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A transfer function is called
realizable if it can be implemented by a causal and stable system. The given frequency response is continuous and doesn't have any impulses, so the corresponding system is stable.
The transfer function (as a function of $s$) is given by
$$H(s)=\frac{a}{b-s^2}e^{-st_0}\tag{1}$$
The given frequency response is obtained form $(1)$ by substituting $s=j\omega$. Since $b>0$, $H(s)$ has two real-valued poles at $\pm\sqrt{b}$, and, consequently, the corresponding impulse response is two-sided (i.e., non-causal).
Partial fraction expansion of $(1)$ gives
$$H(s)=\frac{a}{2\sqrt{b}}\left[\frac{1}{\sqrt{b}-s}+\frac{1}{\sqrt{b}+s}\right]e^{-st_0},\qquad -\sqrt{b}<\textrm{Re}\{s\}<\sqrt{b}\tag{2}$$
From $(2)$ it is straightforward to obtain the impulse response
$$\begin{align}h(t)=\mathcal{L}^{-1}\{H(s)\}&=\frac{a}{2\sqrt{b}}\left[e^{\sqrt{b}(t-t_0)}u(-(t-t_0))+e^{-\sqrt{b}(t-t_0)}u(t-t_0)\right]\\&=\frac{a}{2\sqrt{b}}e^{-\sqrt{b}|t-t_0|}\tag{3}\end{align}$$
No matter how large you choose $t_0$, the impulse response will never be causal. However, since the system is stable, the impulse response decays for $|t|\to\infty$, so you can approximate the actual impulse response well by sufficiently shifting it to the right (i.e., by choosing some large $t_0$), and truncating the non-causal part.
Also take a look at this related question and its answer. |
Consider a $\mathcal{C}^1$ function $V:\Omega\rightarrow\mathbb{R}$ where $\Omega\subset\mathbb{R}^n$. If a random vector $X$ has a parametric density $p_\theta(\textbf{x})$ that's smooth in its parameters and uniformly supported on $\Omega$ we can use the score estimator for a stochastic derivative wrt $\theta\in\mathbb{R}^m$:
$$ \nabla \mathbb{E}\left[V(X)\right] = \mathbb{E}\left[V(X)\nabla \log p_\theta(X)\right] $$
Consider now $p_\theta$ as the density for $X\sim N(f(\theta), \epsilon I)$. Then $\nabla \log p_\theta (X)= \epsilon^{-1}J_f(\theta)^\top(f(\theta)- X)$. This then makes it look like we can get a deterministic approximate derivative for $V$ without access to a gradient oracle for $V$, for small $\epsilon$:
$$ \nabla V(f(\theta))\approx\epsilon^{-1} J_f(\theta)^\top\int d\textbf{x}\;p_\theta(\textbf{x})V(\textbf{x})(f(\theta)-\textbf{x}) $$
By continuity of $\nabla V$ equality should hold in the limit. Since the integral above looks like a Gaussian convolution, perhaps this is some kind of generalization of the standard $m$-dimensional finite difference procedure for generating an approximation to $\nabla V\circ f$.
Does this have a name? If $V$ is, say, $L$-Lipschitz at $f(\theta)$, then would using a quadrature rule for the integral yield a sensible gradient approximation (sensible being as good or better than the $O(m\epsilon^2)$ 2-norm error of symmetric finite differences)?
Edit. Perhaps the integral approximation is distracting. Let's assume we can compute the exact value of the integral and ask the same question about its accuracy for a fixed $\epsilon$ (indeed, for a Monte Carlo approximation, we can make the the integral accurate with high probability by the weak LLN). Though any thoughts on the internal integral would be much-appreciated. |
In order to generate an elliptical orbit, you need to have a force which is equal to the required centripetal force:
$$F=m\frac{v^2}{r}\rightarrow a=\frac{v^2}{r}$$
According to Bertrand's Theorem, this can only be solved with a potential for an inverse square force, or a radial harmonic oscillator potential.
So we cannot attain a circular orbit, is that a problem? No.
I generated a system for our sun, Earth, and moon, dependent on a linear inverse force. What we find is that we need to rescale the Gravitational constant to the negative 22nd order. (For clarity's sake I avoided using astronomical units).
So if we set $G = 6.6740831\times10^{-22}$ we find the following orbit patterns:
We can further decrease the orbital eccentricity when $G \rightarrow 4\times10^{-22}$
Note however, that in the long term, the eccentricity will always increase, even for optimal $G$, take the following radial Sol-Earth distance over 500y:
There are more problems though, for instance, would a star even form with this Gravity configuration?
Note that in this configuration, the acceleration of gravity due to Earth on its surface would be $0.000375m/s^2$ instead of $9.8m/s^2$As the gravity drops off more slowly, but is also significantly more massive, a habitable planet would be much more massive, but such massive planets might also more easily form under these parameters.
And here is where things get really interesting, if we suppose that our planet has a mass of $m_{earth}=5.97237\times10^{28}$, four orders higher than that of the current Earth, gravity at the same radius would be $3.75m/s^2$, and we get the following 1000 year progression:
My suspicion is that the collapse happens 4 orders of magnitude slower, meaning you would have at least $10^5y$ of stable orbit, possible a million (1Ma).
If you could have a planet with a mass of order $O\left(29\right)$, then you might get a near-stable orbit over evolutionary time scales, however getting such a large concentration of Earth (oxyen, quartz, aluminium, lime, iron, magnesium) might be difficult to attain, except maybe in a late-stage galaxy.
I do think the peculiar circumstances would make the formation of large planets more likely as distance is less of a factor for matter to come together. Consequently we would expect fewer planets, but of higher average mass. However, it is also possible this situation would lead to more uniformity in mass distributions.You would have to run some galaxy wide gravity calculations for that one, and recalculate the result of the background radiation. These are things beyond my scope. |
Strictly, correlations between any two or more inputs do not affect a model used in regression during use, however, the model may emerge from the analysis of correlations between input features. This is an important AI concept and one of the times when philosophy is directly applicable to science and technology.
To create a model that may explain observed phenomena, the patterns in observations must be recognized and a plausible relationship between variables representing the dimensions of the observation must be determined. That plausible relationship must be tested by correlating data to the model.
However, once the model is stable in the research and development sense, it can be used to control or predict. At that time, the very patterns that caused the emergence of the model must be decoupled (at least in the batch or real time processing) from the model for the model to have use.
A simple case is that observations of current, resistance, and potential voltage led to the model $V = IR$, but that relation is useless to electrical engineers if it is questioned because of newly acquired data. If the readings from a meter do not match the model, it is the readings that are questioned.
Correlations in a set of $(V, I, R)$ samples were discovered in a noisy quantum domain of electron travel. These correlations were discovered during feature extraction too. The combination of assignment of features macroscopic electrical events and the model that relates those features led to the understanding of proportionality or inverse proportionality in the relationship between any pair of those three dimensions.
Said another way, the model emerged through correlation of pairs (when the third variable was held constant) to $x$ or $1/x$ models. After the model gained acceptance in the scientific community, the correlations between input dimensions were not directly used. Neither were the correlations between the model and the original data of early researchers into electricity reexamined.
Today, the correlation between the model and the two known values and the unknown value is the only correlation in common use. A formula is used instead of regression for two reasons.
Two inputs are known to be proportional or inversely proportional to the output of that model (depending on which is the dependent variable when put to use). The expectation of accuracy is low, so only one or two samples may be used.
When the demand for accuracy increases, regression may become important to gain significant digits in quantification of the unknown property. For instance, an accurate resistance value may require several current-voltage pairs. This is a model that reduces to closed form. Other models do not, so no formula can be generated. Convergence strategies that use iteration or recursion, including those leveraged by regression, artificial networks, and other AI components become useful.
To answer the final question in the question body above, the performance of the model depends on correlation between past observations and the model, but performance in terms of accuracy and reliability at run time only depend on the qualities of the model and the quality of its integration into the AI system. The quality of its integration will also affect speed and conservation of computing resources.
This question touches on statistics because regression is a statistical method. It touches on data science because the philosophic aspect of science pronounced by Newton, Maxwell, Planck, von Neumann, and Heisenberg plays out in the realm of data. This question is directly related to AI because the application of intelligence is clear in the development of a model and then the trusting of it as a generalization to address specific cases.
Interestingly, a paradigm shift occurs when trusted models fail under an increasingly annoying array of conditions. These anomalies then lead to the seeking of a new model to which the larger, more contemporary data set correlates with greater accuracy and reliability. As paradigms shift to serve curiosity and technology, human beings must exhibit the mental features that roughly fall under the umbrella term intelligence. But this kind of intelligence begins with doubt and unlearning, not learning.
The subset of intelligence features that facilitates scientific discovery falls under another umbrella term, cognition. During a paradigm shift, the correlation between input variables again falls under scrutiny because doubt in the accepted model is re-introduced.
Response to Comments
The concept of cognition belongs in any comprehensive answer to this question, since correlation correlates with causality. This is central to AI and central to the reason why many regression methods produce correlation indicators. In the analysis of time series for the purposes of prediction, management, or control (which are the only practical reasons to do any statistics at all) the shift in time is significant.
$$ \mathcal{M}_{\Delta t_1, \Delta t_2, ..., \Delta t_X} \Bigg[ \mathcal{C}(\mathcal{Y}, f(P, \mathcal{X}')) \, \land \, \Big(\forall \; i \in \{1, 2, ..., X\} \, \land \, \mathcal{x}'_i(t) = \mathcal{x}_i(t) - \Delta t_i \Big) \Bigg] \; \text{,} $$
where $\mathcal{M}$ is maximization, $\Delta t_i$ is the shift in time that produces the maximum correlation, $\mathcal{C}$ is the correlation mechanism, $\mathcal{Y}$ is the desired result of applying model $f$, parameterized by $P$, to the component-wise shifted values of the features time series $\mathcal{X}'$.
In this case, the optimal shifts of time in the input parameters is evidence of but not proof of causality $\mathcal{C}$. It is itself a correlation $\mathcal{C}'$.
$$ \mathcal{C}' \Big( \text{Pr} (x_{\alpha} \implies x_{\beta}), \mathcal{C} \Big) $$
The doubt and certainty indicated by $\mathcal{C}'$ is the doubt and certainty in cognition and the basis for belief. If we ignore this abstraction, then the answer to these questions are superficially conformed to basic statistics but absent of the concept of cognition.
My question is what to do if there are correlations among different features of the input data? Does correlation b/w inputs affect the performance of Model? |
Four months ago, we were being persuaded that the Chinese have constructed a quantum radar, something that can inform you about an airplane without any actual reflection of any radiation from the airplane. This is obviously impossible by locality, whether you use any quantum subtleties or not.
Right now, we're told about another wonderful achievement by the Chinese. They have succeeded in teleporting a particle to the outer space. It's basically the same topic – about the same Chinese and about the same transmission of quantum information to the outer space. There are stories everywhere. NPR chose the title
There is nothing illogical, magical, or non-local about quantum teleportation. And there is nothing about quantum teleportation that would become significantly more difficult if the distance between the "parts of the lab" grows bigger.
What is quantum teleportation? It's not a real transmission of "general matter". It's just a transmission of quantum information. The quantum information may have many forms but the simplest toy model almost always involves one qubit.
OK, what is a qubit? It's the information that can be carried by a 2-state Hilbert space, e.g. the electron's spin. The general state of the electron spin is\[
\ket\psi = \alpha\ket\uparrow + \beta \ket\downarrow, \quad \alpha,\beta\in\CC
\] Classically, the spin would objectively be up (=0) or down (=1), so there would be one bit of information. Quantum mechanically, things may be uncertain, \(|\alpha|^2\) and \(|\beta|^2\) are the probabilities – we should assume \(|\alpha|^2+|\beta|^2=1\) – and on top of the separation of "one" into these two probabilities, the relative phase between \(\alpha,\beta\) matters, too. The overall phase change of both \(\alpha,\beta\) is unphysical, however.
How does Alice send the qubit to Bob? Well, if the qubit is carried by one electron, she just sends the whole electron. The electron will carry the same quantum information during the whole trip. Is there a different way that doesn't need to transfer the quantum information directly? Yes. It's enough if there were some quantum entanglement created in advance; and if one transfers some classical information even though you will be able to transmit quantum information.
How does it work? Let's first look at the classical toy model of the same situation.
A source AB of classical bits sends bits to two places, Alice's place and Bob's place. It can send either 00 or 01 or 10 or 11. Let's assume that AB sends 00 to both places. It's enough to assume that it's always AB=00.
Now, Alice wants to transfer the bit C which is either 0 and 1. In this case, we really don't know what the value is. She makes a combined measurement of AC, namely a measurement of "A XOR C". Because A=0, as we said in the previous paragraph, "A XOR C" is simply equal to C. She finds out the result of this "A XOR C" and sends the classical result of this measurement to Bob. Bob has his bit, B, but he can change it to the value of "A XOR C" that he heard from Alice: if Alice says "zero", he doesn't do anything. If Alice says "one", Bob negates his bit from "zero" to "one". At the end of the operation, Bob's bit B is equal to what Alice's C was to start with.
It was a talkative way to copy one damn bit, wasn't it? I formulated it in this awkward way because a straightforward – but somewhat more contrived – generalization exists that transmits C which is a qubit.
Let's redo the operation above in the fully quantum setup.
There is a source of entangled particles AB. Let's assume that they're in the state\[
|\Phi^+\rangle_{AB} = \frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_{B} + |1\rangle_A \otimes |1\rangle_{B})
\] You can see that the probability is 100% that the \(\sigma_z\)-measurements of both qubits end up with the same result. We also had to choose the relative phase or sign of the two terms. Fine. This source of entangled qubits AB is split, A is sent to Alice's place, B is sent to to Bob's place.
Alice wants to "beam" a qubit C to Bob. At this moment, we should realize that the state of all three qubits is\[
|\Phi^+\rangle_{AB} \otimes |\psi\rangle_C = \quad\quad\quad\tag{2017.7}\]\[
=\frac{1}{\sqrt{2}} (|0\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B)\otimes (\alpha |0\rangle_C + \beta|1\rangle_C).
\] Now, Alice will make a measurement that is the quantum counterpart of "A XOR C" in my classical story. She will make a combined measurement on the two qubits A,C which she has access to which will produce one of the 4 results in this orthonormal basis of the 2-qubit Hilbert space. These four states are \(\ket{\Phi^\pm}\) and \(\ket{\Psi^\pm}\). Phi is composed of 00 and 11, Psi is composed of 01 and 10, and the \(\pm\) sign determines the relative sign between the two terms.
Just to make sure that we write something newer, let's invert the relationship between the two bases, the basis of 00,01,10,11, and the basis containing \(\ket{\Phi^\pm}\) and \(\ket{\Psi^\pm}\). The former basis vectors may be written in terms of the latter as:\[
\displaystyle |0\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}} (|\Phi^+\rangle + |\Phi^-\rangle),\\
\displaystyle |0\rangle \otimes |1\rangle = \frac{1}{\sqrt{2}} (|\Psi^+\rangle + |\Psi^-\rangle),\\
\displaystyle |1\rangle \otimes |0\rangle = \frac{1}{\sqrt{2}} (|\Psi^+\rangle - |\Psi^-\rangle),\\
\displaystyle |1\rangle \otimes |1\rangle = \frac{1}{\sqrt{2}} (|\Phi^+\rangle - |\Phi^-\rangle).
\] Let us substitute this for the states 00,01,10,11 of the AC qubits in the expression (2017.7); just try to appreciate how much smarter you have gotten if you can actively use equations such as (2017.7). The state may be rewritten as\[
\begin{align}
|\Phi^+\rangle_{AB} \ \otimes\ | & \psi\rangle_C = \\
\frac{1}{2} \Big \lbrack
\ & |\Phi^+\rangle_{AC} \otimes (\alpha |0\rangle_B + \beta|1\rangle_B) \\
\ + \ & |\Phi^-\rangle_{AC} \otimes (\alpha |0\rangle_B - \beta|1\rangle_B) \\
\ + \ & |\Psi^+\rangle_{AC} \otimes (\beta |0\rangle_B + \alpha|1\rangle_B) \\
\ + \ & |\Psi^-\rangle_{AC} \otimes (\beta |0\rangle_B - \alpha|1\rangle_B) \Big \rbrack . \\
\end{align}
\] Excellent. Now, Alice has obtained one of the four results of her measurements, either \(\Phi^\pm\) or \(\Psi^\pm\), of the AC qubit pair. This result makes one of the four terms above relevant. Consequently, the wave function for the remaining qubit B also collapses to one of the four superpositions above. And for each of the 4 forms of the state, there is always a way to manipulate this qubit B – exchanging 0 and 1 and/or changing the relative sign in between the two terms – that ends up with the state of the B qubit in \(\alpha |0\rangle_B + \beta|1\rangle_B\), i.e. in the original state of the qubit C.
So the qubit C in Alice's place had some quantum state and it was transferred to the state of qubit B that was in Bob's place. We needed previously transmitted entangled AB qubit pairs; and we needed Alice to inform Bob about the result of a classical measurement – one of the four options \(\Phi^\pm\) or \(\Psi^\pm\).
That's it. It's quantum teleportation. It's not too different from simply sending the particle with the qubit C from one place to another. Alice still needs to send some information – which can't be done faster than light in any case – and Alice, Bob still need a path for sending entangled/quantum particles that go all the way from A to B – although the direction that is used is from "the AB source at the center to either A or B".
But regardless of the complex technicalities, quantum teleportation must be considered to be
just one of the specific generalizationsof the simple procedure of copying one bit of information that I started with. It's a generalization that can remember the two complex amplitudes. But because quantum cloning is impossible, the original state of the qubit C was destroyed by Alice's combined measurement of the AC qubit pair.
Aside from these technicalities – that we're talking about qubits that may be in superpositions, that may be manipulated by a more complex set of "unitary transformations" etc. – the quantum teleportation is morally nothing else than a version of a simple copying of one (quantum) bit. What Harry Potter's colleagues and Star Trek charlatans were doing was less elementary.
I think it's correct to say that the hype has a "wrong sign". The media are telling you that quantum mechanics allows you to do incredible things – such as those from the children's or sci-fi books and films – easily. But the actual message of the explanation above is that even seemingly trivial procedures – such as the copying of one bit of information – becomes subtle and hard in quantum mechanics. And quantum mechanics and theoretical physics in general is "cool" when it allows to make seemingly simple things hard or if you force you to be careful or hard-working. If someone isn't excited with this "imposed subtlety", then he simply doesn't like theoretical physics. He may like Harry Potter and Star Trek movies but those aren't the same thing as theoretical physics and from many perspectives, they have the opposite sign. |
I have encountered this old question and since I was bored I figured I use statisical mechanics to derive the relationship $E=\frac{3}{2}kT$ for a single particle in an ideal gas without using outside help. As a disclaimer, it has been a while since I had statistical mechanics courses, so I am a bit rusty. I have encountered some strange results I cannot explain and am curious about.
The problem: Using microcanonical ensemble for a single particle gives $E=(1/2)kT$. Using canonical ensemble for a single particle gives $E=(3/2)kT$.
Using canonical enseble for $N$ particles, the energy per particle $E/N$ is $E/N=(3/2)kT$. Using microcanonical ensemble for $N$ particles, the energy per particles is $E/N=(3/2-1/N)kT$. Clearly the latter goes to $(3/2)kT$ in the $N\rightarrow\infty$ limit.
Questions:
An ideal gas is noninteracting, so there should be no difference between using a single-particle ensemble and a multiparticle ensemble, right? Why is it that the canonical ensemble gives the same results in both cases, but the microcanonical is different?
Why do the microcanonical and canonical ensembles give different results, aside from the $N\rightarrow\infty$ limit? In particular, the microcanonical ensemble, while probably the furtherst removed from experiment, is the "purest" ensemble from a theoretical standpoint, the one that comes directly from the postulates of statistical mechanics. Is the formula $E=(3/2)kT$ valid only when $T$ is fixed? Or what?
The calculations I have used are presented below:
Ideal gas using microcanonical ensemble:
Assume we have an ideal gas in a box of volume $V$, at energy $E$ (at an uncertainty $(E,E+\delta E)$), and since the particles of an ideal gas are noninteracting anyways, we consider $N=1$.
The number of microstates is taken to be the phase volume between $E$ and $E+\delta E$, where $\delta E$ is a first-order "infinitesimal".
This can be calculated by calculating $\Omega(E'<E)$ the number of microstates whose energy is less than $E$. This is the following phase volume: $$ \Omega(E'<E)=\int_Vd^3x\int_Bd^3p, $$ where $B$ is the ball in momentum space given by $B=\{(p_x,p_y,p_z):\ p^2<2mE\}$, so $$ \Omega(E'<E)=Vg(2mE)^{3/2}, $$ where $g$ is an irrelevant constant. The number of microstates (of energy $E$) is then $$ \Omega(E,E+\delta E)=\frac{\partial}{\partial E}\Omega(E'<E)\delta E=\frac{3}{2}gV(2mE)^{1/2}2m\delta E=3gVm(2mE)^{1/2}\delta E. $$
The entropy is $S=k\ln\Omega$ so $$ S=k\left[\ln(3gVm\delta E)+\frac{1}{2}\ln(2m)+\frac{1}{2}\ln E\right], $$ the reciprocal of temperature is then $$ \frac{1}{T}=\frac{\partial S}{\partial E}=\frac{k}{2}\frac{1}{E}, $$ from which we get $$ E=\frac{1}{2}kT. $$
This is missing a factor of 3. Using a canonical ensemble on the other hand gives the proper answer:
Ideal gas using canonical ensemble:
We now consider the ideal gas fixed at temperature $T$, at volume $V$ and at particle number $N=1$.
The canonical partition function is $$ Z=V\left(\frac{2m\pi}{\beta}\right)^{3/2}, $$ the expectation value of energy is $$ E=-\frac{\partial}{\partial\beta}\ln Z=\frac{\partial}{\partial\beta}\frac{3}{2}\ln\beta=\frac{3}{2}\frac{1}{\beta}=\frac{3}{2}kT, $$ which is the result I expect.
Now, I have no idea why the microcanonical ensemble gives only the third of the expected result but I considered using an $N$-particle gas instead of a single particle gas.
The canonical ensemble gave $E=(3N/2)kT$, so the energy of a single particle is $E/N=(3/2)kT$, the same as before, so I will not do this calculation here. The microcanonical ensemble was different however. $N$-particle ideal gas using microcanonical ensemble:
The number of microstates whose energy is less than $E$ is now given as $$\Omega(E'<E)=V^Ng(2mE)^{3N/2},$$ where $g$ is a different constant, but still irrelevant. The number of microstates (of energy $E$) is then $$\Omega(E,E+\delta E)=\frac{\partial\Omega(E'<E)}{\partial E}\delta E=3NmgV^N(2mE)^{(3N/2)-1}\delta E,$$ the entropy is $$S=k\ln\Omega=k\left[\left(\frac{3N}{2}-1\right)\ln(2mE)+c\right],$$ the reciprocal temperature is $$\frac{1}{T}=\frac{\partial S}{\partial E}=k\left(\frac{3N}{2}-1\right)\frac{1}{E}, $$ $$ E=\left(\frac{3N}{2}-1\right)kT. $$ The energy for a single particle is then $$ \frac{E}{N}=\left(\frac{3}{2}-\frac{1}{N}\right)kT. $$ In the limit $N\rightarrow\infty$, this reduces to $E=(3/2)kT$. |
The most similar previous blog post was one about the \(\Delta(27)\) group The first hep-ph paper today is dedicated to heterotic string phenomenology. Michal Tučný, "Everyone is already in Mexico". Buenos días, I am also going. One of his top 20 best country music songs.
The Standard Model of particle physics is usually formulated as a gauge theory based on the \(SU(3)\times SU(2)\times U(1)\) gauge group. The particles carry the color and the electroweak charges. The gauge group is continuous which implies that there are gauge bosons in the spectrum.
However, the Standard Model also requires 3 generations of fermions – quarks and leptons. Because of this repetitive structure, it's natural to imagine that they transform as "triplets" under another, family group as well. However, there are apparently no \(SU(3)_{\rm flavor}\) gauge bosons, at least not available at the LHC yet. For this and other reasons, it's more sensible to assume that the 3 generations of fermions are "triplets" under a discrete, and not continuous, family symmetry.
The family groups that have been tried and that admit three-dimensional representations have been \(\ZZ_3\), \(S_3\), and \(\Delta(27)\). However, there exists one larger discrete group with a three-dimensional representation, the \(\Delta(54)\) group. With some definitions, it's the maximal one – the maximal "exceptional" discrete symmetry with three-dimensional representations, and the previous three cases are all subgroups of \(\Delta(54)\).
Surely many people share the gut feeling that the maximum symmetries of a similar exceptional type are (just like \(E_8\) among the simple Lie groups) the "most beautiful ones" and therefore most promising ones, at least from a certain aesthetic viewpoint. The larger symmetry with 54 elements probably makes the models more constrained and therefore more predictive.
These Mexican heterotic string theorists point out that the heterotic orbifolds of tori haven't led to viable models with this \(\Delta(54)\) symmetry yet. But that's because people were focusing on orbifolds\[
T^6 / \ZZ_3, \quad T^6 / \ZZ_3 \times \ZZ_2.
\] However, they propose different orbifolds instead:\[
T^6 / \ZZ_3 \times \ZZ_3
\] There are various discrete groups here and I need to emphasize that the group \(G\) in the description of the orbifold, \(T^6 / G\), is
notthe same group as the group produced as the family symmetry by the resulting heterotic string model. The group \(G\) is being "gauged" on the world sheet. It means that all the one-string states have to be invariant under \(G\) i.e. there are no charged states or non-singlets under \(G\); and closed string sectors with almost periodic boundary conditions up to the action of \(G\) have to be added, the twisted sectors.
The family group produced by orbifolds is in some way a "dual" group. The \(\ZZ_3\times \ZZ_3\) orbifold produces three fixed points not identified with each other. Dynamics at each of them is the same so there is actually an \(S_3\) symmetry (the full permutation group with 6 elements) exchanging them. A semidirect product of this group with a \(\ZZ_3\times \ZZ_3\) group (different from \(G\)) has to be considered because an additional global symmetry results from the action on localization charges \(m,q\) of the twisted sectors etc. The semidirect product with 54 elements is what we call \(\Delta(54)\).
All the fields and terms in the low-energy Lagrangian should simply be symmetric under this \(\Delta(54)\) and that leads to constraints on the Yukawa couplings and other things. It seems that their models – they found some 700 heterotic vacua using some software – have several great phenomenological properties such as
qualitatively realistic quark and lepton masses in general, including the recently proven nonzero \(\theta_{13}\) neutrino mixing angle the right relationship between the Cabibbo angle and the strange-to-down quark mass ratio (the Gatto-Sartori-Tonin relation) a cool equation relating ratios of both (down-type) quark and (charged) lepton masses: \[ \frac{m_s-m_d}{m_b} = \frac{m_\mu - m_e}{m_\tau} \] Well, this relationship isn't obeyed by the observed masses but corrections could make it true the normal, not inverted, hierarchy of neutrino masses, as preferred by latest experiments PMNS neutrino matrix compatible with the known data
But it's amazing how fine questions about the spectrum and parameters of the Standard Model are already being "almost completely explained" by a theory based on a completely different and much more concise starting point – the hybrid of the only bosonic \(D=26\) string theory and the only \(D=10\) superstring with the maximum family group produced at low energies.
When you combine these conditions, you rather naturally obtain three generations with realistic constraints on the quarks' and leptons' masses and mixing angles. Whoever isn't intrigued by those hints is a cold-blooded animal.
I added the label/category "music" to this blog post because of the extensive discussion of (mostly Czech) country music in the comment section. |
Suppose I had the following periodic 1D advection problem:
$\frac{\partial u}{\partial t} + c\frac{\partial u}{\partial x} = 0$ in $\Omega=[0,1]$
$u(0,t)=u(1,t)$ $u(x,0)=g(x)$ where $g(x)$ has a jump discontinuity at $x^*\in (0,1)$.
It is my understanding that for linear finite difference schemes of higher than first order, spurious oscillations occur near the discontinuity as it is advected over time, resulting in a distortion of the solution from its expected wave shape. According to wikipedia explanation, it seems that these oscillations typically occur when a discontinuous function is approximated with a finite fourier series.
For some reason, I can't seem to grasp how a finite fourier series can be observed in the solution of this PDE. In particular, how can I estimate a bound on the "over-shoot" analytically? |
Linear Regression and ANOVA shaken and stirred (Part 2)Tue, Mar 21, 2017 Updated 2018-03-27 Motivation
In the first part of this entry I did show some
mtcars examples, where
am can be useful to explain ANOVA as its observations are defined as:\[am_i = \begin{cases}1 &\text{ if car } i \text{ is manual} \cr 0 &\text{ if car } i \text{ is automatic}\end{cases}\]
Now I’ll show another example to continue the last example from Part 1 and I’ll move to something involved more variables.
ANOVA with one dummy variable
Consider a model where the outcome is
mpg and the design matrix is \(\renewcommand{\vec}[1]{\boldsymbol{#1}} \newcommand{\R}{\mathbb{R}} X = (\vec{1} \: \vec{x}_2)\).
From the last entry let \[ x_2 = \begin{cases}1 &\text{ if car } i \text{ is automatic} \cr 0 &\text{ if car } i \text{ is manual}\end{cases} \]
This will lead to this estimate: \[ \hat{\vec{\beta}} = \begin{bmatrix}\bar{y}_1 \cr \bar{y}_2 - \bar{y}_1\end{bmatrix} \] Fitting the model gives:
y <- mtcars$mpgx1 <- mtcars$amx2 <- ifelse(x1 == 1, 0, 1)fit <- lm(y ~ x2)summary(fit)
Call:lm(formula = y ~ x2)Residuals: Min 1Q Median 3Q Max -9.3923 -3.0923 -0.2974 3.2439 9.5077 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 24.392 1.360 17.941 < 2e-16 ***x2 -7.245 1.764 -4.106 0.000285 ***---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Residual standard error: 4.902 on 30 degrees of freedomMultiple R-squared: 0.3598, Adjusted R-squared: 0.3385 F-statistic: 16.86 on 1 and 30 DF, p-value: 0.000285
So to see the relationship between the estimates and the group means I need additional steps:
x0 <- rep(1,length(y))X <- cbind(x0,x2)beta <- solve(t(X)%*%X) %*% (t(X)%*%y)beta
[,1]x0 24.392308x2 -7.244939
I did obtain the same estimates with
lm command so now I calculate the group means:
x1 <- ifelse(x1 == 0, NA, x1)x2 <- ifelse(x2 == 0, NA, x2)m1 <- mean(y*x1, na.rm = TRUE)m2 <- mean(y*x2, na.rm = TRUE)beta0 <- m1beta2 <- m2 - m1beta0;beta2
[1] 24.39231
[1] -7.244939
In this case this means that the slope for the two groups is the same but the intercept is different, and therefore exists a negative effect of automatic transmission on miles per gallon in average terms.
Again I’ll verify the equivalency between
lm and
aov in this particular case:
y <- mtcars$mpgx1 <- mtcars$amx2 <- ifelse(x1 == 1, 0, 1)fit2 <- aov(y ~ x2)fit2$coefficients
(Intercept) x2 24.392308 -7.244939
I can calculate the residuals by hand:
fit3 <- lm(mpg ~ am, data = mtcars)mean_mpg <- mean(mtcars$mpg)fitted_mpg <- fit3$coefficients[1] + fit3$coefficients[2]*mtcars$amobserved_mpg <- mtcars$mpgTSS <- sum((observed_mpg - mean_mpg)^2) ESS <- sum((fitted_mpg - mean_mpg)^2)RSS <- sum((observed_mpg - fitted_mpg)^2)TSS;ESS;RSS
[1] 1126.047
[1] 405.1506
[1] 720.8966
Here its verified that \(TSS = ESS + RSS\) but aside from that I can extract information from
aov:
summary(fit2)
Df Sum Sq Mean Sq F value Pr(>F) x2 1 405.2 405.2 16.86 0.000285 ***Residuals 30 720.9 24.0 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
And check that, as expected, \(ESS\) is the variance explained by
x2.
I also can run ANOVA over
lm with:
anova(fit)
Analysis of Variance TableResponse: y Df Sum Sq Mean Sq F value Pr(>F) x2 1 405.15 405.15 16.86 0.000285 ***Residuals 30 720.90 24.03 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
The table provides information on the effect of
x2 over
y. In this case the null hypothesis is rejected because of the large F-value and the associated p-values.
Considering a 0.05 significance threshold I can say, with 95% of confidence, that the regression slope is statistically different from zero or that there is a difference in group means between automatic and manual transmission.
ANOVA with three dummy variables
Now let’s explore something more complex than
am. Reading the documentation I wonder if
cyl has an impact on
mpg so I explore that variable:
str(mtcars)
'data.frame': 32 obs. of 11 variables: $ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ... $ cyl : num 6 6 4 6 8 6 8 4 4 6 ... $ disp: num 160 160 108 258 360 ... $ hp : num 110 110 93 110 175 105 245 62 95 123 ... $ drat: num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ... $ wt : num 2.62 2.88 2.32 3.21 3.44 ... $ qsec: num 16.5 17 18.6 19.4 17 ... $ vs : num 0 0 1 1 0 1 0 1 1 1 ... $ am : num 1 1 1 0 0 0 0 0 0 0 ... $ gear: num 4 4 4 3 3 3 3 4 4 4 ... $ carb: num 4 4 1 1 2 1 4 2 2 4 ...
unique(mtcars$cyl)
[1] 6 4 8
One (wrong) possibility is to write:
y <- mtcars$mpgx1 <- mtcars$cyl; x1 <- ifelse(x1 == 4, 1, 0)x2 <- mtcars$cyl; x2 <- ifelse(x1 == 6, 1, 0)x3 <- mtcars$cyl; x3 <- ifelse(x1 == 8, 1, 0)fit <- lm(y ~ x1 + x2 + x3)summary(fit)
Call:lm(formula = y ~ x1 + x2 + x3)Residuals: Min 1Q Median 3Q Max -6.2476 -2.2846 -0.4556 2.6774 7.2364 Coefficients: (2 not defined because of singularities) Estimate Std. Error t value Pr(>|t|) (Intercept) 16.6476 0.7987 20.844 < 2e-16 ***x1 10.0160 1.3622 7.353 3.44e-08 ***x2 NA NA NA NA x3 NA NA NA NA ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Residual standard error: 3.66 on 30 degrees of freedomMultiple R-squared: 0.6431, Adjusted R-squared: 0.6312 F-statistic: 54.06 on 1 and 30 DF, p-value: 3.436e-08
Here the
NAs mean there are variables that are linearly related to the other variables (e.g. the variable pointed with
NA is an average of one or more of the rest of the variables, like \(x_2 = 2x_1 + x_3\) or another linear combination), then there’s no unique solution to the regression without dropping variables.
My model will include these variables: \[ x_2 = \begin{cases}1 &\text{ if car } i \text{ has 6 cylinders} \cr 0 &\text{ otherwise }\end{cases} \quad \quad x_3 = \begin{cases}1 &\text{ if car } i \text{ has 8 cylinders} \cr 0 &\text{ otherwise }\end{cases} \]
In this particular case regression coefficients are given by this estimate: \[ \hat{\vec{\beta}} = \begin{bmatrix}\bar{y}_1 \cr \bar{y}_2 \end{bmatrix} \]
But R has a command called
as.factor() that is useful in these cases and also can save you some lines of code in other cases:
fit2 <- lm(mpg ~ as.factor(cyl), data = mtcars)summary(fit2)
Call:lm(formula = mpg ~ as.factor(cyl), data = mtcars)Residuals: Min 1Q Median 3Q Max -5.2636 -1.8357 0.0286 1.3893 7.2364 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 26.6636 0.9718 27.437 < 2e-16 ***as.factor(cyl)6 -6.9208 1.5583 -4.441 0.000119 ***as.factor(cyl)8 -11.5636 1.2986 -8.905 8.57e-10 ***---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Residual standard error: 3.223 on 29 degrees of freedomMultiple R-squared: 0.7325, Adjusted R-squared: 0.714 F-statistic: 39.7 on 2 and 29 DF, p-value: 4.979e-09
The
aov version of this is:
fit3 <- aov(mpg ~ as.factor(cyl), data = mtcars)TukeyHSD(fit3)
Tukey multiple comparisons of means 95% family-wise confidence levelFit: aov(formula = mpg ~ as.factor(cyl), data = mtcars)$`as.factor(cyl)` diff lwr upr p adj6-4 -6.920779 -10.769350 -3.0722086 0.00034248-4 -11.563636 -14.770779 -8.3564942 0.00000008-6 -4.642857 -8.327583 -0.9581313 0.0112287
As I said many times in this entry, ANOVA is linear regression. Interpreting the coefficients is up to you. |
The SAG equation
The SAG equation is a Fick diffusion equation of a specie c \displaystyle \frac{\partial c}{\partial t} =\nabla^2 c which can be solved with the reaction–diffusion solver.
Concentration at time t+dt, t and maximum difference between the two.
scalar c[], cold[];double errmax;
We will store the statistics on the diffusion solvers in
mgd.
mgstats mgd;
A “crystal growth” boundary condition of the form \displaystyle \frac{\partial c}{\partial y} = bi\; c \;\;\mathrm{on}\;\; y=0
is imposed for -1 < x < 1 and y = 0, where bi is a kind of Biot number. This mixed condition can be discretised to second-order accuracy as
(c[] - c[bottom])/Delta = bi*(c[] + c[bottom])/2.;
The rest of the wall is covered with a mask so that no growth occurs i.e. \frac {\partial c}{\partial y} = 0 (a Neumann boundary condition). Combining everything then gives
double bi = 1;c[bottom] = fabs(x) < 1 ? neumann(0) : c[]*(2. - bi*Delta)/(2. + bi*Delta);
On the top boundary the species concentration is imposed.
c[top] = dirichlet(1);
And there is no flux on the right and left boundaries.
c[right] = neumann(0);c[left] = neumann(0);
Parameters
The domain is the square box [-5:5]\times[0:10].
Initial conditions
In the absence of a mask (i.e. with the Biot condition along the bottom boundary) an exact stationary solution is
#define cexact(y) ((bi*y + 1)/(bi*L0 + 1))
We use this as initial condition.
Time integration
We first set the timestep according to the timing of upcoming events. We choose a maximum timestep of 0.2 which ensures the stability of the reactive terms for this example.
dt = dtnext (0.2);
We use the diffusion solver to advance the system from t to t+dt.
mgd = diffusion (c, dt);
If the difference between
c and cold is small enough, we stop. Results
The flux along y is {\partial c}/{\partial y}
scalar flux[]; foreach() flux[] = (c[0,0] - c[0,-1])/Delta; boundary ({flux}); for (double x = -L0/2 ; x < L0/2; x += L0/N) { double y = x + L0/2; fprintf (fpx, "%g %g %g\n", x, interpolate (c, x, 0) , interpolate (flux, x, 0)); fprintf (fpy, "%g %g %g\n", y, interpolate (c, 0, y) , interpolate (flux, 0, y)); } fclose(fpx); }
At the end of the simulation, we create an image of the error field, in PNG format.
event pictures (t = end) { scalar dce[]; foreach() dce[] = c[] - cexact(y); boundary ({dce}); output_ppm (dce, file = "c.png", spread = 2, linear = true);}
We compare the exact and computed solutions for a cross-section at x=0.
bi=1L0=10set xlabel "y"set key leftplot 'log'u 1:2 t 'c' w l, '' u 1:3 t 'dc/dy' w l, \ bi/(bi*L0+1.) t 'bi/(bi L0+1)', \ (bi*x+1)/(bi*L0+1) t'(bi y+1)/(bi*L0+1)'
We also display the concentration flux divided by the reference flux bi/(bi L0+1).
set xlabel "x"plot './cutx.txt' u 1:($2*(bi*L0+1)) t 'concentration' w l, \ '' u 1:($3*(bi*L0+1)/bi) t 'flux' w l, 1 not
And the error field.
Bibliography
N. Dupuis, J. Décobert, P.-Y. Lagrée, N. Lagay, D. Carpentier, F. Alexandre (2008): “Demonstration of planar thick InP layers by selective MOVPE”. Journal of Crystal Growth issue 23, 15 November 2008, Pages 4795-4798.
N. Dupuis, J. Décobert, P.-Y. Lagrée , N. Lagay, C. Cuisin, F. Poingt, C. Kazmierski, A. Ramdane, A. Ougazzaden (2008): “Mask pattern interference in AlGaInAs MOVPE Selective Area Growth : experimental and modeling analysis”. Journal of Applied Physics 103, 113113 (2008). |
Deceleration parameter
Part of a series on Physical cosmology q \ \stackrel{\mathrm{def}}{=}\ -\frac{\ddot{a} a }{\dot{a}^2}
where \! a is the scale factor of the universe and the dots indicate derivatives by proper time. The expansion of the universe is said to be "accelerating" if \ddot{a} is positive (recent measurements suggest it is), and in this case the deceleration parameter will be negative.
[1] The minus sign and name "deceleration parameter" are historical; at the time of definition \! q was thought to be positive, now it is believed to be negative.
The Friedmann acceleration equation can be written as
3\frac{\ddot{a}}{a} =-4 \pi G (\rho+3p)=-4\pi G(1+3w)\rho,
where \! \rho is the energy density of the universe, \! p is its pressure, and \! w is the equation of state of the universe.
This can be rewritten as
q=\frac{1}{2}(1+3w)\left(1+K/(aH)^2\right)
The derivative of the Hubble parameter can be written in terms of the deceleration parameter:
\frac{\dot{H}}{H^2}=-(1+q).
Except in the speculative case of phantom energy (which violates all the energy conditions), all postulated forms of matter yield a deceleration parameter \! q \ge -1. Thus, any
expanding universe should have a decreasing Hubble parameter and the local expansion of space is always slowing (or, in the case of a cosmological constant, proceeds at a constant rate, as in de Sitter space).
Observations of the cosmic microwave background demonstrate that the universe is very nearly flat, so:
q=\frac{1}{2}(1+3w)
This implies that the universe is decelerating for any cosmic fluid with equation of state \! w greater than \! -1/3 (any fluid satisfying the strong energy condition does so, as does any form of matter present in the Standard Model, but excluding inflation). However observations of distant type Ia supernovae indicate that \! q is negative; the expansion of the universe is accelerating. This is an indication that the gravitational attraction of matter, on the cosmological scale, is more than counteracted by the negative pressure of dark energy, in the form of either quintessence or a positive cosmological constant.
Before the first indications of an accelerating universe, in 1998, it was thought that the universe was dominated by dust with negligible equation of state, \! w \approx 0. This had suggested that the deceleration parameter was equal to one half; the experimental effort to confirm this prediction led to the discovery of possible acceleration. |
Consider the mixed formulation of the Poisson/Darcy system for a region $\Omega$:
$\alpha \mathbf{v} + \nabla p = f \\ \mathrm{div}[\mathbf{v}] = 0 $
with the boundary conditions
$\mathbf{v}\cdot \mathbf{n} = v_n$ on $\Gamma_v$,
$p = p_0$ on $\Gamma_p$
For the weak form: $a(\mathbf{v}, \mathbf{w}) := \int_\Omega \mathbf{v} \cdot \mathbf{w}$,
while $b_1(\mathbf{w},p) := \int_\Omega p\;\mathrm{div}[\mathbf{w}]$ or
$b_2(\mathbf{w},p) := \int_\Omega \nabla p\cdot\mathbf{w}$.
Depending on the choice for $b_1$ or $b_2$, the solution spaces can be chosen as:
1) $\mathbf{v} \in H(div; \Omega)^n $ while $p \in L^2(\Omega)$ for the choice $b_1$. One can choose for this the BDM elements for the velocity and piecewise constant pressures for each element and get a stable solution. In this formulation, the condition on velocity is essential while the boundary condition for pressure is natural.
2) $\mathbf{v} \in L^2(\Omega)^n, p \in H^1_0(\Omega)$ for the choice $b_2$. In this case, the velocity can be approximated by the discontinous Raviart-Thomas elements of polynomial order $k$ and the pressure by Lagrangian elements of polynomial order $k+1$, to get a stable solution. In this formulation, the condition on velocity is natural while the boundary condition for pressure is essential.
Q1. Is one of 1), 2) to be preferred over the other choice? Will the solutions obtained differ ? Q2. However, $H^1_{0}(\Omega)^n \subset H(div;\Omega)^n$. So suppose I choose $\mathbf{v} \in H^1_0(\Omega)^n, p \in L^2(\Omega)$ for the choice $b_1$ and use the standard Taylor-Hood element, enforcing velocity boundary condition as essential and the pressure boundary condition as natural. Is this a stable approximation ? PS: I am unable to create the tags "Sobolev-spaces", "mixed-formulations". If you agree that they are apt for this question, could you please create them. |
General case. In relativistic thermodynamics, inverse temperature $\beta^\mu$ is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the maximum entropy method, where $\beta^\mu p_\mu$ (in units where $c=1$) replaces the term $\beta H$ of the nonrelativistic canonical ensemble. This is done in
C.G. van Weert, Maximum entropy principle and relativistic hydrodynamics, Annals of Physics 140 (1982), 133-162.
for classical statistical mechanics and for quantum statistical mechanics in
T. Hayata et al., Relativistic hydrodynamics from quantum field theory on the basis of the generalized Gibbs ensemble method, Phys. Rev. D 92 (2015), 065008. https://arxiv.org/abs/1503.04535
For an extension to general relativity with spin see also
F. Becattini, Covariant statistical mechanics and the stress-energy tensor, Phys. Rev. Lett 108 (2012), 244502. https://arxiv.org/abs/1511.05439
Conservative case. One can define a scalar temperature $T:=1/k_B\sqrt{\beta^\mu\beta_\mu}$ and a velocity field $u^\mu:=k_BT\beta^\mu$ for the fluid; then $\beta^\mu=u^\mu/k_BT$, and the distribution function for an ideal fluid takes the form of a Jüttner distribution $e^{-u\cdot p/k_BT}$.
For an ideal fluid (i.e., assuming no dissipation, so that all conservation laws hold exacly), one obtains the format commonly used in relativistic hydrodynamics (see Chapter 22 in the book Misner, Thorne, Wheeler, Gravitation). It amounts to treating the thermodynamics nonrelativistically in the rest frame of the fluid.
Note that the definition of temperature consistent with the canonical ensemble needs a distribution of the form $e^{-\beta H - terms~ linear~ in~ p}$, conforming with the identification of the noncovariant $\beta^0$ as the inverse canonical temperature. Essentially, this is due to the frame dependence of the volume that enters the thermodynamics. This is in agreement with the noncovariant definition of temperature used by Planck and Einstein and was the generally agreed upon convention until at least 1968; cf. the discussion in
R. Balescu, Relativistic statistical thermodynamics, Physica 40 (1968), 309-338.
In contrast, the covariant Jüttner distribution has the form $e^{-u_0 H/k_BT - terms~ linear~ in~ p}$. Therefore the covariant scalar temperature differs from the canonical one by a velocity-dependent factor $u_0$. This explains the different transformation law. The covariant scalar temperature is simply the canonical temperature in the rest frame, turned covariant by redefinition.
Quantum general relativity. In quantum general relativity, accelerated observers interpret temperature differently. This is demonstrated for the vacuum state in Minkowski space by the Unruh effect, which is part of the thermodynamics of black holes. This seems inconsistent with the assumption of a covariant temperature.
Dissipative case. The situation is more complicated in the more realistic dissipative case. Once one allows for dissipation, amounting to going from Euler to Navier-Stokes in the nonrelativistic case, trying to generalize this simple formulation runs into problems. Thus it cannot be completely correct. In a gradient expansion at low order, the velocity field defined above from $\beta^\mu$ can be identified in the Landau-Lifschitz frame with the velocity field proportional to the energy current; see (86) in Hayata et al.. However, in general, this identification involves an approximation as there is no reason for these velocity fields to be exactly parallel; see, e.g.,
P. Van and T.S. Biró, First order and stable relativistic dissipative hydrodynamics, Physics Letters B 709 (2012), 106-110. https://arxiv.org/abs/1109.0985
There are various ways to patch the situation, starting from a kinetic description (valid for dilute gases only): The first reasonable formulation by Israel and Stewart based on a first order gradient expansion turned out to exhibit acausal behavior and not to be thermodynamically consistent. Extensions to second order (by Romatschke, e.g., https://arxiv.org/abs/0902.3663) or third order (by El et al., https://arxiv.org/abs/0907.4500) remedy the problems at low density, but shift the difficulties only to higher order terms (see Section 3.2 of Kovtun, https://arxiv.org/abs/1205.5040).
A causal and thermodynamically consistent formulation involving additional fields was given by Mueller and Ruggeri in their book Extended Thermodynamics 1993 and its 2nd edition, called Rational extended Thermodynamics 1998.
Paradoxes. Concerning the paradoxes mentioned in the original post:
Note that the formula $\langle E\rangle = \frac32 k_B T$ is valid only under very special circumstances (nonrelativistic ideal monatomic gas in its rest frame), and does not generalize. In general there is no simple relationship between temperature and velocity.
One can say that your paradox arises because in the three scenarios, three different concepts of temperature are used. What temperature is and how it transforms is a matter of convention, and the dominant convention changed some time after 1968; after Balescu's paper mentioned above, which shows that until 1963 it was universally defined as being frame-dependent. Today both conventions are alive, the frame-independent one being dominant.
This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier |
Multi-objective optimization is a generalization of the standard optimization problem, such that a decision vector \(x \in X\), must be chosen to solve the minimization problem
$$
\begin{equation*} \min_{x \in X} (f_1(x),f_2(x),\ldots,f_k(x)), \end{equation*} $$
where \(f_i(x)\) are the different objectives one is trying to minimize. The problem can also be written by stacking all objectives into a vector function \(f \colon X \to Y\) that maps from the design space \(X\) to the decision space \(Y \subset \mathbb R^k\).
However, unlike the mono-objective setting, there may not be a single solution. If the objective functions conflict with each other, such that minimizing one increases others, then there isn’t a single solution, but a Pareto set of solutions, each one better than the others in some objective. This makes the optimization considerably harder.
Within this field, the main topics I’m currently tackling are:
How to transform the multi-objective optimization problem into a mono-objective one, which allows standard optimization algorithms to be used. How to approximate the objective functions so that the number of function evaluations, which can be expensive, may be reduced. How to deal with dynamic environments, where the objectives may change through time. Scalarizing objectives
There are two main traditional techniques for transforming multi-objective optimization problems into mono-objective ones.
The first one involves defining a vector of weights \(w \in \mathbb R^k, w_i > 0\), so that the optimization can be written as
$$
\begin{equation*} \min_{x \in X} w^T f(x), \end{equation*} $$
which has a single objective given by the weighted original objectives. If the Pareto frontier is convex, then theoretically we are able to recover it by varying \(w\) and solving multiple optimization problems. However, for concave frontiers, there are regions that won’t be seen.
The other method is the \(\epsilon\)-constraint method, where we optimize one function at a time, while constraining the value of the others, such that the problem becomes:
$$
\begin{equation*} \begin{aligned} \min_{x \in X} & ~f_i(x) \\ \text{s.t.} & ~f_j(x) \le \epsilon_{i,j}, \quad j \in \{1,\ldots,k\} \backslash \{i\}, \end{aligned} \end{equation*} $$
which is able to deal with concave frontiers, but it may be hard to define the constraint parameters \(\epsilon\).
Given these problems, one might be inclined to ask: is it possible to construct some scalar objective to replace multiple objectives? Well, there is an indicator named Hypervolume Indicator (also known as S-measure) that has been used in multi-objective optimization as a target for optimization, such that the estimated Pareto frontier’s hypervolume should be maximized.
Although the hypervolume has some nice theoretical properties, like allowing comparison between sets of points approximating the Pareto frontier, it can be computationally expensive and hard to be optimized directly.
So my search is for some way to improve the hypervolume optimization and new methods for transforming the multi-objective problem into a single-objective one.
Objectives surrogates
Surrogates \(\hat f_i\) are approximations to the real objectives \(f\), also mapping from \(X\) to \(Y\), so that expensive evaluation of the objective can be avoided while searching for a good candidate for real testing, which is then evaluated using the real objectives. There are many papers in the mono-objective literature using high-quality surrogates, which were then just copied to the multi-objective optimization field.
However, these methods optimize each objective separately, which ignores the inherent relationship between objectives given by the Pareto frontier and could lead to worse estimates. As a way to solve this, mono-surrogates have been proposed, where one maps directly from the decision space \(X\) to an indicator that states where the given point is in relation to the estimated Pareto frontier.
Nonetheless, even these direct methods ignore the innate shape of the Pareto frontier, finding surrogates where the estimated frontier may violate considerably the definition of the Pareto frontier itself.
Therefore, my research tries to find a method similar to the mono-surrogate ones, but with a mapping from the objective space \(Y\) instead of the decision space \(X\), which allows integration with the existing surrogate methods more appropriate for each objective, and a Pareto frontier estimate that tries its best not to violate the frontier definition.
Dynamic objectives
The problem with dynamic objectives is one that hasn’t been explored much, in my opinion, but one that I find very important to be tackled in real-world scenarios. If the problem doesn’t have to be solved only once, then it may be possible to extract information from previous iterations to predict the future behavior of the objectives and anticipate some decisions.
For instance, if the objectives have some structure as to how they evolve in time, then surrogate models \(\hat f_i \colon X \times T \to Y\) that take the current time-step into account may be able to learn the time-dependent relationship between objectives and be used to predict the system’s future behavior.
Hence my research tries to find ways to build these surrogates in a way that the estimated future objectives can be useful in the current time, either by creating diversity of solutions beforehand or guiding the choice of a solution among the Pareto set. |
One requirement that you don't have listed is that the generator $g$ needs to generate a subgroup that's of a large prime order; here's what can go wrong if that is not true:
If the order of $g$ (which we call $q$) has a factor $r$, then the attacker can, hearing $g^x$, determine $x \bmod r$ in $O(\sqrt{r})$ time. If $r$ isn't large, this immediately implies that you're leaking data
Worse yet, if the order $r$ is smooth (has no large factors), that makes the DLOG problem easy.
One strategy is, as mandragoe suggests, is look for a prime $p$ with $(p-1)/2$ prime as well. There is certainly a lot to like about this strategy; you can use (say) $g=2$, which can make the computation easier (and if the order of $g$ happens to be $2q$, all that means is that the attacker can learn the lsbit of your DH private values, which really doesn't tell him much, and you can avoid even that by making sure that $p \equiv 7 \pmod{8}$.
On the other hand, it is also a comparatively costly option, as looking for a value $p$ that meets both conditions means going through a lot of candidates. Now, if you're doing this computation only rarely (say, once a day), that's really not that big of a deal; however if you're coining a fresh group for every exchange, it is certainly more expensive that you need.
An alternative approach is:
Search for a 256 bit prime $q$
Once you have that, scan through 2048 bit values $kq+1$ for a prime; when you find it, call it $p$
For $g$, select an arbitrary value $x$ (2 works), and compute $x^k \bmod p$; if that value is not 1, use it for $g$.
The order of the generator $g$ will be $q$, which is a sufficiently large prime that we don't need to worry about attacks that take $O(\sqrt{q})$ time.
The expensive operation is the search for $p$, which is no more expensive than any other prime search for that size. |
We present an approach here that relies only on a standard inequality and the squeeze theorem. To that end, we proceed.
First, note that we can write the limit of interest as
$$\lim_{n\to \infty}\left(\frac{(n+1)^{\sqrt{n+1}}}{n^\sqrt{n}}\right)=\lim_{n\to \infty}e^{\sqrt{n+1}\log(n+1)-\sqrt{n}\log(n)} \tag 1$$
Next, we analyze the exponent on the right-hand side of $(1)$.
Using the identity $\log(n+1)=\log(n)+\log\left(1+\frac1n\right)$, the exponent becomes
$$\begin{align}\sqrt{n+1}\log(n+1)-\sqrt{n}\log(n)&=\left(\sqrt{n+1}-\sqrt{n}\right)\log(n)+\sqrt{n+1}\log\left(1+\frac1n\right)\\\\&=\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}+\sqrt{n+1}\log\left(1+\frac1n\right) \tag 2\\\\\end{align}$$
Then, since for any $\alpha>0$, $\log(n)\le \frac{n^\alpha}{\alpha}$, we find that the term of interest satisfies the inequalities
$$0\le\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}+\sqrt{n+1}\log\left(1+\frac1n\right)\le \frac2{n^{1/4}}+\frac2{\sqrt{n}} \tag 3$$
Applying the squeeze theorem to $(3)$ yields
$$\lim_{n\to \infty} \left(\sqrt{n+1}\log(n+1)-\sqrt{n}\log(n)\right)=0 \tag 4$$
whence using $(4)$ in $(1)$, we find the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{(n+1)^{\sqrt{n+1}}}{n^\sqrt{n}}=1}$$
NOTES:
Note $1$
In this note, we show that the logarithm function satisfies the inequality $\log(n)\le \frac{n^\alpha}{\alpha}$ for any $\alpha >0$. In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le\log(x)\le x-1<x \tag{N1}$$
for $x>0$. Then, using $\log(n^\alpha)=\alpha \log(n)$ in $(N1)$ with $x=n^\alpha$, we find
$$\log(n)\le \frac{n^\alpha -1}{\alpha}<\frac{n^\alpha}{\alpha} \tag {N2}$$
Note $2$
In this note, we show that $0\le \frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\le \frac{2}{n^{1/4}}$ for $n\ge 1$. The left-hand side inequality is trivial. For the right-hand side inequality we use $(N2)$ with $\alpha =1/4$.
Then, we see that $\log(n)\le 4n^{1/4}$. Putting that result together with the inequality $\frac{1}{\sqrt{n+1}+\sqrt{n}}\le \frac{1}{2\sqrt{n}}$, we obtain
$$0\le \frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\le \frac{2}{n^{1/4}}$$
as was to be shown.
Note $3$
In this note, we show $0\le \sqrt{n+1}\log\left(1+\frac1n\right)\le \frac2{\sqrt{n}}$, for $n\ge 1$. The left-hand side inequality is trivial. For the right-hand side inequality we use $(N1)$ with $x=1+\frac1n$.
Then, we see that $\log\left(1+\frac1n\right)\le \frac1n$. Putting that result together with $\sqrt{n+1}\le 2\sqrt{n}$, we obtain
$$0\le \sqrt{n+1}\log\left(1+\frac1n\right)\le \frac2{\sqrt{n}}$$
as was to be shown. |
I am currently working on a physics problem that turns into a non-linear boundary value problem. I need an efficient numerical solver that I could run on my laptop with i5 dual core CPU. I am discretizing my 15 equations on an N x N x N cubic grid using fourth order finite difference derivatives. I am interested in using Newton-Krylov methods since the Jacobian that is generated is sparse. My ultimate goal is to solve this system on a 128 x 128 x 128 cube, which means I will be solving almost 15 million equations for the same number of variables. I am already in the process of learning SUNDIALS packages but I recently found the PETSc library, which many people seem to be praising for quality of the code and efficiency. PETSc also seems to have a large number of preconditioners needed for Krylov methods.
So I would like to ask how do both packages compare for serial computations since I would like to run the solver on my laptop (at least for testing purposes on a bit coarser grid). I might eventually move to my university's HPC cluster to solve the system on a finer grid. Also, my intention is to use these packages on a long term basis.
I am new to high performance computing so kindly excuse my ignorance. I would really appreciate any suggestions.
P.S- The equations I am working on come from Non-Abelian gauge theories and are in Tensor form. Individual components (15) equations are long and complicated...so I actually compute these equations using mathematica and then discretize them. These equations are similar to Maxwell's equations but more complicated and non-linear in the fields. I do not know how much this would help here but the equations are as follows
$(D_\mu F^{\mu \nu})^a = g \epsilon_{abc} (D_\nu \phi)^b \phi^c$, and
$D_\mu (D^\mu \phi)^a = - \lambda (\phi^b \phi^b -v^2) \phi^a$.
Here Einstein's summation convention is implied. The indices $\mu$ and $\nu$ run from 0 to 3. And index $a$ runs from 1 to 3. $D$ represents gauge covariant derivatives and $\phi$ and $F$ represent Higgs and gauge fields strength. You can check this article for more details....page 9. http://www-thphys.physics.ox.ac.uk/people/MaximeGabella/higgs.pdf |
If $x_k$ are the Chebyshev nodes, that is for $n \in \mathbb{N}^*$, we have $x_k = \cos(\pi \frac{2k + 1}{n})$. Now suppose you have approximated values of $x_k$ and $x_{k'}$ for $k,k' \leq n$. In the worst case, what is the number of significant values we may loose when we calculate $x_{k'} - x_k$?
Now, taking into account the fact that each real is represented as $x \approx m \times b^e$, we then have a formula that for $x'$ the approximated value of $x$ and $y'$ approximated value of $y$, $\Delta (x' + y') \leq \epsilon(\Delta| x| + \Delta |y| + |x| + |y|)$ where $\epsilon = b^{1-N}$, where $N$ is there number of significant numbers after the comma.
So for a fixed $N \in \mathbb{N}^*$ we have $\Delta (x_{k'}' - x_k') \leq \epsilon (\Delta |x_{k'}| + \Delta |x_k| + |x_{k'}| + |x_k| ) \leq 4 \epsilon $
Thus in worst case we have a loss of precision of $\frac{4}{b^{N-1}}$. Thus we lose one significant term? |
This is an old revision of the document!
$ c_i \equiv \mbox{Chord length at span station} $
$ b_i \equiv \mbox{Section span} $ $ S_i \equiv \mbox{Section area} $ $ S_{total} \equiv \mbox{Total Area} $ $ \bar{c} \equiv \mbox{Weighted chord sum} $ $ L_{ref} \equiv \mbox{Reference length} $ $ D \equiv \mbox{Diameter} $ $ X_{area} \equiv \mbox{Max cross sectional area} $ $ F \equiv \mbox{Fineness ratio} $ $ Re \equiv \mbox{Reynolds number} $ $ V_{inf} \equiv \mbox{freestream velocity} $ $ \nu \equiv \mbox{kinematic viscosity} $ $ C_{f} \equiv \mbox{friction coefficient} $ $ f \equiv \mbox{flat plate drag} $ $ S_{wet} \equiv \mbox{wetted area} $ $ S_{ref} \equiv \mbox{reference area} $ $ C_D \equiv \mbox{coefficient of drag} $
Wetted areas for all components are taken from comp geom and as such as are appropriately trimmed.
Reference length is calculated according to it's geometry type internal to VSP; either a “Wing” type or a “Body” type. If the result of these calculations are zero, due to rotation or abnormal geometry shape the other equation is attempted as a fail safe. If neither of these methods yield a result greater than zero, a default value of 1.0 is used as to prevent division by zero when calculating fineness ratio.
Wing reference length is found by taking the area weighted chord.
First find the width of the wing section. $$ \Delta_x = |x_{le(1)} - x_{le(end)}| $$ $$ \Delta_y = |y_{le(1)} - y_{le(end)}| $$ $$ \Delta_z = |z_{le(1)} - z_{le(end)}| $$ $$ b_i = \sqrt{ {\Delta_x}^2 + {\Delta_y}^2 + {\Delta_z}^2} $$ $$ S_i = b_i * \frac{c_i + c_{\left(i+1\right)}}{2} $$ $$ S_{total} = \sum\left(S_i\right) $$ $$ \bar{c} = \sum\left(\frac{c_i + c_{\left(i+1\right)}}{2}\right) $$ $$ L_{ref} = \frac{\bar{c}}{S_{total}} $$
Body geometry reference lengths are calculated by taking the distance between to the front and back ends of the degenerated stick. $$ \Delta_x = |x_{le(1)} - x_{le(end)}| $$ $$ \Delta_y = |y_{le(1)} - y_{le(end)}| $$ $$ \Delta_z = |z_{le(1)} - z_{le(end)}| $$ $$ L_{ref} = \sqrt{ {\Delta_x}^2 + {\Delta_y}^2 + {\Delta_z}^2} $$
Thickness to chord takes the max thickness to chord from the degenerate stick created from Degen Geom.
Fineness Ratio is calculated by taking the diameter calculated using the max cross sectional area taken from the degenerate stick from Degen Geom.
$$ D = 2 * \sqrt{\frac{X_{area}}{\pi}} $$ $$ F = \frac{D}{L_{ref}} $$
Depending on the atmosphere input type, the kinematic viscosity is calculated accordingly and used to find the Reynolds number for the geometry.
$$ Re = \frac{V_{inf} * L_{ref}}{\nu} $$
Laminar Percent takes a 0 to 100 value and at default is set to 0% for a fully turbulent flow.
The laminar friction coefficient is calculated using the following equation: $$ C_f = C_{f (100\%Turb)} - \left(C_{f (\% Partial Turb)} * \% Lam\right) + \left(C_{f (\% Partial Lam)} * \% Lam\right) $$
where (using a chosen friction coefficient and it's corresponding inputs) $$ C_{f (\% Partial Turb)} = f\left(Re_{Lam}\right) $$ $$ C_{f (\% Partial Lam)} = f\left(Re_{Lam}\right) $$ where $$ Re_{Lam} = Re * \% Lam $$
$ Q \equiv \mbox{Scale factor applied to drag coefficient} $
$$ f = S_{wet} * Q * C_f * FF $$
$$ C_D = \frac{f}{S_{ref}} $$
Utilizing this feature the user is able to combine the wetted area of any geometry with that of another. Currently, geometries can only be grouped with their ancestors and geometries of the same shape type.
This can be used for example if the gear pod is modeled seperately from the fuselage but the wetted area of the gear pod should be applied with the drag qualities (e.g. length, form factor, etc.) of the fuselage.
By default, subsurfaces are incorporated as a part of the geometry as a whole. In other words, the surfaces do not subtract any wetted area from the geometry or have any of their own unique properties. However, the parasite drag tool let's the user choose these as options if they desire.
The three options are:
Treat as Parent: The default option, incorporates the wetted area of the subsurface as part of a continuous geometry. Separate Treatment: Allows the user, to some extent, control the qualities of the subsurface. However, due to limitations of the methodology used, the geometry based qualities (e.g. L_{ref}, Re, etc.) are derived from the parent geometry. The user is allowed to change the form factor equation type, manually set the laminar percentage, and manually set the interference factor for the subsurface. Zero Drag:The subsurface wetted area is subtracted from the total wetted area of the geometry and no longer contributes to the drag of the component.
$ x \equiv \mbox{distance along chord} $
$ k \equiv \mbox{roughness height} $ $ \gamma \equiv \mbox{specific heat ratio} $ $ r \equiv \mbox{recovery factor} = 0.89 $ $ n \equiv \mbox{viscosity power-law exponent} = 0.67 $
$ l \equiv \mbox{length of component} $
$ d \equiv \mbox{diameter of component} $ $ w \equiv \mbox{width at maximum cross sectional area} $ $ h \equiv \mbox{height at maximum cross sectional area} $ $ FR \equiv \mbox{Covert Fineness Ratio} = \frac{l}{\sqrt{wh}} $ $ l_{r} \equiv \mbox{Length of Fuselage} $ $ A_{x} \equiv \mbox{Cross Sectional Area of Fuselage} $ $ Q \equiv \mbox{Interference Factor} $
$$ C_f = \frac{1.32824}{Re^{1/2}}\ $$
$$ C_f = \frac{0.455}{\ln^2(0.06\ Re)}\ $$
$$ C_f = \frac{0.225}{\log(Re)^{2.32}}\ $$
$$ \frac{1}{\sqrt{C_f}}\ = 3.46\log\left(Re\right) - 5.6 $$
$$ \log\left(Re\ C_f\right) = \frac{0.242}{\sqrt{C_f}}\ $$
$$ \frac{1}{\sqrt{C_f}}\ = 4.15\log\left(Re\ C_f\right) + 1.70 $$
$$ \frac{1}{\sqrt{C_f}}\ = 4.13\log\left(Re\ C_f\right) $$
$$ C_f = \frac{0.0592}{Re^{1/5}}\ $$
$$ C_f = \frac{0.074}{Re^{1/5}}\ $$
$$ C_f = \frac{0.027}{Re^{1/7}}\ $$
$$ C_f = \frac{0.058}{Re^{1/5}}\ $$
$$ C_f = \frac{0.455}{\log\left(Re\right)^{2.58}}\ $$
$$ C_f = \frac{0.472}{\log\left(Re\right)^{2.5}}\ $$
$$ C_f = \frac{1}{\left(2\ \log\left(Re\right) - 0.65\right)^{2.3}}\ $$
$$ C_f = \frac{0.370}{\log\left(Re\right)^{2.584}}\ $$
$$ C_f = \frac{0.427}{\left(\log\left(Re\right) - 0.407\right)^{2.64}}\ $$
$$ C_f = \left(1.89 + 1.62\ \log\left(\frac{l}{k}\right)\right)^{-2.5} $$
$$ C_f = \left(2.87 + 1.58\ \log\left(\frac{x}{k}\right)\right)^{-2.5} $$
$$ C_f = \left(1.4 + 3.7\ \log\left(\frac{x}{k}\right)\right)^{-2} $$
$$ C_f = \frac{\left(1.89 + 1.62\ \log\left(\frac{l}{k}\right)\right)^{-2.5}}{\left(1 + \frac{\gamma - 1}{2}\ M_\infty\right)^{0.467}} $$
$$ f = \frac{\left(1 + 0.22\ r\ \frac{\gamma - 1}{2}\ {M_e}^2\ \frac{Te}{Tw}\right)}{\left(1 + 0.3\ \left(\frac{Taw}{Tw} - 1\right)\right)} $$ $$ C_f = \frac{0.451\ f^2\ \frac{Te}{Tw}}{\ln^2\left(0.056\ f\ \frac{Te}{Tw}^{1+n}\ Re\right)} $$
$$ FF = 1 + \frac{t}{c}\ \left(2.94206 + \frac{t}{c}\ \left(7.16974 + \frac{t}{c}\ \left(48.8876 + \frac{t}{c}\ \left(-1403.02 + \frac{t}{c}\ \left(8598.76 + \frac{t}{c}\ \left(-15834.3\right)\right)\right)\right)\right)\right) $$
$$ FF = 1 + 4.275\ \frac{t}{c} $$
Recreated Data from DATCOM is shown in the Figure and is used to find the Appropriate Scale Factor for use in the DATCOM Equation through interpolation. $$ FF = \left[1 + L\ \left(\frac{t}{c}\right) + 100\ \left(\frac{t}{c}\right)^4\right] * R_{L.S.} $$
$$ FF = 1 + 2\ \left(\frac{t}{c}\right) + 60\ \left(\frac{t}{c}\right)^4 $$
$$ FF = 1 + Z\ \left(\frac{t}{c}\right) + 100\ \left(\frac{t}{c}\right)^4 $$ $$ Z = \frac{\left(2-M^2\right) \cos\left(\Lambda_{\frac{c}{4}}\right)} {\sqrt{1-M^2\cos^2\left(\Lambda_{\frac{c}{4}}\right)}} $$
$$ FF = 1 + \frac{2.2 \cos^2\left(\Lambda_{\frac{c}{4}}\right)} {\sqrt{1-M^2\cos^2\left(\Lambda_{\frac{c}{4}}\right)}} \left(\frac{t}{c}\right) + \frac{4.84 \cos^2\left(\Lambda_{\frac{c}{4}}\right) \left(1 + 5\cos^2\left(\Lambda_{\frac{c}{4}}\right)\right)} {2\left(1-M^2\cos^2\left(\Lambda_{\frac{c}{4}}\right)\right)} \left(\frac{t}{c}\right)^2 $$
$$ FF = 1 + 2.7\ \left(\frac{t}{c}\right) + 100\ \left(\frac{t}{c}\right)^4 $$
$$ FF = 1 + 1.8\ \left(\frac{t}{c}\right) + 50\ \left(\frac{t}{c}\right)^4 $$
$$ FF = 1 + 1.44\left(\frac{t}{c}\right) + 2\left(\frac{t}{c}\right)^2 $$
$$ FF = 1 + 1.68\left(\frac{t}{c}\right) + 3\left(\frac{t}{c}\right)^2 $$
$$ F^* = 1 + 3.3\left(\frac{t}{c}\right) - 0.008\left(\frac{t}{c}\right)^2 + 27.0\left(\frac{t}{c}\right)^3 $$ $$ FF = \left(F^* - 1\right)\left(cos^2\left(\Lambda_{\frac{c}{2}}\right)\right) + 1 $$
$$ F^* = 1 + 3.52\left(\frac{t}{c}\right) $$ $$ FF = \left(F^* - 1\right)\left(cos^2\left(\Lambda_{\frac{c}{2}}\right)\right) + 1 $$ $$ Q = 1.2 $$
$$ FF = 1 + \frac{60}{FR^3} + 0.0025\ FR $$
$$ FF = 1 + \frac{0.35}{FR} $$
$$ FF = 1 + \frac{1.5}{\left(\frac{l}{d}\right)^{1.5}} + \frac{7}{\left(\frac{l}{d}\right)^3} $$
$$ FF = 1 + \frac{2.2}{\left(\frac{l}{d}\right)^{1.5}} + \frac{3.8}{\left(\frac{l}{d}\right)^3} $$
$$ FF = 1 + \frac{2.8}{\left(\frac{l}{d}\right)^{1.5}} + \frac{3.8}{\left(\frac{l}{d}\right)^3} $$
$$ FF = 1.02\left(1.0 + \frac{1.5}{\left(\frac{l}{d}\right)^{1.5}} + \frac{7.0}{\left(\frac{l}{d}\right)^3 \left(1.0 - M^3\right)^{0.6}}\right) $$
$$ \Lambda = \left(\frac{l_{r}}{\frac{4}{\pi}A_{x}}\right)^{0.5} $$ $$ FF = 1 + \frac{2.2}{\Lambda^{1.5}} - \frac{0.9}{\Lambda^3} $$
Jenkinson suggests a constant Form Factor for typical nacelles on wings. $$ FF = 1.25 $$
Jenkinson suggests a constant Form Factor for typical nacelles on aft fuselages. $$ FF = 1.50 $$
The parasite drag tool includes several options for atmosphere models as well as the option for the user to have control over 2 atmospheric qualities between Temperature, Pressure, and Density.
Upper Limit: 82,021 feet
Upper Limit: 84,852 meters
$ \frac{t}{c} \equiv \mbox{thickness to chord ratio of selected geometry} $
$ M \equiv \mbox{freestream Mach number for flight condition} $ $ M_{cr} \equiv \mbox{Critical Mach number, the point at which drag creep begins to occur} $ $ M_{DD} \equiv \mbox{Drag Divergence Mach number, the point at which drag significantly begins to rise} $ $ M_{DD,eff} \equiv \mbox{Effective Drag Divergence Mach number, Drag Divergence Mach number with consideration fro wing sweep} $ $ \Delta_{CD} \equiv \mbox{Additional CD due to Transonic Drag effects} $ $ \phi_{25} \equiv \mbox{Average quarter chord sweep of selected geometry} $ $ \gamma \equiv \mbox{Specific heat ratio; typically 1.4} $ $ C_{L} \equiv \mbox{Design Lift Coefficient} $
Example plots are given showing Drag Divergence Mach number using each equation with contours of quarter chord sweep angle at 0°, 15°, 30°, 45°, 60°, and 75° utilizing the following inputs: $$ C_{L} = 1.0 $$ $$ Airfoil Type = Conventional $$ $$ \gamma = 1.4 $$ $$ A_{F} = 0.8 $$ $$ K_{A} = 0.8 $$
$$ \Delta_{CD} = 20 * \left(M - M_{cr}\right)^4 $$
$ M^{*} \equiv \mbox{1.0, conventional airfoils; maximum t/c at about 0.30c} $
$ M^{*} \equiv \mbox{1.05, high-speed (peaky) airfoils, 1960-1970 technology} $ $ M^{*} \equiv \mbox{1.12 to 1.15, supercritical airfoils [Conservative = 1.12; Optimistic = 1.15]} $ $$ M_{DD,eff} = M_{DD} * \sqrt{\cos{\phi_{25}}} $$ $$ \frac{t}{c} = 0.30\cos{\phi_{25}}\left(\left(1 - \left( \frac{5 + {M_{DD,eff}}^2}{5 + {M^*}^2}\right)^{3.5}\right)\frac{\sqrt{1 - {M_{DD,eff}}^2}}{{M_{DD,eff}}^2}\right)^{\frac{2}{3}} $$
$ M^{*} \equiv \mbox{1.0, conventional airfoils; maximum t/c at about 0.30c} $
$ M^{*} \equiv \mbox{1.05, high-speed (peaky) airfoils, 1960-1970 technology} $ $ M^{*} \equiv \mbox{1.12 to 1.15, supercritical airfoils [Conservative = 1.12; Optimistic = 1.15]} $ $$ M_{DD,eff} = M_{DD} * \sqrt{\cos{\phi_{25}}} $$ $$ \frac{t}{c} = 0.30\cos{\phi_{25}}\left(\left(1 - \left( \frac{5 + {M_{DD,eff}}^2}{5 + \left(k_{M} - 0.25c_{l}\right)^2}\right)^{3.5}\right)\frac{\sqrt{1 - {M_{DD,eff}}^2}}{{M_{DD,eff}}^2}\right)^{\frac{2}{3}} $$
Equation solved for M. $$ A = \frac{M^2\cos^2{\phi_{25}}}{\sqrt{1-M^2\cos^2{\phi_{25}}}}\left(\left(\frac{\gamma + 1}{2}\right)\frac{2.64\frac{t}{c}}{\cos{\phi_{25}}} + \left(\frac{\gamma + 1}{2}\right)\frac{2.64\left(\frac{t}{c}\right)\left(0.34C_{L}\right)}{\cos^3{\phi_{25}}}\right) $$ $$ B = \frac{M^2\cos^2{\phi_{25}}}{1-M^2cos^2{\phi_{25}}}\left(\left(\frac{\gamma + 1}{2}\right)\left(\frac{1.32\frac{t}{c}}{\cos{\phi_{25}}}\right)^2\right) $$ $$ C = M^2cos^2{\phi_{25}}\left(1 + \left(\frac{\gamma + 1}{2}\right)\frac{\left(0.68C_{L}\right)}{\cos^2{\phi_{25}}} + \frac{\gamma + 1}{2}\left(\frac{0.34C_{L}}{\cos^2{\phi_{25}}}\right)^2\right) $$
$$ A + B + C - 1 = 0 $$
If a Peaky Airfoil Type is selected $$ M_{DD} = M + 0.06 $$ otherwise $$ M_{DD} = M $$
$ M_{cc} \equiv \mbox{Crest Critical Mach number} $
$$ a = 0.2 $$ $$ b = 2.131 $$ $$ x = \frac{\frac{t}{c}}{\cos{\phi_{25}}} $$ $$ y = \frac{C_{L}}{{\left(\cos{\phi_{25}}\right)}^2} $$
$$ M_{cc} = \frac{2.8355x^2 - 1.9072x + 0.949 - a\left(1-bx\right)y}{\cos{\phi_{25}}} $$
If Conventional Airfoil Type: $$ \Delta_{CD} = 0.04 $$ If Peaky Airfoil Type: $$ \Delta_{CD} = 0 $$ If Supercritical Conservative: $$ \Delta_{CD} = -0.04 $$ If Supercritical Optimistic: $$ \Delta_{CD} = -0.06 $$
$$ M_{DD} = M_{cc} * \left(1.025 + 0.08\left(1-\cos{\phi_{25}}\right)\right) - \Delta_{CD} $$
$ A_{F} \equiv \mbox{Airfoil Technology Factor, typically between 0.8 and 0.95} $
$$ M_{DD,eff} = M_{DD} * \sqrt{\cos{\phi_{25}}} $$ $$ M_{DD,eff} = A_{F} - 0.1C_{L} - \frac{t}{c} $$
$$ \frac{t}{c} = 0.7185 + 3.107e^{-5}\phi_{25} - 0.1298C_{L} - 0.7210M_{DD} $$
$ K_{A} \equiv \mbox{Airfoil Technology Factor, typically between 0.8 and 0.9} $
$$ M_{DD} = \frac{K_{A}}{cos{\phi_{25}}} - \frac{\frac{t}{c}}{cos^2{\phi_{25}}} - \frac{C_{L}}{10cos^3{\phi_{25}}} $$
$$ a = -1.147 $$ $$ b = 0.2 $$ $$ c = 0.838 $$ $$ d = 4.057 $$ $$ M_{DD} = a (C_{L} - b)d + c + \frac{30}{27}\left(\frac{t}{c} - 0.113\right) + 0.00288\left(\phi_{25} - 29.8\right) $$
$$ M_{cr} = M_{DD} $$
$$ M_{cr} = M_{DD} - 0.02 $$
$$ M_{cr} = M_{DD} - 0.1 $$
$$ M_{cr} = M_{DD} - \left(\frac{1}{80}\right)^{\frac{1}{3}} $$ |
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FAISS, Part 2
Recall that you have a set of database vectors \(\{\textbf{y}_i\}_{i=0}^\ell\), each in \(\mathbb{R}^d\). You can do some prep work to create an index. Then at runtime I ask for the \(k\) closest vectors in \(L^2\) distance.
Formally, we want the set \(L=\text{$k$-argmin}_i\norm{\textbf{x}-\textbf{y}_i}\) given \(\textbf{x}\).
The main paper contributions in this regard were a new algorithm for computing the top-\(k\) scalars of a vector on the GPU and an efficient k-means implementation.
Big Lessons from FAISS
Parsimony is important. Not only does it indicate you’re using the right representation for your problem, but it’s better for bandwidth and better for cache. E.g., see this wiki link, HNSW on 1B vectors at 32 levels results in TB-level index size!
Prioritize parallel-first computing. The underlying algorithmic novelty behind FAISS takes a serially slow algorithm, an \(O(n \log^2 n)\) sort, and parallelizes it to something that takes \(O(\log^2 n)\) serial time. Unlike serial computing, we can take on more work if the span of our computation DAG is wider in parallel settings. Here, speed is proper hardware-efficient vectorization.
The GPU
The paper, refreshingly, reviews the GPU architecture.
Logical compute hierarchy is
grid -> block -> warp -> lane (thread)
Memory hierarchy is
main mem (vram) -> global l2 -> stream multiprocessor (SM) l1 + shared mem, going from multi-GB to multi-MB to about
16+48 KB.
There might be one or more blocks scheduled to a single streaming multiprocessor, which is itself a set of cores. Cores have their own floating point processing units and integer units, but other supporting units like the MMU-equivalent are shared.
My takeaways from this section were the usual “maximize the amount of work each core is doing independently, keeping compute density high and memory accesses low, especially shared memory”, but with two important twists:
GPU warps (gangs of threads) exhibit worse performance when the threads aren’t performing the same instructions on possibly different data ( warp divergence). Each thread is best kept dealing with the memory in its own lane (which typically is a slice of a 32-strided array that the block is processing with multiple warps in a higher granularity of parallelism), but there can be synchronization points through the register file which exchange memory between the threads.
Note there are 32 threads to a warp, we’ll see that come up.
FAISS = IVF + ADC
FAISS answers the question of “what are the closest database points to the query point” by constructing a 2-level tree. Database vectors are further compressed to make the tree smaller.
Given \(n\) database vectors, we cluster with k-means for the top level, getting about \(\sqrt{n}\) centroids. Then, at search time, we use exact search to find the closest centroids, and then among the closest centroid’s clusters we look for the closest clusters overall.
For a 2-level tree, a constant factor of \(\sqrt{n}\) is the optimal cluster size since then the exact search that we do is as small as possible at both levels of the tree.
Since it’s possible the point might be near multiple centroids, FAISS looks at the \(\tau\) closest centroids in the top level of the tree, and then searches all cluster members among the \(\tau\) clusters.
So the larger search occurs when looking at the second level.
Compression reduces I/O pressure as the second-level’s database vectors are loaded. Furthermore, the specific compression algorithm chosen for FAISS, Product Quantization (PQ) enables distance computation on the codes themselves! The code is computed on the
residual of the database vector \(\textbf{y}\) from its centroid \(q_1(\textbf{y})\).
The two-level tree format is the inverted file (IVF), which is essentially a list of records for the database vectors associated with each cluster.
ADC, or asymmetric distance computation, refers to the fact that we’re using the code of the database vector and calculating its distance from the exact query vector. This can be made symmetric by using a code for the query vector as well. We might do this because the coded distance computation can actually be faster than a usual Euclidean distance computation.
FAISS, the easy part
The above overview yields a simple algorithm.
Compute exact distances to top-level centroids Compute ADC in inverted list in probed centroids, generating essentially a list of pairs (index of probed database vector, approximate distance to query point) The smallest-\(\ell\) by the second pair item are extracted, for some \(\ell\) not much larger than \(k\). Then the top \(k\) among those is returned.
The meat of the paper is doing these steps quickly.
Fast ADC via PQ
Product Quantization (PQ) boils down to looking compressing subvectors independently. E.g., we might have a four-dimensional vector \(\textbf{y}=[1, 2, 3, 4]\). We quantize it with \(b=2\) factors as \([(1, 2), (3, 4)]\). Doing this for all our vectors yields \(b\) sets of smaller vectors. The FAISS paper denotes these subvectors as \(\textbf{y}^1=(1, 2), \textbf{y}^2=(3, 4)\).
We then cluster the \(b\) sets independently with 256 centroids. The centroids that these subvectors get assigned to might be \(q^1(\textbf{y}^1)=(1, 1), q^2(\textbf{y}^2)=(4, 4.5)\), which is where the lossy part of the compression comes in. On the plus side, we just encoded 4 floats with 2 bytes!
This compression technique is applied to the
residual of the database vectors for their centroids, meaning we have PQ dictionaries for each centroid.
The key insight here is that we can also break up our query vector \(\textbf{x}=[\textbf{x}^1, \cdots, \textbf{x}^b]\), and create distance lookup tables on the sub-vectors individually, so the distance to a database vector is just a sum of \(b\) looked-up values!
Top-k
OK, so now comes the hard part, we just did steps 1 and 2 really fast, and it’s clear those are super parallelizable algorithms, but how do we get the top (smallest) \(k\) items from the list?
Well, on a CPU, we’d implement this in a straightforward way. Use a max-heap of size \(k\), scan through our list of size \(n\), and then if the next element is smaller than the max of the heap or the heap has size less than \(k\), pop-and-insert or just insert, respectively, into the heap, yielding an \(O(n\log k)\) algorithm.
We could parallelize this \(p\) ways by chopping into \(n/p\)-sized chunks, getting \(k\)-max-heaps, and merging all the heaps, but the intrinsic algorithm does not parallelize well. This means this approach works well when you have lots of CPUs, but is not nearly compute-dense enough for tightly-packed GPU threads, 32 to a warp, where you need to do a lot more computation per byte (having each of those threads maintain its own heap results in a lot of data-dependent instruction divergence).
The alternative approach proposed by FAISS is:
Create an extremely parallel mergesort “Chunkify” the CPU algorithm, taking a big bite of the array at a given time, keeping a “messy max-heap” of a lot more than \(k\) (namely, \(k+32t\)) that includes everything the \(k\)-max-heap would. Every once in a while, do a full sort on the messy max-heap.
Squinting from a distance, this looks similar to the original algorithm, but the magic is in the “chunkification” which enables full use of the GPU.
Highly Parallel Mergesort
As mentioned, this innovation is essentially a serial \(O(n\log^2 n)\) in-place mergesort that has a high computational span.
The money is in the merge operation, which is based on Batcher’s bitonic bit sort. The invariant is that we maintain a list of sorted sequences (lexicographically).
First, we have one sequence of length at most \(n\) [trivially holds] Then, we have 2 sequences of length at most \(n/2\) 4 sequences length \(n/4\) Etc.
Each merge has \(\log n\) steps, where at each step we might have up to \(n\) swaps, but they are disjoint and can happen in parallel. The key is to see that these \(n\) independent swaps ensure lexicographic ordering among the sequences
This is the
odd-merge (Algorithm 1) in the paper. There’s additional logic for irregularly-sized lists to be merged. We’ll come back to this.
Once we have a parallel merge that requires logarithmic serial time, the usual merge sort (Algorithm 2), which itself has a recursion tree of logarithmic depth, results in a \(O(\log^2 n)\) serial time (or depth) algorithm, assuming infinite processors.
Chunkification
This leads to WarpSelect, which is the chunkification mentioned earlier. In essence, our messy max-heap is a combination (and thus superset) of:
The strict size \(k\) max-heap with the \(k\) lowest values seen so far. In fact, this is sorted when viewed as a 32-stride array. 32 thread queues, each maintained in sorted order.
So \(T_0^j\le T_i^j\) for \(i>0\) and \(T_0^j\ge W_{k-1}\) . So if an input is greater than any thread queue head, it can be safely ignored (weak bound).
On the fast path, the next 32 values are read in, and we do a SIMT (single instruction, multiple-thread) compare on each value assigned to each thread. A primitive instruction checks if any of the warp’s threads had a value below the cutoff of the max heap (if none did, we know for sure none of those 32 values are in the top \(k\) and can move on).
If there was a violation, after the per-lane insertion sort the thread heads might be smaller than they were before. Then we do a full sort of the messy heap, restoring the fact that the strict max-heap has the lowest \(k\) values so far.
At this point, it’s clear why we needed a merge sort, which is because the strict max-heap (“warp queue” in the image) is already sorted, so we can avoid re-sorting it by using a merge-based sorting algorithm. Finally, it’s worth pointing out that recasting the fully sorted messy heap into the thread queues maintains the sorted order within each lane. Further, it’s clear why FAISS authors created a homebrew merge algorithm that’s compatible with irregular merge sizes, as opposed to existing power-of-2 parallel merge algorithms: the thread queues are irregularly sized compared to \(k\) and it’d be a lot of overhead to round the array sizes
This leads to the question: why have thread queues at all? Why not make their size exactly 1?
This points to a convenient piece of slack, the thread queue length \(t\), which lets us trade off the cost of the full merge sort against the per-thread insertion sort done every time the new values are read in. The optimal choice depends on \(k\).
Results
Remember, it’s not apples to apples, because FAISS gets a GPU and modern methods use CPUs, but who cares.
Recall from the previous post the
R@1 metric is the average frequency the method actually returns the nearest neighbor (it mayhave the query \(k\) set higher). The different parameters used here don’t matter so much, but I’ll highlight what each row means individually.
HNSW is a modern competitor based on the CPU using an algorithm written 2 years after the paper. Flat is naive search. In this benchmark, the PQ optimization was not used (database vector distances were computed exactly).
Here, for the very large dataset, the authors do use compression (OPQ indicates a preparatory transformation for the compression).
On the whole, FAISS is still the winner since it can take advantage of hardware. On the CPUs, it’s still a contender when it comes to a memory-speed-accuracy tradeoff.
Extensions and Future Work
The authors of the original FAISS work have themselves looked into extensions that combine the FAISS approach with then newer graph-based neighborhood algorithms (Link and Code).
Other future work that the authors have since performed has been in improving the organization of the two-level tree structure. The centroid based approach of the IVF implicitly partitions the space with a Voronoi diagram. As the Inverted Multi-Index (IMI) paper explores, this results in a lot of unnecessary neighbors being probed that are far away from the query point but happen to belong to the same Vornoi cell. One extension that now exists in the code base is to use IMI instead of IVF.
It’s also fun to consider how these systems will be evolving over time. As memory bandwidth increases, single node approaches (like FAISS) grow increasingly viable since they can keep compute dense. However, as network speeds improve, distributed approaches with many, many CPUs look attractive. The latter types of algorithms rely more on hierarchy and less on vectorization and compute density. |
Prove that if $ab \equiv 0 \pmod p$, where p is a prime number, then $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$.
All I have right now is that the prime divisibility property may help with the then part of this problem.
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Prove that if $ab \equiv 0 \pmod p$, where p is a prime number, then $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$.
All I have right now is that the prime divisibility property may help with the then part of this problem.
The ideal $(p) \subset \mathbb Z$ is prime, thus if $ab \in (p)$, then $a \in (p)$ or $b \in (p)$.
In other words:
$ab \equiv 0 \pmod p \implies ab=pk \implies p|a$ or $p|b \implies a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$.
Remember that in the integers, for a prime $\;p\;\;,\;\;\;p|a\iff a=kp\;,\;\;a,k\in\Bbb Z\;$ , and then $\;a=0\pmod p\iff p|a\iff a=kp\;$ , so by what you wrote in your second comment below your question:
$$ab=0\pmod p\iff p|ab\iff p|a\;\;\text{or}\;\;p|b\iff $$
$$a=kp\;\;\text{or}\;\;b=mp\;,\;\;k,m\in\Bbb Z\iff a=0\pmod 0\;\;\text{or}\;\;b=0\pmod p$$ |
You can not compare the two models as they do not model the same variable (as you correctly recognise yourself). Nevertheless AIC should work when comparing both nested and nonnested models.
Just a reminder before we continue: a Gaussian log-likelihood is given by
$$ \log(L(\theta)) =-\frac{|D|}{2}\log(2\pi) -\frac{1}{2} \log(|K|) -\frac{1}{2}(x-\mu)^T K^{-1} (x-\mu), $$
$K$ being the covariance structure of your model, $|D|$ the number of points in your datasets, $\mu$ the mean response and $x$ your dependent variable.
More specifically AIC is calculated to be equal to $2k - 2 \log(L)$, where $k$ is the number of fixed effects in your model and $L$ your likelihood function [1]. It practically compares trade-off between variance ($2k$) and bias ($2\log(L)$) in your modelling assumptions. As such in your case it would compare two different log-likelihood structures when it came to the bias term. That is because when you calculate your log-likelihood practically you look at two terms: a fit term, denoted by $-\frac{1}{2}(x-\mu)^T K^{-1} (x-\mu)$, and a complexity penalization term, denoted by $-\frac{1}{2} \log(|K|)$. Therefore you see that your fit term is completely different between the two models; in the first case you compare the residuals from the raw data and in the other case the residuals of the logged data.
Aside Wikipedia, AIC is also defined to equate: $|D| \log\left(\frac{RSS}{|D|}\right) + 2k$ [3]; this form makes it even more obvious why different models with different dependent variable are not comparable. The RSS is the two case is just incomparable between the two.
Akaike's original paper [4] is actually quite hard to grasp (I think). It is based on KL divergence (difference between two distributions roughly speaking) and works its way on proving how you can approximate the unknown true distribution of your data and compare that to the distribution of the data your model assumes. That's why
"smaller AIC score is better"; you are closer to the approximate true distribution of your data.
So to bring it all together the obvious things to remember when using AIC are three [2,5] :
You can not use it to compare models of different data sets.
You should use the same response variables for all the candidate models.
You should have $|D| >> k$, because otherwise you do not get good asymptotic consistency.
Sorry to break the bad news to you but using AIC to show you are choosing one dependent variable over another is not a statistically sound thing to do. Check the distribution of your residuals in both models, if the logged data case has normally distributed residuals and the raw data case doesn't, you have all the justification you might ever need. You might also want to check if your raw data correspond to a lognormal, that might be enough of a justification also.
For strict mathematical assumptions the game is KL divergence and information theory...
Ah, and some references:
http://en.wikipedia.org/wiki/Akaike_information_criterion Akaike Information Criterion, Shuhua Hu, (Presentation p.17-18) Applied Multivariate Statistical Analysis, Johnson & Wichern, 6th Ed. (p. 386-387) A new look at the statistical model identification, H. Akaike, IEEE Transactions on Automatic Control 19 (6): 716–723 (1974) Model Selection Tutorial #1: Akaike’s Information Criterion, D. Schmidt and E. Makalic, (Presentation p.39) |
Let $\mathbb Q(\alpha)/\mathbb Q$ be an algebraic exntension with $\alpha=\sqrt{2+\sqrt{2}}$.
1) Show that the extension is a galois extension (normal and separable)
2) Show that $Gal(\mathbb Q(\alpha)/\mathbb Q)\cong \mathbb Z_4$
3) Find all fields between $\mathbb Q$ and $\mathbb Q(\alpha)$
My attempt:
1) One can see that $f(x)=(x^2-2)^2-2=X^4-4x^2+2$ is a polynomial with $f(\alpha)=0$. By Eisenstein's theorem with p=2 it is irreducible over $\mathbb Q$, hence its the minimal polynomial of $\alpha$.
The roots of $f$ are $z_{1,2}=\pm\sqrt{2+\sqrt{2}}$ and $z_{3,4}=\pm\sqrt{2-\sqrt{2}}$. So the splitting field of $f$ is $\mathbb Q(z_1,z_2,z_3,z_4)$.But $z_2,z_3$ and $z_4$ can be written in terms of $z_1$:
$z_2=0-z_1$, $z_3=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$, $z_4=-z_3$, hence we have actually $\mathbb Q(z_1,z_2,z_3,z_4)=\mathbb Q(z_1)=\mathbb Q(\alpha)$.
We have shown that $\mathbb Q(\alpha)$ is normal and separabel.
2) We have $[\mathbb Q(\alpha):\mathbb Q]=deg(f)=4$. So its sufficient to find 4 automorphisms with: 1 automorphism has order 2, 2 have order 4 and 1 has order 1.
$id: z_i \mapsto z_i$
$\tau_1: z_1 \mapsto z_3$
with $\tau_1(\sqrt{2})=\tau_1(\sqrt{2})\tau_1(\sqrt{2})-2$ and $\sqrt{2\pm\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2\mp\sqrt{2}}}$ all the other images are determined by the image of $z_1$, we get $z_2\mapsto z_4$, $z_3\mapsto z_2$, $z_4\mapsto z_1$
Ans similarly:
$\tau_2:z_1 \mapsto z_4$, $\tau_2:z_2 \mapsto z_3$, $\tau_2:z_3 \mapsto z_1$, $\tau_2:z_4 \mapsto z_2$
$\tau_3:z_1 \mapsto z_2$, $\tau_3:z_2 \mapsto z_1$, $\tau_3:z_3 \mapsto z_4$, $\tau_3:z_4 \mapsto z_3$
Because of $ord(id)=1, ord(\tau_{1,2})=4$ and $ord(\tau_3)=2$, we have $Gal(\mathbb Q(\alpha)/\mathbb Q)\cong \mathbb Z_2$
3) Only the generated subgroup of $\overline{2}\in \mathbb Z_2$ is not trivial. Its corresponding automorphism is $\tau_3$ which fixes $2+\sqrt{2}$ and $2-\sqrt{2}$, hence a non-trivial field is $\mathbb Q(\sqrt{2})$.
I hope that someone can go trough it and look for mistakes or tips. I am open for any other idea :)
Especially in 3) I am a bit unsure whether my solution is correct or not. |
Let $L$ be the language $\{w \mid w \text{ has equal numbers of \(a\)'s, \(b\)'s and \(c\)'s}\}$. Prove that its complement $\overline{L}$ is context free.
So let us remember the closure properties of context-free languages, gone over in class and stated by the TFS:
* closed under union* closed under concatenation* closed under Kleene Star
Now I will show that my language $L'$ is the same as the language $L$. Let $NE(x,y,z)$ be an operation that generates a language such that the number of x's is not the same as the number of y's, with as many z's as one wants anywhere in the language. Thus my language $L'$ is defined as:
$$ NE(a,b,c) \cup NE(c,a,b) \cup NE(b,c,a) $$
This language is obviously $L$. It is all instances of a not equal to b, b not equal to c, and c not equal to a. The nice thing here is that if we prove $NE()$ is context free, then because of symetry and closure under union operation. Let us represent NE as a PDA here:
$$ (s,\epsilon,\epsilon) \rightarrow (p,\$) $$ $$ (p,a,\epsilon) \rightarrow (p,1) $$ $$ (p,b,1) \rightarrow (p,\epsilon) $$ $$ (p,b,\$) \rightarrow (m,1\$) $$ $$ (p,c,\epsilon) \rightarrow (p,\epsilon) $$ $$ (m,c,\epsilon) \rightarrow (p,\epsilon) $$ $$ (m,b,\epsilon) \rightarrow (m,1) $$ $$ (m,a,1) \rightarrow (m,\epsilon) $$ $$ (m,a,\epsilon) \rightarrow (p,1\$) $$ $$ (p,\epsilon,a) \rightarrow (f,\epsilon) $$ $$ (p,\epsilon,b) \rightarrow (f,\epsilon) $$ $$ (m,\epsilon,a) \rightarrow (f,\epsilon) $$ $$ (m,\epsilon,b) \rightarrow (f,\epsilon) $$
So now that I have represented this language as a PDA, all I need to show is that the language I represented corresponds to $NE()$.
First I will show that my PDA, $X$ is $X \subseteq NE$. This is quite easy to see. The only way that one can enter a final state in $X$ is when, under one of our rules, we hit the end of the string, and there is something left on the stack (meaning that there is an unequal number of either a's or b's). we can see here that this is a subset.
Next I will show that $X \supseteq NE$, or in other words, all strings that are part of $NE()$ can be made by $X$. One can see that one can have as many c's as one wants in this model, by the rules: $ (p/m,c,\epsilon) \rightarrow (p/m,\epsilon) $, so that is not a problem. But how does it promise that all strings of unequal number of a's and b's are not in this language. Again, the grammar only keeps track of how many excess a's or b's there are, and can accept any string of arbitrary length with any permutation of letters as long as one of a or b, has an excess. Thus $X \supseteq NE$.
I have shown that $NE$ is Context Free by building a PDA, and I have shown that $L$ is a finite union of $NE$s, thus $L$ is context-free.
Hint: if $w \notin L$ then for some two characters $\alpha,\beta$, $w$ has more of $\alpha$ than of $\beta$. It is enough to show that each such language $L_{\alpha\beta}$ is context-free. Here the idea is to suitably modify a grammar for $\{ w : w \text{ has the same number of $\alpha$'s as $\beta$'s}\}$. |
I need help finding the radius of convergence for $\frac{x^n}{n\sqrt{n}12^n}$
I tried both the root and ratio tests but neither one is getting me anywhere.
Trying the root test I reduced to $\lim \limits_{n \to \infty}\left(\frac{x}{n^{1/2}n^{1/4}12}\right)$ which is either incorrect or not helpful.
I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence. |
I am responsible for assisting with certain error measurements for an FE program. The idea is to set up some benchmark comparisons which we can use for future development of the program.
To simplify the task I chose to look at a simple finite difference scheme for the equation of a parabola. Consider discrete values of $x_i$ where $\mathbb{x}$ is a vector:
$$\mathbb{x} = \left< \begin{matrix} 0&1&2&3&4&5&6&7\end{matrix}\right>$$
Now consider a mapping of $\mathbb{x}\rightarrow\mathbb{f}$ where $f_i = x_i^2$. It sounds fancy when I say it like that but its really just making a new vector with each value in it equalling the value $x_i^2$. $$\mathbb{f} = \left< \begin{matrix} 0&1&4&9&16&25&36&49\end{matrix}\right>$$
Plotting $\mathbb{f} $ versus $\mathbb{x}$ yields some sort of discretized parabola. The numerical derivative of this would be
$$\mathbb{df}_{num} = \left< \begin{matrix} 1&3&5&7&9&11&13&15\end{matrix}\right>$$
Using a very simple finite difference scheme of: $$df_i = \frac{f_{i+1}-f_i}{x_{i+1}-x_i}$$
It will also be noted that $\mathbb{df}_{num}$ contains one less term than $\mathbb{f}$. This, I believe, is inherent to finite differences.
The analytical counterpart for $\mathbb{df}$ which follows from $\frac {d }{d x}x^2 = 2x$ would be
$$\mathbb{df}_{a} = \left< \begin{matrix} 0&2&4&6&8&10&12&14\end{matrix}\right>$$.
Now subtracting the 2 vectors we get $\mathbb{f}_a-\mathbb{f}_{num} = \left< \begin{matrix} -1&-1&-1&-1&-1&-1&-1&-1\end{matrix}\right>$
Which is a sort of crude error measure.
What would be a better error measure? Something that would be simple to implement for things like this but precise enough that it can spot deviations efficiently in the context of a script? The idea is to have an error vector like this and compare each value in it with some sort of threshhold value, though we have no idea what this threshhold value would be... |
Abstract
In this paper, we investigate the simultaneous equations derived from the Golden Ratio, and use differentiation to find the coordinates where Φ in the x-axis meets Φ in the y-axis in a golden quadratic graph. We demonstrate a proof for the solutions to the five simultaneous equations below by using algebraic and graphic solutions.
Introduction
The Golden Ratio is a number represented by the Greek letter Φ and is approximately equal to 1.6180339887…; it is irrational and is infinitely long, similar to π. Nevertheless, Φ can be represented by a fraction (which contains √5, an irrational number). The fraction which represents Φ is
1 + 2 5 . Φ is an extremely interesting number; it appears everywhere in nature. The number is a ratio, the Golden Ratio, and it represents the “perfect cut” of nature. Everything in nature, from the number of pollen grains on a flower to the number of bonds in the compounds of uranium oxide, to the spiral shape of galaxies can be accurately modeled by Φ. Φ is a ratio, and it is often shown in nature as 1:Φ, such as the ratio of the distance between the shoulder and the elbow, and between the elbow and the fingertips. That is yet another Golden Ratio. However, if the larger section is considered to be 1, the smaller ratio can be Φ, known as the Lesser Golden Ratio.
In this paper, we aim to prove that the only possible solution to the simultaneous equations below is to use the Greater and Lesser Golden Ratios, therefore making the equations mathematically “beautiful.”
Most natural feats which contain the Golden Ratio are considered to be esthetically “beautiful.”
Simultaneous Equations
First, we introduce a set of simultaneous equations, which are at the center of this paper.[1]
[latex]x -1 = y[/latex] [latex]\frac{1}{x} = y\[/latex] [latex]n = x[/latex] [latex]n – 1 = x[/latex] [latex]2 + y = n[/latex]
Obviously, it is possible to rearrange these equations to create a longer string of equations such as: x -1=n – 2 = n − 1 = y, or n=2+y=x
2=x-1. However, these cannot give any numerical value for the constants x, y, and z. This is why, the quadratic formula must be used. The final answer that we want to reach is
[latex]x = \frac{1+\sqrt{5}}{2}[/latex]
therefore, we know that there must be a way of using ax2+bx+c=0. We shall then find the values of
a, b,
and
c which will give us the value of x by using the quadratic formula,
[latex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/latex]
The values of
a, b, and c must eventually make
[latex]x = \frac{1+\sqrt{5}}{2}[/latex]
which is Φ ≈ 1.6180339887… Therefore, we use the formula
[latex]\phi = \frac{1+\sqrt{5}}{2}[/latex]
and substitution to find the numerical values which would go into the formula as
a, b, and c. Algebraic Proof
Since we know that [latex]\phi = \frac{1+\sqrt{5}}{2}[/latex], 1=–b ∴ b=–1. 2
Then, [latex]\sqrt{5} = \sqrt{b^2-4ac}[/latex], but we know that
b = -1 ∴ [latex]sqrt{5} = \sqrt{1-4ac}[/latex] ∴ -1
2 = 1 ∴ -x 2 = x
At this stage, it is much simpler to work out
a because of the 2 in the Φ formula. It is evident that a=1 because 2=2a ∴ a=1. At this point, it becomes clear what the missing constant c is:
[latex]\sqrt{5} = \sqrt{1-4c} \therefore \sqrt{5}[/latex] -1∴ 4c = −4 ∴ c = −1
Another way to determine these constants is to use some of the simultaneous equations. Initially, we calculate:
[latex]x – 1 = \frac{1}{x} \therefore x(x-1) = 1 = x^{2} – x = 1[/latex]
Therefore, we have obtained values for
a, b, and c to use in the quadratic formula.
Now that we have all three of the constants needed for the quadratic formula to work, we must prove that it does;
[latex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/latex]
[latex]x = \frac{1\pm \sqrt{5}}{2}a = 1, b = -1, c=-1[/latex]
We now have x = ϕ ≈ 1 . 6180339887 … ∴ y = x – 1 ≈ 0.6180339887… ∴ n = x + 1 ≈ x + 1 ≈ 2.6180339887 and x = – φ, y = – ϕ, z = 0.382
Graphic Solution
Now that we have sorted out the equations
algebraically, we can also use graphs and differentiation to find the lowest possible point of the curve y = x2 – x – 1, which is derived from our
a, b, and c from the quadratic formula. On a graph we can sketch this curve to find the x-intersects and the minimum point (not maximum because it is quadratic and continues increasing to infinity).
From Figure 1, it is possible to say that the
x-intersects of the curve are Φ and –ϕ, the Golden Ratios.
Differentiation: y = x
2 − x − 1
[latex]\therefore (\frac{dy}{dx}(x^{n})=nx^{n-1}) \rightarrow \frac{dy}{dx} =\frac{d(x^{2}-x-1)}{dx}=2x-1=0[/latex]
[latex]\therefore x=\frac{1}{2} \therefore \frac{1^{2}}{2}-\frac{1}{2}-1=\frac{-5}{4}=y[/latex]
Therefore, we can see that the minimum point of this curve has the coordinates
[latex](\frac{1}{2}, -\frac{5}{4})[/latex]
Figure 2 shows a magnification of the graph at the point (ϕ, 0), showing the exact point of the Golden Ratio.[2]
References Olsen S. The Golden Section: Nature’s Greatest Secret. Wooden Books. Walker & Company; 2006. International Conference on Data Engineering. 16 thInternational Conference on Data Engineering: 29 February-3 March 2000. San Diego, California: Institute of Electrical & Electronics Engineers; 2000. About the Author
Gianamar Giovannetti-Singh is extremely interested in Mathematics and Physics. In March 2012, he was awarded the title of Junior UK Young Scientist of the Year for the National Science and Engineering Competition at the Big Bang Fair 2012 in Birmingham for his research on the orbital and physical dynamics of the Trojan asteroid 2010 TK7. Furthermore, he was awarded the CREST Prize for Enthusiasm and Real-World Context, and was selected as an international delegate to the United Kingdom at Broadcom MASTERS International at the Intel International Science and Engineering Fair; amongst the eighteen winners selected from across the world. He is currently taking part in research on possible new cancer therapies with the use of quantum tunneling to increase the chances of total ionization of tumors. He is 15 years old, and is looking forward to continue studying Mathematics with Physics at University. Gianamar Giovannetti-Singh has published two research papers in the Journal of Modern Physics, and in April 2012, he spoke at the IEEE International Conference on Electric, Information, and Control Engineering about one of his projects in the field of quantum mechanics. In July 2012, he will speak at the Biennial International Quantum Structures Association Meeting, whose proceedings will be published in the International Journal of Theoretical Physics about a predictive method for the annihilation tracks of subatomic particles. |
General case. In relativistic thermodynamics, inverse temperature $\beta^\mu$ is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the maximum entropy method, where $\beta^\mu p_\mu$ (in units where $c=1$) replaces the term $\beta H$ of the nonrelativistic canonical ensemble. This is done in
C.G. van Weert, Maximum entropy principle and relativistic hydrodynamics, Annals of Physics 140 (1982), 133-162.
for classical statistical mechanics and for quantum statistical mechanics in
T. Hayata et al., Relativistic hydrodynamics from quantum field theory on the basis of the generalized Gibbs ensemble method, Phys. Rev. D 92 (2015), 065008. https://arxiv.org/abs/1503.04535
For an extension to general relativity with spin see also
F. Becattini, Covariant statistical mechanics and the stress-energy tensor, Phys. Rev. Lett 108 (2012), 244502. https://arxiv.org/abs/1511.05439
Conservative case. One can define a scalar temperature $T:=1/k_B\sqrt{\beta^\mu\beta_\mu}$ and a velocity field $u^\mu:=k_BT\beta^\mu$ for the fluid; then $\beta^\mu=u^\mu/k_BT$, and the distribution function for an ideal fluid takes the form of a Jüttner distribution $e^{-u\cdot p/k_BT}$.
For an ideal fluid (i.e., assuming no dissipation, so that all conservation laws hold exacly), one obtains the format commonly used in relativistic hydrodynamics (see Chapter 22 in the book Misner, Thorne, Wheeler, Gravitation). It amounts to treating the thermodynamics nonrelativistically in the rest frame of the fluid.
Note that the definition of temperature consistent with the canonical ensemble needs a distribution of the form $e^{-\beta H - terms~ linear~ in~ p}$, conforming with the identification of the noncovariant $\beta^0$ as the inverse canonical temperature. Essentially, this is due to the frame dependence of the volume that enters the thermodynamics. This is in agreement with the noncovariant definition of temperature used by Planck and Einstein and was the generally agreed upon convention until at least 1968; cf. the discussion in
R. Balescu, Relativistic statistical thermodynamics, Physica 40 (1968), 309-338.
In contrast, the covariant Jüttner distribution has the form $e^{-u_0 H/k_BT - terms~ linear~ in~ p}$. Therefore the covariant scalar temperature differs from the canonical one by a velocity-dependent factor $u_0$. This explains the different transformation law. The covariant scalar temperature is simply the canonical temperature in the rest frame, turned covariant by redefinition.
Quantum general relativity. In quantum general relativity, accelerated observers interpret temperature differently. This is demonstrated for the vacuum state in Minkowski space by the Unruh effect, which is part of the thermodynamics of black holes. This seems inconsistent with the assumption of a covariant temperature.
Dissipative case. The situation is more complicated in the more realistic dissipative case. Once one allows for dissipation, amounting to going from Euler to Navier-Stokes in the nonrelativistic case, trying to generalize this simple formulation runs into problems. Thus it cannot be completely correct. In a gradient expansion at low order, the velocity field defined above from $\beta^\mu$ can be identified in the Landau-Lifschitz frame with the velocity field proportional to the energy current; see (86) in Hayata et al.. However, in general, this identification involves an approximation as there is no reason for these velocity fields to be exactly parallel; see, e.g.,
P. Van and T.S. Biró, First order and stable relativistic dissipative hydrodynamics, Physics Letters B 709 (2012), 106-110. https://arxiv.org/abs/1109.0985
There are various ways to patch the situation, starting from a kinetic description (valid for dilute gases only): The first reasonable formulation by Israel and Stewart based on a first order gradient expansion turned out to exhibit acausal behavior and not to be thermodynamically consistent. Extensions to second order (by Romatschke, e.g., https://arxiv.org/abs/0902.3663) or third order (by El et al., https://arxiv.org/abs/0907.4500) remedy the problems at low density, but shift the difficulties only to higher order terms (see Section 3.2 of Kovtun, https://arxiv.org/abs/1205.5040).
A causal and thermodynamically consistent formulation involving additional fields was given by Mueller and Ruggeri in their book Extended Thermodynamics 1993 and its 2nd edition, called Rational extended Thermodynamics 1998.
Paradoxes. Concerning the paradoxes mentioned in the original post:
Note that the formula $\langle E\rangle = \frac32 k_B T$ is valid only under very special circumstances (nonrelativistic ideal monatomic gas in its rest frame), and does not generalize. In general there is no simple relationship between temperature and velocity.
One can say that your paradox arises because in the three scenarios, three different concepts of temperature are used. What temperature is and how it transforms is a matter of convention, and the dominant convention changed some time after 1968; after Balescu's paper mentioned above, which shows that until 1963 it was universally defined as being frame-dependent. Today both conventions are alive, the frame-independent one being dominant.
This post imported from StackExchange Physics at 2016-06-24 15:03 (UTC), posted by SE-user Arnold Neumaier |
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ...
Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-06)
The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ...
Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC
(Springer, 2017-01)
The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ...
Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC
(Springer, 2017-06)
We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ... |
Like-Charged Particles at Liquid Interfaces
Entry by Haifei Zhang, AP 225, Fall 2009
Contents 1 Soft matter keywords 2 Overview 3 Electric-field-induced capillary attraction between like-charged particles at liquid interfaces 4 Soft matter details 5 References Soft matter keywords
Capillary attraction, Liquid interfaces, Colloidal particle
Overview
Nanometre- and micrometre-sized charged particles at aqueous interfaces are typically stabilized by a repulsive Coulomb interaction. If one of the phases forming the interface is a nonpolar substance (such as air or oil) that cannot sustain a charge, the particles will exhibit long-ranged dipolar repulsion; if the interface area is confined, mutual repulsion between the particles can induce ordering and even crystallization. However, particle ordering has also been observed in the absence of area confinement, suggesting that like-charged particles at interfaces can also experience attractive interactions. Interface deformations are known to cause capillary forces that attract neighbouring particles to each other, but a satisfying explanation for the origin of such distortions remains outstanding. Here we present quantitative measurements of attractive interactions between colloidal particles at an oil–water interface and show that the attraction can be explained by capillary forces that arise from a distortion of the interface shape that is due to electrostatic stresses caused by the particles’ dipolar field. This explanation, which is consistent with all reports on interfacial particle ordering so far, also suggests that the attractive interactions might be controllable: by tuning the polarity of one of the interfacial fluids, it should be possible to adjust the electrostatic stresses of the system and hence the interparticle attractions.
Electric-field-induced capillary attraction between like-charged particles at liquid interfaces The origin of the attraction The mismatch in dielectric constants of adjacent fluids can result in asymmetric charging of particles adsorbed at their interface; if one of the fluids is water, then the aqueous surface acquires a charge, which combines with the screening ions in the water to produce an effective dipole moment of the particle, leading to repulsion. By contrast, the origin of an attractive interaction is less well understood, even though interface deformation is known to give rise to capillary forces and a logarithmic attraction between neighbouring particles. For sufficiently large particles, the deformation is caused by gravity; the particle weight is balanced by surface tension, deforming the interface. This results in the clumping of breakfast cereals at the surface of a bowl of milk, and has been harnessed to self-assemble millimetre-sized particles at interfaces. However, the buoyancy mismatch of micrometre-sized colloidal particles is too small to deform the interface significantly and the resulting attractive energies are far smaller than thermal energy.
An alternative candidate for the origin of the attraction is wetting of the particles, which also deforms the interface; however, a spherical particle positions itself exactly to achieve the equilibrium contact angle without distorting the interface unless the particles are constrained to a thin layer of fluid. Another candidate for the origin of the attraction is surface roughness, which might also deform the interface; however, the roughness required is far greater than that which typically exists on colloidal particles. A final candidate for the origin of the attraction is thermal fluctuations; entropic interactions lower the free energy of the interface when two particles approach, resulting in a Casimir type of effective attraction. However, the resulting forces are too small to cause a significant attraction for colloid particles.
Quantify the attractive interaction
To investigate this question further, we quantified the attractive interaction between like-charged particles at a water–oil interface. A typical configuration of colloidal particles at the interface of a large water drop in oil is shown in Fig. 1. The long range of the repulsive interaction is apparent from the large particle separation.Moreover, this repulsion, combined with the geometric confinement resulting from the finite area of the emulsion drop, which is completely covered with particles, causes the pronounced ordering of the particles. But even when the particle coverage is not complete, ordering can still be observed, as illustrated by the inset in Fig. 2, which shows a group of seven particles in a hexagonal crystallite. These were the only particles on the surface of a large water drop, and the crystallite remained stable for more than 30min. The persistence of this structure over long times is clear evidence of a long-ranged attractive interaction (Fig. 3). To extract the interaction potential, we exploit the symmetry of the geometry and measure the probability distribution, P(r), of the centre-to-centre distance, r, of the centre particle to each of the outer particles. To obtain the interparticle potential V(r), we invert the Boltzmann distribution:
The potential has a minimum at an interparticle separation of <math>r_{eq}</math>=5.7 um, and is well described as harmonic with a spring constant of <math>k=23kBT um^{-2}</math> (Fig. 3).
Experimental observations
The motion of any given particle in the crystallite is influenced by the pair interactions with all the other particles; however, by analysing only the radial distance of the outer particles to the centre particle, these many-particle effects are largely cancelled by symmetry. This was confirmed by molecular dynamics simulation of seven particles in the same configuration, interacting with the same pair potential; an uncertainty of about 10–20% in <math>r_{eq}</math> and k was introduced by many-particle effects.
The electrical stresses arise because oil and water have very different dielectric constants: <math>\varepsilon _{oil} \approx 2</math> and <math>\varepsilon _{water} \approx 80</math>: When field lines cross the interface, the intensity of the electric field E and the electrostatic energy density are thus roughly 40 times smaller in water than in oil. The colloidal particle therefore behaves as if pulled into the water by an external force F, as shown schematically in Fig. 4. Thus the particles, which are the source of the electric fields, tend to be surrounded by water, lowering the total electrostatic energy. This effect is analogous to electrostriction and to electrowetting.
The force F can be calculated by the integral of the electric pressure over the free surface. Both the repulsive and the attractive components of the interaction potential depend strongly on the dipole moment, <math>P = (\sigma /\kappa )\pi a_w^2</math>. This is directly proportional to the zeta potential of the particle surface, which in turn sensitively depends on surface charge, surface chemistry, and the screening length, <math>\kappa ^{ - 1}</math>, set by the ion concentration in the water. Thus, for example, if the surface charge remains constant, small changes in <math>\kappa ^{ - 1}</math> change P a great deal, whereas if the surface charge varies and the zeta potential remains constant, there will be no variation of P with <math>\kappa ^{ - 1}</math>. The attractive interaction is sensitively dependent on P; thus, if the surface charge remains constant, the attraction becomes negligible compared to <math>k_BT</math> even for low concentrations of added salt, as shown in the Supplementary Information. To verify the sensitivity of the attraction on P, the measurements were repeated with 5mM NaCl added to the water; only the repulsive interaction remained and its range was reduced, leading to a correspondingly lower particle stability, as shown in the Supplementary Information. These results also confirm that the particles are charged on the aqueous side, rather than the oil side. More generally, the existence of a measurable attractive interaction depends on both P and <math>a_w</math>, and thus is sensitive to surface properties, charge and pH, as well as salt concentration.
Summary
Although the authors' theory does not fully account for the geometry of the wetted region, it nevertheless captures the essential physics: Dipolar electric fields induce surface charges that distort the interface; the dipolar interaction causes repulsion, while the interfacial distortion causes capillary attraction. This behaviour has broader implications for interfacial and colloid chemistry, because adsorption of charged particles at fluid interfaces is a common phenomenon in foods, drugs, oil recovery, and even biology.
Soft matter details The original paper
Nikolaides et al. propose that the puzzling attraction that occurs between micrometre-sized particles adsorbed at an aqueous interface is caused by a distortion of the liquid interface that is due to the dipolar electric field of the particles and which induces a capillary attraction. The authors argue that this effect cannot account for the observed attraction, on the fundamental grounds that it is inconsistent with force balance.
Different opinion raised
Megens et al. think the capillary deformation of the interface contributes a mere <math>1.8*10^{-5}</math> to the interaction potential, so it is insignificant thermodynamically. Thus, the mechanism proposed by Nikolaides et al. does not account for the observations, and the origin of the observed attraction remains enigmatic.
Nikolaides et al. propose that the puzzling attraction that occurs between micrometre-sized particles adsorbed at an aqueous interface is caused by a distortion of the liquid interface that is due to the dipolar electric field of the particles and which induces a capillary attraction. The authors argue that this effect cannot account for the observed attraction, on the fundamental grounds that it is inconsistent with force balance.
To estimate the influence of the surface deformation, the authors assume that the sum total of the electrostatic pressure acting on the liquid interface is equivalent to an external force, <math>F</math>, pushing the particle into the water. The resulting deformation of the interface would give rise to a long-range interparticle interaction energy <math>U(r) = (F^2 /2\pi \gamma )\ln (r/r_0 )</math> where<math>\gamma</math> is the surface tension, <math>r</math> is the distance between particles, and <math>r_0</math> is an arbitrary constant. However, the electrostatic force acts on the particle and the liquid interface simultaneously, so the equation does not apply. The force F on the sphere is balanced by surface tension, creating a dimple in the water surface. The shape of the dimple is governed by the Young–Laplace equation <math>[(1/R_1 ) + (1/R_2 )]\gamma = \Delta p</math> If the force acts only on the particle, then the pressure difference across the interface <math>\Delta p \approx 0</math> and the radii of curvature <math>R_1</math> and <math>R_2</math> of the surface are equal but opposite. The resulting water level around an isolated sphere would be <math>h(r) = F/(2\pi \gamma )\ln (r/r_0 )</math> for small surface slopes.
Owing to the long range of the logarithm, the force on the sphere is passed on to the rim of the vessel containing the liquid. The situation changes fundamentally when the force F is not external but originates from a surface pressure. Now the meniscus profile is very short range. The electrostatic force F pushing the sphere into the liquid is balanced by the liquid interface and there is no force on the rim of the vessel. In contrast, by unquestioningly applying equation 1, one implicitly admits that there is also a force on the vessel, a force not balanced by any other force. This is inconsistent with Newton's third law. Capillary attraction between spheres is caused by the overlap of their dimples, which reduces the total surface area of the water. The resulting attraction energy is <math>U(r) = - (F^2 /\pi \gamma )(r_c /r)^6 </math> for large r. this interaction has a much shorter range than that shown in equation (1), even a shorter range than the dipole-dipole repulsion between like-charged particles that is proportional to 1/r^3, so overall there is no attraction at all. The capillary deformation of the interface contributes a mere <math> 1.8*10^{ - 5} </math> to the interaction potential, so it is insignificant thermodynamically. Thus, the mechanism proposed by Nikolaides et al. does not account for the observations.
The authors think that their experiment provides a clear measure of the interactions between charged particles at fluid-fluid interfaces and demonstrates that there can be a long-range attractive interaction between such particles. However, Megens and Aizenberg raise important points about the authors' interpretation of these results. The authors'calculations account for the electrostatic stresses acting on the fluid–fluid interface, but neglect the force that the electric field exerts on the particle itself. A detailed evaluation of this force, obtained by calculating the change in electrostatic energy as the particle is pushed into the fluid, shows that the interfacial force pulling the particle out of the fluid is exactly cancelled by the electrical force pushing the particle into the fluid, in agreement with the suggestion of Megens and Aizenberg.
This creates a puzzle: the data show unambiguously that there is a long-range attraction, but what is its origin? The particles have a long-range repulsive interaction, owing to their charges; this is dipolar in character, with the electrostatic energy decaying as <math>1/r^3</math>. The attractive interaction must balance this electrostatic repulsion, so, if it decays as a power law, the attractive interaction energy must decay more slowly than <math>1/r^3</math> to create the stable energy minimum observed experimentally. This eliminates possibilities such as asymmetries in the contact line or fluctuation-induced forces, all of which have a power-law decay that is more rapid than <math>1/r^3</math> . The most likely interaction that has sufficient range therefore remains capillary distortion of the interface.
The authors have since confirmed that the particles do have a measurable charge, even when immersed in oil.Measurements of the electrophoretic mobility and pair correlation functions of these particles indicate that their charge can be as high as 200e in oil. In the absence of additional solubilized charges, this would result in a screening length of the order of 20um for the particle volume fraction of about <math>10^{-4}</math> used in the authors' experiments. This is larger than the particle separation, allowing the force imbalance between the particle and the interface to persist far enough for significant interfacial distortion to exist at scales comparable to the interparticle separation. The authors therefore believe that electric-field-induced capillary distortion remains the likely culprit for the attractive interactions between like-charged interfacial particles.
References
[1] “Like-Charged Particles at Liquid Interfaces”, M. Megens, J. Aizenberg, Nature, 2003, 424, 1014.
[2] Nikolaides, M. G. et al. Electric-field-induced capillary attraction between like-charged particles at liquid interfaces, Nature 420, 299–301 (2002). |
I tried to rewrite the Schnorr signature algorithm for elliptic curves but I wanted to be sure to have not done any errors. So I would be very happy if someone could look over this algorithm and tell me if I have done anything wrong, or not precise enough:
Let $E$ be an elliptic curve over a finite field $\mathbb{F}_q$ (first question, is it here as in the ECDSA, that either $q=p$ an odd prime or $q=2^m$?) with parameters such that $E$ is a cryptographically safe curve.
Key Generation:
Choose $P\in E(\mathbb{F}_q)$ of prime order $l$, where $l$ is a large prime.
Choose $1 < a < l$ random and calculate $Q = a*P$. Then the public information is $E, \mathbb{F}_q, Q, P$ and the private signature key is $a$.
Signature Scheme: Choose a random $1 \leq k < l$ Compute $S_0 = kP = (x_0,y_0)$ Compute $s_1 = H(m||x_0)$, where $H$ is a hash function, $x_0$ is the integer value of $x_0$ and $m$ is the message. Compute $s_2 \equiv k + a*s_1 \text{ (mod }l)$
The digital signature is $(S_0,s_2)$ and Alice sends $(m, (S_0,s_2))$ to Bob.
Verification
Bob verifies if $s_2*P=S_0+H(m||x_0)*Q$.
This works since: $$s_2*P=S_0+H(m||x_0)*Q$$ $$\Leftrightarrow s_2*P-H(m||x_0)*Q = S_0$$ $$\Leftrightarrow (k+a*s_1)-H(m||x_0)*a*P = S_0$$ $$\Leftrightarrow kP + a*H(m||x_0)*P-a*H(m||x_0)P=S_0=k*P$$ |
Differential and Integral Equations Differential Integral Equations Volume 13, Number 10-12 (2000), 1429-1444. Fredholm properties of Schrödinger operators in $L^P(\mathbbR^N)$ Abstract
We consider real potentials $V$ such that the Schrödinger operator $-\Delta+V$ maps the Sobolev space $W^{2,p}(\mathbb{R}^{N})$ continuously into $L^{p}(\mathbb{R}^{N})$ for a range of values of $p$ which includes 2. Let $\sigma_{e}$ denote the essential spectrum of $-\Delta+V$ as a self-adjoint operator in $L^{2}(\mathbb{R}^{N}).$ If $\lambda\notin$ $\sigma_{e},$ we show that for all $p$ in the range considered, $-\Delta+V-\lambda:W^{2,p}% (\mathbb{R}^{N})\rightarrow L^{p}(\mathbb{R}^{N})$ is a Fredholm operator of index zero, that ker {$-\Delta+V-\lambda$\} is independent of $p$ and that $L^{p}(\mathbb{R}^{N})=$ker {$-\Delta+V-\lambda$\}$\oplus$\{$-\Delta +V-\lambda$\}$W^{2,p}(\mathbb{R}^{N}).$
Article information Source Differential Integral Equations, Volume 13, Number 10-12 (2000), 1429-1444. Dates First available in Project Euclid: 21 December 2012 Permanent link to this document https://projecteuclid.org/euclid.die/1356061133 Mathematical Reviews number (MathSciNet) MR1787075 Zentralblatt MATH identifier 0989.47036 Subjects Primary: 47F05: Partial differential operators [See also 35Pxx, 58Jxx] (should also be assigned at least one other classification number in section 47) Secondary: 35J10: Schrödinger operator [See also 35Pxx] 35Q40: PDEs in connection with quantum mechanics Citation
Rabier, P. J.; Stuart, C. A. Fredholm properties of Schrödinger operators in $L^P(\mathbbR^N)$. Differential Integral Equations 13 (2000), no. 10-12, 1429--1444. https://projecteuclid.org/euclid.die/1356061133 |
This problem in my linear algebra class is intended to demonstrate that all polynomial parametric curves (in $\mathbb{R^2}$) can fulfill a polynomial Cartesian equation. First, we let $x = p(t)$ and $y = q(t)$ where $p,q \in \mathcal{P}\mathbb{(R)}$ are fixed polynomials dependent on the variable $t$.
Part (a) wants us to find a function, $L$, that takes nonnegative integers $(m,n)$ and returns a nonnegative integer $L(m,n)$ so that if $0 \leq i \leq m$ and $0 \leq j\leq n$, then $x^iy^j = p(t)^iq(t)^j \in \mathcal{P}_{L(m,n)} (\mathbb{R})$. This choice of $L$ also depends on $p$ and $q$ in a way.
Part (b) wants us to choose $m$ and $n$ from above so that the number of monomials $x^iy^j$ with $0 \leq i \leq m$ and $0\leq j \leq n$ exceeds the dimension of $\mathcal{P}_{L(m,n)}(\mathbb{R})$.
Part (c) wants us to show that there exists a nonzero two-variable polynomial $f$ in which $f(x,y) = f(p(t),q(t)) = 0$. That is, $f(x,y)$ in the form $f(x,y) = \sum_{i,j} c_{i,j}x^iy^j (c_{i,j} \in \mathbb{R})$ where the sum is finite.
My attempt at a solution:
First, I called the degree of $p(t)= a$, and the degree of $q(t) = b$. For part (a), I created a function which would, given $m,n$, return a number that would mark the highest possible order of polynomial that could result from $p(t)^iq(t)^j$. So, I let $$L(m,n) = am + bn $$ which comes from exponentiating polynomials $p(t)$ and $q(t)$ and adding exponents for the final polynomial product. All polynomials of the form $p(t)^iq(t)^j$ are now in $\mathcal{P}_{L(m,n)} (\mathbb{R})$.
For part (b), I deduced that the number of possible monomials $x^iy^j$ is $(m+1)(n+1)$ since $0 \leq i\leq m$ and $0 \leq j \leq n$. In addition, dim$\mathcal{P}_{L(m,n)}(\mathbb{R}) = L(m,n) + 1$, or, for my function, $am + bn + 1$. So, I have to come up with $m,n$ such that $(m+1)(n+1) > am + bn + 1$, but this boils down to $mn + m + n > am + bn$. I tried various combinations of letting $m,n$ be $0, 1, a,$ and $b$, but could not satisfy the inequality.
Is there something I'm misinterpreting with the problem, or is my function a poor choice in satisfying the conditions in the problem? What type of function would satisfy the conditions?
I was not able to move on to part (c), but I'm not sure how, once (b) is satisfied, one could deduce the conclusion that part(c) wants us to show. What does the result of part(b) indicate? Does it have to do with the number of polynomials in the basis of $\mathcal{P}_{L(m,n)}(\mathbb{R})$ exceeding the dimension and creating a contradiction or something? Why is it important that $f(x,y) = 0$? |
As the comments suggest there can be confusion over what you call "GI". But the idea here is correct. It is polynomial-time equivalent to find generators of an automorphism group as it is to find an isomorphism between two groups. The idea is "classical" in that it appears in early work such as Luks' Group Isomorphism in bounded valence is in polynomial-time, and even there I think the idea was considered "well-known".
Claim. Let $G$ and $H$ be connected graphs. Then $G\cong H$ if, and only if, every generating set $S$ of $\mathrm{Aut}(G\sqcup H)$ contains an element $g\in S$ such that $G^g=H$.
Remark Important here is that every generating set exchange the graphs as otherwise you would sometimes compute generators that don't solve the problem. So for example, isomorphism of two groups does not so easily yield in this way. That is because not all generating sets of $\mathrm{Aut}(G\times H)$ will interchange $G$ and $H$ when $G\cong H$. instead they can go to diagonal copies. That situation can be fixed, but it requires a stronger argument. So the approach here is not one that applies in all categories.
Proof. For the converse if every (or even if one) generating set of $\mathrm{Aut}(G\sqcup H)$ interchanges $G$ and $H$ then $G\cong H$ by the restriction of that function to $G$. So this is all about the forward direction. (But I mention this because the proof is by contrapositive so it can look like I'm about to go the same direction.)
Suppose $\mathrm{Aut}(G\sqcup H)$ is generated by a set $S$ all of whose elements send $G$ to $G$, and $H$ to $H$, (note by connectivity assumption if one vertex of $G$ is sent to one vertex of $H$ then the entire graph $G$ is sent to $H$ and so by pigeon hole some vertex in $H$ will be sent to $G$ and so $|G|=|H|$ and we will have interchanged the two graphs). Since $S$ sends $G$ to $G$, then every composition of functions in $S$ sends $G$ to $G$, and so do inverses of these functions. So every word in $S$ sends $G$ to $G$ (and also $H$ to $H$). So no element of $\mathrm{Aut}(G\sqcup H)$ interchanges $G$ and $H$.
Finally if $G\cong H$ then an isomorphism $\phi:G\to H$ affords an automorphism $\phi\sqcup \phi^{-1}$ of $G\sqcup H$. So the absence of elements in $\mathrm{Aut}(G\sqcup H)$ to interchange $G$ and $H$ implies $G\not\cong H$. The result follows. $\Box$
But now the point to be clear on is that going from decision (Is $G\cong H$?) to search (Give me $\phi:G\to H$ or a certificate that $G\not\cong H$) has still to be argued (and can be). Also from one isomorphism to generators of automorphisms is another argument (individualize the graphs and repeat the isomorphism test). So all told you have a couple pages of argument to make these equivalences. None will show canonical labeling though. That is much harder (NP-hard if I recall). Even though NAutY and Traces handle many examples quickly. |
Recently I stumbled across this language $L=\{a^{n}a^{{(n + 1)}^2-n^2} \in \Sigma^* \mid n\geq 0\}$ that I can rewrite as $a^{3n + 1}$.
So I applied the pumping lemma to see if it is non regular, and I get:
given a pumping length pI should have a string $z\in L$ such that $\mid z \mid \ge p$ I choose $z=a^pa^pa^pa$ such that $z\in L$ and $\mid z \mid = (3p + 1)\ge p$ now I can slipt the string in 3 parts, $z=uvw$ with $u=a^{p-1}, v=a,w=a^{2p}a$ and it satisfies the two following conditions of the lemma $\mid uv \mid \leq p$ and $\mid v \mid > 0$ then given $z = uv^iw$, if I pump $v$ I see that $z \notin L$ because if for example I choose $i=2$ than $\mid z \mid = (p-1)+i+2p+1=3p+2 \geq 3p+1$
So $z \notin L$ and it means that $L$ is not regular, but I noticed that I can construct a DFA that accepts this language:
So now I'm really confused. Probably I'm applying the pumping lemma in the wrong way but I don't understand where.
UPDATE
Now, thanks to all of you, I think I understood my mistake. So to prove that a language is not regular I have to show that every decomposition fails to satisfy the pumping condition for some pumping number $i$. But in this case there is a decomposition that satisfies the pumping condition so I can't prove that the language is regular (but we know it is regular because we can construct a DFA that accepts it). |
Polya-Hurwitz program.
This may become more interesting in light of the recent progress in the
Polya-Jensen program by Griffin, Ono, Rolen, Zagier.
We will first provide definitions of some functions involved.
The Riemann Xi-function $\Xi(z)$ is related to the Riemann zeta-function $\zeta(s)$ via ([A], [B]): $\Xi(z)=\xi(\tfrac{1}{2}+iz)$,$\xi(s)=\tfrac{1}{2}s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)$.
Riemann $\Xi(z)$ can be expressed as a Fourier transform of a positive, fast decaying, and even kernel $\Phi(t)$ ([A], [B]):
\begin{equation}\Xi(z)=2\int_0^{\infty}\Phi(t)\cos(zt)\mathrm{d}t,\tag{1}\end{equation}
where
\begin{equation}\Phi(t)=\sum_{k\geqslant 1}\phi_k(t)=\Phi(-t),\tag{2}\\\end{equation}
\begin{equation}\phi_k(t)=\left(4\pi^2 k^4 e^{9t/2}-6\pi k^2e^{5t/2}\right)\exp\left(-\pi k^2 e^{2t}\right)\tag{3}.\end{equation}
The Polya aspect of this approach is the following:
Truncate the Kernel $\Phi(t)$ of (2) and/or the integration range in (1) such that the resulting Fourier transform leads to a family of entire functions which only have real roots.
One such candidate is given in [C]:\begin{equation}\Phi_{\color{red}n}(t)=(1/2)\sum_{1\leqslant k\leqslant {\color{red}n}}\left(\phi_k(t)+\phi_k(-t)\right)=\Phi_{\color{red}n}(-t)\tag{4}\end{equation}
\begin{equation}\Xi_{\color{red}n}(z)=2\int_0^{(1/2)\log {\color{red}n}}\Phi_{\color{red}n}(t)\cos(zt)\mathrm{d}t=\Xi_{\color{red}n}(-z),\tag{5}\end{equation}
We refer to [D] and [E] for a near complete review on the zeros of entire functions as Fourier transforms.
The Hurwitz aspect of this approach is the following:
Corollary of Hurwitz's theorem in complex analysis (applied to our case) [F]:
If $\Xi(z)$ and $\{\Xi_n(z)\}$ are analytic functions on a domain $S_{1/2}(z)=\{z: 0<Im(z)<1/2\}$, $\{\Xi_n(z)\}$ converges to $\Xi(z)$ uniformly on compact subsets of $S_{1/2}(z)$, and all but finitely many $\Xi_n(z)$ have no zeros in $S_{1/2}(z)$, then either $\Xi(z)$ is identically zero or $\Xi(z)$ has no zeros in $S_{1/2}(z)$.
The functional equation for $\zeta(s)$ becomes $\Xi(-z)=\Xi(z)$. The candidate of $\Xi_n(z)$ in (5) automatically satisfies this functional equation.
Another benefit of Polya-Hurwitz approach is that the entrance barrier is relatively low (comparing to other approaches that usually require the advance knowledge of analytical number theory).
To get started, one only needs to know Fourier transform, basic complex analysis, some knowledge of entire functions, polynomials etc. So anyone who has math training with the college undergraduate math major may start to work on the Polya-Hurwitz approach and learn other necessary new math as he/she goes.
The most difficult part of Polya-Hurwitz approach seems to be the following:(for example,) proving that all the zeros of $\Xi_n(z)$ in (5) are real in $S_{1/2}(z)$.
One may need to have several iterations: guess one form of the Kernel like $\Phi_n(1,t)$ and complete the integration to get explicit expression for $\Xi_n(1,z)$. If all the zeros of $\Xi_n(1,z)$ are found not to be all real in $S_{1/2}(z)$, then move on to $\Phi_n(2,t)$ and $\Xi_n(2,z)$...
References:
[A] Titchmarsh,"The Theory of the Riemann Zeta-Function",
(1986).
[B] Edwards, "Riemann's Zeta Function", (1974).
[C] Shi, "On the zeros of Riemann Xi-function", (2017) arXiv:1706.08868.
[D] Dimitrov and Rusev, “ZEROS OF ENTIRE FOURIER TRANSFORMS” (2001), 108 page review paper.
[E] Hallum, “ZEROS OF ENTIRE FUNCTIONS REPRESENTED BY FOURIER TRANSFORMS” (2014), Master thesis.
[F] Conway, "Functions of One Complex Variable",(1978) |
I am trying to use a projection method that deals with the viscous effects implicitly to model flow around a cylinder. I'm having trouble figuring out what the boundary conditions should be, particularly on the inflow and outflow.
I think we can consider the Stokes equations without loss of generality: $$ \mathbf{u}_t = \Delta \mathbf{u} - \nabla p $$ $$ \nabla \cdot\mathbf{u} = 0$$
If we discretize explicitly in time (ignoring the pressure term) we get: $$ \frac{\mathbf{u}^{*} - \mathbf{u}^n}{\delta} = \Delta\mathbf{u}^{n}$$ This leads to: $$ \mathbf{u}^{n+1}=\mathbf{u}^* - \delta\nabla p^{n+1}$$ Taking the divergence of this equation yields the pressure Poisson equation. The boundary conditions can be found by dotting with the normal: $$\nabla p^{n+1} \cdot\mathbf{n} = \frac{(\mathbf{u}^* - \mathbf{u}^{n+1})}{\delta}\cdot\mathbf{n}$$ where $\mathbf{u}^*\cdot\mathbf{n}$ can be computed and $\mathbf{u}^{n+1}\cdot\mathbf{n}$ is given as a boundary condition.
Now if we discretize implicitly in time we get: $$ \frac{\mathbf{u}^* - \mathbf{u}^n}{\delta} = \Delta \mathbf{u}^{*}$$ This is the diffusion equation for $\mathbf{u}^*$, and requires boundary conditions on $\mathbf{u}^*$. The most obvious choice is setting $\mathbf{u}^* = \mathbf{u}^{n+1}$ on the boundary. However if we do that, for the pressure Poisson equation we end up with $\nabla p\cdot\mathbf{n} = 0$ everywhere (even at the inflow and outflow). This seems incorrect to me.
What should the correct boundary conditions on the pressure Poisson equation be in this case? |
Suppose we have equation of the form:
$$H \Psi = E \Psi $$
where $H$ is Dirac Hamiltonian (also my question can be answered by people who are not familiar with Dirac Hamiltonian but familiar with numerical solution of different eigenvalue problems):
$$ H = \vec{\alpha} \vec{\nabla} + V(r) = H_0 + V(r)$$
and $\vec{\alpha}$ and $\beta$ are $4 \times 4$ Dirac matrices:
$$\vec{\alpha} = \left( \begin{matrix} 0 & \vec{\sigma} \\ -\vec{\sigma} & 0 \end{matrix} \right) \ , \ \ \beta = \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)$$
$\vec{\sigma}$ are $2 \times 2$ Pauli matrices.
Here $H_0$ is free Dirac Hamiltonian (without potential, only kinetic term is preserved).
I think that the best way to start solving such an equation numerically is to introduce plane-wave basis (which diagonalizes free Hamiltonian $H_0$):
$$\Psi = \sum_n c_n \psi_n $$
Then we have a problem like:
$$H \sum_n c_n \psi_n = E \sum_n c_n \psi_n $$
My question is: what are the most commonly used ways to solve this equation numerically for arbitrary potential? Just to compute matrix elements $H_{nm}$ in this plane-wave basis, to find eigenvalue $E$ and coefficients $C_n$ from this computation? Or maybe there are some better methods? |
One quick way to think about it is that
you're transposing the entire problem. Imagine placing the primal coefficients in a matrix, with the objective at the bottom. This gives $P = \begin{bmatrix} 1 & -6 & 2 \\ 5 & 7 & -4 \\ 8 & 3 & \end{bmatrix}$. Transposing yields $P^T = D = \begin{bmatrix} 1 & 5 & 8 \\ -6 & 7 & 3 \\ 2 & -4 & \end{bmatrix}.$ Thus the dual (coefficients and variables only) looks like $$
\begin{align}
\text{max or min } &2w_1 - 4w_2 \\
\text{subject to } &w_1 + 5w_2 \text{ ? } 8\\
&-6w_1 + 7w_2 \text{ ? } 3.
\end{align}
$$Now we have to figure out whether we are maximizing or minimizing, what should go where the ?'s are, and whether the dual variables are nonnegative, nonpositive, or unrestricted. The table tells us all of this.
Since the primal problem is maximizing, the dual is minimizing. Everything else can be read by pairing rows with columns in the transposition: Row $x$ in $P$ goes with Column $x$ in $D$, and Column $y$ in $P$ goes with Row $y$ in $D$. Then the table says...
Row 1 in $P$ is a $\geq$ constraint in a maximization problem $\Rightarrow$ The variable associated with Column 1 in $D$, $w_1$, has $w_1 \leq 0$. Row 2 in $P$ is a $=$ constraint $\Rightarrow$ $w_2$ is unrestricted. Column 1 in $P$ is for $x_1$, and $x_1 \leq 0$ $\Rightarrow$ The constraint associated with Row 1 in $D$ is a $\leq$ constraint. Column 2 in $P$, with $x_2 \geq 0$ $\Rightarrow$ The second constraint in the dual is a $\geq$ constraint.
Thus we get the complete form of the dual$$
\begin{align}
\min &2w_1 - 4w_2 \\
\text{subject to } &w_1 + 5w_2 \leq 8\\
&-6w_1 + 7w_2 \geq 3 \\
&w_1 \leq 0 \\
&w_2 \text{ unrestricted.}
\end{align}
$$
For remembering how to do this, I prefer something called the "SOB" method instead of memorizing the table. "SOB" here stands for "sensible," "odd," or "bizarre." The SOB table is the following:
Variables Constraints, Maximizing Constraints, Minimizing
Sensible ≥ 0 ≤ ≥
Odd Unrestricted = =
Bizarre ≤ 0 ≥ ≤
Hopefully it makes sense why these are "sensible," "odd," and "bizarre," at least relatively speaking.
The idea then is that
sensible maps to sensible, odd maps to odd, and bizarre maps to bizarre when you're switching from the primal to the dual. Let's take the example problem. It's "bizarre" to have a $\geq$ constraint in a maximization problem, so the variable in the dual associated with the first constraint, $w_1$, must have the "bizarre" nonpositivity restriction $w_1 \leq 0$. It's "odd" to have an equality constraint, and so the variable in the dual associated with the second constraint, $w_2$, must have the "odd" unrestricted property. Then, it's "bizarre" to have a variable $x_1$ with a nonpositivity restriction, so the constraint in the dual associated with $x_1$ must have the "bizarre" $\leq$ constraint for a minimization problem. Finally, it's "sensible" to have a variable $x_2$ with a nonnegativity restriction, so the constraint in the dual associated with $x_2$ must have the "sensible" $\geq$ constraint for a minimization problem.
I've found that after a few practice examples my students generally internalize the SOB method well enough that they can construct duals without needing to memorize anything. |
Let $m$ denote
pole strength. In the diagrams:
(1) Sky blue: Closed Gaussian surface (2) Red: North pole of magnet (3) Green: South pole of magnet (4) Yellow: Part of magnet cutting Gaussian surface
Case 1: When both poles lie inside Gaussian surface
$$\iint_S \vec{B}.\vec{dS}=4 \pi\ m+4 \pi\ (-m)=0$$
Case 2: When one pole lies inside Gaussian surface and other outside
$$\iint_S \vec{B}.\vec{dS}= 4 \pi\left( \iint_S \vec{H}.\vec{dS}+\iint_S \vec{M}.\vec{dS} \right)=4 \pi\ m+4 \pi\ (-m)=0$$
Case 3: When one pole lies on the Gaussian surface and other inside
Due to the inverse square nature of magnetic field intensity , $4 \pi\ \vec{H}$ at a point on the north pole due to that pole must be infinite. Therefore flux due to positive pole must be infinite. On the other hand flux due to negative pole is finite ($-4 \pi\ m$). Hence net flux must be infinite.
But it should be zero. Where have I gone wrong? |
Consider the BVP: find $u = u(x)$, for $x \in (0,1)$ that satisfies \begin{align} u'' + u u' = f, \\ u'(0) = g_n, u(1) = g_d. \end{align} To derive the weak form for this BVP, we multiply the first equation by a suitably smooth test function $\Phi = \Phi(x)$ and integrate both sides. This leads to \begin{align} - \int_\Omega \Phi ' u' + \int_\Omega \Phi u u' = \int_\Omega \Phi f + \Phi(0) g_n - \Phi(1) u'(1). \end{align} From this equation, we choose our test function space $\mathscr{T}$ to be a subspace of the Sobolev space $H^1(\Omega)$ that eliminate that $\Phi(1) u'(1)$ term, i.e., \begin{align} \mathscr{T} = \{w: w \in H^1(\Omega), w(1) = 0 \}. \end{align} Furthermore, we choose our set of trial functions $\mathscr{S}$ to be a subset of $H^1(\Omega)$ that satisfy the essential boundary condition, i.e., \begin{align} \mathscr{S} = \{v:v \in H^1(\Omega), v(1) = g_d \}. \end{align}
We then consider the WP associated with the original BVP: find $u \in \mathscr{S}$ such that \begin{align} -\int_\Omega \Phi'u'+\int_\Omega \Phi u u' = \int_\Omega \Phi f + \Phi(0)g_n, \qquad \forall \Phi \in \mathscr{T}. \end{align}
For Galerkin methods, we fix a positive integer $n$ and determine an n-dimensional subspace of $\mathscr{T}$ by specifying a basis, i.e., \begin{align} \mathscr{T}_n = \text{span}\{\phi_1, ..., \phi_n\} \subset\mathscr{T}. \end{align}
For Bubnov-Galerkin, we determine an n-dimensional subset of $\mathscr{S}$ by taking $\mathscr{S}_n = \mathscr{T}_n \cup \{g(x)\}$, where $g(x) = g_d$ handles the essential boundary. Our ``finite element solution' can then be written $u_h(x) = \sum_{j = 1}^n u_j \phi_j(x) + g(x)$.
Substituting $u_h$ into the weak equation should lead to a nonlinear system of equations ($\forall \Phi \in \mathscr{T}$ becomes $\forall \phi_i \in \mathscr{T}_n$): \begin{align} -\int \phi_i \left( \sum_{j = 1}^n u_j \phi_j + g\right)' + \int \phi_i \left( \sum_{j = 1}^n u_j \phi_j + g \right) \left( \sum_{j = 1}^n u_j \phi_j + g \right)' = \int \phi_i f + \phi_i (0). \end{align} The first term on the left-hand-side can be split up into \begin{align} - \sum_{j = 1}^n \int u_j \phi_i \phi_j' - \int \phi_i g', \end{align} which is just a bilinear form (matrix) and linear form (vector). The entire right hand side consists of just linear forms (vectors).
What do I do with the term in the middle?
If I distribute or ``foil'' the middle term, I get \begin{align} \int \phi_i \left(\sum_{j = 1}^n u_j \phi_j \right)\left(\sum_{j = 1}^n u_j \phi_j' \right) + \int \phi_i \left(\sum_{j = 1}^n u_j \phi_j g' \right) + \int \phi_i \left(g \sum_{j = 1}^n u_j \phi_j' \right) + \int \phi_i g g'. \end{align} Here the second and third terms look like bilinear forms (matrices), the fourth term looks like a linear form (vector), but what about the first term, the "trilinear form"? It seems to me that this first term would end up like $u^T A u$ since the unknown coefficients $u_j$ appear twice, but I am not too sure.
Also, my motivation for this problem is to understand similar looking nonlinearities are handled practically, because they seem to appear a lot (e.g., in Navier-Stokes). |
Evaluate: $$\int_C \frac{e^z+\sin{z}}{z}dz$$ where, $C$ is the circle $|z|=5$ traversed once in the counterclockwise direction.
I can't find an anti-derivative to this function, and I am not sure one exists. I was thinking about using several theorems but each come up short. For instance, I cannot use the
Closed Curve Theorem since the function in the integrand is not entire (at least I don't think it is).
Any hints (to help get me started)? |
The random variables take values in $\mathbb{R}^d$.
I have tried to prove this using characteristic functions. Let $\hat{\mu}_{X_n},\hat{\mu}_{Y_n},\hat{\mu}_{Z_n}$ be the characteristic functions of the corresponding random variables. Let $X$ (resp. $Y$) be the distributional limit of $X_n$ (resp. $Y_n$), with characteristic function $\hat{\mu}_{X}$ (resp. $\hat{\mu}_{Y}$).
It is easy to show using the uniform convergence on compact sets of the $\hat{\mu}_{X_n}$ and $\hat{\mu}_{Y_n}$ that $\lim_{n \rightarrow \infty} \hat{\mu}_{Z_n}$ exists on some small ball $B_{\epsilon} := \{z \in \mathbb{R}^d : |z| \leq \epsilon\}$, and that the limiting function is continuous at $0$.
This is `nearly' enough to conclude, but the problem is, there is no way of checking that $\lim_{n \rightarrow \infty} \hat{\mu}_{Z_n}$ exists on the set $A:=\{z \in \mathbb{R}^d : \hat{\mu}_{Y}(z)=0\}$. This is why I get stuck.
Many thanks for your help.
EDIT: (Proposed solution)
What is said above is enough (c.f. Chung, page 170-171) to conclude that any subsequence of ${\mu}_{Z_n}$ has a further subsequence converging to a probability measure. It is enough to show that the characteristic functions of these potentially different limits are the same.
Suppose $f$ and $g$ are the characteristic functions of limits of subsequences of subsequences of ${\mu}_{Z_n}$. Then they are both continuous (by Arzela-Ascoli and contents of Chung Theorem 6.3.1). Moreover by the considerations above, $f$ and $g$ agree on the set $A^c:= \mathbb{R}^d \setminus A$ where in fact the limit of $\hat{\mu}_{Z_n}$ exists. It is therefore enough to show that any point in $A$ can be written as the limit of some points in $A^c$. But if this were not the case, $\hat{\mu}_Y$ would vanish on some small ball $D:= B(z_o; \epsilon)$ centered on $z_0 \in\mathbb{R}^d$ with radius $\epsilon$. Integrating shows therefore that
$$\int_{\mathbb{R}^d} \int_D 1 - \cos \langle z, x \rangle dz \mu_Y(dx) = 0 $$
and this is impossible unless $Y=0$ almost surely (I think) in which case the claim is trivial. |
Let $\omega$ be a primitive $k$th root of unity, so that for any integer $b$,
$$\frac1k\sum_{j=1}^k \omega^{bj} = \begin{cases}1,&k\mid b, \\ 0,&k\nmid b\end{cases}.$$
This allows us to write $\sum_{i=0}^{\lfloor\frac{n}{k}\rfloor}{n \choose i * k}$ as$$ \sum_{\substack{0\le b\le n,\\b \equiv 0 \pmod k}}{n \choose b} = \sum_{b=0}^n {n \choose b} \frac1k\sum_{j=1}^k \omega^{bj} = \frac1k \sum_{j=1}^k \sum_{b=0}^n {n\choose b} \omega^{bj}.$$
The innermost sum can be recognized as just the binomial expansion for $(1+\omega^j)^n$. So that means the desired sum is simply
$$\frac1k \sum_{j=1}^k (1+\omega^j)^n.$$
Now if you're coding this you might want to put some thought into what number system to evaluate $\omega$ in. You could do it using floating-point complex arithmetic but if $n$ is large you'll get a lot of precision loss in evaluating $(1+\omega^j)^n$ (but at least the end result can be rounded to an integer, albeit one with many digits). You could alternatively evaluate over different finite fields (say $\mathbb F_p$ where $p \equiv 1 \pmod k$ so that admits $k$th roots of unity), and then Chinese remainder the moduli together. |
There is a definition of the geometric product that applies to general multivectors in any Clifford algebra. It follows directly from the definition of the Clifford algebra. To define a Clifford algebra you need a vector space $V$ and a symmetric bilinear form $B(u,v)$ defined for any $u,v\in V$. The Clifford algebra is the quotient of the tensor algebra of $V$ with respect to the two-sided ideal generated by all elements of the form $u\otimes v+v\otimes u -2B(u,v)$ where $u,v\in V$. The geometric product is the product in the quotient algebra. It is standard and you can find the definition of that in any textbook on abstract algebra. Basically, the geometric product is the product in the tensor algebra of $V$ modulo the ideal.
AN EXAMPLE:
To illustrate, consider $\mathbb R^2$ and the bilinear form defined by $B(e_1,e_1)=1$, $B(e_2,e_2)=1$, $B(e_1,e_2)=0$, where $e_1=(1,0)$ and $e_2=(0,1)$. The two-sided ideal generated by $u\otimes v+v\otimes u -2B(u,v)$ is infinite dimensional as the tensor algebra itself. It contain the following elements among others:
$e_1\otimes e_1-1,\quad e_2\otimes e_2-1,\quad \text{and}\quad e_1\otimes e_2+e_2\otimes e_1$.
This can be used to compute the following products:
$e_1e_1 = e_1\otimes e_1=e_1\otimes e_1 -(e_1\otimes e_1-1)=1$,
$e_2e_2 = e_2\otimes e_2=e_2\otimes e_2 -(e_2\otimes e_2-1)=1$,
$e_1e_2=e_1\otimes e_2= \tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)+\tfrac{1}{2}(e_1\otimes e_2+ e_2\otimes e_1)=\tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)$.
In short, $e_1^2=1$, $e_2^2=1$, and $e_1e_2=-e_2e_1$.
Even though the tensor algebra is infinite-dimensional, the quotient algebra is finite-dimensional. See what happens if you try to get to grade 3. For instance, consider this product
$e_1(e_1\wedge e_2)$ where $e_1\wedge e_2=\tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)$.
It is again a straightforward application of the tensor product modulo the ideal:
$e_1(e_1\wedge e_2)=\tfrac{1}{2}(e_1\otimes e_1\otimes e_2 - e_1 \otimes e_2\otimes e_1)$
but $e_1\otimes e_1\otimes e_2=(e_1\otimes e_1-1)\otimes e_2 +e_2=e_2$ and
$e_1\otimes e_2\otimes e_1=(e_1\otimes e_2+e_2\otimes e_1)\otimes e_1 -e_2\otimes e_1\otimes e_1=$
$=e_2\otimes e_1\otimes e_1=e_2+e_2\otimes(e_1\otimes e_1-1)=e_2$.
So, we get $e_1(e_1\wedge e_2)=\tfrac{1}{2}(e_2+e_2)=e_2$, that is we are back to grade 1. Since the quotient algebra is finite-dimensional, every element can be expressed in term of the basis, which consists of $1, e_1, e_2, e_1e_2$ in the example we are considering. So, every multivector $A$ can be expressed as follows:
$A=s+xe_1+ye_2+pe_1e_2$.
If you have two such multivectors you can compute the product simply by using associativity, distributivity, and the properties we have derived above: $e_1^2=1, e_2^2=1, e_1e_2=-e_2e_1$.
You can easily repeat this exercise for other dimensions and for different bilinear forms.
To come back to the definition of the geometric product, here is how you can understand its significance. In geometry, you are dealing with certain geometric structures. For instance, you might want to find a line passing through two points, or you might want to find a point at the intersection of two lines. These kinds of problems can be dealt with efficiently by applying the exterior structure. You also might want to find, say, a line which passes through a given point and is perpendicular to another line. This kind of problem is related to the orthogonal structure. The tensor product is too general. By using the quotient algebra you are effectively eliminating any part of the tensor product which is not related to exterior or orthogonal structure. What is left has a clear geometric significance. In a way, the geometric product does a lot of work for you behind the curtains, so that you can concentrate on the relevant geometric structures. The expression $uv=u\cdot v+u\wedge v$ is not really the definition of the product. It is just a property that the geometric product of two vectors has.
Alan Macdonald does not use the definition of the geometric product I described above because he does not presume his readers are familiar with the tensor algebra, ideals, or quotients. Instead, he wants to concentrate on applications, geometric properties of the algebra, and on computation. If you are not satisfied with his approach, perhaps you need to read another book. This one
Clifford Algebras and Lie Theory, by Meinrenken
is recent and it uses the same definition I used. There are other equivalent ways to define Clifford algebras. If you are interested, check out these books as well
Quadratic Mappings and Clifford Algebras, by Helmstetter and Micali,
Clifford Algebras: An introduction, by Garling.
Perhaps, after trying to read these books you will appreciate Alan's book more.
Clifford algebra is a well-established part of standard mathematics. It is used in differential geometry and Clifford Analysis, not to mention various applications in physics. No one questions its validity. People who refer to it as Geometric algebra simply want to help promote it in engineering, applied mathematics, and physics. The focus is on applications rather than mathematical rigour. As Alan has pointed out, you don't need to know how the algebra is defined in general in order to use it. You can always compute the product in the basis. It gets tedious to do it by hand as the dimension of the underlying vector space increases, but it can be implemented on a computer quite easily. |
Hello,
Let $G$ be an infinite finitely generated discrete group. I call an infinite set $S$ irregular iff for every $g\in G$, $g\neq 1$, we have that $S\cap gS$ is finite. For example $\{z^3|z\in\mathbb{Z}\}$ is irregular in $\mathbb{Z}$. Now my easy to state question: Does every free ultrafilter on $G$ contain at least one irregular subset?
The following is true and might probably be useful: $G$ acts freely on the space of all ultrafilters (the Stone-Cech compactification of $G$ as a discrete space).
free ultrafilters: http://en.wikipedia.org/wiki/Ultrafilter |
Suppose I have a 20 AmpHour battery at 4V (as in, the actual capacity of the battery is 80 Watt hours). And I'm using a 10V 2A charger to charge it (this is a 20 Watt charger). How can the charge time be calculated?
Is it 4 hours?
Note: The charge time is definitely not 20Ah/2A = 20h as a lot of people state, since these numbers are close to the actual values for most fast-charging smartphones today, and the charge time should be of the order of 1 hr).
Note 2: Please explain using the fundamentals of physics, instead of obscure formulae.
Since the charger is at 10V and the battery is at 4V, how does this affect the charging?
Replay
You don't tell us what voltage is required to charge your 4 V battery so let's assume it's 5 V.
Linear regulator / charger If we use a linear voltage regulator or charger we can drop the 10 V down to 5V and can draw 2 A from the power supply. Power into the battery will be 5V x 2A = 10 W. Power dissipated in the regulator will be 5V x 2A = 10 W. Efficiency will be 50%. Charge time would be \$ \frac {Ah~capacity}{current} = \frac {20Ah}{2A} = 10~h \$ if the battery could store all the energy. Since the battery gets warm during charging we know that some of the input power is being lost so 12 h may be a better estimate.
Buck charger
Using a buck voltage converter is more efficient and when reducing the voltage the current can be increased. Let's assume we could get one with 85% efficiency.
$$ V_O \cdot I_O = \eta V_I \cdot I_I $$ where \$ \eta \$ is the efficiency. From this we can work out the maximum output current is
$$ I_O = \eta \frac {V_I \cdot I_I}{V_O} = 0.85 \frac {10 \cdot 2}{5} = 3.4~A $$
Ideal charge time will be \$ \frac {Ah~capacity}{current} = \frac {20Ah}{3.4A} = 5.9~h \$ so say 7 hours allowing for heat losses.
I hope the formulae weren't too obscure. ;^) |
In contemporary US secondary mathematics textbooks, geometric means occasionally make a brief appearance. For example:
In Geometry, students learn that when an altitude is dropped to the hypotenuse of a right triangle, three different geometric relationships are created, each involving a geometric mean. (See the two figures below.) Also in Geometry, students learn that the length of a tangent segment to a circle is equal to the geometric mean of two secant segments. See the figure below, which -- curiously -- does not use the phrase "geometric mean", even though it had been previously taught two chapters earlier.
(All three of the above images are taken from Glencoe / McGraw-Hill
Geometry, 2008.) In Algebra 2, students commonly learn that a "geometric mean" is a number between two given (nonconsecutive) terms of a geometric sequence, as in the image below.
(Source: Glencoe / McGraw-Hill
Algebra 2, 2008.)
One thing that all of these examples have in common is that students never find the geometric mean of
more than two numbers. And yet there are plenty of instances in which it makes sense to compute the geometric mean of three or more numbers! Compound interest is one such context: for example, if the value of an investment increases in three consecutive years by 10%, 15%, and 2%, the correct way to describe the average annual growth rate is not to compute the arithmetic mean $\frac{10+15+2}{3} = 9$, but rather the geometric mean $$\sqrt[3]{1.10\cdot 1.15 \cdot 1.02}\approx 1.0887$$ (so the average annual growth rate is 8.87%, not 9%).
As far as I can tell, the topic of geometric means used to play a larger role in the secondary curriculum. The first "old" Algebra 2-type textbooks I looked at (
Higher Algebra, 1891, by Hall & Knight) included a discussion of general geometric means, as well as a statement of the AM-GM inequality (i.e. that the arithmetic mean is always greater than or equal to the geometric mean). (See figure below).
So (assuming these isolated examples are representative), it seems that once upon a time the GM was part of the curriculum, and now it is not, except in the limited case of the GM of two numbers. I was in high school in the late 1980s, and I seem to recall the general case was still being taught at the time -- but my memory may be wrong.
My question, then: When did the geometric mean (of more than two numbers) disappear from the curriculum? And does anybody have any thoughts as to why?
Just to be clear, I don't expect any one person to be able to give a definitive answer to this. I would be more than happy with some additional data points, of the form "In this 1985 textbook, I find... but in this 1994 textbook from the same publisher, it's gone." |
This is a very basic Lagrangian Field Theory question, it is about a definition convention. It takes much more time to typeset it than answering, but here it is:
Consider a field Lagrangian with only a kinetic term,
$$L = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi $$
Consider the very simple transformation $\phi \rightarrow \phi + \alpha$ ($\alpha$ constant), and so I understand here that $\alpha$ plays the role of $\delta\phi$. I determine the Noether current as $$\frac{\partial L}{\partial[\partial_{\mu}\phi]}\delta\phi$$
and the result is $$\alpha\partial_\mu\phi$$
But in Peskin & Schroeder (just above eq 2.14), the result they give is:
$$\partial_\mu\phi$$
And it doesn't seem to be an erratum. I don't care that "localized" Lagrangian very much (hey, wait before closing, please), but a
very general question arises:
Is $\alpha$ dropped simply because $\partial_\mu\phi$ is too a conserved quantity (and so under "conserved current" one understands the general concept, momentum, energy or whatever, regardless of its value), or am I missing some other very basic detail that is assumed to be known by the reader?
Later edit: I have eventually understood this question and more, by reading the beginning of chapter 22 of Srednicki. I am finding that book (well, the free preprint for the moment) crystal clear, it seems excellent. |
I am studying the Winternitz signature and I describe its algorithms in the next
W-OTS Key Generation. Select the parameter $w\geq 2$ that is the bit size of the partitions of the message to be signed. Compute the number of partitions
$$t_1 = \Big \lceil \dfrac{r}{w} \Big\rceil,$$ and parameter $$t_2 = \Bigg\lceil \dfrac{\lfloor \log_2 t_1 \rfloor + 1 + w}{w}\Bigg\rceil,$$ The signature key is $X=(x_0, x_1 \cdots, x_{t-1}) \in \{0,1\}^{r,t}$, where the bit-strings $x_i$ with $0 \leq i \leq t-1$ are randomly chosen. The verification key $Y$ is computed by applying ${\mathcal{H}}$ $2^w-1$ times in each bit-string $x_i$ of the signature key, i.e. $Y=(y_{0},\cdots, y_{t-1})$ with $y_i = {\mathcal{H}}^{2^w-1}(x_i)$.
W-OTS Signature Generation. Compute the digest $d$ of message $M$ to be signed using $g$, i.e., $g(M) = d = (d_0, \cdots, d_{r-1})$. Add zeros to $d$ until it is divisible by $w$. The partition of $d$ into $t_1$ $w$-sized parts is $$d = b_0|| b_1|| \cdots || b_{t_1-1}.$$Note that each element $b_i$ can be transformed into base-$10$ representation and identified with integers $\{0,\cdots, 2^w-1\}$.
Calculate the checksum $$c=\sum_{i=0}^{t_1-1} 2^w - b_i.$$ The maximum number of bits representing $c$ in binary base is $\lfloor \log_2 t_1 \rfloor + w + 1 $. Indeed, note that $c = t_1 2^w - \sum_{i=0}^{t_1-1}b_i \leq t_1 2^w$ and the number of bits required to represent $t_1 2^w$ is $\lfloor \log_2 t_1 2^w \rfloor + 1 = \lfloor \log_2 t_1\rfloor + w + 1.$
Similarly, split the binary representation of $c$ into $t_2$ $w$-sized parts, i.e., $$c = c_0|| c_1|| \cdots || c_{t_2-1}.$$ Finally, computing the signature as $$\sigma_{\text{W-OTS}} = (\sigma_0,\cdots,\sigma_{t-1})=({\mathcal{H}}^{b_0}(x_0),\cdots,{\mathcal{H}}^{b_{t_1-1}}(x_{t_1-1}),{\mathcal{H}}^{c_0}(x_{t_1}),\cdots, {\mathcal{H}}^{c_{t_2-1}}(x_{t-1})).$$
W-OTS Verification. To verify the signature $\sigma_{\text{W-OTS}}$ of the message $M$, compute the values $b_i$ and $c_i$ in the same way as above described, then compute$$y'=({\mathcal{H}}^{2^w-1-b_0}(\sigma_0),\cdots,{\mathcal{H}}^{{2^w-1-b_{t_1-1}}}(\sigma_{t_1-1}),{\mathcal{H}}^{2^w-1-c_0}(\sigma_{t_1}),\cdots,{\mathcal{H}}^{2^w-1-c_{t_2-1}}(\sigma_{t-1})),$$ and compare $y'$ with $Y$. The signature is valid only if $y'_i=y_i$, $\forall i$. \
I need to understand why this present resistence against adaptative chosen message attack. According my knowledge the adaptative chosen message attack works as follow: an attacker can choose a message on which he learns a signature. Afterwards that attacker can choose the message he wants to forge a signature for. This is called the standard model for secure signatures.
My question is why winternitz signature is secure in the standard model? I make this question because using the fact that
"an attacker can choose a message on which he learns a signature"
an attacker can choose the message 00000..0000 and then He learn $\mathcal{H}(x_i)$ for $i \leq t_1$. With these value he can obtain all next values of the form $\mathcal{H}^v(x_i)$, that is he can forge partially any signature because each $\sigma_i$ his obtained using that form. |
Intuitive Answer:
$\newcommand{\curl}[1]{\nabla \times #1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$There is an emf created because of the Lenz's Law, which means an
electric field is being created. The electric field is circular (thus the force, which is proportional to the electric field, is also circular) and the charged particle would spiral out due to added velocity by the circular force. Depending on the charge of the particle and the direction of the increasing magnetic field, it would go either clockwise or anti-clockwise. Mathematical Answer
Starting with the Faraday's law
$$\curl {\vec E} = - \pdv {\vec B} t$$
and assuming that $\vec B(t)= \beta t \hat z$ for some constant $\beta$, where $\hat z$ is the unit vector along the $z$ direction, we have (I'll drop the vector arrows from now on)
$$(\curl E)_z = - \beta \quad, \tag 1$$
where $(\curl E)_z$ denotes the $z$ component of $\curl E$.
Changing to the cylindrical coordinates $(r, \theta, z)$ we have,
$$(\curl E )_z = \frac 1 r \left( \pdv {(r E_\theta)} r - \pdv {E_r} \theta \right) = \frac {E_\theta} r + \pdv {E_\theta} r - \frac 1 r \pdv {E_r} \theta \quad. \tag 2$$
From the symmetry of the problem it is clear that $\partial E_r/ \partial \theta$ is zero. Therefore substituting equation $(2)$ in the Faraday's law $(1)$ we are left with the differential equation
$$ \pdv {E_\theta} r + \frac {E_\theta} r + \beta = 0$$
One can take the ansatz$^*$ $\tilde E_\theta = \frac {C_1} r - C_2 \beta x$ for some constants $C_1$ and $C_2$. We have
$$\pdv {\tilde E_\theta} r + \frac {\tilde E_\theta} r + \beta = \beta - 2\beta C_2 = 0 \implies C_2= \frac 1 2 \quad .$$
Hence we can write the real solution as
$$E_\theta = \frac C r - \beta \frac r 2 $$
for some constant $C$ that should be determined by the initial conditions. Note that when $C \neq 0$ funny things happen near the origin, in particular the electric field diverges. Now we can avoid this problem by two ways:
We'll look at the solution, which is far a way from the origin, i.e. as $r \to \infty$ the term $C/r \to 0$. Therefore we will proceed by ignoring that term altogether. We'll say that the solution for $C\neq 0$ is not physical at the origin, therefore we'll set $C=0$.
You can choose whichever reason you want but the upshot is that we'll ignore the $C/r$ term.
Then we have for the electric field
$$\fbox{ $E_\theta = - \beta \frac r 2 $}\quad. \tag 3$$
Easily enough the force on the particle is given by
$$\vec F = q \vec E = - \beta \frac {q \cdot r} 2 \, \hat \theta \quad ,$$
where $\hat \theta$ denotes the unit vector in the $\theta$ direction, which you can convert into Cartesian coordinates by substituting
$$\hat \theta = \begin{pmatrix} -\sin \theta \\ \cos \theta \\ 0 \end{pmatrix} \quad \text {with} \quad \theta = \arctan \frac y x $$
and remembering that $r= \sqrt{x^2+y^2}$.
In any event you know that $\theta = \rho$ for some constant $\rho$ give a cylinder in cylindrical coordinates. Thus it is trivial to see that the solution $(3)$ describes a circular electric field, which in return corresponds to an outgoing spiral motion of the particle as we predicted using only our intuition.
$^*$As to why we have chosen that ansatz i.e. an educated guess as to why the solution should look like that, we first start by noting that
$$\pdv {} r \frac 1 r = - \frac 1 {r^2} = \frac 1 r \cdot \frac 1 r$$
Hence a term of the form $C_1/r$ takes care of the term $E_\theta/r$ up to a constant term, so we must include $C_1/r$ in our ansatz $\tilde E_\theta$.
Having taken care of the term $E_\theta/r$ we are left with a pesky constant $\beta$. My intuition says that adding a term of the form $C_2 \beta r $ should clean up the $\beta$ when differentiated with respect to $r$. However adding this term may also screw up the $E_\theta / r $ term. Luckily however what we want to add depends only on $r$ in the first order, hence differentiating this term with respect to $r$ and dividing by $r$ gives the same result! For safety one always adds a dummy constant $C_2$ in front of this term, which you should adjust later on, which is exactly what I did above. |
This article appeared originally in my Day Book.
I was watching the
Mythbusters today, and they were testing a myth that involved a human falling from from great height. They explained the concept of terminal velocity, and stated that the terminal velocity of a human body is approximately 120 mph. They then proceeded with a calculation, according to which it takes ~5.5 seconds for a falling body to reach this velocity.
Except that it takes almost twice as long.
120 mph is about 192 kph, or ~53.3 m/s. At an acceleration of 9.81 m/s
2, it indeed takes about 5.44 seconds to reach this velocity. That is, if you are falling in a vacuum.
But we're talking about falling in air, aren't we. There's no terminal velocity in a vacuum; a falling body would continue to accelerate until it hits the ground. In air, however, air resistance doesn't just kick in at the moment when you reach terminal velocity; it is there all the time, proportional to the square of your velocity, which is how it gets to be equal the to force of gravity eventually when you reach terminal velocity.
So how long would it take for a falling body to reach terminal velocity? Wrong question! A falling body never actually reaches terminal velocity; as air resistance increases, the acceleration of the body decreases, so it'd approach, but never quite get to, that terminal velocity.
So then... how long would it take for it to reach 95% of its terminal velocity?
The acceleration $a$ of a rapidly falling body has two components: a constant acceleration $g$ due to gravity, and a variable acceleration that is a function of the square of the velocity $v$, multiplied by a proportionality factor $\mu$ ($\mu=\rho AC_d/2m$ where $\rho$ is the air density, $A$ is the cross-sectional area, $C_d$ is the drag coefficient, and $m$ is the mass of the body):
\[a=g-\mu v^2.\]
But the acceleration $a$ is none other than the time derivative of the velocity, so we have a simple differential equation for the velocity as a function of time, i.e., $v(t)$:
\[\frac{dv}{dt}=g-\mu v^2.\]
Acceleration becomes zero when terminal velocity is reached. At this point, $g-\mu v_\mathrm{max}^2=0$, from which $\mu=g/v_\mathrm{max}^2$; if $g=9.81~{\rm m}/{\rm s}^2$ and $v_\mathrm{max}=53.3~\rm{m}/\rm{s}$, we get $\mu=0.0035$. (At $\rho=1.3~{\rm kg}/{\rm m}^3$, $m=80~{\rm kg}$, and $A=0.4~{\rm m}^2$, this means $C_d\simeq 1$.)
The solution to the differential equation comes in the form of
\[v=v_\mathrm{max}\tanh(t\sqrt{\mu g}),\]
and we now seek the value of $t$ for which
\[\tanh(t\sqrt{\mu g})=0.95.\]
The answer can be easily calculated:
\[t\simeq 10~{\rm s}.\]
Another question is, how much distance would the body travel in this amount of time? In the Mythbusters episode, they were using a figure of 500-odd feet, but that was again based on the same incorrect math (well, the math was correct, it was just applied incorrectly, so perhaps I should say, incorrect physics.) The distance traveled can be calculated by computing the integral of the velocity:
\[s=\int v_{\rm max}\tanh(t\sqrt{\mu g})~dt=\mu^{-1}\ln\cosh(t\sqrt{\mu g})\simeq 340~{\rm m},\]
which is over 1100 feet, or about twice the height that the Mythbusters have calculated.
But then, who cares about sloppy math (or physics!) when the episode provided an excuse to blow up some 500 pounds of high explosives? Fortunately, they did not make a mistake in
those calculations, so nobody got hurt! |
Problem 8: Mixed Norms
Suggested by Piotr Indyk Source Kanpur 2006 Short link https://sublinear.info/8
For any vector $x$, let $\|x\|_0$ be a norm-like function computing the number of non-zero elements in $x$. Consider the following norm-like function $\|\cdot\|_{2,0}$ over $n \times n$ matrices $A = [a_1 \ldots a_n]$: \[ \|A\|_{2,0} = \left ( \sum_{i=1}^n (\|a_i\|_0)^2 \right )^{1/2}. \]
Assume we are given a stream of $m$ updates $(i,j,\delta)$ to $A$, interpreted as $A[i,j]:=A[i,j]+\delta$, starting from $A=0$. What is the smallest space needed by a streaming algorithm estimating $\|A\|_{2,0}$ up to a factor of $1 \pm \epsilon$? An upper bound of $O(\operatorname{poly}(\epsilon^{-1})\cdot\sqrt{n}\cdot\operatorname{polylog}(n))$ is known as long as $A \ge 0$ [CormodeM-05b]. There are no non-trivial lower bounds known. |
Hi, Can someone provide me some self reading material for Condensed matter theory? I've done QFT previously for which I could happily read Peskin supplemented with David Tong. Can you please suggest some references along those lines? Thanks
@skullpatrol The second one was in my MSc and covered considerably less than my first and (I felt) didn't do it in any particularly great way, so distinctly average. The third was pretty decent - I liked the way he did things and was essentially a more mathematically detailed version of the first :)
2. A weird particle or state that is made of a superposition of a torus region with clockwise momentum and anticlockwise momentum, resulting in one that has no momentum along the major circumference of the torus but still nonzero momentum in directions that are not pointing along the torus
Same thought as you, however I think the major challenge of such simulator is the computational cost. GR calculations with its highly nonlinear nature, might be more costy than a computation of a protein.
However I can see some ways approaching it. Recall how Slereah was building some kind of spaceitme database, that could be the first step. Next, one might be looking for machine learning techniques to help on the simulation by using the classifications of spacetimes as machines are known to perform very well on sign problems as a recent paper has shown
Since GR equations are ultimately a system of 10 nonlinear PDEs, it might be possible the solution strategy has some relation with the class of spacetime that is under consideration, thus that might help heavily reduce the parameters need to consider to simulate them
I just mean this: The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
@ooolb Even if that is really possible (I always can talk about things in a non joking perspective), the issue is that 1) Unlike other people, I cannot incubate my dreams for a certain topic due to Mechanism 1 (consicous desires have reduced probability of appearing in dreams), and 2) For 6 years, my dream still yet to show any sign of revisiting the exact same idea, and there are no known instance of either sequel dreams nor recurrence dreams
@0celo7 I felt this aspect can be helped by machine learning. You can train a neural network with some PDEs of a known class with some known constraints, and let it figure out the best solution for some new PDE after say training it on 1000 different PDEs
Actually that makes me wonder, are the space of all coordinate choices more than all possible moves of Go?
enumaris: From what I understood from the dream, the warp drive showed here may be some variation of the alcuberrie metric with a global topology that has 4 holes in it whereas the original alcuberrie drive, if I recall, don't have holes
orbit stabilizer: h bar is my home chat, because this is the first SE chat I joined. Maths chat is the 2nd one I joined, followed by periodic table, biosphere, factory floor and many others
Btw, since gravity is nonlinear, do we expect if we have a region where spacetime is frame dragged in the clockwise direction being superimposed on a spacetime that is frame dragged in the anticlockwise direction will result in a spacetime with no frame drag? (one possible physical scenario that I can envision such can occur may be when two massive rotating objects with opposite angular velocity are on the course of merging)
Well. I'm a begginer in the study of General Relativity ok? My knowledge about the subject is based on books like Schutz, Hartle,Carroll and introductory papers. About quantum mechanics I have a poor knowledge yet.
So, what I meant about "Gravitational Double slit experiment" is: There's and gravitational analogue of the Double slit experiment, for gravitational waves?
@JackClerk the double slits experiment is just interference of two coherent sources, where we get the two sources from a single light beam using the two slits. But gravitational waves interact so weakly with matter that it's hard to see how we could screen a gravitational wave to get two coherent GW sources.
But if we could figure out a way to do it then yes GWs would interfere just like light wave.
Thank you @Secret and @JohnRennie . But for conclude the discussion, I want to put a "silly picture" here: Imagine a huge double slit plate in space close to a strong source of gravitational waves. Then like water waves, and light, we will see the pattern?
So, if the source (like a Black Hole binary) are sufficent away, then in the regions of destructive interference, space-time would have a flat geometry and then with we put a spherical object in this region the metric will become schwarzschild-like.
if**
Pardon, I just spend some naive-phylosophy time here with these discussions**
The situation was even more dire for Calculus and I managed!
This is a neat strategy I have found-revision becomes more bearable when I have The h Bar open on the side.
In all honesty, I actually prefer exam season! At all other times-as I have observed in this semester, at least-there is nothing exciting to do. This system of tortuous panic, followed by a reward is obviously very satisfying.
My opinion is that I need you kaumudi to decrease the probabilty of h bar having software system infrastructure conversations, which confuse me like hell and is why I take refugee in the maths chat a few weeks ago
(Not that I have questions to ask or anything; like I said, it is a little relieving to be with friends while I am panicked. I think it is possible to gauge how much of a social recluse I am from this, because I spend some of my free time hanging out with you lot, even though I am literally inside a hostel teeming with hundreds of my peers)
that's true. though back in high school ,regardless of code, our teacher taught us to always indent your code to allow easy reading and troubleshooting. We are also taught the 4 spacebar indentation convention
@JohnRennie I wish I can just tab because I am also lazy, but sometimes tab insert 4 spaces while other times it inserts 5-6 spaces, thus screwing up a block of if then conditions in my code, which is why I had no choice
I currently automate almost everything from job submission to data extraction, and later on, with the help of the machine learning group in my uni, we might be able to automate a GUI library search thingy
I can do all tasks related to my work without leaving the text editor (of course, such text editor is emacs). The only inconvenience is that some websites don't render in a optimal way (but most of the work-related ones do)
Hi to all. Does anyone know where I could write matlab code online(for free)? Apparently another one of my institutions great inspirations is to have a matlab-oriented computational physics course without having matlab on the universities pcs. Thanks.
@Kaumudi.H Hacky way: 1st thing is that $\psi\left(x, y, z, t\right) = \psi\left(x, y, t\right)$, so no propagation in $z$-direction. Now, in '$1$ unit' of time, it travels $\frac{\sqrt{3}}{2}$ units in the $y$-direction and $\frac{1}{2}$ units in the $x$-direction. Use this to form a triangle and you'll get the answer with simple trig :)
@Kaumudi.H Ah, it was okayish. It was mostly memory based. Each small question was of 10-15 marks. No idea what they expect me to write for questions like "Describe acoustic and optic phonons" for 15 marks!! I only wrote two small paragraphs...meh. I don't like this subject much :P (physical electronics). Hope to do better in the upcoming tests so that there isn't a huge effect on the gpa.
@Blue Ok, thanks. I found a way by connecting to the servers of the university( the program isn't installed on the pcs on the computer room, but if I connect to the server of the university- which means running remotely another environment, i found an older version of matlab). But thanks again.
@user685252 No; I am saying that it has no bearing on how good you actually are at the subject - it has no bearing on how good you are at applying knowledge; it doesn't test problem solving skills; it doesn't take into account that, if I'm sitting in the office having forgotten the difference between different types of matrix decomposition or something, I can just search the internet (or a textbook), so it doesn't say how good someone is at research in that subject;
it doesn't test how good you are at deriving anything - someone can write down a definition without any understanding, while someone who can derive it, but has forgotten it probably won't have time in an exam situation. In short, testing memory is not the same as testing understanding
If you really want to test someone's understanding, give them a few problems in that area that they've never seen before and give them a reasonable amount of time to do it, with access to textbooks etc. |
I was reading a lemma:
Lemma.Let $G$ be a locally profinite group, and let $H$ be an open subgroup of $G$ of finite index. If $(\pi, V)$ is a smooth representation of $G$, then $V$ is $G$-semisimple if and only if it is $H$-semisimple. Let $(\sigma, W)$ be a semisimple smooth representation of $H$. The induced representation $\operatorname{Ind}_H^G \sigma$ is $G$-semisimple
Proof.Suppose that $V$ is $H$-semisimple, and let $U$ be a $G$-subspace of $V$. By hypothesis, there is an $H$-subspace $W$ of $V$ such that $V = U \oplus W$. Let $f \colon V \to U$ be the $H$-projection with kernel $W$. Consider the map $$ f^G \colon v \mapsto (G : H)^{-1} \sum_{g \in G/H} \pi(g) f( \pi(g)^{-1} v ), \quad v \in V. $$ The definition is independent of the choice of coset representatives and it follows that $f^G$ is a $G$-projection $V \to U$. We then have $V = U \oplus \ker f^G$ and $\ker f^G$ is a $G$-subspace of $V$. Thus $V$ is semisimple ( cf.2.2 Proposition).
Conversely, suppose that $V$ is $G$-semisimple. Thus $V$ is a direct sum of irreducible $G$-subspaces (2.2), and it is enough to treat the case where $V$ is irreducible over $G$. As representations of $H$, the space $V$ is finitely generated and so admits an irreducible $H$-quotient $U$. Suppose for the moment that $H$ is a
normalsubgroup of $G$. By Frobenius Reciprocity (2.4.2), the $H$-map $V \to U$ gives a non-trivial, hence injective, $G$-map $V \to \operatorname{Ind}_H^G U$. As representations of $H$, the induced representation $\operatorname{Ind}_H^G U = c-\operatorname{Ind}_H^G U$ is a direct sum of $G$-conjugates of $U$ ( cf.2.5 Lemma). These are all irreducible over $H$, so $\operatorname{Ind} U$ is $H$-semisimple. Proposition 2.2 then implies that $V \subset \operatorname{Ind} U$ is $H$-semisimple.
In general, we set $H_0 = \bigcap_{g \in G/H} g H g^{-1}$. This is an open normal subgroup of $G$ of finite index. We have just shown that the $G$-space $V$ is $H_0$-semisimple; the first part of the proof shows it is $H$-semisimple.
This completes the proof of 1., and 2. follows readily from the same arguments.
I don't know how $U$ is constructed (after "conversely" in the second paragraph), that is:
Given a locally profinite group $G$, an open subgroup $H$ of finite index, suppose $(\pi, V)$ is an irreducible smooth representation of $G$, we need to construct an $H$-irreducible space $U$ with a nontrivial $H$-homomorphism $V\rightarrow U$.
I had considered $U=V/W$ where $W=$ the $H$-space generated by $\{v-\pi(g_i)v;v\in V,g_i$'s are the right coset representatives of $H$ in $G\}$, but I'm not sure wether $V/W$ is non-trivial or not. Help please. |
This is from an old qualifying examination question.
The premise of the question is as follows: Let $f:\mathbb R \rightarrow \mathbb R$ be continuous. Show that
$$\lim_{n\rightarrow \infty} \int_0^1nx^{n-1}f(x)dx=f(1) $$
I know how to show this without using Weierstrass approximation theorem. But I'd still like to see someone else's take on this.
My questions are
1) Is there a way to show this using just the Weierstrass approximation theorem (i.e., without using Stone's generalization). Weierstrass says that a continuous function defined on a closed bounded interval can be uniformly approximated by polynomials.
2) a sequence of functions is uniformly convergent iff it is convergent in the sup norm. Is there a analogous characterization for uniform continuity interms of suprema?. What about continuity?
Any comments hints are appreciated.
Thanks, |
Given integral = $\iint_R \frac{x-y}{(x+y)^3}dxdy$ :$R=[0,1]\times[0,1]$
Let $u = x+y, v= x-y \implies x = \frac{u+v}{2}, y = \frac{u-v}{2}$
Now x limits are from 0 to 1 and y limits are from 0 to 1. I am not able to understand how to find limits of integration in terms of u, v..
Pls enlighten me. |
Take an infinite hexagonal lattice (or equivalently, an equilateral triangular lattice), with unit spacing between the closest lattice point pairs, and draw a disc of radius $r$ centered on a lattice point at $(0, 0)$. Let $N(r, hex)$ denote the number of hexagonal lattice points at coordinates $(a, b)$ s.t. $(a^2 + b^2) \leq r^2$, i.e. the number of lattice points on or within the aforementioned disc of radius $r$.
Are there any literature references for approximations to $N(r, hex)$ (I haven't been able to find any through a Google search)? What is an exact counting solution for $N(r, hex)$?
Using the exact counting solution for the $Z^2$ integer lattice, (http://mathworld.wolfram.com/GausssCircleProblem.html) I suppose we can guess a lowerbound for the hexagonal lattice of:
Lowerbound $N(r, hex) = 1 + Floor[\frac{r}{2}] + 4*\sum^{Floor[\frac{r}{2}]}_{i=1} Floor[((\frac{r}{2})^2-i^2)^{\frac{1}{2}}] + 2*Floor[r]$
Where we simply overlay the $Z^2$ lattice with (closest) nearest-neighbor spacing $2$ on top of an $A_2$ hexagonal lattice with (closest) nearest-neighbor spacing $1$, and add an additional $2*Floor[r]$ correctional term.
[10/13/12] The OEIS sequences are extremely helpful, but after searching the literature for awhile, I'm still having difficulty finding an exact (counting) solution for the number of lattice points within a circle of real number radius $r$. Any references would be very much appreciated!
[10/14/12] Still no luck finding a reference in the literature. Surely someone has looked at this problem for, say, graphene and other molecular or atomic lattices where one would like to have a precise atom count a certain physical distance away from one atom?
[10/19/12] I managed to find the exact OEIS sequence I was looking for: http://oeis.org/A053416
However, I'd still like to find an exact counting solution, like the one presented above the $Z^2$ integer lattice. |
Thermal engineers often use so called dry bulb and wet bulb temperatures as a measure of humidity in the air. Dry bulb temperature reflects what a dry thermometer measures. A wet thermometer measures a lower temperature due to evaporative cooling, unless the relative humidity is 100%.
The conversion between wet-bulb and dry-bulb temperatures is surprisingly non-trivial. Full details can be found in Chapter 6 of the ASHRAE Fundamentals handbook. (My notation differs slightly from ASHRAE's.)
To convert between the two, you need to know either the dry-bulb or the wet-bulb temperature, in addition to atmospheric pressure and relative humidity. In metric units, the calculation goes as follows. First, absolute temperature ($T$, in Kelvin) is related to the Celsius scale ($t$) as
$$T=t+273.15.$$
You can compute the saturation water pressure for a given absolute temperature:
$$\ln p_{\rm ws}(T)=\sum_{i=-1}^4C_iT^i + C_l\ln T,$$
where the coefficients have different values for $t<0~^\circ{\rm C}$ and $t>0~^\circ{\rm C}$ (i.e., below and above freezing). For the former,
\begin{align}
C_{-1}&=-5.800 220 6\times 10^3,\\ C_0&=1.391 499 3,\\ C_1&=-4.864 023 9\times 10^{-2},\\ C_2&=4.176 476 8\times 10^{-5},\\ C_3&=-1.445 209 3\times 10^{-8},\\ C_4&=0,\\ C_l&=6.545 967 3, \end{align}
whereas for the latter (above freezing),
\begin{align}
C_{-1}&=-5.674 535 9\times 10^{-3},\\ C_0&=6.392 524 7,\\ C_1&=-9.677 843 0\times 10^{-3},\\ C_2&=6.221 570 1\times 10^{-7},\\ C_3&=2.074 782 5\times 10^{-9},\\ C_4&=-9.484 024 0\times 10^{-13},\\ C_l&=4.163 501 9. \end{align}
From $p_{\rm ws}$ and the pressure $p$, the humidity ratio at saturation can be calculated:
$$W_s(T,p)=0.62198\frac{p_{\rm ws}(T)}{p-p_{\rm ws}(T)}.$$
Given a relative humidity $\phi$ at dry-bulb temperature $T_{\rm db}$, the water vapor pressure is given by
$$p_w(T_{\rm db},\phi)=\phi p_{\rm ws}(T_{\rm db}),$$
and if the pressure and water vapor pressure are known, the humidity ratio can be calculated as
$$W(T_{\rm db},p,\phi)=0.62198\frac{p_w(T_{\rm db},\phi)}{p-p_w(T_{\rm db},\phi)}.$$
Finally, the relationship between dry-bulb and wet-bulb temperature is given implicitly by the following equations. First, above freezing:
$$W(T_{\rm db},p,\phi)=\frac{(2501-2.326t_{\rm db})W_s(T_{\rm wb},p)-1.006(t_{\rm db}-t_{\rm wb})}{2501+1.86t_{\rm db}-4.186t_{\rm wb}},$$
and below freezing:
$$W(T_{\rm db},p,\phi)=\frac{(2830-0.24t_{\rm wb})W_s(T_{\rm wb},p)-1.006(t_{\rm db}-t_{\rm wb})}{2830+1.86t_{\rm db}-2.1t_{\rm wb}}.$$
These equations can be solved numerically to obtain the desired temperature from the other value. There is no closed form solution. |
Preprint Series 1998 1998:25 P. Petrushev
We prove in this paper the existence of a Schauder basis for $C[0,1]$consisting of rational functions of uniformly bounded degrees. Thissolves an open question of some years concerning the possible existenceof such bases. This result follows from a more general construction ofbases on $I ...
[Full Abstract
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1998:24 R. Getsadze
Let
$$\mathbb{B}^2:=\left\{|(x,y)\in{R}^2:x^2+y^2\leq1\right\}$$
denote the unit disc on the plane and
$$u _ mt:=\left.\frac{1}{\sqrt{\pi}}\frac{\sin(m+1)\arccos{t}}{\sqrt{1-t^2}}\right.,$$
$m=0,1,\ldots,t\in[-1,1 ...
[Full Abstract
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1998:23 V. Temlyakov
The question of finding an optimal dictionary for nonlinear
m-term approximation is studied in the paper. We consider this problem in the periodic multivariate ( d variables) case for classes of functions with mixed smoothness. We prove that the well known dictionary U which consists of trigonometric polynomials (shifts ... d
[Full Abstract
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1998:22 V. Temlyakov
The paper contains two theorems on approximation of functions withbounded mixed derivative. These theorems give some progress in two oldopen problems. The first one gives, in particular, an upper estimate inthe Bernstein $L-1$-inequalityfor trigonometric polynomials on two variables with harmonics inhyperbolic crosses. The second ...
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1998:21 P. Erdős, K. Rice, L. Székely, and T. Warnow
Reconstruction phylogenetic (evolutionary) trees is a major research problem in biology, but unfortunately the current methods are either inconsistent somewhere in the parapeter space (and hence do not reconstruct the tree even given unboundedly long sequences), have poor statistical power (and hence require extremely long sequences on large or highly ...
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1998:20 F. Shahrokhi, O. Sykora, L. Székely, and I. Vrto
Let
G be a connected bipartite graph. We give a short proof using the variation of Menger's Theorem, for a new lower bound which relates the bipartite crossing number of G, denoted by bcr( G), to the edge connectivity properties of G. The general lower bound implies a weaker ...
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1998:19 F. Shahrokhi, O. Sykora, L. Székely, and I. Vrto
The bipartite crossing number problem is studied, and a connection between this problem and the linear arrangement problem is established. It is shown that when the arboricity is close to the minimum degree and the graph is not too sparse, then the optimal number of crossings has the same order ...
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1998:18 F. Shahrokhi and L. Székely
We study the integral uniform (multicommodity) flow problem in a graph
G and construct a fractional solution whose properties are invariant under the action of a group of automorphisms Γ < Aut( G). The fractional solution is shown to be close to an integral solution (depending on properties of Γ), and ...
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1998:17 P. Erdős, M. Steel, L. Székely, and T. Warnow
Inferring evolutionary trees is an interesting and important problem in biology that is very difficult from a computational point of view as most associated optimization problems are NP-hard. Although it is known that many methods are provably statistically consistent (i.e. the probability of recovering the correct tree converges on ...
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1998:16 P. Erdős, M. Steel, L. Székely, and T. Warnow
A phylogenetic tree (also called an "evolutionary tree") is a leaf–labeled tree which represents the evolutionary history for a set of species, and the construction of such trees is a fundamental problem in biology. Here we address the issue of how many sequence sites are required in order to ...
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1998:15 A. Cohen, W. Dahmen, and R. DeVore
This paper is concerned with the construction and analysis ofwavelet-based adaptive algorithms for the numerical solution ofelliptic equations. These algorithms approximate the solution
u of the equation by a linear combination of N wavelets. Therefore, a benchmark for their performance is provided by the rate of best approximation ...
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1998:14 J. Griggs
We discuss applications of combinatorial arguments to databasesecurity: maximizing the "usabiliby" of a statistical database underthe control of the mechanism Audit Expert of Chin and Ozsoyoglu. Asmodeled by Mirka Miller
et al., the goal is to maximize thenumber of SUM queries from a database of real ...
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1998:13 J. Griggs, M. Simonovits, and G. Thomas
Let $Ex(n,k,\mu)$ denote the maximum number of edges of an
n-vertex graph in which every subgraph of k vertices has at most $\mu$ edges. Here we summarize some known results of the problem determining $Ex(n,k,\mu)$,give simple proofs, and find some new estimates ...
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1998:12 J. Griggs and C. Ho
In "Bulgarian Solitaire," a player divides a deck of
n cardsinto piles. Each move consists of taking a card from each pile to forma single new pile. One is concerned only with how many piles there areof each size. Starting from any division into piles, one always ...
[Full Abstract
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1998:11 J. Griggs and C. Ho
Consider the minimum number $f(m,n)$ of zeroes in a $2m\times2n(0,1)$-matrix
M that contains no $m\times{n}$ submatrix of ones. This special case of the well-known Zarankiewicz problem was studied by Griggs and Ouyang, who showed, for $m\leq{n}$, that $2n+m+1 ...
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1998:10 J. Griggs and G. Rote
n analogue of the Littlewood-Offord problem posed by the first authoris to find the maximum number of subset sums equal to the same vectorover all ways of selecting
n vectors in $R^d$in general position. This note reviews past progress and motivation forthis problem, and presents ...
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1998:09 J. Griggs and K. Reid
Two new elementary proofs are given of Landau's Theorem on necessary and sufficient conditions for a sequence of integers to be the score sequence for some tournament. The first is related to existing proofs by majorization, but it avoids depending on any facts about majorization. The second is natural ...
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1998:08 F. Chudak and J. Griggs
P. L. Erdős and G.O.H. Katona gave an inequality involving binomial coefficients summed over an antichain in the product of two chains. Her we present the common generalization of this inequality and Lubell’s famous inequality for the Boolean lattice to an arbitrary product of chains (lattice of ...
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1998:07 S. Brenner A nonstandard finite element interpolation estimate (file not available) (Numer. Funct. Anal. Optim. 20 (1999), 245-250)
We show that $\|u _ I\|H^{l+e}(\Omega)\leq{C}\|u\|H^{l+e}(\Omega)$, where $\Omega$ is a bounded polygonal domain in $R^20<\epsilon<(\frac{1}{2})$, $u _ 1$ is the piecewise linear nodal interpolant of
u with ...
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1998:06 K. Oskolkov
The goal is to compare
free (non-linear), equispaced ridge and algebraic polynomial approximations $R^{\textrm{fr}} _ N[f]$, $R^{\textrm{eq}} _ N[f]$, $E _ N[f]$ of individual functions $f(x)$ in the norm of $L^2(\mathbb{B}^2),\mathbb{B}^2$ - the unit disc $|x ...
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1998:05 S. Brenner and L. Sung Lower bounds for two-level additive Schwarz preconditioners for nonconforming finite elements (file not available) (Advances in Computational Mathematics (Proceedings of the Guangzhou International Symposium 1997) Lecture Notes in Pure and Appl. Math. 202, pp. 585-604, Dekker, New York, 1999)
Lower bounds for the condition numbers of the preconditioned systems are obtained for two-level additive Schwarz preconditioners for nonconforming nite element methods. They show that ...
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1998:04 S. Brenner and L. Sung Lower bounds for nonoverlapping domain decomposition preconditioners in two dimensions (file not available) (Math. Comp. 69 (2000), 1319-1339)
Lower bounds for the condition numbers of the preconditions systems are obtained for the Bramble-Pasciak-Schatz substructuring preconditioner and the Neumann-Neumann preconditioner in two dimensions. They show that the known upper bounds are sharp.
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1998:03 K. Oskolkov
Our goal is to compare the efficiencies of linear and nonlinear methodsin the problem of ridge approximation. We confine ourselves byfunctions of two variables $f(x)=f(x _ 1,x _ 2)$ and the norm $\|\cdot\|$ of Hilbert space $L^2(\mathbb{B}^2)$, where $\mathbb{B ...
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1998:02 R. DeVore
This is a survey of nonlinear approximation, especially that part of the subject which is important in numerical computation. Nonlinear approximation means that the approximants do not come from linear spaces but rather from nonlinear manifolds. The central question to be studied is what, if any, are the advantages of ...
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1998:01 S. Brenner Lower bounds for two-level additive Schwarz preconditioners with small overlap (file not available) (SIAM J. Sci. Comp. 21 (2000), 1657-1669.)
Lower bounds for the condition numbers of the preconditioned systems are obtained for two-level additive Schwarz preconditioners for both second order and fourth order problems. They show that the known upper bounds are sharp in the case of a small overlap.
[Full Abstract
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I must confess I'm not an educator, but I like this question and at the very least I can answer with the intuitive picture I use in my own head.The $n$-th central moment $\mu_n$ of a random variable $X$ is defined by:$$\mu_n \equiv E[(X-E[X])^n]$$For what follows we will take the measure space of $X$ to be the real numbers, since that's the easiest to ...
I think that the most interesting application of (sum of numerators/sum of denominators) “addition” is to continued fractions. For example, if you want to calculate the continued fraction expansion of the square root of 2,start with 1/0 and 0/1 and “add” them in the following way:In the top row you put the results of the “adding” that are greater then the ...
The problem is with the question, not with the students' answers. The question is ambiguous and I think the students' answer is actually much better than yours.Suppose I drive a thousand miles at 25mpg and you drive one mile at 35mpg. What's the average fuel efficiency? Your answer is 30mpg but I honestly can't think of any situation in which that is a ...
Think of an example with two ratios: 1/3 and 4/5.When you add the numerators, and divide this by the sum of the denominators, you get (1 + 4)/(3 + 5) = 5/8. Now, think about what is happening with the denominators - the denominator of the first ratio should only act on the first numerator. But instead, when you add the ratios in this way, the denominator ...
Allow me to offer another example:Imagine you and your best friend both want to buy a new smart phone. The phone you have chosen will cost you 300€ but your friend chooses a phone that will cost as much as 600€! Luckily, you have two vouchers that will give you a discount:The first voucher will give you the cheaper phone for free, if you buy two phones....
One observation is that (sum of numerators) divided by (sum of denominators) is not well defined.For example, let's work with the two ratios $a=\frac01$ and $b=\frac11$.The ratio of the sum of numerators to sum of denominators is $\frac12$.However, we can also write $a=\frac03$ and $b=\frac22$. Now the ratio is $\frac25$, which is not equal to $\...
I like guest's answer. To elaborate, here is a possible question to ask them.You take two trips in your car:Trip 1 is a 100 mile drive that takes you 2 hours.Trip 2 is a 200 mile drive that takes you 1 hour.(a) What is the average speed of your car?(b) What is the average speed on an average trip?The answer to (a) is $\frac{...
It actually depends on exactly what you're asking. Or even what you SHOULD be asking.If you want the average profitability of all the 500+ operators in the Permian, you could just average all the profit margin percentages. This is taking the ratios (profit/revenue) for each company and averaging them. It corresponds to your expected (mean) profit margin ... |
This is homework and I'm looking for a push in the right direction. Proofs were never something I was properly taught, so now they're kind of a weak point.
Here's the problem:
The following grammar generates numbers in binary notation ($C$ is the start symbol):
$\qquad \begin{align}C &\to C 0 \mid A 1 \mid 0 \\ A &\to B 0 \mid C 1 \mid 1 \\ B &\to A 0 \mid B 1 \end{align}$
Prove that the alternating sums of the digits of the generated numbers are multiples of $3$. The alternating sum of $w=w_0\dots w_n$ is defined as $\sum_{i=0}^n (-1)^i \cdot w_i$. As an example, $C$ generates $1001$ via $C \Rightarrow A1 \Rightarrow B01 \Rightarrow A001 \Rightarrow 1001$ with alternating sum of $0$; clearly, $0$ is a multiple of $3$.
Prove that all such numbers (i.e., numbers whose alternating sum is a multiple of 3) are generated by the grammar.
I'm thinking I need to show that the grammar can only generate strings which are made up of repeated subsequences of digits which always add up to 0, 3, or -3. But I'm not sure how to show that it can only generate those three subsequences.
I also have worked out these thoughts:
Consider that any even number of consecutive 1s is irrelevant, as they cancel each other out.
Consider that all zeros are in of themselves irrelevant, as they add nothing.
Consider then that the only relevant pattern is that of alternating 1s and zeros, and where this pattern starts and ends. |
In 1917, Abram Besicovitch asked the following question:
If \(f:\mathbb{R}^2 \to \mathbb{R}\) is Riemann-integrable, does there exist a pair of orthogonal coordinate axes which would render both \(x \mapsto \int f(x,y) \, dy \) and \(y \mapsto \int f(x,y) \, dx\) Riemann-integrable?
Besicovitch noted that a counterexample could be given if a compact set of measure zero in \(\mathbb{R}^2\) that contains a line segment in every direction could be constructed.
To see this, we assume that \(F\) is such a set. We fix a pair of axes and translate \(F\) so that the line segments in \(F\) parallel to the axes are of irrational distance apart from the axes. Let \(F_0\) be the set of points in \(F\) consisting of at least one rational coordinate with respect to the axes chosen. For each line segment \(l\) in \(F\), we observe that both \(F_0 \cap l\) and \((F \smallsetminus F_0) \cap l\) are dense in \(l\).
This results in severe "one-dimensional discontinuity" of the characteristic function \(\chi_{F_0}\) in each direction. This, in turn, implies that \(x \mapsto \chi_{F_0}(x,y)\) is not Riemann-integrable regardless of the choice of axes or \(y\). \(F\) is of planar measure zero, however, and so \(\chi_{F_0}\) is Riemann-integrable on \(\mathbb{R}^2\) by the Lebesgue criterion for Riemann integrability.
Two years later, Besicovitch succeeded in constructing such a set:
Theorem(Besicovitch, 1919). There exists a compact planar set of measure zero that contains a unit line segment in each direction.
We construct such a set by considering first an equilateral triangle. We slice it up into thin pieces and overlap them in different directions. This yields a set with smaller measure that nevertheless contains a wider variety of unit line segments than the original triangle.
Continuing this way, we end up with a set of measure zero that contains a unit line segment in each direction from, say, 0 to 90 degrees. Piecing together several of these sets yields the desired set. See Terry Tao's interactive applet on the first ten steps of the construction.
The above construct generalizes straightforwardly to higher dimensions. For the sake of efficiency, it is convenient to make the following definition:
Definition.A (or a Kakeya set ) in \(\mathbb{R}^n\) is a compact set in \(\mathbb{R}^n\) that contains a unit line segment in each direction, viz., Besicovitch set
\[\forall e \in \mathbb{S}^{n-1} \, \, \exists x \in \mathbb{R}^n \, \, \forall t \in [-1/2,1/2] \, \, x + te \in E/\]
By considering the products of the two-dimensional Besicovitch sets, we obtain the following results:
Theorem(Besicovitch, 1919). If \(n \geq 2\), then there are Kakeya sets of measure zero in \(\mathbb{R}^n\).
Why the name
? Unbeknownst to Besicovitch, the Japanese mathematician Soichi Kakeya was investigating a similar problem in 1917: Kakeya sets Question(Kakeya needle problem). What is the minimum area of a compact planar set in which a unit line segment can be rotated 180 degrees?
The civil war and the blockade prevented the rest of the world from finding out about Besicovitch's work at the time. For a number of years the three-cusped hypercycloid was considered to be the optimal solution.
Besicovitch eventually learned of the problem, shortly after his departure from Russia in 1924. By slightly modifying his construction of the two-dimensional Kakeya set, Besicovitch was able to establish the following:
Theorem(Besicovitch, 1928). For each \(\varepsilon > 0\), there exists a compact planar set of measure \(\varepsilon\) in which a unit line segment can be rotated 180 degrees.
And so the needle problem can be solved on a
very small set. But just how small is very small? This seemingly innocent question is, in fact, a starting point of a deep theory, which serves as a meeting ground for several major areas of modern mathematics.
In this post, we explore the connections between the Kakeya problem and the
Fourier transform
\[\hat{f}(\xi) = \int_{\mathbb{R}^n} f(x) e^{-2 \pi i x \cdot \xi} \, dx,\]
defined for functions \(f: \mathbb{R}^n \to \mathbb{C}\). Here \(\cdot\) denotes the standard dot product.
If \(f \in L^1\), we can make sense of the above definition. Moreover, the Fourir transform \(\hat{f}:\mathbb{R}^n \to \mathbb{C}\) is continuous, decays at infinity, and satisfies the estimate
\[\|\hat{f}\|_\infty \leq \|f\|_1,\]
which implies that the Fourier transform is a bounded linear operator from \(L^1\) to \(L^\infty\).
How about functions in \(L^p < p < \infty\)?
Clearly, the integral definition makes no sense if the function is not in \(L^1\). We can nevertheless define the Fourier transform operator by considering a dense subset of \(L^p\) on which the integral definition is valid.
The prime candidate is the
\(\mathscr{S}\), which consists of infinitely-differentiable functions on \(\mathbb{R}^n\) such that all of their partials, including the functions themselves, decrease more rapidly than polynomials. Since \(\mathscr{S}\) is dense in \(L^p\) for all \(1 \leq p < \infty\), the definition of Fourier transform is valid on \(\mathscr{S}\). Schwartz space
In fact, the Fourier transform is a homeomorphism of \(\mathscr{S}\) onto itself. The inverse map, called the
, is defined by Fourier inversion formula
\[f^\vee(x) = \int_{\mathbb{R}^n} f(\xi) e^{2\pi i \xi\cdot x} \, d\xi\]
for each \(f \in \mathscr{S}\). The inversion formula establishes the identity
\[\|\hat{f}\|_2 = \|f\|_2\]
for all \(f \in \mathscr{S}\). It follows that we can extend the Fourier transform operator onto all of \(L^2\) such that the above norm estimate holds for all \(f \in L^2\). This is
. Plancherel's theorem
An "interpolation" theorem (specifically, the Riesz–Thorin theorem) now establishes the
Hausdorff–Young inequality
\[\|\hat{f}\|_{p'} \leq \|f\|_p\]
for all \(1 \leq p \leq 2\), where \(p'\) is the
of \(p\), defined by the identity \(p^{-1} + (p')^{-1} = 1\). conjugate exponent
Unfortunately, the Fourier transform is not a bounded operator into another Lebesgue space for \(p>2\). How do we, then, make sense of the equality
\[f = (\hat{f})^\vee\]
for general \(f \in L^p\)? More precisely, if
\[S_Rf(x) = \int_{\vert \xi \vert \leq R} \hat{f}(\xi) e^{2 \pi i \xi \cdot x} \, d\xi,\]
in what sense does \(S_R f\) converge to \(f\) as \(R \to \infty\)?
The \(p=1\) case is hopeless, for Andrey Kolmogorov exhibited an \(L^1\) functino whose Fourier inversion fails to converge at every point. We thus restrict our attention to \(1 < p < \infty\).
For \(n = 1\), there is the pointwise almost-everywhere result, established for \(L^2\) functions by Lennart Carleson in 1966 and extended to \(L^p\) functions for all \(1 < p < \infty\) by Richard Hunt two years later.
As for \(n > 1\), we must first consider the method of summing the integrals. If the integral is summed over rectangles, then the multivariate almost-everywhere convergence follows from the Carleson–Hunt theorem. The pointwise convergence result over balls is still open.
How about \(L^p\) convergence? For \(n=1\), we have the classical theorem of Marcel Riesz:
Theorem(M. Riesz, 1928). Fix \(1 < p < \infty\), let \(f \in L^p(\mathbb{R})\), and assume that \(\hat{f}\) exists. Then \(S_R f \to f\) in $L^p$$.
How about \(n > 1\)? Again, spherical summation is the difficult part of the multivariate case. The \(L^2\) convergence is given by the Plancherel theorem, but the convergence result fails to materialize for all other values of \(p\).
Theorem(Fefferman ball multiplier theorem, 1971). The spherical summation of the Fourier inversion formula converges in the norm topology of \(L^p\) if and only if \(p = 2\).
By a result of de Leeuw, settling the ball multipler problem on \(n = k\) settles it for \(n = k-1\). It thus suffices to consider the ball multipler problem for \(n = 2\).
Moreover, the Riesz representation theorem carries over the ball multipler theorem on \(L^p\) to its conjugate exponent, and so it suffices to consider the \(p > 2\) case.
Finally, the uniform boundedness principle implies that the convergence of spherical summation is equivalent to the norm estimate
\[\|S_R f\|_p \leq k_p \|f\|_p\]
for all sufficiently large \(R\). A scaling argument shows that the above estimate is, in fact, equivalent to the single norm estimate
\[\|S_1 f \|_p \leq k_p \|f\|_p.\]
It thus suffices to exhibit, for each \(N > 0\) and every \(p > 2\), a non-zero function \(f\) such that
\[\|S_1 f \|_p \geq N \|f\|_p.\]
By taking \(f\) to be a sum of characteristic functions on thin needles in different directions, we can minimize \(S_1 f\) while maximizing its \(L^p\) norm. It follows that the above inequality fails.
How badly does the convergence fail? To this end, we consider the
Bochner–Riesz mean
\[S_R^\delta f(x) = \int_{\vert \xi \vert \leq R} \hat{f}(\xi) e^{2 \pi i \xi \cdot x} \left( 1 - \frac{\vert \xi \vert^2}{R^2} \right)^{\delta} \, d\xi\]
for each \(R > 0\) and \(\delta \geq 0\).
\(S_R^0\) is the regular spherical summation, whose lack of convergence we have discussed above. For the \(\delta > 0\) case, a similar reduction argument as above implies that the \(L^p\) convergence of the Bochner–Riesz mean is equivalent to the norm estimate
\(\|S_1^\delta f \|_p \leq k_p\|f\|_p\).
By a result of Herz, the above estimate does not hold unless
\[\left\vert \frac{1}{p} - \frac{1}{2} \right\vert < \frac{2\delta +1}{2n}.\]
The
posits that the \(L^p\) norm estimate does hold for all values of \(p\) that satisfy the above restriction. Bochner–Riesz conjecture
As it turns out, Fefferman's construction could disprove the Bochner–Riesz conjecture if it can be modified to solve the \(n\)-dimensional needle problem on a set
too small to be considered \(n\)-dimensional.
To make this notion precise, we introduce the concept of Hausdorff measure and dimension.
Fix \(m \in \mathbb{N}\), and let \(\omega_m\) be the \(m\)-dimensional Lebesgue measure of the closed unit ball in \(\mathbb{R}^m\). The
\(m\)-dimensional Hausdorff outer measure \(\mathscr{H}^m\) defined for every subset \(E \subseteq \mathbb{R}^n\) by
\[\mathscr{H}^m(E) = \lim_{\delta \to 0} \inf_{\substack{E \subseteq \bigcup S_j \\ \operatorname{diam}(S_j) \leq \delta}} \sum_{j=1}^\infty \omega_m \left( \frac{\operatorname{diam}(S_j)}{2}\right)^m,\]
where the infimum is taken over a countable cover \((S_j)_{n=1}^\infty\) of \(E\) of diameter at most \(\delta\).
More generally, we let
\[\omega_s = \frac{\pi^{s/2}}{\Gamma(s/2+1)}\]
for each \(s > 0\) and define the
\(\mathscr{H}^s\) of \(E \subseteq \mathbb{R}^n\) to be \(s\)-dimensional Hausdorff outer measure
\[\mathscr{H}^s(E) = \lim_{\delta \to 0} \inf_{\substack{E \subseteq \bigcup S_j \\ \operatorname{diam}(S_j) \leq \delta}} \sum_{j=1}^\infty \omega_s \left( \frac{\operatorname{diam}(S_j)}{2}\right)^s.\]
The
of \(E \subseteq \mathbb{R}^n\) is defined to be either the supremum of all \(s\) such that \(\mathscr{H}^s(E) = \infty\) or the infimum of all \(s\)such that \(\mathscr{H}^s(E) = 0\), which always agree. Hausdorff dimension
The Hausdorff measure is defined from the Hausdorff outer measure via the standard Carathéodory construction. Here we have chosen the normalization so that the \(n\)-dimensional Hausdorff measure coincides with the \(n\)-dimensional Lebesgue measure.
With this normalization, \(\mathscr{H}^0\) is the counting measure. \(\mathscr{H}^1\) of a rectifiable curve is its length. In general, the \(m\)-dimensional Hausdorff measure of an \(m\)-manifold coincides with its canonical surface measure.
The Hausdorff dimension \(E \subseteq \mathbb{R}^n\) specifies the
fair dimension of \(E\) in the following sense. While a curve and a surface both have zero \(\mathscr{H}^3\)-value, it would be unfair to say that they are equally small. Indeed, a curve would have zero \(\mathscr{H}^2\)-value as well.
We are now ready to state the following
Kakeya set conjecture.Every Kakeya set in \(\mathbb{R}^n\) is of Hausdorff dimension \(n\).
Let us transform the Kakeya set problem into a problem in harmonic analysis. For each \( \delta > 0\), \(e \in \mathbb{S}^{n-1}\), and \(a \in \mathbb{R}^n\), we let \(T_e^\delta(a)\) be the set of all \(x \in \mathbb{R}^n\) such that \(\vert (x-a) \cdot e \vert \leq 1/2\) and \(\vert x - (x \cdot e) e \vert \leq \delta\). This is essentially the \(\delta\)-neighborhood of the unit line segment in the direction of \(e\) centered at \(a\).
We define the
to be Kakeya maximal function
\[f_\delta^*(e) = \sup_{a \in \mathbb{R}^n} \frac{1}{m(T^\delta_e(a))} \int_{T^\delta_e(a)} \vert f(x) \vert \, dx\]
for each \(e \in \mathbb{S}^{n-1}\). The following conjecture implies the Kakeya set conjecture:
Kakeya maximal function conjecture.For all \(\varepsilon > 0\), there exists a constant \(C_\varepsilon > 0\) such that
\[\|f_\delta^*\|_{L^n(\mathbb{S}^{n-1})} \leq C_\varepsilon \delta^{-\varepsilon } \|f\|_n.\]
Here \(\|\cdot\|_{L^n(\mathbb{S}^{n-1})}\) is the norm of the Lebesgue space on \(\mathbb{S}^{n-1}\), defined to be
\[\|f\|_{L^n(\mathbb{S}^{n-1})} = \int_{\mathbb{S}^{n-1}} \vert f(x) \vert^n \, d \sigma(x).\]
\(d\sigma\) is the canonical surface measure on \(\mathbb{S}^{n-1}\), which coincides with the \((n-1)\)-dimensional Hausdorff measure on \(\mathbb{S}^{n-1}\).
The formulation of the Kakeya maximal function conjecture is, in part, inspired by the problem of restricting the Fourier transform onto a lower-dimensional subset of \(\mathbb{R}^n\). Specifically, we consider the spherical Fourier transform
\[\widehat{f \, d\sigma}(\xi) = \int_{\mathbb{S}^{n-1}} f(\xi) e^{-2 \pi i x \cdot \xi} \, d\sigma(x)\]
for functions \(f:\mathbb{S}^{n-1} \to \mathbb{C}\). The following conjecture implies the Kakeya maximal conjecture.
Stein's restriction conjecture.If \(f \in L^\infty(\mathbb{S}^{n-1})\), then, for each \(q > \frac{2n}{n-1}\), there exists a constant \(C_q > 0\) such that
\[\|\widehat{f \, d\sigma}\|_q \leq C_q \|f\|_\infty.\]
It is known that the Bochner–Riesz conjecture implies the restriction conjecture, whence it follows that the Bochner–Riesz conjecture implies the Kakeya set conjecture.
The Kakey aproblem is an active area of research, with contributions from some of the finest analysts of our time. See (Lab) or (Tao1) for a quick survey. More systematic treatments can be found in (Wol) and (Tao2). |
Yes, you can do symbolic math in Python! The library to take a look at is
SymPy. Its official website is SymPy.org. This article is not a SymPy tutorial, as I only want to walk you through some examples to show you the kinds of things that it can do. A good place to start if you want more information is the SymPy Tutorial from the official documentation.
Now, to pick an example… Given the site that you’re reading this on, what better example to use than the
sinc function, as used in the creation of windowed-sinc filters? The sinc function is defined as
\[\mathrm{sinc}(x)=\frac{\sin(\pi x)}{\pi x}.\]
This is the
normalized version that is typically used in signal processing. It is normalized by multiplying \(x\) with \(\pi\). The effect of this is that the integral becomes one instead of \(\pi\), and that the zeros of the function are (all the non-zero) integer values of \(x\). The central part of the normalized sinc function is shown in Figure 1.
In the code fragment that follows, I’ll define the sinc function in SymPy. There are lots of imported symbols in the first two lines, for the other examples that follow.
>>> from sympy import symbols, sin, pi, pprint, latex, Limit, limit, \ ... Integral, oo, integrate, diff >>> x = symbols('x') >>> sinc = sin(pi * x) / (pi * x) >>> pprint(sinc) sin(π⋅x) ──────── π⋅x
pprint() is a SymPy function that does pretty printing, as you can tell. You can also convert the output to \(\LaTeX\) using the
latex() function, as follows:
>>> print(latex(sinc)) \frac{1}{\pi x} \sin{\left (\pi x \right )}
If I then put this into the current article in the usual way, it renders as
\[\frac{1}{\pi x} \sin{\left (\pi x \right )}.\]
You can then evaluate an expression by subsituting a value for one or more of the symbols. Let’s evaluate our sinc function for some values of \(x\):
>>> sinc.subs(x, 1) 0 >>> sinc.subs(x, -2) 0 >>> sinc.subs(x, 0.5) 2.0/pi >>> sinc.subs(x, 0) nan
The evaluation for \(x=0\) returns
nan. This is because the sinc function has the indeterminate form of \(0/0\) at that point. Of course, since we are looking at a symbolic math toolbox, let’s compute the limit for \(x\to0\). We can first let SymPy generate the \(\LaTeX\) expression, as follows:
>>> print(latex(Limit(sinc, x, 0))) \lim_{x \to 0^+}\left(\frac{1}{\pi x} \sin{\left (\pi x \right )}\right)
This renders as
\[\lim_{x \to 0^+}\left(\frac{1}{\pi x} \sin{\left (\pi x \right )}\right),\]
but remember that you can also print it directly on your terminal screen using
pprint():
>>> pprint(Limit(sinc, x, 0)) ⎛sin(π⋅x)⎞ lim ⎜────────⎟ x─→0⁺⎝ π⋅x ⎠ >>> pprint(Limit(sinc, x, 0), use_unicode=False) /sin(pi*x)\ lim |---------| x->0+\ pi*x /
The formatting with Unicode (the first
pprint() statement) is not entirely correct here, but it looks really good in an actual terminal window. Not that the above code uses
Limit(), with a capital
L. Evaluating the limit can be done by using the
limit() function (with lowercase
l), or by calling the
doit() member on the result of the call to
Limit(). Evaluating the limit results in
>>> print(limit(sinc, x, 0)) 1
Note that this result implies that SymPy is smart enough to apply L’Hôpital’s rule, because the limit expression
also results in the indeterminate form of \(0/0\). If you want to apply L’Hôpital’s rule manually, you can compute the derivative of the numerator and of the denominator and then compute the resulting limit. >>> print(diff(sin(pi * x))) pi*cos(pi*x) >>> print(diff(pi * x)) pi >>> print(limit(diff(sin(pi * x)) / diff(pi * x), x, 0)) 1
Finally, now that we’re doing calculus anyway, let’s evaluate the integral from \(-\infty\) to \(+\infty\) of the sinc function:
>>> pprint(Integral(sinc, (x, -oo, oo))) ∞ ⌠ ⎮ sin(π⋅x) ⎮ ──────── dx ⎮ π⋅x ⌡ -∞ >>> print(integrate(sinc, (x, -oo, oo))) 1
This is the expected result. Note the cute use of
oo for infinity… Anyway, do remember, for when the need arises, that you
can use Python for symbolic math! |
From The design of APX algorithms book by
David P. Williamson and David B. Shmoys, at the bottom of page 21 I saw the following statement (it is about the set cover LP and its dual):
Let $y^*$ be an optimal solution to the following dual LP $$max \sum_{i=1}^n{y_i}$$$$subject \sum_{i:e_i\in{S_j}}{y_i \leq w_j}, y_i \geq0; j=1,...,m;i=1,...,n$$ and consider the solution in which we choose all subsets for which the corresponding dual inequality is
tight; that is, the inequality is met with equality for subset $S_j$, and $\sum_{i:e_i \in{S_j}}{y^* = w_j}$. Let $I^{'}$ denote the indices of the subsets in this solution. We will prove that this algorithm also is an $f$-approximation algorithm for the set cover problem
I would like to know what does the writer mean by
corresponding dual inequality? Is there any correspondence between a solution of the Set Cover problem and a dual LP solution?
I can't understand what does the paragraph mean.
Thanks in advance. |
I have a certain convex optimization problem which depends on the order of a set of $N$ objects. Meaning, for every possible ordering, I get a different convex optimization problem. My ultimate goal is to find the best optimal solution given all possible orderings. One way of doing that is by solving all $n!$ possible problems. A better way (I think) is to write the whole thing as one MILP and let MOSEK take care of the hard work. I did that as follows: I created $N$ binary variables for every object for a total of $N^2$ binary variables. Say the objects are $F_1,\ldots,F_N$ and the binary variables are $b_{kj}$ where $k,j\in\{1,\ldots,N\}$.
I let $$b_{kj} = \left\{ \begin{array}{ll} 1 & \text{if object $F_k$ goes into position $j$;} \\ 0 & \text{otherwise}. \end{array} \right.$$
In order to guarantee that every object goes into exactly one position, I force the following constraint for every $k$: $$\sum_{j=1}^N b_{kj} = 1$$ Also, in order to guarantee that every position receives exactly one object, I force the following constraint for every $j$: $$\sum_{k=1}^N b_{kj} = 1$$
This worked for me and I was able to solve the problem using MOSEK.
My
first question is: is there a better way of doing this? Possibly with fewer than $N^2$ binary variables?
My
second question is: Let's say I want object $p$ to come before object $q$. How do I force that in the constraints? The only idea I was able to come up with is the following. We must have the following implication for every $k \in \{1,\ldots,N-1\}$.$$b_{pk} = 1 \Rightarrow b_{qk} = \ldots = b_{qN} = 0$$Which we translate into:$$b_{pk} = 0 \vee b_{qk} = \ldots = b_{qN} = 0$$which would need additional binary variables to deal with due to the "OR". Is there a better idea?
Thank you |
Say a black hole is travelling at $c/2$, does the shape of the event horizon change? What about the location of the event horizon? If it is travelling at a hypothetical $c$, does the event horizon simply resemble a light cone?
It is convenient to go from Schwarzschild coordinates $$ ds^2 = \left( 1 - \frac{r_s}{r}\right) c^2 dt^2 - \left( 1 - \frac{r_s}{r}\right)^{-1}dr^2 - r^2 d\Omega $$ to Isotropic coordinates $r\rightarrow r^2(1+ r_s/4r)^4$, then to quasi-Minkowski using the usual spherical-cartesian coordinate transformation to get $$ ds^2 = -\left( \frac{1 - r_s/4r}{1 + r_s/4r} \right)^2 c^2 dt^2 + \left( 1 + r_s/4r\right)d\vec{x}^2 $$ Now you can finally use the familiar Lorentz boost formula \begin{align*} ct & \rightarrow \gamma\left(ct - \vec{\beta}\cdot\vec{x}\right) \\ \vec{x} & \rightarrow \vec{x} - \gamma \vec{\beta}ct + \frac{\gamma-1}{\beta^2}\left(\vec{\beta}\cdot\vec{x}\right)\vec{\beta}\\ \Rightarrow r^2 &\rightarrow \left| \vec{x} - \vec{\beta}ct\right|^2 + \gamma^2 \left(\vec{\beta}\cdot\left( \vec{x} - \vec{\beta}ct\right) \right)^2 \end{align*}
Only now you can start asking physical questions using the new coordinate system. For example let's look at the event horizon. In the old coordinates (pre boosted isotropic quasi-minkowski) requiring $g^{00}=0$ gave us the spherical surface surface $r=r_s/4$. Now let's look at our new $g^{00}$ \begin{align*} g^{00} = \gamma^2\left( -\left( \frac{1-r_s/4r'}{1+r_s/4r'}\right)^2 + \beta^2 \left(1+r_s/4r' \right) \right) \end{align*} So using $g^{00}=0$ the horizon in the new coordinate system is found to be $$ r' = \frac{r_s}{4}\frac{1+\beta}{1-\beta} $$ $\Rightarrow$ $$ \left| \vec{x} - \vec{\beta}ct\right|^2 + \gamma^2 \left(\vec{\beta}\cdot\left( \vec{x} - \vec{\beta}ct\right) \right)^2 = \left( \frac{r_s}{4}\frac{1+\beta}{1-\beta} \right)^2 $$ Now since our new coordinates are isotropic the metric is unchanged if we had aligned the $z-$axis with $\vec{\beta}$, in which case you get an ellipsoid with moving center along the $z-$axis $(x_0,y_0,z_0) =(0,0,vt)$ and contracted along the $z$-axis.
Note: proof read for arithmetic mistakes.
The event horizon is by definition a set of points in spacetime that is independent of the observer. Therefore its location is independent of the state of motion of the black hole and the observer relative to each other.
We could ask whether the moving observer thinks the event horizon is spherical. This is not a question that is meaningful in general relativity, unless you specify what kind of observations are involved. When we define spherical symmetry in GR, we take pains to do it in a way that is independent of coordinates or observers. You might think that you could take the intersection of the event horizon with a surface of simultaneity for a particular observer, but GR doesn't define that kind of surface of simultaneity.
If you want to define how the horizon appears to an observer, one natural way to do it would be to use optical observations. In SR, the optical silhouette of a sphere is always a disk, regardless of the state of motion, so I suspect that the same would be true for the silhouette of a black hole. This does seem to be true, as far as I can tell, in figures 25-28 of Riazuelo, https://arxiv.org/abs/1511.06025 , although it's not especially clear. I've done some similar simulations myself, but I only did radial motion, so I haven't actually seen what happens for transverse motion.
But of course, even if we established that optical measurements showed the silhouette of a Schwarzschild black hole as a disk for an observer in any state of motion, that isn't an answer as to the question of whether the event horizon appears length contracted, since the result is the same as the SR result for the silhouette of a sphere. Perhaps there is some other type of observation that you could define that would give a natural sense in which this becomes a meaningful question, but it's not immediately obvious to me what type of observation that would be. |
Degree $n$ : $40$ Transitive number $t$ : $15$ Group : $C_5\times OD_{16}$ Parity: $-1$ Primitive: No Nilpotency class: $2$ Generators: (1,21,3,23,2,22,4,24)(5,26,7,28,6,25,8,27)(9,32,11,30,10,31,12,29)(13,35,16,34,14,36,15,33)(17,37,19,40,18,38,20,39), (1,14,25,37,10,24,33,6,18,30,3,15,27,40,12,21,35,8,20,31,2,13,26,38,9,23,34,5,17,29,4,16,28,39,11,22,36,7,19,32) $|\Aut(F/K)|$: $20$
|G/N| Galois groups for stem field(s) 2: $C_2$ x 3 4: $C_4$ x 2, $C_2^2$ 5: $C_5$ 8: $C_4\times C_2$ 10: $C_{10}$ x 3 16: $C_8:C_2$
Resolvents shown for degrees $\leq 10$
There are no siblings with degree $\leq 10$ A number field with this Galois group has no arithmetically equivalent fields.
There are 50 conjugacy classes of elements. Data not shown.
Order: $80=2^{4} \cdot 5$ Cyclic: No Abelian: No Solvable: Yes GAP id: [80, 24]
Character table: Data not available. |
Strictly speaking, $O(f(n))$ is a set of functions. So the value of $O(f(n))$ is simply the set of all functions that grow asymptotically not faster than $f(n)$. The notation $T(n) = O(f(n))$ is just a conventional way to write that $T(n) \in O(f(n))$.Note that this also clarifies some caveats of the $O$ notation.For example, we write that $(1/2) n^2 + ...
The paragraph is wrong. Unfortunately, it looks exactly like the kind of thing that a student who does not understand the material would write as an answer to an exercise. This sort of nonsense has no place in a textbook. Make no sudden movements. Put the book down. Step away from the book.We say that the "order of growth" of the sequential search ...
You are referring to the Landau notation. They are not different symbols for the same thing but have entirely different meanings. Which one is "preferable" depends entirely on the desired statement.$f \in \cal{O}(g)$ means that $f$ grows at most as fast as $g$, asymptotically and up to a constant factor; think of it as a $\leq$. $f \in o(g)$ is the ...
If you want rigorous proof, the following lemma is often useful resp. more handy than the definitions.If $c = \lim_{n\to\infty} \frac{f(n)}{g(n)}$ exists, then$c=0 \qquad \ \,\iff f \in o(g)$,$c \in (0,\infty) \iff f \in \Theta(g)$ and$c=\infty \quad \ \ \ \iff f \in \omega(g)$.With this, you should be able to order most of the ...
$O$ is a function$$\begin{align}O : (\mathbb{N}\to \mathbb{R}) &\to \mathbf{P}(\mathbb{N}\to \mathbb{R})\\ f &\mapsto O(f)\end{align}$$i.e. it accepts a function $f$ and yields a set of functions that share the asymptotic bound of (at most) $f$. And strictly speaking the correct notation is thus$$(n \mapsto T(n)) \in O(n\mapsto f(n))$$...
Big O: upper bound“Big O” ($O$) is by far the most common one. When you analyse the complexity of an algorithm, most of the time, what matters is to have some upper bound onhow fast the run time¹ grows when the size of the input grows. Basically we want to know that running the algorithm isn't going to take “too long”. We can't express this in actual time ...
Landau notation denotes asymptotic bounds on functions. See here for an explanation of the differences among $O$, $\Omega$ and $\Theta$.Worst-, best-, average or you-name-it-case time describe distinct runtime functions: one for the sequence of highest runtime of any given $n$, one for that of lowest, and so on..Per se, the two have nothing to do with ...
You are right. Notice that the term $O(n+m)$ slightly abuses the classical big-O Notation, which is defined for functions in one variable. However there is a natural extension for multiple variables.Simply speaking, since$$ \frac{1}{2}(m+n) \le \max\{m,n\} \le m+n \le 2 \max\{m,n\},$$you can deduce that $O(n+m)$ and $O(\max\{m,n\})$ are equivalent ...
Believe it or not, it seems (in my experience) that many algorithms people have actually not thought about what the big O notation formally means, and when asked about it, you can get several different answers. Some issues are discussed in the paper On Asymptotic Notation with Multiple Variables by Rodney R. Howell.Curiously, it also seems that most ...
To rationalize how asymptotic notations ignore constant factors, I usually think of it like this: asymptotic complexity isn't for comparing performance of different algorithms, it's for understanding how performance of individual algorithms scales with respect to the input size.For instance, we say that a function that takes $3n$ steps is $O(n)$, because, ...
Typically $O$ is used for stating upper-bounds (an estimate from above), while $\Omega$ is used to state lower-bounds (an estimate from below), and $\Theta$ is used when they match, in which case you can use $\Theta$ in place of them (usually) to state the result.
In mathematics, functions like this are called multilinear functions. But computer scientists probably won't generally know this terminology. This function should definitely not be called linear, either in mathematics or computer science, unless you can reasonably consider one of $m$ and $n$ a constant.
Consider the following algorithm (or procedure, or piece of code, or whatever):Contrive(n)1. if n = 0 then do something Theta(n^3)2. else if n is even then3. flip a coin4. if heads, do something Theta(n)5. else if tails, do something Theta(n^2)6. else if n is odd then7. flip a coin8. if heads, do something Theta(n^4)9. else if ...
Wikipedia says:An algorithm is said to be of polynomial time if its running time is upper bounded by a polynomial expression in the size of the input for the algorithm.$\mathcal{O}(\log n)$ is upper bounded by $\mathcal{O}(n)$, and $\mathcal{O}(n \log n )$ is upper bounded by $\mathcal{O}(n^2)$, therefore they are both in $P$.
Skiena provides a sorted list of the dominance relations between the most common functions in his book, The Algorithm Design Manual:$$n!\gg c^n \gg n^3 \gg n^2 \gg n^{1+\epsilon} \gg n \lg n \gg n \gg n^{1/2}$$$$ \gg \lg^2n \gg \lg n \gg \frac{\lg n}{\lg\lg n} \gg \lg\lg n \gg \alpha(n) \gg 1$$
There are many examples for such functions. Perhaps the easiest way to understand how to get such an example, is by manually constructing it.Let's start with function over the natural numbers, as they can be continuously completed to the reals.A good way to ensure that $f\neq O(g)$ and $g\neq O(f)$ is to alternate between their orders of magnitude. For ...
You need to ignore for a moment the strong feeling that the "$O$" is in the wrong place and plough on with the definition regardless. $f(n) = \log^{O(1)}n$ means that there exist constants $k$ and $n_0$ such that, for all $n\geq n_0$, $f(n) \leq \log^{k\cdot 1}n = \log^k n$.Note that $\log^k n$ means $(\log n)^k$. Functions of the form $\log^{O(1)}n$ are ...
Recall that for $k > 1$, by definition we have $\log^*k = \log^*(\log{k}) + 1$.By applying the definition twice, we see that $\log^*(2^{2^n}) = \log^*n + 2$. Now we can compare $\log^*n + 2$ and $(\log^*n)^2$.
It's certainly not cheating. Think in calculus how substitution may be used to solve a tricky integral. The substitution makes the equation more manageable for manipulation. Additionally, substitution may transform somewhat complex recurrences into familiar ones.This is exactly what occurred in your example. We define a new recurrence $S(m)=T(2^m)$. ...
We first need to clarify what we mean by "does this hold if we have an infinite chain?".We interpret it as an infinite sequence of functions $\{f_i:\mathbb{N}\to\mathbb{N} \mid 1\leq i\}$ such that for all $i$ we have $f_i(n) = O(f_{i+1}(n))$. Such a sequence might not have a last function.We can look at the limit of the functions in the sequence, i.e. $...
It's a variant of the big-O that "ignores" logarithmic factors:$$f(n) \in \widetilde O \left(h(n) \right)$$is equivalent to:$$ \exists k : f(n) \in O \left( h(n) log^k(h(n)) \right) $$From wikipedia:Essentially, it is big O notation, ignoring logarithmic factors because the growth-rate effects of some other super-logarithmic function indicate a ...
Do what you did, but let $a = (3^{0.2})$... that should do it, right?The reason that what you did didn't work is as follows. The big-oh bound is not tight; while the logarithm to the fifth is indeed big-oh of linear functions, it is also big oh of the fifth root function. You need this stronger result (which you can also get from the theorem) to do what ...
Let me start of with a recommendation: treat Landau notation just as you (should) treat rounding: round rarely, round late. If you know something more precise than $O(.)$, use it until you are done with all calculations, and Landauify at the end.As for the question, let's dig through this abuse of notation¹. How would we interpret something like $h \in O(f ...
Tell the lecturer they're wrong. Take the function$$ f(n) = \begin{cases} n & n \text{ is even}, \\ 1 & n \text { is odd}. \end{cases} $$This function is $O(n)$ but neither $o(n)$ nor $\Theta(n)$.Here is a monotone example, which might be more convincing:$$ g(n) = \exp \exp \lfloor \ln \ln n \rfloor. $$
Yes, of course. This is fine and perfectly acceptable. It is common and standard to see algorithms whose running time depends upon two parameters.For instance, you will often see the running time of depth-first search expressed as $O(n+m)$, where $n$ is the number of vertices and $m$ is the number of edges in the graph. This is perfectly valid. The ...
First, let's unpack what $\Theta(1)$ means.Big $O$, and big $\Theta$, are classes of functions. There's a formal definition here, but for the purposes of this question, we say that a function $f$ is in $O(1)$ if there's a constant $c$ where, for all $x$, $f(x) \leq C$. That is, $f$ grows at most as fast as a constant function.Big-$\Theta$ doesn't mean ...
Formally speaking, $O(f(n))$ is a the set of functions $g$ such that $g(n)\leq k\,f(n)$ for some constant $k$ and all large enough $n$. Thus, the most pedantically accurate way of writing it would be $T(n)\in O(f(n))$. However, using $=$ instead of $\in$ is completely standard, and $T(n)=O(f(n))$ just means $T(n)\in O(f(n))$. This is essentially ...
Let's say the final goal is to prove $T(n) = \mathcal{O}(n)$. You start with the induction hypothesis:$T(i) \leq ci$ for all $i < n$.And to complete the proof, you have to show that $T(n) \leq cn$ as well.However, what you're able to deduce is $T(n) \leq (c+1)n$, which is not helpful to complete the proof; you need one constant $c$ for (almost) all ... |
I am trying to numerically evaluate a Greens function for this equation:
$$ \left[\frac{\partial^2}{\partial x^2} + f(x) \right] G(x) = \delta (x-x_0) $$
With Neumann boundary conditions. Here, the function $f(x)$ is known. The difficulty in solving this lies in the numerical approximation of the $\delta$ function. Rather than approximate $\delta \sim \frac{1}{x}$ I attempted to resolve the $\delta$ by integrating both sides in a small interval around $x_0$, but this leaves me with conditions on the derivative of $G$ at $x_0$ that I don't know how to implement in a finite elements method.
I would really appreciate a reference for solving such equations (preferably with a finite elements method), or if someone could lend a guiding hand for solving such an equation. |
This page contains a JavaScript calculator of Hawking radiation and other parameters of a Schwarzschild black hole.
The original idea belongs to Jim Wisniewski, whose page from 2006 (link) appears to be no longer available available again, but since it is not archived by the Wayback Machine, I think my functional clone is still useful.
Wisniewski's original code included a fictitious unit of mass, the "standard industrial neuble", equivalent to a billion metric tons, from Will McCarthy's novel The Collapsium. I kept the unit, but decided to use instead the much more useful value of one solar mass as the initial mass.
As in Wisniewski's version, specifying any quantity causes the others to be recalculated accordingly (see source). The drop-down menus select the units of measure to be used for their corresponding input field.
An added feature is the calculation of the "peak photon" wavelength, corresponding frequency, and photon energy, representing the peak of the blackbody radiation curve per unit logarithm (of wavelength or frequency) that corresponds to the black hole temperature.
Quantity Value Units Expression Mass $M$ Radius $R = M \dfrac{2G}{c^2}$ Surface area $A = M^2 \dfrac{16 \pi G^2}{c^4}$ Surface gravity $\kappa = \dfrac{1}{M} \dfrac{c^4}{4G}$ Surface tides $d\kappa_R = \dfrac{1}{M^2}\dfrac{c^6}{4G^2}$ Entropy (dimensionless) $S = M^2 \dfrac{4\pi G }{\hbar c}$ Temperature $T = \dfrac{1}{M}\dfrac{\hbar c^3}{8\pi k_BG}$ Peak photons $\lambda_{\rm logpeak}=\dfrac{hc}{k_BT[{\rm W}(-4e^{-4})+4]}$ Effective density $\rho = \dfrac{3c^6}{32\pi G^3M^2}$ Luminosity $L = \dfrac{1}{M^2}\dfrac{\hbar c^6}{15360\pi G^2}$ Time to singularity $t_S = \dfrac{\pi GM}{c^3}$ Lifetime $t = M^3 \dfrac{5120\pi G^2}{\hbar c^4}$ Discussion
Wisniewski started his calculations with the standard formula for the Schwarzschild radius of a mass $M$:
$$R = \frac{2G}{c^2}M$$
As per [Hawking 1974], the thermodynamic temperature of such a black hole is
$$T = \frac{\kappa}{2\pi} = \frac{\hbar c^3}{8\pi k_BG}\frac{1}{M}.$$
Its surface area is
$$A = 4\pi R^2 = \frac{16\pi G^2}{c^4} M^2,$$
making the Hawking radiation luminosity at least
$$L = A\sigma T^4 = \frac{\hbar c^6}{15360\pi G^2 }\frac{1}{M^2}.$$
The dimensionless Bekenstein-Hawking entropy of the black hole is defined by
$$S = \frac{c^3A}{4G\hbar}=M^2\frac{4\pi G}{\hbar c}.$$
Multiply this value by the Boltzmann constant $k_B$ to get the entropy in conventional units.
At a distance $r$ from a black hole with mass $M$, the incident radiation flux is, therefore,
$$\Phi = \frac{L}{4\pi r^2} = \frac{\hbar c^6}{61440\pi^2 G^2r^2}\frac{1}{M^2} .$$
The amount of radiation actually intercepted by an object necessarily depends upon its exposed area.
The free-fall time $t_S$ from horizon to singularity is calculated as
$$t_s=\frac{1}{c}\int\frac{1}{\sqrt{\dfrac{2GM}{c^2r}-1}}~dr=\frac{\pi GM}{c^3}\simeq \frac{M}{M_\odot}\times 1.54\times 10^{-5}~{\rm s}.$$
The expression for $L$ makes it possible to calculate the lifetime of a black hole of given initial mass $M_0$, assuming no mass input. Luminosity means energy output, thus
$$-\frac{dE}{dt} = \frac{\hbar c^6}{15360\pi G^2}\frac{1}{M^2}.$$
Since $dE = dM c^2$,
$$-\frac{dM}{dt} = \frac{\hbar c^4}{15360 \pi G^2}\frac{1}{M^2}.$$
Separating variables and integrating, we obtain
$$t = \frac{5120\pi G^2}{\hbar c^4} M^3.$$
Plugging in the various constants, this works out to
$$t = \left(\frac{M}{M_\odot}\right)^3 \times 2.097 \times 10^{67}{\rm yr},$$
where $M_\odot=1.989\times 10^{30}~{\rm kg}$.
The lifetime of a $1~M_\odot$ black hole, therefore, is calculated as more than 57 orders of magnitude longer than the present age of the universe. But that does not take into account the fact that such a black hole is colder than the cosmic microwave background radiation bathing it. Therefore, whatever little energy it radiates, it actually receives more in the form of heat from the cosmos. So rather than shrinking, it would continue to grow. Indeed, any black hole with a mass greater than about 0.75% of the Earth's mass is colder than the cosmic background, and thus its mass increases for now. As the universe expands and cools, however, eventually the black hole may begin to lose mass-energy through Hawking radiation. |
Bagged Trees Algorithm References 2015 http://en.wikipedia.org/wiki/Random_forest#Tree_bagging The training algorithm for random forests applies the general technique of bootstrap aggregating, or bagging, to tree learners. Given a training set
X=
x, …,
1
xwith responses
n
Y=
y, …,
1
y, bagging repeatedly selects a random sample with replacement of the training set and fits trees to these samples … After training, predictions for unseen samples
n
x'can be made by averaging the predictions from all the individual regression trees on
x': :[math]\hat{f} = \frac{1}{B} \sum_{b=1}^B \hat{f}_b (x')[/math] or by taking the majority vote in the case of decision trees.
This bootstrapping procedure leads to better model performance because it decreases the variance of the model, without increasing the bias. This means that while the predictions of a single tree are highly sensitive to noise in its training set, the average of many trees is not, as long as the trees are not correlated. Simply training many trees on a single training set would give strongly correlated trees (or even the same tree many times, if the training algorithm is deterministic); bootstrap sampling is a way of de-correlating the trees by showing them different training sets.
The training algorithm for random forests applies the general technique of bootstrap aggregating, or bagging, to tree learners. Given a training set 2006 (Caruana & Niculescu-Mizil, 2006) ⇒ Rich Caruana, and Alexandru Niculescu-Mizil. (2006). “An Empirical Comparison of Supervised Learning Algorithms.” In: Proceedings of the 23rd International Conference on Machine learning. ISBN:1-59593-383-2 doi:10.1145/1143844.1143865 QUOTE: A number of supervised learning methods have been introduced in the last decade. Unfortunately, the last comprehensive empirical evaluation of supervised learning was the Statlog Project in the early 90's. We present a large-scale empirical comparison between ten supervised learning methods: SVMs, neural nets, logistic regression, naive bayes, memory-based learning, random forests, decision trees, bagged trees, boosted trees, and boosted stumps. We also examine the effect that calibrating the models via Platt Scaling and Isotonic Regression has on their performance. |
In this paper we study a class of fast geometric image inpainting methodsbased on the idea of filling the inpainting domain in successive shells fromits boundary inwards. Image pixels are filled by assigning them a color equalto a weighted average of their already filled neighbors. However, there isflexibility in terms of the order in which pixels are filled, the weights usedfor averaging, and the neighborhood that is averaged over. Varying thesedegrees of freedom leads to different algorithms, and indeed the literaturecontains several methods falling into this general class. All of them are veryfast, but at the same time all of them leave undesirable artifacts such as"kinking" (bending) or blurring of extrapolated isophotes. Our objective inthis paper is to build a theoretical model, based on a continuum limit and aconnection to stopped random walks, in order to understand why these artifactsoccur and what, if anything, can be done about them. At the same time, weconsider a semi-implicit extension in which pixels in a given shell are solvedfor simultaneously by solving a linear system. We prove (within the continuumlimit) that this extension is able to completely eliminate kinking artifacts,which we also prove must always be present in the direct method. Although ouranalysis makes the strong assumption of a square inpainting domain, it makesweak smoothness assumptions and is thus applicable to the low regularityinherent in images.
We consider the problem of convergence to a saddle point of a concave-convexfunction via gradient dynamics. Since first introduced by Arrow, Hurwicz andUzawa in [1] such dynamics have been extensively used in diverse areas, thereare, however, features that render their analysis non trivial. These includethe lack of convergence guarantees when the function considered is not strictlyconcave-convex and also the non-smoothness of subgradient dynamics. Our aim inthis two part paper is to provide an explicit characterization to theasymptotic behaviour of general gradient and subgradient dynamics applied to ageneral concave-convex function. We show that despite the nonlinearity andnon-smoothness of these dynamics their $\omega$-limit set is comprised oftrajectories that solve only explicit linear ODEs that are characterized withinthe paper. More precisely, in Part I an exact characterization is provided to theasymptotic behaviour of unconstrained gradient dynamics. We also show that whenconvergence to a saddle point is not guaranteed then the system behaviour canbe problematic, with arbitrarily small noise leading to an unbounded variance.In Part II we consider a general class of subgradient dynamics that restricttrajectories in an arbitrary convex domain, and show that when an equilibriumpoint exists their limiting trajectories are solutions of subgradient dynamicson only affine subspaces. The latter is a smooth class of dynamics with anasymptotic behaviour exactly characterized in Part I, as solutions to explicitlinear ODEs. These results are used to formulate corresponding convergencecriteria and are demonstrated with several examples and applications presentedin Part II.
In part I we considered the problem of convergence to a saddle point of aconcave-convex function via gradient dynamics and an exact characterization wasgiven to their asymptotic behaviour. In part II we consider a general class ofsubgradient dynamics that provide a restriction in an arbitrary convex domain.We show that despite the nonlinear and non-smooth character of these dynamicstheir $\omega$-limit set is comprised of solutions to only linear ODEs. Inparticular, we show that the latter are solutions to subgradient dynamics onaffine subspaces which is a smooth class of dynamics the asymptotic propertiesof which have been exactly characterized in part I. Various convergencecriteria are formulated using these results and several examples andapplications are also discussed throughout the manuscript.
In this paper, we establish uniqueness of the solution of the Vlasov-Poissonsystem with spatial density belonging to a certain class of Orlicz spaces. Thisextends the uniqueness result of Loeper (which holds for uniformly boundeddensity) and the uniqueness result of the second author. Uniqueness is a directconsequence of our main result, which provides a quantitative stabilityestimate for the Wasserstein distance between two weak solutions with spatialdensity in such Orlicz spaces, in the spirit of Dobrushin's proof of stabilityfor mean-field PDEs. Our proofs are built on the second-order structure of theunderlying characteristic system associated to the equation.
We develop a new technique for establishing quantitative propagation of chaosfor systems of interacting particles. Using this technique we prove propagationof chaos for diffusing particles whose interaction kernel is merely H\"oldercontinuous, even at long ranges. Moreover, we do not require specially preparedinitial data. On the way, we establish a law of large numbers for SDEs thatholds over a class of vector fields simultaneously. The proofs bring togetherideas from empirical process theory and stochastic flows.
We develop a technique of multiple scale asymptotic expansions along meanflows and a corresponding notion of weak multiple scale convergence. These areapplied to homogenize convection dominated parabolic equations with rapidlyoscillating, locally periodic coefficients and $\mathcal{O}(\eps^{-1})$ meanconvection term. Crucial to our analysis is the introduction of a fast timevariable, $\tau=\frac{t}{\eps}$, not apparent in the heterogeneous problem. Theeffective diffusion coefficient is expressed in terms of the average ofEulerian cell solutions along the orbits of the mean flow in the fast timevariable. To make this notion rigorous, we use the theory of ergodic algebraswith mean value.
The problem of approximating the discrete spectra of families of self-adjointoperators that are merely strongly continuous is addressed. It is well-knownthat the spectrum need not vary continuously (as a set) under strongperturbations. However, it is shown that under an additional compactnessassumption the spectrum does vary continuously, and a family of symmetricfinite-dimensional approximations is constructed. An important feature of theseapproximations is that they are valid for the entire family uniformly. Anapplication of this result to the study of plasma instabilities is illustrated.
We study convergence to equilibrium for the kinetic Fokker-Planck equation onthe torus. Solving the stochastic differential equation, we show exponentialconvergence in the Monge-Kantorovich-Wasserstein $\mathcal{W}_2$ distance.Finally, we investigate if such a coupling can be obtained by a co-adaptedcoupling, and show that then the bound must depend on the square root of theinitial distance.
The relativistic Vlasov-Maxwell system describes the evolution of acollisionless plasma. The problem of linear instability of this system isconsidered in two physical settings: the so-called "one and one-half"dimensional case, and the three dimensional case with cylindrical symmetry.Sufficient conditions for instability are obtained in terms of the spectralproperties of certain Schr\"odinger operators that act on the spatial variablealone (and not in full phase space). An important aspect of these conditions isthat they do not require any boundedness assumptions on the domains, nor dothey require monotonicity of the equilibrium. |
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1. Observation of a peaking structure in the J/psi phi mass spectrum from B-+/- -> J/psi phi K-+/- decays
PHYSICS LETTERS B, ISSN 0370-2693, 06/2014, Volume 734, Issue 370-2693 0370-2693, pp. 261 - 281
A peaking structure in the J/psi phi mass spectrum near threshold is observed in B-+/- -> J/psi phi K-+/- decays, produced in pp collisions at root s = 7 TeV...
PHYSICS, NUCLEAR | ASTRONOMY & ASTROPHYSICS | PHYSICS, PARTICLES & FIELDS | Physics - High Energy Physics - Experiment | Physics | High Energy Physics - Experiment | scattering [p p] | J/psi --> muon+ muon | experimental results | Particle Physics - Experiment | Nuclear and High Energy Physics | Phi --> K+ K | vertex [track data analysis] | CERN LHC Coll | B+ --> J/psi Phi K | Peaking structure | hadronic decay [B] | Integrated luminosity | Data sample | final state [dimuon] | mass enhancement | width [resonance] | (J/psi Phi) [mass spectrum] | Breit-Wigner [resonance] | 7000 GeV-cms | leptonic decay [J/psi] | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
PHYSICS, NUCLEAR | ASTRONOMY & ASTROPHYSICS | PHYSICS, PARTICLES & FIELDS | Physics - High Energy Physics - Experiment | Physics | High Energy Physics - Experiment | scattering [p p] | J/psi --> muon+ muon | experimental results | Particle Physics - Experiment | Nuclear and High Energy Physics | Phi --> K+ K | vertex [track data analysis] | CERN LHC Coll | B+ --> J/psi Phi K | Peaking structure | hadronic decay [B] | Integrated luminosity | Data sample | final state [dimuon] | mass enhancement | width [resonance] | (J/psi Phi) [mass spectrum] | Breit-Wigner [resonance] | 7000 GeV-cms | leptonic decay [J/psi] | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
Journal Article
2. Measurement of the ratio of the production cross sections times branching fractions of B c ± → J/ψπ ± and B± → J/ψK ± and ℬ B c ± → J / ψ π ± π ± π ∓ / ℬ B c ± → J / ψ π ± $$ \mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm }{\pi}^{\pm }{\pi}^{\mp}\right)/\mathrm{\mathcal{B}}\left({\mathrm{B}}_{\mathrm{c}}^{\pm}\to \mathrm{J}/\psi {\pi}^{\pm}\right) $$ in pp collisions at s = 7 $$ \sqrt{s}=7 $$ TeV
Journal of High Energy Physics, ISSN 1029-8479, 1/2015, Volume 2015, Issue 1, pp. 1 - 30
The ratio of the production cross sections times branching fractions σ B c ± ℬ B c ± → J / ψ π ± / σ B ± ℬ B ± → J / ψ K ± $$ \left(\sigma...
B physics | Branching fraction | Hadron-Hadron Scattering | Quantum Physics | Quantum Field Theories, String Theory | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory
B physics | Branching fraction | Hadron-Hadron Scattering | Quantum Physics | Quantum Field Theories, String Theory | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory
Journal Article
3. Precise Measurement of the e+e- →π+π-J /ψ Cross Section at Center-of-Mass Energies from 3.77 to 4.60 GeV
Physical Review Letters, ISSN 0031-9007, 03/2017, Volume 118, Issue 9, p. 092001
The cross section for the process e^{+}e^{-}→π^{+}π^{-}J/ψ is measured precisely at center-of-mass energies from 3.77 to 4.60 GeV using 9 fb^{-1} of data...
Journal Article
Physical Review Letters, ISSN 0031-9007, 06/2019, Volume 122, Issue 23, p. 232002
Journal Article
5. The theological method of Stanley J. Grenz
: constructing evangelical theology from Scripture, tradition, and culture
2011, ISBN 9780773415188, iv, 346
Book
Physics Letters B, ISSN 0370-2693, 05/2016, Volume 756, Issue C, pp. 84 - 102
A measurement of the ratio of the branching fractions of the meson to and to is presented. The , , and are observed through their decays to , , and ,...
scattering [p p] | pair production [pi] | statistical | Physics, Nuclear | 114 Physical sciences | Phi --> K+ K | Astronomy & Astrophysics | LHC, CMS, B physics, Nuclear and High Energy Physics | f0 --> pi+ pi | High Energy Physics - Experiment | Compact Muon Solenoid | pair production [K] | Science & Technology | mass spectrum [K+ K-] | Ratio B | Large Hadron Collider (LHC) | Nuclear & Particles Physics | 7000 GeV-cms | leptonic decay [J/psi] | J/psi --> muon+ muon | experimental results | Nuclear and High Energy Physics | Physics and Astronomy | branching ratio [B/s0] | CERN LHC Coll | B/s0 --> J/psi Phi | CMS collaboration ; proton-proton collisions ; CMS ; B physics | Physics | Física | Physical Sciences | hadronic decay [f0] | [PHYS.HEXP]Physics [physics]/High Energy Physics - Experiment [hep-ex] | Physics, Particles & Fields | 0202 Atomic, Molecular, Nuclear, Particle And Plasma Physics | colliding beams [p p] | hadronic decay [Phi] | mass spectrum [pi+ pi-] | B/s0 --> J/psi f0
scattering [p p] | pair production [pi] | statistical | Physics, Nuclear | 114 Physical sciences | Phi --> K+ K | Astronomy & Astrophysics | LHC, CMS, B physics, Nuclear and High Energy Physics | f0 --> pi+ pi | High Energy Physics - Experiment | Compact Muon Solenoid | pair production [K] | Science & Technology | mass spectrum [K+ K-] | Ratio B | Large Hadron Collider (LHC) | Nuclear & Particles Physics | 7000 GeV-cms | leptonic decay [J/psi] | J/psi --> muon+ muon | experimental results | Nuclear and High Energy Physics | Physics and Astronomy | branching ratio [B/s0] | CERN LHC Coll | B/s0 --> J/psi Phi | CMS collaboration ; proton-proton collisions ; CMS ; B physics | Physics | Física | Physical Sciences | hadronic decay [f0] | [PHYS.HEXP]Physics [physics]/High Energy Physics - Experiment [hep-ex] | Physics, Particles & Fields | 0202 Atomic, Molecular, Nuclear, Particle And Plasma Physics | colliding beams [p p] | hadronic decay [Phi] | mass spectrum [pi+ pi-] | B/s0 --> J/psi f0
Journal Article
Physics Letters B, ISSN 0370-2693, 03/2017, Volume 766, Issue C, pp. 212 - 224
Journal Article
2014, ISBN 1610973143, xxi, 460
Book
9. Evidence for a Narrow Near-Threshold Structure in the J/psi phi Mass Spectrum in B+ -> J/psi phi K+ Decays
PHYSICAL REVIEW LETTERS, ISSN 0031-9007, 06/2009, Volume 102, Issue 24
Journal Article
PHYSICAL REVIEW LETTERS, ISSN 0031-9007, 06/2019, Volume 122, Issue 23
Journal Article
SAE technical paper series, Volume 790811.
eJournal |
I type fast enough that for most things it's not a pain, but I have a few big stacks of old course notes I'd like in LaTeX which I'm dreading having to go through. So, I'm just wondering what the best solution for handwriting -> LaTeX is so far, if any.
So, I'm just wondering what the best solution for handwriting -> LaTeX is so far, if any.
There is none, and if there’ll ever be one it’s probably years, if not decades off. I know people who are currently working on recognizing just the
layout of a document, i.e. recognizing that a paper represents a letter, etc.
That works fairly well, but it’s still research level, and going from recognizing the layout to replicating the layout using LaTeX is a big, non-obvious step. And we’re not even talking about text recognition itself.
Just text recognition (i.e. ignoring any layout issue) works fairly well today but only for plain text, not with any formatting.
That said, there’s JMathNotes which recognizes basic formulas and produces LaTeX output. It’s a nice and quite powerful proof of concept.
But it’s important to realize that even though many of the individual building blocks exist, piecing together a working solution is hard.
Very impressed by VisualObjects Web Equation
Screenshot
(of doncherry's clueless scribbling):
Screenshot
(of Aymon's expert scribing):
Just take pictures, and you can export as Latex, PDF, or you can get an Overleaf link (they have a really nice browser based editor). The iOS link is:
and the main website is just http://mathpix.com/.
Disclaimer: I'm the founder of Mathpix. I started working on this as a Stanford grad student in applied math, I hated how long it took to digitize my notes / homework sets. Anyway, Mathpix want to take the pain out of Latex for everyone, I hope this helps!
As the author of write-math.com, I think I can give this question an update.
First of all, there are two types of handwriting recognition: On-line and Off-line. On-line recognition means you can use the information how a symbol is written, whereas in off-line recognition you only have a pixel-map (aka "image"). Imagine on-line recognition as a movie where you get exact information where the tip of the pen was, whereas in off-line recognition you only get the end result. This means on-line recognition is simpler than off-line recognition as you can always just generate the end result.
I am doing research in on-line recognition.
There is an international conference on on-line handwriting recognition called ICDAR (international conference on document analysis and recognition) and a competition called CROHME. In this competition you get a very nice data (meaning: clearly written, no errors in the input as it often occurs in real live) and your classifier has to recognize the recording. The recordings are also very simple: The symbols are written on one line (no
\begin{align}\end{align}, but multiple fractions are possible), a very simple set of 75 allowed symbols (0-9, a-e, i-k, n, x-z, A, B, C, X, Y - you can see that this list was designed to be minimal and not have difficult combinations like 0, O, o or \pi and \prod), no matrices, no delayed strokes (e.g. you write
a < b and then decide to correct it to
a \leq b). And still the best system in 2013 (by VisionObjects, see web demo) only got 60.36% correct.
There are three tasks which have to be solved:
single symbol recognition (quite easy): Given only a single handwritten symbol, find its LaTeX code segmentation (MUCH harder): Given a handwritten equation, find which strokes belong to which symbol (not classifying the symbols but only saying "this is symbol a, this is another symbol b, ...") structural analysis: given a list of symbols
aand
b, say if its
abor
a^bor
a_b. (I didn't try that by now, but I think that's relatively easy)
Why is segmentation so hard? It is the mind-blowing number of possibilities you have to segment. Suppose you have
n=3 strokes. Then you could have the following segmentations:
1: [[0, 1, 2]] 2: [[0, 1], [2]] 3: [[0, 2], [1]] 4: [[0], [1, 2]] 5: [[0], [1], [2]]
Possibility 3 is what makes it so complicated. I've collected a lot of recordings with write-math.com and manually segmented them. About 10% of all multi-symbol recordings have such delayed strokes (see above). The number of possibilities grows a shown in https://oeis.org/A000110 But even without delayed strokes, you still have 2^{n-1} possible ways to segment n stokes.
You can see my progress on this topic here: https://github.com/MartinThoma/hwrt/issues/21
All of my material (papers, presentations, tools) are here: http://martin-thoma.com/write-math/
TL;DR
If you have something non-trivial, you still have to write it yourself. But I try hard to change that :-)
It's not a LaTeX solution, but very useful to me: Get a
new version of a speech recognition programm and read aloud to your computer.
This is a lot faster than typing, even if you were a professionell typewriter. I bought a "premium" version. There you can define your own speech commands. So the command "techenumeration" makes the software type
\begin{enumerate}\item\end{enumerate}
Give it a try, the software works way better than some years ago.
I have summarized the most of the current answers here or below.
I will cover now some papers, work in progress. I understand it so that the Tapio -paper, before preprocessing, uses LP -methods for his formulated QP -puzzle. The Knerr -paper uses discretization of words so one word can have many routes, now getting easily an exponential network-optimization problem. The ON-REC -method is almost the same as the REC-REC -method but some modifications. Knerr has publized a new paper
"Combining diverse systems for handwritten text line recognition" (2011). The Japananese paper contains pretty much no details, mostly programming-biased rhetory or worse marketing of their InftyReader. Academia "Recognition of on-line handwritten mathematical formulas in the e-chalk system"here
Key terms:empirical risk, structural risk, pattern recognition, QP -problem, Lagrange multipliers, theory developed by Vapnik and Chervonenkis (VC),
Perhaps important terms: radial basis functions (RBFs), polynomial kernels, hyperbolic kernels, sequential minimal optimization (SMO), --
II.
Stefan Knerr(CEO of Vision Objects here, over 70 employees) has publications here, they approach the problem differently -- firstly quantifying different segments into markov chains. Then they get some sort of network -optimization problem that I cannot yet fully understand but trying. "Recognition-directed recovering of temporal information from handwriting images"-paper converts words into finite state-machines like the picture here.
Key terms:frame-extraction/vector-quantization/discreate-HMMs here, discrete Hidden Markov Models (HMMs), Tabou method (1984), Baum–Welch training algorithm, ON-REC system, REC–REC system,
"(i) a left–right scan of the word—referred as SCAN–REC further, (ii) a time order of the strokes recovered previously from the static image—referred latter as REC–REC, (iii) a time order of the strokes corresponding to the true online ordering—referred as ON–REC."(the Knerr -paper)
Perhaps important things:IRONOFF database,
III. Japanese researchers such as Masakazu Suzuki, Toshihiro Kanahori, Nobuyuki Ohtake and Katsuhito Yamaguchi -- apparently something to do with Ideal Group -companies such as InftyReader here. Anyway, their most-cited paper below shows a more programming-biased -prototype.
"An Integrated OCR Software for Mathematical Documents and Its Output with Accessibility"(2004)
Perhaps Key terms:Unified Braille Code (UBC) by BANA (Braille Authority of North American), working requires "scanned binary images in either 600 DPI or 400 DPI" Puzzles
"baseline structure analysis method developed by Zanibbi et al [14].
The idea is that mathematical notation can be described as a hierarchical structure of nested baselines."(the Tapia -paper I added the bolding)
The Knerr -paper mentions
"the second optimization process usesand directed graph models" "(I added bolding) The number of possible paths of the ‘‘REC– REC’’approach for a word with N segments is 2N!" Future development
Open-source OCR system for FPGA?. Glen recommends to look for
"dynamic programming algorithms and systolic array processors"here: I think the key is to break the problem into things like subgame perfect equilibriums so it can be parallerized and done fast.
Real-time skeletonization using FPGA, A real-time matching system for large fingerprint databases (almost the speed of ASIC) -- it is non-trivial task to do the cover and skeletonization.
Products Microtask breaks the OCR-detection into games where players identify parts that are too hard for computers to detect. In exchange, the gamers can receive digital currencies.
Prior answer:
On the Samsung Galaxy Note, the proprietary program "S-Note" appears to have a licensed version of VisualObjects included, which it uses to do a very creditable job of converting handwriting to formatted equations. Downside: the LaTeX is not accessible (grrr). I actually was curious so I looked inside the application and decompiled some of the java source files... and I found that internally, it is using LaTeX! So honestly I'm sure that a creditable Java hacker (which I am not) could decompile the app, and with literally just a few lines of changes, make it convert to LaTeX rather than to a bitmap. If you grep the codebase for "EquationRecognition", you'll quickly find the relevant files. You could then recompile just those classes, re-bundle the app, and sign the hacked version. (Which of course would only be legal if you have a legit license to the app.)
I know, this is not actually a useful answer, but I just spend an hour or so finding this out, so I might as well share it.
An iteresting research on Mathematical Information Processing is explained here at the InftyReader project page. It's a Japanese research group.
Given that it hasn't been mention, detexify basically takes handwritten text and produces TeX/LaTeX code (granted on a single symbol scale).
Inlage (http://www.inlage.com) is a Latex editor which offers recognition of handwritten formulas on Windows 7. It makes use of the Windows 7 math input panel and converts the generated MathML to Latex. See a video of how it works at Inlage II feature: Math Input Panel to LaTeX.
Note: I'm in no way affiliated with this program. TexTablet might be a free alternative.
This solution may not answer your question entirely, but if just want a write-on-the-fly solution on Windows with equations as well as handdrawing pictures, then try using
TeXStudio (http://www.texstudio.org/). This software has integrated the Windows write recognization engine into itself after my request a few years ago on the software's open-source development issue tracker. See the screen snapshoot ,
the write LaTeX option is under the
Wizards menu named as Math Assistant. It will then open up an OCR window as shown in the figure below.
Handwrite your equations on the panel, and then you only need to copy the equations shown on top as a LaTeX code to your TeXStudio input window. Or, you can just press the insert button to insert the tex code to your main text. Errors can be fixed by rewriting your equations or just by correcting the generated LaTeX code.
I have used this software to take notes on Physics classes where math and graphics are needed all the time. I use the
Insert Graphic tool under the Wizards menu of the software to insert hand-drawing pictures which can be done easily from OneNote on the fly if you have a tablet (I use Thinkpad X200 tablet, Thinkpad Tablet 2 and Thinkpad Yogo 460 for taking notes electrically). If you really want to handwrite everything, you can use the default writing panel from Windows (Windows 7 and above I guess) to input your text as well, and then use the two tools I have introduced above to input handdrawing diagrams and equations all into LaTeX format.
One real example of compiled output using this technique and this set of tools can be found in my shared Notes on Classical Mechanics on Github.
If you use LyX on Windows 7 or later, there is a Math Input Panel Helper.
It's a little program that converts the Math Panel output to LaTeX or MathML and lets you insert math directly to LyX (or any other LaTeX editor). |
This is a derivation of the deflection of light by a point mass, closely following the derivation in Weinberg's book,
Gravitation and Cosmology (1972).
We start with the Schwarzschild metric:
$$d\tau^2=Bdt^2-Adr^2-r^2(d\theta^2+\sin^2\theta d\phi^2),$$
with
$$B = A^{-1}=\left(1-\frac{2GM}{r}\right).$$
A test particle characterized by 4-velocity $u^\mu$ is following a geodesic trajectory given by the equation of motion $du^\mu/d\tau+\Gamma^\mu_{\alpha\beta}u^\alpha u^\beta=0$. In the Schwarzschild metric, this equation translates into
\begin{align*}
\dfrac{d^2t}{d\tau^2}+\dfrac{B'}{B}\dfrac{dt}{d\tau}\dfrac{dr}{d\tau}&=0,\\ \dfrac{d^2r}{d\tau^2}+\dfrac{B'}{2A}\left(\dfrac{dt}{d\tau}\right)^2+\dfrac{A'}{2A}\left(\dfrac{dr}{d\tau}\right)^2-\dfrac{r}{A}\left(\dfrac{d\phi}{d\tau}\right)^2&=0,\\ \dfrac{d^2\phi}{d\tau^2}+\dfrac{2}{r}\dfrac{dr}{d\tau}\dfrac{d\phi}{d\tau}&=0, \end{align*}
where the prime denotes differentiation with respect to $r$. The first of these equations can be rearranged and integrated:
\begin{align*}
B\dfrac{d^2t}{d\tau^2}+\dfrac{dB}{d\tau}\dfrac{dt}{d\tau}&=0,\\ \dfrac{dt}{d\tau}&=\dfrac{C}{B}, \end{align*}
where $C$ is some constant. Since at infinity, $B=dt/d\tau=1$, we must have $C=1$.
Meanwhile, the third equation of motion can be integrated directly:
$$r^2\dfrac{d\phi}{d\tau}=J,$$
where $J$, another integration constant, can be identified as the angular momentum per unit mass. Using these results in the second equation of motion, we get
$$\dfrac{d^2r}{d\tau^2}+\dfrac{A'}{2A}\left(\dfrac{dr}{d\tau}\right)^2-\dfrac{J^2}{Ar^3}+\dfrac{B'}{2A}\left(\dfrac{1}{B}\right)^2=0.$$
Multiplication by $2Adr/d\tau$ yields
$$2A\dfrac{dr}{d\tau}\dfrac{d^2r}{d\tau^2}+A'\left(\dfrac{dr}{d\tau}\right)^3-2\dfrac{dr}{d\tau}\dfrac{J^2}{r^3}+\dfrac{dr}{d\tau}\dfrac{B'}{B^2}=0,$$
or
$$\dfrac{d}{d\tau}\left[A\left(\dfrac{dr}{d\tau}\right)^2+\dfrac{J^2}{r^2}-\dfrac{1}{B}\right]=0,$$
or
$$A\left(\dfrac{dr}{d\tau}\right)^2+\dfrac{J^2}{r^2}-\dfrac{1}{B}=-{\cal E},$$
where ${\cal E}$ is another integration constant. In the limit of large $r$ ($A=B=1$, $dr/dt=dr/d\tau=v$) we get
$${\cal E}=1-v^2.$$
Using our earlier result for $r^2d\phi/d\tau$ in the preceding equation, we get
$$A\left(\dfrac{dr}{d\phi}\dfrac{J}{r^2}\right)^2+\dfrac{J^2}{r^2}-\dfrac{1}{B}=-{\cal E},$$
or
$$\dfrac{A}{r^4}\left(\dfrac{dr}{d\phi}\right)^2+\dfrac{1}{r^2}=-\dfrac{\cal E}{J^2}+\dfrac{1}{J^2B}.$$
At closest approach, $r=r_0$ and $dr/d\phi=0$. Thus
$$J=r_0\sqrt{\dfrac{1}{B_0}-{\cal E}},$$
where $B_0=1-2GM/r_0$ is the value of $B$ at $r=r_0$. We can solve the preceding equation for $\phi$:
$$\phi=\pm\int\dfrac{\sqrt{A}}{r^2\left[\dfrac{1}{r_0^2}\left(\dfrac{1}{B}-{\cal E}\right)\left(\dfrac{1}{B_0}-{\cal E}\right)^{-1}-\dfrac{1}{r^2}\right]^{1/2}}~dr.$$
For an ultrarelativistic particle, ${\cal E}=1-v^2\sim 0$ and can be omitted, leaving us with
$$\phi=\pm\int\dfrac{\sqrt{A}}{r\left[\dfrac{r^2}{r_0^2}\dfrac{B_0}{B}-1\right]^{1/2}}~dr.$$
The square rooted expression in the denominator can be simplified to first order:
\begin{align*}
\dfrac{r^2}{r_0^2}\dfrac{B_0}{B}-1 &=\dfrac{r^2}{r_0^2}\left[1+2GM\left(\dfrac{1}{r}-\dfrac{1}{r_0}\right)\right]-1\\ &=\left(\dfrac{r^2}{r_0^2}-1\right)\left[1-2GM\left(\dfrac{r}{r_0(r+r_0)}\right)\right] \end{align*}
So to first order, we can write the integral as
$$\phi=\pm\int\dfrac{r_0}{r\sqrt{r^2-r_0^2}}\left[1+\dfrac{GM}{r}+GM\dfrac{r}{r_0(r+r_0)}\right]~dr.$$
This is an elementary integral that can be readily evaluated. Now for a particle coming from infinity and going to infinity, the deflection angle will be given by
$$\Delta\phi=2\left|\int_{r_0}^\infty\dfrac{r_0}{r\sqrt{r^2-r_0^2}}\left[1+\dfrac{GM}{r}+GM\dfrac{r}{r_0(r+r_0)}\right]~dr\right|-\pi=\frac{4GM}{r_0}.$$
After restoring the speed of light, we get therefore
$$\Delta\phi=\dfrac{4GM}{c^2r_0}.$$
Substituting $G=6.674\times 10^{-11}~{\rm m}^3/{\rm s}^2{\rm kg}$, $M=2\times 10^{30}~{\rm kg}$, $r_0=6.95\times 10^8~{\rm m}$ (and of course $c=3\times 10^8~{\rm m}/{\rm s}$), parameters that characterize a ray of light grazing the Sun, we get
$$\Delta\phi\sim 1.76".$$
It is also interesting to contemplate where the factor of 4 is coming from. In the Schwarzschild metric, both time and space have curvature, characterized by the metric coefficients $B$ and $A$. Newtonian gravity amounts to ignoring the spatial curvature; i.e., setting $A=1$. And indeed, if we omit the $\sqrt{A}$ term in the integral expression for $\Delta\phi$, we get exactly half the predicted deflection, $2GM/c^2r_0$. Which demonstrates explicitly that it is because of the presence of spatial curvature, which has no nonrelativistic analog, that the bending of light has the predicted magnitude. Therefore, observations like Eddington's (concerns about the data quality notwithstanding) are indeed genuine tests of general relativity. |
How to prove that $Ax = e^x$ has two solutions when $e < A < \infty$? This is easy to visualise graphically, but how can it be shown with algebra?
Consider the function $f(x)=e^x-Ax$. Then $f(0)=1$. We have $f'(x)=e^x-A$, $f''(x)=e^x$. As the second derivative is always positive, our function is convex. The derivative has a single zero at $x=\log A$, so $f$ has a minimum at that point. This means that $$ f(x)\geq f(\log A)=A-A\log A. $$ As $\lim_{x\to\infty}f(x)=\infty$, if the minimum is negative, then $f$ will have two roots (and none if the minimum is positive).
Assuming $A>0$, we have $A\log A>A$ precisely when $\log A>1$, i.e. $A>e$.
In conclusion, we have
Two points where $Ax=e^x$ when $A>e$; One point where $Ax=e^x$ when $A=e$; No points where $Ax=e^x$ when $A<e$;
This is a proof that uses the intermedate value theorem (I don't know an algebraic proof of this fact):
Let $g(x)=e^x-Ax$. $g(0)=1$ and $g(1)=e-A<0$. Using the intermediate value theorem, we deduce that there is a root $r_1$ of $g(x)$ in the interval $[0,1]$
Since $g$ goes increases without bound after sufficiently large x, thus there exists M>1 such that $g(M)>0$. By the intermediate value theorem, we know that there exists another root $r_2$ of $g$ in the interval $]1,M[$
Here is one with calculus.
If $f(x)=e^x-Ax$ then: $f(0)=1>0.$ $f(1)=e-A<0.$ $f(e)=e^e-Ae>e^e-e^2>0.$
Now use intermediate value theorem.
Since $f$ is differentiable in $\mathbb{R}$ with $f'(x)=e^x-A$ and $f'(x)<0 \iff x<\ln A , \ \ f'(x)>0 \iff x>\ln A \Rightarrow f(x)=0$ has at most two solutions. |
Capillary force
The force by which water ascends in wood, sponge, blotting-paper, and other porous bodies. By the same action the flame of a lamp is fed with oil. The wick is a bundle of threads whose surfaces are nearly in contact, and the oil rises between them in the same way as if they were narrow tubes. Water is supposed to rise from reservoirs and springs below the surface of the ground to the roots or plants in the same way as it rises in fine tubes.
Formula
The height
h of a liquid column is given by: <math>h={2{ \gamma \cos{\theta}}\over{\rho g r}}</math>
where:
<math>\scriptstyle \gamma </math> is the liquid-air surface tension(energy/area) θis the contact angle ρis the density of liquid (mass/volume) gis acceleration due to gravity (length/time 2) ris radius of tube (length).
For a water-filled glass tube in air at sea level, using SI units:
<math>\scriptstyle \gamma </math>is 0.0728 J/m² at 20 °celsius|C θis 20° (0.35 radians|rad) ρis 1000 kg/m 3 gis 9.8 m/s²
therefore, the height of the water column is given by:
<math>h\approx {{1.4 \times 10^{-5}\ \mbox{m}^2}\over r}</math>. |
$$ f(x) = \cases{ x^2\sin(1/x) & if $x \neq 0$ \\ 0 & if $x = 0$} $$ justify that $f(x)$ is differentiable at the origin using $f'(x)=\lim_{h\to0} \frac{f(x+h)-f(x)}h$
closed as off-topic by Hans Lundmark, Aqua, Xander Henderson, jvdhooft, Dave Oct 10 '17 at 13:20
This question appears to be off-topic. The users who voted to close gave this specific reason:
" This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hans Lundmark, Aqua, Xander Henderson, jvdhooft, Dave
$\frac{f(0+h)-f(0)}{h}=h \sin(1/h)$, hence
$|\frac{f(0+h)-f(0)}{h}| \le |h|$.
What can you say about $ \lim_{h \to 0}\frac{f(0+h)-f(0)}{h}$ ?
Hint: $$x^2\sin(1/x)=x\times\frac{\sin(\frac{1}{x})}{(\frac{1}{x})}$$
Now try using limit $x \to 0$ |
I need to show that these two Lagrangians are equivalent:
\begin{align} L(\dot{x},\dot{y},x,y)&=\dot x^2+\dot y + x^2-y ,\\ \tilde{L}(\dot x, \dot y, x, y)&=\dot x^2+\dot y -2y^3. \end{align}
It is the case iff they differ for a total derivation like $\frac{dF}{dt}(x,y)$.
In this case, the difference is $x^2+y^3$ and I can't imagine such an $F(x,y)$ whose total derivative is the one above. How should I move?
I tried with the following $F(x,y)=\frac{x^3}{3\dot x} + \frac{y^4}{4\dot y}$, but it shouldn't have the dotted terms.
Actually, I just proved they don't give rise to the same Lagrange equations, so I can conclude they're not equivalent, right? |
Mathematics - Functional Analysis and Mathematics - Metric Geometry
Abstract
The following strengthening of the Elton-Odell theorem on the existence of a $(1+\epsilon)-$separated sequences in the unit sphere $S_X$ of an infinite dimensional Banach space $X$ is proved: There exists an infinite subset $S\subseteq S_X$ and a constant $d>1$, satisfying the property that for every $x,y\in S$ with $x\neq y$ there exists $f\in B_{X^*}$ such that $d\leq f(x)-f(y)$ and $f(y)\leq f(z)\leq f(x)$, for all $z\in S$. Comment: 15 pages, to appear in Bulletin of the Hellenic Mayhematical Society
Given a finite dimensional Banach space X with dimX = n and an Auerbach basis of X, it is proved that: there exists a set D of n + 1 linear combinations (with coordinates 0, -1, +1) of the members of the basis, so that each pair of different elements of D have distance greater than one. Comment: 15 pages. To appear in MATHEMATIKA |
Global well-posedness and large time behavior of classical solutions to the diffusion approximation model in radiation hydrodynamics
1.
Department of Mathematics, College of Science, Hohai University, Nanjing 210098, China
2.
Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260, USA
We are concerned with the global well-posedness of the diffusion approximation model in radiation hydrodynamics, which describe the compressible fluid dynamics taking into account the radiation effect under the non-local thermal equilibrium case. The model consist of the compressible Navier-Stokes equations coupled with the radiative transport equation with non-local terms. Global well-posedness of the Cauchy problem is established in perturbation framework, and rates of convergence of solutions toward equilibrium, which are algebraic in the whole space and exponential on torus, are also obtained under some additional conditions on initial data. The existence of global solution is proved based on the classical energy estimates, which are considerably complicated and some new ideas and techniques are thus required. Moreover, it is shown that neither shock waves nor vacuum and concentration in the solution are developed in a finite time although there is a complex interaction between photons and matter.
Keywords:Radiation hydrodynamics, diffusion approximation, compressible Navier-Stokes equations, global well-posedness, rate of convergence. Mathematics Subject Classification:76N10, 78A40. Citation:Peng Jiang. Global well-posedness and large time behavior of classical solutions to the diffusion approximation model in radiation hydrodynamics. Discrete & Continuous Dynamical Systems - A, 2017, 37 (4) : 2045-2063. doi: 10.3934/dcds.2017087
References:
[1]
J. W. Bond, K. M. Watson and J. A. Welch,
[2]
C. Buet and B. Després,
Asymptotic analysis of fluid models for the coupling of radiation and hydrodynamics,
[3]
J. A. Carrillo, R. Duan and A. Moussa,
Global classical solution close to equillibrium to the Vlasov-Euler-Fokker-Planck system,
[4] [5]
B. Ducomet and E. Feireisl,
The equation of magnetohydrodynamics: On the interation between matter and radiation in the evlution of gaseous stars,
[6] [7]
Th. Goudon and P. Lafitte,
A coupled model for radiative transfer: Doppler effects, equilibrium and non equilibrium diffusion asymptotics,
[8] [9] [10] [11]
E. Hopf,
[12]
S. Jiang, F. C. Li and F. Xie,
Nonrelativistic limits of the compressible Navier-Stokes-FourierP1 approximation model arising in radiation hydrodynamics,
[13]
S. Jiang, F. Xie and J. W. Zhang, A global existence result in radiation hydrodynamics, Industrial and Applied Mathematics in China,
[14]
S. Kawashima,
[15]
S. Kawashima and S. Nishibata,
A singular limit for hyperbolic-elliptic coupled systems inradiation hydrodynamics,
[16] [17]
R. Kippenhahn and A. Weigert,
[18]
S. Klainerman and A. Majda,
Singular limits of quasilinear hyperbolic systems with largeparameters and the incompressible limit of compressible fluids,
[19] [20] [21]
A. Matsumura and T. Nishida,
The initial value problem for the equations of motion of compressible viscous and heat-conductive fluids,
[22] [23]
D. Mihalas and B. Weibel-Mihalas,
[24]
S. S. Penner and D. B. Olfe,
[25]
G. C. Pomraning,
[26]
C. Rohde and W.-A. Yong,
The nonrelativistic limit in radiation hydrodynamics: I. Weakentropy solutions for a model problem,
[27]
R. N. Thomas,
[28]
W. J. Wang and F. Xie,
The initial value problem for a multi-dimensional radiation hydrodynamics model with viscosity,,
[29]
Y. B. Zeldovich and Y. P. Raizer, Phsics of Shock Waves and High-Temperture Hydrodynamic Phenomenon, Academic Press, 1966.Google Scholar
[30]
show all references
References:
[1]
J. W. Bond, K. M. Watson and J. A. Welch,
[2]
C. Buet and B. Després,
Asymptotic analysis of fluid models for the coupling of radiation and hydrodynamics,
[3]
J. A. Carrillo, R. Duan and A. Moussa,
Global classical solution close to equillibrium to the Vlasov-Euler-Fokker-Planck system,
[4] [5]
B. Ducomet and E. Feireisl,
The equation of magnetohydrodynamics: On the interation between matter and radiation in the evlution of gaseous stars,
[6] [7]
Th. Goudon and P. Lafitte,
A coupled model for radiative transfer: Doppler effects, equilibrium and non equilibrium diffusion asymptotics,
[8] [9] [10] [11]
E. Hopf,
[12]
S. Jiang, F. C. Li and F. Xie,
Nonrelativistic limits of the compressible Navier-Stokes-FourierP1 approximation model arising in radiation hydrodynamics,
[13]
S. Jiang, F. Xie and J. W. Zhang, A global existence result in radiation hydrodynamics, Industrial and Applied Mathematics in China,
[14]
S. Kawashima,
[15]
S. Kawashima and S. Nishibata,
A singular limit for hyperbolic-elliptic coupled systems inradiation hydrodynamics,
[16] [17]
R. Kippenhahn and A. Weigert,
[18]
S. Klainerman and A. Majda,
Singular limits of quasilinear hyperbolic systems with largeparameters and the incompressible limit of compressible fluids,
[19] [20] [21]
A. Matsumura and T. Nishida,
The initial value problem for the equations of motion of compressible viscous and heat-conductive fluids,
[22] [23]
D. Mihalas and B. Weibel-Mihalas,
[24]
S. S. Penner and D. B. Olfe,
[25]
G. C. Pomraning,
[26]
C. Rohde and W.-A. Yong,
The nonrelativistic limit in radiation hydrodynamics: I. Weakentropy solutions for a model problem,
[27]
R. N. Thomas,
[28]
W. J. Wang and F. Xie,
The initial value problem for a multi-dimensional radiation hydrodynamics model with viscosity,,
[29]
Y. B. Zeldovich and Y. P. Raizer, Phsics of Shock Waves and High-Temperture Hydrodynamic Phenomenon, Academic Press, 1966.Google Scholar
[30]
[1]
Zhilei Liang.
Convergence rate of solutions to the
contact discontinuity for the compressible Navier-Stokes equations.
[2]
Bin Han, Changhua Wei.
Global well-posedness for inhomogeneous Navier-Stokes equations with logarithmical hyper-dissipation.
[3]
Daniel Coutand, J. Peirce, Steve Shkoller.
Global well-posedness of weak solutions for the Lagrangian averaged Navier-Stokes equations on bounded domains.
[4]
Weimin Peng, Yi Zhou.
Global well-posedness of axisymmetric Navier-Stokes equations with one slow variable.
[5]
Daoyuan Fang, Ruizhao Zi.
On the well-posedness of inhomogeneous hyperdissipative Navier-Stokes equations.
[6] [7]
Matthias Hieber, Sylvie Monniaux.
Well-posedness results for the Navier-Stokes equations in the rotational framework.
[8]
Boris Haspot, Ewelina Zatorska.
From the highly compressible Navier-Stokes equations to the porous medium equation -- rate of convergence.
[9]
Maxim A. Olshanskii, Leo G. Rebholz, Abner J. Salgado.
On well-posedness of a velocity-vorticity formulation of the stationary Navier-Stokes equations with no-slip boundary conditions.
[10]
Yoshihiro Shibata.
Local well-posedness of free surface problems for the Navier-Stokes equations in a general domain.
[11]
Roberta Bianchini, Roberto Natalini.
Convergence of a vector-BGK approximation for the incompressible Navier-Stokes equations.
[12] [13]
Peixin Zhang, Jianwen Zhang, Junning Zhao.
On the global existence of classical solutions for compressible Navier-Stokes equations with vacuum.
[14] [15]
Chao Deng, Xiaohua Yao.
Well-posedness and ill-posedness for the 3D generalized Navier-Stokes equations in $\dot{F}^{-\alpha,r}_{\frac{3}{\alpha-1}}$.
[16]
Peng Jiang.
Unique global solution of an initial-boundary value problem to a diffusion approximation model in radiation hydrodynamics.
[17] [18] [19]
Bingkang Huang, Lusheng Wang, Qinghua Xiao.
Global nonlinear stability of rarefaction waves for compressible Navier-Stokes equations with temperature and density dependent transport coefficients.
[20]
Yuming Qin, Lan Huang, Zhiyong Ma.
Global existence and exponential stability in
$H^4$ for the nonlinear compressible Navier-Stokes equations.
2018 Impact Factor: 1.143
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When going from the strong form of a PDE to the FEM form it seems one should always do this by first stating the variational form. To do this you multiply the strong form by an element in some (Sobolev) space and integrate over your region. This I can accept. What I don't understand is why one also has to use Green's formula (one or several times).
I've mostly been working with Poisson's equation, so if we take that (with homogenous Dirichlet boundary conditions) as an example, i e
$$ \begin{align} -\nabla^2u &= f,\quad u\in\Omega \\ u &= 0, \quad u\in\partial\Omega \end{align} $$
then it is claimed that the correct way to form the variational form is
$$ \begin{align} \int_\Omega fv\,\mathrm{d}\vec{x} &= -\int_\Omega\nabla^2 uv\,\mathrm{d}\vec{x} \\ &=\int_\Omega\nabla u\cdot\nabla v\,\mathrm{d}\vec{x} - \int_{\partial\Omega}\vec{n}\cdot\nabla u v\,\mathrm{d}\vec{s} \\ &=\int_\Omega\nabla u\cdot\nabla v\,\mathrm{d}\vec{x}. \end{align} $$
But what stops me from using the expression on the first line, isn't that also a variational form that can be used to get a FEM form? Isn't it corresponding to the bilinear and linear forms $b(u,v)=(\nabla^2 u, v)$ and $l(v)=(f, v)$? Is the problem here that if I use linear basis functions (shape functions) then I'll be in trouble because my stiffness matrix will be the null matrix (not invertible)? But what if I use non-linear shape functions? Do I still have to use Green's formula? If I don't have to: is it advisable? If I don't, do I then have a variational-but-not-weak formulation?
Now, let's say that I have a PDE with higher order derivatives, does that mean that there are many possible variational forms, depending on how I use Green's formula? And they all lead to (different) FEM approximations? |
Noob here: first time using Latex and can't seem to write one line of code without getting ! Undefined control sequence. error. Here is the line it has a problem with:
$$ \left \{ x\in\mathbb{Z}:-2\leq x< 7 \right \} $$
And here is the rest of my document for context:
\documentclass{article}\usepackage{amsmath}\usepackage{graphicx}\usepackage{amstext}\begin{document}\section{Chapter 1: Sets}\subsection{Introduction to Sets}Write each of the following sets by listing their elements between braces:\begin{enumerate}\item Problem 3$$ \left \{ x\in\mathbb{Z}:-2\leq x< 7 \right \} $$\end{enumerate}\end{document} |
Kakeya problem
A
Kakeya set in [math]{\mathbb F}_3^r$ is a subset \ltmath\gtA\subset{\mathbb F}_3^n[/math] that contains an algebraic line in every direction; that is, for every [math]d\in{\mathbb F}_3^n[/math], there exists [math]a\in{\mathbb F}_3^n[/math] such that [math]a,a+d,a+2d[/math] all lie in [math]A[/math]. Let [math]k_n[/math] be the smallest size of a Kakeya set in [math]{\mathbb F}_3^n[/math].
Clearly, we have [math]k_1=3[/math], and it is easy to see that [math]k_2=7[/math]. Using a computer, it is not difficult to find that [math]k_3=13[/math] and [math]k_4\le 27[/math]. Indeed, it seems likely that [math]k_4=27[/math] holds, meaning that in [math]{\mathbb F}_3^4[/math] one cannot get away with just [math]26[/math] elements.
General lower bounds
Trivially, we have
[math]k_n\le k_{n+1}\le 3k_n[/math].
Since the Cartesian product of two Kakeya sets is another Kakeya set, we also have
[math]k_{n+m} \leq k_m k_n[/math];
this implies that [math]k_n^{1/n}[/math] converges to a limit as [math]n[/math] goes to infinity.
From a paper of Dvir, Kopparty, Saraf, and Sudan it follows that [math]k_n \geq 3^n / 2^n[/math], but this is superseded by the estimates given below.
To each of the [math](3^n-1)/2[/math] directions in [math]{\mathbb F}_3^n[/math] there correspond at least three pairs of elements in a Kakeya set, determining this direction. Therefore, [math]\binom{k_n}{2}\ge 3\cdot(3^n-1)/2[/math], and hence
[math]k_n\gtrsim 3^{(n+1)/2}.[/math]
One can derive essentially the same conclusion using the "bush" argument, as follows. Let [math]E\subset{\mathbb F}_3^n[/math] be a Kakeya set, considered as a union of [math]N := (3^n-1)/2[/math] lines in all different directions. Let [math]\mu[/math] be the largest number of lines that are concurrent at a point of [math]E[/math]. The number of point-line incidences is at most [math]|E|\mu[/math] and at least [math]3N[/math], whence [math]|E|\ge 3N/\mu[/math]. On the other hand, by considering only those points on the "bush" of lines emanating from a point with multiplicity [math]\mu[/math], we see that [math]|E|\ge 2\mu+1[/math]. Comparing the two last bounds one obtains [math]|E|\gtrsim\sqrt{6N} \approx 3^{(n+1)/2}[/math].
A better bound follows by using the "slices argument". Let [math]A,B,C\subset{\mathbb F}_3^{n-1}[/math] be the three slices of a Kakeya set [math]E\subset{\mathbb F}_3^n[/math]. Form a bipartite graph [math]G[/math] with the partite sets [math]A[/math] and [math]B[/math] by connecting [math]a[/math] and [math]b[/math] by an edge if there is a line in [math]E[/math] through [math]a[/math] and [math]b[/math]. The restricted sumset [math]\{a+b\colon (a,b)\in G\}[/math] is contained in the set [math]-C[/math], while the difference set [math]\{a-b\colon (a,b)\in G\}[/math] is all of [math]{\mathbb F}_3^{n-1}[/math]. Using an estimate from a paper of Katz-Tao, we conclude that [math]3^{n-1}\le\max(|A|,|B|,|C|)^{11/6}[/math], leading to [math]|E|\ge 3^{6(n-1)/11}[/math]. Thus,
[math]k_n \ge 3^{6(n-1)/11}.[/math] General upper bounds
We have
[math]k_n\le 2^{n+1}-1[/math]
since the set of all vectors in [math]{\mathbb F}_3^n[/math] such that at least one of the numbers [math]1[/math] and [math]2[/math] is missing among their coordinates is a Kakeya set.
This estimate can be improved using an idea due to Ruzsa (seems to be unpublished). Namely, let [math]E:=A\cup B[/math], where [math]A[/math] is the set of all those vectors with [math]r/3+O(\sqrt r)[/math] coordinates equal to [math]1[/math] and the rest equal to [math]0[/math], and [math]B[/math] is the set of all those vectors with [math]2r/3+O(\sqrt r)[/math] coordinates equal to [math]2[/math] and the rest equal to [math]0[/math]. Then [math]E[/math], being of size just about [math](27/4)^{r/3}[/math] (which is not difficult to verify using Stirling's formula), contains lines in a positive proportion of directions: for, a typical direction [math]d\in {\mathbb F}_3^n[/math] can be represented as [math]d=d_1+2d_2[/math] with [math]d_1,d_2\in A[/math], and then [math]d_1,d_1+d,d_1+2d\in E[/math]. Now one can use the random rotations trick to get the rest of the directions in [math]E[/math] (losing a polynomial factor in [math]n[/math]).
Putting all this together, we seem to have
[math](3^{6/11} + o(1))^n \le k_n \le ( (27/4)^{1/3} + o(1))^n[/math]
or
[math](1.8207+o(1))^n \le k_n \le (1.8899+o(1))^n.[/math] |
I personally would prefer a textbook recommendation I can download or pick up that is [preferably] not old and does not make trigonometry intimidating to approach (especially one that emphasizes understanding proofs behind properties/theorems).
I don't have textbooks to recommend, but I can recommend an approach to
doing trigonometry that facilitates mathematical understanding of it by crystallizing the logical foundation of trigonometry and algebraic structure of trigonometric expressions. There are two 'levels' to this, depending on whether you want to go straight to complex numbers or stay within real trigonometry. In either case, the focus is on identifying the intrinsic core of trigonometry and reducing everything else to that. Real trigonometry
The key quantities are $\cos(t)$ and $\sin(t)$, which are the $x$ and $y$ coordinates of the point $P_t$ on the unit circle that subtends an arc of length $t$ anti-clockwise from the $x$-axis, as depicted in the image from wikipedia:
Here arc length is measured along the unit circle, and $π$ is
defined as the arc length of the semicircle, so $2π$ is $360°$. (This way of measuring angles is often called measuring them in "radians", but I personally think it is an unnecessary term.) Note that $P_t = P_{t+2πk}$ for any integer $k$, because $2πk$ would be an integer multiple of full rounds. Also note that increasing $t$ moves $P_t$ anti-clockwise, while decreasing $t$ moves $P_t$ clockwise. Related to that, $P_{-t}$ is the reflection of $P_t$ across the $x$-axis.
Note that the signs of $\cos(t)$ and $\sin(t)$ match exactly the signs of the $x$ and $y$ coordinates of the point on the circle. (Do not listen to people who tell you to memorize something to determine which of them is positive in which quadrant.)
And just by definition, $\cos(t)^2+\sin(t)^2 = 1$ for every real $t$. This is the
first key algebraic fact.
Next, $\tan(t)$ is
defined as $\sin(t)/\cos(t)$. (Historically, we also have defined $\sec(t) := 1/\cos(t)$ and $\csc(t) := 1/\sin(t)$ and $\cot(t) := 1/\tan(t)$, but frankly there is little benefit to have so many when $\cos,\sin$ alone suffice.) Whenever you wish to simplify any trigonometric expression involving $\cos,\sin,\tan,\sec,\csc,\cot$, you probably should perform the standard mathematical technique of rewriting in canonical form, which in this case means rewriting in terms of $\cos,\sin$ alone, while taking note of where the original expression is not defined (for instance, $1/\csc(t) = \sin(t)$ for any real $t$ only when $t$ is not a multiple of $π$).
The
other key algebraic facts arise from considering rotation matrices applied to vectors. (If you are unfamiliar with matrices as operators on vectors, please read this first. For an introduction to vectors in euclidean space, see here.) Let $R$ be any rotation about the origin in the plane. Then $R$ satisfies three properties: $R(u+v) = R(u)+R(v)$ for any vectors $u,v$ (i.e. summing two vectors then rotating the result gives the same as rotating the two vectors first before summing them). If $R,S$ are anti-clockwise rotations of angles $t,u$ respectively, then $R∘S$ is an anti-clockwise rotation of angle $t+u$. If $R$ is an anti-clockwise rotation of angle $t$, then: a. $R(⟨x,0⟩) = ⟨x·\cos(t),x·\sin(t)⟩$ for any real $x$. b. $R(⟨0,y⟩) = ⟨-y·\sin(t),y·\cos(t)⟩$ for any real $y$.
We can take these properties as axioms (assumption) about rotations. After all, if $R$ does not satisfy them then we would not call $R$ a rotation to begin with. To see why, property (1) captures the intuition that rotating two connected rods will rotate both rods by the rotation angle while preserving where they connect. Property (2) is only needed in conjunction with property (3). Property (3a) follows from the definition of $\cos,\sin$, and property (3b) follows from the same definition rotated $90°$ anti-clockwise.
Properties (1) and (3) yield the matrix form of a 2d rotation:
If $R$ is an anti-clockwise rotation of angle $t$, then $R = \small \pmatrix{ \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) }$.
And then using property (2) we get:
$\small \pmatrix{ \cos(t+u) & -\sin(t+u) \\ \sin(t+u) & \cos(t+u) } = \pmatrix{ \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) } \pmatrix{ \cos(u) & -\sin(u) \\ \sin(u) & \cos(u) }$ for any reals $t,u$.
Multiplying the matrix product on the right and comparing with the matrix on the left immediately gives the angle-sum identities:
$\cos(t+u) = \cos(t)·\cos(u) - \sin(t)·\sin(u)$ for any reals $t,u$.
$\sin(t+u) = \cos(t)·\sin(u) + \sin(t)·\cos(u)$ for any reals $t,u$.
Whenever you wish to simplify expressions involving trigonometric functions on sums of angles, you should consider using these identities to
reduce the expression to be in terms of $\cos,\sin$ of as few angles as possible.
In fact, all trigonometric identities involving only arithmetic operations and trigonometric functions can be proven using just the above definitions and key algebraic facts. A bit curiously, even the symmetry properties can be proven algebraically as follows.
Given any real $t$:
$1 = \cos(t+(-t)) = \cos(t)·\cos(-t) - \sin(t)·\sin(-t)$. [angle-sum]
$0 = \sin(t+(-t)) = \cos(t)·\sin(-t) + \sin(t)·\cos(-t)$. [angle-sum]
$\cos(t) = \cos(t)^2·\cos(-t) - (\cos(t)·\sin(-t))·\sin(t)$
$ = \cos(t)^2·\cos(-t) + (\sin(t)·\cos(-t))·\sin(t)$
$ = (\cos(t)^2+\sin(t)^2)·\cos(-t)$
$ = \cos(-t)$.
$\sin(t) = (\sin(t)·\cos(-t))·\cos(t) - \sin(t)^2·\sin(-t)$
$ = -(\cos(t)·\sin(-t))·\cos(t) - \sin(t)^2·\sin(-t)$
$ = -\sin(-t)·(\cos(t)^2+\sin(t)^2)$
$ = -\sin(-t)$.
Going on to real analysis, we would need the following facts, which can be taken as axioms for now (and justified separately later):
$\sin' = \cos$. $\cos' = -\sin$.
As before, everything can be reduced to these, so there is no real need to memorize anything more (even though it may be convenient to do so).
Complex trigonometry
Personally, I think it is best to go straight to the complex-valued trigonometric functions, if one desires a complete and rigorous foundation for the mathematical field of
analysis. One simply defines:$\def\rr{\mathbb{R}}\def\cc{\mathbb{C}}\def\lfrac#1#2{{\large\frac{#1}{#2}}}$
$\exp(z) := \sum_{k=0}^∞ \lfrac{z^k}{k!}$ for every complex $z$ (after proving that the sum converges).
$\cos(z) := \lfrac{\exp(iz)+\exp(-iz)}{2}$.
$\sin(z) := \lfrac{\exp(iz)-\exp(-iz)}{2i}$.
$π$ is twice the first positive root of $\cos$ (after proving that it exists).
The motivation is that we want $\exp : \cc→\cc$ such that $\exp' = \exp$ and $\exp(0) = 1$, to be able to solve general linear differential equations, and we want $\cos,\sin : \rr→\rr$ such that $\cos'' = -\cos$ and $\sin'' = -\sin$ and $⟨\cos(0),\cos'(0)⟩ = ⟨1,0⟩$ and $⟨\sin(0),\sin'(0)⟩ = ⟨0,1⟩$, to be able to solve simple harmonic motion, and Taylor expansion brings us to the above definitions for $\exp,\cos,\sin$, which we can prove to converge on the entire complex plane. The above definition of $π$ is the easiest one that I know of that does not depend on any geometry. (For more details on this motivation, see this post.)
Suffice to say that with these definitions, we can prove by basic analysis that $\exp,\cos,\sin$ satisfy the desired motivating properties as well as another
key property of $\exp$:
$\exp(z+w) = \exp(z)·\exp(w)$ for any complex $z,w$.
Using this property, we can prove all the trigonometric identities via algebraic manipulation alone (and they hold for complex variables and not just real variables).
For instance, given any complex $z$:
$\cos(z)^2+\sin(z)^2 = \lfrac{(\exp(iz)+\exp(-iz))^2}{4} - \lfrac{(\exp(iz)-\exp(-iz))^2}{4}$
$ = \exp(iz)·\exp(-iz) = \exp(0) = 1$.
Nevertheless, it is often still easier to first prove the same key algebraic facts for $\cos,\sin$ and then use them to prove other identities, than to reduce everything to $\exp$. |
When multiplying floating points:
we add the (real) exponents, to get the output exponent we multiply the two mantissas (remember the implied 1), round and shift as necessary.
So the new exponent is $E=(19-15) + (3-15)= -8$.
The multiplication of the mantissas give$$1.0011110111_2 \times = 1.0010011000_2 = 1.01101100111010101$$
taking only 10 digits (rounding down, see here) we get $M= 0110110011$.
The sign is a minus.
So the output is$Out= 1\ 00111\ 0110110011$.
For sanity check, let's convert to decimal and see that it makes sense.
$A = (-1)^0 \times 2^{4} \times 1.2412109_{10} = 19.8593744$
$B = (-1)^1 \times 2^{-12} \times 1.1484375_{10} = -0.0002804 $
Then,
$A\cdot B = -0.0055686$
We can convert $Out$ from above to decimal, and get $Out = (-1)^1 \times 2^{-8} \times 1.4248047 = -0.0055656$ and the error is $0.000003 \approx 2^{-18}$ which makes sense (since the real exponent is $2^{-8}$ and we have a 10 digit significand)
See also https://oletus.github.io/float16-simulator.js/ for a calculator, and https://en.wikipedia.org/wiki/Half-precision_floating-point_format or https://en.wikipedia.org/wiki/Floating_point for definition and some more explanations. |
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Bertrand-Chebyshev theorem
Definition from Dictionary, a free dictionary
Life's a bitch, and life's got lots of sisters.Ross Presser
EnglishWikipedia Etymology Noun Bertrand-Chebyshev theorem (Template loop detected: Template:context 1) the theorem that there is at least one prime number between nand 2 nfor every n>1, i.e.: <math>\forall n\in\mathbb{N}:n>1\Rightarrow\exists p\in\mathbb{P}:n<p<2n</math> Elsewhere on the web
Onelook IATE IATE IATE IATE ProZ Dict.cc Wordnik IATE Linguee IATE |
If $A$ is dense in $(X,T)$ and $A$ is an $G_{\delta}$ set $A$ $= \bigcap _{n=1}^{\infty} B_n $ such that $B_n \in T$ $\Rightarrow B_n$ for each n is dense in $(X,T)$.
I have: $A$ $= \bigcap _{n=1}^{\infty} B_n $ $\Rightarrow$ $A$ $\subseteq \bigcap _{n=1}^{\infty} B_n $ $\Rightarrow \forall n \in \Bbb N, A \subseteq B_n$.
And any subset that contains a dense subset in X is dense in X.
I feel like I'm skipping something between the subset and the for all statements. Does this proof make sense or should I have more between the two statements? |
1. Extension of Basis:
Let $V$ be a finite dimensional vector space over a field $k$, and let $S:= \{v_1, \cdots , v_r\}$ be any
linearly independent set of vectors in $V$.
Claim: The set $S$ can be extended to a basis of $V$.
If $\dim(V)=r$, we are done. If not, then there is a non-zero element $v_{r+1}\in V$ which is not in the subspace (of $V$) spanned by $S$. One can show that $S \cup \{v_{r+1}\}$ is also linearly independent. If $S \cup \{v_{r+1}\}$ spans the whole space $V$, we are done. Otherwise we continue this process. This process will terminate because $V$ is finite dimensional, and thus we get a basis of $V$.
2. Idea/Method:
Let $V, W$ be finite dimensional vector spaces over a field $k$. Suppose we are given a subspace $V' \subset V$ with a basis $\{e_1, \cdots, e_d\}$ and a subspace$W'\subset W$ with a basis $\{f_1, \cdots ,f_r\}$. We want to find a linear transformation $T: V \to W$ with $\ker T= V'$ and $\operatorname{Im}T=W'$. If such $T$ exists, then by Rank-nullity theorem, $\dim V=d+r$. If this equality doesn't hold, then there is no such $T$. So assume $\dim V=d+r$. Now extend the basis of $\ker T$ to a basis of $V$, say, $\{e_1, \cdots ,e_d, e_{d+1}, \cdots ,e_{d+r}\}$. Define a map $$T: V \to W$$ by $$T(e_i)=0, 1 \leq i \leq d$$$$T(e_{d+i})=f_i, 1 \leq i \leq r.$$Then $T$ is a linear transformation (
why?), and $\ker T=V'$, $\operatorname{Im}T=W'$.
Note: The linear transformation $T$ is uniques only up to conjugate.
3. Example:
Let $e_1=(1,0,0), e_2=(0,1,0) \in \mathbb{R^3}$ and $f_1=(1,0) \in \mathbb{R}^2$. We want a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ such that $\ker T$ has a basis $\{e_1, e_2\}$ and $\operatorname{Im}T$ is generated by $f_1$. Note that, for $e_3=(0,0,1) \in \mathbb{R}^3,\{e_1, e_2, e_3\}$ is a basis of $\mathbb{R}^3$. Now use the above technique to define a linear transformation. We can define $T$ more concretely as follows:
Any $(x,y,z) \in \mathbb{R}^3$ can be written as $xe_1+ye_2+ze_3$. So $$T(x,y,z)=xTe_1+yTe_2+zTe_3=zf_1=(0,z).$$ |
In this paper the fractional Cox–Ingersoll–Ross process on ${\mathbb{R}_{+}}$ for $H<1/2$ is defined as a square of a pointwise limit of the processes ${Y_{\varepsilon }}$, satisfying the SDE of the form $d{Y_{\varepsilon }}(t)=(\frac{k}{{Y_{\varepsilon }}(t){1_{\{{Y_{\varepsilon }}(t)>0\}}}+\varepsilon }-a{Y_{\varepsilon }}(t))dt+\sigma d{B^{H}}(t)$, as $\varepsilon \downarrow 0$. Properties of such limit process are considered. SDE for both the limit process and the fractional Cox–Ingersoll–Ross process are obtained.
We investigate the pricing of cliquet options in a geometric Meixner model. The considered option is of monthly sum cap style while the underlying stock price model is driven by a pure-jump Meixner–Lévy process yielding Meixner distributed log-returns. In this setting, we infer semi-analytic expressions for the cliquet option price by using the probability distribution function of the driving Meixner–Lévy process and by an application of Fourier transform techniques. In an introductory section, we compile various facts on the Meixner distribution and the related class of Meixner–Lévy processes. We also propose a customized measure change preserving the Meixner distribution of any Meixner process.
with multiplicative stochastic volatility, where Y is some adapted stochastic process. We prove existence–uniqueness results for weak and strong solutions of this equation under various conditions on the process Y and the coefficients a, $\sigma _{1}$, and $\sigma _{2}$. Also, we study the strong consistency of the maximum likelihood estimator for the unknown parameter θ. We suppose that Y is in turn a solution of some diffusion SDE. Several examples of the main equation and of the process Y are provided supplying the strong consistency.
We investigate the convergence of hitting times for jump-diffusion processes. Specifically, we study a sequence of stochastic differential equations with jumps. Under reasonable assumptions, we establish the convergence of solutions to the equations and of the moments when the solutions hit certain sets. |
Finite dimensional smooth attractor for the Berger plate with dissipation acting on a portion of the boundary
1.
Department of Mathematics, University of Nebraska-Lincoln, Lincoln, Nebraska 68588
2.
Department of Mathematics, Faculty of Science, Hacettepe University, Beytepe 06800, Ankara
3.
Haceteppe University, Ankara , Turkey
one boundary conditionon a portion of the boundary. In [24] this type of boundary damping was considered for a Berger plate on the whole boundary and shown to yield the existence of a compact global attractor. In this work we address the issues arising from damping active only on a portion of the boundary, including deriving a necessary trace estimate for $(\Delta u)\big|_{\Gamma_0}$ and eliminating a geometric condition in [24] which was utilized on the damped portionof the boundary.
Additionally, we use recent techniques in the asymptotic behavior of hyperbolic-like dynamical systems [11, 18] involving a ``stabilizability" estimate to show that the compact global attractor has finite fractal dimension and exhibits additional regularity beyond that of the state space (for finite energy solutions).
Keywords:Global attractor, boundary dissipation, dissipative dynamical system., nonlinear plate equation. Mathematics Subject Classification:Primary: 35B41, 74K20; Secondary: 35Q74, 35A0. Citation:George Avalos, Pelin G. Geredeli, Justin T. Webster. Finite dimensional smooth attractor for the Berger plate with dissipation acting on a portion of the boundary. Communications on Pure & Applied Analysis, 2016, 15 (6) : 2301-2328. doi: 10.3934/cpaa.2016038
References:
[1]
J. P. Aubin, Une théorè de compacité,,
[2]
G. Avalos and I. Lasiecka, Exponential stability of a thermoelastic system without mechanical dissipation,,
[3]
G. Avalos and I. Lasiecka, Boundary controllability of thermoelastic plates via the free boundary conditions,,
[4]
A. Babin and M. Vishik,
[5] [6]
H. M. Berger, A new approach to the analysis of large deflections of plates,,
[7]
V. V. Bolotin,
[8] [9]
F. Bucci, I. Chueshov and I. Lasiecka, Global attractor for a composite system of nonlinear wave and plate equations,,
[10]
F. Bucci and I. Chueshov, Long-time dynamics of a coupled system of nonlinear wave and thermoelastic plate equations,,
[11] [12] [13]
I. Chueshov,
[14]
I. Chueshov, M. Eller and I. Lasiecka, Finite dimensionality of the attractor for a semilinear wave equation with nonlinear boundary dissipation,,
[15]
I. Chueshov and I. Lasiecka, Global attractors for von Karman evolutions with a nonlinear boundary dissipation,,
[16] [17]
I. Chueshov and I. Lasiecka, Long-time dynamics of von Karman semi-flows with non-linear boundary/interior damping,,
[18] [19]
I. Chueshov, I. Lasiecka and D. Toundykov, Global attractor for a wave equation with nonlinear localized boundary damping and a source term of critical exponent,,
[20]
I. Chueshov, I. Lasiecka and J. T. Webster, Attractors for delayed, non-rotational von Karman plates with applications to flow-structure interactions without any damping,,
[21]
P. Ciarlet and P. Rabier,
[22]
A. Eden and A. J. Milani, Exponential attractors for extensible beam equations,,
[23]
P. Fabrie, C. Galusinski, A. Miranville and S. Zelik, Uniform exponential attractors for a singularly perturbed damped wave equation,,
[24]
P. G. Geredeli and J. T. Webster, Qualitative results on the dynamics of a Berger plate with nonlinear boundary damping,,
[25]
P. G. Geredeli, I. Lasiecka and J. T. Webster, Smooth attractors of finite dimension for von Karman evolutions with nonlinear frictional damping localized in a boundary layer,,
[26]
P. G. Geredeli and J. T. Webster, Decay rates to eqilibrium for nonlinear plate equations with geometrically constrained, degenerate dissipation,
[27]
J. K. Hale and G. Raugel, Attractors for dissipative evolutionary equations,,
[28]
G. Ji and I. Lasiecka, Nonlinear boundary feedback stabilization for a semilinear Kirchhoff plate with dissipation acting only via moments-limiting behavior,,
[29] [30] [31]
I. Lasiecka and R. Triggiani,
[32] [33] [34]
J. L. Lions,
[35]
J. L. Lions, Contrôlabilité exacte, perturbations et stabilization de systèmes distribués,,
[36] [37]
A. Miranville and S. Zelik, Attractors for dissipative partial differential equations in bounded and unbounded domains,,
[38] [39] [40] [41] [42] [43]
C. P. Vendhan, A study of Berger equations applied to nonlinear vibrations of elastic plates,,
show all references
References:
[1]
J. P. Aubin, Une théorè de compacité,,
[2]
G. Avalos and I. Lasiecka, Exponential stability of a thermoelastic system without mechanical dissipation,,
[3]
G. Avalos and I. Lasiecka, Boundary controllability of thermoelastic plates via the free boundary conditions,,
[4]
A. Babin and M. Vishik,
[5] [6]
H. M. Berger, A new approach to the analysis of large deflections of plates,,
[7]
V. V. Bolotin,
[8] [9]
F. Bucci, I. Chueshov and I. Lasiecka, Global attractor for a composite system of nonlinear wave and plate equations,,
[10]
F. Bucci and I. Chueshov, Long-time dynamics of a coupled system of nonlinear wave and thermoelastic plate equations,,
[11] [12] [13]
I. Chueshov,
[14]
I. Chueshov, M. Eller and I. Lasiecka, Finite dimensionality of the attractor for a semilinear wave equation with nonlinear boundary dissipation,,
[15]
I. Chueshov and I. Lasiecka, Global attractors for von Karman evolutions with a nonlinear boundary dissipation,,
[16] [17]
I. Chueshov and I. Lasiecka, Long-time dynamics of von Karman semi-flows with non-linear boundary/interior damping,,
[18] [19]
I. Chueshov, I. Lasiecka and D. Toundykov, Global attractor for a wave equation with nonlinear localized boundary damping and a source term of critical exponent,,
[20]
I. Chueshov, I. Lasiecka and J. T. Webster, Attractors for delayed, non-rotational von Karman plates with applications to flow-structure interactions without any damping,,
[21]
P. Ciarlet and P. Rabier,
[22]
A. Eden and A. J. Milani, Exponential attractors for extensible beam equations,,
[23]
P. Fabrie, C. Galusinski, A. Miranville and S. Zelik, Uniform exponential attractors for a singularly perturbed damped wave equation,,
[24]
P. G. Geredeli and J. T. Webster, Qualitative results on the dynamics of a Berger plate with nonlinear boundary damping,,
[25]
P. G. Geredeli, I. Lasiecka and J. T. Webster, Smooth attractors of finite dimension for von Karman evolutions with nonlinear frictional damping localized in a boundary layer,,
[26]
P. G. Geredeli and J. T. Webster, Decay rates to eqilibrium for nonlinear plate equations with geometrically constrained, degenerate dissipation,
[27]
J. K. Hale and G. Raugel, Attractors for dissipative evolutionary equations,,
[28]
G. Ji and I. Lasiecka, Nonlinear boundary feedback stabilization for a semilinear Kirchhoff plate with dissipation acting only via moments-limiting behavior,,
[29] [30] [31]
I. Lasiecka and R. Triggiani,
[32] [33] [34]
J. L. Lions,
[35]
J. L. Lions, Contrôlabilité exacte, perturbations et stabilization de systèmes distribués,,
[36] [37]
A. Miranville and S. Zelik, Attractors for dissipative partial differential equations in bounded and unbounded domains,,
[38] [39] [40] [41] [42] [43]
C. P. Vendhan, A study of Berger equations applied to nonlinear vibrations of elastic plates,,
[1]
Francesca Bucci, Igor Chueshov, Irena Lasiecka.
Global attractor for a composite system of nonlinear wave and plate equations.
[2]
Moncef Aouadi, Alain Miranville.
Quasi-stability and global attractor in nonlinear thermoelastic diffusion plate with memory.
[3]
Azer Khanmamedov, Sema Simsek.
Existence of the global attractor
for the plate equation with nonlocal nonlinearity in $
\mathbb{R}
^{n}$.
[4]
Yongqin Liu, Shuichi Kawashima.
Global existence and asymptotic behavior
of solutions for quasi-linear dissipative plate equation.
[5] [6]
Nikos I. Karachalios, Nikos M. Stavrakakis.
Estimates on the dimension of a global attractor for a semilinear dissipative wave equation on $\mathbb R^N$.
[7]
Wided Kechiche.
Regularity of the global attractor for a nonlinear Schrödinger equation with a point defect.
[8] [9] [10]
Vladimir V. Chepyzhov, Monica Conti, Vittorino Pata.
Totally dissipative dynamical processes
and their uniform global attractors.
[11]
Moez Daoulatli, Irena Lasiecka, Daniel Toundykov.
Uniform energy decay for a wave equation with partially supported nonlinear boundary dissipation without growth restrictions.
[12]
Wen Tan.
The regularity of pullback attractor for a non-autonomous
[13]
Sébastien Court.
Stabilization of a fluid-solid system, by the deformation of the self-propelled solid. Part II: The nonlinear system..
[14]
Boyan Jonov, Thomas C. Sideris.
Global and almost global existence of small solutions to a dissipative wave equation in 3D with nearly null nonlinear terms.
[15]
Dominique Blanchard, Nicolas Bruyère, Olivier Guibé.
Existence and uniqueness of the solution of a Boussinesq
system with nonlinear dissipation.
[16] [17] [18]
Jun Zhou.
Global existence and energy decay estimate of solutions for a class of nonlinear higher-order wave equation with general nonlinear dissipation and source term.
[19]
Marek Fila, Kazuhiro Ishige, Tatsuki Kawakami.
Convergence to the Poisson kernel for the Laplace equation with a nonlinear dynamical boundary condition.
[20]
Boling Guo, Zhengde Dai.
Attractor for the dissipative Hamiltonian amplitude equation governing modulated wave instabilities.
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How can I prove that $$\lim_{x\to +\infty}\frac{\left \lfloor{x}\right \rfloor}{x}=1$$
L'Hôpital's rule seems to fail here, since the floor function is not differentiable for integers. What other ways are there to prove this?
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How can I prove that $$\lim_{x\to +\infty}\frac{\left \lfloor{x}\right \rfloor}{x}=1$$
L'Hôpital's rule seems to fail here, since the floor function is not differentiable for integers. What other ways are there to prove this?
Note that $$ x=\lfloor{x}\rfloor+\zeta $$ where $0\leq \zeta<1.$ Then $$ \frac{x}{\lfloor{x}\rfloor}=1+\frac{\zeta}{\lfloor{x}\rfloor}\to1 $$ as $x\to \infty$ since $\zeta/\lfloor{x}\rfloor\to 0$ as $x\to\infty$.
Big hint: $x-1 \le \lfloor x \rfloor \le x$ for all $x \in \Bbb R$. |
Say you've got a number of sets $S_1,...,S_n$ given and are supposed to calculate $\,\,\displaystyle\big|\bigcup_{i=1}^n S_i\big|$ .
The basic approach probably would be to use the classic inclusion-exclusion principle $\displaystyle\sum_{\emptyset \not= I \subseteq \{1, \dotsc, n\}} \left(-1\right)^{|I|+1}|A_I|$ , where $A_I = \,\,\displaystyle\bigcap_{i\in I}^n S_i$ .
However, inclusion-exclusion feels like it's terribly inefficient - after all, we need to sum over $\displaystyle\sum_{i=1}^n\binom{n}{i} = 2^n-1$ possible sets.
While I guess that there's no algorithm with polynomial complexity for this problem, I'm curious about alternatives that are more efficient (i.e. have polynomial run time in most cases), or are simply largely different.
I'd be thankful if you could name alternative algorithms or optimizations to inclusion-exclusion.
The particular problem I'm trying to write an algorithm for is unsatisfiability in Propositional logic for formulas in CNF. I.e. we have a finite set of atomic formulas $\{p_1,...,p_m\}$. Over these atomic formulas, the clauses of the CNF of the formula are built after these rules:
For ever atomic formula $p$, there may either appear $p$, $\lnot p$ or none of the two in the clause.
As the whole problem was either $NP$ or $co-NP$, there most likely isn't a polynomial algorithm, but I am struggling to just make it anyhow just a little bit more efficient than inclusion-exclusion (besides using ideas that have little to do with the algorithm itself like pure-literals or pure-clauses).
Yes, each set itself is discrete with each element in the set being specified. One example would be $\{ \{\bar A, B \},\{B,C \},\{ A,B \} \}$. However, to fit the example, the initially given inclusion-exclusion algorithm would need to be modified.
This is the case as every clause in the CNF is a representative for a bigger clause, e.g. in the above example $\{\bar A, B \}$ stands for
$ \{\bar A, B ,C \} \{\bar A, B ,\bar C \} $ and $\{B,C \}$ stands for $ \{B,C, A \} \{B,C ,\bar A \} $
You get the actual set the representative represents by the following algorithm:
Be $M$ a set of the clause, and $S$ our output set. Add $M$ to $S$. Now repeat as long as possible:
Find an atomic formula $A$ so that $A\notin K$ and $\bar A \notin K$ for some $K\in S$.
Remove $K$ from $S$. Add $K\cup \{A\}$ and $K\cup \{\bar A\}$ to $S$.
Finally, delete all sets in $S$ where $A,\bar A \in S$ for some $A$ and return $S$.
(An atomic formula in this case is simply any letter somewhere in $M$ without the bar (i.e. $A$ for $\bar A$))
This version of the algorithm however is at this point pretty dumb. You can improve it by not explicitly generating the actual sets and calculate it with the representants and applying some arithmetic instead.
It goes like this:Let $A_1,...,A_n$ be our atomic formulas that appear in a set of clauses $\mathcal{K}$. For every $K\in\mathcal{K}$ you can calculate how many sets $K$ actually represents by counting the elements of $K$, let's name it $c$, and calculate $2^{n-c}$.
So, we calculate the sum all the way above, i.e. $$\displaystyle\sum_{\emptyset \not= I \subseteq \{1, \dotsc, n\}} \left(-1\right)^{|I|+1}|A_I|\text{ , where }A_I = \,\,\displaystyle\bigcap_{i\in I}^n S_i$$ just that we calculate $|A_I|$ not by $\big|\,\,\displaystyle\bigcap_{i\in I}^n S_i\big|$, but by calculating how many sets $\,\,\displaystyle\bigcup_{i\in I}^n S_i$ represents (with special case $|A_I|=0$, if $A_I$ contains $A$ and $\bar A$ for some atomic formula). |
My previous answer addressed the question "How can I make sense of the counting types definition of stability outside of a first-order model theory context?"
After discussion in the comments, I realized that you really wanted to know what stability means in the context of an (abstract) projective plane. So I'll try to address that question in this answer.
First some background in model theory, just to make sure if we're on the same page. Let $M$ be an $L$-structure and $C$ a subset of $M$. We write $L(C)$ for the set of all first-order formulas with parameters from $C$. For example, if $L = \{P,L,I\}$ is the language of incidence structures (where $P$ and $L$ are unary relations picking out the points and lines and $I$ is the binary incidence relation), then $\varphi(x_1,x_2): \forall y\, ((L(y) \land I(x_1,y)\land I(x_2,y))\rightarrow \lnot I(c,y))$ is a formula with a single parameter $c\in C$ expressing that no line through $x_1$ and $x_2$ is incident with $c$.
Let $x = (x_1,\dots,x_n)$ be a tuple of variables. A
complete type $p(x)$ over $C$ (in variable context $x$) is a maximal set of $L(C)$-formulas with free variables from $x$ which is consistent with $\mathrm{Th}_{L(C)}(M)$, the set of all $L(C)$-sentences true in $M$. Maximality amounts to saying that for every formula $\varphi(x)\in L(C)$, either $\varphi(x)\in p(x)$ or $\lnot \varphi(x)\in p(x)$. We write $S_x(C)$ for the set of all complete types over $C$ in context $x$.
Let $a = (a_1,\dots,a_n)$ be an element of $M$. Then the
type of $a$ over $C$ is the set of all formulas with parameters from $C$ satisfied by $a$ in $M$: $$\mathrm{tp}(a/C) = \{\varphi(x)\in L(C)\mid M\models \varphi(a)\}.$$
We could equivalently define $S_x(C) = \{\mathrm{tp}(b/C)\mid b\in N, M\preceq N\}$. Note here that in general we have to pass to an elementary extension of $M$: it may be that not every complete type over $C$ is the type of an element in $M$, but every complete type is realized in some elementary extension.
Now a theory $T$ is
stable if there is some infinite cardinal $\kappa$ such that for all models $M\models T$, all sets $C\subseteq M$ with $|C| = \kappa$, and all variable contexts $x$, we have $|S_x(C)| = \kappa$.
In Shelah's slides, he gives a slightly different definition, depending on the following facts:
It suffices to check stability when the context is a single variable $x$ instead of a tuple. It suffices to check stability when $C$ is an (elementary) submodel. If $T$ is stable, then actually we have that for all infinite $\kappa$, $|C| = \kappa$ implies $|S_x(C)| = \kappa^{|L|}$, where $|L|$ is the number of formulas in the language (with no parameters). The fact that there are cardinals $\kappa$ such that $\kappa^{|L|} = \kappa$ gives the converse. Usually we're most interested in the case $|L| = \aleph_0$.
In practice, if you want to prove that a theory is stable by counting types, then you need some way to understand all the possible complete types. This can be hard, because first-order formulas can be very complicated in general. There are essentially two ways of doing this:
Quantifier elimination. A theory $T$ has quantifier elimination if for every first-order formula $\varphi(x)$, there is a quantifier-free formula $\psi(x)$ such that $T\models \forall x\, \varphi(x)\leftrightarrow \psi(x)$. Since quantifier-free formulas are usually much easier to understand than general formulas (they are just Boolean combinations of the atomic formulas - basic relations and equalities between terms), quantifier elimination can be very helpful for counting types. Some theories admit weaker forms of quantifier elimination, where you can show that every formula is equivalent to a Boolean combination of formulas in some restricted class that are still easy to understand. Automorphisms. If $a,a'\in M$ and $\sigma\colon M\to M$ is an automorphism with $\sigma(c) = c$ for all $c\in C$ and $\sigma(a) = a'$, then automatically $\mathrm{tp}(a/C) = \mathrm{tp}(a'/C)$. So upper bounds on the number of orbits of automorphism groups of models of $T$ gives upper bounds on the number of complete types.
OK, now let's look at the example of projective planes. The easiest example of a stable theory of projective planes is the complete theory $T$ of $\mathbb{P}^2(k)$, where $k$ is an algebraically closed field. One way to see that $T$ is stable is to show that it is bi-interpretable with the complete theory of $k$. Any theory of algebraically closed fields is stable, and stability is preserved under bi-interpretation.
$T$ does not have quantifier elimination, but if you add a binary function $f$ to the language such that $f(p_1,p_2)$ is the unique line through the points $p_1$ and $p_2$, and $f(l_1,l_2)$ is the unique point at the intersection of $l_1$ and $l_2$, then $T$ has quantifier elimination in the expanded language (this fact rests heavily on the fact that the theory of algebraically closed fields has quantifier elimination).
Let's think about the counting types criterion. Let $M\models T$ (then $M = \mathbb{P}^2(K)$ where $K$ is algebraically closed of the same characteristic as $k$). Suppose $|M| = \kappa$. How big is $S_x(M)$ (where $x$ is a single variable)?
A type over $M$ could be the type of a point or a line. It suffices to count the types of points, by duality. It turns out that if we coordinatize $M = \mathbb{P}^2(K)$, the type of a point over $M$ is completely determined by the algebraic relationships between the coordinates which are definable over $K$. In short, the types are exactly the scheme-theoretic points of $\mathbb{P}^2(K)$.
The points of dimension $0$ are the types of points in $M$ (there are $\kappa$-many of these). The points of dimension $1$ are the generic types of curves in the projective plane defined over $K$ (there are $\kappa$-many of these, since each can be associated with a homogeneous polynomial over $K$). And there is a unique point of dimension $2$, the generic type of the plane. Altogether, there are only $\kappa$-many types, so $T$ is stable (in fact, it is $\omega$-stable, meaning that $|M| = \kappa$ implies $|S_x(M)| = \kappa$ for
all infinite cardinals $\kappa$).
When the field is not algebraically closed, we do not get such a nice description of types thanks to the loss of quantifier elimination in the field (though understanding types relative to the theory of $\mathbb{P}^2(k)$ can still be reduced to understanding types relative to the theory of $k$).
The situation is even worse for non-Desarguesian projective planes, which are not coordinatized by fields or even division rings. As I alluded to in the comments, however, Hrushovski constructions can be used to produce non-Desarguesian projective planes with stable theories. These structures are extremely homogeneous, which allows stability to be checked using the automorphism criterion I described above. |
A neutrino has rest-mass and travels at (near) $c$, why isn't its mass/ energy (nearly) infinite?
Because it has too low rest mass or still too low speed. Neutrinos are very light particles: Their rest energy is comparable to energy of a hydrogen bond (weaker than typical chemical bound). So you can understand, that full energy must not be infinite.
However, this is not a practical point of view. From the practical point of view, neutrinos have mass-energy as they have (because they got it when they were born (or sometimes in an interaction event afterwards)) and because of this they have speed (that for all
currently observable neutrinos is only slightly less than the speed of light).
The mass-energy can be anything bigger than the rest mass of neutrino. Neutrino detector using a gallium → germanium transformation has detection threshold of 0.233 MeV and this is the lowest value I have found in the Wikipedia article about neutrino detection. Solar neutrinos have energies up to 18 MeV. Famous, but now known to be mistake, faster-than-light neutrino anomaly has something to do with 28 GeV neutrinos. 6.5 TeV per beam (planed for LHC since early in 2015) can become almost wholly energy of neutrino. (Two protons could stop and emit neutrino and antineutrino, but probability is low, digression 1).
The equation for energy of particle with rest mass $m$ is
$E = \gamma m$
where
$\gamma = \frac{1}{\sqrt{1-v^2}}$ so $v = \sqrt{1-\frac{1}{\gamma^2}} \approx 1 - \frac{1}{2 \gamma^2}$.
In the cases about which I write above, the speed is respectively $(1-9.2 \cdot 10^{-14}) c$, $(1-1.5 \cdot 10^{-17}) c$, $(1-3.4 \cdot 10^{-24}) c$ and $(1-1.2 \cdot 10^{-28}) c$.
On the other hand, "relic background neutrinos are estimated to have (...) temperature 1.9 K ($1.7×10^{−4}$ eV) if they are massless, much colder if their mass exceeds 0.001 eV". So even assuming temperature too high for such "heavy" 0.1 eV neutrinos, they are non-relativistic and have average speed (from non-relativistic equations $E = m v^2 / 2$ and $E = \frac{3}{2} k T$) $v = \sqrt{3 k T / m} \approx 0.071 c$. That's still about 21000 km/s, but clearly less than the speed of light.
So if we plug in to find the total energy of the neutrino we find.
ev∼
18keV∼0.03 $m_{electron} Which isn't that big.
Ia this answer correct?
J.J. must have forgotten about square. I have corrected it and $0.03 m_{electron}$ is now even better approximation, but this says something only about neutrinos with speed $0.999999997 c$.
Now, the energy of an electron is .
5 Mega eV, what is then the correct maximum value of the energy of a neutrino?
It is true that "the mass of the neutrino is tiny, but its kinetic energy can be of the same scale as that of the electron" for neutrinos from beta decay and the energy of beta decay is comparable to the mass of electron, so this question makes some sense, but knowing just this, one can only say "of the order of .5 Mega eV". Solar neutrinos are from nuclear reactions and have energies up to 18 MeV - the same order more or less.
You probably wanted some calculations, so I will calculate maximum energy of the neutrinos from decay of tritium. In the first approximation, energy of neutrino is simply difference between rest energy of tritium atom and rest energy of helium-3 atom. (If neutrino takes the whole energy, we could really get neutral helium-3 atom, but probability is low. Digression 2.) Neglecting neutrino mass, its energy is equal to its momentum ($p$), which is equal (with opposite direction) to the helium-3 momentum.
$E_{{^3}\mathrm{He}}^2 = m_{{^3}\mathrm{He}}^2+p^2$
$m_\mathrm{T} = E_{{^3}\mathrm{He}} + p = \sqrt{m_{{^3}\mathrm{He}}^2+p^2} + p$
$m_{{^3}\mathrm{He}}^2 + p^2 = m_\mathrm{T}^2 - 2 m_\mathrm{T} p + p^2$
$p = \frac{m_\mathrm{T}^2 - m_{{^3}\mathrm{He}}^2}{2 m_\mathrm{T}}$
$p = 0.0186 \mathrm{M}e\mathrm{V}$
using
$1 \mathrm{u} = 931.4812 Me\mathrm{V}$
$m_\mathrm{T} = 3.01604927 \mathrm{u}$
$m_{{^3}\mathrm{He}} = 3.01602931 \mathrm{u}$
In this case, we get low energy. From 0.0186 MeV to 18 MeV is something about 0.5 MeV, but writing this I have realised that the rule that nuclear reaction energy is comparable to the mass of electron is not very precise.
Digression 1: Probability is low, but I do not know how low - there are very many events and I do not have any idea, how often this is going to happen - maybe more than once a second, maybe less than once per trillion years.
This should be quite easy to estimate if one knows appropriate rule, but I don't know it and I am afraid that looking for it would take too much time. This has something to do with parton energy distribution and probability of reaction. And what does "almost wholly" mean? Something like 99%, in worst case 99.9%, is enough to get $(1-1.2 \cdot 10^{-28}) c$ with 1.2, not 1.3. Anyway, taking one half of this energy should be much more probable and give comparable neutrino speed.
Digression 2: Again I do not know how improbable this is.
This should be quite easy to estimate if one knows appropriate rule, but I don't know it and I am afraid that looking for it would take too much time. This has something to do with distribution of speed in decays, but also with something like resonances.
General notes:
I do not know more than Wikipedia says about the 0.320 ± 0.081 eV estimation by Planck collaboration, but here I'm assuming that it is correct. Than, oscillation data show that the mass difference is small, and masses of all three neutrinos are about 0.1 eV, but this is rough approximation from error range itself, so I probably underestimate other errors and write to many digits. All not taking into account the possibility that something is very wrong.
Besides I am using here $c=1$ convention. Masses and energy are both measured in electronovolts, and speed is always compared to the speed of light, so this should not cause any problems. |
Alternative Formulation
I came up with an alternative formulation to the below problem. The alternative formulation is actually a special case of the problem bellow and uses bipartite graphs to describe the problem. However, I believe that the alternative formulation is still NP-hard. The alternative formulation uses a disjoint set of incoming and outgoing nodes that simplifies the problem definition.
Given $n$ outgoing and $n$ incoming nodes (the red and blue nodes in the figure respectively), and a set $w_{ij}$'s of size $n \times n$ of edge weights between the outgoing and incoming vertices. The goal of the problem is to color the thick edges in the figure so that for every incoming node, a condition holds.
Given a set $\{ O_i \; | \; i=1 \dots n \}$ of output vertices, a set $\{ I_i\; | \; i=1 \dots n \}$ of input vertices, $n \times n$ weights $w_{ij} \ge 0$ between $O_i$'s and $I_j$'s for $i,j=1 \dots n$, and a positive constant $\beta$, find the minimum number of colors for the edges $e_{ii}$ (thick edges in the above figure) such that for all $j=1 \dots n$,
$$ \frac{w_{jj}}{1+\sum_{c(i)=c(j),i \neq j} w_{ij}} \ge \beta $$
where $c(i)$ shows the color of the edge $e_{ii}$.
Old Formulation
The following problem looks NP-hard to me, but I couldn't show it. Any proof/comment to show the hardness or easiness of it is appreciated.
Assume $K_n=\langle V,E \rangle$ is a complete weighted directed graph with $n$ nodes and $n(n-1)$ edges. Let $w_{ij} \ge 0$ show the weight of the edge $ij$ and $c(ij)$ shows the color of edge $ij$. Given a subset of the edges $T \subseteq E$ and a positive constant $\beta$ the goal is: find the minimum number of colors such that for each $e_{ij} \in T$:
$$ \frac{w_{ij}}{1+\sum_{c(kl)=c(ij),kl \neq ij} w_{kj}} \ge \beta. $$ and $$ c(ij) \neq c(ik) \quad for \quad j \neq k $$
Please note that in the above problem, only the edges in $T$ needs to be colored. That is the problem can be solved in $\mathcal{O}(|T|!)$.
Update:
After Tsuyoshi Ito's comment I updated the problem. The denominator is changed from $1+\sum_{c(kj)=c(ij),k \neq i,e_{kj} \in T} w_{kj}$ to $1+\sum_{c(kl)=c(ij),kl \neq ij} w_{kj}$. Therefore, the denominator contains the weights outside $T$ as well. That's actually why I mentioned the complete graph in the definition.
I also added an additional constraint $c(ij) \neq c(ik) \quad for \quad j \neq k$. That means, the outgoing edges from a node must be of different colors (but the incoming colors can be the same as long as the inequality holds). This puts an intuitive lower bound on the number of colors, which is the maximum out-degree of the nodes in $T$.
As Tsuyoshi mentioned, $w_{ij}$'s, $T$, and $\beta$ are inputs to the problem and the edge colors are the output.
Update 2:
Problem does not enforce the edges $e_{ij}$ and $e_{ji}$ be of a same color. |
I am trying to compute the phase spectrum of the signal $$ s(t)=\frac{A}{\pi}\left[H\left(t+\frac{\tau}{2}\right)-H\left(t-\frac{\tau}{2}\right)\right] $$
The Fourier transform is $$ S(j\omega)=\frac{A\tau}{\pi}\cdot\frac{\sin\left(\omega\displaystyle\frac{\tau}{2}\right)}{\omega\displaystyle\frac{\tau}{2}}$$
My answer for this would be $\phi_s(\omega)=0$, since the imaginary part of $S(j\omega)$ is zero, but according to my textbook, the answer is
$$\phi_s(\omega)=\begin{cases}0&S(j\omega)>0\\\pm\pi&S(j\omega)<0\end{cases}$$
How can I get to this answer and interpret it correctly? |
A forum where anything goes. Introduce yourselves to other members of the forums, discuss how your name evolves when written out in the Game of Life, or just tell us how you found it. This is the forum for "non-academic" content.
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
Ah! Ah! Yoven urge losh = very long house!
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
testitemqlstudop wrote:correct, finally!
Awesome! Who will do mine?
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
testitemqlstudop Posts: 1194 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
3 - lobster, duh
7 - tub with tail, the "twit" gave it away lol Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
testitemqlstudop wrote:3 - lobster, duh
7 - tub with tail, the "twit" gave it away lol
both correct.
Yeah, halibut twit was too good to pass up.
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Hdjensofjfnen Posts: 1338 Joined: March 15th, 2016, 6:41 pm Location: r cis θ
Chimp sale = Camelship
EDIT x4: Solving the easy ones and putting them in code tags to not spoil anything.
Code: Select all
SPOILER ALERT!
2. A call with wit = Claw with tail (similar to the twit thing)
4. Chatters rent = Antstretcher
8. Flaky rehab = Half-bakery
9. Bad plank sheikh fight = Half-baked knightship
11. Hellish lag = Shillelagh
13. Taboo surf = Four boats
14. Jumbo erect floor = Boojum reflector
16. Rephrases plurals = Pre-pulsar hassler
17. Cheapest lightship wig = Lightweight spaceship
Last edited by Hdjensofjfnen
on March 14th, 2019, 7:33 pm, edited 3 times in total.
"A man said to the universe:
'Sir, I exist!'
'However,' replied the universe,
'The fact has not created in me
A sense of obligation.'" -Stephen Crane
Code: Select all
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
Hdjensofjfnen wrote:Chimp sale = Camelship
EDIT:
11. Hellish lag = Shillelagh
13. Taboo surf = Four boats
yes, yes, yes!
It seems that in modern times hellish lag is as painful as being hit by a shillelagh.
Last edited by Moosey
on March 14th, 2019, 7:49 pm, edited 1 time in total.
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Hdjensofjfnen Posts: 1338 Joined: March 15th, 2016, 6:41 pm Location: r cis θ
I just hit you with some more!
"A man said to the universe:
'Sir, I exist!'
'However,' replied the universe,
'The fact has not created in me
A sense of obligation.'" -Stephen Crane
Code: Select all
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact:
#15: Hellish amnesia seeks lag —> shillelagh siamese snake.
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
A for awesome wrote:#15: Hellish amnesia seeks lag —> shillelagh siamese snake.
Yes!
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Hdjensofjfnen Posts: 1338 Joined: March 15th, 2016, 6:41 pm Location: r cis θ
"A man said to the universe:
'Sir, I exist!'
'However,' replied the universe,
'The fact has not created in me
A sense of obligation.'" -Stephen Crane
Code: Select all
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
testitemqlstudop Posts: 1194 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
Very, very hard (increasing order of difficulty):
Code: Select all
yokes noon meme dyke
catastrophe weaned okayed PayPal
unconstitutional reciprocals tomato Lolita
misrepresentations frighteningly chili shit Bork
reforestation lintels ennoble boloney benefit Whitney Bennett
Hints for very, very hard (each hint is a word in the original sentence)
Code: Select all
yokes noon meme dyke - monkey
catastrophe weaned okayed PayPal - apple
unconstitutional reciprocals tomato Lolita - total (hmmm!)
misrepresentations frighteningly chili shit Bork - beer (x2)
reforestation lintels ennoble boloney benefit Whitney Bennett - robin
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
Sorry to bump, but
Code: Select all
Enable an exile pool
Hint: scroll down for hint
Horrible egoistical artillery captain. The answer is “Island name exile _____”
Also, is
STONE CROP = Top censor?
And I hope
yokes noon meme dyke
Is not
MONKEY KNEED MOOSEY
That fourth hard one, testitemqlstudop, sounds like a Swedish chef
recipe.
How did you get those anagrams? By hand or by software, and, if the latter, what software? Ag?
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
testitemqlstudop Posts: 1194 Joined: July 21st, 2016, 11:45 am Location: in catagolue Contact:
I got them from an (Debian super-fast anagram generator.)
"Enable an exile pool" doesn't have an "s" so it can't be "island".
STONE CROP is necropost.
Further hints:
Code: Select all
yokes noon meme dyke - see
catastrophe weaned okayed PayPal - doctor
unconstitutional reciprocals tomato Lolita - cellular
misrepresentations frighteningly chili shit Bork - ninety
reforestation lintels ennoble boloney benefit Whitney Bennett - sir, CGOL
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
testitemqlstudop wrote:I got them from an (Debian super-fast anagram generator.)
"Enable an exile pool" doesn't have an "s" so it can't be "island".
But it does contain the name of an island!
Also, (Re: Stonecrop) can you not give the answers? I’d rather keep guessing
Re: meme dyke noon whatever
Monkey see monkey do?
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Gustone Posts: 462 Joined: March 6th, 2019, 2:26 am
reforestation lintels ennoble boloney benefit Whitney Bennett
Sir Robin You hinted the whole thing
I like making
color palettes for rules Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
Gustone wrote:reforestation lintels ennoble boloney benefit Whitney Bennett
Sir Robin
You hinted the whole thing
That’s only part of the answer
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Hunting
Posts: 1117 Joined: September 11th, 2017, 2:54 am Location: Ponyville, Equestria
Puzzles:
1. passe chips 2. Noke's flaws 3. Got ganla 4. Few or skir
Last edited by Hunting
on May 9th, 2019, 8:39 am, edited 1 time in total.
This post was brought to you by the Element of Magic.
Plz correct my grammar mistakes. I'm still studying English.
Working on:
Nothing.
Favorite gun ever:
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#C Favorite Gun. Found by me.
x = 4, y = 6, rule = B2e3i4at/S1c23cijn4a
o2bo$4o3$4o$o2bo!
Saka
Posts: 3138 Joined: June 19th, 2015, 8:50 pm Location: In the kingdom of Sultan Hamengkubuwono X
Hunting wrote:Puzzle: passe chips
spaceships?
Airy Clave White It Nay
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x = 17, y = 10, rule = B3/S23
b2ob2obo5b2o$11b4obo$2bob3o2bo2b3o$bo3b2o4b2o$o2bo2bob2o3b4o$bob2obo5b
o2b2o$2b2o4bobo2b3o$bo3b5ob2obobo$2bo5bob2o$4bob2o2bobobo!
(Check gen 2)
Sarp
Posts: 203 Joined: March 1st, 2015, 1:28 pm A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact:
3: Tagalong
4: Fireworks
x₁=ηx
V ⃰_η=c²√(Λη)
K=(Λu²)/2
Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt)
$$x_1=\eta x$$
$$V^*_\eta=c^2\sqrt{\Lambda\eta}$$
$$K=\frac{\Lambda u^2}2$$
$$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all
Aidan F. Pierce
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
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Neat Bozo Titan (hint: think chess)
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Moosey Posts: 2490 Joined: January 27th, 2019, 5:54 pm Location: A house, or perhaps the OCA board. Contact:
Code: Select all
REACH PUMA DOG
CAN MY LAC (Obviously the cyan clam)
CHILLS A SLEIGH
VARY EVERY TV-BOY REVERY
INCREASED HONK
BEE WHISTLE STUNT
DRUDGE IODINE FOXILY
DOT EQUINOX
Here’s one of extreme difficulty
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NORTHERN MUFFIN ESCALATED (Hint: The answer will be difficult to find since it isn’t entirely English)
Another:
Here’s a joke:
Also, NO ANAGRAMS
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?" |
The Joule-Thomson effect is also known as the Joule-Kelvin effect. This effect is present in non ideal gasses, where a change in temperature occurs upon expansion.
Introduction
The Joule-Thomson coefficient is given by
\[\mu_{\mathrm JT} = \left. \dfrac{\partial T}{\partial p} \right\vert_H\]
where
Tis the temperature, pis the pressure and His the enthalpy.
In terms of heat capacities one has
\[\mu_{\mathrm JT} C_V = -\left. \dfrac{\partial E}{\partial V} \right \vert_T \]
and
\[\mu_{\mathrm JT} C_p = -\left. \dfrac{\partial H}{\partial p} \right \vert_T \]
In terms of the second virial coefficient at zero pressure one has
\[\mu_{\mathrm JT}\vert_{p=0} = ^0\!\!\phi = B_2(T) -T \dfrac{dB_2(T)}{dT}\]
References Jacques-Olivier Goussard and Bernard Roulet "Free expansion for real gases", American Journal of Physics 61pp. 845-848 (1993) E. Albarran-Zavala, B. A. Espinoza-Elizarraraz, F. Angulo-Brown "Joule Inversion Temperatures for Some Simple Real Gases", The Open Thermodynamics Journal 3pp. 17-22 (2009) Thomas R. Rybolt "A virial treatment of the Joule and Joule-Thomson coefficients", Journal of Chemical Education 58pp. 620-624 (1981) |
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