TITLE: How do we prove that uniform convergence preserves continuity?
QUESTION [2 upvotes]: Suppose that $(f^{(n)})_{n=1}^{\infty}$ is a sequence of functions from one metric space $(X,d_{X})$ to another $(Y,d_{Y})$, and suppose that this sequence converges uniformly to another function $f:X\to Y$. Let $x_{0}$ be a point of $X$. If the functions $f^{(n)}$ are continuous at $x_{0}$ for each $n$, then the limit function $f$ is also continuous at $x_{0}$
My solution (edit)
According to the comments, we have the following solution.
Let us start with the definition of uniform convergence.
For every $\varepsilon/3 > 0$ there is a natural number $N\geq 1$ such that for every $x\in X$ we have that $d_{Y}(f^{(n)}(x),f(x)) < \varepsilon/3$ for every $n\geq N$.
Thus, if we fix $n = N$, since each $f^{(n)}$ is continuous at $x_{0}$, we conclude that for every $\varepsilon/3 > 0$ there is a $\delta > 0$ such that for every $x\in X$ the relation holds
\begin{align*}
d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f^{(N)}(x),f^{(N)}(x_{0})) < \varepsilon/3
\end{align*}
Hence for every $\varepsilon > 0$ there is a $\delta > 0$ such that for every $x\in X$ we have that
\begin{align*}
d_{Y}(f(x),f(x_{0})) \leq d_{Y}(f(x),f^{(N)}(x)) + d_{Y}(f^{(N)}(x),f^{(N)}(x_{0})) + d_{Y}(f^{(N)}(x_{0}),f(x_{0})) < \varepsilon
\end{align*}
whenever $d_{X}(x,x_{0}) < \delta$. Therefore $f$ is continuous at $x_{0}$, just as it has been claimed.

REPLY [4 votes]: Let $\epsilon >0$, then there exists $N\in \mathbb N $ and $\delta >0$ such that $d(x,x_0)<\delta \Rightarrow d(f_N(x),f_N(x_0))< \epsilon /3$ and $d(f_N(y),f(y))< \epsilon /3$ for any chosen $y$, therefore
$$
\begin{align*}
d(f(x),f(x_0))&\leqslant d(f(x),f_N(x))+d(f_N(x),f(x_0))\\
&\leqslant d(f(x),f_N(x))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))\\
&< \epsilon
\end{align*}
$$
that is, we had shown that $d(x,x_0)< \delta \Rightarrow d(f(x),f(x_0))< \epsilon $, so $f$ is continuous at $x_0$.


TITLE: Does it make sense to continue searching for dark matter on Earth?
QUESTION [18 upvotes]: The Large Underground Xenon Detector (LUX) recently released results1 that they have found no signs of dark matter2 after a ~3 month search this spring and summer. The LUX group plans to spend all of 2014 looking for dark matter interactions with the liquid xenon. (I thought I read somewhere that they were planning on doubling the amount of xenon, but cannot find a link that confirms this).
Early last year, researchers from Chile reported that there was no significant amount of dark matter in the solar vicinity3. They studied4 the motions of about 400 stars within 13,000 lightyears radius from the sun, and all the matter is accounted for in what we see: gas, dust, star, etc., and there's little room for dark matter.
Does it make sense to continue searching for dark matter via direct detection method when research suggests there's nothing to find in our solar vicinity?
Would it make more sense to build more sensitive space-based detectors?

1 preprint, Akerib et al 2013
2 article, K Jepsen, Symmetry Magazine, 2013
3 article, ESO, Phys.Org, 2012
4 preprint, Mono Bidin et al 2012

REPLY [3 votes]: It would be remarkable if there were no dark matter in the vicinity of the Earth. As far as we know, the hypothesised properties of dark matter mean that it should be quite smoothly distributed. If we can't find any here on Earth, why should we find it anywhere else? Therefore it makes perfect sense to push sensitivity limits as low as possible here on Earth.
Moni Bidin et al. are sticking to their guns on there being no (negligible) dark matter in the solar vicinity, but the measurements and interpretation are difficult and plenty disagree (e.g. Bovy & Tremaine 2012; Garbari et al. 2011). So from that point of view, the controversial thesis in your question should not be sufficient to shut down an entire research area without a lot more hard evidence that achieves a consensus.


TITLE: Proving that a function is uniformly continuous.
QUESTION [0 upvotes]: Prove thatt $f(x) = x^2 \sin (1/x)$ is uniformly continuous on $(0,1]$.
My attempt:
Let's extend our function $f$ to $[0,1]$ by defining thus
$$
\tilde f(x) =
\begin{cases}
x^2 \sin(1/x) & x \in (0,1] \\
0 & x= 0
\end{cases}
$$
$\color{green}{Claim}$: $\tilde f$ is continuous on $[0,1]$.
$\color{green}{Proof}$: Since, $\tilde f$ is same as $f$ on $(0,1]$, therefore is it is continuous on that interval.
For any $\varepsilon \gt 0$, take $\delta = \sqrt{\varepsilon}$, and thus
$$
|x-0| \lt \delta = \sqrt{\varepsilon}$$
$$
|x^2| \lt \varepsilon $$
$$
|x^2 \sin(1/x)| \lt |x^2| \lt \varepsilon $$
$$|\tilde f(x) - \tilde f(0)| \lt \varepsilon$$
Thus, $\tilde f$ is continuous at $0$, and hence it is continuous on the whole set.

A function continuous function $f$ is uniformly continuous on $(a,b)$, if and only if it can be extended to $\tilde f$ which is continuous on $[a,b]$.

Using the above theorem, we get that $f(x) = x^2 \sin(1/x)$ is continuous on $(0,1]$.

I have two questions to ask:

Is my proof formally written?

Can we do it in any other way? That extension theorem doesn't seem to be of my taste.

REPLY [0 votes]: Your proof looks fine. It might be helpful to mention
at the beginning that $f(x)$ is continuous over $(0, 1]$
since it is a product of 2 functions that are both continuous over $(0, 1]$.
The extension theorem seems to come from 2 facts:

A uniformly continuous function is continuous.
A function that is continuous over a compact set $S$ is uniformly continuous over $S$.

If you don't prefer using any of these facts, you can try proving straight from the definition of uniform continuity:

Let $X, Y$ be metric spaces and $f: X \to Y$. $f$ is uniformly continuous if for any $\epsilon > 0$, there exists $\delta > 0$, such that for any $p,q\in X$ with $d_X(p, q) < \delta$, we have $d_Y(f(p), f(q)) < \epsilon$.


TITLE: Constrained Homogeneous LS Minimization over Time
QUESTION [0 upvotes]: I am new to this forum and would like to ask a question. Let us assume I have a set of matrices $U_{(t)}$ over $t = 1 ... T$ time steps. For each time step I want to find a vector $p_{(t)}$ which satisfies the equation $U_{(t)}p_{(t)}=0$. The solution can easily be found by computing the eigenvector with the smallest eigenvalue. However, I want the adjacent $p_{(t)}$ and $p_{(t-1)}$ to be close to each other for $t>1$, because I know $U_{(t)}$ is subject to little changes over time, $U_{(t)}=U_{(t-1)} + \epsilon$. This amounts to solving
$$argmin_{p_{(t)}} ||U_{(t)}p_{(t)}||^2$$
subject to
$$||p_{(t)}||^2=1$$
$$||p_{(t-1)}-p_{(t)}||^2=0$$
Reformulating this in Lagrangian form is thus:
$$p_{(t)}U_{(t)}^TU_{(t)}p_{(t)} + \lambda(1-p_{(t)}^Tp_{(t)}) + \gamma\bigl((p_{(t-1)}-p_{(t)})^T(p_{(t-1)}-p_{(t)})\bigr)$$
Taking the derivative of the equation $\frac{\partial}{\partial p_{(t)}}$ setting it to zero leads me to:
$$2U_{(t)}^TU_{(t)}p_{(t)} - 2\lambda p_{(t)}-2\gamma p_{(t-1)}+2\gamma p_{(t)}$$
Dropping the constants
$$U_{(t)}^TU_{(t)}p_{(t)} - \lambda p_{(t)}+\gamma(p_{(t)} - p_{(t-1)}) = 0$$
I've made a slight mistake in the original derivation. However, I'm struggling a bit to solve the problem analytically. Thank you in advance.

REPLY [0 votes]: Ok, so I've figured out the question. I am sharing the it with you. My current second side condition is simply $p^T_{(t−1)}p_{(t)}>=0$, which is equivalent to the cosine-similarity equation, as $p$ is unit-length anyways.
The solution can then easily be obtained by scipy.optimize.minimize, which can handle inequalities and equalities: https://www.youtube.com/watch?time_continue=490&v=cXHvC_FGx24


text:
\DOC PURE_ONCE_REWRITE_RULE

\TYPE {PURE_ONCE_REWRITE_RULE : thm list -> thm -> thm}

\SYNOPSIS
Rewrites a theorem once with only the given list of rewrites.

\KEYWORDS
rule.

\DESCRIBE
{PURE_ONCE_REWRITE_RULE} generates rewrites from the list of theorems
supplied by the user, without including the tautologies given in
{basic_rewrites}. The applicable rewrites are employed once, without
entailing in a recursive search for matches over the theorem.
See {GEN_REWRITE_RULE} for more details about rewriting strategies in
HOL.

\FAILURE
This rule does not fail, and it does not diverge.

\SEEALSO
ASM_REWRITE_RULE, GEN_REWRITE_RULE, ONCE_DEPTH_CONV,
ONCE_REWRITE_RULE, PURE_REWRITE_RULE, REWRITE_RULE.

\ENDDOC

subset_name: curated


TITLE: Derive the Hajek pojection of $T_n$.
QUESTION [1 upvotes]: Let $X_1, \dots , X_n$ i.i.d. copies of $X$ with distribution $F$ and density $f$. Let $(X_{1:n}, \dots , X_{i:n}, \dots , X_{n:n})$ be the order statistic. For a given $p \in (0, 1)$ consider the Harrell-Davis estimator defined by
$$T_n = \sum_{i=1}^n c_{ni}X_{n:i}$$
$$c_{ni} = \frac{\Gamma(n + 1)}{\Gamma(k)\Gamma(n − k + 1)} \int^{i/n}_{(i−1)/n} u^{k−1}(1−u)^{n−k} du, i = 1,\dots , n, k = [np].$$
I have three questions regarding this. I cannot seem to find any solid clear answers to them in my textbook
1st question: Which parameter $\theta$ is estimated? I'm not sure if I have completely misinterpreted but it is rather unclear what we are estimating?
2nd question: What is the definition of the Hajek projection? I get that the Hajek projection has to do with the Hoeffding Decomposition, but I would really like a definition of the Hajek projection without delving into Hoeffding Decomposition.
3rd question: how does one derive the Hajek projection of $T_n$ from this example?

REPLY [0 votes]: Answering in an itemized list:

$\theta$ is the expectation of your estimator, $T_n$. I'm not familiar with the Harrel-Davis estimator, but a quick google search mentions it estimates the quantiles of a distribution.
Hajek Projection is a projection of a random variable onto the class $\mathcal{S}$, which is the set of all variables of the form $\sum_{i=1}^n g_i(X_i)$, where $g_i \in L^2$, $X_i$ are iid. If you've had a class in Linear Modeling, then you would have been exposed to the "Projection Theorem," working with vectors and matrices. Hajek is a more general form of this.
To calculate directly, the Hajek Projection Principle can be used:
$$\hat{S} = \sum_{i=1}^n E \left[ T_n | X_i \right] - \left(n-1\right) E\left[ T_n \right].$$ Altneratively, but probably not suggested here, would be to express $T_n$ as a U-statistic, and use the projection formula specific to U-statistics.


TITLE: How to solve the differential equation $y' = \frac{2xy}{x^2-y^2}$?
QUESTION [0 upvotes]: Solve the following differential equation:
$$y' = \frac{2xy}{x^2-y^2}$$
Someone please help me start this problem. This does not look like a regular first-order differential equation in the form  $y' + 2xy = x$. Thank you.

REPLY [1 votes]: Hint: You can write $y' = \dfrac{\frac{2y}{x}}{1 - \left(\frac{y}{x}\right)^2}$ and use $u = \frac{y}{x}$.


TITLE: Bounded Dini derivative
QUESTION [0 upvotes]: I'm so stuck right now. I feel like I lost all my analysis skills. Assume I have a continuous map $s:[0,\infty) \to [0,\infty)$ with bounded Dini derivative for all $t_0 > 0$, in particular $$\limsup\limits_{t\nearrow t_0} \frac{s(t_0)-s(t)}{t_0-t} \le L \cdot s(t_0)$$ for a constant $L>0$ independent of $t_0$.
Assume further that $s(0) = 0$.
Show that $s\le0$.
Of course I tried assuming there was a $t>0$ with $s(t) > 0$, but I were not able to get a contradiction.
Any help is greatly appreciated.

REPLY [0 votes]: In this answer I will, for simplicity, be writing $f'(t_0) = \limsup\limits_{t\nearrow t_0} \frac{f(t_0)-f(t)}{t_0-t}$ for functions $f:[0,\infty) \to [0,\infty)$. So by assumption we know that $$s'(t_0) \le L \cdot s(t_0)$$
for $t_0 > 0 $. Define $g:[0,\infty)\to[0,\infty)$ by $g(t) := s(t) \cdot e^{-Lt}$. Since $s$ is continuous, the same is true for $g$.
We now want to apply the product rule in the case of Dini derivatives. The same exact proof as for the normal product rule goes through, but here we use $\limsup (a_n + b_n) \le (\limsup a_n) + (\limsup b_n)$ and therefore get the "$\le$" instead of "$=$". Also, the Dini derivative of a differentiable function (here $e^{-Lt}$) is of course just its normal derivative.
So we get $$g'(t_0) \le s'(t_0)\cdot e^{-Lt_0} + s(t) \cdot (-L)e^{-Lt_0} \le L \cdot s(t_0) \cdot e^{-Lt_0} + s(t) \cdot (-L)e^{-Lt_0} = 0$$
for all $t_0 >0$. Since $s(0)=0$, we also have $g(0) = 0$.
From $$g'(t_0) = \limsup_{t \nearrow t_0} \frac{g(t_0)-g(t)}{t_0-t} \le 0$$
one immediately gets that $g(t_0)-g(t)\le 0$ for all $t\le t_0$ close to $t_0$, since $t_0-t > 0$ (otherwise we would find a sequence $(t_n)_n$ converging to $t$ from below with $\frac{g(t_0)-g(t)}{t_0-t} > 0$). In particular, we can conclude that $g(t) \ge g(t_0)$ for $t\le t_0$ close to $t_0$. In other words, $g$ has to decrease infinitesimally. Using $g(0)=0$ and continuity of $g$, it follows that $g\le0$.
Finally, we can also conclude that $s\le0$, since $e^{-Lt} > 0$ for all $t$.


TITLE: difficult harmonic conjugate
QUESTION [1 upvotes]: I need to first find c so that u(x,y) = $x^3 + cxy^2$ is harmonic, then create $v(x,y) $ so that $f(z) = u(x,y) + iv(x,y)$ is holomorphic at z.
First, to satisfy laplace,
$ u_x = 3x^2 + cy^2 \rightarrow u_{xx} = 6x$
$ u_y = 2cy \rightarrow u_{yy} = 2c$
$\rightarrow 6x = -2c $
$\rightarrow c = -3x$
This gives me $ u = x^3 - 3x^2y^2$
But then when I use the partials for Cauchy-Rieman and integrate to get v, I keep ending up with $ v = 3x^2y - xy^3 + C$ which does not produce the desired partials!
Help!

REPLY [1 votes]: The unknown $c$ is supposed to be a constant, not a function. Look at your $u_y$ again.


TITLE: Can civilizations orbiting black holes keep time with their interstellar travellers?
QUESTION [4 upvotes]: Due to relativistic-velocity interstellar travel time dilation, travellers would return in the far future of any civilization launching such an effort. A chart from wikipedia indicates typical time dilation ratios for various destinations:

Could an advanced civilization orbiting a black hole position themselves within its gravity well such that they enjoy constant time with their travellers?

REPLY [4 votes]: If we assume everything takes place in the Schwarzschild metric, then the time dilation factor is obtained from the metric as
$$c^2 d\tau^2 = c^2\left(1 - \frac{r_s}{r}\right)^2dt^2 - \left( 1-\frac{r_s}{r}\right)^{-1} dr^2 -r^2 d\theta^2 - r^2\sin^2\theta d\phi^2\ ,$$
where $d\tau$ is  proper time interval and $r_s$ is the Schwarzschild radius.
To answer your question we need to make some simplifications. Let's assume the "stay at home" observer is in a circular orbit around the black hole at fixed $r$. In which case for them, using $v_{\rm orb} = rd\phi/dt$ and $dr=d\theta=0$, we obtain
$$ d\tau = \left( 1 - \frac{r_s}{r} - \frac{v_{\rm orb}^2}{c^2}\right)^{1/2}dt = \left(1 - \frac{3r_s}{2r}\right)^{1/2}dt\ .  $$
For the body racing away and then coming back, I'm going to assume a uniform speed $v$ on a radial trajectory. Of course you can make it more complicated by having say uniform acceleration or soemthing, but that doesn't change the principle of the matter. So here we have $d\theta=d\phi=0$, and because the vast majority of any flight will take place with $r\gg r_s$, then we can just assume the flat spacetime metric and get
$$ d\tau = \left(1 - \frac{v^2}{c^2}\right)^{1/2} dt \  $$
for the proper time interval of the travelling observer.
In order for the proper time interval to be the same for both observers, then we need
$$  1 - \frac{3r_s}{2r} \simeq 1 - \frac{v^2}{c^2}\ , $$
$$ r = \frac{3r_sc^2}{2v^2}\ .$$
In the Schwarzschild metric, the innermost stable circular orbit is at $r=3r_s$. So long as $v < c/\sqrt{2}$ then a solution could be found.
Of course, in your example, speeds of $c/\sqrt{2}$ are actually reached pretty quickly if you accelerate constantly at $1g$. If you want to match more extreme time dilations (any more than a factor of $\sqrt{2}$ in fact), then you would need to have your stay-at-home observer either somehow kept stationary at a a radius not much larger than $r_s$ or in a close orbit around a Kerr black hole for which orbital stable radii could be much smaller, but would require an almost maximally spinning black hole to get very large time dilation factors.
In order to avoid nonsensically large tidal forces, this will also have to be a supermassive black hole.


TITLE: How much energy from air to ice?
QUESTION [0 upvotes]: If...
"A" is the mass of air (standard air composition and pressure)
"I" is the mass of ice
"V" is the volume of the box where the change of heat happens (no heat interaction with out of the box)
"Ta" is the temperature of the air
"Ti" is the temperature of the ice
Ta > Ti
(Ta - Ti) < 10
Which is the formulas to know the balance temperature and which one to know how much ice has melted (if it melt)?
And it's possible to know even how much time it will take to reach balance temperature? (this is not much important to me, just if you'd like to...)
Thanks a lot in advance ^^

REPLY [0 votes]: It’s not clear what you are asking. But if “10” is 10 C, and assuming that the ice starts at 0 C at atmospheric pressure then the heat required to melt all the ice will be
$$Ih_v$$
Where $h_v$ is the latent heat of fusion of water.
That in turn would have to equal
$$AC_{P}\Delta T$$
Where $C_P$ is the specific heat of air at constant pressure.
Setting the two equal will give you the required $\Delta T$ of the air to melt all the ice. If it’s equal to or greater than 10 C, all the ice won’t melt. If it’s less than 10 C you will need to know the actual initial temperature of the air.
Hope this helps.


TITLE: Finding groups with given Automorphism
QUESTION [2 upvotes]: Familiar Exercise: Suppose $G$ is a finite group and $T$ is an automorphism of $G$ which sends more than three quarters of elements of $G$ onto their inverses, then prove that $G$ is abelian.
The group of Quarternions is an example of a group which sends exactly $\displaystyle \frac{3}{4}$ elements onto their inverses.
Is there a finite group with an automorphism $T$ which sends exactly $\displaystyle\frac{4}{5}$ of elements of $G$ onto their inverses?
Similarly can we find groups with Automorphism $T$ which sends exactly $\displaystyle \frac{n}{n+1}$ of elements of $G$ onto their inverses.

REPLY [4 votes]: Prove that
if $G$ is abelian and $\phi$ is an automorphism of $G$
then $\{g\in G : \phi(g)=g^{-1} \}$
is a subgroup of $G$.


TITLE: Placing color boxes on a colored image such that color consistency is maximized
QUESTION [2 upvotes]: I have encountered the following challenging problem that I think to be a non straightforward generalization of the Knapsack problem.
Given an image with black background that contains blobs whose colors are taken from a set C (that does not contain black). Suppose that we have an infinite number of rectangular bounding boxes such that:

Each bounding box has a color taken from C.
Every bounding box of color c has width w_c and height h_c.

Question: How to place these bounding boxes on the image such that there is no overlap between any two boxes, and that n_good - n_bad is maximized, where n_good is the total number of pixels that are inside a box of the same color (as the pixel's color), and n_bad is the total number of pixels that are inside a box of a different color?
Any suggestions on how to solve this problem would be very much appreciated!
Thank you very much in advance for your help!

REPLY [3 votes]: Just an extended note to show that the decision version of your problem ("Is there a placement of the bounding boxes such that $n_{good} - n_{bad} \geq k$?" ) is NP-complete even for two colors $C = \{c_1, c_2\}$ and two bounding boxes $B_1 \equiv \langle col=c_1, w=3, h=1 \rangle$, $B_2 \equiv \langle col=c_2, w=1, h=1 \rangle$.
The reduction is straightforward from: Danièle Beauquier, Maurice Nivat, Eric Rémila, Mike Robson: Tiling Figures of the Plane with Two Bars. Comput. Geom. 5: 1-25 (1995)
They prove that tiling a figure on the plane grid with two horizontal/vertical bars with length at least 3 is NP-complete.
The reduction to your problem is simple: just color the "pixels" of the figure with color $c_1$ and fill the remaining background pixels of the $N+1\times N+1$ square region around it with color $c_2$. You can place the bounding boxes $B_1, B_2$ with $n_{good} - n_{bad} = (N+1)^2$ if and only if the original figure is tileable with $3 \times 1$ horizontal/vertical bars.


TITLE: Is $\frac{1-z}{|1-z|}$ is holomorphic in $|z|<1$?
QUESTION [3 upvotes]: Let $f(z) = \frac{1-z}{|1-z|}$ with $z\in\mathbb{C}$ such that $|z|< 1$. I want to know whether $f(z)$ is holomorphic in $|z|<1.$
Write $\frac{1-z}{|1-z|}=e^{i\arg(1-z)}$, where $\arg$ denotes the principal argument (from $-\pi$ to $\pi$). Then $$f(z)=e^{i\arg(1-z)}.$$
I know $e^w$ is an entire function of $w$. So it suffices to see if $\arg(1-z)$ is holomorphic in $|z|<1.$ Is this true?

REPLY [1 votes]: The argument function has real values. If you think of the Cauchy-Riemann equations, it follows that it must be constant if it is holomorphic.
Your function is also not holomorphic. Assume for a contradiction that it is. By the maximum modulus principle, since $|f(z)|$ is constant, $f(z)$ has to be constant. But your $f(z)$ is not constant.


TITLE: A distributional limit
QUESTION [2 upvotes]: I solved the distributional limit as seen below
$$\lim_{x\rightarrow+0}\left(\int_0^\infty \exp(\frac{-ARx}{2})\cos(R(y-t)) \, dR\right)=\frac{2\pi \delta(t-y)}{A}$$
where  $ -1<y<1 $  ,  $ -1<t<1 $  and $A$ is a positive constant.
I am not sure about the existence of the multiplier $\frac{2}{A}$ on the rigth hand side of the equation. Does it exist? if it does not, what is the correct solution.
Best wishes..

REPLY [1 votes]: Rewrite your integral as
$$
I(A,t,y)=\lim_{x\rightarrow0+}\Re\int_{0}^{\infty}e^{-ARx/2} e^{iR(y-t)} =\lim_{x\rightarrow0+}\Re\frac{1}{Ax/2-i(y-t)}=\lim_{x\rightarrow0+}\frac{Ax/2}{(Ax/2)^2+(y-t)^2}
$$
It's well known that such expressions are a representation of Dirac's Delta distribution: $\delta(y)=\lim_{x\rightarrow0}\frac{x}{x^2+y^2}$
We  obtain
$$
I(A,t,y)=\frac{2 \pi}{A}\delta\left(\frac{4}{A^2}(y-t)\right)
$$
Using the identiy $\delta(ax)=\frac{1}{|a|}\delta(x)$ this simplifies to
$$
I(A,t,y)=\frac{A\pi}{2}\delta\left(y-t\right)
$$
So i think your solution is not correct.


text:
\setcounter{figure}{0}
\renewcommand{\thefigure}{E\arabic{figure}}

\section{Analyzing the similarities using the left basis} \label{sec:appx_similarities}

Here, we take a more unsupervised approach to analyze the similarities of images and to identify isolated images present in a dataset.

Using HO-GSVD, we explained how the vectors in $V$ represent the orthogonal patterns extracted from the entire dataset. We also know that each image in the dataset can be reconstructed by a specific combination of patterns in $V$, defined by the singular values and the left basis vectors. In fact, $V$ is common among all the classes in the dataset $V$ and what distinguishes each image from other images, is its corresponding vector in $U_i \Sigma_i$. The number of rows in $U \Sigma$ correspond to the number of images in the class $i$. Let's consider the $j^{th}$ image in class $i$. We can reconstruct this image using the coefficients in the $j^{th}$ row of $U_i \Sigma_i$ to combine the vectors of $V$.\footnote{Clearly, all of these computations will be in  the wavelet space, and at the end we have to reconstruct them back in the pixel space.} Mathematically, if we have two identical images in the dataset, we will get identical coefficients for them.

Overall, we can say that the HO-GSVD extracts a set of common patterns in the entire dataset, and gives us a set of coefficients in $U_i \Sigma_i$, determining how these patterns should be combined in order to obtain the images. To analyze the similarity of images, we can analyze the similarity of their coefficients, because patterns are common for all of them. In fact, patterns can be considered images in the pixel space, but coefficients are just a vector of real numbers. This way, analyzing the similarity of two images is reduced to basically comparing two vectors.

Figure~\ref{fig:cifar_cat_similarity} shows the similarity matrix of all images in the Cat class. To derive this similarity matrix, we followed these steps:

\begin{enumerate}
    \item Compute the matrix $U_i \Sigma_i$ which has 5,000 rows and 3,000 columns for the Cat class.
    \item Compute the 2-norm distance between each row in this matrix to obtain a distance matrix of 5,000 rows and columns.
    \item Convolve the distance matrix with a Gaussian kernel to obtain the similarity matrix.
\end{enumerate}

\begin{figure}[H]
  \centering
   \includegraphics[width=0.4\linewidth]{fig_cat_similarity.jpg}
  \caption{Similarity matrix of images in the Cat class based on the coefficients in $U \Sigma$.}
  \label{fig:cifar_cat_similarity}
\end{figure}

Once we obtain a similarity matrix, a wide array of analysis can be performed. For example, one can perform clustering to identify communities of similar images. By analyzing the eigenvalues of a graph Laplacian built from the similarity matrix, we identified that there are not any large communities inside this dataset, as previously reported by \cite{yousefzadeh2020using}. However, coarse graining (i.e., organizing the data in large number of clusters) can easily identify small groups of images (usually 2 to 5) that are very similar, and sometimes nearly identical.

Figure~\ref{fig:cifar_dog_anomalies} shows dog images that are most isolated within the dog class. Clearly, we may consider them anomalies or out-of distribution images in the context of this dataset, because picture of a dog with a red bucket on its head, or picture of a dog with red flowers around the perimeter of image are not common. This relates to the broad literature on detecting out-of-distribution images and anomalies in image classification datasets, e.g., \cite{liang2018enhancing,huang2019out,ren2019likelihood}. This is another possible extension of our method.

\begin{figure}[H]
  \centering
   \includegraphics[width=0.1\linewidth]{cifar_dog_anomaly_1.png}
   \includegraphics[width=0.1\linewidth]{cifar_dog_anomaly_2.png}
   \includegraphics[width=0.1\linewidth]{cifar_dog_anomaly_3.png}
   \includegraphics[width=0.1\linewidth]{cifar_dog_anomaly_4.png}
   \includegraphics[width=0.1\linewidth]{cifar_dog_anomaly_5.png}
   \includegraphics[width=0.1\linewidth]{cifar_dog_anomaly_6.png}
   \includegraphics[width=0.1\linewidth]{cifar_dog_anomaly_9.png}
  \caption{Dog images that are most isolated within the dog class, identified by the similarity matrix. We can consider these out-of-distribution images in the context of CIFAR-10 training set.}
  \label{fig:cifar_dog_anomalies}
\end{figure}

We can also use the coefficients in $U_i \Sigma_i$ to visualize the dataset in form of an embedding. Figure~\ref{fig:cifar_cat_tsne} shows the embedding obtained by the t-SNE (t-Distributed Stochastic Neighbor Embedding) algorithm \citep{maaten2008visualizing}. t-SNE is a dimensionality reduction algorithm, specifically designed for visualizing high dimensional data.

This approach of visualizing a set of images in form of an embedding relates to a separate line of research in the literature. Those methods usually suffer from very high computational costs, because they rely on pairwise comparison of images in the pixel space, e.g., \cite{vo2019unsupervised}. We also note that t-SNE is used before for visualizing image datasets, for example by \cite{parde2017face}, but such approaches use a pre-trained VGG model for comparison of images. Our approach does not require a pre-trained model, however, we note that its usefulness requires further experiments and comparison with other methods.

\begin{figure}[H]
  \centering
  \includegraphics[width=0.9\linewidth]{fig_cat_tsne.jpg}
  \caption{t-SNE plot obtained from the $U_i \Sigma_i$ matrix of cat images.}
  \label{fig:cifar_cat_tsne}
\end{figure}

config: arxiv

file: 2006.09879/patterns in image classification/a5-similarities.tex


TITLE: Loopwise expansion of effective action $\Gamma[\phi]$
QUESTION [6 upvotes]: My question is about the loopwise expansion of the effective action $\Gamma(\varphi)$ up to 1-loop contributions. I've understood well the results for both $Z[J]$ and $W[J]$ functionals loopwise expansions.
But then something is missing when I follow the path towards the expansion of the effective action.
Following this excerpt from Zinn-Justin's "Quantum field theory and critical phenomena":

I don't understand the statement:
"Therefore a correction of order $\hbar$ to the relation between $J(x)$ and $\varphi(x)$ will produce a change of order $\hbar^2$ to the r.h.s. of equation (6.47)"
Could someone please write down some more details?
My undestanding at this stage is that the relation between $J(x)$ and $\varphi$ is fixed by the stationarity of the (6.47) (as for any Legendre transformation). But then, how to proceed?

REPLY [4 votes]: The problem is to evaluate $\Gamma[\phi]$ at fixed $\phi$ given an expansion of $W[J]$ in powers of $\hbar$, where $\Gamma[\phi]$ is the Legendre transform of $W[J]$.
By definition,
$$
\Gamma[\phi]=\sup_{J}\Big[\phi\cdot J-W[J]\Big].
$$
Suppose that the expression in the RHS attains its maximum at some $\hat{J}$. Then formally, we can expand $\phi\cdot J-W[J]$ about $\hat{J}$ to obtain
$$
\phi\cdot (\hat{J}+\delta J)-W[\hat{J}+\delta J]=\Gamma[\phi]+\phi\cdot\delta J-\frac{\delta W}{\delta J}\Big|_{\hat{J}}\cdot\delta J+\mathcal{O}(\delta J^2).
$$
Since a maximum is attained at $\hat{J}$, the linear term $(\phi-\delta_J W)\cdot\delta J$ vanishes for arbitrary $\delta J$, and we see that corrections start at order $\delta J^2$.
Now we return to the expansion in $\hbar$. If $W[J]$ has an expansion, then it is reasonable that $\hat J$ has an expansion too after solving $\delta_J W=\phi$. In fact, if $\hat J$ has an expansion in $\hbar$ then the zeroth order term must be the solution to $\delta_J W_0 = \phi$, where $W=W_0+\hbar W_1+\dots$. Hence, we can write $\hat J_0=\hat J-\hbar \delta\hat J$, where $\delta \hat J$ includes all higher order corrections.
From the above argument, evaluating $\phi\cdot J-W[J]$ at $\hat J_0$ will differ from the evaluation at $\hat J$ by terms of order $\hbar^2$. Hence, to order $\hbar$ we can use $\hat J_0$ in evaluating $\Gamma[\phi]$.


TITLE: Is $\operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n, \mathbb{Z})$ a Hausdorff space?
QUESTION [6 upvotes]: The special linear group $\operatorname{SL}(n,\mathbb{R})$ of degree $n$ over $\mathbb{R}$ is the set of $n \times n$ matrices with determinant $1$, with the group operations of ordinary matrix multiplication and matrix inversion. We denote by $\operatorname{SL}(n,\mathbb{Z})$ the group of  $n \times n$ matrices  with integer entries and determinant equals 1. Note that  $\operatorname{SL}(n,\mathbb{Z})$  is a discrete subgroup of $\operatorname{SL}(n,\mathbb{R})$.
Let $G$ to be a topological group and $H$ to be a subgroup of $G$. We will say that a regular Borel measure $\mu$ on the quotient $G/H$ is a left invariant Haar measure if for all Borel sets $E \subseteq G/H$ and all $g \in G$ we have $\mu(gE) = \mu(E)$.
If $G$ is a locally compact Hausdorff group and $\Gamma$ is a discrete subgroup such that $G/H$  carries a finite left G-invariant Harr measure, then we say that  $\Gamma$ is a $\textbf{lattice}$ in $G$.
We have the following results:

$\operatorname{SL}(n,\mathbb{Z})$ is a lattice in $\operatorname{SL}(n,\mathbb{R})$. Moreover, the quotiont  $\operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n,\mathbb{Z})$ is not compact.

(Mahler Criterion) For a sequnece $(g_{m})_{m\in \mathbb{N}}$ of $\operatorname{SL}(n,\mathbb{R})$,  the sequence  $(\pi (g_{m}))_{m\in \mathbb{N}}$ of $\operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n, \mathbb{Z})$  does not have a convergent subsequence if, and only if, there exists a sequence $v_{m} \in \mathbb{Z}^{n}$ with $v_{m} \neq 0$ such that $g_{m}(v_{m})$ tends to $0$.

where  $\pi\colon\operatorname{SL}(n,\mathbb{R}) \to  \operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n,\mathbb{Z})$ is the natural projection.
The second result suggest to me that $\operatorname{SL}(n,\mathbb{R})/\operatorname{SL}(n, \mathbb{Z})$  is a Hausdorff space, but I can't find any reference. So, I do not know if it is true.
Thanks

REPLY [7 votes]: To summarize the comments: Combining Left translation on Lie group of a discrete subgroup is properly discontinuous, Quotient of a topological group by a closed subgroup is Hausdorff, Why is every discrete subgroup of a Hausdorff group closed?
answers your question in affirmative: The quotient is Hausdorff.


TITLE: Algebraic manipulation from notable limit
QUESTION [0 upvotes]: I have no idea of how to solve the following:
$$\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{3x}$$
I know about the notable special limit $$\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{x}=1$$, and I know that I have to do some algebraic manipulation and change what I have above to the notable one, but I can't quite see how.

REPLY [0 votes]: Note, that the limit that you have provided is indeterminate.
$$\begin{align}
\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{3x}
&=\frac{e^0-1}{3(0)}
=\frac{0}{0}
\end{align}$$
Therefore, you can use L'Hopital's Rule which states:
$$\displaystyle \lim_{x \rightarrow 0}\frac{f(x)}{g(x)} = \displaystyle \lim_{x \rightarrow 0}\frac{f'(x)}{g'(x)}$$
Therefore, let $f(x)$ be $e^x -1$ and let $g(x)$ be $3x$ and apply L'Hopital's Rule.
$$\begin{align}
&\displaystyle \lim_{x \rightarrow 0}\frac{f'(x)}{g'(x)} \\
&\displaystyle \lim_{x \rightarrow 0}\frac{e^x}{3}\\
&\displaystyle \lim_{x \rightarrow 0}\frac{e^0}{3}
= \frac{1}{3}
\end{align}$$
Therefore, the limit is $\frac{1}{3}$. Note, that the derivative of $e^x$ is $e^x$.


TITLE: Why is this non-linear transformation not a Lorentz transformations? It does preserve $x^2 + y^2 + z^2 - c^2t^2 = x'^2 + y'^2 + z'^2 - c^2t'^2 $
QUESTION [13 upvotes]: The Lorentz transformation is a mapping of the $x', y',
z',$ and $t'$ coordinates relative to $x, y, z, t$ that preserves the speed of
light:
(1) $x^2 + y^2 + z^2 - c^2t^2 = x'^2 + y'^2 + z'^2 - c^2t'^2 = 0$
If $K'$ is moving with speed $v$ in the $x$ direction relative to $K$ (so that $K$ is moving with speed $-v$ in the $x$ direction relative to $K'$), and we use the Galilean transformation, we get an erroneous
(2) $x'^2 + y'^2 + z'^2 - c^2t'^2 = -2vxt + v^2t^2$
which is NOT equal to zero, and hence does not describe a sphere of
radius $ct$ emanating from the origin of $K'$, but something else (I'm
pretty sure it's a sphere that's been shifted negatively in the $x$
direction by a small amount, relative to $K'$). The Lorentz
transformation says we modify both the $x' = x - vt$ and the $t' = t$
relationships in order to ensure that we end up getting 0.
However, as a naive mathematician, I look at equation (2) and think,
"Ok, so we've got to modify the rule for $x'$, or $t'$, or both, to get
rid of that nasty $-2xvt + v^2t^2$ term in the $x'$ rule. Why not just
set $t'$ up so that instead of coming out to $-c^2t^2$, it actually comes
out to $-c^2t^2 + 2vxt - v^2t^2$, so that the extra terms cancel out?" I
naively went ahead with this idea and I got the following coordinate
transformation:
$$
x' = x - vt\\
y' = y\\
z' = z\\
t' = \frac{1}{c} \sqrt{(v^2 + c^2)t^2 - 2vxt}
$$
If you plug this into (1), you get 0 like you're supposed to.
Now, I haven't been able to find an explanation as to why this transformation is unsatisfactory. Of course, it's no longer linear - but is there some basic reason why the transformation must be linear? Obviously this transformation satisfies Einstein's postulate about the constancy of the speed of light for all inertial observers.
Presumably, then, it must violate the first postulate, regarding the equivalence of all physical phenomena for inertial observers - otherwise we'd need more information in order to derive the Lorentz transformation, which I think is supposed to be derivable from the two postulates alone.
So, in the end, I have one question I'm trying to answer:

Does this transformation violate equivalence for inertial observers? Why? (ideally, someone could give me an example of a force that suddenly appears in my coordinate system that didn't already exist in the other coordinate system)

REPLY [1 votes]: There is a theorem due to Zeeman that under very mild assumptions proves that admissible transformations in special relativity are elements of the Lorentz group. This is treated e.g. in Naber's "The Geometry of Minkowski Spacetime".
To fix definitions, an observer is admissible if it has a 3-dimensional, right-handed, Cartesian spatial coordinate system based on an agreed unit of length and relative to which photons propagate rectilinearly at a fixed speed in any direction, as well as an ideal standard clock based on an agreed unit of time with which to provide a quantitative temporal order to the events on his worldline, in which light speed will be 1.
The mapping from the coordinates of one admissible observer to that of another one constitutes an invertible map
$$F:\mathbb R^4\rightarrow\mathbb R^4$$
The following assumption will be made: any two admissible observers agree on the temporal order of any two events on the worldline of a photon, i.e. if $x$ and $y$ are events in the coordinates of one observer, and $\hat x$ and $\hat y$ those for another one, then $x^0 - y^0$ (time coordinates) and $\hat x^0 - \hat y^0$ have the same sign (it is not demanded that the times are equal).
Let $v$ be the velocity vector of the photon (which is constant). Since in our units $c = 1$, we have $\|v\| = 1$. We have for any two points $x$ and $y$ on the world line of the photon that
$$x^i - y^i = v^i(x^0 - y^0)$$
for $i = 1,2,3$. Consequently, on the world line of the photon
$$(x^0 - y^0)^2 - (x^1 - y^1)^2 - (x^2 - y^2)^2 - (x^3 - y^3)^2 = 0$$
This is a cone in $\mathbb R^4$ with vertex at $y$. Obviously this cone must be preserved in any coordinate system, hence so does the change of coordinates map $F$. Zeeman called such a map a causal automorphism, and proved that $F = T\circ K\circ\Lambda$, where $T$ is a translation and $K$ is a dilatation of $\mathbb R^4$, and $\Lambda\in O(1,3)$ with $\Lambda^0_0\ge 1$.
Note that for this result it is not even assumed that $F$ is continuous.


TITLE: Is there any working strategy for finding the distance between a point and a function/curve in high dimensions?
QUESTION [0 upvotes]: You have a point $P(x_1,...,x_n)$ and an arbitrary function $f(x_1,...,x_n) = c$ . Is finding the minimum distance from the point to the function/curve possible in realistic time?
I mean something like this, but in any dimension:
Shortest distance from point to curve
Minimizing the distance function in 1000 dimension is not a possibility I think.

REPLY [1 votes]: The general from of a curve, surface, etc. in a finite-dimensional space is given by a map $f : U \to \Bbb B^n$, where $U \subseteq \Bbb R^m$ is a region (connected non-empty open set). The "hypersurface" is the image of $U$ under $f$, that is, the set $f(U)$.
Assuming $f$ is differentiable, only a small variation of the technique in the other thread is needed. If $p$ is the point whose distance from the hypersurface is desired, one forms the squared distance function $$D(x) = \|f(x) - p\|^2$$
where $x = (x_1, x_2, ..., x_m)$. $D$ is differentiable because $f$ is, and at the minimum, the derivative with respect to each of the coordinates must be $0$. So you solve the system of equations
$$\frac {\partial D}{\partial x_1} = 0\\\frac {\partial D}{\partial x_2} = 0\\\vdots\\\frac {\partial D}{\partial x_m} = 0$$
This gives you a system of $m$ equations in the $m$ unknowns $x_1, ..., x_m$, which you then solve to minimize $D$.
While this might be impractical to do by hand for a 1000-dimensional problem, it is not impossible.