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Let $$R(\theta)=\begin{bmatrix} \cos\theta &-\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$
Also, $a$ and $b$ are real numbers. We suppose that $b\neq 0$ and we consider the matrix: $$A=\begin{bmatrix} a &-b \\b &a \end{bmatrix}$$
Show that it exist a unique number $\lambda>0$ and a unique number $\theta \in\ ]0,2\pi[$ such as $A=\lambda R(\theta)$.
I don't even understand the question. What is the number $\lambda$? How am I supposed to proceed here? Any help to point me in the right direction would help me a lot.
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Let $X$ be a locally compact and Hausdorff space, and let $\mu$ be a positive measure on the Borel sets of $X$ (here $\mu$ is
not necessarily regular).
Then the linear map $L : C_c(X) \to \Bbb C$ defined by $L : f \mapsto \int_X f \;d\mu$ can be represented, by Riesz theorem, as$$L(f) = \int_X f \;d\nu$$where $\nu$ is a
regular Borel measure (here $C_c(X)$ denotes the set of continuous functions on $X$ that are compactly supported).
My question is:
Is my previous reasoning correct? That is : have I proven that for
everymeasure $\mu$ we can find a regular Borelmeasure $\nu$ with respect to which integration is the same as the integration w.r.t $\mu$?
This would mean (in some sense) that we can always only handle with regular measures. (Or am I misinterpreting something?)
In Folland,
A Guide to Advanced Real Analysis, page 59-60 : "If $\mu$ is a Borel measure on $X$ such that $\mu(K) < \infty$ for every compact $K$ of $ X$, then […] It is a fundamental fact, and a rich source of measures, that every positive linear functional on $C_c(X)$ is of this form, and that the measure can be taken to be regular."
Thank you for your comments!
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Edit: This question is about rejecting the null hypothesis.
Last month my evil twin and I were at a game show.
The rules are as follows: There is a sealed booth with two magic boxes.
Box A has a button that generates a real number according to a normal distribution with mean 5 and standard deviation 1. Box B has a button that generates a real number according to some second fixed probability distribution. We know nothing about the second distribution only it is different from Box A. Edit: If it makes a difference to the answer we can assume some knowledge about Distribution $B$. For example it is normal and we have some bound on the mean and standard deviation. Say Mean from $-5$ to $15$ and standard deviation from $0.5$ to $1.5$.
The host goes into the booth and presses some combination of the two buttons to generate five real numbers $x_1,x_2,x_3,x_4,x_5$. The host then leaves the booth and the contestant enters. There is a camera inside the booth so the audience can see the contestant through an overhead screen, but the contestant cannot see out.
There is a screen showing $x_1 \text{_ _ _ _}$ and a button that says
reveal. The contestant can press the reveal button up to four times to reveal $x_2,x_3,x_4,x_5$. They then write either $A$ or $B$ next to each revealed number and leave the booth.
The contestant wins prize money equal to $\displaystyle \frac{ € 1000 \times \text{|correct answers|}}{\text{|numbers revealed|}}$.
I went into the booth to see $x_1 = 9.132987$. Since I knew it's very unlikely for a normal distribution to throw out something 4 standard deviations from the mean, I wrote $B$ under $x_1$ and left the booth.
Then the host generated $5$ new numbers and my evil twin entered the booth. I saw on the overhead screen that $x_1=5.134998$. I expected my twin to reveal some more numbers. But instead they wrote $B$ next to $x_1$ and left the booth.
I was surprised. "Why did you do that?" I asked my twin. They said, "Well the number $x_1$ is in the range $[5.134997,5.134999]$. Since the interval is very small I know a normal distribution is very unlikely to throw out something in that interval. So the number was probably generated by box $B$".
Unsurprisingly I won $€1000$ that day and my evil twin won nothing.
"Damn, bad luck!" that said.
As a result of their
bad luck my evil twin defaulted on their house payment and has been sleeping on my couch since then.
I know there is something wrong with my twin's reasoning but don't know how to explain it. After all their reasoning is very similar to my own reasoning about Box $A$ throwing out something bigger than $9$. It seems like the difference should somehow involve how $[9,\infty]$ is a very large set and $[5.134997,5.134999]$ is very small.
We're going on the same gameshow next week and I want my evil twin to play better next time. What should I say to convince them their strategy last month was flawed?
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I have the following expression
$$ \sum_{n=-\infty}^{+\infty} \frac{1}{(1-q^{2n-1})(1-q^{m-2n})}$$ where $m$ is an odd positive integer and $|q|<1$.
Given this, the sum should converge to a finite answer, which I would like to compute.
As $|q|<1$ I can expand the reciprocal factors in power series depending on whether $n\in (-\infty,0)$, $n\in (1,\frac{m-1}{2})$ and $n\in(\frac{m+1}{2},+\infty)$. Then by summing over the $n$ index I am left with the following expression:
$$\sum_{a,b=0}^{+\infty}\frac{q^{a+mb}-q^{ma+b}}{q^{2b}-q^{2a}} - \frac{q^{1+a+mb}+q^{1+ma+b}}{1-q^{2(a+b+1)}}$$
What I would like to do now is to compute the sums over $a,b$ but I can't seem to figure out how to do this.
Any help would be greatly appreciated.
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Let us consider a canonical system of N independent, distingusiable harmonic oscillators in 1D. Its partition function $$Z = (Z_{sp})^N = \left(\sum_{n=0}^{\infty} e^{-\beta E_i}\right)^N$$ where, $Z_{sp}$ is the single particle partition function of a simple harmonic oscillator, $E_i$ is the energy of a simple harmonic oscillator in 1D given by $(n+1/2)\hbar\omega$, $n = 0,1,2, \cdots.$
I want to know the fraction of oscillators in a given exicited state of energy $E_j$. I think the answer should be as follows.
$$\frac{e^{-\beta E_i}}{Z}$$ where, $Z$ is given above.
But I have got confused because in Atkin's
Physical Chemistry $Z_{sp}$ has been used instead of $Z$ (Eq. (16.7)). Question: Why is this so? Where am I missing the point?
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For the dynamics of open quantum systems, the Kraus operators $K_\kappa$ can be derived from the unitary orbit $U(t)\rho U(t)^\dagger$ for $\rho=\rho_S\otimes\rho_E$ of the composite system given by the time-propagators $U(t)=\exp(-\mathrm{i} tH)$ via tracing over the environmental states $|\mu\rangle$: $$K_\kappa(t)=K_{\mu\nu}(t):=\langle\mu|U(t)|\nu\rangle$$
I wanted to derive an expression for $K_\kappa$ directly in terms of the interaction Hamiltonian $H_I=\sum_\alpha S_\alpha\otimes E_\alpha$ for the whole Hamiltonian $H=H_S\otimes I_E+I_S\otimes H_E+H_I$ (assuming time-independent $H$). So I wrote $U(t)$ as the exponential $\sum_k (-\mathrm{i} tH)^k/k!$ and employed binomial expansions for the powers of sums of Hamiltonians to obtain
$$K_{\mu\nu}(t)=\sum_{\alpha,k,j\leq k,i\leq j} (-\mathrm{i} t)^k/k!{k\choose j}{j\choose i}\langle\mu|H_S^iI_S^{j-i}S_\alpha^{k-j}\otimes I_E^iH_E^{j-i} E_\alpha^{k-j}|\nu\rangle$$
How can this be simplified to get a clean formula only depending on $H_S$ and $S_\alpha$? It should be possible, as the Kraus operators only act on the reduced Hilbert space alone. In addition, how would one proceed for time-dependent $H(t)$ and their time-ordered products?
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What is conservation of flux linkage ?? Please explain with suitable diagram. I understand charge conservation and textbooks usually say that flux linkage conservation is analogous to that, but they rarely explain this in detail.
I wiil try to go through a simple application of flux conservation using an example.
I will start from something perhaps more familiar to most of us.
When you try to find final status of this circuit using charge conservation principle
you will quickly come to the conclusion that charge at time 0 yields $$ Q(0)=C_1v_1(0)+C_2v_2(0) $$ and will be conserved and shared between the two capacitor giving a final voltage $$ v_1(\infty)=v_2(\infty)=\frac{Q(\infty)}{C_1+C_2}=\frac{Q(0)}{C_1+C_2}=\frac{C_1v_1(0)+C_2v_2(0)}{C_1+C_2} $$
So what have we done? We had a two node circuit made of two capacitors:
this two capacitors were initially charged at some initial voltage initial chargewas calculated considering positive the upper capacitor armatures this chargeis conserved and shared between the two capsin final status. Upper nodewill now remain at constant voltage
just note that at final state
voltage across resistoris zero, otherwise dissipation would change system energy and so we were not at final status yet.
Now le's make it dual and use flux conservation principle.
Now we have a two mesh circuit made of two inductors: this two inductors were initially charged at some initial current
initial fluxis calculated considering positive clockwise mesh current
$$ \Phi(0)=L_1i_1(0)+L_2i_2(0) $$
this fluxis conserved and shared between the two coilsin final status Outer meshwill now remain at constant current
$$ i_1(\infty)=i_2(\infty)=\frac{\Phi(\infty)}{L_1+L_2}=\frac{\Phi(0)}{L_1+L_2}=\frac{L_1i_1(0)+L_2i_2(0)}{L_1+L_2} $$
just note that at final state current through resistoris zero, otherwise dissipation would change system energy and so we were not at final status yet.
I understand this is far from a thorough study but I believe could help to make some clear around that.
Notes:
Both electric charge and magnetic flux conservation priciples could be named toghether
flux conservation, electric flux the first one and magnetic flux the second one.
Resistors in above schematics are there just to avoid singularities in equations. Those could be rigorously coped with Dirac distributions but so far I'd rather spare the subject.
Yes Debajyoti Datta, you are right. Just make sure you do not run throught some circular reference.
I mean KCL proof may come from charge conservation, then you cannot use KCL to proof charge conservation.
The way I understand it is that the term flux linkage is used to describe how much energy is conserved going from magnetic field to current and other similar EM processes
Magnetic flux is the density of magnetic fields going through a plane * the surface area of that plane
A coil would be consider near the ideal case where the total magnetic flux would be the flux of one ring in the coil * # of rings in that coil. In this case the flux linkage would be the same as the total magnetic flux, because the current produce would almost have the same amount of energy, only losing some energy to heat dissipation.
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There is known a lot about the use of groups -- they just really appear a lot, and appear naturally. Is there any known nice use of semigroups in Maths to sort of prove they are indeed important in Mathematics? I understand that it is a research question, but may be somebody can hint me the direction to look on so that I would see sensibility of semigroups, if you see what I mean (so some replies like look for wikipedia are not working as they are anti-answers).
Semigroups provide a fundamental, algebraic tool in the analysis of regular languages and finite automata. This book chapter (pdf) by J-E Pin gives a brief overview of this area.
Am slightly surprised no one has mentioned the Galvin-Glazer proof of Hindman's theorem via the existence of semigroup structure on $\beta{\mathbb N}$, the Stone-Cech compactification of the positive integers (see, for instance, part of this note by Hindman.
The relevance to the original question is that knowing that ``compact right topological semigroups have idempotents'' may sound recondite, but it is just what was needed to answer Galvin's original question about translation-invariant ultrafilters, which was itself motivated by a "concrete" question in additive combinatorics.
On a related note, while it is in general not possible to embed a locally compact group as a dense subgroup of something compact (the map from a group to its Bohr compactification need not be injective), you can always embed it densely into various semigroups equipped with topological structure that interacts with the semigroup action: there are various of these, perhaps the most common being the WAP-compactification and the LUC-compactification. Unfortunately this often says more about the complicated behaviour of compact semitopological semigroups (and their one-sided versions) than about anything true for all locally compact groups, but the compactifications are a useful resource in some problems in analysis, and the semigroup structure gives one some extra grip on how points in this compactification behave. (Disclaimer: this is rather off my own fields of core competence.)
Victor, I don't understand your claim that $C^0$-semigroups aren't really semigroups. You are not free to decide for all the mathematical community what is a semigroup (I guess that you are interested only on discrete semigroups, aren't you ?).
$C^0$-semigroups are fundamental in PDEs (in probability too as mentioned by Steinhurst). The reason is that a lot of evolution PDEs (basically all parabolic ones, like the heat equation, or Navier-Stokes) can be solved only forward but not backward. In linear PDEs, this is a consequence of the Uniform Boundedness Principle (= Banach-Steinhaus Theorem). There is a nice theory relating operators and semigroups, the former being the generator of the latter. In the linear case, a fundamental result is the Hille-Yosida Theorem. Subsequent tools are Duhamel's principle and Trotter's formula. A part of the theory extends to nonlinear semigroups.
Edit. John B. expresses a doubt on the fundamental aspect of semigroups, compared with the evolution equations from which they arise. Let me say that semi-groups say much more, for the following reason. Evolutionary PDEs have classical solutions only when the initial data $u_0$ is smooth enough, typically when $u_0$ belongs to the so-called domain of the generator. This result can never be used to pass from a linear context to a non-linear one via the Duhamel's principle. In other words, in order to have a well-posed Cauchy-problem in Hadamard's sense, we need to invent a notion of weaker solutions ; this is where the semi-group theory comes into play.
An important application of semigroups and monoids is algebraic theory of formal languages, like regular languages of finite and infinite words or trees (one could argue this is more theoretical computer science than mathematics, but essentialy TCS
is mathematics).
For example, regular languages can be characterized using finite state automata, but can also be described by homomorphisms into finite monoids. The algebraic approach simplifies many proofs (like determinization of Buchi automata for infinite words or proving that FO = LTL) and gives deeper insight into the structure of languages.
Semigroups of bounded $L^{2}$ operators are very important in probability. They in fact provide one of the main ways show the very close connection between a self-adjoint operator and a `nice' Markov process (nice can be taken to mean strong Markov, cadlag, and quasi-left continuous.) So how does one get this semigroup from a Markov process? If $X_t$ is your process let $\mu_{t}(x,A)$ be measure with mass $\le 1$ with value $P(X_t \in A | X_0=x)$. Then $\int f(y) \mu_t(x,dy) = T_tf(x)$ gives a semigroup of bounded $L^{2}$ operators.
Why are such constructions important and natural? If $f$ is your initial distribution of something (heat for example) and $X_t$ is Brownian motion. Then $T_t$ acts by letting the heat distribution $f$ diffuse the way heat should. This then gives a nice way to connect PDE and probability theory. I'd end by offering that semigroups are important, in part, because they do arise is so many places and can bridge between disciplines. There are other reasons as well.
Circuit complexity. See
Straubing, Howard, Finite automata, formal logic, and circuit complexity. Progress in Theoretical Computer Science. Birkhäuser Boston, Inc., Boston, MA, 1994.
If you want a research problem relating circuit complexity with (finite) semigroups, there are many in the book and papers by Straubing and others. See also Eilenberg and Schutzenberger (that is in addition to Pin's book mentioned in another answer) - about connections between finite semigroups and regular languages and automata.
(Commutative) semigroups and their analysis shows up in the theory of misère combinatorial games. The "misère quotient" semigroup construction gives a natural generalization of the normal-play Sprague-Grundy theory to misere play which allows for complete analysis of (many) such games. (See http://miseregames.org/ for various papers and presentations.)
Though you say that $C_0$ semigroups are not really semigroups, the structure of compact semitopological semigroups plays an important role in the investigation of their asymptotic behaviour. For example, Glicksberg-DeLeeuw type decompositions or Tauberian theorems are obtained such a way, see Engel-Nagel: One-Parameter semigroups for Linear evolution Equations, Springer, 2000, Chapter V.2.
Unary (1-variable) functions mapping a set X to itself under composition is a semigroup. Cayley's Theorem (one of them) says that every semigroup is isomorphic to one of this kind.
Gerhard "Ask Me About System Design" Paseman, 2011.02.18
Given a group $G$, the Block Monoid $B(G)$ consists of sequences of elements in $G$ that sum to zero. So for example, an element of $B(\mathbb{Z})$ is $(-2,-3,1,1,3)$. The monoid operation is concatenation, and the empty block is the identity element.
Given a Dedekind domain, one can take its ideal class group, and consider the block monoid over that group. Note that in the obvious way, elements of the block monoid can be irreducible or not. One can study irreducible factorization in the Dedekind domain by studying irreducible factorization in the block monoid.
Why nobody explicitly mentioned one of the most natural examples: symmetric semigroup of a set? This is the semigroup anologue of a permutation group (http://en.wikipedia.org/wiki/Transformation_semigroup).
Toric varieties in Algebraic Geometry!! Indeed, the category of normal toric varieties is equivalent with the dual of the category of finitely generated, integral semigroups.
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Evaluating Riemann-Stieltjes Integrals
We will now look at evaluating some Riemann-Stieltjes integrals. Before we do, be sure to recall the results summarized below. Let $f$ and $g$ be Riemann-Stieltjes integrable function with respect to $\alpha$ and $\beta$ on the interval $[a, b]$ and let $c \in \mathbb{R}$.
Additivity of the Integrand:$\displaystyle{\int_a^b [f(x) + g(x)] \: d \alpha (x) = \int_a^b f(x) \: d \alpha (x) + \int_a^b g(x) \: d \alpha (x)}$. Homogeneity of the Integrand:$\displaystyle{\int_a^b cf(x) \: d \alpha (x) = c \int_a^b f(x) \: d \alpha (x)}$. Additivity of the Integrator:$\displaystyle{\int_a^b f(x) \: d [\alpha (x) + \beta(x)] = \int_a^b f(x) \: d \alpha (x) + \int_a^b f(x) \: d \beta (x)}$. Homogeneity of the Integrator:$\displaystyle{\int_a^b f(x) \: d[c \alpha(x)] = c \int_a^b f(x) \: d \alpha (x)}$. Integration by Parts:$\displaystyle{\int_a^b f(x) \: d \alpha (x) + \int_a^b \alpha (x) \: d f(x) = f(b)\alpha(b) - f(a)\alpha(a)}$. Reduction to a Riemann-Integral:If $f$ is bounded on $[a, b]$, $\alpha'$ exists and is continuous on $[a, b]$ then $\displaystyle{\int_a^b f(x) \: d \alpha(x) = \int_a^b f(x) \alpha'(x) \: dx}$
Let's now look at some examples.
Example 1 Evaluate the Riemann-Stieltjes integral $\int_0^1 x \: d x^2$.
We see that the integrand $f(x) = x$ is bounded on $[0, 1]$, the derivative of the integrator $\alpha (x) = x^2$ exists and is $\alpha'(x) = 2x$ and is continuous on $[0, 1]$, so we reduce the Riemann-Stieltjes integral above to get:(1)
Example 2 Evaluate the Riemann-Stieltjes integral $\int_0^{\pi} x \: d \cos x$.
Using integration by parts gives us that:(2)
Example 3 Evaluate the Riemann-Stieltjes integral $\int_0^{\pi} (x + 1) \: d (\sin x + \cos x)$.
Using the additivity of the integrand we have that:(3)
Using the additivity of the integrator and we have that:(4)
The Riemann-Stieltjes integral $\int_0^{\pi} x \: d \cos x = -\pi$ from example 1. The other three integrals can be evaluated by using Integration by parts:(5)
Therefore:(8)
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Tagged: group Problem 343
Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$. Problem 332
Let $G=\GL(n, \R)$ be the
general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices. Consider the subset of $G$ defined by \[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$. The subgroup $\SL(n,\R)$ is called special linear group Problem 322
Let $\R=(\R, +)$ be the additive group of real numbers and let $\R^{\times}=(\R\setminus\{0\}, \cdot)$ be the multiplicative group of real numbers.
(a) Prove that the map $\exp:\R \to \R^{\times}$ defined by \[\exp(x)=e^x\] is an injective group homomorphism.
Add to solve later
(b) Prove that the additive group $\R$ is isomorphic to the multiplicative group \[\R^{+}=\{x \in \R \mid x > 0\}.\]
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Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Centrality dependence of particle production in p-Pb collisions at $\sqrt{s_{\rm NN} }$= 5.02 TeV
(American Physical Society, 2015-06)
We report measurements of the primary charged particle pseudorapidity density and transverse momentum distributions in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV, and investigate their correlation with experimental ...
Measurement of jet quenching with semi-inclusive hadron-jet distributions in central Pb-Pb collisions at ${\sqrt{\bf{s}_{\mathrm {\bf{NN}}}}}$ = 2.76 TeV
(Springer, 2015-09)
We report the measurement of a new observable of jet quenching in central Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV, based on the semi-inclusive rate of charged jets recoiling from a high transverse momentum ...
Centrality dependence of inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Springer, 2015-11)
We present a measurement of inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV as a function of the centrality of the collision, as estimated from the energy deposited in the Zero Degree ...
One-dimensional pion, kaon, and proton femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm {NN}}}$ =2.76 TeV
(American Physical Society, 2015-11)
The size of the particle emission region in high-energy collisions can be deduced using the femtoscopic correlations of particle pairs at low relative momentum. Such correlations arise due to quantum statistics and Coulomb ...
Measurement of jet suppression in central Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2015-06)
The transverse momentum ($p_{\rm T}$) spectrum and nuclear modification factor ($R_{\rm AA}$) of reconstructed jets in 0-10% and 10-30% central Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV were measured. Jets were ...
Coherent $\psi(2S)$ photo-production in ultra-peripheral Pb-Pb collisions at $\sqrt{s_{\rm NN}}$= 2.76 TeV
(Elsevier, 2015-12)
The ALICE Collaboration has performed the first measurement of the coherent $\psi(2S)$ photo-production cross section in ultra-peripheral Pb-Pb collisions at the LHC. This charmonium excited state is reconstructed via the ...
Production of inclusive $\Upsilon$(1S) and $\Upsilon$(2S) in p-Pb collisions at $\mathbf{\sqrt{s_{{\rm NN}}} = 5.02}$ TeV
(Elsevier, 2015-01)
We report on the production of inclusive $\Upsilon$(1S) and $\Upsilon$(2S) in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV at the LHC. The measurement is performed with the ALICE detector at backward ($-4.46< y_{{\rm ...
Measurement of dijet $k_T$ in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2015-06)
A measurement of dijet correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector is presented. Jets are reconstructed from charged particles measured in the central tracking detectors and ...
Measurement of charged jet production cross sections and nuclear modification in p-Pb collisions at $\sqrt{s_\rm{NN}} = 5.02$ TeV
(Elsevier, 2015-10)
Charged jet production cross sections in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV measured with the ALICE detector at the LHC are presented. Using the anti-$k_{\rm T}$ algorithm, jets have been reconstructed in ...
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Linear Maps Examples 3
Recall from the Linear Maps page that a linear map or linear transformation from the vector space $V$ to the vector space $W$ is a function $T : V \to W$ such that for all $u, v \in V$ and for all $a \in \mathbb{F}$ we have that $T(u + v) = T(u) + T(v)$ (additivity property) and $T(av) = aT(v)$ (homogeneity property).
We will now look at some more example questions regarding linear maps.
Example 1 Let $T \in \mathcal L (V, W)$ and suppose that $\{ T(v_1), T(v_2), ..., T(v_m) \}$ is a set of linearly independent vectors in $W$. Prove that $\{ v_1, v_2, ..., v_m \}$ is a set of linearly independent vectors in $V$.
Let $T \in \mathcal L (V, W)$ and suppose that $\{ T(v_1) T(v_2), ..., T(v_m) \}$ is a set of linearly independent vectors in $W$.
Consider the following vector equation for $a_1, a_2, ..., a_m \in \mathbb{F}$:(1)
Now apply the linear transformation $T$ to both sides of the equation above to get that:(2)
The equation above implies that $a_1 = a_2 = ... = a_m = 0$ since $\{ T(v_1), T(v_2), ..., T(v_m) \}$ is a set of linearly independent vectors in $W$. Hence $\{ v_1, v_2, ..., v_m \}$ is a set of linearly independent vectors in $V$.
Example 2 Consider the linear map $T : \wp (\mathbb{R}) \to \mathbb{R}^2$ defined by $T(p(x)) = \left (p(1) + 2p(2) + 3p(3), \int_0^1 x^2 p(x) \: dx \right )$. Determine whether or not $T$ is a linear map from $\wp (\mathbb{R})$ to $\mathbb{R}^2$.
We will first show that the additivity property holds. Let $p(x), q(x) \in \wp (\mathbb{R})$. Then:(3)
So the additivity property holds. We will now show that the homogeneity property holds. Let $p(x) \in \wp (\mathbb{R})$ and let $a \in \mathbb{F}$. Then:(4)
Therefore the homogeneity property holds and so $T$ is indeed a linear map from $\wp (\mathbb{R})$ to $\mathbb{R}$.
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I. As a background, in Traces of Singular Moduli (p.2), Zagier defines the modular form of weight 3/2,
$$g(\tau) = \frac{\eta^2(\tau)}{\eta(2\tau)}\frac{E_4(4\tau)}{\eta^6(4\tau)}=\vartheta_4(\tau)\, \eta^2(4\tau)\,\sqrt[3]{j(4\tau)}$$
which has the nice
q-expansion (A027652, negated terms),
$$g(\tau) = 1/q - 2 + 248q^3 - 492q^4 +(15^3+744)q^7 + \dots + (5280^3+744)q^{67} + \dots + (640320^3+744)q^{163}+\dots$$
However, one can use other Eisenstein series $E_k(\tau)$ as the one below.
II. In a paper by Bruinier (p.6), Borcherds defines a modular form of weight 1/2. First let,
$$K(\tau) = \tfrac{1}{16}\big(\vartheta_3^4(\tau) - \tfrac{1}{8}\vartheta_2^4(\tau)\big)\,\vartheta_2^4(\tau)\vartheta_3(\tau)\vartheta_4^4(\tau)$$
then,
$$b(\tau) = 60\vartheta_3(\tau)+\frac{K(\tau)E_6(4\tau)}{\eta^{24}(4\tau)} = 60\vartheta_3(\tau)+\frac{K(\tau)\sqrt{j(4\tau)-1728}}{\eta^{12}(4\tau)} $$
This has the
q-expansion (A013953),
$$b(\tau) = 1/q^3 + 4 - 240q + (\color{blue}{27000}-240)q^4 - \color{blue}{85995}q^5 + \color{blue}{1707264}q^8 - (\color{blue}{4096000}+240)q^9 + \color{blue}{44330496}q^{12}-91951146q^{13}+\dots$$
I noticed that the blue numbers appear in the degrees of irreducible representations of the
Thompson group $Th$, given by the finite sequence of 48 integers (A003916),
$$1, 248, 4123, \color{blue}{27000}, 30628, 30875, 61256, \color{blue}{85995}, 147250, 767637, 779247, 957125, \color{blue}{1707264}, 2450240, 2572752, 3376737, \color{blue}{4096000}, 4123000, 4881384, 4936750,\dots\color{blue}{44330496},\dots 91171899, 111321000, 190373976.$$
with repeated terms deleted for brevity.
: Questions If it is not coincidence, what is the reason? In Bruinier (p.6) he says one can use $E_k(\tau)$ for $k=4,6,8,10,14$. What are the other functions for higher $k$? : As pointed out by S. Carnahan below, there is already a known moonshine for $Th$. A partial(?) list of others from Griess' Edit "Happy Family"can be found in Monstrous Moonshine, such as for the Higman-Sims $HS$ and so on. (Are there more?)
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Are you sure about the factor of $3$ in Koszul's formula? The computation below does not give such a factor, and it gives a stronger result.
Let $X_i$ be any basis of the left-invariant vector fields (where the indices run from $0$ to $n$), with $[X_i,X_j] = c^k_{ij}X_k$, where, of course, $c^k_{ij}=-c^k_{ji}$ (and where here, as below, we use the summation convention that we sum over repeated indices in any given term if no summation is explicitly given). Let $\langle,\rangle$ be an $\mathrm{ad}$-invariant inner product, with $\langle X_k,X_l\rangle = g_{kl}$. Then $\mathrm{ad}$-invariance is just the equation$$g_{lk}c_{ij}^k + g_{jk}c_{il}^k = 0.$$If we assume that the $X_i$ are $\langle,\rangle$-orthonormal, then $g_{kl} = \delta_{kl}$, so $c^k_{ij} = - c^j_{ik}$, i.e., $c^i_{jk}$ is skew-symmetric in
all its indices.
For simplicity, let's agree to write $c^i_{jk} = c_{ijk}$, when the $X_i$ are orthonormal with respect to an $\mathrm{ad}$-invariant $\langle,\rangle$, which I'll assume from now on.
Now consider the dual $1$-forms $\theta_i$ to the vector fields $X_i$. By the formula relating exterior derivative and Lie bracket, they satisfy$$\mathrm{d}\theta_i = -\tfrac12\,c_{ijk}\,\theta_j\wedge\theta_k= -\sum_{j<k} c_{ijk}\,\theta_j\wedge\theta_k\,.$$By definition, the Cartan form is just$$\omega = -\tfrac16\,c_{ijk}\,\theta_i\wedge\theta_j\wedge\theta_k= -\sum_{i<j<k} c_{ijk}\,\theta_i\wedge\theta_j\wedge\theta_k\,.$$
Let $\alpha = \theta_0$. Then, according to your definition, $$\omega_2 = -\sum_{0<j<k}\,c_{0jk}\,\theta_0\wedge\theta_j\wedge\theta_k= \theta_0\wedge\mathrm{d}\theta_0 = \alpha\wedge\mathrm{d}\alpha\,,$$which is a stronger result than Koszul claimed (although the factor of $3$ in his formula is mysterious), while$$\omega_3 = -\sum_{0<i<j<k}\,c_{ijk}\,\theta_i\wedge\theta_j\wedge\theta_k\,.$$
Remark: By the way, the definition that you give for 'the' Cartan form is a bit odd, because this form definitely depends on choice of the inner product, as you can see, since, if you double the metric, you'll double the Cartan form. This isn't so serious in the simple case because it's just a matter of a scale, but in the general semi-simple case, you run the risk of having many different 'Cartan forms'. However, maybe this isn't such a bad defect, since it's not possible, even in the simple case, to define 'the' Cartan form for all simple groups so that pullback to every simple subgroup $H$ of $G$ yields 'the' Cartan form of $H$. (Just look at the various $\mathrm{SU}(2)$ subgroups of $G$ when there is more than one such subgroup up to conjugacy.)
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Let $A \subset \mathbb{C}$ be an topological annulus, i.e. a region of $\mathbb{C}$ bounded by two disjoint Jordan curves.
Let $B \subset \mathbb{C}$ be a quadrilateral, i.e. a topological disc with four distinct marked points $(z_1,z_2,z_3,z_4)$ arranged anticlockwise on the boundary.
Both annuli and quadrilaterals, as defined above, have a conformal invariant, in both cases known as the modulus. Let $A_R$ be a geometrical annulus with inner boundary a circle of radius $1$ and outer boundary a circle of radius $R$. If $A$ can be mapped conformally and bijectively onto $A_R$ we say that $A$ has
modulus $\ln{R}/2\pi$. Likewise, let $Q_m$ be a geometrical rectangle with vertices $(0, m, i+m, i)$, we say that $B$ has modulus $m$ if there exists a conformal bijection mapping $B$ onto $Q_m$ with $z_1$,$z_2$,$z_3$, and $z_4$ being mapped onto $0$, $m$, $i+m$, and $i$ respectively. The two concepts are linked by the fact that if we take a geometric annulus centred on the origin with modulus $m$, remove all the points on the positive real line from the annulus, and then take the preimage under the exponential map, we obtain (infinitely many copies of) a geometric rectangle also with modulus $m$.
My questions relate to the degree in which the choice of branch cut is arbitrary.
More particularly:
Let $C \subset \mathbb{C}$ be another annulus and represent $C$ as the union $R_1 \cup R_2$ of two simply connected regions with the intersection $R_1 \cap R_2$ consisting of two smooth simple curves $\gamma_1$ and $\gamma_2$, disjoint from each other and each of which has one end point, $x_1$ and $x_2$ respectively, on the inner boundary and one end point, $y_1$ and $y_2$ on the outer boundary. Assume we have a continuous function $f_1$ mapping $R_1$ onto $A$, and such that the restriction of $f_1$ to the interior of $R_1$ is a conformal bijection and such that the image of $\gamma_1$ and $\gamma_2$ is a single simple curve from the inner to the outer boundary of $A$, with $f_1(x_1) = f_1(x_2)$ and $f_1(y_1) = f_1(y_2)$. Similarly assume that $f_2$ is a bijection mapping $R_2$ onto $B$, conformal on the interior, with $\gamma_1$ being mapped onto the component of the boundary joining $z_1$ and $z_2$ and $\gamma_2$ being mapped to the component of the boundary joining $z_3$ and $z_4$. See the diagram below.
My first question is whether the modulus of $C$ is determined entirely by the modulus of $A$ and $B$? (This seems unlikely to me, but no harm in asking)
Secondly, if this is not the case can we give any bound for the modulus of $C$ given the modulus of $A$ and $B$.
Thirdly, are there any conditions we can impose to reduce these bounds? For example specifying that the curves $\gamma_1$ and $\gamma_2$ meet the boundary curves of $C$ orthogonally.
(source: qmul.ac.uk)
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860 42
Hello Forum,
A sound wave intensity (pure frequency) is proportional to the square of the wave pressure amplitude, i.e. ##I \approx p_0^2##, where ##p_0## is the pressure wave amplitude: ##p_0 sin(\omega t \pm kx)##. This means that the (gauge) pressure value goes larger (positive) and lower (negative) than the local atmospheric pressure ##p_atm##. The sound intensity can also be expressed in dB. In that case, the reference pressure is the pressure amplitude ##p_{min}## associated to the faintest, barely audible sound: $$I (db) \approx log \frac{p_0^2}{p_{min}^2}$$ A microphones connected to sound processing software converts the impinging sound pressure into an analog voltage signal which the software displays in terms of its time-varying intensity ##I##. The analog voltage signal is sampled (sampling rate) and converted to a digital signal using a certain a bit depth (8, 16, 24 bits, etc. the higher the bit depth the better I guess). When the software reports the sound intensity ##I## as db values , the dB range goes from a maximum 0 dB down to negative dB values because the intensity is not calculated using the pressure amplitude ##p_{min}## associated to a barely audible sound but it is relative to the loudest sound pressure in the time interval. How can I make sure that the dB intensity values that the software reports are the same dBs commonly calculated and representing the sound loudness (equation above) to get a sense of how truly loud the signal is? What kind of calibration is needed? I know the microphone and software have gains, etc... thanks!!!!
A sound wave intensity (pure frequency) is proportional to the square of the wave pressure amplitude, i.e. ##I \approx p_0^2##, where ##p_0## is the pressure wave amplitude: ##p_0 sin(\omega t \pm kx)##. This means that the (gauge) pressure value goes larger (positive) and lower (negative) than the local atmospheric pressure ##p_atm##.
The sound intensity can also be expressed in dB. In that case, the reference pressure is the pressure amplitude ##p_{min}## associated to the faintest, barely audible sound: $$I (db) \approx log \frac{p_0^2}{p_{min}^2}$$
A microphones connected to sound processing software converts the impinging sound pressure into an analog voltage signal which the software displays in terms of its time-varying intensity ##I##. The analog voltage signal is sampled (sampling rate) and converted to a digital signal using a certain a bit depth (8, 16, 24 bits, etc. the higher the bit depth the better I guess).
When the software reports the sound intensity ##I## as db values , the dB range goes from a maximum 0 dB down to negative dB values because the intensity is not calculated using the pressure amplitude ##p_{min}## associated to a barely audible sound but it is relative to the loudest sound pressure in the time interval.
How can I make sure that the dB intensity values that the software reports are the same dBs commonly calculated and representing the sound loudness (equation above) to get a sense of how truly loud the signal is?
What kind of calibration is needed? I know the microphone and software have gains, etc...
thanks!!!!
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2018-08-25 06:58
Recent developments of the CERN RD50 collaboration / Menichelli, David (U. Florence (main) ; INFN, Florence)/CERN RD50 The objective of the RD50 collaboration is to develop radiation hard semiconductor detectors for very high luminosity colliders, particularly to face the requirements of the possible upgrade of the large hadron collider (LHC) at CERN. Some of the RD50 most recent results about silicon detectors are reported in this paper, with special reference to: (i) the progresses in the characterization of lattice defects responsible for carrier trapping; (ii) charge collection efficiency of n-in-p microstrip detectors, irradiated with neutrons, as measured with different readout electronics; (iii) charge collection efficiency of single-type column 3D detectors, after proton and neutron irradiations, including position-sensitive measurement; (iv) simulations of irradiated double-sided and full-3D detectors, as well as the state of their production process.. 2008 - 5 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 596 (2008) 48-52 In : 8th International Conference on Large Scale Applications and Radiation Hardness of Semiconductor Detectors, Florence, Italy, 27 - 29 Jun 2007, pp.48-52 Detailed record - Similar records 2018-08-25 06:58 Detailed record - Similar records 2018-08-25 06:58
Performance of irradiated bulk SiC detectors / Cunningham, W (Glasgow U.) ; Melone, J (Glasgow U.) ; Horn, M (Glasgow U.) ; Kazukauskas, V (Vilnius U.) ; Roy, P (Glasgow U.) ; Doherty, F (Glasgow U.) ; Glaser, M (CERN) ; Vaitkus, J (Vilnius U.) ; Rahman, M (Glasgow U.)/CERN RD50 Silicon carbide (SiC) is a wide bandgap material with many excellent properties for future use as a detector medium. We present here the performance of irradiated planar detector diodes made from 100-$\mu \rm{m}$-thick semi-insulating SiC from Cree. [...] 2003 - 5 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 509 (2003) 127-131 In : 4th International Workshop on Radiation Imaging Detectors, Amsterdam, The Netherlands, 8 - 12 Sep 2002, pp.127-131 Detailed record - Similar records 2018-08-24 06:19
Measurements and simulations of charge collection efficiency of p$^+$/n junction SiC detectors / Moscatelli, F (IMM, Bologna ; U. Perugia (main) ; INFN, Perugia) ; Scorzoni, A (U. Perugia (main) ; INFN, Perugia ; IMM, Bologna) ; Poggi, A (Perugia U.) ; Bruzzi, M (Florence U.) ; Lagomarsino, S (Florence U.) ; Mersi, S (Florence U.) ; Sciortino, Silvio (Florence U.) ; Nipoti, R (IMM, Bologna) Due to its excellent electrical and physical properties, silicon carbide can represent a good alternative to Si in applications like the inner tracking detectors of particle physics experiments (RD50, LHCC 2002–2003, 15 February 2002, CERN, Ginevra). In this work p$^+$/n SiC diodes realised on a medium-doped ($1 \times 10^{15} \rm{cm}^{−3}$), 40 $\mu \rm{m}$ thick epitaxial layer are exploited as detectors and measurements of their charge collection properties under $\beta$ particle radiation from a $^{90}$Sr source are presented. [...] 2005 - 4 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 546 (2005) 218-221 In : 6th International Workshop on Radiation Imaging Detectors, Glasgow, UK, 25-29 Jul 2004, pp.218-221 Detailed record - Similar records 2018-08-24 06:19
Measurement of trapping time constants in proton-irradiated silicon pad detectors / Krasel, O (Dortmund U.) ; Gossling, C (Dortmund U.) ; Klingenberg, R (Dortmund U.) ; Rajek, S (Dortmund U.) ; Wunstorf, R (Dortmund U.) Silicon pad-detectors fabricated from oxygenated silicon were irradiated with 24-GeV/c protons with fluences between $2 \cdot 10^{13} \ n_{\rm{eq}}/\rm{cm}^2$ and $9 \cdot 10^{14} \ n_{\rm{eq}}/\rm{cm}^2$. The transient current technique was used to measure the trapping probability for holes and electrons. [...] 2004 - 8 p. - Published in : IEEE Trans. Nucl. Sci. 51 (2004) 3055-3062 In : 50th IEEE 2003 Nuclear Science Symposium, Medical Imaging Conference, 13th International Workshop on Room Temperature Semiconductor Detectors and Symposium on Nuclear Power Systems, Portland, OR, USA, 19 - 25 Oct 2003, pp.3055-3062 Detailed record - Similar records 2018-08-24 06:19
Lithium ion irradiation effects on epitaxial silicon detectors / Candelori, A (INFN, Padua ; Padua U.) ; Bisello, D (INFN, Padua ; Padua U.) ; Rando, R (INFN, Padua ; Padua U.) ; Schramm, A (Hamburg U., Inst. Exp. Phys. II) ; Contarato, D (Hamburg U., Inst. Exp. Phys. II) ; Fretwurst, E (Hamburg U., Inst. Exp. Phys. II) ; Lindstrom, G (Hamburg U., Inst. Exp. Phys. II) ; Wyss, J (Cassino U. ; INFN, Pisa) Diodes manufactured on a thin and highly doped epitaxial silicon layer grown on a Czochralski silicon substrate have been irradiated by high energy lithium ions in order to investigate the effects of high bulk damage levels. This information is useful for possible developments of pixel detectors in future very high luminosity colliders because these new devices present superior radiation hardness than nowadays silicon detectors. [...] 2004 - 7 p. - Published in : IEEE Trans. Nucl. Sci. 51 (2004) 1766-1772 In : 13th IEEE-NPSS Real Time Conference 2003, Montreal, Canada, 18 - 23 May 2003, pp.1766-1772 Detailed record - Similar records 2018-08-24 06:19
Radiation hardness of different silicon materials after high-energy electron irradiation / Dittongo, S (Trieste U. ; INFN, Trieste) ; Bosisio, L (Trieste U. ; INFN, Trieste) ; Ciacchi, M (Trieste U.) ; Contarato, D (Hamburg U., Inst. Exp. Phys. II) ; D'Auria, G (Sincrotrone Trieste) ; Fretwurst, E (Hamburg U., Inst. Exp. Phys. II) ; Lindstrom, G (Hamburg U., Inst. Exp. Phys. II) The radiation hardness of diodes fabricated on standard and diffusion-oxygenated float-zone, Czochralski and epitaxial silicon substrates has been compared after irradiation with 900 MeV electrons up to a fluence of $2.1 \times 10^{15} \ \rm{e} / cm^2$. The variation of the effective dopant concentration, the current related damage constant $\alpha$ and their annealing behavior, as well as the charge collection efficiency of the irradiated devices have been investigated.. 2004 - 7 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 530 (2004) 110-116 In : 6th International Conference on Large Scale Applications and Radiation Hardness of Semiconductor Detectors, Florence, Italy, 29 Sep - 1 Oct 2003, pp.110-116 Detailed record - Similar records 2018-08-24 06:19
Recovery of charge collection in heavily irradiated silicon diodes with continuous hole injection / Cindro, V (Stefan Inst., Ljubljana) ; Mandić, I (Stefan Inst., Ljubljana) ; Kramberger, G (Stefan Inst., Ljubljana) ; Mikuž, M (Stefan Inst., Ljubljana ; Ljubljana U.) ; Zavrtanik, M (Ljubljana U.) Holes were continuously injected into irradiated diodes by light illumination of the n$^+$-side. The charge of holes trapped in the radiation-induced levels modified the effective space charge. [...] 2004 - 3 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 518 (2004) 343-345 In : 9th Pisa Meeting on Advanced Detectors, La Biodola, Italy, 25 - 31 May 2003, pp.343-345 Detailed record - Similar records 2018-08-24 06:19
First results on charge collection efficiency of heavily irradiated microstrip sensors fabricated on oxygenated p-type silicon / Casse, G (Liverpool U.) ; Allport, P P (Liverpool U.) ; Martí i Garcia, S (CSIC, Catalunya) ; Lozano, M (Barcelona, Inst. Microelectron.) ; Turner, P R (Liverpool U.) Heavy hadron irradiation leads to type inversion of n-type silicon detectors. After type inversion, the charge collected at low bias voltages by silicon microstrip detectors is higher when read out from the n-side compared to p-side read out. [...] 2004 - 3 p. - Published in : Nucl. Instrum. Methods Phys. Res., A 518 (2004) 340-342 In : 9th Pisa Meeting on Advanced Detectors, La Biodola, Italy, 25 - 31 May 2003, pp.340-342 Detailed record - Similar records 2018-08-23 11:31
Formation and annealing of boron-oxygen defects in irradiated silicon and silicon-germanium n$^+$–p structures / Makarenko, L F (Belarus State U.) ; Lastovskii, S B (Minsk, Inst. Phys.) ; Korshunov, F P (Minsk, Inst. Phys.) ; Moll, M (CERN) ; Pintilie, I (Bucharest, Nat. Inst. Mat. Sci.) ; Abrosimov, N V (Unlisted, DE) New findings on the formation and annealing of interstitial boron-interstitial oxygen complex ($\rm{B_iO_i}$) in p-type silicon are presented. Different types of n+−p structures irradiated with electrons and alpha-particles have been used for DLTS and MCTS studies. [...] 2015 - 4 p. - Published in : AIP Conf. Proc. 1583 (2015) 123-126 Detailed record - Similar records
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Compare these two results:
Theorem (Scott):$ZFC+V=L\vdash \nexists~\text{Measurable cardinal}$
Theorem (Kunen):$ZFC+AC\vdash \nexists~\text{Reinhardt cardinal}$
Now compare these two chains:
$V=L\rightarrow AC\rightarrow \cdots$
$Con(ZF+ \exists~\text{Measurable cardinal})\leftarrow Con(ZF+\exists~\text{Reinhardt cardinal})\leftarrow \cdots$
By an imaginary correspondence it seems we may assume:
Large cardinals are contradictory with choice-like axioms. i.e. Larger large cardinals are contradictory with weaker versions of choice. (See the imaginary diagram below).
In the other words, based on the relation $V=L\rightarrow GCH\rightarrow AC\rightarrow AC_{\omega}\rightarrow \cdots$ we may expect the followings:
(a)There is a large cardinal axiom above the Reinhardt cardinals (e.g. Berkeley cardinals) which is provably inconsistent with the axiom of countable choice, $AC_{\omega}$.
(b)There is a large cardinal axiom between measurable and Reinhardt cardinals which is inconsistent with $GCH$.
In some sense, the statement (b) is true if we assume Shelah's Proper Forcing Axiom (PFA) as a large cardinal axiom, however if we remove PFA from our large cardinal hierarchy, it is not known yet.
Let's focus on the statement (a) which is a interpretation-free statement. We may ask two different questions here:
Question 1:What is the weakest version of the axiom of choice which is necessary to prove inconsistency of Berkeley cardinals?
Question 2:Within $ZF$, what is the weakest large cardinal axiom (above Reinhardt) which is provably inconsistent with $AC_{\omega}$? Remark: For more on equivalent versions and consequences of Berkeley cardinals see here.
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I want to ask for some help with my OR model. What guidance is available to me for doing so?
How to ask for help with your OR model
We love mathematical models and are happy to help you with yours. Before you ask your question, please read these suggestions about how to ask a good question. Doing so will make it easier for us to provide good answers.
Use a descriptive title.Include (some of) the relevant details and be specific. Avoid vague, general titles like "Why isn't my model working?" Good titles will catch the reader's eye and make us want to read the question. Include your algebraic model.Include your entire algebraic formulation, or at least the relevant portion of it. Write your formulation using MathJax. Explain all of the notation and what the objective function and constraints are doing, if it's not obvious. Include minimal, reproducible code.If you are including code (AMPL, GAMS, Python, etc.), use a minimal, complete, verifiable example. That is, give us enough code so that we can run it, reproduce your problem/error, and try to debug it. Don't post very long code listings that contain a lot of stuff that's irrelevant to the problem. Be specific about what's not working.Tell us exactly what's not working the way you want it to. Are you getting error messages? Include them. Are you getting results that don't seem logical or numerically correct? Include them, and tell us what you think is wrong with them. Keep it self-contained.Your question should be as self-contained as possible. Avoid posts that require us to click on external images just to see your formulation, or to read a journal article to understand what you want your model to do. External links are good if they provide additional context, but the question should be able to stand on its own even if we don't click on the links. Use themodeling tag.This tag will let users know that your question is one that needs help with a modeling problem. A good example Title: Why does AMPL/CPLEX give me "nonlinear" error for a knapsack-type problem? Question: I am trying to modify a 0–1 knapsack problem to require that at least half of the items chosen are "priority" items. But I'm getting an error message in AMPL that I don't understand.
Here is my model. We have $n$ items, each with a weight $w_i$ and a value $v_i$. For item $i$, $p_i$ = 1 if the item is a "priority" item and 0 otherwise. The knapsack has a capacity of $W$. (These are all parameters.) The decision variables are $x_i$, which equals 1 if we choose the item, and 0 otherwise. The integer programming formulation is:
$$\begin{alignat}{2} \text{maximize} \quad & \sum_{i=1}^n v_ix_i && \\ \text{subject to} \quad & \sum_{i=1}^n w_ix_i \le W &\quad& \forall i=1,\ldots,n \\ & \frac{\sum_{i=1}^n p_ix_i}{\sum_{i=1}^n x_i} \ge 0.5 \\ & x_i \in \{0,1\} && \forall i=1,\ldots,n \end{alignat}$$
The first and third constraints are standard knapsack constraints. The second constraint says at least half of the items chosen have to be priority items.
I implemented my model in AMPL. Here is my *.mod file.
param n; # number of itemsset ITEMS = 1..n; # set of itemsparam weight{ITEMS}; # weight of each itemparam value{ITEMS}; # value of each itemparam is_priority{ITEMS} binary; # 1/0 if item is priority/notparam capacity; # knapsack capacityvar x{ITEMS} binary; # do we choose the item?maximize TotalValue: sum {i in ITEMS} value[i] * x[i];subj to Capacity: sum {i in ITEMS} weight[i] * x[i] <= capacity;subj to HalfPriority: (sum {i in ITEMS} is_priority[i] * x[i]) / (sum {i in ITEMS} x[i]) >= 0.5;
Here is a simplified version of my *.dat file. (My real *.dat file has
n > 500.)
param n := 4;param: weight value is_priority :=1 20 50 12 15 40 03 15 55 04 20 30 1 ;param capacity := 40;
I am using CPLEX as the solver. When I solve the model, I get the following error message:
CPLEX 12.8.0.0: Constraint _scon[1] is a nonquadratic nonlinear constraint.
What am I doing wrong?
(
Spoiler: The "priority" constraint is nonlinear. Just multiply both sides by the denominator.) A bad example Title: Question about solver Question: I want the knapsack problem to include a constraint that half the items are priority items.
Here is the constraint in AMPL:
subj to HalfPriority: (sum {i in ITEMS} is_priority[i] * x[i]) / (sum {i in ITEMS} x[i]) >= 0.5;
But AMPL gives me an error. What am I doing wrong?
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Pointwise Convergence of Fourier Series Review
Pointwise Convergence of Fourier Series Review
We will now review some of the recent material regarding pointwise convergence of Fourier series.
Recall from the Periodic Functionspage that a function $f$ is said to be $p$-Periodic($p > 0$) if the following equality holds for all $x \in D(f)$:
\begin{align} \quad f(x + p) = f(x) \end{align}
We noted that the trigonometric functions $\sin x$, $\cos x$, and $\tan x$ are all $2\pi$-periodic functions. We also looked at some examples of functions with different periods. On the Dirichlet's Kernelpage we looked at a special collection of functions known as Dirichlet's Kernelwhere for each $n \in \mathbb{N}$, the function $D_n$ is defined as:
\begin{align} \quad D_n(x) = \frac{1}{2} + \sum_{k=1}^{n}\cos kt \end{align}(3)
\begin{align} \quad D_n(x) = \left\{\begin{matrix} n + \frac{1}{2} & \mathrm{if} \: x = 2m\pi, m \in \mathbb{Z} \\ \frac{\sin \left ( \left ( n + \frac{1}{2} \right ) t \right )}{2 \sin \left ( \frac{t}{2} \right )} & \mathrm{if} \: x \neq 2m \pi, m \in \mathbb{Z} \\ \end{matrix}\right. \end{align}
The graphs of the first four functions $D_1$, $D_2$, $D_3$, and $D_4$ of Dirichlet's kernel are graphed below: We then looked at some important properties of the functions in Dirichlet's Kernel. We first noted that for all $n \in \mathbb{N}$, $D_n$ is an even function since each $D_n$ is a finite sum of even functions. Also, each $D_n$ is $2\pi$-periodic. Lastly, we noted that for each $n \in \mathbb{N}$:
\begin{align} \quad \int_0^{2\pi} D_n(t) \: dt = \pi \end{align}
On the Dirichlet's Kernel Representation of the Partial Sums of a Fourier Serieswe began to investigate the usefulness of Dirichlet's kernel. We proved a very important result which said that if $f \in L([0, 2\pi])$ is a $2\pi$-periodic function, then if for each $n \in \mathbb{N}$ we let $s_n$ be the $n^{\mathrm{th}}$ partial sum of the Fourier series of $f$ generated by the trigonometric system, then we can actually obtain an integral representation for $s_n(x)$, namely:
\begin{align} \quad s_n(x) = \frac{2}{\pi} \int_0^{2\pi} \frac{f(x + t) + f(x - t)}{2} D_n(t) \: dt \end{align}
On the The Riemann Localization Theorempage we used this result to prove a strong result regarding convergence of a Fourier series. We proved that if $f \in L([0, 2\pi])$ is a $2\pi$-periodic function then the Fourier series of $f$ generated by the trigonometric system converges at a point $x$ if and only if there exists a $b \in \mathbb{R}$ with $0 < b \leq \pi$ for which the following limit exists:
\begin{align} \quad \lim_{n \to \infty} \frac{2}{\pi} \int_0^{b} \frac{f(x + t) + f(x - t)}{2} \frac{\sin \left (\left ( n + \frac{1}{2} \right ) t \right )}{t} \: dt \end{align}
If this limit does exist for some $x$, then the Fourier series of $f$ generated by the trigonometric series will converge at $x$ to this limit. On the Jordan's and Dini's Tests for Convergence of a Fourier Series at a Pointpage, we then connected previous results to obtain two important convergence tests for Fourier series. Notice that the integral above is a Dirichlet integral if we set (for a fixed $x$):
\begin{align} \quad g(t) = \frac{f(x + t) + f(x - t)}{2} \end{align}
So by Jordan's Theorem we obtain Jordan's test for convergence which says that if $f \in L([0, 2\pi])$ is a $2\pi$-periodic function and if $f$ is of bounded variation on $[x - \delta, x + \delta]$ for some $\delta$ with $0 < \delta \leq b$ then the Fourier series of $f$ generated by the trigonometric system will converge at $x$ to $\displaystyle{g(0+) = \lim_{t \to 0+} \frac{f(x + t) + f(x - t)}{2}}$. Also, by Dini's Theorem we obtain Dini's test for convergence which says that if $f \in L([0, 2\pi])$ is a $2\pi$-periodic function and $g(0+)$ exists and Lebesgue integral $\displaystyle{\int_0^{\delta} \frac{g(t) - g(0+)}{t} \: dt}$ exists for some $\delta$ with $0 < \delta < \pi$, then the Fourier series of $f$ generated by the trignonometric system will converge at $x$ to $\displaystyle{g(0+) = \lim_{t \to 0+} \frac{f(x + t) + f(x - t)}{2}}$.
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What is Wave Optics?
Wave optics stands as a witness for a famous standoff between two great scientific communities who dedicated their lives to understand the nature of light. One supports the particle nature of light; the other supports the wave nature. Sir Isaac Newton stands as a prominent figure that supported the voice of particle nature of light, who proposed corpuscular theory which states that, “light consists of extremely light and tiny particles known as corpuscles which travel with very high speeds from the source of light to create sensation of vision by reflecting on the retina of the eye”. Using this theory Newton was able to explain reflection and refraction but failed to explain the cause of interference, diffraction and polarization. The major failure of Newton’s corpuscular theory is it could not explain why velocity of light is lesser in denser medium compared to vacuum.
Huygens Wave Theory
No one dared to challenge Newton’s corpuscular theory until Christopher Huygens proposed his wave theory of light in early 18
th century. According to Huygens theory, light consists of waves which travel through a very dilute and highly elastic material medium present everywhere in space”. This medium is called as ether. As the medium is supposed to be very dilute and highly elastic, its density would be very low and modulus of elasticity would be very high so that speed of the light would be very large.
Huygens wave theory could able to explain the phenomena like reflection, refraction, interference and diffraction of light. But failed to explain:
Polarization as Huygens assumed light waves to be mechanical disturbances which are longitudinal in nature. Black body radiation, photo electric effect and Compton Effect. Hypothetical medium etherwhich was never discovered and now we know light can propagate in vacuum. Maxwell Electromagnetic Theory
According to Maxwell, light is not a mechanical wave. It is an electromagnetic wave which is transverse in nature which travels with a finite speed given by\(C=\frac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}=3\times {{10}^{8}}{}^{m}/{}_{s}\)
Wavefront and Wave Normal What is Wavefront?
A
wavefront is defined as the laws of all points of the medium which vibrate in the same phase. Depending on the shape of the source of light, wavefronts can be of three types. Spherical Wave Front When light is emerging from a point source, the wavefronts are spherical in shape.
In spherical wavefront,
Amplitude of light waves, \(A\propto \frac{1}{r}\)
and, Intensity of light waves, \(I\propto \frac{1}{{{r}^{2}}}\)
Cylindrical Wave Front
When the source of light is linear, the wavefronts are cylindrical in shape. All the points are equidistant from the source.
In this case, Amplitude of light waves, \(A\propto \frac{1}{\sqrt{r}}\)
Intensity of light waves \(I\propto \frac{1}{r}\)
Plane Wave Front
when the light is coming from a very far-off source, the wavefronts are planar. For a plane wavefront, Amplitude remains constant therefore Intensity remains constant.
What is Wave Normal?
A perpendicular drawn to the surface of a wavefront at any point, in the direction of propagation of light, is called as “wave normal”. The direction in which light travels is called as a ‘ray’ of light. Thus, a wave normal is same as a ray of light.
Shape of Wavefronts
A lens can be used to change the shape of wavefronts. The concept of wavefronts for reflection and refraction is explained below.
Wavefronts For Reflection If light falls on a plane mirror:
If the plane wavefronts are being reflected on the plane mirror, the shape of wavefront of the reflected light is again planar.
If light falls on a concave mirror; convex mirror:
If a plane wavefront falls on a concave mirror, the shape of the reflected light is spherical.
If a plane wavefront falls on a convex mirror, the shape of the reflected light is spherical.
Wavefronts For Refraction If light falls on plane surfaces:
If a plane wave front falls on a plane surface the refracted ray will also have a plane wave front.
If light falls on curved surfaces:
If a plane wavefront falls on a converging (or) diverging lens, the emergent light will have a spherical wavefront.
Check Your Understanding:
A wavefront is represented by the plane \(y=8-\sqrt{3}x\). Find the direction of propagation of the wave.
Solution: Given, \(y=8-\sqrt{3}x,\) which is straight line equation. So, the wavefront can be represented as a straight line having a slope \(\tan \theta =-\sqrt{3}.\)
As it has a negative slope, the wavefront is represented like
∴ wavefront makes an angle of \(150{}^\circ\) with x-axis. As wave must be perpendicular to wavefront, it will make an angle of \(60{}^\circ\) with the x-axis.
Huygen’s Principle
According to Huygen’s principle, Every point on a given wavefront can be regarded as a fresh source of new disturbance and sends out its own spherical wavelets called secondary wavelets. These secondary wavelets spread out in all directions with the velocity of the wave. A surface touching these secondary wavelets tangentially in the forward direction at any instant \(\left( \Delta t \right)\) gives the position and shape of the new wavefront at the instant. This is called “secondary wavefront”.
As discussed in the introduction Huygen’s principle of secondary wavelet’s could able to explain many optical phenomena like reflection, refraction, interference, and diffraction, but could not explain why wavefronts of secondary wavelets are formed in forward direction not in backward direction.
Coherent and Incoherent Sources
Two sources which emit a monochromatic light continuously with a zero (or) constant phase difference between them are called coherent sources. The sources which do not emit light with constant phase difference are called as Incoherent sources.\(S=\frac{D}{d}\left( \mu -1 \right)t\)
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first of all, I need to confess my ignorance with respect to any physics since I'm a mathematician. I'm interested in the physical intuition of the Langlands program, therefore I need to understand what physicists think about homological mirror symmetry which comes from S-duality. This question is related to my previous one Intuition for Homological Mirror Symmetry.
As I have heard everything starts with an $S$-duality between two $N= 4$ super-symmetric Yang-Mills gauge theories of dimension $4$, $(G, \tau)$ and $(^{L}G, \frac{-1}{n_{\mathfrak{g}}\tau})$, where $\tau = \frac{\theta}{2\pi} + \frac{4\pi i}{g^2}$, $G$ is a compact connected simple Lie group and $n_{\mathfrak{g}}$ is the lacing number (the maximal number of edges connecting two vertices in the Dynkin diagram) . And, then the theory would be non-perturbative, since it would be defined "for all" $\tau$, because amplitudes are computed with an expansion in power series in $\tau$
So I need to understand what this would mean to a physicist.
1) First of all, what's the motivation form the Yang-Mills action and how should I understand the coupling constants $\theta$ and $g$?
2) How can I get this so called expansion in power series with variable $\tau$ of the probability amplitude?
3) What was the motivation to start looking at this duality? A creation of an everywhere defined (in $\tau$) gauge theory, maybe?
Thanks in advance.
This post imported from StackExchange Physics at 2015-03-17 04:40 (UTC), posted by SE-user user40276
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
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Criteria for a Set to be Closed in a Metric Space
Recall from the Adherent, Accumulation and Isolated Points in Metric Spaces page that if $(M, d)$ is a metric space and $S \subseteq M$ then:
A point $x \in M$ is said to be an adherent point of $S$ if every ball centered at $x$ contains points of $S$, that is, for all $r > 0$ we have that $B(x, r) \cap S \neq \emptyset$. A point $x \in M$ is said to be an accumulation point of $S$ if every ball centered at $x$ contains points of $S$ different from $x$, that is, for all $r > 0$ we have that $B(x, r) \cap S \setminus \{ x \} \neq \emptyset$. A point $x \in S$ is said to be an isolated point of $S$ if there exists a ball centered at $x$ that contains no other points of $S$, that is, there exists an $r_0 > 0$ such that $B(x, r_0) \cap S = \emptyset$.
We will now look at a nice theorem which gives us criteria for the set $S$ to be closed with regards to adherent and accumulation points.
Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$. Then the following statements are equivalent: a) $S$ is a closed set in $M$. b) $S$ contains all of its adherent points. c) $S$ contains all of its accumulation points. Proof of $a) \implies b)$:Let $S$ be a closed set. Then $S^c = M \setminus S$ is an open set. Suppose that $S$ does not contain all of its adherent points. Then there exists an $x \in S^c = M \setminus S$ such that all ball centered at $x$ contains points of $S$, i.e., for all $r > 0$ we have that: Since $S^c$ is an open set we have that there exists a ball centered at $x$ with radius $r_0 > 0$ such that: But $x \not \in S$ and $B(x, r_0) \subseteq M \setminus S$ so $(B,x, r_0) \cap S = \emptyset$ which is a contradiction to the assumption that $S$ does not contain all of its adherent points. Hence if $S$ is closed then $S$ contains all of its adherent points. Proof of $b) \implies c)$Suppose that $S$ contains all of its adherent points. Every accumulation point is an adherent point, so $S$ also contains all of its accumulation points. Proof of $c) \implies a)$:Suppose that $S$ contains all of its accumulation points and assume that $S$ is not closed. Then $S^c$ is not open. So there exists a point $x \in S^c$ such that $x \not \in \mathrm{int} (S^c)$, i.e., for every ball centered at $x$ with radius $r$ we have that: So for all $r > 0$ we must have that $B(x, r) \cap S \neq \emptyset$. Since $x \in S^c$, we have that $x \not \in S$ and so: But then $x \in S^c$ is an accumulation point of $S$ that is not contained in $S$ which is a contradiction. Hence the assumption that $S$ was not closed is false and so $S$ is closed. $\blacksquare$
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Proofs Regarding Functions
We will now look at some proofs regarding functions, direct images, inverse images, etc… Before we look at such proofs, let's first recall some very important definitions:
Definition: A function $f$ is called an injection, injective function, or one-to-one function if for all $m, n \in A$ where $m \neq n$ satisfies that $f(m) \neq f(n)$.
Definition: A function $f$ is called a surjection, surjective function, or an onto function if $B = R(f)$, that is, the codomain of $f$ is equal to the range of $f$. For every $b \in B$, $b = f(a)$ for some $a \in A$.
Definition: A function $f$ is called a bijection or bijective function if $f$ is both injective and surjective, that is, for all $m, n \in A$ where $m \neq n$ satisfies $f(m) \neq f(n)$, and $B = R(f)$, so for every $b \in B$, $b = f(a)$ for some $a \in A$.
We also want to acknowledge the following implications that will be very useful:
$f(x) \in f(E)$ implies that there exists an element $e \in E$ such that $f(e) = f(x)$. $x \in f^{-1}(H)$ implies that there exists an element $f(x) \in H$ such that $f^{-1}(f(x)) = x$. Example 1 Prove that if $f : A \to B$ is a bijective function then $\forall a \in A$, $f^{-1} ( f(a)) = a$.
We note that from the context of our example that the notation $f^{-1}$ represents the inverse function and NOT an inverse image.
Proof:Let $a \in A$, and let $x = f^{-1}(f(a))$. By the definition of the inverse function we have that $x = f^{-1}(f(a))$ implies that $f(x) = f(a)$. Now since $f$ is a bijective function, it is injective by definition, so then if $f(x) = f(a)$ then $x = a$. So $a = f^{-1}(f(a))$. Example 2 Prove that if $f : A \to B$ is a bijective function then $\forall b \in B$, $f(f^{-1}(b)) = b$.
Once again we note from our context that $f^{-1}$ represents the inverse function.
Proof:Let $b \in B$. Since $f$ is a bijective function it is surjective by definition, so there exists an $a \in A$ such that [$f(a) = b$. Taking the inverse of both sides we get that $a = f^{-1}(b)$. And since $f$ is a function then $f(a) = f(f^{-1}(b))$. But $f(a) = b$ and so $b = f(f^{-1}(b))$. Example 3 Prove that if $f : A \to B$ is a surjective function and $H \subseteq B$ then $f(f^{-1}(H)) = H$.
We note from the context of our example that $f^{-1}$ represents the inverse image of a set and NOT an inverse function. Therefore we have to prove the set $f(f^{-1}(H)) \subseteq H$ and $H \subseteq f(f^{-1}(H))$.
Proof:$\Rightarrow$ We first want to show that $f(f^{-1}(H)) \subseteq H$. Let $f(x) \in f(f^{-1}(H))$. It follows by the definition of the direct image of a set that then $x \in f^{-1} (H)$. And by the definition of the inverse image image of a set we have that $f(x) \in H$. So therefore $f(f^{-1}(H)) \subseteq H$. $\Leftarrow$ We now want to show that $H \subseteq f(f^{-1}(H))$. Let $f(x) \in H$. Now since $f$ is surjective there exists an element $x \in A$ such that $f(x) = y$. So then $y \in H$. This implies that $x \in f^{-1} (H)$ and so $f(x) \in f(f^{-1} (H)$. Therefore we have that $f(f^{-1}(H)) = H$. Example 4 Prove that if $f : A \to B$ is an injective function and $E \subseteq A$ then $f^{-1} (f (E)) = E$.
We note that from the context of our example that the notation $f^{-1}$ represents the inverse image of a set and NOT an inverse function. Therefore we have to prove the set $f^{-1} ( f(E)) = E$ by first showing that $f^{-1} (f (E)) \subseteq E$ and $E \subseteq f^{-1} (f (E))$.
Proof:$\Rightarrow$ We first want to show that $f^{-1} (f (E)) \subseteq E$. Let $x \in f^{-1} (f (E))$. It follows by the definition of the direct image of a set that since $f(x) \in f(E)$, there exists an element $e \in E$ such that $f(x) = f(e)$. Now since $f$ is an injective function we have that $x = e$ and so then $x \in E$. Therefore $f^{-1} (f (E)) \subseteq E$. $\Leftarrow$ We now want to show that $E \subseteq f^{-1} (f (E))$. Let $x \in E$. By the definition of direct image we have that $f(x) \in f(E)$. Furthermore, $f^{-1}(f(x)) = x \in f^{-1}(f(E))$ by the definition of an inverse image of a set. So $x \in f^{-1} (f (E))$ and therefore $E \subseteq f^{-1} (f (E))$. We conclude $f^{-1} (f (E)) = E$ if $f$ is injective.
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I need help with this problem. Some numerical answers would also be very helpful.
I'll augment the correct answers in the comments with the following circuit model. A resistor with noise can be modeled as a noiseless resistor in series with a voltage source representing the Johnson (thermal) noise. More on that here. So we have the following model:
Where \$V_{johnson}\$ is a pure white noise source with spectral density \$ \sqrt{4k_BTR} \$.
Now to get the rms at the voltmeter, we start with the low pass formed by the resistance \$R\$ and the capacitance \$C\$. Find the equivalent noise bandwidth which is \$ \frac\pi{2}\frac1{2 \pi RC} \$.
Take the sqare root of the bandwidth times the spectral density to get the RMS voltage at the input to the amplifier. Then multiply by \$G\$ to get the RMS voltage at the voltmeter
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Let $C$ be a smooth irreducible (complex) curve of genus $g\geq2$.
The
gonality of $C$ is defined as the minimum degree of surjective morphisms $C\rightarrow\Bbb{P}^1$. So $C$ has gonality $d$ if it has a $g^1_d$ but no $g^1_{d-1}$, where $g^r_d$ denotes a linear system of dimension $r$ and degree $d$ on $C$.
Now my question is very basic, I don't understand the following remark:
If $C$ has gonality $d$ then each $g^1_d$ on $C$ is base point free and complete.
First, why are they complete? But even if they are complete, so we have the system $|A|$ with $A$ divisor on $C$, with $\deg A=d\geq2$, and $h^0(A)=2$. Then why is it base point free?
Let $P\in C$. We need to show $h^0(A-P)=h^0(A)-1=1$. Riemann-Roch gives $$h^0(A-P)\geq d-g.$$
Now I'm stuck here. From Hurwitz's formula we get $d=\frac{\deg R}{2}+1-g$, where $R$ is the ramification divisor of the morphism, but I don't see how this may help.
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I am reading the paper "The category of good modules over a quasi-hereditary algebra has almost split sequences", the link is here:https://pub.uni-bielefeld.de/publication/1780235.
In the paper, $A$ is an artin algebra, and $A$-mod the category of (finitely generated) $A$-modules. Let $\Theta=\{\Theta(1), \dots, \Theta(n) \}$ be a finite set of $A$-modules with $Ext^1_A(\Theta(j), \Theta(i))=0$ for $j \geq i$. We denote $\mathcal{F}(\Theta)$ the full subcategory of $A$-mod of direct summands of modules having a filtration with factors in $\Theta$. I have some questions about $\mathcal{F}(\Theta)$:
It says that"$M$ belongs to $\mathcal{F}(\Theta)$ if and only if $M$ has submodules $0=M_0 \subseteq M_1 \subseteq \cdots \subseteq M_t=M$ such that $M_s/M_{s-1}$ is isomorphic to a module in $\Theta $"(By the definition, if $M \in \mathcal{F}(\Theta)$, there is an $A$-module N such that $M \oplus N$ has submodules $0=M'_0 \subseteq M'_1 \subseteq \cdots \subseteq M'_t=M \oplus N$ and $M'_s/M'_{s-1}$ is isomorphic to a module in $\Theta $, then how to get $M$ also has submodules $0=M_0 \subseteq M_1 \subseteq \cdots \subseteq M_t=M$ such that $M_s/M_{s-1}$ is isomorphic to a module in $\Theta $?) It also defines $\mathcal{X}(\Theta)$ the full subcategory of $A$-mod of all modules which are direct summands of modules in $\mathcal{F}(\Theta)$. (By the definition, I think $\mathcal{F}(\Theta)$ has already been closed under direct summands, what is the difference between $\mathcal{X}(\Theta)$ and $\mathcal{F}(\Theta)$?) It gives an example to show $\mathcal{X}(\Theta)$ and $\mathcal{F}(\Theta)$ are not same: $A=\begin {pmatrix} k & k & k \\ 0 &k &0 \\ 0 &0& k\end {pmatrix}$,where k is a field, $\Theta(2)$ is the simple projective $A$-module, $\Theta(1)$ is its injective hull: here the indecomposable modules of length 2 belong to $\mathcal{X}(\Theta)$ but not to $\mathcal{F}(\Theta)$. (I don't know how to get the conclusion)
Thank you for any help.
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Hollow cylinder in plain strain condition¶ Overview¶
The aim of this test case is to validate the following functions:
distributed pressure symmmetry BC nodal displacements strains and nodal stresses
The simulation results of SimScale were compared to the analytical results in [SSLV04_A] and the numerical results in [SSLV04]. The meshes used in (A) and (B) were created with the parametrized-tetrahedralization-tool on the SimScale platform and the meshes used in (C) and (D) were meshed with Salome, resulting in four different modelizations.
Geometry¶
A B E F A’ B’ E’ F’ x [m] 0.1 0.2 2/2 2 0.1 0.2 2/2 2 y [m] 0 0 2/2 2 0 0 2/2 2 z [m] 0 0 0 0 0.01 0.01 0.01 0.01 Analysis type and Domain¶ Tool Type : Code_Aster Analysis Type : Static (linear elastic) Mesh and Element types :
Case Mesh type Number of nodes Element type (A) linear tetrahedral 1542 3D isoparametric (B) quadratic tetrahedral 9623 3D isoparametric (C) linear hexahedral 768 3D isoparametric (D) quadratic hexahedral 2720 3D isoparametric Simulation Setup¶
Material:
isotropic: E = 200 GPa, \(\nu\) = 0.3
Constraints:
Face EFE’F’ zero normal-displacement Face ABA’B’ zero y-displacement Face ABEF and face A’B’E’F’ zero z-displacements fixed
Loads:
Pressure of 60 MPa on Face AEA’E’ Results¶
Case Quantity [SSLV04_A] [SSLV04] SimScale Error (%) (A) \(d_X\) [m] 5.90E-005 5.89E-005 5.72E-005 -3.10 (B) \(d_X\) [m] 5.90E-005 5.90E-005 5.72E-005 -3.05 (C) \(d_X\) [m] 5.90E-005 5.90E-005 5.71E-005 -3.14 (D) \(d_X\) [m] 5.90E-005 5.89E-005 5.72E-005 -3.05 (A) \(\sigma_{XX}\) [Mpa] -6.00E+001 -5.72E+001 -5.61E+001 -6.45 (B) \(\sigma_{XX}\) [Mpa] -6.00E+001 -6.04E+001 -6.00E+001 -0.02 (C) \(\sigma_{XX}\) [Mpa] -6.00E+001 -5.92E+001 -5.12E+001 -14.63 (D) \(\sigma_{XX}\) [Mpa] -6.00E+001 -5.94E+001 -5.99E+001 -0.19 (A) \(\sigma_{YY}\) [Mpa] 1.00E+002 9.79E+001 1.02E+002 1.51 (B) \(\sigma_{YY}\) [Mpa] 1.00E+002 9.92E+001 1.00E+002 0.00 (C) \(\sigma_{YY}\) [Mpa] 1.00E+002 1.00E+002 1.03E+002 3.49 (D) \(\sigma_{YY}\) [Mpa] 1.00E+002 9.96E+001 9.99E+001 -0.06 (A) \(\epsilon_{XX}\) -4.50E-004 -4.33E-004 -4.53E-004 0.71 (B) \(\epsilon_{XX}\) -4.50E-004 -4.49E-004 -4.68E-004 3.98 (C) \(\epsilon_{XX}\) -4.50E-004 -4.47E-004 -4.35E-004 -3.36 (D) \(\epsilon_{XX}\) -4.50E-004 -4.47E-004 -4.67E-004 3.86 (A) \(\epsilon_{YY}\) 5.90E-004 5.75E-004 5.71E-004 -3.14 (B) \(\epsilon_{YY}\) 5.90E-004 5.88E-004 5.72E-004 -3.05 (C) \(\epsilon_{YY}\) 5.90E-004 5.90E-004 5.71E-004 -3.26 (D) \(\epsilon_{YY}\) 5.90E-004 5.87E-004 5.72E-004 -3.13
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Let $f:[0,1]\to[0,1]$ be a continuous function such that its derivative $f'$ exists on $(0,1)$. My question is:
Q1.If $E\subset[0,1]$ is a nowhere dense closed subset, is $f(E)$ also nowhere dense in $[0,1]$?
If the answer is negative, what will happen when we additionaly assume that $f'\ge 0$ on $(0,1)$?
My question originates from the follow question:
Q2.If $g:[0,1]\to[0,1]$ is the Cantor function, can we find homeomorphisms $\varphi$ and $\psi$ both from $[0,1]$ to itself, such that $\psi\circ g\circ \varphi$ is differntiable on $(0,1)$?
If
Q1 with the addtional assumption $f'\ge 0$ has a positive answer, then clearly it gives a negative answer to the second question, because for any $\varphi$ and $\psi$, $f=\psi\circ g\circ \varphi$ maps a nowhere dense closed set onto $[0,1]$. Otherwise, I am still interested in whether $\varphi$ and $\psi$ exist or not. Q2 comes from an attempt in providing a simple counter-example to this question for the case $X=Y=(0,1)$. Update: Thanks to Henno Brandsma's comment below, I realized to add a remark that Q1has a positive answer when $f$ is (piecewise) $C^1$.
Thanks to the discussion with Jim Belk, I realized that my original argument on
Q1under the assumption that $f$ is $C^1$ was incorrect. The following is a corrected argument.
Denote the Lebesgue measure on $[0,1]$ by $|\cdot|$ and denote $C=\{x\in[0,1]:f'(x)=0\}$. Note that $C$ is a closed. Using the fact that for every closed subset $K$ of $[0,1]$, $$|f(K)|\le\int_K|f'(x)|dx,$$ or otherwise, we know that $f(C)$ is closed and $|f(C)|=0$, so $f(C)$, and hence $f(C\cap E)$, are closed and nowhere dense. Note that $[0,1]\setminus C$ is a disjoint union of at most countably many intervals, say $[0,1]\setminus C=\sqcup_n I_n$. Note that $f|_{\overline{I_n}}$ is homeomorphic, so $f(\overline{I_n}\cap E)$ is closed and nowhere dense. Then by Baire category theorem, $$f(E)=f(C\cap E)\cup\big(\cup_n f(\overline{I_n}\cap E)\big)$$ is nowhere dense.
Moreover, I removed another question similar to
Q1in this post, and started a new post for it.
Any hint or suggestion is appreciated. Thanks in advance.
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This is a popular question on this site but I haven't found the answer I'm looking for in other questions. It is often stated that charge conservation in electromagnetism is a consequence of local gauge invariance, or perhaps it is due to some global phase symmetry. Without talking about scalar or spinor fields, the EM Lagrangian that we're familiar with is:$$ \mathcal{L} = -\frac{1}{4}F^2 - A \cdot J + \mathcal{L}_\mathrm{matter}(J) $$The equation of motion for $A$ is simply$$ \partial_\mu F^{\mu \nu} = J^\nu $$From which it follows that $J$ is a conserved current (by the antisymmetry of the field strength). But what symmetry gave rise to this?
I'm not supposing that my matter has any global symmetry here, that I might be able to gauge. Then so far as I can tell, the Lagrangian given isn't gauge invariant. The first term is, indeed, but the second term only becomes gauge invariant on-shell (since I can do some integrating by parts to move a derivative onto $J$). If we demand that our Lagrangian is gauge-invariant even off-shell, then we can deduce that $\partial \cdot J = 0$ off-shell and hence generally. But we can't demand that this hold off-shell, since $J$ is not in general divergenceless!
For concreteness, suppose that $$ \mathcal{L}_\mathrm{matter}(J) = \frac{1}{2} (\partial_\mu \phi) (\partial^\mu \phi) \qquad J^\mu \equiv \partial^\mu \phi $$ Then we find that $\phi$ (a real scalar) satisfies some wave equation, sourced by $A$. The equations of motion here constrain the form of $J$, but off-shell $J$ is just some arbitrary function, since $\phi$ is just some arbitrary function. Then it is clear that the Lagrangian is not gauge-invariant off-shell.
And this is a problem, because when we derive conserved quantities through Noether's theorem, it's important that our symmetry is a symmetry of the Lagrangian for
any field configuration. If it's only a symmetry for on-shell configurations, then the variation of the action vanishes trivially and we can't make any claims about conserved quantities.
So here's my question:
what symmetry does the above Lagrangian have that implies the conservation of the quantity $J$, provided $A$ satisfies its equation of motion? Thank you.
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If the galaxy is axisymmetric and $\Phi = \Phi(R, z)$ is the potential then
$$v_c^2 = \left.R\frac{\partial \Phi(R, z)}{\partial R}\right|_{z = 0}$$
The trick is now in getting the potential $\Phi$. For the Milky Way you have a bunch of components
$$\Phi = \Phi_{\rm halo, ~DM} + \Phi_{\rm disk} + \Phi_{\rm halo, \star} + \Phi_{\rm bulge} + \Phi_{\rm BH}$$
you can even include the gas disk, or the hot gas halo. Now, to answer your question: there's a relationship between mass and potential. For example, if the component is spherical (e.g. the dark halo), then
$$\Phi(r) = -\int_r^{+\infty}{\rm d}r'\frac{G M(r')}{r'^2}$$
where $M(r)$ is the enclosed mass at a given radius
$$M(r) = 4\pi \int_0^r{\rm d}r' r'^2 \rho(r')$$
For the disk the expression is a bit more complicated, but the idea is the same: the circular velocity depends on the gradient of the potential, which in turn depends on the enclosed mass at a given radius.
EDIT The above clearly depends on the election of the model for the components. To give you an example, consider a Hernquist Dark Matter halo with density
$$\rho_{\rm halo,~DM} = \frac{\rho_0}{r / r_{\rm halo} (1 + r/r_{\rm halo})^3}$$
and an exponential razor-thin disk with density
$$\Sigma_{\rm disk}(R) = \Sigma_0 e^{-R / R_{\rm disk}}$$
It is not very complicated to calculate the circular velocity for these two components
$$v_{c, \rm halo}^2 = \frac{GM}{2R_{\rm halo}} \frac{(R / R_{\rm halo})}{(1 + R/R_{\rm halo})^2}$$
and
$$v_{c, \rm disk}^2 = \frac{2 G M_{\rm disk}}{R_{\rm disk}} y^2[I_0(y) K_0(y) - I_1(y)K_1(y)]$$
with $y = R / (2 R_{\rm disk})$. The curve below shows a model with $R_{\rm disk} = 3$ kpc, $R_{\rm halo} = 30$ kpc, $M_{\rm disk} = 10^{10}~M_{\odot}$ and $M_{\rm halo} = 3\times 10^{11}~M_{\odot}$
These are just to give you an example of the numbers
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Algebra and Algebraic Geometry Seminar Spring 2018
The seminar meets on Fridays at 2:25 pm in room B235.
Contents 1 Algebra and Algebraic Geometry Mailing List 2 Fall 2018 Schedule 3 Spring 2018 Schedule 4 Abstracts Algebra and Algebraic Geometry Mailing List Please join the AGS Mailing List to hear about upcoming seminars, lunches, and other algebraic geometry events in the department (it is possible you must be on a math department computer to use this link). Fall 2018 Schedule Spring 2018 Schedule Abstracts Tasos Moulinos Derived Azumaya Algebras and Twisted K-theory
Topological K-theory of dg-categories is a localizing invariant of dg-categories over [math] \mathbb{C} [/math] taking values in the [math] \infty [/math]-category of [math] KU [/math]-modules. In this talk I describe a relative version of this construction; namely for [math]X[/math] a quasi-compact, quasi-separated [math] \mathbb{C} [/math]-scheme I construct a functor valued in the [math] \infty [/math]-category of sheaves of spectra on [math] X(\mathbb{C}) [/math], the complex points of [math]X[/math]. For inputs of the form [math]\operatorname{Perf}(X, A)[/math] where [math]A[/math] is an Azumaya algebra over [math]X[/math], I characterize the values of this functor in terms of the twisted topological K-theory of [math] X(\mathbb{C}) [/math]. From this I deduce a certain decomposition, for [math] X [/math] a finite CW-complex equipped with a bundle [math] P [/math] of projective spaces over [math] X [/math], of [math] KU(P) [/math] in terms of the twisted topological K-theory of [math] X [/math] ; this is a topological analogue of a result of Quillen’s on the algebraic K-theory of Severi-Brauer schemes.
Roman Fedorov A conjecture of Grothendieck and Serre on principal bundles in mixedcharacteristic
Let G be a reductive group scheme over a regular local ring R. An old conjecture of Grothendieck and Serre predicts that such a principal bundle is trivial, if it is trivial over the fraction field of R. The conjecture has recently been proved in the "geometric" case, that is, when R contains a field. In the remaining case, the difficulty comes from the fact, that the situation is more rigid, so that a certain general position argument does not go through. I will discuss this difficulty and a way to circumvent it to obtain some partial results.
Juliette Bruce Asymptotic Syzygies in the Semi-Ample Setting
In recent years numerous conjectures have been made describing the asymptotic Betti numbers of a projective variety as the embedding line bundle becomes more ample. I will discuss recent work attempting to generalize these conjectures to the case when the embedding line bundle becomes more semi-ample. (Recall a line bundle is semi-ample if a sufficiently large multiple is base point free.) In particular, I will discuss how the monomial methods of Ein, Erman, and Lazarsfeld used to prove non-vanishing results on projective space can be extended to prove non-vanishing results for products of projective space.
Andrei Caldararu Computing a categorical Gromov-Witten invariant
In his 2005 paper "The Gromov-Witten potential associated to a TCFT" Kevin Costello described a procedure for recovering an analogue of the Gromov-Witten potential directly out of a cyclic A-inifinity algebra or category. Applying his construction to the derived category of sheaves of a complex projective variety provides a definition of higher genus B-model Gromov-Witten invariants, independent of the BCOV formalism. This has several advantages. Due to the categorical invariance of these invariants, categorical mirror symmetry automatically implies classical mirror symmetry to all genera. Also, the construction can be applied to other categories like categories of matrix factorization, giving a direct definition of FJRW invariants, for example.
In my talk I shall describe the details of the computation (joint with Junwu Tu) of the invariant, at g=1, n=1, for elliptic curves. The result agrees with the predictions of mirror symmetry, matching classical calculations of Dijkgraaf. It is the first non-trivial computation of a categorical Gromov-Witten invariant.
Aron Heleodoro Normally ordered tensor product of Tate objects and decomposition of higher adeles
In this talk I will introduce the different tensor products that exist on Tate objects over vector spaces (or more generally coherent sheaves on a given scheme). As an application, I will explain how these can be used to describe higher adeles on an n-dimensional smooth scheme. Both Tate objects and higher adeles would be introduced in the talk. (This is based on joint work with Braunling, Groechenig and Wolfson.)
Moisés Herradón Cueto Local type of difference equations
The theory of algebraic differential equations on the affine line is very well-understood. In particular, there is a well-defined notion of restricting a D-module to a formal neighborhood of a point, and these restrictions are completely described by two vector spaces, called vanishing cycles and nearby cycles, and some maps between them. We give an analogous notion of "restriction to a formal disk" for difference equations that satisfies several desirable properties: first of all, a difference module can be recovered uniquely from its restriction to the complement of a point and its restriction to a formal disk around this point. Secondly, it gives rise to a local Mellin transform, which relates vanishing cycles of a difference module to nearby cycles of its Mellin transform. Since the Mellin transform of a difference module is a D-module, the Mellin transform brings us back to the familiar world of D-modules.
Eva Elduque On the signed Euler characteristic property for subvarieties of Abelian varieties
Franecki and Kapranov proved that the Euler characteristic of a perverse sheaf on a semi-abelian variety is non-negative. This result has several purely topological consequences regarding the sign of the (topological and intersection homology) Euler characteristic of a subvariety of an abelian variety, and it is natural to attempt to justify them by more elementary methods. In this talk, we'll explore the geometric tools used recently in the proof of the signed Euler characteristic property. Joint work with Christian Geske and Laurentiu Maxim.
Harrison Chen Equivariant localization for periodic cyclic homology and derived loop spaces
There is a close relationship between derived loop spaces, a geometric object, and (periodic) cyclic homology, a categorical invariant. In this talk we will discuss this relationship and how it leads to an equivariant localization result, which has an intuitive interpretation using the language of derived loop spaces. We discuss ongoing generalizations and potential applications in computing the periodic cyclic homology of categories of equivariant (coherent) sheaves on algebraic varieties.
Phil Tosteson Stability in the homology of Deligne-Mumford compactifications
The space [math]\bar M_{g,n}[/math] is a compactification of the moduli space algebraic curves with marked points, obtained by allowing smooth curves to degenerate to nodal ones. We will talk about how the asymptotic behavior of its homology, [math]H_i(\bar M_{g,n})[/math], for [math]n \gg 0[/math] can be studied using the representation theory of the category of finite sets and surjections.
Wei Ho Noncommutative Galois closures and moduli problems
In this talk, we will discuss the notion of a Galois closure for a possibly noncommutative algebra. We will explain how this problem is related to certain moduli problems involving genus one curves and torsors for Jacobians of higher genus curves. This is joint work with Matt Satriano.
Daniel Corey Initial degenerations of Grassmannians
Let Gr_0(d,n) denote the open subvariety of the Grassmannian Gr(d,n) consisting of d-1 dimensional subspaces of P^{n-1} meeting the toric boundary transversely. We prove that Gr_0(3,7) is schoen in the sense that all of its initial degenerations are smooth. The main technique we will use is to express the initial degenerations of Gr_0(3,7) as a inverse limit of thin Schubert cells. We use this to show that the Chow quotient of Gr(3,7) by the maximal torus H in GL(7) is the log canonical compactification of the moduli space of 7 lines in P^2 in linear general position.
Alena Pirutka Irrationality problems
Let X be a projective algebraic variety, the set of solutions of a system of homogeneous polynomial equations. Several classical notions describe how ``unconstrained
the solutions are, i.e., how close X is to projective space: there are notions of rational, unirational and stably rational varieties. Over the field of complex numbers, these notions coincide in dimensions one and two, but diverge in higherdimensions. In the last years, many new classes of non stably rational varieties were found, using a specialization technique, introduced by C. Voisin. This method also allowed to prove that the rationality is not a deformation invariant in smooth and projective families of complex varieties: this is a joint work with B. Hassett and Y. Tschinkel. In my talk I will describe classical examples, as well as the recent progress around these rationality questions. Nero Budur Homotopy of singular algebraic varieties
By work of Simpson, Kollár, Kapovich, every finitely generated group can be the fundamental group of an irreducible complex algebraic variety with only normal crossings and Whitney umbrellas as singularities. In contrast, we show that if a complex algebraic variety has no weight zero 1-cohomology classes, then the fundamental group is strongly restricted: the irreducible components of the cohomology jump loci of rank one local systems containing the constant sheaf are complex affine tori. Same for links and Milnor fibers. This is joint work with Marcel Rubió.
Alexander Yom Din Drinfeld-Gaitsgory functor and contragradient duality for (g,K)-modules
Drinfeld suggested the definition of a certain endo-functor, called the pseudo-identity functor (or the Drinfeld-Gaitsgory functor), on the category of D-modules on an algebraic stack. We extend this definition to an arbitrary DG category, and show that if certain finiteness conditions are satisfied, this functor is the inverse of the Serre functor. We show that the pseudo-identity functor for (g,K)-modules is isomorphic to the composition of cohomological and contragredient dualities, which is parallel to an analogous assertion for p-adic groups.
In this talk I will try to discuss some of these results and around them. This is joint work with Dennis Gaitsgory.
John Lesieutre Some higher-dimensional cases of the Kawaguchi-Silverman conjecture
Given a dominant rational self-map f : X -->X of a variety defined over a number field, the first dynamical degree $\lambda_1(f)$ and the arithmetic degree $\alpha_f(P)$ are two measures of the complexity of the dynamics of f: the first measures the rate of growth of the degrees of the iterates f^n, while the second measures the rate of growth of the heights of the iterates f^n(P) for a point P. A conjecture of Kawaguchi and Silverman predicts that if P has Zariski-dense orbit, then these two quantities coincide. I will prove this conjecture in several higher-dimensional settings, including for all automorphisms of hyper-K\"ahler varieties. This is joint work with Matthew Satriano.
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O.D.E. type behavior of blow-up solutions of nonlinear heat equations
1.
Département de Mathématiques, Université de Cergy-Pontoise and IUF, 2 Ave. Adolphe Chauvin, BP 222, Pontoise, 95 302 Cergy-Pontoise cedex, France
2.
Département de Mathématiques et Applications, CNRS École Normale Supérieure, 45 rue d'Ulm, 75 230 Paris cedex 05, France
$\frac{\partial w}{\partial s}=\Delta w-\frac{1}{2}y \cdot \nabla w -\frac{w}{p-1}+|w|^{p-1}w$
(where $w : \mathbb R^N\times \mathbb R \to \mathbb R^M, p> 1$ and $(N - 2)p < N + 2$) are independent of space and completely explicit. We then derive from this various uniform estimates and a uniform localization property for blow-up solutions of $\partial_t u=\Delta u + |u|^{p-1}u$.
Mathematics Subject Classification:35K40, 35K55, 35A2. Citation:Frank Merle, Hatem Zaag. O.D.E. type behavior of blow-up solutions of nonlinear heat equations. Discrete & Continuous Dynamical Systems - A, 2002, 8 (2) : 435-450. doi: 10.3934/dcds.2002.8.435
[1]
Van Tien Nguyen.
On the blow-up results for a class of strongly perturbed semilinear heat equations.
[2]
Dongho Chae.
On the blow-up problem for the Euler equations and
the Liouville type results in the fluid equations.
[3]
Lan Qiao, Sining Zheng.
Non-simultaneous blow-up for heat equations with positive-negative sources and coupled boundary flux.
[4]
Francesca De Marchis, Isabella Ianni.
Blow up of solutions of semilinear heat equations in non radial domains of $\mathbb{R}^2$.
[5] [6]
Yohei Fujishima.
Blow-up set for a superlinear heat equation and pointedness of the initial data.
[7]
Futoshi Takahashi.
Morse indices and the number of blow up points of blowing-up solutions for a Liouville equation
with singular data.
[8]
Begoña Barrios, Leandro Del Pezzo, Jorge García-Melián, Alexander Quaas.
A Liouville theorem for indefinite fractional diffusion equations and its application to existence of solutions.
[9] [10] [11]
Alexander Gladkov.
Blow-up problem for semilinear heat equation with nonlinear nonlocal Neumann boundary condition.
[12]
Yohei Fujishima.
On the effect of higher order derivatives of initial data on the blow-up set for a semilinear heat equation.
[13]
Xiumei Deng, Jun Zhou.
Global existence and blow-up of solutions to a semilinear heat equation with singular potential and logarithmic nonlinearity.
[14] [15]
Mohamed-Ali Hamza, Hatem Zaag.
Blow-up results for semilinear wave equations in the
superconformal case.
[16]
Huyuan Chen, Hichem Hajaiej, Ying Wang.
Boundary blow-up solutions to fractional elliptic
equations in a measure framework.
[17] [18] [19] [20]
2018 Impact Factor: 1.143
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I'm working through different books about financial mathematics and solving some problems I get stuck.
Suppose you define an arbitrary stochastic process, for example
$ X_t := W_t^8-8t $ where $ W_t $ is a Brownian motion.
The question is, how could I deduce that this stochastic process is a martingale or not using Itô's formula?
The only thing I know is:
Looking at the stochastic integral $ \int K dM $ where $ M=\{M_t\} $ is a martingale, which is right continuous with left limit, null at $0$ and satisfies $ sup_t E[M_t] < \infty$ and $ K $ a stochastic process bounded and predictable, then $ \int K dM $ is a martingale too.
But I'm not sure if this is helpful in this situation. An example of how to solve such types of problems would be appreciated.
Just to be sure, I state Itô's formula which I know so far.
Let $\{X_t\}$ a general $ \mathbb{R}^n $ valued semimartingale and $f: \mathbb{R}^n \to \mathbb{R}$ such that $ f\in C^2 $. Then $ \{f(X_t)\} $ is again a semimartingale and we get Itô's formula (in differential form):
$$ df(X_t) = \sum_{i=1}^n f_{x_i}(X_t)dX_{t,i} + \frac{1}{2}\sum_{i,j=1}^n f_{x_i,x_j}(X_t)d\langle X_i,X_j\rangle_t$$
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The question is as follows:
Let $X_1$ ~ Binomial(3,1/3) and $X_2$ ~ Binomial(4,1/2) be independent random variables. Compute P($X_1$ = $X_2$)
I'm not sure what it means to compute the probability of two
random variables being equal.
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.Sign up to join this community
The question is as follows:
Let $X_1$ ~ Binomial(3,1/3) and $X_2$ ~ Binomial(4,1/2) be independent random variables. Compute P($X_1$ = $X_2$)
I'm not sure what it means to compute the probability of two
random variables being equal.
The probability that $X_1=X_2$ is the probability that both are zero, plus the probability that both are one, plus the probability that both are two, and so on.
Let $Z=X_1-X_2$
$P(Z=z)=\sum_{x_1=z+x_2}^\infty P(X_1=x,X_2=x-z)$ (since $X_1$ and $X_2$ are independent)
$P(Z=z)=\sum_{x_1=z+x_2}^\infty P(X_1=x)P(X_2=x-z)\\=\sum_{x_1=z+x_2}^\infty \binom{3}{x}(\frac{1}{3})^x(\frac{2}{3})^{3-x}\binom{4}{4-x+z}(\frac{1}{2})^{x-z}(\frac{1}{2})^{4-x+z}$
When $X_1=X_2\Rightarrow z=0 $
Therefore
$P(Z=0)=\sum_{x_1=x_2}^\infty \binom{3}{x}(\frac{1}{3})^x(\frac{2}{3})^{3-x}\binom{4}{4-x+0}(\frac{1}{2})^{x-0}(\frac{1}{2})^{4-x+0}\\=\binom{3}{0}(\frac{1}{3})^0(\frac{2}{3})^{3-0}\binom{4}{4-0+0}(\frac{1}{2})^{0-0}(\frac{1}{2})^{4-0+0} \quad (x_1=x_2=0)\\+\binom{3}{1}(\frac{1}{3})^1(\frac{2}{3})^{3-1}\binom{4}{4-1+0}(\frac{1}{2})^{1-0}(\frac{1}{2})^{4-1+0}\quad(x_1=x_2=1)\\+\binom{3}{2}(\frac{1}{3})^2(\frac{2}{3})^{3-2}\binom{4}{4-2+0}(\frac{1}{2})^{2-0}(\frac{1}{2})^{4-2+0}\quad(x_1=x_2=2)\\+\binom{3}{3}(\frac{1}{3})^3(\frac{2}{3})^{3-3}\binom{4}{4-3+0}(\frac{1}{2})^{3-0}(\frac{1}{2})^{4-3+0}\quad(x_1=x_2=3)$
Ok, you can add up them to get the probability.
And remember for binomial distribution the random variable is total number of success. In your case $X_1$ and $X_2$ can be both 0, 1,2,3 for the total number of success of the two distribution.This answer in fact the same as the answer by Daniel and comment by ACE.
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Table of Content: Chemical Equilibrium Ionic Equilibrium – Degree of Ionization and Dissociation Equilibrium Constant – Characteristics and Applications Le Chatelier’s Principle on Equilibrium Solubility and Solubility Product Acid and Base pH Scale and Acidity pH and Solutions Hydrolysis, Salts, and Types Buffer Solutions Introduction to pH Scale
Acid solutions have protons and basic solutions have hydroxide ions.
Concentrations of the ions are low (negative power of ten). pH scale is a convenient way of expressing these low concentrations in simple numbers between 1 and 14.
pH is the negative logarithm to the base ten of hydrogen ion concentration in moles per litre.
pH = – log [H
+]
p(OH) is the negative logarithm to the base ten of hydroxide ion concentration in moles per litre.
p(OH) = – log [OH
–]
In aqueous solutions, pH + p(OH) = 14.
pH scale is based on neutral water, where [H
+] = [OH –] = 10 -7 For a neutral solution pH = = – log [H +] = – log [10 -7] = +7
pH of strong acid decreases with a limit of 1 and pH of a base increases up to 14
Generally, acids and bases will have a pH between.
But negative and greater than14 pH values are also possible.
Limitations of pH Scale pH values does not reflect directly the relative strength of acid or bases.
A solution of pH = 1 has a hydrogen ion concentration 100 times that of a solution of pH = 3 (not three times). A 4 x 10
-5 HCI is twice concentrated of a 2 x 10 N -5 HCI solution, but the pH values of these solutions are 4.40 and 4.70 (not double). N pH value is zero for 1N solution of strong acid. Concentration of 2 , 3 N , 10 N , etc. gives negative pH values.. N A solution of an acid having very low concentration, say 10 -8 , shows a pH 8and hence should be basic, but actual pH value is less than 7. N Periodic Variation of Acidic and Basic Properties (a) Hydracids of the Elements of the Same Periods
Along the period acidic strength increases.. Hydrides become increasingly
acidic from CH 4, NH 3, H 2O and HF. The increase in acidic properties is due to the fact that the stability of their conjugate bases increases in the order
CH
– 3< NH – 2 < OH – < F – (b) Hydracids of the Elements of the Same Group Acidic nature increases down the column. Hydrides of V group elements (NH 3, PH 3, AsH 3, SbH 3, BiH 3) show basic character which decreases due to increase in size and decrease in electronegativity from N to Bi. There is a decrease in electron density in, sp 3-hybrid orbital and thus electron donor capacity decreases. Hydracids of VI group elements (H 20, H 2S, H 2Se, H 2Te) act as weak acids. The strength increases in the order H 20 < H 2S < H 2Se < H 2Te. The increasing acidic propertiesreflects decreasing trend in the electron donor capacity of OH –, HS –, HSe –or HTe –ions. Hydracids of VII group elements (HF, HCI, HBr, HI) show acidic propertieswhich increase from HF to HI. This is explained by the fact that bond energies decrease.
(H-F = 135 kcal/mol, HCI = 103, HBr = 88 and HI = 71 kcal/mol).
Oxyacids
The
acidic properties of oxyacids of the same element which is in different oxidation states increase with an increase in oxidation number.
+ 1 +3 +5 +7
HCIO < HC1O
2 < HC1O 3 < HCIO 4
+4 +6 +3 +5
H
2SO 3 < H 2SO 4; HNO 2 < HNO 3
But this rule fails in
oxyacids of phosphorus.
H
3PO 2 > H 3PO 3 > H 3PO 4
The
acidic properties of the oxyacids of different elements which are in the same oxidation state decreases as the atomic number increases. This is due to increase in size and decrease in electronegativity.
HC1O
4 > HBrO 4 > HIO 4
H
2SO 3 > H 2SeO 3
But there are a number of acid-base reactions in which no proton transfer takes place, e.g.,
SO
2 + SO 2 ↔ SO 2+ + S
Acid
1 Base 2 Acid 2 Base 1
Thus, the protonic definition cannot be used to explain the reactions occurring in non-protonic solvents such as COCl
2, S0 2, N 2O 4, etc. Water – Amphoteric Weak Electrolyte 1) Water can behave like acid or a base. So it is amphoteric.
Water accepts proton from HCl and acts as a base.
Water gives proton to ammonia and can be an acid \(NH_3(aq)+H_2O(l)\rightleftharpoons NH_{4}^{+}(aq)+OH^{-}\)
Molarity of water Molarity = Number of moles per litre of solution = = 55.55 mole l -1 Ionization constant of water\(H_2O\rightleftharpoons H_{4}^{+}+OH^{-}\)
Ka = Kb = \(\frac{[H+[OH-]]}{[H_2O]}\) = \(\frac{10^{-7}\times 10^{-7}}{55.55}=1.8\times10^{-16}\)
Where, Ka is the acid ionization constant and Kb is the base ionization constant.
pKa = pKb = – log[Ka] = \(-\log 1.8\times10^{-16}\) = 15.74
Degree of ionization of water\(H_2O\rightleftharpoons H^{+}+OH^{-}\) Initial concentration moles 55.55 0 0 At equilibrium moles 10 -7 10 -7
Degree of ionization = α =\(\frac{number\;of\;moles\;ionized}{initial\;number\;of\;moles}\) \(=\frac{10^{-7}}{55.55}=1.8\times 10^{-9}\)
Only about 2 parts per billion (ppb) of the water molecules dissociate into ions at room temperature.
Ionic product of water
It is the product of the concentrations of hydrogen and hydroxide ions in water.
Ionic product of water = Kw = [H
+][OH –] = 10 -14
pKw = – log[Kw] = – log10
-14= 14
Ionic product, pKw, pKa and pKb remains the same whether the solution is acidic, neutral or
basic.
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The Partial Summation Formula for Series of Real Numbers
Table of Contents
The Partial Summation Formula for Series of Real Numbers
We will shortly look at two more very useful tests for determining the convergence of series known as Dirichlet's test and Abel's test. Before we look at these two tests though, we will need first look at the following theorem and corollary.
Theorem 1: If $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are two sequences of real numbers and $A_n = a_1 + a_2 + ... + a_n$ then $\displaystyle{\sum_{k=1}^{n} a_kb_k = A_nb_{n+1} - \sum_{k=1}^{n} A_k(b_{k+1} - b_{k})}$. Proof:Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be two sequences of real numbers. Note that $a_k = A_k - A_{k-1}$ where we define $A_0 = 0$. Then:
\begin{align} \quad \sum_{k=1}^{n} a_kb_k &= \sum_{k=1}^{n} [A_k - A_{k-1}]b_k \\ \quad &= \sum_{k=1}^{n} A_kb_k - \sum_{k=1}^{n} A_{k-1}b_k \\ \quad &= \sum_{k=1}^{n} A_kb_k - \sum_{k=1}^{n} A_kb_{k+1} + A_0b_1 + A_nb_{n+1} \\ \quad &= \sum_{k=1}^{n} A_k(b_k - b_{k+1}) + A_nb_{n+1} \\ \quad &= A_nb_{n+1} - \sum_{k=1}^{n} A_k(b_{k+1} - b_k) \quad \blacksquare \end{align}
Corollary 1: If $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are two sequences of real numbers and $A_n = a_1 + a_2 + ... + a_n$ then if $\lim_{n \to \infty} A_nb_{n+1}$ converges and $\displaystyle{\sum_{k=1}^{\infty} A_k(b_{k} - b_{k+1})}$ converges then $\displaystyle{\sum_{k=1}^{\infty} a_kb_k}$ converges. From Theorem 1 we note that:
\begin{align} \quad \sum_{k=1}^{\infty} a_k = \lim_{n \to \infty} \sum_{k=1}^{n} = \lim_{n \to \infty} A_nb_{n+1} - \sum_{k=1}^{\infty} A_k(b_{k+1} - b_{k}) \end{align}
Since the terms on the righthand side of the equation above converge, we have that the lefthand side of the equation also converges. $\blacksquare$
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De Bruijn-Newman constant
For each real number [math]t[/math], define the entire function [math]H_t: {\mathbf C} \to {\mathbf C}[/math] by the formula
[math]\displaystyle H_t(z) := \int_0^\infty e^{tu^2} \Phi(u) \cos(zu)\ du[/math]
where [math]\Phi[/math] is the super-exponentially decaying function
[math]\displaystyle \Phi(u) := \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3 \pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}).[/math]
It is known that [math]\Phi[/math] is even, and that [math]H_t[/math] is even, real on the real axis, and obeys the functional equation [math]H_t(\overline{z}) = \overline{H_t(z)}[/math]. In particular, the zeroes of [math]H_t[/math] are symmetric about both the real and imaginary axes. One can also express [math]H_t[/math] in a number of different forms, such as
[math]\displaystyle H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du[/math]
or
[math]\displaystyle H_t(z) = \frac{1}{2} \int_0^\infty e^{t\log^2 x} \Phi(\log x) e^{iz \log x}\ \frac{dx}{x}.[/math]
In the notation of [KKL2009], one has
[math]\displaystyle H_t(z) = \frac{1}{2} \Xi_{4t}(2z).[/math] Note: there may be a typo in the definition of [math]\Xi_\lambda[/math] in [KKL2009], they may instead have intended to write [math]4\lambda (\log x)^2 + 2 it \log x[/math] in place of [math]\frac{\lambda}{4} (\log x)^2 + \frac{it}{2} \log x[/math] in that definition. If so, the relationship would be [math]H_t(z) = z) = \frac{1}{2} \Xi_{t/4}(z/2)[/math] instead of [math]H_t(z) = z) = \frac{1}{2} \Xi_{4t}(2z)[/math].
De Bruijn [B1950] and Newman [N1976] showed that there existed a constant, the
de Bruijn-Newman constant [math]\Lambda[/math], such that [math]H_t[/math] has all zeroes real precisely when [math]t \geq \Lambda[/math]. The Riemann hypothesis is equivalent to the claim that [math]\Lambda \leq 0[/math]. Currently it is known that [math]0 \leq \Lambda \lt 1/2[/math] (lower bound in [RT2018], upper bound in [KKL2009]). Contents [math]t=0[/math]
When [math]t=0[/math], one has
[math]\displaystyle H_0(z) = \frac{1}{8} \xi( \frac{1}{2} + \frac{iz}{2} ) [/math]
where
[math]\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{s/2} \Gamma(s/2) \zeta(s)[/math]
is the Riemann xi function. In particular, [math]z[/math] is a zero of [math]H_0[/math] if and only if [math]\frac{1}{2} + \frac{iz}{2}[/math] is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of [math]H_0[/math] being real, and Riemann-von Mangoldt formula (in the explicit form given by Backlund) gives
[math]\displaystyle N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})| \lt 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 [/math]
for any [math]T \gt 4[/math], where [math]N_0(T)[/math] denotes the number of zeroes of [math]H_0[/math] with real part between 0 and T.
The first [math]10^{13}[/math] zeroes of [math]H_0[/math] (to the right of the origin) are real [G2004]. This numerical computation uses the Odlyzko-Schonhage algorithm.
[math]t\gt0[/math]
For any [math]t\gt0[/math], it is known that all but finitely many of the zeroes of [math]H_t[/math] are real and simple [KKL2009, Theorem 1.3]
Let [math]\sigma_{max}(t)[/math] denote the largest imaginary part of a zero of [math]H_t[/math], thus [math]\sigma_{max}(t)=0[/math] if and only if [math]t \geq \Lambda[/math]. It is known that the quantity [math]\frac{1}{2} \sigma_{max}(t)^2 + t[/math] is non-decreasing in time whenever [math]\sigma_{max}(t)\gt0[/math] (see [KKL2009, Proposition A]. In particular we have
[math]\displaystyle \Lambda \leq t + \frac{1}{2} \sigma_{max}(t)^2[/math]
for any [math]t[/math].
The zeroes [math]z_j(t)[/math] of [math]H_t[/math] (formally, at least) obey the system of ODE
[math]\partial_t z_j(t) = - \sum_{k \neq j} \frac{2}{z_k(t) - z_j(t)}[/math]
where the sum may have to be interpreted in a principal value sense. (See for instance [CSV1994, Lemma 2.4]. This lemma assumes that [math]t \gt \Lambda[/math], but it is likely that one can extend to other [math]t \geq 0[/math] as well.)
In [KKL2009, Theorem 1.4], it is shown that for any fixed [math]t\gt0[/math], the number [math]N_t(T)[/math] of zeroes of [math]H_t[/math] with real part between 0 and T obeys the asymptotic
[math]N_t(T) = \frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} + t \log T + O(1) [/math]
as [math]T \to \infty[/math] (caution: the error term here is not uniform in t). Also, the zeroes behave like an arithmetic progression in the sense that
[math] z_{k+1}(t) - z_k(t) = (1+o(1)) \frac{4\pi}{\log |z_k|(t)} = (1+o(1)) \frac{4\pi}{\log k} [/math]
as [math]k \to +\infty[/math].
Threads Polymath proposal: upper bounding the de Bruijn-Newman constant, Terence Tao, Jan 24, 2018. Other blog posts and online discussion Heat flow and zeroes of polynomials, Terence Tao, Oct 17, 2017. The de Bruijn-Newman constant is non-negative, Terence Tao, Jan 19, 2018. Lehmer pairs and GUE, Terence Tao, Jan 20, 2018. A new polymath proposal (related to the Riemann hypothesis) over Tao's blog, Gil Kalai, Jan 26, 2018. Code and data Wikipedia and other references Bibliography [B1950] N. C. de Bruijn, The roots of trigonometric integrals, Duke J. Math. 17 (1950), 197–226. [CSV1994] G. Csordas, W. Smith, R. S. Varga, Lehmer pairs of zeros, the de Bruijn-Newman constant Λ, and the Riemann hypothesis, Constr. Approx. 10 (1994), no. 1, 107–129. [G2004] Gourdon, Xavier (2004), The [math]10^13[/math] first zeros of the Riemann Zeta function, and zeros computation at very large height [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. Citeseer [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251. [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is negative, preprint. arXiv:1801.05914
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In a flat toric universe (up connects down, right connects left and front connects back), every points repeats at $size_x$, $size_y$ and $size_z$ intervals.
In such case the Newtonian gravitational acceleration undergone in one point x, y, from a single mass M placed in one corner of a cube, would be undergone from every «image» of this mass M.
eg: lets take one cube of space with a mass (the red dot) in a corner :
Then let's repeat it ad-infinitum in every direction, we see that the mass appears in all corners of every cubes:
In such a situation, the acceleration undergone in the center of any cube would be null and it's general formula would be: $a(x, y, z) = - M G \lim\limits_{N\to\infty}\sum\limits_{i=-N}^{N}\sum\limits_{j=-N}^{N}\sum\limits_{k=-N}^{N} {(x-size_x.i, y-size_y.i, z-size_z.i) \over |{(x-size_x.i, y-size_y.i, z-size_z.i)}|^3}$
To make computing simpler, it would be nice to have a simple equivalent of this formula. which is obviously a periodic function.
In a first approach, I should be able to bring back this problem to a simpler one dimensional function : $\begin{align}f(x) &= \lim\limits_{N\to\infty}\sum\limits_{n=-N}^{N}{x-n\pi \over ((x-n\pi)^2+a^2)^{3/2}} \\\\ &={x \over (x^2+a^2)^{3/2}} + \lim\limits_{N\to\infty}[\sum\limits_{n=-N}^{-1}{x-n\pi \over ((x-n\pi)^2+a^2)^{3/2}} + \sum\limits_{n=1}^{N}{x-n\pi \over ((x-n\pi)^2+a^2)^{3/2}}] \\\\ &={x \over (x^2+a^2)^{3/2}} + \sum\limits_{n=1}^{\infty}[{x+n\pi \over ((x+n\pi)^2+a^2)^{3/2}} + {x-n\pi \over ((x-n\pi)^2+a^2)^{3/2}}]\end{align}$
But then, I wonder how I could compute f(x)... As told above, it will obviously be a periodic function, but it doesn't look like an obvious one. On this simulation I made, one can see various graphs for this function (red for a = 0.6, green for a = 0.3 and blue for a = 0). In addition, cyan is a graph of $-\cot x$, magenta is a graph of $-\cot x/f(x)$ and yellow is $|\sin x|$ (The «V» shape on the magenta graph is caused by the 0/0 situation).
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Segment of Example from Textbook:
$t=\dots$
More usefully, we have:
$t\sim n\log n$
I recall $\sim$ means "similarity" in geometry, same shape but not same size. How is it interpreted here?
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In this context, it means that $$\lim_{n \to \infty}\frac{t}{n\log n}=1$$ That is, the quotient of both sides tends to $1$.
The answer by Daniel Littlewood is absolutely correct in this context. To extend this to the general definition (not tied to the example at hand):
$$f(n) \sim g(n) \iff \lim_{n\to\infty} \left(\frac{f}{g}\right)(n) = 1$$
The symbol $\sim$ does not have a set meaning across all subjects, but it is almost always used to denote an equivalence relation: a relation that is reflexive, symmetric, and transitive.
Daniel Littlewood and anorton have already discussed what $\sim$ means in this instance, and we can verify that it is an equivalent relation between functions on $\Bbb R$.
Clearly for any function $f(n)$, we have that $f\sim f$ since $(f/f)(n)=1$ for all $n$ (with caveats about the zeros of $f$) and $\lim_{n\rightarrow\infty}1=1$.
Also if $f\sim g$ we then have that $g\sim f$. This follows from the fact that if $\lim_{n\rightarrow\infty}h(n)=1$ then $\lim_{n\rightarrow\infty}(1/h)(n)=1$. Applying this fact with $h=f/g$ we tell us that $\sim$ is symmetric.
Finally $\sim$ is transitive. This follows from the fact that if $\lim_{n\rightarrow\infty}h_1(n)=1$ and $\lim_{n\rightarrow\infty}h_2(n)=1$ then $\lim_{n\rightarrow\infty}(h_1h_2)(n)=1$. If we have that $f\sim g$ and $g\sim h$, apply the previous fact with $h_1=f/g$ and $h_2=g/h$, and you will get that $f\sim h$.
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I am interested in the practical method and I like to discover if it is cheap enough to be done as an experiment in a high school.
Method
The method is based on measuring variations in perceived revolution time of Io around Jupiter. Io is the innermost of the four Galilean moons of Jupiter and it takes around 42.5 hours to orbit Jupiter.
The revolution time can be measured by calculating the time interval between the moments Io enters or leaves Jupiter's shadow. Depending on the relative position of Earth and Jupiter, you will either be able to see Io entering the shadow but not leaving it or you will be able to see it leaving the shadow, but not entering. This is because Jupiter will obstruct the view in one of the cases.
You might expect that if you keep looking at Io for a few weeks or months you will see it enter/leave Jupiter's shadow at roughly regular intervals matching Io's revolution around Jupiter.
However, even after introducing corrections for Earth's and Jupiter's orbit eccentricity, you still notice that for a few weeks as Earth moves away from Jupiter the time between observations becomes longer (eventually by a few minutes). At other time of year, you notice that for a few weeks as Earth moves towards Jupiter the time between observations becomes shorter (again, eventually by a few minutes). This few minutes difference comes from the fact that when Earth is further away from Jupiter it takes light more time to reach you than when Earth is closer to Jupiter.
Say you have made two consecutive observations of Io entering Jupiter's shadow at
t 0 and tseparated by 1 nIo's revolutions about Jupiter T. If the speed of light was infinite, one would expect
\begin{equation} t_1 = t_0 + nT \end{equation}
This is however not the case and the difference
\begin{equation} \Delta t = t_1 - t_0 - nT \end{equation}
can be used to measure the speed of light since it is the extra time that light needs to travel the distance equal to the difference in the separation of Earth and Jupiter at
t 1 and t: 0
\begin{equation} c = \frac{\Delta d}{\Delta t} = \frac{d_{EJ}(t_1)-d_{EJ}(t_0)}{\Delta t} \end{equation}
(both numerator and denominator can be negative representing Earth approaching or receding from Jupiter)
In reality more than two observations are needed since
T isn't known. It can be approximated by averaging observations equally distributed around Earth's orbit accounting for eccentricity or simply solved for as another variable. Practical considerations
Note that you will not manage to see Io enter/leave Jupiter's shadow every Io's orbit (i.e. roughly every 42.5 hours) since some of your observation times will fall on a day or will be made impossible by weather conditions. This is of no concern however. You should simply number all Io's revolutions around Jupiter (timed by Io entering/leaving Jupiter's shadow) and note which ones you managed to observe. For successful observations you should record precise time. It might be good to use UTC to avoid problems with daylight saving time changes. After a few weeks you will notice cumulative effect of the speed of light in that the average intervals between Io entering/leaving Jupiter's shadow will become longer or shorter. Cumulative effect is easier to notice. At minimum you should try to make two observations relatively close to each other (separated by just a few Io revolutions) and then at least one more observation a few weeks or months later (a few dozens of Io revolutions). This will let you calculate the average time interval between observations within a short and long time period by dividing the length of the time period by the number of revolutions Io has made around Jupiter in that period. The average computed over the long time period will exhibit cumulative effect of the speed of light by being noticeably longer or shorter than the average computed over the short time period. More observations will help you make a more accurate determination of the speed of light. You must plan all of the observations ahead since you can't make the observations when Earth and Jupiter are close to conjunction or opposition.
Calculations
Once you collected the observations you should determine the position of Earth and Jupiter at the times of the observations (for example using JPL's Horizons system). You can then use the positions to determine the distance between the planets at the time the observations were made. Finally, you can use the distance and the variation in Io's perceived revolution period to compute the speed of light.
You will notice that roughly every 18 millions kms change in the distance of Earth and Jupiter makes an observation happen 1 minute earlier or later.
Cost
The cost of the experiment is largely the cost of buying a telescope that allows you to see Io. Note that the experiment takes a few months and requires measuring time of the observations with the accuracy of seconds.
History
See this wikipedia article for historical account of the determination of the speed of light by Rømer using Io.
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Main Page The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]x[/math] in it, e.g., [math]112x1xx3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. Density Hales-Jewett (DHJ) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers.
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (active) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (final call) (500-599) TBA (600-699) A reading seminar on density Hales-Jewett (active) Unsolved questions
Gowers.462: Incidentally, it occurs to me that we as a collective are doing what I as an individual mathematician do all the time: have an idea that leads to an interesting avenue to explore, get diverted by some temporarily more exciting idea, and forget about the first one. I think we should probably go through the various threads and collect together all the unsolved questions we can find (even if they are vague ones like, “Can an approach of the following kind work?”) and write them up in a single post. If this were a more massive collaboration, then we could work on the various questions in parallel, and update the post if they got answered, or reformulated, or if new questions arose.
IP-Szemeredi (a weaker problem than DHJ)
Solymosi.2: In this note I will try to argue that we should consider a variant of the original problem first. If the removal technique doesn’t work here, then it won’t work in the more difficult setting. If it works, then we have a nice result! Consider the Cartesian product of an IP_d set. (An IP_d set is generated by d numbers by taking all the [math]2^d[/math] possible sums. So, if the n numbers are independent then the size of the IP_d set is [math]2^d[/math]. In the following statements we will suppose that our IP_d sets have size [math]2^n[/math].)
Prove that for any [math]c\gt0[/math] there is a [math]d[/math], such that any c-dense subset of the Cartesian product of an IP_d set (it is a two dimensional pointset) has a corner.
The statement is true. One can even prove that the dense subset of a Cartesian product contains a square, by using the density HJ for k=4. (I will sketch the simple proof later) What is promising here is that one can build a not-very-large tripartite graph where we can try to prove a removal lemma. The vertex sets are the vertical, horizontal, and slope -1 lines, having intersection with the Cartesian product. Two vertices are connected by an edge if the corresponding lines meet in a point of our c-dense subset. Every point defines a triangle, and if you can find another, non-degenerate, triangle then we are done. This graph is still sparse, but maybe it is well-structured for a removal lemma.
Finally, let me prove that there is square if d is large enough compare to c. Every point of the Cartesian product has two coordinates, a 0,1 sequence of length d. It has a one to one mapping to [4]^d; Given a point ((x_1,…,x_d),(y_1,…,y_d)) where x_i,y_j are 0 or 1, it maps to (z_1,…,z_d), where z_i=0 if x_i=y_i=0, z_i=1 if x_i=1 and y_i=0, z_i=2 if x_i=0 and y_i=1, and finally z_i=3 if x_i=y_i=1. Any combinatorial line in [4]^d defines a square in the Cartesian product, so the density HJ implies the statement.
Gowers.7: With reference to Jozsef’s comment, if we suppose that the d numbers used to generate the set are indeed independent, then it’s natural to label a typical point of the Cartesian product as (\epsilon,\eta), where each of \epsilon and \eta is a 01-sequence of length d. Then a corner is a triple of the form (\epsilon,\eta), (\epsilon,\eta+\delta), (\epsilon+\delta,\eta), where \delta is a \{-1,0,1\}-valued sequence of length d with the property that both \epsilon+\delta and \eta+\delta are 01-sequences. So the question is whether corners exist in every dense subset of the original Cartesian product.
This is simpler than the density Hales-Jewett problem in at least one respect: it involves 01-sequences rather than 012-sequences. But that simplicity may be slightly misleading because we are looking for corners in the Cartesian product. A possible disadvantage is that in this formulation we lose the symmetry of the corners: the horizontal and vertical lines will intersect this set in a different way from how the lines of slope -1 do.
I feel that this is a promising avenue to explore, but I would also like a little more justification of the suggestion that this variant is likely to be simpler.
Gowers.22: A slight variant of the problem you propose is this. Let’s take as our ground set the set of all pairs (U,V) of subsets of \null [n], and let’s take as our definition of a corner a triple of the form (U,V), (U\cup D,V), (U,V\cup D), where both the unions must be disjoint unions. This is asking for more than you asked for because I insist that the difference D is positive, so to speak. It seems to be a nice combination of Sperner’s theorem and the usual corners result. But perhaps it would be more sensible not to insist on that positivity and instead ask for a triple of the form (U,V), ((U\cup D)\setminus C,V), (U, (V\cup D)\setminus C, where D is disjoint from both U and V and C is contained in both U and V. That is your original problem I think.
I think I now understand better why your problem could be a good toy problem to look at first. Let’s quickly work out what triangle-removal statement would be needed to solve it. (You’ve already done that, so I just want to reformulate it in set-theoretic language, which I find easier to understand.) We let all of X, Y and Z equal the power set of \null [n]. We join U\in X to V\in Y if (U,V)\in A.
Ah, I see now that there’s a problem with what I’m suggesting, which is that in the normal corners problem we say that (x,y+d) and (x+d,y) lie in a line because both points have the same coordinate sum. When should we say that (U,V\cup D) and (U\cup D,V) lie in a line? It looks to me as though we have to treat the sets as 01-sequences and take the sum again. So it’s not really a set-theoretic reformulation after all.
O'Donnell.35: Just to confirm I have the question right…
There is a dense subset A of {0,1}^n x {0,1}^n. Is it true that it must contain three nonidentical strings (x,x’), (y,y’), (z,z’) such that for each i = 1…n, the 6 bits
[ x_i x'_i ] [ y_i y'_i ] [ z_i z'_i ]
are equal to one of the following:
[ 0 0 ] [ 0 0 ] [ 0, 1 ] [ 1 0 ] [ 1 1 ] [ 1 1 ] [ 0 0 ], [ 0 1 ], [ 0, 1 ], [ 1 0 ], [ 1 0 ], [ 1 1 ], [ 0 0 ] [ 1 0 ] [ 0, 1 ] [ 1 0 ] [ 0 1 ] [ 1 1 ]
?
McCutcheon.469: IP Roth:
Just to be clear on the formulation I had in mind (with apologies for the unprocessed code): for every $\delta>0$ there is an $n$ such that any $E\subset [n]^{[n]}\times [n]^{[n]}$ having relative density at least $\delta$ contains a corner of the form $\{a, a+(\sum_{i\in \alpha} e_i ,0),a+(0, \sum_{i\in \alpha} e_i)\}$. Here $(e_i)$ is the coordinate basis for $[n]^{[n]}$, i.e. $e_i(j)=\delta_{ij}$.
Presumably, this should be (perhaps much) simpler than DHJ, k=3.
High-dimensional Sperner Kalai.29: There is an analogous for Sperner but with high dimensional combinatorial spaces instead of "lines" but I do not remember the details (Kleitman(?) Katona(?) those are ususal suspects.)
Fourier approach
Kalai.29: A sort of generic attack one can try with Sperner is to look at f=1_A and express using the Fourier expansion of f the expression \int f(x)f(y)1_{x<y} where x<y is the partial order (=containment) for 0-1 vectors. Then one may hope that if f does not have a large Fourier coefficient then the expression above is similar to what we get when A is random and otherwise we can raise the density for subspaces. (OK, you can try it directly for the k=3 density HJ problem too but Sperner would be easier;) This is not unrealeted to the regularity philosophy. Gowers.31: Gil, a quick remark about Fourier expansions and the k=3 case. I want to explain why I got stuck several years ago when I was trying to develop some kind of Fourier approach. Maybe with your deep knowledge of this kind of thing you can get me unstuck again.
The problem was that the natural Fourier basis in \null [3]^n was the basis you get by thinking of \null [3]^n as the group \mathbb{Z}_3^n. And if that’s what you do, then there appear to be examples that do not behave quasirandomly, but which do not have large Fourier coefficients either. For example, suppose that n is a multiple of 7, and you look at the set A of all sequences where the numbers of 1s, 2s and 3s are all multiples of 7. If two such sequences lie in a combinatorial line, then the set of variable coordinates for that line must have cardinality that’s a multiple of 7, from which it follows that the third point automatically lies in the line. So this set A has too many combinatorial lines. But I’m fairly sure — perhaps you can confirm this — that A has no large Fourier coefficient.
You can use this idea to produce lots more examples. Obviously you can replace 7 by some other small number. But you can also pick some arbitrary subset W of \null[n] and just ask that the numbers of 0s, 1s and 2s inside W are multiples of 7.
DHJ for dense subsets of a random set
Tao.18: A sufficiently good Varnavides type theorem for DHJ may have a separate application from the one in this project, namely to obtain a “relative” DHJ for dense subsets of a sufficiently pseudorandom subset of {}[3]^n, much as I did with Ben Green for the primes (and which now has a significantly simpler proof by Gowers and by Reingold-Trevisan-Tulsiani-Vadhan). There are other obstacles though to that task (e.g. understanding the analogue of “dual functions” for Hales-Jewett), and so this is probably a bit off-topic.
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Constraining Dark Matter with the Long-Term Variability of Quasars Abstract.
By comparing the results of numerical microlensing simulations to the observed longterm variability of quasars, strong upper limits on the cosmological density of compact objects in the mass range \(10^{-2}-10^{-4} M_\odot\) may be imposed. Using recently developed methods to better approximate the amplification of large sources, we investigate in what way the constraints are affected by assumptions concerning the size of the optical continuum-emitting region of quasars in the currently favored ( \(\Omega_{rm M} = 0.3, \Omega_\Lambda = 0.7\)) cosmology.
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Let $H$ be a Hilbert space, $X$ a topological space, and $\{A_t\}_{t\in X}$ a continuous family of bounded, invertible operators on $H$. Continuous here in the sense that the corresponding map $X\rightarrow B(H)$ is continuous (with respect to the uniform topology on $B(H)$). Then the family of inverses $\{A^{-1}_t\}_{t\in X}$ is also continuous.
Question: what kind of generalizations are there for families of unbounded operators?
To be more specific, suppose that $\{A_t\}_{t\in X}$ is a family of unbounded, self-adjoint operators on $H$ whose domains $D_t$ are (possibly) varying in $H$, but with $D:=\bigcap_t D_t$ dense in $H$. If each $A_t$ has a bounded inverse $A_t^{-1}$, I would like to know if there are situations (i.e. conditions we have to impose) where we can conclude that $\{A^{-1}_t\}_{t\in X}$ is a continuous family of operators. One of the issues (which is part of the question) is what the right notion of continuity would be in this situation. I am aware of some partial answers:
If all the $A_t$ have the same domain, i.e. $D=D_t$ for all $t$, and if the matrix coefficients $\langle A_t u, v\rangle$ are continuous functions of $t\in X$, for every $u\in D$ and $v\in H$, then the answer is yes (this is proved e.g. in: "Kriegl, Michor; Differentiable perturbation of unbounded operators. Math. Ann. 327 (2003), no. 1, 191–201. 47A55")
Example where uniform continuity seems to fail: $H:=L^2(\mathbb{R})$, $A_t:=\mathcal{F}^{-1} (1+|\xi|^2)^{t/2}\mathcal{F}:H\rightarrow H$, where $\mathcal{F}$ denotes the Fourier transform. Then $\{A\}_{t\in [0,1]}$ is a family of unbounded, self-adjoint operators on $H$, with domains given by the Sobolev spaces $D_t=L^2_t$. The inverses are $A_t^{-1}=A_{-t}$, which are bounded on $L^2$, but $\{A_{-t}\}_{t\in [0,1]}$ is not continuous in the uniform topology (it is continuous for the strong topology).
The application I have in mind is this; given a smooth family of pseudo-convex complex structures on a compact manifold with boundary such that the corresponding $\overline{\partial}$-Laplacians are invertible on $L^2(\Lambda^{p,q}T^\ast)$ for some $p,q$, then is the family of inverses uniformly continuous? (In the case without boundary the answer is yes)
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Let $k$ be a $p$-adic field, $G$ a connected reductive group over $k$ with minimal parabolic $P_0$ containing a maximal split torus $A_0$. Let $W = N_G(A_0)(k)/Z_G(A_0)(k)$ be the Weyl group, and $S \subset W$ the simple reflections from $P_0$.
For $\theta, \Omega \subset S$, we have the standard parabolic subgroups $P_\theta, P_\Omega$.
Each double coset $P_{\theta}w P_{\Omega}$, for $w \in W$ with a $k$-rational representative, is a locally closed subvariety of $G$, with $P_\theta wP_\Omega(k) = P_\theta(k)wP_\Omega(k)$.
How do we know that the quotient $P_\theta \backslash P_\theta wP_\Omega$ is an algebraic variety over $k$? This variety and its dimension are considered in Casselman's notes on representation theory, Chapter 6. I do not know the general theory of quotients of algebraic group actions which would make sense out of something like this.
Once this variety is made sense of, can we say that its $k$-rational points coincides with $P_\theta(k) \backslash P_\theta(k) wP_\Omega(k)$?
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This problem:
I was able to solve it doing this:
But I think i have nowhere " exploited " the equivalence given above. How can this be used in this problem?
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The first relation has been derived by using the equivalence given above.
$$\frac {dQ}{dt} = j(\vec r)*4\pi r^2$$ Putting $$\frac {dT}{dr} =k* j(\vec r)$$ Since $$\frac {dV}{dr} = \vec E(\vec r)$$
We will get the same relation
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Suppose I take two independent random samples from a population of size $N$: the first is a simple random sample of size $n$ and the second is a simple random sample of size $m$. Let set $S$ contain all of the records selected from the two samples of size $n$ and $m$ less any duplicates. Thus, $S$ should consist of $n_s$ records where $n_s$ $\le ($$m$+$n$) since $S$ does not contain duplicates.
I am trying to create an unbiased estimator for the population total that only depends on the records in $S$ in terms of only all $y_i$, $n$, $m$, and $N$.
I tried to mimic the unbiased Horvitz Thompson estimator $\hat{t} = \sum_{iϵS}y_i/\pi_i$, where $\pi_i$ is the probability that $y_i$ is in $S$. Here, I believe $\pi_i$ is the probability of the union of the event that $y_i$ is selected in the first sample of $n$ people and the event that $y_i$ is selected in the second sample of $m$ people. However, I am unsure of how to express the upper limit of the sum without using $n_s$.
Is there a way to mimic the Horvitz Thompson estimator in terms only of all $y_i$, $n$, $m$, and $N$? Or should I be taking a different approach to creating an unbiased estimator?
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Here is a beautiful result from numerical analysis. Given any nonsingular $n\times n$ system of linear equations $Ax=b$, an optimal Krylov subspace method like GMRES must necessarily terminate with the exact solution $x=A^{-1}b$ in no more than $n$ iterations (assuming exact arithmetic).
The Cayley-Hamilton theorem provides a simple, elegant proof of this statement. To begin, recall that at the $k$-th iteration, minimum residual methods like GMRES solve the least-squares problem$$\underset{x_k\in\mathbb{R}^n}{\text{minimize }} \|Ax_k-b\|$$by picking a solution from the $k$-th Krylov subspace$$\text{subject to } x_k \in \mathrm{span}\{b,Ab,A^2b,\ldots,A^{k-1}b\}.$$If the objective $ \|Ax_k-b\|$ goes to zero, then we have found the exact solution at the $k$-th iteration (we have assumed that $A$ is full-rank).
Next, observe that $x_k=(c_0 + c_1 A + \cdots + c_{k-1}A^{k-1})b=p(A)b$, where $p(\cdot)$ is a polynomial of order $k-1$. Similarly, $\|Ax_k-b\|=\|q(A)b\|$, where $q(\cdot)$ is a polynomial of order $k$ satisfying $q(0)=-1$. So the least-squares problem from above for each fixed $k$ can be equivalently posed as a polynomial optimization problem with the same optimal objective
$$\text{minimize } \|q_k(A)b\| \text{ subject to } q_k(0)=-1,\; q_k(\cdot) \text{ is an order-} k \text{ polynomial.}$$Again, if the objective $\|q_k(A)b\|$ goes to zero, then GMRES has found the exact solution at the $k$-th iteration.
Finally, we ask: what is a bound on $k$ that guarantees that the objective goes to zero? Well, with $k=n$, and the optimal polynomial $q_n(\cdot)$ for our polynomial optimization problem is just the characteristic polynomial of $A$. According to Cayley-Hamilton, $q_n(A)=0$, so $\|q_n(A)b\|=0$. Hence we conclude that GMRES always terminate with the exact solution at the $n$-th iteration.
This same argument can be repeated (with very minor modifications) for other optimal Krylov methods like conjugate gradients, conjugate residual / MINRES, etc. In each case, the Cayley-Hamilton forms the crux of the argument.
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How do we know that dark matter is dark, in the sense that it doesn't give out any light or absorb any? It is impossible for humans to be watching every single wavelength. For example, what about wavelengths that are too big to detect on Earth?
If dark matter emitted very long wave lengths of electromagnetic radiation it would mean it is composed of charged particles. There is no escape from that conclusion. Somebody might propose that dark matter is some strange configuration of charged particles which acts as a very long wavelength antenna. That might be a good model, but there is a hitch with it. If it emits longer wavelength radiation it must be colder. The Wien displacement law is that $\lambda~=~c/T$, for $c$ a constant, which may be calculated, but is not relevant here. For $T~=~2.7$K the wavelength of radiation is $.1cm$. Let me assume that there does exist some very long wavelength radiation, say $100m$ as the lower bound on this. The ratio of temperatures gives that $$ T_{100m}~=~\frac{10^{-3}}{10^2}2.7K~=~2.7\times 10^{-5}K $$ So all this can be is a very cold gas, which is the hitch. This gas is much colder than the background radiation and not in equilibrium. So from some physical grounds this is not likely, and dark matter is most likely not some ordinary form of matter that interacts by EM.
There is indeed very good reason to believe dark matter is dark - apart from all the evidence from "missing mass" in luminosity counts and gravitational lensing studies.
This comes from theories of large-scale structure formation: That there has always had to be some sort of matter that doesn't interact electromagnetically at all is crucial to most scenarios of large-scale structure formation. The density fluctuations in the present universe would be too large than what would be predicted if there were only ordinary baryonic matter that interacted only electromagnetically. With dark matter, you can have something that gravitates yet decouples from radiation much before baryonic matter does. This allows the dark matter to form gravitational wells (under collapse) which have a much longer time to expand with the universe. By the time ordinary matter decouples from radiation and joins the rest of the expansion flow, the ordinary matter will quickly fall into these large gravitational wells of the dark matter that have had far more time to grow. This, in a way, amplifies density perturbations in the early universe and allows large-scale structure to form. (to the extent that we see it today in the form of clusters and galaxies)
The required amount of dark matter calculated, in this way, in order to observe the present scale of density fluctuations matches very well with the amount of dark matter required to explain galactic rotation curves, gravitational lensing, etc. So there's excellent agreement that all of these are due to the same thing - some sort of matter which doesn't interact electromagnetically at all, viz. dark matter.
Aside from the very good theoretical reasons dbrane gives, I'd say that we don't have observational evidence that DM is truly dark. Some of the largely discredited MACHO candidates, like large numbers of red dwarfs, brown dwarfs, planetissimals etc. have been rejected for theoretical -not observational reasons. DM, that is lightly emitting is not currently detectable.
Nobody is saying that dark matter
necessarily doesn't emit or reflect any EM radiation at all. The reason we call it dark matter is that it doesn't emit radiation in the visible spectrum.
The original discovery was simply that the
stars in galaxies did not have enough mass to account for the measured mass of the galaxy. Astronomers therefore realized that there had to be some stuff that wasn't emitting visible light, hence they called it "dark." Originally, it was thought that brown dwarfs (very small stars without sustained fusion) could be the dark matter; those wouldn't emit visible light, but they would certainly emit infrared radiation. However, later experiments ruled that out, along with other forms of "normal" (baryonic) matter known to exist.
At this point in time, I believe we know from experimental observations that dark matter is not only dark in the visible spectrum, it appears to emit/reflect negligible amounts of radiation in all other frequency bands that we've checked. But that's just an observation. I don't know of any requirement that dark matter
can't interact with EM radiation at all. Most theories about the nature of dark matter do predict that it's totally (or almost totally) decoupled from the EM field. But there is still the possibility that it isn't.
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This is at least feasible if you allow a one-time setup where related keys are distributed among the $n$ devices and the master device. There is a primitive known as aggregated encryption which has the following properties:
The $n$ devices $(D_0, \cdots, D_{n-1})$ each receive a secret key $(s_0, \cdots, s_{n-1})$. The master device receives an "aggregated secret key"; concretely, this will in general be $s = \sum_i s_i$. Each device $D_i$ can encrypt some message $m_i$ with the key $s_i$. Given $n$ encrypted messages $(E_0,\cdots, E_{n-1}) = (E_{s_0}(m_0), \cdots, E_{s_{n-1}}(m_{n-1}))$ and the aggregated secret key $s$, the master device can decrypt $\sum_i s_i$, and cannot learn anything more.
This is relatively easy to build using cryptosystem with appropriate homomorphic properties. The article I link to builds it from the ElGamal cryptosystem, however there also exists constructions from the Paillier cryptosystem, which will be more suited to your problem.
Now, given such a primitive, there is a natural way to ensure that the master device can learn
all messages, but without knowing from which device they come: in the one-time setup phase, also give to each device $D_i$ his secret index $j = \pi^{-1}(i)$ ($\pi$ is a uniformly random permutation of $\{0, \cdots, n-1\}$, $\pi^{-1}$ is the reverse permutation, $\pi^{-1}(i)$ will be the "pseudonymous index" of the device $D_i$). Let $\ell$ be a bound on the size (in bit) of each message $m_i$ (I assume $\ell$ to be publicly known) Then, to send $n$ message while concealing the identity of the device each message came from, each device $D_i$ with message $m_i$ computes and sends:
$E_i = E_{s_i}(2^{\ell\cdot\pi^{-1}(i)}\cdot m_i)$
Given $(E_0, \cdots, E_{n-1})$ and his aggregated key $s$, the master device can only recover
$M = \sum_{i=0}^{n-1} 2^{\ell\cdot\pi^{-1}(i)}\cdot m_i = \sum_{j=0}^{n-1} 2^{\ell \cdot j}\cdot m_{\pi(j)}$
Which is exactly the value $m_{\pi(0)} || m_{\pi(1)} || \cdots || m_{\pi(n-1)}$ (when padding each message $m_i$ with zeroes such that it is exactly $\ell$ bit long),
id est, all the messages of the devices in some fixed secret permuted order.
Note that an ElGamal-based scheme would not work here since you need your aggregated scheme to support a very large message space, but the Joye-Libert scheme has no restriction on the message space, and should work fine here.
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The question is the same as asking when are the solutions to the Schrodinger equation of the form $\psi(x)=A\sin(kx)$ or $\psi(x)=Ae^{\kappa x}$.
Assume $\psi(x)=A\sin(kx)$ is your solution with $k$ constant but an unknown $V(x)$ and plug into the Schrodinger equation\begin{align}-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}A\sin(kx) + V(x)A\sin(kx)&=EA\sin(kx)\, , \\\frac{\hbar^2k^2}{2m}A\sin(kx)+V(x)A\sin(kx)&=EA\sin(kx) \tag{1}\, .\end{align}Since (1) must hold for every $x$, and assuming $A\sin(kx)\ne 0$ for
some $x$, we can cancel the common factor $A\sin(kx)$ for this $x$ and are left with: $$V(x)=E-\frac{\hbar^2k^2}{2m}\, .$$In other words, the potential is constant: $V(x)=V_0$. Reorganizing we get$$\frac{\hbar^2k^2}{2m}=E-V_0\, .$$Since the left hand side is necessarily positive, the $\psi(x)=A\sin(kx)$ canonly occur when $E-V_0>0.$
The same manipulations repeated with $\psi(x)=Ae^{\kappa x}$ yields$$V(x)=E+\frac{\hbar^2\kappa ^2}{2m}\, ,$$also showing the potential is constant. This time however one concludes that$$\frac{\hbar^2\kappa ^2}{2m}=V_0-E$$so that this solution can only occur then the potential is greater than the energy.
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In Yang-Mills theory the field strength tensor $F_{\mu \nu}$ can be calculated as $$ \begin{equation} F_{\mu\nu} \equiv \frac{i}{g} [D_\mu,D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\nu], \end{equation} $$ where covariant derivative is defined as $$ D_\mu = \partial_\mu -ig A_\mu. $$ Can Yang-Mills field strength be defined as covariant derivative squared instead, rather than as the non-commutatibity of the covariant derivatives? The two definitions are essentially the same, but the former affords an alternative way of deriving the covariance properties of field strength (see below).
It's easier to demonstrate the idea in terms of differential forms. The field strength 2-form $F$ (also called curvature 2-form) is defined as (here the coupling constant and $i$ are absorbed into the definition of $A$ and $F$) $$ F = F_{\mu \nu}dx^\mu\wedge dx^\nu = d\wedge A + A\wedge A, $$ gauge field 1-form $A$ is defined as $$ A = A_\mu dx^\mu, $$ and the covariant derivative of the Dirac fermion field $\psi$ (spinor) is defined as $$ D\psi = (d+A)\psi. $$
Now the covariant derivative squared $D\wedge D$ of a spinor is $$ D\wedge D \psi = (d+A)\wedge (d+A)\psi $$ $$ = (d\wedge d + d\wedge A + A \wedge d + A\wedge A) \psi $$ $$ = 0 + d\wedge (A \psi) + A \wedge (d \psi) + (A\wedge A) \psi $$ $$ = (d\wedge A) \psi - A \wedge (d \psi) + A \wedge (d \psi) + (A\wedge A) \psi $$ $$ = (d\wedge A + A\wedge A) \psi $$ $$ = F \psi, $$ or in short $$ F = D\wedge D, $$ where we have used the fact that $d\wedge d = 0$ and 1-forms $d$ and $A$ anti-commute in wedge (outer) $\wedge$ product.
The beauty of the covariant derivative squared definition is that the covariance of field strength 2-form $F$$$F \rightarrow F' = gFg^{-1}$$is an automatic outcome, since by definition of the
covariant derivative we have $$F\psi = D\wedge D \psi \rightarrow g D\wedge D \psi = g F \psi = g F g^{-1}g\psi = F'\psi',$$where$$\psi \rightarrow \psi' = g\psi.$$
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Left-Hand and Right-Hand Limits
Definition: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then the left-hand limit of $f$ at $c$ denoted $\lim_{x \to c^-} f(x) = L_1$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $x < c$ then $\mid f(x) - L_1 \mid < \epsilon$. Similarly, the right-hand limit of $f$ at $c$ denoted $\lim_{x \to c^+} f(x) = L_2$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $x > c$ then $\mid f(x) - L_2 \mid < \epsilon$.
By the definition of a left-hand limit, we are purely interested in what happens to the function to the left of the cluster point $c$, while for right-hand limits, we are purely interested in what happens to the function to the right of the cluster point $c$. There are five different cases that can happen with regards to left-hand and right-hand limits.
Both the left-hand and right-hand limits exist and $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$. Both the left-hand and right-hand limits exists but $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$. Only the left-hand limit exists. Only the right-hand limit exists. Neither the left-hand limit nor the right-hand limit exists.
For example, consider the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \left\{\begin{matrix} x^2 & x > 1\\ \frac{3}{2} - x & x < 1 \end{matrix}\right.$. Notice that $\lim_{x \to 1^-} f(x) = \frac{1}{2}$ and $\lim_{x \to 1^+} f(x) = 1$, and so left-hand and right-hand limits need not be equal.
We will now look at some theorems regarding left-hand and right-hand limits.
Theorem 1 (Uniqueness of One-sided Limits): Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. If $\lim_{x \to c^-} f(x) = L_1$ and $\lim_{x \to c^-} f(x) = L_2$ then $L_1 = L_2$. Similarly if $\lim_{x \to c^+} f(x) = M_1$ and $\lim_{x \to c^+} f(x) = M_2$ then $M_1 = M_2$.
The proof of theorem 1 is analogous to that of The Uniqueness of Limits of a Function Theorem.
Theorem 2: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then $\lim_{x \to c} f(x) = L$ if and only if $\lim_{x \to c^-} f(x) = L = \lim_{x \to c^+} f(x)$. Proof:Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. $\Rightarrow$ Suppose that $\lim_{x \to c} f(x) = L$. Then $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$. Therefore if $x < c$ and $x \in A$ are such that $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$ and so $\lim_{x \to c^-} f(x) = L$. Similarly if $x > c$ and $x \in A$ are such that $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$ and so $\lim_{x \to c^+} f(x) = L$. Therefore $\lim_{x \to c^-} f(x) = L = \lim_{x \to c^+} f(x)$. $\Leftarrow$ Now suppose that $\lim_{x \to c^-} f(x) = L = \lim_{x \to c^+}$. Notice that $x > c$ and $x < c$ is equivalent to saying $x \neq 0$, which is also equivalent to saying $0 < \mid x - c \mid$. Therefore $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$ and so $\lim_{x \to c} f(x) = L$. $\blacksquare$
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The expected number of picks needed equals the sum of the probabilities that at least $t$ picks are needed, which means that $t-1$ subsets left at least one value uncovered. We can use inclusion-exclusion to get the probability that at least one value is uncovered.
The probability that a particular set of $k$ values is uncovered after $t-1$ subsets are chosen is
$$\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1}$$
So, by inclusion-exclusion, the probability that at least one value is uncovered is
$$ \sum_{k=1}^n {n \choose k}(-1)^{k-1}\Bigg(\frac{n-k \choose r}{n \choose r}\Bigg) ^{t-1} $$
And then the expected number of subsets needed to cover everything is
$$ \sum_{t=1}^\infty \sum_{k=1}^n {n \choose k}(-1)^{k-1} \Bigg(\frac{n-k \choose r}{n \choose r}\Bigg)^{t-1} $$
Change the order of summation and use $s=t-1$:
$$ \sum_{k=1}^n {n \choose k}(-1)^{k-1} \sum_{s=0}^\infty \Bigg( \frac{n-k \choose r}{n \choose r}\Bigg)^s$$
The inner sum is a geometric series.
$$ \sum_{k=1}^n {n \choose k} (-1)^{k-1}\frac{n \choose r}{{n \choose r}-{n-k \choose r}}$$
$$ {n \choose r} \sum_{k=1}^n (-1)^{k-1}\frac{n \choose k}{{n \choose r}-{n-k \choose r}}$$
I'm sure that should simplify further, but at least now it's a simple sum. I've checked that this agrees with the coupon collection problem for $r=1$.
Interestingly, Mathematica "simplifies" this sum for particular values of $r$, although what it returns even for the next case is too complicated to repeat, involving EulerGamma, the gamma function at half-integer values, and PolyGamma[0,1+n].
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Can someone please tell me how I get vertical lines in the columns of a matrix like it is shown in the picture?
I started with
... = \begin{pmatrix} \lambda A' & \mu B' & \tau C' \end{pmatrix} ...
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Can someone please tell me how I get vertical lines in the columns of a matrix like it is shown in the picture?
I started with
... = \begin{pmatrix} \lambda A' & \mu B' & \tau C' \end{pmatrix} ...
You could use a \vline
\documentclass{article}\usepackage{amsmath}\begin{document}$ M = \begin{pmatrix}\vline & \vline & \vline \\\lambda A' & \mu B' & \tau C' \\\vline & \vline & \vline\end{pmatrix}$\end{document}
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Every Diagonalizable Nilpotent Matrix is the Zero Matrix Problem 504
Prove that if $A$ is a diagonalizable nilpotent matrix, then $A$ is the zero matrix $O$.
Contents
Definition (Nilpotent Matrix)
A square matrix $A$ is called
nilpotent if there exists a positive integer $k$ such that $A^k=O$. Proof. Main Part
Since $A$ is diagonalizable, there is a nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$.
As we show below, the only eigenvalue of any nilpotent matrix is $0$.
Thus, $S^{-1}AS$ is the zero matrix. Hence $A=SOS^{-1}=O$. The only eigenvalue of each nilpotent matrix is $0$
It remains to show that the fact we used above: the only eigenvalue of the nilpotent matrix $A$ is $0$.
Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$.
That is, \[A\mathbf{v}=\lambda \mathbf{v}, \tag{*}\]
Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$.
Then we use the relation (*) inductively and obtain
\begin{align*} A^k\mathbf{v}&=A^{k-1}A\mathbf{v}\\ &=\lambda A^{k-1}\mathbf{v} && \text{by (*)}\\ &=\lambda A^{k-2}A\mathbf{v}\\ &=\lambda^2 A^{k-2}\mathbf{v} && \text{by (*)}\\ &=\dots =\lambda^k \mathbf{v}. \end{align*}
Hence we have
\[\mathbf{0}=O\mathbf{v}=A^k\mathbf{v}=\lambda^k \mathbf{v}.\]
Note that the eigenvector $\mathbf{v}$ is a nonzero vector by definition.
Thus, we must have $\lambda^k=0$, hence $\lambda=0$. This proves that the only eigenvalue of the nilpotent matrix $A$ is $0$, and this completes the proof. Another Proof of the Fact
Even though the fact proved above is true regardless of diagonalizability of $A$, we can make use that $A$ is diagonalizable to prove the fact as follows.
Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of the $n\times n$ nilpotent matrix $A$.
Then we have \[S^{-1}AS=D,\] where \[D:=\begin{bmatrix} \lambda_1 & 0 & \dots & 0 \\ 0 &\lambda_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \lambda_n \end{bmatrix}.\]
Then we have
\(\require{cancel}\) \begin{align*} D^k&=(S^{-1}AS)^k\\ &=(S^{-1}A\cancel{S})(\cancel{S}^{-1}A\cancel{S})\cdots (\cancel{S}^{-1}AS)\\ &=S^{-1}A^kS\\ &=S^{-1}OS=O. \end{align*}
Since
\[D^k=\begin{bmatrix} \lambda_1^k & 0 & \dots & 0 \\ 0 &\lambda_2^k & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \lambda_n^k \end{bmatrix},\] it yields that $\lambda_i^=0$, and hence $\lambda_i=0$ for $i=1, \dots, n$. Related Question.
The converse of the above fact is also true: if the only eigenvalue of $A$ is $0$, then $A$ is a nilpotent matrix.
See the post ↴
Nilpotent Matrix and Eigenvalues of the Matrix for a proof of this fact.
Add to solve later
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Difference between revisions of "De Bruijn-Newman constant"
(→t>0)
(→Threads)
(6 intermediate revisions by the same user not shown) Line 92: Line 92:
* [https://terrytao.wordpress.com/2018/03/18/polymath15-sixth-thread-the-test-problem-and-beyond/ Polymath15, sixth thread: the test problem and beyond], Terence Tao, Mar 18, 2018.
* [https://terrytao.wordpress.com/2018/03/18/polymath15-sixth-thread-the-test-problem-and-beyond/ Polymath15, sixth thread: the test problem and beyond], Terence Tao, Mar 18, 2018.
* [https://terrytao.wordpress.com/2018/03/28/polymath15-seventh-thread-going-below-0-48/ Polymath15, seventh thread: going below 0.48], Terence Tao, Mar 28, 2018.
* [https://terrytao.wordpress.com/2018/03/28/polymath15-seventh-thread-going-below-0-48/ Polymath15, seventh thread: going below 0.48], Terence Tao, Mar 28, 2018.
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== Other blog posts and online discussion ==
== Other blog posts and online discussion ==
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* [https://github.com/km-git-acc/dbn_upper_bound/tree/master/Writeup Writeup subdirectory of Github repository]
* [https://github.com/km-git-acc/dbn_upper_bound/tree/master/Writeup Writeup subdirectory of Github repository]
+ +
== Test problem ==
== Test problem ==
Latest revision as of 17:37, 30 April 2019
For each real number [math]t[/math], define the entire function [math]H_t: {\mathbf C} \to {\mathbf C}[/math] by the formula
[math]\displaystyle H_t(z) := \int_0^\infty e^{tu^2} \Phi(u) \cos(zu)\ du[/math]
where [math]\Phi[/math] is the super-exponentially decaying function
[math]\displaystyle \Phi(u) := \sum_{n=1}^\infty (2\pi^2 n^4 e^{9u} - 3 \pi n^2 e^{5u}) \exp(-\pi n^2 e^{4u}).[/math]
It is known that [math]\Phi[/math] is even, and that [math]H_t[/math] is even, real on the real axis, and obeys the functional equation [math]H_t(\overline{z}) = \overline{H_t(z)}[/math]. In particular, the zeroes of [math]H_t[/math] are symmetric about both the real and imaginary axes. One can also express [math]H_t[/math] in a number of different forms, such as
[math]\displaystyle H_t(z) = \frac{1}{2} \int_{\bf R} e^{tu^2} \Phi(u) e^{izu}\ du[/math]
or
[math]\displaystyle H_t(z) = \frac{1}{2} \int_0^\infty e^{t\log^2 x} \Phi(\log x) e^{iz \log x}\ \frac{dx}{x}.[/math]
In the notation of [KKL2009], one has
[math]\displaystyle H_t(z) = \frac{1}{8} \Xi_{t/4}(z/2).[/math]
De Bruijn [B1950] and Newman [N1976] showed that there existed a constant, the
de Bruijn-Newman constant [math]\Lambda[/math], such that [math]H_t[/math] has all zeroes real precisely when [math]t \geq \Lambda[/math]. The Riemann hypothesis is equivalent to the claim that [math]\Lambda \leq 0[/math]. Currently it is known that [math]0 \leq \Lambda \lt 1/2[/math] (lower bound in [RT2018], upper bound in [KKL2009]).
The
Polymath15 project seeks to improve the upper bound on [math]\Lambda[/math]. The current strategy is to combine the following three ingredients: Numerical zero-free regions for [math]H_t(x+iy)[/math] of the form [math]\{ x+iy: 0 \leq x \leq T; y \geq \varepsilon \}[/math] for explicit [math]T, \varepsilon, t \gt 0[/math]. Rigorous asymptotics that show that [math]H_t(x+iy)[/math] whenever [math]y \geq \varepsilon[/math] and [math]x \geq T[/math] for a sufficiently large [math]T[/math]. Dynamics of zeroes results that control [math]\Lambda[/math] in terms of the maximum imaginary part of a zero of [math]H_t[/math]. Contents [math]t=0[/math]
When [math]t=0[/math], one has
[math]\displaystyle H_0(z) = \frac{1}{8} \xi( \frac{1}{2} + \frac{iz}{2} ) [/math]
where
[math]\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{-s/2} \Gamma(s/2) \zeta(s)[/math]
is the Riemann xi function. In particular, [math]z[/math] is a zero of [math]H_0[/math] if and only if [math]\frac{1}{2} + \frac{iz}{2}[/math] is a non-trivial zero of the Riemann zeta function. Thus, for instance, the Riemann hypothesis is equivalent to all the zeroes of [math]H_0[/math] being real, and Riemann-von Mangoldt formula (in the explicit form given by Backlund) gives
[math]\displaystyle \left|N_0(T) - (\frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} - \frac{7}{8})\right| \lt 0.137 \log (T/2) + 0.443 \log\log(T/2) + 4.350 [/math]
for any [math]T \gt 4[/math], where [math]N_0(T)[/math] denotes the number of zeroes of [math]H_0[/math] with real part between 0 and T.
The first [math]10^{13}[/math] zeroes of [math]H_0[/math] (to the right of the origin) are real [G2004]. This numerical computation uses the Odlyzko-Schonhage algorithm. In [P2017] it was independently verified that all zeroes of [math]H_0[/math] between 0 and 61,220,092,000 were real.
[math]t\gt0[/math]
For any [math]t\gt0[/math], it is known that all but finitely many of the zeroes of [math]H_t[/math] are real and simple [KKL2009, Theorem 1.3]. In fact, assuming the Riemann hypothesis,
all of the zeroes of [math]H_t[/math] are real and simple [CSV1994, Corollary 2].
It is known that [math]\xi[/math] is an entire function of order one ([T1986, Theorem 2.12]). Hence by the fundamental solution for the heat equation, the [math]H_t[/math] are also entire functions of order one for any [math]t[/math].
Because [math]\Phi[/math] is positive, [math]H_t(iy)[/math] is positive for any [math]y[/math], and hence there are no zeroes on the imaginary axis.
Let [math]\sigma_{max}(t)[/math] denote the largest imaginary part of a zero of [math]H_t[/math], thus [math]\sigma_{max}(t)=0[/math] if and only if [math]t \geq \Lambda[/math]. It is known that the quantity [math]\frac{1}{2} \sigma_{max}(t)^2 + t[/math] is non-increasing in time whenever [math]\sigma_{max}(t)\gt0[/math] (see [KKL2009, Proposition A]. In particular we have
[math]\displaystyle \Lambda \leq t + \frac{1}{2} \sigma_{max}(t)^2[/math]
for any [math]t[/math].
The zeroes [math]z_j(t)[/math] of [math]H_t[/math] obey the system of ODE
[math]\partial_t z_j(t) = - \sum_{k \neq j} \frac{2}{z_k(t) - z_j(t)}[/math]
where the sum is interpreted in a principal value sense, and excluding those times in which [math]z_j(t)[/math] is a repeated zero. See dynamics of zeros for more details. Writing [math]z_j(t) = x_j(t) + i y_j(t)[/math], we can write the dynamics as
[math] \partial_t x_j = - \sum_{k \neq j} \frac{2 (x_k - x_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math] [math] \partial_t y_j = \sum_{k \neq j} \frac{2 (y_k - y_j)}{(x_k-x_j)^2 + (y_k-y_j)^2} [/math]
where the dependence on [math]t[/math] has been omitted for brevity.
In [KKL2009, Theorem 1.4], it is shown that for any fixed [math]t\gt0[/math], the number [math]N_t(T)[/math] of zeroes of [math]H_t[/math] with real part between 0 and T obeys the asymptotic
[math]N_t(T) = \frac{T}{4\pi} \log \frac{T}{4\pi} - \frac{T}{4\pi} + \frac{t}{16} \log T + O(1) [/math]
as [math]T \to \infty[/math] (caution: the error term here is not uniform in t). Also, the zeroes behave like an arithmetic progression in the sense that
[math] z_{k+1}(t) - z_k(t) = (1+o(1)) \frac{4\pi}{\log |z_k|(t)} = (1+o(1)) \frac{4\pi}{\log k} [/math]
as [math]k \to +\infty[/math].
Threads Polymath proposal: upper bounding the de Bruijn-Newman constant, Terence Tao, Jan 24, 2018. Polymath15, first thread: computing H_t, asymptotics, and dynamics of zeroes, Terence Tao, Jan 27, 2018. Polymath15, second thread: generalising the Riemann-Siegel approximate functional equation, Terence Tao and Sujit Nair, Feb 2, 2018. Polymath15, third thread: computing and approximating H_t, Terence Tao and Sujit Nair, Feb 12, 2018. Polymath 15, fourth thread: closing in on the test problem, Terence Tao, Feb 24, 2018. Polymath15, fifth thread: finishing off the test problem?, Terence Tao, Mar 2, 2018. Polymath15, sixth thread: the test problem and beyond, Terence Tao, Mar 18, 2018. Polymath15, seventh thread: going below 0.48, Terence Tao, Mar 28, 2018. Polymath15, eighth thread: going below 0.28, Terence Tao, Apr 17, 2018. Polymath15, ninth thread: going below 0.22?, Terence Tao, May 4, 2018. Polymath15, tenth thread: numerics update, Rudolph Dwars and Kalpesh Muchhal, Sep 6, 2018. Polymath15, eleventh thread: Writing up the results, and exploring negative t, Terence Tao, Dec 28, 2018. Effective approximation of heat flow evolution of the Riemann xi function, and a new upper bound for the de Bruijn-Newman constant, Terence Tao, Apr 30, 2019. Other blog posts and online discussion Heat flow and zeroes of polynomials, Terence Tao, Oct 17, 2017. The de Bruijn-Newman constant is non-negative, Terence Tao, Jan 19, 2018. Lehmer pairs and GUE, Terence Tao, Jan 20, 2018. A new polymath proposal (related to the Riemann hypothesis) over Tao's blog, Gil Kalai, Jan 26, 2018. Code and data Writeup
Here are the Polymath15 grant acknowledgments.
Test problem Zero-free regions
See Zero-free regions.
Wikipedia and other references Bibliography [A2011] J. Arias de Reyna, High-precision computation of Riemann's zeta function by the Riemann-Siegel asymptotic formula, I, Mathematics of Computation, Volume 80, Number 274, April 2011, Pages 995–1009. [B1994] W. G. C. Boyd, Gamma Function Asymptotics by an Extension of the Method of Steepest Descents, Proceedings: Mathematical and Physical Sciences, Vol. 447, No. 1931 (Dec. 8, 1994),pp. 609-630. [B1950] N. C. de Bruijn, The roots of trigonometric integrals, Duke J. Math. 17 (1950), 197–226. [CSV1994] G. Csordas, W. Smith, R. S. Varga, Lehmer pairs of zeros, the de Bruijn-Newman constant Λ, and the Riemann hypothesis, Constr. Approx. 10 (1994), no. 1, 107–129. [G2004] Gourdon, Xavier (2004), The [math]10^{13}[/math] first zeros of the Riemann Zeta function, and zeros computation at very large height [KKL2009] H. Ki, Y. O. Kim, and J. Lee, On the de Bruijn-Newman constant, Advances in Mathematics, 22 (2009), 281–306. Citeseer [N1976] C. M. Newman, Fourier transforms with only real zeroes, Proc. Amer. Math. Soc. 61 (1976), 246–251. [P2017] D. J. Platt, Isolating some non-trivial zeros of zeta, Math. Comp. 86 (2017), 2449-2467. [P1992] G. Pugh, The Riemann-Siegel formula and large scale computations of the Riemann zeta function, M.Sc. Thesis, U. British Columbia, 1992. [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. arXiv:1801.05914 [T1986] E. C. Titchmarsh, The theory of the Riemann zeta-function. Second edition. Edited and with a preface by D. R. Heath-Brown. The Clarendon Press, Oxford University Press, New York, 1986. pdf
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Prove the identity:
$$n(n-1)2^{n-2}=\sum_{k=1}^n {k(k-1) {n \choose k}}$$
I tried using the binomial coefficients identity $2^n = \sum_{k=1}^n {n \choose k}$ but got stuck along the way.
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HINT:
For $k\ge2,$
$$k(k-1)\binom nk=k(k-1)n(n-1)\frac{(n-2)!}{k(k-1)\cdot (k-2)! \{(n-2)-(k-2)\}!}$$
$$=n(n-1)\binom{n-2}{k-2}$$
The $n(n-1)$ and $k(k-1)$ are a signal that differentiating twice should be at hand.
So start with $$ (1+X)^n=\sum_k\binom nkX^k, $$ differentiate twice with respect to $X$ to get$$ n(n-1)(1+X)^{n-2}=\sum_k\binom nk k(k-1)X^k, $$ and finally set $X:=1$.
HINT: Start with a pool of $n$ people. You want to pick a team of at least two people, designate one member of the team as captain, and designate a different member of the team as assistant captain. Both sides count the ways to do this. More details in the spoiler-protected block if you get stuck.
On the left you’re picking the captain and assistant captain and the rest of the team. On the right you’re choosing the team size, $k$, picking the team, and then choosing a captain and assistant captain from that team.
We can make use of the following binomial identity: $$\binom ab\binom bc=\binom ac \binom {a-c}{b-c}$$ Also, we can take the summation from $k=2$ instead of $k=1$, since the latter yields a zero term.
Hence $$\begin{align} \sum_{k=1}^n k(k-1)\binom nk&=\sum_{k=2}^n \binom nk k(k-1)\\ &=2\sum_{k=2}^n \binom nk\binom k2\\ &=2\sum_{k=2}^n \binom n2 {n-2\choose k-2}\\ &=2\binom n2 \sum_{k=2}^n {n-2\choose k-2}\\ &=n(n-1)\sum_{k=0}^{n-2} {n-2\choose k}\\ &=n(n-1)2^{n-2}\qquad \blacksquare \end{align}$$
An interesting point to note:
From the above result, by dividing both sides by 2 and noting that $\frac{r(r-1)}2=\binom r2$we can see that $$\sum_{k=2}^n \binom k2 \binom nk=\binom n22^{n-2}$$
This forms a nice pattern continuing from the commonly known results: $$\sum_{k=1}^n \binom k1 \binom nk=\binom n1 2^{n-1}$$ and $$\sum_{k=0}^n \binom k0 \binom nk=\binom n0 2^{n-0}$$
NB: the last two are equations in their more commonly known forms are
$\sum_{k=1}^n k\binom nk=n\cdot 2^{n-1}$ and $\sum_{k=0}^n \binom nk=2^n$ respectively.
From the pattern it appears that $$\sum_{k=m}^n \binom km \binom nk =\binom nm 2^{n-m}$$
This can easily be proven as follows: $$\begin{align} \sum_{k=m}^n \binom km \binom nk &=\sum_{k=m}^n \binom nk \binom km \\ &=\sum_{k=m}^n \binom nm \binom {n-m}{k-m}\\ &=\binom nm\sum_{k=m}^n \binom {n-m}{k-m}\\ &=\binom nm\sum_{k=0}^{n-m} \binom {n-m}{k}\\ &=\binom nm 2^{n-m} \end{align}$$
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Table of Contents
The Alternating Series Test for Alternating Series of Real Numbers
Recall from the Alternating Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ is a strictly positive sequence then the corresponding alternating sequence is given by $((-1)^{n+1}a_n)_{n=1}^{\infty}$.
We will now look at a very important theorem known as the alternating series test which provides criterion for an alternating series to converge.
Theorem 1: If $(a_n)_{n=1}^{\infty}$ is a decreasing sequence and $\displaystyle{\lim_{n \to \infty} a_n = 0}$ then the alternating series $\displaystyle{\sum_{n=1}^{\infty} (-1)^{n+1} a_n}$ converges. Proof:Let $(a_n)_{n=1}^{\infty}$ be a decreasing sequence that converges to $0$ and let $(a_n^*)_{n=1}^{\infty}$ be the corresponding alternating sequence of terms whose general term is defined for all $n \in \mathbb{N}$ by: Let $(s_n)_{n=1}^{\infty}$ be the corresponding sequence of partial sums for the alternating series. Since $(a_n)_{n=1}^{\infty}$ is a decreasing sequence we note that for all $n \in \mathbb{N}$ that $-a_{2n} > a_{2n+1}$, i.e., $a_{2n} + a_{2n+1} < 0$ and so we see that: This shows that the subsequence of odd partial sums forms a decreasing sequence and so: Furthermore we also have that $a_{2n} > -a_{2n + 1}$, i.e., $a_{2n + 1} + a_{2n} > 0$ and so we see that: This shows that the subsequence of even partial sums forms an increasing sequence and so: So in total, the sequence of partial sums $(s_n)_{n=1}^{\infty}$ is bounded above by $s_1$ and bounded below by $s_2$ as seen by combining $(*)$ and $(**)$: Moreover, the decreasing subsequence of odd partial sums $(s_{2n-1})_{n=1}^{\infty}$ is bounded below and hence converges to some $s_1 \in \mathbb{R}$. Similarly, the increasing subsequence of even partial sums $(s_{2n})_{n=1}^{\infty}$ is bounded above and hence converges to some $s_2 \in \mathbb{R}$. We are given that $\displaystyle{\lim_{n \to \infty} a_n = 0 }$ and since $a_{2n} = s_{2n} - s_{2n-1}$ we see that: So $s_1 = s_2$. So, let $s = s_1 = s_2$. Then $\lim_{n \to \infty} s_n = s$ since every subsequence of $(s_n)_{n=1}^{\infty}$ converges to $s$ as seen from the inequality presented by $(***)$. Thus $\displaystyle{\sum_{n=1}^{\infty} (-1)^n a_n}$ converges. $\blacksquare$
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I tried to prove that if $A$ is the disk algebra then the Gelfand transform is the identity map. The statement can be found here in Theorem 4.4 but it is given without proof. Please can someone check my proof?
Proof:
$\Omega (A)=$ character space of $A$
$\sigma (a) = $ spectrum of $a$
If $A$ is the disk algebra then it is generated by $1$ and $z$ and $\widehat{z}$ is a homeomorphism $\Omega (A) \to \sigma (z)$. We know that $\sigma (z) =\mathbb D$ hence it follows that $\Omega (A) \cong \mathbb D$, that is, every character corresponds to exactly one $\lambda \in \mathbb D$. We now claim that the Gelfand transform which we will denote by $\Gamma: A \to C_0(\Omega (A))$ is the identity map. First of all note that $\Omega (A) $ is compact so that $C_0 (\Omega (A)) = C(\Omega (A))$. Since $\Omega (A) = \mathbb D$, the Gelfand transform is a map $\Gamma : A \to C(\mathbb D)$. It now only remains to be shown that $\Gamma (1) = 1$ and $\Gamma (z) = z$. To this end, note that characters are unital and hence $\Gamma (1) = 1$ is clear. If $f \in A$ denotes the map $f(z) = z$ then for a character $\chi$ corresponding to $\lambda \in \mathbb D$ it then is clear that $\chi (f) = f(\lambda) = \lambda$, that is, $\Gamma (z) = z$ for all $z \in \mathbb D$. Since $1$ and $z$ generate $A$ it follows that $\Gamma$ is the inclusion map $A \hookrightarrow C(\mathbb D)$.
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Just another variant that is somewhat simplistic but I think deliver the message without explicitly using the library
boot that may confuse some people with the syntax it uses.
We have a linear model: $y = X \beta + \epsilon$, $\quad \epsilon \sim N(0,\sigma^2)$
The following is a
parametric bootstrap for that linear model, that means that we do not resample our original data but actually we generate new data from our fitted model. Additionally we assume that the bootstrapped distribution of the regression coefficient $\beta$ is symmetric and that is translation invariant. (Very roughly speaking that we can move the axis of it with affecting its properties) The idea behind is that the fluctuations in the $\beta$ 's are due to $\epsilon$ and therefore with enough samples they should provide a good approximation of the true distribution of $\beta$ 's. As before we test again $H_0 : 0 = \beta_j$ and we defined our p-values as "the probability, given a null hypothesis for the probability distribution of the data, that the outcome would be as extreme as, or more extreme than, the observed outcome" (where the observed outcomes in this case are the $\beta$ 's we got for our original model). So here goes:
# Sample Size
N <- 2^12;
# Linear Model to Boostrap
Model2Boot <- lm( mpg ~ wt + disp, mtcars)
# Values of the model coefficients
Betas <- coefficients(Model2Boot)
# Number of coefficents to test against
M <- length(Betas)
# Matrix of M columns to hold Bootstraping results
BtStrpRes <- matrix( rep(0,M*N), ncol=M)
for (i in 1:N) {
# Simulate data N times from the model we assume be true
# and save the resulting coefficient in the i-th row of BtStrpRes
BtStrpRes[i,] <-coefficients(lm(unlist(simulate(Model2Boot)) ~wt + disp, mtcars))
}
#Get the p-values for coefficient
P_val1 <-mean( abs(BtStrpRes[,1] - mean(BtStrpRes[,1]) )> abs( Betas[1]))
P_val2 <-mean( abs(BtStrpRes[,2] - mean(BtStrpRes[,2]) )> abs( Betas[2]))
P_val3 <-mean( abs(BtStrpRes[,3] - mean(BtStrpRes[,3]) )> abs( Betas[3]))
#and some parametric bootstrap confidence intervals (2.5%, 97.5%)
ConfInt1 <- quantile(BtStrpRes[,1], c(.025, 0.975))
ConfInt2 <- quantile(BtStrpRes[,2], c(.025, 0.975))
ConfInt3 <- quantile(BtStrpRes[,3], c(.025, 0.975))
As mentioned the whole idea is that you have the bootstrapped distribution of $\beta$ 's approximates their true one.(Clearly this code is optimized for speed but for readability. :) )
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The Annals of Mathematical Statistics Ann. Math. Statist. Volume 27, Number 3 (1956), 854-855. On Minimum Variance Among Certain Linear Functions of Order Statistics Abstract
Suppose there are $n$ normal populations $N(\mu_i, 1), i = 1, \cdots, n$ and that one random observation from each of these $n$ populations is given. Let $x_1 \leqq x_2 \leqq \cdots \leqq x_n$ be the observations when arranged in order of magnitude and let the corresponding $n$ random variables be denoted by $X_i, i = 1, \cdots, n.$ The following theorem is proved: THEOREM. \begin{equation*}\operatorname{Var}\big(\sum^n_{i = 1} c_i X_i\big), \text{where}\end{equation*}\begin{equation*}\tag{1}\sum^n_{i = 1} c_i = 1,\end{equation*} is minimum when $c_i = 1/n, i = 1, \cdots, n.$ The above theorem may be applied to provide a direct proof of the result that $\sum^n_{i = 1}X_i$ is the best unbiased linear function of order statistics for estimating the sum $\sum^n_{i = 1}\mu_i.$
Article information Source Ann. Math. Statist., Volume 27, Number 3 (1956), 854-855. Dates First available in Project Euclid: 28 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aoms/1177728196 Digital Object Identifier doi:10.1214/aoms/1177728196 Mathematical Reviews number (MathSciNet) MR79870 Zentralblatt MATH identifier 0074.13702 JSTOR links.jstor.org Citation
Seal, K. C. On Minimum Variance Among Certain Linear Functions of Order Statistics. Ann. Math. Statist. 27 (1956), no. 3, 854--855. doi:10.1214/aoms/1177728196. https://projecteuclid.org/euclid.aoms/1177728196
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Elementary Matrices
Definition: A square $n \times n$ matrix is an Elementary Matrix $E$ if it can be obtained by performing exactly one elementary row operation on the identity matrix $I_n$.
The following three matrices are all considered elementary matrices:(1)
$E_1$ is obtained by taking $I_3$ and multiplying row 2 by 4 ($4R_2 \to R_2$). $E_2$ is obtained by interchanging the two rows of $I_2$ ($R_1 \leftrightarrow R_2$). $E_3$ is obtained by taking the second row of $I_3$ and adding 2 times the first row ($R_2 + 2R_1 \to R_2$).
In all of these cases, $E_1$, $E_2$, and $E_3$ were all obtained from taking an identity matrix and performing one elementary row operation on it to produce $E$.
Theorem 1: Suppose that $E$ is an elementary matrix resulting from a single elementary row operation on $I$. For any matrix $A$ where the product $EA$ is defined, $EA$ will be the matrix that results when applying that same elementary row operation on $A$.
For example, consider the elementary matrix $E = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$ that results from taking $I_2$ and interchanging rows $R_1 \leftrightarrow R_2$. Let's also consider the following matrix $A = \begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$.
The resulting product $EA = \begin{bmatrix} 3 & 4\\ 1 & 2 \end{bmatrix}$, which is the matrix obtained by taking $A$ and interchanging the corresponding rows $R_1 \leftrightarrow R_2$ which is what we should expect from theorem 1.
Inverse Row Operations
Suppose that we create an elementary matrix $E$ by taking $I$ and performing a single row operation on it. If we then take $E$ and
reverse that operation, we will obtain $I$ once again. If we create $E$ by multiplying some row by a constant $k ≠ 0$ ($kR_a \to R_a$), then the inverse operation would be to take $E$ and multiply it by $\frac{1}{k}$ ($\frac{1}{k} R_a \to R_a$). If we create $E$ by interchanging two rows ($R_a \leftrightarrow R_b$), then the inverse operation would be to take $E$ and interchange those same rows again ($R_b \leftrightarrow R_a$). If we create $E$ by adding a multiple of one row to another ($R_a + kR_b \to R_a$), then the inverse operation would be to take $E$ and subtract the multiple of that row off ($R_a - kR_b \to R_a$).
For example, consider the elementary matrix $E = \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}$ that resulted from taking the row operation $3R_2 \to R_2$ on $I$. If we then perform the operation $\frac{1}{3}R_2 \to R_2$ on $E$, we obtain $I$ again.
Elementary Matrix Inverses
Theorem 2: Every elementary matrix $E$ is invertible. Proof:To prove theorem 2, we will look at three cases, each pertaining to the elementary row operations. Case 1:Suppose that $E_{n \times n}$ is obtained by multiplying a row by a constant $k \in \mathbb{R}$, $k \neq 0$. Without loss of generality, suppose we multiply the second row of $I_n$ by $k$. Then $E = \begin{bmatrix} 1 & 0 & \cdots & 0\\ 0 & k & \cdots & 0\\ \vdots & \vdots & \ddots & 0\\ 0 & 0 & \cdots & 1 \end{bmatrix}$. Let $E^{-1} = \begin{bmatrix} 1 & 0 & \cdots & 0\\ 0 & \frac{1}{k} & \cdots & 0\\ \vdots & \vdots & \ddots & 0\\ 0 & 0 & \cdots & 1 \end{bmatrix}$. Then $EE^{-1} = I_{n}$. Case 2and Case 3are left to the reader to show, that is the cases when the elementary row operation performed on $I$ is adding a multiple of one row to another, or interchanging two rows. $\blacksquare$
Theorem 3: If $E$ is an elementary matrix, then $E^{-1}$ is an elementary matrix resulting from performing the inverse row operation from $E$ onto $I$.
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Is there something similar to the Magnus effect for electromagnetism? Would a spinning particle generate a centripetal force moving through something like a circular accelerator? And how would you calculate this?
Electron spin contributes to the centrifugal force on an electron.
Compare the wave function of the hydrogen atom for the Klein-Gordon and the Dirac equation. The KG energy values are
$E_{nl} = \frac{m}{\sqrt{1+[\alpha^2/(n-\delta_l)^2]}}$
with $\delta_l=l+\frac{1}{2}$. The Dirac eigenvalues are obtained by replacing $l\rightarrow j\pm \frac{1}{2}$. See QFT, Itzykson&Zuber, Ch. 2.3.
The reason is that the radial equation in these cases contains the centrifugal potential $\frac{L^2-\alpha^2}{r^2}$ (KG) and $\frac{J^2\alpha^2}{r^2}$ (Dirac). This clearly proves that electron spin like orbital angular momentum exhibits a centrifugal force.
There is a relativistic effect that is similar. If you have an object with a magnetic dipole moment (say a magnet if you want) that is moving straight through the air and encounters an inhomogeneous electric field perpendicular to its motion, then the object will bend its trajectory. Note that is an ELECTRIC field modifying the trajectory of a particle with a MAGNETIC property. The key of the effect is the relative motion. Due to special relativity a moving magnetic moment becomes sort of a electric dipole moment. The magnetic dipole moment may come from neutral electric currents inside the material or intrinsic magnetism.
This effect is a key mechanism to understand the fine structure of hydrogen and spin Hall effect in solids. In such cases, the particle with magnetic moment is an electron with an intrinsic angular momentum AKA spin, some people like to point out an analogy with the Magnus effect.
The Magnus force in electromagnetism is given by $(4\pi c/\mu_0\omega)\,\vec{B}\times\vec{A}$.
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I am attempting to work through "Shahverdiev, Sivaprakasam, and Shore (2002) Lag synchronization in time-delayed systems", but I'm missing something basic up front.
The problem is to take a unidirectionally coupled dynamical system,
$\dot{x} =-\alpha x + (m_1+m_3) \sin x_{\tau_1}$
$\dot{y} =-\alpha y + m_2 \sin y_{\tau_1} + m_3 \sin x_{\tau_2}$
where subscript $\tau$ denotes a time-delay, e.g. $x_{\tau_1} = x(t-\tau_1)$, and determine its stability manifold. The synchronization error is defined to be $\Delta = x_{\tau_2 - \tau_1} - y$, but we are interested in the error dynamics, $\dot{\Delta}$, which the authors list as
$\dot{\Delta}=-\alpha \Delta + m_2 \cos x_{\tau_1} \Delta_{\tau_1}$
My question is about this last step. To differentiate $\Delta$, it seems to me that one could use the expressions for $\dot{x}$ and $\dot{y}$ directly, i.e.
$\dot{\Delta} = \dot{x}(t-\tau_2+\tau_1) - \dot{y}(t)$.
Doing that, however, leaves me with
$\dot{\Delta} = -\alpha \Delta + m_2 \left( \sin x_{\tau_2} - \sin y_{\tau_1} \right)$
I feel as though I'm missing a simple step here. Could anyone point me in the right direction?
Here's a second example, this time from "Senthilkumar, Kurths, and Lakshmanan (2009) Stability of synchronization in coupled time-delay systems using Krasovskii-Lyapunov theory":
The system:
$\dot{x} = -a x(t) + b f(x(t-\tau))$
$\dot{y} = -a y(t) + b f(y(t-\tau)) + K(t)(x(t)-y(t))$
The synchronization error (for small values, the authors state):
$\dot{\Delta} = -(a + K(t))\Delta + b f^\prime(y(t-\tau))\Delta(t-\tau)$
Again, I am unsure of the source of the $f^\prime$ term and why that term isn't simply $b \left( f(x(t-\tau)) - f(y(t-\tau)) \right)$. In both examples, it seems a derivative is being taken, but I don't see why that would be.
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Table of Contents
Indeterminate Forms
We are now going to look a little more thoroughly into evaluating limits before moving forward with calculus. We will soon learn an important limit rule known as L'Hospital's Rule, but before we do, we must first look at some indeterminate forms that may arise in determining the limits of a function.
Indeterminate Form of Type 0/0
Definition: If we have a limit $L = \lim_{x \to a} \frac{f(x)}{g(x)}$ such that as $x \to a$, $f(x) \to 0$ and $g(x) \to 0$, then we say $L$ is of Indeterminate Form of Type $\frac{0}{0}$..
For forms like this, sometimes we are able to cancel factors which are the same. For example:(1)
However, sometimes we cannot cancel our terms to make evaluating the limit easier:(2)
Indeterminate Form of Type ∞/∞
Definition: If we have a limit $L = \lim_{x \to a} \frac{f(x)}{g(x)}$ such that as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$, then we say $L$ is of Indeterminate Form of Type $\frac{\infty}{\infty}$..
For example:(3)
There are some cases such that when $x \to \pm \infty$ then $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$, then we can cancel out problematic spots by dividing each term by the highest power of $x$ in the denominator. For example:(4)
Of course, sometimes we cannot use this division trick to determine the limit of the function.
Indeterminate Form of Type 0 * ∞
Definition: If we have a limit $L = \lim_{x \to a} [f(x)g(x)]$ such that as $x \to a$, $f(x) \to 0$ and $g(x) \to \pm \infty$, then we say $L$ is of Indeterminate Form of Type $0 \cdot \infty$.
Recall that for two functions $f$ and $g$, that the following property holds:(5)
Which is equivalent to the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$. For example:(6)
Once again, these algebraic tricks won't always work. Recognize that this form conversion is very useful nevertheless.
Indeterminate Form of Type ∞ - ∞.
Definition: If we have a limit $L = \lim_{x \to a} [f(x) - g(x)]$ such that as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$, then we say $L$ is of Indeterminate Form of Type $\infty - \infty$.
To solve these limits, we can rewrite the limit over a common denominator and solve it as an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$. For example:(7)
Which is now in indeterminate form $\frac{0}{0}$.
Indeterminate Form of Type 0 0
Definition: If we have a limit $L = \lim_{x \to a} f(x)^{g(x)}$ such that as $x \to a$, $f(x) \to 0$ and $g(x) \to 0$, then we say $L$ is of Indeterminate Form of Type $0^0$. Indeterminate Form of Type ∞ 0
Definition: If we have a limit $L = \lim_{x \to a} f(x)^{g(x)}$ such that as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to 0$, then we say $L$ is of Indeterminate Form of Type $\infty ^0$. Indeterminate Form of Type 1 ∞
Definition: If we have a limit $L = \lim_{x \to a} f(x)^{g(x)}$ such that as $x \to a$, $f(x) \to 1$ and $g(x) \to \pm \infty$, then we say $L$ is of **Indeterminate Form of Type $1^{\infty}$. Importance of Indeterminate Forms
As you can see, there are clearly many indeterminate forms that we have just defined. When we take a limit of an indeterminate form and cannot apply any algebraic tricks, we must then use a rule known as L'Hospital's Rule in order to determine limits that we wouldn't have been able to calculate otherwise.
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High-dimensional Sperner
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High-dimensional Sperner
Kalai.29: There is an analogous for Sperner but with high dimensional combinatorial spaces instead of "lines" but I do not remember the details (Kleitman(?) Katona(?) those are ususal suspects.)
Kalai.29: There is an analogous for Sperner but with high dimensional combinatorial spaces instead of "lines" but I do not remember the details (Kleitman(?) Katona(?) those are ususal suspects.)
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Fourier approach
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Fourier approach
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DHJ for dense subsets of a random set
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DHJ for dense subsets of a random set
Tao.18: A sufficiently good Varnavides type theorem for DHJ may have a separate application from the one in this project, namely to obtain a “relative” DHJ for dense subsets of a sufficiently pseudorandom subset of {}[3]^n, much as I did with Ben Green for the primes (and which now has a significantly simpler proof by Gowers and by Reingold-Trevisan-Tulsiani-Vadhan). There are other obstacles though to that task (e.g. understanding the analogue of “dual functions” for Hales-Jewett), and so this is probably a bit off-topic.
Tao.18: A sufficiently good Varnavides type theorem for DHJ may have a separate application from the one in this project, namely to obtain a “relative” DHJ for dense subsets of a sufficiently pseudorandom subset of {}[3]^n, much as I did with Ben Green for the primes (and which now has a significantly simpler proof by Gowers and by Reingold-Trevisan-Tulsiani-Vadhan). There are other obstacles though to that task (e.g. understanding the analogue of “dual functions” for Hales-Jewett), and so this is probably a bit off-topic.
Revision as of 20:56, 11 February 2009 Contents The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]x[/math] in it, e.g., [math]112x1xx3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. Density Hales-Jewett (DHJ) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers.
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (active) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (final call) (500-599) TBA (600-699) A reading seminar on density Hales-Jewett (active)
A spreadsheet containing the latest lower and upper bounds for [math]c_n[/math] can be found here.
Unsolved questions Gowers.462: Incidentally, it occurs to me that we as a collective are doing what I as an individual mathematician do all the time: have an idea that leads to an interesting avenue to explore, get diverted by some temporarily more exciting idea, and forget about the first one. I think we should probably go through the various threads and collect together all the unsolved questions we can find (even if they are vague ones like, “Can an approach of the following kind work?”) and write them up in a single post. If this were a more massive collaboration, then we could work on the various questions in parallel, and update the post if they got answered, or reformulated, or if new questions arose. IP-Szemeredi (a weaker problem than DHJ) Solymosi.2: In this note I will try to argue that we should consider a variant of the original problem first. If the removal technique doesn’t work here, then it won’t work in the more difficult setting. If it works, then we have a nice result! Consider the Cartesian product of an IP_d set. (An IP_d set is generated by d numbers by taking all the [math]2^d[/math] possible sums. So, if the n numbers are independent then the size of the IP_d set is [math]2^d[/math]. In the following statements we will suppose that our IP_d sets have size [math]2^n[/math].)
Prove that for any [math]c\gt0[/math] there is a [math]d[/math], such that any [math]c[/math]-dense subset of the Cartesian product of an IP_d set (it is a two dimensional pointset) has a corner.
The statement is true. One can even prove that the dense subset of a Cartesian product contains a square, by using the density HJ for [math]k=4[/math]. (I will sketch the simple proof later) What is promising here is that one can build a not-very-large tripartite graph where we can try to prove a removal lemma. The vertex sets are the vertical, horizontal, and slope -1 lines, having intersection with the Cartesian product. Two vertices are connected by an edge if the corresponding lines meet in a point of our [math]c[/math]-dense subset. Every point defines a triangle, and if you can find another, non-degenerate, triangle then we are done. This graph is still sparse, but maybe it is well-structured for a removal lemma.
Finally, let me prove that there is square if [math]d[/math] is large enough compare to [math]c[/math]. Every point of the Cartesian product has two coordinates, a 0,1 sequence of length [math]d[/math]. It has a one to one mapping to [math][4]^d[/math]; Given a point [math]( (x_1,…,x_d),(y_1,…,y_d) )[/math] where [math]x_i,y_j[/math] are 0 or 1, it maps to [math](z_1,…,z_d)[/math], where [math]z_i=0[/math] if [math]x_i=y_i=0[/math], [math]z_i=1[/math] if [math]x_i=1[/math] and [math]y_i=0, z_i=2[/math] if [math]x_i=0[/math] and [math]y_i=1[/math], and finally [math]z_i=3[/math] if [math]x_i=y_i=1[/math]. Any combinatorial line in [math][4]^d[/math] defines a square in the Cartesian product, so the density HJ implies the statement.
Gowers.7: With reference to Jozsef’s comment, if we suppose that the d numbers used to generate the set are indeed independent, then it’s natural to label a typical point of the Cartesian product as (\epsilon,\eta), where each of \epsilon and \eta is a 01-sequence of length d. Then a corner is a triple of the form (\epsilon,\eta), (\epsilon,\eta+\delta), (\epsilon+\delta,\eta), where \delta is a \{-1,0,1\}-valued sequence of length d with the property that both \epsilon+\delta and \eta+\delta are 01-sequences. So the question is whether corners exist in every dense subset of the original Cartesian product.
This is simpler than the density Hales-Jewett problem in at least one respect: it involves 01-sequences rather than 012-sequences. But that simplicity may be slightly misleading because we are looking for corners in the Cartesian product. A possible disadvantage is that in this formulation we lose the symmetry of the corners: the horizontal and vertical lines will intersect this set in a different way from how the lines of slope -1 do.
I feel that this is a promising avenue to explore, but I would also like a little more justification of the suggestion that this variant is likely to be simpler.
Gowers.22: A slight variant of the problem you propose is this. Let’s take as our ground set the set of all pairs (U,V) of subsets of \null [n], and let’s take as our definition of a corner a triple of the form (U,V), (U\cup D,V), (U,V\cup D), where both the unions must be disjoint unions. This is asking for more than you asked for because I insist that the difference D is positive, so to speak. It seems to be a nice combination of Sperner’s theorem and the usual corners result. But perhaps it would be more sensible not to insist on that positivity and instead ask for a triple of the form (U,V), ((U\cup D)\setminus C,V), (U, (V\cup D)\setminus C, where D is disjoint from both U and V and C is contained in both U and V. That is your original problem I think.
I think I now understand better why your problem could be a good toy problem to look at first. Let’s quickly work out what triangle-removal statement would be needed to solve it. (You’ve already done that, so I just want to reformulate it in set-theoretic language, which I find easier to understand.) We let all of X, Y and Z equal the power set of \null [n]. We join U\in X to V\in Y if (U,V)\in A.
Ah, I see now that there’s a problem with what I’m suggesting, which is that in the normal corners problem we say that (x,y+d) and (x+d,y) lie in a line because both points have the same coordinate sum. When should we say that (U,V\cup D) and (U\cup D,V) lie in a line? It looks to me as though we have to treat the sets as 01-sequences and take the sum again. So it’s not really a set-theoretic reformulation after all.
O'Donnell.35: Just to confirm I have the question right…
There is a dense subset A of {0,1}^n x {0,1}^n. Is it true that it must contain three nonidentical strings (x,x’), (y,y’), (z,z’) such that for each i = 1…n, the 6 bits
[ x_i x'_i ] [ y_i y'_i ] [ z_i z'_i ]
are equal to one of the following:
[ 0 0 ] [ 0 0 ] [ 0, 1 ] [ 1 0 ] [ 1 1 ] [ 1 1 ] [ 0 0 ], [ 0 1 ], [ 0, 1 ], [ 1 0 ], [ 1 0 ], [ 1 1 ], [ 0 0 ] [ 1 0 ] [ 0, 1 ] [ 1 0 ] [ 0 1 ] [ 1 1 ]
?
McCutcheon.469: IP Roth:
Just to be clear on the formulation I had in mind (with apologies for the unprocessed code): for every $\delta>0$ there is an $n$ such that any $E\subset [n]^{[n]}\times [n]^{[n]}$ having relative density at least $\delta$ contains a corner of the form $\{a, a+(\sum_{i\in \alpha} e_i ,0),a+(0, \sum_{i\in \alpha} e_i)\}$. Here $(e_i)$ is the coordinate basis for $[n]^{[n]}$, i.e. $e_i(j)=\delta_{ij}$.
Presumably, this should be (perhaps much) simpler than DHJ, k=3.
High-dimensional Sperner
Kalai.29: There is an analogous for Sperner but with high dimensional combinatorial spaces instead of "lines" but I do not remember the details (Kleitman(?) Katona(?) those are ususal suspects.)
Fourier approach
Kalai.29: A sort of generic attack one can try with Sperner is to look at f=1_A and express using the Fourier expansion of f the expression \int f(x)f(y)1_{x<y} where x<y is the partial order (=containment) for 0-1 vectors. Then one may hope that if f does not have a large Fourier coefficient then the expression above is similar to what we get when A is random and otherwise we can raise the density for subspaces. (OK, you can try it directly for the k=3 density HJ problem too but Sperner would be easier;) This is not unrealeted to the regularity philosophy.
Gowers.31: Gil, a quick remark about Fourier expansions and the k=3 case. I want to explain why I got stuck several years ago when I was trying to develop some kind of Fourier approach. Maybe with your deep knowledge of this kind of thing you can get me unstuck again.
The problem was that the natural Fourier basis in \null [3]^n was the basis you get by thinking of \null [3]^n as the group \mathbb{Z}_3^n. And if that’s what you do, then there appear to be examples that do not behave quasirandomly, but which do not have large Fourier coefficients either. For example, suppose that n is a multiple of 7, and you look at the set A of all sequences where the numbers of 1s, 2s and 3s are all multiples of 7. If two such sequences lie in a combinatorial line, then the set of variable coordinates for that line must have cardinality that’s a multiple of 7, from which it follows that the third point automatically lies in the line. So this set A has too many combinatorial lines. But I’m fairly sure — perhaps you can confirm this — that A has no large Fourier coefficient.
You can use this idea to produce lots more examples. Obviously you can replace 7 by some other small number. But you can also pick some arbitrary subset W of \null[n] and just ask that the numbers of 0s, 1s and 2s inside W are multiples of 7. DHJ for dense subsets of a random set
Tao.18: A sufficiently good Varnavides type theorem for DHJ may have a separate application from the one in this project, namely to obtain a “relative” DHJ for dense subsets of a sufficiently pseudorandom subset of {}[3]^n, much as I did with Ben Green for the primes (and which now has a significantly simpler proof by Gowers and by Reingold-Trevisan-Tulsiani-Vadhan). There are other obstacles though to that task (e.g. understanding the analogue of “dual functions” for Hales-Jewett), and so this is probably a bit off-topic.
Bibliography H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem for k=3“, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–241. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem“, J. Anal. Math. 57 (1991), 64–119.
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Difference between revisions of "Main Page"
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== The Problem ==
== The Problem ==
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Let <math>[3]^n</math> be the set of all length <math>n</math> strings over the alphabet <math>1, 2, 3</math>. A ''combinatorial line'' is a set of three points in <math>[3]^n</math>, formed by taking a string with one or more wildcards <math>x</math> in it, e.g., <math>112x1xx3\ldots</math>, and replacing those wildcards by <math>1, 2</math> and <math>3</math>, respectively. In the example given, the resulting combinatorial line is: <math>\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}</math>. A subset of <math>[3]^n</math> is said to be ''line-free'' if it contains no lines. Let <math>c_n</math> be the size of the largest line-free subset of <math>[3]^n</math>.
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Let <math>[3]^n</math> be the set of all length <math>n</math> strings over the alphabet <math>1, 2, 3</math>. A ''combinatorial line'' is a set of three points in <math>[3]^n</math>, formed by taking a string with one or more wildcards <math>x</math> in it, e.g., <math>112x1xx3\ldots</math>, and replacing those wildcards by <math>1, 2</math> and <math>3</math>, respectively. In the example given, the resulting combinatorial line is: <math>\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}</math>. A subset of <math>[3]^n</math> is said to be ''line-free'' if it contains no lines. Let <math>c_n</math> be the size of the largest line-free subset of <math>[3]^n</math>.
'''<math>k=3</math> Density Hales-Jewett (DHJ(3)) theorem:''' <math>\lim_{n \rightarrow \infty} c_n/3^n = 0</math>
'''<math>k=3</math> Density Hales-Jewett (DHJ(3)) theorem:''' <math>\lim_{n \rightarrow \infty} c_n/3^n = 0</math>
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The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular [http://gowers.wordpress.com/2009/02/01/a-combinatorial-approach-to-density-hales-jewett/ combinatorial approach to DHJ, suggested by Tim Gowers.] Some background to this project can be [http://gowers.wordpress.com/2009/01/30/background-to-a-polymath-project/ found here], and general discussion on massively collaborative "polymath" projects can be [http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/ found here].
+
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular [http://gowers.wordpress.com/2009/02/01/a-combinatorial-approach-to-density-hales-jewett/ combinatorial approach to DHJ, suggested by Tim Gowers.] Some background to this project can be [http://gowers.wordpress.com/2009/01/30/background-to-a-polymath-project/ found here], and general discussion on massively collaborative "polymath" projects can be [http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/ found here].
== Threads ==
== Threads ==
Revision as of 06:52, 13 February 2009 The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]x[/math] in it, e.g., [math]112x1xx3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. [math]k=3[/math] Density Hales-Jewett (DHJ(3)) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers. Some background to this project can be found here, and general discussion on massively collaborative "polymath" projects can be found here. A cheatsheet for editing the wiki may be found here.
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (active) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (inactive) (500-599) Possible proof strategies (active) (600-699) A reading seminar on density Hales-Jewett (active)
We are also collecting bounds for Fujimura's problem.
Here are some unsolved problems arising from the above threads.
Bibliography M. Elkin, "An Improved Construction of Progression-Free Sets ", preprint. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem for k=3“, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–241. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem“, J. Anal. Math. 57 (1991), 64–119. B. Green, J. Wolf, "A note on Elkin's improvement of Behrend's construction", preprint. K. O'Bryant, "Sets of integers that do not contain long arithmetic progressions", preprint. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished.
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In
mathematics, is a Steinhaus– Moser notation notation for expressing certain extremely large numbers. It is an extension of Steinhaus's polygon notation. [1] Definitions [ edit ] a number in a n triangle means . n n a number in a n square is equivalent to "the number inside n triangles, which are all nested." n a number in a n pentagon is equivalent with "the number inside n squares, which are all nested." n
etc.:
written in an ( n )-sided polygon is equivalent with "the number m + 1 inside n nested n -sided polygons". In a series of nested polygons, they are m associated inward. The number inside two triangles is equivalent to n inside one triangle, which is equivalent to n n raised to the power of n n . n n
Steinhaus defined only the triangle, the square, and the
circle , which is equivalent to the pentagon defined above. Special values [ edit ]
Steinhaus defined:
mega is the number equivalent to 2 in a circle: megiston is the number equivalent to 10 in a circle: ⑩ Moser's number is the number represented by "2 in a megagon". Megagon is here the name of a polygon with "mega" sides (not to be confused with the polygon with one million sides).
Alternative notations:
use the functions square(x) and triangle(x) let M( be the number represented by the number n, m, p) in n nested m -sided polygons; then the rules are: p M ( n , 1 , 3 ) = n n {\displaystyle M(n,1,3)=n^{n}} M ( n , 1 , p + 1 ) = M ( n , n , p ) {\displaystyle M(n,1,p+1)=M(n,n,p)} M ( n , m + 1 , p ) = M ( M ( n , 1 , p ) , m , p ) {\displaystyle M(n,m+1,p)=M(M(n,1,p),m,p)} and mega = M ( 2 , 1 , 5 ) {\displaystyle M(2,1,5)} megiston = M ( 10 , 1 , 5 ) {\displaystyle M(10,1,5)} moser = M ( 2 , 1 , M ( 2 , 1 , 5 ) ) {\displaystyle M(2,1,M(2,1,5))}
A mega, ②, is already a very large number, since ② =square(square(2)) = square(triangle(triangle(2))) =square(triangle(2
2)) = square(triangle(4)) =square(4 4) =square(256) =triangle(triangle(triangle(...triangle(256)...))) [256 triangles] =triangle(triangle(triangle(...triangle(256 256)...))) [255 triangles] ~triangle(triangle(triangle(...triangle(3.2 × 10 616)...))) [254 triangles] =...
Using the other notation:
mega = M(2,1,5) = M(256,256,3)
With the function
we have mega = f ( x ) = x x {\displaystyle f(x)=x^{x}} where the superscript denotes a f 256 ( 256 ) = f 258 ( 2 ) {\displaystyle f^{256}(256)=f^{258}(2)} functional power, not a numerical power.
We have (note the convention that powers are evaluated from right to left):
M(256,2,3) = ( 256 256 ) 256 256 = 256 256 257 {\displaystyle (256^{\,\!256})^{256^{256}}=256^{256^{257}}} M(256,3,3) = ≈ ( 256 256 257 ) 256 256 257 = 256 256 257 × 256 256 257 = 256 256 257 + 256 257 {\displaystyle (256^{\,\!256^{257}})^{256^{256^{257}}}=256^{256^{257}\times 256^{256^{257}}}=256^{256^{257+256^{257}}}} 256 256 256 257 {\displaystyle 256^{\,\!256^{256^{257}}}}
Similarly:
M(256,4,3) ≈ 256 256 256 256 257 {\displaystyle {\,\!256^{256^{256^{256^{257}}}}}} M(256,5,3) ≈ 256 256 256 256 256 257 {\displaystyle {\,\!256^{256^{256^{256^{256^{257}}}}}}}
etc.
Thus:
mega = , where M ( 256 , 256 , 3 ) ≈ ( 256 ↑ ) 256 257 {\displaystyle M(256,256,3)\approx (256\uparrow )^{256}257} denotes a functional power of the function ( 256 ↑ ) 256 {\displaystyle (256\uparrow )^{256}} . f ( n ) = 256 n {\displaystyle f(n)=256^{n}}
Rounding more crudely (replacing the 257 at the end by 256), we get mega ≈
, using 256 ↑↑ 257 {\displaystyle 256\uparrow \uparrow 257} Knuth's up-arrow notation.
After the first few steps the value of
is each time approximately equal to n n {\displaystyle n^{n}} . In fact, it is even approximately equal to 256 n {\displaystyle 256^{n}} (see also 10 n {\displaystyle 10^{n}} approximate arithmetic for very large numbers). Using base 10 powers we get: M ( 256 , 1 , 3 ) ≈ 3.23 × 10 616 {\displaystyle M(256,1,3)\approx 3.23\times 10^{616}} ( M ( 256 , 2 , 3 ) ≈ 10 1.99 × 10 619 {\displaystyle M(256,2,3)\approx 10^{\,\!1.99\times 10^{619}}} is added to the 616) log 10 616 {\displaystyle \log _{10}616} ( M ( 256 , 3 , 3 ) ≈ 10 10 1.99 × 10 619 {\displaystyle M(256,3,3)\approx 10^{\,\!10^{1.99\times 10^{619}}}} is added to the 619 {\displaystyle 619} , which is negligible; therefore just a 10 is added at the bottom) 1.99 × 10 619 {\displaystyle 1.99\times 10^{619}} M ( 256 , 4 , 3 ) ≈ 10 10 10 1.99 × 10 619 {\displaystyle M(256,4,3)\approx 10^{\,\!10^{10^{1.99\times 10^{619}}}}}
...
mega = , where M ( 256 , 256 , 3 ) ≈ ( 10 ↑ ) 255 1.99 × 10 619 {\displaystyle M(256,256,3)\approx (10\uparrow )^{255}1.99\times 10^{619}} denotes a functional power of the function ( 10 ↑ ) 255 {\displaystyle (10\uparrow )^{255}} . Hence f ( n ) = 10 n {\displaystyle f(n)=10^{n}} 10 ↑↑ 257 < mega < 10 ↑↑ 258 {\displaystyle 10\uparrow \uparrow 257<{\text{mega}}<10\uparrow \uparrow 258} Moser's number [ edit ]
It has been proven that in
Conway chained arrow notation, m o s e r < 3 → 3 → 4 → 2 , {\displaystyle \mathrm {moser} <3\rightarrow 3\rightarrow 4\rightarrow 2,}
and, in
Knuth's up-arrow notation, m o s e r < f 3 ( 4 ) = f ( f ( f ( 4 ) ) ) , where f ( n ) = 3 ↑ n 3. {\displaystyle \mathrm {moser} <f^{3}(4)=f(f(f(4))),{\text{ where }}f(n)=3\uparrow ^{n}3.}
Therefore, Moser's number, although incomprehensibly large, is vanishingly small compared to
Graham's number: m o s e r ≪ 3 → 3 → 64 → 2 < f 64 ( 4 ) = Graham's number . {\displaystyle \mathrm {moser} \ll 3\rightarrow 3\rightarrow 64\rightarrow 2<f^{64}(4)={\text{Graham's number}}.} See also [ edit ] References [ edit ] ^ Hugo Steinhaus, Mathematical Snapshots, Oxford University Press 1969 3, ISBN 0195032675, pp. 28-29 External links [ edit ]
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Let G be a nontrivial finite group. Does there exist an irreducible representation of G of dimension greater than or equal to the cardinality of G?
[Edited for clarity. -- PLC]
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Let G be a nontrivial finite group. Does there exist an irreducible representation of G of dimension greater than or equal to the cardinality of G?
[Edited for clarity. -- PLC]
EDIT: Part 4 added. EDIT2: Second proof of Part 4 added. 1. The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $\leq \left|G\right|$. This is actually obvious: Take any nonzero vector $v\in V$; then, $k\left[G\right]v$ is a nontrivial subrepresentation of $V$ of dimension $\leq\dim\left(k\left[G\right]\right)=\left|G\right|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $\dim V\leq\left|G\right|$. 2. Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $\leq\left|G\right|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step:
If $\dim V=\left|G\right|$, then the map $k\left[G\right]\to V,\ g\mapsto gv$ must be bijective (in fact, it is surjective, since $k\left[G\right]v=V$, and it therefore must be bijective since $\dim\left(k\left[G\right]\right)=\left|G\right|=\dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $V\cong k\left[G\right]$. But $k\left[G\right]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation).
3. Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<\sqrt{\left|G\right|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $\left|G\right|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $\dim V\leq \left|G\right|-1$ can be sharp (take cyclic groups). 4. There is a way to improve 2. so that it comes a bit closer to 3.: Theorem 1. If $V_1$, $V_2$, ..., $V_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $\dim V_1+\dim V_2+...+\dim V_m\leq\dim A$.
(Of course, if $A$ is the group algebra of some finite group $G$, then $\dim A=\left|G\right|$, and we get
2. as a consequence.) First proof of Theorem 1. At first, for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, the (left) representation $V_i^{\ast}$ of the algebra $A^{\mathrm{op}}$ (this representation is defined by $a\cdot f=\left(v\mapsto f\left(av\right)\right)$ for any $f\in V_i^{\ast}$ and $a\in A$) is irreducible (since $V_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{\mathrm{op}}$ (since we can choose some nonzero $u\in V_i^{\ast}$, and then the map $A^{\mathrm{op}}\to V_i^{\ast}$ given by $a\mapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V_i^{\ast}$ and therefore equal to $V_i^{\ast}$ due to the irreducibility of $V_i^{\ast}$). Hence, by duality, $V_i$ is isomorphic to a subrepresentation of the (left) representation $A^{\mathrm{op}\ast}=A^{\ast}$ of $A$. Hence, from now on, let's assume that $V_i$ actually is a subrepresentation of $A^{\ast}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$.
Now, let us prove that the vector subspaces $V_1$, $V_2$, ..., $V_m$ of $A^{\ast}$ are linearly disjoint, i. e., that the sum $V_1+V_2+...+V_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V_1+V_2+...+V_{m-1}$ is already a direct sum. It remains to prove that $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=0$. In fact, assume the contrary. Then, $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=V_m$ (since $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)$ is a nonzero subrepresentation of $V_m$, and $V_m$ is irreducible). Thus, $V_m\subseteq V_1+V_2+...+V_{m-1}$. Consequently, $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ (because the sum $V_1+V_2+...+V_{m-1}$ is a direct sum, according to our induction assumption).
Now, according to Theorem 2.2 and Remark 2.3 of Etingof's "Introduction to representation theory", any subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be a direct sum of the form $r_1V_1\oplus r_2V_2\oplus ...\oplus r_{m-1}V_{m-1}$ for some nonnegative integers $r_1$, $r_2$, ..., $r_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. Since we know that $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$, we conclude that $V_m$ is isomorphic to one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. This contradicts the non-isomorphy of the representations $V_1$, $V_2$, ..., $V_m$. Thus, we have proven that the sum $V_1+V_2+...+V_m$ is actually a direct sum. Consequently, $\dim V_1+\dim V_2+...+\dim V_m=\dim\left(V_1+V_2+...+V_m\right)\leq \dim A^{\ast}=\dim A$, and Theorem 1 is proven.
Second proof of Theorem 1. I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in Crawley-Boevey's "Lectures on representation theory and invariant theory":
Let $0=A_0\subseteq A_1\subseteq A_2\subseteq ...\subseteq A_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $i\in \left\lbrace 1,2,...,k\right\rbrace$, the representation $A_i/A_{i-1}$ of $A$ is irreducible.
Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$).
In fact, let $I$ be the smallest element $i\in \left\lbrace 1,2,...,k\right\rbrace$ satisfying $A_iT\neq 0$ (such elements $i$ exist, because $A_kT=AT=T\neq 0$). Then, $A_IT\neq 0$, but $A_{I-1}T=0$. Now, choose some vector $t\in T$ such that $A_It\neq 0$ (such a vector $t$ exists, because $A_IT\neq 0$), and consider the map $f:A_I\to T$ defined by $f\left(a\right)=at$ for every $a\in A_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A_{I-1}$ to $0$ (because $f\left(A_{I-1}\right)=A_{I-1}t\subseteq A_{I-1}T=0$), it gives rise to a map $g:A_I/A_{I-1}\to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A_I/A_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A_I/A_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $g\left(A_I/A_{I-1}\right)=f\left(A_I\right)=A_It\neq 0$), so that $g$ must be an isomorphism, i. e., we have $T\cong A_I/A_{I-1}$.
So we have just proven that
(1) For every irreducible representation $T$ of $A$, there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$).
Denote this $I$ by $I_T$ in order to make it clear that it depends on $T$. So we have $T\cong A_{I_T}/A_{I_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V_i$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, we see that $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$. Hence, the elements $I_{V_1}$, $I_{V_2}$, ..., $I_{V_m}$ of the set $\left\lbrace 1,2,...,k\right\rbrace$ are pairwise distinct (because $I_{V_i}=I_{V_j}$ would yield $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}=A_{I_{V_j}}/A_{I_{V_j}-1}\cong V_j$, but the representations $V_1$, $V_2$, ..., $V_m$ are pairwise nonisomorphic), and thus
$\sum\limits_{i=1}^{m}\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\sum\limits_{\substack{j\in\left\lbrace 1,2,...,k\right\rbrace ;\ \\ \text{there exists }\\ i\in\left\lbrace 1,2,...,m\right\rbrace \\ \text{ such that }j=I_{V_i}}}\dim\left(A_j/A_{j-1}\right)$ $\leq \sum\limits_{j\in\left\lbrace 1,2,...,k\right\rbrace}\dim\left(A_j/A_{j-1}\right)$ (since $\dim\left(A_j/A_{j-1}\right)\geq 0$ for every $j$, so that adding more summands cannot decrease the sum) $=\sum\limits_{j=1}^{k}\dim\left(A_j/A_{j-1}\right)=\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$.
Since $\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\dim V_i$ for each $i$ (due to $A_{I_{V_i}}/A_{I_{V_i}-1}\cong V_i$) and $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)=\dim A$ (in fact, the sum $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$ is a telescopic sum and simplifies to $\dim A_k-\dim A_0=\dim A-\dim 0=\dim A-0=\dim A$), this inequality becomes $\sum\limits_{i=1}^{m}\dim V_i\leq\dim A$. This proves Theorem 1.
I just want to add that for the $\sqrt{|G|}$ bound, we don't need the field to be algebraically closed, we only require that it be a splitting field for the group (i.e., all irreducible representations over the algebraic closure are realized in the group). For a finite group of exponent
d, any field over which $x^d - 1$ splits completely into distinct linear factors is a splitting field for the group. (Such fields are called "sufficiently large fields" in some places). The converse doesn't hold -- there could be splitting fields that are not sufficiently large.
If the field is not a splitting field for the group, but has a degree
r extension that is a splitting field for the group, then the degrees of irreducible representations are bounded by $r\sqrt{|G|}$ because each irreducible representation over the field can splitinto at most $r$ irreducibles when we pass to the splitting field.
So, over the real numbers, the degree of any irreducible representation is at most $2\sqrt{|G|}$, which is smaller than $|G| - 1$ for groups of size greater than five.
On the other hand, over the rational numbers, the cyclic group of prime order has an irreducible representation of degree equal to the order minus one, so the bound is tight for the rationals.
The $\sqrt{|G|}$ bound is true [for algebraically closed fields -- PLC] in any characteristic. In fact if $R$ is the Jacobson radical of $k[G]$, then $k[G]/R\cong\prod_iM_{n_i}(k)$ where $i$ runs over the irreducible representations and $n_i$ is the degree of the corresponding representation. This gives $\sum_in_i^2=\dim k[G]/R\leq|G|$. In the modular case we always have that $R\neq0$ so that in particular we have strict inequality. Also for every irreducible representation in charateristic $p$ there is an irreducible representation in characteristic $0$ whose degree is $\ge$ than the degree of the characteristic $p$ representation: Choose a number field $K$ which is a splitting field for $G$ and let $R$ be its ring of integers and let further $P$ be a maximal ideal of $R$ dividing $p$. We may filter $K[G]$ by a Jordan-Hölder filtration $W'_i$ so that $W'_i/W' _{i-1}$ is irreducible. Put $W_i:=W'_i\cap R[G]$ so that in particular $W_i/W_{i-1}$ is $R$-torsion free. Hence reducing modulo $P$ we get a filtration $\overline{W}_i$ of $k[G]$. This filtration can be refined to a Jordan-Hölder filtration showing that every irreducible $k[G]$-module which appears in any Jordan-Hölder filtration of $k[G]$ must appear in any Jordan-Hölder filtration of some $\overline{W}_i/\overline{W} _{i-1}$ and thus its degree is $\leq \dim(\overline{W}_i/\overline{W} _{i-1})=\mathrm{rank}(W_i/W _{i-1})=\dim(W'_i/W' _{i-1})$. Hence the degree of any $k[G]$-representation is $\le$ the degree of some $K[G]$-representation. It is rare (but does happen) that $\overline{W}_i/\overline{W} _{i-1}$ is irreducible in the modular case.
In the modular case (
i.e., when the characteristic of the field divides the order of the group), it is more interesting to look for indecomposable representations (because there are very few irreps or, as they are known in this context, simple modules: the 'worst' case is that of a $p$-group, which has one simple module!), and those can be as big as you want, in general, as soon as the group is not of finite representation type. This follows from the first Brauer-Thrall conjecture, proved by Maurice Auslander et al.; see, for example, [Ringel, Claus Michael. On algorithms for solving vector space problems. I. Report on the Brauer-Thrall conjectures: Rojter's theorem and the theorem of Nazarova and Rojter. Representation theory, I (Proc. Workshop, Carleton Univ., Ottawa, Ont., 1979), pp. 104--136, Lecture Notes in Math., 831, Springer, Berlin, 1980. MR0607142] and the references therein.
The smallest non-trivial example is the Klein Klein four group $\mathbb Z_2\oplus\mathbb Z_2$ in characteristic two, which is of infinite tame representation type, so has indecomposable modules of arbitrary high dimension. They have been known for ages, in various forms; they are described nicely, e.g., in [Benson, D. J. Representations and cohomology. I. Basic representation theory of finite groups and associative algebras. Second edition. Cambridge Studies in Advanced Mathematics, 30. Cambridge University Press, Cambridge, 1998. xii+246 pp. MR1644252].
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I think this is a bit odd but I am juggling since hours with $\sin$, $\cos$, $\tan$ and other stuff to proof a formula, but I can't do it. Slowly I am thinking that this formula is wrong. Maybe there is some expert who could tell me if I am right. I have the following problem:
In the end I want to reach the form:
$$ L_{BC} = \frac{L_{AC}\cos{\alpha} - L_{AC'}}{\sin{\alpha}} $$
starting with the formula for similar triangles:
$$ \frac{L_{AC}}{\sin{\theta}} = \frac{L_{AC'}}{\sin{( \theta - \alpha )}} $$
When I combine these two formulas I come to the point that
$$ L_{BC} = L_{AC'} \frac{\cos\theta}{\sin(\theta - \alpha)} $$
Now I don't see any way to replace $ \theta $ so that I am only dependent on the known variables: $$ L_{AC} \hspace{1cm} L_{AC'} \hspace{1cm} \alpha $$
Also expanding the fractions with sin / cos brings me to an deadend. Am I not seeing an obvious connection in these triangles or is there really something wrong about the formula? To make the actual question more clear: I want to calculate $L_{BC}$ using only $\alpha$ , $L_{AC'}$ and $L_{AC}$! And yes, we have $L_{AB}=L_{AB′}$! Thanks!
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My favourite one: $[0, 1]$ is compact, i.e. every open cover of $[0, 1]$ has a finite subcover.
Proof: Suppose for a contradiction that there is an open cover $\mathcal{U}$ which does not admit any finite subcover. Thus, either $\left[ 0, \frac{1}{2} \right]$ or $\left[ \frac{1}{2}, 1 \right]$ cannot be covered with a finite number of sets from $\mathcal{U}$ - name it $I_1$. Again, one of the two $I_1$'s subintervals of length $\frac{1}{4}$ can't be covered with a finite number of sets from $\mathcal{U}$. Continuing, we get a descending sequence of intervals $I_n$ of length $\frac{1}{2^n}$ each, every of which cannot be finitely covered.
By the Cantor Intersection Theorem,
$$\bigcap_{n=1}^{\infty} I_n = \{ x \}$$
for some $x \in [0, 1].$ But there is such $U \in \mathcal{U}$ that $x \in U$ and so $I_n \subseteq U$ for some sufficiently large $n$. That's a contradiction.
But given an arbitrary cover $\mathcal{U}$, I think finding a finite subcover may be a somewhat tedious task. :p
P.S. There actually comes a procedure from the proof above:
See if $[0, 1]$ itself is covered by one set from $\mathcal{U}$. If so, we're done. If not, execute step 1. for $\left[ 0, \frac{1}{2} \right]$ and $\left[ \frac{1}{2}, 1 \right]$ to get their finite subcovers, then unite them.
The proof guarantees you will eventually find a finite subcover (i.e. you'll never end up going downwards infinitely), but you cannot tell how long it will take. So it is not
as constructive as one would expect.
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@HarryGindi So the $n$-simplices of $N(D^{op})$ are $Hom_{sCat}(\mathfrak{C}[n],D^{op})$. Are you using the fact that the whole simplicial set is the mapping simplicial object between cosimplicial simplicial categories, and taking the constant cosimplicial simplicial category in the right coordinate?
I guess I'm just very confused about how you're saying anything about the entire simplicial set if you're not producing it, in one go, as the mapping space between two cosimplicial objects. But whatever, I dunno. I'm having a very bad day with this junk lol.
It just seems like this argument is all about the sets of n-simplices. Which is the trivial part.
lol no i mean, i'm following it by context actually
so for the record i really do think that the simplicial set you're getting can be written as coming from the simplicial enrichment on cosimplicial objects, where you take a constant cosimplicial simplicial category on one side
@user1732 haha thanks! we had no idea if that'd actually find its way to the internet...
@JonathanBeardsley any quillen equivalence determines an adjoint equivalence of quasicategories. (and any equivalence can be upgraded to an adjoint (equivalence)). i'm not sure what you mean by "Quillen equivalences induce equivalences after (co)fibrant replacement" though, i feel like that statement is mixing category-levels
@JonathanBeardsley if nothing else, this follows from the fact that \frakC is a left quillen equivalence so creates weak equivalences among cofibrant objects (and all objects are cofibrant, in particular quasicategories are). i guess also you need to know the fact (proved in HTT) that the three definitions of "hom-sset" introduced in chapter 1 are all weakly equivalent to the one you get via \frakC
@IlaRossi i would imagine that this is in goerss--jardine? ultimately, this is just coming from the fact that homotopy groups are defined to be maps in (from spheres), and you only are "supposed" to map into things that are fibrant -- which in this case means kan complexes
@JonathanBeardsley earlier than this, i'm pretty sure it was proved by dwyer--kan in one of their papers around '80 and '81
@HarryGindi i don't know if i would say that "most" relative categories are fibrant. it was proved by lennart meier that model categories are Barwick--Kan fibrant (iirc without any further adjectives necessary)
@JonathanBeardsley what?! i really liked that picture! i wonder why they removed it
@HarryGindi i don't know about general PDEs, but certainly D-modules are relevant in the homotopical world
@HarryGindi oh interesting, thomason-fibrancy of W is a necessary condition for BK-fibrancy of (R,W)?
i also find the thomason model structure mysterious. i set up a less mysterious (and pretty straightforward) analog for $\infty$-categories in the fappendix here: arxiv.org/pdf/1510.03525.pdf
as for the grothendieck construction computing hocolims, i think the more fundamental thing is that the grothendieck construction itself is a lax colimit. combining this with the fact that ($\infty$-)groupoid completion is a left adjoint, you immediately get that $|Gr(F)|$ is the colimit of $B \xrightarrow{F} Cat \xrightarrow{|-|} Spaces$
@JonathanBeardsley If you want to go that route, I guess you still have to prove that ^op_s and ^op_Delta both lie in the unique nonidentity component of Aut(N(Qcat)) and Aut(N(sCat)) whatever nerve you mean in this particular case (the B-K relative nerve has the advantage here bc sCat is not a simplicial model cat)
I think the direct proof has a lot of advantages here, since it gives a point-set on-the-nose isomorphism
Yeah, definitely, but I'd like to stay and work with Cisinski on the Ph.D if possible, but I'm trying to keep options open
not put all my eggs in one basket, as it were
I mean, I'm open to coming back to the US too, but I don't have any ideas for advisors here who are interested in higher straightening/higher Yoneda, which I am convinced is the big open problem for infinity, n-cats
Gaitsgory and Rozenblyum, I guess, but I think they're more interested in applications of those ideas vs actually getting a hold of them in full generality
@JonathanBeardsley Don't sweat it. As it was mentioned I have now mod superpowers, so s/he can do very little to upset me. Since you're the room owner, let me know if I can be of any assistance here with the moderation (moderators on SE have network-wide chat moderating powers, but this is not my turf, so to speak).
There are two "opposite" functors:$$ op_\Delta\colon sSet\to sSet$$and$$op_s\colon sCat\to sCat.$$The first takes a simplicial set to its opposite simplicial set by precomposing with the opposite of a functor $\Delta\to \Delta$ which is the identity on objects and takes a morphism $\langle k...
@JonathanBeardsley Yeah, I worked out a little proof sketch of the lemma on a notepad
It's enough to show everything works for generating cofaces and codegeneracies
the codegeneracies are free, the 0 and nth cofaces are free
all of those can be done treating frak{C} as a black box
the only slightly complicated thing is keeping track of the inner generated cofaces, but if you use my description of frak{C} or the one Joyal uses in the quasicategories vs simplicial categories paper, the combinatorics are completely explicit for codimension 1 face inclusions
the maps on vertices are obvious, and the maps on homs are just appropriate inclusions of cubes on the {0} face of the cube wrt the axis corresponding to the omitted inner vertex
In general, each Δ[1] factor in Hom(i,j) corresponds exactly to a vertex k with i<k<j, so omitting k gives inclusion onto the 'bottom' face wrt that axis, i.e. Δ[1]^{k-i-1} x {0} x Δ[j-k-1] (I'd call this the top, but I seem to draw my cubical diagrams in the reversed orientation).
> Thus, using appropriate tags one can increase ones chances that users competent to answer the question, or just interested in it, will notice the question in the first place. Conversely, using only very specialized tags (which likely almost nobody specifically favorited, subscribed to, etc) or worse just newly created tags, one might miss a chance to give visibility to ones question.
I am not sure to which extent this effect is noticeable on smaller sites (such as MathOverflow) but probably it's good to follow the recommendations given in the FAQ. (And MO is likely to grow a bit more in the future, so then it can become more important.) And also some smaller tags have enough followers.
You are asking posts far away from areas I am familiar with, so I am not really sure which top-level tags would be a good fit for your questions - otherwise I would edit/retag the posts myself. (Other than possibility to ping you somewhere in chat, the reason why I posted this in this room is that users of this room are likely more familiar with the topics you're interested in and probably they would be able to suggest suitable tags.)
I just wanted to mention this, in case it helps you when asking question here. (Although it seems that you're doing fine.)
@MartinSleziak even I was not sure what other tags are appropriate to add.. I will see other questions similar to this, see what tags they have added and will add if I get to see any relevant tags.. thanks for your suggestion.. it is very reasonable,.
You don't need to put only one tag, you can put up to five. In general it is recommended to put a very general tag (usually an "arxiv" tag) to indicate broadly which sector of math your question is in, and then more specific tags
I would say that the topics of the US Talbot, as with the European Talbot, are heavily influenced by the organizers. If you look at who the organizers were/are for the US Talbot I think you will find many homotopy theorists among them.
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Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors Problem 606
Let $V$ be a vector space and $B$ be a basis for $V$.
Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$.
After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form
\[\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) What is the dimension of $V$? (b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$? (The Ohio State University, Linear Algebra Midterm) Add to solve later
Contents
Solution. (a) What is the dimension of $V$?
Suppose the dimension of $V$ is $n$.
This means that the basis $B$ consists of $n$ vectors.
Then the coordinate of $\mathbf{w}_1$ with respect to $B$ is an $n$-dimensional vector $[\mathbf{w}_1]_B \in \R^n$.
Thus, the number of rows in the matrix $A$ is $n$.
As the elementary row operations do not change the number of rows, we see from the given matrix that the $A$ has four rows.
Thus, the dimension of $V$ is $4$. (b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$?
Note that the dimension of $W:=\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$ is the same as the dimension of
\[W’:=\Span\{[\mathbf{w}_1]_B, [\mathbf{w}_2]_B, [\mathbf{w}_3]_B, [\mathbf{w}_4]_B, [\mathbf{w}_5]_B\}.\]
Since the column vectors of $A$ are these coordinate vectors $[\mathbf{w}_i]_B$ and its reduced row echelon contains the leading 1’s in the first two columns, we conclude that $\{[\mathbf{w}_1]_B, [\mathbf{w}_2]_B\}$ is a basis for $W’$ by the leading 1 method.
It follows that $\{\mathbf{w}_1, \mathbf{w}_2\}$ is a basis for $W$, and its dimension is $2$.
Comment.
This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.
List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017 Vector Space of 2 by 2 Traceless Matrices Find an Orthonormal Basis of the Given Two Dimensional Vector Space Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent? Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$ Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors ←The current problem Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less
Add to solve later
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also, if you are in the US, the next time anything important publishing-related comes up, you can let your representatives know that you care about this and that you think the existing situation is appalling
@heather well, there's a spectrum
so, there's things like New Journal of Physics and Physical Review X
which are the open-access branch of existing academic-society publishers
As far as the intensity of a single-photon goes, the relevant quantity is calculated as usual from the energy density as $I=uc$, where $c$ is the speed of light, and the energy density$$u=\frac{\hbar\omega}{V}$$is given by the photon energy $\hbar \omega$ (normally no bigger than a few eV) di...
Minor terminology question. A physical state corresponds to an element of a projective Hilbert space: an equivalence class of vectors in a Hilbert space that differ by a constant multiple - in other words, a one-dimensional subspace of the Hilbert space. Wouldn't it be more natural to refer to these as "lines" in Hilbert space rather than "rays"? After all, gauging the global $U(1)$ symmetry results in the complex line bundle (not "ray bundle") of QED, and a projective space is often loosely referred to as "the set of lines [not rays] through the origin." — tparker3 mins ago
> A representative of RELX Group, the official name of Elsevier since 2015, told me that it and other publishers “serve the research community by doing things that they need that they either cannot, or do not do on their own, and charge a fair price for that service”
for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing) @EmilioPisanty
> for example, I could (theoretically) argue economic duress because my job depends on getting published in certain journals, and those journals force the people that hold my job to basically force me to get published in certain journals (in other words, what you just told me is true in terms of publishing)
@0celo7 but the bosses are forced because they must continue purchasing journals to keep up the copyright, and they want their employees to publish in journals they own, and journals that are considered high-impact factor, which is a term basically created by the journals.
@BalarkaSen I think one can cheat a little. I'm trying to solve $\Delta u=f$. In coordinates that's $$\frac{1}{\sqrt g}\partial_i(g^{ij}\partial_j u)=f.$$ Buuuuut if I write that as $$\partial_i(g^{ij}\partial_j u)=\sqrt g f,$$ I think it can work...
@BalarkaSen Plan: 1. Use functional analytic techniques on global Sobolev spaces to get a weak solution. 2. Make sure the weak solution satisfies weak boundary conditions. 3. Cut up the function into local pieces that lie in local Sobolev spaces. 4. Make sure this cutting gives nice boundary conditions. 5. Show that the local Sobolev spaces can be taken to be Euclidean ones. 6. Apply Euclidean regularity theory. 7. Patch together solutions while maintaining the boundary conditions.
Alternative Plan: 1. Read Vol 1 of Hormander. 2. Read Vol 2 of Hormander. 3. Read Vol 3 of Hormander. 4. Read the classic papers by Atiyah, Grubb, and Seeley.
I am mostly joking. I don't actually believe in revolution as a plan of making the power dynamic between the various classes and economies better; I think of it as a want of a historical change. Personally I'm mostly opposed to the idea.
@EmilioPisanty I have absolutely no idea where the name comes from, and "Killing" doesn't mean anything in modern German, so really, no idea. Googling its etymology is impossible, all I get are "killing in the name", "Kill Bill" and similar English results...
Wilhelm Karl Joseph Killing (10 May 1847 – 11 February 1923) was a German mathematician who made important contributions to the theories of Lie algebras, Lie groups, and non-Euclidean geometry.Killing studied at the University of Münster and later wrote his dissertation under Karl Weierstrass and Ernst Kummer at Berlin in 1872. He taught in gymnasia (secondary schools) from 1868 to 1872. He became a professor at the seminary college Collegium Hosianum in Braunsberg (now Braniewo). He took holy orders in order to take his teaching position. He became rector of the college and chair of the town...
@EmilioPisanty Apparently, it's an evolution of ~ "Focko-ing(en)", where Focko was the name of the guy who founded the city, and -ing(en) is a common suffix for places. Which...explains nothing, I admit.
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Inside the whale: the structure and dynamics of the isolated Cetus dwarf spheroidal Date2007-02-05 Author
Lewis, G. F.
Ibata, R. A.
Chapman, S. C.
McConnachie, A.
Irwin, M. J.
Tolstoy, E.
Tanvir, N. R.
MetadataShow full item record Abstract
This paper presents a study of the Cetus dwarf, an isolated dwarf galaxy within the Local Group. A matched-filter analysis of the INT/WFC imaging of this system reveals no evidence for significant tidal debris that could have been torn from the galaxy, bolstering the hypothesis that Cetus has never significantly interacted with either the Milky Way or M31. Additionally, Keck/Deimos spectroscopic observations identify this galaxy as a distinct kinematic population possessing a systematic velocity of $-87\pm2{\rm km\ s^{-1}}$ and with a velocity dispersion of $17\pm2{\rm km s^{-1}}$; while tentative, these data also suggest that Cetus possesses a moderate rotational velocity of $\sim8{\rm km s^{-1}}$. The population is confirmed to be relatively metal-poor, consistent with ${\rm [Fe/H]\sim-1.9}$, and, assuming virial equilibrium, implies that the Cetus dwarf galaxy possesses a $M/L\sim70$. It appears, therefore, that Cetus may represent a primordial dwarf galaxy, retaining the kinematic and structural properties lost by other members of the dwarf population of the Local Group in their interactions with the large galaxies. An analysis of Cetus's orbit through the Local Group indicates that it is at apocentre; taken in conjunction with the general dwarf population, this shows the mass of the Local Group to be $\gta2\times10^{12}M_\odot$.
Citation
Lewis, G. F., R. A. Ibata, S. C. Chapman, A. McConnachie, et al. 2007. "Inside the whale: the structure and dynamics of the isolated Cetus dwarf spheroidal." Monthly Notices of the Royal Astronomical Society 375(4): 1364-1370
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In this kind of questions, I believe that most of the attention should be focused on the general approach and the philosophy of attacking. Afterwards, each individual may proceed by writing its own equations and solving them in its prefer way.In our particular case, I guess you examined the whole structure, realized that there are 4 reactions (Ax, Ay, Cx ...
The question has not defined any section properties so we assume all members are rigid. Then by inspection, we see this structure reduces to a 2 member frame AB and BC connected at a hinge at B and supported at pins on ends.AB is at 45 degrees angle and BC is arctan(5/15) = arctan(1/3).EDITAfter the OP correcting my arithmetic error.$$ \Sigma M_a= -...
Brake torque usually means torque at the engine crankshaft, while there is no loss of power or toque due to transmission, powertrain or in your case the chain and gearbox.You would have to subtract those losses if the indicated torque is the wheel torque.
the rate at which heat is added is so fast that the piston has no opportunity to move during the process. Furthermore, when the piston is at the top of its stroke (which is where the heat is added) the crank-and-connecting rod mechanism yields a finite residence time of the piston in that position for a span of crank rotation, which means that even though ...
Hydraulics can be precise, down to ~ 100µm. However:hydraulic pistons [...] used in conjunction with a positive displacement pump and using a large hydraulic transmission ratio the issue of precision could be sorted out, and the incompressibility of the fluid would help to keep backlash to a minimum.This is not how this high precision is achieved in ...
The curve of a spring loaded cam is certain curve called exponential spiral with the unique property that the angle it touches the wall of the crack and a horizon remains the same (13-14 degrees) no matter haw much the cam expands.This small angle means the force applied by the cam to the crack is always $F_{horiz}= W_{weight}*cotan(13)= W*4.3$Which ...
This is a hugely complicated question that's not going to be answered both accurately and briefly, because there's too many variables presented by the real world. So I'm going to make some simplifying assumptions, to give you an idea.What I'm going to present is a simple lever calculation, but it'll almost never apply to the real world.Let's say that ...
According to a couple of textbooks on machine design and assuming that your FBD is correct, the maximum bending moment occurs at the centre of the shaft.For a point load, it is equal to:$WL/4 = M$Where W is the point load and L is the length of your beam.$(200*10^3)*1/4 = 50*10^3 Nm = 50*10^6 Nmm = M$According to the maximum shear stress theory ...
I would calculated the I being allocated to two different materials, foam and paper. one way of getting a sense of the difference between the two is to let the paper beam on its side flatwise, and measure the E of the paper by neglecting the contribution of foam, if it is not paper on both sides glue 2 back to back to make it so. Because in that ...
I think your convergence tolerance is not enough. Since you obtain first few solutions with low convergence requirements you are moving away from the real results for the next solution steps. From the code, it seems you are using unrealistic values. In example, total displacement is 5. The tolerance is 0.01, they are crude numbers. So it is not easy to ...
If you are using standard Gauss quadrature, from the formulation, you determine the coefficients (weights of Gauss location) and their locations. You cannot arbitrarily decide their locations. Here are some examples: http://edwilson.org/BOOK-Wilson/G-inter.pdf
You design the system for a demand that it should meet, 3x 100l/s in your example. If the demand is higher, the sytem will likely fail to deliver the actual demand. If the demand is lower, no problem.On the network side, you know the pressure you want to deliver the water at. You could supply valves at the tap that deliver only (say) 100l/s at 4 bar, but ...
You can define the h and beta constants, either in a cell or as a defined name.After that, you can create a column with the theta parameter and another column with the calculated y parameter.The formula for the y should use the IF and AND functions. Suppose the theta parameters are on column A. The formula should be something like=IF(AND(0<=A1,A1<...
As recommended by @Solar Mike, the most "professional" solution is to let the end-user decide. Give them some input mechanism (a textbox, slider or dropdown, for example) so they can define which value to adopt. This is what you'll see in basically any modeling software you'll find in the wild.However, it's also very important to note the observations made ...
You have a tall skinny skid-steer vehicle, and you're trying to run it on a surface with a high friction coefficient. So the wheels are constantly grabbing and letting go, grabbing and letting go.Try making either the front or rear pair of wheels castoring.
Your connection of the wheels to the robot enclosure (suspension) is flimsy and has too much free play. It starts to jerk and gives enough kick energy to the flimsy box to keep it dancing. And the skidding of the wheels resonates with the vibration of the robot.To improve the stability without having to use electronics self balancing chips:Use an ...
Model airplane motors run at 50000 RPM with ball bearings; I don't think that your limit is due to speed.Basically, the assembly you picture will have the friction of, and be as rugged as, just using the inner bearing. The outer will just be going along for the ride (and adding weight and expense to your assembly).
You will increase the friction due to two sets of balls and lubricants with the races.Also, this will cause more play in the system...If it worked and gave an advantage then it would be used by the F1 teams...
I think the problem is most Americans are too lazy or more interested in having fun ( A K A liberal arts courses). Other than medicine; engineering is the most reliable area to get a good paying job . Now if you want to be a multi-millionaire , you need to invent some new kind of computer chip or develop a new social website. But if you want a good job with ...
Validation: Are we building the right product?Verification: Are we building the product right?Validation is the process of making sure that you have objective evidence that user needs and intended uses are met. ... Verification is typically making sure that you have objective evidence that specified requirements are met. It is usually done by tests, ...
Here's an illustration to explain the difference. Suppose you were creating an accounting system for Enron, before it collapsed. The specification said you need to ensure that the output assumed that 2+2=5.Verification is asking the system "what is 2 + 2" and getting the answer "5".Validation is discovering that the Securities and Exchange Commission ...
You likely did both, but with different processes. You create a mathematical or computer model and you check that it matches the ideas that underlie it. In a computer model you check that you haven't made mistakes in the programming. In a mathematical model you check that you have used the right equations (or whatever) and evaluated them correctly. This is ...
By experience, it is not a good idea to take any standard value or manufacturer value. Just check your chain on 10 degrees and on 25 degrees for a very long distance, on the limit of your chain, with no arrow, so on a flat ground. And just make your numbers.When you need to make a high precition work, you can not just take the values of one specification ...
Why not just use a standard salt spray cabinet ? I am sure it can use different salts other than sea salt. Those we had were roughly 4 ft. cubes which may be big for your work , but may have future uses. And a standard unit results are easy to compare to literature and/or publish. And as a standard unit it would have some resale value. I don't remember the ...
You have the right idea using a shoulder screw.In between the two pieces, use flat washers and a springy Belleville washer.A Belleville washer, also known as a coned-disc spring, conical spring washer, disc spring, Belleville spring or cupped spring washer, is a conical shell which can be loaded along its axis either statically or dynamically. A ...
Search back-pressure regulators. Basic idea is they have a dome with a membrane diaphragm pressurized to the control pressure. If the incoming pressure is higher than the dome it pushes the membrane up and shuts the flow.This is a link to one. back pressure regulator
Here is a trick you can try. Cut the silicone edge so that the edge dimension (for lack of a better term) is a little smaller than you final finished dimension. Now make a temporary casting mold for the cut edge such that the inside edge of the mold adjacent to the cut edge forms a wall , the gap between the cut edge and the smooth mold wall is sized ...
Your raw material will come in the form of a large spool of sheet metal containing a strip of steel hundreds of meters long. You will need an unspooler to feed steel off the spool and a straightener to take the curvature out of the steel strip. Then you need either a punch press or a shear to cut the sheet metal blanks to size and trim their corners. To put ...
Thought experiment -Given a free body $B$ that is being acted on by forces, find its instantaneous center of rotation $P1$ and its instantaneous center of zero acceleration $P2$.Does this let us solve for the center of curvature for any point $P$ on the body?To do so, we need the acceleration vector at $P$ and the velocity vector at $P$. That lets us ...
This is not a statically determinate system. To find how the load splits between the members, start by finding their stiffness.Since the materials and cross-section areas are the same, the stiffness is inversely proportional to the length.Length of AC = $L\cos\theta$.If the stiffness of AB is $K$, the stiffness of AC is $K/\cos\theta$.Now consider a ...
Just bring the wire down to the hinge and have it go inline with the hinge so it rotates with the movement.This is done on car tailgates and works fine. Some wires fail but only after several years...
you can use a small spring loaded roller for window shade. They cost in the range of $20 and many hardware stores custom cut them down to something like 6 inches.You can remove the shade and wind your wire around it.here is a photo.
let's call the tension at the bars at B and D, T1 and at C, T. We see that $T_1=T\frac{ \sqrt3}{2}$$$P = 2*T_1 +T= T +2 T\frac{\sqrt{3}}{2}=2.73205\cdot T$$The share of C bar from P load will be $$T=P\frac{1}{2.73205} $$$$ \delta_c= \frac{Pl_c}{2.73205EA}= \frac{Pl\sqrt{3}/2}{2.73205EA} $$The extended length of the C bar will be $l_c= Pl(\sqrt{3}/2) ...
These phases can be quite ductile due to their layered structure at the atomic scale, naturally as others have suggested here, it depends on compositions and stoichiometry.Under compressive conditions, high levels of performance can be achieved. This can be further enhanced by fabrication of metal-max phase composities. Generally, crack propagation is the ...
I made one for my son's camera.The issue is usually with the bearings. I used 2 pairs of bearings taken from those "spinners" and put them on a threaded rod.Used double nuts to "just" load the bearings to provide some damping and he had good results when he was taking videos while skiing...
I'm going to guess at what's happening from your description. In general, the "moment" on both sides of the pivot (weight x distance) should be close to equal. You can have slightly more weight at the bottom so it wants to stay upright. Inertia is what keeps things from tilting when you start or stop moving. If you have weight just at the bottom (or a ...
the crank handle is connected to the little black box, inside of which there are gears that rotate the metal shaft you can see running through the box and across the back of the table support. the right hand end of that shaft enters the support leg on the right which has another gearbox in it that rotates a shaft leading down the length of the support leg. ...
The noise you hear is the engine running at the idle speed. To provide minumu energy for lubrication, charging the battery, air conditiong, water pump etc. In newer cars it has been eliminated for duration of intermittent stops to save fuel and pollute less. The computer will start the engine the moment you release the brakes.The gear box, manual and ...
Thankfully, I once took an Automotive engineering course. It was two years ago, but I still remember something... Of course, my answer is open for improvement, so if there is any constructive criticism, please go aheadThe gas pedal is the controller for the throttle (think how much air is going into the engine), and this, in turn, controls the fuel ...
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Upon reading about the principal bundle picture of (quantum) field theory I encountered two different definitions of the gauge group:
Local gauge group $G$. Corresponds to the fibers of the $G$-bundle. Local gauge transformations correspond to the change of coordinates in which fields and form are written in trivializations. In other words, as far as I understand, local gauge transformations = transition functions for trivializations of the principal bundle and the associated bundles. Global gauge group $\mathfrak G=Aut(P)$. Is the group of principal bundle diffeomorphisms, and is claimed to be much bigger then $G$.
I would like to understand which notions in physics correspond to these two, apparently very distinct, notions.
First, in physics, there is the notion of
global symmetry, as far as I understand this has nothing to do with gauge/bundles. But on the other hand, the limit of a gauge symmetry where the parameter becomes constant is a global symmetry. Typical $U(1)$ example: $\exp(i\alpha(x))$ = $U(1)$ gauge transformation, but if $\alpha(x)=\alpha=const.$, then it is a global symmetry. Is there any deeper connection between global symmetry and gauge "symmetry"/bundles?
Second, I would appreciate some examples which clarify the distinction between local and global gauge groups mentioned above. If, for example we take a prototype $SU(2)$ gauge theory of the form $$ \mathcal L = -\frac{1}{4}(F_{\mu\nu}^a)^2+|(D_\mu\phi)^a|^2-V(\phi) $$ then we can put it in the bundle picture as a $SU(2)$ principal bundle, the fields $A_\mu^a$ living in the associated bundle $$ (P\times\mathfrak su(2))/\{(p,A)\sim(pg,Ad_{g^{-1}}A)\} $$ and the doublet field $\phi$ in fundamental representation of $SU(2)$ living in the associated bundle $$ (P\times\mathbb C^2)/\{(p,\phi)\sim(pg,\rho(g^{-1})\phi)\}. $$ So, the $SU(2)$ group which we are referring to here is what was the local gauge group $G$ above, is this correct? The gauge transformations apply to coordinate versions of the fields in the trivializations of the associated bundles to change between different coordinates.
Now if, this is the case, what is the global gauge group $\mathfrak G$ then? In particular I am seeking the answer for the following questions:
Do global gauge transformations apply to the model $SU(2)$ theory above? In what sense is the global gauge group much bigger then the local gauge group? Is there a model example where one can clearly write down the local gauge group and the global gauge group? Are there cases where the two groups are the same? How does this discussion relate to quantization of classical fields (if it does at all)? Edit
I'll post here answers to various questions in the comments/answers.
About the definitions I used in my text. I took them from my lecture notes from a course on mathematical gauge theories taught by a mathematician. Although there is no on-line version of this course, I can provide more detailed definitions if needed.
Specifically, concerning the distinction global/local gauge, I have found this piece of information on nLab, which seems to be agreement with my definition: http://ncatlab.org/nlab/show/gauge+group. Although this page provides some explanations, it uses some mathematical terminology which I am not entirely familiar with. The problem both with the nLab page and with the lecture I attended is to make the connection to physics (I am a physicist).
Ok, I'll just write down the definitions from my lecture notes.
Local gauge
Let $P\to M$ a principal $G$-bundle, $\omega$ a connection 1-form on $P$. $U_i, U_j\subset M$ such that $U_i\cap U_j\neq\emptyset$ over which P is trivial: $$ \psi_i:\pi^{-1}(U_i)\to U_i\times G\\ \psi_j:\pi^{-1}(U_j)\to U_j\times G $$ with transition functions $$ \psi_i\circ\psi_j^{-1}:(U_i\cap U_j)\times G\to (U_i\cap U_j)\times G\\ \phantom{\psi_i\circ\psi_j^{-1}:XXX}(m,g)\mapsto (m,\psi_{ij}(m)g) $$ where $\psi_{ij}:U_i\cap U_j\to G$.
Using these trivializations we can construct preferred sections as follows: $$ \sigma_i:U_i\to\pi^{-1}(U_i)\\ \phantom{\sigma_i:x}m\mapsto \psi_i^{-1}(m,e)\,. $$ Define $$ \omega_i:=\sigma_i^*\omega\,, $$ which is a $\mathfrak g$-valued 1-form on $U_i$. Changing between coordinates on $U_i$ and $U_j$ is what is called choice of local gauge. We would like to write down a transformation prescription for $\omega_{i/j}$.
On G we have the canonical 1-form $\theta$ with values in $\mathfrak g$ defined as follows: $$ \theta_g(X_g)=A\in\mathfrak g\quad\text{if }(A^*)_g=X_g $$ where by $A^*$ we mean the fundamental vector field on $G$ corresponding to $A$. One can show that the 1-forms transform in the following way: $$ \omega_j=Ad_{\psi_{ij}}^{-1}\omega_i+\psi_{ij}^*\theta\,. $$ This looks to me like a gauge transformation of guage fields $A=A_\mu dx^\mu$ in physics where we write $$ A'=gAg^{-1}+gdg^{-1} $$ and I even believe that the above transformation for $\omega_j$ can be rewritten as something like this $$ \omega_j=Ad_{\psi_{ij}}^{-1}\omega_i+\psi_{ij}^{-1}d\psi_{ij}\,, $$ Then it looks like the transformation for $A$ with $g=\psi_{ij}^{-1}$.
So much about the local gauge transformation.
Global gauge group
The group of global gauge transformation $\mathfrak G$ is the set of automorphisms of the principal bundle P: $$ \mathfrak G = Aut(P)\,. $$ This must not be confused with the structure group $G$, which is sometimes called gauge group in leterature, but is much smaller.
The (global) gauge group can be described in three ways:
$\mathfrak G=\{\phi:P\to P|\phi\text{ is a diffeo., }\pi\circ\phi=\pi, \phi(pg)=\phi(p)g\,\forall g\in G\}$ $\mathfrak G=\{u:P\to G|u\text{ smooth s.t. }u(pg)=g^{-1}u(p)g\,\forall g\in G\}$. Note that $\phi(p)=p\cdot u(p)$. $\mathfrak G=\{\text{sections }s:M\to F\}$, where $F=(P\times G)/\sim$ with $(p,h)\sim(pg,g^{-1}pg)\forall g\in G$.
If $\omega$ is a connection 1-form which corresponds to a choice to the horizontal tangent space $H$ on $P$, then $\phi^*\omega$ is also a connection 1-form, defining the pull-back connection $$ (\phi^*H)_p:=(D_p\phi)^{-1}H_{\phi(p)}\quad\Leftrightarrow\quad D_p\phi((\phi^*H)_p):=H_{\phi(p)} $$ We conclude that $\mathfrak G$ acts on connections, but the explicit form of this action if fairly complicated. In contrast, the action of $\mathfrak G$ on the curvature is easy to understand.
From the last equation above we conclude that $\phi$ maps $\phi^*H$ to $H$, and so it maps $\tilde\Omega$, the curvature of $\phi^*H$, to $\Omega$, the curvature of $H$: $$ \phi(\tilde\Omega(X,Y))=\Omega(X,Y)\quad\forall X,Y\in T_mM\,. $$ One can show that the following transformation relation holds: $$ \tilde\Omega=Ad_{u^{-1}}\Omega\,. $$ where $u:P\to G$ is the map from the second definition of $\mathfrak G$ above corresponding to $\phi$.
Two connections $H_1$ and $H_2$ on P are called gauge equivalent if there is a $\phi\in\mathfrak G$ with $\phi^*H_2=H_1$.
One of the things which I am curious about is that from the local gauge transformations the transformation law for the 1-form $\omega$ seems to be what we call the gauge transformation of gauge fields $A$ (see above), but from the global gauge group transformation, the transformation of the curvature is what we see in physics, namely if $F$ is the field strength corresponding to $A$, and $g$ an element of the gauge group then $$ F\to F'=gFg^{-1}\,. $$ Also, as I said at the very beginning, I would like to pin down what these mathematical definitions correspond to in what we learn in physics. In physics, when we discuss a certain gauge theory, there is THE gauge group, such as $U(1)$, $SU(2)$, etc. under which fields transform in certain representations, as well as the gauge fields themselves, which transform in the adjoint representation. Now in mathematics I see the distinction local/global gauge group. The essence of my question is do understand this distinction and to relate it to phyics.
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In the theory of relativistic wave equations, we derive the Dirac equation and the Klein-Gordon equation by using representation theory of the Poincare algebra.
For example, in this paper
http://arxiv.org/abs/0809.4942
the Dirac equation in momentum space (equation [52], [57] and [58]) can be derived from the 1-particle state of irreducible unitary representation of the Poincare algebra (equation [18] and [19]). The ordinary wave function in position space is its Fourier transform (equation [53], [62] and [65]).
Note that at this stage, this Dirac equation is simply a classical wave equation. i.e. its solutions are classical Dirac 4-spinors, which take values in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$.
If we regard the Dirac waves $\psi(x)$ and $\bar{\psi}(x)$ as 'classical fields', then the quantized Dirac fields are obtained by promoting them into fermionic harmonic oscillators.
What I do not understand is that when we are doing the path-integral quantization of the Dirac fields, we are, in fact, treating $\psi$ and $\bar{\psi}$ as Grassmann numbers, which are counter-intuitive for me. As far as I understand, we do path-integral by summing over all 'classical fields'. While the 'classical Dirac wave $\psi(x)$' we derived in the beginning are simply 4-spinors living in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$. How can they be treated as Grassmann numbers instead?
As I see it, physicists are trying to construct a 'classical analogue' of Fermions that are purely quantum objects. For instance, if we start from a quantum anti-commutators
$[\psi,\psi^{\dagger}]_{+}=i\hbar1$ and $[\psi,\psi]_{+}=[\psi^{\dagger},\psi^{\dagger}]_{+}=0$,
then we can obtain the Grassmann numbers in the classical limit $\hbar\rightarrow0$. This is how I used to understand the Grassmann numbers. The problem is that if the Grassmann numbers are indeed a sort of classical limit of anticommuting operators in Hilbert space, then the limit $\hbar\rightarrow0$ itself does not make any sense from a physical point of view since in this limit $\hbar\rightarrow0$, the spin observables vanish totally and what we obtain then would be a $0$, which is a trivial theory.
Please tell me how exactly the quantum Fermions are related to Grassmann numbers.
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Limits of Functions on Metric Spaces Review
Limits of Functions on Metric Spaces Review
We will now review some of the recent material regarding limits of functions on metric spaces.
Let $(S, d_S)$ and $(T, d_T)$ be metric spaces with $A \subseteq S$.
On the Limits of Functions on Metric Spacespage we that a Functionfrom the metric spaces $S$ and $T$ is a rule $f : A \to T$ defined for all $x \in A$ by $f(x) = y$ for some $y \in T$. The set $A$ is called the Domainof $f$ often written $D(f) = A$ and the set $T$ is called the Codomainof $f$ and is often written $C(f) = T$. The image of $A$ under $f$, $f(A)$, is called the Rangeof $f$. Furthermore, if $f : A \to T$ and $p \in S$ is an accumulation point on $A$ and $b \in T$ then we say that the Limit of $f$ as $x$ Approaches $p$ is $b$written $\displaystyle{\lim_{x \to p} f(x) = b}$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in D(f) \setminus \{ p \} = A \setminus \{p \}$ and $d_S(x, p) < \delta$ then:
\begin{align} \quad d_T(f(x), b) < \epsilon \end{align}
An equivalent definition is that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in [D(f) \setminus \{ p \}] \cap B_S(p, \delta)$ then $f(x) \in B_T(b, \epsilon)$. Regardless, it is important to note that we require $p \in S$ to be an accumulation point of $A$ because then every open ball centered at $p$ contains points in $A$ and so we can "approach" $p$ by taking $x$ sufficiently closed to $p$ in terms of the metric defined on $S$. On The Uniqueness of Limits of Functions on Metric Spacespage we saw that if the limit of $f$ as $x$ approaches $p$ exists then that limit is unique - as expected. On the Sequential Criterion for the Limit of a Function on Metric Spacespage we looked at a very useful theorem which told us that the limit of $f$ as $x$ approaches $p$ is equal to $b$ if and only if for every sequence $(x_n)_{n=1}^{\infty}$ in $A \setminus \{ p \}$ that converges to $p$ in $S$ we have that the sequence $(f(x_n))_{n=1}^{\infty}$ converges to $b$ in $T$. We then turned out attention to limits of complex-valued functions. On the Limits of Sums and Differences of Complex-Valued Functions, Limits of Products of Complex-Valued Functions, and the Limits of Reciprocals and Quotients of Complex-Valued Functionspages we looked at a bunch of theorems regarding the limits of complex-valued functions. We said that if $(S, d_S)$ and $(\mathbb{C}, d)$ are metric spaces where $d$ is the usual metric on $\mathbb{C}$ defined for all $x, y \in \mathbb{C}$ by $d(x, y) = \mid x - y \mid$, $A \subseteq S$, $p \in S$ is an accumulation point of $A$, and $a, b \in T$ with $\displaystyle{\lim_{x \to p} f(x) = a}$ and $\displaystyle{\lim_{x \to p} g(x) = b}$ then we can conclude that:
\begin{align} \quad \lim_{x \to p} [f(x) + g(x)] = a + b \end{align}(3)
\begin{align} \quad \lim_{x \to p} [f(x) - g(x)] = a - b \end{align}(4)
\begin{align} \quad \lim_{x \to p} f(x)g(x) = ab \end{align}(5)
\begin{align} \quad \lim_{x \to p} \frac{1}{g(x)} = \frac{1}{b} \quad \mathrm{if \:} b \neq 0 \end{align}(6)
\begin{align} \quad \lim_{x \to p} \frac{f(x)}{g(x)} = \frac{a}{b} \quad \mathrm{if \:} b \neq 0 \end{align}
We then turned our attention to vector-valued functions. On the Existence of Limits of Vector-Valued Functions, Limits of Sums and Differences of Vector-Valued Functions, Limits of Scalar Multiples of Vector-Valued Functions, Limits of Dot Products of Vector-Valued Functions, and Limits of Norms of Vector-Valued Functionspages we saw that if $(S, d_S)$ and $(\mathbb{R}^n, d)$ are metric spaces where $d$ is the usual metric on $\mathbb{R}^n$ given for all $\mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n$ by $d(\mathbf{x}, \mathbf{y}) = \| \mathbf{x} - \mathbf{y} \|$, $A \subseteq S$, $p \in S$ is an accumulation point of $A$, $\mathbf{a}, \mathbf{b} \in \mathbb{R}^n$, $\mathbf{f}, \mathbf{g} : A \to \mathbb{R}^n$, and $\lim_{x \to p} \mathbf{f}(x) = \mathbf{a}$, $\lim_{x \to p} \mathbf{g}(x) = \mathbf{b}$, then:
\begin{align} \quad \lim_{x \to p} \mathbf{f}(x) = \mathbf{a} \quad \mathrm{if \: and \: only \: if} \quad \lim_{x \to p} f_j(x) = a_j \: , \forall j \in \{1, 2, ..., n \} \end{align}(8)
\begin{align} \quad \lim_{x \to p} [\mathbf{f}(x) + \mathbf{g}(x)] = \mathbf{a} + \mathbf{b} \end{align}(9)
\begin{align} \quad \lim_{x \to p} [\mathbf{f}(x) - \mathbf{g}(x)] = \mathbf{a} - \mathbf{b} \end{align}(10)
\begin{align} \quad \lim_{x \to p} \lambda \mathbf{f}(x) = \lambda \mathbf{a} \:, \forall \lambda \in \mathbb{R} \end{align}(11)
\begin{align} \quad \lim_{x \to p} \mathbf{f}(x) \cdot \mathbf{g}(x) = \mathbf{a} \cdot \mathbf{b} \end{align}(12)
\begin{align} \quad \lim_{x \to p} \| \mathbf{f}(x) \| = \| \mathbf{a} \| \end{align}
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I honestly don't recall learning the inverse function theorem until first year uni to be fair (and basically never used it from 2nd year and later) , so I doubt the majority of the high-school students would even consider it.Please share some more rants pls this is hilarious.
Wow this is definitely harder than the previous years. I see that Carrotsticks having extra assistance from former BOS members (who are also math whizzes :P) definitely revoluntionised the content.However, with the new syllabus being implemented, I wonder what mathematical theorem do you plan...
You can't compare the BOS trials with the HSC. It's just at a completely different level since it's specifically aimed to challenge the students who have mastered virtually every single topics, i.e people who have the potential to easily achieve 95+ HSC. Even if you did do badly, it wouldn't...
When you're finding the rate, or how fast or slow something is, you use the derivative expression.so for c)you find dS/dt and then sub in t = 5 and your value of k from part b), that's your rate.d)sub S = 4 (as that's the amount that's left undissolved ) and your k from b)Then you solve...
The problem with checking your answers over, is that sometimes you would think it's right in the back of your mind so it could blind you from seeing the mistake.What I would do is that I would read the question thoroughly a few times (even though it's something I'm familiar with), and then do...
I'm not sure what you mean by "New" HSC Math courses (you mean progressing to senior or a complete revamp in the content?)Anyhow, all I can say is that you general math is basically learning applied "real life" math like finding areas and doing some basic finance stuff you might encounter in...
My parents and relatives would talk about how my mum and dad did really well in math back at school in China. I'm not sure if it's legit or not haha but I was able to gasp at most concepts a bit better than the average student so I guess it's a factor.
That's not even true since you're squaring a number less than 10<ln(2)<ln(e) = 1 \\ \\ ln(2) > (ln(2))^22ln(2) > (ln(2))^2 \Rightarrow 2 > ln(2)At the moment I still can't figure the lower bound either.
In order to find the horizontal asymptote, you need to take the limit as the x-value approaches infinity. You also need use this concept.lim_{x \rightarrow 0} \frac{sinx}{x} =lim_{x \rightarrow \infty} \frac{sin(\frac{1}{x})}{\frac{1}{x}} = 1lim_{x \rightarrow \infty} \frac {x^2...
Note that is says distinct points so that already implies p \neq qYou can easily see that's true if you simply observe the equation mQRmPR=-1. which leads to (p+1)(q+1) = -4 and you'll understand that p and q can't be -1 as you'll get zero.However also observe that both p+1 and q+1 can't...
Yeah that's because this not your straight forward "top minus bottom" kind of question.First you should draw out the curves, then find the point of intersection (which is x = 1 by inspection)The area given above is\int_0^1 \ x^3 \ \ dx + \int_1^{\frac{4}{3}} \ \ -3x + 4 \ \ dx = \frac{5}{12}
Alright zID 501Issued at Tue Dec 4 16:47:02 2018Received the email at 5:39 PMHehe a pretty crap semester for me, but I'm just happy that I survived all of them and thus my undergraduate life is officially over.
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Sreedhar, K and Ganguly, P (1990)
Evolution and the concomitant disappearance of high-$T_c$ superconductivity with carrier concentration in the $La_{2-x}Sr_xCuO_{4-\delta}$ system $(0.0\leq x \leq 1.2)$: Crossover from a Mott insulator to a band metal. In: Physical Review B, 41 (1). 371 - 382.
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Abstract
Electrical resistivity, magnetic susceptibility, thermogravimetric analysis, and infrared absorption spectra of the compound $La_{2-x}Sr_xCuO_{4-\delta}$ have been studied for a wide range of $Sr$ concentrations $(0.0\leq x \leq 1.2)$. The samples annealed at an oxygen pressure of 1 bar were stoichiometric $(\delta =0.0)$ in the range 0.0<x<0.33. In this range the compounds are characterized by a decrease in the a parameter, an increase in the c parameter, and a maximum in the c/a ratio (at x =0.33) typical of the formation of low-spin $Cu^{3+}$ ions. In the range 0.15<x<0.33, the compounds show a positive temperature coefficient of resistivity, decrease in the magnitude of the Pauli magnetic susceptibility, infrared oscillator strengths, thermopower S, as well as the Hall coefficient $R_H$. The superconducting transition temperature $T_c$, as well as the percentage Meissner fraction also decrease with x in this range. In particular the stoichiometric x=0.33 composition having a hole concentration of $\sim 0.33$ holes/Cu shows a minimum in the Pauli magnetic susceptibility and disappearance of all infrared absorption bands. The superconductivity also disappears down to 4.2 K at this composition, even though it is more metallic. These results have been attributed to the occurrence of a transition from a highly correlated narrow-band "Mott conductor" to a broadband metal at high carrier concentrations. At still higher $Sr$ concentrations (0.33 <x < 1.0), holes and oxygen vacancies coexist. The decrease in the c/a ratio, increase in resistivity, reappearance of the infrared bands, and the Curietype magnetic susceptibility observed in the range $0.66\leq x\leq 1.2$ indicate the dominating role of oxygen vacancies which induce disorder and localization.
Item Type: Journal Article Additional Information: Copyright of this article belongs to The American Physical Society. Department/Centre: Division of Chemical Sciences > Solid State & Structural Chemistry Unit Depositing User: Ramya Krishna Date Deposited: 30 Jun 2007 Last Modified: 18 Jan 2012 06:19 URI: http://eprints.iisc.ac.in/id/eprint/11335 Actions (login required)
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You are currently browsing the tag archive for the ‘springer’ tag.
This happens a lot: I am reading a paper, as usual going directly to the results and skipping the introduction, related literature, discussion, preliminaries, formal model etc. And then there is some which I have no idea what it stands for. I would like to search for `\alpha’ in the pdf document, but if there is a way to do it then I have never heard about it.
So, imagine my delight when I heard of Springer’s LaTeX Search tool, which does something that I never even dared to wish — search in their database for an equation that contains a given latex code. Pretty awesome, isn’t it ?
I tried some arbitrary code
i\hbar\frac{\partial}{\partial t}\Psi=\hat H\Psi
(which translates to )
but apparently nobody has used this equation before.
So I tried something else: E=mc^2. Again no exact matches but this time there are a couple of similar results
Well, as Jeffrey Shallit said, it is, at least, a start.
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Evaluating Double Integrals in Polar Coordinates
So we have looked at evaluating double integrals over general domains, however, sometimes it may be rather difficult to compute double integrals over certain domains due to the nature of integrating with the rectangular coordinate system. For example, computing a double integral over a circle with equation $(x - h)^2 + (y - k)^2 = r^2$.
Sometimes, it may be easier to use the polar coordinate system to compute a double integral. We will see that we can often times do this using the following formula to convert our double integral for a continuous function $f$ over the rectangle $R = \{ (x, y) : a ≤ r ≤ b, \alpha ≤ \theta ≤ \beta \}$ and where $0 ≤ \beta - \alpha ≤ 2 \pi$:(1)
Recall that if $(x, y)$ is a point in $\mathbb{R}^2$, then the relationship between $(x, y)$ and the corresponding polar coordinates $(r, \theta)$ come from the equations:(2)
Furthermore, $r^2 = x^2 + y^2$. Sometimes it is easier to represent certain regions $R$ as $R = \{ (r, \theta) : a ≤ r ≤ b , \alpha ≤ \theta ≤ \beta \}$.
Now let's see how we would compute $\iint_R f(x, y) \: dA$. We start by taking the interval $[a, b]$ and dividing it into $m$ equal width subintervals $[ r_{i-1}, r_i ]$ where the width of each subinterval is $\Delta r = \frac{b - a}{m}$. We will also divide the interval $[ \alpha, \beta ]$ up into $n$ equal width subintervals $[\theta_{j-1}, \theta_j]$ where the width of each subinterval is $\Delta \theta = \frac{\beta - \alpha}{n}$. $R_{ij}$ will denote the polar sub-rectangle for which $r_{i-1} ≤ r ≤ r_{i}$ and $\theta_{j-1} ≤ \theta ≤ \theta_{j}$. The center of each polar subrectangle $R_{ij}$ which we will denote as $(r_i^*, \theta_j^*)$ is such that $r_i^* = \frac{1}{2} \left ( r_{i-1} + r_i \right )$ and $\theta_j = \frac{1}{2} \left ( \theta_{j-1}, \theta_j \right )$.
Now recall that the area of a sector of a circle is given by the formula:(3)
Therefore we can compute the areas of each $R_{ij}$ as:(4)
Using the conversion equations above, we have that the centers of each $R_{ij}$ in rectangular coordinates is $(r_i^* \cos \theta_j^*, r_i^* \sin \theta_j^*)$. Now the Riemann sum for setting up our double integral will be:(5)
Now let $g(r, \theta) = f(r \cos \theta, r \sin \theta) \cdot r$ then we have that:(6)
Thus we can evaluate our double integral as:(7)
Another type of region that we may need to integrate over is one such as $D = \{ (r, \theta) : h_1 (\theta) ≤ r ≤ h_2 (\theta) , \alpha ≤ \theta ≤ \beta \}$:
The double integral of $f(x, y)$ over $D$ can be computed as:(8)
What's nice is that we can also compute the area of polar region using double integrals.
Theorem 1: The area enclosed in a polar region $D$ bounded by $\alpha ≤ \theta ≤ \beta$ and $0 ≤ r ≤ h(\theta)$ is given by $\mathrm{Area} = \iint_D 1 \: dA$. Proof:Suppose that $D$ is bounded by $\theta = \alpha$, $\theta = \beta$, $r = 0$, and $r = h(\theta)$. Then: From the Areas Enclosed by Polar Curves page, we see that this is precisely the area enclosed by a general polar curve. $\blacksquare$
Now the following Theorem gives us the value for a very important single integral that is widely used in areas such as probability.
Theorem 2: $\int_{-\infty}^{\infty} e^{-x^2} \: dx = \sqrt{\pi}$. Proof:Let $I = \int_{-\infty}^{\infty} e^{-x^2} \: dx$. Note that we can also write $I = \int_{-\infty}^{\infty} e^{-y^2} \: dy$ as the choice of variable does not matter. Thus: We now express $\mathbb{R}^2$ as a polar region. We have that $\mathbb{R}^2 = \{ 0 ≤ r ≤ \infty, 0 ≤ \theta ≤ \pi \}$, and so:
Let's now look at some examples of calculating integrals with polar coordinates.
Example 1 Evaluate the double integral $\iint_R x + 2y \: dA$ where $R$ is the region of area between the semicircles above the $x$-axis of $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.
The region $R$ can be described as $R = \{ (r, \theta) : 1 ≤ r ≤ 2, 0 ≤ \theta ≤ \pi \}$:
We note that $f(x, y) = x + 2y$ and so $f(r \cos \theta, r \sin \theta) = r \cos \theta + 2r \sin \theta$. Using the formula above, we can convert our double integral into the following iterated integrals:(12)
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I am reading the following paper by Slater:
https://journals.aps.org/pr/pdf/10.1103/PhysRev.81.385
On page 5 they write above equation (12) the following:
"If we now average over all wave functions, we find that the properly weighted average of $F(\eta)$ is 3/4."
Now, $F(\eta)=1/2 + \frac{1-\eta^2}{4\eta}\ln((1+\eta)/(1-\eta))$.
I don't understand what does it mean to average over wave functions, I thought that they calculated: $\lim_{T\to \infty} \frac{1}{T}\int_0^T F(x)dx$, but I have given maple to calculate this limit (for the additive part without 1/2), and it didn't gave me 1/4.
So I don't understand which average of this function did they calculate?
ANyone knows?
Thanks!
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Sign-changing solutions for some nonhomogeneous nonlocal critical elliptic problems
Departamento de Matemática, Universidad Técnica Federico Santa María, Avenida España 1680, Valparaíso, Chile
$ (-\Delta_{\Omega})^{s} u = \left| u\right| ^{\frac{4}{N-2s}}u +\varepsilon f(x)\quad \mbox{in }\Omega, $
$ \Omega $
$ \mathbb{R}^{N} $
$ N>4s $
$ s\in (0,1] $
$ f\in L^{\infty}(\Omega) $
$ f\geq 0 $
$ f\neq0 $
$ \varepsilon>0 $
$ (-\Delta_{\Omega})^{s} $ spectralfractional Laplacian. We show that the number of sign-changing solutions goes to infinity as
$ \varepsilon\rightarrow 0 $
$ \Omega $
$ f $ Keywords:Fractional Laplacian, sign-changing solution, nonlinear elliptic equation, critical exponent, reduction method. Mathematics Subject Classification:Primary: 35B20, 35B40; Secondary: 35J60, 35B38. Citation:Salomón Alarcón, Jinggang Tan. Sign-changing solutions for some nonhomogeneous nonlocal critical elliptic problems. Discrete & Continuous Dynamical Systems - A, 2019, 39 (10) : 5825-5846. doi: 10.3934/dcds.2019256
References:
[1]
S. Alarcón,
Double-spike solutions for a critical inhomogeneous elliptic problem in domains with small holes,
[2] [3]
B. Barrios, E. Colorado, A. de Pablo and U. Sánchez,
On some critical problems for the fractional Laplacian operator,
[4]
H. Brezis and L. Nirenberg,
Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents,
[5] [6] [7]
D. Cao and H. Zhou,
On the existence of multiple solutions of nonhomogeneous elliptic equations involving critical Sobolev exponents,
[8]
A. Capella,
Solutions of a pure critical exponent problem involving the half-Laplacian in annular-shaped domains,
[9]
A. Capella, J. Dávila, L. Dupaigne and Y. Sire,
Regularity of radial extremal solutions for some non-local semilinear equations,
[10] [11]
W. Choi, S. Kim and K. Lee,
Asymptotic behavior of solutions for nonlinear elliptic problems with the fractional Laplacian,
[12]
M. Clapp, M. del Pino and M. Musso,
Multiple solutions for a non-homogeneous elliptic equation at the critical exponent,
[13] [14]
J. Dávila, M. del Pino and Y. Sire,
Nondegeneracy of the bubble in the critical case for nonlocal equations,
[15] [16]
M. del Pino, P. Felmer and M. Musso,
Two-bubble solutions in the super-critical Bahri-Coron's problem,
[17] [18]
A. A. Kilbas, H. M. Srivastava and J. J. Trujillo, Fractional differential equations: a emergent field in applied and mathematical sciences,
[19]
N. Krall and A. W. Trivelpiece,
Principles of Plasma Physics, Academic Press, New York, London, 1973.
Google Scholar
[20]
M. Musso,
Sign-changing blowing-up solutions for a non-homogeneous elliptic equation at the critical exponent,
[21] [22]
X. Shang, Y. Yang and J. Zhang,
Positive solutions of nonhomogeneous fractional laplacian problem with critical exponent,
[23] [24] [25]
G. Tarantello,
On nonhomogeneous elliptic equations involving critical Sobolev exponent,
[26]
A. Upadhyaya, J.-P. Rieu, J. A. Glazier and Y. Sawada,
Anomalous diffusion and non-Gaussian velocity distribution of Hydra cells in cellular aggregates,
show all references
References:
[1]
S. Alarcón,
Double-spike solutions for a critical inhomogeneous elliptic problem in domains with small holes,
[2] [3]
B. Barrios, E. Colorado, A. de Pablo and U. Sánchez,
On some critical problems for the fractional Laplacian operator,
[4]
H. Brezis and L. Nirenberg,
Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents,
[5] [6] [7]
D. Cao and H. Zhou,
On the existence of multiple solutions of nonhomogeneous elliptic equations involving critical Sobolev exponents,
[8]
A. Capella,
Solutions of a pure critical exponent problem involving the half-Laplacian in annular-shaped domains,
[9]
A. Capella, J. Dávila, L. Dupaigne and Y. Sire,
Regularity of radial extremal solutions for some non-local semilinear equations,
[10] [11]
W. Choi, S. Kim and K. Lee,
Asymptotic behavior of solutions for nonlinear elliptic problems with the fractional Laplacian,
[12]
M. Clapp, M. del Pino and M. Musso,
Multiple solutions for a non-homogeneous elliptic equation at the critical exponent,
[13] [14]
J. Dávila, M. del Pino and Y. Sire,
Nondegeneracy of the bubble in the critical case for nonlocal equations,
[15] [16]
M. del Pino, P. Felmer and M. Musso,
Two-bubble solutions in the super-critical Bahri-Coron's problem,
[17] [18]
A. A. Kilbas, H. M. Srivastava and J. J. Trujillo, Fractional differential equations: a emergent field in applied and mathematical sciences,
[19]
N. Krall and A. W. Trivelpiece,
Principles of Plasma Physics, Academic Press, New York, London, 1973.
Google Scholar
[20]
M. Musso,
Sign-changing blowing-up solutions for a non-homogeneous elliptic equation at the critical exponent,
[21] [22]
X. Shang, Y. Yang and J. Zhang,
Positive solutions of nonhomogeneous fractional laplacian problem with critical exponent,
[23] [24] [25]
G. Tarantello,
On nonhomogeneous elliptic equations involving critical Sobolev exponent,
[26]
A. Upadhyaya, J.-P. Rieu, J. A. Glazier and Y. Sawada,
Anomalous diffusion and non-Gaussian velocity distribution of Hydra cells in cellular aggregates,
[1] [2]
Mateus Balbino Guimarães, Rodrigo da Silva Rodrigues.
Elliptic equations involving linear and superlinear terms and critical
Caffarelli-Kohn-Nirenberg exponent with sign-changing weight functions.
[3]
Yohei Sato, Zhi-Qiang Wang.
On the least energy sign-changing solutions for a nonlinear elliptic system.
[4]
Yanfang Peng, Jing Yang.
Sign-changing solutions to elliptic problems with two critical Sobolev-Hardy exponents.
[5]
Tsung-Fang Wu.
On semilinear elliptic equations involving critical Sobolev exponents and sign-changing weight function.
[6]
Gabriele Cora, Alessandro Iacopetti.
Sign-changing bubble-tower solutions to fractional semilinear elliptic problems.
[7]
Yohei Sato.
Sign-changing multi-peak solutions for nonlinear Schrödinger equations with critical frequency.
[8]
Wei Long, Shuangjie Peng, Jing Yang.
Infinitely many positive and sign-changing solutions for nonlinear fractional scalar field equations.
[9]
Jun Yang, Yaotian Shen.
Weighted Sobolev-Hardy spaces and sign-changing solutions of
degenerate elliptic equation.
[10] [11] [12]
M. Ben Ayed, Kamal Ould Bouh.
Nonexistence results of sign-changing solutions to a supercritical nonlinear problem.
[13]
Guirong Liu, Yuanwei Qi.
Sign-changing solutions of a quasilinear heat equation with a source term.
[14]
Yuxin Ge, Monica Musso, A. Pistoia, Daniel Pollack.
A refined result on sign changing solutions for a critical elliptic problem.
[15]
Yuanxiao Li, Ming Mei, Kaijun Zhang.
Existence of multiple nontrivial solutions for a $p$-Kirchhoff type
elliptic problem involving sign-changing weight functions.
[16]
Wen Zhang, Xianhua Tang, Bitao Cheng, Jian Zhang.
Sign-changing solutions for fourth order elliptic equations with Kirchhoff-type.
[17]
Huxiao Luo, Xianhua Tang, Zu Gao.
Sign-changing solutions for non-local elliptic equations with asymptotically linear term.
[18]
Bartosz Bieganowski, Jaros law Mederski.
Nonlinear SchrÖdinger equations with sum of periodic and vanishing potentials and sign-changing nonlinearities.
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Teodora-Liliana Dinu.
Entire solutions of the nonlinear eigenvalue logistic problem with sign-changing potential and absorption.
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Angela Pistoia, Tonia Ricciardi.
Sign-changing tower of bubbles for a sinh-Poisson equation with asymmetric exponents.
2018 Impact Factor: 1.143
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I have to solve the differential equation for radial disks, and therefore I need the shear force distribution. The disk is rotationally symmetric.
I want to use the superpositions principle and set 3 areas.
$0 \le r \le D_i / 2$
$D_i / 2 < r \le k / 2$
$k/2 < r \le D / 2$
The shear forces are for area 1
$\sum F_z = 0$
$F_{rz1} \pi r - \frac{\pi r^2}{2}p = 0$
$F_{rz1} = \frac{p r}{2}p$
Area 2
$0 = -F_{rz2} \pi r + F_s \pi k $
$F_{rz2} = F_s \frac{k}{r} $
Area 3
$0 = F_{rz3} $
The solution confirms $F_{rz1}$ and $F_{rz3}$ for me. However it states that only after $F_{rz3}$ is known one can solve for $F_{rz2}$ for the area $D_1/2 < r \le D / 2$ The solution should be $F_{rz2} = F_s \frac{k}{2r}$
Did I cut correctly?
What am I missing here?
I know realized I had to insert $k/2$ instead of $k$ and then I get the correct solution for
$0 = -F_{rz2} \pi r + F_s \pi k/2 $
$F_{rz2} = F_s \frac{k}{2r} $
However the question that remains is, why the hint to extend the area? Why is it not possible to calculate $F_{rz2}$ before $F_{rz3}$ when it apparantly is possible?
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I came across the following statement in my course notes:
Consider $R = \left\{f \in \mathbb{Q}[X] \mid f(\mathbb{Z}) \subset \mathbb{Z}\right\}$ the
ring of integer-valued polynomials. Then the ideal generated by $X$ is nota prime ideal.
Isn't this incorrect? I'd argue that $R/(X)$ (where $(X)$ denotes the ideal generated by $X$) is isomorphic to $\mathbb{Z}$, which is an integral domain and thus $(X)$ must be prime.
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Journal of Symbolic Logic J. Symbolic Logic Volume 57, Issue 1 (1992), 172-178. Lusin-Sierpinski Index for the Internal Sets Abstract
We prove that there exists a function $f$ which reduces a given $\Pi^1_1$ subset $P$ of an internal set $X$ of an $\omega_1$-saturated nonstandard universe to the set $\mathbf{WF}$ of well-founded trees possessing properties similar to those possessed by the standard part map. We use $f$ to define the Lusin-Sierpinski index of points in $X$, and prove the basic properties of that index using the classical properties of the Lusin-Sierpinski index. An example of a $\Pi^1_1$ but not $\Sigma^1_1$ set is given.
Article information Source J. Symbolic Logic, Volume 57, Issue 1 (1992), 172-178. Dates First available in Project Euclid: 6 July 2007 Permanent link to this document https://projecteuclid.org/euclid.jsl/1183743898 Mathematical Reviews number (MathSciNet) MR1150932 Zentralblatt MATH identifier 0753.03031 JSTOR links.jstor.org Citation
Zivaljevic, Bosko. Lusin-Sierpinski Index for the Internal Sets. J. Symbolic Logic 57 (1992), no. 1, 172--178. https://projecteuclid.org/euclid.jsl/1183743898
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I am reading the following paper by Slater:
https://journals.aps.org/pr/pdf/10.1103/PhysRev.81.385
On page 5 they write above equation (12) the following:
"If we now average over all wave functions, we find that the properly weighted average of $F(\eta)$ is 3/4."
Now, $F(\eta)=1/2 + \frac{1-\eta^2}{4\eta}\ln((1+\eta)/(1-\eta))$.
I don't understand what does it mean to average over wave functions, I thought that they calculated: $\lim_{T\to \infty} \frac{1}{T}\int_0^T F(x)dx$, but I have given maple to calculate this limit (for the additive part without 1/2), and it didn't gave me 1/4.
So I don't understand which average of this function did they calculate?
ANyone knows?
Thanks!
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Difference between revisions of "Main Page"
(Fixing location.)
(Collected background materials)
Line 5: Line 5:
'''<math>k=3</math> Density Hales-Jewett (DHJ(3)) theorem:''' <math>\lim_{n \rightarrow \infty} c_n/3^n = 0</math>
'''<math>k=3</math> Density Hales-Jewett (DHJ(3)) theorem:''' <math>\lim_{n \rightarrow \infty} c_n/3^n = 0</math>
−
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular [http://gowers.wordpress.com/2009/02/01/a-combinatorial-approach-to-density-hales-jewett/ combinatorial approach to DHJ, suggested by Tim Gowers.] Some background to
+
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular [http://gowers.wordpress.com/2009/02/01/a-combinatorial-approach-to-density-hales-jewett/ combinatorial approach to DHJ, suggested by Tim Gowers.]
+ + + +
Some background to project can be [http://gowers.wordpress.com/2009/01/30/background-to-a-polymath-project/ found here]discussion on massively collaborative "polymath" projects can be [http://gowers.wordpress.com/2009/01/27/is-massively-collaborative-mathematics-possible/ found here]. A cheatsheet for editing the wiki may be [http://meta.wikimedia.org/wiki/File:MediaWikiRefCard.png found here].
== Threads ==
== Threads ==
Line 32: Line 36:
# K. O'Bryant, "[http://arxiv.org/abs/0811.3057 Sets of integers that do not contain long arithmetic progressions]", preprint.
# K. O'Bryant, "[http://arxiv.org/abs/0811.3057 Sets of integers that do not contain long arithmetic progressions]", preprint.
# R. McCutcheon, “[http://www.msci.memphis.edu/~randall/preprints/HJk3.pdf The conclusion of the proof of the density Hales-Jewett theorem for k=3]“, unpublished.
# R. McCutcheon, “[http://www.msci.memphis.edu/~randall/preprints/HJk3.pdf The conclusion of the proof of the density Hales-Jewett theorem for k=3]“, unpublished.
− − − − Revision as of 07:17, 13 February 2009 The Problem
Let [math][3]^n[/math] be the set of all length [math]n[/math] strings over the alphabet [math]1, 2, 3[/math]. A
combinatorial line is a set of three points in [math][3]^n[/math], formed by taking a string with one or more wildcards [math]x[/math] in it, e.g., [math]112x1xx3\ldots[/math], and replacing those wildcards by [math]1, 2[/math] and [math]3[/math], respectively. In the example given, the resulting combinatorial line is: [math]\{ 11211113\ldots, 11221223\ldots, 11231333\ldots \}[/math]. A subset of [math][3]^n[/math] is said to be line-free if it contains no lines. Let [math]c_n[/math] be the size of the largest line-free subset of [math][3]^n[/math]. [math]k=3[/math] Density Hales-Jewett (DHJ(3)) theorem: [math]\lim_{n \rightarrow \infty} c_n/3^n = 0[/math]
The original proof of DHJ(3) used arguments from ergodic theory. The basic problem to be consider by the Polymath project is to explore a particular combinatorial approach to DHJ, suggested by Tim Gowers.
Useful background materials
Some background to the project can be found here. General discussion on massively collaborative "polymath" projects can be found here. A cheatsheet for editing the wiki may be found here. Finally, here is the general Wiki user's guide
Threads (1-199) A combinatorial approach to density Hales-Jewett (inactive) (200-299) Upper and lower bounds for the density Hales-Jewett problem (active) (300-399) The triangle-removal approach (inactive) (400-499) Quasirandomness and obstructions to uniformity (inactive) (500-599) Possible proof strategies (active) (600-699) A reading seminar on density Hales-Jewett (active)
We are also collecting bounds for Fujimura's problem.
Here are some unsolved problems arising from the above threads.
Bibliography M. Elkin, "An Improved Construction of Progression-Free Sets ", preprint. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem for k=3“, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–241. H. Furstenberg, Y. Katznelson, “A density version of the Hales-Jewett theorem“, J. Anal. Math. 57 (1991), 64–119. B. Green, J. Wolf, "A note on Elkin's improvement of Behrend's construction", preprint. K. O'Bryant, "Sets of integers that do not contain long arithmetic progressions", preprint. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished.
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4:28 AM
@MartinSleziak Here I am! Thank you for opening this chat room and all your comments on my post, Martin. They are really good feedback to this project.
@MartinSleziak Yeah, using a chat room to exchange ideas and feedback makes a lot of sense compared to leaving comments in my post. BTW. Anyone finds a
\oint\frac{1}{1-z^2}dz expression in old posts? Send to me and I will investigate why this issue occurs.
@MartinSleziak It is OK, don't feel anything bad. As long as there is a place that comes to people's mind if they want to report some issue on Approach0, I am willing to come to that place and discuss. I am really interested in pushing Approach0 forward.
4:57 AM
Hi @WeiZhong thanks for joining the room. I will write a bit more here when I have more time. For now two minor things.
I just want to make sure that you know that the answer on meta is community wiki. Which means that various users are invited to edit it, you can see from revision history who added what to the question.
You can see in revision history that this bullet point was added by Workaholic: "I searched for
\oint $\oint$, but I only got results related to
\int $\int$. I tried for
\oint \frac{dz}{1-z^2} $\oint \frac{dz}{1-z^2}$ which is an integral that appears quite often but it did not yield any correct results."
So if you want to make sure that this user is notified about your comments, you can simply add @Workaholic. Any of the editors can be pinged.
And I noticed also this about one of the quizzes (I did not check whether some of the other quizzes have similar problem.)
I suppose that the quizzes are supposed to be chosen in such way that Approach0 indeed helps to find the question. I.e., each quiz was created with some specific question in mind, which should be among the search results. Is that correct?
I guess the quiz saying "Please list all positive integers $i,j,k$ such that $i^5 + j^6 = k^7$." was made with this question in mind: Find all positive integers satisfying: $x^5+y^6=z^7$.
However when I try the query from this quiz, I get completely different results.
I vaguely recall that I tried some quizzes, including this one, and they worked. (By which I mean that the answer to the question from the quiz could be found among the search results.) So is this perhaps due to some changes that were made since then? Or is that simply because when I tried the quiz last time, less questions were indexed. (And now that question is still somewhere among the results, but further down.)
I was wondering whether to add the word to my last message, but it is probably not a bug. It is simply that search results are not exactly as I would expect.
My impression from the search results is that not only x, y, z are replaced by various variables, but also 5,6,7 are replaced by various numbers.
5:40 AM
I think that this implicitly contains a question whether when searching for $x^5+y^6=z^7$ also the questions containing $x^2+y^2=z^2$ or $a^3+b^3=c^3$ should be matches.
For the sake of completeness I will copy here the part of quiz list which is relevant to the quiz I mentioned above:
"Q": "Please list all positive integers [imath]i,j,k[/imath] such that [imath]i^5 + j^6 = k^7[/imath]. ",
Hmm, I should have posted this as a single multiline message. But now I see that it is already too late to delete the above messages. Sorry for the duplication:
{ /* 4 */
"Q": "Please list all positive integers [imath]i,j,k[/imath] such that [imath]i^5 + j^6 = k^7[/imath]. ", "hints": [ "This should be easy, the only thing I need to do is do some calculation...", "I can use my computer to enumerate...", "... (10 minutes after) ...", "OK, I give up. Why borther list them <b>all</b>?", "Is that possible to <a href=\"#\">search it</a> on Internet?" ], "search": "all positive integers, $i^5 + j^6 = k^7$" },
"Q": "Please list all positive integers [imath]i,j,k[/imath] such that [imath]i^5 + j^6 = k^7[/imath]. ",
"hints": [
"This should be easy, the only thing I need to do is do some calculation...",
"I can use my computer to enumerate...",
"... (10 minutes after) ...",
"OK, I give up. Why borther list them <b>all</b>?",
"Is that possible to <a href=\"#\">search it</a> on Internet?"
],
"search": "all positive integers, $i^5 + j^6 = k^7$"
},
8 hours later…
1:19 PM
@MartinSleziak OK, I get it. So next time I would definitely reply to whom actually makes the revision.
@MartinSleziak Yes, remember the first time when we talk in a chat room? At that version of approach0, when a very limited posts have been indexed, you can actually get relevant posts on $i^5+j^6=k^7$. However, when I has enlarged the index (now almost the entire MSE), that piece of quiz (in fact, some quiz I selected earilier like [this one]()) does not find relevant posts anymore.
I have noticed that "quiz" does not work, but I am really lazy and have not investigated it. Instead of change that "quiz", I agree to investigate on why that relevant result has gone. As far as I can guess, there can be two reasons:
1) the crawler missed that one (I did the crawling in China, the network condition is not always good, sometimes crawler fails to fetch random posts and have to skip them) 2) there is a bug in approach0 that I am not aware
1) the crawler missed that one (I did the crawling in China, the network condition is not always good, sometimes crawler fails to fetch random posts and have to skip them)
2) there is a bug in approach0 that I am not aware
In order to investigate this problem, I am trying to find the original posts that you and me have seen (as you remember vaguely) which is relevant to $i^5+j^6=k^7$ quiz, if you find that post, please send me the URL.
@MartinSleziak It can be a bug, but I need to know if my index does contain a relevant post, so first let us find that post we think relevant. And I will have a look whether or not it is in my index, perhaps the crawler just missed that one. If it is in our index currently, then I should spend some time to find out the reason.
@MartinSleziak As for you last question, I need to illustrate it a little more. Approach0 will first find expressions that are
structurallyrelevant to query. So $x^5+y^6=z^7$ will get you $x^2+y^2=z^2$ or $a^3+b^3=c^3$, because they (more specifically, their operator tree representation) are considered structurally identical.
After filtering out these structurally relevant expressions, Approach0 will evaluate their symbolic relevance degree with regarding to query expression. Suppose $x^5+y^6=z^7$ gives you $x^2+y^2=z^2$, $a^3+b^3=c^3$ and also $x^5+y^6=z^7$, expression $x^5+y^6=z^7$ will be ranked higher than $x^2+y^2=z^2$ and $a^3+b^3=c^3$, this is because $x^5+y^6=z^7$ has higher symbolic score (in fact, since it has identical symbol set to query, it has the highest possible symbolic score).
I am sorry, I should use "and" instead of "or". Let me repeat the message before previous one below:
As for you last question, I need to illustrate it a little more. Approach0 will first find expressions that are
structurallyrelevant to query. So $x^5+y^6=z^7$ will get you both$x^2+y^2=z^2$ and$a^3+b^3=c^3$, because they (more specifically, their operator tree representation) are considered structurally identical.
Now the next things for me to do is to investigate some "missing results" suggested by you.
1. Try to find `\oint` expression in an old post (by old I mean at least 5 weeks old, so that it is possible been indexed)
1. Try to find `\oint` expression in an old post (by old I mean at least 5 weeks old, so that it is possible been indexed)
2:23 PM
Unfortunately, I fail to find any relevant old post in neither case 1 nor case 2 after a few tries (using MSE default search). So the only thing I can do now is to do an "integrated test" (see the new code I have just pushed to Github: github.com/approach0/search-engine/commit/…)
An "integrated test" means I make a minimal index with a few specified math expressions and search a specified query, and see if the results is expected. For example, the test case tests/cases/math-rank/oint.txt specified the query $\oint \frac{dz}{1-z^2}$, and the entire index has just two expressions: $\oint \frac{dz}{1-z^2}$ and $\oint \frac{dx}{1-x^2}$, and the expected search result is both these two expressions are HIT (i.e. they should appear in search result)
10 hours ago, by Martin Sleziak
I guess the quiz saying "Please list all positive integers $i,j,k$ such that $i^5 + j^6 = k^7$." was made with this question in mind: Find all positive integers satisfying: $x^5+y^6=z^7$.
2:39 PM
For anyone interested, I post the screenshot of integrated test results here: imgur.com/a/xYBD5
3:04 PM
For example like this: chat.stackexchange.com/transcript/message/32711761#32711761 You get the link by clicking on the little arrow next to the message and then clicking on "permalink".
I am mentioning this because (hypothetically) if Workaholic only sees your comment a few days later and then they come here to see what the message you refer to, they might have problem with finding it if there are plenty of newer messages.
However, this room does not have that much traffic, so very likely this is not going to be a problem in this specific case.
Another possible way to linke to a specific set of messages is to go to the transcript and then choose a specific day, like this: chat.stackexchange.com/transcript/46148/2016/10/1
Or to bookmark a conversation. This can be done from the room menu on the right. This question on meta.SE even has some pictures.
This is also briefly mentioned in chat help: chat.stackexchange.com/faq#permalink
3:25 PM
@MartinSleziak Good to learn this. I just posted another comment with permalink in that meta post for Workaholic to refer.
I just checked the index on server, yes, that post is indeed indexed. (for my own reference, docID = 249331)
2 hours later…
5:13 PM
Update: I have fixed that quiz problem. See: approach0.xyz/search/…
That is not strictly a bug, it is because I put a restriction on the number of document to be searched in one posting list (not trying to be very technical). I have pushed my new code to GitHub (see commit github.com/approach0/search-engine/commit/…), this change gets rid of that restriction and now that relevant post is shown as the 2nd search result.
2 hours later…
6:57 PM
« first day (2 days earlier) next day → last day (1104 days later) »
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I am reading the following paper by Slater:
https://journals.aps.org/pr/pdf/10.1103/PhysRev.81.385
On page 5 they write above equation (12) the following:
"If we now average over all wave functions, we find that the properly weighted average of $F(\eta)$ is 3/4."
Now, $F(\eta)=1/2 + \frac{1-\eta^2}{4\eta}\ln((1+\eta)/(1-\eta))$.
I don't understand what does it mean to average over wave functions, I thought that they calculated: $\lim_{T\to \infty} \frac{1}{T}\int_0^T F(x)dx$, but I have given maple to calculate this limit (for the additive part without 1/2), and it didn't gave me 1/4.
So I don't understand which average of this function did they calculate?
ANyone knows?
Thanks!
|
This Lecture summarizes some well known facts about $\#P$ completeness of permanent.
Given a CNF formula $\phi$ on $n$ variables, they construct matrix $A$ such that:
$$perm(A)=4^{3m} \#SAT(\phi)$$
This gives easy upper bound on $perm(A)$.
$\phi$ is unsatisfiable iff $perm(A)=0$, so the decision problem "Is the permanent zero (ZP)" is NP-complete.
The special case "Is the permanent of (0,1) matrix zero? (ZP01)" is polynomial.
The permanent of (0,1) matrix counts the number of vertex disjoint cycle covers of the digraph defined by this adjacency matrix, so the permanent is zero iff if the digraph doesn't have vertex disjoint cycle covers, which is polynomial.
Then on p.5 they give reduction:
Thus, for an integer matrix $A$, there exists an (0, 1)-matrix $B$, such that, $perm(B) = perm(A)\pmod{Q}$.
Soon after, remarks stops a naive attack of ZP01.
$perm(B)=0$ doesn't imply $perm(A)=0$, it implies $perm(A)=QN$ for integer $N$.
Choose $Q$ larger than the upper bound for $perm(A)$ and check if $perm(B)=0$. If this doesn't hold (cycle cover exists), then $perm(A) \ne 0$ so $\phi$ is satisfiable.
If it holds, $perm(A)=QN$ for integer $N$.
Since $Q$ is larger than the upper bound, this means $N=0$ and $perm(A)=0$, so $\phi$ is unsatisfiable.
Everything is polynomial.
What is wrong with this?
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Let's assume that we have measured the height of each member in our population below:
Figure 1 - The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm. Images taken from http://www.mathsisfun.com/data/standard-deviation.html
We'll put these in a numpy array so we can use them later on:
import numpy as npheights = np.array([600, 470, 170, 430, 300])
Numpy has many useful built-in functions. One of these calculates the standard deviation for any set of data, below you can see this in action:
print("standard deviation is {}".format(np.std(heights)))
standard deviation is 147.32277488562318
As always, it's recommended you use these well-maintained functions instead of writing your own. However, with this population size being so small, we can get a better understanding of how the standard deviation is calculated if we do it ourselves.
Let's have a look at the formula: $$\sigma = \sqrt{\frac{\Sigma{(x - \mu)^2}}{N}}$$
To those of you who aren't comfortable with mathematics, this can be overwhelming. But let's break it down and calculate everything step-by-step:
First we want to find $\mu$ which is just the mean of the population values. The mean is just all the values summed together and then divided by the population size. For our population, this would be:
population_mean = np.sum(heights) / heights.sizeprint("population mean is: {}".format(population_mean))
population mean is: 394.0
We can find the mean using a built-in numpy function too:
population_mean = np.mean(heights)print("population mean is: {}".format(population_mean))
population mean is: 394.0
Great - now we have the mean, let's plot it on our graph in green:
Figure 2 - The population mean (green) plotted on our graph.
Now we want to calculate $(x - \mu)$, which is difference between each height and the mean:
Figure 3 - Difference between each height and the mean.
In Python + numpy, we can calculate this:
height_differences = heights - population_meanprint("height differences: {}".format(height_differences))
height differences: [ 206. 76. -224. 36. -94.]
Now to get the variance which is this part: $$\frac{\Sigma{(x - \mu)^2}}{N}$$
To do this, we first square all our height differences above, and get the average of them.
variance = (height_differences**2) variance = np.sum(variance)variance = variance / height_differences.sizeprint("variance: {}".format(variance))
variance: 21704.0
Again, we can find the variance using a built-in numpy function too:
variance = np.var(height_differences)print("variance: {}".format(variance))
variance: 21704.0
Finally, the last part is to find the square root of our variance, which is the standard deviation:$$\sigma = \sqrt{\frac{\Sigma{(x - \mu)^2}}{N}}$$
standard_deviation = np.sqrt(variance)print("standard deviation is {}".format(standard_deviation))
standard deviation is 147.32277488562318
When we plot this standard deviation on our graph we get the following:
Figure 4 - Standard deviation (purple) plotted on our graph.
With the standard deviation, we can suggest which heights are within our standard deviation (147mm) of the mean. So now there's an approach to knowing what is normal, what is extra large, or extra small.
If there data we're working with is a sample taken from a larger population, then there's a slight difference in calculating the standard deviation.
Instead of: $$\sigma = \sqrt{\frac{\Sigma{(x - \mu)^2}}{N}}$$
We use: $$s = \sqrt{\frac{\Sigma{(x - \bar{x})^2}}{n - 1}}$$
where $\bar{x}$ is just the sample mean, and $n - 1$ is the sample size minus 1.
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We write the group operation multiplicatively.Let $x, y\in T(A)$. Then $x, y$ have finite order, hence there exists positive integers $m, n$ such that $x^m=e, y^n=e$, where $e$ is the identity element of $A$. Then we have\begin{align*}(xy)^{mn}&=x^{mn}y^{mn} \qquad \text{ (since $A$ is abelian)}\\&=(x^m)^n(y^m)^n=e^me^n=e.\end{align*}Therefore the element $xy$ has also finite order, hence $xy \in T(A)$.
Also, we have\begin{align*}(x^{-1})^m=(x^m)^{-1}=e^{-1}=e.\end{align*}Hence the inverse $x^{-1}$ of $x$ has finite order, hence $x^{-1}\in T(A)$.
Therefore, the subset $T(A)$ is closed under group operation and inverse, hence $T(A)$ is a subgroup of $A$.
(b) $A/T(A)$ is a torsion-free abelian group
Since $A$ is an abelian group, the quotient $G=A/T(A)$ is also an abelian group.For $a\in A$, let $\bar{a}=aT(A)$ be an element of $G=A/T(A)$. Suppose that $\bar{a}$ has finite order in $G$. We want to prove that $\bar{a}=\bar{e}$ the identity element of $G$.
Since $\bar{a}$ has finite order, there exists a positive integer $n$ such that\[\bar{a}^n=\bar{e}.\]This implies that\[a^nT(A)=T(A)\]and thus $a^n\in T(A)$.
Since each element of $T(A)$ has finite order by definition, there exists a positive integer $m$ such that $(a^n)^m=e$.It follows from $a^{nm}=e$ that $a$ has finite order, and thus $a\in T(A)$.Therefore we have\[\bar{a}=aT(A)=T(A)=\bar{e}.\]
We have proved that any element of $G=A/T(A)$ that has finite order is the identity, hence $G$ is the torsion-free abelian subgroup of $G$.
Quotient Group of Abelian Group is AbelianLet $G$ be an abelian group and let $N$ be a normal subgroup of $G$.Then prove that the quotient group $G/N$ is also an abelian group.Proof.Each element of $G/N$ is a coset $aN$ for some $a\in G$.Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]
Pullback Group of Two Group Homomorphisms into a GroupLet $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms.Define the subset $M$ of $G_1 \times G_2$ to be\[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\]Prove that $M$ is a subgroup of $G_1 \times G_2$.[…]
Elements of Finite Order of an Abelian Group form a SubgroupLet $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is,\[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\]Prove that $H$ is a subgroup of $G$.Proof.Note that the identity element $e$ of […]
Normal Subgroups, Isomorphic Quotients, But Not IsomorphicLet $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.Proof.We give a […]
Commutator Subgroup and Abelian Quotient GroupLet $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.Let $N$ be a subgroup of $G$.Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.Definitions.Recall that for any $a, b \in G$, the […]
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Table of Contents
Internal Direct Products Isomorphic To External Direct Products
Recall from The Internal Direct Product of Two Groups page that if $(G, *)$ is a group and $(H, *)$, $(K, *)$ are two subgroups of $(G, *)$ then we say that $(G, *)$ is the internal direct product of $(H, *)$ and $(K, *)$ if the following three conditions are met:
$G = \{ h * k : h \in H, k \in K \}$. $H \cap K = \{ e \}$ where $e$ is the identity element in $G$. $h * k = k * h$ for all $h \in H$ and for all $k \in K$.
We will now look at a theorem which tells us that if $(G, *)$ is the internal direct product of $(H, *)$ and $(K, *)$ then $(G, *)$ is isomorphic to the external direct product $(H \times K, \cdot)$.
Theorem 1: Let $(G, *)$ be a group and let $(H, *)$, $(K, *)$ be subgroups of $(G, *)$. If $(G, *)$ is the internal direct product of $(H, *)$ and $(G, *)$ then $G \cong H \times K$. Proof:Define a function $f : H \times K \to G$ for all $(h, k) \in H \times K$ by: We need show that $f$ is bijective. For injectivity, let $(h_1, k_1), (h_2, k_2) \in H \times K$ and suppose that $f(h_1, k_1) = f(h_2, k_2)$. Then: Thus: But $(H, *)$ and $(K, *)$ are groups and are closed under $*$ so $k_2^{-1} * k_1 \in K$ and $h_2 * h_1^{-1} \in H$. Since $H \cap K = \{ e \}$ this must mean that the equality at $(*)$ implies that $k_2^{-1} * k_1 = e$ and $h_2 * h_1^{-1} = e$. So $h_1 = h_2$ and $k_1 = k_2$ so $(h_1, k_1) = (h_2, k_2)$, i.e., $f$ is injective. We now show that $f$ is surjective. Let $g \in G$. Since $G$ is an interal direction product of $H$ and $K$ we have that $G = \{ h * k : h \in H, k \in K \}$ so $g = h * k$ for some $h \in H$ and for some $k \in K$. So $(h, k) \in H \times K$ is such that: Hence $f$ is surjective. Lastly, we see that: Since $G$ is an internal direct product of $H$ and $K$ we see that $h_2 * k_1 = k_1 * h_2$ so: Therefore $f$ is an isomorphism between $(G, *)$ and $(H \times K, \cdot)$ so $(G, *) \cong (H \times K, \cdot)$. $\blacksquare$
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How big of an engine would I need, if the original is 65hp? What would be the weight of the up-scaled plane? Simply twice the weight? What would be the stall speed? How fast would it go? 75mph? Is there a software program to test aircraft design?
Unfortunately a simple scale-up would fall victim to the square-cube law: all dimensions times two means the volume and weight are 2x2x2 = 8 times the original, while the wing area is 2x2 = 4 times the original. It is the wing area that keeps the aeroplane up. It is the reason why birds cannot get any larger than they already are.
Someone in the 1920s analysed this and predicted that the maximum size for aeroplanes would be about that of the DC-3, but improved technology has proved that prediction wrong. A B777 has a way higher wing loading than a DC-3 of course. The square-cube law is
not a law, it can be defeated by engineering. Point is that linearly amplifying all dimensions does not work.
How big of an engine would I need, if the original is 65hp?
Minimal: 8 x 65hp = 520 hp. We need a serious amount of extra thrust.
From this answer:
$$D = C_{D0} \cdot \frac{1}{2} \rho V^2 \cdot S + \frac{2 W^2}{\pi A e \cdot \rho V^2 \cdot S} \cdot $$
$C_{D0}$ would stay the same if everything is scaled up S = wing area = 4 times as high W is 8 times as high.
$D_2 = D_1 \cdot (4 + 8^2/4) = D_1 \cdot 20$
So you need 20 times as much thrust at the same speed.
Increase V and the first factor increases, second factor decreases. There are of course many optimisations possible (weight saving, wing loading increase etc), but that goes beyond the scope of the question.
What would be the weight of the up-scaled plane? Simply twice the weight?
Eight times that of the original.
What would be the stall speed?
Higher than that of the original because the wing loading is higher. Depends on the flaps & slats installed.
How fast would it go? 75mph?
Yeah it might..but it would probably need to go faster as @xxavier mentions.
Is there a software program to test aircraft design?
FlightGear is an open source flight simulator that allows you to link your own flight dynamics model into it.
In general, the power required by an airplane is proportional to its weight times its speed. The weight scales with the cube of the linear dimension, so the bigger J3 would weigh $ 2^3 = 8 $ times more. Since the airspeed is squared in the formula for lift, the exponent for the proportionality of lift should be 1/2.
Now you have two powers for the scaling, 3 for the weight (or mass, in this case) and 1/2 for the speed. As the weight and speed are multiplied in order to get the proportionality of the power required, I add the exponents, and the result is that the bigger J3 plane would need an engine of $ 2^{3.5} = 11.3 \times 65 = 735 $ hp.
Concerning the stall speed, and as airspeed scales with a power of 1/2, and the scale factor is 2, if the 'small' J3 has a stall speed of –say– 50 mph, you'll have a stall speed of $ 50 \times \sqrt{2} = 71 $ mph for the 'big one'... The cruise and maximum speeds will also scale with $ \sqrt{2} $...
Of course, all these are rough estimations... To start with, the weight does not grow
exactly with the 3rd power of the linear dimension...
Yes, it would fly.
Scaling up a successful design has worked well in the past. Just two examples:
When Howard Hughes wanted a fast airliner for his TWA airline, he turned to Lockheed and asked them to design one on the basis of the wing of their P-38 fighter. The result was the Lockheed Constellation (the picture below shows the military version C-69). In the summer of 1940, Willy Messerschmitt was asked to design a cargo glider with the payload bay the size of a Reichsbahn cargo car. He used the largest aircraft he had designed so far, the M-20, and scaled this up. The result was the Me-321, which was a considerably better design than the analog Junkers design Ju-322, which was designed from scratch and turned out to be uncontrollable.
What would be the weight of the up-scaled plane? Simply twice the weight?
No, it will be much heavier. The structure will have to be beefed up for the same reason that an elephant needs big, sturdy legs while an ant can use spindly legs and still carry a multiple of its own weight. Normally, this factor is the length factor by the power of 2.3 or 2.4. Since the aircraft will be hollow on the inside, simple cube laws which apply for solids will give too high a value.
If you look at existing aircraft, you will find that wing loading increases with size. That makes sense: The wing of an aircraft twice as big as the original will have four times its area. Its mass, however, will be five times bigger according to the exponential weight growth.
Engine size should be selected so the power to weight ratio is maintained. Your twice-as-big Piper J-3 will need 320 hp to be useable.
Since it is bigger and has a higher wing loading, the upscaled J-3 will also fly a bit faster. Stall speed will be 10 - 12% higher due to the higher wing loading. Friction drag will be smaller relative to its wetted area since it flies at a higher Reynolds number. The higher wing loading will again result in higher maximum flight speed, if only because the optimum point of lowest drag is at a higher dynamic pressure. But don't expect too much: I would expect that the top speed is just 8% higher. Maybe 10% if you are lucky.
Is there a software program to test aircraft design?
Back in 1973, Air Progress magazine published an article on a proposed three times scaled-up Piper J-3 Cub with 500 hp, original top speed, 200 mile range, etc. You can find the article at:
The Piper Cub J3 is an excellent place to start in aircraft design. It represents all that was learned in the first 40 years of aircraft design as a tractor driven high wing mono plane with a properly proportioned and placed vertical and horizontal control surfaces. It is no accident this was the first trainer aircraft for many aviators.
Undoubtedly this design can be scaled up, but beyond the actual build it may be beneficial to look for the J3's attributes in other larger aircraft. The high wing gives improved stability in pitch and roll. This is a plane that wants to stay "upright". Now look at the 132 foot wingspan C130 Hercules cargo plane. Yes, a lot of Piper Cub there. And that large tail keeps it steady and can handle a wide range of CG.
So, for your scale up, plan on needing more power, and high lift devices such as slats and flaps to keep landing speed at a minimum. Also, evaluate strengthening requirements for fuselage and especially the wing to bear the heavier loads. And check out existing designs in the size range you desire.
And if you are really motivated, here is another Hercules, the Hughes H-4, aka "Spruce Goose". At 320 feet, it's wingspan is more than 9 times that of the J3. See if you can pick out the similarities.
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