TITLE: There exists a divisor linearly equivalent to $P$ not containing $P$
QUESTION [4 upvotes]: Let $X$ be a non-singular projective curve over $\mathbb C$ and let $P\in  X$ be a closed point. Which is, in your opinion,  the easiest way to prove that there exists a divisor $D$ of $X$ satisfying the following conditions:

$D$ is linearly equivalent to $P$.
$D$ doesn't contain $P$.

Thank you in advance.

REPLY [2 votes]: I have it on good authority that Relapsarian is suffering an identity crisis at the moment, so let me post the comment above as an answer as per @GeorgesElencwajg's impassioned suggestion.
Let $H_1$ and $H_2$ be nonsingular hyperplane sections of $X$ such that $P$ lies in $H_1$ but not $H_2$. Then the divisor you want is $D=P−H_1+H_2$. Why? "Nonsingular" means that each hyperplanes is intersecting $X$ transversely; that implies in particular that the divisor $H_1$ contains $P$ with coefficient 1. On the other hand $P$ is not in $H_2$, so in the divisor $P-H_1+H_2$ we end up with no term involving $P$.
Finally, (exercise!) any two hyperplanes in projective space are linearly equivalent, and linear equivalence is preserved by restriction, so $-H_1+H_2$ is a principal divisor on $X$.


TITLE: Sets equivalence relations
QUESTION [1 upvotes]: I seem to be having a hard time understanding some basic sets concepts. In week 5 of my class, I learnt about the cross product of 2 sets to be the following $A \times B = \{(a,b) : a \in A, b \in B \} $
To give an example, this will look something like
$A = \{1,2\} \text{ B } = \{x, y,z \} \text{ then } A  \times B = \{ (1, x), (1, y), (1, z), (2, x), (2, y), (2, z) \}$
Now fast forward to the middle of the semester where we are talking about equivalence relation and partitions. He made the case that for a set $S$, $R \subseteq S \times S$
I am having a bit a of hard time understanding why we have $S \times S$ if we are only given one set. Partly because I only see one set $S$ and not two sets.
Any help in breaking down the concepts would be greatly appreciated. Basically how do you explain something like this to your 10 year old.

REPLY [3 votes]: The notion of "equivalence" comes to generalize "equality". If two objects are equal if and only if they are the same, equivalence relations come to relax that strict demand. For example, two people are equivalent if and only if they have the same height; or the same car; or the same biological mother.
These are all properties by which we can compare people and liken them to one another, even if they are not the same person.
Going back to the mathematical world, equivalence relation generalize the notion of equality, but for that we need a fixed domain of discourse. And we want to compare two elements from that domain of discourse. If "similar" means that $(x,y)\in R$ then we want both $x$ and $y$ to be from our domain. In particular this means that $R$ is a subset of $S\times S$.
And if what bothers you is that we take the product of $S$ with itself, and not with another set, remember that $2\times 2$ is a well-defined notion in the natural number. Cartesian product, much like multiplication, is defined for any two [suitable] objects, not necessary distinct two.


TITLE: Why can't I find the value of years in compound interest, when there is a compound frequency?
QUESTION [0 upvotes]: I'm trying to figure out why when I do not have an 'a' value (Compound Frequency) I can no longer use logarithms to solve for 'n' (number of years). Let the formula look like this:
FV=PV(1+(i/100)/a)^na

So by using the next formula:
(log(FV/PV))/(log(1+(i/100)/a))=n

It only appears to work when a is equal to 1. When I change it to something like 52, I end up get the incorrect number. Try using these variables:

FV = 122.12
PV = 100
i = 10
a = 52

I know that n=2, but using the second formula I constantly get 104. Why is this happening?
Apologies for poor formatting, I'm just in a rush.

REPLY [0 votes]: I think it is the way the formula is formatted.

now plug in the variables


TITLE: $f(x) = e^{-{1\over x^2}}+\int_0^{\pi x\over2}(1+\sin t)^{1\over2}dt$ for $x\in(0,\infty)$
QUESTION [2 upvotes]: Let

$$f(x) = e^{-{1\over x^2}}+\int_0^{\pi x\over2}(1+\sin t)^{1\over2}dt$$

for $x\in(0,\infty)$
Then which of the following are true?
(A) $f′$ exists and is continuous.
(B) $f′′$ exists for all x.
(C) $f′$ is bounded.
(D) there exists $α > 0$ such that $|f(x)| > |f′(x)|$ for every $x$ in $(α,1)$
My approach: I found out $f'$ and $f''$. Since $x>0$, $f''\ne 0$ so (C) is not true.
(A) & (B) seems true. But I can't understand (D).

REPLY [0 votes]: Use Lagrange mean value theorem.you will get f(c)=1+c-1f'(c) c=alpha. Find f(c)and f'(c). c>0. Put these values in the above and compare.


TITLE: Microscope: relation between numearical aperture and working distance
QUESTION [0 upvotes]: I am looking for help in understanding the relation between the numercial aperture and the working distance of a microscope objective.
I've searched for many hours but couldn't find any paper / book or other literature that could tell me the relation between these two parameters.
Are there any formulas on how to calculate the working distance of an microscope objective?
I would also really appreciate literature recommendations.

REPLY [0 votes]: The working distance is a direct consequence of the actual optical design, and less so in regard to the numerical aperture (though one can indirectly affect the other).
Either way, as far as I know, there's no direct analytical relation between them in a real-world optical design (one may design different lenses with the same focal length and numerical aperture, but different working distances).


TITLE: Determine cubic function from 2 roots and a maximum.
QUESTION [1 upvotes]: If I am trying to find a cubic function with 3 real roots, and I know two of them, and one local maximum, is it possible?
Assuming my roots are $0.05$, $0.95$ and $u$, and my local maximum is $(i, j)$.
Since a cubic takes the form $y = ax^3 + bx^2 + cx + d$:
$$ 0 = a(0.05^3) + b(0.05^2) + c(0.05) + d$$
$$ 0 = a(0.95^3) + b(0.95^2) + c(0.95) + d$$
$$ 0 = a(u^3) + b(u^2) + c(u) + d$$
My local maximum is where the derivative = $0$;
$$ 0 = 3ax^2 + 2bx + c $$
Of course, if I could calculate $u$, I could write the cubic in the form:
$$y = (x-0.05)(x-0.95)(x-u)$$

REPLY [1 votes]: Given any cubic equation
$$f(x) = ax^3 + bx^2 + cx + d = a(x-\alpha)(x-\beta)(x-\gamma)$$
which you know $\alpha$ and $\beta$. If $f(x)$ reaches local maximum at $x_m$ with value $f(x_m) = m$, one has
$$f'(x_m) = 0 \quad\implies\quad \frac{f'(x_m)}{f(x_m)} = 0 \quad\implies\quad
\frac{1}{x_m-\alpha}+\frac{1}{x_m-\beta}+\frac{1}{x_m-\gamma} = 0$$
This give us $\quad\displaystyle \gamma = x_m + \frac{1}{\frac{1}{x_m-\alpha} + \frac{1}{x_m-\beta}}$.
Once you figure out what $\gamma$ is, then $f(x)$ is given by:
$$f(x) = m \frac{(x-\alpha)(x-\beta)(x-\gamma)}{(x_m-\alpha)(x_m-\beta)(x_m-\gamma)}$$

REPLY [0 votes]: Start with an equation of the form:
$$y = a(x-r_{1})(x-r{_{2}})(x-r{_{3}})$$
or
$$y = a(x^3 -(r_{1}+r_{2}+r_{3})x^2+(r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1})x-r_{1}r_{2}r_{3})$$
You have $r_{1}$ and $r_{2}$, and want to find $a$ and $r_{3}$.
The derivative condition at the known maximum $(x_{max},y_{max})$ will be
$$a(3x_{max}^2-2(r_{1}+r_{2}+r_{3})x_{max}+(r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}))\equiv0$$
Which is linear in $r_{3}$, so use your known $x_{max}$, $r_{1}$ and $r_{2}$ values to determine $r_{3}$.
Once you have $r_{3}$, substitute $(x_{max},y_{max})$ into the first equation to find $a$.


TITLE: Angles satisfying $(\cos\theta,\sin\theta)=(\frac{\sqrt a }{\sqrt{a+b}},\frac{\sqrt b}{\sqrt{a+b}})$ for integer $a$, $b$
QUESTION [1 upvotes]: If we list the commonly used angles in the first quadrant $\theta_0=0$, $\theta_1=\dfrac{\pi}{6}$, $\theta_2=\dfrac{\pi}{4}$, $\theta_3=\dfrac{\pi}{3}$, and $\theta_4=\dfrac{\pi}{2}$,
then $$(\cos \theta_i, \sin \theta_i)=\left(\dfrac{\sqrt{4-i}}{2},\dfrac{\sqrt{i}}{2}\right) \qquad\mbox{for}\qquad i=0,\dots,4.$$
The angles $\dfrac{\pi}{6}, \dfrac{\pi}{4},\dfrac{\pi}{3} $ are obtained by bisecting or trisecting the first quadrant.

Question: Are there special constructions for angles $\theta$ such that
$$(\cos \theta, \sin \theta)=\left(\dfrac{\sqrt{a}}{\sqrt{a+b}},\dfrac{\sqrt{b}}{\sqrt{a+b}}\right)$$
where $a,b$, and $\sqrt{a+b}\ $ are positive integers? Or specifically, is there a rational number $r=r(a,b)$ such that $\theta=r\pi$?

Note that if $a,b$ are nonnegative integers such that a+b=4, then we are describing the angles $\theta_0,\dots,\theta_4$.

REPLY [2 votes]: Regarding the second question, there are no other possible values of $\theta\in[0,\pi/2]$. Squaring, we have
$$\cos^2\theta=\frac{a}{a+b}\Leftrightarrow\frac{1+\cos 2\theta}{2}=\frac{a}{a+b}\Leftrightarrow\cos 2\theta=\frac{a-b}{a+b} $$
This means that $\cos2\theta$ is rational and $2\theta$ is a rational multiple of $\pi$ (since $\theta$ is a rational multiple of $\pi$). By Niven's theorem, $\cos2\theta=0,\pm\frac 12$ or $\pm 1$, which gives us $\theta\in\left\{0,\frac{\pi}{6},
\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2}\right\}$.


TITLE: How can I prove that if $2^x+1$ is prime, then $x$ is a power of $2$?
QUESTION [0 upvotes]: How can I prove that if $2^x+1$ is prime, then $x$ is a power of $2$? Can someone help me prove this?

REPLY [1 votes]: Hint. You can probably show that
$$
A-B \text{ divides } A^n - B^n .
$$
Then try to show that
$$
A+B \text{ divides } A^n + B^n
$$
when $n$ is odd.


text:
\section{Links with domain adaptation}\label{sec:link_DA}
In this part we revisit a standard bound from domain adaptation. In this setting, $P$ and $Q$ stand for the source and target distributions over a joint space $\measSpace = \mathcal X\times \mathcal Y$ where $\mathcal X$ is called the sample space and $\mathcal Y$ the label space and is a finite set of classes. Given a classifier $h$ (a.k.a an hypothesis) one would like to control the error $\epsilon_Q(h)$ in the target domain which is not directly observable (for lack of supervision) with a bound relying one the corresponding error $\epsilon_P(h)$ in the source domain (where supervision is available). The simplest bound proposed in the literature is the following \cite{ben2010theory}:
\begin{equation}
    \label{eq:da-tv}

    \epsilon_Q(h):=\int \1_{h(x)\neq y} dQ(x,y)

    \leq
    \epsilon_P(h) + \Vert P-Q\Vert_{TV}

\end{equation}

In the common {\em covariate-shift} assumption (where the distribution of labels given samples is the same in both domains), the bound can be expressed in terms of marginal distribution over $\mathcal X$.
It was noted directly by \cite{ben2010theory} that the previous bound suffers from two major issues in practice. The first one is that the bound cannot be estimated efficiently from finite i.i.d samples because of its reliance on the TV norm (which involves a class of measurable sets that is not restricted enough for such purpose).
And the second issue lies in the fact that the bound does not imply the class of hypothesis functions where $h$ lives (e.g. a class of small VC dimension or of small Rademacher capacity).

As such, the bound in question is over-pessimistic, because it takes for granted that error set $\{h(x)\neq y\}$ may be spread as an arbitrary measurable set, which may be far from the case depending on restrictions applying to $h$.
The authors then derived a more adapted bound based on the so-called $\mathcal H \Delta \mathcal H$-divergence, answering the previous two points, and many other bounds were later proposed. The interested reader may refer to \cite{redko2020survey} for a very up-to-date and comprehensive review of such works.

\subsection{A refined bound}
Despite its shortcomings, we would like to revisit Eq.~\ref{eq:da-tv} and provide an optimized version of it which has an intuitive interpretation.
We will first derive a bound, based on the Lorenz curve between $P$ and $Q$ and then express it in a form closer to Eq~\ref{eq:da-tv} by relying on the duality between Lorenz and PR curves.

\begin{proposition}\label{prop:error_bounds_Lorenz}
The Lorenz curve provides a lower bound for domain adaptation.
\begin{equation}
    \epsilon_Q(h)\leq 1-F(1-\epsilon_P(h))
\end{equation}
\end{proposition}
\begin{proof}

\begin{equation*}
    \begin{split}
    F(1-\epsilon_P(h))
        & = \inf_{\substack{g \text{ measurable}\\0\leq g \leq 1 \\\int g d\S \ge 1 - \int  \1_{h\neq f} d\S}} \int g d\T
        \\
        & = \inf_{\substack{g \text{ measurable}\\0\leq 1-g \leq 1 \\\int (1 - g) d\S \le \int  \1_{h\neq f} d\S}} \int 1 - (1-g) d\T
        \\
        & = 1 - \sup_{\substack{g' \text{ measurable}\\0\leq g' \leq 1 \\\int g' d\S \le \int  \1_{h\neq f} d\S}} \int g' d\T
    \end{split}
\end{equation*}
Considering the particular case: $g'=\1_{h\neq f}$ one gets
\begin{equation*}
    \begin{split}
    F(1-\epsilon_P(h))
        & \le 1 - \int  \1_{h\neq f} d\T
        = 1 - \epsilon_Q(h)\\
    \end{split}
\end{equation*}
\end{proof}

The exact same bound can be expressed with the PR curve parametrization thanks to Theorem \ref{thm:duality}.
\begin{proposition}
\label{prop:pr-bound}
\begin{equation}
\label{eq:pr-bound}
    \epsilon_Q(h)\leq \lambda^* \epsilon_P(h)+(1-\alpha_{\lambda^*})
\end{equation}
With $\lambda^*=\argmax\limits_{\lambda\in\mathbf{R^+}}\lbrace\alpha_\lambda - \lambda \epsilon_P(h)\rbrace$
\end{proposition}

\begin{proof}
This result follows from Thm.~\ref{thm:duality}.
\end{proof}

The first part of our upper-bound corresponds to the errors occurring in the common support of $\S$ and $\T$. There, the error rate is controlled in the source domain, and is therefore also controlled in the target domain. The amplification factor $\lambda^*$ accounts for the fact that the common mass between $\S$ and $\T$ is present in different ratios in the two domains. The second part corresponds to errors occurring in the target domain, within mass that is not present in the source domain. We do not have control over this error, and must account for it by considering the worst case where $h$ is always wrong. As a result, the only way to keep this term under control is to make some assumptions on the class of admissible functions for $h$ and the distribution of labels, that is to say the hypothesis class and the concept class.

\subsection{Discussion}
The latter form of our bound (Eq.~\eqref{eq:pr-bound}) highlights  a strong tie with Eq.~\eqref{eq:da-tv}. In particular, if $\lambda^*=1$ then, noting that $\alpha_1=1-\frac 12 \Vert P-Q\Vert_{TV}$, then the two bounds are almost identical.
The only difference resides in a factor $\frac 12$ in our favor, which stems from the fact that we explicitly leverage the non-negativity of the error. Besides in general, $1$ is not the optimal $\lambda^*$ and our bound is even tighter.
This is where the reliance on a trade-off curve comes in handy: we obtain virtually one bound for each value of $\lambda$ and we can pick the sharpest one. Let us consider the simple example of Fig.~\ref{fig:complex} where we can read the value of $\alpha_\lambda$ on the y-axis at the location where the PR-curve meets the line of equation $\alpha = \lambda \beta$. In this example $\alpha_1\approx 0.38$ which means that $\Vert P -Q\Vert_{TV} = 2(1-\alpha_1) \approx 1.24$. Therefore the bound of Eq.~\eqref{eq:da-tv} is larger than $1$ and is hence non informative.
On the other hand, Eq.~\eqref{eq:dprd2} with $\lambda=1$ gives $\epsilon_Q(h)\leq \epsilon_P(h) + (1-\alpha_1)\approx \epsilon_P(h)+0.62$ which is informative as soon as $\epsilon_P(h) < 0.38$. This condition is easily meet in concrete cases because $\epsilon_P(h)$ is the error in the source domain, which is under control thanks to supervision.  More importantly, we can examine how the optimal trade-off between the two terms of the error bound can yield a much sharper bound.
In fact, given the convexity of $1-\alpha_\lambda$, the optimal $\lambda$ is characterized by the first-order critical point condition, that is $\epsilon_P(h)\in \partial_\lambda \alpha_\lambda$.
Note that it is not trivial to read the derivative of $\alpha_\lambda$ from the PR curve directly, but in this example this derivative is much larger than $1$ and then larger than $\epsilon_P(h)$. This means that the optimal $\lambda$ is far from $1$.

Besides, as discussed earlier, our bound can be easily understood in terms of shared vs separate mass between $P$ and $Q$.
Despite those advantages, one must keep in mind that it suffers from the same limitations when considering its practical estimation from finite i.i.d. samples and its pessimistic nature with regard to the actual regularity of the error set.
That being said, Lorenz and PR curves draw several similarities with tools developed in domain adaptation, which calls for further scrutiny.
For instance, the proof of Thm~\ref{thm:pr-div} allows to express PR-curves as trade-off curves computed from weight ratios as defined in \cite{ben2012hardness}.
Inspired by their work, it would be natural to restrict the class of ``admissible'' measurable sets in the PR-curves and get more useful bounds while retaining the notion of an optimal trade-off.
A similar step could be taken on the dual representation by restricting the class of functions in the Lorenz diagram.
In that way, one would leverage the restrictions on the hypothesis class and get similar bounds as many of those derived from Integral Probability Metrics \cite{redko2020survey}.
Nonetheless, given the proven ability of deep neural nets to overfit random labels \cite{zhang2016understanding}, leveraging classical notions of complexity is probably not sufficient to get bounds that are representative of the  current domain adaptation problems.
It seems inevitable to leverage some kind of ''implicit bias'' related to the optimization procedure.
Doing so while relying on trade-off curves is an exciting research avenue : it is on-going and will certainly yield more practical bounds for domain adaptation.

\iffalse
\section{Error Bounds}\label{sec:domain_adaptation}
Let $P_S$ and $P_T$ be two distributions over $(\Omega, \mathcal{A})$, in this case the sample space. Given labeled samples of $P_S$ and unlabeled samples of $P_T$, we want to minimize the target generalization error $\epsilon_T(h,f)$ for some hypothesis $h\in \mathcal{H}$ w.r.t a labeling function $f$. We hence look for an upper bound over this quantity.

\begin{comment}
One of the most cited source exposing a useful upper bound for domain adaptation is \cite{ben2010theory}. The bound reads as follow:

\begin{equation*}
    \begin{split}
        \epsilon_T(h,f)\leq&\epsilon_S(h,f)+\gamma +\delta_{H\Delta H}(\S,\T)
    \end{split}
\end{equation*}

where, $\gamma:=\inf_{h'} \epsilon_S(h', f)+\epsilon_T(h', f)$ is the optimal combined error on source and target, and
$\delta_{H\Delta H}(\S,\T):=\sup_{h,h'\in H}\epsilon_T(h,h')-\epsilon_S(h,h')$ is called the $H\Delta H$-divergence, and accounts for how much two
hypotheses can deviate from one another and in a different way in the two domains.

As noted, in \cite{zhang2019bridging}, it is possible to improve their bound in several directions.
In particular, they propose to adapt the bound by replacing the $L^0$ error by differentiable loss like a margin loss.
They also propose to improve the bounds by keeping it adaptive w.r.t the current hypothesis $h$.
Inspired by the derivations \cite{zhang2019bridging}, one can extract a refined version of the bound of \cite{ben2010theory}, as follows:

\begin{proposition}
\begin{equation}
    \begin{split}
        \epsilon_T(h,f)\leq&\epsilon_S(h,f)+\gamma+\delta(h,\S,\T)
    \end{split}
\end{equation}
where $\delta(h,\S,\T):=\sup_{h'}\epsilon_T(h,h')-\epsilon_S(h,h')$ is a kind of adaptive divergence, that accounts for
how differently on $\S$ and $\T$ another hypothesis $h'$ can deviate from our current hypothesis $h$.
\end{proposition}
\begin{proof}
The proof is given mainly to highlight its use of the triangle identity $\epsilon_D(h,g)\leq \epsilon_D(h,f)+\epsilon_D(f,g)$ which
is a pretty loose inequality.
\begin{equation*}
    \begin{split}
        \epsilon_T(h,f)\leq& \epsilon_T(h',f) +\underbrace{\epsilon_T(h,h')}_{=\epsilon_S(h,h')+\epsilon_T(h,h')-\epsilon_S(h,h')}\\
        \leq& \epsilon_T(h',f) + \underbrace{\epsilon_S(h,h')}_{\leq\epsilon_S(h,f)+\epsilon_S(h',f)} +\delta(h,\S,\T)\\
        \leq& \epsilon_S(h,f)+ \epsilon_S(h',f)+\epsilon_T(h',f) +\delta(h,\S,\T)
    \end{split}
\end{equation*}
This inequality being true for any arbitrary choice of $h'$, one can take the infimum of the right hand side w.r.t $h'$, which is the desired bound.
\end{proof}
\begin{remark}
The proof relies twice on the use of the triangle inequality, leading to a potential overestimate by a factor $3$. Indeed, if the best hypothesis is the same in the two domains, then $\gamma=2\inf_{h'}\epsilon_s(h',f)$.  And $h$ is approximately equal to this common best hypothesis, then the bound is at least 3 times as large as what is needed.
\end{remark}
\end{comment}

\subsection{Lorenz bound}
From the results of  Sec.~\ref{sec:equivalence}, we derive a new, tighter bound to control the target risk.
\begin{proposition}\label{prop:error_bounds_Lorenz}
The Lorenz curve provides a lower bound for domain adaptation.
\begin{equation}
    \epsilon_T(h,f)\leq 1-F(1-\epsilon_S(h,f))
\end{equation}
\end{proposition}
\begin{proof}

\begin{equation*}
    \begin{split}
    F(1-\epsilon_S(h,f))
        & = \inf_{\substack{g \text{ measurable}\\0\leq g \leq 1 \\\int g d\S \ge 1 - \int  \1_{h\neq f} d\S}} \int g d\T
        \\
        & = \inf_{\substack{g \text{ measurable}\\0\leq 1-g \leq 1 \\\int (1 - g) d\S \le \int  \1_{h\neq f} d\S}} \int 1 - (1-g) d\T
        \\
        & = 1 - \sup_{\substack{g' \text{ measurable}\\0\leq g' \leq 1 \\\int g' d\S \le \int  \1_{h\neq f} d\S}} \int g' d\T
    \end{split}
\end{equation*}
Considering the particular case: $g'=\1_{h\neq f}$ one gets
\begin{equation*}
    \begin{split}
    F(1-\epsilon_S(h,f))
        & \le 1 - \int  \1_{h\neq f} d\T
        = 1 - \epsilon_T(h,f)\\
    \end{split}
\end{equation*}
\end{proof}

\begin{remark}
\label{rk:f-determinism}
In a degenerate feature-space, $f$ might become non-deterministic. In such cases, the bound remains valid if $P_S$ and $P_T$ stand for the joint sample/label distribution in all definitions and $f$ merely extracts the label component. In that case, we also need target labels to compute the bound.
\end{remark}

\subsection{PR bound}
The exact same bound can be expressed with the PR curve parametrization thanks to Theorem \ref{thm:duality}.

\begin{proposition}
\label{prop:pr-bound}
\begin{equation}
\label{eq:pr-bound}
    \epsilon_T(h,f)\leq \lambda^* \epsilon_S(h,f)+(1-\alpha_{\lambda^*})
\end{equation}
With $\lambda^*=\argmax\limits_{\lambda\in\mathbf{R^+}}\lbrace\alpha_\lambda - \lambda \epsilon_S(h, f)\rbrace$
\end{proposition}

\begin{proof}
This result follows trivially from Thm.~\ref{thm:duality}.
\if0
Starting from the Lorenz bound and applying Theorem \ref{thm:duality}:\\

\begin{equation}
\label{eq:pr-bound-v2}
\begin{split}
    \epsilon_T(h, f) &\leq 1 - F(1 - \epsilon_S(h, f))\\
&=1 - \sup_{\lambda\in\mathbf{R^+}}\lbrace\alpha_\lambda - \lambda \epsilon_S(h, f)\rbrace\\
&=\lambda^* \epsilon_S(h,f)+(1-\alpha_{\lambda^*})
\end{split}
\end{equation}
\fi

\end{proof}

The first part of our upper-bound corresponds to the errors occurring in the common support of $\S$ and $\T$. There, the error rate is controlled in the source domain, and is therefore also controlled in the target domain. The amplification factor $\lambda^*$ accounts for the fact that the common mass between $\S$ and $\T$ is present in different ratios in the two domains. The second part corresponds to errors occurring in the target domain, within mass that is not present in the source domain. We do not have control over this error, and must account for it by considering the worst case where we are always wrong. As a result, the only way to keep this term under control is to make some assumption on the class of admissible functions for $h$ and $f$, namely the hypothesis class and the concept class.
\fi

config: arxiv


TITLE: estimate Gamma parameters based on mean and variance
QUESTION [0 upvotes]: I am following these two approaches (which are the same)this and  this, to estimate the two parameters of Gamma dist based on mean and var. I am not sure why I cannot get the same mean and var from rgamma based on the estimated parameters in R:
 mean1=20
 var1=30

 shapeg = (mean1*2)/var1
 scaleg = var1/mean1

 nn=(rgamma(1000, shape = shapeg, scale = scaleg))
 mean(nn);
 var(nn);

 > mean(nn);
 [1] 1.975715
 > var(nn);
 [1] 3.6593

Based on https://stat.ethz.ch/R-manual/R-devel/library/stats/html/GammaDist.html
The Gamma distribution with parameters shape = a and scale = s
The mean and variance are E(X) = as and Var(X) = as^2.
Based on this: https://stats.stackexchange.com/questions/280459/estimating-gamma-distribution-parameters-using-sample-mean-and-std
$\mbox{Var}[X] = \alpha \theta^2$
therefore, $\theta$ must be equal to scale in  R. SO I am replacing the parameters correctly.

REPLY [1 votes]: I think the mistake is in your code. It looks like you tried to square with *2 instead of ^2. Maybe you were thinking of Python, which would use **2.


TITLE: Conclude that $A^TA=I$, so $A^{-1}=A^T$.
QUESTION [2 upvotes]: A matrix $A$ is said to be orthogonal if for every vector $\textbf{x}$ we have
$||A\textbf{x}||_2=||\textbf{x}||_2$.
Show that such a matrix will necessarily have the property that
$\left<A\textbf{x},A\textbf{y}\right>=\left<\textbf{x},\textbf{y}\right>$.
From here, conclude that $A^TA=I$, so $A^{-1}=A^T$.
$\textbf{Hint:}$ Make use of the identity $\left<\textbf{x},\textbf{y}\right>=\frac{1}{4}||\textbf{x}+\textbf{y}||^2_2 - \frac{1}{4}||\textbf{x}-\textbf{y}||^2_2$.

I am not sure how to even begin this question. If someone could at the least provide definitions of the notation. How do you read the first line? Is it read, "The norm of the matrix A with vector x in $\mathbb{R}^2$"? Then what does the hint say? Why is there a 2 as a superscript and subscript?

REPLY [2 votes]: The 2 in the superscript means squared. The 2 in the subscript means the L-2 norm, aka $\Vert x \Vert_2=\sqrt {\sum_{i=1}^n x_i^2}$. Usually the subscript is dropped when you know it’s L-2 because it’s most common. Other norms include 1 norm, infinity norm, etc. Notice that after squaring the norm there’s no overarching square root.
So
$$
\begin{split}
(Ax)^T(Ay)
 &= 1/4(Ax+Ay)^T(Ax+Ay)-1/4(Ax-Ay)^T(Ax-Ay)\\
 &= 1/4(A(x+y))^T(A(x+y))-1/4(A(x-y))^T(A(x-y)) \\
 &= 1/4(x+y)^T(x+y)-1/4(x-y)^T(x-y) \\
 &= x^Ty
\end{split}
$$
Using the notation you’re given,
$$
\begin{split}
\langle Ax, Ay\rangle
 &= 1/4\Vert Ax+Ay\Vert ^2-1/4\Vert Ax-Ay \Vert^2 \\
 &= 1/4\Vert A(x+y)\Vert^2 -1/4\Vert A(x-y) \Vert^2 \\
 &= 1/4\Vert x+y\Vert^2 -1/4\Vert x-y\Vert^2 \\
 &= \langle x, y\rangle \ldots
\end{split}
$$
For the second part, You showed already that $(Ax)^T(Ay)=x^Ty$. But $(Ax)^T(Ay)=x^TA^TAy$. Looking at both these, It’s sandwiched in between, so it must be that $A^TA=I$.
Because $AA^{-1}=I$, A^T is the inverse. Learn the moves involving matrix distribution, inner products, norms squared, and transposing a matrix-vector product and you’re good to go.


TITLE: Let $f$ be a periodic holomorphic function in some neighborhood of $\infty$ , prove $f$ is a constant.
QUESTION [0 upvotes]: I have trouble understanding the part about in some neighborhood of $\infty$.
To my understanding, $f(z) = f(z+n)$ for some large enough $z$. If so, then naively I want to say that by taking a closed bounded neighborhood of $\infty$, the due to the holomorphicity, $f$ has to attain its maximum on the boundary. Say it is $f(z*)$, then $z*+n$, will be also in the neighborhood, and by maximum modulus principle, $f$ must be constant.
However, I don’t believe that’s how neighborhood of $\infty$ works.
To prove $f$ is constant, I only know $0$ derivative or maximum modulus principle. Can anyone correct me on my misconception about $\infty$ neighborhood, and point me a strategy on solving this problem? Thanks.

REPLY [1 votes]: EDIT: Actually the claim is not true as $f(z) = e^z$ satisfies the hypotheses (it is entire and $f(z+2\pi i) = f(z)$) and is not constant.
Below we only prove that $f$ may be considered entire.

Fix $R>0$ and $\{ z \in \mathbb{C} \mid |z| > R\}$ a neighborhood of $\infty$ where $f$ is defined. Being periodic means that there exist some complex number $\tau$ such that $f(z+\tau) = f(z)$ if $|z|>R$ and $|z+\tau|>R$.
Now let $w\in \mathbb{C}$. There exists a natural $n$ such that $|w+n\tau|>R$. Hence we define a function $g\colon \mathbb{C} \to \mathbb{C}$ by $g(w)= f(w+n\tau)$.
It follows that $g$ is well defined by the periodicity of $f$, it extends $f$ tho the whole $\mathbb{C}$ and $g$ is entire.

To give an affirmative answer we need $f$ to be doubly-periodic i.e to have $\tau_1$ and $\tau_2$ such that $$f(z + \tau_1) = f(z+\tau_2) = f(z)$$
in this case the values of $f$ are determined in the parallelogram with vertices $\{0, \tau_1, \tau_2, \tau_1+\tau_2\}$. In particular $f$ is bounded.


TITLE: Changing index of summation. $\sum_{y=1}^\infty (1-\theta)^{y-1}\theta$
QUESTION [0 upvotes]: I am always confused how to change the index of summation.
$$\sum_{y=1}^\infty (1-\theta)^{y-1}\theta$$
The above is supposed to be a geometric sum and sum up to $1-(1-\theta)^x$? But how?

REPLY [2 votes]: $$\sum_{y=1}^x(1-\theta)^{y-1}\theta$$
more precisely
$$\sum_{y=1}^{y=x}(1-\theta)^{y-1}\theta \tag{1}$$
Now the equation of lower bound
$$y=1 $$
can be transformed to
$$y-1=0$$
and the equation of the upper bound
$$y=x$$
to
$$y-1=x-1$$
by subtracting $1$ from the equation.
We get
$$\sum_{y-1=0}^{y-1=x-1}(1-\theta)^{y-1}\theta \tag{2}$$
In this expression we substitute $y-1$ by $y$ to get
$$\sum_{y=0}^{y=x-1}(1-\theta)^{y}\theta$$
This substitution is only correct if all old $y$ are substituted by new $y$. This is more clear if in $(2)$ you first substitute $y-1$ by $z$
$$\sum_{z=0}^{z=x-1}(1-\theta)^{z}\theta$$
check if all $y$ have disappeared and now substitute $z$ by $y$
$$\sum_{y=0}^{y=x-1}(1-\theta)^{y}\theta$$
A sligthly other way to process $(1)$.
We want to change $y-1$ to $z$ so we set
$$y-1=z$$
and from this we get
$$y=z+1$$
Substituting this in $(1)$ we get
$$\sum_{z+1=1}^{z+1=x}(1-\theta)^{(z+1)-1}\theta $$
and this can be transformed to
$$\sum_{z+1=1}^{z+1=x}(1-\theta)^{(z+1)-1}\theta $$
or better
$$\sum_{z=0}^{z=x-1}(1-\theta)^{z}\theta $$
You can change the variable name from $z$ to $y$ and get
$$\sum_{y=0}^{y=x-1}(1-\theta)^{y}\theta$$


TITLE: Check me- Limits and differentiation
QUESTION [2 upvotes]: Evaluate the following limit:
$\lim\limits_{x \to 0^{+}} [\ln{(1+x)}]^{x}$
This is what I have, and I get the "correct" answer- just want to make sure my reasoning is valid:
let $y=[\ln{(1+x)}]^{x}$
$\ln{y}=x\ln{(\ln{(1+x)})}$
$\ln{y}=x\ln{(1+\ln{(1+x)}-1)}$
Taylor expand $\ln{(1+x)}$ (the outermost one, where $x=(1+\ln{(1+x})-1)$
$\ln{y}=x(\ln{(1+x)}-1)-\frac{(\ln{(1+x)}-1)^{2}}{2} + H.O.T.$
$\ln{y}=x((x-\frac{x^{2}}{2}+H.O.T.)-1) - \frac{((x-\frac{x^{2}}{2} + H.O.T.)-1)^{2}}{2} + H.O.T.$
At this point, hopefully it is clear that everything will be multiplied out leaving
$\ln{y}=O{(x)}$
with this then
$e^{ln{y}}=e^{O{(x)}}$
$\lim\limits_{x \to 0^{+}} y=e^{O{(x)}}=1$
Anyone see a mistake I'm making or want to poke a hole in this method?

REPLY [0 votes]: You have taken the right approach and the only issue is to calculate $$L = \lim_{x \to 0^{+}}x\log(\log(1 + x))$$ which can be done very easily as follows $$\begin{aligned}L &= \lim_{x \to 0^{+}}x\log(\log(1 + x))\\
&= \lim_{x \to 0^{+}}x\log(\log(1 + x))\\
&= \lim_{x \to 0^{+}}x\log\left(x\cdot\frac{\log(1 + x)}{x}\right)\\
&= \lim_{x \to 0^{+}}x\log x + x\log \left(\frac{\log(1 + x)}{x}\right)\\
&= \lim_{x \to 0^{+}}x\log x + 0\cdot\log 1\\
&= \lim_{x \to 0^{+}}x\log x\\
&= \lim_{t \to \infty}-\frac{\log t}{t}\text{ where } t = 1/x\\
&= 0\end{aligned}$$ Here we have used two standard logarithmic limits namely $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\text{ and }\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0\text{ for any }a > 0$$ Now as you have done $\lim_{x \to 0^{+}} y = e^{L} = 1$.


TITLE: Subspaces of $\Bbb R^3$
QUESTION [0 upvotes]: Given
$U=\{(1, -1, 3)^t\}$ and $V_a=\{(x, 3x-az,z)^t\}$ for any $a\in\mathbb{R}$, how to determine $a$, such that $U \cap V_a = \{o\} \land U+V_a=\mathbb{R}^3$?
I already found out that $V_a$ always describes a plane in this 3D space, because its base is $\{(1,3,0)^t,(0,-a,1)^t\}$ ($\dim(V)=2$) and that in order to satisfy both conditions, the line determined by $U$ should not lie within this plane and the line should cross the plane in $o$.
However, I forgot much of the algebra of my senior year and don't remember how exactly I am supposed to find out if a line is contained in the plane or not. I tried to set a vector equation:
$$(1,-1,3)^t = l_1 (1,3,0)^t + l_2 (0,-a,1)^t + l_3 (0, 0, 0)$$
but that yields a solution for the $l$'s and not for my $a$.

REPLY [0 votes]: We have $V_a = \operatorname{sp} \{(1,3,0)^T, (0,-a,1)^T \}$.
Then $(1,-1,3)^T \in V_a$ iff $a = \frac{4}{3}$.
Since $\dim V_a = 2$, then $\dim (V_a + \operatorname{sp} ( U)) = 2 $ iff $(1,-1,3)^T \in V_a$.


text:
\section{Derivatives of the conditional entropy: Proof of \eqref{eq:perturbed_f:derivative1a}--\eqref{eq:perturbed_f:derivative3b}}
\label{app:entropy-derivatives}
A large part of this appendix is an adaptation of \cite{Mac:2007,KudM:2009}. We recall that $h_{t,s;\epsilon,\delta} = n^{-1}\mathbb{E}\ln \bar{\mathcal{Z}}_{t, s; \bar{\mathcal{E}}}$ where
$\bar{\mathcal{Z}}_{t, s; \bar{\mathcal{E}}}$ is the partition function associated to the hamiltonian
 \eqref{eq:Htse-abstract}. Therefore, as explained in section \eqref{sec:thermal-and-quenched}, the free entropy only depends on $(\epsilon, \delta)$ through the combination \eqref{eq:E-epsilon-bar},
 with an explicit dependence as well as (possibly) an implicit one through the choice of $\underline \sfx$.
 To alleviate the notations in this appendix we  drop the subscripts $t,s;\bar{\mathcal{E}}$ in the Gibbs brackets.

\subsection{Proof of \eqref{eq:perturbed_f:derivative1a}}
\label{app:entropy-derivatives:1a}
Let $\sH_{t,s; \bar{\mathcal{E}}}^{\sim i}(\underline{\sigma}, \underline{\tilde{J}}, \underline{\bar{H}})$ be the Hamiltonian
$\sH_{t,s; \bar{\mathcal{E}}}(\underline{\sigma}, \underline{\tilde{J}}, \underline{\bar{H}})$ with $\bar{H}_i^{(t,s)}=0$.
Let ${\cal Z}_{t,s; \bar{\mathcal{E}}}^{\sim i}$ and $\< - \>_{\sim i}$ be the partition function and the Gibbs expectation
associated with $\sH_{t,s; \bar{\mathcal{E}}}^{\sim i}(\underline{\sigma}, \underline{\tilde{J}}, \underline{\bar{H}})$.
The identities

\begin{align}
\ln \biggl\{\frac{{\cal Z}_{t,s; \bar{\mathcal{E}}}}{{\cal Z}_{t,s; \bar{\mathcal{E}}}^{\sim i}} \biggr\}&= \ln \< e^{\bar{H}_i^{(t,s)} (\sigma_i-1)} \>_{\sim i}, \nonumber\\
e^{\bar{H}_i^{(t,s)} (\sigma_i-1)} & = \frac{1 + \sigma_i\tanh \bar{H}_i^{(t,s)} }{1+\tanh \bar{H}_i^{(t,s)}},
\label{eq:entropy-derivatives1}
\end{align}

imply

\begin{align}
h_{t,s; \epsilon,\delta}
	& = \frac{1}{n} \mathbb{E} \ln {\cal Z}_{t,s; \bar{\mathcal{E}}}^{\sim i}
	+ \frac{1}{n} \mathbb{E}  \ln \biggl\{\frac{1+\< \sigma_i \>_{\sim i}\tanh \bar{H}_i^{(t,s)}}{1+\tanh \bar{H}_i^{(t,s)}}\biggr\}.  \label{eq:entropy-derivatives2}
\end{align}

As $\tanh \bar{H}_i^{(t,s)}$ and $\< \sigma_i \>_{\sim i}$ equal either $0$ or $1$, \eqref{eq:entropy-derivatives2} simplifies to

\begin{align}
h_{t,s; \epsilon,\delta}
	& = \frac{1}{n} \mathbb{E} \ln {\cal Z}_{t,s; \bar{\mathcal{E}}}^{\sim i} - \frac{1}{n} \mathbb{E} [ \bar{\epsilon}_i^{(t,s)} ] \ln 2 \left ( 1 - \mathbb{E} \< \sigma_i \>_{\sim i} \right ).
\label{eq:entropy-derivatives2a}
\end{align}

Therefore, we have

\begin{align}
\frac{d}{d\mathbb{E}[\bar{\epsilon}^{(t,s)}]} h_{t,s; \epsilon, \delta}
	& = \sum_{i=1}^{n} \frac{d}{d\mathbb{E}[\bar{\epsilon}_i^{(t,s)}]} h_{t,s; \bar{\mathcal{E}}}\biggr\vert_{ \mathbb{E}[\bar{\epsilon}_1^{(t,s)}]=\dots=\mathbb{E}[\bar{\epsilon}_n^{(t,s)}]= \mathbb{E}[\bar{\epsilon}^{(t,s)}]}
	= -\frac{\ln 2}{n} \sum_{i=1}^{n} \left ( 1 - \mathbb{E} \< \sigma_i \>_{\sim i} \right ), \label{1_25}
\end{align}

which is the first equality in \eqref{eq:perturbed_f:derivative1a}.

To obtain the second equality, simply notice that as $1 - \< \sigma_i \>_{t,s;\epsilon, \delta} = 0$ when $\bar{H}_i^{(t,s)}=+\infty$
(which happens with probability $\bar{\epsilon}_i^{(t,s)}$). Performing the expectation over $\bar{H}_i^{(t,s)}$ in the following expression we get
\begin{align}
1-\mathbb{E}\<\sigma_i\>_{t,s;\epsilon,\delta}
	& = \mathbb{E} [ 1 - \<\sigma_i\>_{t,s;\epsilon,\delta} ] \nonumber \\
	& = \mathbb{E} [ (1-\bar{\epsilon}_i^{(t,s)})(1-\mathbb{E}\<\sigma_i\>_{\sim i}) + \bar{\epsilon}_i^{(t,s)} (1 - \mathbb{E}\<\sigma_i\>_{\bar{H}_i^{(t,s)}=\infty}) ] \nonumber \\
	& = (1-\mathbb{E}[\bar{\epsilon}^{(t,s)}])(1-\mathbb{E}\<\sigma_i\>_{\sim i}). \label{eq:recover-original-gibbs}
\end{align}
Replacing in \eqref{1_25} yields the second equality in \eqref{eq:perturbed_f:derivative1a}.

\subsection{Proof of \eqref{eq:perturbed_f:derivative1b}}
\label{app:entropy-derivatives:1b}
Let $\sH_{t,s; \bar{\mathcal{E}}}^{\sim i,j}(\underline{\sigma}, \underline{\tilde{J}}, \underline{\bar{H}})$ be the Hamiltonian
$\sH_{t,s; \bar{\mathcal{E}}}(\underline{\sigma}, \underline{\tilde{J}}, \underline{\bar{H}})$ with $\bar{H}_i^{(t,s)}=\bar{H}_j^{(t,s)}=0$. Let ${\cal Z}_{t,s; \bar{\mathcal{E}}}^{\sim i,j}$ and $\< - \>_{\sim i,j}$ be the partition function and the Gibbs expectation associated with $\sH_{t,s; \bar{\mathcal{E}}}^{\sim i,j}(\underline{\sigma}, \underline{\tilde{J}}, \underline{\bar{H}})$.

Using again \eqref{eq:entropy-derivatives1} on the identity
\begin{align*}
\ln \biggl\{\frac{{\cal Z}_{t,s;\epsilon,\delta}}{{\cal Z}_{t,s; \bar{\mathcal{E}}}^{\sim i,j}} \biggr\}
	& = \ln \<e^{\bar{H}_i^{(t,s)}(\sigma_i-1) + \bar{H}_j^{(t,s)}(\sigma_j-1)} \>_{\sim i, j},
\end{align*}
we have
\begin{align}
h_{t,s; \bar{\mathcal{E}}}
	& = \frac{1}{n} \mathbb{E} \ln {\cal Z}_{t,s; \bar{\mathcal{E}}}^{\sim i,j}
\nonumber\\
&\qquad+ \frac{1}{n} \mathbb{E}  \ln \biggl\{\frac{1 + \< \sigma_i \>_{\sim i,j}\tanh \bar{H}_i^{(t,s)}  + \< \sigma_j \>_{\sim i,j}\tanh \bar{H}_j^{(t,s)}
+ \< \sigma_i \sigma_j \>_{\sim i,j}\tanh \bar{H}_i^{(t,s)} \tanh \bar{H}_j^{(t,s)} }{1 + \tanh \bar{H}_i^{(t,s)} + \tanh \bar{H}_j^{(t,s)} + \tanh \bar{H}_i^{(t,s)} \tanh \bar{H}_j^{(t,s)}}\biggr\} \nonumber \\
	& = \frac{1}{n} \mathbb{E} \ln {\cal Z}_{t,s; \underline{\epsilon}}^{\sim i,j}
+ \frac{\mathbb{E}[\bar{\epsilon}_i^{(t,s)} ] \mathbb{E}[\bar{\epsilon}_j^{(t,s)} ]}{n} \mathbb{E} \ln \biggl\{\frac{1 + \< \sigma_i \>_{\sim i,j} + \< \sigma_j \>_{\sim i,j} + \< \sigma_i \sigma_j \>_{\sim i,j}}{4}\biggr\}
\nonumber \\
	& \qquad + \frac{\mathbb{E}[\bar{\epsilon}_i^{(t,s)} ] (1-\mathbb{E}[\bar{\epsilon}_j^{(t,s)} ])}{n} \mathbb{E} \ln \biggl\{\frac{1 + \< \sigma_i \>_{\sim i,j}}{2}\biggr\}
	+ \frac{(1-\mathbb{E}[\bar{\epsilon}_i^{(t,s)} ]) \mathbb{E}[\bar{\epsilon}_j^{(t,s)} ]}{n} \mathbb{E} \ln \biggl\{\frac{1 + \< \sigma_j \>_{\sim i,j}}{2}\biggr\},
\label{eq:entropy-derivatives3}
\end{align}
where \eqref{eq:entropy-derivatives3} follows from taking the expectation over $\bar{H}_i^{(t,s)}$ and $\bar{H}_j^{(t,s)}$. From \eqref{eq:entropy-derivatives2a}
one can deduce that $\frac{d^2}{d \mathbb{E}[\bar{\epsilon}_i^{(t,s)}]^2} h_{t,s; \epsilon,\delta} = 0$. Therefore
\begin{align*}
  \frac{d^2}{d \mathbb{E}[\bar{\epsilon}^{(t,s)}]^2} h_{t,s; \epsilon, \delta}
	& =  \sum_{i,j=1}^{n} \frac{d^2}{d \mathbb{E}[\bar{\epsilon}_j^{(t,s)}] d \mathbb{E}[\bar{\epsilon}_i^{(t,s)}]} h_{t,s; \bar{\mathcal{E}}}\biggr\vert_{\epsilon_1=\dots=\epsilon_n=\epsilon}
	\nonumber \\ &
	=  \sum_{i \neq j} \frac{d^2}{d\mathbb{E}[\bar{\epsilon}_j^{(t,s)}] d\mathbb{E}[\bar{\epsilon}_i^{(t,s)}]}
	h_{t,s; \bar{\mathcal{E}}}\biggr\vert_{\mathbb{E}[\bar{\epsilon}_i^{(t,s)}]=\dots=\mathbb{E}[\bar{\epsilon}_n^{(t,s)}] = \mathbb{E}[\bar{\epsilon}^{(t,s)}]}.
\end{align*}
The derivatives $\frac{d^2}{d \mathbb{E}[\bar{\epsilon}_j^{(t,s)}] d \mathbb{E}[\bar{\epsilon}_i^{(t,s)}]} h_{t,s; \epsilon,\delta}$
can be readily obtained from \eqref{eq:entropy-derivatives3}. This provides
\begin{align}
\frac{d^2}{d \mathbb{E}[\bar{\epsilon}^{(t,s)}]^2} h_{t,s; \epsilon, \delta}
	& = \frac{1}{n} \sum_{i \neq j}
	\mathbb{E}\ln \biggl\{\frac{1 + \< \sigma_i \>_{\sim i,j} + \< \sigma_j \>_{\sim i,j} + \< \sigma_i \sigma_j \>_{\sim i,j}}{1 + \< \sigma_i \>_{\sim i,j}
	+ \< \sigma_j \>_{\sim i,j} + \< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j}}\biggr\} . \label{eq:entropy-derivatives6a}
\end{align}

We now simplify each term in the sum \eqref{eq:entropy-derivatives6a}.
Given that $\< \sigma_S \>_{\sim i,j}$ equals either $0$ or $1$ for any subsets $S \subset \{ 1 \dots n\}$, one can verify that the numerator
and denominator of \eqref{eq:entropy-derivatives6a} can be written as

\begin{align}
	& \hspace{-0.5cm} \ln \bigl( 1 + \< \sigma_i \>_{\sim i,j} + \< \sigma_j \>_{\sim i,j} + \< \sigma_i \sigma_j \>_{\sim i,j} \bigr)
	=  \bigl(\< \sigma_i \>_{\sim i,j} + \< \sigma_j \>_{\sim i,j} + \< \sigma_i \sigma_j \>_{\sim i,j}\bigr) \ln 2
	\nonumber \\
	& +\bigl(\< \sigma_i \>_{\sim i,j}\< \sigma_j \>_{\sim i,j} + \< \sigma_i \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j} + \< \sigma_j \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j}\bigr)
	\bigl(\ln 3 - 2\ln 2\bigr)
	\nonumber \\
	&
	+  \< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j} \bigl(5\ln2 - 3\ln 3\bigr),
	\label{eq:entropy-derivatives7}
\end{align}

and

\begin{align*}
\ln \bigl( 1 + \< \sigma_i \>_{\sim i,j} + \< \sigma_j \>_{\sim i,j} + \< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j} \bigr)
	& =  \bigl(\< \sigma_i \>_{\sim i,j} + \< \sigma_j \>_{\sim i,j}\bigr) \ln 2.
\end{align*}

Special cases of the Nishimori identities \eqref{eq:Nishimori},

\begin{align*}
\mathbb{E}[\< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j}]
	& = \mathbb{E}[\< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j}], \\
\mathbb{E}[\< \sigma_i \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j}]
	& = \mathbb{E}[\< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j}], \\
\mathbb{E}[\< \sigma_j \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j}]
	& = \mathbb{E}[\< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j} \< \sigma_i \sigma_j \>_{\sim i,j}],
\end{align*}

can now be used to simplify \eqref{eq:entropy-derivatives7} so that each term in the sum \eqref{eq:entropy-derivatives6a} becomes

\begin{align}
\ln (2) \, \mathbb{E} [\< \sigma_i \sigma_j \>_{\sim i,j} - \< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j} ].
\label{eq:entropy-derivatives8}
\end{align}

Moreover, as $\< \sigma_i \sigma_j \> - \< \sigma_i \> \< \sigma_j \> = 0$ when $\bar{H}_i^{(t,s)}$ and/or $\bar{H}_j^{(t,s)}$ equal $+\infty$, we obtain

\begin{align}
\mathbb{E} [\< \sigma_i \sigma_j \> - \< \sigma_i \> \< \sigma_j \> ]
	& = (1- \mathbb{E}[\bar{\epsilon}_i^{(t,s)}]) (1-\mathbb{E}[\bar{\epsilon}_j^{(t,s)}]) \mathbb{E} [\< \sigma_i \sigma_j \>_{\sim i,j} - \< \sigma_i \>_{\sim i,j} \< \sigma_j \>_{\sim i,j} ].
	\label{eq:entropy-derivatives9}
\end{align}
Finally, from \eqref{eq:entropy-derivatives6a}, \eqref{eq:entropy-derivatives8} and \eqref{eq:entropy-derivatives9} we obtain \eqref{eq:perturbed_f:derivative1b}.

\subsection{Derivation of \eqref{eq:perturbed_f:derivative4a} and \eqref{eq:perturbed_f:derivative4b}}
The derivation of \eqref{eq:perturbed_f:derivative4a} is the same as Sec.~\ref{app:entropy-derivatives:1a} except that the steps should be done on $\tilde{H}_i$ instead of $\bar{H}_i^{(t,s)}$. The derivation of \eqref{eq:perturbed_f:derivative4a} is the same as Sec.~\ref{app:entropy-derivatives:1b} except that the steps should be done on $\tilde{H}_i, \tilde{H}_j$ instead of $\bar{H}_i^{(t,s)}, \bar{H}_j^{(t,s)}$.

\subsection{Proof of \eqref{eq:perturbed_f:derivative3a} and \eqref{eq:perturbed_f:derivative3b}}
For $\underline{\sfx}(\epsilon)$ independent of $\delta$, from \eqref{eq:E-epsilon-bar} we have
\begin{align*}
\frac{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]}{d\delta} = \frac{1}{n^{\theta}}(1-\epsilon) e^{-\frac{K}{RT}(s \tilde{x}^{t}
+ \sum_{t'=1}^{t-1} \tilde{x}^{t'})} = \frac{1-\mathbb{E} [\bar{\epsilon}^{(t,s)}]}{n^{\theta}-\delta}
\quad\quad{\rm and}\quad\quad\frac{d^2 \mathbb{E} [\bar{\epsilon}^{(t,s)}]}{d\delta^2} = 0.
\end{align*}
Together with \eqref{eq:perturbed_f:derivative1a} and \eqref{eq:perturbed_f:derivative1b} we can immediately derive
\begin{align*}
\frac{d}{d \delta} h_{t,s; \epsilon, \delta}
	& = \frac{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]}{d\delta} \frac{d h_{t,s; \epsilon, \delta} }{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]}
	\- = - \frac{\ln 2}{n^{1+\theta}(1-\delta/n^{\theta})} \sum_{i=1}^n (1 - \mathbb{E}\<\sigma_i\>_{t,s;\epsilon,\delta}) \\
\frac{d^2}{d \delta^2} h_{t,s; \epsilon}
	& = \frac{d}{d \delta} \bigg( \frac{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]}{d\delta} \frac{d h_{t,s; \epsilon, \delta} }{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]} \bigg)
	\- = \bigg( \frac{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]}{d\delta} \bigg)^2 \frac{d^2 h_{t,s; \epsilon,\delta}}{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]^2} + \frac{d^2 \mathbb{E} [\bar{\epsilon}^{(t,s)}]}{d\delta^2} \frac{d h_{t,s; \epsilon, \delta}}{d \mathbb{E} [\bar{\epsilon}^{(t,s)}]} \\
	& = \frac{\ln 2}{n^{1+2\theta}(1-\delta/n^{\theta})^2} \sum_{i \neq j} \mathbb{E} [\< \sigma_i \sigma_j \>_{t,s;\epsilon,\delta}
- \< \sigma_i \>_{t,s;\epsilon,\delta} \< \sigma_j \>_{t, s;\epsilon,\delta}].
\end{align*}
The first equality of \eqref{eq:perturbed_f:derivative3a} follows from applying the same argument in \eqref{eq:recover-original-gibbs} to $\tilde{H}_i$.

\subsection{Proof of \eqref{eq:Q1:delta-der}}
We rearrange \eqref{eq:perturbed_f:derivative3a} to obtain
\begin{align}
\frac{1}{n} \sum_{i=1}^{n} \mathbb{E}\<\sigma_i\>_{t,s;\epsilon,\delta} = \frac{n^{\theta}(1-\delta/n^{\theta})}{\ln 2} \frac{d}{d\delta} h_{t,s;\epsilon,\delta}.
\end{align}
Then using \eqref{eq:perturbed_f:derivative3a} and \eqref{eq:perturbed_f:derivative3b} we have
\begin{align*}
	& \frac{d}{d\delta} \big ( \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}\<\sigma_i\>_{t,s;\epsilon,\delta} \big )
	\- = \frac{n^{\theta}(1-\delta/n^{\theta})}{\ln 2} \frac{d^2}{d\delta^2} h_{t,s;\epsilon,\delta} - \frac{1}{\ln 2} \frac{d}{d\delta} h_{t,s;\epsilon,\delta} \\
	& = \frac{1}{n^{1+\theta}(1-\delta/n^{\theta})} \big ( \sum_{i \neq j} \mathbb{E} [\< \sigma_i \sigma_j \>_{t,s;\epsilon,\delta}
- \< \sigma_i \>_{t,s;\epsilon,\delta} \< \sigma_j \>_{t, s;\epsilon,\delta}] - \sum_{i=1}^{n} (1 - \mathbb{E}\<\sigma_i\>_{t,s;\epsilon,\delta}) \big ) \\
	& = \frac{1}{n^{1+\theta}(1-\delta/n^{\theta})} \sum_{i, j=1}^{n} \mathbb{E} [\< \sigma_i \sigma_j \>_{t,s;\epsilon,\delta}
- \< \sigma_i \>_{t,s;\epsilon,\delta} \< \sigma_j \>_{t, s;\epsilon,\delta}]
\end{align*}
where the last equality uses one of the Nishimori identities \eqref{eq:Nishimori}, namely $\mathbb{E} \< \sigma_i \> = \mathbb{E} [ \< \sigma_i \>^2 ]$.


TITLE: $ \sum\limits_{n=1}^\infty \dfrac{(1+i)^n}{n^2}$ is divergent, and no idea about $\sum\limits_{n=1}^\infty \dfrac{(3+4i)^n}{5^n\,\sqrt[999]{n}}$
QUESTION [1 upvotes]: How can one see that $ \sum\limits_{n=1}^\infty \dfrac{(1+i)^n}{n^2}$ is divergent, and by which criterion? I was using a binomial theorem for $ (1+i)^n $ as $ \sum\limits_{k=0}^n \dbinom{n}{k} i^n$, but it seems to get very complicated and not clear for me.
Question 2:
What is the result in this case: $\sum\limits_{n=1}^\infty \dfrac{(3+4i)^n}{5^n\,\sqrt[999]{n}}$?
The root criterion gives me 1 in this case...

REPLY [0 votes]: This is for your second sum:
$$|3+4i|=5$$
$$\arg(3+4i)=\arctan(4/3)=:\alpha \approx 1$$
$$\sum\limits_{n=1}^\infty \frac{(3+4i)^n}{5^n\,\sqrt[999]{n}} =
\sum\limits_{n=1}^\infty \frac{\exp(\alpha n i)}{\sqrt[999]{n}}$$
$$\frac{1}{\sqrt[999]{n}} \downarrow 0$$
I'm pretty much certain that this sum will converge by something like Leibniz' alternating series test, but this is not exactly an alternating series so I can't say 100%.


TITLE: If $\kappa$ is weakly inaccessible and $A\subset\kappa$, can $L[A]$ violate $\kappa^{\lt\kappa}=\kappa$?
QUESTION [10 upvotes]: In some current work, my co-authors and I had wanted in a certain
argument to appeal to $\kappa^{\lt\kappa}=\kappa$ in $L[A]$, in a
situation where $A\subset\kappa$ and $\kappa$ was weakly
inaccessible in $V$, but $\kappa$ was below the continuum in $V$
(and so $\kappa^{\lt\kappa}\neq\kappa$ in $V$). But we have lost
confidence in this statement. Is it consistent that there is a
counterexample?
Question. Is it (relatively) consistent that $\kappa$ is
weakly inaccessible, but there is $A\subset\kappa$ with
$L[A]\models\kappa^{\lt\kappa}>\kappa$?
The model $L[A]$ here is the relative constructible universe, defined as in the constructible universe, but with a predicate for the set $A$.
We had at first thought a condensation argument might show
$\kappa^{\lt\kappa}=\kappa$ in every $L[A]$, but upon looking at it closely, I
can't quite make it work. Condensation works fine above $\kappa$, but I don't see why, for example, every real number in $L[A]$ must be added by a stage before $\kappa$. It would be interesting to me even to resolve the question of how big $2^\omega$ must be or can be in comparison with $\kappa$ in $L[A]$, where $A\subset\kappa$ and $\kappa$ is a regular limit cardinal in $V$.
(Meanwhile, we've saved our application by finding another route
for our argument that does not rely on this issue.)

REPLY [12 votes]: By Lemma 2.2. of my paper Shelah's strong covering property and CH in V[r], we can show that:

Claim. If $A\subseteq \kappa,$ and if $Y\in L[A]$ is a bounded subset of $\kappa,$ then there exists a proper initial segment $A'$ of $A$ such that $Y\in L[A'].$

From this, $\kappa^{<\kappa}=\kappa$ in $L[A].$


TITLE: Matrix multiplied by itself n times equals identity
QUESTION [3 upvotes]: Given a $A \in \mathbb{R}^{100\times 100}$ and $A^{6} = I_{100}$ and $A^{14} = I_{100}$, ist say $A^{2} = I_{100}$ too?
And what would then be the relationship governing whether $A^{2} = I_{100}$ or even $A^{n}$  is valid given two different conditions as above?

REPLY [3 votes]: $\mathbb Z\to \mathbb R^{100\times 100}$, $n\mapsto A^n$ is a group homomorphism and its kernel must be a subgroup of $\mathbb Z$ and it contains $14$ and $6$. The smallest such subgroup is $2\mathbb Z$.


TITLE: Prove that if two matrices commute iff their inverse also commute
QUESTION [1 upvotes]: Let A, B be two invertible matrices
es; prove that A and B
commute if and only if $A^{−1}$ and $B^{−1}$ Commute.
Any suggestions?

REPLY [2 votes]: Let $a,b$ be elements of an arbitrary group $G$. Then $ab=ba$ is equivalent to $a^{-1}b^{-1}=b^{-1}a^{-1}$, because it is equivalent to $(ab)^{-1}=(ba)^{-1}$, and
$(ab)^{-1}=b^{-1}a^{-1}$, $(ba)^{-1}=a^{-1}b^{-1}$.
Here we have the special case $G=GL_n(K)$, where $K$ probably is meant to be a field. It could be more general of course, e.g., $G=GL_n(\mathbb{Z})$.

REPLY [1 votes]: Assuming you know that $$(AB)^{-1}=B^{-1}A^{-1},$$ then you get by interchanging the roles of $A,B$ that also $$(BA)^{-1}=A^{-1}B^{-1}.$$
Now the left hand sides of these equations are equal if and only if their right hand sides are equal. And inverses of two matrices are equal if and only if the matrices themselves are equal.


TITLE: How to linearize z <-> x == y?
QUESTION [1 upvotes]: I have a constraint that reads
$z \iff x = y$
where $z$ is a 0-1 variable and $x,y$ are non-zero, positive integer variables. I'm managed to formulate equivalence going in the right direction, but not in the left. Does anyone have any idea here?

REPLY [2 votes]: Here's a straightforward big-M approach.  Introduce binary variables $u$ and $v$, and impose linear constraints
\begin{align}
-M(1-z) \le x - y &\le M(1-z) \tag1 \\
x - y + 1 &\le M (1-u) \tag2 \\
y - x + 1 &\le M (1-v) \tag3 \\
1 - z &\le u + v \tag4
\end{align}
Constraint $(1)$ enforces $z \implies x=y$.
Constraint $(2)$ enforces $u \implies x<y$.
Constraint $(3)$ enforces $v \implies x>y$.
Constraint $(4)$ enforces $\lnot z \implies (u \lor v)$.

A stronger formulation is:
\begin{align}
-Mu + 0z + 1v \le x - y &\le -1u + 0z + Mv \tag5 \\
u + z + v &= 1 \tag6
\end{align}
The three cases are:

$u \implies -M \le x - y \le -1$, that is $x < y$.
$z \implies 0 \le x - y \le 0$, that is $x = y$.
$v \implies 1 \le x - y \le M$, that is $x > y$.


text:
\begin{document}

\title[Inverse mean curvature flow in hyperbolic space]{Inverse mean curvature flows in the hyperbolic 3-space revisited}
\author{Pei-Ken Hung and Mu-Tao Wang}
\begin{abstract}
This note revisits the inverse mean curvature flow in the 3-dimensional hyperbolic space. In particular, we show that the limiting shape is not necessarily round after scaling, thus resolving an inconsistency in the literature.
\end{abstract}

\address{Department of Mathematics \\ Columbia University \\ 2990 Broadway \\ New York, NY 10027}
\thanks{ The second author was supported by the National Science Foundation under grant DMS-1105483 and DMS-1405152. The authors would like to thank Andre Neves for his interest in this work. }

\maketitle

\section{Introduction}
 Let $(\mathbb{H}^3,\bar{g})$ be the 3-dimensional hyperbolic space with the metric \[\bar{g}=dr^2+(\sinh^2 r)\sigma\] in the $(r, \theta)$ coordinates, where $\sigma=\sigma_{ij}d\theta^i d\theta^j$ is the standard metric on $S^2$. We show that there exists a star-shaped mean convex 2-surface $\Sigma_0$ in  $\mathbb{H}^3$ such that the inverse mean curvature flow with  $\Sigma_0$ as the initial surface does not converge to a round sphere (of constant curvature) after scaling. Such an example on an asymptotic hyperbolic 3-space with positive mass was constructed
 by Neves \cite{Neves} to demonstrate the impossibility of proving the hyperbolic Penrose inequality by the method of the inverse mean curvature
 flow. However, in \cite[Theorem 1.2 and 6.11]{Gerhardt2} it was claimed that the inverse mean curvature behaves better on the hyperbolic 3-space and deforms the induced metric on the surface to a round one after scaling. We show by a concrete example in this note that the claim does not hold true.  More precisely, we prove:
 \begin{theorem}
 There exists a star-shaped mean convex closed surface $\Sigma_0$ in $\mathbb{H}^3$  that has the following property. Let $\Sigma_t$ be the inverse mean curvature
 flow of $\Sigma_0$, and $|\Sigma_t|$ and $g_t$ be the area of $\Sigma_t$ and the induced metric on $\Sigma_t$, respectively. As $t\rightarrow \infty$, $|\Sigma_t|^{-1}g_t$ converges to a metric on $S^2$ that is not of constant curvature.
 \end{theorem}

 The construction in \cite{Neves} relies on the positivity of the limit of the Hawking mass. However, in the hyperbolic case, the Hawking mass always limits
 to zero along the inverse mean curvature flow. We introduce a new geometric quantity (modified Hawking mass) to handle this degenerate phenomenon.
 A new monotonicity formula for the inverse mean curvature flow is discovered along the way.

The result in this paper affirms that the limiting behavior of the inverse mean curvature flow
depends distinctively on the structure of the end at infinity. For an (asymptotically) Euclidean end, the limiting shape is always round and this fact plays a critical role in the inverse mean curvature flow proof of both the Riemannian Penrose inequality \cite{Huisken-Ilmanen} and quermassintegral inequalities \cite{Guan-Li}. In the asymptotically hyperbolic case, it is already demonstrated by Neves \cite{Neves} that the limiting shape is not necessarily round. We show that even in the hyperbolic case when the Hawking mass degenerates, this phenomenon persists. A new strategy relying on the conformal structure and the Sobolev type inequality
on the sphere at infinity was devised in \cite{Brendle-Hung-Wang, Brendle-Wang}  to tackle this difficulty and to obtain Penrose-Gibbon inequalities in the asymptotically flat case. Such a strategy has been further developed by several authors \cite{Ge-Wang-Wu, Lopes-de-Lima-Girao} to solve related problems.

\section{A modified Hawking mass and its monotonicity along the inverse mean curvature flow}
We recall the Hawking mass for a closed embedded surface $\Sigma$ in $\mathbb{H}^3$:
\[m_{H}(\Sigma)=\sqrt{\frac{|\Sigma|}{16\pi}}\left(1-\frac{1}{16\pi}\int_\Sigma (H^2-4)d\mu\right).\]
By the Gauss equation in $\mathbb{H}^3$, we have $H^2-4=4K+2|\mathring{A}|^2$, where $K$ is the Gauss curvature of $\Sigma$ and $\mathring{A}$ is the traceless part of the second fundamental form $A$ with $|A|^2=|\mathring{A}|^2+\frac{1}{2} H^2$. Therefore, one rewrites
\begin{equation}\label{traceless}
 1-\frac{1}{16\pi}\int_\Sigma (H^2-4)d\mu=-\frac{1}{8\pi}\int_\Sigma |\mathring{A}|^2d\mu.
 \end{equation}

 In this article, we consider a modified quantity
 \begin{equation}
 \tilde{m}(\Sigma)=-|\Sigma| \int_\Sigma |\mathring{A}|^2d\mu.
 \end{equation}
 Though the scaling of $\tilde{m}(\Sigma)$ is no longer the same as a mass, we call it the modified Hawking mass in this note.

Consider the inverse mean curvature flow $\Sigma_t$ of a closed embedded surface $\Sigma_0$, which deforms the surface in the normal direction with speed $\frac{1}{H}$. In the following, we compute the evolution equation of the modified Hawking mass $\tilde{m}(\Sigma_t)$. The evolution equation for mean curvature is
\[  \frac{\partial H}{\partial t}=\frac{\Delta H}{H^2}-2\frac{|\nabla H|^2}{H^3}-\frac{|A|^2}{H}+\frac{2}{H}. \]

We compute
\begin{align*}
\frac{\partial H^2}{\partial t}&=2\frac{\Delta H}{H}-4\frac{|\nabla H|^2}{H^2}-2|A|^2+4\\
                               &=2(\Delta \log H)-2\frac{|\nabla H|^2}{H^2}-2(H^2-4)+4K.
\end{align*}
Taking into account of $\frac{\partial}{\partial t} d\mu_t =d\mu_t$ and integrating by parts, we obtain
\begin{align*}
\frac{d}{dt}\left(1-\frac{1}{16\pi}\int_{\Sigma_t} (H^2-4)d\mu_t\right)&=-\left(1-\frac{1}{16\pi}\int_{\Sigma_t} (H^2-4)d\mu_t\right)+\frac{1}{8\pi}\int_{\Sigma_t}\frac{|\nabla H|^2}{H^2}d\mu_t.
\end{align*}

In view of \eqref{traceless},
\begin{align*}
\frac{d}{dt} \int_{\Sigma_t} |\mathring{A}|^2d\mu_t &=-\int_{\Sigma_t} |\mathring{A}|^2d\mu_t -\int_{\Sigma_t}\frac{|\nabla H|^2}{H^2}d\mu_t.
\end{align*}

We thus obtain the following proposition:
\begin{proposition} Along an inverse mean curvature flow $\Sigma_t$, the modified Hawking mass $\tilde{m}(\Sigma_t)$ satisfies the following evolution equation:
\[\frac{d}{dt} \tilde{m}(\Sigma_t) = |\Sigma_t| \int_{\Sigma_t}\frac{|\nabla H|^2}{H^2}d\mu_t.\]
\end{proposition}

\section{The construction}

  We consider a family of star-shaped surfaces $\tilde{\Sigma}_s \subset \mathbb{H}^3$ that are described in the $(r, \theta)$ coordinates by
\begin{equation}\label{family_surface} \tilde{\Sigma}_s=\{ (\tilde{r}(\theta, s ),\theta):\ \theta\in S^{2} \} \end{equation}
for a smooth function $\tilde{r}(\theta, s)$ on $S^{2}\times \mathbb{R}^+$.  We assume $\tilde{r}(\theta, s)$ is of the following asymptotics as $s\rightarrow \infty$:
\begin{equation}\label{asymptotic} \tilde{r}(\theta, s)=c s+ f(\theta)+o(1)\end{equation} where $c$ is a constant and $f$ is a function on $S^2$.

Consider a new function
\[ \varphi(\theta, s):=-\int_{\tilde{r}(\theta, s)}^\infty\frac{dx}{\sinh(x)}\] such that $D_i \varphi=(\sinh^{-1} \tilde{r}) D_i\tilde{r}$.
Let $\varphi_i=D_i\varphi$ and $\varphi_{ij}=D_j D_i\varphi$ be  covariant derivatives with respect to the round metric $\sigma$. Let $|\cdot |_{\sigma}$ be the norm corresponding to $\sigma$.  Let
 \[g_{ij}=\sinh^2 r(\sigma_{ij}+\varphi_i\varphi_j)\]
  be the induced metric of $\tilde{\Sigma}_s$ and $h_{ij}$ be the second fundamental form of $\tilde{\Sigma}_s$.
Denote $v=\sqrt{1+|D\varphi|_{\sigma}^2}$ and the area form of $g_{ij}$  is \[\sqrt{\det g}=\sinh^2\tilde{r} \sqrt{1+|D\varphi|_{\sigma}^2}\sqrt{ \det \sigma}=(\sinh^2\tilde{r}) v \sqrt{ \det \sigma}. \]

$h_j^i=g^{ik}h_{kj}$ can be expressed in terms of $\varphi$:
\begin{equation}\label{2nd_fund} h_j^i=\frac{\cosh \tilde{r}}{v\sinh \tilde{r} }\delta_j^i-\frac{1}{v\sinh \tilde{r} }(\sigma^{ik}-\frac{\varphi^i\varphi^k}{v^2})\varphi_{kj}. \end{equation}

\begin{proposition} \label{asymptotics}Let $\tilde{\Sigma}_s$ be a family of surfaces in $\mathbb{H}^3$ that are radial graphs of the function $\tilde{r}(\theta, s)=cs+f(\theta)+o(1)$ and $g_{ij}$ be the induced metric on $\tilde{\Sigma}_s$. Then the limit of the rescaled metric $e^{-2cs} g_{ij}$ as $s\rightarrow \infty$ is round if and only if  $\lim_{s\rightarrow \infty} \tilde{m}(\tilde{\Sigma}_s)=0$.

\end{proposition}

\begin{proof}
From \eqref{2nd_fund}, we compute the traceless part $|\mathring{A}|^2=|A|^2-\frac{1}{2}H^2$:
\begin{align*}
&|\mathring{A}|^2=\frac{1}{v^2\sinh^2\tilde{r}}\left( \tilde{\sigma}^{ik}\varphi_{kj}\tilde{\sigma}^{jl}\varphi_{li}-\frac{1}{2}(\tilde{\sigma}^{ij}\varphi_{ij})^2 \right),
\end{align*} where $\tilde{\sigma}^{ik}=\sigma^{ik}-\frac{\varphi^i\varphi^k}{v^2}$ is the inverse of $\sigma_{ij}+\varphi_i\varphi_j$.

It is more convenient to express all terms in $\tilde{r}$ now. We have
\[v^2=1+(\sinh^{-2} \tilde{r}) |D \tilde{r}|^2_\sigma=1+O(e^{-2cs})\] and
\[\sigma^{ik}-\frac{\varphi^i\varphi^k}{v^2}=\sigma^{ik}-\frac{\tilde{r}^i\tilde{r}^k}{ \sinh^2 \tilde{r}+|D \tilde{r}|^2_\sigma}=\sigma^{ik}+O(e^{-2cs}).\]
On the other hand, we compute \[\varphi_{ij}=(\sinh^{-1} \tilde{r}) B_{ij}\] where
\[B_{ij}=\tilde{r}_{ij}-\frac{\cosh \tilde{r}}{\sinh \tilde{r} }\tilde{r}_i\tilde{r}_j =f_{ij}-f_if_j+o(1)\] as $s\rightarrow \infty$.

Therefore,
\begin{align*}
&\lim_{s\to\infty}|\tilde{\Sigma}_s|\int_{\tilde{\Sigma}_s}|\mathring{A}|^2 d\mu_s\\
&=\lim_{s\to\infty}\left(\int_{S^2}(\sinh^2 \tilde{r}) d\mu_{\sigma}\right)\left( \int_{S^2} (\sinh^{-2} \tilde{r}) |\mathring{B}_{ij}|_{\sigma}^2d\mu_{\sigma}  \right)\\
&=\lim_{s\to\infty} (\int_{S^2}\frac{e^{2f}}{4}d\mu_{\sigma})(\int_{S^2}\frac{4}{e^{2f}}|\mathring{B}_{ij}|^2_{\sigma}d\mu_{\sigma})\\
&=\int_{S^2}e^{2f} d\mu_{\sigma}\int_{S^2}|\mathring{D}^2 e^{-f}|^2_{\sigma}d\mu_{\sigma}.
\end{align*}
In particular, $\lim_{s\rightarrow \infty} \tilde{m}(\tilde{\Sigma}_s)\leq 0$ and the equality holds if and only if $e^{-f}$ is a linear combination of constants and first eigenfunctions of $S^2$. On the other hand, the limit of the rescaled induced metric is
\[\lim_{s\to \infty} e^{-2cs} g_{ij}=\lim_{s\to \infty} e^{-2cs} (\sinh^2\tilde{r}) \sigma_{ij}=e^{2f} \sigma_{ij}\]
In the view of the following lemma, the proof is complete.
\end{proof}

\begin{lemma} Let $\sigma$ be the standard round metric on $S^{n-1}$.
The conformal metric $e^{2f}\sigma$ has constant section curvature if and only if $e^{-f}$ is a linear combination of constants and first eigenfunctions of $S^{n-1}$.
\end{lemma}
\begin{proof}\label{conf_factor}
From \cite[Chapter 6 Lemma 1.2]{Schoen-Yau} the conformal group of $S^{n-1}$ is $SO(n,1)$. By direct computation, for each $\phi$ in the conformal group we have
\[ \phi^*\sigma=e^{2u}\sigma,\ \ e^{-u}=a_0+\sum_{i=1}^n a_iX^i \]
where $a_0,\ \dots ,\ a_n$ are some constants and $X^i, i=1\cdots n$ are the first $n$ eigenfunctions of $S^{n-1}$. On the other hand, if $e^{2f}\sigma$ has constant sectional curvature, it can be realized as $\phi^*\sigma$ for a conformal diffeomorphism $\phi$ \cite[Theorem 6.1.2]{Petersen}. The assertion follows.
\end{proof}

\subsection{Proof of Theorem 1}
Pick a function $\bar{f}$ on $S^2$ such that
\[\int_{S^2}e^{2\bar{f}}d\mu_{\sigma}\int_{S^2}|\mathring{D}^2 e^{-\bar{f}}|^2_{\sigma}d\mu_{\sigma}= c_0> 0\] where $\mathring{D}^2 e^{-\bar{f}}$ is the traceless
part of the Hessian of $e^{-\bar{f}}$.
Let $\tilde{\Sigma}_s$ be the family of surface described by $r=s+\bar{f}(\theta)$ in the $(r, \theta)$ coordinates. From the analysis in the last section, we have
\begin{equation}\label{limit}\lim_{s\to\infty}\tilde{m}(\tilde{\Sigma}_s)=-c_0.\end{equation}

Consider the inverse mean curvature flow $\Sigma_t^s$ with $\tilde{\Sigma}_s$ as the initial surface, where $t$ is the flow parameter.
Neves \cite[Theorem 3.1]{Neves} proved that for $s$ large enough, the following estimates hold for $\Sigma_t^s$:
\begin{equation}\label{pinching}\begin{split} |\tilde{\Sigma}_s||H^2-4|&\leq Ce^{-t}\\
 |\tilde{\Sigma}_s|^{3}|\nabla A|^2&\leq Ce^{-3t} \end{split}\end{equation}
where the constant $C$ does not depends on $s$.
Therefore,
\begin{equation}\label{monotone} \frac{d}{dt}\tilde{m}(\Sigma^s_t)=|\Sigma^s_t|\int_{\Sigma^s_t}\frac{|\nabla H|^2}{H^2}d\mu^s_t \leq C|\tilde{\Sigma}_s|^{-1}e^{-t},\end{equation} where  $|\Sigma_t^s|=
|\tilde{\Sigma}_s| e^t$ is used.
Pick $s_0$ large enough such that:

(1) $\tilde{\Sigma}_{s_0}$ is mean-convex.

(2) $\tilde{m}(\tilde{\Sigma}_{s_0})<-\frac{c_0}{2}$, which is possible by \eqref{limit}.

(3) $C|{\tilde{\Sigma}}_{s_0}|^{-1}\leq \frac{c_0}{4}$.

Let $\Sigma_t$ be the inverse mean curvature flow with $\tilde{\Sigma}_{s_0}$ as the initial surface, \eqref{monotone} implies  \[\lim_{t\rightarrow \infty} \tilde{m}(\Sigma_t)< -\frac{c_0}{4}<0.\]

On the other hand, by the analysis in \cite{Ding, Gerhardt2} we know that starting from a mean-convex star-shaped surface, the solution of inverse mean curvature flow $\Sigma_t$ exists for all time and the surface $\Sigma_t$ is given as the graph of  \[\tilde{r}(\theta ,t)=\frac{t}{2}+f(\theta) +o(1)\] for a smooth function $f(\theta)$ on $S^2$ as $t\rightarrow \infty$. We can apply Proposition \ref{asymptotics} to conclude that
$\Sigma_t$ does not converge to a round sphere after scaling.

\qed

\subsection{Inverse mean curvature flows in the ball model}
We consider the ball model of the hyperbolic metric
\[\frac{1}{(1-\frac{1}{4} \rho^2)^2}(d\rho^2+\rho^2 \sigma)\] which can be turned into the form
\[ dr^2+(\sinh^2r) \sigma\] by the change of variable \[\rho=2-\frac{4}{e^r+1}.\]

Therefore, a solution of the inverse mean curvature flow defined by the radial function $r=\tilde{r}(\theta, t)=\frac{t}{2}+f(\theta)+o(1)$ in the $(r, \theta)$ coordinates
is the same the family of surfaced defined by $\rho=u(\theta, t)$ in the $(\rho, \theta)$ coordinates, where
\[u(\theta, t)=2-\frac{4}{e^{\tilde{r}(\theta, t)}+1}.\]
In particular,
\[\lim_{t\rightarrow \infty} (u-2) e^{\frac{t}{2}}=-4 e^{-f(\theta)}.\] Our result indicates that $f$ does not have to be a linear combination of constants and
first eigenfunctions of $S^2$. This should be compared with the claim in  \cite[Theorem 1.2 and 6.11]{Gerhardt2}.

\section{The higher dimensional case}
In this section, we show that the same conclusion holds in higher dimensions, i.e. there exists a star-shaped mean convex hypersurface $\Sigma_0$ in the $n$ dimensional hyperbolic space $\mathbb{H}^n$ for $n\geq 4$ such that the inverse mean curvature flow with $\Sigma_0$ as the initial data does not converge to a round sphere (of constant sectional curvature) after scaling.
Let $(\mathbb{H}^n,\bar{g})$ be the hyperbolic space with the metric \[\bar{g}=dr^2+(\sinh^2 r)\sigma,\] where $\sigma$ is the standard metric on $S^{n-1}$.
For a closed hypersurface $\Sigma$ in $\mathbb{H}^n$, we consider the quantity:
\[ Q(\Sigma):=|\Sigma|^{-\frac{n-5}{n-1}}\int_{\Sigma}|\mathring{A}|^2 d\mu,\] where $\mathring{A}$ is the traceless part of the second fundamental form $A$. We have the following proposition analogous to Proposition  \ref{asymptotics}

\begin{proposition}\label{asymptotics_n}
Let $\tilde{\Sigma}_s$ be a family of hypersurfaces in $\mathbb{H}^n$ that are radial graphs of the functions $\tilde{r}(\theta, s)=cs+f(\theta)+o(1)$. Then
\begin{align*}
&\lim_{s\to\infty}Q({\tilde{\Sigma}}_s)=(\int_{S^{n-1}}e^{(n-1)f}\mu_{\sigma})^{-\frac{n-5}{n-1}}\int_{S^{n-1}}e^{(n-3)f}|\mathring{D^2} e^{-f}|^2_{\sigma}\mu_{\sigma}
\end{align*}
The limit of the rescaled induced metric is $e^{2f}\sigma$. It is a round metric if and only if $\lim_{s\to\infty} Q({\tilde{\Sigma}_s})=0$.
\end{proposition}
Now we turn to the inverse mean curvature flow. We need the following lemma concerning the evolution of $|\mathring{A}|^2$.
\begin{lemma} \label{evolution_of_A}The following equation holds along the inverse mean curvature flow of a hypersurface in $\mathbb{H}^n$:
\[
\frac{\partial |\mathring{A}|^2}{\partial t}=2\nabla^j\left(\frac{\nabla_iH}{H^2} \right)\mathring{h}^i_j-\frac{4}
{n-1}|\mathring{A}|^2-2\frac{\mathring{h}^k_i\mathring{h}^j_k\mathring{h}^i_j}{H},
\] where $\mathring{h}^i_j=h^i_j-\frac{1}{n-1}H\delta^i_j$.
\end{lemma}
\begin{proof} We compute
\begin{align*}
\frac{\partial h_i^j}{\partial t}
&= \frac{1}{H^2} \, \nabla^j \nabla_i H - 2 \, \frac{\nabla_i H \, \nabla^j H}{H^3} - \frac{h_i^kh_k^j}{H} - \frac{1}{H} \, g^{mj} \, R_{\nu i\nu m} \\
&=\nabla^j\left(\frac{\nabla_iH}{H^2} \right)-\frac{1}{H}\left(\mathring{h}^k_i+\frac{H}{n-1}\delta^k_i\right)\left(\mathring{h}^j_k+\frac{H}{n-1}\delta^j_k\right)-\frac{1}{H}g^{mj}(-g_{im})\\
&=\nabla^j\left(\frac{\nabla_iH}{H^2} \right)-\frac{2}{n-1}\mathring{h}^j_i-\frac{\mathring{h}^k_i\mathring{h}^j_k}{H}+\left(\frac{1}{H}-\frac{H}{(n-1)^2}\right)\delta_i^j.
\end{align*}
Therefore,
\begin{align*}
\frac{\partial \mathring{h}_i^j}{\partial t}=\nabla^j\left(\frac{\nabla_iH}{H^2} \right)-\frac{2}{n-1}\mathring{h}^j_i-\frac{\mathring{h}^k_i\mathring{h}^j_k}{H}+\left(\frac{1}{H}-\frac{H}{(n-1)^2}-\frac{1}{n-1}\frac{\partial H}{\partial t}\right)\delta_i^j.
\end{align*}
Contracting the above equation with $\mathring{h}^i_j$ gives the desired formula.\end{proof}

 We know that starting from a mean-convex star-shaped surface, the solution of inverse mean curvature flow exists for all time and $\tilde{r}(.,t)-\frac{t}{n}$ converges to a smooth function $f$.

The construction is similar to the $n=3$ case. Pick a function $\bar{f}$ on $S^{n-1}$ such that
\[\left(\int_{S^{n-1}}e^{(n-1)\bar{f}}\mu_{\sigma}\right)^{-\frac{n-5}{n-1}}\int_{S^{n-1}}e^{(n-3)\bar{f}}|\mathring{D^2} e^{-\bar{f}}|^2_{\sigma}\mu_{\sigma}=c_0> 0\]
Consider $\tilde{\Sigma}_s=\{(s+\bar{f}(\theta),\theta)\}$ and let $\Sigma^s_t$ be the solution of the inverse mean curvature flow with initial data $\tilde{\Sigma}_s$. The pinching estimate \eqref{pinching} can be generalized to higher dimensions, and for  $s$ large enough, we have the following estimate on $\Sigma^s_t$:
\begin{equation}\label{pinching2}\begin{split} |\tilde{\Sigma}_s|^{\frac{4}{n-1}}\left(|H-(n-1)|^2+|\mathring{A}|^2\right)&\leq Ce^{-\frac{4t}{n-1}}\\
 |\tilde{\Sigma}_s|^{\frac{6}{n-1}}|\nabla A|^2&\leq Ce^{-\frac{6t}{n-1}}\end{split}, \end{equation}
where $C$ is a constant independent of $s$.

With this pinching estimate and Lemma \ref{evolution_of_A}, we deduce:
\begin{align*}
\frac{d}{dt}\int_{\Sigma^s_t}|\mathring{A}|^2 d\mu^s_t&=\frac{n-5}{n-1}\int_{\Sigma^s_t}|\mathring{A}|^2 d\mu^s_t+\int_{\Sigma^s_t}\left(-2\frac{\nabla_iH}{H^2} \nabla^j\mathring{h}^i_j-2\frac{\mathring{h}^k_i\mathring{h}^j_k\mathring{h}^i_j}{H} \right)d\mu^s_t\\
&\geq\frac{n-5}{n-1}\int_{\Sigma^s_t}|\mathring{A}|^2 d\mu^s_t-\tilde{C}|\tilde{\Sigma}_s|^{\frac{n-7}{n-1}}e^{\frac{n-7}{n-5}t},
\end{align*} where $\tilde{C}$ is independent of $s$.
Thus $\frac{d}{dt}Q(\Sigma^s_t)\geq -\tilde{C}|\tilde{\Sigma}_s|^{\frac{-2}{n-1}}e^{\frac{-2t}{n-1}}$.

Pick $s_0$ large enough such that:

(1) $\tilde{\Sigma}_{s_0}$ is mean-convex.

(2) $Q(\tilde{\Sigma}_{s_0}) > c_0/2$.

(3) $-\tilde{C}|\tilde{\Sigma}_{s_0}|^{\frac{-2}{n-1}}>-\frac{1}{2(n-1)}c_0$.

(4) The pinching estimate \eqref{pinching2} holds on $\Sigma^{s_0}_t$.

Denote by $\Sigma_t$ the inverse mean curvature flow with $\tilde{\Sigma}_{s_0}$ as the initial data. The flow exists for all time and for $t$ large, $\Sigma_t$ is given as the graph of $\frac{t}{n-1}+f(\theta)+o(1)$ for another smooth function $f$ on $S^{n-1}$.

It follows from the above conditions on $s_0$ that $Q(\Sigma_0)>\frac{c_0}{2}$ and $\frac{d}{dt}Q(\Sigma_t)\geq -\frac{1}{2(n-1)}e^{\frac{-2t}{n-1}}c_0$. Therefore $\lim_{t\rightarrow \infty} Q(\Sigma_t)> c_0/4>0$. From Proposition \ref{asymptotics_n} the limit of the rescaled induced metric on $\Sigma_t$ is not a round one.


TITLE: Let $\{E_k\}$ be a sequence of measurable sets in X. Show that $A\in\mathfrak{M}$, where $A$ is the set of all x that lie in infinitely many $E_k$'s.
QUESTION [1 upvotes]: The problem arises from Walter Rudin's Real & Complex Analysis:

The highlighted texts, however, would NOT make any sense if $A$ is not measurable. All my attempts to show that $A\in\mathfrak{M}$ have failed; any hints would be greatly appreciated.

REPLY [1 votes]: $$
A =\bigcap_{N=1}^\infty \bigcup_{n\ge N} E_k
$$
By definition, this means any point of $A$ belongs to infinitely many of $E_k$. By the fact that $E_k$ are all measurable, you conclude $A$ is as well.


TITLE: Fock-space-related combinatorial problem v2
QUESTION [3 upvotes]: This problem comes from counting degrees of freedom in a system of bosons.
The integers $k_n$, $n=1..m$ are allowed to take values on the interval $-\Lambda \leq k_n\leq\Lambda,\,k_n\neq0$. I need to calculate the total number of $m$-tuples of $k_n$'s whose elements sum to zero:
$$
\underbrace{
\{k_1,k_2,\ldots,k_m\}:
\quad
k_1+k_2+\ldots+k_m = 0}_{\text{How many of those?}}
$$
Actually, even an asymptotic estimate would work for me. Can this be reduced to some well-known problem?
My idea is that we could probably shift everything by a constant and use the number of integer partitions?..
UPDATE
I forgot to mention that the order DOES matter in this problem ($m$-tuples differing only by order count as different ones), physically those correspond to different momentum modes.

REPLY [3 votes]: Let $X_1,X_2,\dots,X_m$ be iid random variables uniformly distributed over the nonzero integers in $[-\Lambda,\Lambda]$, and let $S_m=X_1+\dots+X_m$. You want the probability that $S_m=0$, times $(2\Lambda)^m$.
As long as $m$ is large enough, then by the central limit theorem, $S_m/\sqrt{m} \approx N(0,\sigma^2)$, where $\sigma^2$ is the variance of $X_1$. Therefore, letting $\Phi(x)$ be the standard normal cdf, where $\Phi'(x)=(2\pi)^{-1/2}\exp(-x^2/2)$,  we have
\begin{align}
P(S_m=0)
  &=P\left(-\frac{1/2}{\sigma\sqrt m}<\frac{S_m}{\sigma\sqrt{m}}<\frac{1/2}{\sigma\sqrt m}\right)
\\&\approx\Phi\left(\frac{1/2}{\sigma\sqrt m}\right)-\Phi\left(\frac{1/2}{\sigma\sqrt m}\right)
\\&\approx\frac1{\sigma\sqrt{m}}\cdot \Phi'(0)
\\&=\frac1{\sigma\sqrt{2\pi m}}
\end{align}
Now, what is $\sigma$? Since $X_1$ is approximately a discrete uniform on the interval $[-\Lambda,\Lambda]$, which has variance $((\Lambda-(-\Lambda)+1)^2-1)/12$ (see Wikipedia: discrete uniform), it will be true that $\sigma\approx \Lambda/\sqrt{3}$. Therefore,
$$
\text{# of solutions}\approx \frac1{\Lambda \sqrt{\frac23 \pi m}}\cdot (2\Lambda )^m
$$
Furthermore, you can quantify the error in this approximation. The error in the approximating $S_m$ by a normal distribution is described by the Edgeworth series. Fortunately, since $X_1$ is symmetric and therefore has zero skewness, the error in this approximation is $O(1/m)$, so
$$
\text{# of solutions}=(2\Lambda )^m\left(\frac1{\Lambda \sqrt{\frac23 \pi m}}+O(1/m)\right)
$$

You can also given an exact answer to this problem in terms of a double sum of binomial coefficients. This may be useful if you want to verify the quality of the previous asymptotic answer.
Let us first count tuples without the constraint $k_n\neq 0$. If you add the quantity $\Lambda+1$ to each entry, you get a tuple of positive integers between $1$ and $2\Lambda+1$ summing to $m(\Lambda+1)$. Referring to here, the number of solutions is
$$
\sum_{i=0}^m (-1)^i\binom{m}i\binom{m(\Lambda+1)-(2\Lambda+1)i -1 }{m-1}
$$
Here, we use the convention that $\binom{r}k=0$ when $r$ is negative.
Now, to add back in the constraint $k_i\neq 0$, we use the principle of inclusion exclusion. Take all the solutions counted by the last equation, subtract the ones where some variable is zero, add back in the doubly subtract solutions, etc. The result is
$$
\sum_{k=0}^m (-1)^{m-k}\binom{m}k
\sum_{i=0}^k (-1)^i\binom{k}i
\binom{k(\Lambda+1)-(2\Lambda+1)i -1 }{k-1}
$$


TITLE: Coboundary operators, 1-cocycles and computing cohomology
QUESTION [0 upvotes]: My question is about the compatibility and consistency between two definitions of cohomology in two books.
I asked this question about 10 days ago on MathSE
and I set a bounty on it, but I didn't receive an answer.
I was reading cohomology from Neukirch's book, and there he referenced Hall's book. The two approaches are almost the same (are they not?), and they should give us the same results (cohomology groups). But I can not see their compatibility and consistency, and I can not recover them from each other. My problem is that: I can not propose the compatibility between these two books.

Hall's book:
15.7. Cochains, Coboundaries, and Cohomology Groups.
Given a double $\Omega$-modulo $A$ we define $C^n = C^n(A, \Omega)$ to be the additive group of all functions $f$ of $n$ variables which range independently over $\Omega$ and taking values in $A$, subject to the condition
$$ (15.7.1) \ \ \ \ \ \ \ \ \ \ \ \ \ \
f(\xi_1, \dots, \xi_n) = 0,$$
whenever at least one of the $\xi_i = 1$. The elements of $C_n$ are called
$n$-dimensional cochains. $C^0 = A$ by definition and a zero dimensional
cochain is simply any element of A.
The coboundary operator $\delta$ maps each $C_n$ into the next, $C_{n+1}$ in
accordance with the rule
$$(\delta f)(\xi_0, \xi_1, \dots, \xi_n)\\
=\xi_0f(\xi_1, \dots, \xi_n) \\
+\sum_{t=1}^{n} (-1)^t (\xi_0, \xi_1, \dots, \xi_{t-2}, \xi_{t-1}\xi_{t}, \xi_{t+1}, \dots, \xi_{n}) \\
+f(\xi_0\xi_1, \dots, \xi_{n-1})\xi_n . \textit{ something that I cannot read},
$$
And let $Z^n=Z^n(A, \Omega)= \ker (C^n\longrightarrow C^{n+1})$, and let $B^n=B^n(A, \Omega)= Im (C^{n-1}\longrightarrow C^{n})$. The quotient group $Z^n/B^n$ is called the $n$-dimensional cohomology group of the double $\Omega$-module $A$. We write it $$H^n(A, \Omega) = Z^n/B^n.$$
In the part something that I can not read, in the above formula, probably it is written $(-1)^{n+1}$, but I am not sure.

I have some confusions about coboundary operators and 1-cocycles and computing cohomology, and I can not find where the bugs are.
In Neukirch's Class field theory, page 13, does the last summand of $d_q(\sigma_1, \dots, \sigma_q)$ written truly? Or should it be equal to
$$d_q(\sigma_1, \dots, \sigma_q)=
\sigma_1(\sigma_2, \dots, \sigma_{q})
+
\sum_{i=1}^{q-1} (-1)^i (\sigma_1, \dots, \sigma_{i-1}, \sigma_{i}\sigma_{i+1}, \sigma_{i+2}, \dots, \sigma_{q})
+
(-1)^q(\sigma_1, \dots, \sigma_{q-1})\sigma_q ?$$
I've seen something closely similar to these relations in Hall's book (The theory of groups), but I think at least one of them should have a typo. I know that they are not exactly the same, but I think they are not compatible unless for instance we replace the definition of Neukirch as above.
Also I have ambiguity about cocycles and 1-cocycles. If the things in Neukirch's book are correct, then I expected the cocycle should be $f(\sigma\tau)=\sigma f(\tau)+f(\sigma)$, and I am confused with $\delta(f(\sigma, \tau))=\sigma f(\tau)-f(\sigma\tau)+f(\sigma)\tau$, and I can not see the compatibility.

REPLY [4 votes]: Let me just put my comments as an answer: (i) To compute the cohomology groups $H^n(G,A)$, for $A$ a (left) $G$-module and where
$$H^n(G,A)=\text{Ext}^n_{{\mathbb Z}G}({\mathbb Z},A)$$
one needs a projective resolution of the $G$-module ${\mathbb Z}$, say
$$\cdots\stackrel{d_3}\rightarrow P_2\stackrel{d_2}\rightarrow P_1\stackrel{d_0}\rightarrow P_0\stackrel{\varepsilon}\rightarrow {\mathbb Z}\rightarrow 0$$
such that
$$H^n(G,A)=\text{Ext}^n_{{\mathbb Z}G}(
{\mathbb Z},A)=\ker(d^*_{n+1})/\text{Im}(d_n^*).$$
(ii) One choice of projective resolution of ${\mathbb Z}$ is given by the free ${\mathbb Z}$-modules $P_n$ with basis
the $(n+1)$-tuples $(g_0,\ldots,g_n)$ of elements of the group $G$ and where the action of $G$ is by (left) translation: $g(g_0,\ldots,g_n)=(gg_0,\ldots,gg_n)$. It follows that the $P_n$ are free $G$-modules with basis the $(n+1)$-tuples of the form $(1,g_1,\ldots,g_n)$. The $G$-morphisms
$$d_n:P_n\rightarrow P_{n-1}$$
are defined on the generators by the formula
$$d_n(g_0,\ldots,g_n)=\sum_{j=0}^n(-1)^n(g_0,\ldots,\widehat{g_j},\ldots,g_n)$$
where $\widehat{g_j}$ means that this term is omitted.
The morphism $\varepsilon$ sends a generator $(g_0)$ to $1$. One shows that the $(P_n,d_n)$  form indeed an exact sequence and upon appliying the functor $\text{Hom}_{G}(-,A)$ one obtains a complex whose cohomology groups are $H^n(G,A)$. All of these is pretty standard fare in any homological algebra textbook. The important point is now to observe that an $n$-cochain $f\in \text{Hom}_G(P_n,A)$ is determined by its values in the generators $(g_0,\ldots,g_n)$ of $P_n$ and since the cochain is a $G$-module morphism (we say that it is $G$-covariant), then
$$f(g_0,g_1,\ldots,g_n)=f(g_0(g_0^{-1}g_1,\ldots,g_0^{-1}g_n))=g_0f(1,g_0^{-1}g_1,\ldots,g_0^{-1}g_n)$$
i.e., $f$ is determined by its values on the generators of $P_n$ of the form $(1,g_1,\ldots,g_n)$ and then one may also consider another projective resolution of ${\mathbb Z}$ given by the free $G$-modules $Q_n$  with generators $[g_1,\ldots,g_n]$ (and where $Q_0$ is the free $G$-module generated by a symbol $[\;]$. One then shows that $P_n\simeq Q_n$ as $G$-modules and defining
$$\delta_n:Q_n\rightarrow Q_{n-1}$$
by
$$\delta_n[g_1,\ldots,g_n]=g_1[g_2,\ldots,g_n]+\sum_{j=1}^{n-1}(-1)^j[g_1,\ldots,g_jg_{j+1},\ldots,g_n]+(-1)^n[g_1,\ldots,g_{n-1}]$$
the $(Q_n,\delta_n)$ form a free $G$-resolution of ${\mathbb Z}$ and there is an isomorphism of chain complexes $\varphi:(P_n,d_n)\rightarrow (Q_n,\delta_n)$, answering your question. Let just add some details: First, the morphisms $\varphi_n:P_n\rightarrow Q_n$ are given by $\varphi_n(g_0,\ldots,g_n)=g_0[g_0^{-1}g_1,g_1^{-1}g_2,\ldots, g_{n-1}^{-1}g_n]$ and the corresponding $\psi_n:Q_n\rightarrow P_n$ are $\psi_n[g_1,\ldots,g_n]=(1,g_1,g_1g_2,g_1g_2g_3,\ldots, g_1g_2\cdots g_n)$. A lenghty, but straightforward, computation shows that $\delta_n=\varphi_{n-1}\circ d_n\circ \psi_n$.


TITLE: Find number of $n$-tuples of naturals greater than 1 such that their product is at most $K$.
QUESTION [1 upvotes]: I have to find the number of different solutions to: $$\begin{align}0<\prod_{i=1}^nk_i\leqslant K\in\Bbb N,\ k_i \in \Bbb N^{+}\setminus\{1\}&\  \forall i\in\{1,\ldots,n\},\end{align}$$
It seems trivial to enumerate this in a recursive manner as ${k_i}_\max=\frac{K}{2^{n-1}}$. However, I would like to get the number of different solutions as a function of $K$ and $n$? Any hints or suggestions appreciated.

REPLY [1 votes]: With all the useful comments from above, I devised an answer. The number of n
-tuples of naturals greater than 1 such that their product is at most K equals to:
\begin{equation}
f(K, n) = \sum_{k=2}^{\lfloor \frac{K}{2^{n-1}}\rfloor} f(\lfloor \frac{K}{k} \rfloor, n - 1)
\end{equation}
\begin{equation}
f(K, 0) = 1
\end{equation}
Similarly, in the previous question: Number of $n$-tuples with bounded product in which naturals greater or equal 1 are considered the count is obtained recursively as well. This corresponds to divisor summatory function: https://en.wikipedia.org/wiki/Divisor_summatory_function. However, in this case naturals in tuples are greater than one.


TITLE: Find a function $f: \mathbb{R} \to \mathbb{R}$ that is continuous at precisely one point?
QUESTION [24 upvotes]: Is the function $f(x) = \lim_{k \to +\infty} \tan{kx}$ a right example?
But I do not know whether this is a function at first. Is this a function?
Could you give a correct real-valued function that is continuous at precisely one point?
Thanks.

REPLY [0 votes]: Your example is an expression that defines a function, but the domain of said function is not the real line, unless you allow extended real numbers as the output; but, by the criteria of your problem statement, you do not allow that.If you use the extended real line as a codomain, then the formula that you gave describes a function that maps the entire real line onto the three point set $\{-\infty,0,\infty\}$.  In this case, the function would be continuous at every real number except $0$.
However, at the outset, your problem statement requires that the codomain and range both be the real line.  This is incompatible with the above discussion, because the real line includes neither $\infty$ nor $-\infty$.


TITLE: Property of angle bisectors in a triangle
QUESTION [1 upvotes]: Let $ABC$ be a triangle having circumcenter $O$. Suppose $AH$ is the altitude from vertex $A$ and $AT$ bisects angle $A$. I would like a simple geometric proof that $AT$ also bisects angle $OAH$.

REPLY [4 votes]: Without loss of generality, assume that $\angle B \geq \angle C$. Now suppose that $\angle B \leq \pi / 2$. Drop a perpendicular from $O$ to $D$ on $\overline{AC}$.

By the Inscribed Angle Theorem, $\angle AOC = 2 \angle B$, so $\angle AOD = \angle B$. Therefore, $\angle BAH$ and $\angle OAD$ are congruent complements of $\angle B$, whereupon
$$\angle HAT = \angle BAT - \angle BAH = \angle CAT - \angle OAD = \angle OAT$$ so that $\overline{AT}$ bisects $\angle OAH$.
Now suppose that $\angle B > \pi / 2$.

We have $\angle AOC + 2\angle B = 2\pi$, whence $\angle AOD = \pi - \angle B$. Therefore, $\angle BAH$ and $\angle OAD$ are congruent complements of $\angle AOD$, whereupon $$\angle HAT = \angle BAT + \angle BAH= \angle CAT + \angle OAD = \angle OAT.$$


TITLE: Is $Pr(A \cap C) = Pr(B \cap C)$ given that $Pr(A) = Pr(B)$?
QUESTION [0 upvotes]: Suppose that $Pr(A) = Pr(B)$. For any event $C$, is it hold that $Pr(A \cap C) = Pr(B \cap C)$? When $A$ and $C$ are independent and $B$ and $C$ are independent the proof is easy but, is it hold in the general case?

REPLY [5 votes]: Does this diagram answer your question?


TITLE: $\sum a_{n}^{2}$ converges $\Rightarrow\sum \frac{a_{n}}{n}$ converges
QUESTION [2 upvotes]: $\sum a_{n}^{2}$ converges $\Rightarrow\sum \frac{a_{n}}{n}$ converges.
any hints?

REPLY [4 votes]: If these are sequences of real numbers, note that, for any $a,b\in \mathbb{R}$ we have
$$ 0 \leq (a-b)^2, $$
and so
$$ ab \leq \frac{1}{2}(a^2 + b^2). $$
Can you pick $a$ and $b$ so that this gives
$$ \frac{a_n}{n} \leq \text{stuff} $$
where $\sum \text{stuff}$ is a convergent series?


TITLE: Unitarity of Quantum Mechanical Systems
QUESTION [2 upvotes]: I was reading this lecture notes "Black holes from A to Z" by Andrew Strominger. In the first chapter Introduction the following statement is made:

"If we know the present, there are laws that determine the future and can be used to predict the past. This is true in Classical Physics. It is also true in Quantum Physics, where we determine the state, instead of classical variables, in the future or the past. We refer to this as the Unitarity of Quantum Mechanical systems."

I haven't understood how unitarity is related to the system being deterministic? How unitarity is related to time evolution of the system?
The link to the lecture notes is "https://scholar.harvard.edu/files/yshi/files/physics_211r_-_black_holes_final.pdf"

REPLY [1 votes]: In Quantum Mechanics if you know the state of the system at a given time, then the quantum equations of motion tells you what the state is for any future as well as past time.
The state of the system at instant $t_0$ is the vector $|\psi(t_0)\rangle$ and we postulate that it evolves in time through a linear operator $U(t-t_0)$ (the time evolution operator),
$$|\psi(t)\rangle = U(t-t_0)|\psi(t_0)\rangle.$$
You can think of $U(t-t_0)$ as a matrix depending on the parameter $t$. If you know the state of the system at any instant $t_0$, then you act with that matrix on the vector state and you obtain the vector state at time $t$. Totally deterministic!
Notice that this operator being unitary is not necessary to a deterministic time evolution of a vector $|\psi\rangle$. In fact, what makes sure that time evolution is deterministic is the existence of a time evolution operator satisfying the above equation. The fact that $U$ is unitary is necessary in QM in order that the relation between states is preserved though. For instance, if $\langle\psi(0)|\phi(0)\rangle=0$, then $\langle\psi(t)|\phi(t)\rangle=\langle\psi(0)U(t)^\dagger U(t)|\phi(0)\rangle=\langle\psi(0)|\phi(0)\rangle=0$.


TITLE: Integral domain and ascending chain condition proof
QUESTION [2 upvotes]: Show that an integral domain $A$ is a principal ideal domain if every ideal $I$ of $A$ is principal, that is, of the form $I=(a)$. Show directly that the ideals in a PID satisfy the a.c.c.

REPLY [2 votes]: Let $I_1 \subseteq I_2 \subseteq \ldots$ be an infinite ascending chain of ideals. Now, note that $I := \bigcup_{k \ge 1} I_k$ is again an ideal, so there is some $a \in I$ such that $I = (a)$. But by the definition of $I$, there must be some $k \ge 1$ such that $a \in I_k$. Obviously, $(a) \subseteq I_k \subseteq I = (a)$, so $I_k = (a) = I$. Furthermore, for $j \ge k$, we have $I = I_k \subseteq I_j \subseteq I$, so for all $j \ge k$, $I_j = I_k$.
Edit: This seems answer the question (cf. comments).


TITLE: Optics viewing angle and distance
QUESTION [1 upvotes]: Almost all optics like microscope, telescope.. require observer putting their eyes very close to the optics to see.
Can the same optics designed so observer can see like viewing TV screen, i.e. from a larger distance like 30cm from optics? Or is it impossible?

REPLY [0 votes]: There are a few laws of optics that come into play here.  Probably the most significant one is simply this: if you step back 30 cm from the exit aperture, the aperture now subtends a very small angular field of view. Since all the light coming from the telescope must pass through that aperture, it can't take up a large part of the space your eye can see! (There are additional optics laws -- look up terms like "exit pupil" and "angular magnification" -- which explain why moving even a cm away makes the image look tiny inside the eyepiece)
Now, by comparison, if you look at the optics of a projector, that lens is designed to produce an image at some distance.  You, the viewer, can look at the entire image on the screen because it takes up a large part of your field of view.  But if you were standing at the image screen, looking at the projector, all you would be able to see is that small piece of the image which hits your eyeball.


TITLE: Does this situation indicate that $g(x)>f(x)$?
QUESTION [1 upvotes]: I have two functions $f(x)$ and $g(x)$ for $0<x<5$. By calculating the first and second derivatives I see that both functions are increasing and concave (not strictly concave), i.e. $f''<0$ and $g''<0$. Then If I have $$f(0)=g(0)=0$$ and $$f(5)=10\quad ,\quad g(5)=11.$$ Then, does this prove that $g$ is always grater than $f$?

REPLY [0 votes]: No. Take $f(x)=10\sqrt[3]{x/5}$ and $g(x)=11\sqrt{x/5}$.
You get something like this

You can easily come up with the same type of plot if $f''$ and $g''$ are positive


TITLE: The composition of soft curves is not always soft.
QUESTION [2 upvotes]: An exercise I found in one of my text books:
(In my book a curve that is "soft by pieces" is a curve $\gamma :[a,b]\subset \mathbb{R}\to\mathbb{R}^n$ such that there is a partition $P=\{a=t_0<t_1<...t_k=b\}$ of $[a,b]$ such that $\gamma$ has a continuous derivative on each $[t_{i-1},t_i]$ for $i \in \{1,2,...k\}$; I believe "soft by pieces" may not be a universal term to refer to such curves, this is a direct translation of the term used in the book, which is in Spanish)
''If $\gamma :[a,b]\subset \mathbb{R}\to\mathbb{R}^n$ is $\textit{soft by pieces}$ and $\alpha:[c,d]\subset\mathbb{R}\to[a,b]\subset\mathbb{R}$, surjective on $[a,b]$ and $C^1$ on $[c,d]$. Show, with an example, that $\gamma \circ \alpha:[c,d]\subset\mathbb{R}\to \mathbb{R}^2$ is not always soft by pieces"
I took $\gamma:[-1,1]\to\mathbb{R}^2$, $\gamma(t)=(t,|t|)$, I know that this function is soft by pieces (I need only to take $P=\{-1,0,1\}$, $\gamma$'s derivative is definitely continuous in $[-1,0]$ and $[0,1]$).
The problem I'm faced with lies in finding the adequate $\alpha$, at first I thought about proposing a function similar to $sin(\frac{1}{x})$ such that this function is surjective, but has problems on 0, the problem with this function is that is is not differentiable on zero. Then I thought about $x^2\sin\left(\frac{1}{x}\right)$ and $x^3\sin\left(\frac{1}{x}\right)$, however these don't work either because even though I could make them continuous on zero (by defining a function that is 0 if $x=0$ and $x^2\sin\left(\frac{1}{x}\right)$ or $x^3\sin\left(\frac{1}{x}\right)$) everywhere else) and their derivative is continuous, they are not surjectve on $[-1,1]$
I believe that my intuition about my choice if $\gamma$ is sound and the idea of choosing an $\alpha$ that "has problems on zero" could work. Still, I'm stuck and unable to find an $\alpha$ that "has problems on zero", but is surjective and has a continuous derivative.

REPLY [3 votes]: That's a fine idea. Let$$f(x)=\begin{cases}x^2\sin\left(\frac1x\right)&\text{ if }x\ne0\\0&\text{ if }x=0.\end{cases}$$Note that $\lim_{x\to\infty}f(x)=\infty$. Let $\alpha\in(0,\infty)$ be the smallest number greater than $0$ such that $f(\alpha)=1$ ($\alpha\approx1.14$). And now consider the minimum of $f(x)$ when $x\in(0,\alpha]$; let us call $\beta$ the point at which that minimum is reached ($\beta\approx0.234$) and let $\eta=f(\beta)$ ($\eta\approx-0.05$). For each $x\in[\eta,1]$, let $\gamma(x)=|x|$. Then $f$ maps $[0,\alpha]$ onto $[\eta,1]$, $f|_{[0,\alpha]}$ is a $C^1$ function, $\gamma$ is piecewise differentiable, but $\gamma\circ f|_{[0,\alpha]}$ isn't: there are infinitely many points at which it is not differentiable.


TITLE: How can I find the maximum value of $x_6-x_1$ subject to the two constraints $\sum_{j=1}^{6} x_j^2$ and $\sum_{j=1}^{6} x_j = 0$
QUESTION [3 upvotes]: I currently have six variables $x_1, x_2, x_3, x_4, x_5, x_6$. I am trying to determine how large I can make the difference $x_6-x_1$ while satisfying the constraints:
$\sum_{j=1}^{6} x_j^2 \leq 1$
$\sum_{j=1}^{6} x_j = 0$
So far I have tried to use an optimization package in R but to no avail. Is there a way I can program this out or would it be too much to handle? Thanks!

REPLY [3 votes]: Note that
$$
(x_6-x_1)^2\leq 2(x_6^2+x_1^2)\le2
$$
with all inequalities becoming equalities when $x_6=\pm1/\sqrt{2}$ and $x_1=-x_6$. Thus,
$$
\max(x_6-x_1)=\sqrt{2}
$$
which realizes when $x_1=-1/\sqrt{2}$, $x_2=x_3=x_4=x_5=0$, and $x_6=1/\sqrt{2}$.


TITLE: What is the connection between this hint and the question?
QUESTION [0 upvotes]: The picture is a question and its hint, I am not asking for solutions, but how whether a differential equation is linear can help me solve this question?
It's obvious that it is not supposed to be solved in an ordinary way.

REPLY [0 votes]: You can convert each of the equations into a characteristic equation and then have a general solution in terms of $e$ (and in some cases, $\sin x$ and $\cos x$).
The characteristic equation for (a) is $$c^5 + 4c^3 -5c = 0$$ which can be factored into $$c(c^4+4c^2-5) \rightarrow c(c^2-1)(c^2+5) \rightarrow c(c-1)(c+1)(c^2+5)$$
The characteristic equation for (b) is $$c^3 +2c^2 + c = 0$$ which can be factored into $$c(c^2 + 2c + 1) \rightarrow c(c+1)^2$$
Once you have the general solutions, you can use either undetermined coefficients or variation of parameters to find the particular solution.


TITLE: Simple divisibility problem in elementary number theory
QUESTION [0 upvotes]: Find all positive integers $n$ such that $n+2009$ divides $n^2+2009$ and $n+2010$ divides $n^2+2010$.
I'm kind of new in number theory and got stuck in this simple problem. I'm almost sure that the only solution is $n=1$, but I can't show that in a simple way.
Thanks in advance!

REPLY [1 votes]: If $n+2009$ divides $n^2+2009$, it also divides $n^2-n=n(n-1)$  Similarly, $n+2010$ divides $n(n-1)$  If $n \gt 1$, one of them would have to divide $n$ and the other $n-1$, but that is not possible, so $n=1$


TITLE: possibly found a "counterexample" to a multilinear algebra problem
QUESTION [3 upvotes]: Edit: Exact Question. my question is b part
$\phi:E\times F\to G$ be bilinear
$\psi:E\times F\to H$ be bilinear
Given $N_1(\phi)\subset N_1(\psi)$ and $N_2(\phi)\subset N_2(\psi)$ Show that there exist a linear function $f:G\to H$ such that $f\cdot\phi=\psi$ where $N_1(\phi) = \{x|\phi(x,y)=0\forall y\in F\}$ and similarly $N_2$ for the second coordinate
My "counterexample"
$F=\mathbb R^3$
$H=E=\mathbb R^2$
$G=\mathbb R^5$
$\phi(e,f)=(e_1f_1,e_1f_2,e_1f_3,e_2f_1,e_2f_2)$
$\psi(e,f)=(e_2f_2,e_2f_3)$
$N_1(\phi)=\{(0,0)\}\subset\{(x,0)\}=N_1(\psi)$
$N_2(\phi)=\{(0,0,0)\}\subset\{(x,0,0)\}=N_2(\psi)$
Please point out the mistake in the "counterexample". Also please provide hints for the problem

REPLY [1 votes]: Your counterexample is correct, it shows the claim made before cannot be correct (there is no way to get $e_2f_3$ as a linear combination of the components of $\phi(e,f)$). Are you sure you copied the problem correctly, especially the definition of $N_1, N_2$?