text
stringlengths 256
16.4k
|
|---|
Recently I am reading the book
The Ricci Flow: techniques and applications: Part I: geometric aspects by Bennett Chow et al.. Here I have encountered a result (Proposition 1.13) due to Hamilton which states that "Any expanding or steady Ricci soliton $(g,X)$ on a closed manifold $M^n$ is Einstein. Any shrinking Ricci solution on a closed $n$-manifold has positive scalar curvature."In the proof, the general Ricci soliton equation is given by $$\Delta R + 2|\mathrm{Rc}|^2=\mathcal{L}_X(R)-\epsilon R,$$ which can be written as $$\Delta R + 2|\mathrm{Rc}|^2-\langle\nabla R,X\rangle+\epsilon R=0.$$ The proof says that the last equation can be rewritten as $$\Delta\left(R+\frac {n\epsilon} {2}\right)+2\,\left|\mathrm{Rc}+\frac {\epsilon} {2}g\right|^2-\left\langle\nabla\left(R+\frac {n\epsilon} {2}\right),X\right\rangle-\epsilon\left(R+\frac {n\epsilon} {2}\right)=0.$$ I can not understand the last equation. Please some one help me to understand this.
Recently I am reading the book
It is a direct simple computation We use the followings :
$$\Delta C=\nabla C=0$$ for constant $C$
$$|Rc +Cg|^2=|Rc|^2+ 2C(Rc,g)+C^2 |g|^2=|Rc|^2+2CR +nC^2$$
|
Probably an application of Glivenko-Cantelli's theorem as I suggested in the comments will work.
Assume that the $\xi_t$ are defined on a probability space $(\Omega,\mathcal F,\mu)$. We can assume that $G$ is the cumulative distribution function of a real valued random variable. Let $(\eta_t,t\in\Bbb Z)$ be a collection of independent random variables with cumulative distribution function $G$, defined on $(\Omega',\mathcal G,\nu)$. Using Glivenko-Cantelli's theorem for the product space and the family of i.i.d. random variables $(\xi_t-\eta_t,t\in\Bbb Z)$ gives $$\sup_{u\in\Bbb R}\left|\frac 1n\sum_{i=1}^n\chi_{\{\eta_i(\omega')-\xi_i(\omega)\leqslant u\}}-\mu\otimes\nu\{\eta_i(\omega')-\xi_i(\omega)\leqslant u\}\right|\to 0$$ for $\mu\otimes\nu$ almost every $(\omega,\omega')$. Then integrate with respect to $\nu$.
An alternative approach would use the result of the paper
Ramon van Handel,
The universal Glivenko-Cantelli property, Probab. Th. Rel. Fields 155, 911-934 (2013)
which is available here. We consider $\mathcal F:=\{t\mapsto G(x+t), x\in\Bbb R)\}$, which is a Glivenko-Cantelli class as $G$ is bounded. But it seems quite overkill for this problem, since their result is more general.
|
As far as I know, the first Diophantine problem (over a number field) that was solved using Spec and other tools of algebraic geometry was the following result (proved by Mazur and Tate in a paper from
Inventiones in the early 1970s):
If $E$ is an elliptic curve over $\mathbb Q$, then $E$ has no rational point of order 13.
The proof as it's written uses quite a bit more than you can learn just from reading Hartshorne; I don't know if there is any way to significantly simplify it. [Added: Rereading the first page of the Mazur--Tate paper, I see that theyrefer to another proof of this fact by Blass, which I've never read, but whichseems likely to be of a more classical nature.]
There is another result, which goes back to Billing and Mahler, of the same nature:
If $E$ is an elliptic curve over $\mathbb Q$, then $E$ has no rational point
of order $11$.
This was proved by elementary (if somewhat complicated) arguments. An analogousresult with $11$ replaced by $17$ wasproved by Ogg againusing elementary arguments.
These results were all generalized by Mazur (in the mid 1970s) as follows:
If $E$ is an elliptic curve over $\mathbb Q$, then $E$ has no rational point of any order other than $2,\ldots,10$, or $12$.
Mazur's paper doing this (the famous
Eisenstein ideal paper) was the one whichreally established the effectiveness of Grothendieck's algebro-geometric tools for solving classical number theory problems. For example, Wiles's work on Fermat's Last Theorem fits squarely in the tradition established by Mazur's paper.
As far as I know, no-one has found an elementary proof of Mazur's theorem; theelementary techniques of Billing--Mahler and Ogg don't seem to be extendable to the general case. So this is an interesting Diophantine problem which seems to require modern algebraic geometry to solve.
Often when a Diophantine problem is solved by algebro-geometric methods, it is not as simple as the way you suggest in your question.
For example, in the results described above, one does not work with one particular elliptic curve at a time. Rather, for each $N \geq 1$, there is a Diophantine equation, whose solutions over $\mathbb Q$ correspond to ellipticcurves over $\mathbb Q$ with a rational solution of order $N$.
This is the so-called modular curve $Y_1(N)$; although it was in some sense known to Jacobi, Kronecker, and the other 19th century developers of the theory of elliptic and automorphic functions, its precise interpretation as a Diophantine equation over $\mathbb Q$ is hard to make precise without moderntechniques of algebraic geometry. (As its name suggests, it is a certain
moduli space.)
An even more important contribution of modern theory is that this Diophantine equation even has a canonical model over $\mathbb Z$, which continues to havea moduli-space interpretation. (Concretely, this means that one starts withsome Diophantine equation --- or better, system of Diophantine equations --- over $\mathbb Q$, and then clears the denominators in a canonical fashion,to get a particular system of Diophantine equations with integral coefficientswhose solutions have a conceptual interpretation in terms of certain data relatedto elliptic curves.)
The curve $Y_1(N)$ is affine, not projective, and it is more natural to study projective curves. One can naturally complete it to a projective curve,called $X_1(N)$. It turns out that $X_1(N)$ can have rational solutions --- some of the extra points we added in going from $Y_1(N)$ to $X_1(N)$ might be rational --- and so we can rephrase Mazur's theorem as saying thatthe only rational points of $X_1(N)$ (for any $N \neq 2,\ldots,10,12$) lie inthe complement of $Y_1(N)$.
In fact, there are related curves $X_0(N)$, and what he proves is that $X_0(N)$ has only finitely many rational points for each $N$. He is then able to deduce the result about $Y_1(N)$ and $X_1(N)$ by further arguments.
The reason for giving the preceding somewhat technical details is that I wantto say something about how Mazur's proof works in the particular case $N = 11$(recovering the theorem of Billing and Mahler).
The curve $X_0(11)$ is an elliptic curve. One can write down itsexplicit equation easily enough; it is (the projectivization of)
$$y^2 +y = x^3 - x^2 - 10 x - 20.$$
(There is one point at infinity, which serves as the origin of the group law.)
Mazur wants to show it has only finitely many solutions. It's not clear how the explicit equation will help. (In the sense that if you begin with this equation, it's not clear how to directly show that it has only finitely many solutions over $\mathbb Q$.)
Instead, he first notes that it has a subgroup of rational points of order $5$:$$\{\text{ the point at infinity}, (5,5), (16,-61), (16,60), (5,-6) \}.$$
One knows from the general theory of elliptic curves that the full $5$-torsion subgroup of $X_0(11)$ is of order $25$, a product of two cyclic groups of order $5$.We have one of them above, while the other factor is not given byrational points.
In fact, the other $5$-torsion points have coordinates in the field $\mathbb Z[\zeta_5]$. (I don't know their explicit coordinates, unfortunately.)
Mazur doesn't need to know their exact values; instead, what is important forhim is that he is able to show (by conceptual, not computational, arguments)that the full $5$-torsion subgroup of $X_0(11)$, now thought of not just as a Diophantine over $\mathbb Q$ but as a scheme over Spec $\mathbb Z$,is a product of two group schemes of order $5$: namely$$\mathbb Z/ 5\mathbb Z \times \mu_5.$$The first factor is the subgroup of order $5$ determined by the points withinteger coordinates; the second factor is a subgroup of order $5$ generated bya $5$-torsion point with coefficients in Spec $\mathbb Z[\zeta_5]$.
What does it mean that this second factor is $\mu_5$?
Well, $X^5 - 1$ is a Diophantine equation, whose solutions are defined over$\mathbb Z[\zeta_5]$, and have a natural (multiplicative) group structure, and this is what $\mu_5$ is.What Mazur says is that an isomorphic copy of this "Diophantine group" (more precisely, this
group scheme) lives inside $X_0(11)$.
Note that the classical theory of Diophantine equations is not very well set upto deal with concepts like "isomorphisms of Diophantine equations whose solutions admits a natural group structure". (One already sees this if one tries to develop the theory of elliptic curves, including the group structure, in an elementary way.) So this is already a place where scheme theory provides new and important expressive power.
In any event, once Mazur has this formula for the $5$-torsion, he can make an infinite descent to prove that there are no other rational points besides the $5$ that we already wrote down. He doesn't phrase this infinite descent inthe naive way, with equations, as Fermat did with his descents (although itis the same underlying idea): rather, he argues as follows:
The curve $X_0(11)$ stays non-singular modulo every prime except $11$ (as you can check directly from the above equation). Modulo $11$ it becomes singular:you can check directly that reduced modulo $11$, the above equation becomes$$(y-5)^2 = (x-2)(x-5)^2,$$which has a singular point (a node) at $(5,5)$.
Note now that all our rational solutions $(5,5), (16,-61),$ etc. (other than the point at infinity) reduceto the node when you reduce them modulo $11$.
Using this (plus a little more argument) what you can show is that if$(x,y)$ is any rational point of $X_0(11)$, then after subtracting off (in the group law) a suitable choice of one of our $5$ known points, you obtaina point which
does not reduce to the node upon reduction modulo $11$.
So what we have to show is that if $(x,y)$ is any rational solution on $X_0(11)$which does not map to the node mod $11$, it is trivial (i.e. the point atinfinity).
Suppose it is not: then Mazur considers a point $(x',y')$ (no longer necessarily rational,just defined over some number field) which maps to $(x,y)$ under multiplicationby $5$ (in the group law). (This is the descent argument.)
Now this point is not uniquely determined, but it is determined up to addition (in the group law) of a $5$-torsion point. Because we know the precisestructure of the $5$-torsion (even over Spec $\mathbb Z$) we see that thispoint would have to have coordinates in some compositum of fields of the following type: (a) an everywhere unramified cyclic degree $5$ extension of $\mathbb Q$ (this relates to the $\mathbb Z/5\mathbb Z$ factor); and (b) an everywhereunramifed extension of $\mathbb Q$ obtained by extracting the $5$th root ofsome number (this relates to the $\mu_5$ factor). Now no such extension of $\mathbb Q$ exist (e.g. because$\mathbb Q$ admits
no non-trivial everywhere unramified extension), and hence$(x',y')$ again has to be defined over $\mathbb Q$. Now we repeat the aboveprocedure ad infitum, to get a contradiction (via infinite descent).
I hope that the above sketch gives some idea of how more sophisticated methods can help with the solution of Diophantine equations. It is not just that one writes down Spec and magically gets new information. Rather, the introduction of a more conceptual way of thinking gives whole new ways of transferring information around and making computations which are not accessible when working in a naive manner.
A good high-level comparison would be the theory of solutions of algebraic equations before and after Galois's contributions.
A more specific analogy would be the difference between studying surfaces in space (say) with the tools of an undergraduate multi-variable calculus class,compared to the tools of manifold theory. In undergraduate calculus, one has toat all times remember the equation for the surface, work with explicit coordinates, make explicit coordinate changes to reduce computations from the curved surface to the plane, and so on. In manifold theory, one has a conceptual apparatus which lets one speak of the surface as an object independent of the equation cutting it out; one can say "consider a chart in the neighbourhood of the point $p$" without having to explicitly writedown the functions giving rise to the chart. (The implicit function theoremsupplies them, and that is often enough; you don't have to concretely determine the outputof that theorem every time you want to apply it.)
So it goes with the scheme-theoretic point of view. One can use the modularinterpretation to write down points of $X_0(11)$ without having to give theircoordinates. In fact, one can show that it has a node when reduced modulo $11$without ever having to write down an equation. The determination of the $5$-torsion group is again made by conceptual arguments, without having to write down the actual solutions in coordinates. And as the above sketch of the infinite descent (hopefully) makes clear, it is any case the abstract natureof the $5$-torsion points (the fact that they are isomorphic to$\mathbb Z/5\mathbb Z \times \mu_5$) which is important for the descent, not any information about their explicit coordinates.
I hope this answer, as long and technical as it is, gives some hint as to the utility of the scheme-theoretic viewpoint.
References: A nice introduction to $X_0(11)$ is given in this expository article of Tom Weston.
As for Mazur's theorem, I don't know of any expositions which are not at a much higher level of sophistication. (There are simpler proofs of his main technical results now, e.g. here,but these are simpler only in a relative sense; they are still not accessibleto non-experts in this style of number theory.)
|
Search
Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
|
Search
Now showing items 1-9 of 9
Measurement of $J/\psi$ production as a function of event multiplicity in pp collisions at $\sqrt{s} = 13\,\mathrm{TeV}$ with ALICE
(Elsevier, 2017-11)
The availability at the LHC of the largest collision energy in pp collisions allows a significant advance in the measurement of $J/\psi$ production as function of event multiplicity. The interesting relative increase ...
Multiplicity dependence of jet-like two-particle correlations in pp collisions at $\sqrt s$ =7 and 13 TeV with ALICE
(Elsevier, 2017-11)
Two-particle correlations in relative azimuthal angle (Δ ϕ ) and pseudorapidity (Δ η ) have been used to study heavy-ion collision dynamics, including medium-induced jet modification. Further investigations also showed the ...
The new Inner Tracking System of the ALICE experiment
(Elsevier, 2017-11)
The ALICE experiment will undergo a major upgrade during the next LHC Long Shutdown scheduled in 2019–20 that will enable a detailed study of the properties of the QGP, exploiting the increased Pb-Pb luminosity ...
Azimuthally differential pion femtoscopy relative to the second and thrid harmonic in Pb-Pb 2.76 TeV collisions from ALICE
(Elsevier, 2017-11)
Azimuthally differential femtoscopic measurements, being sensitive to spatio-temporal characteristics of the source as well as to the collective velocity fields at freeze-out, provide very important information on the ...
Charmonium production in Pb–Pb and p–Pb collisions at forward rapidity measured with ALICE
(Elsevier, 2017-11)
The ALICE collaboration has measured the inclusive charmonium production at forward rapidity in Pb–Pb and p–Pb collisions at sNN=5.02TeV and sNN=8.16TeV , respectively. In Pb–Pb collisions, the J/ ψ and ψ (2S) nuclear ...
Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions
(Elsevier, 2017-11)
Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
Jet-hadron correlations relative to the event plane at the LHC with ALICE
(Elsevier, 2017-11)
In ultra relativistic heavy-ion collisions at the Large Hadron Collider (LHC), conditions are met to produce a hot, dense and strongly interacting medium known as the Quark Gluon Plasma (QGP). Quarks and gluons from incoming ...
Measurements of the nuclear modification factor and elliptic flow of leptons from heavy-flavour hadron decays in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 and 5.02 TeV with ALICE
(Elsevier, 2017-11)
We present the ALICE results on the nuclear modification factor and elliptic flow of electrons and muons from open heavy-flavour hadron decays at mid-rapidity and forward rapidity in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ ...
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
|
Alan Jeffrey tweeted the following in reply to the previous post:
@jeremysiek wouldn't it be easier to change the defn of application to be ⟦MN⟧σ = { W | T ∈ ⟦M⟧σ, V ∈ ⟦N⟧σ, (V′,W) ∈ T, V′ ⊆ V }?
The idea is that, for higher order functions, if the function \(M\)is expecting to ask all the questions in the table \(V'\),then it is OK to apply \(M\) to a table \(V\) that answers
more questionsthan \(V'\). This idea is quite natural, it is like Liskov's subsumption principlebut for functions instead of objects. If this change can help us with theself application problem, then it will be preferable to the graph-o-tables approach described in the previous post because it retains the simpleinductive definition of values. So let's see where this takes us!
We have the original definition of values\[ \begin{array}{lrcl} \text{values} & v & ::= & n \mid \{ (v_1,v'_1),\ldots,(v_n,v'_n) \} \end{array} \]
and here is the denotational semantics, updated with Alan's suggestion to include the clause \(v'_2 \sqsubseteq v_2\) in the case for application.\begin{align*} E[\!| n |\!](\rho) &= \{ n \} \\ E[\!| x |\!](\rho) &= \{ \rho(x) \} \\ E[\!| \lambda x.\; e |\!](\rho) &= \{ T \mid \forall v v'. (v,v') \in T \Rightarrow v' \in E[\!|e|\!](\rho(x:=v)) \} \\ E[\!| e_1\;e_2 |\!](\rho) &= \left\{ v \middle| \begin{array}{l} \exists T v_2 v'_2. T {\in} E[\!| e_1 |\!](\rho) \land v_2 {\in} E[\!| e_2 |\!](\rho) \\ \land v'_2 \sqsubseteq v_2 \land (v'_2,v) {\in} T \end{array} \right\} \end{align*}
The ordering on values \(\sqsubseteq\) used above is just equality on numbers and subset on function tables.
The first thing to check is whether this semantics can handle self application at all, such as \[ (\lambda f. f \; f) \; (\lambda g. \; 42) \]
Example 1. \( 42 \in E[\!| (\lambda f. f \; f) \; (\lambda g. \; 42) |\!](\emptyset) \) The main work is figuring out witnesses for the function tables. We're going to need the following tables: \begin{align*} T_0 &= \emptyset \\ T_1 &= \{ (\emptyset, 42)\} \\ T_2 &= \{ (T_1, 42) \} \end{align*} Here's the proof, working top-down, or goal driven. The important use of subsumption is the \( \emptyset \sqsubseteq T_1 \) below. \( T_2 \in E[\!| (\lambda f. f \; f)|\!](\emptyset)\) So we need to show: \( 42 \in E[\!| f \; f|\!](f:=T_1) \) \( T_1 \in E[\!| f |\!](f:=T_1) \) \( T_1 \in E[\!| f |\!](f:=T_1) \) \( \emptyset \sqsubseteq T_1 \) \( (\emptyset, 42) \in T_1 \) \( T_1 \in E[\!| (\lambda g. \; 42) |\!](\emptyset)\) So we need to show \( 42 \in E[\!| 42 |\!](g:=\emptyset)\), which is immediate. \( T_1 \sqsubseteq T_1 \) \( (T_1,42) \in T_2 \)
Good, so this semantics can handle a simple use of self application. How about factorial? Instead of considering factorial of 3, as in the previous post, we'll go further this time and consider factorial of an arbitrary number \(n\).
Example 2. We shall compute the factorial of \(n\)using the strict version of the Y combinator, that is, the Zcombinator. \begin{align*} M & \equiv \lambda x. f \; (\lambda v. (x\; x) \; v) \\ Z & \equiv \lambda f. M \; M \\ F & \equiv \lambda n. \mathbf{if}\,n=0\,\mathbf{then}\, 1\,\mathbf{else}\, n \times r \,(n-1)\\ H & \equiv \lambda r. F \\ \mathit{fact} & \equiv Z\, H\end{align*}We shall show that \[ n! \in E[\!|\mathit{fact}\;n|\!](\emptyset)\]For this example we need very many tables, but fortunately there are justa few patterns. To capture these patterns, be define the followingtable-producing functions.\begin{align*} T_F(n) &= \{ (n,n!) \} \\ T_H(n) &= \{ (\emptyset,T_F(0)), (T_F(0), T_F(1)), \ldots ,(T_F(n-1), T_F(n)) \} \\ T_M(n) &= \begin{cases} \emptyset & \text{if } n = 0 \\ \{ (T_M(n'), T_F(n')) \} \cup T_M(n') & \text{if } n = 1+n' \end{cases} \\T_Z(n) &= \{ (T_H(n), T_F(n) )\}\end{align*} \(T_F(n)\) is a fragment of the factorial function, for the one input \(n\).\(T_H(n)\) maps each \(T_F(i)\) to \(T_F(i+1) \) for up to \(i+1 = n\).\(T_M(n)\) is the heart of the matter, and what makes the self application work.It maps successively larger versions of itself to fragments of the factorialfunction, that is\[ T_M(n) = \left\{ \begin{array}{l} T_M(0) \mapsto T_F(0) \\ T_M(1) \mapsto T_F(1) \\ \vdots & \\ T_M(n-1) \mapsto T_F(n-1) \end{array} \right\}\]For example, here is \(T_M(4)\): \( T_M(n) \sqsubseteq T_M(1+n) \) \( (T_M(n), T_F(n)) \in T_M(1+n) \) Lemma 1. If \(n \le k\), then \(T_M(1+n) \in E[\!| M |\!](f:=T_H(k)) \). Lemma 2. \( T_Z(n) \in E[\!| Z |\!](\emptyset) \)
If you're curious about the details for the complete proof of \( n! \in E[\!|\mathit{fact}\;n|\!](\emptyset) \) you can take a look at the proof in Isabelle that I've written here.
This is all quite promising! Next we look at the proof of soundness with respect to the big step semantics.
Soundness with Respect to the Big-Step Semantics
The proof of soundness is quite similar to that of the first version, as the relation \(\approx\) between the denotational and big-step values remains the same. However, the following two technical lemmas are needed to handle subsumption.
Lemma (Related Table Subsumption)If \(T' \subseteq T\) and \(T \approx \langle \lambda x.e, \rho \rangle\),then \(T' \approx \langle \lambda x.e, \rho \rangle\). The proof is by induction on \(T'\). Lemma (Related Value Subsumption)If \(v_1 \approx v'\) and \(v_2 \sqsubseteq v'\), then \(v_2 \approx v'\). The proof is by case analysis, using the previous lemma when the values are function tables. Theorem (Soundness). If \(v \in E[\!| e |\!](\rho) \) and \( \rho \approx \rho' \), then \( \rho' \vdash e \Downarrow v' \) and \(v \approx v'\) for some \(v'\).
The mechanization of soundness in Isabelle is here.
|
I need help finding the formula for this summation notation: $$\sum_{k=1}^n{k^{2k} }$$ or $$1^2 + 2^4 +3^6 +.....+n^{2n} $$
And preferably not involving calculus.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
This probably doesn't have a nice closed form, since the simpler sum of $k^k$ (called the "hypertriangular function of $n$") doesn't.
If $$a_n = \sum\limits_{k=1}^{n} k^k$$
Then
$$\lim_{n\to\infty}\left(\frac{1}{n}\cdot \frac{a_{n+1}}{a_n}\right)=e$$
At best, you can hope for a similar asymptotic result for your sum.
|
Let $p$ be a prime number.
1) What is the degree over $\Bbb Q_p$ of 'the' algebraic closure $\overline{\Bbb Q_p}$ of $\Bbb Q_p$ ? 2) What is the order (the cardinality) of the absolute Galois group $G$ of $\Bbb Q_p$ ? 3) What is the degree over $\overline{\Bbb Q_p}$ of the completion $\Bbb C_p$ of $\overline{\Bbb Q_p}$ ?
$ $
Here are some comments and thoughts :
1)
I know that $[\overline{\Bbb Q_p} : \Bbb Q_p] \geq \aleph_0$, since $X^n-p$ is irreducible in the UFD $\Bbb Z_p[X]$ for any $n \geq 1$. According to an answer here, we have $[\overline{\Bbb Q_p} : \Bbb Q_p] = \aleph_0$, but there is no proof of this claim.
Since $\Bbb Q_p$ has a unique unramified extension of degree $n$ for any $n \geq 1$, we have $[\Bbb Q_p^{\rm unr} : \Bbb Q_p] = \aleph_0$.
In general, we have $|k| =|k^{\rm sep}| = |\bar k|$, when $k$ is infinite (see here, exercise 4.5). But it doesn't help for knowing $[\bar k : k]$ (nor $[k^{\rm sep} : k]$, think of the separably closed but not algebraically closed field $\Bbb F_p(T)^{\rm sep}$). As mentioned here, we can have fields $k$ with $[\bar k : k] > \aleph_0$. In general, I think that we have $[\bar k : k] \leq |k|$ when $k$ is infinite. For any infinite field $k$ and any set $I$, we have $|k^{(I)}| = |k| \cdot |I| = \max(|k|, |I|)$ (where $\dim_k(k^{(I)}) = |I|$, while $\dim_k(k^{I}) = |k^I| = |k|^{|I|}$).
2)
Since $G$ surjects onto the absolute Galois group $\widehat{\Bbb Z}$ of $\Bbb F_p$, it is uncountable — I don't know if one can use a similar proof as for the absolute Galois group of $\Bbb Q$, using permutation of the square roots of primes, using this (maybe the fact that $\Bbb Q_p^{\times, 2}$ has finite index in $\Bbb Q_p^{\times}$ [$8$ if $p=2$ and $4$ if $p>2$] could be a problem?).
Moreover, $G$ is at most of cardinality $|\mathrm{Bij}(\overline{\Bbb Q_p})| = 2^{2^{\aleph_0}}$. So how to choose (assuming GCH...) between $2^{\aleph_0}$ and $2^{2^{\aleph_0}}$ for the cardinality of $G$ ?
3)
My question here asked about the degree of $\Bbb C_p$ over a completion of some subfield of $\overline{\Bbb Q_p}$, which is different from here. I know that $[\Bbb C_p : \overline{\Bbb Q_p}]>1$, but it is not obvious to me that this degree is $\geq \aleph_0$. (By the way, we have $|\Bbb C_p| = 2^{\aleph_0}$ by exercise 2.30.5 here).
According to
Arithmetic of Algebraic Curves (Stepanov), p. 239(d), the transcendence degree of $\Bbb C_p$ over $\overline{\Bbb Q_p}$ is already $2^{\aleph_0}$, without proof. I'm not sure how to prove this, and actually I should think to the relation between $\mathrm{tr.deg}_k(L)$ and $[L:k]$ : if $B$ is a transcendence basis of $L$ over an infinite field $k$, then we have$$|k| \cdot |B| = [k(B) : k] \leq [L:k] \leq [\overline{k(B)} : k] \leq |k| \cdot |B| \cdot |k(B)| = (|k| \cdot |B| ) \cdot (|k| \cdot |B|)$$In particular, if $|B| = \mathrm{deg.tr}_k(L) = 2^{\aleph_0} = |k|$, then we get $[L:k] = 2^{\aleph_0}$ (see also 11.4 here...).
Thank you for your various comments!
|
Assume that the uniformly distributed bound surface and bound volume current, $\vec{K}_{b}$,$\vec{J}_{b}$ respectively, are in the azimuthal direction $\phi$ within an infinitely long solenoid with a current passing through the wounded coils centered on the z-axis.
Clearly, the magnetic field field has no z and $\phi$ dependence due to the symmetry of current about the z-axis and the infinitely long solenoid. Hence,
$\vec{B}=\left \langle B_{s}\left ( s,\phi,\theta \right ),B_{\phi}\left ( s,\phi,\theta \right ),B_{z}\left ( s,\phi,\theta \right ) \right \rangle\rightarrow \vec{B}=\left ( B_{s}\left ( s \right ),B_{\phi}\left ( s \right ) ,B_{\theta}\left ( s \right )\right )$
Evidently, $B_{\phi}$ cannot exists for if it did it would violate the fact that a magnetic field cannot in a parallel to the current that generates it.
I know that using the Right hand rule in context of a current carrying solenoid, if the current runs in a direction 'towards' me, the thumb points in the positive z-direction. If the current runs in a direction 'away' from me, the thumb points in the negative z-direction. That thumb tells me the direction of the magnetic field. In both cases, it points outward/ away from the solenoid.
Here is where a bit of explanation would help:
How does this ties in with the fact that it violates the Biot-Savard law?
|
Cryptology ePrint Archive: Report 2002/031 A Parallelizable Design Principle for Cryptographic Hash Functions Palash Sarkar and Paul J. Schellenberg Abstract: We describe a parallel design principle for hash functions. Given a secure hash
function $h:\{0,1\}^n\rightarrow \{0,1\}^m$ with $n\geq 2m$, and a binary tree of
$2^t$ processors we show how to construct a secure hash function $h^{*}$ which can hash
messages of lengths less than $2^{n-m}$ and a secure hash function $h^{\infty}$ which can
hash messages of arbitrary length. The number of parallel rounds required to hash a
message of length $L$ is $\lfloor\frac{L}{2^t}\rfloor+t+2$. Further, our algorithm is incrementally
parallelizable in the following sense : given a digest produced using a binary tree of $2^t$
processors, we show that the same digest can also be produced using a binary tree of
$2^{t^{\prime}}$ $(0\leq t^{\prime}\leq t)$ processors.
Category / Keywords: foundations / hash function, parallel algorithm Publication Info: earlier version appeared in the proceedings of Indocrypt 2001. Date: received 12 Mar 2002, last revised 2 Aug 2002 Contact author: palash at isical ac in Available format(s): Postscript (PS) | Compressed Postscript (PS.GZ) | BibTeX Citation Note: Revision prepared on the basis of comments from an anonymous referee. Version: 20020802:112007 (All versions of this report) Short URL: ia.cr/2002/031 [ Cryptology ePrint archive ]
|
There is a "high temperature" limit, it is actually the same point. 5K is hotter than 1K, and 1000K is hotter than both. The temperature $\pm \infty$ is actually the same temperature, and for negative temperatures you get hotter the closer you get to zero (from below).
Most systems cannot reach negative temperatures, but even if you can you can't get all the way to zero from either direction.
If this seems weird, note that $$\frac{1}{kT}=\frac{d S}{d E},$$
So if giving a little bit of energy $E$ gives a large increase in entropy $S$, then $1/kT$ is large (and positive), so $T$ is small (and positive). If a little bit of energy gives no change in entropy (not possible for most systems) then $1/kT$is zero, so $T=\pm\infty$. If a little bit of energy gives a large decrease in entropy $S$, then $1/kT$ is large (and negative), so $T$ is small (and negative).
These changes in entropy can never be infinite because entropy is a real and finite thing and we don't make vanishingly small changes in energy, so there is no infinite rate $\frac{d S}{d E}$. And that's why temperature is never zero.
edit to discuss when negative temperatures can occur
All you need for negative energy is that the range of small-and-practically-accessible energy exchanges can only cause the entropy to get smaller when it accepts energy. This makes it statistically favorable to give energy to positive temperature things (then both entropies can go up), so it is hotter than all of them (so negative temperature is hotter than positive temperature). This might not be because the negative temperature system has components with a strict maximum energy, just that those higher entropy states are not accessible. For instance if your collection of two state systems don't have enough kinetic energy to combine to create electron-positron pairs it doesn't matter that with a large amount of additional energy they could increase their entropy by doing so, as long as that isn't going to happen. Statistical physics is about what will happen on time scales long enough for quasi-equilibrium to occur.
|
February 3rd, 2019, 04:04 AM
# 1
Newbie
Joined: Feb 2019
From: Uppsala
Posts: 1
Thanks: 0
Solve x^120 + x^3 + 2x^2 + x +3 ≡ 0 (mod 7)
Hi,
I'm having trouble with this one:
x^120 + x^3 + 2x^2 + x +3 ≡ 0 (mod 7)
What is x?
February 3rd, 2019, 05:19 AM
# 2
Senior Member
Joined: Aug 2017
From: United Kingdom
Posts: 313
Thanks: 112
Math Focus: Number Theory, Algebraic Geometry
You can just try each number from 0 to 6 to see which are solutions. The only obstacle in doing this is computing the 120-th powers mod 7. I'll explain a common trick for computing large powers modulo a relatively small number, and use it to compute $3^{120} \bmod 7$.
Say you have natural numbers $a, n$ and $m$ and you want to compute $a^n \bmod m$. Write $n$ in its binary form: $n = n_0 + n_1 2 + n_2 2^2 + \dots + n_k 2^k$, with each $n_i$ here being either $0$ or $1$. We now compute $a^{2^i} \bmod m$ for $i = 0, 1 \dots, k$. But this is particularly easy if $m$ is small: once we know $a^{2^i} \bmod m$ for some $i$, we just need to square this value and reduce mod $m$ to get $a^{2^{i+1}} \bmod 7$. (Do you see why this is the case?) Then we're pretty much done in light of the equation $a^n = a^{n_0} a^{n_1 2} a^{n_2 2^2} \dots a^{n_k 2^k}$.
Let's see this in practice with $a = 3, n = 120, m = 7$. We have $120 = 2^6 + 2^5 + 2^4 + 2^3$. Now:
$\begin{align*}
3 &\equiv 3 \bmod 7 \\
3^2 &= 9 \equiv 2 \bmod 7 \\
3^{2^2} &\equiv 2^2 = 4 \bmod 7 \\
3^{2^3} &\equiv 4^2 = 16 \equiv 2 \bmod 7 \\
3^{2^4} &\equiv 2^2 = 4 \bmod 7 \\
3^{2^5} &\equiv 4^2 \equiv 2 \bmod 7 \\
3^{2^6} &\equiv 2^2 = 4 \bmod 7
\end{align*}$
(Aside: as soon as we hit a value we've seen before in the list, we'll enter a cycling pattern. More precisely, let $x_i$ denote $a^{2^i} \bmod m$. If $x_{r + s} = x_s$ for some naturals $r$ and $s$, we will have $x_{dr + s + i} = x_{s+i}$ for all natural numbers $d$ and $i$. And we're guaranteed to have such $r, s$ with $s \leq m+1$ by the pigeonhole principle.)
Now we just compute
$\begin{align*}
3^{120} &= 3^{2^3} 3^{2^4} 3^{2^5} 3^{2^6} \\
&\equiv 2 \times 4 \times 2 \times 4 \\
&= 8 \times 8 \equiv 1 \times 1 = 1 \bmod 7
\end{align*}$
Last edited by cjem; February 3rd, 2019 at 05:25 AM.
Tags 2x2, ≡, mod, solve, x120
Search tags for this page
Click on a term to search for related topics.
Thread Tools Display Modes
Similar Threads Thread Thread Starter Forum Replies Last Post Please Solve this 4 EPIC Problems.... i couldn't solve it myself gen_shao Algebra 12 November 2nd, 2014 06:11 AM
|
Above-Average Smoothing of Impulsive Noise
In this blog I show a neat noise reduction scheme that has the high-frequency noise reduction behavior of a traditional moving average process but with much better impulsive-noise suppression.
In practice we may be required to make precise measurements in the presence of highly-impulsive noise. Without some sort of analog signal conditioning, or digital signal processing, it can be difficult to obtain stable and repeatable, measurements. This impulsive-noise smoothing trick, originally developed to detect microampere changes in milliampere signals, describes a smoothing algorithm that improves the repeatability of signal amplitude measurements in the presence of impulsive noise[1].
Traditional Moving Average
Practical noise reduction methods often involve an
N-point moving-average process performed on a sequence of measured x( n)
values, to compute a sequence of
N-sample
arithmetic means,
M( q). As such, the moving-average sequence
M( q)
is defined by:
$$ M(q) = \frac{1}{N} \sum_{k=q}^{q+(N-1)} x(k) \tag{1} $$
where the time index of this
averaging process is
q = 0,1,2,3,
... .
The Smoothing Algorithm
The proposed impulsive-noise smoothing
algorithm is implemented as follows: collect
N samples of an x( n)
input sequence and compute the arithmetic mean,
M( q), of the N samples. Each sample in the block is
then compared to the mean. The direction of each
x( n) sample
relative to the mean (greater than, or less than) is accumulated, as well as
the cumulative magnitude of the deviation of the samples in one direction
(which, by definition of the mean, equals that of the other direction). This
data is used to compute a correction term that is added to the
M( q)
mean according to the following formula:
$$ A(q) = M(q) + \frac{(P_{os}-N_{eg}) \cdot \lvert D_{total} \rvert }{N^2} \tag{2}$$
Variable
A( q) is the corrected mean, M( q) is the arithmetic mean (average) from Eq. (1), P os is the number of samples greater than M( q), and N eg is the number of samples less than M( q). Variable D total is the sum of deviations from the mean (absolute values and one direction only). Variable | D total,| then, is the magnitude of the sum of the differences between the P os samples and M( q). A more detailed description of | D total| is given in Appendix A.
For a simple example, consider a system acquiring 10 measured samples of 10, 10, 11, 9, 10, 10, 13, 10, 10, and 10. The mean is
M = 10.3, the total number of samples positive is P os = 2, and the total number of samples negative is N eg = 8 (so P os‑ N eg = ‑6). The total deviation in either direction from the mean is 3.4 [using the eight samples less than the mean, (10.3‑10) times 7 plus (10.3‑9); or using the two samples greater than the mean, (13‑10.3) plus (11‑10.3)]. With D total = 3.4, Eq. (2) yields the "smoothed" result of A = 10.096.
The Eq. (2) smoothing algorithm's performance, relative to traditional moving-averaging, can be illustrated by a more realistic example. Figure 1(a) shows a measured 500-sample
x( n) input signal sequence comprising a step signal of amplitude one contaminated with random noise and three large impulsive-noise spike samples.
Figure 1: Impulsive
noise smoothing for
N = 16: (a) input
x(n)
signal; (b) moving average output; (c) smoothing
(b) moving average output; (c) smoothing
algorithm output.
Figure 1(b) shows the Eq. (1) moving average output while Figure 1(c) shows the Eq. (2) smoothing algorithm's output.
We compare the outputs of the moving average and
smoothing algorithms in a more concise way in Figure 2. There we see that the input
x( n)'s
high-frequency noise has been reduced in amplitude by the Eq. (2) smoothing
algorithm (in a way very similar to moving averaging) but the large unwanted
impulses in Figure 1(a) have been significantly reduced at the output of the
Eq. (2) smoothing algorithm.
Figure 2: Comparison of the moving average and smoothing algorithm outputs.
If
computational speed is important to you, notice in Eq. (2) that if
N is an integer power of two then the
compute-intensive division operation can be accomplished by binary arithmetic
right shifts to reduce the computational workload.
References
[1] Dvorak, R. “Software Filter Boosts Signal-Measurement Stability, Precision,”
Electronic Design Magazine, February 3, 2003.
Appendix A: Definition of | D total | total
The definition of the variable
|D , for a given sequence, can be illustrated with a simple example. have a look at the simple 8-sample total| x( n) sequence in Figure A-1(a) whose mean (average) is M= 5.19, where
$$ M = \frac{1}{8} \sum_{n=0}^7 x(n) \tag{A-1}$$
For this
x( n)
input, five samples are greater than
M
and three samples are less than M.
For the sequence
Figure A1: Values used to compute | D total | for the example x( n) sequence. x( n) the variable |D is the absolute value of the sum of positive deviations from total| M, | D +(1)+ D +(2)+ D +(3)+ D +(4)+ D +(5)|, as shown in Figure A1(b). And the nature of the arithmetic mean given in Eq. (A-1) is such that variable |D is also equal to the absolute value of the sum of negative deviations from total| M, | D -(1)+ D -(2)+ D -(3)|. So we can express variable |D as: total|
$$ \lvert D_{total} \rvert = \lvert D_+(1) + D_+(2) + D_+(3) + D_+(4) + D_+(5) \rvert $$ $$ = \lvert D_-(1) + D_-(2) + D_-(3) \rvert \tag{A-2}$$
Previous post by Rick Lyons:
Looking For a Second Toolbox? This One's For Sale
Next post by Rick Lyons:
Errata for the book: 'Understanding Digital Signal Processing'
Rick
Any idea how this smoother might compare with a N-point median filter? Thanks, for all the books and ideas that you have shared with the community over the years!
Wayne
With my simple test signals, constant amplitude signals contaminated with wideband random noise and occassional unwanted narrow impulsive amplitude spikes, I found that the proposed scheme and median filtering (median smoothing) produced VERY similar results. The two smoothing schemes reduced the unwanted impulsive noise to a very similar degree and the variances of the two schemes' outputs were VERY similar. So I'll "stick my neck out" and say, "It appears to me that there's no significant performance difference between the proposed smoothing scheme and your every-day garden-variety median filtering."
Wayne, your question illustrates my laziness. What I didn't do is compare the computational workload of the proposed smoothing scheme to the computational workload of median filtering. :-(
Rick
There are many whom I might call lazy but you are not one of them! I have learned a great deal from your books over the years. Writing clear English prose is often harder than writing code.
My very dog eared copy of "Understanding Digital Signal Processing (third edition) bears witness to your efforts.
Thanks for your insights concerning this filtering scheme. The filter is not hard to implement, I might do my own comparison between a median filter and your smoother.
Wayne
Hi Wayne.
Thanks again. I can send you the errata for the 3rd edition of my "Understanding DSP" book. If you're interested, send my a private e-mail at:
R.Lyons@ieee.org
Wayne, if you model the two smoothing schemes I'd be interested to learn about your modeling results.
On behalf of Wayne (and by the way it is first time I see his posts) I did the following matlab modelling and you can see the median is far more powerful on (single isolated) spikes. I hope I done it right
*******************************************
x = randn(1,1024);
x(36) = 17; x(178) = -20; x(455) = -33;
p = 0; n = 0; s = 0;
for i = 1+10:1024
m(i) = mean(x(i-10:i));
if x(i) > m(i),
p = p+1;
s = s + x(i);
elseif x(i) < m(i),
n = n + 1;
end
m(i) = m(i) + (p - n)*s/i^2;
test(i) = median(x(i-10:i));
end
plot(m);hold;
plot(test,'r--');
******************************
s = s + x(i);
should be: s = s + x(i) -m(i);
Also, I think your command:
m(i) = m(i) + (p - n)*s/i^2;
should be: m(i) = m(i) + (p - n)*s/11^2;
where the denominator of the second term is eleven squared. (That denominator must be a constant.)
However, when I make those changes to your MATLAB code my "proposed smoothing" computational results are obviously incorrect. So Kaz, ...I haven't yet figured out what's going on here, but I'm still working on it.
Hi Rick,
Your first correction of:
s = s + x(i) -m(i);
is ok but second one is wrong as it should be I^2 (i.e. N^2) since filter is in running mode.
I can see why median filter is more potent for spikes as it totally ignores peaks if peaks with other samples fall within the taps segment while your formula gives weight to every sample.
Hi Kaz. I'm not sure what your last reply's variable 'I' (uppercase 'i') is. In your code:
when i = 11, you process samples x(1) -to- x(11),
when i = 12, you process samples x(2) -to- x(12),
when i = 13, you process samples x(3) -to- x(13), and so on.
So your value for 'N' in my blog's Eq. (2) is: N = 11. And that's why I said I thought your command:
m(i) = m(i) + (p - n)*s/i^2; should be: m(i) = m(i) + (p - n)*s/11^2; where the denominator of the second term is eleven squared.
I think for each new value of your loop counter 'i' you need to reset variables 'p', 'n', and 's' to zero. I say that because, for example when N = 11 the maximum value for 'p' can never be greater than 10. But your maximum value for 'p' often increases in value to be greater than 500!
Hi Rick (and Kaz),
You are correct.
I tried the code from Kaz, but I included a small function to reset 'p', 'n', and 's' to zero. It worked ok. In fact, seems like it improves the median filter. I've also changed "I" to "N". Here is the code:
**********************************
x = 5+randn(1,1024);
x(36) = 17; x(178) = -20; x(455) = -33; L=length(x); N=10; p = 0; n = 0; s = 0; s2=0; for i = N+1:L m(i) = mean(x(i-N:i)); [s,p,n]=Dtotal(m(i),x(i-N:i)); m(i) = m(i) + (p - n)*s/N^2; test(i) = median(x(i-N:i)); end subplot(2,1,1) plot(m,'b-');hold; plot(test,'r--'); title("Smoothers") legend('AboveAverage','meanfilter') subplot(2,1,2) plot(x,'k') title("original")
******************************
The function is:
+++++++++++++++++++++++++++++++
function [s,p,n]=Dtotal(m,x)
p = 0; n = 0; s = 0; for i=1:length(x) if x(i) > m, s = s + x(i)-m; p = p+1; elseif x(i) < m, n = n + 1; end end endfunction
+++++++++++++++++++++++++++++
Sorry if this is too simple, but I had to try it :)
Thanks gmiram. Model now looks ok to me. I added more spikes as below and it does seem that Rick's algorithm is better than median but not so if dc = 0.
%%%%%%%%%%%%%%%%%%
x = 5*0+randn(1,2^14);
x(36:100:end) = 17; x(178:200:end) = -20; x(455:50:end) = -33;
%%%%%%%%%%%%%%%%%
and the rest of code is same as yours above
Hi Kaz. That's a clever way to add multiple impulses. Neat!!
Hi gmiram. Your code looks correct to me (except I think the legend title: "meanfilter" should be "median filter"). Thanks for posting your code.
Sure!
that should say "median filter".
Thanks.
As I read through the description of the filter, I came to the conclusion that it must actually be a "conceptually easy" way to
estimate the median. It would be interesting to develop the math to compare them and compute the error, but I'll leave that work to someone else :-).
I think this approach offers some smoothing whereas the median tends to stick to the quantum levels of the original inputs (or a half-way point between them). It would also be interesting to compare computational load -- and for that, I'm guessing the most efficient way to keep track of median across a sliding window would be by managing two heaps: a min-heap for the values above median and a max-heap for the values below median. I've implemented such a thing before -- its a nice exercise, and the resulting data structure is probably about as efficient as you can get for the job. My first guess is that computing the median that way (with the heaps) would require fewer comparisons at each advance of the window. But I still like the idea of what this median-estimate can do.
Of course, another filter type to compare is the LULU, and I've found that combining different sizes can give good results (like averaging the results of a 5x3 and and a 3x5 for a small-ish window size) -- I expect LULU is similar to this approach in terms of computational load.
Hello zimbot.
What does "LULU" mean?
[-Rick-]
Think of it as "lower" and "upper". You take a neighborhood, break it into so many chunks of a given size, take the max and min of each chunk, then get the max of the mins and the min of the maxes. You can average those two values or something even more clever with them. There's a nice overview on wikipedia of Lulu smoothing.
Hi Rick,
This algorithm is amazing. This helped in comparing the running average filter for my application.
Could this algorithm be extended for specific cutoff frequency by adjusting the window size or any other parameters. This could be interesting if you have any further comments regarding this.
Pranav
Hi Yottabyte. Regarding the proposed smoothing technique, the only other comments I have are the following:
[1] That larger the magnitude of the unwanted impulses, the larger must be 'N' to achieve a given amount of impulse smoothing.
[2] The wider in time (measured in samples) are the unwanted impulses, the larger must be 'N' to achieve a given amount of impulse smoothing.
[3] Although my statistical analysis comparing the outputs of a traditional median filter and the proposed smoothing filter is correct, that analysis doesn't indicate what I've noticed to be a performance difference between those two smoothing techniques. Follow-on experimentation leads me to believe that median filtering is superior to the smoothing scheme I've described in this blog.
[4] Again, I have not compared the computational workload (arithmetic operations per output sample) of the proposed smoothing scheme to the computational workload of median filtering. :-(
Hi Rick,
Thank you for the comments. However as an observation from my results, there is always an inherent time delay which increases larger the N.
Pranav
Hi Pranav,
you can use impulse input for x, get output from above code and do fft to study frequency response such as cutoff, attenuation etc.
Hi Kaz,
Ya! That would give more details.
Pranav
Hi Rick
Thanks to your skillful and practical idea
I had a question. The idea of implementing it was much better than Translation Invariant Wavelet Denoising
Is there any difference in noise removal with this method?
Hello kafshvandi. I must tell you that the proposed impulsive-noise reduction scheme described in my blog is not my idea. I learned of that scheme from the Reference given in my blog.
I hate to admit it but I have never heard of the "Translation Invariant Wavelet Denoising" method before. Being almost totally ignorant of wavelet transforms, sadly, I'm unable to answer your question. Sorry about that.
Thanks for sharing.
To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
|
This question concerns a system of equations that arise in the study of one-soliton solutions to the Davey-Stewartson equation.
In what follows, $f(z)$ denotes a function which depends smoothly (but not necessarily analytically!) on $z=x+iy$. Thus $f:\mathbb{C} \rightarrow \mathbb{R}$ or equivalently $f:\mathbb{R}^2 \rightarrow \mathbb{R}$. We denote by $\overline{\partial}$ and $\partial$ the usual operators$$ \overline{\partial} = \frac{1}{2} \left( \partial_x + i \partial_y \right) $$and$$\partial = \frac{1}{2} \left( \partial_x - i \partial_y \right). $$
The system is:
$$\overline{\partial} n_1(z) = (1+|z|^2)^{-1} n_2(z)$$$$\partial n_2(z) = -(1+|z|^2)^{-1} n_1(z)$$
and the question is as follows. Suppose that
$$\lim_{|z|\rightarrow \infty} |z| n_1(z) = \lim_{\|z| \rightarrow \infty} |z| n_2(z) = 0$$
Can one prove that $n_1(z)=n_2(z)=0$ if one assumes
a priori that $n_1$ and $n_2$ belong to $L^p(R^2)$ for all $p>2$ (including $p=\infty$)? For this purpose one can assume that the limits above exist.
Thanks in advance for any help.
Peter Perry, University of KentuckyThis post imported from StackExchange MathOverflow at 2014-09-20 22:30 (UCT), posted by SE-user Peter Perry
|
Jean Duchon
Retired from CNRS, Institut Fourier (Grenoble).
Postal address: 297 Grande Rue, 38160 Saint-Antoine, France.
My professional email address is no longer active. My new email (not to be spread, please) is through gmail dot com, it is my name dot my wife's name (mesdag).
Saint-Antoine l'Abbaye (Isère), France
Member for 4 years, 3 months
2 profile views
Last seen Feb 22 at 10:29 Communities (1) Top network posts 10 Completion of $C_0^{\infty}(\mathbb{R}^N)$ with norm $\|u\|= \Bigg(\int_{{\mathbb{R}}^N} |\Delta u |^2 \, \mathrm{d}x \Bigg)^{\frac{1}{2}}. $ 8 Structure of the Cantor part of the derivative of a BV function 7 Is there a measure on the sphere with positive Fourier transform? 6 Bound on the joint distribution of three real random variables with given two dimensional marginals 6 Continuous right inverse to the Laplacian operator on $C^\infty$ 5 Obstruction to Navier-Stokes blowup with cylindrical symmetry 5 Is there an $\infty$ version of the Wasserstein distance between two distributions? View more network posts →
|
There is also one question directly relating to this question, that is, how to define the sum of uncountably many numbers (not necessarily positive numbers). The difficulty lies in the fact that there could not be any order of this summation, since there are uncountably many of terms. So, when we talk about the sum of $x_\alpha$, namely, $\sum_{\alpha\in A}x_\alpha$, we are actually saying the following,
For every countable subset of $I\subset A$ with arbitrarily given order, the sequence $(x_\alpha)_{\alpha\in I}$ should be convergent. In other words, the sequence $(x_\alpha)_{\alpha\in I}$ should be absolutely convergent.
A proper definition is given in Paul Halmos' book, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, as follows:
$x=\sum_{\alpha\in A}x_\alpha$ means that for any positive number $\varepsilon$ there is some finite set $I_0$ such that for any finite set (or more generally, countable set) $I\supset I_0$, we have $|x-\sum_{\alpha\in I}x_\alpha|<\varepsilon$.
Note that, the set $\{1,-\frac{1}{2},\frac{1}{3},\cdots,(-1)^{k-1}\frac{1}{k},\cdots, 0,\cdots\}$, where in the end there are uncountably many $0$'s is not convergent any more. But the sequence $\{1,-\frac{1}{2},\cdots\}$ is convergent in the standard sense.
Now we invoke Zorn's lemma, on all countable subsets $I$, with respect to which, the sequence $x_\alpha$ is absolutely convergent, with the inclusion as the order. Note that for any $I_1\subset I_2\subset I_3\subset \cdots$ a chain of countable subsets of $A$, the set $I^*=\bigcup_iI_i$ is also a countable subset of $A$ and by the definition, the sequence with index in $I^*$ is also absolutely convergent. By Zorn's lemma, there should exists a maximal countable subset $I_{max}$. This means that any number $x_\alpha$ with $\alpha\notin I_{max}$ should be $0$, otherwise we can construct another strictly larger countable subset on which the number sequence is absolutely convergent.
|
Question:
If two identical masses are attached to two walls facing each other and the two masses themselves are connected by a third spring (all springs have same initial length and spring constant k). Now we apply two driving forces to these two masses respectively: one on the left with $F_dcos{2\omega t}$ and the one on the left with $2F_dcos{2\omega t}$. Find a particular solution for $x_1(t)$ and $x_2(t)$.
My solution:
$$m\frac{d^2x_1}{dt^2}=-kx_1+k(x_2-x_1)+F_dcos(2\omega t)$$ $$m\frac{d^2x_2}{dt^2}=-kx_2-k(x_2-x_1)+2F_dcos(2\omega t)$$ $$\rightarrow\frac{d^2}{dt^2}(x_1+x_2)+\omega^2(x_1+x_2)=3\frac{F_d}{m}cos(2\omega t)$$ $$\frac{d^2}{dt^2}(x_2-x_1)+3\omega(x_2-x_1)=\frac{F_d}{m}cos(2\omega t)$$ $$Let ~z_1=x_1+x_2~and~z_2=x_2-x_1~\rightarrow~Guess ~z_1=Ccos(2\omega t)~and ~z_2=Dcos(2\omega t)$$ $$\rightarrow~C=D=-\frac{F_d}{m\omega^2}$$ $$\rightarrow x_1(t)=0,~x_2(t)=Acos(2\omega t)$$
where $A=C=D=-\frac{F_d}{m\omega^2}$.
This solution seems very weird since mass 1 is not moving at all. Could somebody give a hint what is wrong with this anti-intuitive solution?
|
I'm trying to read through and understand this proof from Rudin, but am a bit confused at just a few steps. I'm going to try to replicate the proof and pause at those particular steps.
Theorem. Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$. Proof. Let $B \subset S$, where $S$ is an ordered set and $B$ is non-empty and bounded below. Define $L$ to the set of all lower bounds of $B$, which is non-empty due to the fact that $B$ is bounded below.
(1) This is where my first question comes in: Rudin argues here that $L$ consists of all elements $y \in S$ s.t. $y \leq x$ for every element $x \in B$. I understand why this holds for all elements of $B$, as this is the definition of lower bounds. But, we aren't given the size or nature of $S$ (finite, infinite, etc.). If $S\neq \mathbb{R}$, would it noe be possible for us to have other element not in $S$ which bounds $B$ below? We also don't know whether $S$ is bounded below; if so, there exist lower bounds of $B$, and thus elements of $L$, which are in $\mathbb{R}$ but not $L$. So, might it be more accurate to say that $L = \{y \in \mathbb{R} : y \leq x, \forall x \in B\}$?
Continuing the proof, and assuming Rudin's last step as to the definition of $L$, it's clear that every $x$ is an upper bound of $L$, by definition, and thus every element of $B$ bounds $L$ above.
(2) Here, Rudin argues that the properties we assumed about $S$ tell us that $L$ has a supremum, $\alpha$, in $S$. Here, I am not quite following. Clearly, $S$ is ordered and has a least-upper bound. We don't know whether $B$ contains that least upper bound, as we don't know the nature of the subset. Does this step require -- or perhaps embed, without proof -- a lemma that a subset of a ordered set with a least upper bound also has a least upper bound, which is less than or equal to the least upper bound of the one contained in its superset? I'm also struggling a bit to see why this element must exist in $S$, but this may be that not only is every element of $B$ an upper bound of $L$, but that $B$ contains all such upper bounds of $L$. Is this correct? Would this not presuppose that $B$, and thus $L$ (as the largest element of $L$ and the smallest element of $B$ are the same), are composed of real numbers, and that this list is "complete"? Otherwise, we can find a real that is in neither set but bounds $B$ below.
So, again assuming this last step, we define the supremum of $L$ as some element $\alpha \in S$. We then consider some arbitrary element $\gamma$, which has the property that $\gamma < \alpha$. (Rudin does not state that this element is in $S$, but I'm not sure that it quite matters here.) Since $\gamma$ is the least-upper-bound of $L$, $\alpha$ cannot be an upper bound of $L$. But, all elements of $B$ are upper bounds of $L$, so $\gamma \not \in B$. Hence, $B$ cannot contain any element less than $\alpha$, and we have $\alpha \leq x, \forall x \in B$.
(3) I understand everything from the above except, perhaps, the final line that I have for the moment omitted. Rudin concludes from this that $\alpha \in L$. Is the argument here, again, that because $L$ contains all elements $y \leq x$ for $x \in B$, that $\alpha$ is simultaneously in line and $B$ due to the fact that we could have $y = x$? (In other words, assuming some notion of 'completeness,' the largest element of $L$ is simultaneously the smallest element of $B$? I hope I'm not assuming the conclusion of the proof here.)
So, again assuming the above that $\alpha \in L$, we consider some element $\beta$. (Rudin again makes no assumptions on whether $\beta$ lives in $S$, so I assume we don't need to make any assumptions, and can just take $\beta$ to be a real number.) Then, we let $\beta > \alpha$.
(4) From here, I am having difficulty seeing how Rudin concludes that $\beta \not \in L$. It seems that the argument is that, since $\alpha$ is the least upper bound of $S$, and every element of $L$ (per Rudin, though I questioned why this is the case above) is also in $S$, that clearly $\alpha$ must also bound this subset of $S$ above. So, clearly, a greater element, $\beta$, can't be in this set that's bounded above by $\alpha$. Is this correct? If so, this suggests that the argument that $L$ contains elements only in $S$ is even more crucial, though I still am struggling to see why this is the case.
From here, the remainder of the proof is easy: $\alpha \in L$, as we showed above, but any greater element than $\alpha$, such as $\beta$ from above, cannot be in $L$, so $\alpha$ is not only an upper bound of $L$, but the greatest such upper bound. Since $L$ contains all of the lower bounds of $B$, this implies that $\alpha$ is the greatest lower bound, so $\alpha = \inf B$.
I'd appreciate any help and insights on the questions I raised above. Hopefully I have understood and correctly framed at least parts of this proof.
Thanks in advance.
|
To understand what is going on lets examine your initial plot. We assign it to the variable
g:
g = Show[
Plot[1/2 x + 1/2, {x, -6, 7}],
ListPlot[{{4, -2.5}, {2, 1.5}}, Joined -> True, PlotMarkers -> Automatic]
]
g // ImageDimensions
{360, 224}
The figure's aspect ratio is therefore:
#2/#1 & @@ ImageDimensions[g] // N
0.6222222222
which happens to be close to $\frac{2}{\sqrt{5}+1}\approx 0.6180339887$ or the reciprocal of the golden ratio. This ratio is widely considered as being aesthetically pleasing and is therefore the default aspect ratio for Mathematica's
Plot function:
Options[Plot, AspectRatio]
{AspectRatio -> 1/GoldenRatio}
With
AspectRatio->Automatic Mathematica sets the aspect ratio to the ratio of the plot ranges in x and y directions:
h = Show[
Plot[1/2 x + 1/2, {x, -6, 7}],
ListPlot[{{4, -2.5}, {2, 1.5}}, Joined -> True, PlotMarkers -> Automatic],
AspectRatio -> Automatic
];
#2/#1 & @@ ImageDimensions[h] // N
0.5083333333
#2/#1 & @@ (Subtract @@@ PlotRange[h])
0.4999999796
(small difference caused by the discrete pixel size).
This choice causes the the plot space to be isotropic which is needed for a correct display of angles.
h
Note that the default aspect ratio for
Graphics is:
Options[Graphics, AspectRatio]
{AspectRatio -> Automatic}
This makes sense as in most drawings we want an isotropic space.
|
Your score is simply the sum of difficulties of your solved problems. Solving the same problem twice does not give any extra points. Note that Kattis' difficulty estimates vary over time, and that this can cause your score to go up or down without you doing anything.
Scores are only updated every few minutes – your score and rank will not increase instantaneously after you have solved a problem, you have to wait a short while.
If you have set your account to be anonymous, you will not be shown in ranklists, and your score will not contribute to the combined score of your country or university. Your user profile will show a tentative rank which is the rank you would get if you turned off anonymous mode (assuming no anonymous users with a higher score than you do the same).
The combined score for a group of people (e.g., all users from a given country or university) is computed as a weighted average of the scores of the individual users, with geometrically decreasing weights (higher weights given to the larger scores). Suppose the group contains $n$ people, and that their scores, ordered in non-increasing order, are $s_0 \ge s_1 \ge \ldots \ge s_{n-1}$ Then the combined score for this group of people is calculated as \[ S = \frac{1}{f} \sum_{i=0}^{n-1} \left(1-\frac{1}{f}\right)^i \cdot s_i, \] where the parameter $f$ gives a trade-off between the contribution from having a few high scores and the contribution from having many users. In Kattis, the value of this parameter is chosen to be $f = 5$.
For example, if the group consists of a single user, the score for the group is 20% of the score of that user. If the group consists of a very large number of users, about 90% of the score is contributed by the 10 highest scores.
Adding a new user with a non-zero score to a group always increases the combined score of the group.
Kattis has problems of varying difficulty. She estimates the difficulty for different problems by using a variant of the ELO rating system. Broadly speaking, problems which are solved by many people using few submissions get low difficulty scores, and problems which are often attempted but rarely solved get high difficulty scores. Problems with very few submissions tend to get medium difficulty scores, since Kattis does not have enough data about their difficulty.
The difficulty estimation process also assigns an ELO-style rating to you as a user. This rating increases when you solve problems, like your regular score, but is also affected by your submission accuracy. We use your rating to choose which problems to suggest for you to solve. If your rating is higher, the problems we suggest to you in each category (trivial, easy, medium, hard) will have higher difficulty values.
|
We define the thermal density operator as
$$\tau(\beta) = \frac{e^{-\beta H}}{\mathrm{Tr}(e^{-\beta H})}$$
where $H$ is the systems Hamiltonian. The thermal state is characterized by the fact that it maximizes the entropy for a given, fixed energy.
Now when we consider the second law of thermodynamics which states that
$$\Delta S \geq 0$$
which states that over time the entropy of systems either remains the same or increases.
Now I know that equilibration/thermalization is a process which is not yet well understood on the quantum scale. But if we would forget about this fact and just say that yes, quantum systems equilibrate aswell would this mean that they equilibrate
towards the thermal state?
|
Colloquia/Fall18 Contents 1 Mathematics Colloquium 1.1 Spring 2018 1.2 Spring Abstracts 1.2.1 January 29 Li Chao (Columbia) 1.2.2 February 2 Thomas Fai (Harvard) 1.2.3 February 5 Alex Lubotzky (Hebrew University) 1.2.4 February 6 Alex Lubotzky (Hebrew University) 1.2.5 February 9 Wes Pegden (CMU) 1.2.6 March 2 Aaron Bertram (Utah) 1.2.7 March 16 Anne Gelb (Dartmouth) 1.2.8 April 6 Edray Goins (Purdue) 1.3 Past Colloquia Mathematics Colloquium
All colloquia are on Fridays at 4:00 pm in Van Vleck B239,
unless otherwise indicated. Spring 2018
date speaker title host(s) January 29 (Monday) Li Chao (Columbia) Elliptic curves and Goldfeld's conjecture Jordan Ellenberg February 2 (Room: 911) Thomas Fai (Harvard) The Lubricated Immersed Boundary Method Spagnolie, Smith February 5 (Monday, Room: 911) Alex Lubotzky (Hebrew University) High dimensional expanders: From Ramanujan graphs to Ramanujan complexes Ellenberg, Gurevitch February 6 (Tuesday 2 pm, Room 911) Alex Lubotzky (Hebrew University) Groups' approximation, stability and high dimensional expanders Ellenberg, Gurevitch February 9 Wes Pegden (CMU) The fractal nature of the Abelian Sandpile Roch March 2 Aaron Bertram (University of Utah) Stability in Algebraic Geometry Caldararu March 16 Anne Gelb (Dartmouth) Reducing the effects of bad data measurements using variance based weighted joint sparsity WIMAW April 4 (Wednesday) John Baez (UC Riverside) TBA Craciun April 6 Edray Goins (Purdue) Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups Melanie April 13 Jill Pipher (Brown) TBA WIMAW April 16 (Monday) Christine Berkesch Zamaere (University of Minnesota) TBA Erman, Sam April 20 Xiuxiong Chen (Stony Brook University, CANCELLED) TBA Bing Wang April 25 (Wednesday) Hitoshi Ishii (Waseda University) Wasow lecture TBA Tran date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty date person (institution) TBA hosting faculty Spring Abstracts January 29 Li Chao (Columbia)
Title: Elliptic curves and Goldfeld's conjecture
Abstract: An elliptic curve is a plane curve defined by a cubic equation. Determining whether such an equation has infinitely many rational solutions has been a central problem in number theory for centuries, which lead to the celebrated conjecture of Birch and Swinnerton-Dyer. Within a family of elliptic curves (such as the Mordell curve family y^2=x^3-d), a conjecture of Goldfeld further predicts that there should be infinitely many rational solutions exactly half of the time. We will start with a history of this problem, discuss our recent work (with D. Kriz) towards Goldfeld's conjecture and illustrate the key ideas and ingredients behind these new progresses.
February 2 Thomas Fai (Harvard)
Title: The Lubricated Immersed Boundary Method
Abstract: Many real-world examples of fluid-structure interaction, including the transit of red blood cells through the narrow slits in the spleen, involve the near-contact of elastic structures separated by thin layers of fluid. The separation of length scales between these fine lubrication layers and the larger elastic objects poses significant computational challenges. Motivated by the challenge of resolving such multiscale problems, we introduce an immersed boundary method that uses elements of lubrication theory to resolve thin fluid layers between immersed boundaries. We apply this method to two-dimensional flows of increasing complexity, including eccentric rotating cylinders and elastic vesicles near walls in shear flow, to show its increased accuracy compared to the classical immersed boundary method. We present preliminary simulation results of cell suspensions, a problem in which near-contact occurs at multiple levels, such as cell-wall, cell-cell, and intracellular interactions, to highlight the importance of resolving thin fluid layers in order to obtain the correct overall dynamics.
February 5 Alex Lubotzky (Hebrew University)
Title: High dimensional expanders: From Ramanujan graphs to Ramanujan complexes
Abstract:
Expander graphs in general, and Ramanujan graphs , in particular, have played a major role in computer science in the last 5 decades and more recently also in pure math. The first explicit construction of bounded degree expanding graphs was given by Margulis in the early 70's. In mid 80' Margulis and Lubotzky-Phillips-Sarnak provided Ramanujan graphs which are optimal such expanders.
In recent years a high dimensional theory of expanders is emerging. A notion of topological expanders was defined by Gromov in 2010 who proved that the complete d-dimensional simplical complexes are such. He raised the basic question of existence of such bounded degree complexes of dimension d>1.
This question was answered recently affirmatively (by T. Kaufman, D. Kazdhan and A. Lubotzky for d=2 and by S. Evra and T. Kaufman for general d) by showing that the d-skeleton of (d+1)-dimensional Ramanujan complexes provide such topological expanders. We will describe these developments and the general area of high dimensional expanders.
February 6 Alex Lubotzky (Hebrew University)
Title: Groups' approximation, stability and high dimensional expanders
Abstract:
Several well-known open questions, such as: are all groups sofic or hyperlinear?, have a common form: can all groups be approximated by asymptotic homomorphisms into the symmetric groups Sym(n) (in the sofic case) or the unitary groups U(n) (in the hyperlinear case)? In the case of U(n), the question can be asked with respect to different metrics and norms. We answer, for the first time, one of these versions, showing that there exist fintely presented groups which are not approximated by U(n) with respect to the Frobenius (=L_2) norm.
The strategy is via the notion of "stability": some higher dimensional cohomology vanishing phenomena is proven to imply stability and using high dimensional expanders, it is shown that some non-residually finite groups (central extensions of some lattices in p-adic Lie groups) are Frobenious stable and hence cannot be Frobenius approximated.
All notions will be explained. Joint work with M, De Chiffre, L. Glebsky and A. Thom.
February 9 Wes Pegden (CMU)
Title: The fractal nature of the Abelian Sandpile
Abstract: The Abelian Sandpile is a simple diffusion process on the integer lattice, in which configurations of chips disperse according to a simple rule: when a vertex has at least 4 chips, it can distribute one chip to each neighbor.
Introduced in the statistical physics community in the 1980s, the Abelian sandpile exhibits striking fractal behavior which long resisted rigorous mathematical analysis (or even a plausible explanation). We now have a relatively robust mathematical understanding of this fractal nature of the sandpile, which involves surprising connections between integer superharmonic functions on the lattice, discrete tilings of the plane, and Apollonian circle packings. In this talk, we will survey our work in this area, and discuss avenues of current and future research.
March 2 Aaron Bertram (Utah)
Title: Stability in Algebraic Geometry
Abstract: Stability was originally introduced in algebraic geometry in the context of finding a projective quotient space for the action of an algebraic group on a projective manifold. This, in turn, led in the 1960s to a notion of slope-stability for vector bundles on a Riemann surface, which was an important tool in the classification of vector bundles. In the 1990s, mirror symmetry considerations led Michael Douglas to notions of stability for "D-branes" (on a higher-dimensional manifold) that corresponded to no previously known mathematical definition. We now understand each of these notions of stability as a distinct point of a complex "stability manifold" that is an important invariant of the (derived) category of complexes of vector bundles of a projective manifold. In this talk I want to give some examples to illustrate the various stabilities, and also to describe some current work in the area.
March 16 Anne Gelb (Dartmouth)
Title: Reducing the effects of bad data measurements using variance based weighted joint sparsity
Abstract: We introduce the variance based joint sparsity (VBJS) method for sparse signal recovery and image reconstruction from multiple measurement vectors. Joint sparsity techniques employing $\ell_{2,1}$ minimization are typically used, but the algorithm is computationally intensive and requires fine tuning of parameters. The VBJS method uses a weighted $\ell_1$ joint sparsity algorithm, where the weights depend on the pixel-wise variance. The VBJS method is accurate, robust, cost efficient and also reduces the effects of false data.
April 6 Edray Goins (Purdue)
Title: Toroidal Belyĭ Pairs, Toroidal Graphs, and their Monodromy Groups
Abstract: A Belyĭ map [math] \beta: \mathbb P^1(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] is a rational function with at most three critical values; we may assume these values are [math] \{ 0, \, 1, \, \infty \}. [/math] A Dessin d'Enfant is a planar bipartite graph obtained by considering the preimage of a path between two of these critical values, usually taken to be the line segment from 0 to 1. Such graphs can be drawn on the sphere by composing with stereographic projection: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq \mathbb P^1(\mathbb C) \simeq S^2(\mathbb R). [/math] Replacing [math] \mathbb P^1 [/math] with an elliptic curve [math]E [/math], there is a similar definition of a Belyĭ map [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C). [/math] Since [math] E(\mathbb C) \simeq \mathbb T^2(\mathbb R) [/math] is a torus, we call [math] (E, \beta) [/math] a toroidal Belyĭ pair. The corresponding Dessin d'Enfant can be drawn on the torus by composing with an elliptic logarithm: [math] \beta^{-1} \bigl( [0,1] \bigr) \subseteq E(\mathbb C) \simeq \mathbb T^2(\mathbb R). [/math]
This project seeks to create a database of such Belyĭ pairs, their corresponding Dessins d'Enfant, and their monodromy groups. For each positive integer [math] N [/math], there are only finitely many toroidal Belyĭ pairs [math] (E, \beta) [/math] with [math] \deg \, \beta = N. [/math] Using the Hurwitz Genus formula, we can begin this database by considering all possible degree sequences [math] \mathcal D [/math] on the ramification indices as multisets on three partitions of N. For each degree sequence, we compute all possible monodromy groups [math] G = \text{im} \, \bigl[ \pi_1 \bigl( \mathbb P^1(\mathbb C) - \{ 0, \, 1, \, \infty \} \bigr) \to S_N \bigr]; [/math] they are the ``Galois closure
of the group of automorphisms of the graph. Finally, for each possible monodromy group, we compute explicit formulas for Belyĭ maps [math] \beta: E(\mathbb C) \to \mathbb P^1(\mathbb C) [/math] associated to some elliptic curve [math] E: \ y^2 = x^3 + A \, x + B. [/math] We will discuss some of the challenges of determining the structure of these groups, and present visualizations of group actions on the torus.
This work is part of PRiME (Purdue Research in Mathematics Experience) with Chineze Christopher, Robert Dicks, Gina Ferolito, Joseph Sauder, and Danika Van Niel with assistance by Edray Goins and Abhishek Parab.
|
How many digits are there in the number $200^{2010}$? I have tried to re-write it as $(2\cdot 100)^{2010} = (2\cdot 10^2)^{2010} = 2^{2010} \cdot 10^{4020} = 1024^{201} \cdot 10^{4020} = 1.024^{201} \cdot 10^{4623}$. But how do I write $1.024^{201}$ as a power of base 10? I would like to solve the problem without logarithms or a calculator.
$\log 200^{2010}=2010\times\log 200=2010\times(2+0.3010)=4625.01$
$\therefore$ $4625\leq \log 200^{2010} <4626$
$\therefore$ $10^{4625}\leq 200^{2010}<10^{4626}$
$\therefore$ $4626$ digits
Think about some very simple cases $10$ has $2$ digits as does $99$, but $100$ has $3$ digits.$1\le \log_{10} 99 \lt 2$. A few explorations should convince you that the number of decimal digits of a positive integer $N$ is $\lfloor \log_{10} N\rfloor +1$.
You can prove this by writing $N=a\times 10^k$ where $1\le a\lt 10$ and $k$ is an integer and taking logs to base 10.
Now $\log_{10} (200^{2010})=2010 \log_{10}(200)=2010 \cdot(2+\log_{10}2)$ so you just need to have a sufficiently accurate estimate of $\log_{10}2$.
$1024^{201} \approx (10^3)^{201}=10^{603}$ is a rough estimate for that term.
We can do slightly better by writing $1024=10^3 + 24$, so
$1024^{201} = (10^3 + 24)^{201} = 10^{603} + 201\cdot24\cdot10^{600} + \tbinom {201}2\cdot24^2\cdot10^{597} + \cdots$
After just a few terms, they should be small enough so as not to add digits, and you can probably work that out without too much trouble.
This is kind of clunky, but it does completely avoid logarithms, if that's important.
You cannot really solve it without logarithms since this is exactly what logarithm is – it measures how many digits you need to write a number in a certain base. The number of decimal digits is $⌊\log_{10}(n)⌋ + 1$. $\log(200^{2010}) = 2010 * \log(200) = 2010 * \log(2 * 100) = 2010 * (\log(2) + 2) = 4625.07…$. So the exact number of digits is 4626. If you don't want to use calculator, you can estimate $\log_{10}(2)$ by another method.
|
If you could travel to the center of the Earth (or any planet), would you be weightless there?
Correct. If you split the earth up into spherical shells, then the gravity from the shells "above" you cancels out, and you only feel the shells "below" you. When you are in the middle there is nothing "below" you.
{I am using some simplistic terms, but I don't want to break out surface integrals and radial flux equations}
Edit: Although the inside of the shell will have zero gravity classically, it will also have non zero gravity relativistically. At the perfect center the forces may balance out, yielding an unstable solution, meaning that a small perturbation in position will result in forces that exaggerate this perturbation.
The simplest way to think about it is that there is mass all around you in the center of the Earth so you get an equal gravitational "pull" from all directions. The pulls cancel out so you get no acceleration.
If one assumes constant density for the Earth (which isn't strictly speaking true but it is close enough for this illustration) the gravitational acceleration drops linearly from 1g at the surface to 0 at the center of the Earth. So you'd get a zero if you stepped on a scale at the center of the Earth.
The more complicated explanation is that acceleration due to gravity is the derivative of the gravitational potential. This potential is a minimum at the center of the Earth and grows quadratically up to the surface. It then continues to increase at a lower rate. Since at the exact center is flat (like the bottom of a valley), the derivative which is a measure of the rate of change is zero, and there is no acceleration.
Interestingly, even though you would be weightless there, the effects of gravity are highest at the center of the Earth. You get more gravitational time dilation, for example, than you do at the surface.
I like answers that appeal to symmetry, so I answer this one with a question: If you were at the center, which way would you fall? That tells us you could stay floating there.
In the following the term "charge" refers either to mass or to electric charge and the term "Inverse Square Law" refers either to Newton's Gravitational Law or to Coulomb's Law, respectively.
SECTION 1
A. The Inverse Square Law for Spheres with uniform surface charge density
Proposition A: Let a sphere of radius $\:\rm{R}\:$ with uniform surface charge density $\:\rho_{s}\:$ and empty interior. Then:
(a1) the force exerted upon a point charge $\:\xi\:$ in the interior or on the surface of the sphere, as in Fig. 01, is
zero (cancels out). In terms of potentials, the whole sphere (surface + interior) is an equi-potential region.
(a2) the force exerted upon a point charge $\:\xi\:$ in the exterior of the sphere, as in Fig. 03, is equal to the force exerted by a point particle at the center of the sphere with charge equal to its total surface charge $\:\Xi_{s}=\rho_{s}\cdot4\pi{\rm{R}}^{2}\:$. In terms of potentials, the potential outside the sphere equals that created by its total surface charge $\:\Xi_{s}\:$ concentrated on its center.
An intermediate conclusion in the proof of this Proposition is that the magnitude of the force exerted by the "cup" AKBMA of the sphere on the point charge $\:\xi\:$ in Fig. 02 is proportional to $\:\sin^{2}\omega\:$, where $\:\omega\:$ is the angle by which any point of the cyclic edge AMBA of the cup observes the line segment $\:b\:$ (that between the charge $\:\xi\:$ and the center of the sphere). More exactly this force is by magnitude:
\begin{equation} \vert \mathbf{f}_{AKBMA}\vert=k \cdot \dfrac{\Xi_{s}\cdot \xi}{{\rm{b}}^{2}}\sin^{2}\left(\dfrac{\omega}{2}\right)=\left(k \cdot \dfrac{4\pi\rho_{s}\xi{\rm{R}}^{2}}{{\rm{b}}^{2}}\right)\sin^{2}\left(\dfrac{\omega}{2}\right)=constant\cdot \sin^{2}\left(\dfrac{\omega}{2}\right) \tag{A-01} \end{equation}
But this force is cancelled by the force exerted by the "cup" CLDNC of the sphere which is equal in magnitude, but opposite direction:
\begin{equation} \mathbf{f}_{CLDNC}=\;-\;\mathbf{f}_{AKBMA} \tag{A-02} \end{equation}
So, if we remove these two "cups" the force doesn't change. But if we enlarge "cup" AKBMA by moving its cyclic edge AMBA to the left, this last will coincide with the cyclic edge CNDC of the left cup CLDNC. Then removing the two cups is as if we remove the whole sphere leaving the net force unchanged, that is, zero.
Also in Fig. 02 we have
\begin{equation} \mathbf{f}_{AMBDNCA}=\;\mathbf{0} \tag{A-03} \end{equation}
B. The Inverse Square Law for Spheres with uniform volume charge density
Proposition B: Let a sphere of radius $\:\rm{R}\:$ with uniform volume charge density $\:\rho_{v}\:$. Then:
(b1) the force exerted upon a point charge $\:\xi\:$ in the interior of the sphere located at a radial distance $\:\rm{r}\:$ from is center is equal, according to
Proposition A, to that exerted by the volume charge density of a sphere of radius $\:\rm{r}\:$, $\:\Xi_{v}\left(\rm{r}\right)=\rho_{v}\cdot \dfrac{4}{3}\pi{\rm{r}}^{3}\:$, concentrated on the center. The magnitude of this force is:
\begin{equation} \vert f_{inside} \vert =k \cdot \dfrac{\Xi_{v}\left(\rm{r}\right)\cdot \xi}{{\rm{r}}^{2}}=k \cdot \dfrac{\rho_{v}4\pi\xi{\rm{r}}^{3}}{3{\rm{r}}^{2}}=constant \cdot \rm{r}\:,\quad \rm{r}\le \rm{R} \tag{B-01} \end{equation}
(b2) the force exerted upon a point charge $\:\xi\:$ in the exterior of the sphere and at radial distance $\:\rm{r}\:$ from is center is equal, according to
Proposition A, to that exerted by the volume charge density of a sphere of radius $\:\rm{R}\:$, $\:\Xi_{v}\left(\rm{R}\right)=\rho_{v}\cdot \dfrac{4}{3}\pi{\rm{R}}^{3}\:$, concentrated on the center. The magnitude of this force is:
\begin{equation} \vert f_{outside} \vert =k \cdot \dfrac{\Xi_{v}\left(\rm{R}\right)\cdot \xi}{{\rm{r}}^{2}}=k \cdot \dfrac{\rho_{v}4\pi\xi{\rm{R}}^{3}}{3{\rm{r}}^{2}}=constant \cdot \rm{r}^{-2}\:,\quad \rm{r}>\rm{R} \tag{B-02} \end{equation}
SECTION 2
Suppose that Earth is a perfect sphere with uniform volume mass density. Then:
Proposition C:
(c1) A body located at the center of the Earth is
weightless.
(c2) Imagine a tunnel of small cross section running along a whole diameter, so passing through the center of the Earth. A body placed in the tunnel at a radial distance $\:{\rm{r}}_{0}\:$ from the center will execute
a simple rectilinear harmonic oscillation with center the center of the Earth, since in the case of gravity the force is always attractive to the center and according to equation (B-01) proportional in magnitude to the distance from this center of attraction.
You would
not be weightless at the center of the Earth. In other words, the Earth does not follow a geodesic. Let me explain.
The Earth is not spherical, it is an oblate spheroid. The acceleration of a uniform non-spherical body in a spherical gravitational field does not follow an inverse square law. The acceleration
of the center of mass does not equal the acceleration at the center of mass. An accelerometer fixed at the center of the Earth would read approx 1.75 pgal (1.75e-14 m/$\mathrm{s^2}$), not zero.
protected by Qmechanic♦ Apr 24 '13 at 16:05
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
|
I'm a mathematician interested in abstract QFT. I'm trying to undersand why, under certain (all?) circumstances, we must have $T^2 = -1$ rather than $T^2 = +1$, where $T$ is the time reversal operator. I understand from the Wikipedia article that requiring that energy stay positive forces $T$ to be represented by an anti-unitary operator. But I don't see how this forces $T^2=-1$. (Or maybe it doesn't force it, it merely allows it?)
Here's another version of my question. There are two distinct double covers of the Lie group $O(n)$ which restrict to the familiar $Spin(n)\to SO(n)$ cover on $SO(n)$; they are called $Pin_+(n)$ and $Pin_-(n)$. If $R\in O(n)$ is a reflection and $\tilde{R}\in Pin_\pm(n)$ covers $R$, then $\tilde{R}^2 = \pm 1$. So saying that $T^2=-1$ means we are in $Pin_-$ rather than $Pin_+$. (I'm assuming Euclidean signature here.) My question (version 2): Under what circumstances are we forced to use $Pin_-$ rather than $Pin_+$ here?
(I posted a similar question on physics.stackexchange.com last week, but there were no replies.)
EDIT: Thanks to the half-integer spin hint in the comments below, I was able to do a more effective web search. If I understand correctly, Kramer's theorem says that for even-dimensional (half integer spin) representations of the Spin group, $T$ must satisfy $T^2=-1$, while for the odd-dimensional representations (integer spin), we have $T^2=1$. I guess at this point it becomes a straightforward question in representation theory: Given an irreducible representation of $Spin(n)$, we can ask whether it is possible to extend it to $Pin_-(n)$ (or $Pin_+(n)$) so that the lifted reflections $\tilde R$ (e.g. $T$) act as an anti-unitary operator.
|
Let's assume that the vapor pressure of both components can be modeled with the Antoine equation. This gives us a convenient way to address the question algebraically. The vapor pressure of most chemicals is excellently modeled by the correct Antoine equation for that chemical. The equation is:$$\log_{10}{p} = A-\frac{B}{C+T}$$
where the the parameters $A$, $B$, and $C$ are compound specific parameters. The equation can be solved for temperature explicitly yielding $T = \frac{B}{A-\log_{10}\, p} - C$.
So at $\pu{1 atm}$, the equation for pure species $i$ and also a different pure species $j$ that is higher boiling by $\pu{50 ^\circ C}$ is:
\begin{align}T_1 &= \frac{B_i}{A_i-\log_{10}\, p_{\pu{1 atm}}} - C_i\\T_1 &= \frac{B_j}{A_j-\log_{10}\, p_{\pu{1 atm}}} - C_j - \pu{50 ^\circ C}\end{align}
where $T_1$ is the boiling point of $i$ at $\pu{1 atm}$. To make the math easier, let's suppose the equation is in units of atmospheres, in which case $\log_{10}\, p_{\pu{1 atm}} = 0$.
\begin{align}T_1 &= \frac{B_i}{A_i} - C_i\\T_1 &= \frac{B_j}{A_j} - C_j - \pu{50 ^\circ C}\end{align}
The question is, what happens when $p$ changes, in this case, when it is lowered by ~760-fold, let's call that 1000-fold to keep the math easy.
\begin{align}T_{0,i} &= \frac{B_i}{A_i-\log_{10}0.001} - C_i = \frac{B_i}{A_i+3}-C_i\\T_{0,j} &= \frac{B_j}{A_j-\log_{10}0.001} - C_j = \frac{B_j}{A_j+3}-C_j\end{align}
Now we have four equations but
six unknown parameters, $A_i$, $A_j$, $B_i$, $B_j$, $C_i$, and $C_j$. So in general, there is no generally valid constraint for the temperature difference $T_{0,i}-T_{0,j}$: It could be higher, lower, or about the same as the difference at room temperature.
However, we could make some further assumptions. A good one here could be that both species in question have a constant heat of vaporization, which means that $C_i=C_j=0$.
If that assumption is true, then the four equations become:
\begin{align}T_1 &= \frac{B_i}{A_i}\\T_1 &= \frac{B_j}{A_j} - \pu{50 ^\circ C}\\T_{0,i} &= \frac{B_i}{A_i+3} =\frac{\frac{B_i}{A_i}}{1+\frac{3}{A_i}} =\frac{T_1}{1+\frac{3}{A_i}}\\T_{0,j} &= \frac{B_j}{A_j+3} =\frac{T_1+50 °C}{1+\frac{3}{A_j}}\end{align}
Now, the temperature difference of interest $T_{0,j}-T_{0,i}$ is
$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{1+\frac{3}{A_j}} - \frac{T_1}{1+\frac{3}{A_i}}$$
To keep the formulas easy let me define new parameters $a_i$ and $a_j$ such that $a_x=1+\frac{3}{A_x}$. Then the temperature difference is
$$T_{0,j}-T_{0,i} = \frac{T_1+\pu{50 ^\circ C}}{a_j} - \frac{T_1}{a_i}= \frac{a_i(T_1+\pu{50 ^\circ C})-a_j T_1}{a_i a_j}=\frac{(a_i-a_j)T_1 + a_i(\pu{50 ^\circ C}) }{a_i a_j}$$
Now we must introduce further assumptions about the Antoine $A$ parameters. They are positive and thus the $a$ parameters must also be positive, and further positive $A$ implies $a>1$.
If we further assume $a_j=a_i$, then
$$T_{0,j}-T_{0,i} =\frac{(\pu{50 ^\circ C}) }{a_j}$$
Since we know already that $a_j>1$, then the temperature difference (i.e. the difference in boiling points) at the lower pressure is
less than the temperature difference of $\pu{50 ^\circ C}$ at \pu{1 atm}. Now, even if $a_i$ is not precisely equal to $a_j$, then the temperature difference will still be less than $\pu{50 ^\circ C}$ as long as the $a_i$ and $a_j$ parameters are similar enough to keep $(a_i-a_j)T_1 << a_i(\pu{50 ^\circ C})$. It would take a very unusual choice of chemicals $i$ and $j$ for that to happen. I don't think I could identify such a pair.
So, in summary, to get to the result that I think your instructor was asking for, we required
many assumptions: Both chemicals obey the Antoine equation in the temperature range of interest (probably an OK assumption). Both chemicals have approximately constant enthalpy of vaporization in the T range of interest so that we can neglect the Antoine $C$ parameters. The Antoine $A$ parameters for both chemicals must be positive. (A good assumption but it still must be made.) The chemicals have Antoine $A$ parameters (and thus also $a$ parameters as I defined them here) that are not too different from each other.
It might be possible to weaken some of the assumptions and still find that the lower-pressure boiling point difference must be lower than $\pu{50 ^\circ C}$, but I'm confident that the result is not necessarily generally true. It isn't a thermodynamic law or anything. It definitely depends on several key assumptions about how vapor pressure "usually" or "typically" behaves for most chemicals.
|
Possible Duplicate: Why can ALL quadratic equations be solved by the quadratic formula?
How to derive this: $x = \frac{-b + {\sqrt{b^2 + 4ac}}}{2a}$
From this:
$ax^2 + bx + c = 0$
I know this may be a little elementary :)
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
Possible Duplicate: Why can ALL quadratic equations be solved by the quadratic formula?
How to derive this: $x = \frac{-b + {\sqrt{b^2 + 4ac}}}{2a}$
From this:
$ax^2 + bx + c = 0$
I know this may be a little elementary :)
The usual approach is to complete the square:
$$\begin{align*} ax^2+bx+c&=a\left(x^2+\frac{b}ax+\frac{c}a\right)\\ &=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}a\right)\\ &=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c\;, \end{align*}$$
which equals $0$ if and only if $$a\left(x+\frac{b}{2a}\right)^2=\frac{b^2}{4a}-c=\frac{b^2-4ac}{4a}\;.$$ Now divide both sides by $a$ to get
$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\;,$$ and take square roots:
$$x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\pm\frac1{2a}\sqrt{b^2-4ac}\;.$$ Hence
$$x=-\frac{b}{2a}\pm\frac1{2a}\sqrt{b^2-4ac}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;.$$
Since $a\neq0$ we can divide the equation by $a$ and have it as $x^{2}+px+q=0$ where $p=\frac{b}{a},q=\frac{c}{a}$.
Note $$x^{2}+px+q=(x+\frac{p}{2})^{2}+q-\frac{p^{2}}{4}$$ so we need to solve $(x+\frac{p}{2})^{2}+q-\frac{p^{2}}{4}=0$
so we have it that $(x+\frac{p}{2})^{2}=q-\frac{p^{2}}{4}$. now take the square root, reduce $\frac{p}{2}$ from both sides and substitute $p,q$.
from $ax^2 + bx + c = 0 \Leftrightarrow 4a^2x^2 + 4abx + b^2 = b^2 - 4ac$.
Hence $(2ax + b)^2 = b^2 - 4ac$ then $2ax + b = \pm \sqrt{b^2 - 4ac}$.
from this, we have $ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
First off, note that you can easily transform the equation to $x^2-2dx+e=0$ by simply dividing out $a$ and a factor $-2$, then we're looking to explain the solution $x=d\pm\sqrt{d^2-e}$.
Since $x^2-2dx+e=(x^2-2dx+d^2)-(d^2-e)=(x-d)^2-(d^2-e)=0$, taking the square root yields $x-d=\pm\sqrt{d^2-e}$, which was requested.
|
@egreg It does this "I just need to make use of the standard hyphenation function of LaTeX, except "behind the scenes", without actually typesetting anything." (if not typesetting includes typesetting in a hidden box) it doesn't address the use case that he said he wanted that for
@JosephWright ah yes, unlike the hyphenation near box question, I guess that makes sense, basically can't just rely on lccode anymore. I suppose you don't want the hyphenation code in my last answer by default?
@JosephWright anway if we rip out all the auto-testing (since mac/windows/linux come out the same anyway) but leave in the .cfg possibility, there is no actual loss of functionality if someone is still using a vms tex or whatever
I want to change the tracking (space between the characters) for a sans serif font. I found that I can use the microtype package to change the tracking of the smallcaps font (\textsc{foo}), but I can't figure out how to make \textsc{} a sans serif font.
@DavidCarlisle -- if you write it as "4 May 2016" you don't need a comma (or, in the u.s., want a comma).
@egreg (even if you're not here at the moment) -- tomorrow is international archaeology day: twitter.com/ArchaeologyDay , so there must be someplace near you that you could visit to demonstrate your firsthand knowledge.
@barbarabeeton I prefer May 4, 2016, for some reason (don't know why actually)
@barbarabeeton but I have another question maybe better suited for you please: If a member of a conference scientific committee writes a preface for the special issue, can the signature say John Doe \\ for the scientific committee or is there a better wording?
@barbarabeeton overrightarrow answer will have to wait, need time to debug \ialign :-) (it's not the \smash wat did it) on the other hand if we mention \ialign enough it may interest @egreg enough to debug it for us.
@DavidCarlisle -- okay. are you sure the \smash isn't involved? i thought it might also be the reason that the arrow is too close to the "M". (\smash[t] might have been more appropriate.) i haven't yet had a chance to try it out at "normal" size; after all, \Huge is magnified from a larger base for the alphabet, but always from 10pt for symbols, and that's bound to have an effect, not necessarily positive. (and yes, that is the sort of thing that seems to fascinate @egreg.)
@barbarabeeton yes I edited the arrow macros not to have relbar (ie just omit the extender entirely and just have a single arrowhead but it still overprinted when in the \ialign construct but I'd already spent too long on it at work so stopped, may try to look this weekend (but it's uktug tomorrow)
if the expression is put into an \fbox, it is clear all around. even with the \smash. so something else is going on. put it into a text block, with \newline after the preceding text, and directly following before another text line. i think the intention is to treat the "M" as a large operator (like \sum or \prod, but the submitter wasn't very specific about the intent.)
@egreg -- okay. i'll double check that with plain tex. but that doesn't explain why there's also an overlap of the arrow with the "M", at least in the output i got. personally, i think that that arrow is horrendously too large in that context, which is why i'd like to know what is intended.
@barbarabeeton the overlap below is much smaller, see the righthand box with the arrow in egreg's image, it just extends below and catches the serifs on the M, but th eoverlap above is pretty bad really
@DavidCarlisle -- i think other possible/probable contexts for the \over*arrows have to be looked at also. this example is way outside the contexts i would expect. and any change should work without adverse effect in the "normal" contexts.
@DavidCarlisle -- maybe better take a look at the latin modern math arrowheads ...
@DavidCarlisle I see no real way out. The CM arrows extend above the x-height, but the advertised height is 1ex (actually a bit less). If you add the strut, you end up with too big a space when using other fonts.
MagSafe is a series of proprietary magnetically attached power connectors, originally introduced by Apple Inc. on January 10, 2006, in conjunction with the MacBook Pro at the Macworld Expo in San Francisco, California. The connector is held in place magnetically so that if it is tugged — for example, by someone tripping over the cord — it will pull out of the socket without damaging the connector or the computer power socket, and without pulling the computer off the surface on which it is located.The concept of MagSafe is copied from the magnetic power connectors that are part of many deep fryers...
has anyone converted from LaTeX -> Word before? I have seen questions on the site but I'm wondering what the result is like... and whether the document is still completely editable etc after the conversion? I mean, if the doc is written in LaTeX, then converted to Word, is the word editable?
I'm not familiar with word, so I'm not sure if there are things there that would just get goofed up or something.
@baxx never use word (have a copy just because but I don't use it;-) but have helped enough people with things over the years, these days I'd probably convert to html latexml or tex4ht then import the html into word and see what come out
You should be able to cut and paste mathematics from your web browser to Word (or any of the Micorsoft Office suite). Unfortunately at present you have to make a small edit but any text editor will do for that.Givenx=\frac{-b\pm\sqrt{b^2-4ac}}{2a}Make a small html file that looks like<!...
@baxx all the convertors that I mention can deal with document \newcommand to a certain extent. if it is just \newcommand\z{\mathbb{Z}} that is no problem in any of them, if it's half a million lines of tex commands implementing tikz then it gets trickier.
@baxx yes but they are extremes but the thing is you just never know, you may see a simple article class document that uses no hard looking packages then get half way through and find \makeatletter several hundred lines of trick tex macros copied from this site that are over-writing latex format internals.
|
You want a table that models "stacked" probabilities.
For example, if the chance of any individual attempt's success is 40%, your table might look like this:
\begin{array}{cc}n\text{ to success} & \text{d}100\text{ roll}\\ \hline1 & 01-40\\2 & 41-64\\3 & 65-78\\4 & 79-87\\5 & 88-92\\6 & 93-95\\7 & 96-97\\8 & 98\\9 & 99\\10 & 100\\\end{array}
But how do you generate this table?
Given the probability \$p\$ of succeeding on any given attempt, then the probability of succeeding
on
the \$n^\text{th}\$ attempt is given by $$P(n)=(1-p)^{n-1}\cdot p$$
This is because in order to succeed on the \$n^\text{th}\$ attempt we must first
fail, with a probability \$(1-p)\$, \$n-1\$ times, then succeed (with probability \$p\$).
Thus in the example above we see that \begin{align*}P(1)&=p=0.4\\P(2)&=(1-p)\cdot p =0.24\\P(3)&=(1-p)^2\cdot p \approx0.14\\P(4)&=(1-p)^3\cdot p \approx 0.09\\\vdots\quad& \hspace{2cm}\vdots\end{align*}
But in order to
stack the probabilities we recognize that the "break points"--the highest number in each of the percentile ranges--are given by the sum of all probabilities up to the \$n^\text{th}\$.Luckily, this is just a geometric series: $$\sum_{i=0}^{n-1}{(1-p)^i} = \frac{1-(1-p)^n}{p}$$
With this in hand it's easy to generate a table of "break points" for any given \$p\$:
\begin{array}{c|ccccccc}& \text{d}100\text{ roll}\\n\text{ to success} & p=0.1 & p=0.2 & p=0.3 & p=0.4 & p=0.5 &p=0.6 &p=0.7\\ \hline1 & 01-10 & 01-20 & 01-30 & 01-40 & 01-50 & 01-60 & 01-70\\2 & 11-19 & 21-36 & 31-51 & 41-64 & 51-75 & 61-84 & 71-91\\3 & 20-27 & 37-49 & 52-66 & 65-78 & 76-88 & 85-94 & 92-97\\4 & 28-34 & 50-59 & 67-76 & 79-87 & 89-94 & 95-97 & 98-99\\ \hline5 & 35-41 & 60-67 & 77-83 & 88-92 & 95-97 & 98-99 & 100\\6 & 42-47 & 68-74 & 84-88 & 93-95 & 98 & 100\\7 & 48-52 & 75-79 & 89-92 & 97-97 & 99\\8 & 53-57 & 80-83 & 93-94 & 98 & 100\\\hline9 & 58-61 & 84-87 & 95-96 & 99 & \\10 & 62-65 & 88-89 & 97 & 99 & \\11 & 66-69 & 90-91 & 98 & 100 & \\12 & 70-72 & 92-93 & 99 & & \\\hline13 & 73-75 & 94-95 & 99 & & \\14 & 76-77 & 96 & 99 & & \\15 & 78-79 & 96 & 100 & & \\16 & 80-81 & 97 & & & \\\vdots & \vdots & \vdots\\\end{array}
Notes:
1. In a few places the same single number appears twice. ( p=0.2, n=14,15, for instance.) In these cases multiple break-points round to the same percent-value. You could randomly choose from among n values if this is rolled, simply take the lower, or devise some other scheme.
2. Obviously, there is a non-zero probability that success would take longer than I've indicated, ending my tables at the first appearance of a rounded-to-100 percent-value. However, by construction there is less than a 1/2% chance that any n greater than the last presented could occur. I felt fine leaving the long tail off the table.
|
Coefficients of Cascaded Discrete-Time Systems
In this article, we’ll show how to compute the coefficients that result when you cascade discrete-time systems. With the coefficients in hand, it’s then easy to compute the time or frequency response. The computation presented here can also be used to find coefficients of mixed discrete-time and continuous-time systems, by using a discrete time model of the continuous-time portion [1].
Consider the cascade of two systems, H
1(z) and H 2(z), where
$$H_1(z)=K_1\frac{b_0+b_1z^{-1}+...+b_Nz^{-N}}{1+a_1z^{-1}+...+a_Nz^{-N}} = K_1\frac{b_0z^N+b_1z^{N-1}+...+b_N}{z^N+a_1z^{N-1}+...+a_N }$$
$$H_2(z)=K_2\frac{d_0+d_1z^{-1}+...+d_Mz^{-M}}{1+c_1z^{-1}+...+c_Mz^{-M}} = K_2\frac{d_0z^M+d_1z^{M-1}+...+d_M}{z^M+c_1z^{M-1}+...+c_M }$$
H
1(z) and H 2(z) are assumed to have the same sample rate. H(z) = H 1(z)H 2(z) can be found by multiplying-out the numerator and denominator polynomials of H 1 and H 2. This can be done by hand, but it turns out there is a rule for multiplying polynomials [2]. The rule is as follows: Given two polynomials with coefficients b = [b 0 b 1 … b n] and d = [d 0 d 1 … d n], the coefficients of the product of the polynomials are given by the convolution of b and d.Thus H(z) = H 1(z)H 2(z) has coefficients: numerator coeffs = b ⊛ d (1)
denominator coeffs = a ⊛ c (2)
where ⊛ indicates convolution. The overall gain constant is K
1K 2. For FIR systems, the denominator is unity, so the coefficients of the cascaded systems are b⊛d (which is perhaps no surprise). Example
Here we’ll cascade 2
nd-order lowpass and 1(z) and H 2(z) as follows:
$$H_1(z)=K_1\frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}} $$
where H
1(z) is lowpass with
a= [1 .3695 .1958] b= [1 2 1] K1= .3913
$$H_2(z)=K_2\frac{d_0+d_1z^{-1}+d_2z^{-2}}{1+c_1z^{-1}+c_2z^{-2}} $$
where H
2(z) is band-reject with
c= [1 -1.1074 .8841] d= [1 -1.1756 1] K2= .942
The frequency responses of the two filters are computed using Matlab as follows:
fs= 100; % Hz sample frequency [h1,f]= freqz(K1*b,a,2048,fs); H1= 20*log10(abs(h1)); % dB magnitude response [h2,f]= freqz(K2*d,c,2048,fs); H2= 20*log10(abs(h2)); % dB magnitude response
The responses are plotted in Figure 1. The cascade coefficients are computed using Equations 1 and 2:
bb= conv(b,d) aa= conv(a,c) K= K1*K2
This gives coefficients:
aa = 1.0000 -0.7379 0.6707 0.1098 0.1731 bb = 1.0000 0.8244 -0.3512 0.8244 1.0000 K = 0.3686
The length of aa is 5, so the cascade transfer function is 4
th order. The transfer function is:
$$H(z)=K\frac{bb_0+bb_1z^{-1}+bb_2z^{-2}+bb_3z^{-3}+bb_4z^{-4}}{1+aa_1z^{-1}+aa_2z^{-2}+aa_3z^{-3}+aa_4z^{-4}} $$
(I apologize for the ungainly notation). The frequency response is:
[h,f]= freqz(K*bb,aa,2048,fs); H= 20*log10(abs(h)); % dB magnitude response
The cascaded response is shown in the lower plot of Figure 1.
In case you are interested, the lowpass and band-reject coefficients were calculated using the following Matlab code:
fs= 100; fc= 30; [b,a]= butter(2,2*fc/fs) % lowpass coeffs f0= 15; fU= 16; [d,c]= br_synth1(1,f0,fU,fs) % band-reject coeffs
Here, br_synth1 is a Matlab function developed in an earlier post [3].
Figure 1. Magnitude response of cascaded lowpass and Band-reject filters.
top: Lowpass response |H
1| middle: Band-reject response |H 2| bottom: cascade response
Neil Robertson March, 2018
Appendix Convolution Formulas
The output y[n] of an FIR filter is the discrete convolution of its coefficients b with an input x[n]. We call each of these terms a “sequence”. Discrete convolution is defined as [4]:
$$y[n]=\sum_{k=-\infty}^{\infty}b[k]\cdot x[n-k]$$
$$=\sum_{k=0}^N b[k]\cdot x[n-k]$$
where N= length(b)-1 and indices of x that are negative or greater than length(x)-1 are ignored.
Cascading two FIR filters uses the same convolution formula. However, the two sequences are the coefficients of the filters. When multiplying polynomials, the two sequences are the polynomial coefficients. In these cases, since we are not dealing with an input signal x, we replace x with another symbol. Thus the convolution of two discrete sequences b and d is:
$$y[n]=\sum_{k=0}^N b[k]\cdot d[n-k]$$
Where, again, N= length(b)-1. Convolution is commutative, so we can also write the above equation as:
$$y[n]=\sum_{k=0}^M d[k]\cdot b[n-k]$$
where M= length(d)-1.
References
1.Robertson, Neil, https://www.dsprelated.com/showarticle/1055.php
2.Smith, Julius O.,
Mathematics of the Discrete Fourier Transform (DFT),
3.Robertson, Neil, https://www.dsprelated.com/showarticle/1131.php
4.Lyons, Richard G.,
Understanding Digital Signal Processing, 2 nd Ed., Prentice Hall, 2004, section 5.9.1.
Previous post by Neil Robertson:
Design IIR Filters Using Cascaded Biquads
Next post by Neil Robertson:
Phase or Frequency Shifter Using a Hilbert Transformer
To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
|
I know the answer to the above question, but I have a question on some of the reasoning.
The way I know how to solve it is $$\lim_{x \rightarrow 0}f(x) = \lim_{x \rightarrow 0}\left(f(x)\cdot \frac{x^2}{x^2}\right) = \lim_{x \rightarrow 0}\left(\frac{f(x)}{x^2}\cdot x^2\right) = \left(\lim_{x \rightarrow 0}\frac{f(x)}{x^2}\right)\left(\lim_{x \rightarrow 0}x^2\right) = 5\cdot0 = 0.$$
I saw another solution elsewhere that gets the right answer, but I am unsure if the steps are actually correct. \begin{align*} &\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = 5 \\ \Longrightarrow &\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5 \\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot \lim_{x \rightarrow 0}x^2\\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot 0 = 0. \end{align*}
My issue is with that first step. I know that $\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a}f(x)}{\lim_{x \rightarrow a}g(x)}$, but only when $\lim_{x \rightarrow a}g(x) \neq 0$. Since $\lim_{x \rightarrow 0}x^2 = 0$, wouldn't this invalidate the above work? However, it still got the same answer, so my real question is why did it work and when will it work in general?
EDIT: Does anyone have a nice example for when the logic in the second method doesn't work?
|
I have the feeling, that this is a superbly stupid question. And I also have the feeling that it is quite possible, that my answer will be wrong.
First of all the parts-per-notation should be avoided as it is not compliant with SI and highly ambiguous. Secondly, it generally may only refer to a unitless number - a molar concentration has a unit ($\mathrm{mol/L}$).
The here used unit may refer to a couple of things. In general its meaning should most likely refer to $\mathrm{PPTV = 10^{-12}L/L}$. Sometimes the use of $\mathrm{PPTV = 10^{-12}g/L}$
From here you can probably make your best guesses, determining how many moles of compound do you have:\begin{align}pV &=n\mathcal{R}T\\n &= \frac{pV}{\mathcal{R}T}\\n &= \frac{101325~\mathrm{Pa}\cdot 261\cdot10^{-12}~\mathrm{L}}{8.314~\mathrm{\frac{J}{mol\cdot K}}283.15~\mathrm{K}}\\n &= 1.12\cdot10^{-8}~\mathrm{mol} &\implies c&= 1.12\cdot10^{-8}~\mathrm{\frac{mol}{L}}\end{align}
If you have the mass given, then you need to know which gas it is, as $n=\frac{m}{M}$.
Another possible use could be the particles by volume notation, so it could also refer to $\mathrm{PPTV = 10^{-12}/m^3}$. (Apparently this is somewhat common in ecological air measurements.) Here you simply have to use $n=\frac{N}{\mathcal{N}_\mathrm{A}}$ to figure out the concentration in $\mathrm{mol/L}$.
|
In Poincare Gauge Theories spin connections take the role of the Levi-Civitta connection of GR for defining covariant derivatives. The GR can be formulated as a Poincare Gauge Theory which is called the Teleparallel equivalent of GR. In this theory the spin connection is decomposed into a flat spin connection $A_{ab\mu}=-A_{ba\mu}$ and a contortion $K_{ab\mu}=1/2(T_{ba\mu}+T_{\mu ab}-T_{ab\mu})$.
The Weitzenböck connection $\Gamma^{\beta}_{\phantom{\beta}\mu\nu} = h^{\beta}_i\delta_{\nu}h^i_{\mu}$ is normally used for the flat spin connection. However, a general flat spin connection is only defined by the condition that its curvature is zero \begin{equation} R^a_{\phantom{a}b\mu\nu} = \partial_{\mu}A^a_{\phantom{a}b\nu} - \partial_{\nu}A^a_{\phantom{a}b\mu} + A^a_{\phantom{a}c\mu}A^c_{\phantom{c}b\nu} - A^a_{\phantom{a}c\nu}A^c_{\phantom{c}b\mu} = 0. \end{equation} These equations do not define a single connection. Instead, a set of spin connections comply with this condition.
To understand how large this subset is, the number of independent curvature components have to be counted. The curvature tensor has at most 36 independent components. However, the Bianchi identities show that not all are independent. The first Bianchi identities \begin{equation} D_{\rho}T^a_{\phantom{a}\mu\nu} + D_{\mu}T^a_{\phantom{a}\nu\rho} + D_{\nu}T^a_{\phantom{a}\rho\mu} = R^a_{\phantom{a}\rho\mu\nu} + R^a_{\phantom{a}\mu\nu\rho} + R^a_{\phantom{a}\nu\rho\mu} \end{equation} sets constraints for the Torsion components, but the second Bianchi identities \begin{equation} D_{\rho}R^a_{\phantom{a}b\mu\nu} + D_{\mu}R^a_{\phantom{a}b\nu\rho} + D_{\nu}R^a_{\phantom{a}b\rho\mu} = 0 \end{equation} with $D_{\mu}$ a Weitzenböck covariant derivative, do set constraints in the components of the curvature.
Which is the final number of independent curvature components? How large is the resulting subset of flat spin connections?
Thank you for your help!
|
Linear models describe a continuous response variable as a function of one or more predictor variables. They can help you understand and predict the behavior of complex systems or analyze experimental, financial, and biological data.
Linear regression is a statistical method used to create a linear model. The model describes the relationship between a dependent variable \(y\) (also called the response) as a function of one or more independent variables \(X_i\) (called the predictors). The general equation for a linear model is:
\[y = \beta_0 + \sum \ \beta_i X_i + \epsilon_i\]
where \(\beta\) represents linear parameter estimates to be computed and \(\epsilon\) represents the error terms.
There are several types of linear regression:
Simple linear regression:models using only one predictor Multiple linear regression:models using multiple predictors Multivariate linear regression:models for multiple response variables Generate predictions Compare linear model fits Plot residuals Evaluate goodness-of-fit Detect outliers
To create a linear model that fits curves and surfaces to your data, see Curve Fitting Toolbox. To create linear models of dynamic systems from measured input-output data, see System Identification Toolbox. To create a linear model for control system design from a nonlinear Simulink model, see Simulink Control Design.
|
Some existence results for dynamical systems on non-complete Riemannian manifolds
DOI: http://dx.doi.org/10.12775/TMNA.1999.008
Abstract
Let $\mathcal M^*$ be a non-complete Riemannian manifold with bound-ed topological
boundary and $V: \mathcal M \to \mathbb R$ a $C^2$ potential function subquadratic at infinity. In this paper we look for curves $x: [0,T]\to\mathcal M$ having prescribed period $T$ or joining two fixed points of $\mathcal M$, satisfying the system $$ D_t (\dot x(t))=-\nabla_R V(x(t)), $$ where $D_t(\dot x(t))$ is the covariant derivative of $\dot x$ along the direction of $\dot x$ and $\nabla_R V$ the Riemannian gradient of $V$. We assume that $V(x) \to -\infty$ if $d(x,\partial\mathcal M)\to 0$ and, in the periodic case, suitable hypotheses on the sectional curvature of $\mathcal M$ at infinity. We use variational methods in addition with a penalization technique and Morse index estimates.
boundary and $V: \mathcal M \to \mathbb R$ a $C^2$ potential function subquadratic
at infinity.
In this paper we look for curves $x: [0,T]\to\mathcal M$ having prescribed period $T$ or joining two
fixed points of $\mathcal M$, satisfying the system
$$
D_t (\dot x(t))=-\nabla_R V(x(t)),
$$
where $D_t(\dot x(t))$ is the covariant derivative of $\dot x$ along the direction
of $\dot x$ and $\nabla_R V$ the Riemannian gradient of $V$.
We assume that $V(x) \to -\infty$ if $d(x,\partial\mathcal M)\to 0$ and, in the periodic case,
suitable hypotheses on the sectional curvature of $\mathcal M$ at infinity.
We use variational methods in addition with a penalization technique and Morse
index estimates.
Keywords
Variational methods; Riemannian manifold; Morse index; sectional curvature
Full Text:FULL TEXT Refbacks There are currently no refbacks.
|
I would proceed as follows:\begin{align*}x_t &= x_{t-\delta t} + \alpha (\beta - x_{t-\delta t}) \delta t + \sigma x_{t-\delta t}^\gamma (w_t - w_{t-\delta t}),\\x_t &= \max (x_t, \ 0),\end{align*}where $w_t - w_{t-\delta t}$ is a normal random variable with mean $0$ and variance $\delta t$, which can be obtained by an independent draw for each ...
For GBM you can show it by theory (see Kloeden) or show it empirically as follows. You can observe that EM has strong order of convergence equal to 0.5, whereas Milstein's is 1.0. Not always the latter is superior. E.g. simple EM beats simple Mistein for Heston. In this case is good to use adapted schemes (Giles) for instance, especially dealing with jumps....
In arithmetic brownian, drift does not depend on the previous price, so it is simply $\mu \Delta t$ as you have done. It depends on the previous price in geometric brownian though. Let’s recall the GBM equation:$dS_t=\mu S_t dt +\sigma S_t dB_t$Discretising:$\Delta S_t=\mu S_t \Delta t + \sigma S_t \sqrt{\Delta t} N[0,1]$$S_{t+1}-S_t=\mu S_t \Delta t ...
I figured it out. The problem is just, that I am not taking the same driving brownian motion. That would mean, if I am calculating $E[\bar X^{\delta}]$ and $E[\bar X^{1/2\delta}]$. The sample of path is complete different and thus not comparable. For weak convergence, one need to be sure, that the sample path are almost surely the same. As a result, the ...
first, there is a formula for the continuously monitored case.second, if you use log coordinates the Euler discretization is exact so this should be done.third, the convergence for discretely monitored to continuously is actually very slow so you will need a lot of steps.fourth, it's actually better to draw the hitting time to the barrier rather than ...
$r-\frac{\sigma^2}{2}$ for the drift only applies to the log-returns. The Euler discretisation simply discretises the SDE directly. You'd use the risk-free ratefor you drift under the risk-neutral measure for your question.For your reference:Please read the wikipedia for more details.
|
ISSN:
1531-3492
eISSN:
1553-524X
All Issues
Discrete & Continuous Dynamical Systems - B
June 2014 , Volume 19 , Issue 4
Select all articles
Export/Reference:
Abstract:
RS feedback models have been successful in explaining the observed phenomenon of clustering in autonomous oscillation in yeast, but current models do not include the biological reality of dynamical delay and do not have the related property of quorum sensing. Here an RS type ODE model for cell cycle feedback, including an explicit term modeling a chemical feedback mediating agent, is analyzed. New dynamics include population dependent effects: subcritical pitchfork bifurcations, and quorum sensing occur. The model suggests new experimental directions in autonomous oscillation in yeast.
Abstract:
This survey focuses on the most important aspects of the mathematical theory of population genetic models of selection and migration between discrete niches. Such models are most appropriate if the dispersal distance is short compared to the scale at which the environment changes, or if the habitat is fragmented. The general goal of such models is to study the influence of population subdivision and gene flow among subpopulations on the amount and pattern of genetic variation maintained. Only deterministic models are treated. Because space is discrete, they are formulated in terms of systems of nonlinear difference or differential equations. A central topic is the exploration of the equilibrium and stability structure under various assumptions on the patterns of selection and migration. Another important, closely related topic concerns conditions (necessary or sufficient) for fully polymorphic (internal) equilibria. First, the theory of one-locus models with two or multiple alleles is laid out. Then, mostly very recent, developments about multilocus models are presented. Finally, as an application, analysis and results of an explicit two-locus model emerging from speciation theory are highlighted.
Abstract:
This article was accidentally posted online but only to be discovered that the same article had been published (see [1]) in the previous issue of the same journal. Thus this publication is retracted. The Editorial Office offers apologies for the confusion and inconvenience it might have caused.
Abstract:
We consider sensor array imaging for simultaneous noise blended sources. We study a migration imaging functional and we analyze its sensitivity to singular perturbations of the speed of propagation of the medium. We consider two kinds of random sources: randomly delayed pulses and stationary random processes, and three possible kinds of perturbations. Using high frequency analysis we prove the statistical stability (with respect to the realization of the noise blending) of the scheme and obtain quantitative results on the image contrast provided by the imaging functional, which strongly depends on the type of perturbations.
Abstract:
In this paper, we consider a class of epidemic models described by five nonlinear ordinary differential equations. The population is divided into susceptible, vaccinated, exposed, infectious, and recovered subclasses. One main feature of this kind of models is that treatment and vaccination are introduced to control and prevent infectious diseases. The existence and local stability of the endemic equilibria are studied. The occurrence of backward bifurcation is established by using center manifold theory. Moveover, global dynamics are studied by applying the geometric approach. We would like to mention that in the case of bistability, global results are difficult to obtain since there is no compact absorbing set. It is the first time that higher (greater than or equal to four) dimensional systems are discussed. We give sufficient conditions in terms of the system parameters by extending the method in Arino et al. [2]. Numerical simulations are also provided to support our theoretical results. By carrying out sensitivity analysis of the basic reproduction number in terms of some parameters, some effective measures to control infectious diseases are analyzed.
Abstract:
A damped stochastic beam equation driven by a Non-Gaussian Lévy process is studied. Under appropriate conditions, the existence theorem for a unique global weak solution is given. Moreover, we also show the existence of a unique invariant measure associated with the transition semigroup under mild conditions.
Abstract:
We study in this article the stochastic Zakharov-Kuznetsov equation driven by a multiplicative noise. We establish, in space dimensions two and three the global existence of martingale solutions, and in space dimension two the global pathwise uniqueness and the existence of pathwise solutions. New methods are employed to deal with a special type of boundary conditions and to verify the pathwise uniqueness of martingale solutions with a lack of regularity, where both difficulties arise due to the partly hyperbolic feature of the model.
Abstract:
A virus dynamics model for HIV or HBV is studied, which incorporates saturation effects of immune responses and an intracellular time delay. With the aid of persistence theory and Liapunov method, it is shown that the global stability of the model is totally determined by the reproductive numbers for viral infection, for CTL immune response, for antibody immune response, for antibody invasion and for CTL immune invasion. The results preclude the complicated behaviors such as the backward bifurcations and Hopf bifurcations which may be induced by saturation factors and a time delay.
Abstract:
In this paper, using an approach of Lyapunov functional, we establish the complete global stability of a multi-group SIS epidemic model in which the effect of population migration among different regions is considered. We prove the global asymptotic stability of the disease-free equilibrium of the model for $R_0\leq 1$, and that of an endemic equilibrium for $R_0>1$. Here $R_0$ denotes the well-known basic reproduction number defined by the spectral radius of an irreducible nonnegative matrix called the next generation matrix. We emphasize that the graph-theoretic approach, which is typically used for multi-group epidemic models, is not needed in our proof.
Abstract:
We justify some characterizations of the ground states of spin-1 Bose-Einstein condensates exhibited from numerical simulations. For ferromagnetic systems, we show the validity of the single-mode approximation (SMA). For an antiferromagnetic system with nonzero magnetization, we prove the vanishing of the $m_F=0$ component. In the end of the paper some remaining degenerate situations are also discussed. The proofs of the main results are all based on a simple observation, that a redistribution of masses among different components will reduce the kinetic energy.
Abstract:
We provide sufficient conditions for the existence of limit cycles for the Floquet differential equations $\dot {\bf x}(t) = A{\bf x}(t)+ε(B(t){\bf x}(t)+b(t))$, where ${\bf x}(t)$ and $b(t)$ are column vectors of length $n$, $A$ and $B(t)$ are $n\times n$ matrices, the components of $b(t)$ and $B(t)$ are $T$--periodic functions, the differential equation $\dot {\bf x}(t)= A{\bf x}(t)$ has a plane filled with $T$--periodic orbits, and $ε$ is a small parameter. The proof of this result is based on averaging theory but only uses linear algebra.
Abstract:
We consider a finite discrete nonlinear Schrödinger equation with localized forcing, damping, and nonautonomous perturbations. In the autonomous case these systems are shown numerically to have multiple attracting spatially localized solutions. In the nonautonomous case we study analytically some properties of the pullback attractor of the system, assuming that the origin of the corresponding autonomous system is hyberbolic. We also see numerically the persistence of multiple localized attracting states under different types of nonautonomous perturbations.
Abstract:
Much recent work has focused on persistence for epidemic models with periodic coefficients. But the case where the infected compartments satisfy a delay differential equation or a partial differential equation does not seem to have been considered so far. The purpose of this paper is to provide a framework for proving persistence in such a case. Some examples are presented, such as a periodic SIR model structured by time since infection and a periodic SIS delay model.
Abstract:
In this paper, we consider a time-delayed and nonlocal population model with migration and relax the monotone assumption for the birth function. We study the global dynamics of the model system when the spatial domain is bounded. If the spatial domain is unbounded, we investigate the spreading speed $c^*$, the non-existence of traveling wave solutions with speed $c\in(0,c^*)$, the existence of traveling wave solutions with $c\geq c^*$, and the uniqueness of traveling wave solutions with $c>c^*$. It is shown that the spreading speed coincides with the minimal wave speed of traveling waves.
Abstract:
Stochastic averaging for a class of stochastic differential equations (SDEs) with fractional Brownian motion, of the Hurst parameter $H$ in the interval $(\frac{1}{2},1)$, is investigated. An averaged SDE for the original SDE is proposed, and their solutions are quantitatively compared. It is shown that the solution of the averaged SDE converges to that of the original SDE in the sense of mean square and also in probability. It is further demonstrated that a similar averaging principle holds for SDEs under stochastic integral of pathwise backward and forward types. Two examples are presented and numerical simulations are carried out to illustrate the averaging principle.
Abstract:
This paper is devoted to the existence of pullback attractors for the process $\{U(t,\tau)\}_{t\geq \tau}$ associated with the three dimensional non-autonomous planetary geostrophic viscous equations of large-scale ocean circulation. We first prove the existence of pullback absorbing sets in $H$ and $V$ for the process $\{U(t,\tau)\}_{t\geq \tau}$ associated with (1)-(8), and then we prove the existence of a pullback attractor in $H$ by the Sobolev compactness embedding theorem. Finally, we obtain the existence of a pullback attractor in $V$ for the process $\{U(t,\tau)\}_{t\geq \tau}$ associated with (1)-(8) by verifying the pullback $\mathcal{D}$ condition $(PDC)$.
Readers Authors Editors Referees Librarians More Email Alert
Add your name and e-mail address to receive news of forthcoming issues of this journal:
[Back to Top]
|
Search
Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
|
Trigonometry Complex Numbers Geometric Series Integrals of Complex Functions and Integration by parts
A series of the form $a b^k, a b^{k+1}, \ldots, a b^{l}$, where $a$ and $b$ can be any {\em complex} number is called a geometric series with $l-k+1$ terms. For example, $1,\frac{1}{2},\frac{1}{4},\ldots$ is an infinite geometric series with $a=1$, $b=\frac12$. You may have seen these before, but in this class often we will be interested in the case when $b$ (and $a$) are complex numbers. Luckily, nothing changes from when $a$ and $b$ are just real numbers.We will particularly be interested in writing a closed form expression for the sum of consecutive terms of a geometric series. The most general result that you should(1)
A few special cases of the above general result are important. Just convince yourself that these are true(2)
The following identity is also true, although we will not use this often in this class.(3)
Here are couple of examples to try out
1. For any two given integers $k$ and $M$, what is$\displaystyle{\sum_{n=0}^{M-1} e^{\frac{j 2 \pi k n}{M}}}$?
2. Just for intellectual curiosity - Can you prove the results in Equation 1 and Equation 3 ?
|
Distribution of charge
within the region where the field is located, is obviously uniquely defined, because it's just
$$\rho=\epsilon_0\nabla \vec{E}$$
However, if you cut out a region of space, and want to predict the contents of this region based only on the field outside, you can't do it in a unique way.
The reason is that you have too many degrees of freedom. If you decompose the field outside into the inhomogeneous contribution of the charges outside, plus the extra that was supposed to be caused by the cut out region (and obeys the Laplace equation outside the region), then the second contribution (the one with $\nabla \vec{E}=0$ outside the region) can be exactly reproduced simply by putting the correct charges
on the surface of the boundary! That's what many call the holographic principle. You see now that there is one unique surface solution for every bulk distribution - but however differently you choose to distribute the charges inside the region, you can always rearrange the surface charge to accomodate the same external field. That's how conductive objects "mask" whatever you put inside them.
So... charges that occupy the space where the field is known, are uniquely defined. Charges in regions where the field is unknown, are arbitrary. Charges on the boundary of the known region are uniquely defined for each particular choice of distribution inside - assuming no charge is inside, and all field is caused by the surface, is simply one of the infinite number of solutions.
|
Infinite product of complex numbers Set
context $(z_i)$ … Sequence($\mathbb C$) definition $\prod_{i=1}^\infty z_i\equiv \mathrm{lim}_{n\to\infty}\prod_{i=1}^n z_i$ Discussion todo: Interestingly, I think I see the nLab doesn't want to allow e.g. $\prod_{n=1}^\infty (17-n)$ to be zero. (Reference below)
I recon infinite products may arise when x is written as $111\cdots11x$ and 1 is represented as something.
see the context of the “prod-step to exp-sum” formula blow. can Euler products arise in this way?
And they may arise ans determinants of infinite dimensional operators, i.e. products of eigenvalues.
Theorems Limit of a sequence
For nonzero $a_n$ and any $N,M,K$, we have
$\lim_{n\to\infty}a_n=a_N\cdot\prod_{n=N}^\infty\frac{a_{n+1}}{a_n}$ $\lim_{n\to\infty}a_n=a_M+\sum_{n=M}^\infty(a_{n+1}-{a_{n}})=a_M+\sum_{n=M}^\infty\left(\frac{a_{n+1}}{a_n}-1\right)\,a_n$
So both are equal. This might be useful for rewriting and computing $\prod_{n=d}^\infty f(n)$, or even find the limit. Mathematica can find the $a$-sequence with $a_{n+1}=f(n)\cdot a_n$, see (“RSolve”).
Also, from the sum formula with $a_n=\prod_{k=K}^{n-1}b_k$, we get
$\prod_{k=K}^\infty b_k=\lim_{n\to\infty}a_n=\prod_{k=K}^{M-1}b_k+\sum_{n=M}^\infty(b_n-1)\,\prod_{k=K}^{n-1}b_k$
$\prod_{k=0}^\infty b_k = \prod_{k=0}^{M-1}b_k + \sum_{n=M}^\infty(b_n-1)\,\prod_{k=0}^{n-1}b_k$ b[n_] = 1 + 1/n!; Ks = 0; Ke = 100; Mid = 4; Product[b[k], {k, Ks, Ke}] Ks = 0; Ke = 100; Mid = 4; Product[b[k], {k, Ks, Mid - 1}] + Sum[(b[k] - 1) Product[b[k2], {k2, Ks, k - 1}], {k, Mid, Ke}] todo: I think this formula holds also for finite products, i.e. when $\infty$ is replaces by some natural number $K_2>M>K$. prod-step to exp-sum Note that this isn't of the form as in this entry, because the terms depend on the upper limit $N$ that goes to infinite
$\lim_{N\to\infty} \prod_{n=0}^{N-1} \left(1+\frac{x_n}{N}\right)= \exp\left(\lim_{N\to\infty}\frac{1}{N}\sum_{n=0}^{N-1}x_n\right)$
I've seen this pop up in the derivation of the basic path integral. Where else does it pop up? Why, if $x_n$ doesn't depend on $N$, does this turn to an $\exp(\int something)$? I think they join some limits to a single one. I think that relates to what Ron said about quantization procedures. h = (b - a)/m; f[X_] = c X^2; p = Product[ 1 + f[i] h , {i, a, b, h}] // simple a = 0; b = 7; c = 3; Limit[p, m -> \[Infinity]] Exp[Integrate[c X^2, {X, 0, 7}]] Jacobi triple product
$\prod_{m=1}^\infty \left( 1 - q^{2m}\right)\left( 1 + w^{2}q^{2m-1}\right)\left( 1 + w^{-2}q^{2m-1}\right)= \sum_{n=-\infty}^\infty w^{2n}q^{n^2}.$
If the sum is a partition function, what does $\prod_{m=1}^\infty$ represent in terms of (algebraic) operations on state spaces?
Also
$\prod_{n=0}^{\infty}\left(1 + x^{2^n}\right) = \frac{1}{1-x}$
|
What your book states is not generally true. Two counter-examples:
Ammonia ($\ce{NH3}$) will liquefy at room temperature and a pressure of approximately $\pu{10bar}$. (
CRC Handbook of Chemistry and Physics 44th ed; information as cited on Wikipedia’s data page).
Butane ($\ce{C4H10}$) will liquefy at room temperature and a pressure of just over $\pu{2bar}$. (Same source as cited on Wikipedia’s data page).
Especially the second is well-known to all those who have a butane lighter of transparent material. If it is half-full, a clear distinction between the liquid and gaseous phases can be seen inside at room temperature. Naturally, the inside of the lighter must be under positive pressure for gas to flow out if a flame is required.
On the other hand, there are a great many gases that cannot be liquefied at room temperature by applying pressure. One of these is nitrogen ($\ce{N2}$). In this case, applying enough pressure will make the sample transform directly from the gaseous to a solid phase.
The difference between nitrogen and ammonia/butane is their thermodynamic data, most importantly their critical point. The critical point defines the last position of the liquid/solid separation of a phase diagram. If the temperature is higher than the critical temperature, no liquid phase will be observed under any compression. Likewise, if the pressure is higher than the critical pressure, a substance will not liquefy even under extreme cooling (it will go straight to the solid phase).
The critical points of the compounds mentioned herein are:
$$\begin{array}{lrr}\hline\text{Compound} & \vartheta_\text{crit} [\mathrm{^\circ C}] & p_\text{crit} [bar] \\\hline\ce{NH3} & 132.4 & 112.8\phantom{0} \\\ce{C4H10} & 152\phantom{.0} & 37.96\\\ce{N2} & -146.9 & 33.90\\ \hline \end{array}$$
As you can see, the critical pressures of all compounds are well above standard pressure (hence they can be liquefied under reduced temperature). Nitrogen’s critical temperature is below room temperature, so it cannot be liquefied by compression; the others can.
|
This question already has an answer here:
I have to show that for any $b >1$, we have $$ b^n > n$$ for all $n$ sufficiently large, using only very basic analysis (no calculus). My attempt is as follows.
We know that $b^{n+1} - b^n = b^n(b-1)$. For $n$ sufficiently large, say $$n \geq N = \left\lceil \frac{\ln(2/(b-1))}{\ln b} \right\rceil + 1,$$ we have $$ b^{n+1} - b^n > 2.$$
Now let $\Delta = N - b^N$. Then for any $j\geq 1$, we have $$ b^{N+j} = (b^{N+j} - b^{N+j-1}) + \ldots + (b^{N+1} - b^N) + b^N > 2j + b^N = 2j + N - \Delta = N+j + (j - \Delta).$$ Thus we have $b^n > n$ for any $n \geq N + |\Delta|$.
This works, but it seems messy.
Is there is better way? I know induction is usual for this type of problem, but establishing the base case for generic $b$ seems difficult.
|
I am writing this answer to try and learn something about the various scattering processes for myself, so hopefully another more detailed and more sophisticated answer will also be posted, that both I (and the OP, of course!) can learn from.
Bhabha scattering is the electron-positron scattering interaction involving an electron and a positron :
${ e^{+}e^{-}\rightarrow e^{+}e^{-}}$
The two-order Feynman diagrams describing this interaction are an annihilation process and a scattering process.
The Bhabha scattering rate is used as a luminosity monitor in electron-positron collider experiments. In scattering theory and accelerator physics, luminosity (L) is the ratio of the number of events detected (N) in a certain time (t) to the interaction cross-section (σ):
${ L={\frac {1}{\sigma }}{\frac {dN}{dt}}}$
Luminosity values are useful in determine the performance and efficency of a particle accelerator, as the greater the integrated luminosity, the more data emerges from the (often expensive) experiment.
Ionization, (as I'm sure you already know, apologies) is the mechanism by which an atom or a molecule acquires a negative or positive charge by gaining or losing electrons to form ions. Ionization can result from the loss of an electron after collisions with subatomic particles, collisions with other atoms, molecules and ions, or through the interaction with light.
In contrast to Bhabha scattering , ${ e^{+}e^{-}\rightarrow e^{+}e^{-}}$, Møller scattering ${\ e^{-}e^{-}\longrightarrow e^{-}e^{-}}$ denotes electron-electron scattering. The electron interaction that is idealized in Møller scattering forms the theoretical basis of many familiar phenomena such as the repulsion of electrons in the helium atom.
Again, as I am sure you are aware, we need to include the both diagrams to complete the calculation of Møller scattering amplitudes. These illustrations give an indication of some of the terms included in the scattering amplitude, but to calculate a measurable quantity, we need to pick a reference frame, usually the COM frame, and assign helicities to the particles or average/sum over all possible spin states.
|
I am thinking about the deletion error correcting codes for quantum information.
In classical information theory, there exist some deletion error correcting codes. An easy example is the following situation:
Alice prepares a set $\{ 00,11 \}$ and sends an element of the set to Bob. Here Bob knows the set. One deletion error occurs at the message so $00$ becomes $0$ and $11$ becomes $1$. Then Bob receives the message which is $0$ or $1$. Bob can correctly get the message Alice sent because he can restore $0$ to $00$ and $1$ to $11$.
If Alice prepared a set $\{ 01,11 \}$, Bob could not correctly get the messsage when he received $1$. That is why the set $\{ 00,11 \}$ is called a deletion error correcting code.
I would like to construct the quantum version of deletion error correcting codes by replacing the deletion error in the classical theory to the partial trace in the quantum information theory. The situation I expect is as follows:
Alice prepares a quantum state $\left | \psi \right >\in \mathbb{C}^{2 \otimes n}$. This state is represented as a density matrix $\rho=\left | \psi \right >\left < \psi \right |$. One deletion error occurs at $\rho$ so that the state becomes $\rho'=Tr_i(\rho)$. Then Bob can perform some unitary operations and some measurements to obtain $\rho$. I wonder this is possible or not.
If Alice prepares a separable state such as $\left | \psi \right >=\left | 0\right >\otimes \left | 0\right >\in \mathbb{C}^{2 \otimes 2}$, it can obviously be restored by preparing $\left | 0\right >$. This is not interesting. So I would like to construct some non-trivial solutions, which means Alice prepares an entangled state. I tried to find a specific example, but I could not find it at all.
Would someone give me an example to satisfy the situation above? Perhaps, such states do not exist. If so, I would like to know the reason.
|
I've read that any real skew-symmetric matrix $A$, where $A^T = -A$, can be brought into block diagonal form
$ A = Q \, \Sigma \, Q^T = \left( \begin{array}{ccccl} \vec{q}_1 & \vec{q}_2 & \vec{q}_3 & \vec{q}_4 & \dots \end{array} \right) \, \left( \begin{array}{ccccl} 0 & \lambda_1 & 0 & 0 & \dots \\ -\lambda_1 & 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & \lambda_2 & \dots \\ 0 & 0 & -\lambda_2 & 0 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right) \, \left( \begin{array}{c} \vec{q}^T_1 \\ \vec{q}^T_2 \\ \vec{q}^T_3 \\ \vec{q}^T_4 \\ \vdots \end{array} \right)$
where $\vec{q}_i$ are real, orthogonal column vectors.
But also, the non-zero eigenvalues of $A$ are purely imaginary and occur in pairs $\pm i \lambda_i$, which are the same $\lambda_i$ as occur in the blocks of $\Sigma$. We could then also write $A$ in complex diagonal form
$ A = U \, D \, U^\dagger = \left( \begin{array}{ccccl} \vec{u}_1 & \vec{u}_2 & \vec{u}_3 & \vec{u}_4 & \dots \end{array} \right) \, \left( \begin{array}{ccccl} i\lambda_1 & 0 & 0 & 0 & \dots \\ 0 & -i\lambda_1 & 0 & 0 & \dots \\ 0 & 0 & i\lambda_2 & 0 & \dots \\ 0 & 0 & 0 & -i\lambda_2 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right) \, \left( \begin{array}{c} \vec{u}^\dagger_1 \\ \vec{u}^\dagger_2 \\ \vec{u}^\dagger_3 \\ \vec{u}^\dagger_4 \\ \vdots \end{array} \right)$
where $\vec{u}_i$ are complex, orthogonal column vectors.
My question is what is the relationship between the complex eigenvectors $\vec{u}_i$ and the real vectors $\vec{q}_i$ that bring $A$ into block diagonal form? And how can I show the link?
|
Let $\mu_4 = E(X-\mu)^4$. Then, the formula for the SE of $s^2$ is:
$$se(s^2) = \sqrt{ \frac{1}{n}\left(\mu_4 -\frac{n-3}{n-1} \sigma^4\right)}$$This is an exact formula, valid for any sample size and distribution, and is proved on page 438, of Rao, 1973, assuming that the $\mu_4$ is finite. The formula you gave in your question applies only to Normally distributed data.
Let $\hat{\theta} = s^2$. You want to find the SE of $ g(\hat{\theta})$, where $g(u) = \sqrt{u}$.
There is no general exact formula for this standard error, as @Alecos Papadopoulos pointed out. However, one can drive an approximate (large sample) standard error by means of the delta method. (See Wikipedia entry for "delta method").
Here's how Rao, 1973, 6.a.2.4 put it. I include the absolute value indicators, which he incorrectly omitted.
$$se(g(\hat{\theta})) \approx |g'(\hat\theta)|\times se(\hat{\theta})$$where $g'$ is the first derivative.
Now for the square root function $g$
$$g'(u) = \frac{1}{2\thinspace u^{1/2}}$$
So:
$$se(s)\approx \frac{1}{2 \sigma} se(s^2)$$
In practice I would estimate the standard error by the bootstrap or jackknife.
Reference:
CR Rao (1973) Linear Statistical Inference and its Applications 2nd Ed, John Wiley & Sons, NY
|
What if we make a tunnel through the earth and drop a ball or any other body into it? How would the gravity behave on the ball, and , will the ball pop-up on the other side? If yes, then how much time will it take? A question, which is usually asked in competitive exams and for general interest.
This article is a descriptive analysis of bodies inside the tunnel through earth. To do so, we assume no- frictional forces between the earth’s inner surface and the body — so there is no dissipation in the energy of the body thrown to the tunnel. It’s also assumed that the body is leaved at rest from one end of the tunnel (i.e., from the surface of earth) and the density of Earth is uniform (it’s not, actually).
When the body is on the surface of Earth, the acceleration due to gravity
g is about $ 9.8 m/s^2$ derived from the formula $ g= G \dfrac{M}{R^2}$ , where G represents the universal gravitational constant (with value $ G= 6.67 \times 10^{-11} \mathrm{Newton-metre^2-kg^{-2}}$ ) M denotes the total mass of the earth and R, radius of earth. Now as body leaves the surface and enters into the Earth — the value of g decreases linearly by the formula $ g= G \dfrac{M}{R^2} \left( {1-\dfrac{r}{R}} \right)$ where, r denotes the depth traveled by the body vertically. It’s clear that as the body reaches at half of the radius of Earth (i.e., r=R/2 ) and to the center of earth, ( r =R), the gravitational acceleration becomes half and then ceases to zero . In other words, the weight reduces to a half and then zero at these points.
Now, the force acting on the body is
F=mass * acceleration = m *(-g’). Here g’ being the reduced gravitational acceleration working on the body. Now, as g’=g'(r) ( g’ is a function of r), we can easily have
$ F = m \times -g'(r) = -kr$ where
k being a constant which depends on m, M & R. Now, as F=-kr, the force working on the body is attractive and directly proportional to the displacement from the reference point. Such a body executes Simple Harmonic Motion. So the body we put inside, will do a simple harmonic motion through the tunnel.
The time period for this system can be given by $ T= 2 \pi \sqrt{\dfrac{m}{k}} = 2 \pi \sqrt{\dfrac{mR}{mg}}$ (Remember; –
mg=-kR & -mg’=-kr)
Thus, $ T=2 \pi \sqrt{\dfrac{R}{g}}= 5068 \mathrm{seconds}$ or 84.47 minutes. This tells that the body will pop-up on the other side after 42.235 minutes after it was released from one side and will return back after another 42.235 minutes.
Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word.
How to write better comments?
|
We present state sums for quantum link invariants arising from the representation theory of U_q(\mathfrak{gl}_{N|M}). We investigate the case of the N-th exterior power of the standard...
Pages
We provide a finite dimensional categorification of the symmetric evaluation of \mathfrak{sl}_N-webs using foam technology. As an output we obtain a symmetric link homology theory categorifying...
Using a modified foam evaluation, we give a categorification of the Alexander polynomial of a knot. We also give a purely algebraic version of this knot homology which makes it appear as the...
Recently, we constructed the first-principle derivation of the holographic dual of N=4 SYM in the double-scaled γ-deformed limit directly from the CFT side. The dual fishchain model is a novel...
We study various non-perturbative approaches to the quantization of the Seiberg-Witten curve of {\cal N}=2, SU(2) super Yang-Mills theory, which is closely related to the modified Mathieu...
We explore the heat current in the quantum Hall edge at filling factors \nu = 1 and \nu = 2 in the presence of dissipation. Dissipation arises in the compressible strip forming at the edge in...
Proteins are a matter of dual nature. As a physical object, a protein molecule is a folded chain of amino acids with multifarious biochemistry. But it is also an instantiation along an...
By using the notion of a rigid R-matrix in a monoidal category and the Reshetikhin--Turaev functor on the category of tangles, we review the definition of the associated invariant of long knots...
Future galaxy clustering surveys will probe small scales where non-linearities become important. Since the number of modes accessible on intermediate to small scales is very high, having a...
Recently, string theory on \text{AdS}_3 \times \text{S}^3 \times \mathbb{T}^4 with one unit of NS-NS flux (k=1) was argued to be exactly dual to the symmetric orbifold of \mathbb{T}^4, and in...
We study metastable behavior in a discrete nonlinear Schrödinger equation from the viewpoint of Hamiltonian systems theory. When there are n < \infty sites in this equation, we consider...
In this short paper we determine the effects of structure on the cosmological consistency relation which is valid in a perfect Friedmann Universe. We show that within \LambdaCDM the consistency...
|
I'm working on an exercise from functional analysis.
Let $E$ be a vector space and $\|\cdot\|_1$ and $\|\cdot\|_2$ be two complete norms on $E$. Now suppose that $E$ satisfies the following property:
$\bullet$ if $(x_n)$ is a sequence in $E$ and $x,y\in E$ such that $\|x_n-x\|_1\to 0$ and $\|x_n-y\|_2\to 0$, then $x=y$.
Now we want to show that the norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.
My idea is as follows: If for any $n>0$, there is an element $x_n\in E$ such that $\|x_n\|_1>n\|x_n\|_2$. Then consider $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$. Clearly, $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$ converges to $0$. However, I cann't get a contradicition from this. Maybe my idea is wrong.
In fact, I even don't konw how to show that a Cauchy sequence in norm $\|\cdot\|_1$ is also a Cauchy sequence in norm $\|\cdot\|_2$.
Anyone can give me some hints or a counter example? Thank you very much.
|
How can I get the closed-form of the above expression.I tried to find it but I can't do.
Let $x=\tan(\pi/9)$.
First, $$\tan 3a=\frac{\tan a+\tan 2a}{1-\tan a\tan 2a}=\frac{\tan a+\frac{2\tan a}{1-\tan^2 a}}{1-\frac{2\tan^2 a}{1-\tan^2a}}=\frac{3\tan a-\tan^3a}{1-3\tan^2a}$$
Since $\tan(\pi/3)=\sqrt 3$ you have
$$\sqrt 3=\frac{3x-x^3}{1-3x^2}$$
Therefore, $\tan(\pi/9)$ is one of the roots of the equation $$X^3-3\sqrt 3X^2-3X+\sqrt 3=0$$
If you are able to solve it, you will eventually get that 'closed form'. Cubic equations have a method to solve them, but it is far from simple. Alternatively you may use a computer program to solve the equation, like Maxima, Mathematica, Maple, etc.
Do you allow complex radicals in your "closed form"? (If you apply Cardano's formulas in ajotatxe's solution you will get them.) If so, why not just use radicals like $$ (-1)^{1/9} $$ so that you can write $$ \tan\frac{\pi}{9} = \frac{-2 \left( -1 \right) ^{5/9}+4 \left( -1 \right) ^{4/9}-2 \left( -1 \right) ^{2/9}-2 (-1)^{1/9}}{\sqrt{3}}+\sqrt {3} $$ ??
|
Last Updated: May 4, 2019
Several methods have been developed in order to save computational time and cost for unsteady flow computations compared to LES.
Keywords Detached-Eddy Simulation (DES), Scale Adaptive Simulation (SAS), Grey area problem Detached Eddy Simulation (DES) Detached-Eddy Simulation (DES) is a hybrid Reynolds Averaged Navier-Stokes/Large-Eddy Simulation model. The DES model was first proposed in 1997 by Spalart et al. [1] based on the Spalart-Allmaras RANS model and it is commonly referred to as DES97. Spalart-Allmaras Based DES Formulation (DES97)
Its formulation is briefly described in [2]:
The driving length scale of the RANS S-A model is the distance to the closest wall, \(d\). This makes a modification to this model for DES mode quite straightforward (exactly for this reason it was used as a basis of DES in the first publication [1]). The modification consists in substituting for \(d\), everywhere in the equations, the new DES length scale, \(\tilde{d}\). This length is also based on the grid spacing \(\Delta\) and is defined as:
\begin{equation} \tilde{d} = {\rm min}\left( d, C_{DES}\Delta \right), \tag{1} \label{eq:dTilda} \end{equation} where \(C_{DES}\) is the only new adjustable model constant, and \(\Delta\) is based on the largest dimension of the local grid cell \begin{equation} \Delta = {\rm max}\left( \delta_{x}, \delta_{y}, \delta_{z} \right). \tag{2} \label{eq:delta} \end{equation} Here we assume for simplicity that the grid is structured and that the coordinates \(\left(x, y, z\right)\) are aligned with the grid cell, but the generalizations are obvious.
For wall-bounded separated flows, the above formulation results in a bybrid model that functions as the standard RANS S-A model inside the whole attached boundary layer, and as its subgrid-scale version in the rest of the flow including the separated regions and near wake. Indeed, in the attached boundary layer, due to the significant grid anisotropy \(\left(\delta_{x} \approx \delta_{z} \gg \delta_{y}\right)\) typical of this flow region, in accordance with \eqref{eq:dTilda}, \(\tilde{d} = d\), and the model reduces to the standard S-A RANS model. Otherwise, once a field point is far enough from walls \(\left( d > C_{DES}\Delta \right)\), the length scale of the model becomes grid-dependent, i.e., the model performs as a subgrid-scale version of the S-A model. Note that at “equilibrium” (meaning a balance of production and destruction terms) this model reduces to an algebraic miximg-length Smagorinski-like subgrid model.
The “DES limiter” defined by Eq. \eqref{eq:dTilda} that is used in DES97 switches between the RANS length scale and LES length scale so that the model behaves in RANS-like and LES-like manners as illustrated in Figure 1. The length scale \eqref{eq:dTilda} depends only on the grid used in the simulation and it is solution-independent.
Menter’s SST Based DES Formulation k-\(\omega\) SST DES model k-\(\omega\) SST DDES model k-\(\omega\) SST IDDES model
There is a project called
DESider ( Detached Eddy Simulation for Industrial Aerodynamics) and many other DES models have been developed based on different RANS models, including RSM ones. Other Models \(v^2-f\) DES model
We can check the RANS and LES regions using the
function objects in OpenFOAM. DESModelRegions Scale Adaptive Simulation (SAS) k-\(\omega\) SST SAS model Useful Links FLOMANIA(Flow Physics Modelling An Integrated Approach) Project: 2002-2004 DESider(Detached Eddy Simulation for Industrial Aerodynamics) Project: 2004-2007 ATAAC(Advanced Turbulence Simulation for Aerodynamic Application Challenges) Project References
[1] P. R. Spalart, W.-H. Jou, M. Strelets and S. R. Allmaras, Comments on the feasibility of LES for wings, and on a hybrid RANS/LES approach. 1st AFOSR Int. Conf. on DNS/LES, Aug. 4-8, 1997, Ruston, LA. In “Advances in DNS/LES”, C. Liu and Z. Liu Eds., Greyden Press, Columbus, OH.
[2] M. Strelets, Detached Eddy Simulation of Massively Separated Flows. AIAA, 2001-0879. [3] P. R. Spalart, S. Deck, M. L. Shur, K. D. Squires, M. Kh. Strelets, and A. Travin, A new version of detached-eddy simulation, resistant to ambiguous grid densities. Theoretical and Computational Fluid Dynamics, 20(3), 181-195, 2006. [4] M. L. Shur, P. R. Spalart, M. Kh. Strelets, and A. Travin, A hybrid RANS-LES approach with delayed-DES and wall-modelled LES capabilities. International Journal of Heat and Fluid Flow, 29, 1638–1648, 2008. [5] Detached eddy simulation (DES). Available at: http://www.cfd-online.com/Wiki/Detached_eddy_simulation_(DES) [Accessed: 03 May 2019]. [6] OpenFOAM® v3.0+: New Solver and Physical Modelling Functionality. Available at: https://www.openfoam.com/releases/openfoam-v3.0+/solvers-and-physics.php#physics-kOmegaSSTDES [Accessed: 03 May 2019]. [7] C. M. Winkler, A. J. Dorgan and M. Mani, Scale Adaptive Simulations of Turbulent Flows on Unstructured Grids. AIAA, 2011-3559.
|
Alright, I have this group $\langle x_i, i\in\mathbb{Z}\mid x_i^2=x_{i-1}x_{i+1}\rangle$ and I'm trying to determine whether $x_ix_j=x_jx_i$ or not. I'm unsure there is enough information to decide this, to be honest.
Nah, I have a pretty garbage question. Let me spell it out.
I have a fiber bundle $p : E \to M$ where $\dim M = m$ and $\dim E = m+k$. Usually a normal person defines $J^r E$ as follows: for any point $x \in M$ look at local sections of $p$ over $x$.
For two local sections $s_1, s_2$ defined on some nbhd of $x$ with $s_1(x) = s_2(x) = y$, say $J^r_p s_1 = J^r_p s_2$ if with respect to some choice of coordinates $(x_1, \cdots, x_m)$ near $x$ and $(x_1, \cdots, x_{m+k})$ near $y$ such that $p$ is projection to first $m$ variables in these coordinates, $D^I s_1(0) = D^I s_2(0)$ for all $|I| \leq r$.
This is a coordinate-independent (chain rule) equivalence relation on local sections of $p$ defined near $x$. So let the set of equivalence classes be $J^r_x E$ which inherits a natural topology after identifying it with $J^r_0(\Bbb R^m, \Bbb R^k)$ which is space of $r$-order Taylor expansions at $0$ of functions $\Bbb R^m \to \Bbb R^k$ preserving origin.
Then declare $J^r p : J^r E \to M$ is the bundle whose fiber over $x$ is $J^r_x E$, and you can set up the transition functions etc no problem so all topology is set. This becomes an affine bundle.
Define the $r$-jet sheaf $\mathscr{J}^r_E$ to be the sheaf which assigns to every open set $U \subset M$ an $(r+1)$-tuple $(s = s_0, s_1, s_2, \cdots, s_r)$ where $s$ is a section of $p : E \to M$ over $U$, $s_1$ is a section of $dp : TE \to TU$ over $U$, $\cdots$, $s_r$ is a section of $d^r p : T^r E \to T^r U$ where $T^k X$ is the iterated $k$-fold tangent bundle of $X$, and the tuple satisfies the following commutation relation for all $0 \leq k < r$
$$\require{AMScd}\begin{CD} T^{k+1} E @>>> T^k E\\ @AAA @AAA \\ T^{k+1} U @>>> T^k U \end{CD}$$
@user193319 It converges uniformly on $[0,r]$ for any $r\in(0,1)$, but not on $[0,1)$, cause deleting a measure zero set won't prevent you from getting arbitrarily close to $1$ (for a non-degenerate interval has positive measure).
The top and bottom maps are tangent bundle projections, and the left and right maps are $s_{k+1}$ and $s_k$.
@RyanUnger Well I am going to dispense with the bundle altogether and work with the sheaf, is the idea.
The presheaf is $U \mapsto \mathscr{J}^r_E(U)$ where $\mathscr{J}^r_E(U) \subset \prod_{k = 0}^r \Gamma_{T^k E}(T^k U)$ consists of all the $(r+1)$-tuples of the sort I described
It's easy to check that this is a sheaf, because basically sections of a bundle form a sheaf, and when you glue two of those $(r+1)$-tuples of the sort I describe, you still get an $(r+1)$-tuple that preserves the commutation relation
The stalk of $\mathscr{J}^r_E$ over a point $x \in M$ is clearly the same as $J^r_x E$, consisting of all possible $r$-order Taylor series expansions of sections of $E$ defined near $x$ possible.
Let $M \subset \mathbb{R}^d$ be a compact smooth $k$-dimensional manifold embedded in $\mathbb{R}^d$. Let $\mathcal{N}(\varepsilon)$ denote the minimal cardinal of an $\varepsilon$-cover $P$ of $M$; that is for every point $x \in M$ there exists a $p \in P$ such that $\| x - p\|_{2}<\varepsilon$....
The same result should be true for abstract Riemannian manifolds. Do you know how to prove it in that case?
I think there you really do need some kind of PDEs to construct good charts.
I might be way overcomplicating this.
If we define $\tilde{\mathcal H}^k_\delta$ to be the $\delta$-Hausdorff "measure" but instead of $diam(U_i)\le\delta$ we set $diam(U_i)=\delta$, does this converge to the usual Hausdorff measure as $\delta\searrow 0$?
I think so by the squeeze theorem or something.
this is a larger "measure" than $\mathcal H^k_\delta$ and that increases to $\mathcal H^k$
but then we can replace all of those $U_i$'s with balls, incurring some fixed error
In fractal geometry, the Minkowski–Bouligand dimension, also known as Minkowski dimension or box-counting dimension, is a way of determining the fractal dimension of a set S in a Euclidean space Rn, or more generally in a metric space (X, d). It is named after the German mathematician Hermann Minkowski and the French mathematician Georges Bouligand.To calculate this dimension for a fractal S, imagine this fractal lying on an evenly spaced grid, and count how many boxes are required to cover the set. The box-counting dimension is calculated by seeing how this number changes as we make the grid...
@BalarkaSen what is this
ok but this does confirm that what I'm trying to do is wrong haha
In mathematics, Hausdorff dimension (a.k.a. fractal dimension) is a measure of roughness and/or chaos that was first introduced in 1918 by mathematician Felix Hausdorff. Applying the mathematical formula, the Hausdorff dimension of a single point is zero, of a line segment is 1, of a square is 2, and of a cube is 3. That is, for sets of points that define a smooth shape or a shape that has a small number of corners—the shapes of traditional geometry and science—the Hausdorff dimension is an integer agreeing with the usual sense of dimension, also known as the topological dimension. However, formulas...
Let $a,b \in \Bbb{R}$ be fixed, and let $n \in \Bbb{Z}$. If $[\cdot]$ denotes the greatest integer function, is it possible to bound $|[abn] - [a[bn]|$ by a constant that is independent of $n$? Are there any nice inequalities with the greatest integer function?
I am trying to show that $n \mapsto [abn]$ and $n \mapsto [a[bn]]$ are equivalent quasi-isometries of $\Bbb{Z}$...that's the motivation.
|
This blog discusses a problematic situation that can arise when we try to implement certain digital filters. Occasionally in the literature of DSP we encounter impractical digital IIR filter block diagrams, and by impractical I mean block diagrams that cannot be implemented. This blog gives examples of impractical digital IIR filters and what can be done to make them practical.
Implementing an Impractical Filter: Example 1
Reference [1] presented the digital IIR bandpass filter...
I've recently encountered a digital filter design application that astonished me with its design flexibility, capability, and ease of use. The software is called the "ASN Filter Designer." After experimenting with a demo version of this filter design software I was so impressed that I simply had publicize it to the subscribers here on dsprelated.com.What I Liked About the ASN Filter Designer
With typical filter design software packages the user enters numerical values for the...
This blog describes a general discrete-signal network that appears, in various forms, inside so many DSP applications.
Figure 1 shows how the network's structure has the distinct look of a digital filter—a comb filter followed by a 2nd-order recursive network. However, I do not call this useful network a filter because its capabilities extend far beyond simple filtering. Through a series of examples I've illustrated the fundamental strength of this Swiss Army Knife of digital networks...
Recently I've been thinking about the process of envelope detection. Tutorial information on this topic is readily available but that information is spread out over a number of DSP textbooks and many Internet web sites. The purpose of this blog is to summarize various digital envelope detection methods in one place.
Here I focus on envelope detection as it is applied to an amplitude-fluctuating sinusoidal signal where the positive-amplitude fluctuations (the sinusoid's envelope)...
I just discovered a useful web-based source of signal processing information that was new to me. I thought I'd share what I learned with the subscribers here on DSPRelated.com.
The Home page of the web site that I found doesn't look at all like it would be useful to us DSP fanatics. But if you enter some signal processing topic of interest, say, "FM demodulation" (without the quotation marks) into the 'Search' box at the top of the web page
and click the red 'SEARCH...
This blog discusses a not so well-known rule regarding the filtering in multistage decimation and interpolation by an integer power of two. I'm referring to sample rate change systems using half-band lowpass filters (LPFs) as shown in Figure 1. Here's the story.
Figure 1: Multistage decimation and interpolation using half-band filters.Multistage Decimation – A Very Brief Review
Figure 2(a) depicts the process of decimation by an integer factor D. That...
Recently I've been thinking about digital differentiator and Hilbert transformer implementations and I've developed a processing scheme that may be of interest to the readers here on dsprelated.com.
This blog presents a novel method for simultaneously implementing a digital differentiator (DD), a Hilbert transformer (HT), and a half-band lowpass filter (HBF) using a single tapped-delay line and a single set of coefficients. The method is based on the similarities of the three N =...
This blog proposes a novel differentiator worth your consideration. Although simple, the differentiator provides a fairly wide 'frequency range of linear operation' and can be implemented, if need be, without performing numerical multiplications.Background
In reference [1] I presented a computationally-efficient tapped-delay line digital differentiator whose $h_{ref}(k)$ impulse response is:$$ h_{ref}(k) = {-1/16}, \ 0, \ 1, \ 0, \ {-1}, \ 0, \ 1/16 \tag{1} $$
and...
This blog discusses a little-known filter characteristic that enables real- and complex-coefficient tapped-delay line FIR filters to exhibit linear phase behavior. That is, this blog answers the question:What is the constraint on real- and complex-valued FIR filters that guarantee linear phase behavior in the frequency domain?
I'll declare two things to convince you to continue reading.
Declaration# 1: "That the coefficients must be symmetrical" is not a correct
If you need to compute inverse fast Fourier transforms (inverse FFTs) but you only have forward FFT software (or forward FFT FPGA cores) available to you, below are four ways to solve your problem.
Preliminaries To define what we're thinking about here, an N-point forward FFT and an N-point inverse FFT are described by:$$ Forward \ FFT \rightarrow X(m) = \sum_{n=0}^{N-1} x(n)e^{-j2\pi nm/N} \tag{1} $$ $$ Inverse \ FFT \rightarrow x(n) = {1 \over N} \sum_{m=0}^{N-1}...
Some time ago I was studying various digital differentiating networks, i.e., networks that approximate the process of taking the derivative of a discrete time-domain sequence. By "studying" I mean that I was experimenting with various differentiating filter coefficients, and I discovered a computationally-efficient digital differentiator. A differentiator that, for low fequency signals, has the power of George Foreman's right hand! Before I describe this differentiator, let's review a few...
This blog presents two very easy ways to test the performance of multistage cascaded integrator-comb (CIC) decimation filters [1]. Anyone implementing CIC filters should take note of the following proposed CIC filter test methods.
Introduction
Figure 1 presents a multistage decimate by D CIC filter where the number of stages is S = 3. The '↓D' operation represents downsampling by integer D (discard all but every Dth sample), and n is the time index.
If the Figure 3 filter's...
There are two code snippets associated with this blog post:
and
This blog discusses an accurate method of estimating time-domain sinewave peak amplitudes based on fast Fourier transform (FFT) data. Such an operation sounds simple, but the scalloping loss characteristic of FFTs complicates the process. We eliminate that complication by...
There are so many different time- and frequency-domain methods for generating complex baseband and analytic bandpass signals that I had trouble keeping those techniques straight in my mind. Thus, for my own benefit, I created a kind of reference table showing those methods. I present that table for your viewing pleasure in this blog.
For clarity, I define a complex baseband signal as follows: derived from an input analog xbp(t)bandpass signal whose spectrum is shown in Figure 1(a), or...
Recently I was on the Signal Processing Stack Exchange web site (a question and answer site for DSP people) and I read a posted question regarding Goertzel filters [1]. One of the subscribers posted a reply to the question by pointing interested readers to a Wikipedia web page discussing Goertzel filters [2]. I noticed the Wiki web site stated that a Goertzel filter:"...is marginally stable and vulnerable tonumerical error accumulation when computed usinglow-precision arithmetic and...
I just learned a new method (new to me at least) for computing the group delay of digital filters. In the event this process turns out to be interesting to my readers, this blog describes the method. Let's start with a bit of algebra so that you'll know I'm not making all of this up.
Assume we have the N-sample h(n) impulse response of a digital filter, with n being our time-domain index, and that we represent the filter's discrete-time Fourier transform (DTFT), H(ω), in polar form...
I've recently encountered a digital filter design application that astonished me with its design flexibility, capability, and ease of use. The software is called the "ASN Filter Designer." After experimenting with a demo version of this filter design software I was so impressed that I simply had publicize it to the subscribers here on dsprelated.com.What I Liked About the ASN Filter Designer
With typical filter design software packages the user enters numerical values for the...
Most of us are familiar with the process of flipping the spectrum (spectral inversion) of a real signal by multiplying that signal's time samples by (-1)n. In that process the center of spectral rotation is fs/4, where fs is the signal's sample rate in Hz. In this blog we discuss a different kind of spectral flipping process.
Consider the situation where we need to flip the X(f) spectrum in Figure 1(a) to obtain the desired Y(f) spectrum shown in Figure 1(b). Notice that the center of...
Earlier this year, for the Linear Audio magazine, published in the Netherlands whose subscribers are technically-skilled hi-fi audio enthusiasts, I wrote an article on the fundamentals of interpolation as it's used to improve the performance of analog-to-digital conversion. Perhaps that article will be of some value to the subscribers of dsprelated.com. Here's what I wrote:
We encounter the process of digital-to-analog...
This blog explains why, in the process of time-domain interpolation (sample rate increase), zero stuffing a time sequence with zero-valued samples produces an increased-length time sequence whose spectrum contains replications of the original time sequence's spectrum.
Background
The traditional way to interpolate (sample rate increase) an x(n) time domain sequence is shown in Figure 1.
Figure 1
The '↑ L' operation in Figure 1 means to...
There are so many different time- and frequency-domain methods for generating complex baseband and analytic bandpass signals that I had trouble keeping those techniques straight in my mind. Thus, for my own benefit, I created a kind of reference table showing those methods. I present that table for your viewing pleasure in this blog.
For clarity, I define a complex baseband signal as follows: derived from an input analog xbp(t)bandpass signal whose spectrum is shown in Figure 1(a), or...
This blog discusses two ways to determine an exponential averager's weighting factor so that the averager has a given 3-dB cutoff frequency. Here we assume the reader is familiar with exponential averaging lowpass filters, also called a "leaky integrators", to reduce noise fluctuations that contaminate constant-amplitude signal measurements. Exponential averagers are useful because they allow us to implement lowpass filtering at a low computational workload per output sample.
Figure 1 shows...
If you've read about the Goertzel algorithm, you know it's typically presented as an efficient way to compute an individual kth bin result of an N-point discrete Fourier transform (DFT). The integer-valued frequency index k is in the range of zero to N-1 and the standard block diagram for the Goertzel algorithm is shown in Figure 1. For example, if you want to efficiently compute just the 17th DFT bin result (output sample X17) of a 64-point DFT you set integer frequency index k = 17 and N =...
This blog presents several interesting things I recently learned regarding the estimation of a spectral value located at a frequency lying between previously computed FFT spectral samples. My curiosity about this FFT interpolation process was triggered by reading a spectrum analysis paper written by three astronomers [1].
My fixation on one equation in that paper led to the creation of this blog.
Background
The notion of FFT interpolation is straightforward to describe. That is, for example,...
This blog is not about signal processing. Rather, it discusses an interesting topic in number theory, the magic of the number 9. As such, this blog is for people who are charmed by the behavior and properties of numbers.
For decades I've thought the number 9 had tricky, almost magical, qualities. Many people feel the same way. I have a book on number theory, whose chapter 8 is titled "Digits — and the Magic of 9", that discusses all sorts of interesting mathematical characteristics of the...
It works like this: say we have a real xR(n) input bandpass...
This blog describes a general discrete-signal network that appears, in various forms, inside so many DSP applications.
Figure 1 shows how the network's structure has the distinct look of a digital filter—a comb filter followed by a 2nd-order recursive network. However, I do not call this useful network a filter because its capabilities extend far beyond simple filtering. Through a series of examples I've illustrated the fundamental strength of this Swiss Army Knife of digital networks...
There have been times when I wanted to determine the z-domain transfer function of some discrete network, but my algebra skills failed me. Some time ago I learned Mason's Rule, which helped me solve my problems. If you're willing to learn the steps in using Mason's Rule, it has the power of George Foreman's right hand in solving network analysis problems.
This blog discusses a valuable analysis method (well known to our analog control system engineering brethren) to obtain the z-domain...
I just encountered what I think is an interesting technique for multiplying two integer numbers. Perhaps some of the readers here will also find it interesting.
Here's the technique: assume we want to multiply 18 times 17. We start by writing 18 and 17, side-by-side in column A and column B, as shown at the top of Figure 1. Next we divide the 18 at the top of column A by two, retaining only the integer part of the division, and double the 17 at the top of column B. The results of those two...
I have read, in some of the literature of DSP, that when the discrete Fourier transform (DFT) is used as a filter the process of performing a DFT causes an input signal's spectrum to be frequency translated down to zero Hz (DC). I can understand why someone might say that, but I challenge that statement as being incorrect. Here are my thoughts.
Using the DFT as a Filter It may seem strange to think of the DFT as being used as a filter but there are a number of applications where this is...
|
Tex2im tex2im is a simple tool that converts LaTeX formulas into high resolution pixmap graphics for inclusion in text processors or presentations. With tex2im you can write files containing only the formula in latex mathmode and transform them to many different graphic formats.
latex2html can do something similar, but only on whole documents. This is the direct solution.
Contents Usage tex2im [options] inputfile
The content of input file should be plain latex mathmode code. Alternatively, a string containing the latex code can be specified.
Options
-v
show version
-h
show help
-o file
specifies output filename, default is inputfile with new extension
-f expr
specifies output format; possible examples: gif, jpg, tif (all formates supported by "convert" should work; default: png
-r expr
specifies desired resolution in dpi; possible examples: 100x100, 300x300, 200x150; default is 150x150
-b expr
specifies the background colour; default: white
-t expr
specifies the text colour; default: black
-x file
file containing extra header lines of latex; default:
~/.tex2im_header
-z
transparent background; default: off
-n
no-formula mode (do not wrap in
eqnarray*environment); default: off
-a
change status of antialiasing; default is on for normal mode and off for transparent mode
For user specific default values,
tex2im reads the file '
.tex2imrc' in your home-directory. To change a value, put inside one or more of the following 5 commands:
resolution="150x150" format="png" color1="white" color2="black" extra_header="~/.tex2im_header" trans=0 aa=1
In the file '
~/.tex2im_header', additional latex header lines can be put, for example '
\usepackage{amsmath}' or your own definitions.
The transparency implementation of ImageMagick has some problems in combination with antialiasing. Therefore, by default, antialiasing is off in transparent mode. If you switch it on via the
-a flag you can expect pixels with something between background and text colour around the letters. This can look good if the background used in tex2im and the one of your document are identical, but also very bad if this is not the case. True antialising with transparency seems not to be possible. Just play around a bit to find the optimal settings.
Examples Note: The following examples were taken directly from the official tex2im website with some slight modifications. Example 1
Generate a file, called "
formula1.tex", containing the following:
\psi_{tot}(x,-t_0,r) = \frac{1}{(2\pi)^2} \int\!\!\!\int \tilde\Psi_{tot}\left(k_x,\frac{c}{2}\sqrt{k_x^2 + k_r^2},r=0\right)
Note that you should not put any
\begin{document} or
\begin{equation} in the file. Just plain latex mathmode code is required.
Then type on the command line:
% tex2im formula1.tex Example 2 Q = 1 + \sum_{i=1}^n\sum_{k=1}^n\sigma\left(\prod_{j=i}^kS_j\right) \exp\left\lbrace\frac{-K}{RT}\left(\Delta Lk - \left[\sum_{j=i}^k0.36j\right] - 0.8\right)^2\right\rbrace Example 3
You can also use colour. Generate a file "
formula2.txt", containing the following:
[C_3] = {\color{red} \langle \vec k_{3L} \vec k_{3L}^{\:\adj} \rangle }= \left[ \begin{array}{rrrr} S_{11} & S_{21} & S_{31} \\ S_{12} & S_{22} & S_{32} \\ S_{13} & S_{23} & S_{33} \end{array} \right]
Then type on the command line:
% tex2im -b yellow -t blue formula2.tex
And you'll get this:
Example 4
Finally, you can use tex2im without input file by specifying the latex-code directly on the command line:
% tex2im "\sum_{i=0}^5 x_i^2"
generates a file "
out.png" in the local directory (as long as you do not specify another output filename), containing:
Convert to SVG
Once one has the TeX code, one can produce an SVG file via the following, which assume the TeX file is called comm.tex:
latex comm.tex dvips -E -y 2500 -o comm.eps comm.dvi eps2eps -dNOCACHE comm.eps comm2.eps pstoedit -f sk comm2.eps comm.sk inkscape -z -f comm.sk -l comm.svg Download and install
To install tex2im, download the archive, uncompress it, and place the contained script "tex2im" in a directory in your path and make it executable.
For example:
tar xzf tex2im-1.8.tar.gz mv tex2im/tex2im /path/of/your/choice chmod 755 /path/of/your/choice/tex2im
|
My question refers to differential equations in Sobolev spaces. It is as follows:
Let $\Omega \subset \mathbb{R}^n$ be a bounded open set. Let $a: H_0^1(\Omega) \times H_0^1(\Omega) \rightarrow \mathbb{R}$ be a bilinear symmetric form, such that:
(continuity in $H^1$) $\forall u,v \in H^1(\Omega) \; \; \; a(u,v) \leq c_1 \|u\|_{H^1(\Omega)} \|v\|_{H^1(\Omega)}$, ($L_2$ coercivity in $H^1_0$) $\forall \eta \in H_0^1(\Omega) \; \; \; a(\eta,\eta) \geq c_2 \|\eta\|_{L_2(\Omega)}^2$.
Let $u,w$ be solutions of equations $$ a(u,v) = \int_{\Omega} g(x)v(x) dx \quad \forall v \in H_0^1(\Omega); u|_{\partial \Omega} = \bar{u}|_{\partial \Omega},$$ $$ a(w,v) = \int_{\Omega} g(x)v(x) dx \quad \forall v \in H_0^1(\Omega); w|_{\partial \Omega} = \bar{w}|_{\partial \Omega},$$ where $\bar{u}, \bar{w} \in \mathcal{C}^2(\overline{\Omega})$ and $g \in L_2(\Omega)$ are given.
Let us define $\varphi := u - \bar{u}$, $\psi := w - \bar{w}$.
My question is:
Under the following assumptions, is it true that $$a(\varphi - \psi, \varphi - \psi) \leq c_3 \|\bar{u} - \bar{w}\|_{H^1(\Omega)}^2.$$ Rationale:
If we pick some $a$, for example $a(u,v) := \int_\Omega \nabla u \cdot \nabla v \; dx$, then $u$ and $w$ may be treated as weak solutions of a given differential equation with a Dirichlet boundary values determined by the functions $\bar{u}$, $\bar{w}$. The aim of this question is to estimate the difference between these solutions in some manner.
This result seems easy if the coercivity in assumption 2 will be in $\|\cdot\|_{H^1(\Omega)}$ or even in $|\cdot|_{H_0^1(\Omega)}$ instead of $\|\cdot\|_{L_2(\Omega)}$. Then we could take $v = \varphi - \psi$ in both equations, subtract them and use assumptions 1,2 to obtain the above estimate.
Is the coercivity in $L_2$ norm enough?
|
I'm considering a symmetry transformation on a Lagrangian
$$ \delta A = \int L(q +\delta q, \dot{q} + \delta \dot{q} , \ddot{q} + \delta \ddot{q}) dt $$
the general variation takes the form
$$ \delta A = \int \frac{ \partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q} + \frac{\partial L}{\partial \ddot{q}} \delta \ddot{q} dt $$
Now, the second term inside the integral is normally handled as:
$$ \int \frac{\partial L}{\partial \dot{q}} \delta \dot{q} = \frac{\partial L}{\partial \dot{q}} \delta q - \int \frac{\partial}{\partial t}( \frac{\partial L}{\partial \dot{q}} ) \delta q $$
the third terms requires some more work, I'm having it as:
$$ \int \frac{\partial L}{\partial \ddot{q}} \delta \ddot{q} = \frac{\partial L}{\partial \ddot{q}} \delta \dot{q} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \ddot{q}} ) \delta q + \int \frac{\partial^2}{\partial^2 t}( \frac{\partial L}{\partial \ddot{q}} ) \delta q $$
So my variation (that in the case of symmetry must be zero up to boundary terms is)
$$ \delta A = \int \Big \{ \frac{\partial L}{\partial q} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \dot{q}} ) + \frac{\partial^2}{\partial^2 t}( \frac{\partial L}{\partial \ddot{q}} ) \Big \} \delta q dt + \Big \{ \frac{\partial L}{\partial \dot{q}} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \ddot{q}} ) \Big \} \delta q + \frac{\partial L}{\partial \ddot{q}} \delta \dot{q} $$
Now, I'm taking
both boundary terms to be conserved currents:
$$ \frac{\partial L}{\partial \dot{q}} - \frac{\partial}{\partial t}( \frac{\partial L}{\partial \ddot{q}} ) $$
and
$$ \frac{\partial L}{\partial \ddot{q}} $$
But if the second is a conserved current, then its derivative is zero, and the conserved current becomes trivially identical to the first-order case
What the error in my derivation?
|
User:Nikita2 Pages of which I am contributing and watching
Analytic function | Cauchy criterion | Cauchy integral | Condition number | Continuous function | D'Alembert criterion (convergence of series) | Dedekind criterion (convergence of series) | Derivative | Dini theorem | Dirichlet-function | Ermakov convergence criterion | Extension of an operator | Fourier transform | Friedrichs inequality | Fubini theorem | Function | Functional | Generalized derivative | Generalized function | Geometric progression | Hahn-Banach theorem | Harmonic series | Hilbert transform | Hölder inequality | Lebesgue integral | Lebesgue measure | Leibniz criterion | Leibniz series | Lipschitz Function | Lipschitz condition | Luzin-N-property | Newton-Leibniz formula | Newton potential | Operator | Poincaré inequality | Pseudo-metric | Raabe criterion | Riemann integral | Series | Sobolev space | Vitali theorem |
TeXing
I'm keen on turning up articles of EoM into better appearance by rewriting formulas and math symbols in TeX.
Now there are
2103 (out of 15,890) articles with Category:TeX done tag.
$\quad \rightarrow \quad$ $\sum_{n=1}^{\infty}n!z^n$ Just type $\sum_{n=1}^{\infty}n!z^n$. Today You may look at Category:TeX wanted. How to Cite This Entry:
Nikita2.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Nikita2&oldid=34319
|
Let $X$ be an abelian variety.
In "Mumford, Abelian Varieties" the Riemann-Roch Theorem has the following form:
For all line bundles $\mathcal{L}$ on $X$, if $\mathcal{L}\cong\mathcal{O}_X(D)$, we have $\chi(\mathcal{L})=\frac{(D^g)}{g!}$, $\chi(\mathcal{L})^2=\deg\phi_\mathcal{L}$, where $(D^g)$ is the $g$-fold self-intersection number of $D$.
$\mathcal{L}$ ample implies $\dim H^0(X,\mathcal{L})>0$; $\mathcal{L}$ ample implies $i(\mathcal{L})=0$.
Where $i(\mathcal{L})$ comes from the Vanishing Theorem in Mumford:
Let $\mathcal{L}$ be a line bundle on $X$ such that $K(\mathcal{L})$ is finite. There exists a unique integer $i=i(\mathcal{L})$, $0\leq i(\mathcal{L})\leq g=\dim X$, such that $H^p(X,\mathcal{L})=(0)$ for $p\neq i$ and $H^i(X,\mathcal{L})\neq (0)$.
Can someone give me more details of how to do that? Thanks!
|
What are some alternative ways to represent the golden ratio? I already know the relatively boring ones compared to the complex ones as well as:
$\displaystyle \frac{1+\sqrt 5}{2},$ $\displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1+}} \dots},$ $\displaystyle \phi + 1 = \phi ^{-1},$
and also the multiplier of consecutive Fibonacci terms. As some current answers have given, I would
not like any formulas that reproduce the above. They are not classed as interesting, as they include repetition. I am looking for formulas that are interesting, and I am hoping to find some without repetition.
|
This answer focuses on identifying families of solutions to the problem described in the question.
I've made two provisional conjectures in order to make progress with the problem:
The result can be stated for three $2n$-gons rather than two $n$-gons and one $2n$-gon.
Solutions have mirror symmetry. Or equivalently, in any solution there are two pairs of $2n$-gons which have the same degree of overlap. [This turns out to be false - see 'Solution family 5' below. However, this condition is assumed in Solution families 1-4.]
[
Continuation 6: in an overhaul of the notation I've halved $\phi$ and doubled $m$ so that $m$ is always an integer.]
If we define the degree of overlap, $j$, between two $2n$-gons $(n>3)$ as the number of edges of one that lie wholly inside the other, then $1 < j < n$.
If $$\phi = \frac{\pi}{2n}$$is half the angle subtended at the centre of the $2n$-gon by one of its edges, then the distance between the centres of two overlapping $2n$-gons is $$D_{jn} = 2\cos{j\phi}$$Consider a $2n$-gon P which overlaps a $2n$-gon O with degree $j$. Now bring in a third $2n$-gon, Q, which also overlaps O with degree $j$ but is rotated about the centre of O by an angle $m\phi$ with respect to P, where $m$ is an integer.
The distance between the centres of P and Q, which I'll denote by $D_{kn}$ for a reason that will become apparent, is$$D_{kn} = 2D_{jn}\sin{\tfrac{m}{2}\phi} = 4\cos{j\phi} \, \sin{\tfrac{m}{2}\phi}$$
We now demand that P and Q should overlap by an integer degree, $k$, so that$$D_{kn} = 2\cos{k\phi}$$This will ensure that all points of intersection coincide with vertices of the intersecting polygons, and thus provide a configuration satisfying the requirements of the question (with the proviso that the condition does not guarantee that there is a common area of overlap shared by all three polygons).
We have omitted mention of the orientation of the polygons, but it is easily shown that this is always such as to achieve the desired overlap.
Combining the two expressions for $D_{kn}$ gives the condition
$$2\cos{j\phi}\, \sin{\tfrac{m}{2}\phi} = \cos{k\phi}$$or (since $n\phi=\pi/2$)$$2\cos{j\phi}\, \cos{(n-\tfrac{m}{2})\phi} = \cos{k\phi} \tag{1}$$
The configurations we seek are solutions of this equation for integer $n$, $j$, $k$ and $m$.
In the first example in the question $n = 12, j = 8, k = 6, m = 12$.
In the second example $n = 15, j = 6, k = 10, m = 6$.
[
Continuation 6: for solutions under the constraint of conjecture 2, $m$ is always even, but in the more general case $m$ may be odd.]
I'll now throw this open to see if anyone can provide a general solution. It seems likely that $j$, $k$ and $m/2$ must be divisors of $2n$ [this turns out to be incorrect], and I have a hunch that the solution will involve cyclotomic polynomials [this turns out to be correct].
Continuation (1)
I've now identified 3 families of solutions consistent with conjecture 2 (mirror symmetry), all involving angles of 60 degrees. There may be others.
Solution family 1
This family is defined by setting $j=2n/3$. This means that half the angle subtended at the centre of O by its overlapping edges is $\tfrac{\pi}{3}$ radians or 60 degrees. Since $\cos{\tfrac{\pi}{3}} = \tfrac{1}{2}$ it reduces equation 1 to$$\cos{(n-\tfrac{m}{2})\phi} = \cos{k\phi}$$so there are solutions with$$n-\tfrac{m}{2} = k$$(where $\tfrac{m}{2}$ is an integer) subject to $2 \le k \le n-1\,\,$, $1 \le \tfrac{m}{2} \le n-2\,\,$ and $3|n$.
The first example in the question belongs to this family. The complete set of solutions for $n=12$ combine to make this pleasing diagram:
Solution family 2
This family has $m=2n/3$. This makes $\cos{(n-\tfrac{m}{2})\phi}=\cos{(\pi/3)} = \tfrac{1}{2}$, which reduces equation 1 to$$\cos{j\phi} = \cos{k\phi}$$so (given that $j<n$ and $k<n$)$$j = k$$These solutions have threefold rotational symmetry. The only restriction is that $n$ must be divisible by 3. Example ($n=9, j=k=4, m=6$):
Solution family 3
This family is the most interesting of the three, but yields only one solution. It is defined by setting $k=2n/3$ so that $\cos{k\phi}=\cos{\tfrac{\pi}{3}} = \tfrac{1}{2}$. Equation 1 then becomes
$$2\cos{j\phi}\,\cos{(n-\tfrac{m}{2})\phi} = \tfrac{1}{2}$$which may be written in the following equivalent forms:$$\cos{(n+\tfrac{m}{2}-j)\phi} + \cos{(n+\tfrac{m}{2}+j)\phi} = -\tfrac{1}{2} \tag{2}$$$$\cos{(n-\tfrac{m}{2}-j)\phi} + \cos{(n-\tfrac{m}{2}+j)\phi} = \tfrac{1}{2} \tag{3}$$Solutions to these equations can be found using the following theorem relating the roots $z_i(N)$ of the $N$th cyclotomic polynomial to the Möbius function $\mu(N)$:
$$\sum_{i=1}^{\varphi(N)} {z_i(N)} = \mu(N)$$where $\varphi(N)$ is the Euler totient function (the number of positive integers less than $N$ that are relatively prime to $N$) and $z_i(N)$ are a subset of the $N$th roots of unity.Taking the real part of both sides and using symmetry this becomes:$$\sum_{i=1}^{\varphi(N)/2} { \cos{(p_i(N) \frac{2\pi}{N})} } = \tfrac{1}{2} \mu(N) \tag{4}$$where $p_i(N)$ is the $i$th integer which is coprime with $N$.
The Möbius function $\mu(N)$ takes values as follows:
$\mu(N) = 1$ if $N$ is a square-free positive integer with an even number of prime factors.
$\mu(N) = −1$ if $N$ is a square-free positive integer with an odd number of prime factors.
$\mu(N) = 0$ if $N$ has a squared prime factor.
Equation 4 thus provides solutions to equations 2 and 3 if $\varphi(N) = 4$, $\mu(N)$ has the appropriate sign and the cosine arguments are matched.
The first two conditions are true for only two integers:
$N=5$, with $\mu(5)=-1$, $p_1(5) = 1, p_2(5) = 2$
$N=10$, with $\mu(10)=1$, $p_1(10) = 1, p_2(10) = 3$.
We first set $N=5$ and look for solutions to equation 2.
Matching the cosine arguments requires firstly that$$2j \frac{\pi}{2n} = (p_2(5)-p_1(5))\frac{2\pi}{5}$$from which it follows that$$5j = 2n$$
$n$ must be divisible by 3 to satisfy $k=2n/3$, so the smallest value of $n$ for which solutions are possible is $n=15$, with $k=10$ and $j=6$. All other solutions will be multiples of this one.Matching the cosine arguments also requires that$$(n+\tfrac{m}{2}-j) \frac{\pi}{2n} = p_1(5) \frac{2\pi}{5}$$which implies $m=6$.
This is the solution illustrated by the second example in the question.
Setting $N=10$ and looking for solutions to equation 3 yields the same solution.
Continuation (2)
Solution family 4
A fourth family of solutions can be obtained by writing equation 1 as
$$\cos{(n+\tfrac{m}{2}-j)\phi} + \cos{(n+\tfrac{m}{2}+j)\phi} + \cos{k\phi} = 0 \tag{5}$$
and viewing this as an instance of equation 4 with $\varphi(N)/2 = 3$ and $\mu(N) = 0$. There are two values of N which satisfy these conditions, $N = 9$ and $N = 18$, which lead to three solutions:
For $N = 9$:$$n=9, j=6, k=8, m=2\\n=9, j=4, k=4, m=6$$
For $N=18$:$$n=9, j=2, k=2, m=6$$
However, these are not new solutions. The first is a member of family 1 and the last two are members of family 2.
Continuation (3)
Solution family 5
Rotating a $2n$-gon about a vertex by an angle $m\phi$ moves its centre by a distance $$2\sin{ \tfrac{m}{2}\phi} = 2\cos{(n-\tfrac{m}{2})\phi} = D_{n-m/2,n}.$$If $m$ is even the rotated $2n$-gon thus overlaps the original $2n$-gon with integer degree $n-\tfrac{m}{2}$, and a third $2n$-gon with a different $m$ may overlap both of these, providing another type of solution to the problem.
Solutions of this kind may be constructed for all $n \ge 3$. The diagram below includes the complete set of such solutions for $n=5$. A similar diagram with $n=12$ (but with a centrally placed $2n$-gon of the same size which can only be added when $3|n$) is shown above under Solution family 1.
This family of solutions provides exceptions to conjecture 2: not all groups of three $2n$-gons overlapping in this way show mirror symmetry.
Continuation (4)
If we relax the condition set by conjecture 2, allowing solutions without mirror symmetry, we need an additional parameter, $l$, to specify the degree of overlap between O and P (which is now no longer $j$).
The distances between the centres of the three $2n$-gons are now related by the cosine rule:
$$D_{nk}^2 = D_{nj}^2 + D_{nl}^2 - 2 D_{nj}D_{nl}\cos{m_k\phi},$$where a subscript $k$ has been added to $m$ to acknowledge the fact that $j$, $l$ and $k$ can be cycled to generate three equations of this form. These can be written$$\\ \cos^2{J} + \cos^2{L} - 2 \cos{J} \cos{L} \cos{M_k} = \cos^2{K} \\ \cos^2{K} + \cos^2{J} - 2 \cos{K} \cos{J} \cos{M_l} = \cos^2{L} \\ \cos^2{L} + \cos^2{K} - 2 \cos{L} \cos{K} \cos{M_j} = \cos^2{J} $$where$$J = j\phi,\, L = l\phi,\, K = k\phi,\\M_j = m_j\phi,\, M_l = m_l\phi,\, M_k = m_k\phi$$
The same result in a slightly different form is derived in the answer provided by @marco trevi.
$M_j$, $M_l$ and $M_k$ are the angles of the triangle formed by the centres of the three polygons. Since these sum to $\pi$ we have$$m_j + m_l + m_k = 2n$$
The sine rule gives another set of relations:$$\frac{\cos{J}}{\sin{M_j}} = \frac{\cos{L}} {\sin{M_l}} = \frac{\cos{K}}{\sin{M_k}} $$
In general the $m$ parameters are limited to integer values (as can be seen by considering the symmetry of the overlap between a $2n$-gon and each of its two neighbours). But they are now not necessarily even.
|
Unlike usual spin the isospin is an
internal degree of freedom, the particle type that doesn't have any relation to spacetime. The similarity between ordinary spin and isospin originates from the $2\times 2$ unitary matrices group $SU(2)$ being an universal covering of the 3d rotation group $SO(3)$. That means that they have same Lie algebras and their representations are related to each other.
Indeed if you look at the commutation relations of the Pauli matrices $\{\tau_k\}$ they are the same as the commutation relations of the 3d rotation generators $\{J_k\}$. Because of that if you take some 2-dimensional column and $2\times 2$ Hermitian traceless matrix,\begin{equation}\begin{pmatrix}u\\v\end{pmatrix},\quad \hat{T}=T_1\tau_1+T_2\tau_2+T_3\tau_3\end{equation}and act with some $2\times 2$ unitary matrix on them they will transform as a spinor and a 3d vector $\vec{T}=(T_1,T_2,T_3)$.
The symmetry under 3d rotations correspond to the conserved 3d vector - angular momentum. Similarly because of the aforementioned relation the symmetry under $SU(2)$ tranformations correspond to the conserved 3d vector - isospin. However by construction the isospin has no relation to our space. We could live in the flat world but if we had any $SU(2)$ symmetry we could get conserved isospin 3d vector.
So any time you have two similar particle species, you get $SU(2)$ symmetry and from there you get some conserved isospin-like vector.
In the nuclear physics this appears in the following way. We have proton and neutron,\begin{equation}\begin{pmatrix}p\\n\end{pmatrix},\quad m_p\simeq m_n\end{equation}Because strong interaction doesn't differentiate proton and neutron we can consider act on that column with arbitrary $2\times 2$ unitary matrix and all strong processes will remain almost the same. Thus we have $SU(2)$ symmetry and from that we have a isospin vector conserved by strong interaction. However weak interaction differentiates them and thus doesn't conserve isospin.
Where this structure comes from? The deepest explanation at present comes from the Standard model. There we have three gauge fields. One of them is $B_mu$ with $U(1)$ gauge group that you may think of as just electromagnetic field (indeed it is one of the precursors of the electromagnetic field). Another two are non-abelian Yang-Mills fields that you may think of as generalizations of the electromagnetism - instead of simple $A_mu$ you have some traceless $n\times n$ matrix. Those two are $\hat{W}_\mu=W^{k}_\mu\tau_k$ with $SU(2)$ gauge group and $g_\mu$ with $SU(3)$ gauge group. All the fermions are organized according to those gauge groups. So with lightest generation of quarks we get,\begin{equation}\begin{pmatrix}(u_r,u_g,u_b)_{SU(3)}\\(d_r,d_g,d_b)_{SU(3)}\end{pmatrix}_{SU(2)}\end{equation}
The $SU(2)$ symmetry acts on the columns whereas $SU(3)$ acts on the rows. Each one of them doesn't care about the structure associated with the other. The initial $SU(2)$ symmetry is associated with the so-called
weak isospin $\vec{T}_W$ with $u$ quarks having $T_{W,3}=+\frac{1}{2}$ and $d$ quarks $T_{W,3}=-\frac{1}{2}$.
$SU(3)$ gauge field corresponds to the strong interaction. Its associated structure will not be important for us now because of the confinement. What is important is that it doesn't touch the structure associated with $SU(2)$.
At low energies when the Higgs field becomes nonzero the remaining $U(1)\times SU(2)$ symmetry breaks down to $U(1)$ giving rise to the electromagnetic and weak interactions. The quarks (and many other fields) get different masses. But what's important the strong interaction doesn't care about all that and interacts with all quarks in the same way, so strong interaction doesn't differentiate e.g. $uud$ and $udd$.
However $u$ and $d$ quarks for some unknown reason are very light to the point when they often can safely assumed to be massless. The weak interaction is, well, very weak compared to the strong so most of the mass of the hadrons comes from the strong interaction that doesn't differentiate $u$ and $d$. Thanks to that $p=uud$ and $n=udd$ are very close in mass. That's how we get low energy $SU(2)$ symmetry of $(p,n)$ and fundamental weak isospin structure gives rise to the isospin of the nuclear physics.
Now if you ask, why is it so that $SU(2)$ and $SU(3)$ are independent of each other? Why $u$ and $d$ are so light? No one knows that right now. There are various hypotheses about beyond Standard model physics that explain that but unfortunately we haven't got yet evidence for any of them.
|
Talk:Absolute continuity
Could I suggest using $\lambda$ rather than $\mathcal L$ for Lebesgue measure since
it is very commonly used, almost standard it would be consistent with the notation for a general measure, $\mu$ calligraphic is being used already for $\sigma$-algebras
--Jjg 12:57, 30 July 2012 (CEST)
Between metric setting and References I would like to type the following lines. But for some reason which is misterious to me, any time I try the page comes out a mess... Camillo 10:45, 10 August 2012 (CEST)
if for every $\varepsilon$ there is a $\delta > 0$ such that, for any $a_1<b_1<a_2<b_2<\ldots < a_n<b_n \in I$ with $\sum_i |a_i -b_i| <\delta$, we have \[ \sum_i d (f (b_i), f(a_i)) <\varepsilon\, . \] The absolute continuity guarantees the uniform continuity. As for real valued functions, there is a characterization through an appropriate notion of derivative. Theorem 1A continuous function $f$ is absolutely continuous if and only if there is a function $g\in L^1_{loc} (I, \mathbb R)$ such that\begin{equation}\label{e:metric}d (f(b), f(a))\leq \int_a^b g(t)\, dt \qquad \forall a<b\in I\,\end{equation}(cp. with ). This theorem motivates the following Definition 2If $f:I\to X$ is a absolutely continuous and $I$ is compact, the metric derivative of $f$ is the function $g\in L^1$ with the smalles $L^1$ norm such that \ref{e:metric} holds (cp. with ) OK, I found a way around. But there must be some bug: it seems that whenever I write the symbol "bigger" then things gets messed up (now even on THIS page). Camillo 10:57, 10 August 2012 (CEST) But I did not understand what is the problem. Messed up? On this page? Where? And what was the way around? --Boris Tsirelson 13:06, 10 August 2012 (CEST) > How to Cite This Entry:
Absolute continuity.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Absolute_continuity&oldid=27473
|
I have this assignement for my discreet math course, and I would like a opinion on my answer:
Let $a = p_1^{e_1} + p_2^{e_2} + ... + p_s^{e_s}$ and $b = q_1^{f_1} + q_2^{f_2} + ... + q_t^{f_t}$ be positive integers expressed as sum of powers of prime numbers. In $\Bbb N$ , let these two binary relations be defined:
$$aR_1b \Leftrightarrow e_1 + e_2 + ... e_s \le f_1 + f_2 + ... + f_t$$ $$aR_2b \Leftrightarrow e_1 + e_2 + ... e_s \lt f_1 + f_2 + ... + f_t$$
Which of these relations is an order relation?
What I came up with:
For $R_1$ to be a large ordering it needs to be reflexive, antisymmetrical and transtive.
Reflexivity: $aRa$ so $e_1 + e_2 + ... e_s \le e_1 + e_2 + ... + e_s$ must be true. And it is, the sum of the exponents is $\le$ to itself, so $R_1$ is reflexive.
Antisymmetry: If $aRb$ and $bRa$, then $a=b$.
$$aRb \Leftrightarrow e_1 + e_2 + ... e_s \le f_1 + f_2 + ... + f_s$$ $$bRa \Leftrightarrow f_1 + f_2 + ... + f_s \le e_1 + e_2 + ... e_s$$
This can happen, and it implies that the values of the two sums are equal. But the fact that the sums are equal, does not imply that $a=b$, because it doesn't take into consideration the bases of the powers. We could have $A=5^3+7^2$ and $B=3^2+2^3$ and they would have the same sum for the exponents, but A and B themselves would be different numbers. So $aRb$ and $bRa$ can both happen even if $a\neq b$, so the relation is not antisymmetric.
Transitivity: This is straightforward. If $$e_1 + e_2 + ... e_s \le f_1 + f_2 + ... + f_s$$
and
$$f_1 + f_2 + ... + f_s \le g_1 + g_2 + ... g_h$$
of course
$$e_1 + e_2 + ... e_s \le g_1 + g_2 + ... + g_h$$
So the relation is transitive.
$R_1$ is not a large ordering because it's not antisymmetric.
To prove that it is a strict ordering we need to prove that $R_1$ is irreflexive and transitive. We already know that it is transitive, and we also know that it is reflexive, so $R_1$ can't be irreflexive, thus it is not an order relation.
For $R_2$ :
Reflexivity: $R$ can't be reflexive because the sum of the exponents of $a$ can't be lesser than itself.
Antisymmetry: We already know that if $$e_1 + e_2 + ... e_s \lt f_1 + f_2 + ... + f_s$$ is true, the opposite can't be. So if $aRb$ we can never have $bRa$, be $a=b$ or not. So the relation is antisymmetric.
Transitivity: Same as $R_1$ but with $\lt$ instead of $\le$.
Since the relation is not reflexive, it is not a large ordering.
Is the relation irreflexive? It is because $\forall x \in S : a\lnot Ra$, the reason being the counter-proof of reflexivity. So $R_2$ is a strict ordering relation.
Thank you for your help
|
Does the following integral converge?
$$\int_1^\infty \sin^2 (x^2) \, dx$$
I tried $$\int_1^\infty \sin^2(x^2) \, dx=\int_1^\infty \frac{1-\cos(2x^2)}{2} \, dx = \frac{\sqrt{2}}{4} \int_1^\infty\frac{1-\cos(u)}{2\sqrt{u}} \, du$$
The idea is that, I want to compare the original integral to a divergent $p$-integral. But I am not sure how to proceed from here.
|
Given that $\textbf{F} = \langle z,x,y \rangle$,
The plane $ z=2x+2y-1$ and the paraboloid $ z= x^2 +y^2$ intersect in a closed curve.
I'm trying to use stokes theorem to find the line integral.
Attempt:
We know that Stokes Theorem is given by: $\iint_S (\nabla \times \textbf{F}) \cdot \textbf{N}\: d\textbf{S} $
$\nabla \times \textbf{F} = \vec{\imath} + \vec{\jmath} + \vec{k} = \langle 1 , 1, 1\rangle$
Now, the problem I have is parameterizing. This is what I did.
$$\vec{r} = \langle x,y,2x+2y-1\rangle$$ $R_x \times R_y = 2\vec{\imath} - 2\vec{\jmath} + 0\vec{k} $
Now, $\iint_S \langle 1 , 1, 1\rangle \cdot \langle 2 , -2, 0\rangle \: d\textbf{S} = 0 $. But this is not right!
Perhaps what I'm struggling most is getting the parameterization right for the surface. Also would I have to convert to polar coordinates to complete the integral?
Any help would be appreciated! Thank you
|
I've had to rewrite the question as i made agrievious errors in the first go-round.
Given a (pseudo)Riemannian metric g, we can identify its components with the symmetric product of gamma matrices:
$$\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}I$$
The gammas act as basis vectors and their products and linear combinations form a clifford algebra. Under the geometric algebra our basis vectors (matrices now) transform in a two-sided way:
$$\bar{\gamma}^{\mu}=\psi^{\dagger}\gamma^{\mu}\psi$$
for a general non-unitary transformation (i,e not just a change of coordinate basis) $\psi^{\dagger}\neq\psi^{-1}$ how does the metric g and its determinant Det(g) transform??
I had initially thought $g\longrightarrow\psi^{\dagger}g\psi$ however that is silly since we have the non-unitary generalization.
$$g^{\mu\nu}I=\frac{1}{2}\psi^{\dagger}\left(\gamma^{\mu}\psi\psi^{\dagger}\gamma^{\nu}+\gamma^{\nu}\psi\psi^{\dagger}\gamma^{\mu}\right)\psi$$
For my problem set $\psi\psi^{\dagger}$ should be a scalar (in case you havent guessed the $\psi$ are spinors). this should leave me with:
$$g^{\mu\nu}I=\frac{1}{2}\psi^{\dagger}\left(\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}\right)\psi\left(\psi\psi^{\dagger}\right)$$
|
While I’ve been quite happy with the performance of my Predictaball football rating system, one thing that that’s bothered me since its inception last summer is the reliance on hard-coded parameters.
Similar to many other football rating methods, it’s an adaptation of the Elo system that was designed for Chess matches by Arpad Elo in the 1950s. His aim was to devise an easily implementable system to rate competitors in a 2-person zero-sum game.
Predictaball, on the other hand, is more concerned with accurately predicting future matches; model interpretability and portability is far less of a requirement. Furthermore, we have the advantage of an abundance of historical performance data and cheap computation to fit our models.
This post summarises the implementation of a rating system that simultaneously optimises parameters used in both the match prediction model and the rating update function in order to maximise match prediction accuracy.
Current Elo implementation
I’ll firstly run a quick refresher on how the standard Elo method is used by Predictaball.
Elo expected outcome
The formula for calculating the expected outcome $E$ of a match in Elo’s original formulation is shown below, where $\delta$ is the difference in rating between the home and away team, i.e. $\delta = R_{H} - R_{A}$.
$$E = \frac{1}{1+10^{- \delta /400}}$$
Note that this is the inverse logit function in base 10 rather than the standard natural base, and is therefore equivalent to logistic regression with 1 predictor ($\delta$) and the coefficient $\beta = \frac{1}{400}$. The version used by Predictaball (and others including 538) also adds in a factor to account for home advantage, i.e. $\delta = R_{H} - R_{A} + \alpha$, where $\alpha$ is estimated from the historical data as the rating difference associated with the proportion of matches that are home wins.
This expected outcome framed as logistic regression is therefore
$$\text{logit}(P(\text{home win})) = \alpha + \beta(R_{H} - R_{A})$$
However, this model isn’t overly useful on its own as it doesn’t tell us whether an away win or draw is more likely, and we might be more interested in other outcomes such as the number of goals scored (what Predictaball currently outputs). Furthermore, we might want to include additional predictor variables, such as if a team is missing a key player, or has had a new manager recently. We definitely have sufficient training data to fit more complex predictive models and not rely upon hardcoded simplifications.
Rating update equation
The Elo rating update equation is a function of the difference between the actual outcome $O$ ($O \in \left\{0, 0.5, 1\right\}$ for away win, draw, and home win respectively), the expected outcome $E$ from the above equation, and a gain factor $K$ that controls how much rating is won/lost per game. $K$ is traditionally fixed at 20 for a long-running competition or set higher for a short tournament.
$$R^{\prime}_{H} = R_{H} + K(O - E)$$
Predictaball also scales the output to account for the margin of victory (MoV), similar to the method used by 538
$$R^{\prime}_{H} = R_{H} + KG(\text{MoV})(O - E)$$
The functional form of G is shown below, where the multiplier decays as the margin of victory increases. Also note the adjustment by $\delta$ (labelled $\text{dr}$ in the figure) to stop higher rating teams getting further and further ahead. The parameters of $G$ were derived by hand by tweaking them until the resulting shape looked appropriate, rather than using a data-driven approach.
Secondary prediction network
The final limitation with my implementation is that currently the ratings are retrofitted to my dataset using the above (hardcoded) parameters, and then a secondary predictive model is fitted using the $\delta$ as input predictors to a Bayesian model that predicts how many goals each team will score. This model is far more flexible and useful than the logistic regression used in the Elo system, but it still seems a waste to fit a second predictive-model separately from the rating system.
Rating System Optimised for Predictive Accuracy (RSOPA)
The properties of my ideal rating system are:
Be entirely data driven, i.e. parameters fitted using historical data Flexible to choice of predictive model Primary objective is predictive accuracy
I’ll call this method Rating System Optimised for Predictive Accuracy (RSOPA), just because there aren’t enough acronyms in the world already, and will now detail its implementation.
Predictive model
The crux of the Elo system is the $O-E$ term that is in $[-1, 1]$. Any predictive model for $E$ and any measure of outcome $O$ can be used provided a statistic in this range can be calculated.
For an initial proof-of-concept I’m going to use an ordinal multinomial regression for the match prediction model. This estimates probabilities for each of the 3 possible outcomes while taking the natural ordering into account (away win < draw < home win). The model is specified as follows
$$\mu = \alpha_{league} + \beta \delta$$ $$\text{logit}(q_1) = \eta_1 - \mu$$ $$\text{logit}(q_2) = \eta_2 - \mu$$ $$P(\text{away win}) = q_1$$ $$P(\text{draw}) = q_2 - q_1$$ $$P(\text{home win}) = 1 - q_2$$
Where $\alpha_{league}$ is a league specific intercept, while the coefficient $\beta$ and the two thresholds $\eta_1$ and $\eta_2$ are constant across leagues.
A value for $E$ in $[0,1]$ can be obtained by either scaling the linear predictor $\mu$, or by using a more hacky solution as the normalised weighted average of the 3 outcomes.
$$E = \frac{(1P(\text{away win}) + 2P(\text{draw}) + 3P(\text{home win}) - 1)}{3-1}$$
Using the same output formulation as before ($O \in \left\{0, 0.5, 1\right\}$), we have a more useful predictive model that can still be incorporated into the Elo rating system.
With 4 leagues in this dataset, this is a total of
7 parameters to be estimated for the predictive model. Rating system
The rating system is much as before
$$R^{\prime}_{H} = R_{H} + KG(\text{MoV})(O - E)$$
Where $G(x)$ now takes the general form $G(x) = A(1-exp(-Bx))+1$, where $A$ controls the maximum output and $B$ the decay rate, as shown in the plot below.
Alongside $A$ and $B$, the gain parameter $K$ can also be treated as free, and I’ll let it vary by league.Therefore, there are
6 free parameters to optimise from the rating update model. Parameter optimisation using Genetic Algorithm
In total there are 13 parameters to be solved: 7 from the match prediction model and 6 from the rating update equation. For an initial proof-of-concept I’m going to solve for these using a Genetic Algorithm - in particular the R interface to CMA-ES. As I’m primarily interested in obtaining the most accurate match predictions, my fitness function is the Ranked Probability Score (RPS), the specific case of the Brier score when the output is ordered.
I’m training the network on the 13,014 games in the top 4 European leagues (Premiership, Bundesliga, La Liga, Serie A) between the 2005-2006 and 2013-2014 seasons, although I’m allowing 3 seasons for the ratings to settle before the RPS is counted. To obtain an idea of how well the model works on unseen data I’ll form a test set comprising the seasons from 2014-2015 up to and including 2018-2019.
The hardcoded aspects of this model are the functional forms of the 2 models and the starting rating, which I’m setting at 2,000 completely arbitrarily.
Final model
The table below compares the RPS from RSOPA against two Bayesian models that were fitted to the original hand-implemented rating system. The ‘Goals’ method predicts the goals scored by each team and was used in the Predictaball forecasts for the 2018-2019 season while ‘Ordinal’ is an ordinal multinomial regression and was used prior to 2018.
Unsurprisingly, RSOPA does far better on the training set - an improvement of 0.001 rps would be enough to move me up to 5th from 11th in 2017-2018 season - but there is less of an difference on the test set. This could be one area for future work, adding in regularisation methods or early-stopping to minimise the risk of over-fitting.
Model Training data Test data RSOPA 0.1998 0.1974 Bayesian Ordinal 0.2007 0.1977 Bayesian Goals 0.2009 0.1978
The plot below shows the modelled outcome probabilities for a range of $\delta$ across the 4 leagues. It immediately highlights the benefits of using a more complex match prediction model than the standard Elo logistic regression as the away win and draw probabilities aren’t linearly related. Incidentally, the peak in the draw probability at around $\delta = -200$ suggests that the home advantage is worth 200 rating points. Finally, the known fact that the Bundesliga has a lower proportion of home wins can be seen by the red curve lying lower than the others, resulting from the league specific $\alpha$.
Moving onto the rating network, and the fitted $G(\text{MoV})$ function is shown below. It allows for very high multipliers which concerns me slightly as it could allow for stronger teams’ ratings to skyrocket. If this becomes a problem I could reshape $G$ to include a penalty term for the rating difference to reduce this auto-correlation, like what 538 use and I’ve borowed for the current version of Predictaball.
The team ratings at the end of the 2018-2019 season are shown below for both the current implementation (red) and RSOPA (blue). RSOPA definitely produces more dispersed teams due to the large multiplier effect of $G$, which is around 3x for a 5-goal win in the current method but 4.4x in the new system.
This has some advantages, in that it’s clear to see who the best and worst-rated teams are, for example the top 6 in the Premier League are far easier to identify than under the old system. But it could be easy to overinterpret these ratings by reading too much into the large differences between the teams, when really these need to be taken into consideration alongside the $\beta$ coefficient.
For example, if we were to predict the outcome of Liverpool vs Everton under the old system we would get H/D/A of 72% / 17% / 11%, which isn’t that much less than the 78% / 14% / 8% under RSOPA, so even though Liverpool’s rating is far closer to Everton’s under the old method the $\beta$ was larger to compensate.
Further work
I’ve run out of time to tinker with this concept further this Summer, but the next thing I’d be looking at improving would be to remove $G$ entirely and encapsulate the predicted scale of victory in $E$, so that the rating update is simply $K(O-E)$ as before.
In particular, I’d like to use my current goal prediction model rather than the ordinal multinomial regression I’m using here and I’d like to go back to using a Bayesian model rather than just using point estimates of parameters. However, this is computationally more challenging to fit so I’d need to find time to sit down and properly think about it.
As for now, I’m going to use these ratings on The Predictaball’s website for the coming 2019-2020 season and see how they get on.
|
Title Minimization of functionals of the gradient by Baire's theorem Publication Type Journal Article Year of Publication 2000 Authors Zagatti, S Journal SIAM J. Control Optim. 38 (2000) 384-399 Abstract
We give sufficient conditions for the existence of solutions of the minimum problem $$ {\mathcal{P}}_{u_0}: \qquad \hbox{Minimize}\quad \int_\Omega g(Du(x))dx, \quad u\in u_0 + W_0^{1,p}(\Omega,{\mathbb{R}}), $$ based on the structure of the epigraph of the lower convex envelope of g, which is assumed be lower semicontinuous and to grow at infinity faster than the power p with p larger than the dimension of the space. No convexity conditions are required on g, and no assumptions are made on the boundary datum $u_0\in W_0^{1,p}(\Omega,\mathbb{R})$.
URL http://hdl.handle.net/1963/3511 DOI 10.1137/S0363012998335206 Minimization of functionals of the gradient by Baire's theorem
Research Group:
|
Search
Now showing items 1-1 of 1
Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
|
Hedonic Evaluation Approach
The
hedonic approach to economic assessment can be used for evaluating the economic value of environmental goods such as noise, air or water quality, landscape and similar goods.
The
hedonic approach is based on the assumption that goods can be considered aggregates of different attributes, some of which, as they cannot be sold separately, do not have an individual price. On real estate markets for example, it is not possible to purchase separately the room, the preferred location, the panoramic qualities, quality of air or of surrounding landscape. It is one frequently applied methodology for the assessment of natural values based on market values.
The hedonic approach to evaluation attempts to estimate the economic value using implicit prices of single characteristics of a good on the basis of market values of the whole good. The use of this approach is of particular interest in the field of environmental valuation, as it can be assumed that the values attributed to natural resources are attributes of commodities which are sold on the market. Obviously, the more closely the market good is related to the use of a natural resource, the more suitable is this approach for the evaluation of the natural resource. Real estate properties are very interesting in this context, as their values are strongly influenced by locational characteristics.
The approach assumes that the economic value of each attribute influences the total value of the commodity and can thus be revealed as a difference in price, for instance real estate properties, assuming all other characteristics to remain constant. It is based on the identification of the part of value which regards the characteristic to be evaluated, and in case of variations of the characteristic, on the possibility of estimating its demand function. The theory at the basis of hedonic methods have been formulated by Rosen (1974
[1]) and successively improved with regards to the valuation of environmental goods by Freeman (1979 [2]). Rosen’s model simulates a competitive market and foresees the simultaneous estimation of demand and supply function. In the particular case of real estate markets, with a very rigid supply, the model can be simplified and traced back to the neoclassic scheme of the theory of consumer’s demand (Diamond and Smith [3]).The correct application of the hedonic approach requires that the following hypothesis have been positively verified: the market must offer a continuous range of choices, that means, all combinations between private good and environmental conditions; purchasers must be able to behave according to the principle of decreasing marginal utility with respect to the environmental characteristics; purchasers must have the same opportunity of access to the market. In other words, they must have the same cost of information, transaction and transfer, the same income available and the same mobility; a perfect transparency of prices and characteristics must be given; prices must adapt immediately to changes in the demand for environmental goods.
These hypothesis are of course restrictive, as they describe a market which is perfectly transparent on the side of the offer, and, is homogeneous and perfectly competitive on the demand side.
On an operative level the approach proceeds in two phases:
estimate the value function of the private good; estimate of the demand function of the environmental characteristics of the surplus.
The estimation of the function of the prices of private goods is generally obtained applying statistical methods of linear (or linearized) multiple regression: [math]V=\beta_0+\sum_i\cdot x_i+ \epsilon[/math]
In this way each single characteristic of the good is associated with a parameter
ß i which represents, if the above conditions are satisfied, according to the specification of the variable, the implicit price.The hedonic approach shows, besides its doubtless advantages, some limits which add to the general one of respecting the hypothesis on the characteristics of real estate markets: it can be applied only in presence of a good number of market exchanges, as the model representing the market requires a certain number of good quality data; the market must be sufficiently transparent; the valuation might be biased if there are expectations with regards to changes in environmental qualities; It is not possible to estimate the total economic value of the environmental good, but only the value connected to present and, with some caution, future uses. References Rosen, S. (1974). "Hedonic Prices and Implicit Markets: Product Differentiation in Pure Competition." Journal of Political Economy 82 Freeman, A. M., III, (1979). "Hedonic Prices, Property Values and Measuring Environmental Benefits: A Survey of the Issues." Scandinavian Journal of Economics 81(2) Diamond, D. B. and B. A. Smith (1985). "Simultaneity in the Market for Housing Characteristics." Journal of Urban Economics 17(280-292) See also Multifunctionality and Valuation in coastal zones: concepts, approaches, tools and case studies Multifunctionality and Valuation in coastal zones: introduction Total Economic Value Economic value
Please note that others may also have edited the contents of this article.
Paolo Rosato (2019): Hedonic Evaluation Approach. Available from http://www.coastalwiki.org/wiki/Hedonic_Evaluation_Approach [accessed on 23-09-2019]
|
We consider implicit signatures over finite semigroups determined by sets of
pseudonatural numbers. We prove that, under relatively simple hypotheses on a
pseudovariety V of semigroups, the finitely generated free algebra for the
largest such signature is closed under taking factors within the free pro-V
semigroup on the same set of generators. Furthermore, we show that the natural
analogue of the Pin-Reutenauer descriptive procedure for the closure of a
rational language in the free group with respect to the profinite topology
holds for the pseudovariety of all finite semigroups. As an application, we
establish that a pseudovariety enjoys this property if and only if it is full.
Section: Automata, Logic and Semantics
When nodes can repeatedly update their behavior (as in agent-based models
from computational social science or repeated-game play settings) the problem
of optimal network seeding becomes very complex. For a popular
spreading-phenomena model of binary-behavior updating based on thresholds of
adoption among neighbors, we consider several planning problems in the design
of \textit{Sticky Interventions}: when adoption decisions are reversible, the
planner aims to find a Seed Set where temporary intervention leads to long-term
behavior change. We prove that completely converting a network at minimum cost
is $\Omega(\ln (OPT) )$-hard to approximate and that maximizing conversion
subject to a budget is $(1-\frac{1}{e})$-hard to approximate. Optimization
heuristics which rely on many objective function evaluations may still be
practical, particularly in relatively-sparse networks: we prove that the
long-term impact of a Seed Set can be evaluated in $O(|E|^2)$ operations. For a
more descriptive […]
Section: Distributed Computing and Networking
Word $W$ is an instance of word $V$ provided there is a homomorphism $\phi$
mapping letters to nonempty words so that $\phi(V) = W$. For example, taking
$\phi$ such that $\phi(c)=fr$, $\phi(o)=e$ and $\phi(l)=zer$, we see that
"freezer" is an instance of "cool".
Let $\mathbb{I}_n(V,[q])$ be the probability that a random length $n$ word on
the alphabet $[q] = \{1,2,\cdots q\}$ is an instance of $V$. Having previously
shown that $\lim_{n \rightarrow \infty} \mathbb{I}_n(V,[q])$ exists, we now
calculate this limit for two Zimin words, $Z_2 = aba$ and $Z_3 = abacaba$.
Section: Combinatorics
We study the configuration space of rectangulations and convex subdivisions
of $n$ points in the plane. It is shown that a sequence of $O(n\log n)$
elementary flip and rotate operations can transform any rectangulation to any
other rectangulation on the same set of $n$ points. This bound is the best
possible for some point sets, while $\Theta(n)$ operations are sufficient and
necessary for others. Some of our bounds generalize to convex subdivisions of
$n$ points in the plane.
Section: Combinatorics
In this paper, we study the behaviour of the generalized power domination number of a graph by small changes on the graph, namely edge and vertex deletion and edge contraction. We prove optimal bounds for $\gamma_{p,k}(G-e)$, $\gamma_{p,k}(G/e)$ and for $\gamma_{p,k}(G-v)$ in terms of $\gamma_{p,k}(G)$, and give examples for which these bounds are tight. We characterize all graphs for which $\gamma_{p,k}(G-e) = \gamma_{p,k}(G)+1$ for any edge $e$. We also consider the behaviour of the propagation radius of graphs by similar modifications.
Section: Graph Theory
Closed monopolies in graphs have a quite long range of applications in
several problems related to overcoming failures, since they frequently have
some common approaches around the notion of majorities, for instance to
consensus problems, diagnosis problems or voting systems. We introduce here
open $k$-monopolies in graphs which are closely related to different parameters
in graphs. Given a graph $G=(V,E)$ and $X\subseteq V$, if $\delta_X(v)$ is the
number of neighbors $v$ has in $X$, $k$ is an integer and $t$ is a positive
integer, then we establish in this article a connection between the following
three concepts:
- Given a nonempty set $M\subseteq V$ a vertex $v$ of $G$ is said to be
$k$-controlled by $M$ if $\delta_M(v)\ge \frac{\delta_V(v)}{2}+k$. The set $M$
is called an open $k$-monopoly for $G$ if it $k$-controls every vertex $v$ of
$G$.
- A function $f: V\rightarrow \{-1,1\}$ is called a signed total
$t$-dominating function for $G$ if $f(N(v))=\sum_{v\in N(v)}f(v)\geq t$ […]
Section: Graph Theory
We find surprisingly simple formulas for the limiting probability that the
rank of a randomly selected vertex in a randomly selected phylogenetic tree or
generalized phylogenetic tree is a given integer.
Section: Combinatorics
The notion of Shared Risk Link Groups (SRLG) captures survivability issues when a set of links of a network may fail simultaneously. The theory of survivable network design relies on basic combinatorial objects that are rather easy to compute in the classical graph models: shortest paths, minimum cuts, or pairs of disjoint paths. In the SRLG context, the optimization criterion for these objects is no longer the number of edges they use, but the number of SRLGs involved. Unfortunately, computing these combinatorial objects is NP-hard and hard to approximate with this objective in general. Nevertheless some objects can be computed in polynomial time when the SRLGs satisfy certain structural properties of locality which correspond to practical ones, namely the star property (all links affected by a given SRLG are incident to a unique node) and the span 1 property (the links affected by a given SRLG form a connected component of the network). The star property is defined in a multi-colored […]
Section: Distributed Computing and Networking
The new finite state machine package in the mathematics software system
SageMath is presented and illustrated by many examples. Several combinatorial
problems, in particular digit problems, are introduced, modeled by automata and
transducers and solved using SageMath. In particular, we compute the asymptotic
Hamming weight of a non-adjacent-form-like digit expansion, which was not known
before.
Section: Analysis of Algorithms
A graph is an efficient open domination graph if there exists a subset of
vertices whose open neighborhoods partition its vertex set. We characterize
those graphs $G$ for which the Cartesian product $G \Box H$ is an efficient
open domination graph when $H$ is a complete graph of order at least 3 or a
complete bipartite graph. The characterization is based on the existence of a
certain type of weak partition of $V(G)$. For the class of trees when $H$ is
complete of order at least 3, the characterization is constructive. In
addition, a special type of efficient open domination graph is characterized
among Cartesian products $G \Box H$ when $H$ is a 5-cycle or a 4-cycle.
Section: Graph Theory
The partial sum of the states of a Markov chain or more generally a Markov
source is asymptotically normally distributed under suitable conditions. One of
these conditions is that the variance is unbounded. A simple combinatorial
characterization of Markov sources which satisfy this condition is given in
terms of cycles of the underlying graph of the Markov chain. Also Markov
sources with higher dimensional alphabets are considered.
Furthermore, the case of an unbounded covariance between two coordinates of
the Markov source is combinatorically characterized. If the covariance is
bounded, then the two coordinates are asymptotically independent.
The results are illustrated by several examples, like the number of specific
blocks in $0$-$1$-sequences and the Hamming weight of the width-$w$
non-adjacent form.
Section: Analysis of Algorithms
Ruskey and Savage in 1993 asked whether every matching in a hypercube can be
extended to a Hamiltonian cycle. A positive answer is known for perfect
matchings, but the general case has been resolved only for matchings of linear
size. In this paper we show that there is a quadratic function $q(n)$ such that
every matching in the $n$-dimensional hypercube of size at most $q(n)$ may be
extended to a cycle which covers at least $\frac34$ of the vertices.
Section: Graph Theory
The objective of this paper is to find in a setting of n sequential observations of objects a good online policy to select the k bestof these n uniquely rankable objects. This focus is motivated by the fact that it is hard to find closed form solutions of optimalstrategies for general k and n. Selection is without recall, and the idea is to investigate threshold functions which maintain allpresent information, that is thresholds which are functions of all selections made so far. Our main interest lies in the asymptoticbehaviour of these thresholds as n -> infinity and in the corresponding asymptotic performance of the threshold algorithm.
Section: Analysis of Algorithms
We prove geometric Ramsey-type statements on collections of lines in 3-space.
These statements give guarantees on the size of a clique or an independent set
in (hyper)graphs induced by incidence relations between lines, points, and
reguli in 3-space. Among other things, we prove that: (1) The intersection
graph of n lines in R^3 has a clique or independent set of size Omega(n^{1/3}).
(2) Every set of n lines in R^3 has a subset of n^{1/2} lines that are all
stabbed by one line, or a subset of Omega((n/log n)^{1/5}) such that no
6-subset is stabbed by one line. (3) Every set of n lines in general position
in R^3 has a subset of Omega(n^{2/3}) lines that all lie on a regulus, or a
subset of Omega(n^{1/3}) lines such that no 4-subset is contained in a regulus.
The proofs of these statements all follow from geometric incidence bounds --
such as the Guth-Katz bound on point-line incidences in R^3 -- combined with
Turán-type results on independent sets in sparse graphs and […]
Section: Combinatorics
A right ideal (left ideal, two-sided ideal) is a non-empty language $L$ over
an alphabet $\Sigma$ such that $L=L\Sigma^*$ ($L=\Sigma^*L$,
$L=\Sigma^*L\Sigma^*$). Let $k=3$ for right ideals, 4 for left ideals and 5 for
two-sided ideals. We show that there exist sequences ($L_n \mid n \ge k $) of
right, left, and two-sided regular ideals, where $L_n$ has quotient complexity
(state complexity) $n$, such that $L_n$ is most complex in its class under the
following measures of complexity: the size of the syntactic semigroup, the
quotient complexities of the left quotients of $L_n$, the number of atoms
(intersections of complemented and uncomplemented left quotients), the quotient
complexities of the atoms, and the quotient complexities of reversal, star,
product (concatenation), and all binary boolean operations. In that sense,
these ideals are "most complex" languages in their classes, or "universal
witnesses" to the complexity of the various operations.
Section: Automata, Logic and Semantics
A narrow connection between infinite binary words rich in classical
palindromes and infinite binary words rich simultaneously in palindromes and
pseudopalindromes (the so-called $H$-rich words) is demonstrated.
The correspondence between rich and $H$-rich words is based on the operation
$S$ acting over words over the alphabet $\{0,1\}$ and defined by
$S(u_0u_1u_2\ldots) = v_1v_2v_3\ldots$, where $v_i= u_{i-1} + u_i \mod 2$.
The operation $S$ enables us to construct a new class of rich words and a new
class of $H$-rich words.
Finally, the operation $S$ is considered on the multiliteral alphabet
$\mathbb{Z}_m$ as well and applied to the generalized Thue--Morse words. As a
byproduct, new binary rich and $H$-rich words are obtained by application of
$S$ on the generalized Thue--Morse words over the alphabet $\mathbb{Z}_4$.
Section: Combinatorics
In this paper, we confirm conjectures of Laborde-Zubieta on the enumeration
of corners in tree-like tableaux and in symmetric tree-like tableaux. In the
process, we also enumerate corners in (type $B$) permutation tableaux and
(symmetric) alternative tableaux. The proof is based on Corteel and Nadeau's
bijection between permutation tableaux and permutations. It allows us to
interpret the number of corners as a statistic over permutations that is easier
to count. The type $B$ case uses the bijection of Corteel and Kim between type
$B$ permutation tableaux and signed permutations. Moreover, we give a bijection
between corners and runs of size 1 in permutations, which gives an alternative
proof of the enumeration of corners. Finally, we introduce conjectural
polynomial analogues of these enumerations, and explain the implications on the
PASEP.
Section: Combinatorics
We study two different objects attached to an arbitrary quadrangulation of a regular polygon. The first one is a poset, closely related to the Stokes polytopes introduced by Baryshnikov. The second one is a set of some paths configurations inside the quadrangulation, satisfying some specific constraints. These objects provide a generalisation of the existing combinatorics of cluster algebras and nonnesting partitions of type A.
Section: Combinatorics
For every interval map with finitely many periodic points of periods 1 and 2, we associate a word by taking the periods of these points from left to right. It is natural to ask which words arise in this manner. In this paper we give two different characterizations of the language obtained in this way. We study Markov chains for $\alpha$-orientations of plane graphs, these are
orientations where the outdegree of each vertex is prescribed by the value of a
given function $\alpha$. The set of $\alpha$-orientations of a plane graph has
a natural distributive lattice structure. The moves of the up-down Markov chain
on this distributive lattice corresponds to reversals of directed facial cycles
in the $\alpha$-orientation. We have a positive and several negative results
regarding the mixing time of such Markov chains.
A 2-orientation of a plane quadrangulation is an orientation where every
inner vertex has outdegree 2. We show that there is a class of plane
quadrangulations such that the up-down Markov chain on the 2-orientations of
these quadrangulations is slowly mixing. On the other hand the chain is rapidly
mixing on 2-orientations of quadrangulations with maximum degree at most 4.
Regarding examples for slow mixing we also revisit the case of 3-orientations
of triangulations […]
Section: Graph Theory
|
Probability Seminar Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 14, Timo Seppäläinen, UW-Madison
Title:
Geometry of the corner growth model
Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah).
February 21, Diane Holcomb, KTH
Title:
On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Title: Quantitative homogenization in a balanced random environment
Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison).
Wednesday, February 27 at 1:10pm Jon Peterson, Purdue
Title:
Functional Limit Laws for Recurrent Excited Random Walks
Abstract:
Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina.
March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the
character ratio:
$$ \text{trace}(\rho(g))/\text{dim}(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant
rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM). April 4, Philip Matchett Wood, UW-Madison
Title:
Outliers in the spectrum for products of independent random matrices
Abstract: For fixed positive integers m, we consider the product of m independent n by n random matrices with iid entries as in the limit as n tends to infinity. Under suitable assumptions on the entries of each matrix, it is known that the limiting empirical distribution of the eigenvalues is described by the m-th power of the circular law. Moreover, this same limiting distribution continues to hold if each iid random matrix is additively perturbed by a bounded rank deterministic error. However, the bounded rank perturbations may create one or more outlier eigenvalues. We describe the asymptotic location of the outlier eigenvalues, which extends a result of Terence Tao for the case of a single iid matrix. Our methods also allow us to consider several other types of perturbations, including multiplicative perturbations. Joint work with Natalie Coston and Sean O'Rourke.
April 11, Eviatar Procaccia, Texas A&M Title: Stabilization of Diffusion Limited Aggregation in a Wedge.
Abstract: We prove a discrete Beurling estimate for the harmonic measure in a wedge in $\mathbf{Z}^2$, and use it to show that Diffusion Limited Aggregation (DLA) in a wedge of angle smaller than $\pi/4$ stabilizes. This allows to consider the infinite DLA and questions about the number of arms, growth and dimension. I will present some conjectures and open problems.
April 18, Andrea Agazzi, Duke
Title:
Large Deviations Theory for Chemical Reaction Networks
Abstract: The microscopic dynamics of well-stirred networks of chemical reactions are modeled as jump Markov processes. At large volume, one may expect in this framework to have a straightforward application of large deviation theory. This is not at all true, for the jump rates of this class of models are typically neither globally Lipschitz, nor bounded away from zero, with both blowup and absorption as quite possible scenarios. In joint work with Amir Dembo and Jean-Pierre Eckmann, we utilize Lyapunov stability theory to bypass this challenges and to characterize a large class of network topologies that satisfy the full Wentzell-Freidlin theory of asymptotic rates of exit from domains of attraction. Under the assumption of positive recurrence these results also allow for the estimation of transitions times between metastable states of this class of processes.
April 25, Kavita Ramanan, Brown
Title:
Beyond Mean-Field Limits: Local Dynamics on Sparse Graphs
Abstract: Many applications can be modeled as a large system of homogeneous interacting particle systems on a graph in which the infinitesimal evolution of each particle depends on its own state and the empirical distribution of the states of neighboring particles. When the graph is a clique, it is well known that the dynamics of a typical particle converges in the limit, as the number of vertices goes to infinity, to a nonlinear Markov process, often referred to as the McKean-Vlasov or mean-field limit. In this talk, we focus on the complementary case of scaling limits of dynamics on certain sequences of sparse graphs, including regular trees and sparse Erdos-Renyi graphs, and obtain a novel characterization of the dynamics of the neighborhood of a typical particle. This is based on various joint works with Ankan Ganguly, Dan Lacker and Ruoyu Wu.
Friday, April 26, Colloquium, Van Vleck 911 from 4pm to 5pm, Kavita Ramanan, Brown
Title:
Tales of Random Projections
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
Tuesday , May 7, Van Vleck 901, 2:25pm, Duncan Dauvergne (Toronto)
Title:
The directed landscape
Abstract: I will describe the construction of the full scaling limit of (Brownian) last passage percolation: the directed landscape. The directed landscape can be thought of as a random scale-invariant `directed' metric on the plane, and last passage paths converge to directed geodesics in this metric. The directed landscape is expected to be a universal scaling limit for general last passage and random growth models (i.e. TASEP, the KPZ equation, the longest increasing subsequence in a random permutation). Joint work with Janosch Ormann and Balint Virag.
|
There is no way to calculate the Starting Current or Locked Rotor Current (LRA) without more information!
Single-phase or three-phase? NEMA Motor Design B, C or D?
What does academic education's sake mean? A voltage of 15V with a power of 132kW is meaningless for an induction motor. You just can't make up numbers. You are also using \$P = V\ I\$, which is DC power.
You'd be better off looking up a motor nameplate and going from there.
Take a 150hp, 1789rpm, 460V, Design B, Code G, 3-phase induction motor. So rated current is 163A, with a power factor of 0.897 lagging and an efficiency of 96.2%.Code G gives you locked rotor kVA on a per hp basis. Locked rotor kVA will allow you to calculate LRA. Code G = 5.6 up to but not including 6.3. Worst case = 6.3.$$150hp \times 6.3 = 945 kVA$$
$$ S = \sqrt {3}\ V_{Line}\ I_{Line} $$$$ I_{Line} = \frac {S} {\sqrt {3}\ V_{Line}} = \frac {945 kVA} {\sqrt {3} \times 460V} = 1,186A $$
LRA will be between 1,102A and < 1,186A vs 163A or 676% to 728% of full-load current.
|
Separable Differential Equations: Exponential Decay
When I was in high school, my chemistry teacher presented me with a radioactive decay problem, and a formula that read:
$$Q=Q_{0} * e^{-r t}$$
Where \(Q\) represented the current amount of radioactive material, \(Q_0\) represented the starting amount of material, and \(r\) was some unknown “decay rate”. I was greatly confused by this. “Why do we raise the time function to the power of e?” I asked. “Why not raise the time function to the power of \(\frac{1} {2}\) ? We are working with half-lives after all”. It turns out that the answer to this question is surprisingly nuanced, and requires a deep dive into differential equations. So let’s take a look at a radioactive decay problem through the lens of differential equations.
Now I know this problem has four parts to it, but don’t panic—you won’t be needing your brown pants today. In fact, don’t even read Parts A-D yet. Let’s just look at the differential equation, and solve it before doing anything else:
$$\frac{d Q}{d t}=-r Q$$
You get bonus points if you recognized this as a
Separable Differential Equation, which means that you can move the two variables (Q and t) to opposite sides of the equation. The best way to succeed in a differential equations class is by always knowing what type of equation you’re working with. If you’re not sure what type of equation you’re looking at, you can always try to separate it. In this case, we multiply both sides by \(dt\), and divide both sides by \(Q\).
$$\frac{d Q}{Q}=-r * d t$$
The point of separating the equation is so we can take the integral of both sides, thereby getting rid of the differential terms \(dQ\) and \(dt\).
$$\int \frac{d Q}{Q}=-r * \int d t$$
Note that we can leave the \(-r\) term outside the integral for \(t\), since \(r\) is a constant, and does not depend on \(t\). We solve both integrals to get
$$\ln (Q)=-r t+C_{1}$$
Never forget that \(+C\). Hopefully you kept those good habits from MAT21B. Also don’t forget why we’re doing this. “Solving” a differential equation means finding detailed temporal information about the time-sensitive variable, \(Q(t)\). When I write \(Q(t)\), I’m saying that “\(Q\) is a function of \(t\)”. Let’s finish up this equation:
$$Q(t)=e^{\left(-r t+C_{1}\right)}$$
Note that \(e^{(-r t+C)}\) can be written as: [\(e^{-r t} * e^{C_{1}}\)]. Also note that \(e^{C_{1}}\) is just another constant, which I’ll call \(C_2\). That was a bit fast—let me do it again in slow motion:
$$Q(t)=e^{\left(-r t+C_{1}\right)}=e^{-r t} * e^{c_{1}}=C_{2} e^{-r t}$$
Great, now we’re almost done! The last thing that you have to deal with in any differential equation is getting rid of any lingering \(C\) The way that we do this is through
Initial Conditions. Initial conditions allow us to plug in predetermined values for \(Q\) and \(t\) that reveal the value of \(C_2\). If you take a close look at the problem, you might notice that the wording for Part A gives us an initial condition right away. Let’s use it to find \(C_2\):
From this phrasing, and our equation: \(Q(t)=C_{2} e^{-r t}\), we get:
$$Q(0)=C_{2} e^{-r(0)}=C_{2}=100 \mathrm{mg}$$
And there you have it. Now that we know \(C_2\), we can write our solved differential equation.
$$\boldsymbol{Q}(\boldsymbol{t})=\mathbf{1} \mathbf{0} \mathbf{0} \boldsymbol{m} \boldsymbol{g} * \boldsymbol{e}^{-\boldsymbol{r} t}$$
I know I haven’t solved any of Parts A-D yet, but now that you have the finished equation, they will basically solve themselves! Here we go:
Part A
Here we are asked to find \(r\) if \(Q(1 week) = 84 \mathrm(mg)\). Let’s plug it in:
$$\begin{array}{c}{Q(1)=84 m g=100 m g * e^{-r}} \\ {e^{-r}=0.84} \\ {r=-\ln (0.84)=0.1744}\end{array}$$
To make sure that units are consistent, the units of \(r\) must be \(\frac {1} {weeks}\). Otherwise, \(e^{-rt}\) would have non-cancelling units, and that would be awkward.
Part B
We practically already solved this one; just plug in \(r\) from Part A:
$$Q(t)=100 m g * e^{-0.1744 t}$$
Part C
This time, we’re looking for \(t\) when \(Q(t) = 50mg \). Again, just plug it in:
\begin{array}{c}{Q(t)=50 m g=100 m g * e^{-0.1744 t}} \\ {e^{-0.1744 t}=0.5} \\ {t=-\frac{\ln (0.5)}{0.1744}=3.974}\end{array}
Quick, what units does \(t\) have? Hopefully you said “weeks”. We have to remain consistent with our given values.
Part D
The only tricky thing going on here is labelling the two points. We know 3 points easily: \(Q(0)=100 m g, Q(1)=84 m g\), and \(Q(3.974)=50 m g\). If you want to be fancy you include: \(Q\left(\frac{1}{0.1774}\right)=\frac{100}{e} m g\), or \(Q(2 * 3.974)=25 m g\). The following graph is a reasonable solution:
What Did We Learn?
Aside from how to solve the above problem, we learned two important things today: first of all, that solving a differential equation should always be your first step in tackling a problem, regardless of how many parts the question has. Additionally, we learned that the equation [ \(Q=Q_{0} * e^{-r t}\) ] is the ultimate solution to exponential decay problems—problems in which the rate of change of a substance, \(\frac{d Q}{d t}\), is proportional to how much of that substance you have. When I was given this equation in high school, I didn’t understand why we used the natural number \(e\). But knowing that the equation represents the
general case of this differential equation completely explains why mathematicians prefer it over all other potential solutions for such a problem.
|
What questions led to the invention of complex differentiation/integration? How were their definitions agreed upon?
Real differentiation/integration has an obvious meaning. To extend calculus to the complex numbers, why would this be done, is it even meaningful to call it 'differentiation'/'integration'?
What questions led to the invention of complex differentiation/integration? How were their definitions agreed upon?
There is no difference in differentiation. Derivative of a complex function of a complex variable is defined by the same formula $$f'(z)=\lim_{h\to 0} (f(z+h)-f(z))/h$$ as for the real variable. Concerning integration, one reason was the desire to investigate integrals of real functions. One of the consequences (and motivations) of Cauchy theory is that it evaluates integrals of real functions on real intervals by manipulating with integrals in the complex domain.
But there is a more important and more profound reason. Many equations (differential and functional) can be solved in the form of power series. This was essentially discovered by Newton, and he considered this his main discovery in Calculus. A power series, if it converges anywhere except $x=0$, converges in a disk in the complex plane, and it is impossible to understand its properties while staying in the real domain only. This is the most important motivation of extending Calculus to complex domain.
Here is a simple example: we have a power series expansion $$f(x)=\frac{1}{1+x^2}=1-x^2+x^4-x^6+\ldots.$$ It converges in the real domain for $-1<x<1$ only. But the left hand side $f(x)=1/(1+x^2)$ is a nice function on the whole real line. By looking at this function on the real line only it is impossible to understand what happens at $\pm 1$, why the series suddenly stops to converge.
Remark. I did not like the sentence "definition agreed upon". These definitions are essentially forced on us by the nature of things. There is nothing here to agree or disagree upon:-) Most of these things were discovered by Cauchy in the beginning of 19th century, with important contributions of Gauss and Abel. But they were discovered, not "agreed upon".
|
Can the centralizer of Inn(G) in Aut(G), where G is preferably any non-abelian finite one, equal to Inn(G) itself? Clearly, such centralizer contains all $f$ in Aut(G) where $f(g)g^{-1}$ are in Z(G).
To summarize (thanks to Steve D. for the article pointer):
If $G$ is a group, an automorphism $\sigma\in\mathrm{Aut}(G)$ is said to be a "central automorphism" if $\sigma$ commutes with every automorphism in $\mathrm{Inn}(G)$, or equivalently if $\sigma(g)g^{-1}\in Z(G)$ for all $g\in G$. The central automorphisms form a normal subgroup of $\mathrm{Aut}(G)$, denoted by $\mathrm{Aut}_c(G)$.
You are asking for examples with $\mathrm{Inn}(G)=\mathrm{Aut}_c(G)$. In particular, this requires $\mathrm{Inn}(G)\cong G/Z(G)$ to be abelian, hence $G$ is nilpotent of class at most $2$.
Case 1. $G$ abelian.
If $G$ is abelian, then $\mathrm{Inn}(G)=\{1\}$ and $\mathrm{Aut}_c(G)=\mathrm{Aut}(G)$. For the equality between the group of central automorphisms and inner automorphisms to hold, we would need the group of automorphisms to be trivial. If $G$ is finite, the only possibilities are $|G|=1$ and $|G|=2$. As was recently discussed in this very forum, whether there are any infinite abelian groups with this property depends on your Set Theory. If we assume the Axiom of Choice (in fact, weaker axioms are sufficient here), then there are no infinite abelian examples. But there are models of $ZF$ without choice where there are such infinite groups, as Asaf Karagila showed in that question.
Case 2. Finite nonabelian groups.
Moving on to groups of nilpotency class exactly $2$. A finite nilpotent group is the products of its Sylow $p$-subgroups, which are characteristic and have no nontrivial homomorphisms between them; so the group of automorphisms of a finite nilpotent group naturally breaks up into a product of automorphisms groups of $p$-groups. Thus, the problem reduces to characterizing the finite $p$-groups of nilpotency class two in which every central automorphism is inner. A theorem of M.J. Curran and D.J. McCaughan (
Central automorphisms that are almost inner, Comm. Algebra 29 (no. 5), 2081-2087, 2001) completely solves the problem then: Theorem. If $G$ if a finite nonabelian p-group, then $\mathrm{Aut}_c(G)=\mathrm{Inn}(G)$ if and only if $G'=Z(G)$ and $Z(G)$ is cyclic.
There are many examples of such groups: all nonabelian groups of order $p^3$, all extra-special $p$-groups, etc.
So a finite group satisfies the condition if and only:
$G$ is trivial; or $G$ is a direct product of its Sylow $p$-subgroups, $Z(G)=G'$ and $Z(G)$ is cyclic; or $G\cong C_2\times H$, where $C_2$ is the cyclic group of order two, and $H$ is a group of odd order of types 1 or 2. Case 3. Infinite nonabelian groups. I can't answer it completely, except to say that examples certainly exist.
This one seems more complicated; I'll take a look at the Curran-MacCaughan paper later on, but there are certainly examples with $G'=Z(G)$ cyclic. For example, consider the relatively free nilpotent group of class $2$. This is $$F = \Bigl\langle x,y\Bigm| [x,y,x]=[x,y,y]=1\Bigr\rangle,$$ and is isomorphic to the group of $3\times 3$ upper triangular matrices with integer coefficients and $1$s in the diagonal, by the map that sends $x$ to $I+E_{12}$ and $y$ to $I+E_{21}$. Every element has normal form $x^ay^b[y,x]^c$ with $a,b,c\in\mathbb{Z}$, and multiplication given by $$(x^ay^b[y,c]^c)(x^{\alpha}y^{\beta}|[y,x]^{\delta}) = x^{a+\alpha}y^{b+\beta}[y,x]^{c+\delta+\alpha b}.$$
Any two elements of $F$ determine an endomorphism by mapping $x$ to the first element and $y$ to the second; thus, the automorphisms of $F$ correspond to maps of the form $$\begin{align*} x &\longmapsto x^a y^b [y,x]^{k}\\ y &\longmapsto x^c y^d [y,x]^{\ell} \end{align*}$$ with $ad-bc = \pm 1$. An inner automorphism given by conjugation by $x^ry^s$ maps $x$ to $x[x,x^ry^s] = x[y,x]^{-s}$ and $y$ to $y[y,x^ry^s]=y[y,x]^r$. Thus, inner automorphisms correspond to $b=c=0$ and $a=d=1$.
If an automorphism of $F$ is central, then $(x^ay^b[y,x]^k)x^{-1} = x^{a-1}y^b[y,x]^{k-b}$ and $(x^cy^d[y,x]^{\ell})y^{-1}=x^cy^{d-1}[y,x]^{\ell}$ are both central; since the center is the commutator subgroup, this requires $a-1=b=c=d-1=0$, showing that the automorphism is inner. Hence $\mathrm{Inn}(F)=\mathrm{Aut}_c(F)$.
I
think that the conditions $G'=Z(G)$ and $Z(G)$ cyclic will still be at least sufficient in the infinite case.
I was fiddling with this, but never had time to finish. Maybe it is interesting to someone.
We define two bijections that help to describe the centralizer of the inner automorphism group in the full automorphism group.
exp : Hom(G,Z(G)) → C Aut(G)(Inn(G)) : ζ ↦ ( g ↦ gζ(g) ) log : C Aut(G)(Inn(G)) → Hom(G,Z(G)) : φ ↦ ( g ↦ g −1φ(g) )
We need to show these are well-defined (that is they have the specified ranges).
Proof: If ζ is a homomorphism from G to Z(G), then exp(ζ) : G → G : g ↦ gζ(g) is indeed an automorphism, with inverse exp(−ζ) defined by −ζ, the homomorphism from G to Z(G) that takes g to ζ(g)
−1. The inner automorphism defined by h takes g to h −1gh = g h. If one applies exp(ζ) and then conjugation by h, one gets g hζ(g), since ζ(g) is central and so ζ(g) h = ζ(g). On the other hand, if one conjugates by h and then applies exp(ζ), then one gets g hζ(g h), but since ζ is a homomorphism into an abelian group, ζ is constant on conjugacy classes, and g hζ(g h) = g hζ(g). Hence every exp(ζ) is in C Aut(G)(Inn(G)). Conversely, if φ is an automorphism of G commuting with all inner automorphisms, then φ(x) φ(h) = φ(x h) = φ(x) h, so φ(h) and h define the same automorphism of G, and so φ(h) and h differ by some element of the center of G. Define log(φ):G→Z(G) implicitly from φ(h) = h log(φ)(h). It is not hard to check that log(φ) is a homomorphism from G to Z(G).
Notice that log(exp(ζ)) = ζ and exp(log(φ))=φ, so these are mutually inverse bijections.
Suppose then that Inn(G) = C
Aut(G)(Inn(G)). This means that exp(Hom(G,Z(G))) = Inn(G). This means that log : Inn(G) → Hom(G,Z(G)) is bijective! However, log(∧h) = ( g ↦ [g,h] ) has a very simple form for the inner automorphism ∧h defined by conjugation by h. In particular, [G,G] ≤ Z(G) for log(∧h) to even make sense, and so G has nilpotency class at most 2. Notice that the kernel of log(∧h) contains all of Z(G), for any h. Indeed, the kernel of log(∧h) is C G(h), and so all centralizers are normal, which is weird.
Another sick thing: log(∧hk)(g) = [g,hk] = [g,k][g,h] = [g,h][g,k] = (log(∧h) + log(∧k))(g), so in fact log is an isomorphism of abelian groups G/Z(G) ≅ Hom(G,Z(G)).
Unfortunately, I cannot find a good element of Hom(G,Z(G)). The transfer is useless, and the central series induction arguments give inequalities pointing the wrong way.
For finite groups, presumably this is taken care of by counting as in theorem 1 of:
but it didn't seem to address infinite (nilpotent) groups. The way it is done in this (very nice) paper mentioned by Steve certainly doesn't work out for infinite groups. In particular, Lemma I would be wonderful to have more concretely with homomorphisms rather than dualizing finite abelian groups a zillion times.
Yes, this can for instance happen when $Inn(G) = Aut(G)$ which is the case for the symmetric groups $S_n$ when $n\neq 6$
|
Let $$U_1 = \left\{ \left( \begin{array}{cc} x_1\\ x_2\\ x_3\\ x_4\\ \end{array} \right) \in \mathbb{R} :-x_1-x_2+x_3=0 \right\}$$
$$U_2 = span \left( \left( \begin{array}{cc} -1 \\ 1 \\ 1 \\ 2 \\ \end{array} \right), \left( \begin{array}{cc} 2 \\ -1 \\ -1 \\ 2 \\ \end{array} \right) \right) $$
, find the bases for $U_1, U_2, U_1 \cap U_2, U_1 + U_2$ and give the dimension of each subspace.
Now $U_1$ can be written as $$\left\{ \left( \begin{array}{cc} x_1\\ x_2\\ x_1 + x_2\\ x_4\\ \end{array} \right) \right\}$$ and so a basis would be $$B= \left( \left( \begin{array}{cc} 1 \\ 0 \\ 1 \\ 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 \\ 1 \\ 1 \\ 0 \\ \end{array} \right), \left( \begin{array}{cc} 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right) \right)$$
the vectors in the span of $U_2$ form a basis since they are linearly independent.
Now to find the basis of $U_1 + U_2$ I could take the span of all the basis vectors in $U_1$ and $U_2$, but since there are only four linearly independent vectors in that span, only four of those vectors form the basis of $U_1 + U_2$.
I cannot find a basis for the intersection of $U_1$ and $U_2$. I can only determine the dimension of that subspace by using the dimension formula for subspaces. So $U_1 \cap U_2$ must have dimension 1, but I cannot find a spanning vector. Can anybody help please?
|
The primal assignment problem that I have is:
$\begin{align} & \max \sum\limits_{i=i}^N\sum\limits_{j=1}^N c_{ij}x_{ij} \\ &\text{subject to} \\ &\sum_{i=1}^N x_{ij} = 1,~~j=1,\dots,N \\ &\sum_{j=1}^N x_{ij} = 1, ~~i=1,\dots,N \\ &x_{i,j} ~\in \{ 0,1 \} \end{align} $
and I would like to obtain its dual, which should be
$$ \min_{p_j} \bigg\{ \sum_{i=1}^N p_j + \sum_{i=1}^N \max \{ a_{ij} - p_j \} \bigg\} $$
Can you give me any hints on how to approach this derivation? I have tried with Lagrange multipliers but I have not been able to reach this expression.
|
Given a large sparse matrix $\Sigma$, I want to find a (block) diagonal matrix $\Omega$ which approximates $\Sigma^{-1}$.
In this specific problem, we know that $\Sigma$ is a block tridiagonal matrix. The blocks are of size $3 \times 3$. The blocks on the diagonal of $\Sigma$ stem from covariance matrices, i.e. they are symmetric positive semi-definite. The block $\Sigma_{i,i}$ represents the measurement uncertainty of the parameter block $i$ in my nonlinear least squares problem. The blocks $\Sigma_{i,i+1}$ on the upper diagonal represent the covariance matrix between parameter $i$ and $i+1$. (In general, they are smaller than the blocks on the diagonal, if that helps.) Therefore, $\Sigma$ is symmetric, positive semi-definite, and contains only real values.
As stated above, I am interested in finding the blockdiagonal matrix $\Omega$ which (best) approximates $\Sigma^{-1}$ in the sense that $\Omega \cdot \Sigma \approx I$. In the literature, I've often seen that the off-diagonal entries of $\Sigma$ are ignored which makes it easy to construct $\Omega$. I'd think there should be better approximations for finding $\Omega$.
I've also found this question, which is interested in the diagonal of the exact inverse. Again, I'd intuitively expect that this is not optimal as the off-diagonal entries of $\Omega$ are not small compared to its diagonal entries. This question aims to find any sparse matrix that approximates the inverse, but I am specifically interested in an approximation of a block diagonal form.
What is the best way to approach this problem?
|
Let us call any commutative semigroup $(S, +)$ strongly homogeneous if it satisfies the following three properties:
P-1. Every endomorphism on $S$ is a bijection.
P-2. Any two endomorphisms on $S$ commute.
P-3. For every $x, y \in S$ there exist one and only one endomorphism $\psi$ on $S$ such that $\psi(x) = y$.
Is the following true?
Corresponding to any element $1 \in S$ there exist one a̶n̶d̶ ̶o̶n̶l̶y̶ ̶o̶n̶e̶ associative binary operation $\times$ satisfying, for all $x,y,z \in S$,
$$\tag 1 1 \times 1 = 1$$
$$\tag 2 x \times y = y \times x$$
$$\tag 3 x \times (y+z) = x \times y + x \times z$$
Moreover, for every $x \ S$ there exists $y \in S$ with $x \times y = 1$.
My Work
I sketched out a proof for a special case here, and I think the argument can be lifted, but I'm not completely confident in using this logic and would like verification/confirmation.
If the theory is false, a counter-example would be appreciated.
|
Can anyone please help me to prove $ X=M$ using the following set of the equation(first-order logic)in Isabelle/HOL?
$ N>=M$
$ \forall n. 0\leq n<N \rightarrow n<M$
$ X=N$
where $ N, M, X$ are integers constant. $ n$ integer variable.
Can anyone please help me to prove $ X=M$ using the following set of the equation(first-order logic)in Isabelle/HOL?
$ N>=M$
$ \forall n. 0\leq n<N \rightarrow n<M$
$ X=N$
where $ N, M, X$ are integers constant. $ n$ integer variable.
How to get the site users and site groups in SPFx using PnP JS?
Thanks in advance.
I want to add a user group to sharepoint subsite using javascript. I have used follwing code and it added a user group to subsite(ABS). and it also added user group to the parent site(OMSTest) which is a root sitecollection. why is that?
function CreateUserGroups(url) { var siteUrl = 'https://xxxx.sharepoint.com/sites/OMSTest/ABS'; var context = new SP.ClientContext(siteUrl); var currentWEB = context.get_web(); var membersGRP = new SP.GroupCreationInformation(); var membershipArray = []; var collContribute = SP.RoleDefinitionBindingCollection.newObject(context); var rdContribute = currentWEB.get_roleDefinitions().getByName("Contribute"); collContribute.add(rdContribute); membersGRP.set_title("GROUPP1"); membersGRP.set_description("GROUPP1"); var oMembersGRP = currentWEB.get_siteGroups().add(membersGRP); oMembersGRP.set_allowMembersEditMembership(false); oMembersGRP.set_onlyAllowMembersViewMembership(false); var assignments = currentWEB.get_roleAssignments(); var roleAssignment = assignments.add(oMembersGRP, collContribute); oMembersGRP.update(); membershipArray.push(oMembersGRP); context.load(membershipArray[membershipArray.length - 1]); //clientContext.executeQueryAsync(Function.createDelegate(this, this.onQuerySucceeded2), Function.createDelegate(this, this.onQueryFailed2)); context.executeQueryAsync( function () { alert("Group Created Successfully"); }, function (sender, args) { alert("Failed to create groups " + args.get_message()); }); }
I’m at a loss with an SSIS package I inherited. It contains: 1 Script Task 3 Execute SQL tasks 5 Data flow tasks (each contains a number of merges, lookups, data inserts and other transformations) 1 file system task of the package.
All of these are encapsulated in a Foreach loop container. I’ve been tasked with modifying the package so that if any of the steps within the control/data flow fails, the entire thing is rolled back. Now I’ve tried two different approaches to accomplish this:
I. Using Distributed Transactions.
I ensured that:
MSDTC was running on target server and executing client (screenshot enclosed) msdtc.exe was added as an exception to server and client firewall Inbound and outbound rules were set for both server and client to allow DTC connections. ForeachLoop Container TrasanctionLevel: Required All other tasks TransactionLevel: Supported My OLEDB Connection has RetainSameConnection set to TRUE and I’m using SQL Server Authentication with Save Password checked
When I execute the package, it fails right after the script task (first step) After spending an entire week trying to figure out a workaround, I decided to try SQL Tasks to try to accomplish my goa using 3 Execute SQL Tasks:
BEGIN TRAN before the foreach loop container COMMIT TRAN After the ForeachLoop Container with a Success Constraint ROLLBACK TRAN After the ForeachLoop Container with a Failure constraint
In this case, foreachloop container and all other tasks have TransactionLevel property set to Supported. Now here, the problem is that the package executes up to the fourth data flow task and hangs there forever. After logging into SQL Server and verifying the running sessions, I noticed sys.sp_describe_first_result_set;1 as a headblocker session
Doing some research, I found it could be related to a few TRUNCATE statements in some of my Data flow tasks which could cause a schema lock. I went ahead and changed the ValidateExternalMetaData property to False for all tasks within my data flow and changed my truncate statements to DELETE statements instead. Re-ran package and still hangs in the same spot with the same headblocker. As an alternative, I tried creating a second OLEDB connection to the same database, assigned that new OLEDB Connection to my BEGIN, ROLLBACK and COMMIT SQL tasks with RetainSameConnectionProperty set to TRUE and changed the RetainSameConnectionProperty to FALSE (and tried it with TRUE as well) in the original OLEDB connection (the one used by the data flow tasks). This worked in the sense that the package appeared to execute (It ran and Commit Tran executed fine) and then I ran it again with a forced error to cause it to fail and the Rollback TRAN task executed successfully, however, when I queried the affected tables, the transaction hadn’t rolled back, all new records were inserted and old ones were updated (the begin tran was clearly started in a different connection and hence didn’t affect the package’s workflow). I’m not sure what else to try at this point. Any help would be truly appreciated, I’m about to go nuts with this!
P.S. additionally, all objects have “DelayValidation” set to true on everything and SQL Server version is 2012.
Until Python 3.4 you were able to determine target’s operating system with Python as follows:
import nmap nm = nmap.PortScanner() scanner = nm.scan(IP, port, arguments='-O') print(scanner['scan'][IP]['osmatch'])
I’m using Python 3.6 and
osmatch returns nothing. Is there a way how to go about this ?
I am having a problem: I connect my Airpods,i start playing a song, and suddenly, it stops playing and I cant find the solution. Ive been trying a couple of solutions, but none of them really worked out.
I’m trying to setup an approval workflow for new and edited items in Document Library.
I created an approval workflow that is set to start when a new item is created and when an item is edited. The workflow updates the approval status on completion.
Now when I create a new item or edit an existing item the workflow kicks off as expected. However when I approve the workflow the workflow completes and then starts a new workflow. It would seem that updating the approval status is causing the workflow to activate again.
Is there any workaround to stop this looping?
My requirement is:
To pass listnames as an array to a method and concatenate the result after each success.
The problem is I’m not able to concatenate the result from the previous ajax call
handleClick = (ListName) => { var reactHandler = this; jquery.ajax({ url: this.props.siteUrl+ "/_api/web/lists/getbytitle('" + ListName + "')/items" , type: "GET", headers:{'Accept': 'application/json; odata=verbose;'}, success: function(resultData) { ///logic??? }); } }, error : function(jqXHR, textStatus, errorThrown) { } }); }
After asking this question, I realized that the true slowness comes from SQL Server Engine, and I used tons of things and still no success.
However, I created a
connection.udl file on desktop and tried to connect to my localhost via different providers and via different connection string combinations.
I realized that when I use
localhost instead of
. and use
Microsoft OLE DB Provider for SQL Server, the connection is almost instantaneous and truly fast.
Now I want to be able to connect to SQL Server via SSMS using that provider. But I don’t know how. Could you please help.
How can I login in SSMS, using ‘Microsoft OLE DB Provider for SQL Server’?
I am trying to become good at web design, but feel I am missing that ‘visual eye’ that so many of you have.
The general flow of what I’m trying to make is a top level subject (ex. Math), will bring you to a page of its categories (ex, Linear Algebra, Calculus, Discrete Math… and so on). You can then click on one of those categories to bring you to the actual content/tutorial/blog page itself (which will have a left side nav for different parts of the category). Some example on the side-left nav might if you click on Calculus might be .. integrals, etc etc).
I want to make a page where I will have multiple top-level subjects, all in their own card with an icon (and maybe a description). I’ve been stuck on this for a long time.
What comes next when you click on a major subject (ex. Math)? Another page with 5-15 minor-subject (Calculus, Linear algebra etc..). Can they go in cards ? Surely they must look a little bit different. I feel like clicking on a page with 4 cards, and going to a page with 10-15 cards would look repetitive and bad.
I only have 4 “Main subjects” so there is a lot of white space below. Maybe I need a a light gray/blue background image on part of the page?
I am thinking about making those 4 cards a little bit bigger… and the next page they lead to have 10-15 cards about the original size of the 4 cards. Any feedback would be greatly appreciated!
|
As I said in the comments I think is fine the way you did it. I would suggest you to use "congruent to" instead of "equal to" as in my comment from the beginning, or add it at the end of the conversions you did as a last step.
Said that, maybe your question is more related with this point:
how you could make a solution starting backwards? from $14m+r$.
Let us start with:
$$n=14m+r$$
Then, let us suppose three possibilities:
$r \lt 7$
$r = 7$
$7 \lt r \lt 14$ so we can define $r=7+r'$, where $r' \lt 7$
$$n=14m+r \pmod{7} \equiv 2\cdot 7m\pmod{7} + r \pmod{7} \equiv 0+r \equiv 3$$
Thus:
$$r=3$$
But that is not possible because it means that $n = 14m + 3$, but $14m+3$
is not even, and we know that $n$ is even, so that solution is not possible.
$$n=14m+7 \pmod{7} \equiv 2\cdot 7\pmod{7} + 7 \pmod{7} \equiv 0 \equiv 3$$
So it is impossible for the residue to be $r=7$ because $14m+7 \not \equiv 3 \pmod {7}$,
the residue does not comply the premise, being $\equiv {3} \pmod{7}$, it is indeed $\equiv {0} \pmod {7}$.
$$n=14m+r \pmod{7} \equiv 2\cdot 7\pmod{7} + 7 + r' \pmod{7} \equiv 0+0+r' \equiv 3$$
Thus finally the unique remaining valid option is $7 \lt r = 10 \lt 14$:
$$r'=3, r=7+r'=7+3=10$$
|
We have to point out some simple remark of capital importance. First of all, the lagrangian
is not defined as $L = T - V$. This turns out to be true only on a riemannian manifold; the fact this is almost always true in classical mechanics is an "accident" due to the postulates of classical (non relativistic) mechanics. For the most precise definition of the lagrangian, see my answer to this question.
Here we are interested to the fact the lagrangian is
by definition the function that satisfies the Euler-Lagrange equations along the curves of motions and so reproduce the equations of motion.
In fact, if you follow, for example, Landau's treatment (see References), you can see how the form of lagrangian is established from a physical point of view. Assume the
action principle and denote with $S$ the action for our system and with $q$ the set of generalized coordinates we are using. Then the lagrangian is, by definition, the function such that the integral
$$S = \int_{t_0}^{t_1} L( \dot q, q, t ) \mbox{d}t$$
get an extremal value along the trajectories of the motion. By varying $S$, we find Euler-Langrange equations and the fact that $L$ is defined up to a total derivative with respect to time of an arbitrary differentiable function of time and coordinates, $f(q,t)$. Why is $L$ a function of $\dot q, q$ and $t$ but not, for example, of $\ddot q$? Because newtonian mechanics postulates state that the state of a system is completely determined by the knowledge of all positions and velocities of its constituents at a given instant. Now we would establish the functional form of $L$. In order to do this, we consider the galilean postulate of relativity. We can rephrase it in a more suitable form for our goals:
Postulate(Galilean relativity) There exists a class of frame of reference respect to which the space is homogeneus and isotropic. In those frames, equations of motion retain the same form.
Consider a free-moving particle in an inertial frame of reference. Let's indicate with $\mathbf{r}$ its position and with $\mathbf{v}$ its velocity in such a frame. Since there are no preferred points in the space (nor in time: time flows in the same manner for
all observers - i.e. in all frames - in newtonian mechanics), $L$ can't depend on $\mathbf{r}$ or $t$. So it is a function of $\mathbf{v}$ only. Moreover, in virtues of absence of preferred directions, it depends only on $v^2$. Hence $L = L(v^2)$. It follows that
$$ \frac{\partial L}{\partial{\mathbf{r}}} = 0 \implies \frac{\mbox{d}}{\mbox{d}t} \frac{\partial L}{\partial{\mathbf v}} = 0$$.
This implies that $\mathbf{v} = \mbox{constant}$ also. From here we obtain Galilei's transformations. Take two inertial frames of reference, $K$ and $K'$, moving one respect to other with a small velocity $\mathbf{a}$. In $K'$ lagrangian is $L'(v'^2)$, while in $K$ is $L(v^2)$. Since the form of equation of motion must be the same,
$$L'(v'^2) = L(v^2) = L(v^2 + 2\mathbf{a \cdot v} + a^2)$$
Expanding the last member in series of powers of $\mathbf{a}$ and retaining only the first order term, we get
$$L'(v'^2) = L(v^2) + \frac{\partial L}{\partial{v^2}} 2 \mathbf{a \cdot v}.$$
The last term in RHS is a total derivative with respect to time. From the above remark, this means that $\frac{\partial L}{\partial{v^2}}$ can't depend on $v^2$ and hence
$$L = a v^2.$$
We reproduce known equation of motion for a free moving particle if we set $a=m/2$.
Now, if we take a compound system and we separate its constituents in such a way that interactions turn out to be negligible, then for each constituent we can write a free-body lagrangian and the lagrangian for the whole system will the sum of all of them. This means that the free-body lagrangian (in an inertial frame of reference) enjoys the expected characteristics for a kinetic energy.
Finally, consider a system in which interactions between particles are not negligible. We make the following ansatz:
$$L = \sum_\alpha \frac{m_\alpha v_\alpha^2}{2} - V( \mathbf{r_1,r_2 \dots} ).$$
($V$ is a differentiable function of coordinate.)Using Euler-Langrange equations, we get
$$m_\alpha \frac{\mbox{d}\mathbf{v}_\alpha}{\mbox{d}t} = -\frac{\partial V}{\partial \mathbf{r_\alpha}},$$
that is Newton's Second Law. (at least when masses are fixed.) So our ansatz works.
We have to note that in order to derive the form of $L$, we have used the postulates of newtonian mechanics. So our result is no longer valid when those do not apply. In particular, $L = T - V$ is only a
particular case and is not a good definition for $L$.
A side question is: does $L$ have a physical significance?
No. We measure the equations of motion, not the lagrangian! So the form of $L$ is not important, provided that it reproduces the equations of motion.
Now, the energy. I think you would mean that the hamiltonian, often written as $H = T + V$, makes more sense. In effect, energy is measured quantity, but the hamiltonian is not defined as $H = T + V$. It is, by definition, the Legendre transform of $L$ with respect to $\mathbf{v}$ ($\dot q$ in general). It turns out to be the energy only in the case $L = T - V$ and,
in addition, $T$ takes the form $T = \sum_{i,j} a_{ij}(q, t) \dot q_i \dot q_j$, i.e. is a symmetric quadratic form in $\dot q$ whose coefficients are function of $q$ and $t$ only and are symmetric in $i,j$.
References
L. Landau, E. Lifshitz,
Mechanics
V. Arnold,
Mathematical methods of classical mechanics
|
In Introduction to Algorithms (CLRS) 3rd Edition, page 299, the section attempts to prove:
The expected height of a randomly built binary search tree on $ n$ distinct keys is $ O(\lg n)$ .
We define “randomly built binary search tree on $ n$ keys” as:
a binary search tree that arises from inserting the keys in random order into an initially empty tree, where each of the $ n!$ permutations of the input keys is equally likely.
In the proof, we defined some random variables:
Let $ X_n$ denotes the height of a randomly built binary search tree on $ n$ keys and the exponential height $ Y_n = 2^{X_n}$ . Of the $ n$ keys, we choose one key as the root of the tree, and we let $ R_n$ denotes the random variable that holds the position that this key would occupy if the set of keys were sorted (also know as ‘rank’ in the text). If we know that $ R_n=i$ , it follows that $ Y_n=2\cdot max(Y_{i-1},Y_{n-1})$ .
We also define indicator random variables $ Z_{n,1}, Z_{n,2}, …, Z_{n,n}$ , $ Z_{n,i} = I\{R_n=i\}$ .
\begin{equation} \begin{split} E[Y_n] & = E\Big[\sum_{i=1}^{n} Z_{n,i} (2\cdot max(Y_{i-1}, Y_{n-i}))\Big] \ & = \sum_{i=1}^{n} E[Z_{n,i} (2\cdot max(Y_{i-1}, Y_{n-i}))] && \text{(by linearity of expectation)}\ & = \sum_{i=1}^{n} E[Z_{n,i}] E[(2\cdot max(Y_{i-1}, Y_{n-i}))] && \text{(by independence)}\ \end{split} \end{equation}
I would like to ask why is the last equality correct? In other words, why can we assume independence of $ Z_{n,i}$ and $ max(Y_{i-1},Y_{n-i})$ ?
In the proof, there was a brief explanation:
Having chosen $ R_n = i$ , the left subtree (whose ranks are less than $ i$ . This subtree is just like any other randomly built binary search tree on $ i-1$ keys. Other than the number of keys it contains, this subtree’s structure is not affected at all by the choice of $ R_n=i$ , and hence the random variables $ Y_{i-1}$ and $ Z_{n,i}$ are independent. Likewise, the right subtree, whose exponential height is $ Y_{n-i}$ , is randomly built on the $ n-i$ keys whose ranks are greater than $ i$ .
However, since $ R_n$ does affect the number of keys $ Y_{i-1}$ and $ Y_{n-i}$ contain (as acknowledged by the explanation above), wouldn’t it mean that $ Y_{i-1}$ and $ Y_{n-i}$ are dependent on $ Z_{n,i}$ ?
A note regarding similar questions posted on CS stackexchange
I have read the answers for the following questions which are very similar to mine:
Proof that a randomly built binary search tree has logarithmic height Average height of a BST with n Nodes. Randomized BST height analysis : How $ Z_{n,i}$ and $ Y_{k-1}$ are independent?
For 1 & 2, the question was a more general one which asked for an explanation for the entire proof. Also, the answer for both questions did not attempt to justify the independence and simply used the result.
For 3, the question was based on a MIT lecture, https://www.youtube.com/watch?v=vgELyZ9LXX4 and it lacked some of the details which I have included above. Also, the answer was also not clear.
|
The covariance of two random variables $X$ and $Y$ is defined by: $$\mathrm{Cov}(X,Y）= \operatorname{E}(X-\operatorname{E}(X))(Y-\operatorname{E}(Y))=\operatorname{E}(XY)-\operatorname{E}(X)\operatorname{E}(Y)$$
Another related definition is correlation coefficient $$\rho(X,Y) = \frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}}$$
It can be proved that the correlation coefficient $\rho(X,Y)$ always lies between −1 and +1. $X$ and $Y$ are two independent standard normal random variables. We now define another random variable $Z$ by $Z=\rho X+\sqrt{1-\rho^2}\cdot Y$ where $\rho \in [−1,1]$.
How can one prove hat $\rho(X,Z) = \rho$?
|
Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision classical_phase_density [2015/08/18 20:27]
nikolaj
classical_phase_density [2015/08/18 20:29] (current)
nikolaj
Line 25: Line 25: ^ $ \frac{\mathrm d}{\mathrm dt}{\hat\rho}(\pi(t),t)=0 $ ^ ^ $ \frac{\mathrm d}{\mathrm dt}{\hat\rho}(\pi(t),t)=0 $ ^ where $\pi$ is the solution of the [[Hamiltonian equations]]. where $\pi$ is the solution of the [[Hamiltonian equations]]. - - === Volume in statistical physics === - A characteristic volume $V$ may be given by an integral over the spatial part of ${\mathcal M}$. This is e.g. how $V$ arises in the statistical mechanics derivation in the classical setting of the ideal gas law $p := -\frac{\partial}{\partial V}\langle{H}\rangle = \frac{N}{V}\cdot k_B T$. See also [[https://en.wikipedia.org/wiki/Cluster_expansion|Cluster expansion]]. - Introducing the density $n=\frac{N}{V}$, this holds true for infinite volumes. - In the derivation via quantum gases in an infinite volume, a volume parameter is introduced in when the momenta are quantized (see [[Classical density of states]]). - - A remark on the latter case: Note that the physical constants $\hbar$ and $c$ can be used to translate energy to frequency (or time) and further translates time to length. Using this, we can write down models involving a volume parameter $V$, defining a characteristic energy $\frac{(\hbar c)^3}{V}$. This may then e.g. be embedded via (unitless!) expressions as complicated as - - $\frac{V}{(\hbar c)^3}\int {\mathrm d}E\, f(\frac{V}{(\hbar c)^3}E)$. === Reference === === Reference ===
|
With careful use of the
\phantom family of commands, you can get proper alignment inside and outside of the
cases (i.e., the second portion of your equation) as well:
This is a general solution that I often use and will work across different environment as well. We fix a size for portions of the equations, and use
\makebox to set that text in the specified fixed width. You can specify this fixed width as in:
\newcommand{\FixedSize}[1]{\makebox[1.5in][l]{\ensuremath{#1}}}%
This allows you to easily adjust this width to get the display that you want for the appropriate portions of the equation.
A better solution is to automatically compute the required width using the
calc package's
\widthof command. For this we select what is the widest portion that we need to be able to accomodate and then compute the width of that. You can do this all in one line, but for readability I have defined
\WidestPart separately, and used it to compute the width.
One slight complication in this specific situation is the left brace from the
cases. To adjust for this I defined
\PhantomBrace which produces a horizontal space equivalent to the width of the brace used for a three line cases, and include that in the adjustment of the first equation.
I added a
\quad to separate these two portions but any value can be used here.
\documentclass{article}
\usepackage{calc}
\usepackage{amsmath}
\newcommand{\WidestPart}{\ensuremath{aT\left(\frac{n}{b}\right) + n^{c},}}%
\newcommand{\FixedSize}[1]{\makebox[\widthof{\WidestPart}][l]{\ensuremath{#1}}}%
\newcommand{\PhantomBrace}{\hphantom{\left\{\vphantom{\begin{cases}\\\\\end{cases}}\right.}}%
\begin{document}
\begin{align*}
T_{in\ general}\left(n\right)&=\FixedSize{aT\left(\frac{n}{b}\right) + n^{c},}\PhantomBrace\qquad a\geq1, b\geq1, c>0\\
T_{cases}\left(n\right) &=
\begin{cases}
\FixedSize{\Theta\left(n^{\log_{b}a}\right)} \qquad a>b^{c}\\
\FixedSize{\Theta\left(n^{c}\log_{b}n\right)} \qquad a=b^{c}\\
\FixedSize{\Theta\left(n^{c}\right)} \qquad a<b^{c}
\end{cases}
\end{align*}
\end{document}
|
NULL CONTROLLABILITY OF DEGENERATE NONAUTONOMOUS PARABOLIC EQUATIONS DOI Number First page Last page Abstract
$$
u_{t}-M(t)(a(x)u_{x})_{x}=h\chi_{\omega},\qquad (x,t)\in Q=(0,1)\times(0,T),$$ where $\omega=(x_{1},x_{2})$ is a
small nonempty open subset in $(0,1)$, $h\in L^{2}(\omega\times(0,T))$, the diffusion coefficients $a(\cdot)$ is
degenerate at $x=0$ and $M(\cdot)$ is non degenerate on $[0,T]$. Also the boundary conditions are considered to
be Dirichlet or Neumann type related to the degeneracy rate of $a(\cdot)$. Under some conditions on the functions
$a(\cdot)$ and $M(\cdot)$, we prove some global Carleman estimates which will yield the observability inequality
of the associated adjoint system and equivalently the null controllability of our parabolic equation.
Keywords Keywords Full Text:PDF References
Acquistapace P, Terreni B. 1987. A unified approach to abstract linear nonautonomous parabolic equations. Rend. Sem. Mat. Univ. Padova. 78, 47-107.
Acquistapace P. 1988. Evolution operators and strong solutions of abstract linear parabolic equations. Differential Integral Equations. 1, 433-457.
Ait Ben Hassi E, Ammar-Khodja F, Hajjaj A , Maniar L. 2011. Null controllability of degenerate parabolic cascade systems. Portugaliae Mathematica. 68, 345-367.
Ait Ben Hassi E, Ammar-Khodja F, Hajjaj A, Maniar L. 2013. Carleman estimates and null controllability of coupled degenerate systems. Evolution Equations and Control Theory. to appear.
Alabau-Boussouria F, Cannarsa P, Fragnelli G. 2006. Carleman estimates for degen- erate parabolic operators with applications to null controllability. J. Evol. Equ. 6, 161-204.
Cabanillas V. R, Menezes S. B, Zuazua E. 2001. Null controllability in unbounded domains for the semilinear heat equation with nonlinearities involving gradient terms. Journal of Optimization Theory and Applications. 110, 245-264.
Cannarsa P, Martinez P, Vancostenoble J. 2005. Null Controllability of degenerate heat equations. Adv. Differential Equations. 10, 153-190.
Cannarsa P, Martinez P, Vancostenoble J. 2008. Carleman estimates for a class of degenerate parabolic operators. SIAM, J. Control Optim. 47, 1-19.
Cannarsa P, Tort J, Yamamoto M. 2012. Unique continuation and approximate con- trollability for a degenerate parabolic equation. newblock Applicable Analysis. 91, 1409- 1425.
Gobbino M. 1999. Quasilinear degenerate parabolic equations of Kirchhoff type, Math. Methods Appl. Sci. 22 (5), 375-388.
De Teresa L, Zuazua E. 1999. Approximate controllability of the semilinear heat equa- tion in unbounded domains. Nonlinear Analysis TMA. 37, 1059-1090.
Fattorini H. O, Russell D. L. 1971. Exact controllability theorems for linear parabolic equations in one space dimension. Arch. Rat. Mech. Anal. 4, 272-292.
Fernandez-Cara E. 1997. Null controllability of the semilinear heat equation. ESAIM: Control, Optim, Calv. Var. 2, 87-103.
Fernandez-Cara E, Limaco J, De Menezesc S. B. 2012. Null controllability for a parabolic equation with nonlocal nonlinearities. Systems and Control Letters. 61, 107-111.
Fernandez-Cara E, Zuazua E. 2000. Controllability for weakly blowing-up semilinear heat equations. Annales de l’Institut Henry Poincar´e, Analyse non lin´eaire. 17, 583- 616.
Fursikov A. V, Yu Imanuvilov O. 1996. Controllability of evolution equations, Lecture Notes Series 34, Seoul National University, Seoul, Korea.
Lebeau G, Robbiano L. 1995. Controle exact de l’´equation de la chaleur. Comm. in PDE. 20, 335-356.
Lopez A, Zhang X, Zuazua E. 2000. Null controllability of the heat equation as singular limit of the exact controllability of dissipative wave equations. J. Math. Pures Appl. 79, 741–808.
Martinez P, Vancostenoble J. 2006. Carleman estimates for one-dimensional degener- ate heat equations. J. Evol. Equ. 6, 325-362.
Micu S, Zuazua E. 2001. On the lack of null controllability of the heat equation on the half-line. Trans. Amer. Math. Soc.52, 1635-1659.
Fotouhi M, Salimi L. 2012. Controllability results for a class of one dimensional degen- erate/singular Parabolic Equations. Journal of Dynamical and Control Systems, 18, 573–602.
Russell D. L. 1973. A unified boundary controllability theory for hyperbolic and parabolic partial differential equations. Studies in Applied Mathematics. 52, 189-221.
Tataru D. 1994. Apriori estimates of Carleman’s type in domains with boundary. Journal de Maths. Pures et Appliqu´ees. 73, 355-387.
Vancostenoble J. 2011. Improved Hardy-Poincare inequalities and sharp Carleman es- timates for degenerate-singular parabolic problems. Discrete.Contin. Dyn. Syst. Ser.S 4. 761-790.
DOI: https://doi.org/10.22190/FUMI1902311B
Refbacks There are currently no refbacks.
ISSN 0352-9665 (Print)
|
X
Search Filters
Format
Subjects
Library Location
Language
Publication Date
Click on a bar to filter by decade
Slide to change publication date range
Physical Review Letters, ISSN 0031-9007, 11/2017, Volume 119, Issue 19, pp. 191802 - 191802
We report the first evidence for isospin violation in B -> K*gamma and the first measurement of the difference of CP asymmetries between B+->.K*(+)gamma and...
BREAKING | SYMMETRY | PHYSICS, MULTIDISCIPLINARY | ANNIHILATION | K-ASTERISK-GAMMA | TO-LEADING ORDER | DECAYS | Physics - High Energy Physics - Experiment | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
BREAKING | SYMMETRY | PHYSICS, MULTIDISCIPLINARY | ANNIHILATION | K-ASTERISK-GAMMA | TO-LEADING ORDER | DECAYS | Physics - High Energy Physics - Experiment | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS
Journal Article
Physical Review Letters, ISSN 0031-9007, 03/2017, Volume 118, Issue 11, pp. 111801 - 111801
We present a measurement of angular observables and a test of lepton flavor universality in the B→K^{*}ℓ^{+}ℓ^{-} decay, where ℓ is either e or μ. The analysis...
Journal Article
3. Observation of $$\Xi _{c}(2930)^0$$ Ξc(2930)0 and updated measurement of $$B^{-} \rightarrow K^{-} \Lambda _{c}^{+} \bar{\Lambda }_{c}^{-}$$ B-→K-Λc+Λ¯c- at Belle
: Belle Collaboration
The European Physical Journal C, ISSN 1434-6044, 3/2018, Volume 78, Issue 3, pp. 1 - 8
We report the first observation of the $$\Xi _{c}(2930)^0$$ Ξc(2930)0 charmed-strange baryon with a significance greater than 5$$\sigma $$ σ . The $$\Xi...
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Nuclear Physics, Heavy Ions, Hadrons | Measurement Science and Instrumentation | Nuclear Energy | Quantum Field Theories, String Theory | Physics | Elementary Particles, Quantum Field Theory | Astronomy, Astrophysics and Cosmology
Journal Article
4. Observation of $$\Xi _{c}(2930)^0$$ Ξc(2930)0 and updated measurement of $$B^{-} \rightarrow K^{-} \Lambda _{c}^{+} \bar{\Lambda }_{c}^{-}$$ B-→K-Λc+Λ¯c- at Belle
European Physical Journal C: Particles and Fields, ISSN 1434-6044, 03/2018, Volume 78, Issue 3, pp. 1 - 8
Abstract We report the first observation of the $$\Xi _{c}(2930)^0$$ Ξc(2930)0 charmed-strange baryon with a significance greater than 5$$\sigma $$ σ . The...
Journal Article
PHYSICAL REVIEW LETTERS, ISSN 0031-9007, 03/2012, Volume 108, Issue 13, pp. 134801 - 134801
Journal Article
6. Ingestion of hyaluronans (molecular weights 800 k and 300 k) improves dry skin conditions: a randomized, double blind, controlled study
Journal of Clinical Biochemistry and Nutrition, ISSN 0912-0009, 2014, Volume 56, Issue 1, pp. 66 - 73
Hyaluronan (HA) has been increasingly used as a dietary supplement to improve the skin. However, the effect of ingested HA may depend on its molecular weight...
molecular weight | skin moisture content | ingestion | dry skin | hyaluronan | Hyaluronan | Dry skin | Skin moisture content | Molecular weight | Ingestion | RESPONSES | CELLS | NUTRITION & DIETETICS | ACID | OLIGOSACCHARIDES | Original
molecular weight | skin moisture content | ingestion | dry skin | hyaluronan | Hyaluronan | Dry skin | Skin moisture content | Molecular weight | Ingestion | RESPONSES | CELLS | NUTRITION & DIETETICS | ACID | OLIGOSACCHARIDES | Original
Journal Article
PHYSICAL REVIEW LETTERS, ISSN 0031-9007, 01/2019, Volume 122, Issue 2
Journal Article
Physical review letters, 04/2016, Volume 116, Issue 16, pp. 161801 - 161801
We measure the decay B_{s}^{0}→K^{0}K[over ¯]^{0} using data collected at the ϒ(5S) resonance with the Belle detector at the KEKB e^{+}e^{-} collider. The data...
Journal Article
Physical Review Letters, ISSN 0031-9007, 01/2019, Volume 122, Issue 2
Journal Article
12/2015
Phys. Rev. Lett. 116, 161801 (2016) We measure the decay $B_s^0\rightarrow K^0\overline{K}^0$ using data collected at the $\Upsilon(5S)$ resonance with the...
Physics - High Energy Physics - Experiment
Physics - High Energy Physics - Experiment
Journal Article
Physical Review Letters, ISSN 0031-9007, 04/2016, Volume 116, Issue 16
Journal Article
12. In situ X-ray photoelectron spectroscopy using a conventional Al-Kα source and an environmental cell for liquid samples and solid-liquid interfaces
Applied Physics Letters, ISSN 0003-6951, 04/2019, Volume 114, Issue 17, p. 173702
X-ray photoelectron spectroscopy (XPS), which intrinsically requires vacuum, was used to characterize chemical species in a liquid using laboratory XPS...
PHYSICS, APPLIED | XPS | PHOTOEMISSION | Organic chemistry | X rays | Aqueous solutions | Liquid-solid interfaces | X ray spectroscopy | Spectrum analysis | Silicon nitride | Cesium | X ray photoelectron spectroscopy | Photoelectrons | Quantitative analysis
PHYSICS, APPLIED | XPS | PHOTOEMISSION | Organic chemistry | X rays | Aqueous solutions | Liquid-solid interfaces | X ray spectroscopy | Spectrum analysis | Silicon nitride | Cesium | X ray photoelectron spectroscopy | Photoelectrons | Quantitative analysis
Journal Article
13. Measurement of D 0 -D 0 mixing and search for CP violation in D 0 →K + K - ,π + π - decays with the full Belle data set
Physics Letters, Section B: Nuclear, Elementary Particle and High-Energy Physics, ISSN 0370-2693, 02/2016, Volume 753, pp. 412 - 418
Journal Article
Astronomical Journal, ISSN 0004-6256, 07/2017, Volume 154, Issue 1, p. 1
We report the discovery and the analysis of the planetary microlensing event, OGLE-2013-BLG-1761. There are some degenerate solutions in this event because the...
gravitational lensing: micro | planetary systems | NEPTUNE-MASS | OGLE-2005-BLG-169 | MICROLENSING OPTICAL DEPTH | CONFIRMATION | GALACTIC BULGE | EVENT | DISCOVERY | COLD NEPTUNE | ASTRONOMY & ASTROPHYSICS | LATE-TYPE STAR | EXTINCTION | Physics - Earth and Planetary Astrophysics
gravitational lensing: micro | planetary systems | NEPTUNE-MASS | OGLE-2005-BLG-169 | MICROLENSING OPTICAL DEPTH | CONFIRMATION | GALACTIC BULGE | EVENT | DISCOVERY | COLD NEPTUNE | ASTRONOMY & ASTROPHYSICS | LATE-TYPE STAR | EXTINCTION | Physics - Earth and Planetary Astrophysics
Journal Article
2015, ISBN 0674598474, vi, 388
Book
16. Observation of Ξ c (2930) 0 and updated measurement of B- → K- Λc+Λ¯c- at Belle: Belle Collaboration
European Physical Journal C, ISSN 1434-6044, 03/2018, Volume 78, Issue 3
Journal Article
|
But if you don't want to have a Google account: Chrome is really good. Much faster than FF (I can't run FF on either of the laptops here) and more reliable (it restores your previous session if it crashes with 100% certainty).
And Chrome has a Personal Blocklist extension which does what you want.
: )
Of course you already have a Google account but Chrome is cool : )
Guys, I feel a little defeated in trying to understand infinitesimals. I'm sure you all think this is hilarious. But if I can't understand this, then I'm yet again stalled. How did you guys come to terms with them, later in your studies?
do you know the history? Calculus was invented based on the notion of infinitesimals. There were serious logical difficulties found in it, and a new theory developed based on limits. In modern times using some quite deep ideas from logic a new rigorous theory of infinitesimals was created.
@QED No. This is my question as best as I can put it: I understand that lim_{x->a} f(x) = f(a), but then to say that the gradient of the tangent curve is some value, is like saying that when x=a, then f(x) = f(a). The whole point of the limit, I thought, was to say, instead, that we don't know what f(a) is, but we can say that it approaches some value.
I have problem with showing that the limit of the following function$$\frac{\sqrt{\frac{3 \pi}{2n}} -\int_0^{\sqrt 6}(1-\frac{x^2}{6}+\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$equal to $1$, with $n \to \infty$.
@QED When I said, "So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0." I guess what I'm saying is that (f(x+h)-f(x))/h is not continuous since it's not defined at h=0.
@KorganRivera There are lots of things wrong with that: dx=0 is wrong. dy/dx - what/s y? "dy/dx is by definition not continuous" it's not a function how can you ask whether or not it's continous, ... etc.
In general this stuff with 'dy/dx' is supposed to help as some kind of memory aid, but since there's no rigorous mathematics behind it - all it's going to do is confuse people
in fact there was a big controversy about it since using it in obvious ways suggested by the notation leads to wrong results
@QED I'll work on trying to understand that the gradient of the tangent is the limit, rather than the gradient of the tangent approaches the limit. I'll read your proof. Thanks for your help. I think I just need some sleep. O_O
@NikhilBellarykar Either way, don't highlight everyone and ask them to check out some link. If you have a specific user which you think can say something in particular feel free to highlight them; you may also address "to all", but don't highlight several people like that.
@NikhilBellarykar No. I know what the link is. I have no idea why I am looking at it, what should I do about it, and frankly I have enough as it is. I use this chat to vent, not to exercise my better judgment.
@QED So now it makes sense to me that the derivative is the limit. What I think I was doing in my head was saying to myself that g(x) isn't continuous at x=h so how can I evaluate g(h)? But that's not what's happening. The derivative is the limit, not g(h).
@KorganRivera, in that case you'll need to be proving $\forall \varepsilon > 0,\,\,\,\, \exists \delta,\,\,\,\, \forall x,\,\,\,\, 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$ by picking some correct L (somehow)
Hey guys, I have a short question a friend of mine asked me which I cannot answer because I have not learnt about measure theory (or whatever is needed to answer the question) yet. He asks what is wrong with \int_0^{2 \pi} \frac{d}{dn} e^{inx} dx when he applies Lesbegue's dominated convergence theorem, because apparently, if he first integrates and then derives, the result is 0 but if he first derives and then integrates it's not 0. Does anyone know?
|
Here's a common definition of a function (for example, Wiki follows this definition):
A
relationbetween sets $A$ and $B$ is any subset $R \subseteq A \times B$. We say that this relation is a functionif it satisfies the property $$ (a,b_1) \in R \text{ and }(a,b_2) \in R \implies b_1 = b_2 $$
This definition of a function is good for most purposes. Certainly, we can find the domain and range of a function defined in this way. This also allows us to answer such questions as
Is the set $\{(1,2),(2,5),(3,5)\}$ a function? (answer: yes)
With such a definition, we could certainly go on to define a function's domain and range. The problem with this definition, however, is that it includes no notion of a codomain. This presents a problem when we want to answer a question such as
Are the functions $f:\Bbb R \to \Bbb R$ given by $f(x) = x^2$ and $g:\Bbb R \to [0,\infty)$ given by $g(x) = x^2$ the same function? (answer: no)
On the one hand, the "graphs" of the function are the same. If we are to believe that these functions are merely subsets of a Cartesian product, then we should say that $f = g$ since both are merely the set $\{(x,x^2) : x \in \Bbb R\}$. On the other hand, we would like to say that "the function $g$ is surjective, but the function $f$ is not". If surjectivity is a property of functions, then the fact that $f$ and $g$ do not share this property should mean that $f \neq g$.
So what gives? Is there a setting in which both of these questions are well-posed? If someone has a reference that handles all this well, I would appreciate it.
Edit: Wiki apparrarently has a discussion of this issue here
|
Definition:Greatest Common Divisor/Integral Domain Definition
Let $a, b \in D: a \ne 0 \lor b \ne 0$.
Let $d \divides a$ denote that $d$ is a divisor of $a$.
Let $d \in D$ have the following properties: $(1): \quad d \divides a \land d \divides b$ $(2): \quad c \divides a \land c \divides b \implies c \divides d$
Then $d$ is called
a greatest common divisor of $a$ and $b$ (abbreviated GCD or gcd) and denoted $\gcd \set {a, b}$. We see that, trivially: $\gcd \set {a, b} = \gcd \set {b, a}$
so the set notation is justified.
The
greatest common divisor is often seen abbreviated as GCD, gcd or g.c.d.
Some sources write $\gcd \set {a, b}$ as $\tuple {a, b}$, but this notation can cause confusion with ordered pairs.
The notation $\map \gcd {a, b}$ is frequently seen, but the set notation, although a little more cumbersome, can be argued to be preferable.
Highest common factor when it occurs, is usually abbreviated as HCF, hcf or h.c.f.
It is written $\hcf \set {a, b}$ or $\map \hcf {a, b}$.
Also see
However, see Elements of Euclidean Domain have Greatest Common Divisor where it is shown that the same does not apply to a Euclidean domain.
|
Version 17 (modified by 7 months ago) (diff), 'perturbative' and 'real' particles; the perturbative weight 'perturbative' and 'real' particles
(following text is taken - slightly modified - from: O.Buss, PhD thesis, pdf, Appendix B.1)
Reactions which are so violent that they disassemble the whole target nucleus can be treated only by explicitly propagating all particles, the ones in the target and the ones produced in the collision on the same footing.
For reactions which are not violent enough to disrupt the whole target nucleus, e.g. low-energy πA, γA or neutrino A collisions at not too high energies, the target nucleus stays very close to its ground state. Henceforth, one keeps as an approximationthe phase-space density of the target nucleus constant in time ('frozen approximation'). In GiBUU this is controlled by the switch
freezeRealParticles. The test-particles which represent this constant target nucleus are called
real test-particles. However, one also wants to consider the final state particles. Thus one defines another type of test-particles which are called perturbative.The perturbative test-particles are propagated and may collide with real ones, the products are perturbative particles again. However, perturbative particles may not scatter among each other.Furthermore, their feedback on the actual densities is neglected. One can simulate inthis fashion the effects of the almost constant target on the outgoing pparticles without modifyingthe target. E.g. in πA collisions we initialize all initial state pions as perturbative test-particles.Thus the target automatically remains frozen and all products of the collisions of pions andtarget nucleons are assigned to the perturbative regime.
Furthermore, since the
perturbative particles do not react among themselves or modify the realparticles in a reaction, one can also split a perturbative particle into \(N_{test}\) pieces (several perturbativeparticles) during a run. Each piece is given a corresponding weight \(1/N_{test}\) and one simulates likethis \(N_{test}\) possible final state scenarios of the same perturbative particle during one run. The perturbative weigth 'perWeight'
Usually, in the cases mentioned above, where one uses the seperation into
real and perturbative particles, one wants to calculate some final quantity like \(d\sigma^A_{tot}=\int_{nucleus}d^3r\int \frac{d^3p}{(2\pi)^3} d\sigma^N_{tot}\,\times\,\dots \). Here we are hiding all medium modifications, as e.g. Pauli blocking, flux corrections or medium modifications of the cross section in the part "\(\,\times\,\dots \)". Now, solving this via the testparticle ansatz (with \(N_{test}\) being the number of test particles), this quantity is calulated as \(d\sigma^A_{tot}=\frac{1}{N_{test}}\sum_{j=1}^{N_{test}\cdot A}d\sigma^j_{tot}\,\times\,\dots \), with \(d\sigma^j_{tot}\) standing for the cross section of the \(j\)-th test-particle.
The internal implementation of calculations like this in GiBUU is, that a loop runs over all \(N_{test}\cdot A\) target nucleons and creates some event. Thus all these events have the same probability. But since they should be weighted according \(d\sigma^j_{tot}\), this is corrected by giving all (final state) particles coming out of event \(j\) the weight \(d\sigma^j_{tot}\).
This information is stored in the variable
perWeight in the definition of the particle type.
Thus, in order to get the correct final cross section, one has to
sum the perWeight, and not the particles.
As an example: if you want to calculate the inclusive pion production cross section, you have to loop over all particles and sum the perWeights of all pions. Simply taking the number of all pions would give false results.
The weights can also be negative. This happens, e.g., in the case of pion production on nucleons. In this case the cross section is determined by the square of a coherent sum of resonance and background amplitudes and as such is positive. In the code the resonance contribution is separated out as the square of the resonance amplitude and as such is positive as well. The remainder, i.e. the sum of the square of the background amplitude and the interference term of resonance and background amplitudes, can be negative, however. This latter contribution is just the event type labeled 32 and 33 in the code that describes the 1pi bg plus interference. How to compute cross sections from the perturbative weights for neutrino-induced reactions
The output file
FinalEvents.dat contains all the events generated. Per event all the four-momenta of final state particles are listed together with the incoming neutrino energy, the 'perWeight' and various other useful properties (see documentation for
FinalEvents.dat). In each event there is one nucleon with perWeight=0 which represents the hit nucleon; for 2p2h processes the second initial nucleon is not written out.
The final state nucleons may have masses which are spread out around the physical mass in a very narrow distribution. There are two reasons for that: 1. nucleons may still be inside the potential well and thus have lower masses. These nucleons can be eliminated from the final events file by imposing a condition that they are outside the nuclear potential (the spatial coordinates of all particles are also given in the
FinalEvents.dat file). 2. For numerical-practical reasons the nucleons are given a Breit-Wigner mass distribution with a width of typically 1 MeV around the physical mass when calculating the QE cross section.
As an example we consider here the calculation of the CC inclusive differential cross section dsigma/dE_mu for a neutrino-induced reaction on a nucleus; E_mu is the energy of the outgoing muon. In
FinalEvents.dat the lines with the particle number 902 contain all the muon kinematics as well as the perweight. In order to produce a spectrum one first has to bin the muon energies into energy bins. This binning process must preserve the connection between energy and perweight. Then all the perweights in a given energy bin are summed and divided by the bin width to obtain the differential cross section. If the GiBUU run used - for better statistics - a number of runs >1 at the same energies, then this number of runs has to be divided out to obtain the final differential cross section.
All cross sections in GiBUU, both the precomputed ones and the reconstructed ones, are given per nucleon. The units are 10
-38 cm 2 for neutrinos and 10 -33 cm 2 for electrons.
|
it is gained by: start with a torus — a shape like the surface of a doughnut — and remove a slice. Attach two interlocking smaller tori, one on each side of the gap left by the slice, and repeat the process, slicing each torus and inserting an interlocking pair of smaller tori that you will subsequently slice and insert even smaller tori into. Perform the process infinitely many times.
@AaronStevens I just had to google what +ve and -ve mean (I'm sure I've seen them, but probably in poorly written context so I ignore it). If you're going to be that lazy you might as well just leave it at "+" and "-" instead of adding some weird letters that seem like acronyms.
@AaronStevens I could see like one teacher coming up with it and using it with their students; but I'm surprised that it wasn't something that was quickly questioned and corrected by others.
I've had teachers try to teach nonsense conventions before. My favourite example was my "Grade 11 Advanced Math" teacher. She asked us how we wanted to round, and we agreed as a class with the standard "0.5 rounds up in every case" instead of "0.5 rounds to the nearest even number". For some reason, she concluded that this means that "0.45 rounds up to 1 because 0.45 rounds up to 0.5 which rounds up to 1". I never managed to convince her that it defeated the purpose of rounding.
@AaronStevens It was extremely frustrating to me. To the extent that she considered writing me up to administration for being belligerent or something because I kept pushing that issue. I wouldn't drop it until she pushed it that far, because it was just so illogical. I was bringing up graphing calculators and showing that they never did that. I just tried explaining the purpose of rounding; but she wouldn't even budge. It still blows my mind, rounding is so logical.
I got a good laugh out of it one time when I got an extra % because of it though. (countered the % I lost for not rounding that way on a different test)
@JMac Cursory googling seems to suggests it's used a lot in handwritten medical records, where writing "positive" and "negative" is too long but especially - alone would be in danger of looking like a smudge
Back in the 1970s (ask your father) this was a standard notation in electronics, and since there were no computers in those days us nerds used to dabble in electronics instead. I picked up the notation then and it has stayed with me.
@JohnRennie Means the same as long as the person understands the notation. Personally, I would understand "+" to mean positive quicker than I associate "+ve" to mean only positive. It's always evoked a sense of being an abbreviation to me, like +V____ Energy and - V_____ Energy". IDK, it just personally confuses me more than it clears things up.
Well at least I understand a bit about the notation. I definitely would suggest people avoid it though. It makes sense in the context of handwritten medical records; but with typed responses it does more harm that just using +/- or positive/negative IMO. Especially since younger people don't really see it and it's not exactly intuitive.
Yeah, that's why it's confusing me I guess. Typing a full word is barely any more effort than the shorthand, and a symbol like +/- doesn't look like a mistaken dash as much. I'll use weird shorthand in my own hand notes if I think it helps, but obviously I wouldn't try to use that notation to explain things to others.
@JohnRennie I think I've seen that before. I always just automatically associated "+ve" and "-ve" as if the "ve" part was some sort of relic notation for voltage, not the "ve" part of positive and negative.
Miller indices form a notation system in crystallography for planes in crystal (Bravais) lattices.In particular, a family of lattice planes is determined by three integers h, k, and ℓ, the Miller indices. They are written (hkℓ), and denote the family of planes orthogonal tohb1+kb2+ℓb3...
on that note, there was a paper I've used a lot in my own work lately, a stats paper from 1937. setting aside the content of the paper for a moment, the paper starts like this:
"In the case of “correlations”, as indeed is often the case, many discussions arise from confusion between different concepts. For a long time we have seen papers demonstrating and repeating that is necessary to distinguish between the concept of “correlation” that is measured by the “correlation coefficient” r (of Bravais), and the concept of “stochastic dependence” defined in probability theory."
The quantum model for the hydrogen atom is a Hamiltonian dynamical system based on Schrödinger operator $H$. When compared with the physical atom it has important shortcomings. First and foremost the quantum system has no trajectories joining stationary electron configurations at different energy...
I couldn't understand how this transformation works $$ \left[i \gamma^{\mu} \partial_{\mu}-m\right] \psi(x) \rightarrow\left[i \gamma^{\mu}\left(\Lambda^{-1}\right)_{\mu}^{\nu} \partial_{\nu}-m\right] \Lambda_{\frac{1}{2}} \psi\left(\Lambda^{-1} x\right) $$ under this law of transformation $$ \Lambda_{\frac{1}{2}}^{-1} \gamma^{\mu} \Lambda_{\frac{1}{2}}=\Lambda_{\nu}^{\mu} \gamma^{\nu} $$ ?
@EmilioPisanty one thing this makes me curious about: Given the above limit, it should be the case that $\displaystyle \sum_{n=0}^\infty \frac{K_n(z)}{n!}t^n$ should have radius of convergence $|t|<1/z$
When I read physics explanations of "quantization", I am confused, because they talk about particles, momentum, and other specific things. It seems to me that quantum formalism is much more general than this (e.g. in quantum computing there are no "particles").What is the most general statemen...
Image of problem: https://ibb.co/6HPQxHJ My question is: if i need to write equation Module of forces about this mass that rotate in a circle with radius $R$, why that way: $ma_n=mgcos(t)+T$ it isn't correct? My book show this equation $ma_n=mg+T$.
$a_n$ is normal acceleration. $a_t$ is tangetinal acceleration.
Oh sorry wait. Now I'm going to explain better the situation
In general, I'll explain what happens. I have this ball of mass $m$ running through this circumference. I need to set the minimum speed to travel around the circumference. The problem is perfectly clear to me I don't have a problem with it. To find the critical speed I set $T = 0$. Well I say that my book writes that equation because the result it writes is $v = \sqrt {gR}$ while to me $v = \sqrt {gRcos (t)}$. Is it clearer now?
Well, I’m not sure I understand how the book is getting its answer (I’m having to visualize) but your answer doesn’t make much sense: if the ball is to keep moving around the circle, then having an angular dependence to the minimum speed doesn’t make sense to me
For one, when t=180 degrees your answer would require v^2 = -g R
In another exercise it's write:"The limit condition to be able to make the turn around the pin along the circular trajectory of the radius R is obtained by imposing that the thread tension is zero at the moment when the mass is on the vertical above the pin: "write initial equation""
I have caculated the transformed left hand Weyl spinor and found it correct, but the plus sign of right hand Weyl spinor in front of $$\beta$$ is fuzzy $${\psi_{R} \rightarrow\left(1-i \boldsymbol{\theta} \cdot \frac{\sigma}{2}+\boldsymbol{\beta} \cdot \frac{\sigma}{2}\right) \psi_{R}}$$
|
In this vignette, we illustrate on a synthetic dataset how to perform post-selection inference (PSI) for a set of kernels using our R package
kernelPSI (Slim et al. 2019). The kernels are selected in a forward fashion according to quadratic kernel association scores, leading to the modeling of the selection event as a set of quadratic constraints. For valid inference, we need to correct for the bias introduced by the prior selection of the kernels. Determining the exact post-selection distribution of our test statistics under the selection event was impossible. To overcome that, we use sampling in order to derive empirical \(p\)-values.
We first start by giving the setup for our simulation. We associate \(10\) gaussian kernels to \(10\) independent groups of variables of size \(5\) each. Within each group, the variables are drawn from a multivariate normal distribution with mean \(0\) and a correlation matrix \(V_{ij} = \rho^{\lvert i-j\rvert}\), where \(\rho = 0.6\) and \(i,j\in[1\mathrel{{.}\,{.}} 5]\). The dataset we consider here consists of \(100\) independent samples.
require("MASS")#> Loading required package: MASSn_kernels <- 10 # total number of kernelssize_kernels <- 5 # dimensionality of the data associated to each kerneln <- 100 # sample sizerho <- 0.6 # correlation parameter# intra-group correlation matrixcorr <- outer(seq_len(size_kernels), seq_len(size_kernels), function(i, j) return(rho^(abs(i-j))))# design matrixX <- replicate(n_kernels, mvrnorm(n, mu = rep(0, size_kernels), Sigma = corr), simplify = FALSE)
To each group, we associate a local gaussian kernel \(K\) of variance \(\sigma^2 = 5\),
i.e. \(K(x_i, x_j) = \exp(\lvert\lvert x_i-x_j\rvert\rvert^2/\sigma^2)\).
require("kernlab")#> Loading required package: kernlabK <- replicate(n_kernels, rbfdot(sigma = 1 / size_kernels)) # full list of Gaussian kernels# List of Gram matricesKmat <- sapply(seq_len(n_kernels), function(i) {kMatrix <- kernelMatrix(K[[i]], X[[i]]); return(as.kernelMatrix(kMatrix, center = TRUE))}, simplify = FALSE)
We select the first three kernels as causal kernels, and simulate the outcome \(Y\) in the following way: \(Y\sim 0.1 * U + \epsilon\). \(U\) is the first eigenvector of the similarity matrix of the sum kernel of the first three kernel and \(\epsilon\) is a reduced gaussian random error. For the outcome, we consider the linear kernel.
m_kernels <- 3 # number of causal kernelstheta <- 0.1 # amplitude of size effectKsum <- Reduce(`+`, Kmat[seq_len(m_kernels)]) # sum kernel of the causal kernelsdecompK <- eigen(Ksum) # eigenvalue decomposition of the sum kernel KsumY <- as.matrix(theta * decompK$values[1] * decompK$vectors[, 1] + rnorm(n), ncol = 1) # response vectorLmat <- kernelMatrix(new("vanillakernel"), Y) # linear response vector
The first step in PSI is to select the kernels. For that purpose, we use the function
FOHSIC for the fixed variant and
adaFOHSIC for the adaptive variant. Afterwards, to generate the list of matrices associated to the quadratic constraint of each selection event, we respectively apply
forwardQ and
adaQ.
require("kernelPSI")#> Loading required package: kernelPSIcandidate_kernels <- 3 # number of kernels for the fixed variantselectFOHSIC <- FOHSIC(Kmat, Lmat, mKernels = candidate_kernels) # fixed variantconstraintFO <- forwardQ(Kmat, selectFOHSIC) # list of quadratic constraints modeling the selection event
If the number of causal kernels is not available beforehand, we resort to the adaptive version:
Finally, using the obtained constraints, we can derive PSI significance values for three statistics: the log-likelihood ratio for the ridge prototype, the log-likelihood ratio for the kernel PCA prototype and the HSIC score. The \(p\)-values are computed by comparing the statistics of the original response to those of replicates sampled within the acceptance region of the selection event. Most often, because of the difference in their statistical power, the methods yield different \(p\)-values.
n_replicates <- 5000 # number of replicates (statistical power and validity require a higher number of samples)burn_in <- 1000 # number of burn-in iterations# Fixed variant ------------------# selected methods: 'ridge' for the kernel ridge regression prototype # and 'pca' for the kernel principal component regression prototypekernelPSI(Y, K_select = Kmat[selectFOHSIC], constraintFO, method = c("ridge", "pca"), n_replicates = n_replicates, burn_in = burn_in)#> $ridge#> [1] 0.0382#> #> $pca#> [1] 0.2502# Adaptive variant ------------------# selected methods: 'hsic' for the unbiased HSIC estimatorkernelPSI(Y, K_select = Kmat[adaS], constraintFO, method = "hsic", n_replicates = n_replicates, burn_in = burn_in)#> $hsic#> [1] 0.0912
Slim, Lotfi, Clément Chatelain, Chloe-Agathe Azencott, and Jean-Philippe Vert. 2019. “KernelPSI: A Post-Selection Inference Framework for Nonlinear Variable Selection.” In
Proceedings of the 36th International Conference on Machine Learning, edited by Kamalika Chaudhuri and Ruslan Salakhutdinov, 97:5857–65. Proceedings of Machine Learning Research. Long Beach, California, USA: PMLR. http://proceedings.mlr.press/v97/slim19a.html.
|
tl;dr: I've found a fatal gap in this proof that I've been unable to close. I'll leave this answer up in case either: a) I figure out how to fix it or b) it inspires someone else to figure out how to fix it.
Let $G = (X \cup Y, E)$ be a bipartite graph without a perfect matching. We'll say that a subset $S$ is
deficient if $|N(S)| \lt |S|$. We are looking for a minimum, deficient subset of $X$. The general approach will be to identify potential minimum, deficient sets by characterizing (and finding) all the mini mal, deficient sets, i.e.: deficient sets $S\subset X$ that contains no deficient subsets. Let's make a few observations on the properties of these minimal, deficient sets.
Observation 1: A subset $S$ is a minimal deficient subset of $X$ iff for all $s \in S$, the set $S\setminus \{s\}$ has a perfect matching in $G$. This is just Hall's Theorem.
Observation 2: If $S$ is a minimal deficient subset of $X$, then for all $s_1, s_2 \in S$, there exists a path in $G$ from $s_1$ to $s_2$. Otherwise, we could decompose $S$ into two (or more) components, at least one of which would have to be deficient, thus contradicting minimality.
Now, let us fix $M$, some maximum matching in $G$. Let $X' \subset X$ and $Y' \subset Y$ be the vertices that are matched by $M$ and let $U = X\setminus X'$ be the subset of unmatched vertices in $X$. For any subset $S$ of $X$, we will also denote $m(S)$ as the set of vertices in $G$ reachable from $S$ via $M$-alternating paths.
In an answer to the question linked in the OP, we see a proof that if we take $S = U \cup (m(U) \cap X)$ then $S$ is deficient. Careful reading of that proof reveals that it works not just for $U$ but any subset of $U$. That is to say, if we take any subset $U_1 \subseteq U$, then $U_1 \cup (m(U_1) \cap X)$ is a deficient subset of $X$. In particular, we may take $U_1$ to be a singleton set. For any $u \in U$, let's define $D_u = \{u\} \cup (m(\{u\}) \cap X)$.
Lemma 1: $D_u$ is a minimal, deficient set for all $u \in U$.
Proof: We'll take for granted that $D_u$ is deficient via proof given in the previously referenced answer. To show that $D_u$ is minimal w.r.t. deficiency, we observe that $D_u \setminus \{u\}$ is simply a subset of $X'$, hence there exists a perfect matching for it inside of $G$ (just take the restriction of $M$ to $D_u\setminus \{u\}$). For any other $y \in D_u$, we follow the $M$-alternating path from $y$ to $u$, flip all the edges along this path, and obtain a perfect matching of $D_u \setminus \{y\}$ in $G$. So, by Observation 1, $D_u$ is a minimal, deficient set. $\square$
Ok, now that we've identified one collection of minimal, deficient subsets of $X$, we need to ask: what about any others?
To add a little structure, let us consider any set $S \subseteq X$ to be in the form $S = U_1 \cup Z_1 \cup Z_2$ where $U_1 \subseteq U$, $Z_1 \subseteq m(U_1)$ and $Z_2 \subseteq X' \setminus m(U_1)$. In other words, we break $S$ into the portion that's unmatched by $M$ ($U_1$), the portion that's reachable from $U_1$ via $M$-alternating paths ($Z_1$), and the portion that's not reachable from $U_1$ via $M$-alternating paths ($Z_2$). It's trivial to observe that if $S$ is a deficient set, then $U_1$ must be non-empty.
Via Lemma 1, we have covered the case where $Z_1 = m(U_1)$ and $Z_2$ is empty. This leaves three cases to examine:
$Z_2$ is non-empty $|U_1| > 1$ and $Z_1 \subsetneq m(U_1)$ $Z_1$ and $Z_2$ are both empty (i.e.: $S \subseteq U$).
Lemma 2: If $S = U_1 \cup Z_1 \cup Z_2 \subseteq X$ is such that $Z_2 \neq \emptyset$, then $S$ is not a minimal, deficient subset of $X$.
Proof: Let $M(Z_2)$ be the elements of $Y$ that are matched with $Z_2$ in $M$. By definition, there can be no edges from $U_1$ nor $Z_1$ to $M(Z_2)$ since that would imply an $M$-alternating path from $U_1$ to vertices in $Z_2$.
If $S$ is a minimal, deficient set, then every subset of $S$ has a complete matching. In particular, $U_1 \cup Z_1$ has a complete matching, say $M_1$. By our previous observation, we note that this complete matching $M_1$ does not use any of the vertices in $M(Z_2)$. Thus, the matching formed by using $M_1$ to match $U_1 \cup Z_1$ and $M$ to match $Z_2$ is a complete matching for $S$, contradicting the assumption that $S$ was deficient. $\square$
In a previous version of this answer, I had neglected case 2), assuming that it was somehow covered during the proof of Lemma 1. However, this is not the case. There can exist minimal, deficient sets which do not look like $D_u$. The following diagram shows such an example. Taking the bolded edges as the matching $M$, we can see that $S = \{A, B, C\}$ is a minimal, deficient set and is not of the form $D_u$. I haven't yet been able to find an effective characterization of minimal deficient sets that fall into case 2, so I am currently unable to complete this proof.
|
This Integral came up while attempting another question:
$$f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y^2 \cos^2x+ \sin^2x) .dx$$
The suggested solution was as follows: $$f'(y) = 2y \int_{0}^{\pi/2}\frac{cos^{2}x}{sin^{2}x + y^{2}cos^{2}x}dx$$ $$= 2y \int_{0}^{\pi/2}\frac{dx}{tan^{2}x + y^{2}}$$ $$= 2y \int_{0}^{\pi/2}\frac{sec^{2}x - tan^{2}x }{tan^{2}x + y^{2}}dx$$ $$= 2y . \frac{1}{y} tan^{-1}( \frac{1}{y}) |_{0}^{\infty} -2y\frac{\pi}{2} + y^{2}f'(y)$$ $$f'(y) = \frac{\pi}{1 + y}$$ Unfortunately, I was unable to understand the how to get the last 3 steps. As per my understanding,
$$f'(y)= 2y \int_{0}^{\pi/2}\frac{sec^{2}x - tan^{2}x }{tan^{2}x + y^{2}}dx$$ Splitting into 2 integrals, $$= 2y\int_0^{\pi/2}\dfrac{\sec^2 x}{\tan^2 x + y^2}dx-2y \int_{0}^{\pi/2}\frac{tan^{2}x }{tan^{2}x + y^{2}}dx$$ Substituting $\tan(x)=u$ in the first integral, $$ 2y\int_0^{\pi/2}\dfrac{\sec^2 x}{\tan^2 x + y^2}dx$$ $$= 2y\int_0^{\infty}\dfrac{du}{u^2 + y^2}$$ $$=2y\times \dfrac{1}{y}\tan^{-1}(\dfrac{u}{y})|_0^\infty$$ However, this does not seem to agree with the given solution. Also, I have no idea as to how to integrate $-2y \int_{0}^{\pi/2}\frac{tan^{2}x }{tan^{2}x + y^{2}}dx$ ould somebody please be so kind as to point out my error in computing the first integral and also help me integrate the second integral? Many, many thanks in advance!
|
What you're looking for is a cross spectral analysis. These notes show how it works for a bivariate case, but it's easy to apply this to multivariate series.
You start with multivariate time series:$x_t=(x_{1t},\dots,x_{nt})$. You define the matrix of autocovariate functions
$\Gamma(j)=\gamma_{mk}(j)\equiv cov(x_{mt},x_{k,t-j})$,
then the cross spectrum matrix is
$f(\omega)=f_{mk}(\omega)\equiv\frac{1}{2\pi}\sum_{j=-\infty}^\infty e^{-i\omega j}\gamma_{mk}(j)$.
From here you go on and define cross periodogram, cross spectral densities and all the usual stuff. For instance, the squared coherence is:
$\rho_{mk}=\frac{|f_{mk}(\omega)|^2}{f_{mm}(\omega)f_{kk}(\omega)}$
Similarly, you can get cross phases.
|
Let $n\in \mathbb N$ and $\{x_n\}$ is a monotone sequence. Prove that: $$ \{y_n\} = {1\over x_1 + x_2 + \dots + x_n} $$ is also a monotone sequence.
Given $\{x_n\}$ is monotone then by definition: $$ \forall n\in\mathbb N:x_n \le x_{n+1} \tag1 $$ or: $$ \forall n\in\mathbb N:x_n \ge x_{n+1} \tag2 $$ Let's prove for $(1)$ first. Using definition of a monotone sequence: $$ x_1 \le x_2 \le x_3 \le \dots \le x_n $$ So from this: $$ x_1 + x_2 + x_3 + \dots + x_n \le x_1 + x_2 + x_3 + \dots + x_n + x_{n+1} \tag 3 $$ Thus taking the reciprocal of $(3)$ we get that:
$$ {1\over x_1 + x_2 + x_3 + \dots + x_n} \ge {1\over x_1 + x_2 + x_3 + \dots + x_n + x_{n+1}} $$ But LHS is $y_n$ and RHS of the inequality is $y_{n+1}$, therefore by definition of a monotone sequence we conclude that $y_n$ is also monotone.
Case $(2)$ is obtained similarly. What bugs me is that the following still holds (at least for $x_n \ge 0$):
$$ x_1 + x_2 + x_3 + \dots + x_n \le x_1 + x_2 + x_3 + \dots + x_n + x_{n+1} $$
And we get once again that:
$$ {1\over x_1 + x_2 + x_3 + \dots + x_n} \ge {1\over x_1 + x_2 + x_3 + \dots + x_n + x_{n+1}} $$
But how is this possible? From the above it looks like $y_n$ is always monotonically decreasing, which seems false. For example when $\sum_{k=1}^nx_k < 1$.
Where have i taken the wrong road?
Update
As shown in answers and comments monotonicity is only preserved assuming all $x_n$ have the same sign.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.