TITLE: Error for graph Theory proof
QUESTION [1 upvotes]: I am looking for an error in the proof but I am not certain about it. Pretty sure it has something to do with how there is not always a cycle of length 3.

Theorem 1. For every (undirected) graph $G = (V; E)$ without loops, if $|V| \ge  3$ and for every
  vertex has a degree of at least $2$, there exists a cycle of length $3$.
Proof. We proceed by induction on $|V|$. As a base case, consider $|V| = 3$. For any graph on three
  vertices, if all vertices have degree $2$, then every vertex is connected to both other vertices. Thus
  there is always a cycle of length $3$. To prove the inductive step, let $G$ be a graph with $k$ vertices,
  and construct a new graph $G_0$ on $k + 1$ vertices by adding one new vertex to $G$ and $2$ edges incident
  to the new vertex. Since $G$ has $k$ vertices, by the inductive hypothesis it has a cycle of length $3$.
  Thus the graph $G_0$
  contains the same cycle of length $3$. This completes the proof.

REPLY [2 votes]: The way you proceed in your inductive proof doesn't work. Not every graph (of your form) can be constructed in that manner. As in mlo105's answer, $C_4$ is a counterexample: you can't make $C_4$ by starting with a $3$-cycle and adding a vertex and two edges (which is the only kind of $4$-vertex graph your method can create).
In general, when you're doing an inductive proof in graph theory, you usually start with a graph on $k+1$ vertices and somehow reduce it to the case of a graph with $k$ vertices - not the other way around.


TITLE: Every finitely generated module is the quotient of a finitely generated projective module.
QUESTION [0 upvotes]: Every finitely generated module is the quotient of a finitely generated projective module.
I already find a proof of this but it uses tensor functor and flat modules propositions so im looking for a simpler proof

REPLY [1 votes]: In fact, it is a quotient of a finitely generated free module. Suppose that $R$ is your ring, and that $M$ is your module. Choose a generating set $\{m_1,...,m_n\}$. Define a module map $\varphi:R^n\rightarrow M$ by $$(r_1,...,r_n)\mapsto r_1m_1+\ldots+r_nm_n.$$ This map is surjective. By the First Isomorphism Theorem, $M$ is isomorphic to the quotient of $R^n$ by the kernel.


TITLE: Best intuitive metaphors for math concepts (of any level)
QUESTION [17 upvotes]: Frequently, we introduce a new concept with a formal definition, then immediately say "Intuitively, what this means is..." What are the absolute best metaphors you've seen (for concepts of any level)?
For example, I think the comparison of modular arithmetic to arithmetic with times, on a clock face, is fantastic, because it corresponds so perfectly and because clocks are so widely known.

REPLY [1 votes]: I find the concept of a digital sundial quite intriguing.  From wikipedia:

Roughly speaking, there exists a set
  with prescribed projections in almost
  all directions.


TITLE: Is it possible for an object to stop sinking after a while?
QUESTION [1 upvotes]: Suppose you drop an object into a body of fluid (on a planet with constant gravity). Suppose the fluid is pure (as in, there are no solid granulates or bubbles present in it), and finally suppose that there are no waves in the fluid, and that its density is constant everywhere.
Is it possible for said object to start sinking, but eventually slow down and completely stop sinking further (as in, it will stop and remain suspended in the liquid without hitting the bottom)?

REPLY [5 votes]: I see this is a follow-up post to Suppose a hollow metal sphere filled with helium is dropped in a body of water
Well, the situation you are describing is possible if the object in question can change its average density while in the water. It will stop sinking when $\rho_\text{average} = \rho_w$.
In fact, there's a vessel that uses this exact principle: the submarine. When it wants to submerge it fills up its ballast tanks with water, becoming more dense in the process. Then it sinks. When it wants to stop (and/or rise) it forces air into the ballast tanks which pushes the water out, so it becomes less dense, and it stops sinking/rises.


TITLE: Proving by definition that sequence is Cauchy
QUESTION [0 upvotes]: I need to prove by using the definition that sequence is Cauchy.
Sequence in question is
$$a_n=\sum_{k=1}^n {(\arccos{1 \over k} +3k-1)\over (2k-1)^4}$$
I was confused by sum and don't know how to proceed and what to compare it with.

REPLY [1 votes]: Hint. We have that
$$\lim_{k\to +\infty} \frac{{\arccos(1/k) +3k-1\over (2k-1)^4}}{1/k^3}=\frac{3}{16}<1$$
which implies that there exist $N>0$ such that for $k> N$,
$$0\leq {\arccos(1/k) +3k-1\over (2k-1)^4}\leq \frac{1}{k^3}$$
Hence for $n\geq m> N$,
$$ 0\leq a_n-a_m\leq \sum_{k=m+1}^n\frac{1}{k^3}=b_n-b_m.$$
where $b_n:=\sum_{k=1}^{n}\frac{1}{k^3}$. Now note that $(b_n)_n$ is a convergent sequence.


TITLE: Why does $ \frac{\partial}{\partial\theta}f(re^{i\theta})=i\frac{\partial}{\partial r}f(re^{i\theta})$?
QUESTION [1 upvotes]: I'm reading something, and don't understand why a certain equality comes up.
Suppose $u:\mathbb{C}\setminus\{0\}\to\mathbb{R}$ given by $u(z)=\ln(|z|^2)$ is a harmonic function. We want to see if $u$ has a harmonic conjugate $v$. If it does, then $f=u+iv$ is holomorphic.
But then I read that since the differential of $f$ is complex linear, then
$$
\frac{\partial}{\partial\theta}f(re^{i\theta})=i\frac{\partial}{\partial r}f(re^{i\theta}).
$$
I don't follow this. Even writing it out with the chain rule in terms of $u$ and $v$ doesn't make it clear to me. Why does this equality follow? Thanks.

REPLY [4 votes]: I think this is an error in what you're reading. You can see that without going into details simply because $\partial/\partial r$ has units of inverse length whereas $\partial/\partial \theta$ doesn't. The derivatives are
$$\frac{\partial}{\partial\theta}f(r\mathrm e^{\mathrm i\theta})=\frac{\partial(r\mathrm e^{\mathrm i\theta})}{\partial\theta}f'(r\mathrm e^{\mathrm i\theta})=r\mathrm i\mathrm e^{\mathrm i\theta}f'(r\mathrm e^{\mathrm i\theta})$$
and
$$\frac{\partial}{\partial r}f(r\mathrm e^{\mathrm i\theta})=\frac{\partial(r\mathrm e^{\mathrm i\theta})}{\partial r}f'(r\mathrm e^{\mathrm i\theta})=\mathrm e^{\mathrm i\theta}f'(r\mathrm e^{\mathrm i\theta})\;,$$
so the correct relationship is
$$\frac{\partial}{\partial\theta}f(r\mathrm e^{\mathrm i\theta})=r\mathrm i\frac{\partial}{\partial r}f(r\mathrm e^{\mathrm i\theta})\;.$$


TITLE: When to use the Functional Determinant in Polar Coordinate Transformation
QUESTION [0 upvotes]: I am currently learning about polar coordinate transformation, especially for integrating over certain regions. Let's say we have to calculate
$\int_{n}{xy \; dx dy}$
Then I think the correct transformation would be (note the functional determinant $r$)
$\int_{m}{r cos \phi \cdot r sin \phi \cdot r \; d\phi dr}$
First question: Is that correct?
My second question is: I have some examples a friend of mine once wrote for an exam. The task was for the following region
$ B = \{ (x,y) \in R : 0 \leq x^2 + y^2 \leq 1, x \leq y \}  $
to calculate the integral
$\int_{B}{x^2+y^2 \; dx dy}$
And he began with
$\int_{B}{x^2+y^2 \; dx dy} = \int_{0}^{\frac{5\pi}{4}} \int_{0}^{1} r^2 sin^2 \phi + r^2 cos^2 \phi \; d\phi dr$
Second question: Where is the functional determinant here? Is it missing or am I wrong assuming there should belong one?
I'd be happy if anyone could help me with some insight on this. Thanks!

REPLY [2 votes]: Yes, the determinant is $r$. Your friend made a mistake, there should be an $r$ there too (and the lower limit for $\phi$ should be $\pi/4$).


TITLE: Calculating the acceleration of a car
QUESTION [2 upvotes]: I'm trying to calculate the maximum acceleration a car can achieve with the current gear ratio. I am ignoring drag forces and friction to keep it simple.
I'm doing this by:

calculating the torque on the wheels: $T_e/G$ (engine torque / Gear ratio)
Calculate the Force exerted on the ground: T_w/r (wheel torque / wheel radius)
then finding the acceleration using $F=ma$

However acceleration values are too small. Is there something wrong with my reasoning?
My car's engine Torque is 500Nm, gear ratios: [2.97,2.08,1,0.86,0.56],
car's mass = 2500Kg

REPLY [2 votes]: The gear ratios are gearing down, i.e. they reduce the angular velocity at the wheels and increase the torque. The correct formula is:
$$ T_{wheels} = T_{engine} G $$
Plus don't forget that there is a differential gear as well as the gearbox, and this reduces the angular velocity and increases the torque as well. The value of $G$ in your equation is the gearbox ratio multiplied by the differential ratio.
Response to comment:
I don't know what the differential gearing is in your car, but differentials are generally around 4:1 so let's take this value. In that case the torque at the wheels in first gear is:
$$T_{wheels} = 500 \times 2.97 \times 4 = 5940Nm $$
I would guess the tyre radius is about 0.3m, so the force at the contact between the tyre and ground is:
$$ F = \frac{T_{wheels}}{r} = \frac{5940}{0.3} = 19800N $$
So the acceleration is:
$$ a = \frac{F}{m} = \frac{19800}{2500} = 7.92ms^{-2} $$
That seems pretty good to me. 0.8g!


TITLE: Projective curve branch point
QUESTION [1 upvotes]: Could someone explain why the projective curve $y^2=x$ has a branch point at infinity like the one at 0? I don't understand why this is true, and supposedly it helps probe that the curve is topologically a circle, which I don't understand how it follows from having a branch point at infinity.

REPLY [0 votes]: Assuming you want $x,y \in \mathbb{R}$, let  $[u,v,w]\in\mathbb{R}\mathbb{P}^2$. The affine point $(x,y)\in\mathbb{R}^2$ (away from infinity) corresponds to the point $[u/w, v/w,1]$ for $w\neq 0$. Then $y^2=x \Leftrightarrow v^2=uw$. Clearly this curve has a branch point at $w=0$.
(Another easier way to do the computation is, formally, let $x=1/t, y=y/t$ then letting $t\to 0$ corresponds to letting $x\to \infty$. This agrees with the above if you set $u=1$ which is allowed for $u\neq 0$ in projective coordinates.)
The compactified curve is topologically a circle since the graph is "on the positive branch" above the x-axis and switches branches (goes below the x-axis) at both $0$ and $\infty$.
Of course if $x,y\in \mathbb{C}$, everything is the same here except topologically you have something different than a circle, namely you have a Riemann surface which topologically looks like $S^3$.


TITLE: Parametric Equation of a Surface
QUESTION [0 upvotes]: If $\left\{P_{k,l}^{0}\right\}_{k,l=0}^{n}$ is a set of $\left(n+1\right)^2$ three-dimensional points, and
$$
P_{k,l}^{r+1}\left(t,s\right)=\left(P_{k+1,l+1}^{r}+P_{k+1,l}^{r}+P_{k,l+1}^{r}+P_{k,l}^{r}\right)ts,\tag{$*$}
$$
then, is $P_{0,0}^{n}(t,s)$ the parametric equation of a surface?
I tried plotting this with a set of test points and obtained a spatial, straight line, when I was instead aiming to obtain a surface (since I am making use of the two parameters $t$ and $s$).
What is the problem with $(*)$? I am just interested in knowing whether this is a surface or not. Thanks in advance!

REPLY [2 votes]: As far as I read your notation, the formulka does not describe a surface. You work with two parameters, but only ever use the product of these. So the set of points $P_{0,0}^n(t,s)$ can be described using a single parameter, e.g. $P_{0,0}^n(u,1)$. You can see that by taking $u=ts$, there is a one-to-one correspondence between your two-parameter form any my one-parameter version. Therefore, the best you can hope for is a curved line, not a surface.


TITLE: What can i say about $\sum_{n=1}^\infty p^n\int_0^\pi 2\sin(mx)\cos(nx)\mathrm dx$ ??
QUESTION [0 upvotes]: I'm trying to solve the following sum:
$$\sum_{n=1}^\infty p^n\underbrace{\int_0^\pi  2\sin(mx)\cos(nx)\mathrm dx}_{\kappa_{n,m}}$$
where for $\kappa_{n,m}$ I did:
\begin{align*}
\kappa_{n,m}&=\int_0^\pi\left\{\sin\left[(n+m)x\right]-\sin\left[(n-m)x\right]\right\}\mathrm dx\\
&=\frac{\cos\left[(n-m)\pi\right]}{n-m}-\frac{\cos\left[(n+m)\pi\right]}{n+m}-\frac1{n-m}+\frac1{n+m}\\
&=\frac{(-1)^{n-m}}{n-m}-\frac{(-1)^{n+m}}{n+m}-\frac{2m}{n^2-m^2}\\
&=\frac{2m}{n^2-m^2}\big((-1)^{n-m}-1\big).
\end{align*}
Thus
\begin{align*}
\sum_{n=1}^\infty p^n\kappa_{n,m}=\sum_{n=1}^\infty p^n\frac{2m}{n^2-m^2}\big((-1)^{n-m}-1\big)
\end{align*}
What I'm able to say from this last point is that the sum above is equal to $\displaystyle i\pi
p^m$ when $n=m$ for $n,m\in\mathbb{N}$. Also, it is equal to zero for the case where $n-m$ is an even number for $n,m\in\mathbb{N}$. I would like to shed some light on this last step in particular.

REPLY [1 votes]: Just for your curiosity.
There is an explicit expression for the infinite summation but it involves several special functions
$$F(p,m)=\sum_{n=1}^\infty p^n\frac{2m}{n^2-m^2}\big((-1)^{n-m}-1\big)
=p \Big[\Phi (p,1,1+m)-\Phi (p,1,1-m)\Big]+$$ $$\frac{(-1)^m }{m^2-1}p\Big[(m+1) \, _2F_1(1,1-m;2-m;-p)+(m-1) \, _2F_1(1,1+m;2+m;-p)\Big]$$ where, beside the gaussian hypergeometric function appears the Hurwitz-Lerch transcendent function.
The contour plots of $\Re [F(p,m)]$ and $\Im [F(p,m)]$ are very interesting to look at; may be, they could give you some ideas.


text:
\section{Sharing or Pricing the Privacy Budget?}
Theorem \ref{th:convergence rate of inexact AMA} provides a relationship between sub-optimality and the cumulative variance of the Laplacian noise inserted by all subsystems.
In real scenarios, a system designer or operator may specify a desired level of (sub-)optimality $ \mathcal S_K$ achieved after $K$ iterations, that is $\mathbb{E}[D(\lambda^{K}) - D(\lambda^{\star})] \le \mathcal S_K$.
Rewriting Equation~\eqref{eq: upper-bound of the distributed optimization algorithm}, we can compute a bound on the amount of cumulative Laplacian noise allowed at run-time
\begin{equation}
\sum^{M}_{i=1} \sigma^{2}_i \le  \frac{1}{4} \rho^{2}_{\phi} K \mathcal S_K - \sum^{M}_{i=1} G^{2}_i \triangleq \Sigma_{\text{budget}}  \ .
\end{equation}
Hence, once $\mathcal S_K$ is specified, a \emph{cumulative privacy budget} $\Sigma_{\text{budget}}$ is set.
Remember that each individual agent may define a different distance metric, or different weights, which may lead to different sensitivities and hence different noise levels required by the various agents.
Since the privacy budget is limited, a fair and transparent allocation procedure is required to divide the allowable noise over all agents.
Here, we propose two approaches for going about such allocation.

The first approach, would entail a proportional division of the pie. This could be done in two ways. The first way involves splitting the noise variance budget by the number of agents:~$\sigma_i^2 = \frac{1}{M} \Sigma_{\text{budget}} \,, \forall i = 1,\hdots,M$. In this case, the allowed noise variance $\sigma_i^2$ determines the maximum level of local differential privacy~$\epsilon_i$ that can be maintained, given a set sensitivity~$\Theta_i$ (or vice versa), as outlined in~\eqref{eq:privacy_level}), which will differ from agent to agent. The second way involves setting all local differential privacy levels~$\epsilon_i$ equal, and, given set sensitivities~$\Theta_i$, splitting the noise among all agents. This is equivalent to equating Equation \eqref{eq:privacy_level} for all agents $i = 1, \hdots,M$, leading to $M$ equations for~$\sigma_1,\hdots,\sigma_M$:
\begin{equation}
 \frac{\Theta_1}{\sigma_1} =  \frac{\Theta_2}{\sigma_2} = \cdots =
 \frac{\Theta_M}{\sigma_M}, \quad \sum^{M}_{i=1} \sigma^{2}_i = \Sigma_{\text{budget}} \,,
\end{equation}
where we assume that the noise variances are constant for all time steps~$k = 1,\hdots,K$.

The second approach we anticipate is a pricing scheme, in which the value of privacy is left to a market or negotiation.
In the context of d-OPF, it is natural to assume that different DER owners with varying privacy levels will have varying degrees of willingness to pay or incur a deduction on their revenue for preserving privacy of their local parameters.
Here, we propose two scenarios to perform allocation via the so-called \emph{Kelly mechanism}~\cite{kelly_rate_1998}.
We assume a one-directional bid $w_i$ done by all agents after seeing a price $\pi^{\sigma}_i$ given by the network operator to the agent. With the bid, the operator constructs a surrogate utility function~$w_i \log \sigma_i^2$. The operator then determines the allocation and payment by maximizing the sum of surrogate utility functions as:
\begin{equation}
\sigma^2_{\text{Kelly}}(w) = \arg \displaystyle \max_{ \Sigma_{i=1}^M \sigma_i^2 \le \Sigma_{\text{budget}}} \sum_{i=1}^M w_i \log \sigma_i^2 \, .
\end{equation}
The authors of~\cite{yang_vcg-kelly_2007} show that this mechanism works neatly if one assumes that buyers are \emph{price-takers}.
In the case buyers are strategic, realizing that the eventual price is influenced by all bids, i.e. $\pi^{\sigma}_i(w)$, the Kelly mechanism may not yield an efficient Nash equilibrium.
To account for such behavior, the same authors combined the Kelly mechanism with the celebrated Vickrey-Clark-Groves (VCG) mechanism~\cite{vickrey_counterspeculation_1961,clarke_multipart_1971,groves_incentives_1973}, proposing the VCG-Kelly mechanism.
The VCG mechanism is lauded because it incentivizes all participating buyers to report truthfully about the value they assign to the goods sold. It has an intuitive payment scheme, which says that each player $i$ should pay the difference in utility of the other players $j \neq i$ between the scenarios that player $i$ does and does not participate, which leads to each player caring about both its own utility and that of others, leading to truthful reporting.
The VCG-Kelly mechanism extends the VCG mechanism to problems with divisible goods~\cite{yang_vcg-kelly_2007}. It considers a more general class of surrogate utility functions~$V_i(w_i,\sigma^2_i) = w_i f_i(\sigma^2_i)$, with $f_i$'s strictly increasing, strictly concave, and twice differentiable.
The allocation rule is then similar in form:
\begin{equation}
\sigma^2_{\text{VCGK}}(w) = \arg \displaystyle \max_{ \sum_{i=1}^M \sigma_i^2 \le \Sigma_{\text{budget}}} \sum_{i=1}^n V_i(w_i,\sigma^2_i) \, ,
\end{equation}
and the payment scheme reads:
\begin{equation}
\begin{array}{rcl}
m_{i,\text{VCGK}}(w) &=& \displaystyle \left( \max_{\sum_{i=1}^M \sigma_i^2 \le \Sigma_{\text{budget}}, \sigma^2_i=0} \displaystyle \Sigma_{j=1,j \neq i}^M V_j(w_j,\sigma^2_j) \right) \\
&&- \displaystyle \sum_{j=1,j \neq i}^M V_i(w_i,\sigma^2_{i,\text{VCGK}})\, .
\end{array}
\end{equation}
The actual form of the utility functions in the context of our application is left as an open problem, which requires further investigation. Readers interested in further details and examples of the VCG-Kelly mechanism are directed to~\cite{yang_vcg-kelly_2007}.

We propose these allocation schemes to trigger a discussion about what a fair and workable division of the allowable noise variance in our privacy-preserving looks like.
This may vary based on the application.
Given that privacy in engineered systems is a value of increasing importance, it may be wise to consider the impact of pricing privacy for participants with varying socioeconomic background. In situations where certain agents have significantly less resources but privacy is equally important, a proportional scheme may be the more ethical approach to take.
If pricing is still used and it is anticipated that some participants outbid others, one may think about allowing each participant to have a minimum amount of privacy, which can be translated into an extra linear constraint in the Kelly mechanisms.

config: arxiv


TITLE: Proving that $\sum_{j=0}^n 2^j=2^{n+1}-1$ for $n\geq 0$ by induction
QUESTION [1 upvotes]: Solving $S=$ $\displaystyle \sum_{j=0}^n$ $2$$^j$ = $2$$^n$$^+$$^1$ $-1$
So I was able to find  the basis and the the RHS but I'm not sure how I should go about solving the LHS. Since I have K+1 in the sigma
Solving  $2$$^K$$^+$$^1$ $-1$ +  $\displaystyle \sum_{j=0}^{n+1}2^j$ = $2$$^K$$^+$$^2$ $-1$

REPLY [2 votes]: See if you can follow the meat of the induction proof below (getting from the LHS to the RHS):
\begin{align}
\sum_{j=0}^{k+1}2^j &= \sum_{j=0}^k 2^j+2^{k+1}\tag{by defn. of $\Sigma$}\\[0.5em]
&= (2^{k+1}-1)+2^{k+1}\tag{by inductive hypothesis}\\[0.5em]
&= 2\cdot2^{k+1}-1\tag{group like terms}\\[0.5em]
&= 2^{k+2}-1\tag{exponent law}\\[0.5em]
&= 2^{(k+1)+1}-1.\tag{simplify}
\end{align}


TITLE: Are there logics without modus ponens?
QUESTION [5 upvotes]: The question doesn't go beyond the title.
And I don't mean logics that merely just don't have it as a primitive rule - I'm interested in logic where you can't actually use it.
I've searched around and looked at the more exotic logics that I know, but all use modus ponens. Are there logic that do have implication but go without that rule?
But maybe I'm confused and if I take away that tool to syntactically go from knowing $P$ to knowing $Q$, then there isn't anything to $\to$ left.

REPLY [3 votes]: It was observed by Lewis & Langford that modus ponens fails in Lukasiewicz 3-valued logic.  What they did not make clear is how it fails and why it should fail. It fails because Lukasiewicz allows conditionals to have the third logical value. Unrestricted use of modus ponens would allow the use of doubtful premises (a and a→b) to draw a conclusion b that is false. This is not and should not be valid reasoning. It seems to have escaped their attention that the deficiency can be repaired by restricting conditionals to those that are definitely true: (a & L(a→b)) → b is a tautology.
The suggestion of ((a→b) & (b→c)) → (a→c) also fails for the same reason and can be repaired the same way.
Modus ponens as well as most other rules of inference, such as (P $\rightarrow $ P) also fail in Kleene's 3-valued logic. This is because the conditional used in it assigns the third logical value whenever the antecedent and consequent both have that value.


TITLE: Multi-keyhole contour integral with branch cut
QUESTION [3 upvotes]: How to construct a contour to calculate complex line integral

$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\arctan\left(\frac{1}{z}\right)\arctan\left(\frac{1}{s-z}\right)\,\mathrm{d}z$$

This integral is derived from the Laplace transform of the square of the Sinc function.
$$\mathscr{L}[f^2(t);s]=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F\left(z\right)F\left(s-z\right)\,\mathrm{d}z$$
As shown in the expression above, Wikipedia gives the Laplace transform formula of function product as. I want to verify this formula by taking function as the Sinc function. I've found six singularities $z=\pm\,\!i,0,s,s\pm\,\!i$ so far, but I don't know how to construct the contour.
I already know how to calculate $\displaystyle\int_{0}^{+\infty}\!\left(\frac{\sin(x)}{x}\right)^{\!\!2}\,e^{-sx}\,dx$ through the parametric integral, but now I don't know how to construct the contour and use the residue theorem to calculate the arctangent complex integral.

REPLY [1 votes]: We want to evaluate $\displaystyle I(S)=\int_{0}^{+\infty}\!\left(\frac{\sin(x)}{x}\right)^{\!\!2}\,e^{-sx}\,dx$ via contour integration.
In order to ensure that our result is correct, let's first evaluate the integral using an easier method. After that we will accomplish a complex integration program.
$$\frac{\partial^2}{\partial s^2}I(s)=\int_0^\infty\sin^2xe^{-sx}dx=\frac{1}{2s}-\frac{1}{2}\Re \int_0^\infty e^{2ix-sx}dx=\frac{1}{2s}-\frac{1}{2}\frac{s}{s^2+4}$$
Integrating two times
$$I(s)=\frac{s}{2}\,\log s-\frac{s}{4}\,\log(s^2+4)-\frac{i}{2}\log\frac{s+2i}{s-2i}+As+B$$
where $A$ and $B$ are some constants to be defined.
We will also use $\tan^{-1}(z)\,$ representation: $$\tan^{-1}(z)=-\frac{i}{2}\log\frac{i+z}{i-z}=-\frac{i}{2}\log\frac{1+iz}{1-iz}$$
$\text{At}\,s\to\infty\,\,\, I(s)\to 0 \,\,\Rightarrow\,\,\, A=B=0$
$\text{At}\,s\to0\,\,\, I(s)\to  \int_0^{\infty}\frac{\sin^2x}{x^2}dx=\frac{\pi}{2}\,\,\Rightarrow\,\, -\frac{i}{2}\log\frac{s+2i}{s-2i}\to \frac{\pi}{2}$,  and we should interpret
$$-\frac{i}{2}\log\frac{s+2i}{s-2i}=-\frac{i}{2}\log\big(-\frac{i+s/2}{i-s/2} \big)=\frac{\pi}{2}-\tan^{-1}\Big(\frac{s}{2}\Big)\to 0 \,\text{at} \,\,s\to 0\,$$
Finally,
$$I(s)=\frac{s}{2}\,\log s-\frac{s}{4}\,\log(s^2+4)+\frac{\pi}{2}-\tan^{-1}\Big(\frac{s}{2}\Big)$$

Now, let's get this result via integration in the complex plane.
$$I(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F\left(z\right)F\left(s-z\right)\,\mathrm{d}z,\,\,\text{where}\,\, F(z)=\int_0^\infty\frac{\sin x}{x}e^{-xz}dx$$
It is not difficult to find  $\,F(z)=\tan^{-1}(\frac{1}{z})=\frac{i}{2}\log\frac{z-i}{z+i}\,$, and $\,\,F(s-z)=\frac{i}{2}\log\frac{z-s+i}{z-s-i}$
At $|z|\to\infty \,\,\,F(z)F(s-z)\,\sim\frac{1}{4}\log\frac{1+i/z}{1-i/z}\log\frac{1+(s-i)/z}{1+(i-s)/z}\sim\frac{i(s-i)}{z^2}$
The integrand declines rapidly enough, and we can close the contour by a segment of a big circle with radius $R$; integral along this circle does not contribute at $R\to\infty$.
$$I(S)=-\frac{1}{8\pi i}\int_{c-i\infty}^{c+i\infty}\log\frac{z-i}{z+i}\log\frac{z-s+i}{z-s-i}dz=-\frac{1}{8\pi i}\oint\log\frac{z-i}{z+i}\log\frac{z-s+i}{z-s-i}dz$$
We should note that this formula is applicable for $c>p$ and $s>p+c$, where $p$ is the growth power for $f(z)$. For $\,\frac{\sin x}{x} \,\,\,p=0$, so the line of integration in the complex plane goes between the points $z=0$ and $z=s$. This line position can be also defined by the requirement for the integral convergence for any $s>0$.
We can close the contour by the circle segment either in the right or left side of the complex plane. Let's choose the left side (you can check that closing the contour in the right side gives the same result). To make the integrand single-valued we have to add cuts connecting branch points of logarithms: $[-i;i]$ and $[s-i;s+i]$.
Our contour looks like

Apart from the cut, there no special points inside our closed contour. Therefore, we can squeeze the contour, pulling it over the cut. Due to our choice the main logarithm branch is at $0<z<s$; when going around $z=i$ to the left bank of the cut $\log\frac{z-i}{z+i}\to \log\frac{z-i}{z+i}+2\pi i$.
So, the integral around the cut
$$I(s)=-\frac{1}{8\pi i}\int_{-i}^i\log\frac{z-i}{z+i}\log\frac{z-s+i}{z-s-i}dz-\frac{1}{8\pi i}\int_{i}^{-i}\Big(2\pi i+\log\frac{z-i}{z+i}\Big)\log\frac{z-s+i}{z-s-i}dz$$
$$=-\frac{1}{4}\int_{i}^{-i}\log\frac{z-s+i}{z-s-i}dz=-\frac{1}{4}\Big(s\log\frac{2i-s}{-s}+s\log\frac{-2i-s}{-s}+2i\log\frac{-2i-s}{2i-s}\Big)$$
$$=\frac{s}{2}\log s-\frac{s}{4}\log(s^2+4)-\frac{i}{2}\log\frac{1+s/2}{-i+s/2}$$
Considering the boundary conditions (at $s\to0$ and $s\to\infty$) we have to choose the right branch of logarithm and identify the third term as $\,\,\frac{\pi}{2}-\tan^{-1}(\frac{s}{2})$.
Finally,
$$I(s)=\frac{s}{2}\,\log s-\frac{s}{4}\,\log(s^2+4)+\frac{\pi}{2}-\tan^{-1}\Big(\frac{s}{2}\Big)$$


TITLE: Simple differential equation solution mismatches given answer
QUESTION [3 upvotes]: Solve $$\left(y-x\frac{dy}{dx}\right)=a\left(y^2+\frac{dy}{dx}\right)$$

I solved it by dividing both sides with $y^2/dx$: $$\frac{(ydx-xdy)}{y^2}=a\left(dx+\frac{dy}{y^2}\right)$$ and then integrate to get $$x/y=ax-a/y+c$$But that doesn't match the answer given by SymbolLab:
$$y=\frac{ac_1+xc_1}{a\left(-1+xc_1+ac_1\right)}$$
What's the problem here? Why doesn't my simple differential equation solution match with the given answer?

REPLY [0 votes]: In fact, if you do it by hand, distribute, agrupate, and order in the standar form of a L.D.E. you'll find that It´s a Bernoulli D.E. do all the subs. to reduce it and the answer is what hypernova wrote. he deserves all the credit


TITLE: Bases for two linear subspaces in the sets of all the functions $f:\mathbb{R} \to \mathbb{R}$ and $f:\mathbb{R}^n \to \mathbb{R}^n$, respectively.
QUESTION [0 upvotes]: Let $X$ the set of the all the functions $f:\mathbb{R} \to \mathbb{R}$.
Let $Y:=\{f\in X : \frac{d^2 f}{d x^2}+f=0\}$. Is that a linear subspace (I would say yes)? How to write a base for $Y$?
Furthermore, let $A$ the set of all the functions $f:\mathbb{R}^n \to \mathbb{R}^n$. Let $B:\{f\in A: ||\frac{\partial f}{\partial x}||\le 1, \forall x \in \mathbb{R}^n \}$. Is that a subspace (I am attempted to say yes)? How to write a base for such subspace?

REPLY [1 votes]: Regarding your first question, the space $Y$ is the space of all solutions to the linear differential equation $f'' + f = 0$ or $f'' = -f$. This is a second order linear differential equation which, if you will solve using the regular techniques, will yield two linearly independent solutions which will form a basis for $Y$. Since this equation is so simple, you don't even need to really solve it (can you think of two linearly independent functions $f$ such that their second derivative $f''$ is $-f$?).
Regarding your second question, $B$ won't be a subspace. To see this clearly, take $n = 1$ and ask yourself whether $B$ is closed under scalar multiplication or addition.


TITLE: Effective Serre Vanishing
QUESTION [7 upvotes]: Suppose that $X = \mathbb{P}^n_k$ and $G$ is a coherent sheaf on $X$.
Question: Is there a way to determine some integer $n_0$ such that $H^1(X, G \otimes O_X(n)) = 0$ for all $n \geq n_0$?  (obviously $n_0$ depends on $G$)
Of course, the higher cohomologies would be very interesting as well, but perhaps the above is simpler.  One can also ask this for the relative case as well $f : Y \to X$ a projective morphism.
What I'd be particularly interested in is something that can be implemented into a computer.
In my particular case, I know I only have to compute some other things until Serre vanishing hits, at that point, I know I can stop computing.  Unfortunately, I don't know how to tell when I've arrived.  You can assume I have a presentation of my module $G$ if it helps.
PS: Actually, what I really want is vanishing of $$R^1 f_* (G(-nE))$$ for a blow-up $f : Y \to X$ with $E$ the pullback of the blownup ideal.

REPLY [7 votes]: Take $n_0$ as (Castelnuovo–Mumford) regularity of $G$ minus $1$.
This generalizes to higher cohomology: for $H^i$ take the regularity minus $i$
Let $M$ be a module representing $G$.
Then, the (Castelnuovo–Mumford) regularity of $M$ is an upper bound of the regularity of $G$.
So for an implementation you can take the regularity of $M$, which can be computed using a free resolution of $M$.
If you need a better bound then you can compute a "better" module for $G$ by approximating its module of global sections.
However, this is rather expensive to my experience.


TITLE: "Exactly Two" - Are these statements logically equivalent?
QUESTION [2 upvotes]: Let P(x,y) = "x loves y"
Let the Domain of Discourse be all people
"Someone loves exactly two people"
I found that this can be expressed as:
∃x∃y∃z( y≠z ∧ P(x,y) ∧ P(x,z) ∧ ∀w(P(x,w) → (w=y ∨ w=z)))
(Person y is not person z, and person x loves y and person x loves z, and for all people, if person x loves person w, then person w must be person y, or person w must be person z)
However I also see this expression in many places and in similar circumstances:
∃x∃y∃z( y≠z ∧ ∀w(P(x,w) ↔ (w=y ∨ w=z)))
(Person y is not person z, and for all people, person x loves person w if and only if person w is person y and person w is person z)
Are these logically equivalent?

REPLY [1 votes]: Yes, they are equivalent, because $$\forall w ((w=y \lor w=z) \rightarrow P(x,w)) \;\leftrightarrow\; P(x,y) \land P(x,z)$$
Now try to prove this, and try to complete the equivalence proof using this.


TITLE: To what extent can you manipulate differentials like dy and dt like actual values?
QUESTION [8 upvotes]: I have been thinking about the differentials that we use in derivatives and integrals.
For example, I have an equation:
$${\int{w}{dr}} = \text{other stuff}$$
The context for this strange equation was:
$$\begin{align*}
q'(t) &= a - \frac{b}{r(t)},\\
r'(t) &= c - d r(t) - e q(t)\\
r''(t)&= d r'(t) - e q'(t)\\
r''(t)&= d r' - e a + \frac{eb}{r}\\
w&=r'\\
w' &= d w - ea + \frac{eb}{r}\\
\frac{dw}{dr} \frac{dr}{dt}&= d w - ea + \frac{eb}{r}\\
w \frac{dw}{dr} &= dw - ea + \frac{eb}{r}
\end{align*}$$
Integrate both sides with respect to $r$:
$$\frac{w^2}{2} = \int{d*w dr} - ea r + eb log(r)$$
Shift the integral to the left hand side:
$$\int d * w dr = - \frac{w^2}{2} - ea * r - eb * log(r)$$
Divide by $d$:
$$\int w dr = \frac{w^2}{2d} - \frac{ear}{d}-\frac{eb}{d} log(r)$$
And I get to that $w\, dr$.
So if I can replace $w$ with ${\frac{dr}{dt}}$, and differentials can be multiplied like normal, I would get ${\frac{dr^2}{dt}}$.
Then I have the idea of multiplying by ${\frac{dt}{dt}}$. Can I go ahead and do this? That would give me
$${\int{\frac{dr^2}{dt^2} dt}}$$
Which looks like ${w^2}$ to me. But can I do that?

REPLY [0 votes]: You can manipulate differentials in the following manner:

You can add differentials: if $z=x+y,$ then $\mathrm{d}z=\mathrm{d}x+\mathrm{d}y.$
You can multiply differentials by functions: if $y=f(x),$ then $\mathrm{d}y=f'(x)\mathrm{d}x$
You are allowed to "divide" differentials, but only in the context of appliying the chain rule. So if $y=f\circ{g},$ where $\circ$ denotes composition, then $\frac{\mathrm{d}y}{\mathrm{d}s}=\frac{\mathrm{d}(f\circ{g})}{\mathrm{d}g}\frac{\mathrm{d}g}{\mathrm{d}s},$ which looks like the differential $\mathrm{d}g$ is being divided out, at least visually (but there genuinely is no actual division happening here).

Everything else is not permitted.


TITLE: Understanding quantum stochastic master equations
QUESTION [6 upvotes]: I'm teaching myself open quantum systems and the concept of a stochastic master equation has arisen. As someone who has studied classical stochastic processes a fair bit, this seems, at least to my naive intuition, like a completely nonsensical quantity. Can someone explain the purpose/functioning of such a thing?
Specifically, my intuition is as follows:
For classical system one has:
Classical quantity subject to uncertain dynamics $\rightarrow$ Noisy/random state (modelled by SDEs for example) $\rightarrow$ Deterministic probability dynamics (Fokker planck or master equation description)
Removing the noise in the SDE means removing higher order terms in the deterministic FP-eqn leaving us with something like the Liouville eqn.
For quantum systems:
A quantum state description subject to uncertain dynamics $\rightarrow$ e.g. a Lindblad equation in the reduced density matrix (which could be unravelled into a stochastic schrodinger equation)
Removing the noise in the Schrodinger eqn removes the (deterministic) non-coherent terms in the Linblad eqn leaving us with the Quantum Liouville eqn.
So we have
Stochastic Dif. eq./jump process $\leftrightarrow$ Deterministic Fokker Planck/Master equation $\rightarrow_{noise free}$ Liouville equation
Stochastic Schrodinger equation $\leftrightarrow$ Deterministic Lindblad equation $\rightarrow_{noise free}$ Quantum Liouville equation
I.e. the Lindblad equation, describing the evolution of the ensemble of states under uncertain dynamics, is deterministic. What then, should I make of things like this (https://en.wikipedia.org/wiki/Belavkin_equation) where there are additional stochastic terms in the Lindblad equation? They claim to average to the Lindblad equation.
But this seems to be getting the levels of description/sources of uncertainty extremely muddled: no one adds stochastic terms to a classical master equation (or FP eqn), the entire point is that they describe the ensemble behaviour that results from the underlying stochastic dynamics.
It implies one would need/could build a 'super-master equation' which would be deterministic in the dynamics of distributions of (apparently stochastic) density matrices...
But density matrices are perfectly capable of describing mixed states (indeed that's the point), so isn't the above just pathological and unnecessary?

REPLY [3 votes]: The Belavkin equation and other stochastic master equations describe the evolution of a system which is being continuously measured. Since quantum measurement is inherently stochastic, this is what leads to the 'extra' layer of stochasticity that you have noticed.
As you correctly noted, the Lindblad equation is deterministic, as it represents an 'averaging out' of the system-environment interaction. Mathematically, this is achieved by performing a partial trace over the environment following a unitary system-environment interaction. There is no measurement involved in the dynamics that lead to Linblad equation.
Stochastic master equations (SME), on the other hand, do involve continuous measurements on some part of the system. The outcomes of these measurements are stochastic, so the exact dynamics will change depending on what outcomes you get. For example, the SME for a continuously measured continuous variable $X$ is:
$$
d \rho = -k[X,[X,\rho]]dt + \sqrt{2k} (X \rho + \rho X - 2 \langle X\rangle \rho)dW
$$
where $\rho$ is the density operator of the system, $k$ is a determines the strength of the measurement and $dW$ is a Wiener Process-a continuous stochastic process. If you evolve a state using this equation multiple times, starting from the same state each time, you will find that the dynamics each time will be different, due to the inherent randomness of the quantum measurement process captured by $dW$. These different evolutions are sometimes known as 'unravellings' of the master equation.
It is important to note that in the above equation, the density operator acts as a'state of knowledge' of the observer. In other words, it assumes that the observer is keeping track of the measurement result. If the measurement result is discarded, then one ends up with deterministic evolution of the density operator, as the stochastic nature of the measurement is 'averaged out'. Mathematically, this is achieved by setting the the term proportional to $dW$ equal to 0 and retaining only the term proportional to $dt$.
For a derivation of the above equation and further discussion, I recommend 'A Straightforward Introduction of Continuous Quantum Measurement' by Jacobs and Steck.


TITLE: Photon absorption probability for a given molecule in gas phase
QUESTION [2 upvotes]: So I'm pretty sure I'm approaching this problem in the wrong way and I need some guidance (my first hint is that I think I'm thinking about a quantum mechanical problem too classically)
Suppose there is an isolated molecule in the gas phase with an average cross-sectional area to be exposed to radiation of $A$.  (For my specific problem, the molecule is trapped in a superfluid Helium droplet, but I think the calculation should be roughly the same).  If the radiation source has a flux $f$ (in units of photons/second/square area/0.1% BW) at energy $E$, what is the probability of the molecule absorbing a photon within a given interaction time $t$ if the absorption probability at a given energy is $P(E)$?
This is pretty easy to calculate if I treat the whole problem classically, i.e. like a ball and a target model.  For some reason, though, I get numbers that seem to be way too low if I do this.  I know it has to be more complicated than that, since light is a wave also.  What am I missing?
I understand that transition probabilities are related to wavefunction overlap, etc.  Also, I should note that the radiation in my specific problem is in the hard x-ray region, though I don't think that should change the answer.
Thanks in advance for your help.

REPLY [1 votes]: It is not unusual in spectroscopy to define an absorption cross-section of a molecule in $cm^2$. Although it has no real physical meaning (it can even be larger that the molecule itself) and can be derived from Einstein coefficients, it is what you can directly measure by experiment $$cross-section\;\;\alpha= \frac{total\;energy\;absorbed\;/s}{total\;incident\;intensity\;(energy/s/area)}$$
The relation to the Einstein coefficient if simply $\alpha=\frac{\hbar\omega}{c}B_{21}$
Basically, transition probability is Einstein coefficient times photon density. You problem could be to correctly define a fraction of photons that are resonant with the transition that you study. This would greatly depend on the kind of light source that you use.


TITLE: Are these particular kinds of matrices well known?
QUESTION [0 upvotes]: Given two positive integers $n$ and $a \leq \frac{n}{2}$ consider a $n \times n$ matrix $A$ such that,

all the diagonal entries are either $a$ or $a+1$
all the non-zero off-diagonal entries are $\pm 1$
if $A_{ii} =a$ then the $i^{th}$ row and the $i^{th}$ column have $a$ non-zero off-diagonal entries.
(..and similarly for $A_{ii} = a+1$..)

Is something known about such matrices?
Are they characterized somehow?
Like what would their maximum eigenvalue be?

REPLY [7 votes]: Gerschgorin's circle theorem gives the upper bound $2(a+1)$ for the eigenvalues. This bound should be sharp if I'm not mistaken. I think the family of $n\times n$ circulant matrices
$$\begin{pmatrix}
2 & -1 & 0 & \cdots & 0 & -1 \\
-1 & 2 & -1 & 0 & \cdots & 0 \\
\vdots & & \ddots & &  & \vdots \\
0 & \cdots & 0 & -1 & 2 & -1 \\
-1 & 0 & \cdots & 0 & -1 & 2
\end{pmatrix}$$
should have maximal eigenvalue arbitrarly close to $2\cdot (1+1)$ (here $a=1$) for $n\gg 0$. If I didn't make any mistakes the eigenvalues of this matrix are $2-2\cos(\frac{2\pi j}{n})$ for $j=0,1,\ldots,n-1$. For $j=\lfloor \frac{n}{2} \rfloor$ and $j=\lceil \frac{n}{2} \rceil$ this is maximal and arbitrary close to 4, even equal to 4 if $2\mid n$.
EDIT:
And just as I finished writing the first part of this post I remembered the Perron-Frobenius theorem. Considere the matrices with constant diagonal $a+1$ and exactly $a+1$ times an entry equal to +1 off the diagonal in each row. These are positive matrices and have constant row sum $2(a+1)$. Therefore this row sum is an eigenvalue with eigenvector $(1,1,\cdots,1)^T$. It coincides with the $\infty$-operator norm (=row sum norm) of the matrix and is therefore equal to the spectral radius. These matrices therefore attain the upper bound exactly.


TITLE: If a function is defined by a taylor series, how can you derive its inverse in the form of a taylor series?
QUESTION [0 upvotes]: Say you want to find the inverse of a function, and it's not expressible in terms of known functions, so you want to find the taylor series for that inverse. But, you at least know the taylor series for your original function f(x), defined as $$f(x)=\sum_{n=0}^{\infty} \frac{g(x,n)h(n)}{n!}$$ and it converges for a sufficient radius.
Since Taylor series stem from derivatives, and there is a theorem called the inverse function theorem that relatives derivatives to a functions' inverse
$$(f^{-1})'(f(x))=\frac{1}{f'(x)},$$
is there any consequential theorem where you can derive a new Taylor series for the inverse of a function by using the original Taylor series for that original function?
In other words, $$f^{-1}(x)=\sum_{n=0}^{\infty} \frac{???}{n!}$$

REPLY [1 votes]: Considering $y=\sin(x)$, let us develop it as a truncated Taylor series
$$y=x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots\tag 1$$ Now, following the steps given here, let $$x=a_1y+a_2y^2+a_3y^3+a_4y^4+a_5y^5+\cdots\tag 2$$ and replace.
Expand all terms and group the terms for a given power of $y$. You will get
$$y=a_1 y+a_2 y^2+\left(a_3-\frac{a_1^3}{6}\right) y^3+\left(a_4-\frac{1}{2} a_1^2
   a_2\right) y^4+\frac{1}{120} \left(a_1^5-60 a_3 a_1^2-60 a_2^2 a_1+120
   a_5\right) y^5+\cdots\tag 3$$ Now, identify the powers to get $$a_1=1$$ $$a_2=0$$ $$a_3-\frac{a_1^3}{6}=0\implies a_3=\frac 16 $$ $$a_4-\frac{1}{2} a_1^2
   a_2=0\implies a_4=0$$ $$a_1^5-60 a_3 a_1^2-60 a_2^2 a_1+120
   a_5=0\implies a_5=\frac 3 {40}$$


TITLE: $\zeta (s+c)/\zeta (s)$ is increasing
QUESTION [3 upvotes]: It appears that for the Riemann zeta function
$$\zeta (s) = \sum_{n\ge 1} \frac{1}{n^s} \quad (s > 1)$$
and any constant $c > 0$, the function
$$\frac{\zeta (s+c)}{\zeta (s)}$$
(considered for real $s > 1$) is increasing.
Could anybody give a proof of this fact?
Thank you very much!

Just in case, here's what I tried: we have
$$\left(\frac{\zeta (s+c)}{\zeta (s)}\right)' = \frac{\zeta (s) \zeta' (s+c) - \zeta (s+c) \zeta'(s)}{\zeta (s)^2},$$
and $\zeta' (s) = -\sum_{n\ge 2} \frac{\log n}{n^s}$, but after playing around with series I don't see how to deduce that $\zeta (s) \zeta' (s+c) > \zeta (s+c) \zeta'(s)$.

REPLY [5 votes]: Since $s>1$, $\zeta(s)>0$ and it's easier to consider the logarithmic derivative: $f'(x)/f(x)$ has the same sign as $f'(x)$. In particular, for $s>1$ we can use the Euler product
$$ \zeta(s) = \prod_{p} (1-p^{-s})^{-1} : $$
from this we have
$$ \frac{d}{ds} \log{\zeta(s)} = \frac{d}{ds} \sum_p -\log{(1-p^{-s})} = -\sum_p \frac{1}{p^s-1} \log{p} . $$
Then
$$ \frac{d}{ds} \log{\frac{\zeta(s+c)}{\zeta(s)}} = \frac{d}{ds} \log{\zeta(s+c)} - \frac{d}{ds} \log{\zeta(s)} = \sum_p \left( \frac{1}{p^{s}-1} - \frac{1}{p^{s+c}-1} \right) \log{p} . $$
Since $s \mapsto 1/(p^s-1)$ is a decreasing function, every term in the sum is positive, so the whole sum is positive. Therefore $s \mapsto \zeta(s+c)/\zeta(s)$ is in fact strictly increasing when $c>0$ and $s>1$.
(All the interchanges of logarithms and derivatives with sums and products can be justified using continuity and local uniform convergence, as usual: there's nothing weird going on here.)


TITLE: Nearly Free Electron Model and the Reduced Zone Scheme
QUESTION [7 upvotes]: When for example studying the vibrational modes of a one dimensional diatomic chain we find that the dispersion relation $\omega(k)$ is periodic in the one dimensional reciprocal lattice vector $\frac{2\pi}{a}$, and so the dispersion relation can all be displayed in the first Brillouin zone in the reduced zone scheme.
When studying free electrons perturbed by a weak periodic potential we find that the dispersion relation of the free electrons $E=\frac{\hbar^2k^2}{2m}$ develops gaps at the Brillouin zone boundaries. However this (almost) parabolic dispersion does not have a functional dependence that is periodic (i.e for a single value of $k$ there is only one possible energy). Books seem to suggest we can display this in a reduced zone scheme though - why is this?
Edit: Possibly related to Bloch's theorem? I guess Bloch's theorem says that we can always reduce things into the first Brillouin zone - however if this is the case why doesn't it naturally drop out of the calculations rather than us having to force it?

REPLY [4 votes]: Preamble:
The dispersion relation for a periodic potential is always periodic itself. That is, if $E_k$ is an energy eigenvalue for wavevector $k$ in a lattice with cell size $a$, then
$$
E_{k + G} = E_k, \;\;\;\forall k,\;G = \frac{2\pi}{a}n\;\;\text{(G vector of the reciprocal lattice)}
$$
which follows directly from the translational symmetry of the Hamiltonian.  To see this, let $\psi(x)$ be an eigenfunction of the Hamiltonian. Translational symmetry means that (Bloch's theorem)
$$
\psi(x+a) = e^{ika}\psi(x)
$$
for some wavevector $k$, and that any eigenfunction behaving as above under translation corresponds to the same eigenvalue $E_k$. So the corresponding eigenfunctions are also indexed by $k$ (and other quantum numbers, e.g. spin), $\psi_k(x)$.
Now let $\psi_{k+G}(x)$ be an eigenfunction corresponding to energy eigenvalue $E_{k+G}$. Then by translational symmetry again,
$$
\psi_{k+G}(x+a) = e^{i(k+G)a}\psi_{k+G}(x) = e^{ika}\psi_{k+G}(x)
$$
which means that $\psi_{k+G}(x)$ must correspond to eigenvalue $E_k$, and so $E_{k+G} = E_k$. As a side note, from time reversal symmetry it also follows that $E_k = E_{-k}$.
The first Brillouin zone singles out a first period in $E_k$, by setting $-\pi \le ka \le \pi$. If there are $N$ particles in the lattice, then there are $N$ distinct values of $k$ in the first Brillouin zone. It can be shown that relations $E_{k+G} = E_k$, $E_k = E_{-k}$ imply $\frac{dE_k}{dk} = 0$ at $k=\frac{\pi}{a}n$, $\;n=0,\pm1,\pm2,\dots $, that is, $E_k$ has extrema in the center and on the edges of Brillouin zones, where the dispersion relation can be approximated as a parabolic one, $E_k \sim \pm k^2$. In the nearly free electron model, this parabolic approximation becomes, around the center of the first Brillouin zone, $E_k = \frac{\hbar^2 k^2}{2m}$.
Short answer to the question:
The reason why this periodicity is not explicit in expressions such as $E_k = \frac{\hbar^2 k^2}{2m}$ is that the complexity of energy levels, or energy bands, in the periodic system can be described in 3 different schemes. The representation described above, using $E_k = E_{k+G}$ and showing all energy levels (bands) in all regions of the wavevector space, is the so-called periodic zone scheme. The representation using $E_k = \frac{\hbar^2 k^2}{2m}$ shows all bands in the first Brillouin zone only and is known as the reduced zone scheme. There is also the extended zone scheme, shown in the other answer, which shows different bands in different Brillouin zones, with discontinuities at zone edges.
You can find a very good depiction of the 3 representation schemes in Fig.8, pg.25 of these notes on the Bloch Theorem and Energy Band[s]. The lecture also contains a great introduction to energy bands in periodic systems from symmetry principles (translation, parity, time reversal, etc).


text:
\begin{document}

\title[]
{An extension problem for convex functions.}

\author[]{ Bo Berndtsson}

\address{B Berndtsson :Department of Mathematics\\Chalmers University
  of Technology
  and the University of G\"oteborg\\S-412 96 G\"OTEBORG\\SWEDEN,\\}

\email{ bob@math.chalmers.se}

\begin{abstract}
{ We give a statement on extension with estimates of convex functions defined on a linear subspace, inspired by similar extension results concerning metrics on positive line bundles. }
\end{abstract}

\bigskip

\maketitle

\section{Introduction}
The aim of this paper is to prove the following theorem on extension with estimates of convex functions.

\begin{thm} Let $\phi(t,x)$ be a convex function in $\R^{n+m}=\R^m_t\times\R^n_x$ and let $\psi(x)$ be a convex function in $\R^n$. Assume
\be
\int_{R^n}e^{\psi(x)-\phi(0,x)} dx =1.
\ee
Then $\psi$ can be extended to a convex function, $\Psi$, on all of $\R^{n+m}$
in such a way that
\be
\int_{R^n}e^{\Psi(t,x)-\phi(t,x)} dx \leq 1
\ee
for all $t$ in $\R^m$.
\end{thm}

\bigskip

One motivitation for studying this extension problem comes from the analogy with certain extension problems in complex analytic geometry, in particular the problem of invariance of plurigenera, see e g \cite{Siu} and \cite{Paun}. In these complex analytic extension problems one seeks to extend holomorphic sections to certain line bundles from  hypersurfaces in a bigger manifold.  The main point in the proofs is to extend a positively curved metric on the line bundle, initially defined only over the hypersurface, to a positively curved metric over the ambient manifold. The analogy to the situation discussed in the theorem lies in the parallellism between positively curved metrics and convex functions on $\R^n$. The convex situation is however (and of course) much simpler and allows for more complete  results, and no theorem as precise as theorem 1.1 is known in the complex analytic setting.

\noindent The theorem can also be viewed as a generalization of Prekopa's theorem, \cite{Prekopa},\cite{B-Lieb}, which says that the function $\tilde\phi$ defined by
$$
e^{-\tilde\phi(t)}=\int_{\R^n} e^{-\phi(t,x)} dx
$$
is convex. Indeed, Prekopa's theorem says that if the function $\psi$ is identically equal to zero, then we may take $\Psi(t,x)=\tilde\phi(t)$.

\noindent It is not hard to see that, conversely, the case $\psi=0$ in theorem 1 implies Prekopa's theorem. This is so because, since $\Psi(t,x)$ is convex with respect to all the variables,
$$
\Psi_0(t):=\inf_x \Psi(t,x)
$$
is also convex, and satisfies (1.2) as well. Hence
$$
\tilde \phi(t)\geq\Psi_0(t)
$$
with equality for $t=0$. In particular, the graph of $\tilde\phi$ has a supporting hyperplane at the origin. Replacing $t=0$ by any other value of $t$, we see that $\tilde\phi$ is convex.

\bigskip

\noindent In the next section we will prove Theorem 1 by a reduction to the case of $\psi=0$. We will also give a simple corollary on the convexity with respect to parameters of a certain extremal convex function.

\bigskip

\noindent
 I would like to thank Mihai P\u aun for many very stimulating discussions on these matters. As mentioned above this little note was inspired by (joint work with him on) similar extension problems for positive metrics on line bundles. Thanks also to the Mittag-Leffler institute where this work was carried out.

\section{Proof of Theorem 1}

To avoid some issues of convergence we  will first prove a version of
Theorem 1 where, instead of integrating over $\R^n$, we integrate over a ball $B_R$ in $\R^n$ of radius $R$ and center 0. Let us call a function $\psi$ convex in $B_R$ ``good'', if $\psi$ satisfies the conclusion of Theorem 1 for any choice of $\phi$, convex in $\{|t|<R|\}\times B_R$. More precisely, $\psi$ is good if for any convex function $\phi$ in $\{|t|<R|\}\times B_R$ such that
$$
\int_{B_R} e^{\psi(x)-\phi(0,x)} dx \leq 1,
$$
there is a convex extension $\Psi(t,x)$ such that

$$
\int_{B_R} e^{\Psi(t,x)-\phi(t,x)} dx \leq 1,
$$
for all $t$  with $|t|<R$.

\bigskip

\noindent By the discussion in the introduction, the function $\psi$ which is identically equal to 0 is good - this is one way of stating Prekopa's theorem. We next claim that any affine $\psi(x)=a\cdot x+b$ is also good. To see this,
write
$$
1=\int_{B_R} e^{a\cdot x+b-\phi(0,x)} dx =\int_{B_R}
e^{-(\phi(0,x)-a\cdot x-b)} dx .
$$
Since 0 is good there is a function $\Psi(t,x)$, (or actually $\Psi(t)$), such that $\Psi(0,x)=0$ and
$$
\int_{B_R} e^{\Psi(t,x)-(\phi(t,x)-a\cdot x-b)} dx\leq 1.
$$
Then clearly $\Psi(t,x)+a\cdot x+b$ extends $\psi=a\cdot x+b$ and satisfies the required estimate.

\bigskip

\noindent The next step is to note that if $\psi_\xi(x)$ are good for any $\xi$ in $\R^n$, then $\psi$ defined by
$$
e^\psi =\int e^{\psi_\xi}d\mu(\xi),
$$
where $\mu$ is a positive measure, is also good. This is evident since $\Psi$ defined by
$$
e^{\Psi}=\int e^{\Psi_\xi}d\mu(\xi)
$$
extends $\psi$ if $\Psi_\xi$ extend $\psi_\xi$.

\bigskip

\noindent  The main step of
the proof involves H\"older's inequality. We claim that if $\psi$ is good and $\lambda\geq 1$, then
$\psi/\lambda$ is good. This is proved by an iterative procedure, imitating an argument from \cite{B-P}. We can first clearly find an extension $\Psi_0(t,x)$
of $\psi$ such that
$$
\int_{B_R} e^{\Psi_0(t,x)/\lambda-\phi(t,x)} dx\leq A,
$$
for {\it some} finite constant $A$. This is at least clear if we shrink $R$ slightly, since we may then take $\Psi_0$ independent of $t$. Write
$$
1=\int_{B_R} e^{\psi(x)/\lambda -\phi(0,x)} dx =\int_{B_R} e^{\psi(x) -
(\phi(0,x)+
(1-1/\lambda)\Psi_0(0,x))}dx.
$$
Since $\psi$ is good, there is a convex extension $\Psi_1$ with
$$
\int_{B_R} e^{\Psi_1(t,x) -(\phi(t,x)+  (1-1/\lambda)\Psi_0(t,x))} dx\leq 1.
$$
By H\"older's inequality with exponents $\lambda$ and $\lambda/(\lambda-1)$
$$
\int_{B_R} e^{\Psi_1(t,x)/\lambda-\phi(t,x)} dx=
\int_{B_R} e^{\Psi_1(t,x)/\lambda- (1-1/\lambda)\Psi_0(t,x)/\lambda + (1-1/\lambda)\Psi_0(t,x)/\lambda -\phi(t,x)} dx\leq
$$
$$
\leq
\left(\int_{B_R} e^{\Psi_1(t,x)-(\phi(t,x)+(1-1/\lambda)\Psi_0(t,x))}dx\right)^{1/\lambda} \left(\int_{B_R} e^{\Psi_0(t,x)/\lambda-\phi(t,x)} dx\right)^{(\lambda-1)/\lambda }\leq A^{(\lambda-1)/\lambda}.
$$
If $A>1$ this is strictly smaller than $A$. Iterating the procedure we get extensions $\Psi_k$ of $\psi$ with corresponding integrals bounded by
$$
A^{((\lambda-1)/\lambda))^k}.
$$
A simple compactness argument shows then that a limit of a subsequence of $\Psi_k$ satisfies the desired estimate.

\bigskip

\noindent
With this, we can at last prove that any convex function $\psi$ in $\R^n$ is good. Let $\psi^*$ be the Legendre transform
$$
\psi^*(\xi)=\sup_x (x\cdot \xi -\psi(x)).
$$
Then, by the involutivity of the Legendre transform
$$
\psi(x)=\sup_\xi (x\cdot \xi -\psi^*(\xi)).
$$
Hence
$$
\psi=\lim \psi_\lambda/\lambda,
$$
as $\lambda$ tends to infinity, where
$$
e^{\psi_\lambda(x)}=\int_{\R^n}e^{\lambda(x\cdot\xi-\psi^*(\xi))}d\xi.
$$
By the arguments above, each $\psi_\lambda/\lambda$ is good, so by a simple passage to the limit, $\psi$ is good. Finally, we can let $R$ tend to infinity, so the theorem is proved.

\section{ An extremal convex function}

Given a convex function $\phi$ on $\R^n$ we put
$$
E(\phi)(x):=\sup \{\psi(x); \int e^{\psi(x)-\phi(x)}dx \leq 1\}
$$
We then have the following corollary to Theorem 1.1.
\begin{cor} Let $\phi(t,x)$ be convex in $\R^m_t\times \R^n_x$. Let
$$
\hat\phi(t,x)= E_x(\phi)
$$
where $E_x$ indicates that $E$ is taken with respect to the $x$-variable for $t$ fixed. Then $\hat\phi$ is convex.
\end{cor}

\bigskip

Let us show how the Corollary follows from Theorem 1.1. It is  enough
to show that for any point $p=(t_0,x_0)$ there is a convex function $\phi_p$ such
that
$$
\phi_p\leq\hat \phi,
$$
with equality at $p$. (That implies that the graph of $\hat\phi$ has a supporting hyperplane at every point.) Assume without loss of generality that $p=0$, and let $\psi(x)$ be a function realising the supremum, so that
$$
\psi(0)=\hat\phi(0,0),
$$
and
$$
\int e^{\psi(x)-\phi(0,x)}dx\leq 1.
$$
By Theorem 1.1 there is a convex function $\Psi(t,x)$ such that $\Psi(0,x)=\psi(x)$ and
$$
\int e^{\Psi(t,x)-\phi(t,x)}dx\leq 1,
$$
for any $t$. Then

$$
\Psi(t,x)\leq \hat\phi(t,x)
$$
for any $(t,x)$ with equality at the origin. This finishes the proof.

\bigskip

\def\listing#1#2#3{{\sc #1}:\ {\it #2}, \ #3.}


TITLE: How to calculate P(A∩B), P(A∩C), P(B∩C) and P(A∪C), P(B∪C)?
QUESTION [0 upvotes]: A particular computer program outputs a number in {0,2,3} with probabilities as follows:
$0 $ has a probability of $\frac{1}{3}$
$2$ has a probability of $\frac{1}{2}$
$3$ has a probability of $\frac{1}{6}$
The program is run twice and its first and second output are recorded.
Let A be the event that the  first output is $0$.
Let B be the event that the second output is $2$.
Let C be the event that the sum of the two outputs is exactly $5$.

(a) Find $P(A)$, $P(B)$ and $P(C)$.
(b) Find $P(A∩B)$, $P(A∩C)$ and $P(B∩C)$.
(c) Are $A$ and $C$ independent events? Are $B$ and $C$ independent events?
(d) Find $P(A∪C)$ and $P(B∪C)$.

This is how I attempted this:
For part a) $P(A) = \frac{1}{3}$, $P(B) = \frac{1}{2}$, $P(C) = (2,3) or (3,2) = \frac{1}{6}$
For part b) This is where I have a problem I am not sure how to find $P(A∩B)$, $P(A∩C)$ and $P(B∩C)$.
For part c) A and C independent only if P(A∩C) = P(A)P(C)
For part d) This is also where I have a problem, if I knew the answers to the part b I would be able to find this.

REPLY [2 votes]: $(b)$
The runs of the program are independent.
$$P(A \cap B) = \frac{1}{3} \cdot \frac{1}{2}$$
$C$ can not occur with $A$ because $A$ has the value $0$.
$$P(A \cap C) = \frac{1}{2} \cdot 0$$
For $B$ and $C$ to occur we must have a $3$ on the first run.
$$P(B \cap C) = \frac{1}{2} \cdot \frac{1}{6} $$
$(c)$
You have the probabilities of the intersection of the events
and the individual events, check if
$P(U \cap V) = P(U)P(V)$
for $A$ and $C$, $B$ and $C$.
$(d)$
Use the formula
$P(U \cup V) = P(U) + P(V) - P(U \cap V)$
for $A$ and $C$, $B$ and $C$. You have already calculated what you need.


TITLE: Give an example of a nested sequence of closed but unbounded intervals which does not have a point in its intersection.
QUESTION [0 upvotes]: I'm finding nested intervals hard to understand and I'm really stuck on this homework question. First to make sure, a closed but unbounded interval is something like this right?
$[n,\infty)$
And if so, could we show that $I_n=[n,\infty)$ is an example with no point in its intersection? I think we could use the Archimedean Property to show that for any $x \in [1,\infty)$, there exists an $m \in \mathbb{N}$ greater than $x.$ So $x \not\in I_m$. Like a proof by contradiction?
I'd like to know if my reasoning makes sense and if this is a "concrete" counterexample.
Thanks!
[EDITED]

REPLY [3 votes]: Concretely consider $I_n := [n, \infty)$. By the existence of $\lfloor x+1 \rfloor$ for all real $x$, you have that for $n>\lfloor x+1 \rfloor$, $x\notin I_n$ (wich is basically the archimedian property you mentioned), thus
$$\bigcap_{n\in\mathbb  N} [n,\infty) = \emptyset$$
So it is a valid example. A more general class of examples is given by this proposition:

Let $(a_n)_n$ be a sequence with $\lim\limits_{n\to\infty} a_n = \infty$ then the sequences of sets
  $$I^1_n := [a_n, \infty)\\I^2_n := (-\infty, -a_n]\\I^3_n := \mathbb R \setminus (-a_n, a_n)$$
  all fulfill $I^j_{n+1} \subset I^j_n$ and
  $$\bigcap_{n\in\mathbb N} I^j_n = \emptyset$$


TITLE: Does a mirror help a near-sighted persion see at a distance clearer?
QUESTION [22 upvotes]: A near-sighted person without eye-glasses can not clearly see things at distance.
If he takes a photo of the things at distance, he can see the things from the photo much clearer, because he can place the photo much closer to his eyes.
If he turns his back at the things at distance, and holds a mirror close to his eyes in a position so that the mirror reflects the things at distance behind him, will he see the things much clearer than if he looked at the things at distance directly?

REPLY [0 votes]: Any optical system that can form an image on the retina is well suited. The eye has deficiency of few dioptres and if the optical system can provide them at the right distance one will see clearly.
A mirror with curvature can also be used however it wound not be convenient for every day use. It would be sort of adapted mirror telescope (regular telescope is afocal system).


TITLE: A differential inclusion relating to the slope of a convex function
QUESTION [3 upvotes]: This question is concerned with a possible lemma which would be very useful in one of my current research projects, but which I am currently unable to prove. The project as a whole relates to the asymptotic growth rates of certain matrix products, but once all the dynamical parts are finished with, I need to obtain some quantitative information about the behaviour of the subderivatives of a certain convex function. This would be greatly simplified if the following lemma were true:

Lemma A. Let $f \colon [0,1] \to \mathbb{R}$ be a continuous convex function, and let $\varepsilon>0$. Suppose that for every $t \in (0,1)$, for every subderivative $f'(t)$ of $f$ at $t$ the inequality
  \[   f'(t)\leq \frac{f(t)-f(0)}{t} + \varepsilon\]
  holds. Then there exist subderivatives $f'(0)$ and $f'(1)$ of $f$ (at 0 and 1 respectively) such that $|f'(0)-f'(1)| \leq C\varepsilon$, where $C>0$ is a constant which does not depend on $f$ or on $\varepsilon$.

In visual terms, this means that for any $t \in (0,1)$ the slope of the straight line connecting $(0,f(0))$ and $(t,f(t))$ is constrained to be close to the "gradient" of $f$ at $t$, and I would like to deduce from this that the entire graph of $f$ does not admit very many different "slopes".
This problem can be re-stated in the following form, which I personally find somewhat easier to think about. Let us define $F \colon (0,1] \to \mathbb{R}$ to be the function which describes the left derivatives of the above function $f$. This $F$ is a well-defined monotone increasing function which is continuous on the left.

Lemma B. Fix $\varepsilon>0$. Let $F \colon (0,1] \to [0,\infty)$ be a monotone increasing function, continuous with respect to limits from the left, such that for every $t>0$ we have $F(t) \leq \frac{1}{t}\int_0^tF(s)ds + \varepsilon$. Is it necessarily the case that $\sup F  - \inf F \leq C\varepsilon$ for some constant $C>0$ which does not depend on $F$ or on $\varepsilon$?

I've spent a couple of weeks thinking about this on-and-off without making very much headway, and I am beginning to get a little frustrated at not being able to solve a problem in one-dimensional real analysis! Of course, this does not preclude the possibility that the problem is actually very simple for someone equipped with the correct tools. So my question is: does anyone know whether either of the above two lemmas is true, or have any pointers which might be useful in establishing a proof?

REPLY [4 votes]: Unfortunately the lemma is false.  Given a candidate $C$, let $\varepsilon = 1$ and $F(t) = \ln(te^C+1)$.  Then the hypotheses of Lemma B hold but the conclusion fails.


TITLE: Stream function, velocity potential/field from complex potential
QUESTION [3 upvotes]: Have this question, but have no answers in the book, so just wanted to check my progress.
Consider the following complex potential:
$$\Omega(z) = -\frac{k}{2\pi}\frac{1}{z}$$
Calculate the corresponding velocity potential, stream function and velocity field. Sketch the stream function.
Parameterising $z = re^{i\theta}$, I get: $\Omega(r,\theta) = -\frac{k}{2\pi r}(cos\theta - i\sin\theta)$; so I get the velocity potential and stream function as:
$\phi(r,\theta) = -\frac{kcos\theta}{2\pi r}$;
$\psi(r,\theta) = \frac{ksin\theta}{2\pi r}$
I'm not very sure about the velocity field. Is it just the two directions of the velocity potential?
Also not too sure about sketching the stream function. I converted back into Cartesian coords to get $\psi (x,y) = \frac{ky}{2\pi (x^2+y^2)}$ and started plugging in different y values and solving for different constants (setting $k=1$). Seems very long-winded? I got a rough sketch of:

Any feedback appreciated!
(And I'm no relation to Picasso in case anybody wondered)

REPLY [1 votes]: The velocity field is the gradient of the velocity potential:
$$\nabla \frac{-k\cos \theta}{2\pi r} = \nabla \frac{-k x}{2\pi(x^2+y^2)}=\frac{k}{2\pi(x^2+y^2)^2}\left(x^2-y^2, 2xy\right).$$
Notice that this vector field is divergence-free, as needed, since the potential is harmonic.
To sketch the stream function, you can solve for $r$:
$$r = \frac{k \sin \theta}{2\pi c}$$
which is now in standard polar form (once you specify $k/c$.)


TITLE: Integrals in analysis and category theory
QUESTION [12 upvotes]: Are integrals in analysis special cases of coends in category theory? They are both seen as weighted sums, denoted by $\displaystyle\int $ and share the same formal properties (for example Fubini's Theorem). Also notice that there is a very beautiful reformulation of the Yoneda Lemma
$$F(a) = \int^{p \in I} \mathrm{Hom}(a,p) \otimes F(p)$$
for functors $F : I^{op} \to \mathcal{C}$ into a cocomplete category $\mathcal{C}$ which looks like the equation
$$\mu(A) = \int_{p \in I} \chi_A(p) ~ d \mu(p)$$
for a measure $\mu$ on a measurable space $I$.
Notice that this question is not as silly as it seems: Limits in analysis can be seen as limits in category theory, see  MO/9951 or Wikipedia (Topological limits).
If the measurable space $I$ is discrete, the question is about (infinite) series - are they special cases of coends?

REPLY [6 votes]: Not an answer, but it was too long for a comment.
Lawvere "Metric spaces" paper relates the inner and outer measure on a probability space to Kan extensions (which can be written as ends/coends). This is p. 162:

As an example of the last corollary [see the paper], we could take for $X$ the space of nonnegative step functions of a probability space $S$ and for $Y$ the space of all nonnegative functions with the natural sup metric, and consider as $\varphi$ the elementary integral; thus in general we might call $y\in Y$ $\varphi$-integrable if [a certain right Kan extension, the inner measure] = [a certain left Kan extension, the outer measure].