TITLE: Cosmological redshift in a lab?
QUESTION [2 upvotes]: I want to consider a thought experiment. Lets ignore technical problems of actually performing such an experiment.
Consider two photons having the same wavelength. We send 1 photon to a distant galaxy (millions of light years away). The photon would hit a mirror there and return to Earth. On Earth (I assume) a cosmological redshift would be detected. At the same time we keep the second photon on Earth in small container with perfect mirrors where it bounces back and forth. Would the photon in the container exhibit the same cosmological redshift as the photon that traveled to the distant galaxy and back? If so, (now back to technical problems) could the redshift be ever measured in the lab or is the effect many orders of magnitude too small to be ever measured in a lab on Earth, not to mention the problem of constructing a perfect mirror to keep the photon?

REPLY [1 votes]: This experiment is essentially being conducted at some level by the Gravitational Wave detectors of LIGO and VIRGO.
Laser light is fed into a Fabry Perot interferometer, so that it effectively bounces up and down the 4km arms about 280 times.
The mirrors are suspended and the gravitational wave has the effect of expanding and contracting the space between them (in the laboratory frame).
If the gravitational wave has a wavelength smaller than $2 \times 280 \times 4$ km, then the response of the interferometer is much reduced because the laser light is "stretched" by a similar amount to the arms and the phase difference engendered by the wave is diminished.
If instead we deal with low frequency, long wavelength gravitational waves, the disturbance of the mirrors lasts much longer than the time spent by the laser light in the instrument arms and the phase difference response is maximised.
Whilst this is not a "cosmological redshift" it is pretty much the same physics. The phase difference accumulated by the laser "in the lab" is caused by the metric perturbation caused by the passing gravitational wave and is similar to the accumulated redshift of a photon travelling across the universe.


TITLE: Understanding the definition of Sylow $p$-subgroups
QUESTION [2 upvotes]: Here is the definition of Sylow $p$-group (source: wikipedia)

For a prime number $p$, a Sylow $p$-subgroup of a group $G$ is a maximal $p$-subgroup of $G$, i.e. a subgroup of $G$ that is a $p$-group (so that the order of every group element is a power of $p$) that is not a proper subgroup of any other $p$-subgroup of $G$.

Also here is the definition of a maximal subgroup (source: wikipedia)

In group theory, a maximal subgroup $H$ of a group $G$ is a proper subgroup, such that no proper subgroup $K$ contains $H$ strictly.

So, in accordance with the definition of maximal subgroup, a maximal $p$-subgroup $H$ should be a proper $p$-subgroup of the group $G$ such that no other proper $p$-subgroup $K$ contains $H$ strictly.
But the definition of Sylow $p$-subgroup does not say that it is a proper subgroup of the group $G$. According to its definition, if say, $G$ is a group of order $2^3$, then the Sylow $p$-subgroup is the group itself, but if I follow the definition as given by me, the Sylow $p$-subgroup should be a proper subgroup of $G$ i.e. a subgroup of order $2^2$.
I know I am misunderstanding the definition somewhere. Please help me rectify myself.

REPLY [3 votes]: As I mentioned in comments, I think the issue here is actually that we misname "maximal subgroup".
Of course, given a partially ordered set $P$, a maximal element (or $P$) is an element $p\in P$ with the property that for all $x\in P$, if $p\leq x$ then $p=x$.
If we have a class $\mathfrak{X}$ of subgroups, partially ordered by inclusion, we often talk about "maximal $\mathfrak{X}$-subgroups" (also "minimal $\mathfrak{X}$-groups", though that is less common). Thus, "maximal $p$-subgroup" means "maximal among the $p$-subgroups".
We say "maximal subgroup" (and "maximal normal subgroup"), but we really mean "maximal proper subgroup" (that is, the class $\mathfrak{X}$ is the class of proper subgroups, not the class of all subgroups). (And we likewise really  mean "maximal proper normal subgroup").
So we should say:

A maximal proper subgroup of $G$ is a proper subgroup $H$ of $G$ that is not properly contained in any proper subgroup of $G$.

If we used that terminology and definition, you would likely have no problems. But it is an unfortunate fact that nobody says "maximal proper subgroup", we just say "maximal subgroup" and elide the "proper" clause, on the (flimsy) excuse that of course we must mean proper subgroup, as otherwise the concept would just be "the group $G$" and it would be silly to introduce such a nomenclature for the whole  group.


TITLE: Unable to find coordinates of a point on $\sin(x - k) + c$ graph.
QUESTION [0 upvotes]: For a)i) I have found the coordinates of $A$ by finding $x$ which is equal to $\frac{\frac{4\pi}{3}}{2} = \frac{2\pi}{3}$
To find $y$ I first find $k$ and then $c$ in the equation $\sin(x - k) + c$ and then use the fact that $1 + c = y$.
I get the right answer which is $A(\frac{2\pi}{3}, \frac{3}{2})$.
However the question uses the word hence in the part a)ii). This seems to imply that it is possible to find $A$ without finding $k$ or $c$.
If this is possible, how do you find $A$ without first finding $k$ or $c$.

REPLY [1 votes]: Yes, there is a way. You already found the $x$ coordinate of $A$.
To find the $y$ coordinate, first draw a line parallel to the axis $Ox$ that "symmetrizes" the graph vertically.
You may find out geometrically that this line is precisely $(0,\sin(a))+Ox$, where $a$ is such that $2a+\pi=4\pi/3$. Then $a=\pi/6$ and the line is $(0,1/2)+Ox$.
Then, from that, you may find out that the $y$ coordinate of $A$ is $1+ 1/2$.


TITLE: Difference of numbers in a unit interval
QUESTION [1 upvotes]: $x,y,z\in \mathbb{R}$ are chosen at random from the unit interval $[0, 1]$.
What is the probability that
$$\max(x,y,z) - \min(x,y,z) \leq \frac{2}{3}$$
EDIT- Solutions not using calculus would be appreciated, as this problem appeared on a test where calculus was not meant to be used.

REPLY [4 votes]: Let $E$ be the event $\{ \max(x, y, z) - \min(x, y, z) \le \frac{2}{3}\}$. Our goal is to find $\Pr[E]$. Now let's split $E$ into 6 events $E_{xy}, E_{yx}, E_{xz}, E_{zx}, E_{yz}, E_{zy}$, where
$$E_{xy} = \{ (x = \max(x, y, z)) \land (y = \min(x, y, z)) \land (x - y \le 2/3)\},$$
and the rest are defined similarly. Note that $E$ is the union of these six events and the probability of intersection of any two of these events is $0$.   Also, due to symmetry, all these 6 events have the same probability. Hence $\Pr[E] = 6\Pr[E_{xy}]$.
It remains to  compute $\Pr[E_{xy}]$. Note that $E_{xy}$ happens iff $y\le x\le y + 2/3$ and $z\in[y, x]$. I.e.
\begin{align*}
\Pr[E_{xy}] &= \int\limits_0^1 dy \int\limits_y^{\min(y + \frac{2}{3}, 1)} dx \int\limits_y^x dz
=  \int\limits_0^1 dy \int\limits_y^{\min(y + \frac{2}{3}, 1)} (x - y) dx \\
&= \int\limits_0^1 dy \cdot \frac{(x - y)^2}{2} \Bigg|_{y}^{\min(y + \frac{2}{3}, 1)}
= \int\limits_0^1 \frac{\left(\min(y + \frac{2}{3}, 1) - y\right)^2}{2} dy\\
&= \int\limits_0^{\frac{1}{3}} \frac{\left(\frac{2}{3}\right)^2}{2}dy + \int\limits_{\frac{1}{3}}^1 \frac{(1 - y)^2}{2}dy \\
&= \frac{2}{27} -\frac{(1 - y)^3}{6} \Bigg |_{\frac{1}{3}}^1 = \frac{10}{81}.
\end{align*}
And thus the answer is $ 6\cdot \frac{10}{81} = \frac{20}{27}$


TITLE: Probability that one beta random variable is greater than other beta random variables
QUESTION [0 upvotes]: Suppose we have n independent beta random variables $p_1 \sim beta(\alpha_1, \beta_1), p_2 \sim beta(\alpha_2, \beta_2)...$, the parameters are all known. I want to calculate the probability that $p_i$ is greater than all the others.
\begin{align}
{\rm Pr}(p_i \geq max(p_1, p_2,...,p_n)) &=
\int_0^1\int_0^{p_i}...\int_0^{p_i}f_{p_1}f_{p_2}...f_{p_n}dp_1dp_2...dp_ndp_i\\
&=\int_0^1I_{p_i}(\alpha_1, \beta_1)I_{p_i}(\alpha_2, \beta_2)...I_{p_i}(\alpha_n, \beta_n)f_{p_i}dp_i
\end{align}
This is what I got so far. Is there any closed-form solution for this question? If not, is there an efficient way to solve the integral? It can be assumed that the $\alpha, \beta$ parameters are very large if necessary, but it would be great if it can be solved without this assumption. Thank you.
If the integral can not be efficiently solved, can there be any approximation?
note:
parameters $\alpha, \beta$ are different for different random variables.

REPLY [3 votes]: Let $M=max(p_1,\dots p_{i-1},p_{i+1},\dots,p_n)$
$$
\begin{split}
P(p_i\geq y, y\geq M) &= P(M\leq y)-P(p_i\leq y, y\geq M)\\ & = P(M\leq y)-I_i(\alpha_i,\beta_i)(y)P(M\leq y)
\end{split}
$$
$$\begin{equation}
\begin{split}P(M\leq y) & = P(p_1 \leq y, \dots , p_n \leq y)\\ & =\prod_{k=1, k\neq i}^{n}P(p_k \leq y)\\& = \prod_{k=1,k\neq i}^{n}I_i(\alpha_k,\beta_k)(y)
\end{split}
\end{equation}
$$
This happens $\forall y$. So you would need to integrate this over $(0,1)$. I do not think you can do better than this.


TITLE: Is the following function concave? (or log of it)
QUESTION [0 upvotes]: I have a function
$f(x_1, x_2, ..., x_M) = \displaystyle \prod_{i = 1}^N \frac{(\sum_{j = 1}^{M} a_jx_jI_{ij})^2}{\sum_{j = 1}^{M} a_jx_j}$ in domain $\{{\bf x} \in {\bf R}^m \setminus {\bf 0} \ \mid 0 \leq x_j \leq 1\}$.
Here $a_j$ are some positive integer constants and $I_{ij}$ are indicator values (0 or 1).
I was wondering if $f$ or log $f$ is concave in this domain.
For log $f$, using trivial composition rule led me nowhere, so I was wondering if there are some other techniques I may be able to use to prove that it is indeed concave. Of course, it may not turn out to be concave after all.
Thanks!

REPLY [0 votes]: It seems that the function $f(x,y)=\frac{x^2}{x+y}$ is not concave even on the set $$\{(x,y): 0<x,y<1\mbox{ and }x+y=1\}$$ and its logarithm is  not concave even on the set $$\{(1/2,y): 0<y<1\}.$$


TITLE: Average of a dataset with Ranges of Numbers - A "Simple" high school standarized test question
QUESTION [1 upvotes]: This is similar to a question on a standardized test for high school students: The heights of students are listed below. What is the mean of this dataset? The dataset is: 1 student 55-56 inches, 1 student 57-58 inches. The first thing that confuses me is that 55-56 and 57-58 are not adjacent meaning that the numbers between 56 and 57 are not part of the dataset? Or is the "-" symbol inclusive meaning 55-56 includes 56.5? This doesn't seem mathematically correct because inclusive means that the point of the upper bound would be included, not a whole inch. Or were the heights of students measured to the nearest inch? Also what is the average of ranges? Would the answer be 56-57? Or 56.5? This is further complicated by the confusing ranges that I mentioned above. I find it absurd that the standardized test would contain a question that does not have a standardized answer. Often I get perfect scores on standardized tests but I probably got this one wrong. If anyone could explain the reasoning behind the right answer it would clear things up for me.

REPLY [1 votes]: If all the height ranges had a gap like this then the interpretation that makes most sense is the height range $55 - 56$ inches is in fact $54.5$ to $<56.5$, and  $57 - 58$ is $56.5$ to $<58.5$ etc. It is continuous data made somewhat discrete to simplify the calculation.
If it didn't, that is, some taller ranges went from $60 - 61, 61 - 62$ etc., then the interpretation would be that there was no student in the $56 - 57$ range. Either way, to calculate the overall mean would mean taking the middle value which would be $55.5$ for the $55 - 56$ range
Also, a question that leaves the student guessing the correct interpretation is a somewhat poorly conceived question.


TITLE: Injectivity/Complex Mapping
QUESTION [1 upvotes]: In $\mathbb{C}$, consider the neighborhood $O$ $:=$ {${x+iy; (x+1)^2 + (y-1)^2 < 4}$}. Show that $w=z^2$ is not one-to-one when its domain of definition is restricted to $O$.
I'm not sure where to start since I don't think I properly understand how the domain of definition works for maps. Thanks.

REPLY [0 votes]: This means "show there are two points of $O$ ('two $z$ values') that are sent to the same value (i.e., 'same $w$-value'). Since $z^2  = (-z)^2$, one way to do this is to show that some point $z$ and its negative are both in $O$. Drawing a picture of $O$ should help.


TITLE: Questions about Hamiltonicity of random graphs.
QUESTION [1 upvotes]: I’m reading a proof on the Hamiltonicity of a random graph, and there’s a few details that I’m not clear about. Here’s the setup and argument:

Let $G$ be a graph on $[n]$, $\alpha \in (0,1)$, and $R = \{e_1,e_2,\dots,e_{|R|}\}$ be a collection of random edges chosen uniformly and randomly from all subsets of $[n] \choose 2$. We assume:
$\ \ \bullet$ the longest path in $G$ has length at least $l(n)$
$\ \ \bullet$ $(|R|\cdot\alpha^2) \to \infty$
$\ \ \bullet$ $|R| \geq \frac{100(n - l(n))}{\alpha^2}$
$\ \ \bullet$ $G$ is connected.
For $t \in [|R|]$, let the random variable $Y_t = 1$ if $e_t = \{x,y\}$ and there exists a longest path in $G_t := G_{t-1} \cup e_t$, with endpoints $x,y$ (and $Y_t = 0$ otherwise). Some further analysis (which is omitted here) shows that $\mathbb{P}(Y_t = 1) \geq \alpha^2$.
Observation: if $\sum_{t=1}^{|R|} Y_t \geq n - l(n)$, then $G \cup R$ is Hamiltonian.
Also, since $\sum_{t=1}^{|R|} Y_t$ is dominated by a $Bin(|R|,\alpha^2)$ random variable, we have:
\begin{align}
&\mathbb{P}(\text{$G \cup R$ not Hamiltonian}) \\
\leq \ &\mathbb{P}(Bin(|R|,\alpha^2) \leq n - l(n)) \\
\leq \ &\mathbb{P}(Bin(|R|,\alpha^2) \leq \frac{|R| \cdot \alpha^2}{100}) = o(1)
\end{align}

My questions are as follows:

I want to make sure I understand the statement:
$$\sum_{t=1}^{|R|} Y_t \geq n - l(n) \Rightarrow G \cup R \text{ is Hamiltonian}$$
If we look at the sequence of graphs $G := G_0, G_1, G_2,\dots, G_{|R|}$, i.e. we throw the edges of $R$ on $G$ one by one, then the LHS condition implies that we have closed the longest paths in those graphs at least $n - l(n)$ times, with each iteration lengthens the length of the longest path by $1$?

Why do we have: $\sum_{t=1}^{|R|} Y_t$ is dominated by a $Bin(|R|,\alpha^2)$ random variable?

Lastly, how do we get the $o(1)$ at the end?

REPLY [2 votes]: Yes, you are right about what the proof means by
$$
   \sum_{t=1}^{|R|} Y_t \geq n - l(n) \implies G \cup R \text{ is Hamiltonian}.
$$
Each time $Y_t = 1$, we either increase the length of the longest path by $1$, or turn a path of length $n-1$ into a Hamiltonian cycle. (Or, if $Y_t = 1$ and $G_{t-1}$ is already Hamiltonian, we do nothing.) Taking at least $n - l(n)$ such steps must result in a Hamiltonian cycle.

The sum $\sum_{t=1}^{|R|} Y_t$ is "dominated by $\text{Bin}(|R|,\alpha^2)$" in the following sense. (I would actually say "dominates" rather than "is dominated by".)
It's not just true that $\mathbb P(Y_t = 1) \ge \alpha^2$.It's also true that for any event $A$ defined in terms of $Y_1, \dots, Y_{t-1}$, $\mathbb P(Y_t = 1 \mid A) \ge \alpha^2$. This conditional probability is also key to the proof, and I hope that it is explained in the step you omitted. (Essentially, as long as $G$ is an expander graph with certain parameters, then $G_1, \dots, G_t$ must have this property, regardless of $R$.)
So we can define a sequence $(Y_t')$ with the following properties:

Independently of $Y_1', \dots, Y_{t-1}'$, $\mathbb P(Y_t' = 1) = \alpha^2$.
If $Y_t = 0$, then $Y_t' = 0$ as well.

Essentially, whenever $Y_t = 1$, set $Y_t' = 1$ with the correct probability (depending on all our previous history) so that the overall $\mathbb P(Y_t' = 1)$ in this situation is $\alpha^2$.
Now we have $\sum_{t=1}^{|R|} Y_t \ge \sum_{t=1}^{|R|} Y_t'$ by property 2 above, and this second sum is $\text{Bin}(|R|,\alpha^2)$ exactly. Therefore
$$
   \mathbb P\left(\sum_{t=1}^{|R|} Y_t < n - l(n)\right) \le  \mathbb P\left(\sum_{t=1}^{|R|} Y_t' < n - l(n)\right)
$$
and for this second sum, we can use binomial probability techniques.

Let $Y' = \sum_{t=1}^{|R|} Y_t'$; by the above, we have $Y' \sim \text{Bin}(|R|,\alpha^2)$. We want to bound $\mathbb P(Y' < n - l(n))$.
Well, we have $\mu := \mathbb E(Y') = \alpha^2 |R| \ge 100 (n - l(n))$ by one of the bulleted assumptions made at the beginning, and so it's enough to bound $\mathbb P(Y' < 0.01 \mu)$. By a Chernoff bound,
$$
   \mathbb P(Y' < 0.01 \mu) \le \left(\frac{e^{-0.99}}{0.01^{0.01}}\right)^\mu \le 0.39^\mu.
$$
Since $\mu = \alpha^2 |R| \to \infty$ (another one of the bulleted assumptions), $0.39^\mu \to 0$, and so $\mathbb P(Y' < n - l(n)) = o(1)$.


TITLE: Show that $e^x\ge x^e$ for all $x\ge 0$
QUESTION [2 upvotes]: Show that $e^x\ge x^e$ for all $x\ge0.$
I was playing around with splitting a number into parts (that add to the original number) and then multiplying the parts to get the result, and I found that the product is highest when the parts tended towards $e$. This inequality is a consequence of that. Is there a proof or intuitive explanation for this?

REPLY [0 votes]: Having that $\dfrac{\log x}{x}$ is a increasing function with $x\leq e$
$$\therefore\frac{\log e}{e}\geq\frac{\log x}{x}$$
Excluding the logarithm signs in both side, we have what we need to show.


TITLE: Number of terms of sequence under $n$ with no arithmetic bounds.
QUESTION [2 upvotes]: I mentioned to a friend recently the result that for any $n > 1$, there is an $N$ beyond which there is always a prime between $m$ and $nm$ (for $m>N$, if it wasn't clear).
They proclaimed "This is very intuitive, since by choosing a large enough $m$, the gap between $m$ and $nm$ will be also large and so with an infinite number of primes, it only makes sense that this would be the case." This heuristic makes sense for things like primes but I, not thinking this was intuitive in general, wondered if there were infinite sequences for which no such bounds exist...
Can you give examples of sequences where either:
a) $\exists n$ for which there is no $N$ satisfying the above criteria or preferably
b) $\nexists n$ for which there is an $N$ satisfying the above criteria?
I was thinking that at least a) could be shown for some doubly exponential sequences via generating function magic, but was not sure.

REPLY [0 votes]: It depends on how large a gap
you consider to be intuitive.
Wikipedia's article on prime gaps
(https://en.wikipedia.org/wiki/Prime_gap)
has a number of interesting
results and conjectures
including this:
"In 1931, E. Westzynthius proved that maximal prime gaps grow more than logarithmically. That is,[9]
${\displaystyle \limsup _{n\to \infty }{\frac {g_{n}}{\log p_{n}}}=\infty .}
$"