
Theoretically, magnetism is caused by aligning electrons such that they stay more or less in the same place.
If i heat the electrons,  they're more energetic,  and thus likely won't be in the same place. But could i theoretically use the same system that an Air Conditioning unit uses ( compressed air can more readily exchange heat than uncompressed air ),  by aligning the electrons when they're more energetic, and cooling them down while still aligning them,  could that create a stronger magnet than just aligning them while cold?

REPLY [1 votes]: One needs to be aware that if you get a magnetic material too hot, thermal agitation will prevent the atomic dipoles from maintaining alignment.  Also, when an external field is removed, the dipoles tend to relax into alignment with the crystal axes.  The “remnant” field (after saturation) depends more on the nature of the material than on its history.


TITLE: Is surjectivity preserved in open neighborhoods?
QUESTION [2 upvotes]: Let $X,S$ be schemes of finite type over a field and let $f:X\times S\to S$ be the projection. Suppose we have a morphism of coherent sheaves $\phi:\mathscr E\to \mathscr F$ on $X\times S$.

Is it true that if the restriction $\phi|_{X\times\{s\}}$ to the fiber
  over $s$ is surjective (or injective), then there is an open
  neighborhood $U\subset S$ of $s$ such that $\phi|_{X\times U}$ is
  surjective (or injective)?

A similar statement is true for stalks: for a locally noetherian scheme $S$, if a morphism of coherent $\mathscr O_S$-modules is such that $\phi_s:\mathscr E_s\to \mathscr F_s$ is surjective at $s$, then it is surjective in a neighborhood of $s$.
Thanks for any help!

REPLY [2 votes]: It's simpler to consider the support of the kernel or cokernel sheaf, call it $\mathscr G$ -- we wish to show that it vanishes appropriately.
Edit: the original argument I posted was incorrect (I'll leave it for posterity; see below and in the comments...). Here is a simpler argument:
The statement is false in general without some kind of properness assumption. For example, if $X = S = \mathbb{A}^1$, then the structure sheaf of the hyperbola $\{xs = 1\}$ is zero on the fiber $s = 0$ but is nonzero on every other fiber.
However, if $X$ is complete, then $f : X \times S \to S$ is proper (more generally, if $f : \mathcal{X} \to S$ is any proper map), then we can make the following argument: let $Z$ be the support locus of $\mathscr G$. Then $Z$ is closed because $\mathscr G$ is coherent. By properness, $f(Z)$ is closed in $S$, and by hypothesis $s \notin f(Z)$. Consequently, $U = S - f(Z)$ is an open neighborhood of $s$ and $\mathscr G|_{f^{-1}(U)} = 0$.
(Original argument:)
We can assume $S$ is affine; and we can actually just have $\mathcal{X} \to S$ be any finite type $S$-scheme.
Now the statement is: if a sheaf vanishes on the fiber over $s \in S$, we wish to show that it vanishes on the preimage of a neighborhood of $s$. At this point we can also assume $X$ is affine (reason: cover with finitely-many affine opens, solve the problem there, and intersect the corresponding opens of $S$...)
Now we're reduced to an easy commutative algebra statement! For $B$ a finitely-generated $A$-algebra and $M$ a finite $B$-module, if $M \otimes A_P = 0$, then there exists $f\notin P$ ($f \in A$) such that $M \otimes A[1/f] = 0$. I leave this part to you.


TITLE: Length of perimeter of a right triangle is ‘x’ inches and the length of the altitude perpendicular to the hypotenuse is ‘y’ inches. Find the sides.
QUESTION [1 upvotes]: I would prefer the method rather than a direct solution please. I tried solving for the hypotenuse but couldn't get it in terms of x and y.

REPLY [2 votes]: Always start by drawing a diagram. Make sure you note which angles are right angles.
Then you'll find that the altitude of a right triangle actually makes 2 smaller triangles similar to the big triangle! Look at the diagram below (different names)
Triangle ADC is also a right triangle (angle at D) and one of the other angles (angle at C) is the same as one of the angles of ABC (also the angle at C). So triangles ADC and ABC are similar, but kinda flipped and rotated (make sure to keep good track of how the similar triangles are oriented). Similarly with triangle ABD.

Then it's a matter of setting up a bunch of equations and solving.