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Products of Linear Spaces
Definition: Let $X$ and $Y$ be linear spaces. The Product $X \times Y$ is defined to be the space with the operators of addition and scalar multiplication defined for all $(x_1, y_1), (x_2, y_2) \in X \times Y$ and $\alpha \in \mathbb{R}$ by $(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2)$ and $\alpha (x_1, y_1) = (\alpha x_1, \alpha y_1)$.
It can be shown without too much trouble that if $X$ and $Y$ are linear spaces then their product $X \times Y$ is also a linear space.
Proposition 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. Define a norm $\| \cdot \| : X \times Y \to [0, \infty)$ for all $(x, y) \in X \times Y$ by $\| (x, y) \| = \| x \|_X + \| y \|_Y$. Then $(X \times Y, \| \cdot \|)$ is a normed linear space. There are a few other ways in which we can define a norm on $X \times Y$. For example, we could define $\| (x, y) \| = \max \{ \| x \|_X, \| y \|_Y \}$. Proof:We only need to show that $\| \cdot \|$ is a norm on $X$. First, suppose that $\| (x, y) \| = 0$. Then $\| x \|_X + \| y \|_Y = 0$. A sum of two nonnegative values equaling zero implies that both values equal $0$. That is, $\| x \|_X = 0$ and $\| y \|_Y = 0$. But this implies that $x = 0$ and $y = 0$. So $(x, y) = (0, 0)$. Conversely, suppose that $(x, y) = (0, 0)$. Then $\| (x, y) \| = \| (0, 0) \| = \| 0 \|_X + \| 0 \|_Y = 0$. Now let $\alpha \in \mathbb{R}$. Then: Lastly, let $(x, y), (z, w) \in X \times Y$. Then: Therefore, $\| \cdot \|$ is a norm on $X \times Y$. $\blacksquare$
Proposition 1: Let $(X, \| \cdot \|_X)$ be a normed linear space. Then: a) The function of addition $+ : X \times X \to X$ is continuous on $X \times X$. b) The function of scalar multiplication $\cdot : \mathbb{R} \times X \to X$ is continuous on $\mathbb{R} \times X$. Recall the sequential criterion for continuity: A function $f : X \to Y$ is continuous at $x$ if and only if for all sequences $(x_n)$ in $X$ that converges to $x$ we have that $(f(x_n))$ converges to $f(x)$. Proof of a)Let $(x_n, x_n')$ be a sequence in $X \times X$ that converges to $(y, y')$. Then $(x_n)$ and $(x_n')$ are sequences in $X$ that converge to $y$ and $y'$ respectively. So $(x_n + x_n')$ converges to $y + y'$. So $+$ is continuous on $X \times X$. Proof of b)Let $(\alpha_n, x_n)$ be a sequence in $\mathbb{R} \times X$ that converges to $(\alpha, x)$. Then $(\alpha_n)$ is a sequence in $\mathbb{R}$ that converges to $\alpha$ and $(x_n)$ is a sequence in $X$ that converges to $X$. So $(\alpha_nx_n)$ converges to $\alpha x$, which shows that $\cdot$ is continuous on $\mathbb{R} \times X$. $\blacksquare$
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Question
A heart defibrillator being used on a patient has an $RC$ time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an $8.00 \mu \textrm{F}$ capacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to decline to $6.00 \times 10^2 \textrm{ V}$?
Final Answer
$1.25 \textrm{ k}\Omega$ $30.0 \textrm{ ms}$ Calculator Screenshots
Video Transcript
This is College Physics Answers with Shaun Dychko. The time constant for this heart defibrillator is ten milliseconds, that’s ten times ten to the minus three seconds. And it has a capacitance of eight micro Farads. Now the final resistance of the RC Circuit in this defibrillator will be the time constant equals resistance multiplied by the capacitance and we’ll divide both sides by
Cto solve for R. So Ris tau over C. So that’s ten times ten to the minus three seconds divided by six times ten to the minus six Farads giving us 1.25 kilo Ohms. Part b says given some initial voltage of 12 kilo Volts, how long will it take to reach the voltage of six times ten to the two Volts? So, this is the formula for the discharging capacitor. And we’ll divide both sides by V naught, the starting voltage. And so eto the power negative time over time constant equals the final voltage divided by the initial voltage. And then take the natural log rhythm of both sides and in which case the left side becomes negative tover tau. On the right side is natural log rhythm of Vover V naught. And then multiplied both sides by negative tau and you get the time. So it’s the negative of the time constant, ten times ten to the minus three seconds times the natural log rhythm of times six times ten to the two Volts divided by 12 times ten to the three Volts, giving us 30.0 milliseconds.
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I am studying numerical methods for differential equations. I came accros the trapezoidal method in two forms, an explicit and an iterative one. I would like to know the advantages and disadvantages of each of those methods. Furthermore, how can I study the stability for the iterative method? Which stability definition is the better one and why?. I explain both methods below.
Consider an initial value problem given by $y' = f(t, y)$ and $y(a) = t_0$, where $f$ is defined in $[a, b]\times[\alpha, \beta]$ and satisfies the Lipschitz property with Lipschitz's constant $L$.
Given a natural number $n$ and $h = \frac{b-a}{n}$, we are trying to approximate the unique solution of the problem at $t_i = a + ih \ \forall i = 0, 1 \ldots n$. If $y$ is the solution, we call $y_i = y(t_i)$ and we denote $w_i$ to the approximations obtained by the applied method.
One of the methods studied is the explicit trapezoidal method. It follows the following rule:
$w_{i+1} = w_i + \frac{h}{2} \left[f(t_i,w_i) + f(t_{i}+h, w_i + h f(t_i,w_i))\right]$
We have proved that it has a local error of order 3 and, hence, a global error of order 2.
Then, reading some books I came accros the iterative trapezoidal method, which solves the following implicit equation:
$ w_{i}= w_{i-1} + \frac{h}{2} \left[f(t_{i-1}, w_{i-1}) + f(t_i, w_{i})\right] $
The idea is taking an initial approximation $w_i^{(0)}$ and defining the following sequence:
$w_{i} ^{(j+1)} = w_{i-1} + \frac{h}{2} \left[f(t_{i-1}, w_{i-1}) + f(t_i, w_{i}^{(j)})\right]$
The limit of that sequence is taken as $w_i$. I have proved that the sequence converges if $hL/2 < 1$ and that if we use $w_i$, then the local error is $O(h^3)$. However, why is this method useful and how can I study its stability?
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Di Filippo, Domenico (2013)
The Charged anti-counter for the NA62experiment at CERN SPS. [Tesi di dottorato]
Item Type: Tesi di dottorato Lingua: English Title: The Charged anti-counter for the NA62experiment at CERN SPS Creators:
Creators Email Di Filippo, Domenico domenicodifilippo@gmail.com Date: 2 April 2013 Number of Pages: 117 Institution: Università degli Studi di Napoli Federico II Department: Fisica Scuola di dottorato: Scienze fisiche Dottorato: Fisica fondamentale ed applicata Ciclo di dottorato: 25 Coordinatore del Corso di dottorato:
nome email Velotta, Raffaele velotta@na.infn.it Tutor:
nome email Ambrosino, Fabio Fabio.Ambrosino@na.infn.it Date: 2 April 2013 Number of Pages: 117 Uncontrolled Keywords: NA62 CHANTI design MonteCarlo Inelastic Background [error in script] [error in script] Settori scientifico-disciplinari del MIUR: Area 02 - Scienze fisiche > FIS/01 - Fisica sperimentale Area 02 - Scienze fisiche > FIS/04 - Fisica nucleare e subnucleare Date Deposited: 07 Apr 2013 14:24 Last Modified: 21 Oct 2014 09:19 URI: http://www.fedoa.unina.it/id/eprint/9390
Abstract
The NA62 experiment at CERN is intended to measure, with 10% relative error, the Branching Ratio of the K decay $K^{+}\rightarrow\pi^{+}\nu\bar{\nu}$ which is expected to occur only in about 8 out $10^{11}$ kaon decays.The measurement strategy is to detect the decaying kaon and its product $\pi^{+}$ and to reject all the background channels, which are up to ten orders of magnitude more abundant than the signal, using both kinematical constraints and particle identification systems as veto.The exceptionally low signal yield makes it necessary to prevent also background which is not due to kaon decays, but is rather connected to the interaction of the intense hadron beam with the residual material enconuntered along the decay volume including the beam spectrometer itself: in order to reduce to an acceptable level this last background we designed the Charged Anti Counter (CHANTI).CHANTI is designed to be a compact, efficient and fast response detector to be operated in vacuum and is made up assembling FNAL-NICADD scintillator bars and fast wavelength shifter fibers. It is read out via silicon photomultipliers coupled to fast electronics. It is composed by 6 stations with x-y views and tracking capability; we already constructed 90% of the bars needed, and we characterized more than 50% of them.Finally, the first station was fully assembled and it was tested in the NA62 framework during a testbeam in November 2012. The data collected allowed us to estimate the time resolution of the station, which is a crucial parameter of every veto system.A first MonteCarlo estimation of the amount of the potentially dangerous inelastic interactions was performed showing that it could be a primary source of background, also taking into account the rejection factor ensured by the CHANTI and the other detectors. It is thus indispensable to plan a strategy towards a data-driven estimation of this backgrund source.
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Use Sylow’s theorem. To review Sylow’s theorem, check out the post Sylow’s Theorem (summary). Read the corollary there as well to understand the proof below.
Proof.
Let $n_p, n_q$ be the number of $p$-Sylow subgroups and $q$-Sylow subgroups, respectively.Then by Sylow’s theorem, we have\[n_p\equiv 1 \pmod p, \qquad n_p|q \tag{*}\]and\[n_q \equiv 1 \pmod q, \qquad n_q|p. \tag{**}\]Since $q \equiv 1 \pmod p$, we have $q>p$. Thus $n_q$ must be $1$ from (**).Hence, $G$ has a unique $q$-Sylow subgroup, and it is normal.
From (*), the possibilities for $n_p$ are either $1$ or $q$.We eliminate the possibility of $n_p=1$ as follows.If $n_p=1$, then $G$ has a unique $p$-Sylow subgroup, and hence it is normal.Let $P, Q$ be the unique normal $p$-Sylow subgroup and $q$-Sylow subgroup of $G$, respectively. Then $P, Q$ are normal subgroup of order $p$ and $q$, and hence $P \cap Q=\{e\}$, where $e$ is the identity element of $G$.Since the order $|PQ|=pq=|G|$, these conditions imply that we have\[G\cong P \times Q.\]Since $P$ and $Q$ are groups of prime order, hence it is cyclic, in particular abelian.The direct product of abelian group is abelian, hence $G$ is abelian.This is a contradiction.
Group of Order $pq$ Has a Normal Sylow Subgroup and SolvableLet $p, q$ be prime numbers such that $p>q$.If a group $G$ has order $pq$, then show the followings.(a) The group $G$ has a normal Sylow $p$-subgroup.(b) The group $G$ is solvable.Definition/HintFor (a), apply Sylow's theorem. To review Sylow's theorem, […]
Sylow Subgroups of a Group of Order 33 is Normal SubgroupsProve that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.Hint.We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]
A Group of Order $20$ is SolvableProve that a group of order $20$ is solvable.Hint.Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.See the post summary of Sylow’s Theorem to review Sylow's theorem.Proof.Let $G$ be a group of order $20$. The […]
Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.Hint.Use Sylow's theorem.(See Sylow’s Theorem (Summary) for a review of Sylow's theorem.)Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]
Subgroup Containing All $p$-Sylow Subgroups of a GroupSuppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.Then show that $N$ contains all $p$-Sylow subgroups of […]
Every Group of Order 72 is Not a Simple GroupProve that every finite group of order $72$ is not a simple group.Definition.A group $G$ is said to be simple if the only normal subgroups of $G$ are the trivial group $\{e\}$ or $G$ itself.Hint.Let $G$ be a group of order $72$.Use the Sylow's theorem and determine […]
Abelian Group and Direct Product of Its SubgroupsLet $G$ be a finite abelian group of order $mn$, where $m$ and $n$ are relatively prime positive integers.Then show that there exists unique subgroups $G_1$ of order $m$ and $G_2$ of order $n$ such that $G\cong G_1 \times G_2$.Hint.Consider […]
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Nets in a Normed Algebra
Definition: A Directed Set is a set $\Lambda$ with an relation $\leq$ that satisfies the three conditions: 1) $\lambda \leq \lambda$ for all $\lambda \in \Lambda$ (Reflexivity). 2) If $\lambda_1 \leq \lambda_2$ and $\lambda_2 \leq \lambda_3$ then $\lambda_1 \leq \lambda_3$ (Transitivity). 3) For all $\lambda_1, \lambda_2 \in \Lambda$ there exists a $\lambda \in \Lambda$ such that $\lambda_1 \leq \lambda$ and $\lambda_2 \leq \lambda$ (The Existence of an Upper Bound Between Pairs in $\Lambda$).
Definition: Let $\mathfrak{A}$ be a normed algebra. A Net in $\mathfrak{A}$ is a function $m : \Lambda \to \mathfrak{A}$ where $\Lambda$ is a directed set. Nets are often denoted by $\{ m(\lambda) \}_{\lambda \in \Lambda}$.
Definition: Let $\mathfrak{A}$ be a normed algebra. A net $\{ x(\lambda) \}_{\lambda \in \Lambda}$ in $\mathfrak{A}$ is said to Net Converge (or simply Converge) to $x \in \mathfrak{A}$ denoted $x(\lambda) \overset{\alpha} \to x$ if for every open neighbourhood $U$ of $x$ there exists a $\lambda_0 \in \Lambda$ such that if $\lambda \geq \lambda_0$ then $x(\lambda) \in U$.
The concept of a net is not exclusive to normed algebras. Nets are defined similarly for general topological spaces. Of course, $\mathfrak{A}$ with the norm topology is a topological space!
Note that since $\mathfrak{A}$ is a normed space and $x \in \mathfrak{A}$ then every open ball centered at $x$ is an open neighbourhood of $x$. Thus, if $\{ x(\lambda) \}_{\lambda \in \Gamma}$ net converges to $x \in \mathfrak{A}$ then for all $\epsilon > 0$ there exists a $\lambda_0 \in \Lambda$ such that if $\lambda \geq \lambda_0$ then $x(\lambda) \in B(x, \epsilon)$, or equivalently, for all $\lambda \geq \lambda_0$ we have that:(1)
\begin{align} \quad \| x - x(\lambda) \| < \epsilon \end{align}
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Please note that the recommended version of Scilab is 6.0.2. This page might be outdated.
See the recommended documentation of this function damp
Natural frequencies and damping factors.
Syntax [wn,z] = damp(sys) [wn,z] = damp(P [,dt]) [wn,z] = damp(R [,dt]) Parameters sys
A linear dynamical system (see syslin).
P
An array of polynomials.
R
An array of real or complex floating point numbers.
dt
A non negative scalar, with default value 0.
wn
vector of floating point numbers in increasing order: the natural pulsation in rad/s.
z
vector of floating point numbers: the damping factors.
Description
The denominator second order continuous time transfer function with complex poles can be written as
s^2 + 2*z*wn*s + wn^2 where
z is the damping factor and
wn the natural pulsation.
If
sys is a continuous time system,
[wn,z] = damp(sys) returns in
wn the natural pulsation (in rad/s) and in
z the damping factors of the poles of the linear dynamical system
sys. The
wn and
z arrays are ordered according to the increasing pulsation order.
If
sys is a discrete time system
[wn,z] = damp(sys) returns in
wn the natural pulsation (in rad/s) and in
z the damping factors of the continuous time equivalent poles of
sys. The
wn and
z arrays are ordered according to the increasing pulsation order.
[wn,z] = damp(P) returns in
wn the natural pulsation (in rad/s) and in
z the damping factors of the set of roots of the polynomials stored in the
P array. If
dt is given and non 0, the roots are first converted to their continuous time equivalents. The
wn and
z arrays are ordered according to the increasing pulsation order.
[wn,z] = damp(R) returns in
wn the natural pulsation (in rad/s) and in
z the damping factors of the set of roots stored in the
R array. If
dt is given and non 0, the roots are first converted to their continuous time equivalents.
wn(i) and
z(i) are the the natural pulsation and damping factor of
R(i).
Examples s = %s; num = 22801 + 4406.18*s + 382.37*s^2 + 21.02*s^3 + s^4; den = 22952.25 + 4117.77*s + 490.63*s^2 + 33.06*s^3 + s^4 h = syslin('c', num/den); [wn,z] = damp(h)
The following example illustrates the effect of the damping factor on the frequency response of a second order system.
s = %s; wn = 1; clf(); Z = [0.95 0.7 0.5 0.3 0.13 0.0001]; for k=1:size(Z,'*') z = Z(k) H = syslin('c', 1 + 5*s + 10*s^2, s^2 + 2*z*wn*s + wn^2); gainplot(H, 0.01, 1) p = gce(); p = p.children; p.foreground = k; end title("$\frac{1+5 s+10 s^2}{\omega_n^2+2\omega_n\xi s+s^2}, \quad \omega_n=1$") legend('$\xi = '+string(Z)+'$') plot(wn/(2*%pi)*[1 1], [0 70], 'r') // Natural pulsation
Computing the natural pulsations and daping ratio for a set of roots:
[wn,z] = damp((1:5)+%i)
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What are the open big problems in algebraic geometry and vector bundles?
More specifically, I would like to know what are interesting problems related to moduli spaces of vector bundles over projective varieties/curves.
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What are the open big problems in algebraic geometry and vector bundles?
More specifically, I would like to know what are interesting problems related to moduli spaces of vector bundles over projective varieties/curves.
A few of the more obvious ones:
* Resolution of singularities in characteristic p
*Hodge conjecture * Standard conjectures on algebraic cycles (though these are not so urgent since Deligne proved the Weil conjectures). *Proving finite generation of the canonical ring for general type used to be open though I think it was recently solved; I'm not sure about the details.
For vector bundles, a longstanding open problem is the classification of vector bundles over projective spaces.
(Added later) A very old major problem is that of finding which moduli spaces of curves are unirational. It is classical that the moduli space is unirational for genus at most 10, and I think this has more recently been pushed to genus about 13. Mumford and Harris showed that it is of general type for genus at least 24. As far as I know most of the remaining cases are still open.
Let me mention a couple of problems related to vector bundles on projective spaces.
The Hartshorne conjecture. In its weak form it says that any rank 2 vector bundle on $\mathbf{P}^n_{\mathbf{C}},n>6$ is a direct sum of line bundles, which implies that any codimension 2 smooth subvariety whose canonical class is a multiple of the hyperplane sectionis a complete intersection. In a stronger form Hartshorne's conjecture says that any codimension $>\frac{2}{3}n$ subvariety of $\mathbf{P}^n_{k},k$ an algebraically closed field is a complete intersection. See Hartshorne, Varieties of small codimension in a projective space, Bull AMS 80, 1974. The weak conjecture fails for $n=3$ and $4$ -- there are examples (due to Horrocks and Mumford) of non-split vector bundles of rank 2 on $\mathbf{P}^4_{\mathbf{C}}$, but so far as I know the question if any such examples exist for $n>4$ is open. See here Evidences on Hartshorne's conjecture? References? for a discussion including some references.
The existence of non-algebraic topological vector bundles on $\mathbf{P}^n_{\mathbf{C}}$. It is a classical result that any topological complex vector bundle on $\mathbf{P}^n_{\mathbf{C}}, n\leq 3$ is algebraic, see e.g. Okonek, Schneider, Spindler, Vector bundles on complex projective spaces, chapter 1, \S 6. It is strongly suspected that for $n>3$ there are topological complex vector bundles that are not algebraic. Good candidates are nontrivial rank 2 vector bundles on $\mathbf{P}^n_{\mathbf{C}}, n\geq 5$ all of whose Chern classes vanish which were constructed by E. Rees, see MR0517518. It is claimed there that these bundles do not admit a holomorphic structure, but later a gap was found in the proof. See here Complex vector bundles that are not holomorphic for some more information.
Linearization Conjecture. Every algebraic action of $\mathbb{C}^*$ on $\mathbb{C}^n$ is linear in some coordinates of $\mathbb{C}^n$. Open for $n>3$.
Cancellation Conjecture. If $X\times \mathbb{C}\cong \mathbb{C}^{m+1}$ then $X\cong \mathbb{C}^m$. Open for $m>2$.
Coolidge-Nagata Conjecture. A rational cuspidal curve in $\mathbb{P}^2$ is rectifiable, i.e. there exists a birational automorphism of $\mathbb{P}^2$ which transforms the curve into a line.
There's also the big open question (I think it's still open) about whether rationally connected varieties are always unirational. I think people believe the answer is
NO, but they don't know an example.
Joe Harris had some slides a few years ago with regards to this Seattle 2005
We can also mention two other major open problems :
The abundance conjecture, stating that if a $K_X+\Delta$ is klt and nef, then it is semi-ample (a multiple has no base-point)
The Griffith's conjecture : if $E$ is an ample vector bundle over a compact complex manifold, then it is Griffith-positive. (this is known for line bundles of course)
There's also Fujita's conjecture.
Conjecture: Suppose $X$ is a smooth projective dimensional complex algebraic variety with ample divisor $A$. Then
It's also often stated in the complex analytic world.
Also there are many refinements (and generalizations) of this conjecture. For example, the assumption that $X$ is smooth is probably more than you need (something close to rational singularities should be ok). It also might even be true in characteristic $p > 0$.
It's known in relatively low dimensions (up to 5 in case 1. I think?)
In connection to vector bundles over $\mathbb{P}^n$, Hartshorne's paper from 1979 provides a list of open problems. The paper is "Algebraic vector bundles on projective spaces: A problem list"
Topology, 18:117–128, 1979.
I don't know which of those problems are still open, but I would be interested in knowing how much progress has been made on those problems, since 1979.
Open problems in Algebraic Geometry, S.J. Edixhoven (editor), B.J.J. Moonen (editor), F. Oort (editor).
The
Tate conjecture: Let $k$ be a finitely generated field, $X/k$ a smooth projective geometrically integral variety and $\ell$ invertible in $k$. Then the cycle class map $$\mathrm{CH}^r(X) \otimes_\mathbf{Z} \mathbf{Q}_\ell \to \mathrm{H}^{2r}(\bar{X},\mathbf{Q}_\ell(r))^{G_k}$$ is surjective.
It is e.g. proved for $r=1$ and Abelian varieties, a deep theorem. See http://www.math.harvard.edu/~chaoli/doc/TateConjecture.html.
This is analogous to the Hodge conjecture for complex varieties.
The Maximal Rank Conjecture is a major outstanding problem in Brill-Noether theory, although recent advances in tropical techniques might point the way to a solution; see https://arxiv.org/abs/1505.05460.
EDIT: The Maximal Rank Conjecture was proved by Eric Larson in his PhD thesis; see: https://arxiv.org/abs/1711.04906
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on behalf of the CMS Collaboration
Articles written in Pramana – Journal of Physics
Volume 69 Issue 6 December 2007 pp 1069-1074 Calorimetry and Muons
The performance of the CMS hadron calorimeter is studied using test beam facilities at CERN. Two wedges of brass-scintillator calorimeter are exposed to negative and positive beams with momenta between 3 and 300 GeV/c. Light produced in the scintillators are collected using wavelength shifting fibres and read out using hybrid photo-diodes. Each of the wedges has 17 layers of scintillators. In one of these wedges signal from all 17 layers are grouped together while in the other each layer is read out separately. The response, energy resolution, longitudinal and lateral shower profiles are measured.
Volume 79 Issue 4 October 2012 pp 817-819 Poster Presentations
The compact muon solenoid (CMS) is one of the four experiments which is getting and analysing the results of the collision of protons at LHC. The CMS trigger system is divided into two stages, the level-1 trigger and high-level triggers, to handle the large stream of data produced in collision. The information transmitted from the three muon subsystems (DT, CSC and RPC) are collected by the Global Muon Trigger (GMT) Board and merged. A method for evaluating the RPC system trigger efficiency with data from $pp$ collision was developed using the features of GMT. The results of the study with the real data of 2011 are shown and discussed here along with the comparison of Monte Carlo results
Volume 79 Issue 4 October 2012 pp 839-843 Poster Presentations
The results from various dijet distributions in proton-proton collisions at a centre-of-mass energy of 7 TeV, with 2010 and 2011 data from the CMS experiment, are presented. The measurements of the dijet mass spectra, centrality ratio, azimuthal decorrelation and angular distribution are shown. Sensitivity of the phenomenological parameters used to model different event generators is also investigated. Prospects for observing evidence for new physics in these distributions are presented.
Volume 79 Issue 4 October 2012 pp 845-848 Poster Presentations
Search for the Standard Model Higgs boson in the decay mode $H \rightarrow ZZ \rightarrow 2l2\tau$, where $l = \mu, e$, is presented based on CMS data corresponding to an integrated luminosity of 1.1 fb
-1 at $\sqrt{s} = 7$ TeV. No evidence is found for a significant deviation from Standard Model expectations anywhere in the $ZZ$ mass range considered in this analysis. An upper limit at 95% CL is placed on the product of the cross-section and decay branching ratio for the Higgs boson decaying with Standard Model-like couplings, which excludes cross-sections of about ten times the expected value for Higgs boson masses in the range 200 < $m_{H}$< 400 GeV/c 2.
Volume 79 Issue 4 October 2012 pp 871-874 Poster Presentations
A search for heavy resonance production decaying to tau pair is performed in the proton–proton collisions at the center-of-mass energy $\sqrt{s} = 7$ TeV using 36 pb
-1 of data collected with CMS detector at the LHC during the year 2010. The number of observed events are in good agreement with the predictions of Standard Model background processes. Therefore, an upper limit on the resonance cross-section times the branching ratio to tau pair is obtained as a function of the resonance mass. Ditau resonance Z' with Standard Model couplings having mass less than 468 GeV/c 2 is excluded at 95% confidence level.
Volume 79 Issue 4 October 2012 pp 929-932 Poster Presentations
A search for excited leptons, carried out with the CMS detector in $pp$ collision at the LHC with $\sqrt{s} = 7$ TeV, is presented. The search has been performed for an associated production of a lepton and an oppositely charged excited lepton $pp \rightarrow ll^{*}$, followed by the decay $l^{*} \rightarrow l\gamma$ , resulting in $ll\gamma$ final state, where $l = e,\mu$. No excess above the Standard Model expectation is observed in the data. Interpreting the findings in the context of $l^{∗}$ production through four-fermion contact interactions and $l^{∗}$ decay via electroweak processes, upper limits are reported for $l^{∗}$ production at this collision energy and the exclusion region in the $\Lambda - M(l^{*})$ parameter space is extended beyond the previously established limits.
Volume 79 Issue 4 October 2012 pp 937-940 Poster Presentations
The Drell–Yan differential cross-section is measured in proton–proton collisions at $\sqrt{s} = 7$ TeV, from a data sample collected with the CMS detector at the LHC and corresponding to an integrated luminosity of $36 \pm 1.4$ pb
-1. The measured cross-section is normalized to the cross-section of the 𝑍-peak region, for both dimuon and dielectron final state, in the dilepton invariant mass range of 15–600 GeV/c 2. The normalized cross-section values are quoted in the full phase-space and within the detector acceptance. The effect of final-state radiation is also studied and the measurements are correted for this. The measurements are compared to the theoretical predictions and are found to be in good agreement.
Volume 79 Issue 5 November 2012 pp 1223-1226 Poster Presentations
The measurement of 𝑡-channel single top cross-section in proton–proton collisions at the Large Hadron Collider (LHC) at a centre-of-mass energy of 7 TeV, using data collected with the Compact Muon Solenoid (CMS) experiment during the year 2010 is presented. Both the electronneutrino and muon-neutrino decay channels of 𝑊 boson from top decay are considered. Two complementary multivariate analysis methods to separate signal and background and to extract the cross-section for the single top produced in 𝑡-channel are explored. The result is compared with the most precise Standard Model theory predictions.
Volume 79 Issue 5 November 2012 pp 1263-1266 Poster Presentations
A search is reported for the Higgs boson decaying to $W^{+} W^{−}$ in pp collisions at $\sqrt{s} = 7$ TeV. The analysis is performed using LHC data recorded by the CMS detector, corresponding to an integrated luminosity of 1.55 fb
-1. No significant excess above the Standard Model background expectation is observed, and upper limits on Higgs boson production are derived, excluding the presence of a Higgs boson with a mass in the range of 147–194 GeV/c 2 at the 95% confidence level (CL) using the CLs approach.
Volume 79 Issue 5 November 2012 pp 1285-1288 Poster Presentations
The measurements of the cross-sections are presented for the production of jets in association with 𝑊 and 𝑍 bosons reconstructed in their decays to electrons and muons. The results are based on 36 pb
-1 of $pp$ collisions at $\sqrt{s} = 7$
Volume 79 Issue 5 November 2012 pp 1321-1324 Poster Presentations
A search for the Standard Model (
-1 recorded by the 2. Twenty-one events are observed, while $21.2 \pm 0.8$ events are expected from Standard Model ( 2.
Volume 79 Issue 5 November 2012 pp 1341-1344 Poster Presentations
A search for a Higgs boson decaying into two photons in $pp$ collisions at the
-1 of data recorded in 2011 by the
Current Issue
Volume 93 | Issue 5 November 2019
Click here for Editorial Note on CAP Mode
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I'm attempting to emulate the modulation and demodulation of a PAL video frame in software, but I'm having trouble understanding how the AM (luma) and QAM (UV) components of the signal are separated during the demodulation process.
Background:
The software I'm writing takes an RGB bitmap, translates it to Y'UV using the BT.601 standard, generates (modulates) a signal as a series of samples, then saves those samples to a wave file. Those samples can then be loaded back in and demodulated back into a bitmap, with the H-sync pulses honoured against the H-pos ramp generator as an old CRT TV would do, and finally translated back into an RGB bitmap.
There are two motivations behind this project: first, to create some cool looking analog fuzz in software; and second, to better understand the TV standards and basic modulation / demodulation techniques in a domain (software) where I am more comfortable.
What I've got so far:
The code performs the modulation fine, and when the result is viewed in Audacity (with a sample rate faked at 1/1000th the real frequency) it shows a signal that I recognise to be a series of PAL picture lines. The full process works if I put it in black-and-white mode, thus omitting the QAM signal from the modulation step and leaving only the amplitude modulated luma channel. I've also tested the QAM modulation and demodulation code against a simple audio file, and it works great. The only bit I haven't got working is the image demodulation with both the AM luma and QAM colour components included.
The confusion:
Given a pure untouched QAM signal, it seems relatively trivial to demodulate the signal back to its original composite signals, by multiplying with the carrier sine and its cosine individually:
$$I_t = s_t \cos(2\pi f_c t)$$ $$Q_t = s_t \sin(2\pi f_c t)$$
The two components are then low-passed at the carrier frequency to remove high-frequency terms. This seems to work great, and as I noted above my implementation of QAM demodulation works just fine.
Similarly, I can implement simple AM demodulation, even when the signal is accompanied by other signals in separate parts of the frequency domain - it's just a case of high-pass and/or low-pass until you've got the bit you want.
However, from what I can see, the QAM modulated UV component is added to the base AM luma signal, with both parts overlapping in the frequency domain, as the following diagram from Wikipedia appears to show:
At this point I'm a little stumped. It doesn't look like I can low-pass to get the AM luma signal due to the overlap, and I can't see how I'd subtract the QAM signal from the underlying AM.
What am I missing? What are the logical steps that should be taken to separate the luma and UV components?
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I'm trying to solve $$\operatorname{Arg}(z-2) - \operatorname{Arg}(z+2) = \frac{\pi}{6}$$ for $z \in \mathbb{C}$.
I know that $$\operatorname{Arg} z_1 - \operatorname{Arg} z_2 = \operatorname{Arg} \frac{z_1}{z_2},$$ but that's only valid when $\operatorname{Arg} z_1 - \operatorname{Arg} z_2 \in (-\pi,\pi]$, so I'm not sure how to even begin solving this.
I'm not familiar with modular arithmetic so if it is possible to solve this without using it then that would be great! (not that I know whether it is required to solve this in the first place)
Thank you in advance.
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Overview¶
We’ve already generated quite a few figures in these lectures using Matplotlib.
Matplotlib is an outstanding graphics library, designed for scientific computing, with
high-quality 2D and 3D plots output in all the usual formats (PDF, PNG, etc.) LaTeX integration fine-grained control over all aspects of presentation animation, etc. Matplotlib’s Split Personality¶
Matplotlib is unusual in that it offers two different interfaces to plotting.
One is a simple MATLAB-style API (Application Programming Interface) that was written to help MATLAB refugees find a ready home.
The other is a more “Pythonic” object-oriented API.
For reasons described below, we recommend that you use the second API.
But first, let’s discuss the difference.
import matplotlib.pyplot as plt%matplotlib inlineimport numpy as npx = np.linspace(0, 10, 200)y = np.sin(x)plt.plot(x, y, 'b-', linewidth=2)plt.show()
This is simple and convenient, but also somewhat limited and un-Pythonic.
For example, in the function calls, a lot of objects get created and passed around without making themselves known to the programmer.
Python programmers tend to prefer a more explicit style of programming (run
import this in a code block and look at the second line).
This leads us to the alternative, object-oriented Matplotlib API.
The Object-Oriented API¶
Here’s the code corresponding to the preceding figure using the object-oriented API
fig, ax = plt.subplots()ax.plot(x, y, 'b-', linewidth=2)plt.show()
Here the call
fig, ax = plt.subplots() returns a pair, where
figis a
Figureinstance—like a blank canvas.
axis an
AxesSubplotinstance—think of a frame for plotting in.
The
plot() function is actually a method of
ax.
While there’s a bit more typing, the more explicit use of objects gives us better control.
This will become more clear as we go along.
fig, ax = plt.subplots()ax.plot(x, y, 'r-', linewidth=2, label='sine function', alpha=0.6)ax.legend()plt.show()
We’ve also used
alpha to make the line slightly transparent—which makes it look smoother.
The location of the legend can be changed by replacing
ax.legend() with
ax.legend(loc='upper center').
fig, ax = plt.subplots()ax.plot(x, y, 'r-', linewidth=2, label='sine function', alpha=0.6)ax.legend(loc='upper center')plt.show()
If everything is properly configured, then adding LaTeX is trivial
fig, ax = plt.subplots()ax.plot(x, y, 'r-', linewidth=2, label='$y=\sin(x)$', alpha=0.6)ax.legend(loc='upper center')plt.show()
Controlling the ticks, adding titles and so on is also straightforward
fig, ax = plt.subplots()ax.plot(x, y, 'r-', linewidth=2, label='$y=\sin(x)$', alpha=0.6)ax.legend(loc='upper center')ax.set_yticks([-1, 0, 1])ax.set_title('Test plot')plt.show()
More Features¶
Matplotlib has a huge array of functions and features, which you can discover over time as you have need for them.
We mention just a few.
Multiple Plots on One Axis¶
Here’s an example that randomly generates three normal densities and adds a label with their mean
from scipy.stats import normfrom random import uniformfig, ax = plt.subplots()x = np.linspace(-4, 4, 150)for i in range(3): m, s = uniform(-1, 1), uniform(1, 2) y = norm.pdf(x, loc=m, scale=s) current_label = f'$\mu = {m:.2}$' ax.plot(x, y, linewidth=2, alpha=0.6, label=current_label)ax.legend()plt.show()
Multiple Subplots¶
Here’s an example that generates 6 histograms
num_rows, num_cols = 3, 2fig, axes = plt.subplots(num_rows, num_cols, figsize=(10, 12))for i in range(num_rows): for j in range(num_cols): m, s = uniform(-1, 1), uniform(1, 2) x = norm.rvs(loc=m, scale=s, size=100) axes[i, j].hist(x, alpha=0.6, bins=20) t = f'$\mu = {m:.2}, \quad \sigma = {s:.2}$' axes[i, j].set(title=t, xticks=[-4, 0, 4], yticks=[])plt.show()
from mpl_toolkits.mplot3d.axes3d import Axes3Dfrom matplotlib import cmdef f(x, y): return np.cos(x**2 + y**2) / (1 + x**2 + y**2)xgrid = np.linspace(-3, 3, 50)ygrid = xgridx, y = np.meshgrid(xgrid, ygrid)fig = plt.figure(figsize=(8, 6))ax = fig.add_subplot(111, projection='3d')ax.plot_surface(x, y, f(x, y), rstride=2, cstride=2, cmap=cm.jet, alpha=0.7, linewidth=0.25)ax.set_zlim(-0.5, 1.0)plt.show()
A Customizing Function¶
Perhaps you will find a set of customizations that you regularly use.
Suppose we usually prefer our axes to go through the origin, and to have a grid.
Here’s a nice example from Matthew Doty of how the object-oriented API can be used to build a custom
subplots function that implements these changes.
Read carefully through the code and see if you can follow what’s going on
def subplots(): "Custom subplots with axes through the origin" fig, ax = plt.subplots() # Set the axes through the origin for spine in ['left', 'bottom']: ax.spines[spine].set_position('zero') for spine in ['right', 'top']: ax.spines[spine].set_color('none') ax.grid() return fig, axfig, ax = subplots() # Call the local version, not plt.subplots()x = np.linspace(-2, 10, 200)y = np.sin(x)ax.plot(x, y, 'r-', linewidth=2, label='sine function', alpha=0.6)ax.legend(loc='lower right')plt.show()
The custom
subplots function
calls the standard
plt.subplotsfunction internally to generate the
fig, axpair,
makes the desired customizations to
ax, and
passes the
fig, axpair back to the calling code.
Exercise 1¶
Plot the function$$ f(x) = \cos(\pi \theta x) \exp(-x) $$
over the interval $ [0, 5] $ for each $ \theta $ in
np.linspace(0, 2, 10).
Place all the curves in the same figure.
The output should look like this
θ_vals = np.linspace(0, 2, 10)x = np.linspace(0, 5, 200)fig, ax = plt.subplots()for θ in θ_vals: ax.plot(x, np.cos(np.pi * θ * x) * np.exp(- x))plt.show()
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The Annals of Statistics Ann. Statist. Volume 25, Number 2 (1997), 818-850. Asymptotic expansion of M-estimators with long-memory errors Abstract
This paper obtains a higher-order asymptotic expansion of a class of
M-estimators of the one-sample location parameter when the errors form a long-memory moving average. A suitably standardized difference between an M-estimator and the sample mean is shown to have a limiting distribution. The nature of the limiting distribution depends on the range of the dependence parameter $\theta$. If, for example, $1/3 < \theta < 1$, then a suitably standardized difference between the sample median and the sample mean converges weakly to a normal distribution provided the common error distribution is symmetric. If $0 < \theta < 1/3$, then the corresponding limiting distribution is nonnormal. This paper thus goes beyond that of Beran who observed, in the case of long-memory Gaussian errors, that M-estimators $T_n$ of the one-sample location parameter are asymptotically equivalent to the sample mean in the sense that $\Var(T_n)/\Var(\bar{X}_n) \to 1$ and $T_n = \bar{X}_n + o_P(\sqrt{\Var(\bar{X}_n)}).$ Article information Source Ann. Statist., Volume 25, Number 2 (1997), 818-850. Dates First available in Project Euclid: 12 September 2002 Permanent link to this document https://projecteuclid.org/euclid.aos/1031833675 Digital Object Identifier doi:10.1214/aos/1031833675 Mathematical Reviews number (MathSciNet) MR1439325 Zentralblatt MATH identifier 0885.62101 Subjects Primary: 62M10: Time series, auto-correlation, regression, etc. [See also 91B84] Secondary: 65G30: Interval and finite arithmetic Citation
Koul, Hira L.; Surgailis, Donatas. Asymptotic expansion of M -estimators with long-memory errors. Ann. Statist. 25 (1997), no. 2, 818--850. doi:10.1214/aos/1031833675. https://projecteuclid.org/euclid.aos/1031833675
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Difference between revisions of "De Bruijn-Newman constant"
(→Code and data)
(→Threads)
(12 intermediate revisions by the same user not shown) Line 35: Line 35:
where
where
−
:<math>\displaystyle \xi(s) := \frac{s(s-1)}{2} \pi^{s/2} \Gamma(s/2) \zeta(s)</math>
+
:<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 [https://en.wikipedia.org/wiki/Riemann%E2%80%93von_Mangoldt_formula Riemann-von Mangoldt formula] (in the explicit form given by Backlund) gives
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 [https://en.wikipedia.org/wiki/Riemann%E2%80%93von_Mangoldt_formula Riemann-von Mangoldt formula] (in the explicit form given by Backlund) gives
Line 50: Line 50:
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>.
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>.
+ +
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)>0</math> (see [KKL2009, Proposition A]. In particular we have
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)>0</math> (see [KKL2009, Proposition A]. In particular we have
Line 89: Line 91:
* [https://terrytao.wordpress.com/2018/03/02/polymath15-fifth-thread-finishing-off-the-test-problem/ Polymath15, fifth thread: finishing off the test problem?], Terence Tao, Mar 2, 2018.
* [https://terrytao.wordpress.com/2018/03/02/polymath15-fifth-thread-finishing-off-the-test-problem/ Polymath15, fifth thread: finishing off the test problem?], Terence Tao, Mar 2, 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/2018/03/18/polymath15-sixth-thread-the-test-problem-and-beyond/ Polymath15, sixth thread: the test problem and beyond], Terence Tao, Mar 18, 2018.
+ + + + + +
== Other blog posts and online discussion ==
== Other blog posts and online discussion ==
Line 104: Line 112:
* [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 ==
See [[Polymath15 test problem]].
See [[Polymath15 test problem]].
+ + + +
== Wikipedia and other references ==
== Wikipedia and other references ==
Line 129: Line 143:
* [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. [https://arxiv.org/abs/1801.05914 arXiv:1801.05914]
* [RT2018] B. Rodgers, T. Tao, The de Bruijn-Newman constant is non-negative, preprint. [https://arxiv.org/abs/1801.05914 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. [http://plouffe.fr/simon/math/The%20Theory%20Of%20The%20Riemann%20Zeta-Function%20-Titshmarch.pdf pdf]
* [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. [http://plouffe.fr/simon/math/The%20Theory%20Of%20The%20Riemann%20Zeta-Function%20-Titshmarch.pdf pdf]
+ + 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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Let $A$ be a linear operator on $L$ with domain $D(A)$. To show that $A$ is closable, i.e., it has a closed linear extension, why does it suffice to show that if $\{f_n\}\subset D(A), f_n\to 0,$ and $Af_n\to g\in L,$ then $g=0$? I would greatly appreciate any help.
So we say that an operator is closed if it has a closed graph. This means that given an unbounded operator $T \in \mathcal{B}(H,K)$ that $\{x,Tx)\}$ is closed in $H \times K$. This explicitly means that given a sequence $x_n \to x $ in H such that $Tx_n \to y$ in K that $y=Tx$. Now, if you can show that this holds at a point in H then you can extend to the whole domain of T in the typical way.
Given any operator $T$ on $L$, denote by $D(T)$ its domain and $\Gamma(T)$ its graph, i.e.
$$\Gamma(T)=\{(x,Tx):x\in D(T)\}.$$
Note that $\Gamma(T)$ is a subspace of $L\times L$, and $T$ is closed if and only if $\Gamma(T)$ is a closed subspace. With that in mind, $T$ is closable if and only if $\overline{\Gamma(T)}=\Gamma(T')$ for some operator $T'$ on $L$ (and clearly $T'$ must be closed). I will prove this upon request, but it is a good exercise.
Back to the problem at hand: assume that $\{f_n\}\subset D(A)$, $f_n\to0$, $Af_n\to g$ implies $g=0$. We want to show $\overline{\Gamma(A)}=\Gamma(A')$ for some operator $A'$ on $L$. Let $(f,g),(f,h)\in\overline{\Gamma(A)}$; it remains to show $g=h$. There exist sequences $\{(f^1_n,g_n)\},\{(f^2_n,h_n)\}\subset\Gamma(A)$ such that $(f_n^1,g_n)\to(f,g)$ and $(f_n^2,h_n)\to(f,h)$. Hence $f_n^1-f_n^2\to0$ and $A(f_n^1-f_n^2)=g_n-h_n\to g-h$. By assumption, this implies $g-h=0$, and we are done.
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Refering to Theorem 8.2 in Baby Rudin
8.2 TheoremSuppse $\sum c_n$ converges. Put $$f(x)=\sum_{n=0}^{\infty} c_nx^n \ \ \ \ (-1<x<1)$$ Then $$\lim_{x\rightarrow1}f(x)=\sum_{n=0}^{\infty}c_n$$
The proof in Rudin is that outlined in Wikipedia
However, Factoring out $(1-x)$ seems not natural for me.
The theorem looks like a extension from Theorem 8.1.
Indeed, by imitating the proof of Rudin 7.11 $$|f(t)-\sum^{\infty}_{n=0}c_n|\leq|\sum_{n=0}^{\infty}c_n-\sum_{n=0}^{N}c_n|+|\sum_{n=0}^{N}c_n-\sum_{n=0}^{N}c_nt^n|+|\sum_{n=0}^{N}c_nt^n-f(t)|$$ where $t \in(-1,1)$
1.The first term can be arbitrary small for large N since $\sum c_n$ converges.
2.The second term can be small when $t\rightarrow1$ since polynomial is continuous.
3.Since f(x) is uniform convergent on $(-1,1)$, the third term can also be arbitrary small.
Then, I can also conclude the same result as in Theorem 8.2.
Could anyone kindly explain why Rudin uses a different approach (like factoring out (1-x)) or my reasoning has flaws, if there is any ?
Summary of the discussion :
The problem in my proof arises from the third term in the inequality, where I mistake the order to take limits (I implicitly make $t \rightarrow 1$ and then $ N \rightarrow \infty$. This is wrong).
The motivation (I guess) in Rudin's proof is from Rudin
Theorem 3.42, where we study the criteria to test conditional convergent series. Theorem 8.2 has a similar situation, the sum $\sum c_n$ may very well be not absolutely convergent.
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A few preliminary remarks:
1) Complexifications of Banach lattices are in fact a special case of the more general concept of
complexifications of real Banach spaces.
2) Most books and articles about complex Banach lattices which contain spectral theoretic results focus on
positive operators (or, say, generators of positive semigroups) which are, of course, real.
3) One reason for 2) might be that real operators can actually be defined on every complexification of a real Banach space (no matter whether a lattice structure is present), so the question about "spectral properties of real operators" belongs to the realm of complexifications of real Banach spaces rather than to the realm of complex Banach lattices.
4) While most books on Banach lattices contain a chapter on complex Banach lattices (which is, unfortunately, very brief in most cases), there does not seem to exist much literature on complexifications of real Banach spaces (although the topic often seems to be considered as some kind of "standard folklore" among functional analysts).
The only article I'm aware of which contains a rather extensive treatment of complexifications of real Banach spaces is "G. A. Muñoz, Y. Sarantopoulos, and A. Tonge: Complexifications of real Banach spaces, polynomials and multilinear maps. Studia Math., 134(1):1–33, 1999." Unfortunately, though, spectral theory is not treated there.
References to learn about spectral properties of real operators:
If you primarily want to learn about the spectral properties of real operators, you can find some information in the following references (I apologize for the self-advertisement, but these are really the only references I know for this topic):
[1] Appendix C in my PhD thesis contains a thorough treatment of complexifications. Spectral theoretic aspects are discussed in Section C.3, and a few spectral theoretic properties of real operators are discussed in Proposition C.3.2 (the fact that the operators considered in this proposition are exactly the class of real operators follows from Proposition C.1.6). However, I should add that Appendix C contains almost no proofs, for the following reason (which reflects, of course, only my personal point of view): when it comes to complexfications, it is in most cases more difficult to find the correct statements rather than to prove them (proving them is very straightforward in most cases).
[2] You can also find similar information in Appendix A of the arXiv version 1 of my paper "Spectral and Asymptotic Properties of Contractive Semigroups on Non-Hilbert Spaces" (but in version 2, which was eventually published, I removed this appendix and replaced it with a very brief discussion of complexifications at the beginning of the paper in response to a request in a referee's report).
References to quote:
The above references [1] and [2] are certainly not very well-suited in case that you need a reference to quote in a paper. Indeed, [2] is only a preprint which was eventually published without the information relevant here, and [1] contains almost no proofs (moreover, one might also question whether it is a particularly good idea to refer to the appendix of a PhD thesis for information about a quite classical topic in functional analysis - even though this topic does not seem to be particularly well represented in the literature).
Unfortunately, I do not know any other references about the spectral theory of real operators (and I have severe doubts whether such a reference exists).
So my suggestion is: If you need spectral properties of real operators in a paper, simply state them either in the introduction or in an appendix of the paper. You may include proofs if you feel they are necessary; but in many cases those properties are such elementary that you probably won't need to include proofs.
Additional remark (biased by my personal opinion):
I think it is reasonable to conlcude that a survey article about complexifications of real Banach spaces which complements the aspects discussed in the article of Muñoz, Sarantopoulos and Tonge would be a very useful source of reference. Now, all that is needed is a volunteer to write it and a journal willing to publish it...
Edit in response to the OP's comment:
Here are a few facts with proofs. Throughout, let $E$ be a complexification of a real Banach space $E_{\mathbb{R}}$.
Note that an
everywhere defined linear operator $T$ on $E$ is real if and only if $T(E_{\mathbb{R}}) \subseteq E_{\mathbb{R}}$.
Proposition 1. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real. If $z \in D(A)$, then $\overline{z}$ is also in $D(A)$ and we have $A\overline{z} = \overline{Az}$.
Proof. We may write $z$ as $z = x+iy$, where $x,y \in E_{\mathbb{R}}$. As $A$ is real, we have $x,y \in D(A)$ and $Ax$ as well as $Ay$ is in $E_{\mathbb{R}}$. Hence,\begin{align*} A\overline{z} = A(x-iy) = Ax - iAy = \overline{Ax+ iAy} = \overline{Az}.\end{align*}Note that the third equality above uses the fact that both $Ax$ and $Ay$ are elements of $E_{\mathbb{R}}$.
Proposition 2. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real. If $\lambda$ is a real number in the resolvent set of $A$, then the resolvent $R(\lambda,A) := (\lambda - A)^{-1}$ is real, too.
Proof. Let $x \in E_{\mathbb{R}}$ and set $y := R(\lambda,A)x$. Then $y$ is an element of $D(A)$ and it can be written as $y = a + ib$ for unique $a,b \in E_{\mathbb{R}}$. Since $A$ is real, we conclude that $a,b \in D(A)$.Clearly, $x = (\lambda - A)y = (\lambda-A)a + i (\lambda - A)b$. Since $A$ is a real operator and $\lambda$ is a real number, both vectors $(\lambda - A)a$ and $(\lambda - A)b$ are elements of $E_{\mathbb{R}}$. As the decomposition of each vector in $E$ into its real and imaginary part is unqie, we conclude that $x = (\lambda - A)a$. Hence, $R(\lambda,A)x = a \in E_{\mathbb{R}}$.This shows that $R(\lambda,A)$ is indeed real.
Proposition 3. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real.
(a) If $\lambda \in \mathbb{C}$ is an eigenvalue of $A$ with eigenvector $z \in E$, then $\overline{\lambda}$ is an eigenvalue of $A$ with eigenvector $\overline{z}$.
(b) If $\lambda \in \mathbb{C}$ is a spectral value of $A$, then so is $\overline{\lambda}$.
Proof. (a) According to Proposition 1 we have $\overline{z} \in D(A)$ and $A\overline{z} = \overline{Az} = \overline{\lambda z} = \overline{\lambda} \overline{z}$.
(b) Assume that $\overline{\lambda}$ is not a spectral value of $A$. Then $A$ is closed, so we only have to show that $\lambda - A$ is a bijective mapping from $D(A)$ to $E$. Clearly $\lambda - A$ is injective, since otherwise it would follow from (a) that $\overline{\lambda} - A$ was not injective.In order to show that $\lambda - A$ is surjective, let $z \in E$. Since $\overline{\lambda} - A$ is surjective, we can find a vector $c \in E$ such that $(\overline{\lambda} - A)c = \overline{z}$. Hence, we obtain\begin{align*} \overline{(\lambda - A)\overline{c}} = \overline{\lambda} c - Ac = (\overline{\lambda} - A) c = \overline{z}\end{align*}(where we used Proposition 1 for the first equality), so $(\lambda - A)\overline{c} = z$. This proves that $\lambda - A$ is indeed surjective.
Remark 4. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real and let $\lambda$ be a complex number. Then the following assertions hold:
(a) The number $\lambda$ is an eigenvalue of $A$ if and only if $\overline{\lambda}$ is an eigenvalue of $A$.
(b) The number $\lambda$ is a spectral value of $A$ if and only if $\overline{\lambda}$ is a spectral value of $A$.
Proof. The "only if" implications are the content of Proposition 3, and the converse implications also follow from Proposition 3 by applying the proposition to the number $\overline{\lambda}$ instead of $\lambda$.
Remark 5. The arguments in the proof of Proposition 3 actually show that we have\begin{align*} \overline{R(\lambda,A)z} = R(\overline{\lambda},A) \overline{z}\end{align*}for each $\lambda$ in the resolvent set of $A$ and each $z \in E$. This can be considered as a generalisation of Proposition 2.
Theorem 6. Let $A: E \supseteq D(A) \to E$ be a linear operator which is real and let $\sigma$ be a compact subset of the spectrum $\sigma(A)$ such that $\sigma(A) \setminus \sigma$ is closed. Let $P$ denote the spectral projection associated to $\sigma$.
If $\sigma$ is invariant under complex conjugation (i.e. $\overline{\lambda} \in \sigma$ for all $\lambda \in \sigma$), then the operator $P$ is real.
Sketch of the proof. By Proposition 3 (or Remark 4) the entire spectrum $\sigma(A)$ is conjugation invariant. Since $\sigma$ is compact and $\sigma(A) \setminus \sigma$ is closed, we can find a closed (and smooth) cycle $\gamma$ in $\mathbb{C}$ which circumvents no element of $\sigma(A) \setminus \sigma$ but each element of $\sigma $ exactly once. As $\sigma$ and $\sigma(A) \setminus \sigma$ are conjugation invariant, we can choose $\gamma$ to be conjugation invariant, too, meaning that walking along $\overline{\gamma}$ is the same as walking along $\gamma$ in the converse direction. By definition of the spectral projection we have\begin{align*} P = \frac{1}{2\pi i} \int_{\gamma} R(\mu,A) \, d\mu\end{align*}Applying this to a vector $z \in E$ and employing Remark 5 we conclude that\begin{align*} \overline{Pz} = - \frac{1}{2\pi i} \int_\gamma R(\overline{\mu}, A) \overline{z} \, \overline{d\mu} = \frac{1}{2\pi i} \int_\gamma R(\mu,A) \overline{z} \, d\mu = P\overline{z}.\end{align*}If $z \in E_{\mathbb{R}}$ this implies that $\overline{Pz} = P\overline{z} = Pz$, so $Pz \in E_{\mathbb{R}}$. Hence, $P$ is real.
Of course, one can prove many related results, but I hope that the above arguments suffice to give you an idea of how things usually work in the spectral theory of real operators.
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Books by Independent Authors Advanced Real Analysis 2017, 434-544 Chapter X. Introduction to Wavelets Abstract
This chapter introduces the relatively recent subject of wavelets, which is an outgrowth of Fourier analysis in mathematics and signal processing in engineering. Except in one case, construction of examples of wavelets tends to be difficult. Much of the chapter is devoted to construction of some of the better known examples and lists of their most important properties.
Section 1 defines wavelets and discusses three features of traditional Fourier analysis: the Uncertainty Principle, Gibbs phenomenon, and the Shannon Sampling Theorem. It ends with a brief essay on the need for wavelets in various applications.
Section 2 establishes that the Haar system is an orthonormal basis of $L^{2}(\mathbb {R})$. The Haar wavelet predates the general theory of wavelets by many decades but provides a prototype for some of the development. The section ends with some discussion of convergence of one-sided Haar expansions for function spaces besides $L^{2}(\mathbb {R})$.
Section 3 begins the general theory of wavelets, introducing the notion of multiresolution analysis to abstract the construction in Section 2 of the Haar wavelet. The ingredients of a multiresolution analysis are a scaling function, traditionally called $\varphi$, and an increasing sequence of closed subspaces $V_{j}$ of $L^{2}(\mathbb {R})$ with certain properties. The wavelet that is constructed is traditionally called $\psi$.
Section 4 introduces the Shannon wavelet, whose construction is immediate from the theory of multiresolution analyses. The new ingredient here, beyond the ideas used for the Haar wavelet, is the careful use of the generating function of the scaling function to obtain a formula for the wavelet.
Section 5 supplements the theory of Section 3 by showing how to build a multiresolution analysis out of a candidate for the scaling function.
Section 6 introduces the Meyer wavelets, each of which is smooth and has Fourier transform of a prescribed order of differentiability. The full theory of Sections 3 and 5 is used in their construction.
Section 7 introduces splines, examines one example, and sees the need for more theory. It develops one further aspect of the general theory, showing how to replace a “Riesz system” with an orthonormal set. It therefore allows one to relax the conditions needed in Section 5 for a function to be a scaling function. In addition, it uses elementary complex analysis to prove a series expansion for $\pi^{2}/\sin^{2}\pi z$ that is needed in Section 8.
Section 8 continues the discussion of the role of splines in the theory of wavelets, introducing the Battle–Lemarié wavelets. As with the Meyer wavelets each is smooth and has Fourier transform of a prescribed order of differentiability.
Section 9 develops the Daubechies wavelets. These have compact support, but except for the first one, neither they nor their Fourier transforms have known formulas in closed form. The construction begins by pinpointing necessary conditions on the generating function.
Section 10 deals with smoothness questions. It contains three results. The first gives an estimate for the decay of the Fourier transform of the Daubechies scaling function of each order. The second deduces a certain amount of differentiability of a scaling function from the estimate in the first result. The third shows in the converse direction that a Daubechies wavelet can never be of class $C^{∞}$. The section concludes with a table summarizing properties of the specific wavelets that have been constructed in Sections 2–9.
Section 11 gives a quick introduction to applications. It discusses the discrete wavelet transform and its use in storage and compression of data, it identifies some applications of wavelets in one and two dimensions, and it makes brief remarks about some of the applications.
Chapter information Source Advanced Real Analysis, Digital Second Edition, Corrected version (East Setauket, NY: Anthony W. Knapp, 2017) Dates First available in Project Euclid: 21 May 2018 Permanent link to this document https://projecteuclid.org/euclid.bia/1526871324 Digital Object Identifier doi:10.3792/euclid/9781429799911-10 Rights Copyright © 2017, Anthony W. Knapp Citation
Knapp, Anthony W. Chapter X. Introduction to Wavelets. Advanced Real Analysis, 434--544, Anthony W. Knapp, East Setauket, New York, 2017. doi:10.3792/euclid/9781429799911-10. https://projecteuclid.org/euclid.bia/1526871324
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If Two Subsets $A, B$ of a Finite Group $G$ are Large Enough, then $G=AB$
Problem 493
Let $G$ be a finite group and let $A, B$ be subsets of $G$ satisfying\[|A|+|B| > |G|.\]Here $|X|$ denotes the cardinality (the number of elements) of the set $X$.Then prove that $G=AB$, where\[AB=\{ab \mid a\in A, b\in B\}.\]
Since $A, B$ are subsets of the group $G$, we have $AB\subset G$.Thus, it remains to show that $G\subset AB$, that is any element $g\in G$ is of the form $ab$ for some $a\in A$ and $b\in B$.This is equivalent to finding $a\in A$ and $b\in B$ such that $gb^{-1}=a$.
Consider the subset\[B^{-1}:=\{b^{-1} \mid b \in B\}.\]Since taking the inverse gives the bijective map $B \to B^{-1}$, $b \mapsto b^{-1}$, we have $|B|=|B^{-1}|$.
Also consider the subset\[gB^{-1}=\{gb^{-1} \mid b\in B\}.\]Note that multiplying by $g$ and by its inverse $g^{-1}$ give the bijective maps\[B^{-1} \to gB^{-1}, b^{-1} \mapsto gb^{-1} \text{ and } gB^{-1} \to B^{-1}, gb^{-1} \mapsto b^{-1}.\]Hence we have\[ |B|=|B^{-1}|=|gB^{-1}|.\]
Since $A$ and $gB^{-1}$ are both subsets in $G$ and we have by assumption that\[|A|+|gB^{-1}|=|A|+|B| > |G|,\]the intersection $A\cap gB^{-1}$ cannot be empty.
Therefore, there exists $a \in A\cap gB^{-1}$, and thus $a\in A$ and $a=gb^{-1}$ for some $b\in B$.As a result we obtain $g=ab$.It yields that $G\subset AB$, and we have $G=AB$ as a consequence.
Related Question.
As an application, or use the similar technique, try the following
Problem.Every element in a finite field $F$ is the sum of two squares in $F$.
Finite Group and a Unique Solution of an EquationLet $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that\[b^m=a.\]We give two proofs.Proof 1.Since $m$ and $n$ are relatively prime […]
Group Homomorphism, Preimage, and Product of GroupsLet $G, G'$ be groups and let $f:G \to G'$ be a group homomorphism.Put $N=\ker(f)$. Then show that we have\[f^{-1}(f(H))=HN.\]Proof.$(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$.It follows that there exists $h\in H$ […]
The Product of a Subgroup and a Normal Subgroup is a SubgroupLet $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$.The product of $H$ and $N$ is defined to be the subset\[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\]Prove that the product $H\cdot N$ is a subgroup of […]
Nontrivial Action of a Simple Group on a Finite SetLet $G$ be a simple group and let $X$ be a finite set.Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.Proof.Since $G$ acts on $X$, it […]
Normal Subgroup Whose Order is Relatively Prime to Its IndexLet $G$ be a finite group and let $N$ be a normal subgroup of $G$.Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.(a) Prove that $N=\{a\in G \mid a^n=e\}$.(b) Prove that $N=\{b^m \mid b\in G\}$.Proof.Note that as $n$ and […]
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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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Assume that:
$P(X): \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $Q(X): \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $E(K, X): \{0, 1\}^{\ell_{K}} \times \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $D(K, X): \{0, 1\}^{\ell_{K}} \times \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $\widetilde{E}(K, T, X): \{0, 1\}^{\ell_{K}} \times \{0, 1\}^{\ell_{T}} \times \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $\widetilde{D}(K, T, X): \{0, 1\}^{\ell_{K}} \times \{0, 1\}^{\ell_{T}} \times \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $E_{*}(K, IV, T, X): \{0, 1\}^{\ell_{K}} \times \{0, 1\}^{\ell_{IV}} \times \{0, 1\}^{\ell_{T}} \times \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $D_{*}(K, IV, T, X): \{0, 1\}^{\ell_{K}} \times \{0, 1\}^{\ell_{IV}} \times \{0, 1\}^{\ell_{T}} \times \{0, 1\}^{\ell_{X}} \rightarrow \{0, 1\}^{\ell_{X}}$. $\ell_{X}$, $\ell_{K}$, $\ell_{T}$, and $\ell_{IV}$ are equal. $P$ and $Q$ are each other's inverses. $E(K, X) = P(X \oplus K) \oplus K$. $D(K, X) = Q(X \oplus K) \oplus K$. $\widetilde{E}(K, T, X) = E(E(K, T), X)$. $\widetilde{D}(K, T, X) = D(E(K, T), X)$. $E_{*}(K, IV, T, X) = \widetilde{E}(E(K, IV), T, X)$. $D_{*}(K, IV, T, X) = \widetilde{D}(E(K, IV), T, X)$.
$P$ and $Q$ are inverses and are both publicly known pseudorandom permutations that form the core of a single-key Even-Mansour block cipher. $\widetilde{E}$/$\widetilde{D}$ and $E_{*}$/$D_{*}$ are two modes of operation that I've constructed from it. With the mode of operation built around $\widetilde{E}$/$\widetilde{D}$, $CT_{i} = \widetilde{E}(K, i, PT_{i})$ and $PT_{i} = \widetilde{D}(K, i, CT_{i})$. With the mode of operation built around $E_{*}$/$D_{*}$, $CT_{i} = E_{*}(K, IV, i, PT_{i})$ and $PT_{i} = D_{*}(K, IV, i, CT_{i})$.
As I understand the Even-Mansour scheme, an attacker who possesses $2^{0.5 \times \ell_{X}}$ pairs of plaintext and ciphertext blocks can break the security of it. In the case of the first mode of operation, the key used for each block is different and is derived by encrypting the block's index with $K$ as the key. In the case of the second, an instance vector (as opposed to "initialization vector") is encrypted with $K$ as the key and this new value is used as the key the same way as in the first mode of operation. In either mode of operation does the birthday bound apply? Does security disappear after knowing $2^{0.5 \times \ell_{X}}$ pairs of plaintexts and ciphertexts? Or does the security become optimal?
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Under the auspices of the Computational Complexity Foundation (CCF)
In 1994, Y. Mansour conjectured that for every DNF formula on $n$ variables with $t$ terms there exists a polynomial $p$ with $t^{O(\log (1/\epsilon))}$ non-zero coefficients such that $\E_{x \in \{0,1\}}[(p(x)-f(x))^2] \leq \epsilon$. We make the first progress on this conjecture and show that it is true for several natural subclasses of DNF formulas including randomly chosen DNF formulas and read-$k$ DNF formulas for constant $k$.
Our result yields the first polynomial-time query algorithm for agnostically learning these subclasses of DNF formulas with respect to the uniform distribution on $\{0,1\}^n$ (for any constant error parameter).
Applying recent work on sandwiching polynomials, our results imply that a $t^{-O(\log 1/\epsilon)}$-biased distribution fools the above subclasses of DNF formulas. This gives pseudorandom generators for these subclasses with shorter seed length than all previous work.
The read-k section has been fixed.
In 1994, Y. Mansour conjectured that for every DNF formula on $n$ variables with $t$ terms there exists a polynomial $p$ with $t^{O(\log (1/\epsilon))}$ non-zero coefficients such that $\E_{x \in \{0,1\}}[(p(x)-f(x))^2] \leq \epsilon$. We make the first progress on this conjecture and show that it is true for randomly chosen DNF formulas and read-once DNF formulas.
Our result yields the first polynomial-time query algorithm for agnostically learning these subclasses of DNF formulas with respect to the uniform distribution on $\{0,1\}^n$ (for any constant error parameter).
Applying recent work on sandwiching polynomials, our results imply that a $t^{-O(\log 1/\epsilon)}$-biased distribution fools the above subclasses of DNF formulas. This gives pseudorandom generators for these subclasses with shorter seed length than all previous work.
There was an error in the previous version where similar results were shown for read-k DNF formulas.
In 1994, Y. Mansour conjectured that for every DNF formula on $n$ variables with $t$ terms there exists a polynomial $p$ with $t^{O(\log (1/\epsilon))}$ non-zero coefficients such that $\E_{x \in \{0,1\}}[(p(x)-f(x))^2] \leq \epsilon$. We make the first progress on this conjecture and show that it is true for several natural subclasses of DNF formulas including randomly chosen DNF formulas and read-$k$ DNF formulas for constant $k$.
Our result yields the first polynomial-time query algorithm for agnostically learning these subclasses of DNF formulas with respect to the uniform distribution on $\{0,1\}^n$ (for any constant error parameter).
Applying recent work on sandwiching polynomials, our results imply that a $t^{-O(\log 1/\epsilon)}$-biased distribution fools the above subclasses of DNF formulas. This gives pseudorandom generators for these subclasses with shorter seed length than all previous work.
In 1994, Y. Mansour conjectured that for every DNF formula on $n$ variables with $t$ terms there exists a polynomial $p$ with $t^{O(\log (1/\epsilon))}$ non-zero coefficients such that $\E_{x \in \{0,1\}}[(p(x)-f(x))^2] \leq \epsilon$. We make the first progress on this conjecture and show that it is true for several natural subclasses of DNF formulas including randomly chosen DNF formulas and read-$k$ DNF formulas for constant $k$.
Our result yields the first polynomial-time query algorithm for agnostically learning these subclasses of DNF formulas with respect to the uniform distribution on $\{0,1\}^n$ (for any constant error parameter).
Applying recent work on sandwiching polynomials, our results imply that a $t^{-O(\log 1/\epsilon)}$-biased distribution fools the above subclasses of DNF formulas. This gives pseudorandom generators for these subclasses with shorter seed length than all previous work.
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The solutions of the Schrödinger equation for a hydrogen atom have definite energy. Does this mean that they could be written as a superposition of plane waves of a single frequency - corresponding to that energy - with only the phases and directions differing?
Eigenstates of any Hamiltonian by definition have definite energy.
Let's consider a simpler system: 1D harmonic oscillator, i.e. system with potential $U=x^2$. The only plane waves with different directions and phases you can sum in 1D case is $e^{ikx}$ and $e^{-i(kx+\phi)}$. Now any way you try to sum these two plane waves, this will only give you a function, which oscillates at infinity. But there's no eigenstate here, which oscillates at infinity — all are bound. Thus they can't be represented as a linear combination of plane waves with single frequency.
In hydrogen there's also infinitely many bound states, which don't oscillate at infinity, and summing plane waves over sphere will give you spherical Bessel functions and the like, which do oscillate. So, the answer is no.
What you seem to confuse is frequency versus wave length. Plane waves are eigenstates of Hamiltonian
only for a free particle. If a particle in state of plane wave is put in an inhomogeneous potential, it will no longer be in an eigenstate: its state will be a superposition of different energy states, thus there'll be no single frequency. What is a characteristic for a plane wave is its wavelength, which doesn't depend on potential (although the wave will be distorted in time evolution if it's not an eigenstate, and cease to be a plane wave).
Since hydrogen atom is characterized by Coulomb potential, which is non-constant, plane wave can't have single frequency in such a system — it'll distort in the next moment of time.
The solutions to the time-independent Schrödinger equation are independent of frequency, so your answer is trivially yes.
You might also ask whether the spatial parts of the wavefunction can be written as a superposition of plane waves $e^{ i\vec k \cdot \vec x}$ (they can, by Fourier decomposition) and whether the wavevectors $\vec k$ all have the same magnitude.
I had originally written that not all the plane wavevectors can have the same magnitude, since the bound wavefunctions are localized. A comment by Sebastian Henckel has made me realize this is not correct for plane wave which all have the same phase at the origin. However, I think it'd be quite challenging to produce the hydrogen radial wavefunctions using only plane waves which are coherent at the origin — in particular the feature that the $n$th radial wavefunction has exactly $n-1$ nodes at different values for $r$. A set of plane waves, in phase at the origin, will have infinitely many nodes in $r$, in spherical shells.
The physics here is that the states of the hydrogen which have definite energy do not have describe electrons with definite momentum, or definite wavenumber $|k|$.
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2019-09-04 12:06
Soft QCD and Central Exclusive Production at LHCb / Kucharczyk, Marcin (Polish Academy of Sciences (PL)) The LHCb detector, owing to its unique acceptance coverage $(2 < \eta < 5)$ and a precise track and vertex reconstruction, is a universal tool allowing the study of various aspects of electroweak and QCD processes, such as particle correlations or Central Exclusive Production. The recent results on the measurement of the inelastic cross section at $ \sqrt s = 13 \ \rm{TeV}$ as well as the Bose-Einstein correlations of same-sign pions and kinematic correlations for pairs of beauty hadrons performed using large samples of proton-proton collision data accumulated with the LHCb detector at $\sqrt s = 7\ \rm{and} \ 8 \ \rm{TeV}$, are summarized in the present proceedings, together with the studies of Central Exclusive Production at $ \sqrt s = 13 \ \rm{TeV}$ exploiting new forward shower counters installed upstream and downstream of the LHCb detector. [...] LHCb-PROC-2019-008; CERN-LHCb-PROC-2019-008.- Geneva : CERN, 2019 - 6. Fulltext: PDF; In : The XXVII International Workshop on Deep Inelastic Scattering and Related Subjects, Turin, Italy, 8 - 12 Apr 2019 Úplný záznam - Podobné záznamy 2019-08-15 17:39
LHCb Upgrades / Steinkamp, Olaf (Universitaet Zuerich (CH)) During the LHC long shutdown 2, in 2019/2020, the LHCb collaboration is going to perform a major upgrade of the experiment. The upgraded detector is designed to operate at a five times higher instantaneous luminosity than in Run II and can be read out at the full bunch-crossing frequency of the LHC, abolishing the need for a hardware trigger [...] LHCb-PROC-2019-007; CERN-LHCb-PROC-2019-007.- Geneva : CERN, 2019 - mult.p. In : Kruger2018, Hazyview, South Africa, 3 - 7 Dec 2018 Úplný záznam - Podobné záznamy 2019-08-15 17:36
Tests of Lepton Flavour Universality at LHCb / Mueller, Katharina (Universitaet Zuerich (CH)) In the Standard Model of particle physics the three charged leptons are identical copies of each other, apart from mass differences, and the electroweak coupling of the gauge bosons to leptons is independent of the lepton flavour. This prediction is called lepton flavour universality (LFU) and is well tested. [...] LHCb-PROC-2019-006; CERN-LHCb-PROC-2019-006.- Geneva : CERN, 2019 - mult.p. In : Kruger2018, Hazyview, South Africa, 3 - 7 Dec 2018 Úplný záznam - Podobné záznamy 2019-05-15 16:57 Úplný záznam - Podobné záznamy 2019-02-12 14:01
XYZ states at LHCb / Kucharczyk, Marcin (Polish Academy of Sciences (PL)) The latest years have observed a resurrection of interest in searches for exotic states motivated by precision spectroscopy studies of beauty and charm hadrons providing the observation of several exotic states. The latest results on spectroscopy of exotic hadrons are reviewed, using the proton-proton collision data collected by the LHCb experiment. [...] LHCb-PROC-2019-004; CERN-LHCb-PROC-2019-004.- Geneva : CERN, 2019 - 6. Fulltext: PDF; In : 15th International Workshop on Meson Physics, Kraków, Poland, 7 - 12 Jun 2018 Úplný záznam - Podobné záznamy 2019-01-21 09:59
Mixing and indirect $CP$ violation in two-body Charm decays at LHCb / Pajero, Tommaso (Universita & INFN Pisa (IT)) The copious number of $D^0$ decays collected by the LHCb experiment during 2011--2016 allows the test of the violation of the $CP$ symmetry in the decay of charm quarks with unprecedented precision, approaching for the first time the expectations of the Standard Model. We present the latest measurements of LHCb of mixing and indirect $CP$ violation in the decay of $D^0$ mesons into two charged hadrons [...] LHCb-PROC-2019-003; CERN-LHCb-PROC-2019-003.- Geneva : CERN, 2019 - 10. Fulltext: PDF; In : 10th International Workshop on the CKM Unitarity Triangle, Heidelberg, Germany, 17 - 21 Sep 2018 Úplný záznam - Podobné záznamy 2019-01-15 14:22
Experimental status of LNU in B decays in LHCb / Benson, Sean (Nikhef National institute for subatomic physics (NL)) In the Standard Model, the three charged leptons are identical copies of each other, apart from mass differences. Experimental tests of this feature in semileptonic decays of b-hadrons are highly sensitive to New Physics particles which preferentially couple to the 2nd and 3rd generations of leptons. [...] LHCb-PROC-2019-002; CERN-LHCb-PROC-2019-002.- Geneva : CERN, 2019 - 7. Fulltext: PDF; In : The 15th International Workshop on Tau Lepton Physics, Amsterdam, Netherlands, 24 - 28 Sep 2018 Úplný záznam - Podobné záznamy 2019-01-10 15:54 Úplný záznam - Podobné záznamy 2018-12-20 16:31
Simultaneous usage of the LHCb HLT farm for Online and Offline processing workflows LHCb is one of the 4 LHC experiments and continues to revolutionise data acquisition and analysis techniques. Already two years ago the concepts of “online” and “offline” analysis were unified: the calibration and alignment processes take place automatically in real time and are used in the triggering process such that Online data are immediately available offline for physics analysis (Turbo analysis), the computing capacity of the HLT farm has been used simultaneously for different workflows : synchronous first level trigger, asynchronous second level trigger, and Monte-Carlo simulation. [...] LHCb-PROC-2018-031; CERN-LHCb-PROC-2018-031.- Geneva : CERN, 2018 - 7. Fulltext: PDF; In : 23rd International Conference on Computing in High Energy and Nuclear Physics, CHEP 2018, Sofia, Bulgaria, 9 - 13 Jul 2018 Úplný záznam - Podobné záznamy 2018-12-14 16:02
The Timepix3 Telescope andSensor R&D for the LHCb VELO Upgrade / Dall'Occo, Elena (Nikhef National institute for subatomic physics (NL)) The VErtex LOcator (VELO) of the LHCb detector is going to be replaced in the context of a major upgrade of the experiment planned for 2019-2020. The upgraded VELO is a silicon pixel detector, designed to with stand a radiation dose up to $8 \times 10^{15} 1 ~\text {MeV} ~\eta_{eq} ~ \text{cm}^{−2}$, with the additional challenge of a highly non uniform radiation exposure. [...] LHCb-PROC-2018-030; CERN-LHCb-PROC-2018-030.- Geneva : CERN, 2018 - 8. Úplný záznam - Podobné záznamy
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Normal Lines on a Surface
We have just looked at Tangent Planes to Surfaces to a point on a surface of a two variable real-valued function $z = f(x, y)$. We will now look at what a normal line on a surface $S$ at point $P$ is.
First, let $z = f(x, y)$ be a two variable real-valued function that generates the smooth surface $S$, and let $(x_0, y_0) \in D(f)$. Let $P$ be the point that lies on the surface $S$ with coordinates $P(x_0, y_0, z_0)$, and let $\Pi$ be the tangent plane to $S$ at $P$.
Recall that the vertical plane $y = y_0$ intersects $S$ forming a curve of intersection $C_1$. The tangent line $T_1$, of this curve at point $P$ has slope $\frac{\partial}{\partial x} f(x_0, y_0)$. Also recall that the tangent plane $\Pi$ contains $T_1$ by definition, and more precisely, the tangent plane $\Pi$ intersects the vertical plane $y = y_0$ to form $T_1$. Therefore $T_1$ is parallel to the vector we define as $\vec{T_1} = \vec{i} + 0 \vec{j} + \frac{\partial}{\partial x} f(x_0, y_0) \vec{k}$.
Similarly, the vertical plane $x = x_0$ intersects $S$ forming a curve of intersection $C_2$. The tangent line $T_2$, of this curve at point $P$ has sloe $\frac{\partial}{\partial y} f(x_0, y_0)$. Similarly, the tangent plane $\Pi$ contains $T_2$ by definition, and more precisely, the tangent plane $\Pi$ intersects the vertical plane $x = x_0$ to form $T_2$. Therefore $T_2$ is parallel to the vector we define as $\vec{T_2} = 0\vec{i} + \vec{j} + \frac{\partial}{\partial y} f(x_0, y_0) \vec{k}$.
Since $\vec{T_1}$ and $\vec{T_2}$ are parallel to both $T_1$ and $T_2$ respectively, then the cross product $\vec{T_2} \times \vec{T_1}$ produces a vector that is perpendicular to both of these tangent lines. Since $\Pi$ contains both of these tangent lines, it follows that the cross product $\vec{T_2} \times \vec{T_1}$ produces a vector that is perpendicular to the tangent plane $\Pi$, or rather, produces a normal vector for $\Pi$.(1)
Notice that we can once again obtain the equation of the tangent plane of $P$ by expanding the dot product $(\vec{T_2} \times \vec{T_1} ) \cdot (x - x_0, y - y_0, z - z_0) = 0$.
However, now that we have this normal vector $\vec{T_2} \times \vec{T_1}$ to the tangent plane at $P$, we note that $\vec{T_2} \times \vec{T_1}$ will be parallel to the normal line corresponding to $P$. We have all we need to construct the equation of a normal line according to the Equations of Lines in Three-Dimensional Space page. Since we have a point ($P(x_0, y_0, z_0)$) that passes through this line and a vector ($\vec{T_2} \times \vec{T_1}$) that is parallel to this line, then we obtain the following parametric equations for this normal line.(2)
Definition: Let $z = f(x, y)$ be a two variable real-valued function, and let $P(x_0, y_0, z_0)$ be a point on the surface generated by $f$. The Normal Line at $P$ is the line that passes through $P$ and is perpendicular to the tangent plane at $P$ and perpendicular to the surface $S$ at $P$.
Let's look at some examples of finding normal lines on a surface.
Example 1 Let $z = f(x, y) = -2xe^xy^2 - 2y$. Find the equation of the normal line at $(-1, -1) \in D(f)$.
We first find the partial derivatives of $f$ as $\frac{\partial z}{\partial x} = [-2e^x - 2xe^x]y^2$, and $\frac{\partial z}{\partial y} = -4xe^xy - 2$. If we evaluate both of these partial derivatives at $x = -1$ and $y = -1$, we get that:(3)
We also note that $f(-1, -1) = \frac{2}{e} + 2$. Therefore we have that the equation of the normal line has the following parametric equations:(5)
The graph of $f$ as well as the tangent plane and normal line are depicted below:
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To amplify Yehuda Lindell's answer even within a single family of groups, here is an example of a 2046-bit modulus $p$ for which discrete logs in $(\mathbb Z/p\mathbb Z)^\times$, as in standard modular multiplication Diffie–Hellman, are extraordinarily easy to compute:
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.
2046-bit moduli should be secure, right? No, not at all. First of all, it helps if they're prime (or at least if you don't know their prime factorization, but for public DH or DSA parameters, you would like to know this sort of thing about them). This number $p$ is not prime; rather, it is the product of the first 232 prime numbers except 2. So, given fixed $h$, to solve $$g^x \equiv h \pmod p$$ for $x$, it suffices to compute the discrete logs $x_3$, $x_5$, $x_7$,
etc., satisfying
\begin{align*} g^{x_3} &\equiv h \pmod 3, \\ g^{x_5} &\equiv h \pmod 5, \\ &\vdots\end{align*}
independently, and then assemble them into a solution for $x$ with the Chinese remainder theorem:
\begin{align*} x &\equiv x_3 \pmod{\phi(3)}, \\ x &\equiv x_5 \pmod{\phi(5)}, \\ &\vdots\end{align*}
where $\phi(3) = 3 - 1$, $\phi(5) = 5 - 1$,
etc., since they're all prime.
This algorithm is so cheap that you could do it with
pen and paper using schoolbook long division and schoolbook multiplication in a few hours. But the easiness of this algorithm means nothing about how easy or hard it is to compute discrete logs in $(\mathbb Z/q\mathbb Z)^\times$ where $q$ is the RFC 3526 Group #14 modulus
0xffffffffffffffffc90fdaa22168c234c4c6628b80dc1cd129024e088a67cc74020bbea63b139b22514a08798e3404ddef9519b3cd3a431b302b0a6df25f14374fe1356d6d51c245e485b576625e7ec6f44c42e9a637ed6b0bff5cb6f406b7edee386bfb5a899fa5ae9f24117c4b1fe649286651ece45b3dc2007cb8a163bf0598da48361c55d39a69163fa8fd24cf5f83655d23dca3ad961c62f356208552bb9ed529077096966d670c354e4abc9804f1746c08ca18217c32905e462e36ce3be39e772c180e86039b2783a2ec07a28fb5c55df06f4c52c9de2bcbf6955817183995497cea956ae515d2261898fa051015728e5a8aacaa68ffffffffffffffff.
The same algorithm using the Chinese remainder theorem doesn't work because $q$ is prime, so you can't solve the problem independently modulo all its factors and then recombine the results into a solution.
The choice of field, curve shape, curve parameters,
etc., has the same kind of impact on the ease or difficulty of computing elliptic-curve discrete logs as the choice of modulus has on the ease or difficulty of computing modular multiplication discrete logs.
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Please let me assume that you wonder about control surface effectiveness rather than their efficiency. Both are closely related, but I prefer to address their effectiveness - doing what the pilot demands them to do.
The aerodynamic forces are proportional to the dynamic pressure $q$ of the flow, which is density times velocity squared, as in$$q = \frac{\rho}{2}\cdot v^2$$
To create the same amount of lift at a higher altitude, the airplane has to fly faster, so in effect the forces can stay the same. Only the true air speed will be different, which helps to get where you want faster. The control surface effectivity remains unchanged.
If we dig a little deeper, the atmosphere cools down with increasing altitude until you reach the Tropopause. This cooling makes the engines more efficient but also reduces the Reynolds number of the flow around the aircraft. A lower Reynolds number translates into a thicker boundary layer, which will reduce the effectiveness of control surfaces a tiny bit and will reduce the range of their best effectivity. But this effect is so small that it has no practical consequence.
Airliners fly at transsonic speeds at altitude, so now sonic effects become more pronounced. For the control surfaces this means that they will lose their effectiveness at higher deflection angles. For small corrections all is very similar to low level, slow flight, but if you need full deflections the control surfaces are indeed less effective at altitude.
Another effect will be rather more pronounced, and this is aerodynamic damping. Damping is the tendency of a system to create additional forces from movements which run against this movement. Take a wing: If the aircraft rolls, the wingtips will move up on one side and down on the other. Both movements will add a small angle of attack, lowering the total angle of attack on the up moving tip and increasing it on the down moving tip. Both tips will see a change in lift which will counteract the rolling motion. If the rolling speed is $\omega_x$, the resulting change in angle of attack $\alpha$ at a wing station $y$ is$$\Delta\alpha = arctan \left(\frac{\omega_x \cdot y}{v}\right)$$As you can see, the flight speed is in the denominator, so a higher flight speed will create a smaller angle change for the same roll rate. The same holds true for the other axes of movement, and in consequence the higher-flying aircraft needs either more pilot attention or an artificial damper.
I guess it is the lower damping which made you think that the control surfaces are less effective. They aren't.
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My country has major economical headaches each 10 years or so, like hyperinflation or defaults or things like that. It's been 17 years since the last major crisis (a larger crisis than the 1930's crisis) but it looks like we are aiming again towards the same. Usually we have trade imbalances or fiscal deficits which causes my country being unable to sustain the current money exchange rate, the goverment devaluates the currency and that leads people to increase the price of the goods in the same rate "just in case other prices increase the same rate", that usually is the cause of a snowball inflation. I was taking a look at our current imports. Usually I can't get a detailed list but we have a trade imbalance deficit record of 8,47 billions last year (2017) and the car imports increased in 2 billions in one year, being 6,3 billion total the amount of imported cars. My question is, being the trade imbalance a major cause in the inflation we have, why we just can't stop importing cars, at least in the proportion it creates imbalances? Why we can't live with new cars or imported (non national) cars? Could there be any serious consequences to this or we are just paying the consequences of some people wanting to live with goods/cars which signals their high social status?
1- Some basic macroeconomic fundamentals
I highly doubt a trade imbalance is causing high inflation. The trade deficit is likely happening because of economists call "Twin Deficits". I'll spare you the accounting that leads to this equation and present you a setorial balances identity. Let $S$ be private savings, $I$ be capital formation, $G$ be government spending, $T$ be government revenue (in taxes), $X$ be exports and $M$, imports. We have:
$$(S-I) + (T-G) = (X-M)$$
In intuitive terms, the private sector balance plus the public sector balance, which is basically the domestic balance, must equal the trade balance. This happens due to sheer accounting, it's not a "theory". This is the same in the US, Venezuela, China and in a tribe of herdsman.
When $(T-G)$ is negative, we have a budget deficit - your country's case. Assuming people are saving the same amount, investment must fall or imports must go up. The more realistic scenario is that both things happen.
And we also have the Long-Run Fiscal Equilibrium condition. Let $M^s$ be money supply, $B$ be government debt, $P$ the price level, $r$ be the real interest rate and $t$ be time, we have:
$$ \frac{M^s + B}{P} = \sum_{t=0}^{\infty} \frac{(T_t - G_t)}{(1+r)^t}$$
This basically means that real government passive must equal the liquid present value of expected government savings. If people expect the government to continue to have budget deficits, either interest rates or the price level must go up. That's what's causing inflation. This also means that if the government increases the money supply without credibly signalling it will save more in the future, it will also implicate an increase in the price level, inflation.
2- Some basic microeconomic fundamentals
If importing cars gets prohibited, the supply of cars will go down. The supply curve shifts to the left. This means less cars available and more expensive ones.
Higher car prices signal domestic producers to make more cars. Supply goes up again and prices tend to converge to a new equilibrium. The problem is making cars domesctically might no be very efficient. Your country employs more capital and labor in other activities because it is more productive to do so. If it weren't and there was an easy opportunity to make money by producing cars, people would just do it.
Restricting imports is basically making your country poorer by forcing it to have less goods and services available and allocating precious capital and labor to activities that aren't as productive. It's a
very bad idea.
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Let $K_n$ be the sets of vectors $x \in \mathbb{Z}^d $ with each coordinates $x_i$ between $1$ and $n$. For any subset $A$ of $K_n$, let $S(A)$ be the set of points $x \in K_n$ which are on some line containing at least two points of $A$ (in other words, $S(A)$ is the union of the lines passing through - at least - two points of $A$).
Such a set $A$ is said to
generate $K_n$ if $S(A) = K_n$. Now let $r_d(n)$ be the smallest size of a generating subset of $K_n$.
Question : What are the best known bounds on $r_d(n)$ ? (The first non trivial case is $d=2$)
This problem may be "well-known" ; I'm almost sure this question has already been studied, but I didn't find any reference, and Google gives nothing.
The trivial bound is $r_d(n) \gg_d n^{\frac{d}{2}- \frac{1}{2}} $ : taking a generating subset of size $r_d(n)$, there are at most $O(r_d(n)^2)$ lines to consider, each one intersecting $K_n$ in at most $n$ points, so that $|K_n| \ll r_d(n)^2 \times n$.
A refinement of this argument (a typical line contains much less than $n$ points of $K_n$) gives a lower bound $r_d(n) \gg_d n^{\frac{d}{2}- \frac{1}{4} - \frac{1}{4(2d-1)} } $.
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Limits to Infinity and Negative Infinity
Limits to Infinity and Negative Infinity
Definition: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then we say that the limit as $x \to c$ of $f$ is $\infty$ written $\lim_{x \to c} f(x) = \infty$ if $\forall \alpha \in \mathbb{R}$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $f(x) > \alpha$. Similarly, we say that the limit as $x \to c$ of $f$ is $-\infty$ written $\lim_{x \to c} f(x) = -\infty$ if $\forall \beta \in \mathbb{R}$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x -c \mid < \delta$ then $f(x) < \beta$.
The following diagrams represent the definition of a limit to infinity (left) and to negative infinity (right).
We will now look at a comparison theorem (analogous to that of sequences) of limits to infinity and negative infinity.
Theorem 1: Let $f : A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ be functions and let $c$ be a cluster point of $A$. Then: 1) If $\forall x \in A$, $x \neq c$ we have that $f(x) ≤ g(x)$ and $\lim_{x \to c} f(x) = \infty$ then $\lim_{x \to c} g(x) = \infty$. 2) If $\forall x \in A$, $x \neq c$ we have that $f(x) ≤ g(x)$ and $\lim_{x \to c} g(x) = -\infty$ then $\lim_{x \to c} f(x) = -\infty$. Proof:Suppose that $\lim_{x \to c} f(x) = \infty$ and let $\alpha \in \mathbb{R}$ be given. Then $\exists \delta > 0$ such that if $x \in A$, and $0 < \mid x - c \mid < \delta$ then $f(x) > \alpha$. But $g(x) ≥ f(x)$ $\forall x \in A$ and so for this $\delta > 0$, if $x \in A$ and $0 < \mid x - c \mid < \delta$ we have that $g(x) > \alpha$, and so $\lim_{x \to c} g(x) = \infty$. Similarly, suppose that $\lim_{x \to c} g(x) = -\infty$ and let $\beta \in \mathbb{R}$ be given. Then $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $g(x) < \beta$. But $f(x) ≤ g(x)$ $\forall x \in A$ and so for this $\delta > 0$, if $x \in A$ and $0 < \mid x - c \mid < \delta$ we have that $f(x) < \beta$, and so $\lim_{x \to c} f(x) = -\infty$. $\blacksquare$
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Difference between revisions of "Main Page"
(→Unsolved questions)
(→Threads)
Line 18: Line 18:
[http://blogsearch.google.com/blogsearch?hl=en&ie=UTF-8&q=polymath1&btnG=Search+Blogs Here is a further list of blog posts related to the Polymath1 project]. [http://en.wordpress.com/tag/polymath1/ Here is wordpress's list].
[http://blogsearch.google.com/blogsearch?hl=en&ie=UTF-8&q=polymath1&btnG=Search+Blogs Here is a further list of blog posts related to the Polymath1 project]. [http://en.wordpress.com/tag/polymath1/ Here is wordpress's list].
−
A spreadsheet containing the latest
+
A spreadsheet containing the latest upper bounds for <math>c_n</math> can be found [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DsU-uZ1tK7VEg here].
Here are some [[unsolved problems]] arising from the above threads.
Here are some [[unsolved problems]] arising from the above threads.
− −
== Bibliography ==
== Bibliography ==
Revision as of 18:53, 12 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.
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)
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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I'm typesetting a bunch of source code, and I really like the way the double-nested logical-and and logical-or symbols look for denoting bitwise logical and/or.
Unfortunately, LaTeX doesn't seem to have these symbols, so I resorted to rolling my own using a combination of
\land and
\lor with carefully rotated and aligned rules. The results are acceptable to my eye at 10pt, but the placement of the rules is fragile for a couple of reasons. Here's how they look magnified:
The first reason they're fragile is because (against my better judgment, but not knowing a better way), I'm making assumptions about spacing:
\newcommand{\bland}{% bitwise logical and \hbox{% $\land$% \hspace{-.47em}% \raisebox{-.04em}{% \rotatebox{66}{% \vrule width .3em height .45pt depth 0pt% }% \hspace{.065em}% \rotatebox{-66}{% \hspace{-.3em}% \vrule width .3em height .45pt depth 0pt% }% }% \hspace{.20em}% }%}
Thus, when shown in a smaller size—for example when stacked atop
= or
\Leftarrow—the internal lines are off by a small but noticeable amount.
The second reason is more insidious and seems to have to do with rounding error in PDF conversion or display: the inner wedge "wobbles" half a pixel or so (as compared to the outer wedge) at various sizes.
I'm wondering what the best solution is here. I'm not opposed (in theory) to diving into XeLaTeX to see how
U+2A53 and
U+2A54 look, but if possible I'd prefer to avoid dependencies on XeTeX/XeLaTeX for this. Also, the first thing I tried was nesting a
\tiny\land inside a
\land but the slopes of the strokes were different. Basically I need something with the same slopes and thicknesses, and it would also be nice if they had rounded edges, but I haven't the TeXpertise to make that happen.
Here's how the symbols look at a more regular size (top line is regular boolean logical operators; second two lines are bitwise boolean operators):
By the way, I did come across a double-nested less-than and greater-than in a standard LaTeX package, which I think I could
maybe use—if I rotated them 90 degrees—but the problem I have with those is that they override/redefine
\ll and
\gg rather than adding new control sequences, and I have places elsewhere where I want to use the real
\ll and
\gg.
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Difference between revisions of "Main Page"
(→Bibliography)
(→Threads)
Line 25: Line 25:
A spreadsheet containing the latest upper and lower bounds for <math>c_n</math> can be found [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DsU-uZ1tK7VEg here]. Here are the proofs of our [[upper and lower bounds]] for these constants.
A spreadsheet containing the latest upper and lower bounds for <math>c_n</math> can be found [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DsU-uZ1tK7VEg here]. Here are the proofs of our [[upper and lower bounds]] for these constants.
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We are also collecting bounds for [[Fujimura's problem]].
+
We are also collecting bounds for [[Fujimura's problem]].
Here are some [[unsolved problems]] arising from the above threads.
Here are some [[unsolved problems]] arising from the above threads.
Revision as of 18:14, 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 considered 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 (final call) (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) (700-799) Bounds for the first few density Hales-Jewett numbers, and related quantities (arriving at station)
Here are some unsolved problems arising from the above threads.
Here is a tidy problem page.
Bibliography
Density Hales-Jewett
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. R. McCutcheon, “The conclusion of the proof of the density Hales-Jewett theorem for k=3“, unpublished.
Behrend-type constructions
M. Elkin, "An Improved Construction of Progression-Free Sets ", preprint. 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.
Triangles and corners
M. Ajtai, E. Szemerédi, Sets of lattice points that form no squares, Stud. Sci. Math. Hungar. 9 (1974), 9--11 (1975). MR369299 I. Ruzsa, E. Szemerédi, Triple systems with no six points carrying three triangles. Combinatorics (Proc. Fifth Hungarian Colloq., Keszthely, 1976), Vol. II, pp. 939--945, Colloq. Math. Soc. János Bolyai, 18, North-Holland, Amsterdam-New York, 1978. MR519318 J. Solymosi, A note on a question of Erdős and Graham, Combin. Probab. Comput. 13 (2004), no. 2, 263--267. MR 2047239
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Subspace Spanned by Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$.Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$.(a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ […]
Cosine and Sine Functions are Linearly IndependentLet $C[-\pi, \pi]$ be the vector space of all continuous functions defined on the interval $[-\pi, \pi]$.Show that the subset $\{\cos(x), \sin(x)\}$ in $C[-\pi, \pi]$ is linearly independent.Proof.Note that the zero vector in the vector space $C[-\pi, \pi]$ is […]
Determine the Values of $a$ so that $W_a$ is a SubspaceFor what real values of $a$ is the set\[W_a = \{ f \in C(\mathbb{R}) \mid f(0) = a \}\]a subspace of the vector space $C(\mathbb{R})$ of all real-valued functions?Solution.The zero element of $C(\mathbb{R})$ is the function $\mathbf{0}$ defined by […]
Vector Space of Functions from a Set to a Vector SpaceFor a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by\[ (f+g)(s) = f(s) + […]
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We recall here the definition of a nilpotent group.Let $G$ be group. Define $G^0=G$,\[ G^1=[G, G]=\langle [x,y]:=xyx^{-1}y^{-1} \mid x, y \in G\rangle,\]and inductively define\[G^i=[G^{i-1},G]=\langle [x,y] \mid x \in G^{i-1}, y \in G \rangle.\]Then we obtain so called the lower central series of $G$:\[ G^{0} \triangleright G^{1} \triangleright \cdots \triangleright G^{i} \triangleright \cdots. \]
If there exists $m\in \Z$ such that $G^m=\{e\}$, then the group $G$ is called nilpotent.
Proof.
Consider the natural projection $p:G \to G/H$.Then we have $p(G^i)=(G/H)^i$.
Since $G/H$ is nilpotent, there exists $m \in \Z$ such that $(G/H)^m=\{eH\}$.Thus we obtain\[p(G^m)=(G/H)^m=\{eH\}.\]Thus for any $g \in G^m$, $g \in H \subset Z(G)$.
It follows that for any $g \in G^m$, $x \in G$ we have $gxg^{-1}x^{-1}=e$.Since the elements $gxg^{-1}x^{-1}$ are generators of $G^{m+1}=[G^m, G]$, we conclude that $G^{m+1}=\{e\}$ and $G$ is nilpotent.
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 […]
A Simple Abelian Group if and only if the Order is a Prime NumberLet $G$ be a group. (Do not assume that $G$ is a finite group.)Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.Definition.A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]
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 […]
A Condition that a Commutator Group is a Normal SubgroupLet $H$ be a normal subgroup of a group $G$.Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$.In particular, the commutator subgroup $[G, G]$ is a normal subgroup of […]
Infinite Cyclic Groups Do Not Have Composition SeriesLet $G$ be an infinite cyclic group. Then show that $G$ does not have a composition series.Proof.Let $G=\langle a \rangle$ and suppose that $G$ has a composition series\[G=G_0\rhd G_1 \rhd \cdots G_{m-1} \rhd G_m=\{e\},\]where $e$ is the identity element of […]
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 […]
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There are two primary dust populations near 1 AU, interplanetary dust (IPD) and interstellar dust (ISD) [
Mann, 2010]. I also discussed dust observations in detail at https://physics.stackexchange.com/a/160627/59023. Interplanetary Dust
IPD of ~1 $\mu$m size drift sunward due to Poynting-Robertson drag while following roughly Keplerian orbits [e.g.,
Malaspina et al., 2014]. Closer to the sun, these particles break up due to collisions, sublimation/ablation, and/or sputtering.
Dust grains of ~0.1 $\mu$m size are the so-called "$\beta$ meteorites", which travel away from the sun due to the imbalance of radiation pressure over gravity [
Mann, 2010].
The smallest dust grains with $\ll$ 0.1 $\mu$m size, the so-called
nanograins or nanodust, act like large pickup ions, which are carried anti-sunward by the frame-dependent convective electric field (i.e., just the Lorentz force) produced when the dust grain moves relative to the solar wind flow (i.e., $\mathbf{E}_{conv} = - \mathbf{V}_{sw} \times \mathbf{B}_{sw}$, where the subscript $conv$($sw$) stands for convective(solar wind), and $\mathbf{V}$ and $\mathbf{B}$ are the bulk flow velocity and quasi-static magnetic field). These particles can reach speeds in excess of 100 km/s relative to the sun [ Meyer-Vernet et al., 2009]. Interstellar Dust
ISD was first discovered by the
Ulysses spacecraft, which is ~1 $\mu$m size and moves at ~26 km/s relative to the solar system barycenter. More recent work [ Malaspina et al., 2014] has found a relationship between dust impact count rates and ecliptic longitude.
The reason can be seen from the following. The Earth's transverse speed about the sun is ~29 km/s. Thus, when the Earth moves anti-parallel(parallel) to the ISD flow direction the relative dust-spacecraft speed is ~55(~3) km/s, which produced an enhanced(depressed) dust count rate. This occurs because there is a threshold impact speed necessary to produce a large enough plasma cloud (i.e., $\gtrsim$5-10 km/s depending on dust size) [
Meyer-Vernet et al., 2009; 2014].
What is the origin of the dust near the sun?
The primary sources of ~1 $\mu$m size near 1 AU are cometary debris trails, asteroids, planets, moons, and ISD [
Mann, 2010; Zaslavsky, 2015].
Simply put - wouldn't the gravity pull it in or the solar wind blow it away?
Some are attracted by a combination of gravity and Poynting-Robertson drag while the smaller grains are either "pushed" out by radiation pressure (i.e., $\beta$ meteorites) or "picked-up" by the solar wind Lorentz force (i.e., nanodust).
References D.M. Malaspina et al., "Interplanetary and interstellar dust observed by the Wind/WAVES electric field instrument," Geophys. Res. Lett. 41, pp. 266-272, doi:10.1002/2013GL058786, 2014. Mann, I. "Interstellar Dust in the Solar System," Annual Review of Astronomy and Astrophysics 48, pp. 173-203, doi:10.1146/annurev-astro-081309-130846, 2010. Meyer-Vernet, N., et al. "Dust detection by the Wave instrument on STEREO: Nanoparticles picked up by the solar wind?," Sol. Phys. 256, pp. 463-474, doi:10.1007/s11207-009-9349-2, 2009. Meyer-Vernet, N., et al. "The importance of monopole antennas for dust observations: Why Wind/WAVES does not detect nanodust," Geophys. Res. Lett. 41, pp. 2716-2720, doi:10.1002/2014GL059988, 2014. Zaslavsky, A. "Floating potential perturbations due to micrometeoroid impacts: Theory and application to S/WAVES data," J. Geophys. Res. 120, pp. 855-867, doi:10.1002/2014JA020635, 2015.
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Hi,
Suppose we have a (real, separable) Banach space $V$ and a (linear) set $A\subseteq V$. I presume in general it might not be possible to write every element of the closed span of $A$ as an infinite linear combination $\sum_{i=1}^\infty\beta_i a_i$ of elements of $A$. Are there simple (non-trivial) conditions guaranteeing that the closed linear span of $A$ coincides with its infinite linear span (perhaps with unconditional/absolute convergence)?
My example of interest is the following: Let $X$ be a compact metric space and $F:X\to X$ a continuous map. My space $V$ is the space $C(X)$ of continuous (real-valued) functions on $X$, and $A$ is the subset of functions that can be written as $\varphi\circ F - \varphi$ for some $\varphi\in C(X)$.
Thank you for any help.
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Dirichlet's Kernel Representation of the Partial Sums of a Fourier Series
Recall from the Dirichlet's Kernel page that Dirichlet's kernel is the collection of functions $D_n$ where for each $n \in \mathbb{N}$:(1)
We also looked at some properties of the functions in Dirichlet's kernel. We saw that for each $n \in \mathbb{N}$, $D_n$ is even, $2\pi$-periodic, and $\displaystyle{\int_0^{2\pi} D_n(t) \: dt = \pi}$. We will now look at a nice theorem which tells us that any function $f \in L([0, 2\pi])$ that is $2\pi$-periodic can have the partial sums of the Fourier series generated by $f$ represented in terms of an integral involving a function in Dirichlet's kernel.
Theorem 1: Let $f \in L([0, 2\pi])$ be a $2\pi$-periodic function. For each $n \in \mathbb{N}$ let $\displaystyle{s_n(x) = \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos kx + b_k \sin kx)}$ denote the partial sum of the Fourier series generated by $f$. Then for each $n \in \mathbb{N}$, $\displaystyle{s_n(x) = \frac{2}{\pi} \int_0^{\pi} \frac{f(x + t) + f(x - t)}{2} D_n(t) \: dt}$. Recall that if $f \in L([0, 2\pi])$ then the Fourier series "generated" by $f$ is the Fourier series obtained with respect to the trigonometric system $\displaystyle{\left \{ \frac{1}{\sqrt{2\pi}}, \frac{\cos x}{\sqrt{\pi}}, \frac{\sin x}{\sqrt{\pi}}, \frac{\cos 2x}{\sqrt{\pi}}, \frac{\sin 2x}{\sqrt{\pi}}, ... \right \}}$ and $\displaystyle{f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)}$ where the $a_n$'s and $b_n$'s are the Fourier coefficients of the Fourier series given by $\displaystyle{a_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \cos nt \: dt}$ and $\displaystyle{b_n = \frac{1}{\pi} \int_0^{2\pi} f(t) \sin nt \: dt}$. Be sure to review the content on the The Fourier Series of Functions Relative to an Orthonormal System page as a refresher! Proof:We know that for all $n \in \mathbb{N}$ that $\displaystyle{s_n(x) = \frac{a_0}{2} + \sum_{k=1}^{n} (a_k \cos kx + b_k \sin kx)}$. Compacting this and we get: We now use the following trigonometric identity: Therefore: Notice that $f$ and $D_n$ are both $2\pi$ periodic. Therefore we have that: Let $u = t - x$. Then $du =dt$ and thus: Since $D_n$ is an even function and $[-\pi, \pi]$ is centered at the origin we have that: Replacing the dummy variable $u$ with $t$ yields:
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We will be offering mothur and R workshops throughout 2019. Learn more. Difference between revisions of "Sharedace"
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This file contains the frequency of sequences from each group found in each OTU. Each row consists of the distance being considered, group name, number of OTUS, and the abundance information separated by tabs. The abundance information is as follows. Each subsequent number represents a different OTU so that the number indicates the number of sequences in that group that clustered within that OTU. Note that OTU frequencies can only be compared within a distance definition. Below is a link to the
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This file contains the frequency of sequences from each group found in each OTU. Each row consists of the distance being considered, group name, number of OTUS, and the abundance information separated by tabs. The abundance information is as follows. Each subsequent number represents a different OTU so that the number indicates the number of sequences in that group that clustered within that OTU. Note that OTU frequencies can only be compared within a distance definition. Below is a link to the used in the calculations.
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Revision as of 16:46, 16 January 2009 Example Calculations *.sharedAce Example calculations below will be performed using data from the Eckburg 70.stool_compare files with an OTU definition of 0.03. Estimating the richness of shared OTUs between two communities. A Non-parametric richness estimator of the number of shared OTUs between two communities has been developed that is analogous to the ACE (3) single community richness estimator. The <math>S_{A,B ACE},</math> (9), estimator is calculated as: <math>S_{A,B ACE} = S_{12 \left ( abund \right )} + \frac {S_{12 \left ( rare \right )}}{c_{12}} + \frac {1}{C_{12}} \left [ f_{\left ( rare \right )1+} {\Gamma}_1 + f_{\left ( rare \right )+1} {\Gamma}_2 + f_{11}{\Gamma}_3 \right ]</math> where,
<math>C_{12} = 1 - \frac {\sum_{i=1}^{S_{12\left ( rare \right )}} {\left \{Y_i I \left ( X_i = 1 \right ) + X_iI \left ( Y_i = 1 \right ) - I \left ( X_i = Y_i = 1 \right ) \right \}}} {T_{11}}</math>
<math>{\Gamma}_1 = \frac{S_{12 \left (rare \right )} n_{rare} T_{21}}{C_{12}\left( n_{rare} - 1\right)T_{10}T_{11}} - 1</math>, <math>{\Gamma}_2 = \frac{S_{12 \left (rare \right )} m_{rare} T_{12}}{C_{12}\left( m_{rare} - 1\right)T_{01}T_{11}} - 1</math>
<math>{\Gamma}_3 = \left[ \frac{S_{12\left( rare \right)}}{C_{12}}\right ]^2 \frac{n_{rare}m_{rare}T_{22}}{\left(n_{rare}-1\right)\left(m_{rare}-1\right)T_{10}T_{01}T_{11}} - \frac{S_{12 \left( rare \right)}T_{11}}{C_{12}T_{01}T_{10}}-{\Gamma}_1-{\Gamma}2</math>
<math>T_{10} = \sum_{i=1}^{S_{12\left( rare \right)}} X_i </math>, <math>T_{01} = \sum_{i=1}^{S_{12\left( rare \right)}} Y_i </math>, <math>T_{11} = \sum_{i=1}^{S_{12\left( rare \right)}} X_i Y_i </math>, <math>T_{21} = \sum_{i=1}^{S_{12\left( rare \right)}} X_i \left( X_i - 1 \right) Y_i </math>
<math>T_{12} = \sum_{i=1}^{S_{12\left( rare \right)}} X_i \left( Y_i - 1 \right) Y_i </math>, <math>T_{22} = \sum_{i=1}^{S_{12\left( rare \right)}} {X_i \left( X_i - 1 \right) Y_i \left( Y_i - 1 \right)} </math>
where,
<math>f_{11}</math> = number of shared OTUs with one observed individual in A and B
<math>f_{1+}, f_{2+}</math> = number of shared OTUs with one or two individuals observed in A
<math>f_{+1}, f_{+2}</math> = number of shared OTUs with one or two individuals observed in B
<math>f_{\left(rare \right)1+}</math> = number of OTUs with one individual found in A and less than or equal to 10 in B.
<math>f_{\left(rare \right)+1}</math> = number of OTUs with one individual found in B and less than or equal to 10 in A.
<math>n_{rare}</math> = number of sequences from A that contain less than 10 sequences.
<math>m_{rare}</math> = number of sequences from B that contain less than 10 sequences.
<math>S_{12\left(rare\right)}</math> = number of shared OTUs where both of the communities are represented by less than or equal to 10 sequences.
<math>S_{12\left(abund\right)}</math> = number of shared OTUs where at least one of the communities is represented by more than 10 sequences.
<math>S_{12\left(obs\right)}</math> = number of shared OTUs in A and B.
Calculation of <math>S_{A,B ACE}.</math> is considerably complicated to evaluate. First, we determine that there are 23 rare shared OTUs and 37 abundant shared OTUs. Next, considering only the rare OTUs, we calculate <math>C_{12}</math> as 0.845878. We obtained the following T-values:
<math>T_{10} = 93</math>
<math>T_{01} = 64</math>
<math>T_{11} = 279</math>
<math>T_{21} = 1444</math>
<math>{T_{12}} = 988</math>
<math>T_{22} = 5440</math>
Next, calculation of the Γ-values requires knowing <math>f_{\left(rare \right)1+}, f_{\left(rare \right)+1} \mbox{ and } f_{\left(rare \right)11}</math>, which were 5, 8, and 2. Also, <math> n_{rare} \mbox{ and } m_{rare}</math> were 185 and 167, respectively. Finally, calculation of the Γ-values gives <math>{\Gamma}_1=0.530409, {\Gamma}_2 = 0.523308 \mbox{ and } {\Gamma}_3 = 0.151840</math>. This gives a <math>S_{A,B ACE}.</math> value of 72.3024 as seen below.
File Samples on the Eckburg 70.stool_compare Dataset .shared
This file contains the frequency of sequences from each group found in each OTU. Each row consists of the distance being considered, group name, number of OTUS, and the abundance information separated by tabs. The abundance information is as follows. Each subsequent number represents a different OTU so that the number indicates the number of sequences in that group that clustered within that OTU. Note that OTU frequencies can only be compared within a distance definition. Below is a link to the files used in the calculations.
.sharedAce
The first line contains the labels of all the columns. First sampled which shows the frequency of the <math>S_{A,B ACE}.</math> calculations. The frequency was set to 500, so after each 500 selected the <math>S_{A,B ACE}.</math> is calculated at each of the distances, with a calculation done after all are sampled. The following labels in the first line are the distances at which the calculations were made and the names of the groups compared. Each additional line starts with the number of sequences sampled followed by the <math>S_{A,B ACE}.</math> calculation at the column's distance. For instance, at distance 0.01, after 4392 samples <math>S_{A,B ACE}.</math> was 136.599.
sampled 0.01tissuestool 0.02tissuestool 0.03tissuestool 0.04tissuestool 1 0 0 0 0 500 44.2676 52.4249 43.9391 26.2499 1000 86.2691 53.7864 55.2556 60.1921 1500 114.238 106.452 45.6638 50.0418 2000 180.391 99.0382 57.2304 47.1769 2500 124.966 92.2403 48.1031 48.5068 3000 114.838 94.2194 56.2644 59.6396 3500 126.609 102.88 59.8571 71.1169 4000 134.213 98.837 56.6823 68.317 4392 136.599 86.5079 72.3024 62.117
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Let $T:X\to X$ be a continuous function on a compact metric space $X.$ Let $\mu$ be a $T$ invariant and ergodic probability measure on $X$ with strictly positive Sinai entropy $h_{\mu}(T).$ Let $F:X\to X$ be a continuos transformation that commutes with $T.$
Define $F_{*}\mu(A)=\mu(F^{-1}A)$ for every Borel set $A.$
I am looking for an example in ergodic theory in which
$\frac{1}{N}\sum_{k=1}^NF_{*}^k\mu$ does not converge in the weak* topology as $N\to \infty.$
Pd: It is trivial to find examples in which $F_{*}^N\mu$ does not converge but $\frac{1}{N}\sum_{k=1}^NF_{*}^k\mu$ does.
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You are currently browsing the monthly archive for October 2012.
A quite amusing post I stumbled upon describes pricing algorithms used by book retailers at Amazon.com, and what happens when two retailers use such algorithms. I am pretty sure that economists can say much about the use of these algorithms, optimal pricing algorithms, whether the customer gains or loses when retailers use them, and other similar questions. Alas, I am a mathemtician, so all I can do is read the post and smile.
Symbolab is a new search engine that allows one to search for equations. A great idea for those of us who, for example, forgot how the model of “stochastic game” is called, but remember that in stochastic games one is interested in the discounted sum of payoffs. The search engine presumably searches through articles to look for the equation that you look for.
Excited about the prospect that forgetting words is no longer devastating, I immediately searched for the formula of the discounted sum of payoffs:
$\sum _{t=1}^{\infty }\left(\lambda \left(1-\lambda \right)^{t-1}r\left(s_t,a_t\right)\right)$
Unfortunately 0 results were found. I did not give up and change the payoff function from “r” to “c” and then to “u”. Maybe the engine prefers costs or utilities. When I searched for the equation with costs, I got one result; a sum related to convex sets which is not related whatsoever to discounted sum. When I searched for the equation with utilities the engine gave up. I got 174 results! but they were of various sums, none of them (at least, not the 30 top ones which I looked at) involved discounted sums. There were references to statistics, ergodic theory, and various other topics, but no economics or game theory.
Maybe stochastic games are not that common on the net. Let’s look for something simpler:
$u_i\left(\sigma \right)\ge \:u_i\left(\sigma _{-i},\sigma ‘_i\right)$
which is the condition that defines Nash equilibrium. 0 results again. I gave up.
Unfortunately, it seems that we will have to continue remembering names of concepts, and will have to consult with colleagues when we need results in topics we are not familiar with. Magic has not been found yet.
It was a pleasure to awaken to the news that this years Nobel (memorial) in Economics went to Shapley and Roth (Schwartz lecture series hits bullseye again). Even more had it gone to me…..but, have yet to see pigs in the sky.
Pleasure turned to amusement as I heard and read the attempts of journalists to summarize the contributions honored. NPR suggested that it was about applying statistics. Forbes had a piece that among Indians would be described as putting shit in milk. This always makes me wonder about the other things they get wrong in the subjects I have no knowledge of. Nevertheless, I will plug one outlet for reasons that will become obvious upon reading it.
In this post, I set myself the task of seeing if I can do a better job than the fourth estate of conveying the nature of the contribution that was honored, Oct 15th, 2012. Here goes.
The fictional decentralized markets of the textbook, like the frictionless plane in a vacuum used in physics, are a useful device for establishing a benchmark. Real markets, however, must deal with frictions and the imperfections of their participants. One such market is for College Admissions in the US that is largely decentralized This decentralization increases uncertainty and raises costs. Students, for example, must forecast which colleges are likely to accept them. The greater the uncertainty in these forecasts the more `insurance’ is purchased either by applying to a large set of colleges or aiming `low’. On the college side, this insurance makes yields difficult to forecast. Increasing acceptance rates to increase yields has the effect of driving application numbers in the future down, so, waiting lists grow. These problems could be eliminated were one to switch to a centralized admissions market. How would one design such a centralized process? This is the question that the work of Shapley (and the late David Gale) and Alvin Roth addresses.
A major hurdle that a centralized market for college admissions must overcome is that it must match students with colleges in a way that respects the preferences and incentives of both parties. A centralized market cannot force a student to attend a college she does not want to, or require a college to accept a particular student. There is always the threat that the participants can pick up their marbles and walk away. If enough do, the incentives for the students and colleges that participate in the centralized market decline. Would a student participate in an centralized market if certain brand name colleges opted out? The work of (Gale &) Shapley was the first to formalize this concern with designing centralized markets that would be immune defections on the part of participants, i.e., stable. Their seminal paper articulated a model and a mechanism for matching students to colleges that would be stable in this sense. Alvin Roth’s own work builds on this in a number of ways. The first is to use the notion of stability to explain why some centralized markets fail. Second, to highlight the importance of other sources of instability associated with, say timing. Participants may wish to `jump the clock’. Colleges, for example, offering admission to high school students in their junior year when there might be less competition for that student. The third, is to use the ideas developed in other contexts to allocate scarce resources where money as a medium of exchange is ruled out, most notably kidneys.
The work honored this year has its roots in a specialty of game theory, long considered unfashionable but one Shapley made deep contributions to: co-operative game theory. One can trace an unbroken line between the concern for stability in the design for markets and the abstract notions of stability discussed, for example, in the first book on game theory by von-Neuman and Morgenstern. It proves, once again, Keynes’ dictum:
“The power of vested interests is vastly exaggerated compared with the gradual encroachment of ideas.’’
Le Monde tells us (in French) that researchers found 47-million-year-old fossils of nine mating pairs of turtles of species
Allaeochelys crassesculpta in a lake in Germany. These fossils, which, according to the researchers, are the only fossils of mating pairs of animals to be found, taught the researchers a lot on this extinct species. But what caught my eye is a probabilistic statement made by one of the researchers: “Des millions d’animaux vivent et meurent chaque année, et nombre d’entre eux se fossilisent par hasard, mais il n’y a vraiment aucune raison que ça arrive lorsque vous êtes en train de vous reproduire. Il est hautement improbable que les deux partenaires meurent en même temps, et les chances que les deux soient fossilisés à la fois sont encore plus maigres”, a expliqué Walter Joyce, de l’université allemande de Tübingen. Avec plusieurs couples, les probabilités s’amoindrissent encore.
Since the name Walter Joyce sounds English speaking, I searched for a version in English. I found this:
“No other vertebrates have ever been found like these, so these are truly exceptional fossils,” Joyce said. “The chances of both partners dying while mating are extremely low, and the chances of both partners being preserved as fossils afterward even lower. These fossils show that the fossil record has the potential to document even the most unlikely event if the conditions are right.”
Introduction to Probability. I will ask them to read this article and spot the mistakes. I wonder how many will find what is wrong in this article.
One reason given for the value of an MBA degree is the relationships that one develops with other students as well as the connection to the larger alumni network. Such relationships can eventually be used to open doors, secure a place at `the table’ and traded with others. While I’ve long since replaced the belief in `
res ipsa loquitur‘ for `who you know matters’, I’m still not convinced by the relationship story.
Suppose introductions to gatekeepers and decision makers are scarce resources. When handing them out, why should I favor someone just because we attended the same institution? Presumably what matters is what good turn the other might do for me. Surely, this will depend on the position held rather than the school attended. Furthermore, why should the other’s academic pedigree suggest anything about the likelihood of the other returning the favor in the future? I know of no B-school that claims that it is selecting a class of future Cato’s.
True, favors are not requested or granted until a bond is established between the parties. Having something in common assists the formation of such bonds. But, why should having attended the same school be any more useful in this regard than a common interest in wine, golf or stamps?
Finally, if the alumni network is valuable, then merging two small networks should increase value for members of either network. Thus, in much the same way that some airlines share their frequent flyer programs (eg star alliance), we should see certain schools merging their networks. Haas and Anderson? Johnson School and Olin? One sees something like this at the executive masters level but not at the full time MBA level.
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Boundedness of a Subset of C(X)
Recall from The Set of Real-Valued Continuous Functions on a Compact Metric Space X, C(X) page that if $(X, d)$ is a compact metric space the set $C(X)$ is defined to be the set of all real-valued continuous functions on $X$, that is:(1)
We then defined a very important metric $\rho : C(X) \times C(X) \to [0, \infty)$ given for all $f, g \in C(X)$ by:(2)
We verified that $\rho$ was indeed a metric and so $(C(X), \rho)$ is a metric space.
We will now define the concept of boundedness of a subset of $C(X)$.
Definition: Let $(X, d)$ be a compact metric space and let $\Gamma \subseteq C(X)$. Then $\Gamma$ is said to be Bounded if there exists an $M \in \mathbb{R}$, $M > 0$ such that $\mid f(x) \mid \leq M$ for all $f \in \Gamma$ and for all $x \in X$.
For example, consider the metric space $([-2\pi, 2\pi], d)$ where $d$ is the usual Euclidean metric defined for all $x, y \in [-2\pi, 2\pi]$ by $d(x, y) = \mid x - y \mid$. Also consider the following subcollection:(3)
We claim that $\Gamma$ is bounded. To show this, take $M = 1 > 0$. Then since for all $a \in \mathbb{R}$ we have that $\mid \cos (a) \mid \leq 1$ we see that for all $f_r \in Gamma$ and for all $x \in X$ that then:(4)
Therefore $\Gamma$ is bounded.
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Normally, if an object of mass $m$ is inclined to the horizontal at an angle $b$, we set the reaction force of the object on the inclined plane as $R = mg\cos{b}$ (if we resolve the force of gravity so the line of action coming out of the plane is perpendicular to it).
However in circular motion*. it's assumed that $R\cos{b} = mg$. In the example above, one would have to do this in order to arrive to the correct answer, instead of $R = mg\cos{b}$. Using $R = mg\cos{b}$ seems natural enough, as I am resolving vertically, however, both equations would produce two different values for $R$. Why is this?
To show what I mean: If we set the reaction force in this question as $mg\cos{a}$, then the centripetal force will be $mg\cos{b}\cos(\pi/2-b) = mg\cos{b}\sin{b} = \frac{1}{2}mg\sin(2b)$
Whereas If we use $R\cos{b} = mg$, $R = mg\sec{b}$ and the centripetal force will be $mg\sec{b}\sin{b} = mg\tan{b}$. This will end up with two different values for the radius of the circular motion, and hence two different final answers.
*In the circular motion questions I've seen in my mechanics module
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While scientists have devoted much research to models of neural activity in the brain, they have paid little attention to modeling drugs that target the brain. Development of this class of drugs is very challenging and necessitates an understanding of the highly complex processes that govern the concentration profile of a drug in the brain over time. Since access to the brain for measurement purposes is very limited, a mathematical model is a helpful tool. But before we present a model, we must introduce some of the brain’s physiology and the processes that occur after medicine consumption.
The brain is interlaced with a network of blood capillaries (see Figure 1). Following intravenous or oral administration and subsequent intestinal absorption, the drug in question begins to circulate in the blood and primarily enters the brain from the arterial network by crossing the blood-brain barrier (BBB). One of the BBB’s principal functions is to limit transport into the brain and protect it from harmful substances, thereby preventing brain damage. When the drug does enter the brain through the BBB it circulates in brain fluids, such as the extracellular (ECF) and cerebrospinal fluids. It then binds to receptors on cells in ECF (see Figure 2). When a drug binds to a receptor, it leads to an effect in the body. Here we will focus on drug transport in ECF and the subsequent binding to receptors.
Figure 1. The brain and its interlacing capillary network. 1a. The brain. 1b. The network of capillaries that intertwines the brain. 1c. Brain capillaries from the human cerebellar cortex. 1a and 1b are public domain images, and 1c is courtesy of [1].
Compartmental models are widely used in pharmacology, and have also been developed to model drug concentration in the brain. For example, [5] presents a general compartmental model of the central nervous system. Unfortunately, these models do not account for drug transport in ECF and other tissues, which mainly occurs via diffusion and bulk flow. Moreover, compartmental models do not consider receptor binding. A diffusion-advection equation can model drug transport where the drug is administered directly into the brain [2]. As a first step towards a full model of the brain, our model incorporates diffusion and flow in ECF, inflow through the BBB, and receptor binding.
Figure 2. Cells in the brain lie in extracellular fluid (ECF). The fluid is coloured red. Image courtesy of [3].
Though the brain is three-dimensional, we start by creating a model on a two-dimensional domain, which represents a tissue unit of brain ECF. This square domain is surrounded by brain capillaries and can be considered the smallest building block of the brain, in terms of drug distribution (see Figure 3). In the human brain, the distance between capillaries is on average 50µm. Cells with receptors are located inside the domain. We model drug transport in the unit by diffusion and bulk flow, assuming that the latter occurs in the \(x\)-direction. One can consider ECF a porous medium, as it is filled with many obstacles—such as cells and proteins—that limit diffusion. This leads to an effective diffusion smaller than normal (in a medium without obstacles). We model this with the so-called tortuosity \(\lambda\), thereby dividing the normal diffusion \(d\) by \(\lambda^2\), which results in a smaller diffusion coefficient. Tortuosity differs between drugs due to their varying sizes and deformabilities.
To formulate the model, we denote the concentration of free (unbound) drug by \(D\) and the concentration of receptor-bound drug by \(B\). This yields
\[\frac{\partial D}{\partial t} = \frac{d}{\lambda^2}\Delta L - \upsilon \triangledown L -\: k_{on}D(B_{max} - B) + k_{off}\:B, \\ \frac{\partial B}{\partial t} = k_{on}D(B_{max} - B) + k_{off}\:B,\]
where \(\upsilon\) is the speed of the flow. When a drug binds to a receptor, it forms a drug-receptor complex until it dissociates (unbinds) into drug and receptor again. We model this by the final terms in the \(D\)-equation, which represent receptor binding with a rate \(k_{on}\) and unbinding with a rate \(k_{off}\). The maximum concentration of receptors \(B_{max}\) limits the binding.
Figure 3. The two-dimensional unit of extracellular fluid, which contains cells with receptors and is bounded by blood capillaries. Concept for figure provided by Vivi Rottschäfer.
We assume that no drug is present in the brain at \(t=0\), and hence \(D(t=0)=0\) and \(B(t=0)=0\). We use boundary conditions to model the concentration of the drug in the blood and its crossing through the BBB. While drugs can cross the BBB via several mechanisms, including passive and active transport, we only analyse passive transport resulting from diffusion. At \(x=0\), this leads to
\[d\frac{\partial D}{\partial x} = P (D - D_{blood}(t)),\]
and similar conditions at the other boundaries [4]. \(P\) is a measure of the permeability—transport through the BBB—and \(D_{blood}(t)\) describes the drug concentration in the surrounding capillaries’ blood. This can and will vary with time since the drug enters the blood and is thereafter eliminated from it.
The time dynamics of the concentrations is of interest, and this presents an important mathematical challenge as it differs from the “standard” question of behaviour of solutions as \(t\) becomes large; at larger \(t\), all of the drug is eliminated from the brain.
We perform simulations, study the free drug concentration and the bound complex concentration in the domain over time [4], and choose all coefficients in physiologically-relevant ranges. Many of the coefficients vary widely among different drugs; therefore, we examine the influence of changing various parameters on the concentration. As an example, we show results of the impact of changing the BBB’s permeability \(P\) on the concentration. After fixing the rest of the parameters and only changing \(P\), we plot the concentrations of the free drug \(D\) and bound drug \(B\) versus time in the middle of the domain (see Figure 4). We also plot the concentration of drug in the blood \(D_{blood}(t)\) (in red).
Figure 4. Influence of permeability through the blood-brain barrier (BBB). The effect of changing permeability P on the log concentration-time profiles of D and B for low, intermediate, and high P. Figure courtesy of [4].
We vary \(P\) from the lowest possible physiological value to an intermediate, followed by a larger value. The lowest value of \(P\) corresponds to drugs that have difficulty crossing the BBB. When \(P\) is larger, the drug easily moves through the BBB and \(D\)’s profile strongly follows the profile of \(D_{blood}(t)\). In contrast, \(D\) increases and decreases more slowly when \(P\) is smaller because the drug both enters and leaves brain ECF more slowly. In Figure 4 (right), we plot \(B\) and observe that when \(P\) is of higher value, \(B\) rapidly increases to a maximum before quickly decreasing again. This decrease in \(B\) starts when there is not enough of the free drug present to bind to all the free receptors because it has flowed back through the more permeable BBB. In contrast, \(B\) increases more slowly and limits to a certain value when \(P\) is lower. \(B\) only decreases after time periods longer than those shown in the simulation.
As a next step, we are currently working on a three-dimensional model for a unit of the brain. We can form an entire brain by combining several of these units. This will yield a simplified model that will allow us to assign non-identical parameter values to different units, thus accounting for brain heterogeneity. For example, receptors are not distributed evenly in the brain; drugs target different regions, and receptor concentrations can vary per region. A local disease can also greatly influence the parameters.
A broad range of opportunities exists for mathematicians to collaborate with pharmacologists in various areas, even beyond brain modelling. Among the challenges for modelers is the need for a combination of biological processes with drug influence. We strongly believe that this calls for the continued development of mathematical pharmacology.
Acknowledgments: This is based on joint research with Esmée Vendel (Mathematical Institute) and Liesbeth de Lange (Leiden Academic Centre for Drug Research), both of Leiden University.
References [1] Ferber, D. (2007). Bridging the Blood-Brain Barrier: New Methods Improve the Odds of Getting Drugs to the Brain Cells That Need Them. PLoS Bio., 5(6), e169. [2] Nicholson, C. (2001). Diffusion and related transport mechanisms in brain tissue. Rep. Prog. in Phys., 64(7), 815. [3] Perkins, K., Arranz, A., Yamaguchi, Y., & Hrabetova, S. (2017). Brain extracellular space, hyaluronan, and the prevention of epileptic seizures. Rev. Neurosci., 28(8), 869-892. [4] Vendel, E., Rottschäfer, V., & de Lange, E.C.M. (2018). Improving the prediction of local drug distribution profiles in the brain with a new 2D mathematical model. Special Issue of Bull. Math. Bio.: Mathematics to Support Drug Discovery and Development (submitted). [5] Yamamoto, Y., Välitalo, P.A., van den Berg, D.-J., Hartman, R., van den Brink, W., Wong, Y.C.,…,de Lange, E.C.M. (2017). A Generic Multi-Compartmental CNS Distribution Model Structure for 9 Drugs Allows Prediction of Human Brain Target Site Concentrations. Pharm. Res., 34(2), 333-351.
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An
incidence geometry is a set $P$ (the "points"), a set $L$ (the "lines"), and a relation $I\subseteq P\times L$ ("incidence"). Equivalently, a bipartite graph with the halves of the partition labeled by $P$ and $L$.
A
projective plane is an incidence geometry satisifying the following axioms: For every pair of distinct points, there is a unique line incident to both of them. For every pair of distinct lines, there is a unique point incident to both of them. (Non-degeneracy) There are $4$ points, no $3$ of which are collinear.
Equivalently, a bipartite graph with no subgraph isomorphic to the complete bipartite graph $K_{2,2}$, but such that any subgraph isomorphic to a subgraph of $K_{2,2}$ is contained in a subgraph isomorphic to a
maximal proper subgraph of $K_{2,2}$. Plus non-degeneracy. Replacing $2$ and $2$ with $m$ and $n$ gives the following natural generalization:
A
$(m,n)$-pseudoplane is an incidence geometry satisfying the following axioms: For every $m$ distinct points, there are exactly $(n-1)$ distinct lines which are incident to all of them. For every $n$ distinct lines, there are exactly $(m-1)$ distinct points which are incident to all of them.
It's possible that there is a natural non-degeneracy axiom to include, but the correct formulation is not clear to me.
(See update below)
I recently wrote a paper with Gabe Conant studying the model theory of existentially closed projective planes, and we were led to define $(m,n)$-pseudoplanes when we realized that almost all of our results generalized easily to these structures. But we have had trouble locating them in the combinatorics literature. For example, they do not appear to be mentioned explicitly in the
Handbook of Combinatorial Designs (though it's entirely possible I overlooked them in that massive tome!). Question: What is known about $(m,n)$-pseudoplanes? Do they appear under a different name in the literature? I would appreciate any and all references. Update: In Summer 2018, my REU student Matisse Peppet and I investigated the combinatorics of $(m,n)$-pseudoplanes (when $m,n\geq 2$). I hope you will forgive me bumping this question and advertising Matisse's results below; we are preparing a paper on the topic, and I am still looking for references to $(m,n)$-pseudoplanes in the combinatorics literature. Of course, I especially want to know if any of the results below are already known, or if there is any similar relevant work out there (I am a logician, combinatorics is not my area of expertise).
Matisse identified the following non-degeneracy axiom, generalizing the non-degeneracy axiom for projective planes:
There exist $mn$ points, no $(m+1)$ of which are incident to a single line, and there exist $mn$ lines, no $(n+1)$ of which are incident to a single point.
She then proved the following theorems:
In any non-degenerate $(m,n)$-pseudoplane, there are constants (possibly infinite cardinals) $P_a$ and $L_b$ for $0\leq a \leq n$ and $0\leq b \leq m$ such that for any set of $a$ lines, there are exactly $P_a$ points incident to these lines, and for any set of $b$ points, there are exactly $L_b$ lines incident to these points. (Recall that in projective planes, we have $P_0 = L_0 = k^2+k+1$, $P_1 = L_1 = k+1$, and $P_2 = L_2 = 1$, where $k$ is the order of the plane.)
There are no finite non-degenerate $(m,n)$-pseudoplanes when $m\geq 3$ or $n\geq 3$. The proof involves deriving conflicting identities involving the constants $P_a$ and $L_b$.
Nevertheless, there are infinite non-degenerate $(m,n)$-pseudoplanes. In any such structure, there is a single infinite cardinal $\kappa$ such that $P_a = L_b = \kappa$ for all $0\leq a < n$ and $0\leq b < m$.
As a consequence of 2, the theory of existentially closed $(m,n)$-pseudoplanes has no prime model when $m\geq 3$ or $n\geq 3$. This question is still open for projective planes (when $m = n = 2$); Gabe and I showed that it is equivalent to a longstanding open problem: does every finite $K_{2,2}$-free configuration embed in a finite projective plane? Of course, 2 shows that this question trivializes when $m\geq 3$ or $n\geq 3$, since the non-degeneracy configuration does not embed in any finite $(m,n)$-pseudoplane.
It remains open to classify the (finite) degenerate $(m,n)$-pseudoplanes when $m\geq 3$ or $n\geq 3$ (it may be that there is no satisfying classification).
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According to Wikipedia the Sun's "power density" is "approximately 276.5 $W/m^3$, a value that more nearly approximates that of reptile metabolism or a compost pile than of a thermonuclear bomb." My question is, so why is the Sun's core so hot (15.7 million K)? Using a gardener's (not a physicist's intuition) it seems apparent that you can't keep on increasing the temperature of a compost heap just by making the heap larger.
Your (gardener's) intuition is wrong. If you increase the size of your compost heap to the size of a star, then its core would be as hot as that of the Sun. All other things being equal (though compost heaps are not hydrogen plus helium), the temperature of a spherical compost heap would just depend on its total mass divided by its radius$^{*}$.
To support the weight of all the material above requires a large pressure gradient. This in turn requires that the interior pressure of the star is very large.
But why this particular temperature/density combination?
Nuclear reactions actually stop the core from getting hotter. Without them, the star would radiate from its surface and continue to contract and become even hotter in the centre. The nuclear reactions supply just enough energy to equal that radiated from the surface and thus prevent the need for further contraction.
The nuclear reactions are initiated once the nuclei attain sufficient kinetic energy (governed by their temperature) to penetrate the Coulomb barrier between them. The strong temperature dependence of the nuclear reactions then acts like a core thermostat. If the reaction rate is raised, the star will expand and the core temperature will cool again. Conversely, a contraction leads to an increase in nuclear reaction rate and increased temperature and pressure that act against any compression.
$*$ This relationship arises from the virial theorem, which says that for a fluid/gas that has reached mechanical equilibrium, that the sum of the (negative) gravitational potential energy and twice the internal kinetic energy will equal zero. $$ \Omega + 2K = 0$$
The internal kinetic energy can be approximated as $3k_BT/2$ per particle (for a monatomic ideal gas) and the gravitational potential energy as $-\alpha GM^2/R$, where $M$ is the mass, $R$ the radius and $\alpha$ is a numerical factor of order unity that depends on the exact density profile. The virial theorem then becomes total kinetic energy $$ \alpha G\left(\frac{M^2}{R}\right) \simeq 2\left(\frac{3k_BT}{2}\right) \frac{M}{\mu}\ ,$$ where $\mu$ is the mass per particle. From this, we can see that $$ T \simeq \frac{\alpha G\mu}{3k_B}\left( \frac{M}{R}\right)$$
Another way of seeing that the Sun's power density must be rather low (or that very large compost heaps get very hot) and also to see the temperature-radius relationship is to calculate what the
surface temperature should be, for a given power density.
So, we can model the Sun as a sphere with radius $R$, and a power density of $\rho$. The total volume of the Sun is then
$$V =\frac{4\pi R^3}{3}$$
and the total power output is
$$P =\rho V = \rho\frac{4\pi R^3}{3}$$
The surface area of the Sun is then
$$A = 4 \pi R^2$$
So the power flux through the surface of the Sun is
$$f = \frac{P}{A} = \rho \frac{R}{3}$$
(Notice that this is going like $R$: the bigger the star, the higher the flux for a given power density.)
And we can use the Stefan-Boltzmann law which relates surface temperature to flux, assuming the Sun is a black body
$$f = \sigma T^4$$
And putting this together we get
$$T = \left(\rho\frac{R}{3\sigma}\right)^{1/4}$$
Where $T$ is the surface temperature of the Sun. This formula tells you that large objects which are generating power get much hotter than small ones with the same power density: the surface temperature goes as the fourth root of the radius.
Plugging in the numbers this gives a surface temperature which is much too high, from which I conclude that the power density of the Sun is in fact much
lower than the Wikipedia figure: that figure is probably for the part of the Sun where fusion is occurring only, not the whole volume.
If the heap was self-gravitating then the larger the heap, the warmer its center. The idea behind is that at the core the material will feel the pressure exerted by all the layers above, and you probably know that if you put pressure on something, it will heat up.
What is interesting here is that all the outer layers of your self-gravitating "heap" will tend to collapse towards the center (that's what gravity does best), and as a result the pressure will continue increasing and so will the temperature at the core. If the mass of it is large enough, the temperature reached through this mechanism is so high that molecules split, atoms ionize, and nuclei merge, thus generating thermonuclear energy.
This energy, generated at the center, will travel outwards and generate a pressure on the collapsing material. The result is that the system reaches equilibrium, where the hydrodynamical pressure is balanced by the radiation pressure coming from fusion at the center. It is possible to do the math and calculate what is the temperature required to merge Hydrogen into Helium: the result is 15.7 million K.
protected by Qmechanic♦ Nov 27 '17 at 10:04
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linear poolor weighted averageof Adila's and Benoit's credence functions. That is, we assign a weight to Adila ($\alpha$) and a weight to Benoit ($1-\alpha$) and we take the linear combination of their credence functions with these weights to be our credence function. So my credence in $X$ will be $\alpha c_A(X) + (1-\alpha) c_B(X)$, while my credence in $\overline{X}$ will be $\alpha c_A(\overline{X}) + (1-\alpha)c_B(\overline{X})$.
But now suppose that either Adila or Benoit or both are probabilistically incoherent -- that is, either $c_A(X) + c_A(\overline{X}) \neq 1$ or $c_B(X) + c_B(\overline{X}) \neq 1$ or both. Then, it may well be that the linear pool of their credence functions is also probabilistically incoherent. That is,
$(\alpha c_A(X) + (1-\alpha) c_B(X)) + (\alpha c_A(\overline{X}) + (1-\alpha)c_B(\overline{X})) = $
$\alpha (c_A(X) + c_A(\overline{X})) + (1-\alpha)(c_B(X) + c_B(\overline{X})) \neq 1$
But, as an adherent of Probabilism, I want my credences to be probabilistically coherent. So, what should I do?
A natural suggestion is this: take the aggregated credences in $X$ and $\overline{X}$, and then take the closest pair of credences that are probabilistically coherent. Let's call that process the
coherentizationof the incoherent credences. Of course, to carry out this process, we need a measure of distance between any two credence functions. Luckily, that's easy to come by. Suppose you are an adherent of Probabilism because you are persuaded by the so-called accuracy dominance arguments for that norm. According to these arguments, we measure the accuracy of a credence function by measuring its proximity to the ideal credence function, which we take to be the credence function that assigns credence 1 to all truths and credence 0 to all falsehoods. That is, we generate a measure of the accuracy of a credence function from a measure of the distance between two credence functions. Let's call that distance measure $D$. In the accuracy-first literature, there are reasons for taking $D$ to be a so-called Bregman divergence. Given such a measure $D$, we might be tempted to say that, if Adila and/or Benoit are incoherent and our linear pool of their credences is incoherent, we should notadopt that linear pool as our credence function, since it violates Probabilism, but rather we should find the nearest coherent credence function to the incoherent linear pool, relative to $D$, and adopt that. That is, we should adopt credence function $c$ such that $D(c, \alpha c_A + (1-\alpha)c_B)$ is minimal. So, we should first take the linear pool of Adila's and Benoit's credences; and then we should make them coherent.
But this raises the question: why not first make Adila's and Benoit's credences coherent, and then take the linear pool of the resulting credence functions? Do these two procedures give the same result? That is, in the jargon of algebra, does linear pooling commute with our procedure for making incoherent credences coherent? Does linear pooling commute with coherentization? If so, there is no problem. But if not, our judgment aggregation method faces a dilemma: in which order should the procedures be performed: aggregate, then make coherent; or make coherent, then aggregate.
It turns out that whether or not the two commute depends on the distance measure in question. First, suppose we use the so-called
squared Euclidean distancemeasure. That is, for two credence functions $c$, $c'$ defined on a set of propositions $X_1$, $\ldots$, $X_n$,$$SED(c, c') = \sum^n_{i=1} (c(X_i) - c'(X_i))^2$$ In particular, if $c$, $c'$ are defined on $X$, $\overline{X}$, then the distance from $c$ to $c'$ is $$(c(X) -c'(X))^2 + (c(\overline{X})-c'(\overline{X})^2$$ And note that this generates the quadratic scoring rule, which is strictly proper: $\mathfrak{q}(1, x) = (1-x)^2$ $\mathfrak{q}(0, x) = x^2$ Theorem 1For all $\alpha$, $c_A$, $c_B$, $$\alpha c^*_A + (1-\alpha)c^*_B = (\alpha c_A + (1-\alpha)c_B)^*$$
Second, suppose we use the
generalized Kullback-Leibler divergenceto measure the distance between credence functions. That is, for two credence functions $c$, $c'$ defined on a set of propositions $X_1$, $\ldots$, $X_n$,$$GKL(c, c') = \sum^n_{i=1} c(X_i) \mathrm{log}\frac{c(X_i)}{c'(X_i)} - \sum^n_{i=1} c(X_i) + \sum^n_{i=1} c'(X_i)$$ Thus, for $c$, $c'$ defined on $X$, $\overline{X}$, the distance from $c$ to $'$ is $$c(X)\mathrm{log}\frac{c(X)}{c'(X)} + c(\overline{X})\mathrm{log}\frac{c(\overline{X})}{c'(\overline{X})} - c(X) - c(\overline{X}) + c'(X) + c'(\overline{X})$$ And note that this generates the following scoring rule, which is strictly proper: $\mathfrak{b}(1, x) = \mathrm{log}(\frac{1}{x}) - 1 + x$ $\mathfrak{b}(0, x) = x$ does notcommute with our procedure for making incoherent credences coherent. Given a credence function $c$, let $c^+$ be the closest coherent credence function to $c$ relative to $GKL$. Then: Theorem 2For many $\alpha$, $c_A$, $c_B$, $$\alpha c^+_A + (1-\alpha)c^+_B \neq (\alpha c_A + (1-\alpha)c_B)^+$$ Proofs of Theorems 1 and 2. With the following two key facts in hand, the results are straightforward. If $c$ is defined on $X$, $\overline{X}$: $c^*(X) = \frac{1}{2} + \frac{c(X)-c(\overline{X})}{2}$, $c^*(\overline{X}) = \frac{1}{2} - \frac{c(X) - c(\overline{X})}{2}$. $c^+(X) = \frac{c(X)}{c(X) + c(\overline{X})}$, $c^+(\overline{X}) = \frac{c(\overline{X})}{c(X) + c(\overline{X})}$.
Thus, Theorem 1 tells us that, if you measure distance using SED, then no dilemma arises: you can aggregate and then make coherent, or you can make coherent and then aggregate -- they will have the same outcome. However, Theorem 2 tells us that, if you measure distance using GKL, then a dilemma does arise: aggregating and then making coherent gives a different outcome from making coherent and then aggregating.
Perhaps this is an argument against GKL and in favour of SED? You might think, of course, that the problem arises here only because SED is somehow naturally paired with linear pooling, while GKL might be naturally paired with some other method of aggregation such that that method of aggregation commutes with coherentization relative to GKL. That may be so. But bear in mind that there is a very general argument in favour of linear pooling that applies whichever distance measure you use: it says that if you do not aggregate a set of probabilistic credence functions using linear pooling then there is some linear pool that each of those credence functions expects to be more accurate than your aggregation. So I think this response won't work.
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Can someone give a proof of the "technical part" of the following proof?
I'm not sure whether this approach works, but you can try it:
Suppose $\left\langle g\right\rangle H\neq G$. Then there exists an $x\in G\setminus \left\langle g\right\rangle H$. Let $H'$ denote the smallest subgroup of $G$ containing $H$ and $x$. By maximality of $H$, $H'\cap \left\langle g\right\rangle\neq \left\{e\right\}$. Hence there exists a non-trivial $y\in G$ such that $y\in H'\cap \left\langle g\right\rangle$. Since $G$ is commutative, this implies that we can write $y=hx^i$ for some $h\in H$ and some $i\neq 0$ (otherwise $y\in H\cap \left\langle g\right\rangle=\left\{e\right\}$), and $y=g^n$ for some $n$. Hence $g^n=hx^i$ and thus $x^i=g^nh^{-1}\in \left\langle g\right\rangle H$.
Can you conclude from this that $x\in \left\langle g\right\rangle H$? That would yield a contradiction.
Edit: Here is a another try: Notice that $\left\langle g\right\rangle\cap H=\left\{e\right\}$. Now consider $G/H$.For each $x\in G$ we denote by $\bar{x}$ the corresponding class in $G/H$. We claim that $|\bar{g}|=|g|=p^m$ (here $|x|$ is the order of an element). Suppose that $|\bar{g}|<p^m$. Then $\bar{g}^{p^{m-1}}=\bar{e}$, hence $g^{p^{m-1}}\in H$. But then $g^{p^{m-1}}\in \left\langle g\right\rangle\cap H=\left\{e\right\}$ contradicts the fact that $|g|=p^m$. Thus $|\bar{g}|=p^m$. It follows that the order of $\bar{g}$ is maximal in $G/H$.
By the induction hypothesis $G/H\cong \mathbb{Z}_{p^m}\times K$ where $K$ is a product of cyclic groups. Now consider the subgroup $L:=\left\{x\in G\mid \bar{x}\in K\right\}$. Then $|L|=p^i|K|$ for some $i$, in fact $|L|=|H||K|$ (Why?). Suppose that $x\in \left\langle g\right\rangle\cap L$, then $\bar{x}\in \left\langle \bar{g}\right\rangle\cap K=\left\{\bar{e}\right\}$ (Why?), and thus $x\in H$, hence $x\in \left\langle g\right\rangle\cap H=\left\{e\right\}$. We just proved that $\left\langle g\right\rangle\cap L=\left\{e\right\}$.
Thus $|\left\langle g\right\rangle L|=|\left\langle g\right\rangle||L|=p^i(|\left\langle \bar{g}\right\rangle||K|)=|H||G/H|=|G|$. By maximality of $H$ we definitely need that $|\left\langle g\right\rangle H|=|G|$ which then shows the claim.
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$\int z^2 \log [(z+1)/(z-1)] dz$ taken around circle $|z|=2$
I am taking residues at $\pm 1$.
This gives me 0 as the value of integral. Is this correct.
How do I modify the integral to get value over half the circle?
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$$\int_C f(z)dz=-2\pi i \operatorname{Res}_{z=\infty}f(z)$$ But we have $$f(z)=2z+\frac23z^{-1}+o(z^{-1})$$ as $z\to\infty$, therefore the integral equals $\frac43 \pi i$. $f$ is not meromorphic in $|z|<2$ , so we can not apply residue theorem inside the circle, but we can apply it outside the circle.
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Tagged: cyclic group Problem 613
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective.
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(d) Determine the group structure of the kernel of $\phi$. Problem 529
Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$. (a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have? (b) Let $ax+b+I$ be a nonzero element of the field $\F_3[x]/(x^2+1)$, where $a, b \in \F_3$. Find the inverse of $ax+b+I$. (c) Recall that the multiplicative group of nonzero elements of a field is a cyclic group.
Confirm that the element $x$ is not a generator of $E^{\times}$, where $E=\F_3[x]/(x^2+1)$ but $x+1$ is a generator.Add to solve later
Problem 434
Let $R$ be a ring with $1$.
A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$. (It is also called a simple module.) (a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.
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(b) Determine all the irreducible $\Z$-modules. Problem 420
In this post, we study the
Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
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Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$. Problem 302
Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by
\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\] where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring homomorphism, called the augmentation map and the kernel of $\epsilon$ is called the augmentation ideal. (a) Prove that the augmentation ideal in the group ring $RG$ is generated by $\{g-e \mid g\in G\}$.
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(b) Prove that if $G=\langle g\rangle$ is a finite cyclic group generated by $g$, then the augmentation ideal is generated by $g-e$. Read solution
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Sec2.1 Lorentz Invariance
\(\quad\)QFT is based on quantum mechanics, so we provide only the briefest version of summaries, in the generalized version of Dirac:
Def(Ray):
A \emph{ray} \(\mathscr{R}\) is a set of normalized vectors, i.e., \(\{\psi|\langle\psi,\psi\rangle=1\}\). Here \(\langle\cdot,\cdot\rangle\) is the inner product of Hilbert space, satisfying \(\forall\phi\psi\in H,\xi,\eta\in\mathbb{C}\),\par
1) \(\langle\phi,\psi\rangle=\langle\psi,\phi\rangle^{*}\); 2) \(\langle\phi,\xi_{1}\psi_{1}+\xi_{2}\psi_{2}\rangle=\xi_{1}\langle\phi,\psi_{1}\rangle+\xi_{2}\langle\phi,\psi_{2}\rangle\); 3) \(\langle\eta_{1}\phi_{1}+\eta_{2}\phi_{2},\psi\rangle=\eta_{1}^{*}\langle\phi_{1},\psi\rangle+\eta_{2}\langle\phi_{2},\psi\rangle\); 4) \(\langle\psi,\psi\rangle\geqslant0\quad\) and \(\langle\psi,\psi\rangle=0\Rightarrow\psi=0.\) Def:
The \emph{self-adjoint} operator of one linear operator \(A\) of Hilbert space, denoted \(A^{\dagger}\), is defined as \(\forall\phi,\psi\in H, \langle\phi,A^{\dagger}\psi\rangle:=\langle A\phi,\psi\rangle\).
Axiom(QM): 1) Physical states are represented by rays in Hilbert space. 2) Observables are represented by Hermitian operators\footnote{More seriously, ’cause the domain of operators involve in some complicated problems, observables in quantum mechanics are essentially \emph{unbounded self-adjoint} operators in Functional Analysis.}, that is, \(A=A^{\dagger}\). 3) The probability of finding a state represented by \(\mathscr{R}\) in the mutually orthogonal states \(\mathscr{R}_{n}\) is \(P(\mathscr{R}\rightarrow\mathscr{R}_{n})=|\langle\psi,\psi_{n}\rangle|^{2}\). Def:
A \emph{symmetry transformation} \(T\) is a change in the point of view that does not change the results of possible experiments. That is, observer \(\mathcal{O}\) sees a system in a state represented by ray \(\mathscr{R}_{i}\), and observer \(\mathcal{O}’\) looking at the same system will observe it in a different state, represented by ray \(\mathscr{R}’_{i}\). Then we always have
$$P(\mathscr{R}\rightarrow\mathscr{R}_{n})=P(\mathscr{R}’\rightarrow\mathscr{R}’_{n})$$
\(\quad\)Obviously the set of symmetry transformation \(T_{1}:\mathscr{R}\mapsto\mathscr{R}’\) form a group if we naturally define the trivial transformation \(\mathscr{R}\mapsto\mathscr{R}\) as its identity and the group product is defined as \(T_{2}T_{1}:\mathscr{R}\mapsto\mathscr{R}”\), where \(T_{2}:\mathscr{R}’\mapsto\mathscr{R}”\), and the inverse is \(T_{1}^{-1}:\mathscr{R}’\mapsto\mathscr{R}\).
Theorem(Wigner):
Any symmetry transformation \(T\) of rays can be represented as operators on Hilbert space, with \(U(T)\) either \emph{unitary and linear}:
$$\langle U\phi,U\psi\rangle=\langle\phi,\psi\rangle,\quad U(\eta\phi+\xi\psi)=\eta U\phi+\xi U\psi,$$ or \emph{antiunitary and antilinear}: $$\langle U\phi,U\psi\rangle=\langle\phi,\psi\rangle^{*},\quad U(\eta\phi+\xi\psi)=\eta^{*} U\phi+\xi^{*} U\psi,$$ Prooof:
skipped.
\(\quad\)Since the set of symmetry transformation \(\{T\}\) is a group \(G\), we naturally find a map \(U:G\rightarrow GL(H), U=U(T)\), in the sense of Wigner theorem. But this map may not be \emph{homomorphism}. In fact, we can only get
$$U(T_{1})U(T_{2})=e^{\phi(T_{1},T_{2})}U(T_{1}T_{2})$$ from our former discussion. We call this non-homomorphism map as \emph{projective representation} of the symmetry transformation group. In order to avoid tedious discussion about the phase of projective representation(we will return to this topic in the end of this chapter), we directly admitted that Assertion(non-projective representation):
Any representations of symmetry transformation group with phases \(\phi\) can be canceled through group enlarging(without changing its physical implications). That is, we choose such a homomorphism map of \(U:G\rightarrow GL(H)\) that
$$U(T_{1})U(T_{2})=U(T_{1}T_{2}).$$ Sec2.2 Lorentz Transformation
\(\quad\)By the \emph{Special equivalence principle} of Einstein, the unit distance holds in any inertia coordinates, i.e.,
\begin{equation}\label{2.2.1}\eta_{\mu\nu}\mathrm{d} x’^{\mu}\mathrm{d} x^{\nu}=\eta_{\mu\nu}\mathrm{d} x^{\mu}\mathrm{d} x^{\nu},\end{equation}
where the metric in our notation is
$$\eta_{\mu\nu}=\begin{pmatrix}+1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}.$$ Any transformations \(T:x^{\mu}\mapsto x’^{\mu}\) satisfying \eqref{2.2.1} have the linear form that \begin{equation}\label{2.2.2}x’^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}+a^{\mu},\end{equation} such that \begin{equation}\label{2.2.3}\eta_{\mu\nu}\Lambda_{\rho}^{\mu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}.\end{equation} Precisely denote the transformation as \(T(\Lambda,a)\), then from our former discussion, the set of \(T\) forms a group(the existence of inverse can be seen from \((\mathrm{det}\Lambda)^{2}=1)\), and its group product is Assertion: \begin{equation}\label{2.2.4}T(\bar{\Lambda},\bar{a})T(\Lambda,a)=T(\bar{\Lambda}\Lambda,\bar{\Lambda}a+\bar{a}).\end{equation} Proof: We firstly perform a transformation \(x^{\mu}\rightarrow x’^{\mu}\) as \eqref{2.2.2}, then continue to perform another transformation \(x’^{\mu}\rightarrow x”^{\mu}\), which gives $$x”^{\mu}=\bar{\Lambda}_{\rho}^{\mu}\Lambda_{\nu}^{\rho}x^{\nu}+\left(\bar{\Lambda}_{\rho}^{\mu}a^{\rho}+\bar{a}^{\mu}\right).$$ And thus \(T(\Lambda,a)\) satisfy the claimed relation.
According to the discussion in the former section, transformation group \(\{T(\Lambda,a)\}\) induce a unitary representation \(U(T)\) acting on Hilbert space. Homomorphism transfer the product rule of transformation group \eqref{2.2.4} to \(GL(H)\), giving
\begin{equation}\label{2.2.5}U(\bar{\Lambda},\bar{a})U(\Lambda,a)=U(\bar{\Lambda}\Lambda,\bar{\Lambda}a+\bar{a}).\end{equation} From \eqref{2.2.5}, we can easily write down the inverse map \(U^{-1}(\Lambda,a)\) \begin{equation}\label{2.2.6}U^{-1}(\Lambda,a)=U(\Lambda^{-1},-\Lambda^{-1}a)\end{equation} since \(U(\Lambda,a)U(\Lambda^{-1},-\Lambda^{-1}a)=U(1,0)=\mathbbold{1}\). \(\quad\)Then we are to introduce an important symmetry group in physics. Def:
The group of transformation \(T(\Lambda,a)\) is called \emph{inhomogeneous Lorentz group}, or \(\textit{Poincar}\acute{e} \textit{group}\).
\(\quad\)One crucial subgroup \(L\) of the \(\mathrm{Poincar\acute{e}}\) group’s is one that \(a^{\mu}=0\). Specifically,
$$L=\{T|T(\bar{\Lambda},0)T(\Lambda,0)=T(\bar{\Lambda}\Lambda,0).\}.$$ We call this \emph{homogeneous Lorentz group}, denoted as \(\mathrm{O}(1,3)\). It is easy to show that group \(\mathrm{O}(1,1)\) has four distinct components of the form $$\Lambda=\begin{pmatrix}\pm\cosh\theta&\mp\sinh\theta\\ \mp\sinh\theta&\pm\cosh\theta\end{pmatrix}.$$ Since \(\cosh\theta\) is always more than zero, these four components are disjoint with each other, indicating that group \(\mathrm{O}(1,1)\) is topologically not connected. Basing on our discussion in low-dimension conditions, most of physical textbooks directly claim as their please that so dose group \(\mathrm{O}(1,3)\). However, we should not randomly promote the disconnectedness of general Lorentz group because the topology structure of \(\mathrm{O}(1,n)\) is entirely different from the simple \(\mathrm{O}(1,1)\) case. Theorem(disconnectedness of \(\mathrm{O}(1,3)\)) The Lorentz group \(L\) has four connected components. They are \begin{align*}L_{+}^{\uparrow}&=\{\Lambda|\det\Lambda=1,\Lambda_{00}\geqslant1\},\\ L_{-}^{\uparrow}&=\{\Lambda|\det\Lambda=-1,\Lambda_{00}\geqslant1\},\\ L_{+}^{\downarrow}&=\{\Lambda|\det\Lambda=1,\Lambda_{00}\leqslant-1\},\\ L_{-}^{\downarrow}&=\{\Lambda|\det\Lambda=-1,\Lambda_{00}\leqslant-1\}.\end{align*} Especially, component \(L_{+}^{\uparrow}\) is called the \emph{proper chronological Lorentz group}, or mathematically, \(\mathrm{SO}_{+}(1,3)\). Proof:
Take the determinate of \eqref{2.2.3} immediately gives \(\det\Lambda=\pm1\). Also, let indices \(\rho=\sigma=0\), \eqref{2.2.3} gives \(1=\Lambda_{00}^{2}-\Lambda_{10}^{2}-\Lambda_{20}^{2}-\Lambda_{30}^{2}\). Thus, we have either \(\Lambda_{00}\geqslant1\) or \(\Lambda_{00}\leqslant-1\), which follows a disjoint union of open set:
$$L=L_{+}^{\uparrow}\bigcup L_{-}^{\uparrow}\bigcup L_{+}^{\downarrow}\bigcup L_{-}^{\downarrow}.$$ Moreover, since any Lorentz transformation in the three other set \(L_{-}^{\uparrow}, L_{+}^{\downarrow}, L_{-}^{\downarrow}\) can be written as product of discrete transformation(\(\mathscr{P}\equiv\mathrm{diag}\{-1,1,1,1\}\) or \(\mathscr{T}\equiv\mathrm{diag}\{1,-1,-1,-1\}\)) and one element of \(L_{+}^{\uparrow}\), thus it suffices to show the connectedness of \(L_{+}^{\uparrow}\). Let $$H=\{x=(x^{0},\cdots,x^{3})^{T}\in\mathbb{R}^{4}|x^{\mu}x_{\mu}=1, x^{0}\geqslant1\},$$ certainly \((x^{0},\cdots,x^{3})^{T}\mapsto(x^{1},x^{2},x^{3})^{T}\) defines a diffeomorphism(note that \(x^{\mu}x_{\mu}\) confines \(x^{0}\)) of \(H\) with \(\mathbb{R}^{3}\). If \(v_{0}\in H\), we can complete \(v_{0}\) to an orthogonal normalized basis of \(\mathbb{R}^{4}\) through Gram-Schmidt procedure. Denote the super-vector(as an operator) as \(\mathcal{V}=(v_{0},\cdots,v_{3})\). Because by definition \(v_{0}^{0}>1\) and all basis are orthogonal to each other, implying that \(\det\mathcal{V}=1\), \(\mathcal{V}\) should belongs to \(L_{+}^{\uparrow}\). And for \(e_{0}=(1,0,0,0)^{T}\in H\), $$(v_{0},\cdots,v_{3})e_{0}=\sum_{k=0}^{3}v_{i}e_{0}^{i}=v_{1},$$ thus the map \(\pi:L_{+}^{\uparrow}\rightarrow H\) given by \(\pi(\Lambda)=\Lambda e_{0}\) is onto(for \(v_{0}\) is arbitrary). It can be seen that the element of \(\pi^{-1}(e_{0})=\{\Lambda\in L_{+}^{\uparrow}|\Lambda e_{0}=e_{0}\}\) has the form of $$\Lambda=\left(v_{0},v_{1},v_{2},v_{3}\right),$$ where \(v_{0}=(1,0,0,0)^{T}\). On the other hand, since the orthogonality demands \(\langle v_{0},v_{i}\rangle=0\Rightarrow v_{i}^{0}=0\) and \(\langle v_{i},v_{j}\rangle=0\), \(\Lambda\) must be an element of \(\mathrm{SO}(3)\) and therefore \(\pi^{-1}(e_{0})\cong\mathrm{SO}(3)\). It is also easy to see that \(\pi^{-1}(v^{0})=\mathcal{V}\pi^{-1}(e^{0})\).
Indeed, note that \(L_{+}^{\uparrow}\) is Lie subgroup of \(\mathrm{O}(1,3)\), \(\mathrm{SO}(3)\) acts on \(L_{+}^{\uparrow}\) to the right in such a way that \(\pi:L_{+}^{\uparrow}\rightarrow H\) is a principle fiber bundle over \(H\cong\mathbb{R}^{3}\) with group \(\mathrm{SO}(3)\). Any bundle over \(\mathbb{R}^{n}\) is trivial, and so
\(L_{+}^{\uparrow}\) is topologically \(\mathbb{R}^{3}\times\mathrm{SO}(3)\), which is obviously connected.
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Under the auspices of the Computational Complexity Foundation (CCF)
Let $X$ be randomly chosen from $\{-1,1\}^n$, and let $Y$ be randomly
chosen from the standard spherical Gaussian on $\R^n$. For any (possibly unbounded) polytope $P$ formed by the intersection of $k$ halfspaces, we prove that $$\left|\Pr\left[X \in P\right] - \Pr\left[Y \in P\right]\right| \leq \log^{8/5}k \cdot \Delta,$$ where $\Delta$ is a parameter that is small for polytopes formed by the intersection of ``regular'' halfspaces (i.e., halfspaces with low influence). The novelty of our invariance principle is the polylogarithmic dependence on $k$. Previously, only bounds that were at least linear in $k$ were known.
We give two important applications of our main result:
\begin{itemize}
\item A bound of $\log^{O(1)}k \cdot {\epsilon}^{1/6}$ on the Boolean noise sensitivity of intersections of $k$ ``regular'' halfspaces (previous work gave bounds linear in $k$). This gives a corresponding agnostic learning algorithm for intersections of regular halfspaces. \item A pseudorandom generator (PRG) with seed length $O(\log n\,\poly(\log k,1/\delta))$ that $\delta$-fools {\em all} polytopes with $k$ faces with respect to the Gaussian distribution. \end{itemize}
We also obtain PRGs with similar parameters that fool polytopes formed by intersection of regular halfspaces over the hypercube. Using our PRG constructions, we obtain the first deterministic quasi-polynomial time algorithms for approximately counting the number of solutions to a broad class of integer programs, including dense covering problems and contingency tables.
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The Riemann-Lebesgue Lemma
Recall from the Lebesgue Integrable Functions with Arbitrarily Small Integral Terms page that if $f \in L(I)$ then for all $\epsilon > 0$ there exists upper functions $u^*, v^* \in U(I)$ where $f = u^* - v^*$, $v^*$ is nonnegative almost everywhere on $I$, and $\displaystyle{\int_I v^*(x) \: dx < \epsilon}$.
We also saw that there exists $g \in L(I)$ and $s \in S(I)$ where $f = g + s$ and $\displaystyle{\int_I \mid g(x) \mid \: dx < \epsilon}$.
We will now use the latter result to prove an extremely important result called the Riemann-Lebesgue lemma.
Lemma (The Riemann-Lebesgue Lemma): If $f \in L(I)$ and $\beta \in \mathbb{R}$ then $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$. Proof:We begin by showing that the Riemann-Lebesgue lemma holds for step functions. Let $[a, b] \subseteq I$ and define a function $f$ on $I$ by: Then we have that: However we see that: So by the Squeeze theorem we see that indeed, $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$ for functions $f$ as defined above. However, every step function is the sum of functions of this type, so summing the limits of these functions shows that $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$ holds for every $f \in S(I)$. Now by the theorem mentioned at the top of this page, since $f \in L(I)$, for $\displaystyle{\epsilon_1 = \frac{\epsilon}{2} > 0}$ there exists functions $g \in L(I)$ and $s \in S(I)$ such that $f = g + s$ and where: And since we have already verified the Riemann-Lebesgue lemma to be true for step functions we have that $\displaystyle{\lim_{n \to \infty} \int_I s_n(t) \sin (\alpha t + \beta) \: dt = 0}$ so for $\displaystyle{\epsilon_2 = \frac{\epsilon}{2} > 0}$ there exists an $M \in \mathbb{R}$ such that if $\alpha \geq M$ then: Therefore for $\alpha \geq M$ we have that: So for all $\epsilon > 0$ there exists an $M \in \mathbb{R}$ such that if $\alpha \geq M$ then $\displaystyle{\biggr \lvert \int_I f(t) \sin (\alpha t + \beta) \: dt \biggr \rvert < \epsilon}$ so for any $f \in L(I)$ and for all $\beta \in \mathbb{R}$, $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin (\alpha t + \beta) \: dt = 0}$. $\blacksquare$
Corollary 1: If $f \in L(I)$ then: a) $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \sin \alpha t \: dt = 0}$. b) $\displaystyle{\lim_{\alpha \to \infty} \int_I f(t) \cos \alpha t \: dt = 0}$. Proof of a)Set $\beta = 0$. Then by the Riemann-Lebesgue lemma we have that: Proof of b)Set $\beta = \frac{\pi}{2}$. Once again, by the Riemann-Lebesgue lemma we have that:
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What is Energy?
In recent days, we often hear about
energy. Every invention, civilization is based upon acquiring and effectively using energy. This is possible by the unique property of our Universe energy that it can be transformed and transferred, but the total amount is always the same (conserved). One fundamental focus of physics is to investigate energy.
Energy can be generally defined as a scalar quantity associated with the state or condition of one or more objects. In this chapter, we will mainly focus on two forms of mechanical energy:
Kinetic and Potential energy. Kinetic energy
It is associated with the state of motion of an object. The greater the kinetic energy, the faster it moves. For an object of mass
m and velocity v, it is given by the expression,
Unit of energy (all types) is joule (J)
Example:
A 3kg block moving past you at 2.0m/s has a kinetic energy of 6.0J.
Potential energy
It is associated with the configuration (arrangement) of the system of objects that exert forces on one another.
Example:
When you throw a ball down from a height h above the surface of the earth, the increase in kinetic energy of the ball is accounted by defining
gravitational potential energy. This energy is associated with the state of separation (configuration) between two objects (ball and earth) attracted by gravitational force. Work and Potential energy
In the above example, the work done by the gravitational force is negative if the ball is thrown up. As the ball rises, its velocity (kinetic energy) decreases and reaches zero when it attains the maximum height. Now, we can tell that the energy the ball possess at the maximum height is
gravitational potential energy. The work done by the gravitational force is stored as the potential energy. Hence, for any case, the change in potential energy is defined as being equal to the negative of the work done on the system.
Where F is the constant force and \(\Delta x\) is the displacement due to the force.
More generally, if the force is a conservative force, then the change in potential energy is given by,\(\Delta U = \int_{r_1}^{r_2}F(r)dr\)
Conservative force from potential energy can be found by,\(F(r) = -\frac{dU}{dx} = 0\)
Potential energy and equilibrium
The object is said to be in equilibrium if the net force acting on is equal to zero. \(F(r) = -\frac{dU}{dx} = 0\)
Stable equilibrium: A body is in stable equilibrium if it comes back to its normal position on slight displacement. For slight displacement, the restoring force should act on it.\(\frac{d^{2}U}{dx^{2}}> 0\)
Unstable equilibrium: A body is in unstable equilibrium if it does come back to its normal position on slight displacement. For slight displacement, the restoring force should take away the object.\(\frac{d^{2}U}{dx^{2}}< 0\)
Neutral equilibrium: We can move the particle slightly away from such a point and it will still remain in equilibrium (
i.e., it will neither attempt to return to its initial state, nor will it continue to move) \(\frac{d^{2}U}{dx^{2}}= 0\)
Problem:
The potential energy of a conservative system is given by U = ax
2.bx where a and b are positive constants.
This problem is an application of potential energy and conservative force relation
Find the equilibrium , position and discuss whether the equilibrium is stable, unstable or neutral.
In a conservation field \(F = \frac{dU}{dx}\)
∴ \(F = -\frac{d}{dx}(ax^{2}-bx) = b-2ax\)
For equilibrium F = 0 or b – 2ax = 0
∴ \(x = \frac{b}{2a}\)
From the given equation we can see that \(\frac{d^{2U}}{dx^{2}} = 2a\) (positive)., U is minimum
Therefore, x = \(x= \frac{b}{2a}\) is the stable equilibrium position.
Gravitational Potential Energy
The work done by the gravitational force for displacing the object from height 0 to h is given by,\(W = \int_{r_1}^{r_2}F.dr = \int_{0}^{h}mg \;dy\) \(\Delta U = -\int_{0}^{h}-mg\;dy\) \(\Delta U = mgh\)
Elastic Potential Energy
Let’s consider a spring-mass system with a spring of spring constant k and a block of mass, m displaced through a distance x from its equilibrium position (x=0). As the block moves from x=0 to x, the spring force, F
s=-kx does work on the block. The corresponding change in elastic potential energy is given by, Problem:
A block of mass m is attached to two unstretched springs of spring constant k₁ and k₂ as shown in figure . The block is displaced towards right through a distance of x and released. Find the speed of the block as it passes through a distance x/4 from its mean position.
Applying conservation energy
\(\frac{1}{2}K_1x^{2}+\frac{1}{2}K_2x^{2} = \frac{1}{2}mv^{2}+\frac{1}{2}k_1(x/4)^{2}+\frac{1}{2}k_2(x/4)^{2}\) \(v= \frac{x}{4m}\sqrt{15(k_1+k_2)}\)
Work- Energy theorem
This theorem states that work done by all the forces acting on a particle or body is equal to the change in its kinetic energy.
Let us take an example in which a block of mass m kept on a rough horizontal surface is acted upon by a constant force F parallel to the surface and it is displaced through x. The initial velocity and the final velocity are respectively, v
0 and v.
Now, we apply Newton’s law:\(F = ma = m\left ( \frac{v^{2}-v_{o}^{2}}{2x} \right )\)
Finally, the work done is\(W = Fx = m\left ( \frac{v^{2}-v_{o}^{2}}{2x} \right )x =\frac{1}{2}mv^{2}-\frac{1}{2}mv_{o}^{2}\)
We now define the Kinetic energy(K) of the object:\(K = \frac{1}{2}mv^{2}\)
By this definition, the work done on the object is simply equal to the change in the object’s\(W = K-K_o = \Delta K\)
Problem:
The displacement of a body in meter is a function of time according to x = 2t
4+5. Mass of the body is 2kg. What is the increase in its kinetic energy one second after the start of motion ? 8J b)16J
c) 32J d)64J
D)
X = 2t
4 +5
Increase in the kinetic energy results from the work done by the applied force\(\Delta KE = \int_{0}^{1}48\times8t^{5}dt = \frac{48\times8}{6}t^{6} = \frac{48\times8}{6} = 64J\)
A 2kg block is placed on a frictionless horizontal surface. A force shown in the F – x graph is applied to the block horizontally. The change in kinetic energy is
a) 15J b) 20J
c) 25J d) 30J
b)
Work done =Area under F-x graph
W = 1/2 x(10-2)x5 = 20J
Work done = change in kinetic energy = 20J
Law of conservation of energy
Conservation of energy means conservation of all forms of energy together. Accounting all forms of energy within an isolated system, the total energy remains constant. The mechanical energy accounts for only two forms of energy, namely kinetic energy, K and potential energy, U.
If only conservative forces act on a system, then total mechanical energy of the system remains constant. i.e. \(K+U = constant\)
Example
Consider the case of a simple pendulum of massive bob hanging at the end of a massless rod that is pivoted at a fixed point. If the pendulum is given a velocity from its equilibrium position, it executes a simple harmonic motion where energy is conserved at all points.
As the bob is raised from the equilibrium point, its position above the surface of the earth is increased and hence the potential energy is increased. It is accounted for the decrease in kinetic energy as the bob slows down while raising.
When kinetic energy is maximum (at equillibrium position), potential energy is minimum. When potential energy is maximum (extreme positions), kinetic energy is minimum. Thus at all points, total mechanical energy is conserved.
Problem:
Just before striking the ground, a 2.0kg mass has 400J of kinetic energy. If friction can be ignored, from what height was it dropped?(g = 9.8m/s
2)
By conservation of mechanical energy,
U
f+K f = U i+K i
or,0+K
f = mgh+0
An idea massless spring ‘S’ can be compressed 1.0m by a 100N force. It is placed as shown at the bottom of a frictionless inclined plane which makes an angle of \(\theta = 30^{o}\) with the horizontal. A 10 kg block is released from the top of the incline and is brought to rest momentarily after compressing the spring 2.0m. Through what distance does the mass slide before coming to rest ?
Spring constant = 100 N\m
Conserving energy,\(mg(x+0) \sin 30^{o} = \frac{1}{2}K.(2)^{2}\) \(10\times10(x+2)\times \frac{1}{2} = \frac{1}{2}\times100\) \(x+2 = 4\)
x = 2 along the inclined surface
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The Twin Paradox is undoubtedly one of the most discussed things in special relativity and have a tendency to confuse most of us.
Classically, it's resolved by either stick to one of the three reference frames in question – one with the Earth at rest, one following the ship as it goes outwards and one following the ship as it comes back – or by invoking general relativity equalizing acceleration being in a strong gravitational field.
I, however, would like to find a solution where the ship is in rest
all the time, tracking its "view" on the world.
Let's take real numbers. The binary star system Alpha Centauri is about 4.39 light years away making it the perfect candidate for this thought experiment in special relativity. To make calculations more simple we also choose the velocity of the ship such that $\gamma = 2$ is true: we find that $v = 0.87c$.
Now, the ship takes off from Earth heading towards Alpha Centauri with $v = 0.87c$. Since we are aboard on the ship, we instead see Alpha Centauri closing in on us with $v = -0.87c$ from a distance of $\frac{4.39 \; \text{ly}}{\gamma} = 2.20 \; \text{ly}$. This makes the duration of our journey only $\frac{2.20 \; \text{ly}}{0.87c} = 2.53 \; \text{years}$, while $1.26 \; \text{years}$ have passed on Earth.
We have reached our destination and look back at Earth, concluding that in a coordinate frame with our present ship as the origin $(0, 0, 0, 0)$ the Earth has the coordinates $(0, -2.20 \; \text{ly}, 0, 0)$.
We then
instantaneously change our Earthly relative velocity from $v=0.87c$ to $v=-0.87c$ but we still want a reference frame in which we are at rest. We find that a such coordinate system has a relative velocity of $v = -0.99c$ to our reference frame making $\gamma = 7.09$.
Where is the Earth we looked back at in this new reference system?
Well, we had the event $(0, -2.20 \; \text{ly}, 0, 0)$, which we write $\begin{cases} t = 0 \\ x = 2.20 \; \text{ly} \end{cases}$
The coordinates are then transformed according to $\begin{cases} t' = \gamma(t - \frac{vx}{c^2}) \\ x' = \gamma(x-vt) \end{cases}$ yielding
$\begin{cases} t' = -15.44 \; \text{years} \\ x' = -15.60 \; \text{ly}\end{cases}$
We therefore conclude that our instantaneous turn have resulted in $15.44 \; \text{years}$ passed on Earth. But where is the simultaneous Earth?
Since the simultaneous Earth now has the relative velocity $v=0.87c$ its coordinates must therefore be
$\begin{cases} t' = -15.44 \; \text{years} + \Delta t\\ x' = -15.60 \; \text{ly} + 0.87 c \cdot \Delta t \end{cases}$
For $t' = 0$ to be satisfied we find $x' = -15.60 \; \text{ly} + 0.87 c \cdot 15.44 \; \text{years} = -2.2 \; \text{ly}$.
Perfect! That's the result we expected. Our ride back to Earth takes $\frac{2.20 \; \text{ly}}{0.87c} = 2.53 \; \text{years}$.
But here's my question.
For a stationary person on Earth we travelled a total distance of 8.78 light years with the speed of 0.87c. This translates into 10.1 years of Earth time.
But just our turn back caused the axis of simultaneously to shift the time on Earth 15.44 years in the future.
How is this possible?
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Current browse context:
math.DG
Change to browse by: References & Citations Bookmark(what is this?) Mathematics > Differential Geometry Title: Darboux-Weinstein theorem for locally conformally symplectic manifolds
(Submitted on 1 Nov 2015)
Abstract: A locally conformally symplectic (LCS) form is an almost symplectic form $\omega$ such that a closed one-form $\theta$ exists with $d\omega = \theta \wedge \omega$. We present a version of the well-known result of Darboux and Weinstein in the LCS setting and give an application concerning Lagrangian submanifolds. Submission historyFrom: Alexandra Otiman [view email] [v1]Sun, 1 Nov 2015 11:10:49 GMT (80kb,D)
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I found in this article a straightforward way to calculate the eigenvalues of the hamiltonian of an electron under the influence of an homogenous magnetic field (p. 5): http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-06.pdf
$$\vec{B}=B\hat{z}$$
For this, they express the magnetic potential as $$\vec{A}=\frac{1}{2}B(-y,x,0)$$, which gives rise to the magnetic field using the well-known relation $\vec{\nabla}\times \vec{A}$. Using the Pauli hamiltonian, this leads to the eigenvalue equation:
$$H_0\Psi(r)+\frac{\mu_B B}{\hbar}(L_z+\sigma_z)\Psi(r)=E\Psi(r)$$
which allows a straightforward calculation of the energies.
However, I noticed that one can also calculate the same magnetic field using the potential $$\vec{A}=B(-y,0,0)$$, which is no surprise since the magnetic potential is not unique.
Nevertheless, when trying to perform the same calculations I do not find the same energies as before. In particular, I arrived at the following eigenvalue equation:
$$H_0\Psi(r)+\frac{\mu_B B}{\hbar}(L_z+\sigma_z+i\hbar x\partial_y)\Psi(r)=E\Psi(r)$$
That is, it only differs by an additive term $i\hbar^2x\partial_y$, which is weird since my intuition says I should find the same energies (since the magnetic field is exactly the same in both cases). Why does this happen? Or could I somehow reduce this equation to the previous one?
Additional information: to be more clear on how I arrived at the result, I started by calculating, for $\vec{A}=B(-y,0,0)$ and $e=-e$:
$$\vec{A}\cdot\vec{p}=A_xp_x=B(i\hbar y\partial_x)=B(L_z+i\hbar x\partial_y)$$
$$\vec{\nabla}\cdot\vec{A}=0$$
The hamiltonian is, in it's expanded form:
$$\hat{H}=H_0+\frac{1}{2m}(-e\vec{A}\cdot\vec{p}+i\hbar\vec{\nabla}\cdot\vec{A}+e^2\vec{A}^2)+e\phi+\mu_b \vec{\sigma}\cdot\vec{B}$$
Substituting, with $\phi=0$, and neglecting second order terms in $\vec{A}$:
$$\hat{H}=H_0+\frac{1}{2m}(eB(L_z+i\hbar x\partial_y))+e\phi+\mu_b \vec{\sigma}\cdot\vec{B}$$
$$\hat{H}=H_0+\frac{\mu_B B}{\hbar}(L_z+\sigma_z+i\hbar x\partial_y)$$
For stationary states, we can use the time-independent Schrodinger-equation:
$$\hat{H}\Psi(r)=E\Psi(r)$$
$$H_0\Psi(r)+\frac{\mu_B B}{\hbar}(L_z+\sigma_z+i\hbar x\partial_y)\Psi(r)=E\Psi(r)$$
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Roscoff is a village at the north-west corner of France, located on a small piece of land that protrudes into the English canal. Right here, in 1548, the six-year-old Mary, Queen of Scots, having been betrothed to the Dauphin François, disembarks.
As far as I understood, the most common sights in the area are tourists and sea food. As far as I can tell, the main advantage of Roscoff is the Laboratoire Biologique, which is used to host conferences. Every now and then the French game theory group makes use of this facility and organizes a conference in this secluded place. The first week of July was one of these nows and thens. This is my third time to attend the Roscoff conference, and I enjoyed meeting colleagues, the talks, and the vegetarian food that all non-sea-food eaters got.
Here I will tell you about one of the talks by Roberto Cominetti.
Brouwer’s fixed point theorem states that every continuous function $f$ that is defined on a compact and convex subset $X$ of a Euclidean space has a fixed point. When the function $f$ is a contraction, that is, when there is $ρ ∈ [0,1)$ such that $d(f(x),f(y)) ≤ ρ d(x,y)$ for every $x,y \in X$, then Banach’s fixed point theorem tell us that there is a unique fixed point $x*$ and there is an algorithm to approximate it: choose an arbitrary point $x_0 ∈ X$ and define inductively $x_{k+1} = f(x_k)$. The sequence $(x_k)$ converges to $x*$ at an exponential rate.
When the function $f$ is non-expansive, that is, $d(f(x),f(y)) \leq d(x,y)$ for every $x,y \in X$, there may be more than a single fixed point (e.g., $f$ is the identity) and the sequence defined above need not converge to a fixed point (e.g., a rotation in the unit circle).
In his talk, Roberto talked about a procedure that does converge to a fixed point when $f$ is non-expansive. Let $(α_k)$ be a sequence of numbers in $(0,1)$. Choose $x_0 ∈ X$ in an arbitrary way and define inductively $x_{k+1} = α_{k+1} f(x_k) + (1-α_{k+1}) x_k$. Surprisingly enough, under this definition the distance $d(x_k,f(x_k))$ is bounded by
d(x_k,f(x_k)) ≤ C diameter(X) / \sqrt( α_1 (1-α_1) + α_2 (1-α_2) + … + α_n (1-α_n) ),
where C = 1/\sqrt(π).
In particular, if the denominator goes to infinity, which happens, for example, if the sequence $(α_k)$ is constant, then the sequence $(x_k)$ converges to a fixed point. Since the function that assigns to each two-player zero-sum strategic-form game its value is non-expansive, this result can become handy in various situations.
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The possibility though remote, is intriguing as we may be able in the future to actually "see" our own planet's history. Though sounding science fiction, if we are able to detect bodies in space that are able of reflecting light emitted from our planet earth, using amplification systems and filters,this may - if possible, give us a tool of utmost importance.
Not in any practical sense. First of all, the intensity will be very low: inverse-r-squared going out to the reflector, and inverse-r-squared coming back. As @WillCross pointed out, it will be at very low levels. Secondly, the resolution will be very poor, as there's nothing to focus it and the subtended angle is very very small (in both directions).
Consider that we already
do have such a reflector, about 1.5 light seconds away: the dark side of the Moon, illuminated by Earthshine. It somewhat lights up the lunar surface, and variations in the total illumination can be detected that correlate to cloud cover (increased reflection) on Earth. Even being so close, there's no way to see any sort of useful images as you envision. Maybe you could detect a massive nuclear war by reflections from the Moon or nearby planets, but at light year distances, it would be hard to sort out from the noise.
The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it).
The amount of incident radiation from the Sun at any moment is the solar constant ($F \sim 1.3 \times 10^3$ Wm$^{-2}$) integrated over a hemisphere. Thus the total reflected light from the Earth is $L=5\times 10^{16}$ W.
Let's assume this has the same spectrum as sunlight and let's assume that this light gets reflected from something which is positioned optimally - i.e. it sees the full illuminated hemisphere. In that case, roughly speaking, the incident flux on any reflecting body will be $L/2\pi d^2$ (because it is scattered roughly into a hemisphere of the sky).
Now we have to explore two divergent scenarios.
There just happens to be a large object a 1000 light years away that is highly reflective.
Let's be generous and say it is a perfect reflector, but we can't assume specular reflection. Instead let's assume the reflected light is also scattered isotropically into a $2\pi$ solid angle. Thus the radiation we get back will be $$ f = \frac{L}{2\pi d^2} \frac{\pi r^2}{2\pi d^2} = \frac{L r^2}{4\pi d^4},$$ where $r$ is the radius of the thing doing the reflecting.
To turn a flux into an astronomical magnitude we note that the Sun has a visual magnitude of $-26.74$. The apparent magnitude of the reflected light will be given by $$ m = 2.5\log_{10} \left(\frac{F}{f}\right) -26.74 = 2.5 \log_{10} \left(\frac{4F \pi d^4}{L r^2}\right) -26.74 $$
So let's put in some numbers. Assume $r=R_{\odot}$ (i.e. a reflector as big as the Sun) and let $d$ be 1000 light years. From this I calculate $m=85$.
To put this in context, the Hubble space telescope ultra deep field has a magnitude limit of around $m=30$ (http://arxiv.org/abs/1305.1931 ) and each 5 magnitudes on top of that corresponds to a factor of 100 decrease in brightness. So $m=86$ is 22 orders of magnitude fainter than detectable by HST. What's worse, the reflector also scatters all the light from the rest of the universe, so picking out the signal from the earth will be utterly futile.
A big, flat mirror 1000 light years away.
How did it get there? Let's leave that aside. In this case we would just be looking at an image of the Earth as if it were 2000 light years away (assuming everything gets reflected). The flux received back at Earth in this case: $$ f = \frac{L}{2\pi [2d]^2} $$ with $d=1000$ light years, which will result in an apparent magnitude at the earth of $m=37$.
OK, this is more promising, but still 7 magnitudes below detection with the HST and perhaps 5 magnitudes fainter than might be detected with the James Webb Space Telescope if and when it does an ultra-deep field. It is unclear whether the sky will be actually full of optical sources at this level of faintness and so even higher spatial resolution than HST/JWST might be required to pick it out even if we had the sensitivity.
Just send a telescope to 1000 light years, observe the Earth, analyse the data and send the signal back to Earth.
Of course this doesn't help you see into the past because we would have to send the telescope there. But it could help those in the future see into
their past.
Assuming this is technically feasible, the Earth will have a maximum brightness corresponding to $m \sim 35$ so something a lot better than JWST would be required and that ignores the problem of the brightness contrast with the Sun, which would be separated by only 0.03 arcseconds from the Earth at that distance.
Note also that these calculations are merely to detect the light. To extract anything meaningful would mean collecting a spectrum at the very least!
There are several reflectors that were placed on the moon during the Apollo mission and the number of photons detected (after being reflected) is very small compared to the number of photons the laser emits. See the section "Details"
Rather late, but as LDC3 resurrected this question from 2013....
This has already happened. Radio waves reflected from what is presumed to be an asteroid swarm about 25 LY away have returned to earth and been identified as early TV broadcasts. Quite appropriately, the first identified signals were episodes of Dr. Who.
However, radio and visible light are rather different things. If you are expecting to watch Julius Caesar walk up the steps of the Portico, it won't happen. We can barely do that from low orbit.
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Bessel's Inequality for Inner Product Spaces
Recall from the The Pythagorean Identity for Inner Product Spaces page that if $H$ is an inner product space and $\{ x_1, x_2, ..., x_n \}$ is an orthonormal subset of $H$ then for all $c_1, c_2, ..., c_n \in \mathbb{R}$ we have that:(1)
We now use the Pythagorean identity to prove the very important Bessel's inequality. We first need the following lemma.
Lemma 1: Let $H$ be an inner product space. If $\{ e_1, e_2, ..., e_n \}$ is an orthonormal set then for all $h \in H$, $\displaystyle{\sum_{k=1}^{n} \langle e_k, h \rangle^2 \leq \| h \|^2}$. Proof:Let $\displaystyle{g = \sum_{k=1}^{\infty}\langle e_k, h \rangle e_k}$. Observe that: For each $h \in H$ we have that: Therefore $\| g \|^2 \leq \| h \|^2$ which implies that $\| g \| \leq \| h \|$. In other words:
Theorem 2 (Bessel's Inequality): Let $H$ be an inner product space and let $(e_k)_{k=1}^{\infty}$ be an orthonormal sequence. Then for all $h \in H$ we have that $\displaystyle{\sum_{k=1}^{\infty} \langle e_k, h \rangle^2 \leq \| h \|^2}$. Proof:For each $n \in \mathbb{N}$ we have by lemma 1 that: Taking the limit as $n \to \infty$ gives us Bessel's inequality. $\blacksquare$
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Consider a theory $$\mathcal{L}=(\partial_\mu\Phi^\dagger)(\partial^\mu\Phi)-\mu^2(\Phi^\dagger\Phi)-\lambda(\Phi^\dagger\Phi)^2$$ where $\Phi=\begin{pmatrix}\phi_1+i\phi_2\\ \phi_0+i\phi_3\end{pmatrix}$ is a complex $SU(2)$ doublet. After symmetry breaking there is no residual symmetry and hence there are $(2^2-1)=3$ goldstone bosons. The same Lagrangian can also be written as $$\mathcal{L}=\frac{1}{2}\sum\limits_{i=0}^{3}(\partial_\mu\phi_i)^2-\mu^2(\sum\limits_{i=0}^{3}\phi_i^2)-\lambda(\sum\limits_{i=0}^{3}\phi_i^2)^2$$ which is nothing but the Lagrangian of linear sigma model. After symmetry breaking the symmetry of the Lagrangian reduces from $O(4)$ to $O(3)$. Therefore, there are $3$ goldstone bosons once again and the results match. However, I'm having a confusion with the following. Consider the theory $$\mathcal{L}=(\partial_\mu\xi^\dagger)(\partial^\mu\xi)-\mu^2(\xi^\dagger\xi)-\lambda(\xi^\dagger\xi)^2$$ where $\xi=\begin{pmatrix}\xi_1+i\xi_2\\ \xi_3+i\xi_4\\ \xi_0+i\xi_5\end{pmatrix}$ is a complex $SU(2)$ triplet. The Lagrangian is again $SU(2)$ invariant. Right? After SSB there is no residual symmetry and umber of goldstone boson is 3. However, if we write it as $$\mathcal{L}=\frac{1}{2}\sum\limits_{i=0}^{5}(\partial_\mu\xi_i)^2-\mu^2(\sum\limits_{i=0}^{5}\xi_i^2)-\lambda(\sum\limits_{i=0}^{5}\xi_i^2)^2$$ then $O(6)$ symmetry breaks down to $O(5)$ and number of Goldstone bosons is $=5$. So it doesn't match. Then where am I making the mistake? What is the correct number of Goldstone bosons in this case?
Your first two theories,
Φ in the spinor rep of SU(2), and φ in the vector rep of O(4), are dealt with correctly, with 3 generators broken in both cases, so 3 goldstons and one massive field.
You have completely messed up the counting and symmetry structure of your latter theory. The first form, with a complex triplet
ξ, is SU(3)-, not just SU(2), invariant, and this SU(3) breaks down to the residual SU(2) by the v.e.v., so 8-3=5 broken generators, and thus 5 goldstons, and one residual massive field, just as in the language of your O(4)/O(3) vector representation model.
I am unclear as to how you concluded, erroneously, that "there is no residual SU(2)". There is: it mixes up the components not involving the v.e.v. So, for example, if the v.e.v. is dialed to the 3rd component, the SU(2) subgroup mixing up the upper two components ($\lambda_1, \lambda_2,\lambda_3$ Gell-Mann matrices) is unbroken. You ought to brush up on the standard elementary SSB counting arguments, which your teacher must have assigned to you, Ling-Fong Li (1974)
You are right that that the symmetry breaking breaks all three symmetries of $SU(2)$. Thus the $SU(2)$ generators give you three goldstone bosons in the theory with broken symmetry.
However, we have not yet considered all of the symmetries of the original theory. We know that the full symmetry group has six generators and that five of them must be broken. Thus there must be two additional generators of the symmetry group of the original theory (besides the three $SU(2)$ generators we have already counted) which get broken.
Once we include these two additional broken generators we get $5$ goldstone bosons.
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I was watching a video on "How Does a Quantum Computer Work?".
I'm confused about what they mean by: "
Although the qubits can exist in any combination of states, when they are measured they must fall into one of the basis states."
From what I know about linear algebra, if we represent the state of a qubit by $|\psi\rangle$ it can be written like $\alpha|x\rangle + \beta |y\rangle$ (where $|x\rangle$ and $|y\rangle$ form a basis) or $\gamma (|x\rangle+|y\rangle) + \delta (|x\rangle-|y\rangle)$ or even $A(|x\rangle+|100y\rangle) + B |y\rangle$! What I mean is that no set of basis states is
unique.
So, in reality which set of basis states can a qubit (or more generally a quantum system of qubits)
actually collapse to? Can an actual measurement land us with a basis state like $(|0\rangle + |1\rangle)$ or $(|0\rangle - |1\rangle)$ ? Or is only $|0\rangle$ and $|1\rangle$ possible? Also does the basis vector which a qubit can land up in have to have norm $1$ (i.e. must it be an element of an orthonormal basis)?
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Isomorphisms Between Normed Linear Spaces
Recall from the Isomorphisms Between Normed Linear Spaces page that if $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ are normed linear spaces then an isomorphism between $X$ and $Y$ is a bijective bounded linear operator $T : X \to Y$ such that there exists constants $m > 0$ and $M > 0$ for which:(1)
for every $x \in X$. We now give an alternative criterion for determining if a bounded linear operator $T$ is an isomorphism from $X$ to $Y$.
Theorem 1: Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed linear spaces. A bounded linear operator $T \in \mathcal B(X, Y)$ is an isomorphism from $X$ to $Y$ if and only if there exists a bounded linear operator $S \in \mathcal B(Y, X)$ such that $S(T(x)) = x$ for all $x \in X$ and $T(S(y)) = y$ for all $y \in Y$. Proof:$\Rightarrow$ Suppose that the bounded linear operator $T \in \mathcal B(X, Y)$ is an isomorphism from $X$ to $Y$. Let $S = T^{-1}$. Then $S : Y \to X$. We aim to show that $S$ is linear. Let $y_1, y_2 \in Y$. Then: Since $T$ is injective, we have that $S(y_1 + y_2) = S(y_1) + S(y_2)$. Let $\alpha \in \mathbb{R}$ and let $y \in Y$. Then: Again, since $T$ is injective we have that $S(\alpha y) = \alpha S(y)$. Hence $S$ is linear. We now show that $S = T^{-1}$ is bounded. Since $T$ is an isomorphism there exists $m > 0$ and $M > 0$ such that for every $x \in X$ we have that: Since $T$ is surjective for each $y \in Y$ there is a $x \in X$ for which $y = T(x)$. So for all $y \in Y$ we have that: From the first inequality we get that for all $y \in Y$: So $S$ is a bounded linear operator. Lastly, we have that $S(T(x)) = x$ for all $x \in X$ and $T(S(y)) = y$ for all $y \in Y$ since $S = T^{-1}$. $\Rightarrow$ Suppose that there exists a bounded linear operator $S \in \mathcal B(Y, X)$ such that $S(T(x)) = x$ for all $x \in X$ and $T(S(y)) = y$ for all $y \in Y$. Then $T$ and $S$ are bijective. Since $S$ is bounded, for all $y \in Y$ we have that that for some $m > 0$ and for all $y \in Y$ that: Since $S$ is bijective, for all $x \in X$ there is a $y \in Y$ such that $x = S(y)$. So: Hence for all $x \in X$ we have that: And from the boundedness of $T$, for $M > 0$ we have that for all $x \in X$ that: So $T$ is an isomorphism from $X$ to $Y$. $\blacksquare$
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Hyperelastic materials are the special class of materials which tends to respond elastically when they are subjected to very large strains. They shows both a nonlinear material behavior as well as large shape changes. They posses certain characteristics:
All the polymers tends to show the hyperelastic behaviour such as elastomers, rubbers, sponges and other soft flexible materials.
Hyperelastic materials are mostly used where high flexibility on a long run is required under large loads. The typical examples of there uses are as elastomeric pads in brigdes, rail pads, car door seal, car tyres etc.
In finite element analysis, hyperelasticity theory is used to represent the non-linear response of hyperelastic materials at large strains. Hyperlasticity is popular due to it’s ease of use in finite element models. Normally stress-strain curve data from experiments is used to find the constants of theoretical models to fit the material response. There are some good available choices of hyperelasticity models which are available in SimScale platform including:
The stress-strain relation for hyperelastic materials is normally calculated through strain energy density function. The formulation of which is shown below:
Suppose a solid is subjected to a displacement which can be represented by u ( i
Before giving the appropriate material parameters to define a specific hyperelastic materials, one should know the strain energy density forms of the hyperelasticity models. In this section, we have mentioned the strain energy density forms of all the available hyperelasticity models on SimScale platform (already mentioned in
Hyperelasticity).
The basic strain energy density form of reduced polynomial is represented as:
$$U = \displaystyle\sum_{i=1}^{N} C_{i0}(\bar I_1 – 3)^i + \displaystyle\sum_{i=1}^{N} \frac {1}{D_i}(J – 1)^{2i}$$ where, C ( i0
For
N = 1 in above formulation, one obtains the most basic form known as Neo Hookean, represented as: $$U = C_{10}(\bar I_1 – 3) + \frac {1}{D_1}(J – 1)^2$$
For
N = 2 in above formulation, one obtains: $$U = C_{10}(\bar I_1 – 3) + C_{20}(\bar I_1 – 3)^2 + \frac {1}{D_1}(J – 1)^2 + \frac {1}{D_2}(J – 1)^4$$
For
N = 3 in above formulation, one obtains Yeoh, represented as: $$U = C_{10}(\bar I_1 – 3) + C_{20}(\bar I_1 – 3)^2 + C_{30}(\bar I_1 – 3)^3 + \frac {1}{D_1}(J – 1)^2 + \frac {1}{D_2}(J – 1)^4 + \frac {1}{D_3}(J – 1)^6$$
The basic strain energy density form of polynomial is represented as:
$$U = \displaystyle\sum_{i+j=1}^{N}C_{ij}(\bar I_1 – 3)^i(\bar I_2 – 3)^j + \displaystyle\sum_{i=1}^{N} \frac {1}{D_i}(J – 1)^{2i}$$ As a first approximation for D , same eq. eq2 can be used with i
For
N = 1 in above formulation, one obtains the enhancement to Neo Hookean form known as Mooney Rivlin, represented as: $$U = C_{10}(\bar I_1 – 3) + C_{01}(\bar I_2 – 3) + \frac {1}{D_1}(J – 1)^2$$
For
N = 2 in above formulation, one obtains: $$U = C_{10}(\bar I_1 – 3) + C_{01}(\bar I_2 – 3) + C_{20}(\bar I_1 – 3)^2 + C_{02}(\bar I_2 – 3)^2 + C_{11}(\bar I_1 – 3)(\bar I_2 – 3) + \frac {1}{D_1}(J – 1)^2 + \frac {1}{D_2}(J – 1)^4$$
Strain energy density form for Signorini is represented as:
$$U = C_{10}(\bar I_1 – 3) + C_{01}(\bar I_2 – 3) + C_{20}(\bar I_1 – 3)^2 + \frac {1}{D_1}(J – 1)^2$$
Compared to the previous forms, Ogden is represented by the principal stretches \(\bar \lambda_i (i = 1,2,3)\) compared to the deviatoric strain invariants \(\bar I_i (i = 1,2,3)\). The basic strain energy density form of Ogden is represented as:
$$U = \displaystyle\sum_{i=1}^{N} \frac {2 \mu_i}{\alpha_{i}^{2}}(\bar \lambda_{1}^{\alpha_i} + \bar \lambda_{2}^{\alpha_i} + \bar \lambda_{3}^{\alpha_i} – 3) + \displaystyle\sum_{i=1}^{N} \frac {1}{D_i}(J – 1)^{2i}$$ where, μ and i
The strain energy density of Arruda-Boyce has the following form:
$$U = \mu \displaystyle\sum_{i=1}^{5} \frac {C_i}{\lambda_{m}^{2i-2}} \left(\bar I_{1}^{i} – 3 \right) + \frac {1}{D} \left(\frac {J^2 – 1}{2} – lnJ \right)$$ where, \(C_1 = \frac {1}{2}\), \(C_2 = \frac {1}{20}\), \(C_3 = \frac {11}{1050}\), \(C_4 = \frac {19}{7000}\) and \(C_5 = \frac {519}{673750}\). For the compressibility of a material, eq. eq2 can be used as a first approximation for D with \(G_o = \mu (1 + \frac {3}{5 \lambda_{m}^{2}} + \frac {99}{175 \lambda_{m}^{4}} + \frac {513}{875 \lambda_{m}^{6}} + \frac {42039}{67375 \lambda_{m}^{8}})\) i
In experiments, the response of a material for a specific load is normally deduced by the stress-strain relationship. As an example, a basic stress-strain curve of an elastomer for Uniaxial (Tension), Biaxial and Shear load (provided by Axel Product, Inc.) is shown below:
In order to fit the material properties to the experimental stress-strain curve, one has to take in to account eq. eq1 above. This equation shows the stress as a function of invariants containing principal stretches (1 + “strain values”) and strain energy density (of any of the above model). Using this equation, one can calculate the constants of the undertaken model through numerical scheme to get an accurate fit to the stress-strain material curve. In most of the cases, the accuracy of the fit increases with the increasing order of the model.
As an example, let us consider a stress-strain relation of an incompressible uniaxial case for a general polynomial model: The principal stretches of uniaxial case are: $$\lambda_1 = \lambda, \quad \lambda_2 = \lambda_3 = \lambda^\frac{-1}{2} \\ (where, \lambda = 1 + \epsilon)$$ and the deviatoric strain invariants are: $$\bar I_1 = \lambda^2 + 2 \lambda^{-1}, \quad \bar I_2 = \lambda^{-2} + 2 \lambda$$ From principle of virtual work, it follows that: δ U = σ δ λ Than stress can be represented as: $$\sigma = \frac {\partial U}{\partial \lambda} = \frac {\partial U}{\partial \bar I_1} \frac {\partial \bar I_1}{\partial \lambda} + \frac {\partial U}{\partial \bar I_2} \frac {\partial \bar I_2}{\partial \lambda}\\ = 2(\lambda – \lambda^{-2}) \left(\frac {\partial U}{\partial \bar I_1} + \frac {\partial U}{\partial \bar I_2} \right)$$ Therefore, for the general polynomial case: $$\sigma = 2(\lambda – \lambda^{-2}) \displaystyle\sum_{i+j=1}^{N}[iC_{ij}(\bar I_1 – 3)^i(\bar I_2 – 3)^j + jC_{ij}(\bar I_1 – 3)^i(\bar I_2 – 3)^j]$$ By using the experimental stress-strain data in the above equation, one can calculate the material constants for a specific order of polynomial through a numerical scheme e.g. Least square method.
Once the material constants of a specific model are obtained for a good fit to experimental data, one can input these constants to the specified material model in SimScale under material properties. In case a polynomial of order 2 is used, the following can be interpreted:
Same cases are applied to all other models.
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Under the auspices of the Computational Complexity Foundation (CCF)
In this paper we give a new upper bound on the minimal degree of a nonzero Fourier coefficient in any non-linear symmetric Boolean function.
Specifically, we prove that for every non-linear and symmetric $f:\{0,1\}^{k} \to \{0,1\}$ there exists a set $\emptyset\neq S\subset[k]$ such that $|S|=O(\Gamma(k)+\sqrt{k})$, and $\hat{f}(S) \neq 0$, where $\Gamma(m) \leq m^{0.525}$ is the largest gap between consecutive prime numbers in $\{1,\ldots,m\}$. As an application we obtain a new analysis of the PAC learning algorithm for symmetric juntas, under the uniform distribution, of Mossel et al. Namely, we show that the running time of their algorithm is at most $n^{O(k^{0.525})} \cdot \mathrm{poly}(n,2^{k},\log(1/\delta))$ where $n$ is the number of variables, $k$ is the size of the junta (i.e. number of relevant variables) and $\delta$ is the error probability. In particular, for $k\geq\log(n)^{1/(1-0.525)}\approx \log(n)^{2.1}$ our analysis matches the lower bound $2^k$ (up to polynomial factors).
Our bound on the degree greatly improves the previous result of Kolountzakis et al. who proved that $|S|=O(k/\log k)$.
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Answer
The solution set is $$\{0,\frac{2\pi}{3},\frac{4\pi}{3}\}$$
Work Step by Step
$$\cos2x-\cos x=0$$ over interval $[0,2\pi)$ 1) In this case, only the interval for $x$, which is $[0,2\pi)$, is necessary, as you will see in step 2 that $\cos2x$ would be changed to a function of $x$ only. $$x\in[0,2\pi)$$ 2) Now consider back the equation $$\cos2x-\cos x=0$$ Here we see that $\cos x$ is a trigonometric function of $x$, but $\cos2x$ is that of $2x$. Thus it is essential to change $\cos2x$ to a trigonometric function of $x$ by using the identity $\cos2x=2\cos^2 x-1$ $$2\cos^2x-1-\cos x=0$$ $$(2\cos^2x-2\cos x)+(\cos x-1)=0$$ $$(2\cos x+1)(\cos x-1)=0$$ $$\cos x=-\frac{1}{2}\hspace{1cm}\text{or}\hspace{1cm}\cos x=1$$ For $\cos x=1$, over the interval $[0,2\pi)$, there are 1 value whose $\cos$ equals $1$, which is $\{0\}$ For $\cos x=-\frac{1}{2}$, over the interval $[0,2\pi)$, there are 2 values whose $\cos$ equals $-\frac{1}{2}$, which are $\{\frac{2\pi}{3},\frac{4\pi}{3}\}$ Combining the solutions of 2 cases where $\cos x=1$ or $\cos x=-\frac{1}{2}$, we end up with the solution set: $$x=\{0,\frac{2\pi}{3},\frac{4\pi}{3}\}$$
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Answer
$(x,y) = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$, which matches up with option (D)
Work Step by Step
$x = cos~t$ $y = sin~t$ When $t = \frac{\pi}{4}$: $x = cos~(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ $y = sin~(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$ Then $(x,y) = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$, which matches up with option (D)
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Lunar Lander bug?
06-10-2014, 09:02 PM
Post: #1
Lunar Lander bug?
I was playing with the lunar lander game on my 29C and took a look at the source code. I can't figure out the terminal velocity calculation and I was wondering if someone could walk me through it. The program is here.
Lines 46-50 indicate that a burn b accelerates the craft by a=2b-5. This indicates that gravity is -5. It's also easy to deduce that each burn lasts 1 second.
So when you burn the last of the fuel, what is the crash velocity? To figure this out, recognize that you get a 1 second burn consisting of the all the fuel (a=2b-5), followed by a free fall that ends in a crash landing (a=-5). Using the equations provided:
\begin{equation}a_0=2b-5\\
v_1 = v_0+a_0t = v_0+2b-5\\
x_1 = x_0+v_0t+\frac{1}{2}a_0t^2 = x_0+v_0+b-2.5\\
v_f^2 = v_1^2 + 2a_1(x_f-x_1) = v_1^2 + 2a_1(0-x_1) = v_1^2+2*(-5)*(-x_1)\\
v_f = -\sqrt{v_1^2+10x_1}\end{equation}
Lines 41-44 compare the burn amount to the remaining fuel. If you're burning the last of the fuel it goes to label 6 at line 68.
Lines 69-74 add b-2.5 to x
Lines 75-77 add 2b-5 to v
Lines 78-86 compute vf from x
I don't understand lines 69-74. It seems to me that it should subtract v from x also. Am I missing something or is this a bug?
Thanks,
Dave
06-11-2014, 07:28 AM
Post: #2
RE: Lunar Lander bug?
(06-10-2014 09:02 PM)David Hayden Wrote: I don't understand lines 69-74. It seems to me that it should subtract v from x also. Am I missing something or is this a bug?
But in some cases that might turn out negative and then you have a problem when taking the square root in line 85. You could change the program that in lines 41-44 you just use the remaining fuel if the entry was bigger than that (no branch to LBL 6), calculate the new values, check whether we crashed and check whether we still have fuel. Only then loop back. Otherwise calculate the final crash velocity. With this approach the redundant calculations after LBL 6 could be avoided.
Cheers
Thomas
PS: The LBL 9 in line 40 could probably be removed.
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You can still estimate parameters by using the likelihood directly. Let the observations be $x_1, \dots, x_n$ with the exponential distribution with rate $\lambda>0$ and unknown.The density function is $f(x;\lambda)= \lambda e^{-\lambda x}$, cumulative distribution function $F(x;\lambda)=1-e^{-\lambda x}$ and tail function $G(x;\lambda)=1-F(x;\lambda) = e^{-\lambda x}$. Assume the first $r$ observations are fully observed, while for $x_{r+1}, \dots, x_n$ we only know that $x_j > t_j$ for some known positive constants $t_j$. As always, the likelihood is the "probability of the observed data", for the censored observations, that is given by $P(X_j > t_j) = G(t_j;\lambda)$, so the full likelihood function is$$ L(\lambda) = \prod_{i=1}^r f(x_i;\lambda) \cdot \prod_{i=r+1}^n G(t_j;\lambda)$$The loglikelihood function then becomes$$ l(\lambda) = r\log\lambda -\lambda(x_1+\dots+x_r+t_{r+1}+\dots+ t_n)$$which has the same form as the loglikelihood for the usual, fully observed case, except from the first term $r\log\lambda$ in place of $n\log\lambda$. Writing $T$ for the mean of observations and censoring times, the maximum likelihood estimator of $\lambda$ becomes $\hat{\lambda}=\frac{r}{nT}$, which you yourself can compare with the fully observed case.
EDIT
To try to answer the question in comments: If
all observations were censored, that is, we did not wait long enough to observe any event (death), what can we do? In that case, $r=0$, so the loglikelihood becomes$$ l(\lambda) = -nT \lambda$$that is, it is linear decreasing in $\lambda$. So the maximum must be for $\lambda=0$! But, zero is not a valid value for the rate parameter $\lambda$ since it do not correspond to any exponential distribution. We must conclude that in this case the maximum likelihood estimator do not exist! Maybe one could try to construct some sort of confidence interval for $\lambda$ based on that loglikelihood function? For that, look below.
But, in any case, the real conclusion from the data in that case is that we should wait more time until we get some events ...
Here is how we can construct a (one-sided) confidence interval for $\lambda$ in case all observations get censored. The likelihood function in that case is $e^{-\lambda n T}$, which has the same form as the likelihood function from a binomial experiment where we got all successes, which is $p^n$ (see also Confidence interval around binomial estimate of 0 or 1). In that case we want a one-sided confidence interval for $p$ of the form $[\underset{\bar{}}{p}, 1]$. Then we get an interval for $\lambda$ by solving $\log p = -\lambda T$.
We get the confidence interval for $p$ by solving $$ P(X=n) = p^n \ge 0.95 ~~~~\text{(say)}$$so that $ n\log p \ge \log 0.95 $. This give finally the confidence interval for $\lambda$:$$ \lambda \le \frac{-\log 0.95}{n T}.$$
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After the excellent post by JD Long in this thread, I looked for a simple example, and the R code necessary to produce the PCA and then go back to the original data. It gave me some first-hand geometric intuition, and I want to share what I got. The dataset and code can be directly copied and pasted into R form Github.
I used a data set that I found online on semiconductors here, and I trimmed it to just two dimensions - "atomic number" and "melting point" - to facilitate plotting.
As a caveat the idea is purely illustrative of the computational process: PCA is used to reduce more than two variables to a few derived principal components, or to identify collinearity also in the case of multiple features. So it wouldn't find much application in the case of two variables, nor would there be a need to calculate eigenvectors of correlation matrices as pointed out by @amoeba.
Further, I truncated the observations from 44 to 15 to ease the task of tracking individual points. The ultimate result was a skeleton data frame (
dat1):
compounds atomic.no melting.point
AIN 10 498.0
AIP 14 625.0
AIAs 23 1011.5
... ... ...
The "compounds" column indicate the chemical constitution of the semiconductor, and plays the role of row name.
This can be reproduced as follows (ready to copy and paste on R console):
dat <- read.csv(url("http://rinterested.github.io/datasets/semiconductors"))
colnames(dat)[2] <- "atomic.no"
dat1 <- subset(dat[1:15,1:3])
row.names(dat1) <- dat1$compounds
dat1 <- dat1[,-1]
The data were then scaled:
X <- apply(dat1, 2, function(x) (x - mean(x)) / sd(x))
# This centers data points around the mean and standardizes by dividing by SD.
# It is the equivalent to `X <- scale(dat1, center = T, scale = T)`
The linear algebra steps followed:
C <- cov(X) # Covariance matrix (centered data)
$\begin{bmatrix} &\text{at_no}&\text{melt_p}\\\text{at_no}&1&0.296\\\text{melt_p}&0.296&1\end{bmatrix}$
The correlation function
cor(dat1) gives the same output on the non-scaled data as the function
cov(X) on the scaled data.
lambda <- eigen(C)$values # Eigenvalues
lambda_matrix <- diag(2)*eigen(C)$values # Eigenvalues matrix
$\begin{bmatrix} &\color{purple}{\lambda_{\text{PC1}}}&\color{orange}{\lambda_{\text{PC2}}}\\&1.296422& 0\\&0&0.7035783\end{bmatrix}$
e_vectors <- eigen(C)$vectors # Eigenvectors
$\frac{1}{\sqrt{2}}\begin{bmatrix} &\color{purple}{\text{PC1}}&\color{orange}{\text{PC2}}\\&1&\,\,\,\,\,1\\&1&-1\end{bmatrix}$
Since the first eigenvector initially returns as $\sim \small [-0.7,-0.7]$ we choose to change it to $\small [0.7, 0.7]$ to make it consistent with built-in formulas through:
e_vectors[,1] = - e_vectors[,1]; colnames(e_vectors) <- c("PC1","PC2")
The resultant eigenvalues were $\small 1.2964217$ and $\small 0.7035783$. Under less minimalistic conditions, this result would have helped decide which eigenvectors to include (largest eigenvalues). For instance, the relative contribution of the first eigenvalue is $\small 64.8\%$:
eigen(C)$values[1]/sum(eigen(C)$values) * 100, meaning that it accounts for $\sim\small 65\%$ of the variability in the data. The variability in the direction of the second eigenvector is $35.2\%$. This is typically shown on a scree plot depicting the value of the eigenvalues:
We'll include both eigenvectors given the small size of this toy data set example, understanding that excluding one of the eigenvectors would result in dimensionality reduction - the idea behind PCA.
The
score matrix was determined as the matrix multiplication of the scaled data (
X) by the
matrix of eigenvectors (or "rotations"):
score_matrix <- X %*% e_vectors
# Identical to the often found operation: t(t(e_vectors) %*% t(X))
The concept entails a linear
combination of each entry (row / subject / observation / superconductor in this case) of the centered (and in this case scaled) data weighted by the rows of each eigenvector, so that in each of the final columns of the score matrix, we'll find a contribution from each variable (column) of the data (the entire
X), BUT only the corresponding eigenvector will have taken part in the computation (i.e. the first eigenvector $[0.7, 0.7]^{T}$ will contribute to $\text{PC}\,1$ (Principal Component 1) and $[0.7, -0.7]^{T}$ to $\text{PC}\,2$, as in:
Therefore each eigenvector will influence each variable differently, and this will be reflected in the "loadings" of the PCA. In our case, the negative sign in the second component of the second eigenvector $[0.7, - 0.7]$ will change the sign of the melting point values in the linear combinations that produce PC2, whereas the effect of the first eigenvector will be consistently positive:
The eigenvectors are scaled to $1$:
> apply(e_vectors, 2, function(x) sum(x^2))
PC1 PC2
1 1
whereas the (
loadings) are the eigenvectors scaled by the eigenvalues (despite the confusing terminology in the in-built R functions displayed below). Consequently, the loadings can be calculated as:
> e_vectors %*% lambda_matrix
[,1] [,2]
[1,] 0.9167086 0.497505
[2,] 0.9167086 -0.497505
> prcomp(X)$rotation %*% diag(princomp(covmat = C)$sd^2)
[,1] [,2]
atomic.no 0.9167086 0.497505
melting.point 0.9167086 -0.497505
It is interesting to note that the rotated data cloud (the score plot) will have variance along each component (PC) equal to the eigenvalues:
> apply(score_matrix, 2, function(x) var(x))
PC1 PC2
53829.7896 110.8414
> lambda
[1] 53829.7896 110.8414
Utilizing the built-in functions the results can be replicated:
# For the SCORE MATRIX:
prcomp(X)$x
# or...
princomp(X)$scores # The signs of the PC 1 column will be reversed.
# and for EIGENVECTOR MATRIX:
prcomp(X)$rotation
# or...
princomp(X)$loadings
# and for EIGENVALUES:
prcomp(X)$sdev^2
# or...
princomp(covmat = C)$sd^2
Alternatively, the singular value decomposition ($\text{U}\Sigma \text{V}^\text{T}$) method can be applied to manually calculate PCA; in fact, this is the method used in
prcomp(). The steps can be spelled out as:
svd_scaled_dat <-svd(scale(dat1))
eigen_vectors <- svd_scaled_dat$v
eigen_values <- (svd_scaled_dat$d/sqrt(nrow(dat1) - 1))^2
scores<-scale(dat1) %*% eigen_vectors
The result is shown below, with first, the distances from the individual points to the first eigenvector, and on a second plot, the orthogonal distances to the second eigenvector:
If instead we plotted the values of the score matrix (PC1 and PC2) - no longer "melting.point" and "atomic.no", but really a change of basis of the point coordinates with the eigenvectors as basis, these distances would be preserved, but would naturally become perpendicular to the xy axis:
The trick was now to
recover the original data. The points had been transformed through a simple matrix multiplication by the eigenvectors. Now the data was rotated back by multiplying by the inverse of the matrix of eigenvectors with a resultant marked change in the location of the data points. For instance, notice the change in pink dot "GaN" in the left upper quadrant (black circle in the left plot, below), returning to its initial position in the left lower quadrant (black circle in the right plot, below).
Now we finally had the original data restored in this "de-rotated" matrix:
Beyond the change of coordinates of rotation of the data in PCA, the results must be interpreted, and this process tends to involve a
biplot, on which the data points are plotted with respect to the new eigenvector coordinates, and the original variables are now superimposed as vectors. It is interesting to note the equivalence in the position of the points between the plots in the second row of rotation graphs above ("Scores with xy Axis = Eigenvectors") (to the left in the plots that follow), and the
biplot (to the right):
The superimposition of the original variables as red arrows offers a path to the interpretation of
PC1 as a vector in the direction (or with a positive correlation) with both
atomic no and
melting point; and of
PC2 as a component along increasing values of
atomic no but negatively correlated with
melting point, consistent with the values of the eigenvectors:
PCA$rotation
PC1 PC2
atomic.no 0.7071068 0.7071068
melting.point 0.7071068 -0.7071068
This interactive tutorial by Victor Powell gives immediate feedback as to the changes in the eigenvectors as the data cloud is modified.
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Tagged: norm Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.
Add to solve later
(b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. Problem 381
Consider the matrix
\[A=\begin{bmatrix} 3/2 & 2\\ -1& -3/2 \end{bmatrix} \in M_{2\times 2}(\R).\] (a) Find the eigenvalues and corresponding eigenvectors of $A$. (b) Show that for $\mathbf{v}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}\in \R^2$, we can choose $n$ large enough so that the length $\|A^n\mathbf{v}\|$ is as small as we like.
(
University of California, Berkeley, Linear Algebra Final Exam Problem) Read solution Problem 188
Denote by $i$ the square root of $-1$.
Let \[R=\Z[i]=\{a+ib \mid a, b \in \Z \}\] be the ring of Gaussian integers. We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to \[N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.\]
Here $\bar{\alpha}$ is the complex conjugate of $\alpha$.
Then show that an element $\alpha \in R$ is a unit if and only if the norm $N(\alpha)=\pm 1$. Also, determine all the units of the ring $R=\Z[i]$ of Gaussian integers.
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Probability Seminar Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 14, Timo Seppäläinen, UW-Madison
Title:
Geometry of the corner growth model
Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah).
February 21, Diane Holcomb, KTH
Title:
On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Title: Quantitative homogenization in a balanced random environment
Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison).
Wednesday, February 27 at 1:10pm Jon Peterson, Purdue
Title:
Functional Limit Laws for Recurrent Excited Random Walks
Abstract:
Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina.
March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the
character ratio:
$$ \text{trace}(\rho(g))/\text{dim}(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant
rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM). April 4, Philip Matchett Wood, UW-Madison
Title:
Outliers in the spectrum for products of independent random matrices
Abstract: For fixed positive integers m, we consider the product of m independent n by n random matrices with iid entries as in the limit as n tends to infinity. Under suitable assumptions on the entries of each matrix, it is known that the limiting empirical distribution of the eigenvalues is described by the m-th power of the circular law. Moreover, this same limiting distribution continues to hold if each iid random matrix is additively perturbed by a bounded rank deterministic error. However, the bounded rank perturbations may create one or more outlier eigenvalues. We describe the asymptotic location of the outlier eigenvalues, which extends a result of Terence Tao for the case of a single iid matrix. Our methods also allow us to consider several other types of perturbations, including multiplicative perturbations. Joint work with Natalie Coston and Sean O'Rourke.
April 11, Eviatar Procaccia, Texas A&M Title: Stabilization of Diffusion Limited Aggregation in a Wedge.
Abstract: We prove a discrete Beurling estimate for the harmonic measure in a wedge in $\mathbf{Z}^2$, and use it to show that Diffusion Limited Aggregation (DLA) in a wedge of angle smaller than $\pi/4$ stabilizes. This allows to consider the infinite DLA and questions about the number of arms, growth and dimension. I will present some conjectures and open problems.
April 18, Andrea Agazzi, Duke
Title:
Large Deviations Theory for Chemical Reaction Networks
Abstract: The microscopic dynamics of well-stirred networks of chemical reactions are modeled as jump Markov processes. At large volume, one may expect in this framework to have a straightforward application of large deviation theory. This is not at all true, for the jump rates of this class of models are typically neither globally Lipschitz, nor bounded away from zero, with both blowup and absorption as quite possible scenarios. In joint work with Amir Dembo and Jean-Pierre Eckmann, we utilize Lyapunov stability theory to bypass this challenges and to characterize a large class of network topologies that satisfy the full Wentzell-Freidlin theory of asymptotic rates of exit from domains of attraction. Under the assumption of positive recurrence these results also allow for the estimation of transitions times between metastable states of this class of processes.
April 25, Kavita Ramanan, Brown
Title:
Beyond Mean-Field Limits: Local Dynamics on Sparse Graphs
Abstract: Many applications can be modeled as a large system of homogeneous interacting particle systems on a graph in which the infinitesimal evolution of each particle depends on its own state and the empirical distribution of the states of neighboring particles. When the graph is a clique, it is well known that the dynamics of a typical particle converges in the limit, as the number of vertices goes to infinity, to a nonlinear Markov process, often referred to as the McKean-Vlasov or mean-field limit. In this talk, we focus on the complementary case of scaling limits of dynamics on certain sequences of sparse graphs, including regular trees and sparse Erdos-Renyi graphs, and obtain a novel characterization of the dynamics of the neighborhood of a typical particle. This is based on various joint works with Ankan Ganguly, Dan Lacker and Ruoyu Wu.
April 26, Colloquium, Kavita Ramanan, Brown
Title:
Tales of Random Projections
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
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What is Second Order Cone Programming (SOCP)?
Second Order Cone Programming (SOCP) problems are a type of optimisation problem that have applications in many areas of science, finance and engineering. A summary of the type of problems that can make use of SOCP, including things as diverse as designing antenna arrays, finite impulse response (FIR) filters and structural equilibrium problems can be found in the paper ‘Applications of Second Order Cone Programming’ by Lobo et al. There are also a couple of examples of using SOCP for portfolio optimisation in the GitHub repository of the Numerical Algorithms Group (NAG).
A large scale SOCP solver was one of the highlights of the Mark 27 release of the NAG library (See here for a poster about its performance). Those who have used the NAG library for years will expect this solver to have interfaces in Fortran and C and, of course, they are there. In addition to this is the fact that Mark 27 of the NAG Library for
Python was released at the same time as the Fortran and C interfaces which reflects the importance of Python in today’s numerical computing landscape.
Here’s a quick demo of how the new SOCP solver works in Python. The code is based on a notebook in NAG’s PythonExamples GitHub repository.
NAG’s
handle_solve_socp_ipm function (also known as e04pt) is a solver from the NAG optimization modelling suite for large-scale second-order cone programming (SOCP) problems based on an interior point method (IPM).
$$
\begin{array}{ll} {\underset{x \in \mathbb{R}^{n}}{minimize}\ } & {c^{T}x} \\ \text{subject to} & {l_{A} \leq Ax \leq u_{A}\text{,}} \\ & {l_{x} \leq x \leq u_{x}\text{,}} \\ & {x \in \mathcal{K}\text{,}} \\ \end{array} $$
where $\mathcal{K} = \mathcal{K}^{n_{1}} \times \cdots \times \mathcal{K}^{n_{r}} \times \mathbb{R}^{n_{l}}$ is a Cartesian product of quadratic (second-order type) cones and $n_{l}$-dimensional real space, and $n = \sum_{i = 1}^{r}n_{i} + n_{l}$ is the number of decision variables. Here $c$, $x$, $l_x$ and $u_x$ are $n$-dimensional vectors.
$A$ is an $m$ by $n$ sparse matrix, and $l_A$ and $u_A$ and are $m$-dimensional vectors. Note that $x \in \mathcal{K}$ partitions subsets of variables into quadratic cones and each $\mathcal{K}^{n_{i}}$ can be either a quadratic cone or a rotated quadratic cone. These are defined as follows:
$$
\mathcal{K}_{q}^{n_{i}} := \left\{ {z = \left\{ {z_{1},z_{2},\ldots,z_{n_{i}}} \right\} \in {\mathbb{R}}^{n_{i}} \quad\quad : \quad\quad z_{1}^{2} \geq \sum\limits_{j = 2}^{n_{i}}z_{j}^{2},\quad\quad\quad z_{1} \geq 0} \right\}\text{.} $$
$$
\mathcal{K}_{r}^{n_{i}} := \left\{ {z = \left\{ {z_{1},z_{2},\ldots,z_{n_{i}}} \right\} \in {\mathbb{R}}^{n_{i}}\quad\quad:\quad \quad\quad 2z_{1}z_{2} \geq \sum\limits_{j = 3}^{n_{i}}z_{j}^{2}, \quad\quad z_{1} \geq 0, \quad\quad z_{2} \geq 0} \right\}\text{.} $$
For a full explanation of this routine, refer to e04ptc in the NAG Library Manual
Using the NAG SOCP Solver from Python
This example, derived from the documentation for the
handle_set_group function solves the following SOCP problem
minimize $${10.0x_{1} + 20.0x_{2} + x_{3}}$$
from naginterfaces.base import utils
from naginterfaces.library import opt
# The problem size:
n = 3
# Create the problem handle:
handle = opt.handle_init(nvar=n)
# Set objective function
opt.handle_set_linobj(handle, cvec=[10.0, 20.0, 1.0])
subject to the bounds
$$ \begin{array}{rllll} {- 2.0} & \leq & x_{1} & \leq & 2.0 \\ {- 2.0} & \leq & x_{2} & \leq & 2.0 \\ \end{array} $$ # Set box constraints
opt.handle_set_simplebounds(
handle,
bl=[-2.0, -2.0, -1.e20],
bu=[2.0, 2.0, 1.e20]
)
the general linear constraints
\begin{array}{lcrcrcrclcl}
& & {- 0.1x_{1}} & – & {0.1x_{2}} & + & x_{3} & \leq & 1.5 & & \\ 1.0 & \leq & {- 0.06x_{1}} & + & x_{2} & + & x_{3} & & & & \\ \end{array} # Set linear constraints
opt.handle_set_linconstr(
handle,
bl=[-1.e20, 1.0],
bu=[1.5, 1.e20],
irowb=[1, 1, 1, 2, 2, 2],
icolb=[1, 2, 3, 1, 2, 3],
b=[-0.1, -0.1, 1.0, -0.06, 1.0, 1.0]
);
and the cone constraint
$$\left( {x_{3},x_{1},x_{2}} \right) \in \mathcal{K}_{q}^{3}\text{.}$$
# Set cone constraint
opt.handle_set_group(
handle,
gtype='Q',
group=[ 3,1, 2],
idgroup=0
);
# Set some algorithmic options.
for option in [
'Print Options = NO',
'Print Level = 1'
]:
opt.handle_opt_set(handle, option)
# Use an explicit I/O manager for abbreviated iteration output:
iom = utils.FileObjManager(locus_in_output=False)
Finally, we call the solver
# Call SOCP interior point solver
result = opt.handle_solve_socp_ipm(handle, io_manager=iom)
------------------------------------------------
E04PT, Interior point method for SOCP problems
------------------------------------------------
Status: converged, an optimal solution found
Final primal objective value -1.951817E+01
Final dual objective value -1.951817E+01
result.x
array([-1.26819151, -0.4084294 , 1.3323379 ])
and the objective function value is
result.rinfo[0]
-19.51816515094211
Finally, we clean up after ourselves by destroying the handle
# Destroy the handle:
opt.handle_free(handle)
As you can see, the way to use the NAG Library for Python interface follows the mathematics quite closely.
NAG also recently added support for the popular cvxpy modelling language that I’ll discuss another time. Links
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To me the unification of logistic, linear, poisson regression etc... has always been in terms of specification of the mean and variance in the Generalized Linear Model framework. We start by specifying a probability distribution for our data, normal for continuous data, Bernoulli for dichotomous, Poisson for counts, etc...Then we specify a link function that describes how the mean is related to the linear predictor:
$g(\mu_i) = \alpha + x_i^T\beta$
For linear regression, $g(\mu_i) = \mu_i$.
For logistic regression, $g(\mu_i) = \log(\frac{\mu_i}{1-\mu_i})$.
For Poisson regression, $g(\mu_i) = \log(\mu_i)$.
The only thing one might be able to consider in terms of writing an error term would be to state:
$y_i = g^{-1}(\alpha+x_i^T\beta) + e_i$ where $E(e_i) = 0$ and $Var(e_i) = \sigma^2(\mu_i)$. For example, for logistic regression, $\sigma^2(\mu_i) = \mu_i(1-\mu_i) = g^{-1}(\alpha+x_i^T\beta)(1-g^{-1}(\alpha+x_i^T\beta))$. But, you cannot explicitly state that $e_i$ has a Bernoulli distribution as mentioned above.
Note, however, that basic Generalized Linear Models only assume a structure for the mean and variance of the distribution. It can be shown that the estimating equations and the Hessian matrix only depend on the mean and variance you assume in your model. So you don't necessarily need to be concerned with the distribution of $e_i$ for this model because the higher order moments don't play a role in the estimation of the model parameters.
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Table of Contents: How Capacitors are Connected?
We are also interested to calculate the amount of charge stored in a system where two or more capacitors are connected. The combination can be made in many ways. The combination is connected to a battery to apply a potential difference (V) and charge the plates (Q). We can define the equivalent capacitance of the combination between two points to be
\(C=\frac{Q}{V}\)
Two frequently used methods of combination are: Parallel combination and Series combination
Parallel Combination of Capacitors
When capacitors are connected in parallel, the potential difference V across each is the same and the charge on C
1, C 2 is different i.e., Q 1 and Q 2.
The total charge is Q given as:
\(Q={{Q}_{1}}+{{Q}_{2}}\) \(Q={{C}_{1}}V+{{C}_{2}}V\) \(\frac{Q}{V}={{C}_{1}}+{{C}_{2}}\)
Equivalent capacitance between a and b is:
C = C
1 + C 2
The charges on capacitors is given as:
\(Q1=\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}Q\) \(Q2=\frac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}Q\)
In case of more than two capacitors, C = C
1 + C 2 + C 3 + C 4 + C 5 + ………… Watch this Video for More Reference Series Combination of Capacitors
When capacitors are connected in series, the magnitude of charge Q on each capacitor is same. The potential difference across C
1 and C 2 is different i.e., V 1 and V 2.
Q = C 1 V1 = C2 V2
The total potential difference across combination is:
V = V
1 + V 2
\(V=\frac{Q}{{{C}_{1}}}+\frac{Q}{{{C}_{2}}}\)
\(\frac{V}{Q}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\)
The ratio Q/V is called as the equivalent capacitance C between point a and b.
The equivalent capacitance C is given by: \(\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\)
The potential difference across C
1 and C 2 is V 1 and V 2 respectively, given as follows:
\({{V}_{1}}=\frac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}};\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{V}_{2}}=\frac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}}V\)
In case of more than two capacitors, the relation is:
\(\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+\frac{1}{{{C}_{4}}}+……\)
Important Points: If N identical capacitors of capacitance C are connected in series, then effective capacitance = C/N If N identical capacitors of capacitance C are connected in parallel, then effective capacitance = CN Problems on Combination of Capacitors Problem 1: Two capacitors of capacitance C 1 = 6 μ F and C 2 = 3 μ F are connected in series across a cell of emf 18 V. Calculate: The equivalent capacitance The potential difference across each capacitor The charge on each capacitor Sol: (a) \(\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\) \(\Rightarrow \frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{6\times 3}{6+3}=2\mu F\) (b) \({{V}_{1}}=\frac{C{}_{2}}{{{C}_{1}}+{{C}_{2}}}V=\frac{3}{6+3}\times 18=6\,volts\)
\({{V}_{2}}=\frac{C{}_{1}}{{{C}_{1}}+{{C}_{2}}}V=\frac{6}{6+3}\times 18=12\,volts\)
(c) Q 1 = Q 2 = C 1 V 1 = C 2 V 2 = CV
Charge on each capacitor = C
eq V = 2μF x 18 volts = 36μC
In the above problem, note that the smallest capacitor has the largest potential difference across it.
Example 2: 3 capacitors are arranged as shown in the figure. Find the equivalent capacitance between A and B.
Sol: All the 3 capacitors are in parallel C 4 + C 1 + C 2 + C 3 = 3C Example 3: Find the equivalent capacitance between points A and B capacitance of each capacitor is 2 μF.
Sol: In the system given, 1 and 3 are in parallel. 5 is connected between A and B. So, they can also be represented as follows. As 1 and 3 are in parallel, their effective capacitance is 4μF 4μF and 2μF are in series, their effective capacitance is 4/3μF 4/3μF and 2 10/3μF are in parallel, their effective capacitance is 10/3μF 10/3μF and 2μF are in series, their effective capacitance is 5/4μF 5/4μF and 2μF are in parallel, their effective capacitance is 13/4μF
Therefore the equivalent capacitance of the given system is 13/4μF.
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Concluding Progressive Remarks
The Salmon Prize (photo of the prize here) is offered by mathematician/biologist Elizabeth Allman, and can be appreciated in the broad mathematical context that is provided by Topics in Tensors: A Summer School by Shmuel Friedland.
Until such time as further comments are offered, a working answer is:
An introduction to multilinear varieties considered as algebraic/geometric objects may be found on page 99ff of Joe Harris'
Algebraic Geometry: a First Course.
A recent comprehensive survey is chapter 7 of Joseph Landsberg's brand-new book
Tensors: Geometry and Applications(December 2011), which begins (encouragingly)
"This chapter includes nearly all that is known about defining equations for secant varieties of Segre and Veranese varieties …"
At the conclusion of the meta thread that was started by Andy Putman (who wondered why this question was being downvoted) concrete examples now are given of higher-order 'almost-Hilbert' varieties. The following intent is posted there:
Last night I discovered a brand-new monograph by Joseph Landsberg that encompasses more-or-less the answer sought, and so I have amended the beginning of the question to be a pointer to Landsberg's monograph.
Sometime in the next week or two I will post a concrete mathematical question — framed within the context that Landsberg's monograph supplies — asking for a classification of all multilinear varieties having unit-dimension defect with respect to their natural Segre embedding.
At that time I will request closure of the original question, to be supplanted by this classification question.
A shorter, accessibly written, and free-as-in-freedom introduction to these multilinear varieties is Joseph Landsberg's relatively recent — and much-cited —
Bulletin of the AMSsurvey article "Geometry and the complexity of matrix multiplication" (2008).
Please let me apologize for the deficiencies of original question in conveying to MOF readers the mathematical depth and breadth of these multilinear varieties, their implications for fundamental quantum physics, and their numerous practical applications in engineering fields ranging from the computational complexity matrix multiplication to the simulation of quantum transport.
Fortunately, Joseph Landsberg's recent writings have done an immensely better job of this than the original draft of my MOF question did!
My appreciation and thanks are extended to all who have provided comments, and in particular, sincere congratulations are extended to Theo Johnson-Freyd for providing an answer that has received MOF's first-ever Gold Reversal Medal. It has been great fun to help this happen!
The question asked (as clarified per Joseph Landsberg)
For the theorem of algebraic geometry that is specified below, please provide a reference (or references) that:
states the theorem rigorously, proves the theorem explicitly, within a framework that extends naturally to multi-linear algebraic varieties The theorem for which mathematical references are sought
Let $k\ge1$ be an integer and let $\boldsymbol{\psi}=\{\psi_{(mn)}\}$ and $\boldsymbol{\xi}=\{\xi_{(srm)}\}$ be vectors in $\mathbb{C}^{k^2}$ and $\mathbb{C}^{2(k-1)k}$ respectively. Here ${(}\dots{)}$ is a multi-index, repeated indices are summed, and the indices $\{s,r,m,n\}$ range over $s \in \{1,2\}$, $r \in \{1,\dots,k-1\}$, and $m,n \in \{1,\dots,k\}$. Then we have:
The Second-Hand Lion Theorem (SHLT)$$ %\forall\ \boldsymbol{\psi}\ \colon\ \ \det_{mn}\ \psi_{(mn)} = 0 \quad\Longleftrightarrow\quad \exists\ \boldsymbol{\xi}\ \colon\ \ \psi_{(mn)} = \xi_{(1rm)}\,\xi_{(2rn)}. $$ Context of the question in multilinear algebraic geometry
Primary consideration should be given to references that prove the theorem and/or discuss related theorems within a mathematical framework that extends naturally to multilinear algebraic varieties.
Secondary consideration should be given to references that are reasonably accessible to the (many) engineers and physicists for whom these multilinear varieties increasingly are finding practical applications.
In the context of algebraic geometry, $r$ may be regarded as an index over $(k{-}1)$ order-2 Segre varieties that enter in a rank-$(k{-}2)$ secant join having the natural Segre embedding in the tensor product space $\mathcal{H}_1 \otimes \mathcal{H}_2$.
In the notation of Joe Harris'
Algebraic Geometry: a First Course, the theorem asserts the identity of the preceding Segre embedding with what is called the generic determinantal variety $\mathcal{M}^{(kk)\!}_{k{-}1}$ that comprises (by definition) the set of $k\times k$ complex matrices having matrix rank $k-1$.
Attention is directed particularly to a passage in Harris (page 100) that states:
"We should draw a fundamental and important distinction between bi- and tri- or multilinear objects […] whose invariants are far from being completely understood."
Thus, although the theorem stated can be solved via specialized techniques that apply uniquely to bilinear varieties, a broader and deeper grounding is sought for this theorem within the context that modern algebraic geometry provides, with regard especially to techniques that extend naturally to generic multilinear algebraic varieties. To borrow a phrase from Richard Hamming, "The purpose of the question is insight, not theorems."
Quantum physics and engineering applications
In the context of quantum physics, $s$ may be regarded as an index over two $k$-dimensional Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$, each equipped with an $k$-element orthonormal basis, such that $(mn)$ is a multi-index over the quantum amplitudes $\psi_{(mn)}$ that are naturally associated to the $k^2$ orthonormal basis vectors of the bipartite Hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_2$.
In quantum systems engineering (QSE), determinantal varieties are the bread-and-butter state-spaces of large-scale quantum simulations, because they support both the natural geometric pullback of thermodynamical relations and conservation laws and the numerically efficient integration of dynamical trajectories that respect these relations.
The origin of the SHLT name
The name "SHLT" is a homage to the following dialog line in the film
Second Hand Lions:
Uncle Garth (Michael McCaine):This lion's no good … it's … it's … defective.
The word
defective refers specifically to a $\mathcal{M}^{(kk)\!}_{k{-}1}$ determinantal variety's one-dimensional (nonlinear) rank-defect as a quantum state-space, relative to the $k^2$-dimensional Hilbert space in which it is immersed (see below).
As a starting-point, the SHLT is mentioned — but regrettably only in passing and with no derivation given — in the paragraphs following Example 9.2 on page 99 of Harris.
Two quantum physics conjectures
By definition, let a quantum state $\psi \in \mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$ (so that $\dim \mathcal{H} = k^2$) be called
$k$-Lion iff $\psi \in \mathcal{M}^{(kk)\!}_{k{-}1}$ (in Harris' notation for $\mathcal{M}\,$).
Then we have:
The Weak $k$-Lion Hypothesis
There exists a finite integer $k\lesssim 2^5$ such that no practicable quantum experiment can observationally disprove the hypothesis that the state-space of a symmetrically bipartite dynamical system is $k$-Lion rather than Hilbert.
Physically, the limit $k\lesssim 2^5$ corresponds to the case of quantum entanglement in a bipartite dynamical system having $5+5=10$ qubits in total.
The Weak $k$-Lion Hypothesis is sufficiently difficult to test — although by construction the required tests are far easier than demonstrating fault-tolerant quantum computing (FTLC) — that it is reasonable to suppose that Weak $k$-Lion Hypothesis cannot be feasibly be disconfirmed even for for quite small values of $k$. Hence it is both mathematically and physically natural to conjecture:
The Strong $k$-Lion Hypothesis
There exists a constant of Nature in the form of a finite integer $k$, such that no experiment can observationally disprove the hypothesis that the state-space of a symmetrically bipartite dynamical system is $k$-Lion rather than Hilbert, for the fundamental reason that the dynamical state-space of Nature is a determinantal variety rather than a Hilbert space.
This question's three four five-level reward structure
Associated to this question is a
three four five-level reward structure:
Gain MathOverflow reputation by providing good math literature references.
Contribute to the (wonderful) ongoing GLL debate between Aram Harrow and Gil Kalai, and thereby help also to accelerate the medical goals of the UW/ISH Naturality and Guidance Seminar.
Demonstrate either $k$-Lion Hypothesis to $100K wager on quantum computing">win $100,000 from Scott Aaronson.
Receive MOF's first-ever award of the Gold Reversal Medal. Congratulations, Theo!
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Here's a nice couple of paragraphs from Bohr & Mottelson:
Since time reversal $T$ anticommutes with the total angular momentum, it is convenient to combine $T$ with a rotation $R$ through the angle $\pi$ about an axis perpendicular to the $z$-axis (the axis of space quantization). Such a rotation also inverts [angular momentum] $I_z$ and thus
\begin{align*}
[ RT, I_z ] &= 0 \\
[ RT, (\mathbf I)^2] &= 0.
\end{align*}
It is therefore possible to construct a set of basis states with quantum numbers $IM$, which are also eigenvectors of $RT$. By suitably choosing the phases of these states, the eignvalues of $RT$ may be set equal to unity,
$$
RT \left| \alpha IM\right> = \left| \alpha IM \right>,
$$
where $\alpha$ represents a set of additional quantum numbers specifying the internal structure of the states. The conventional phasing corresponds to choosing the rotation axis of $R$ to be the $y$ axis.
and later
In the representation in which $j_z$ is diagonal, the nonvanishing matrix elements of the angular momentum operators are
\begin{align*}
\left< jm \middle| j_z \middle| jm\right> &= m \\
\left< jm\pm1 \middle| j_x\pm ij_y \middle| jm\right> &= \sqrt{(j\mp m)(j\pm m+1)}
\end{align*}
The nondiagonal matrix elements of $j_x\pm ij_y$ involve arbitrary phase factors associated with the choice of relative phases for the states with different $m$. The phase convention [above] implies that the matrix elements of $j_x$ are real while those of $j_y$ are purely imaginary, since $j_x$ commutes with $RT$ while $j_y$ anticommutes with $RT$. We are thus left with arbitrary real phase factors (±1), which are conventionally fixed by the requirement that the matrix elements of $j_x ± i j_y$ be positive (Condon and Shortley, 1935).
Nowadays these phase "choices" are swallowed up in the standard definitions of the spherical harmonics (particularly the relationship between phase and $m$) and in the typical representation for the three spin axes (one diagonal, one purely real and one purely imaginary).
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The Orthogonality Theorem for Characters of Irreducible Group Representations
Definition: Let $G$ be a group and let $\varphi, \psi : G \to \mathbb{C}$. The Inner Product of $\varphi$ and $\psi$ will be defined as $\displaystyle{\langle \varphi, \psi \rangle = \frac{1}{|G|} \sum_{g \in G} \overline{\varphi(g)} \psi(g)}$.
If $G$ is a group and $\varphi : G \to \mathbb{C}$ has the property that for all $g \in G$ and for all $h \in H$ that:(1)
Then $\varphi$ is said to be invariant under conjugation. If $\varphi$ and $\psi$ are both invariant under conjugation, then the inner product of $\varphi$ and $\psi$ is given by:(2)
Where the sum above is over all conjugacy classes $C \subseteq G$ and where $\overline{\varphi(C)} := \varphi(c)$ and $\psi(C) = \psi(c)$ where $c \in C$.
Lemma 1:Let $G$ be a group and let $(V, \rho_V)$ and $(W, \rho_W)$ be representations of $G$. Let $\phi : V \to w$ be a linear map and let $\phi_0 : V \to W$ be defined for all $v \in V$ by $\displaystyle{\phi_0(v) = \frac{1}{|G|} \sum_{g \in G} \rho_W(g) \circ \phi \circ \rho_V(g)(v)}$. Then: a) $\phi_0$ is a $G$-equivariant map. b) If $V$ and $W$ are irreducible representations of $G$ that are not isomorphic, then $\phi_0 = 0$. c) If $(V, \rho_V) = (W, \rho_W)$ are irreducible representations of $G$ then $\phi_0 = \frac{\mathrm{trace}(\phi)}{\mathrm{dim}(V))} \cdot \mathbf{id}_V$.
The lemma above will be used in order to prove the following major theorem:
Theorem 2 (The Orthogonality Theorem): Let $G$ be a group. a) If $V$ and $W$ are irreducible group representations of $G$ that are non-isomorphic, then $\langle \chi_V, \chi_W \rangle = 0$. b) If $V$ is an irreducible group representation of $G$ then $\langle \chi_V, \chi_V \rangle = 1$. Proof of a):Equip $V$ and $W$ with inner products and let $\{ v_1, v_2, ..., v_m \}$ be an orthonormal basis for $V$ and let $\{ w_1, w_2, ..., w_n \}$ be an orthonormal basis for $W$. Let $\varphi : V \to V$ be a linear operator. The trace of $\varphi$ is then: Look at the inner product of the characters $\chi_W$ and $\chi_V$: Now, since $W \otimes V^* \cong \mathrm{Hom}(V, W)$, we see that if $w \in W$ and $f \in V^*$, $f : V \to \mathbb{C}$ then $w \otimes f$ can be viewed as a linear operator from $V$ to $W$ defined for all $v \in V$ by: For each $v_0 \in V$, we can let $v_0^* : V \to \mathbb{C}$ be defined for all $v \in V$ by: So for each $w \in W$ and each $v_0 \in V$ we have that $w \otimes v_0^*$ can be viewed as a linear operator from $V$ to $W$ defined for all $v \in V$ by: For each $1 \leq i \leq m$ and for each $1 \leq j \leq n$, consider the linear operator $w_j \otimes v_i^* : V \to W$. For each $g \in G$ and for each $v \in V$ we see that: So in the sum at $(*)$ we have that: If $V \not \cong W$ then by the above lemma the above sum is zero. $\blacksquare$ Proof of b)If $V = W$ then we have by the above lemma that:
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One thing you might want to do is write the vector $\vec{a}$ as some magnitude $\theta$ times a
unit vector
{a1, a2, a3}. Then calculate $\exp(i \theta \,a\cdot\sigma)$. This will allow you to get rid of those factors of $\sqrt{a_1^2+a_2^2+a_3^2}$, which are just some $\theta$ anyway.
Here it is in Mathematica:
a = \[Theta] {a1, a2, a3};
b = a.Array[PauliMatrix, 3];
c = MatrixExp[I b];
c//FullForm;
c = c /. Plus[Power[a1, 2], Power[a2, 2], Power[a3, 2]] -> 1;
c // ExpToTrig // MatrixForm
$\left(\begin{array}{cc} \cos (\theta )+\text{a3} i \sin (\theta ) & \text{a2} \sin (\theta )+\text{a1} i \sin (\theta ) \\ i \text{a1} \sin (\theta )-\text{a2} \sin (\theta ) & \cos (\theta )-i \text{a3} \sin (\theta ) \\\end{array}\right)$
I used
FullForm to peek at the matrix
c so I would know how to make the substitution that eliminates the magnitude of the unit vector.
To answer your second question, you can expand any 2x2 matrix $M$ in terms of the set $S$ consisting of the identity matrix and the 3 Pauli matrices, but the coefficients in the expansion will only be real numbers if $M$ is Hermitian. It is fairly easy to prove this. You know that the set $T$ of 4 matrices each with a 1 in one position and 0's in the other positions is a basis for any 2x2 matrix. You can write each matrix in $T$ in terms of the matrices in $S$, but some of the coefficients are complex. If $S$ is basis for $T$ and $T$ is a basis for all 2x2 matrices ...
Thanks to your comment, we now seek coefficients $c_0, c_1, c_2, c_3$ such that matrix $\hat{c} = c_0 \hat{1} + i\,c_1\hat{\sigma}_1 +i\,c_2\hat{\sigma}_2 +i\,c_3\hat{\sigma}_3$. Here is one way to do it in Mathematica. First, create a basis set and a list of coefficients. Note that we have $i$ times the coefficients of the Pauli matrices, so we explicitly put the $i$ into the list of coefficients. The dot product of the coefficients and the basis is the matrix $m$, which we want to to be equal to the given matrix $c$. So, we form the equations and solve for the coefficients. A quick way to display the solution gives a "fair" result, but the Pauli matrices are not in the proper order. A "better" display can be had using the
Row function. Here is the code that takes the previous result for $c$ from above, finds the expansion coefficients and displays the result:
basis = Join[{IdentityMatrix[2]}, Array[PauliMatrix, 3]];
coeff = {c0, I c1, I c2, I c3};
m = coeff.basis // Expand;
m // MatrixForm;
eqns = Flatten[Thread /@ Thread[m == c]];
soln = Solve[eqns, {c0, c1, c2, c3}] // First;
fair = coeff.(MatrixForm /@ basis) /. soln
Table[coeff[[k]] MatrixForm[basis[[k]]], {k, 1, 4}];
better = Row[% /. soln, "+"]
$$\cos (\theta )\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\\end{array}\right) +i \text{a1} \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \\\end{array}\right) \sin (\theta )+i \text{a2} \left(\begin{array}{cc} 0 & -i \\ i & 0 \\\end{array}\right) \sin (\theta )+i \text{a3} \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \\\end{array}\right) \sin (\theta )$$
Factoring out the $i$ and the $\sin\theta$ and putting the Pauli matrix terms in parentheses is a bit trickier. I don't have a solution for that bit of finesse.
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496 results for "fraction".
Three graphs are given with areas underneath them shaded. The student is asked to calculate their areas, using integration. Q1 has a polynomial. Q2 has exponentials and fractional functions. Q3 requires solving a trig equation and integration by parts.
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Factorise $x^2+cx+d$ into 2 distinct linear factors and then find $\displaystyle \int \frac{ax+b}{x^2+cx+d}\;dx,\;a \neq 0$ using partial fractions or otherwise.
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This is a set of questions designed to help you practice adding, subtracting, multiplying and dividing fractions.
All of these can be done without a calculator.
Question Needs to be tested
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This is a homework assignment. Students may complete this assignment during a maths lesson with the support from the teacher. Passing mark 60% Difficulty level 3 Moderat/Average (C+, B-, B Level). Exam (1 question) Draft
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The Cauchy Product of Power Series
The Cauchy Product of Power Series
Recall from Convergence of Cauchy Products of Two Series of Real Numbers page that:
If $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$ are both absolutely convergent series that converge to $A$ and $B$ respectively, then the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges absolutely to $AB$. If $\displaystyle{\sum_{n=0}^{\infty} a_n}$ is absolutely convergent and converges to $A$, while $\displaystyle{\sum_{n=0}^{\infty} b_n}$ is conditionally convergent and converges to $B$, then the Cauchy product $\displaystyle{\sum_{n=0}^{\infty} c_n}$ converges (not necessarily absolutely) to $AB$.
We will now look at a nice consequence of the first part of this theorem with regards to power series.
Theorem 1: Let $\displaystyle{\sum_{n=0}^{\infty} a_nx^n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_nx^n}$ be two power series that are absolutely convergent and converge to $A(x)$ and $B(x)$ respectively. Then the Cauchy product of these power series is $\displaystyle{\sum_{n=0}^{\infty} c_nx^n}$ and it converges to $A(x)B(x)$. Proof:Suppose that $\displaystyle{A(x) = \sum_{n=0}^{\infty} a_nx^n}$ and $\displaystyle{B(x) = \sum_{n=0}^{\infty} b_nx^n}$. Let $\displaystyle{\sum_{n=0}^{\infty} C_n}$ denote the Cauchy product for these two series. Then for each $n \in \mathbb{N}$ we have that:
\begin{align} \quad C_n = \sum_{k=0}^{n} (a_kx^k)(b_{n-k}x^{n-k}) = \sum_{k=0}^{n} a_kb_{n-k}x^n = x^n \sum_{k=0}^{n} a_kb_{n-k} \end{align}
Let $\displaystyle{\sum_{n=0}^{\infty} c_n}$ denote the Cauchy product for the series $\displaystyle{\sum_{n=0}^{\infty} a_n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_n}$. Then:
\begin{align} \quad \sum_{n=0}^{\infty} C_n = \sum_{n=0}^{\infty} x^n \left ( \sum_{k=0}^{n} a_kb_{n-k} \right ) = \sum_{n=0}^{\infty} c_nx^n \end{align}
From the absolute convergence of $\displaystyle{\sum_{n=0}^{\infty} a_nx^n}$ and $\displaystyle{\sum_{n=0}^{\infty} b_nx^n}$, from the first result mentioned at the top of this page we must have that then:
\begin{align} \quad \sum_{n=0}^{\infty} c_nx^n = A(x)B(x) \quad \blacksquare \end{align}
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The Supremum and Infimum of The Bounded Set (a + S)
The Supremum and Infimum of The Bounded Set (a + S)
Recall from The Supremum and Infimum of a Bounded Set page the following definitions:
Definition: Let $S$ be a set that is bounded above. We say that the supremum of $S$ denoted $\sup S = u$ is a number $u$ that satisfies the conditions that $u$ is an upper bound of $S$ and $u$ is the least upper bound of $S$, that is for any $v$ that is also an upper bound of $S$ then $u \leq v$.
Definition: Let $S$ be a set that is bounded below. We say that the infimum of $S$ denoted $\inf S = w$ is a number $w$ that satisfies the conditions that $w$ is a lower bound of $S$ and $w$ is the greatest lower bound of $S$, that is for any $t$ that is also a lower bound of $S$ then $t \leq w$.
Now let $a \in \mathbb{R}$ and define the set $(a + S) := \{ a + x : x \in \mathbb{S} \}$. We will now look at some important supremum and infimum theorems regarding this set. Be sure to check out the The Supremum and Infimum of The Bounded Set (aS) as well for similar proofs.
Theorem 1: Let $S$ be a nonempty bounded subset of $\mathbb{R}$. Then $\sup (a + S) = a + \sup S$. Proof: Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $u = \sup S$. Then we know that $\forall x \in S$, $x ≤ u = \sup S$. By adding $a$ to both sides of this inequality we get that $a + x ≤ a + u$ and so $a + u$ is an upper bound for the set $(a + S)$ and so $\sup (a + S) ≤ a + u$. Now suppose that $v$ is any upper bound to the set $(a + S)$ and then so $\forall x \in S$, $a + x ≤ v$ which implies that $x ≤ v - a$. Therefore $v - a$ is an upper bound to the set $S$ and so $\sup S = u ≤ v - a$ or rather just $u ≤ v - a$ which implies that $a + u ≤ v$. Since $v$ is any upper bound of the set $(a + S)$ we can replace it in this inequality to get that $a + u ≤ \sup (a + S)$. Since $\sup (a + S) ≤ a + u$ and $a + u ≤ \sup (a + S)$ then it follows that $\sup (a + S) = a + u = a + \sup S$. $\blacksquare$.
Theorem 2: Let $S$ be a nonempty bounded subset of $\mathbb{R}$. Then $\inf (a + S) = a + \inf S$. Proof: Let $S \subset \mathbb{R}$ that is nonempty and bounded and let $w = \inf S$. Then we know that $\forall x \in S$, $\inf S = w ≤ x$. By adding $a$ to both sides of this inequality we get that $a + w ≤ a + x$ and so $a + w$ is a lower bound for the set $(a + S)$ and so $a + w ≤ \inf (a + S)$. Now suppose that $t$ is any lower bound to the set $(a + S)$ and then so $\forall x \in S$, $t ≤ a + x$ which implies that $t - a ≤ x$. Therefore $t - a$ is a lower bound to the set $S$ and so $t - a ≤ w = \inf S$ or rather just $t - a ≤ w$ which implies that $t ≤ a + w$. Since $t$ is any lower bound to the set $(a + S)$ we can replace it in this inequality to get that $\inf (a + S) ≤ a + w$. Since $a + w ≤ \inf (a + S)$ and $\inf (a + S) ≤ a + w$ then it follows that $\inf (a + S) = a + w = a + \inf S$. $\blacksquare$
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You asked for a qualitative picture, so here goes.
Consider a simplified example: the quantum harmonic oscillator.
Its ground state is given by
$$ \Psi(x) = \text{const} \cdot \exp \left( - m \omega_0 x^2 / 2 \hbar \right). $$
Now suppose that we are measuring the position of this oscillator in the ground state. We could get any real value, with probability density $|\Psi|^2$. In reality, because of the exponential decay, most of the values are distributed within the window of width
$$ \Delta x \sim \sqrt{\frac{2 \hbar}{m \omega_0}}, $$
with the mean concentrated at $x = 0$.
Because measuring an individual oscillator is a complicated process which results in it getting entangled with the measurement device, let's simplify the problem – say we have an ensemble of non-interacting oscillators all in ground states, and we measure them all independently. The distribution of values $\{x_i\}$ is expected to mostly lie within the mentioned above window, but the actual values are unknown. We usually say that those are due to
quantum fluctuations of the position operator.
The same thing happens with the quantum field, which upon inspection is nothing more than a collection of weakly interacting harmonic oscillators. If we take an ensemble of vacuum quantum field configurations (say, independent experiments at a particle accelerator), and we measure a value of the field at a point, we will see that it is not equal to zero (as it would be in the classical theory), but instead the values are distributed within an error window and are otherwise random. This are quantum fluctuations of the QFT vacuum.
These fluctuations are sometimes attributed to "virtual particles", or "virtual pairs", which are said to be "born from the vacuum". Sometimes it is also said that they can "borrow energy from vacuum for a short period of time". AFAIK these are just analogies, appealing to the consequence of Erenfest's theorem (the so-called time-energy uncertainty relation).
But the fluctuations undisputably have very real, measurable effects. Qualitatively, those effects come from a difference between the physical picture of the same thing painted by classical fields and quantum fields. You can say that quantum fields reproduce classical fields on certain scales (measured in the field value), which are much greater than the size of the error window. But once the precision with which you measure field values becomes comparable to the size of the error window, quantum effects kick in. Those who like painting intuitive pictures in their heads say that this is caused by quantum fluctuations, or virtual particles.
UPDATE
Belief that observed Casimir effect has something to do with vacuum fluctuations of the
fundamental QFT is misguided. In fact, in the calculation of the Casimir force we use an effective field theory – free electromagnetism in the 1D box, bounded by the two plates. Then we look at the effective vacuum state of this effective QFT, and we interpret the Casimir force as a consequence of the dependence of its properties on the displacement between the plates, $d$.
From the point of view of the fundamental QFT however (Standard Model, etc.)
there is no external conducting plates in the first place. If there were, it would violate Lorentz invariance. Real plates used in real experiments are made of the same matter described by the fundamental QFT, thus the state of interest is extremely complicated. What we observe as Casimir force is really just a complicated interaction of the fundamental QFT, which describes the time evolution of the complicated initial state (which describes the plates + electromagnetic field in between).
It is hopeless to try to calculate this in the fundamental QFT, just like it is hopeless to calculate the properties of the tennis ball by studying directly electromagnetic interactions holding its atoms together. Instead, we turn to the effective description, which captures all the interesting properties of our setup. In this case it is free electromagnetic effective QFT in the 1D box.
So to summarize: we are looking at the vacuum state of the effective QFT and the dependence of its properties on $d$. Alternatively, we are observing an extremely complicated fundamental system in a state which we can't hope to describe.
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In one of his seminal papers [1], Moser proved a result, which in the simplest setting, still capturing the gist, states: Given a positive continuous smooth function h on a compact, connected domain \(D\subset R^n\) with the average \([h] = 1\), there exists a diffeomorphism \(\varphi\) of \(D\) onto itself with the Jacobian \(h\):
\[\mbox{det}\,\varphi'(x)=h(x). \qquad (1)\] Solving this nonlinear PDE for the components of \(\varphi\) may seem like a difficult problem, but a physical analogy leads to a solution at once, as follows.
Interpreting \(h\) as the initial density of a chemical dissolved in a medium occupying the domain \(D\), we imagine that the chemical diffuses, equalizing its density as \(t \to \infty\) (the limiting density has to be \(1\) since \([h] = 1\)). The map \(\varphi\), which sends each particle from \(t =0\) to its position at \(t \to \infty\), then satisfies \((1)\).
In a bit more detail, let the density \(\rho = \rho(x,t)\) evolve according to the heat equation
\[\rho_t=\Delta\rho \qquad (2) \] with Neumann boundary conditions (no diffusion through ∂D), starting with \(r(x, 0) = h(x)\). Assume that each particle \(z = z(t)\) diffuses according to
\[\rho\dot{z} = -\nabla\rho; \qquad (3) \] such evolution preserves the mass \(\int\Omega_t\) \(\rho\)dV of any region \(\Omega_t\). Thus, \(h\,dV_0=\rho(x,t)dV_t,\) i.e. \(\frac{dV_t}{dV_0}=\frac{h}{\rho}.\) In the limit \(t \to \infty\) this turns into \((1)\). The “diffusing particle” map \(\varphi\) solves the nonlinear PDE
. The missing details of this proof are not hard to fill in, or to find in [2]. 1
There has been a lot of work on this problem since Moser’s original paper, in particular on the regularity (references can be found in, e.g., [3]), but my modest goal here was to give a simple basic idea rather than a review of the latest results.
Indeed, the mass enters an infinitesimal patch \(dV\) at the rate \(-\mathrm{div}\rho\dot{z}\,dV\stackrel{(3)}{=}\Delta\rho\,dV,\) precisely in agreement with \((2)\). Formally, differentiating the mass integral gives two terms which cancel each other. 1
Acknowledgments: The work from which these columns are drawn is funded by NSF grant DMS-1412542.
References [1] Moser, J. On the volume elements on a manifold, Trans. Amer. Math. Soc. 120, 286-294 (1965).
[2] Levi, M. On a problem by Arnold on periodic motions in magnetic fields,
Comm. Pure and Applied Mathematics. 56 (8), 1165-1177 (2003).
[3] Dacorogna, B and Moser, J. On a partial differential equation involving the Jacobian determinant.
Ann. l’inst. H. Poincaré Anal. non linéaire. 7(1), 1-26 (1990).
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Classification of continuous-time and discrete-time signals deals with the type of independent variable. If the signal amplitude is defined for every possible value of time, the signal is called a continuous-time signal. However, if the signal takes values at specific instances of time but not anywhere else, it is called a discrete-time signal. Basically, a discrete-time signal is just a sequence of numbers.
Consider a football (soccer) player participating in a 20-match tournament. Suppose that his running speed is recorded at each instant of time in the 90-minute duration of a particular match and plotted against time. The result shown in Figure below is clearly a continuous-time signal.
On the other hand, Figure below shows the number of goals he scored during those 20 matches, which is defined only for each match and not in between. Hence, it is a discrete-time signal.
Discrete-time signals usually arise in two ways:
Inherently discrete-time: By recording the number of events over finite time periods. For example, number of trees cut every year in a city for housing and development projects. Conversion from continuous-time to discrete-time: By acquiring the values of a continuous-time signal at fixed time instants. This process is called sampling and is discussed in detail in another article. For example, the actual temperature outside varies continuously throughout the day, but weather stations log the data after specific intervals, say every 30 minutes.
Mathematically, we represent a continuous-time signal as $s(t)$ and a discrete-time signal as $s[n]$, where $t$ is a real number while $n$ is an integer. For example, a discrete-time signal $s[n] = 3n^2$ can be plotted by finding $s[n]$ for various values of $n$. Each member $s[n]$ of a discrete-time signal is called a sample.
\begin{align*} n = -5 \quad \rightarrow \quad s[n] &= 75 \nonumber \\ n = -4 \quad \rightarrow \quad s[n] &= 48 \nonumber \\ n = -3 \quad \rightarrow \quad s[n] &= 27 \nonumber \\ n = -2 \quad \rightarrow \quad s[n] &= 12 \nonumber \\ n = -1 \quad \rightarrow \quad s[n] &= 3 \nonumber \\ n = 0 \quad \rightarrow \quad s[n] &= 0 \nonumber \\ n = +1 \quad \rightarrow \quad s[n] &= 3 \nonumber \\ n = +2 \quad \rightarrow \quad s[n] &= 12 \nonumber \\ n = +3 \quad \rightarrow \quad s[n] &= 27 \nonumber \\ n = +4 \quad \rightarrow \quad s[n] &= 48 \nonumber \\ n = +5 \quad \rightarrow \quad s[n] &= 75 \nonumber \end{align*} Plotting each value of $s[n]$ against every $n$ is then straightforward as shown in Figure below.
Another way of representing a discrete-time signal is in the form of a sequence with an underline indicating the time origin ($n=0$), such as
\begin{equation*}
s[n] = \{…,75, 48, 27, 12, 3, 0, 3, 12, 27, 48, 75, … \} \end{equation*}
Finally, it is incorrect to assume that a discrete-time signal is zero between two values of $n$. It is simply not defined for non-integer values.
In electrical engineering, the time-varying quantity is usually voltage (or sometimes current). Therefore, when we work with a signal, just think of it as a voltage changing over time. After we convert this signal from continuous-time to discrete-time, it becomes a sequence of numbers.
Unit Impulse and Unit Step Signals
Some basic signals play an important role in discrete-time signal processing. We discuss two of them below.
A unit impulse is a signal that is 0 everywhere, except at $n=0$, where its value is 1. Mathematically, it is denoted as $\delta[n]$ and defined as \begin{equation} \delta[n] = \left\{ \begin{array}{l} 1, \quad n = 0 \\ 0, \quad n \neq 0 \\ \end{array} \right. \end{equation} The unit impulse signal is shown in Figure below. A unit step signal is 0 for past ($n<0$) values, while 1 for present ($n=0$) and future ($n>0$) values. To be precise, it is denoted as $u[n]$ and defined as \begin{equation} u[n] = \left\{ \begin{array}{l} 1, \quad n \geq 0 \\ 0, \quad n < 0 \\ \end{array} \right. \end{equation} The unit step signal is shown in Figure below.
Energy of a Signal
Plotting a discrete-time signal or writing it as a sequence of numbers seems good enough, but then how do we make a comparison between two signals? For example, in the case of soccer player above, we need to know how this player stands against other players participating in the tournament. It therefore seems that we should devise some measure of “strength” or “size” of a signal.
One simple method can be just adding all the values on the amplitude axis, which is our target dependent variable. So from the figure depicting player 1 goals, our player has a total number of 10 goals in the tournament. Now we can easily compare his performance with others, provided that the signal always stays positive.
Imagine another footballer who is so “energetic” that he just cannot resist possessing the ball during the game and sometimes even scores own goals, as shown in Figure below. From his team’s point of view, his net total by adding all goals is just 7. However, from an energy perspective in the field, he is much different than player 1 before.
Therefore, a simple addition does not work for signals that assume both positive and negative values, such as a voltage varying with time because addition of both positive and negative values cancels and diminishes the calculated signal strength. This suggests that strength of a signal can be measured by taking the absolute value of the signal and then adding all the values. Or square of the absolute value, or the fourth power of the absolute value, and so on. Due to its mathematical tractability (an exciting must-have for communication theorists), square of the absolute value is the preferred choice.
In light of the above, the energy of a discrete-time signal is defined as
\begin{align} E_s &= \cdots + |s[-2]|^2 + |s[-1]|^2 + |s[0]|^2 + |s[1]|^2 + |s[2]|^2 + \cdots \nonumber \\ &= \sum \limits _n |s[n]|^2 \end{align} where the term $\sum \limits _n$ denotes summation over all values of $n$.
As an example, signal energy in the figure depicting player 1 goals is
\begin{equation*} E_{P1} = 2^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 = 14 \end{equation*} and for the figure showing player 2 goals, it is \begin{align*} E_{P2} &= 2^2 + -1^2 + 1^2 + 1^2 + -1^2 + 1^2 + … \\ &\quad -1^2 + 2^2 + 1^2 + 1^2 + 2^2 + -1^2 = 21 \end{align*} Using such a definition, the difference between the energy levels of the two players is truly reflected.
Similarly, energy in the quadratic signal above is infinite as the signal values extend from $-\infty$ to $+\infty$.
Even and Odd Signals
Some signals have specific properties which make analysis and computations simpler for us later. For example, a signal is called even (or symmetric) if
\begin{equation} s[-n] = s[n] \quad \text{or} \quad s(-t) = s(t) \end{equation}
Flipping an even signal around amplitude axis results in the same signal. An example of an even signal is $s(t) = \cos(2\pi f_0 t)$ shown in the left Figure below. The quadratic signal $s[n] = n^2$ illustrated before is also an even signal.
On the other hand, a signal is called odd (or antisymmetric) if
\begin{equation} s[-n] = -s[n] \quad \text{or} \quad s(-t) = -s(t) \end{equation} An odd signal has symmetry around the origin. An example of an odd signal is $s(t) = \sin(2\pi f_0 t)$ drawn in the right Figure above, as well as $s[n] = -n$. Periodic and Aperiodic Signals
A signal is periodic if it repeats itself after a certain period $N$.
\begin{equation} s[n\pm N] = s[n] \quad \text{for all} ~n \end{equation} Both $s[n] = \cos[2\pi f_0 n]$ and $s[n] = \sin[2\pi f_0 n]$ are examples of periodic signals if $f_0$ is a rational number.
If a signal does not repeat itself forever, there is no value of $N$ that satisfies the above periodicity equation. Such a signal is known as aperiodic, an example of which is a unit step signal $u[n]$ and most other signals encountered in these articles.
An example each of a periodic and an aperiodic signal is shown in Figure below.
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The framework
Recall from last time:
$\mathcal{F}$ is a finite set of propositions. (It is the set of propositions about which our agent has an opinion.) A credence function is a function $c : \mathcal{F} \rightarrow [0, 1]$.
\[
\langle c(X_1), \ldots, c(X_n) \rangle
\]
This is essentially the representation we used last time when we plotted credence functions as points on the Euclidean plane. To avoid prolixity, I'll use $c$ to refer to the credence function and to the vector that represents it.
Under this representation, we have the following:
Let $\mathcal{B}$ be the set of credence functions. That is, $\mathcal{B} := [0, 1]^n$. Let $\mathcal{P}$ be set of probability functions. So $\mathcal{P} \subseteq \mathcal{B} = [0, 1]^n$. The function\[ Q(c, c') := \sum_{X \in \mathcal{F}} (c(X) - c'(X))^2\] genuinely measures the squared Euclidean distance between the vectors that represent $c$ and $c'$.
Recall:
$\mathcal{W}$ is the set of classically consistent assignments of truth-values to the propositions in $\mathcal{F}$. (Thus, we can think of $\mathcal{W}$ as the set of possible worlds grained as finely as the propositions in $\mathcal{F}$ will allow.) Given $w$ in $\mathcal{W}$, define $v_w : \mathcal{F} \rightarrow [0, 1]$ as follows:\[ v_w(X) = \left \{ \begin{array}{ll} 0 & \mbox{if $X$ is false at $w$} \\ 1 & \mbox{if $X$ is true at $w$} \end{array} \right. \] Let $\mathcal{V} := \{v_w : w \mbox{ in } \mathcal{W}\}$. Thus, under the vector representation, $\mathcal{V} \subseteq [0, 1]^n$. The core theorem
With that in hand, we can state Theorem 1, the mathematical core of Joyce's argument for Probabilism. (In fact, the version we state here is slightly stronger than Joyce's; clause (2) is stronger.)
Theorem 1 If $c \not \in \mathcal{P}$, then there is $c^* \in \mathcal{P}$ such that $Q(v, c^*) < Q(v, c)$ for all $v \in \mathcal{V}$. If $c \in \mathcal{P}$, then there is no $c^* \in \mathcal{B}$ such that $Q(v, c^*) \leq Q(v, c)$ for all $v \in \mathcal{V}$.
The first step in the proof of Theorem 1 is the following Lemma, which is due to (de Finetti, 1974). It gives an extremely useful characterization of the credence functions that satisfy Probabilism: they are precisely those that are convex combinations of the omniscient credence functions in $\mathcal{V}$.
Definition 1Let $\mathcal{V}^+$ be the convex hull of $\mathcal{V}$. That is, $\mathcal{V}^+$ is the smallest convex set that contains $\mathcal{V}$. Lemma 1$\mathcal{P} = \mathcal{V}^+$.
Proof: Suppose, to begin with, that $\mathcal{F}$ is an algebra.
First, we prove $\mathcal{V}^+ \subseteq \mathcal{P}$. To do this, we note two things: first, each $v \in \mathcal{V}$ is a probability function, so $\mathcal{V} \subseteq \mathcal{P}$; second, $\mathcal{P}$ is convex. Second, we prove $\mathcal{P} \subseteq \mathcal{V}^+$. Since $\mathcal{F}$ is a finite algebra, it has atoms. Let $\mathcal{A} \subseteq \mathcal{F}$ be the set of atoms of $\mathcal{F}$. Then $\mathcal{A}$ and $\mathcal{V}$ stand in one-one correspondence: if $v \in \mathcal{V}$, there is exactly one $A_v \in \mathcal{A}$ such that $v(A_v) = 1$ and $v(A_{v'}) = 0$ for $v \neq v'$. Now suppose $c \in \mathcal{P}$. And suppose $X \in \mathcal{F}$. Then $X$ is equivalent to the disjunction of the atoms $A_v$ such that $v(X) = 1$. Thus\[c(X) = c\left (\bigvee_{v : v(X) = 1} A_v \right ) = \sum_{v : v(X) = 1} c(A_v) = \sum_{v} c(A_v)v(X)\] So\[c = \sum_{v \in \mathcal{V}} c(A_v) v\] Thus, $c$ is a convex combination of elements of $\mathcal{V}$. That is, $c \in \mathcal{V}^+$.
$\Box$
This characterization of $\mathcal{P}$, the set of probability functions on $\mathcal{F}$, makes our lives a lot easier, since there is a vast array of results about the behaviour of convex sets. Here are the two that are important for our purposes:
Lemma 2Suppose $\mathcal{X} \subseteq \mathbb{R}^n$ is convex. Then if $x \not \in \mathcal{X}$, there is $x^* \in \mathcal{X}$ such that $Q(y, x^*) < Q(y, x)$ for all $y \in \mathcal{X}$. Lemma 3Suppose $\mathcal{X} \subseteq \mathbb{R}^n$. Then, if $x, y \in \mathcal{X}^+$ and $x \neq y$, there is $z \in \mathcal{X}$ such that $Q(z, x) < Q(z, y)$.
We can now see how Lemmas 1, 2, and 3 combine to give Theorem 1: By Lemma 1, $\mathcal{P} = \mathcal{V}^+$. But, by Lemma 2, if there is a point $c$ outside $\mathcal{V}^+$, there is a point $c^*$ inside $\mathcal{V}^+$ such that $c^*$ is closer to all points in $\mathcal{V}^+$ than $c$ is; thus, in particular, $c^*$ is closer to all $v$ in $\mathcal{V}$ than $c$ is. This gives Theorem 1(1). By Lemma 3, if $c, c'$ are in $\mathcal{V}^+$, and $c \neq c'$, then there is $v$ in $\mathcal{V}$ such that $c$ is closer to $v$ than $c'$ is. Thus, $c$ is not even weakly dominated. This gives Theorem 1(2). And we're done!
So far, we've been assuming that distance between credence functions is measured by Squared Euclidean Distance. Does Theorem 1 depend on that? That is, is there a broader class of alternative measures of the distance between credence functions such that Theorem 1 holds for every distance measure in that class? The first thing to say is that Squared Euclidean Distance is not itself a distance measure, strictly speaking: that is, it isn't a \[ d_F(y, x) := F(x) - F(y) - \langle \nabla F(x), (y-x)\rangle \] The core theorem generalized
So far, we've been assuming that distance between credence functions is measured by Squared Euclidean Distance. Does Theorem 1 depend on that? That is, is there a broader class of alternative measures of the distance between credence functions such that Theorem 1 holds for every distance measure in that class? The first thing to say is that Squared Euclidean Distance is not itself a distance measure, strictly speaking: that is, it isn't a
metric. It doesn't satisfy the triangle inequality. It is rather what statisticians call a divergence. In this section, we show that Theorem 1 holds for any Bregman divergence. Squared Euclidean Distance is a Bregman divergence, but so is Kullback-Leibler divergence, the squared Mahalanobis distance, and Itakura-Saito distance. And, as we will see, if an inaccuracy measure is generated by a proper scoring rule there is a Bregman divergence that differs from that inaccuracy measure by a constant. Definition 2Suppose $\mathcal{X} \subseteq \mathbb{R}^n$ is convex. Suppose $F : \mathcal{X} \rightarrow \mathbb{R}$ is strictly convex and $\nabla F$ is defined on $\mathrm{int}(\mathcal{X})$ and extends to a bounded, continuous function on $\mathcal{X}$. Then the Bregman divergence generated by$F$ is
\[
d_F(y, x) := F(x) - F(y) - \langle \nabla F(x), (y-x)\rangle
\]
Essentially, $d_F(y, x)$ is the difference between $F(y)$ and the first-order Taylor expansion around $x$ evaluated at $y$, as the following diagram illustrates:
Given the strict convexity of $F$, it follows that $d_F(y, x) \geq 0$, with equality iff $x = y$.
What follows are some of the crucial theorems concerning Bregman divergences. Each shows that Bregman divergences share many important geometrical features with Squared Euclidean Distance. We will be appealing to these various properties over the coming weeks.
First: Lemmas 2 and 3 holds for any Bregman divergence $d_F$:
Lemma 4Suppose $\mathcal{X} \subseteq \mathbb{R}^n$ is convex. Then if $x \not \in \mathcal{X}$, there is $x^* \in \mathcal{X}$ such that $d_F(y, x^*) < d_F(y, x)$ for all $y \in \mathcal{X}$. Proof: This is Proposition 3 in (Predd, et al., 2009). Lemma 5Suppose $\mathcal{X} \subseteq \mathbb{R}^n$. Then, if $x, y \in \mathcal{X}^+$ and $x \neq y$, there is $z \in \mathcal{X}$ such that $d_F(z, x) < d_F(z, y)$.
Proof: This is proved as part (a) of Theorem 1 in (Predd, et al., 2009).
Together, these are enough to show that Theorem 1 holds for any Bregman divergence. Thus, if we can establish that distance between credence functions ought to be measured by a Bregman divergence, we can run Joyce's argument.
Second: Suppose the inaccuracy of a credence function $c$ at world $w$ is given by $d_F(v_w, c)$ for some Bregman divergence $d_F$. Then, if credence function $c_1$ is further from $c$ than $c_2$ is, then $c$ will expect $c_1$ to have greater inaccuracy than it will expect $c_2$ to have.
Lemma 6Recall that, if $v$ is in $\mathcal{V}$, then $A_v$ is the unique atom of $\mathcal{F}$ such that $v(A_v) = 1$. Then, if $c \in \mathcal{P}$ and $d_F(c, c_1) < d_F(c, c_2)$, then
\[
\sum_{v \in \mathcal{V}} c(A_v)d_F(v_w, c_1) < \sum_{v \in \mathcal{V}} c(A_v)d_F(v_w, c_2)
\]
Proof: This is proved on page 32 of (Pettigrew, 2013).
One consequence of this is that each probabilistic credence function expects itself to be most accurate. The following result gives a partial converse to this.
Notice that $B(c, w)$ is obtained by taking the squared difference between each credence $c(X)$ and the corresponding ideal credence $v_w(X)$ and summing the results. In this sort of situation, we say that $B$ is generated by a scoring rule.
Definition 3A scoring ruleis a function $s : \{0, 1\} \times [0, 1] \rightarrow [0, \infty]$.
The idea is that $s(0, x)$ measures the inaccuracy of having credence $x$ in a false proposition, whereas $s(1, x)$ measures the inaccuracy of having credence $x$ in a true proposition.
Definition 4Given a scoring rule $s$, we say that $I_s$ is the inaccuracy measure generated by$s$, where
\[
I_s(c, w) = \sum_{X \in \mathcal{F}} s(v_w(X), c(X))
\]
We say that a scoring rule is proper if a particular credence expects itself to be more accurate than it expects any other credence to be. More precisely:
Definition 5We say that a scoring rule $s$ is properif
\[
ps(1, x) + (1-p)s(0, x)
\]
is minimal at $x = p$.
Then we have the following theorem, which connects the inaccuracy measures generated by proper scoring rules and Bregman divergences.
Theorem 2Suppose $s$ is a continuous proper scoring rule. Then there is a Bregman divergence $d_F$ such that
\[
I_s(c, w) = d_F(v_w, c) + \sum_{X \in \mathcal{F}} s(v_w(X), v_w(X))
\]
Proof: This is proved as Equation (8) in (Predd, et al.).
References de Finetti, B. (1974) A Theory of Probability(vol 1) (New York: Wiley) Pettigrew, R. (2013) 'A New Epistemic Utility Argument for the Principal Principle' Episteme10(1): 19-35 Predd, et al. (2009) 'Probabilistic Coherence and Proper Scoring Rules' IEEE Transactions on Information Theory55(10): 4786-4792.
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Let X be a continuous random variable with pdf
$f(x) = \begin{cases} 1/4, & \text{if $x$ $\in$ (-2,2)} \\ 0, & \text{otherwise} \end{cases}$
Find the probability density functiions of i) Y=$X^3$ and ii) Z= $X^4$
So far I have done this for i)
$F_Y(Y) = P(Y \le y) = P(X \le y^{1/3}) = F_X(y^{1/3}) = \int1/4.dt $
Carrying this through, I get an answer that is 1/2 of what I should be getting.
For the integral (the last part I was up to) I am using $y^{1/3}$ and $-y^{1/3}$ as my upper and lowers but I am pretty sure I should be using $y^{1/3}$ and 0. Can someone tell me why?
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I follow the hint there and adapt the notations.For the statement (c), you can take the long exact sequence of the sequence of the complex $0\to C^\bullet(\mathfrak{U},\mathcal{F})\to C^\bullet(\mathfrak{U},\mathcal{G})\to D^\bullet(\mathfrak{U})\to 0$. This gives you
$$0\to \check{H}^0(\mathfrak{U},\mathcal{F})\to \check{H}^0(\mathfrak{U},\mathcal{G})\to H^0(D^\bullet(\mathfrak{U}))\to \check{H}^1(\mathfrak{U},\mathcal{F})\to 0.$$
$\check{H}^1(\mathfrak{U},\mathcal{G})=0$ since $\mathcal{G}$ is flasque.
Also you have isomorphisms $\check{H}^0(\mathfrak{U},\mathcal{F})\cong\Gamma(X,\mathcal{F})$ and $\check{H}^0(\mathfrak{U},\mathcal{G})\cong\Gamma(X,\mathcal{G})$ as well as a natural map $H^0(D^\bullet(\mathfrak{U}))\to \check{H}^0(\mathfrak{U},\mathcal{R})=\Gamma(X,\mathcal{R})$. These fit together and give the following commutative diagram:
$\require{AMScd}$\begin{CD}0@>>> \check{H}^0(\mathfrak{U},\mathcal{F})@>>> \check{H}^0(\mathfrak{U},\mathcal{G})@>>> H^0(D^\bullet(\mathfrak{U}))@>>> \check{H}^1(\mathfrak{U},\mathcal{F})@>>> 0\\@. @VVV @VVV @VVV \\0@>>>\Gamma(X,\mathcal{F})@>>> \Gamma(X,\mathcal{G})@>>> \Gamma(X,\mathcal{R})@>>> H^1(X,\mathcal{F})@>>> 0\end{CD}
So you have an induced map $\check{H}^1(\mathfrak{U},\mathcal{F})\to H^1(X,\mathcal{F})$. Now take the direct limit (with respect to the open cover) in the first line. The result will follow if we can show $$\varinjlim_{\mathfrak{U}} H^0(D^\bullet(\mathfrak{U}))\cong H^0(X,\mathcal{R}).$$The injectivity is obvious. For surjectivity, from the definition of quotient sheaves, for any $r\in \Gamma(X,\mathcal{R})$, $r$ can be described by giving an open cover $\mathfrak{V}=\{V_i\}$ of $X$ and elements $g_i\in \Gamma(V_i,\mathcal{G})$ such that $g_i-g_j\in\Gamma(V_i\cap V_j,\mathcal{F})$. Hence the sujectivity follows.
For the last exercise, you can imitate the proof of
Theorem 4.5. Instead of the affineness of the open sets, the condition on the vanishing of higher cohomologies in the exercise implies that you still have the short exact sequence$$0\to \mathcal{F}(U_{i_0\cdots i_p})\to \mathcal{G}(U_{i_0\cdots i_p})\to \mathcal{R}(U_{i_0\cdots i_p})\to 0$$and hence the short exact sequence of Cech complexes. The same argument will work. You need the vanishing assumption here again for the induction step.
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Series Convergence and Divergence Practice Examples 2
We will now look at some more examples of applying the various convergence/divergence tests we have looked at so far to some series without being given what test to apply specifically.
More examples of evaluating series can be found on the following page:
Series Convergence and Divergence Practice Examples 1 Series Convergence and Divergence Practice Examples 2 Series Convergence and Divergence Practice Examples 3 Series Convergence and Divergence Practice Examples 4 Series Convergence and Divergence Practice Examples 5 Example 1 Does the series $\sum_{n=1}^{\infty} \frac{100^n}{n!}$ converge or diverge?
Our first though is to apply the ratio test, that is:(1)
Since $0 ≤ L < 1$, we have that $\sum_{n=1}^{\infty} \frac{100^n}{n!}$ converges by the ratio test.
Example 2 Does the series $\sum_{n=1}^{\infty} \frac{1+n}{2+n}$ converge or diverge?
Notice that $\lim_{n \to \infty} \frac{1+n}{2+n} = \lim_{n \to \infty} \frac{\frac{1}{n} + 1}{\frac{2}{n} + 1} = 1$, so $\sum_{n=1}^{\infty} \frac{1+n}{2+n}$ diverges by the divergence test.
Example 3 Does the series $\sum_{n=1}^{\infty} \frac{1 + \sin n}{n^3}$ converge or diverge?
We note that $0 ≤ 1 + \sin n ≤ 2$ for all $n \in \mathbb{N}$. So we have that $\frac{1 + \sin n}{n^3} ≤ \frac{2}{n^3}$. By the comparison test, we know that $\sum_{n=1}^{\infty} \frac{2}{n^3}$ is convergent, and since $\sum_{n=1}^{\infty} \frac{1 + \sin n}{n^3} ≤ \sum_{n=1}^{\infty} \frac{2}{n^3}$, then $\sum_{n=1}^{\infty} \frac{1 + \sin n}{n^3}$ is also convergent.
Example 4 Does the series $\sum_{n=1}^{\infty} \frac{1}{2^n(n+1)}$ converge or diverge?
Applying the ratio test we get that:(2)
Since $0 ≤ L < 1$, by the ratio test we get that $\sum_{n=1}^{\infty} \frac{1}{2^n(n+1)}$ converges.
Note that we could have also applied the root test as follows (omitting to show that $\lim_{n \to \infty} (1 + n)^{1/n} = 1$):(3)
Once again since $0 ≤ L < 1$, by the root test we get that $\sum_{n=1}^{\infty} \frac{1}{2^n(n+1)}$ converges.
Example 5 Does the series $\sum_{n=1}^{\infty} \frac{1}{\pi^n + 5}$ converge or diverge?
We note that $\frac{1}{\pi^n + 5} < \frac{1}{\pi^n} = \left ( \frac{1}{\pi} \right)^n$. We note that $\sum_{n=1}^{\infty} \left ( \frac{1}{\pi} \right)^n$ is a convergent geometric series since the common ratio $\mid r \mid = \biggr \rvert \frac{1}{\pi} \biggr \rvert < 1$, and so by the comparison test since $\sum_{n=1}^{\infty} \frac{1}{\pi^n + 5} < \sum_{n=1}^{\infty} \left ( \frac{1}{\pi} \right)^n$ we have that $\sum_{n=1}^{\infty} \frac{1}{\pi^n + 5}$ converges.
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Series Convergence and Divergence Practice Examples 3
We will now look at some more examples of applying the various convergence/divergence tests we have looked at so far to some series without being given what test to apply specifically.
More examples of evaluating series can be found on the following page:
Series Convergence and Divergence Practice Examples 1 Series Convergence and Divergence Practice Examples 2 Series Convergence and Divergence Practice Examples 3 Series Convergence and Divergence Practice Examples 4 Series Convergence and Divergence Practice Examples 5 Example 1 Does the series $\sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n^4}$ converge or diverge?
We first might be tempted to use the alternating series test since we notice the term $(-1)^n$ is the numerator, however, this series is NOT alternating signs. Notice that every odd term is positive, and every even term is zero. Instead, let's use the comparison test.
We note that $\frac{1 - (-1)^n}{n^4} ≤ \frac{2}{n^4}$, and we know that $\sum_{n=1}^{\infty} \frac{2}{n^4}$ converges by the p-series test. Since $\sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n^4} ≤ \sum_{n=1}^{\infty} \frac{2}{n^4}$, we have that $\sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n^4}$ converges by the comparison test.
Example 2 Does the series $\sum_{n=3}^{\infty} \frac{1}{n \ln n \sqrt{ \ln \ln n}}$ converge or diverge?
This example will be difficult to evaluate using some of the simpler tests, so let's try the integral test. Let $f(x) = \frac{1}{x \ln x \sqrt{ \ln \ln x}}$ and evaluate the following integral:(1)
Let's first figure out what this indefinite integral is. Let $u = \ln \ln x$. Therefore $du = \frac{1}{x} \cdot \frac{1}{\ln x} \: dx = \frac{1}{x \ln x} \: dx$, and so:(2)
Therefore we have that:(3)
By the integral test $\sum_{n=3}^{\infty} \frac{1}{n \ln n \sqrt{ \ln \ln n}}$ diverges.
Example 3 Does the series $\sum_{n=1}^{\infty} \frac{n!}{n^2e^n}$ converge or diverge?
Applying the ratio test we get that:(4)
So by the ratio test $\sum_{n=1}^{\infty} \frac{n!}{n^2e^n}$ diverges.
Example 4 Does the series $\sum_{n=1}^{\infty} \frac{n + 4}{n^3 - 2n + 3}$ converge or diverge?
We note that for large $n$ that $\frac{n + 4}{n^3 - 2n + 3}$ behaves like $\frac{n}{n^3} = \frac{1}{n^2}$, and using the limit comparison test:(5)
So since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges as a p-series, it follows that $\sum_{n=1}^{\infty} \frac{n + 4}{n^3 - 2n + 3}$ also converges.
Example 5 Does the series $\sum_{n=1}^{\infty} \frac{n}{\sin ^2 n + \cos ^2 n}$ converge or diverge?
Using the trigonometric identity that $\sin ^2 n + \cos ^2 n = 1$ we have that $\sum_{n=1}^{\infty} \frac{n}{\sin ^2 n + \cos ^2 n} = \sum_{n=1}^{\infty} n$, and we note that $\lim_{n \to \infty} n = \infty$, so by the divergence test we have that $\sum_{n=1}^{\infty} \frac{n}{\sin ^2 n + \cos ^2 n}$ diverges.
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Thesis BELLE2-MTHESIS-2019-006 Untagged Exclusive Analysis of the Semileptonic Decay $B\rightarrow\pi\ell\nu$ from Belle II Data in Preparation for |$V_{ub}$| Extraction
Svenja Granderath ; Prof. Dr. Jochen Dingfelder ; Dr. Peter Lewis
2019 Physikalisches InstitutBonn, Germany
Abstract: In this thesis an untagged exclusive analysis of the charmless semileptonic decay $B\rightarrow\pi\ell\nu$ using an integrated luminosity of 5.6 $fb^{−1}$ of early Belle II data was discussed. The analysis was performed in 5 bins of momentum transfer, which makes it sensitive to form factor parametrisations. This method was chosen to prepare for later |$V_{ub}| extraction using a larger dataset. The signal selections were separately optimised in the momentum transfer bins using a figure of merit (FOM) based on the statistical significance of signal on top of background. First, selections based on detector properties and the kinematics of the signal decay were performed. After that, boosted decision trees (BDTs) were employed to separate signal from continuum background, as well as from BB background. After selecting on the BB BDT output classifier the shapes in signal and background of the signal extraction variables, $M_{bc}$ and $\Delta E$, were very similar and the correlations between the yields were high. Therefore, in the end, a selection was only placed on the continuum suppression BDT output classifier. Finally, we set up and validated a simultaneous extended binned maximum likelihood fit on the $M_{bc}$ and $\Delta E$ distributions to extract signal in each momentum transfer bin.bFitting to data was not possible, due to very low statistics and high correlations in the yields of the templates. We performed toy studies to predict the luminosities necessary to achieve statistical significance of 3$\sigma$ and 5$\sigma$ in the total branching fraction for each of the modes. In the $B^{0}$ modes a luminosity of $\sim$ 15 $fb^{−1}$ and $\sim$ 20 $fb^{−1}$ is needed for 3$\sigma$ and 5$\sigma$ significance, respectively. For the $B^{\pm}$ modes higher luminosities are needed: ∼ 30 $fb^{−1}$ for 3$\sigma$ significance and ∼ 50 $fb^{−1}$ for 5$\sigma$ significance. Note: Presented on 19 09 2019 Note: MSc
The record appears in these collections: Books, Theses & Reports > Theses > Masters Theses
Record created 2019-10-04, last modified 2019-10-04
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Problem
Let $X_1, \ldots, X_n$ be a sample from normal distribution $N(\theta, \theta^2)$, where $\theta > 0$. I am to find minimal sufficient statistic and prove that it is not complete.
Finding the minimal sufficient statistic
First of all I did some transformations which led me to the following expression $$f(x|\theta) = \gamma(\theta) \exp \bigg \{ -\bigg(\frac{1}{2\theta^2} \sum_{i=1}^nx_i^2 - \sum_{i=1}^nx_i + \frac{n}{2} \bigg) \bigg \} \tag{1}.$$ Using $(1)$ I can write that my sufficient statistic would be $$T(X) = \bigg(\sum_{i=1}^nx_i^2, \sum_{i=1}^nx_i \bigg) \tag{2}.$$ To show that $(2)$ is minimal I calculate the following fraction $$\frac{f(x|\theta)}{f(y|\theta)} \tag{3}.$$ It's easy to notice that $(3)$ is independent from $\theta$ if and only if $x_i = y_i$. Thus T(X) is minimal.
Doubts
Are my calculation and reasoning correct? How can I prove that $(2)$ is not complete?
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So i have a question which asks to find the fourier series of $\left\vert\,\sin\left(x\right)\,\right\vert\,$. I have worked out the solution as $$ {2 \over\pi} - {4 \over \pi}\sum_{k = 1}^{\infty}{\cos\left(2kx\right) \over 4k^{2} - 1} $$
Which i am pretty sure is correct as i have the solution in my book.
The second part of the question asks to work out the sums of $$ \sum_{k = 1}^{\infty}{1 \over 4k^{2} - 1}\qquad\mbox{and}\qquad \sum_{k = 1}^{\infty}{\left(-1\right)^{k} \over 4k^{2} - 1} $$
Im sure this is probably very simple but i have no solution for this and I am struggling to search for an explanation of how to do this on google. Could someone please tell me know it is done ?.
Many thanks.
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Consider the following experiment. I take a spin-$\frac{1}{2}$ particle and make a $\sigma_x$ measurement (measure the spin in the $x$ direction), then make a $\sigma_y$ measurement, then another $\sigma_x$ one, then $\sigma_y$, and so on for $n$ measurements. The formalism of quantum mechanics tells us that the outcomes of these measurements will be random and independent. I now have a string of completely random bits, of length $n$. Briefly, my question is
where does the information in this string come from?
The obvious answer is "quantum measurements are fundamentally indeterministic and it's simply a law of physics that you get a random string when you do that". The problem I have with this is that it can be shown that unitary evolution of quantum systems conserves von Neumann entropy, just as Hamiltonian evolution of a classical system conserves Shannon entropy. In the classical case this can be interpreted as "no process can create or destroy information on the microscopic level." It seems like the same should be true for the quantum case as well, but this seems hard to reconcile with the existence of "true" randomness in quantum measurement, which does seem to create information.
It's clear that there are some interpretations for which this isn't a problem. In particular, for a no-collapse interpretation the Universe just ends up in a superposition of $2^n$ states, each containing an observer looking at a different output string.
But I'm not a big fan of no-collapse interpretations, so I'm wondering how other quantum interpretations cope with this. In particular, in the "standard" interpretation (by which I mean the one that people adhere to when they say quantum mechanics doesn't need an interpretation), how is the indeterminacy of measurement reconciled with the conservation of von Neumann entropy? Is there an interpretation other than no-collapse that can solve it particularly well?
addendum
It seems worth summarising my current thinking on this, and having another go at making clear what I'm really asking.
I want to start by talking about the classical case, because only then can I make it clear where the analogy seems to break down. Let's consider a classical system that can take on one of $n$ discrete states (microstates). Since I don't initially know which state the system is in, I model the system with a probability distribution.
The system evolves over time. We model this by taking the vector $p$ of probabilities and multiplying it by a matrix T at each time step, i.e. $p_{t+1} = Tp_t$. The discrete analogue of Hamiltonian dynamics turns out to be the assumption that $T$ is a permutation matrix, i.e. it has exacly one 1 on each rown and column, and all its other entries are 0. (Note that permutation matrices are a subset of unitary matrices.) It turns out that, under this assumption, the Gibbs entropy (aka Shannon entropy) $H(p)$ does not change over time.
(It's also worth mentioning, as an aside, that instead of representing $p$ as a vector, I could choose to represent it as a diagonal matrix $P$, with $P_{ii}=p_i$. It then looks a lot like the density matrix formalism, with $P$ playing the role of $\rho$ and $T$ being equivalent to unitary evolution.)
Now let's say I make a measurement of the system. We'll assume that I don't disturb the system when I do this. For example, let's say the system has two states, and that initially I have no idea which of them the system is in, so $p=(\frac{1}{2},\frac{1}{2})$. After my measurement I know what state the system is in, so $p$ will become either $(1,0)$ or $(0,1)$ with equal probability. I have gained one bit of information about the system, and $H(p)$ has reduced by one bit. In the classical case these will always be equal, unless the system interacts with some other system whose state I don't precisely know (such as, for example, a heat bath).
Seen from this point of view, the change in von Neumann entropy when a quantum measurement is performed is not surprising. If entropy just represents a lack of information about a system then of course it should decrease when we get some information. In the comments below, where I refer to "subjective collapse" interpretations, I mean interpretations that try to interpret the "collapse" of a wavefunction as analogous to the "collapse" of the classical probability distribution as described above, and the von Neumann entropy as analogous to the Gibbs entropy. These are also called "$\psi$-epistemic" interpretations.
But there's a problem, which is this: in the experiment described at the beginning of this question, I'm getting one bit of information with every measurement, but the von Neumann entropy is remaining constant (at zero) instead of decreasing by one bit each time. In the classical case, "the total information I have gained about the system" + "uncertainty I have about the system" is constant, whereas in the quantum case it can increase. This is disturbing, and I suppose what I really want to know is whether there's any known interpretation in which this "extra" information is accounted for somehow (e.g. perhaps it could come from thermal degrees of freedom in the measuring apparatus), or in which it can be shown that something other than the von Neumann entropy plays a role analogous to the Gibbs entropy.
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I have some troubles understanding one statement in the book "Nonlinear Optics" by R.Boyd. In chapter 7.1, it is derived that in a nonlinear material a light beam undergoes self-focusing due to the dependence of the refractive index on the intensity of the beam itself, and the value of the corresponding self-focusing angle is obtained.
A few lines below, it is stated that, in the case of small laser power, the convergence angle is reduced by diffraction; the total angle is said to be given by \begin{equation} \theta =(\theta_\mathrm{sf}^2-\theta_\mathrm{dif}^2)^{1/2} \end{equation}
I can't get to understand why this equality holds.. can someone help me? In the figure below you can see the original paragraph from the book (equation 7.1.4 is just the expression for $\theta_\mathrm{sf}$, which I think is unnecessary)
The derivation to the result given by Eq. (7.1.4) ignores the effects of diffraction, and thus might be expected to be valid when self-action effects overwhelm those of diffraction$-$that is, for $P\gg P_\mathrm{cr}$. For smaller laser powers, the self-focusing distance can be estimated by noting that the beam convergence angle is reduced by diffraction effects and is given approximately by $\theta =(\theta_\mathrm{sf}^2-\theta_\mathrm{dif}^2)^{1/2}$, where $$ \theta_\mathrm{dif} = 0.61\lambda_0/n_0d \tag{7.1.5} $$ is the diffraction angle of a beam of diameter $d$ and vacuum wavelength $\lambda_0$.
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I'm trying to understand a part of N.J.Hitchin's paper "The moduli space of complex Lagrangian submanifolds" https://arxiv.org/pdf/math/9901069.pdf . In particular I would like to understand how any given holomorphic function $\mathcal{F}$ will naturally give a special Kahler manifold.
In the following I will try my best to summarize his Remark 1 (page 11). Let $V$ be a real symplectic vector space with coordinates $\{x_1,...,x_{2n}\}$ and a symplectic form $\omega = \sum_{i = 1}^n dx_{i} \wedge dx_{n+i}$. Consider $V \times V$, let's denote a coordinates on the second factor by $\{\xi_1,...,\xi_{2n}\}$ then we introduce a complex structure $I$ on $V\times V$ by \begin{equation} I\frac{\partial}{\partial x_i} = \frac{\partial}{\partial \xi_{n+i}}, \qquad I\frac{\partial}{\partial \xi_i} = \frac{\partial}{\partial x_{n+i}}. \end{equation} So we can define a holomorphic complex coordinates $\{v_i = x_i + i\xi_{n+i}, w_i = \xi_i + ix_{n+i}|i = 1,...,n\}$ and the canonical complex symplectic form $\omega^c = \sum_{i = 1}^n dv_i \wedge dw_i = \omega_1 + i\omega_2$ where \begin{equation} \omega_1 = \sum_{i = 1}^{2n}dx_i\wedge d\xi_i, \qquad \omega_2 = \sum_{i=1}^n (dx_i \wedge dx_{n+i} - d\xi_i \wedge d\xi_{n+i}). \end{equation}
The claim is that given any holomorphic function $\mathcal{F} = \mathcal{F}(w_1,...,w_n)$ a complex Lagrangian submanifold $M = \{z_i = \partial\mathcal{F}/\partial w_i\} \subset V\times V$ is special Kahler with complex structure $I|_M$.
According to the paper this follows since $\omega^c|_M = 0 \implies \omega_i|_M = 0, i = 1,2$ hence $M$ is bi-Lagrangian, then
"From Theorem 2 this is all we need for a special Kahler manifold" (page 12).
I'm having a difficult time understanding this because Theorem 2 (page 7) also required that $M\subset V\times V$ is transversal to two projections onto $V$ in addition to the bi-Lagrangian condition. I interpreted the transversal condition as the two projections of $M$ are surjective to $V$, which seems to be quite an important part in proving Theorem 2. However I don't see why \begin{equation} M = \Big\{v_i = \frac{\partial \mathcal{F}}{\partial w_i}\Big\} = \Big\{x_i = \frac{\partial \Re \mathcal{F}}{\partial \xi_i} = \frac{\partial \Im \mathcal{F}}{\partial x_{n+i}},\ \xi_{n+i} = \frac{\partial \Im \mathcal{F}}{\partial \xi_i} = -\frac{\partial \Re \mathcal{F}}{\partial x_{n+i}}\Big\} \end{equation} would satisfies such condition for an arbitrary $\mathcal{F}$. For example I could take $\mathcal{F} = 0$ then obviously $x_i$ and $\xi_{n+i}$ would be fixed and transversality condition would failed.
Question : what exactly did I miss in trying to follow his arguments above? Or, if not every holomorphic functions $\mathcal{F}$ would yield a special Kahler Lagrangian then what is a criterion to determine which $\mathcal{F}$ would works?
Thank you!
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To give an example where convergence of Cauchy sequences is important: time-evolution is typically calculatedas$$ |\psi(t)\rangle = e^{i\hbar^{-1} \hat{H}\cdot t}|\psi_0\rangle$$now, the exponential of an operator is defined
1 by$$ e^{\hat{A}} = \sum_{i=0}^\infty\frac{\hat{A}^i}{i!}$$where the sum in turn is defined by$$ \Bigl(\sum_{i=0}^b\frac{\hat{A}^i}{i!}\Bigr)|\psi\rangle = \sum_i\frac{\hat{A}^i|\psi\rangle}{i!} =:S_b$$which is simply a superposition of vectors, no problem there – for finite sums. But if the sum is infinite, it is defined as the limit of the sequence of partial sums. Is that a Cauchy sequence? It can be shown under quite reasonable assumptions (essentially, finite energy) that it is. So in a Hilbert space, we normally have a nice well-defined expression for time evolution, which is obviously quite handy if you want to verify any theoretical model with experimental results. In a non-Hilbert inner product space we can't be sure that the result is well-defined. Bad!
Another thing that I imagine to be a big problem: the Riesz representation theorem doesn't hold on general inner product space. Though this theorem is seldom explicitly mentioned in physics, it is the reason that you can do lots of things that are often taken for granted; in particular it's required for uniqueness of the Hermitian adjoint. That, I expect, might cause considerable havoc if you're working with stuff like ladder operators.
1As V Moretti remarks, this definition of the exponential isn't really well-defined for unbounded operators. Nevertheless it's typically used by physicists... and indeed ok if simply taken as a shorthand for the expression applied to a suitable state.
It is an idealization that allows one to do calculations that are otherwise impossible, and in this way accounts for the power of quantum mechanics.
It is the same sort of idealization that makes physicists work with real numbers in place of rational numbers, although all raw numbers measured are rational. However, the rational numbers lack most of the useful properties of the real numbers. Restricted to rationals you dont even have an exponential or trigonometric function. Thus rational numbers can express only very limited physics.
In the same way, working in a Hilbert space allows one to do many operations (such as talking about $e^{itH}$) that don't make sense in an incomplete vector space.
Hilbert spaces (or variants of it such as rigged Hilbert spaces) are therefore absolutely necessary for having conceptual precision in quantum mechanics.
There are many physically fundamental properties that would not hold if the completeness requirement were dropped. First of all the spectral theorem for self-adjoint operators would not hold. So, an observable (let us assume to deal with observables with pure point spectrum) would not have a complete set of eigenstates. There is a fundamental idea in quantum physics that, given an observable and a pure state, the state can always be realized as a coherent superposition of states where the observable is defined. Without completeness this basic idea of quantum theory would not be valid any more.
I think your question is "why wasn't quantum mechanics formulated on normed vector spaces?" i.e "why was the completeness criterion required?"
I don't know a rigorous answer, but it seems reasonable for the following reason:
Completeness means that every Cauchy sequence of elements of H converges to an element of H. The QM postulate says that physical states are represented by vectors (strictly speaking rays) in H, so if I had an infinite sequence of physical states which were getting "physically" closer and closer together - in the sense that the characteristics of the physical quantities encoded in the states were converging, then it seems reasonable to require that the thing they're converging to is also a physical state. Mapping this over to H, then the Cauchy completeness criterion will take care of this.
The reason I'm worried that this is a bit of a weak answer, however, is that not all elements of H necessarily represent realizable physical states. For example the sums of vectors in different superselection sectors certainly doesn't. So maybe the Hilbert space criterion where
every Cauchy sequence converges to an element of H is sufficient but not necessary.
If you can accept that the reals are more
physical than the rationals, then by analogy a complete vector space is more physical than an incomplete one; all finite-dimensional spaces are complete by construction, an infinite-dimensional one isn't and requires completion by hand.
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Differential and Integral Equations Differential Integral Equations Volume 26, Number 7/8 (2013), 757-780. Asymptotic behavior of fractional order semilinear evolution equations Abstract
Fractional calculus is a subject of great interest in many areas of mathematics, physics, and sciences, including stochastic processes, mechanics, chemistry, and biology. We will call an operator $A$ on a Banach space $X$ $\omega$-sectorial ($\omega\in\mathbb R$) of angle $\theta$ if there exists $\theta \in [0,\pi/2)$ such that $S_\theta:=\{\lambda\in\mathbb C\setminus\{0\} : |\mbox{arg} (\lambda)| < \theta+\pi/2\}\subset\rho(A)$ (the resolvent set of $A$) and $\sup\{|\lambda-\omega|\|(\lambda-A)^{-1}\| : \lambda\in \omega +S_\theta\} < \infty$. Let $A$ be $\omega$-sectorial of angle $\beta\pi/2$ with $\omega < 0$ and $f$ an $X$-valued function. Using the theory of regularized families, and Banach's fixed-point theorem, we prove existence and uniqueness of mild solutions for the semilinear fractional-order differential equation \begin{align*} & D_t^{\alpha+1}u(t) + \mu D^{\beta}_t u(t) \\ & = Au(t) + \frac{t^{-\alpha}}{\Gamma(1-\alpha)}u'(0) + \mu \frac{t^{-\beta}}{\Gamma(1-\beta)} u(0) + f(t,u(t)), \,\, t > 0, \end{align*} $0 < \alpha \leq \beta \leq 1,\,\, \mu >0$, with the property that the solution decomposes, uniquely, into a periodic term (respectively almost periodic, almost automorphic, compact almost automorphic) and a second term that decays to zero. We shall make the convention $\frac{1}{\Gamma(0)}=0.$ The general result on the asymptotic behavior is obtained by first establishing a sharp estimate on the solution family associated to the linear equation.
Article information Source Differential Integral Equations, Volume 26, Number 7/8 (2013), 757-780. Dates First available in Project Euclid: 20 May 2013 Permanent link to this document https://projecteuclid.org/euclid.die/1369057816 Mathematical Reviews number (MathSciNet) MR3098986 Zentralblatt MATH identifier 1299.35309 Subjects Primary: 34A08: Fractional differential equations 35R11: Fractional partial differential equations 47D06: One-parameter semigroups and linear evolution equations [See also 34G10, 34K30] 45N05: Abstract integral equations, integral equations in abstract spaces Citation
Keyantuo, Valentin; Lizama, Carlos; Warma, Mahamadi. Asymptotic behavior of fractional order semilinear evolution equations. Differential Integral Equations 26 (2013), no. 7/8, 757--780. https://projecteuclid.org/euclid.die/1369057816
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The usual representation of a free electromagnetic wave in vacuum looks like this:
The blue parts are the local electric field, while the green parts are the local magnetic field.
The
circularly polarized wave is also another standard representation of an EM travelling wave, but it is a bit less well known.
Here are the wave functions describing the above usual linearly polarized wave:
\begin{align} \vec{\boldsymbol{E}} &= \vec{\boldsymbol{a}} \sin{(k \, x - \omega \, t + \phi)}, \tag{1} \\[12pt] \vec{\boldsymbol{B}} &= \vec{\boldsymbol{b}} \sin{(k \, x - \omega \, t + \phi)}, \tag{2} \end{align} where $\vec{\boldsymbol{a}}$ and $\vec{\boldsymbol{b}}$ are two orthogonal constant polarisation vectors, transverse to the propagation (the $x$ axis here).
Now, I believe that I remember (I'm not sure!) that there's a
special superposition of travelling waves that gives a non-standard representation of the travelling EM plane wave, with the magnetic parts relative to the electric parts. I'm unable to find which superposition exactly can do this, but I know that this superposition (if it exists !) is actually shifted by a quarter of wavelength very simple.
I'm NOT talking about standard
standing wave here, and I'm not talking about the circularly polarized travelling wave neither. However, the superposition I'm looking for may be related to such waves, I don't know.
So somebody could tell which superposition of waves can shift the magnetic crests (as seen above) by
one quarter of a wavelength, relative to the electric crests ?
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In the Heisenberg picture (using natural dimensions): $$ O_H = e^{iHt}O_se^{-iHt}. \tag{1} $$ If the Hamiltonian is independent of time then we can take a partial derivative of both sides with respect to time: $$ \partial_t{O_H} = iHe^{iHt}O_se^{-iHt}+e^{iHt}\partial_tO_se^{-iHt}-e^{iHt}O_siHe^{-iHt}. \tag{2} $$ Therefore, $$ \partial_t{O_H} = i[H,O_H]+(\partial_tO_s)_H \, , \tag{3} $$ but this is not equivalent to what many textbooks list as the Heisenberg equation of motion. Instead they state that $$ \frac{d}{dt}{O_H} = i[H,O_H]+(\partial_tO_s)_H. \tag{4} $$ Why, in general, is this true and not the former statement? Am I just being pedantic with my use of partial and total derivatives?
With some definitions to make time dependences explicit, your equation (4) can be made sense of. Let's take the following:
Let $O_s$ be an operator depending on time and other parameters $O_s:\mathbb{R}\times S\rightarrow \mathrm{Op}$, where $S$ is the space of the other parameters and $\mathrm{Op}$ is the space of operators on the Hilbert space. Let $\phi:\mathbb{R}\times\mathrm{Op}\rightarrow\mathrm{Op}$ denote time evolution of operators in the Heisenberg picture, given by $\phi_t(O)=e^{iHt}Oe^{-iHt}$.
Note that $(\partial_t \phi)_t(O)=i[H,\phi_t(O)]$ and $\partial_O\phi=\phi$ (because $\phi$ is linear in $O$). Now, given a parameter $p\in S$ we can define the function of time: $O_H:\mathbb{R}\rightarrow \mathrm{Op}$ with $O_H(t)=\phi_t(O_s(t,p))$. Our function $O_H$ is a one-parameter one, so it only makes sense to take its total derivative: \begin{align} \frac{dO_H}{dt}(t)=&(\partial_t\phi)_t(O_s(t,p))+(\partial_O\phi)_t\left[(\partial_tO_s)(t,p)\right]\\ =& i[H,\phi_t(O_s(t,p))]+\phi_t\left[(\partial_tO_s)(t,p)\right]\\=& i[H,O_H(t)]+e^{iHt}(\partial_tO_s)(t,p)e^{-iHt}, \end{align}
where in the first step I have applied the chain rule and in the others, the equalities we already had.
No, you are not "just" being pedantic with your misuse of partial derivatives: your Eqns (2) and (3) are flat wrong. You simply did not apply the definitions right, as @WeinEld has been pointing out. (You might have spared yourself grief if you illustrated your very question for a simple system, such as the SHO.)
$$O_H \equiv e^{iHt}O_se^{-iHt} ,$$so for $$O_S=f(x,p;t) \qquad \Longrightarrow \qquad O_H=f(x(t),p(t);t),$$where $x(t)= e^{iHt}xe^{-iHt} $ and likewise for
p.
The time-derivative of $O_H$ consists of the partial derivative w.r.t.
t after the semicolon, plus the convective derivative due to the flow of x and p in the Heisenberg picture,$$\frac{\partial O_H}{\partial x(t)} \dot{x} + \frac{\partial O_H}{\partial p(t)} \dot{p} = i[H,O_H] = e^{iHt}(i[H,O_S])e^{-iHt}.$$(Prove this! Unless you did, the discussion is all vapor.)
The partial derivative is $$ \frac{\partial O_H}{\partial t}=e^{iHt} \frac{\partial O_S}{\partial t}e^{-iHt}=\left (\frac{\partial O_S}{\partial t}\right ) _H. $$ (Some express this as $ \frac{\partial O_H}{\partial t}$, trusting the reader would properly understand the evident differentiation of only the argument after the semicolon, but this very question may make them think twice. Now, to be sure, since $O_S$ has vanishing convective derivative, $dO_S/dt=\partial O_S/\partial t$, as raised in a comment, so this is a non-issue.)
In any case, putting the two pieces together nets the conventional $$ \frac{d}{dt}{O_H} = i[H,O_H]+(\partial_tO_s)_H. $$
Monitor the evident behavior of a simple observable such as $O_S=tx$ in the SHO, $H=(p^2+x^2)/2 $, the celebrated rigid classical-like rotation in phase space, $x(t)=x\cos t +p \sin t$, $p(t)=p\cos t - x \sin t$; thus $O_H=tx(t)$. Hence $dO_H/dt= t p(t)+x(t)$: now appreciate efficiencies and differences of the respective pictures. (Such as $$dO_H/dt=\exp(itH) (it[p^2/2,x] + x)\exp(-itH)=e^{it~[(x^2+p^2)/2,}~ (tp + x)~,$$ with the physicists' customary avoidance of the mathematician's
ad map notation.)
You might find your bearings by thinking of the S picture as the Eulerian frame, and the H picture as the Lagrangian, comoving frame.
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