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[Updated in light of some of the comments and answers below]
This is a question about the relationship between higher-order logic and topoi. It's well-known that every topos gives a model of higher-order logic, in which the type in which sentences inhabit is taken to be the subobject classifier $\Omega$.
As far as I can see $\Omega$, qua interpretation of higher-order logic, is playing two distinct roles and I can't see why they're being identitified:
$\Omega$ plays a special role in turning subobjects into elements of a power object in the topos. $Sub(-)\cong Hom(-,\Omega)$ $\Omega$ plays the role of being the truth values of sentences.
There is, for example, another natural choice for the second role, namely: $2$ (the coproduct of the terminal object with itself). Given a choice of arrow $true: 1\to 2$ we can interpret higher-order logic within a topos in an analogous way.
I'd like to know if there's any reason not to split these two roles. Is there a principle of higher-order logic that forces these two roles to coincide? A natural candidate would be the schema:
$\forall f(\forall xy(fx=fy \to x=y) \to \exists G \forall z(Gz \leftrightarrow \exists xfx=z))$
where $f:A\to B$, $x,y: A$, $z:B$ and $G:B\to \Omega$. This schema is schematic in the types $A$ and $B$. Does this sentence fail if I use $2$ in a topos instead of $\Omega$? Or does $\forall xy(fx=fy \to x=y)$ fail to express the fact that $f$ is mono? Or does $\forall z(Gz \leftrightarrow \exists xfx=z)$ fail to express the fact that $G$ is a characteristic function of $f$. Or what?
Lastly, just to clarify where I'm coming from: I'm mainly interested in topos theory as an interpretation of higher-order logic. I realise there are lots of interesting differences between topoi that lead to the same sentences of higher-order logic being true, but these aren't the sorts of differences I'm interested in (at least for these purposes). |
I'm not quite sure that I understand how the generalized likelihood ratio test works for composite hypotheses; observe the example below:
Let $X_1,...,X_n$ be a random sample from an exponential distribution, $X_i\sim EXP(\theta) \implies E(X_i)=\theta$. Derive the generalized likelihood ratio test of $H_0:\theta=\theta_0$ vs. $H_a: \theta>\theta_0$.
I've been able to to a good portion of the work; we know that, in this case, $\bar{X}$ is the maximum likelihood estimator of $\theta$. But here's where I'm confused. Suppose instead we had that $H_0: \theta=\theta_0$ vs. $H_a:\theta\ne\theta_0$.
If this were true instead, we would have that the likelihood ratio is given by: $$\lambda(\vec{X})=\frac{\bar{x}^n e^{-n\bar{x}/\theta_0+n}}{\theta_0^n}$$And then we would reject the null hypothesis if this value was less than some constant $c$. However, considering that a composite hypothesis is given instead, I believe that the decision rule needs to change somehow; the problem is that I don't understand how. The textbook lists the following as the final answer to the original question:
Reject $H_0$ if $2n\bar{x}/\theta_0 \ge \chi^2_{1-\alpha}(2n))$
where $\chi^2_{1-\alpha}(2n)$ represents the percentile function of the chi square distribution with $2n$ degrees of freedom. Can somebody show how to work to this solution; I don't know how they get there because this is a composite hypothesis and I don't understand what needs to be changed. |
This question already has an answer here:
Suppose I have a $d \times n$ matrix $\mathbf X$ (each entry point has $d$ dimensions) and after some manipulation of data (i.e. summarizing the data $\mathbf X$) I get its $d \times d$ symmetric, quadratic correlation matrix $\rho$ (defined by Pearson).
Then by
SVD definition we know that matrix $\mathbf X$ can be decomposed in three matrices:
$\mathbf X = \mathbf U \mathbf \Sigma \mathbf V^\top$
Hence here is the question:
I want to know if its correct to suppose that
$\mathbf X \mathbf X^\top = \rho$
then doing some algebra we can get:
$\rho = (\mathbf U \mathbf \Sigma \mathbf V^\top)(\mathbf V \mathbf \Sigma \mathbf U^\top)$
so
$\mathbf X \mathbf X^\top = \rho = \mathbf U \mathbf \Sigma^2 \mathbf U^\top$
I want to know if this equivalence is correct, or in which cases is correct? Can anyone give an explanation about it?.
Note I am using correlation Matrix instead of covariance matrix, (we know we can get eigenvalues and eigenvectors from a correlation matrix, and that would solve PCA) |
Seminar der Arbeitsgruppe Algebra: Wintersemester 2018/19 Relations between Fourier coefficients of Siegel modular forms
Dienstag, der 16. Oktober 2018, 15:15 – 16:15 Uhr, Raum S2|15 401
Referentin: Jolanta Marzec (TU Darmstadt) Abstract: Fourier coefficients of Siegel modular forms have played a prominent role in number theory. For example, in case of classical modular forms (degree $n=1$), the coefficients are equal to Hecke eigenvalues and are linked to a number of rational points on suitable elliptic curves. In the case of Siegel modular forms of degree $2$ they carry information about certain central critical values of quadratic twists of associated spinor $L$-functions (B\"ocherer's conjecture). Of special importance are \emph{fundamental Fourier coefficients}: their non-vanishing is related to existence of `nice' models for the associated global automorphic representations, and consequently enabled to prove analytic and algebraic properties for $L$-functions for $\mathrm{GSp}_4\times\mathrm{GL}_2$.
Even though a lot of work has been devoted to Siegel modular forms of degree $2$, their Fourier coefficients – especially in case of non-trivial level – lack good understanding. Therefore, motivated by the properties described above and the problem of determination of Siegel modular forms by an interesting subset of Fourier coefficients, we derive relations between `simple' and `more complicated' Fourier coefficients, and indicate their applications.
CM values, cycle integrals, and polyharmonic Maass forms
Donnerstag, den 25. Oktober 2018, 15:20 – 17:00 Uhr, Raum S2|15 244
Referent: Toshiki Matsusaka (Kyushu University / Universität zu Köln) Abstract: In 2011 Duke-Imamoglu-Toth showed the generating functions of traces of cycle integrals and CM values of modular functions are (poly)harmonic Maass forms. I give a generalization of their work to polyharmonic Maass forms. Applications of sieve methods
Dienstag, der 30. Oktober 2018, 15:15 – 16:15 Uhr, Raum S2|15 401
Referent: Christian Elsholtz (TU Graz) Abstract: Apart from the ancient sieve of Eratosthenes, modern sieve theory started in 1915, when Viggo Brun showed that there are not too many twin primes, so that the sum of reciprocals 1/5+1/7+1/11+1/13 +1/17+1/19+… is finite, or convergent. In fact, it can be proved that for any fixed (a_1, …, a_k) the number of k-tuples of primes (n+a_1,n+a_2,…, n+a_k) with n <x is at most C_{k,a_1,…,a_k} x/(log x)^k. Using the large sieve, one can analyze what happens, when k is not a constant. If for example k=log x, one can give the following upper bound for the number of such tuples: O(x/exp( log x log log log x/log log x)). When k is (log x)2, onbe gets the dramatically better bound O(x^(2/3)), and when k increases this approaches x^(1/2). From this one can conclude that the set of primes cannot be written as a sumset P=A+B+C, even if one allows finitely many mistakes. In this talk we give a survey of such upper bounds for the prime k-tuple problem in its full range, from very small k to very large k. Newform theory from Jacobi forms of lattice index
Dienstag, den 27. November 2018, 14:00 – 15:00 Uhr, Raum S2|15 401
Referentin: Andreea Mocanu (University of Nottingham) Abstract: I will give a brief introduction to Jacobi forms, including some examples and their relation to other types of modular forms. After that, I will discuss some of the ingredients that go into developing a theory of newforms for Jacobi forms of lattice index, namely Hecke operators, level raising operators and orthogonal groups of discriminant modules. Non-existence of reflective modular forms
Tuesday, December 18, 2018, 15:15 – 16:15, Room S2|15 401
Speaker: Haowu Wang (Université Lille) Abstract: Reflective modular forms are holomorphic functions whose divisors are determined by rational quadratic divisors associated to reflective vectors. This type of modular forms has applications in arithmetic, geometry and Lie algebra. The classification of reflective modular forms is an old open problem and has been investigated by several mathematicians. In this talk, I introduce an approach based on the theory of Jacobi forms to classify reflective modular forms on lattices of general level. I give an explicit formula to express the weight of reflective modular forms. I also prove two non-existence results. The first one is that there is no lattice of signature (2, n) having a 2-reflective modular form when n > 14 with three exceptions. The second is that there is no lattice of signature (2, n) admitting a reflective modular form when n >22 except the unimodular lattice of signature (2,26). Automorphic representations of symplectic groups
Tuesday, January 8, 2019, 15:15 – 16:15, Room S2|15 401
Speaker: Jolanta Marzec (TU Darmstadt) Abstract: How to associate an automorphic representation of GSp(2n) to a Siegel modular form of degree n. Tomorrow we will focus on spherical representations. p-adic dynamics of Hecke operators
Friday, January 11, 2019, 14:15 -15:15, Raum S2|15 401
Speaker: Eyal Goren (McGill University) Abstract: Let p and q be distinct primes. In a joint work with Payman Kassaei (King’s College) we are studying questions concerning the dynamics of the Hecke operator T_q in its action on the modular curves X_1(N), where (N, pq) = 1, in the p-adic topology. Using Serre-Tate coordinates and isogeny volcanoes for points of X_1(N) with ordinary reduction, and using the graphs of supersingular elliptic curves and the Gross-Hopkins period map for points with supersingular reduction, we are able to get rather precise understanding of orbits of points. If time allows, I will also discuss the work of Herrero, Menares and Rivera-Letelier that is complementary to ours. Automorphic representations of symplectic groups, II
Monday, March 11, 2019, 15:15 – 16:15, Room S2|15 401
Speaker: Jolanta Marzec (TU Darmstadt) Abstract: Part III: We continue studying an automorphic representation associated to a Siegel modular form of degree n. This time we will construct a component at the archimedean place. Generalized Kontsevich-Zagier series via knots
Tuesday, April 23, 2019, 15:15 – 16:15, Room S2|15 401
Speaker: Robert Osburn (University College Dublin) Abstract: Over the past two decades, there has been substantial interest in the overlap between quantum knot invariants, q-series and modular forms. In this talk, we discuss one such instance, namely an explicit q-hypergeometric expression for the colored Jones polynomial for double twist knots. As an application, we generalize a duality at roots of unity between the Kontsevich-Zagier series and the generating function for strongly unimodal sequences. This is joint work with Jeremy Lovejoy (Paris 7 and Berkeley). Non-Archimedean integrals on the Hitchin Fibration
Tuesday, May 14, 2019, 15:15 – 16:15, Room S2|15 401
Speaker Dimitri Wyss (Université Paris 6) Abstract:
Based on mirror symmetry considerations, Hausel and Thaddeus conjectured an equality between 'stringy' Hodge numbers for moduli spaces of SL_n/PGL_n Higgs bundles. We prove this conjecture using non-archimedean integration in the sense of Denef-Loeser and Batyrev and the duality of generic Hitchin fibers.
In a more arithmetic context, similar ideas lead to a new proof of the geometric stabilization theorem for anisotropic Hitchin fibers, a key ingredient in the proof of the fundamental lemma by Ngô. This is joint work with Michael Groechenig and Paul Ziegler.
tba
Tuesday, May 21, 2019, 15:15 – 16:15, Room S2|15 401
Speaker Mark Feldmann (Universität Münster) Abstract: We study Weil group representations over the coefficient feld Qp and establish certain equivalences of categories in the flavor of Fontaine's classification of p-adic representations of the absolute Galois group. If we restrict to crystalline (or de-Rham) Weil group representations, we can describe the category of these Weil group representations in terms of generators. More precisely it is generated as an abelian tensor category by the full subcategory of Galois group representations and finite unramified inductions of the character Qp(| · |) given by Artin's reciprocity law. Fourier coefficients of automorphic forms: p-adic analysis and arithmetic at the cusps
Tuesday, June 25, 2019, 15:15 – 16:15, Room S2|15 401
Speaker Andrew Corbett (University Exeter, UK) Abstract: The information buried in the Fourier coefficients of automorphic forms is not often readily accessible; that is part of their endear. We tackle a selection of problems in number theory whose solution is related a careful understanding of such Fourier coefficients at different cusps. We bring forth a p-adic framework which may be utilised in the various problems. Purity for the Brauer group of singular schemes A Geometric Theta Correspondence for Picard Modular Surfaces
Tuesday, July 9, 2019, 15:15 – 16:15, Room S2|15 401
Speaker Robert Little (Oxford University) Abstract: For a noncompact Picard Modular Surface X, the machinery of Kudla & Millson give us a degree 2 cohomology class $\theta_L(\tau)$ on X, which is a holomorphic modular form in the variable $\tau$ . Following work of Cogdell, Hirzebruch-Zagier and Funke-Millson, we aim to extend this cohomology class to the boundary components of a compactification of X, which will allow us to create a holomorphic, modular, non-trivial and compactly supported degree 2 cohomology class on X. This work uses techniques from representation theory, arithmetic geometry and real analysis; we will also be able to give concrete examples of holomorphic modular forms given by pairing non-compact homology on X. Estimates of holomorphic cusp forms associated to co-compact arithmetic subgroups
Tuesday, November 26, 2019, 15:15 – 16:15, Room S2|15 401
Speaker: Anilatmaja Aryasomayajula (IISER Tirupati) Abstract: In 1995, Iwaniec and Sarnak computed estimates of Hecke eigen Mass forms associated to co-compact arithmetic subgroups of SL_{2}(R). Adapting the arguments of Iwaniec and Sarnak to the holomorphic setting, Das and Sengupta derived sub-convexity bounds of holomorphic cusp forms associated the co-compact arithmetic subgroups considered by Iwaniec and Sarnak. In a recent work of ours, we have combined the approach of Das and Sengupta with a heat kernel approach to improve the sub-convexity bounds of Dad and Sengupta. tba
Tuesday, February 11, 2020, 15:15 – 16:15, Room S2|15 401
Speaker: Dr. Hongbo Yin (MPIM Bonn) |
I think I'm missing something quite basic here but consider the process:
$$ e^- + e^+ \rightarrow 2\gamma$$
Fermions have opposite parity to antifermions so the parity quantum number before the process is $P=-1 \times (-1)^L$ where $L$ is the relative orbital angular momentum, which should vanish in the zero momentum frame where the collision is head on. So we start with $P=-1$. The photons on the other hand have the same parity as each other so have parity $P=+1$ (after again noting that they must be heading in opposite directions to each other in the ZMF so have $L=0$).
I think my reasoning about the angular momentum must be wrong, but I don't see why. (Additionally, the same argument says that if this is $L:0\rightarrow 0$ then charge congugation $C$ is not a conserved quantity either.) |
Write out the simple equations
$$\begin{align}Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3}\end{align}$$
There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started.
Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem$$X\perp Y \implies E[XY] = E[X]E[Y]$$Note that the converse of this statement is not true.
Proof
Assertion $a_1a_0 + a_2a_1 = 0 \iff Y_j \perp Y_{j-1}$
Define $\mu = E[Z]$
($\implies$) Suppose $a_1a_0 + a_2a_1 = 0$. There are two cases where this is possible. Case 1, suppose $a_1 = 0$. The equations become
$$\begin{align}Y_j &= a_0 Z_j + a_2 Z_{j-2}\\Y_{j-1} &= a_0 Z_{j-1} + a_2 Z_{j-3}\end{align}$$
Their $\sigma$-algebras are given by $\sigma(Y_j) = \sigma(Z_j)\cup\sigma(Z_{j-2})$ and $\sigma(Y_{j-1}) = \sigma(Z_{j-1})\cup \sigma(Z_{j-3})$. Thus $Y_j \perp Y_{j-1}$. This could be more gruesomely detailed but I take some for granted. See this for more details including definitions etc.
Case 2, suppose $a_2 = 0$ and $a_0 = 0$. The equations become
$$\begin{align}Y_j &= a_1 Z_{j-1} \\Y_{j-1} &= a_1 Z_{j-2}\end{align}$$
The same $\sigma$-algebra argument applies more easily but a more elegant solution presents itself in the form of the CDF.
$$\begin{align}F_{Y_j, Y_{j-1}}(y_j, y_{j-1}) &= P(Y_j \leq y_j \text{ and } Y_{j-1} \leq y_{j-1}) \\& = F_{Z_j}(y_j/a_1)F_{Z_{j-1}}(y_{j-1}/a_1)\\& = F_{Y_j}(y_j)F_{Y_{j-1}}(y_{j-1})\end{align}$$
($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,
$$\begin{align}E[Y_jY_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \\& + (a_1a_0 + a_2a_1)E[Z^2]\end{align}$$
$$\begin{align}E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2)\mu^2\\&+ (a_1a_0 + a_2a_1)\mu^2\end{align}$$
In the non-degenerate case when the distribution of $Z$ is not a constant, the variance is strictly positive so that $E[Z^2] - \mu^2 > 0$ and so $E[Z^2] > \mu^2$ and more importantly $E[Z^2] \neq \mu^2$ Thus for the equality $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ to hold, it must be the case that $a_1a_0 + a_2a_1 = 0$. |
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Change to browse by: Bookmark(what is this?) Astrophysics > High Energy Astrophysical Phenomena Title: Amplifying magnetic fields of a newly born neutron star by stochastic angular momentum accretion in core collapse supernovae
(Submitted on 20 Mar 2019 (v1), last revised 22 Aug 2019 (this version, v2))
Abstract: I present a novel mechanism to boost magnetic field amplification of newly born neutron stars in core collapse supernovae. In this mechanism, that operates in the jittering jets explosion mechanism and comes on top of the regular magnetic field amplification by turbulence, the accretion of stochastic angular momentum in core collapse supernovae forms a neutron star with strong initial magnetic fields but with a slow rotation. The varying angular momentum of the accreted gas, which is unique to the jittering jets explosion mechanism, exerts a varying azimuthal shear on the magnetic fields of the accreted mass near the surface of the neutron star. This, I argue, can form an amplifying effect which I term the stochastic omega (S$\omega$) effect. In the common $\alpha \omega$ dynamo the rotation has constant direction and value, and hence supplies a constant azimuthal shear, while the convection has a stochastic behavior. In the S$\omega$ dynamo the stochastic angular momentum is different from turbulence in that it operates on a large scale, and it is different from a regular rotational shear in being stochastic. The basic assumption is that because of the varying direction of the angular momentum axis from one accretion episode to the next, the rotational flow of an accretion episode stretches the magnetic fields that were amplified in the previous episode. I estimate the amplification factor of the S$\omega$ dynamo alone to be $\approx 10$. I speculate that the S$\omega$ effect accounts for a recent finding that many neutron stars are born with strong magnetic fields. Submission historyFrom: Noam Soker [view email] [v1]Wed, 20 Mar 2019 05:55:18 GMT (160kb,D) [v2]Thu, 22 Aug 2019 03:44:10 GMT (154kb,D) |
Research Open Access Published: Global existence and uniqueness of solutions to the three-dimensional Boussinesq equations Boundary Value Problems volume 2016, Article number: 85 (2016) Article metrics
1127 Accesses
1 Citations
Abstract
In this paper, we study the three-dimensional Boussinesq equations and obtain the global existence and uniqueness of a suitable weak solution in unbounded exterior domain.
Introduction
Let \(D \subset{\mathbf {R}^{3}}\) be a bounded domain with a connected \(C^{2}\)-smooth boundary
S, and \(D':={\mathbf {R}^{3}} \backslash D \) be the unbounded exterior domain. We consider following three-dimensional Boussinesq equations:
where \(p=p(x,t)\) is the scalar pressure, \(u=u(x,t)\) is the velocity vector field, and \(\theta=\theta(x,t)\) is a scalar quantity such as the concentration of a chemical substance or the temperature variation in a gravity field, in which case \(\theta e_{3}\) represents the buoyancy force. The nonnegative parameters
μ and ν denote the molecular diffusion and the viscosity, respectively; \(u\cdot\nabla u:=u_{a}\partial_{a} u\), \(\partial_{a} u:=\frac{\partial u}{\partial x_{a}}:=u_{;a}\), \(u\cdot\nabla\theta:=u_{a}\partial_{a} \theta\), \(\nabla\cdot u:=u_{a;a}=0\), \(\frac{\partial u^{2}}{\partial x_{a}}:=(u^{2})_{;a}\), \(u^{2}=u_{b}u_{b}\). Over the repeated indices a and b summation is understood, \(1\leq a\), \(b\leq3\).
The Boussinesq system is one of the most commonly used fluid models since it has a vortex stretching effect similar to that in the 3D incompressible flow. The Boussinesq system has an important roles in the atmospheric sciences [1] and is a model in many geophysical applications [2]. For this reason, this system is studied systematically by scientists from different domains.
Before starting and proving our main results, let us first briefly recall the related results in the literature. For the three-dimensional case, the progress on the local well-posedness and regularity criteria on the inviscid Boussinesq equations has been made (see,
e.g., [3–11]). Since there is no global well-posedness for the standard high-dimensional Boussinesq equations for large initial data, it is natural to consider the blow-up (or regularity) criterion. Various criteria for the Boussinesq equations have been obtained by considerable works (see, e.g., [3–5, 7–9, 12–17] and the references therein).
In this paper, we establish the global existence and uniqueness of a suitable weak solution to the three-dimensional Boussinesq equations in unbounded exterior domains.
We assume that \(u\in G\) and \(\theta\in J\), where
with arbitrary \(T>0\).
In this paper, \((f,g):=\int_{D'}f_{a}g_{a}\,dx\) denotes the inner product in \(L^{2}(D')\), \(\Vert f\Vert :=(f,f)^{\frac{1}{2}}\), \(f_{ja}\) denotes the
ath component of the vector function \(f_{j}\), and \(f_{ja;b}\) is the derivative \(\frac{\partial f_{ja}}{\partial x_{b}}\).
Here we took into account that \(-(\Delta u, v)=(\nabla u, \nabla v)\), \(-(\Delta\theta, w)=(\nabla\theta, \nabla w)\), and \((\nabla p, v)=-(p, v_{a;a})=0\), if \(v\in G\), \(w\in J\).
Equation (1.6) is equivalent to the integrated equation
We also assume that
Now we are in position to state our main results.
Theorem 1.1 If assumption (1.8) holds, \(u_{0}\in H_{0}^{1}(D)\) satisfies (1.3), and \(\theta_{0}\in H_{0}^{1}(D)\), then for all \(t\geq0\), there exist weak solutions \(u\in G\), \(\theta\in J\) to (1.6), and the solutions are unique, provided that \(|\theta(\cdot,t)|\), \(\Vert \nabla u\Vert ^{4}\in L_{\mathrm{loc}}^{1}(0,\infty)\). Remark 1.1 Proof of Theorem 1.1 Lemma 2.1 Proof
See,
e.g., [20]. □ Lemma 2.2 Under assumption (1.8), the following estimate holds: where \(C>0\) is a constant. Proof
Take \(v=u\) and \(w=\theta\) in (1.6). Then
Thus, equation (1.6) with \(v=u\) and \(w=\theta\) implies
Denote \(\sup_{s\in[0,t]}\Vert u(s)\Vert =\alpha(t)\). Then (2.4) implies
Since \(\alpha(t)\geq0\), inequality (2.5) yields
Remark 2.3 A priori estimate (2.2) implies that, for every \(T\in [0,\infty)\), \(u\in L^{\infty}(0,T; L^{2}(D'))\cap L^{2}(0,T; H_{0}^{1}(D'))\), \(\theta\in L^{2}(0,T; H_{0}^{1}(D'))\). This and equation (1.6) also imply that \(u_{t}, \theta_{t}\in L^{2}(D'\times[0,T])\). Lemma 2.4 Proof
The idea of the proof is to reduce the problem to the existence of the solution to a Cauchy problem for ordinary differential equations (ODE) of finite order and then to use an
a priori estimate to establish the convergence of these solutions of ODE to the solutions of equations (1.6) and (1.7).
Let us look for the solutions to equation (1.6) of the form \(u^{n}:=\sum_{j=1}^{n}c_{j}^{n}(t)\phi_{j}(x)\), \(\theta^{n}:=\sum_{k=1}^{n}d_{k}^{n}(t)\phi _{k}(x)\), where \(\{\phi_{i}(x)\}_{i=1}^{\infty}\) is an orthonormal basis of the space \(L^{2}(D')\) of divergence-free vector functions belonging to \(H_{0}^{1}(D')\), and in the expression \(u^{n}\), \(\theta^{n}\), the upper index
n is not a power. If we substitute \(u^{n}\), \(\theta^{n}\) into (1.6), take \(v=\phi_{v}\), \(w=\phi_{w}\), and use the orthonormality of the system \(\{\phi_{i}(x)\} _{i=1}^{\infty}\) and relation \((\nabla\phi_{j}, \nabla\phi _{v})=\lambda_{v}\delta_{jv}\), \((\nabla\phi_{k}, \nabla\phi_{w})=\lambda _{w}\delta_{kw}\), where \(\lambda_{v}\), \(\lambda_{w}\) are the eigenvalues of the vector Dirichlet Laplacian in D on the divergence-free vector fields, then we obtain a system of ODE for the unknown coefficients \(c_{v}^{n}\), \(d_{w}^{n}\):
Consider the set \(\{u^{n}=u^{n}(t)\}_{n=1}^{\infty}\), \(\{\theta^{n}=\theta ^{n}(t)\}_{n=1}^{\infty}\). Inequalities (2.2) and (2.8) for \(u=u^{n}\), \(\theta=\theta^{n}\) imply the existence of the weak limits \(u^{n}\rightharpoonup u\), \(\theta^{n}\rightharpoonup\theta\) in \(L^{2}(0,T; H_{0}^{1}(D'))\). This allows us to pass to the limit in equation (1.7) in all the terms except the term \(\int_{0}^{t} [(u_{s}, v)+(\theta_{s}, w) ]\,ds\). The weak limits of the terms \((u^{n}_{a}u_{b;a}^{n}, v_{b})\), \((u^{n}_{a} \theta^{n}_{b;a}, w_{b})\) exist and are equal to \((u_{a}u_{b;a}, v_{b})\), \((u_{a} \theta_{b;a}, w_{b})\), respectively, because
Note that \(v_{b;a},w_{b;a}\in L^{2}(D')\) and \(u^{n}_{a}u_{b}^{n}, u^{n}_{a}\theta _{b}^{n}\in L^{4}(D')\). Applying the interpolation inequality and the Schwarz inequality, we get
where \(\epsilon>0\) is an arbitrary small number. We have \(u^{n}_{a}u_{b}^{n} \rightharpoonup u_{a}u_{b}\) in \(L^{2}(D')\) as \(n\rightarrow\infty\) because bounded sets in a reflexive Banach space \(L^{4}(D')\) are weakly compact. Consequently, \((u^{n}_{a}u_{b;a}^{n}, v_{b})\rightarrow(u_{a}u_{b;a}, v_{b})\) as \(n\rightarrow\infty\). Similarly, we have \((u^{n}_{a}\theta_{b;a}^{n}, w_{b})\rightarrow(u_{a}\theta _{b;a}, w_{b})\) as \(n\rightarrow\infty\).
The weak limit of the \(\int_{0}^{t} [(\nabla u, \nabla v)+(\nabla \theta, \nabla w) ]\,ds\) exists because of the
a priori estimate (2.2) and the weak compactness of the bounded sets in a Hilbert space. Since equation (1.7) holds and the limits of all its terms, except \(\int_{0}^{t} [(u_{s}^{n}, v)+(\theta_{s}^{n}, w) ]\,ds \), do exist, there exists the limit \(\int_{0}^{t} [(u_{s}^{n}, v)+(\theta _{s}^{n}, w) ]\,ds\rightarrow\int_{0}^{t} [(u_{s}, v)+(\theta_{s}, w) ]\,ds\) for all \(v, w\in G\). By passing to the limit as \(n\rightarrow \infty\) we prove that the limits u, θ satisfy (1.7). Differentiating (1.7) with respect to t yields (1.6) almost everywhere. □ Lemma 2.5 The global weak solutions u, θ from Lemma 2.4 are unique. Proof
Suppose that there are two solutions to (1.6), \(u\in G\), \(\theta\in J\) and \(u'\in G\), \(\theta'\in J\), and let \(U=u-u'\), \(\Theta=\theta-\theta'\). Then
Since \(U\in G\), \(\Theta\in J\), we may set \(v=U\), \(w=\Theta\) in (2.11) and have
where we have used the following conclusions:
Thus, (2.12) implies
From Young’s inequality we get the following estimates:
Denoting \(z:=\Vert U\Vert ^{2}+\Vert \Theta \Vert ^{2}\) and choosing \(\epsilon=\frac {1}{2\Vert \nabla u\Vert }\) in inequalities (2.14), we obtain
Assume that \(|\theta(\cdot,t)|\), \(\Vert \nabla u\Vert ^{4}\in L^{1}_{\mathrm{loc}}(0,\infty)\). Then we get \(z=0\) for all \(t\geq0\). □
The proof of Theorem 1.1 is complete.
References 1.
Majda, A: Introduction to PDEs and Waves for the Atmosphere and Ocean. Courant Lecture Notes in Mathematics, vol. 9. AMS/CIMS, Providence (2003)
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Pedlosky, J: Geophysical Fluid Dynamics. Springer, New York (1987)
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Chae, D, Kim, S-K, Nam, H-S: Local existence and blow-up criterion of Hölder continuous solutions of the Boussinesq equations. Nagoya Math. J.
155, 55-80 (1999) 4.
Chae, D, Nam, HS: Local existence and blow-up criterion for the Boussinesq equations. Proc. R. Soc. Edinb., Sect. A
127, 935-946 (1997) 5.
Cui, X, Dou, C, Jiu, Q: Local well-posedness and blow up criterion for the inviscid Boussinesq system in Hölder spaces. J. Partial Differ. Equ.
25, 220-238 (2012) 6.
Danchin, R: Remarks on the lifespan of the solutions to some models of incompressible fluid mechanics. Proc. Am. Math. Soc.
141, 1979-1993 (2013) 7.
Ishimura, N, Morimoto, H: Remarks on the blow-up criterion for the 3-D Boussinesq equations. Math. Models Methods Appl. Sci.
9, 1323-1332 (1999) 8.
Liu, X, Wang, M, Zhang, Z: Local well-posedness and blowup criterion of the Boussinesq equations in critical Besov spaces. J. Math. Fluid Mech.
12, 280-292 (2010) 9.
Shu, WEC: Small-scale structures in Boussinesq convection. Phys. Fluids
6, 49-58 (1994) 10.
Taniuchi, Y: A note on the blow-up criterion for the inviscid 2D Boussinesq equations. Lect. Notes Pure Appl. Math.
223, 131-140 (2002) 11.
Xu, X, Ye, Z: The lifespan of solutions to the inviscid 3D Boussinesq system. Appl. Math. Lett.
26, 854-859 (2013) 12.
Chae, D: Local existence and blow-up criterion for the Euler equations in the Besov spaces. Asymptot. Anal.
38, 339-358 (2004) 13.
Fan, J, Zhou, Y: A note on regularity criterion for the 3D Boussinesq system with partial viscosity. Appl. Math. Lett.
22, 802-805 (2009) 14.
Liu, X, Li, Y: On the stability of global solutions to the 3D Boussinesq system. Nonlinear Anal. TMA
95, 580-591 (2014) 15.
Qiu, H, Du, Y, Yao, Z: Local existence and blow-up criterion for the generalized Boussinesq equations in Besov spaces. Math. Methods Appl. Sci.
36, 86-98 (2012) 16.
Ramm, AG: Existence and uniqueness of the global solution to the Navier-Stokes equations. Appl. Math. Lett.
49, 7-11 (2015) 17.
Ye, Z: Blow-up criterion of smooth solutions for the Boussinesq equations. Nonlinear Anal. TMA
110, 97-103 (2014) 18.
Ladyzhenskaya, O: The Mathematical Theory of Viscous Incompressible Flow. Gordon and Breach, New York (1969)
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Temam, R: Navier-Stokes Equations. Theory and Numerical Analysis. North Holland, Amsterdam (1984)
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Brandolese, L, Schonbek, ME: Large time decay and growth for solutions of a viscous Boussinesq system. Trans. Am. Math. Soc.
364, 5057-5090 (2012) Acknowledgements
The work was in part supported by the NNSF of China (Nos. 11326158, 11271066, and 11571227) and the Innovation Program of Shanghai Municipal Education Commission (No. 13ZZ048).
Additional information Competing interests
The author declares that they have no competing interests. |
What is the relationship between $G_\infty$ (homotopy Gerstenhaber) and $B_\infty$ algebras?
In Getzler & Jones "Operads, homotopy algebra, and iterated integrals for double loop spaces" (a paper I don't well understand) a $B_\infty$ algebra is defined to be a graded vector space $V$ together with a dg-bialgebra structure on $BV = \oplus_{i \geq 0} (V[1])^{\otimes i}$, that is a square-zero, degree one coderivation $\delta$ of the canonical coalgebra structure (stopping here, we have defined an $A_\infty$ algebra) and an associative multiplication $m:BV \otimes BV \to BV$ that is a morphism of coalgebras and such that $\delta$ is a derivation of $m$.
A $G_\infty$ algebra is more complicated. The $G_\infty$ operad is a dg-operad whose underlying graded operad is free and such that its cohomology is the operad controlling Gerstenhaber algebras. I believe that the operad of chains on the little 2-discs operad is a model for the $G_\infty$ operad. Yes?
It is now known (the famous Deligne conjecture) that the Hochschild cochain complex of an associative algebra carries the structure of a $G_\infty$ algebra. It also carries the structure of a $B_\infty$ algebra. Some articles discuss the $G_\infty$ structure while others discuss the $B_\infty$ structure. So I wonder: How are these structures related in this case? In general? |
Surely there are many: these are all polynomials in one variable, so every two of them are algebraically dependent because of the transcendence degree argument :-)
However, I am sure that this is not what you wanted to hear, so here you are a nice argument showing how to guess your formula and obtain other formulas somewhat similar to it. Note that there is a remarkable symmetry property $P_k(-1-N)=(-1)^{k+1} P_k(N)$ for $k>0$. (Basically, for $k>0$ the polynomial $P_k(x)$ is the only polynomial of degree $k+1$ solving the functional equation $f(x)-f(x-1)=x^k$ together with the condition $f(0)=0$, and then you can show that $Q_k(x)=(-1)^{k+1} P_k(-1-x)$ satisfied exactly the same conditions, which proves the symmetry property without any annoying computations.) If we re-define $P_0(N)=N+\frac12$ (and assume $P_{-1}=1$), this symmetry will hold in general.
Now, the polynomial $P_1^2$, as a polynomial of degree $4$, should be a rational combination of $P_0$, $P_1$, $P_2$ and $P_3$ (and such a combination is clearly unique - you yourself observed that they form a basis), and because of the type of symmetry it possesses, it is actually a combination of $P_1$ and $P_3$ (because other polynomials change sign under the symmetry $N\mapsto -1-N$, and this would contradict the linear independence), and looking at it carefully we observe that the $P_1$-coefficient is equal to zero, and the $P_3$-coefficient is equal to~$1$, which is your formula.
For the same reason, the product $P_mP_n$ is expressed as a linear combination of $P_l$ where $l\le m+n+1$, $l\equiv m+n+1\pmod{2}$, - half of the terms disappear for free! (And, because of vanishing at~$0$, the redefined $P_0=N+\frac12$ and $P_{-1}=1$ will not show up in such a combination if $m+n>0$.)
Some examples: $6P_1P_2=5P_4+P_2$, $3P_2^2=2P_5+P_3$, $12P_2P_3=7P_6+5P_4$, $2P_3^2=P_7+P_5$, $60P_3P_4=27P_8+35P_6-2P_4$ (this last one is a bit disappointing!) etc. |
How to Couple a Full-Wave Simulation to a Ray Tracing Simulation
Welcome back to our discussion on multiscale modeling in high-frequency electromagnetics. Multiscale modeling is a simulation challenge that arises when there are vastly different scales in a single simulation, such as the size of an antenna compared to the distance between the antenna and its target. Today, in Part 4 of the series, we will examine how we can construct a multiscale model by coupling a
Full-Wave antenna simulation with a geometrical optics simulation using the Ray Optics Module.
Using the Ray Optics Module for Multiscale Modeling
In Part 2 of the blog series, we used the
Electromagnetic Waves, Frequency Domain interface, which we call a Full-Wave simulation, and a Far-Field Domain node to determine the electric field in the far field. We then coupled a Full-Wave simulation to the Electromagnetic Waves, Beam Envelopes interface (or a Beam-Envelopes simulation) in order to precisely calculate fields in any region, regardless of the distance from the source.
The
Far-Field Domain and Beam-Envelopes solutions that we looked at in the previous blog post are effective, but they share one noteworthy restriction. In each case, we assumed that a homogeneous domain surrounded the antenna in all directions. For many situations, this information is sufficient. In other simulations, you may not have a homogeneous domain surrounding your antenna and you need to account for issues like atmospheric refraction or reflection off of nearby buildings. These simulations require a different approach. A model of several hotels in Las Vegas. A directional antenna emits rays toward the ARIA® Resort & Casino.
The
Geometrical Optics interface in the Ray Optics Module, an add-on product to the COMSOL Multiphysics® software, regards EM waves as rays. This interface can account for spatially varying refractive indices, reflection and refraction from complicated geometries, and long propagation distances. However, these features come with a tradeoff. Since waves are treated as rays, this approach neglects diffraction. In other words, we are assuming that the wavelength of light is much smaller than any geometric features in our environment. You can read a more thorough description of ray optics in a previous blog post. Modeling Coupled Antennas: Preparing for Ray Optics
As you may recall, we introduced an approach to coupling a radiating and receiving antenna in Part 3 of this series. When incorporating ray optics into our multiscale modeling, we are required to use a similar but more generalized approach. Before we show you how to set up a geometrical optics simulation in COMSOL Multiphysics, let’s first review this alternate method.
Mapping the Radiated Fields
As a quick refresher, we are interested in calculating the fields at the location of the receiving antenna using the following equation:
We previously used an integration operator on a single point to calculate this along the line directly between the two antennas. We now wish to retain the angular dependence, so we need to recalculate this equation for each point in the receiving antenna’s domain. Since it is impractical to add numerous points and integration operators, we need to establish a more general technique.
To do so, we replace the integration operator with a General Extrusion operator. As before, we create a variable for the magnitude of
r. We then use the General Extrusion operator to evaluate the scattering amplitude at a point in the geometry that shares the same angular coordinates, (\theta,\phi), as the point in which we are actually interested.
To demonstrate this concept, we use a figure that is slightly more involved than that from the previous post. Note that the subscripts 1, 2, and
r in \vec{r}_j=\left(x_j,y_j,z_j\right) represent a vector in component 1, a vector in component 2, and the offset between the antennas, respectively. Image showing where the scattering amplitude should be calculated and how the coordinates of that point can be determined.
As we previously outlined, the primary complication is determining where to calculate the scattering amplitude. We want the fields at the point \vec{r}_r + \vec{r_2}, which requires calculating the scattering amplitude at \vec{r}_1. The complication, of course, is that each point in the domain around the receiving antenna (each vector \vec{r}_2) will have its own evaluation location \vec{r}_1. We evaluate this by again rescaling the Cartesian coordinates, but instead of doing it for a single point, we define it inside of the general operator so that it can be called from any location. From the above figure, we know that this point is x_1 = \left(x+x_r\right) \frac{|\vec{r}_1|}{|\vec{r}_2+\vec{r_r}|}, with corresponding equations for
y and z. The operator is defined in component 1, so the source will be defined in that component. It will be called from component 2, so the x, y, z in the following expressions refer to x 2, y 2, z 2 in the above figure. The General Extrusion operator used for the scattering amplitude calculation. Note that this is defined in component 1. Storing the Radiated Fields in a Dummy Variable
As a bookkeeping step, we store the calculated fields in a “dummy” variable. By a dummy variable, we mean that we add in an extra dependent variable that takes the value of a calculation determined elsewhere. We do this for two reasons.
The first reason is that most variables in COMSOL Multiphysics are calculated on demand from the dependent variables. In an RF simulation, for example, the dependent variables are the three Cartesian components of the electric field: Ex, Ey, and Ez. These are determined when computing the solution. In postprocessing, every other value (electric current, magnetic field, etc.) is calculated from the electric field when required. In most cases, this is a fast and seamless process. In our case, each field evaluation point requires a general extrusion of a scattering amplitude, and each scattering amplitude point requires a surface integration as defined in the
Far-Field Domain node. This can take a while and we want to ensure that we perform this calculation only once.
The second reason why we do this has to do with the element order. The
Scattered Field formulation requires a background electric field. COMSOL Multiphysics then calculates the magnetic field using the differential form of Faraday’s law (also known as the Maxwell-Faraday equation). This requires taking spatial derivatives of the electric field. There are no issues when taking the spatial derivatives of an analytical function like a plane wave or Gaussian beam, but it can cause a discretization issue when applied to a solved-for variable. This is a rather advanced topic, which you can find out more about in an archived webinar on equation-based modeling.
By using a cubic dummy variable to store the electric field, we can take a spatial derivative of the electric field and still obtain a well-resolved magnetic field for use in the
Scattered Field formulation. Without the increased order of the dummy variable, the magnetic field used would be underresolved. Below, you can see what it looks like to put the General Extrusion operator together with the dummy variable setup. The variable r is identical to the one used in Part 3 of this blog series and is defined in component 2. The dummy variable implementation. Notice that the dummy variable components are called Ebx, Eby, and Ebz.
The only remaining step is to use the dummy variables — Ebx, Eby, and Ebz — in a background field simulation of the half-wavelength dipole discussed in Part 1 and Part 3.
This technique isn’t actually very good for this particular problem. There may be situations where it is useful, but the technique from Part 3 is preferred in the vast majority of cases. The received power from the two simulations is extremely close, but this method takes much longer to calculate and the file size increases drastically. In the demo examples for this post, this method took several times longer than the previous simulation method. While you may conclude that this is not a terribly useful step overall, it is useful when we incorporate ray optics into our multiscale modeling, as discussed in the next section.
Setting Up a Geometrical Optics Simulation in the COMSOL® Software
A geometrical optics simulation implicitly assumes that every ray is already in the far field. Earlier in the blog series, we saw that the
Far-Field Domain feature correctly calculates the electric field at arbitrary points in the far field. Here, we use that information as the input for rays in a geometrical optics simulation. The simulation geometry, symmetry, and electric dipole point source used are the same as in Part 2. The domain assignments for the simulation. The Full-Wave simulation is performed over the entire domain, with the outer region set as a perfectly matched layer (PML). The geometrical optics simulation is only performed in this outer region. Note that this image is not to scale.
With the domains assigned, we select the
Geometrical Optics interface, change the Intensity computation to Compute intensity, and select the Compute phase check box. These steps are required to properly compute the amplitude and phase of the electric field along the ray trajectory. Settings for the Geometrical Optics interface. The Intensity computation is set to Compute intensity and the Compute phase check box is selected.
We also apply an Inlet boundary condition to the boundary between the
Full-Wave simulation domain and Geometrical Optics domain. The inlet settings can be seen in the image below, but let’s walk through them one at a time. First, the Ray Direction Vector section is configured. This will launch the rays normal to the curved surface we’ve selected for the inlet — in other words, radially outwards. The variables Etheta and Ephi are calculated from the scattering amplitude according to
with a similar assignment for Ephi.
This equation comes from our previous blog post about using the
Far-Field Domain node to calculate the fields at an arbitrary location. These variables are used to specify the initial phase and polarization of the rays. The GOP\_I0 variable specifies the correct spatial intensity distribution for the rays (as antennas generally do not emit uniformly) and is calculated according to GOP\_I0 = (|Etheta|^2 + |Ephi|^2)/Z/2, where Z is the impedance of the medium.
The initial radius of curvature has two factors. The parameter dipole\_sim\_r is the radius of the spherical boundary that we are launching the rays from and will correctly initialize the curvature of the ray wavefront.
Finally, we use the Cartesian components of our spherical unit vector \hat{\theta} to specify the initial principal curvature direction. This ensures that the correct polarization orientation is imparted to the rays. The wavefront shape here must be set to
Ellipsoid — even though the surface is technically a sphere — because we need to be able to specify a preferred direction for polarization. If we choose Spherical, then each orientation is degenerate and we cannot make that specification.
Beyond setting the correct frequency, the only other setting here is the placement of a Freeze Wall condition on the exterior boundary to stop the rays. Let’s take a look at the results vs. theory. As before, we express the full solution for a point dipole as a sum of two contributions, which we have labeled near field (NF) and far field (FF).
\overrightarrow{E} & = \overrightarrow{E}_{FF} + \overrightarrow{E}_{NF} \\
\overrightarrow{E}_{NF} & = \frac{1}{4\pi\epsilon_0}[3\hat{r}(\hat{r}\cdot\vec{p})-\vec{p}](\frac{1}{r^3}+\frac{jk}{r^2})e^{-jkr}\\
\overrightarrow{E}_{FF} & = \frac{1}{4\pi\epsilon_0}k^2(\hat{r}\times\vec{p})\times\hat{r}\frac{e^{-jkr}}{r}\\
\end{align}
The electric fields from a geometrical optics simulation compared against theory. Geometrical optics is always in the far field, so we see excellent agreement as the distance from the source increases. For reference, the far-field domain results from the previous post would overlap exactly with the ray optics and FF theory lines.
As mentioned before, the
Geometrical Optics interface is necessarily in the far field, so we do not expect to be able to correctly capture the near-field information as we did in the Beam-Envelopes solution in Part 2. This can also be seen because we seeded the ray tracing simulation with data from the Far-Field Domain node calculation. It is therefore unsurprising that there is disagreement near the source, but we can clearly see that the results match with theory as the distance from the source increases. Summary of Multiscale Modeling Techniques
From looking solely at the above plot, we have to ask ourselves: “What have we actually gained here?”
This is a fair question, because the plot shown above could have been constructed directly from any of the techniques covered in the series so far. To make this clear, let’s review each of them.
Multiscale Technique Regime of Validity Modules Used Notes Far-Field Domain node Far field RF or Wave Optics Requires the antenna to be completely surrounded by a homogeneous domain. Beam-Envelopes Any field Wave Optics Requires specification of the phase function or wave vector. Geometrical Optics Far field Ray Optics Can account for a spatially varying index as well as reflection and refraction from complex geometries. Diffraction is neglected. A summary of the multiscale modeling techniques we have covered in this blog series.
Note that any of these techniques will require a
Full-Wavesimulation of the radiation source. This generally requires the RF Module, although there is a subset of radiation sources that can be modeled using the Wave Optics Module instead. The Far-Field Domainnode is available in both the RF and Wave Optics modules.
We originally motivated this discussion by talking about signal transmission from one antenna to another, and solved that simulation using the
Far-Field Domain node in the last post. In the next blog post in this series, we’ll redo that simulation using the Geometrical Optics interface introduced here.
Access the model discussed in this blog post and any of the model examples highlighted throughout this blog series by clicking on the button above.
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I have peacefully derived geometric Brownian motion by applying Ito's formula to the process $Y(t) = e^{\alpha B(t) - \frac{t}{2}}$ and letting $\alpha = 1$.
The differential form of $dY$ is clearly equal to
$$dY = \alpha YdB + \left(\frac{\alpha^2}{2} - \frac{1}{2}\right)Ydt \qquad (1)$$
and I understand this to be the SDE that governs a geometric Brownian motion with drift.
I'm now trying to look at the issue from another perspective.
Say I have the following system of SDEs
$$\left\{\begin{matrix} dX(t) = Y(t)dB^{(1)}(t)\\ dY(t) = Y(t)dB^{(2)}(t) \end{matrix}\right.$$
where $B^{(1)}(t),t>0$ and $B^{(2)}(t),t>0$ are two standard and independent Brownian motions. I also have the following initial conditions
$$X(0) = 0 \qquad Y(0) = 1$$
I would like to obtain $Y(t)$ as a solution to this system and eventually find out also $X(t)$, but I'm having a hard time.
I can see that the second equation is exactly equal to $(1)$ for $\alpha =1$, however I would like to actually see how the initial conditions play a role.
My attempt was the following.
I take the second equation and treat as follows
$$dY(t) = Y(t)dB^{(2)}(t) \iff \\ \frac{dY(t)}{Y(t)} = dB^{(2)}(t) \iff \\\ln(Y(t)) = \int_0^tdB^{(2)}(t) = B^{(2)}(t) - B^{(2)}(0) \iff \\ Y(t) = e^{B^{(2)}(t)}$$
since $B^{(2)}(t)$ is a standard Brownian motion. Nowhere near the wanted result. And applying the initial condition only gives me a triviality.
What am I missing? I clearly need to reobtain $Y(t)$ to then solve $X(t)$... |
When we write complex numbers we write in $a+ib$ form why we don't write it as we write in Cartesian coordinate system like $(x,y)$ and how the idea of a complex plane emerge
There is a book called "A History of Vector Analysis" by Crowe which addresses this and many more topics.
It is also published as a book by Dover.
The Form $a+ib$ is easier to work with. For example, the product $z_1\cdot z_2$ and the ratio $z_1/z_2$ for $z_1,z_2 \in \mathbb{C}$ is better to calculate using $a+ib$ or $r\mathrm{e}^{i\phi}$.
The idea of the complex plane probalby emerged because of the following formula:
$$z=a+ib=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}+i\frac{b}{\sqrt{a^2+b^2}}\right).$$
In which $\sqrt{a^2+b^2}\neq 0$ can be interpreted as the distance from the origin / hypotenuse of triangle and $a/\sqrt{a^2+b^2}$ and $b/\sqrt{a^2+b^2}$ can be interpreted as the cosine and sine of an angle (see geometric definition of trig functions). So the similarity in the description of the complex numbers and the geometrical interpretation (e.g. $z_1,z_2 \in \mathbb{C}$ then the sum $z_1+z_2$ can be interpreted as a vector sum) made it a useful interpretation of the complex numbers. |
Let $K = \mathbb Q(\mu_m)$ and $\zeta_K$ it's Dedekind zeta function. We know from the class number formula that, around $0$: $$\zeta_K(s) \sim s^{r_1+r_2-1}h(K)R(K)/w(K) $$ where $h,R,w$ stand for the size of the class group, the regulator and the size of the subgroup of roots of unity.
On the other hand, we have the decomposition: $$\zeta_K(s) = \prod_\chi L(s,\chi)$$ where the product is over suitable Dirichlet characters and moreover, $L(0,\chi) = -B_{1,\chi}$ - the generalized Bernoulli numbers.
On the other hand, $R(K)$ should not be algebraic and there should be a corresponding transcendental contribution from $L(0,\chi)$ and the explicit formula for $B_{1\chi}$ shows that this comes only from those $\chi$ such that $L(s,\chi) = 0. $
Question 1: In the case that $L(s,\chi) = 0$, is it possible to say what the first non zero Taylor coefficient is? Question 2: Can we decompose (even if it is only conjecturally) $h(K),R(K),w(K)$ into factors corresponding to each Dirichlet character appearing in the decomposition of $\zeta_K$ . What about the order of vanishing of $L(0,\chi)$? |
The biggest point is of course that all motion is relative, as many have pointed out. Even barring that, pretending that we live in Copernican times and assuming that the Sun is the center of the universe, the question posits that the force of gravity might be proportional to mass and velocity. This would mean something like$$\vec F\left( \vec v, \hat r, \frac{Mm}{r^2} \right)$$Terms like $\vec v \cdot \hat r$ and $\vec v \times \hat r$ are ruled out by the observation that $\vec F$ is parallel to $\hat r$. So, we have $$\vec F = -G' \frac{v}{c} \frac{Mm}{r^2} \hat r$$where $G'$ isn't necessarily the $G$ measured in the Cavendish experiment (constraints will be derived). Since the sun and the orbiting object might be moving in our Copernican/Newtonian "universe", for the solar system this becomes$$\vec F = -G' \frac{\left| \vec v + \vec v_{orb} \right|}{c} \frac{Mm}{r^2} \hat r$$There are two simplifying limits, where the sun's velocity is 0 or much greater than the planet's orbital velocity.
Case 1: Sun's "universe" velocity is zero
Then, for an orbiting planet, the acceleration is$$\vec F = - \frac{G'Mm}{c} \frac{|\vec v|}{r^2} \hat r$$This is still a centripetal force (with no torque, $\vec r \times \vec F = 0$). So, it works out that angular momentum $L$ is conserved. So, the effective one dimensional DE is, since $L = m r^2 \omega$ and $v = r \omega$,$$\frac{d^2 r}{dt^2} = - \frac{G'M}{c} \frac{r \omega}{r^2} + \frac{L^2}{mr^3} = \left(L^2 - \frac{G'ML}{mc} \right) \frac{1}{r^3}$$This is directly integrable! Defining$$A = L^2 - \frac{G'ML}{mc},$$the solution (starting from a point where dr/dt = 0) is$$r(t) = r_0 \sqrt{1 + \frac{At^2}{r_0^4}}$$Three subclasses:
Subcase 1: $L > G'M/mc$ so that $A > 0$
This describes an orbit spiraling away to infinity, horribly contradicting observation.
Subcase 2: $L < G'M/mc$ so that $A < 0$
This describes an orbit spiraling in. Also bad.
Subcase 3: $L = G'M/mc$ so that $A = 0$
Fine, a falling straight trajectory.
So, this horribly fails.
Case 2: The Sun's "universe" velocity is large
The orbital velocity is much less than the speed of light, but the sun's might not beso expanding in series$$\frac{\left| \vec v_{sun} + \vec v_{orb} \right|}{c}\approx \frac{v_{sun}}{c} \left(1 + \frac{\hat v_{sun} \cdot \vec v_{orb}}{v_{sun}} \right)$$This leads to two terms, the first of which is just Newton's laws (absorbing $G'v_{sun}/c$ into $G$) and the second of which is a perturbation:$$\vec F = - \left(1 + \frac{\hat v_{sun} \cdot \vec v_{orb}}{v_{sun}} \right) \frac{GMm}{r^2}$$I don't intend on solving this more complicated equation, but I'll point out that if you did so (analytically or numerically), you could use observational data to put constraints on $v_{sun}$, which I believe would be severe.
Edit: This one sneaks through to lowest order if you assume that the sun's motion is normal to the solar system's plane; you'd have to look at nasty second order terms. |
RC timing
Ignoring the BJT and LED portions of the circuit, the following shows you how to approximate the RC charging process. The left side is from your schematic. The right side is the Thevenin equivalent (same thing, just slightly simplified):
simulate this circuit – Schematic created using CircuitLab
Looking at the right side, find that \$V_\text{TH}=9\:\text{V}\cdot\frac{R_2}{R_1+R_2}\$ and that \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$. It's common practice to invent (create) a new variable: \$\tau=R_\text{TH}\cdot C_1\$ (multiplying resistance in Ohms times capacitance in Farads yields units in seconds.) This is done because much is
from this simplification. In \$1\tau\$ the voltage will rise to about 63% of the supply voltage (the Thevenin value in this case); in \$2\tau\$ it's 86%; and in \$3\tau\$ it's about 95%. Most of the work has been done in about \$3\tau\$ and this fact can be used as a guide to determine when there's very little change remaining. known
(This ignores the fact that the base of the BJT will also require some current and therefore affects the timing just mentioned. But if \$R_\text{TH}\$ is low enough in value, the base current of the BJT won't matter enough to worry about.)
If the voltage across \$C_1\$ starts at \$0\:\text{V}\$ when \$t=0\$, then \$V_{\text{base}\left(t\right)}=V_\text{TH}\cdot\left(1-e^{\left[\frac{-t}{\tau}\right]}\right)\$.
(This allows you to work out what values of \$V_\text{TH}\$, \$R_\text{TH}\$ and \$C_1\$ you want, once you have decided how much current your LEDs will require and what kind of base current will be needed for that.)
Adding the emitter-follower BJT
Let's add the so-called emitter-follower BJT arrangement, now:
simulate this circuit
Here, the BJT's emitter will simply "follow" the base voltage, less one diode drop. So as the voltage at the base follows the \$V_{\text{base}\left(t\right)}\$ equation and rises upward, so also will the emitter rise upward too. This arrangement, though, uses the collector to source a lot of added current, if needed, because the collector can source up to \$\beta\$ times the base current. So a tiny base current permits a very large, added collector current. This means that the emitter can source a great deal of current if the load attached to the emitter wants it, without significantly affecting the emitter voltage which is
the base voltage. At this point, you can consider adding some kind of following to the emitter, knowing that the BJT will boost the current compliance. load
The only caveat here is that the emitter follows the base and cannot exceed a voltage that is higher than one diode drop below the base voltage. And since the base voltage cannot exceed \$V_\text{TH}\$, the emitter voltage cannot exceed \$V_\text{TH}-V_\text{BE}\$. Luckily, you can arrange for different values of \$V_\text{TH}\$ by modifying the relationship between \$R_1\$ and \$R_2\$. But this is still a limitation to be aware of.
Emitter-follower BJT with added load
Here is the effective circuit you'd started with:
simulate this circuit
So what happens now?
Well, the LED diode drops must be met before anything happens. For old-style, typical
red LEDs operating at \$20\:\text{mA}\$, this is about \$2\:\text{V}\$ per LED. For high intensity modern LEDs, this can easily reach \$3.8\:\text{V}\$ per LED, or more (when the currents are rather high, too.) So the base voltage exceed the LED requirements (you show two of them, so twice the voltage requirement of just one.) It must also exceed the \$V_\text{BE}\$ required by the BJT, itself, as well. Any residual voltage that is left, after accounting for the two LEDs and \$V_\text{BE}\$, will appear across \$R_3\$ and this will set the current. must
If you think about this for a moment, you'll realize that it will take some time for the base voltage to reach a point where the sum of the two LED voltages and \$V_\text{BE}\$ is finally met. Up until this point, nothing much happens and the LEDs will not be lit up. Once it is reached, \$R_3\$ will begin to limit the current as the base voltage continues upward (assuming it does so.) So the LED currents will "follow" this rise, once the voltage requirements have been met.
Ultimately, the final current in the LEDs will be \$I_\text{LED}=\frac{V_\text{TH}-2\cdot V_\text{LED}-V_\text{BE}}{R_3}\$.
A significant problem in this design is that the final current depends a great deal on (mostly) \$V_\text{LED}\$ and the BJT's \$V_\text{BE}\$ (less so.) And since the value of \$V_\text{LED}\$ can vary quite a bit from one LED to another (at the same current), this is a source of management error in the circuit. It's not terrible, since most folks won't really notice too much. But if \$V_\text{TH}-2\cdot V_\text{LED}-V_\text{BE}\$ doesn't leave much margin for \$R_3\$ to do its work, then the results can vary quite a lot from one circuit's part selections to another circuit with different parts.
This means this isn't really a
circuit. It may works after a fashion, assuming \$V_\text{TH}\ge 2\cdot V_\text{LED}+V_\text{BE}\$. But you may not get consistent results, time after time. Also, we've completely neglected what happens when the power supply is turned off (how does \$C_1\$ discharge itself so that it starts again where it left off, the last time?) good Simple improvement
One way to improve the circuit is to remove the dependency of the LED current upon the LED voltage. To do that, simply move the two LEDs into the collector of the BJT:
simulate this circuit
This only works well if your BJT collector can remain at a voltage that is above \$V_\text{TH}\$. But since you can design things so that \$V_\text{TH}\lt 9\:\text{V}-2\cdot V_\text{LED}\$, it doesn't pose an insurmountable problem.
Here, the only delay until the LEDs start experiencing current is the short delay required until the voltage on \$C_1\$ (the base of \$Q_1\$) exceeds \$V_\text{BE}\$. (This is a shorter delay than before.)
Also, this removes a dependence upon the variability of your LED voltages. So that's a good thing for the circuit. So this is actually an improvement over the prior design.
Final commentary
There are many additional ideas you could apply, if you wanted to make the circuit still better.
One problem is how to deal with what happens when the power supply is turned off. So far, we've assumed at the capacitor is completely discharged when the power is turned on and that the power supply is "instantly" turned on (it's fast, relative to the circuit shown.)
Perhaps also you'd like the LED to gradually decline in intensity when the power is turned off. This may require some temporary storage of charge to supply current to the LEDs once the power is removed, to allow a gradual decline. Or, perhaps, you'd like an instant turn-off of the LEDs.
There are a lot of such details that may complicate a practical circuit. But each of them can be resolved, if you clearly specify the behaviors you'd like in all cases. It's just a matter of carefully detailing what you expect in all cases and then working out how to achieve those details, one by one. |
I am trying to figure out whether the following is true: a function $f$ has limit $L$ at the point $p$ if and only if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $$ s,t \in \mathbb{B}_{\delta}(p) \implies |f(s) - f(t)| < \epsilon. $$ I believe that it is true because by definition of limit $$ |t - p| < \delta \implies |f(t) - L| < \frac{\epsilon}{2} $$ and $$ |s - p| < \delta \implies |f(s) - L| < \frac{\epsilon}{2} $$ which can be combined to give $$ s,t \in \mathbb{B}_{\delta}(p) \implies |f(t) - L| + |f(s) - L| < \epsilon \implies |f(s) - f(t)| < \epsilon $$ Since every step is reversible then this proves the statment is true. Is this the right way to think about this? This is the first time I've seen the limit characterized in this way and it makes intuitive sense. So, am I on the right track here?
Note that your proposed definition makes no mention of $L$. That means that, using your definition, I could conclude that the limit of $f(x)=x$ as $x\to 0$ is $1$, or $2$, or $3$, or whatever I want. That's a bit of a problem.
Rather, what you give is a definition for the
existence of the limit (provided you disallow $s,t=p$). So you would have:
Definition.Let $f$ be a function defined on a punctured neighborhood of $p$. Then the limit of $f(x)$ as $x\to p$ exists if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $s,t\in \mathbb{B}_{\delta}(p)-\{p\}$, then $|f(s)-f(t)|\lt \epsilon$.
The idea is fine, and works over the real numbers; but the presentation leaves something to be desired. It is not true that "every step is reversible": for example while it is true that if $|f(t)-L|+|f(s)-L|\lt \epsilon$, then $|f(s)-f(t)|\lt \epsilon$,
that implication is not reversible in general.
Your characterization is just a "Cauchy-like" definition; you ask the values to approach one another instead of a specific number. So the definition doesn't tell you
what the limit is, it just tells you that a limit exists (provided you are working over the reals, or some other Cauchy-complete field).
Now, your argument shows that if the limit exists in the usual sense, then it exists in this sense. But it does not establish the converse! For example, every function $\mathbb{Q}\to\mathbb{Q}$ that has a limit will satisfy your definition, but there are functions $\mathbb{Q}\to\mathbb{Q}$ that satisfy your definition and don't have a limit (because the limit is not a rational).
In order to show that your definition implies the limit exists in the usual sense, you need to invoke completeness of the real numbers. The argument is similar as for proving that every Cauchy sequence converges. |
For Discrete Time
There are two quantities you should be careful not to mix up.
One is the number of individuals who will die during a given interval: $d_x = N_x - N_{x+1}$. One is the fraction, out of those alive at the beginning of a given interval, who will die during the interval: $q_x = \frac{d_x}{N_x}$
($N_x$ being the number of survivors at age $x$.)
On a
regular plot, if $d_x$ is constant then survivorship will decrease linearly; if $q_x$ is constant then survivorship will decrease geometrically. On a semilog plot, when $q_x$ is constant then this geometric decrease will look like a straight line. This is, as far as I know, the main reason to even use semilog plots in the first place; it makes geometric decrease easy to recognize. $d_x$ is sometimes called the death rate and $q_x$ is sometimes called the mortality rate.
(In my opinion this usage is careless; they aren't rates, but just numbers. Compare: there is a distinction between travelling 10 miles in an hour, and travelling
at 10 miles per hour.)
$q_x$ can also be thought of as estimating the probability that someone who is age $x$ will die before age $(x+1)$. It is therefore sometimes also called the
"Age-Specific Probability of Death". Notice that it is bounded between 0 and 1.
(Probabilities can't be lower than 0 or higher than 1; and the fraction dying has to be somewhere in between "none of them" and "all of them"!)
Here's an example survivorship curve, on a regular plot, I made for constant $q_x = 0.1$.
This much should be enough to answer your question. But, for completeness...
For Continuous Time
I complained about calling $d_x$ the death rate. That's because rates have a unit of $time^{-1}$ ("...
per second", "... per hour" etc.) Of course, dividing $d_x$ by the length of the interval, does give you the average death rate over the interval. And, The slope of the survivorship curve (multiplied by -1; we want a positive rate of decrease, instead of a negative rate of increase) gives you the instantaneous death rate at age $x$: $-N^{'}_{x}$.
(For instantaneous rates to be useful, we have to "smooth-out" the survivorship curve instead of letting it be a stepped curve. If it was stepped, then the slope would just always be horizontal or vertical.)
This brings us to another quantity: the "
Force of Mortality": The instantaneous, age-specific, per-capita death rate; or, the death rate at age $x$, divided by the cohort size at age $x$: $\mu_x = -\frac{N^{'}_{x}}{N_x}$
To illustrate, if the death rate at age $x$ is "
60 deaths per minute", and if at that instant there are 60 people, then the per-capita death rate is " 1 death per person per minute". Because choice of time unit is arbitrary, this is the same as saying " 60 deaths per person per hour" or " 525600 deaths per person per year".
The Force of Mortality is unintuitive ("per-capita" death rate? - but everybody only dies once!) but the formula is coherent. The point of having a "per-capita rate" is to enable comparison between cohorts of different sizes, or the
same cohort at different ages when it was bigger vs. smaller, and so on; in general the idea is that death rate has something to do with the number of members, but also something to do with the condition of each member, and $\mu_x$ is trying to get at the latter. It can be visualized as the slope of the survivorship curve (multiplied by -1 to make it positive), divided by the height of the survivorship curve.
There is a similarity between these two quantities, and the two which were defined for discrete time. If $-N^{'}_{x}$ is constant, then so is $d_x$; and if $\mu_x$ is constant, then so is $q_x$. And, again: on a regular plot, if $-N^{'}_{x}$ is constant then survivorship will decrease linearly; and if $\mu_x$ is constant then survivorship will decrease exponentially.
But they are distinct. It is the $time^{-1}$ dimension, and the fact that time units are interchangeable, which makes the difference.
If there are only 60 people, then it's obviously not possible for more than 60 people to die in the next year $(0 \le d_x \le 60)$; and it's not possible for the fraction dying to be higher than "all of them" $(0 \le q_x \le 1)$.
But suppose the
rate of death at this instant is " 60 deaths per hour". That's the same thing as saying it's " 1 death per minute" or " 525600 deaths per year". There's no implication that that many people will in fact die, because there's no implication that the rate will be kept up for the whole year. If this particular death rate were held constant as long as possible, then the cohort would all be dead in exactly an hour and then the rate would then hit 0. The instantaneous death rate can be as high as you like; it just can't stay high forever.
(If a death rate of "
60 deaths per hour" is measured at an instant when there are 60 people, then the per-capita death rate is " 1 death per person per hour" - or, equivalently, " $\frac{1}{60}^{th}$ of a death per person per hour", or " 8760 deaths per person per year".)
Although the absolute death rate can't stay high forever, the per-capita death rate ($\mu_x$)
can. That's what happens during constant $\mu_x$ (exponentially decreasing $N_x$), for example. The reason why this is not paradoxical is that the slope ($-N^{'}_{x}$; the numerator) is constantly decreasing; it's just that so is the height ($N_x$; the denominator).
The Force of Mortality $\mu_x$ has
also sometimes been referred to as the " mortality rate". The equivocal use of language is unfortunate. It is also, especially in reliability engineering, known as the "hazard rate".
By the way
$q_x$ and $\mu_x$ have often been conflated, at least in gerontology (with which I am most familiar). But they are not the same. The first estimates a probability, and is bounded between 0 and 1. The second is
not a probability, and has no upper bound.
One way they have been conflated is with the "
Gompertz equation".
A Type II Survivorship Curve is one which is decreasing exponentially. When this is the case, both $q_x$ and $\mu_x$ are constant. We would call such a species "non-aging" or "non-senescing": your age makes no difference to your vulnerability.
But many species do age.
Benjamin Gompertz proposed that for many species including humans $\mu_x$ grows exponentially throughout adulthood:
Gompertz' Law of Mortality: $\mu_x = \mu_0 \cdot e^{Gx}$ Or, in its logarithmic form: $ln(\mu_x) = ln(\mu_0) + Gx$
(Where $\mu_0$ is the "initial mortality" and $G$ is the exponential "Gompertz parameter". In the logarithmic form, $ln(\mu_0)$ is the intercept, and $G$ is the slope.)
Whether or not $\mu_x$ follows such a law
indefinitely is an empirical matter. (It's actually debated whether it decelerates in late life, and if so why.) But notice that $q_x$ couldn't possibly grow exponentially forever, since $q_x$ couldn't possibly ever exceed 1. And yet, due to the conflation, in recent times it has often been in terms of $q_x$ that the Gompertz equation has been presented and discussed.
Further Reading
Unfortunately, as far as I know the literature (both professional and pedagogical) is sorely lacking in treatments which are both intuitive and accurate!
Peter Medawar's "The Definition and Measurement of Senescence" (Chapter 1 in
The CIBA Foundation Colloquia on Ageing, Vol. 1) is probably still the best introduction to the rationale behind why we care about survivorship curves and mortality, from the perspective of biogerontology. Medawar gives the correct definition and formula for $\mu_x$ (though he doesn't stress its distinction from $q_x$; indeed, he doesn't mention $q_x$ at all).
The distinction between $q_x$ and $\mu_x$ is discussed in Gavrilov and Gavrilova's
The Biology of Life Span: A Quantitative Approach, as well as in some of their papers. They present the correct forumula for $\mu_x$, but do not really try to explain it. |
Shoka function
Shoka function is holomorphic function,
\( \mathrm{Shoka}(z)=z+\ln\!\Big( \exp(-z) +\mathrm e -1\Big)\) Contents Range of holomorphizm
The Shoka function is holomorphic at the complex plane with cuts
\( x+ (1\!+\!2n) \mathrm i \pi ~\) at real \(x\) and integer \(n\).
In such a way, the countable set of cut lines are directed to the left.
Inverse function
The inverse function \(\mathrm{ArcShoka}=\mathrm{Shoka}^{-1}\) can be expressed as elementary function; for \(|\Im(z)| < \pi\),
\(\displaystyle \mathrm{ArcShoka}(z)= z + \ln \!\left( \frac{1\!-\!\mathrm e^{-z}}{\mathrm e \!-\!1} \right)\)
Some properties of the inverse function are collected in the special article ArcShoka.
Complex maps of Shoka and ArcShoka look similar. The following relation takes place:
\(\mathrm{ArcShoka}(z)=\mathrm{Shoka}\big(z-\ln(\mathrm e\! -\!1) - \mathrm i \pi\big) -\ln(\mathrm e\!-\!1) +\mathrm i \pi\)
that can be verified with the direct substitution.
Shoko function \( \mathrm{Shoko}(z) = \ln\!\Big( 1+ \exp(z)(\mathrm e -1)\Big)\) Applications in Laser science
As the Shoko function, the Shoka function is superfunction of the Keller function and describes the evolution of the short pulse of light in a two-level laser medium. Properties and expansions of the Shoka function are similar to those of the Shoko function and are not repeated here.
Tania function Origin of the name
Initially, the Shoko function had been implemented; that function is named after Shoko san. Then in happened, that the cut lines of the Shoko function go to the right hand side direction, that makes difficult its comparison with the Tania function in the complex plane. Name Shoka function is created as small modification of name Shoko function.
References |
A $54.0\ \mathrm{mL}$ sample of oxygen is collected over water at $20\ \mathrm{^\circ C}$ and $770\ \mathrm{Torr}$ pressure. What is the volume of the dry gas at STP?
By using the combined gas law, I came up with the answer $\pu{51.2 mL}$.
$770 - 21.1 = 748.9\ \mathrm{Torr}$ pressure for oxygen
$21.1$ = vapor pressure of water at $23\ \mathrm{^\circ C}$
$\frac{P_1V_1}{ T_1 }= \frac{P_2V_2 }{ T_2}$
$\frac{748.9V_1}{273 }= \frac{770\times54}{296}$
$V_1 = \pu{51.2 mL}$
Is this how it's supposed to be done? |
Dear MO community,
let $F$ be a totally real field and $K$ be an imaginary quadratic field of class number one, and $M=FK$ be the quadratic CM extension of $F$ obtained by adjoining $K$ to $F$. Let $p$ be a prime that splits in $K$ and fix an embedding of $K$ into $\mathbb{C}$. It will determine a $p$-ordinary CM type $\Sigma$ of $M$. With these assumptions, Katz and Hida-Tilouine showed that there is a period \begin{align}\Omega_\infty=(\Omega_{\infty,\sigma})_{\sigma \in \Sigma} \in (M\otimes\mathbb{C})^\times \end{align} so that for Hecke character $\tau$ with infinity type $m_0\Sigma +2d$ in certain range, the number \begin{align} L(0,\tau)\frac{\pi^d}{\Omega_{\infty}^{m_0+2d}} \end{align} is algebraic. On the other hand, we know from theory of CM elliptic curves that \begin{align} \frac{L(1,\psi)}{\Omega_{E/K}^+} \end{align} is algebraic where $\psi$ is the Hecke character associated to and elliptic curve $E/\mathbb{Q}$ with CM by $\mathcal O_K$ and $\Omega_{E/K}^{\pm}$ is the periods of Neron differential of minimal Weierstrass model of $E$. My question:
Is there any relations between $\Omega_\infty$ and $\Omega_{E/K}^{\pm}$? Is the ratio $\pi$ times a $p$-adic unit always?
I suspect there should be some relation because the periods of $L(s,\psi\circ N_{M/K})$ are powers of $\Omega_{E/K}^{\pm}$, but do not see how to prove the claims in my question above. One related result about periods was shown by Bouganis and Dokchitser[Algebraicity of L-values for elliptic curves in a false Tate curve tower, 2007), but it seems they do not consider the integrality. |
One disadvantage of the fact that you have posted 5 identical answers (1, 2, 3, 4, 5) is that if other users have some comments about the website you created, they will post them in all these place. If you have some place online where you would like to receive feedback, you should probably also add link to that. — Martin Sleziak1 min ago
BTW your program looks very interesting, in particular the way to enter mathematics.
One thing that seem to be missing is documentation (at least I did not find it).
This means that it is not explained anywhere: 1) How a search query is entered. 2) What the search engine actually looks for.
For example upon entering $\frac xy$ will it find also $\frac{\alpha}{\beta}$? Or even $\alpha/\beta$? What about $\frac{x_1}{x_2}$?
*******
Is it possible to save a link to particular search query? For example in Google I am able to use link such as: google.com/search?q=approach0+xyz Feature like that would be useful for posting bug reports.
When I try to click on "raw query", I get curl -v https://approach0.xyz/search/search-relay.php?q='%24%5Cfrac%7Bx%7D%7By%7D%24' But pasting the link into the browser does not do what I expected it to.
*******
If I copy-paste search query into your search engine, it does not work. For example, if I copy $\frac xy$ and paste it, I do not get what would I expect. Which means I have to type every query. Possibility to paste would be useful for long formulas. Here is what I get after pasting this particular string:
I was not able to enter integrals with bounds, such as $\int_0^1$. This is what I get instead:
One thing which we should keep in mind is that duplicates might be useful. They improve the chance that another user will find the question, since with each duplicate another copy with somewhat different phrasing of the title is added. So if you spent reasonable time by searching and did not find...
In comments and other answers it was mentioned that there are some other search engines which could be better when searching for mathematical expressions. But I think that as nowadays several pages uses LaTex syntax (Wikipedia, this site, to mention just two important examples). Additionally, som...
@MartinSleziak Thank you so much for your comments and suggestions here. I have took a brief look at your feedback, I really love your feedback and will seriously look into those points and improve approach0. Give me just some minutes, I will answer/reply to your in feedback in our chat. — Wei Zhong1 min ago
I still think that it would be useful if you added to your post where do you want to receive feedback from math.SE users. (I suppose I was not the only person to try it.) Especially since you wrote: "I am hoping someone interested can join and form a community to push this project forward, "
BTW those animations with examples of searching look really cool.
@MartinSleziak Thanks to your advice, I have appended more information on my posted answers. Will reply to you shortly in chat. — Wei Zhong29 secs ago
We are open-source project hosted on GitHub: http://github.com/approach0Welcome to send any feedback on our GitHub issue page!
@MartinSleziak Currently it has only a documentation for developers (approach0.xyz/docs) hopefully this project will accelerate its releasing process when people get involved. But I will list this as a important TODO before publishing approach0.xyz . At that time I hope there will be a helpful guide page for new users.
@MartinSleziak Yes, $x+y$ will find $a+b$ too, IMHO this is the very basic requirement for a math-aware search engine. Actually, approach0 will look into expression structure and symbolic alpha-equivalence too. But for now, $x_1$ will not get $x$ because approach0 consider them not structurally identical, but you can use wildcard to match $x_1$ just by entering a question mark "?" or \qvar{x} in a math formula. As for your example, enter $\frac \qvar{x} \qvar{y} $ is enough to match it.
@MartinSleziak As for the query link, it needs more explanation, technologically the way you mentioned that Google is using, is a HTTP GET method, but for mathematics, GET request may be not appropriate since it has structure in a query, usually developer would alternatively use a HTTP POST request, with JSON encoded. This makes developing much more easier because JSON is a rich-structured and easy to seperate math keywords.
@MartinSleziak Right now there are two solutions for "query link" problem you addressed. First is to use browser back/forward button to navigate among query history.
@MartinSleziak Second is to use a computer command line 'curl' to get search results from particular query link (you can actually see that in browser, but it is in developer tools, such as the network inspection tab of Chrome). I agree it is helpful to add a GET query link for user to refer to a query, I will write this point in project TODO and improve this later. (just need some extra efforts though)
@MartinSleziak Yes, if you search \alpha, you will get all \alpha document ranked top, different symbols such as "a", "b" ranked after exact match.
@MartinSleziak Approach0 plans to add a "Symbol Pad" just like what www.symbolab.com and searchonmath.com are using. This will help user to input greek symbols even if they do not remember how to spell.
@MartinSleziak Yes, you can get, greek letters are tokenized to the same thing as normal alphabets.
@MartinSleziak As for integrals upper bounds, I think it is a problem on a JavaScript plugin approch0 is using, I also observe this issue, only thing you can do is to use arrow key to move cursor to the right most and hit a '^' so it goes to upper bound edit.
@MartinSleziak Yes, it has a threshold now, but this is easy to adjust from source code. Most importantly, I have ONLY 1000 pages indexed, which means only 30,000 posts on math stackexchange. This is a very small number, but will index more posts/pages when search engine efficiency and relevance is tuned.
@MartinSleziak As I mentioned, the indices is too small currently. You probably will get what you want when this project develops to the next stage, which is enlarge index and publish.
@MartinSleziak Thank you for all your suggestions, currently I just hope more developers get to know this project, indeed, this is my side project, development progress can be very slow due to my time constrain. But I believe its usefulness and will spend my spare time to develop until its publish.
So, we would not have polls like: "What is your favorite calculus textbook?" — GEdgar2 hours ago
@GEdgar I'd say this goes under "tools." But perhaps it could be made explicit. — quid1 hour ago
@quid I think that the type of question mentioned in GEdgar's comment is closer to book-recommendations which are valid questions on the main. (Although not formulated like that.) I also think that his comment was tongue-in-cheek. (Although it is a bit more difficult for me to detect sarcasm, as I am not a native speaker.) — Martin Sleziak57 mins ago
"What is your favorite calculus textbook?" is opinion based and/or too broad for main. If at all it is a "poll." On tex.se they have polls "favorite editor/distro/fonts etc" while actual questions on these are still on-topic on main. Beyond that it is not clear why a question which software one uses should be a valid poll while the question which book one uses is not. — quid7 mins ago
@quid I will reply here, since I do not want to digress in the comments too much from the topic of that question.
Certainly I agree that "What is your favorite calculus textbook?" would not be suitable for the main. Which is why I wrote in my comment: "Although not formulated like that".
Book recommendations are certainly accepted on the main site, if they are formulated in the proper way.
If there will be community poll and somebody suggests question from GEdgar's comment, I will be perfectly ok with it. But I thought that his comment is simply playful remark pointing out that there is plenty of "polls" of this type on the main (although ther should not be). I guess some examples can be found here or here.
Perhaps it is better to link search results directly on MSE here and here, since in the Google search results it is not immediately visible that many of those questions are closed.
Of course, I might be wrong - it is possible that GEdgar's comment was meant seriously.
I have seen for the first time on TeX.SE. The poll there was concentrated on TeXnical side of things. If you look at the questions there, they are asking about TeX distributions, packages, tools used for graphs and diagrams, etc.
Academia.SE has some questions which could be classified as "demographic" (including gender).
@quid From what I heard, it stands for Kašpar, Melichar and Baltazár, as the answer there says. In Slovakia you would see G+M+B, where G stand for Gašpar.
But that is only anecdotal.
And if I am to believe Slovak Wikipedia it should be Christus mansionem benedicat.
From the Wikipedia article: "Nad dvere kňaz píše C+M+B (Christus mansionem benedicat - Kristus nech žehná tento dom). Toto sa však často chybne vysvetľuje ako 20-G+M+B-16 podľa začiatočných písmen údajných mien troch kráľov."
My attempt to write English translation: The priest writes on the door C+M+B (Christus mansionem benedicat - Let the Christ bless this house). A mistaken explanation is often given that it is G+M+B, following the names of three wise men.
As you can see there, Christus mansionem benedicat is translated to Slovak as "Kristus nech žehná tento dom". In Czech it would be "Kristus ať žehná tomuto domu" (I believe). So K+M+B cannot come from initial letters of the translation.
It seems that they have also other interpretations in Poland.
"A tradition in Poland and German-speaking Catholic areas is the writing of the three kings' initials (C+M+B or C M B, or K+M+B in those areas where Caspar is spelled Kaspar) above the main door of Catholic homes in chalk. This is a new year's blessing for the occupants and the initials also are believed to also stand for "Christus mansionem benedicat" ("May/Let Christ Bless This House").
Depending on the city or town, this will be happen sometime between Christmas and the Epiphany, with most municipalities celebrating closer to the Epiphany."
BTW in the village where I come from the priest writes those letters on houses every year during Christmas. I do not remember seeing them on a church, as in Najib's question.
In Germany, the Czech Republic and Austria the Epiphany singing is performed at or close to Epiphany (January 6) and has developed into a nationwide custom, where the children of both sexes call on every door and are given sweets and money for charity projects of Caritas, Kindermissionswerk or Dreikönigsaktion[2] - mostly in aid of poorer children in other countries.[3]
A tradition in most of Central Europe involves writing a blessing above the main door of the home. For instance if the year is 2014, it would be "20 * C + M + B + 14". The initials refer to the Latin phrase "Christus mansionem benedicat" (= May Christ bless this house); folkloristically they are often interpreted as the names of the Three Wise Men (Caspar, Melchior, Balthasar).
In Catholic parts of Germany and in Austria, this is done by the Sternsinger (literally "Star singers"). After having sung their songs, recited a poem, and collected donations for children in poorer parts of the world, they will chalk the blessing on the top of the door frame or place a sticker with the blessing.
On Slovakia specifically it says there:
The biggest carol singing campaign in Slovakia is Dobrá Novina (English: "Good News"). It is also one of the biggest charity campaigns by young people in the country. Dobrá Novina is organized by the youth organization eRko. |
(@Gijs is right that this is just algebra, but perhaps it will help you to see it worked out.)
Consider that "$\exp(b_0+b_1RS)$" is an odds of some event, and that "${\rm Pr}(RW_i|RS_i,b_0,b_1)$" is the probability of the same event (cf., Interpretation of simple predictions to odds ratios in logistic regression). To make it easier to see what's going on, we can replace these complicated expressions with "$o$" and "$p$", respectively.
Now, recognize that:
$$p=\frac{o}{1+o} \qquad\qquad o=\frac{p}{1-p}$$If you substitute the full expressions into the formula on the left, you will get your third equation.
On the other hand, if we substitute the full expressions into the right formula, you would get this:
$$\exp(b_0+b_1RS) = \bigg(\frac{{\rm Pr}(RW_i|RS_i,b_0,b_1)}{1-{\rm Pr}(RW_i|RS_i,b_0,b_1)}\bigg)$$which isn't quite the same. To get to your fourth formula, take the log of both sides, and then just switch the right hand side and the left hand side: \begin{align}\log(\exp(b_0+b_1RS)) &= \log\bigg(\frac{{\rm Pr}(RW_i|RS_i,b_0,b_1)}{1-{\rm Pr}(RW_i|RS_i,b_0,b_1)}\bigg) \\[8pt]\log\bigg(\frac{{\rm Pr}(RW_i|RS_i,b_0,b_1)}{1-{\rm Pr}(RW_i|RS_i,b_0,b_1)}\bigg) &= b_0+b_1RS\end{align}
Thus, all we need to establish is the fact that the "$o$"s in the right equation above are the same as the "$o$" in the left equation, and likewise for the "$p$"s between the two equations. What is potentially unintuitive is that you will need to take the reciprocal of both sides at a couple of points. It's a bit tedious to write out the algebra, but it doesn't take too many steps:
\begin{align}p &= \frac{o}{1+o} &\text{right formula}& \\[8pt]\frac 1 p &= \frac{1+o}{o} &\text{taking reciprocals}& \\[8pt]\frac 1 p &= \frac 1 o + \frac o o &\text{separate out }\frac o o& \\[8pt]\frac 1 p &= \frac 1 o + 1 &\frac o o = 1& \\[8pt]\frac 1 p - 1 &= \frac 1 o &\text{subtracting 1}& \\[8pt]\frac 1 p - \frac p p &= \frac 1 o &1 = \frac p p& \\[8pt]\frac{1-p}{p} &= \frac 1 o &\text{simplifying}& \\[8pt]\frac{p}{1-p} &= o &\text{taking reciprocals}& \\[8pt]o &= \frac{p}{1-p} &\text{left formula}&\end{align}
You might also find What is the difference between logistic and logit regression? helpful. |
Tweedie Distributions: mle estimation of p
Maximum likelihood estimation of the Tweedie index parameter \(p\).
Keywords models Usage
tweedie.profile(formula, p.vec=NULL, xi.vec=NULL, link.power=0, data, weights, offset, fit.glm=FALSE, do.smooth=TRUE, do.plot=FALSE, do.ci=do.smooth, eps=1/6, control=list( epsilon=1e-09, maxit=glm.control()$maxit, trace=glm.control()$trace ), do.points=do.plot, method="inversion", conf.level=0.95, phi.method=ifelse(method == "saddlepoint", "saddlepoint", "mle"), verbose=FALSE, add0=FALSE)
Arguments formula p.vec
a vector of
pvalues for consideration. The values must all be larger than one (if the response variable has exact zeros, the values must all be between one and two). If
NULL(the default),
p.vecis set to
seq(1.2, 1.8, by=0.1)if the response contains any zeros, or
seq(1.5, 5, by=0.5)if the response contains no zeros. See the DETAILS section below for further details.
xi.vec
the same as
p.vec; some authors use the \(p\) notation for the index parameter, and some use \(\xi\); this function detects which is used and then uses that notation throughout
link.power
the power link function to use. These link functions \(g(\cdot)\) are of the form \(g(\eta)=\eta^{\rm link.power}\), and the special case of
link.power=0(the default) refers to the logarithm link function. See the documentation for
tweediealso.
data
an optional data frame, list or environment (or object coercible by
as.data.frameto a data frame) containing the variables in the model. If not found in
data, the variables are taken from
environment(formula), typically the environment from which
glmis called.
weights
an optional vector of weights to be used in the fitting process. Should be
NULLor a numeric vector.
offset
this can be used to specify an
a prioriknown component to be included in the linear predictor during fitting. This should be
NULLor a numeric vector of length either one or equal to the number of cases. One or more
offsetterms can be included in the formula instead or as well, and if both are specified their sum is used. See
model.offset.
fit.glm
logical flag. If
TRUE, the Tweedie generalized linear model is fitted using the value of \(p\) found by the profiling function. If
FALSE(the default), no model is fitted.
do.smooth
logical flag. If
TRUE(the default), a spline is fitted to the data to smooth the profile likelihood plot. If
FALSE, no smoothing is used (and the function is quicker).
Notethat
p.vecmust contain
at least five pointsfor smoothing to be allowed. do.plot
logical flag. If
TRUE, a plot of the profile likelihood is produce. If
FALSE(the default), no plot is produced.
do.ci
logical flag. If
TRUE, the nominal 100*
conf.levelis computed. If
FALSE, no confidence interval is computed. By default,
do.ciis the same value as
do.smooth, since a confidence interval will only be accurate if smoothing has been performed. Indeed, if
do.smooth=FALSE, confidence intervals are never computed and
do.ciis forced to
FALSEif it is given as
TRUE.
eps
the offset in computing the variance function. The default is
eps=1/6(as suggested by Nelder and Pregibon, 1987). Note
epsis ignored unless the
method="saddlepoint"as it makes no sense otherwise.
control
a list of parameters for controlling the fitting process; see
glm.controland
glm. The default is to use the maximum number of iterations
maxitand the
tracesetting as given in
glm.control, but to set
epsilonto
1e-09to ensure a smoother plot
do.points
plot the points on the plot where the (log-) likelihood is computed for the given values of
p; defaults to the same value as
do.plot
method
the method for computing the (log-) likelihood. One of
"series",
"inversion"(the default),
"interpolation"or
"saddlepoint". If there are any troubles using this function, sometimes a change of method will fix the problem. Note that
method="saddlepoint"is only an approximate method for computing the (log-) likelihood. Using
method="interpolation"may produce a jump in the profile likelihood as it changes computational regimes.
conf.level
the confidence level for the computation of the nominal confidence interval. The default is
conf.level=0.95.
phi.method
the method for estimating
phi, one of
"saddlepoint"or
"mle". A maximum likelihood estimate is used unless
method="saddlepoint", when the saddlepoint approximation method is used. Note that using
phi.method="saddlepoint"is equivalent to using the mean deviance estimator of
phi.
verbose
the amount of feedback requested:
0or
FALSEmeans minimal feedback (the default),
1or
TRUEmeans some feedback, or
2means to show all feedback. Since the function can be slow and sometimes problematic, feedback can be good; but it can also be unnecessary when one knows all is well.
add0
if
TRUE, the value
p=0is used in forming the profile log-likelihood (corresponding to the normal distribution); the default value is
add0=FALSE
Details
For each value in
p.vec, the function computes an estimate of
phi and then computes the value of the log-likelihood for these parameters. The plot of the log-likelihood against
p.vec allows the maximum likelihood value of
p to be found. Once the value of
p is found, the distribution within the class of Tweedie distribution is identified.
Value
The main purpose of the function is to estimate the value of the Tweedie index parameter, \(p\), which is produced by the output list as
p.max. Optionally (if
do.plot=TRUE), a plot is produced that shows the profile log-likelihood computed at each value in
p.vec (smoothed if
do.smooth=TRUE). This function can be temperamental (for theoretical reasons involved in numerically computing the density), and this plot shows the values of \(p\) requested on the horizontal axis (using
rug); there may be fewer points on the plot, since the likelihood some values of \(p\) requested may have returned
NaN,
Inf or
NA.
A list containing the components:
y and
x (such that
plot(x,y) (partially) recreates the profile likelihood plot);
ht (the height of the nominal confidence interval);
L (the estimate of the (log-) likelihood at each given value of
p);
p (the
p-values used);
phi (the computed values of
phi at the values in
p);
p.max (the estimate of the mle of
p);
L.max (the estimate of the (log-) likelihood at
p.max);
phi.max (the estimate of
phi at
p.max);
ci (the lower and upper limits of the confidence interval for
p);
method (the method used for estimation:
series,
inversion,
interpolation or
saddlepoint);
phi.method (the method used for estimation of
phi:
saddlepoint or
phi).
If
glm.fit is
TRUE, the list also contains a component
glm.obj, a
glm object for the fitted Tweedie generalized linear model.
Note
The estimates of
p and
phi are printed. The result is printed invisibly.
If the response variable has any exact zeros, the values in
p.vec must all be between one and two.
The function is sometimes unstable and may fail. It may also be very slow. One solution is to change the method. The default is
method="inversion" (the default); then try
method="series",
method="interpolation" and
method="saddlepoint" in that order. Note that
method="saddlepoint" is an approximate method only. Also make sure the values in
p.vec are suitable for the data (see above paragraph).
It is recommended that for the first use with a data set, use
p.vec with only a small number of values and set
do.smooth=FALSE,
do.ci=FALSE. If this is successful, a larger vector
p.vec and smoothing can be used.
References
Dunn, P. K. and Smyth, G. K. (2008). Evaluation of Tweedie exponential dispersion model densities by Fourier inversion.
Statistics and Computing, 18, 73--86. 10.1007/s11222-007-9039-6
Dunn, Peter K and Smyth, Gordon K (2005). Series evaluation of Tweedie exponential dispersion model densities
Statistics and Computing, 15(4). 267--280. 10.1007/s11222-005-4070-y
Dunn, Peter K and Smyth, Gordon K (2001). Tweedie family densities: methods of evaluation.
Proceedings of the 16th International Workshop on Statistical Modelling, Odense, Denmark, 2--6 July
Jorgensen, B. (1987). Exponential dispersion models.
Journal of the Royal Statistical Society, B, 49, 127--162.
Jorgensen, B. (1997).
Theory of Dispersion Models. Chapman and Hall, London.
Nelder, J. A. and Pregibon, D. (1987). An extended quasi-likelihood function.
Biometrika 74(2), 221--232. 10.1093/biomet/74.2.221
Tweedie, M. C. K. (1984). An index which distinguishes between some important exponential families.
Statistics: Applications and New Directions. Proceedings of the Indian Statistical Institute Golden Jubilee International Conference (Eds. J. K. Ghosh and J. Roy), pp. 579-604. Calcutta: Indian Statistical Institute. See Also Aliases tweedie.profile Examples
# NOT RUN {library(statmod) # Needed to use tweedie.profile# Generate some fictitious datatest.data <- rgamma(n=200, scale=1, shape=1)# The gamma is a Tweedie distribution with power=2;# let's see if p=2 is suggested by tweedie.profile:# }# NOT RUN { out <- tweedie.profile( test.data ~ 1, p.vec=seq(1.5, 2.5, by=0.2) ) out$p.max out$ci# }
Documentation reproduced from package tweedie, version 2.3.2, License: GPL (>= 2) |
(Notes added by Enrico Scalas on 19.08.2009)
What is the common notion of equilibrium in economics?
The concept of equilibrium referred to in General Equilibrium Theory is taken from Physics. It coincides with mechanical equilibrium.
When looking for mechanical equilibrium one minimizes a potential function
subject to boundary conditions, in order to find equilibrium positions; when looking for standard (micro)economic equilibrium, one maximizes a utility function subject to budget constraints (this is the consumer side, in other words: demand) and maximizes the profit subject to cost constraints (this is producer side, in other words, supply); then one equates supply and demand, and finds equilibrium quantities and prices. In both cases, the mathematical tool is optimization with constraints using the method of Lagrange multipliers.
Walras and Pareto explicitly inspired their pioneering work on General Equilibrium Theory to Physics and mechanical equilibrium. This was made clear by Ingrao and Israel (1990).
What is statistical equilibrium?
Statistical equilibrium is another notion of equilibrium in Physics. It was defined by Maxwell and Boltzmann in their early work on the theory of gases, trying to reconcile mechanics and thermodynamics. In order to better understand this notion, it is useful to make use of a Markovianist approach to statistical equilibrium as discussed by Oliver Penrose (the brother of Roger Penrose) in his 1970 book. By the way, a similar
approach was promoted by Richard von Mises (the brother of Ludwig von Mises) in a book reprinted in 1945 (actually the book was written by R. von Mises before WWII).
A finite Markov chain is a stochastic process defined as a sequence of random variables
$X_1,\ldots,X_n$ on the same probability space that assume values in a finite set $S$, known as the state space. For a Markov chain, the predictive probability $\mathbb{P}(X_n = x_n|X_{n-1}=x_{n-1},\ldots,X_1=x_1)$ has the following simple form
In other words, the predictive probability does not depend on al the past states, but only on the last state occupied by the chain. As a consequence of the multiplication theorem, one gets that the finite-dimensional distribution $\mathbb{P}(X_1 = x_1,\ldots,X_n=x_n)$ is given by(2)
As a consequence of Kolmogorov's representation theorem, this means that a Markov chain is fully characterized by the knowledge of the functions $\mathbb{P}(X_m=x_m|X_{m-1}=x_{m-1})$, also known as
transition probabilities and $\mathbb{P}(X_1=x_1)$, also known as initial probabilitydistribution. If the transition probabilities do not depend on the index $m$, but only on the initial and on the final state, than the Markov chain is called homogeneous. In the following, only homogeneous Markov chains will be considered. For the sake of simplicity, it is useful to introduce the notation
for the transition probability and(5)
for the initial probability distribution. Note that $P(x,y)$ is nothing else than
a matrix in the finite case under scrutiny, with the property that
in other words the rows of the matrix sum up to 1 as a consequence of the addition axiom. Such matrices are called
stochastic matrices (to be distinguished from random matrices which are matrices with random entries). Note that the initial distribution can be written as a row vector, so that one can obtain the marginal distribution of the random variable $X_n$ as
Now suppose there is a distribution $p(x)$ satisfying the equation(8)
then $p(x)$ is called a
stationary distribution or invariant measure. If the chain starts with states distributed according to $p(x)$, this distribution does not change as time goes by. Note that the states are jumping from one to another one, but the probability of finding the system in a specific state does not change. This is exactly the idea of statistical equilibrium put forward by Ludwig Boltzmann.
However, more can and should be said. First of all, the stationary distribution may not exist; secondly the chain usually starts from a specific state, so that the initial distribution is a vector full of 0's and with a single 1 in the initial state. The latter state of affairs can be represented by a Kronecker delta $\pi(x) = \delta(x,x_0)$, where $x_0$ is the specific initial state. Usually, this is not a stationary distribution and the convergence of the chain to the stationary distribution is not granted at all.
Fortunately, it turns out that under some rather mild conditions: the stationary distribution exists and is unique; the chain always converges to the stationary distribution irrespective of its initial distribution.
It is indeed sufficient to consider a finite chain that is
aperiodic and irreducible or ergodic. If $x$ is a state of the Markov chain, its period $d_x$is defined as the greatest common divisor of the set $\{n \geq 1: \, P^n (x,x) > 0\}$. A chain is irreducible if any state leads to any other state with finite probability and in a finite number of steps. If a state $x$ leads to a state $y$, then one can prove that also $y$ leads to $x$ and the set of communicating states have a common period. In an irreducible chain all the states communicate and they have a common period $d$. The chain is aperiodic if $d=1$.
If the finite Markov chain is irreducible and aperiodic, then it has a unique stationary distribution $p(x)$ and(9)
irrespective of the initial state $x$. This means that, after a transient period, the distribution of chain states reaches a stationary distribution, which can then be interpreted as an equilibrium distribution in the statistical sense.
Why and where statistical equilibrium may be useful in economics?
There are several possible domains of application of the concept of statistical equilibrium in Economics. Incidentally, note that many agent-based models used in Economic theory are intrinsically Markov chains (or Markovian processes). Therefore the ideas discussed above naturally apply. Up to now, with my coworkers I have used these concepts:
to discuss some toy models for the distribution of wealth (not of income!) as in Scalas et al. (2006) and in Garibaldi et al. (2007). to study a simple agent-based model of financial market with heterogeneous knowledge as in Toth et al. (2007). to generalize a sectoral productivity model originally due to Aoki and Yoshikawa, in Scalas and Garibaldi (2009).
In Scalas et al. (2006), Garibaldi et al. (2007), and Scalas and Garibaldi (2009), we promote the use of a finitary approach to combinatorial stochastic processes which will be the main topic of my presentation in Reykjavik.
References
B. Ingrao and G. Israel (1990),
The Invisible Hand. Economic Equilibrium in the History of Science, MIT Press, Cambridge MA.
O. Penrose (1970),
Foundations of Statistical Mechanics: A Deductive Treatment, Dover, NY.
R. von Mises (1945),
Wahrscheinlichkeitsrechnung und ihre Anwendung in der Statistik und theoretischen Physik, Rosenberg, NY.
E. Scalas, U. Garibaldi and S. Donadio (2006),
Statistical equilibrium in simple exchange games I - Methods of solution and application to the Bennati-Dragulescu-Yakovenko (BDY) game, European Physical Journal B, 53(2), 267-272.
U. Garibaldi, E. Scalas and P. Viarengo (2007),
Statistical equilibrium in simple exchange games II. The redistribution game, European Physical Journal B, 60(2), 241-246.
B. Toth, E. Scalas, J. Huber and M. Kirchler (2007),
The value of information in a multi-agent market model - The luck of the uninformed, European Physical Journal B, 55(1), 115-120.
E. Scalas and U. Garibaldi (2009),
A Dynamic Probabilistic Version of the Aoki–Yoshikawa Sectoral Productivity Model, Economics, The Open-Access, Open-Assessment E-Journal, 3, 2009-15. http://www.economics-ejournal.org/economics/journalarticles/2009-15. |
As part of an optional assignment, we asked our students to create a pendulum, measure its frequency with phyphox and submit the results via a web form. I just picked up the data and got thrilled as the result is amazing:
I just had to clean out some obvious cases in which students did not use the correct units (no, they did not build a 69m pendulum).
They did not even know anything about the math of oscillations. Instead, I will introduce this topic in tomorrow’s lecture and will then use their own data to verify the result for the frequency of a pendulum:
$latex f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{g}{l}}&bg=404040&fg=ffffff&s=3$ |
We can prove this if you dont understand the formula.
Suppose $M$ is a point object at an actual depth $MA$ below the free surface of water $XY$ in a tank. A ray of light incident on $XY$ normally along $MA$ passes straight along $MAA'$.Another ray of light from $M$ incident at $\angle i$ on $XY$,along $MB$ gets deviated away from normal and is refracted at $\angle r$ along $BC$.If we produce $BC$ we will find that it meets $OA$ at $L$.Therefore $L$ is virtual image of $M$ which appears when we see from $C$.Now the apparent depth is $AL$.
$$\angle AMB=\angle MBN'$$$$\angle ALB= \angle NBC$$ In ∆$AMB$, $$\sin i= \frac{AB}{MB}$$
In ∆$IAB$,$$\sin r=\frac{AB}{LB}$$
Now, $$^a\mu_w =\frac{AB}{LB}×\frac{MB}{AB}=\frac{MB}{LB}$$
Suppose that $\angle i\rightarrow0$ then B will near A
Therefore, $$^a\mu_w=\frac{MA}{LA} =\frac{real~depth}{apparent~depth}$$ |
How do you construct a Bayes classifier for a binary target where it is assumed: $p(y=1)=\alpha$ and $p(x|y)$ both multivariate gaussian?
Assumption:I assume that this question is asking for the Bayes classifier given that $Y$ is a Bernoullirandom variable with parameter $\alpha$ and the conditionaldistributions of the observation $X$ are multivariate Gaussian distributions.
If both conditional distributions of $X$ given $Y$ are
non-degeneratemultivariate Gaussian distributions, that is, they possess densities $f_i(x) = f_{X\mid Y=i}(x\mid Y=i), i = 0, 1$, then the Bayes classifier can be expressed as$$\frac{\alpha}{1-\alpha}\cdot \frac{f_1(x)}{f_0(x)}~~\begin{array}{c}\hat Y=1\\\gtrless\\\hat Y=0\end{array}~~ 1.$$The decision boundary can also be stated as the hypersurface defined by the equation $${\alpha}\cdot{f_1(x)} - {1-\alpha}\cdot{f_0(x)} = 0$$which does not have a simple solution in general. The two-dimensional case is treated in this question where the hypersurface is found to be a conic section. |
Motivation:
Many interesting
irrational numbers (or numbers believed to be irrational) appear as answers to natural questions in mathematics. Famous examples are $e$, $\pi$, $\log 2$, $\zeta(3)$ etc. Many more such numbers are described for example in the wonderful book "Mathematical Constants" by Steven R. Finch. The question:
I am interested in theorems where a "special"
rational number makes a surprising appearance as an answer to a natural question. By a special rational number I mean one with a large denominator (and preferably also a large numerator, to rule out numbers which are simply the reciprocals of large integers, but I'll consider exceptions to this rule). Please provide examples.
For illustration, here are a couple of nice examples I'm aware of:
The average geodesic distance between two random points of the Sierpinski gasket of unit side lengths is equal to $\frac{466}{885}$. This is also equivalent to a natural discrete math fact about the analysis of algorithms, namely that the average number of moves in the Tower of Hanoi game with $n$ disks connecting a randomly chosen initial state to a randomly chosen terminal state with a shortest number of moves, is asymptotically equal to $\frac{466}{885}\times 2^n$. See here and here for more information.
The answer to the title question of the recent paper ""The density of primes dividing a term in the Somos-5 sequence" by Davis, Kotsonis and Rouse is $\frac{5087}{10752}$.
Rules:
1) I won't try to define how large the denominator and numerator need to be to for the rational number to qualify as "special". A good answer will maximize the ratio of the number's information theoretic content to the information theoretic content of the statement of the question it answers. (E.g., a number like 34/57 may qualify if the question it answers is simple enough.) Really simple fractions like $3/4$, $22/7$ obviously do not qualify.
2) The question the number answers needs to be natural. Again, it's impossible to define what this means, but avoid answers in the style of "what is the rational number with smallest denominator solving the Diophantine equation [some arbitrary-sounding, unmotivated equation]".
Edit: a lot of great answers so far, thanks everyone. To clarify my question a bit, while all the answers posted so far represent very beautiful mathematics and some (like Richard Stanley's and Max Alekseyev's answers) are truly astonishing, my favorite type of answers involve questions that are conceptual in nature (e.g., longest increasing subsequences, tower of Hanoi, Markov spectrum, critical exponents in percolation) rather than purely computational (e.g., compute some integral or infinite series) and to which the answer is an exotic rational number. (Note that someone edited my original question changing "exotic" to "special"; that is fine, but "exotic" does a better job of signaling that numbers like 1/4 and 2 are not really what I had in mind. That is, 2 is indeed quite a special number, but I doubt anyone would consider it exotic.) |
Equations Learn easily with Video Lessons and Interactive Practice Problems Solving Equations Equations containing one or more variables are algebra equations. Variables represent unknown amounts, and any letter or symbol can be used as a variable.
Algebra equations may need
one-step, two-steps, or multiple-steps to solve for the value of the variable. Equations may have variables on one or both sides of the equal sign. To solve algebra equations, students combine like terms and use inverse operations to isolate the variable. Equations in the format of variable word problems may help students learn to apply algebra skills to real world situations. One-Step Equation
To
isolate the variable to solve these one-step equations, undo the equation: Example 1: Undo the constant by subtracting from both sides of the equation.
$\begin{array}{lcl} x + 3 & = & 8 \\ x + 3 - 3 & = & 8 - 3 \\ x & = & 5 \end{array}$
Example 2: Divide both sides of the equation by the coefficient.
$\begin{array}{lcl} 2x & = & 16\\ \frac{2}{2}x & = & \frac{16}{2} \\ x& = &8 \end{array}$
Two-Step Equation
Use
reverse PEMDAS to solve this two-step equation. Step 1: Subtract the constant from both sides of the equation. Step 2: Divide both sides of the equation by the coefficient.
$\begin{array}{lcl} 3x + 5 &= & 17 \\ 3x + 5 -5 &= &17 -5\\ 3x&=&12 \\ \frac{3}{3}x&=&\frac{12}{3} \\ x&=&4 \end{array}$
Multi-Step Equation with Variable on One Side
Use the
Distributive Property to solve this multi-step equation: Step 1: Apply the Distributive Property. Step 2: Add the constant to both sides of the equation. Step 3: Divide both sides of the equation by the coefficient.
$\begin{array}{lcl} 8 &= &2(x - 2)\\ 8&=&2x -4 \\ 8 +4&=&2x -4 +4 \\ 12&=&2x \\ \frac{12}{2}&=&\frac{2}{2}x \\ x&=&6 \end{array}$
Multi-Step Equation with Variables on Both Sides
To solve multi-step equations such as this,
combine like terms to make the problem easier to solve: Step 1: Use opposite operations to combine like terms. Step 2: Divide both sides of the equation by the coefficient.
$\begin{array}{lcl} 11x&=& -16 + 3x\\ 11x - 3x &=& -16 + 3x- 3x \\ 8x &=& -16 \\ \frac{8}{8}x&=&\frac{-16}{8}\\ x&=& -2 \end{array}$
Word Problems
Write a variable equation to solve this word problem. Let x represent the number of people attending the party.
Husni bakes cakes for a party. He doesn’t know how many people will attend, but he does know he needs 3 eggs per cake and a cake serves 8 people. Write an expression to determine the number of eggs he will need.
$(\frac{x}{8})\times3$
If 24 people attend the party, how many eggs will he need?
$(\frac{24}{8})\times3 = 3\times3=9$
He needs 9 eggs.
Distance Rate Time
Use the
Distance Rate Time (DRT) formula to solve problems of how far, how long, or how fast. Using the formula, calculate problems for travel in the same direction or different directions. The following triangles will help you to remember:
Distance Rate Time – Same Direction Example 1:Use the DRT formula to solve this problem: Alex rode his bike for 2 hours travelling 23 miles. What is his rate?
D = 23 miles
T = 2 hours R = x
$\begin{array}{lcl} 23&=&2x\\ x&=&11.5 \end{array}$
Alex’s rate is 11.5 miles per hour.
Example 2:After school, Kara and Cyrus ride bikes with the riding club. Kara leaves school at 3:00 pm, riding 10 miles per hour. Cyrus stayed a few minutes late to clean out his locker and left at 3:10 pm, riding at a rate of 12 miles per hour. When will he catch up with Kara?
$\begin{array}{lcl} 10t &=& 12(t-\frac{1}{6}) \\ 10t&=&12t -2 \\ 10t -12t&=&12t -12t -2 \\ -2t&=&-2 \\ \frac{-2}{-2}t&=&\frac{-2}{-2} \\ 2t&=&2 \\ t&=&1 \end{array}$
After one hour, Cyrus will catch up with Kara.
Distance Rate Time – Different Directions
For this problem, the travel is from
two different directions. Use the DRT formula to solve:Carly and Pete leave school traveling in opposite directions on a straight road. Pete rides his electric bike 12 mi/h faster than Carly walks. After 2 hours, they are 36 miles apart. Find Carly’s rate and Pete’s rate.
$\begin{array}{lcl} 2(12+x) + 2x &=&36\\ 24 +2x +2x&=&36\\ 24 +4x&=&36\\ 24 -24 +4x&=&36 -24\\ 4x&=&12\\ \frac{4}{4}x&=&\frac{12}{4} \\ x&=&3 \end{array}$
Carly’s rate is 3 mph, and Pete’s rate is 15. |
In this tutorial you will learn how to set up and execute Purpose:
calculations using the
exciting
interface, which allows to obtain full stress tensors of crystal systems for
STRESS-exciting anycrystal structure. In addition, the application of
to the determination of stress tensor for the
STRESS-exciting
is explicitly presented.
hexagonal Be
Table of Contents
0. Define relevant environment variables Read the following paragraphs before starting with the rest of this tutorial!
Before starting, be sure that relevant environment variables are already defined as specified in
. Here is a list of the scripts which are relevant for this tutorial with a short description. How to set environment variables for tutorials scripts
: Python script for generating structures at different strains.
STRESS-exciting-setup.py
: (Bash) shell script for running a series of
STRESS-exciting-submit.sh
calculations.
exciting
: Python script for fitting the energy-vs-strain and CVe-vs-strain curves.
STRESS-exciting-analyze.py
: (Bash) shell script for cleaning unnecessary files.
STRESS-exciting-clean.sh
: Python script for calculating the stress components.
STRESS-exciting-result.py
:
exciting2sgroup.xsl
script for converting the
xsl
input to an
exciting
input file.
sgroup
: File containing
Grace.par
parameters needed by visualization tools.
xmgrace Requirements: Bash shell. Python numpy, lxml, math, sys and os libraries.
(visualization tool). xmgrace
From now on the symbol
will indicate the shell prompt. $
Extra requirement: Tool for space-group determination
The scripts in this tutorial use the
tool. If you have not done before, this tool should be downloaded and installed. sgroup
1. Set up and perform a reference calculation
In order to test the calculation of the stress tensor presented in the next sections, we set up and perform a calculation for the system we have chosen in a reference configuration. The example which is shown here is the hexagona structure of Be under an applied pressure. First, create a working directory
and move inside it. Be-reference
$ mkdir Be-reference$ cd Be-reference
Here, copy the following lines inside a file
. input.xml
<input> <title>Be: Stress calculation</title> <structure speciespath="$EXCITINGROOT/species"> <crystal scale="4.300"> <basevect> 1.00000000 0.00000000 0.00000000 </basevect> <basevect> -0.50000000 0.86602540 0.00000000 </basevect> <basevect> 0.00000000 0.00000000 1.50000000 </basevect> </crystal> <species speciesfile="Be.xml" rmt="1.95"> <atom coord="0.66666667 0.33333333 0.75000000"/> <atom coord="0.33333333 0.66666667 0.25000000"/> </species> </structure> <groundstate ngridk="6 6 4" xctype="GGA_PBE_SOL"> </groundstate> <relax/> </input> Be sure to set the correct path for the
root directory (indicated in this example by exciting
) to the one pointing to the place where the
$EXCITINGROOT
directory is placed. In order to do this, use the command
exciting
$ SETUP-excitingroot.sh
Now, we procede as alredy seen in
in order to generate new structures and determine the corressponding pressure. General lattice optimization
Now, you should have a file which called
inside the current directory. It contains all information which is related to the Birch-Murnaghan fit including the hydrostatic pressure corresponding to the calculated structures. The content of this file should be similar to the following: BM_eos.out
=== Birch-Murnaghan eos ========================= Fit accuracy: Log(Final residue in [Ha]): -6.04 Final parameters: E_min = -29.36096 [Ha] V_min = 106.5871 [Bohr^3] B_0 = 121.951 [GPa] B' = 3.553 ================================================= Volume E_dft-E_eos Pressure [GPa]94.2633 +0.00000009 +18.615 98.7043 -0.00000041 +10.735103.2857 +0.00000065 +4.057108.0004 -0.00000046 -1.569112.8597 +0.00000012 -6.298
As can be seen, the last 5 lines of this file show the volumes of the calculated structures (
coloumn) and the corresponding hydrostatic pressure ( first
coloumn). The first of the 5 structures which are calculated, (with a volume of
third Bohr
94.2633
3) has an hydrostatic pressure of GPa. In the following sections, this structure will be denoted as
+18.615
.
reference configuration
We notice here that the pressure is related to the components of the stress tensor, $\sigma_{\alpha \beta}$, as following:(1)
Next, we will calculate directly the stress tensor of this reference configuration and will use the equation above to (re)calculate the corresponding hydrostatic pressure.
2. Set up and perform a calculation of the stress for the reference configuration
Before continuing, we remind here that all strains considered in this tutorial are
physical strains. To start the preparation of the calculation of the stress we move to the parent directory, create a new directory
and move inside it. Be-stress
$ cd ../$ mkdir Be-stress$ cd Be-stress
Inside this directory, we copy the input file corresponding to the reference configuration:
$ cp ../Be-reference/VOL/vol_01/input.xml ./
i) Generation of input files for distorted structures
In order to generate input files for a series of distorted structures, you have to run the script
, which will produce the following output on the screen. STRESS-exciting-setup.py
$ STRESS-exciting-setup.py >>>> Please enter the exciting input file name: Be_stress.xml Number and name of space group: 194 (P 63/m m c) Hexagonal I structure in the Laue classification. This structure has 2 independent stress components.>>>> Please enter the maximum amount of strain The suggested value is between 0.0010 and 0.0100 : 0.002 The maximum amount of strain is 0.002>>>> Please enter the number of the distorted structures [odd number > 4]: 21 The number of the distorted structures is 21$
Entry values must be typed on the screen when requested. In this case, the entries are the following.
input.xml, the name of the input file. 0.002, the absolute value of the maximum strain for which we want to perform the calculation. 21, the number of deformed structures equally spaced in strain, which are generated between the maximum negative strain and the maximum positive one. ii) Execute the calculations
To execute the series of calculations, run the script
, which will produce an output on the screen similar to the following. STRESS-exciting-submit.sh
$ STRESS-exciting-submit.sh +--------------------------------------+ | SCF calculation of "Dst01_01" starts | +--------------------------------------+ Elapsed time = 0m7sThu Jul 24 13:52:44 CEST 2014 +--------------------------------------+ | SCF calculation of "Dst01_02" starts | +--------------------------------------+ Elapsed time = 0m7sThu Jul 24 13:52:51 CEST 2014 +--------------------------------------+ | SCF calculation of "Dst01_03" starts | +--------------------------------------+... +--------------------------------------+ | SCF calculation of "Dst02_21" starts | +--------------------------------------+ Elapsed time = 0m7sThu Jul 24 13:57:37 CEST 2014$
After the complete run, stress components can be calculated.
iii) Analyzing the calculations
Stress components are obtained from energy calculations using a similar method to the one illustrated in
. Within this approach, first derivatives of the energy curves are evaluated with the help of a curve fitting. Calculations performed above were producing Energy vs. strain calculations
21points for each energy curve. The quality of the fitting procedure can be improved by increasing the number of data points per energy curve.
Now, we analyze our calculations. The script
allows the analysis of the dependence of the calculated derivatives of the energy-vs-strain curve on STRESS-exciting-analyze.py
the range of distortions included in the fitting procedure (the
axis in
x
plots),
xmgrace the degree of the polynomial fit used in the fitting procedure (different color curves in
plots).
xmgrace
The
script is executed as follows. STRESS-exciting-analyze.py
$ STRESS-exciting-analyze.py
At this point, two
plots will appear on your screen automatically (for more information on how to deal with xmgrace
plots, see
xmgrace
).
Xmgrace: A Quickstart
The previous plots can be used to determine the best range of deformations and order of polynomial fit for each distortion.
As an example, we analyze the first plot, corresponding to the distortion
. Distortion types are listed in the file Dst01
. By examining this file, we can see that the
Distorted_Parameters
distortion corresponds to an applied physical strain in the
Dst01
with the form
Voigt notation
, where
(η,η,η,0,0,0)
is a strain parameter. This deformation type is directly connected with the
η
. For each distorsion type, a plot appear. This plot contains the first derivative of the energy with respect to the strain parameter
hydrostatic pressure
as a function of the maximum strain and of the order of polynomial fit.
η
Analyzing the plot for the
distortion, we notice that curves corresponding to the lower order of the polynomial used in the fit show a Dst01
horizontal plateau at about -555 kbar. This can be assumed to be the
value for the first derivative,
converged
(further information on this topic can be found
from the point of view of the fit
). In a similar way, the results for the
here
distortion can be analyzed, too.
Dst02
The script
generates the file STRESS-exciting-analyze.py
, which will be discussed and used in the next section.
STRESS.IN iv)Numerical values of the stress components
In order to obtain the numerical values of the stress tensor, you should use the following procedure. The first step is to edit the file
, which should have the form STRESS.IN
Dst01 eta_max Fit_orderDst02 eta_max Fit_order
In each row of this file, you should insert a value for
and eta_max
. According to the result of the visual analysis of the previous figures, these two values are extracted by considering the
Fit_order plateau regionsof the corresponding plot:
is a value of maximum distortion located in the plateau region of the curve representing the polynomial fit of order
eta_max
. In our case, we can choose for each distortion
Fit_order 0.002and 1as the best distortion range and polynomial fit, respectively. Thus, the
file will now contain
STRESS.IN
Dst01 0.002 1Dst02 0.002 1
Now, you run the
script. STRESS-exciting-result.py
$ STRESS-exciting-result.py
Finally, numerical values of the stress components are written in the output file
. In this case the content of STRESS.OUT
should be similar to
STRESS.OUT
Thu Aug 11 01:29:45 2016 The hydrostatic pressure and stress tensor in kilobar (kbar) P = 184.841222 [kbar] -150.864206 0.000000 0.000000 0.000000 -150.864206 0.000000 0.000000 0.000000 -252.795254 The hydrostatic pressure and stress tensor in gigapascal (GPa) P = 18.484122 [GPa] -15.086421 0.000000 0.000000 0.000000 -15.086421 0.000000 0.000000 0.000000 -25.279525
From this output, we can extract a value of
GPa for the hydrostatic pressure of the reference configuration, which is in very good agreement with the value obtained in +18.484
.
Section 1 Exercise Repeat all calculations using a denser -point grid and a larger number of strain points. k |
Research talks;Partial Differential Equations;Mathematical Physics
In this talk we present recent results on the Hall-MHD system. We consider the incompressible MHD-Hall equations in $\mathbb{R}^3$.
$\partial_tu +u \cdot u + \nabla u+\nabla p = \left ( \nabla \times B \right )\times B +\nu \nabla u,$ $\nabla \cdot u =0, \nabla \cdot B =0, $ $\partial_tB - \nabla \times \left (u \times B\right ) + \nabla \times \left (\left (\nabla \times B\right )\times B \right ) = \mu \nabla B,$ $u\left (x,0 \right )=u_0\left (x\right ) ; B\left (x,0 \right )=B_0\left (x\right ).$ Here $u=\left (u_1, u_2, u_3 \right ) = u \left (x,t \right ) $ is the velocity of the charged fluid, $B=\left (B_1, B_2, B_3 \right ) $ the magnetic field induced by the motion of the charged fluid, $p=p \left (x,t \right )$ the pressure of the fluid. The positive constants $\nu$ and $\mu$ are the viscosity and the resistivity coefficients. Compared with the usual viscous incompressible MHD system, the above system contains the extra term $\nabla \times \left (\left (\nabla \times B\right )\times B \right ) $ , which is the so called Hall term. This term is important when the magnetic shear is large, where the magnetic reconnection happens. On the other hand, in the case of laminar ows where the shear is weak, one ignores the Hall term, and the system reduces to the usual MHD. Compared to the case of the usual MHD the history of the fully rigorous mathematical study of the Cauchy problem for the Hall-MHD system is very short. The global existence of weak solutions in the periodic domain is done in [1] by a Galerkin approximation. The global existence in the whole domain in $\mathbb{R}^3$ as well as the local well-posedness of smooth solution is proved in [2], where the global existence of smooth solution for small initial data is also established. A refined form of the blow-up criteria and small data global existence is obtained in [3]. Temporal decay estimateof the global small solutions is deduced in [4]. In the case of zero resistivity we present finite time blow-up result for the solutions obtained in [5]. We note that this is quite rare case, as far as the authors know, where the blow-up result for the incompressible flows is proved.
In this talk we present recent results on the Hall-MHD system. We consider the incompressible MHD-Hall equations in $\mathbb{R}^3$. $\partial_tu +u \cdot u + \nabla u+\nabla p = \left ( \nabla \times B \right )\times B +\nu \nabla u,$ $\nabla \cdot u =0, \nabla \cdot B =0, $ $\partial_tB - \nabla \times \left (u \times B\right ) + \nabla \times \left (\left (\nabla \times B\right )\times B \right ) = \mu \nabla B,$ $u\left (x,0 \right ...
35Q35 ; 76W05
... Lire [+] |
Volume by rings, also known as volume by disks or volume by washers (if the area between two functions is being rotated around an axis), is a method of finding the volume of a solid of revolution. This method involves splitting the shape into infinitely small circular rings and summing them up. The formula for the volume of any solid of rotation is $ V=\int\limits_a^b A(x)dx $, where $ A(x) $ denotes an area function.In the case of volume by rings, the formula is $ V=\pi\int\limits_a^b f(x)^2dx $
assuming the rotation is around the x-axis. If the rotation is of an area between two functions $ f(x) $ and $ g(x) $, the formula is
$ V=\pi\int\limits_a^b\bigl(f(x)^2-g(x)^2\bigr)dx $ Examples
To find the volume of the resulting solid when $ f(x)=\sqrt{x} $ is rotated around the $ x $-axis on the interval $ (0,4) $, substitute into the formula.
$ \begin{align} &V=\pi\int\limits_a^b f(x)^2dx=\pi\int\limits_0^4(\sqrt{x})^2dx \\&V=\pi\int\limits_0^4 xdx \\&V=\pi\left[\frac{x^2}{2}\right]_0^4=\pi\left(\frac{4^2}{2}-\frac{0^2}{2}\right)=\pi(8-0)=8\pi\end{align} $
This method can also be used to find the formula for the volume of shapes. Take for example the formula for the volume of a sphere, $ V=\frac{4\pi}{3}r^3 $. A sphere is a graph of $ x^2+y^2=r^2 $ rotated around an axis (here we will assume it is the $ x $-axis). Begin by isolating $ y $. It is important to keep in mind that $ r $ is not a variable and must not be treated as one.
$ \begin{align}y&=\sqrt{r^2-x^2} \\V&=\pi\int\limits_a^b f(x)^2dx=\pi\int\limits_{-r}^r\big(\sqrt{r^2-x^2}\big)^2dx \\&=\pi\int\limits_{-r}^r(r^2-x^2)dx \\&=\pi\left[r^2x-\frac{x^3}{3}\right]_{-r}^r \\&=\pi\left(\left(r^2(r)-\frac{r^3}{3}\right)-\left(r^2(-r)-\frac{(-r)^3}{3}\right)\right)=\pi\left(\left(r^3-\frac{r^3}{3}\right)-\left(-r^3+\frac{r^3}{3}\right)\right) \\&=\pi\left(2r^3-\frac{2r^3}{3}\right)=\pi\left(\frac{4r^3}{3}\right)=\frac{4\pi}{3}r^3\end{align} $ |
deterrant
Every day one sees politicians on TV assuring us that nuclear deterrence works because there no nuclear weapon has been exploded in anger since 1945. They clearly have no understanding of statistics.
With a few plausible assumptions, we can easily calculate that the time until the next bomb explodes could be as little as 20 years.
Be scared, very scared.
The first assumption is that bombs go off at random intervals. Since we have had only one so far (counting Hiroshima and Nagasaki as a single event), this can’t be verified. But given the large number of small influences that control when a bomb explodes (whether in war or by accident), it is the natural assumption to make. The assumption is given some credence by the observation that the intervals between wars are random [download pdf].
If the intervals between bombs are random, that implies that the distribution of the length of the intervals is exponential in shape, The nature of this distribution has already been explained in an earlier post about the random lengths of time for which a patient stays in an intensive care unit. If you haven’t come across an exponential distribution before, please look at that post before moving on.
All that we know is that 70 years have elapsed since the last bomb. so the interval until the next one must be greater than 70 years. The probability that a random interval is longer than 70 years can be found from the cumulative form of the exponential distribution.
If we denote the true mean interval between bombs as $\mu$ then the probability that an intervals is longer than 70 years is
\[ \text{Prob}\left( \text{interval > 70}\right)=\exp{\left(\frac{-70}{\mu_\mathrm{lo}}\right)} \]
We can get a lower 95% confidence limit (call it $\mu_\mathrm{lo}$) for the mean interval between bombs by the argument used in Lecture on Biostatistics, section 7.8 (page 108). If we imagine that $\mu_\mathrm{lo}$ were the true mean, we want it to be such that there is a 2.5% chance that we observe an interval that is greater than 70 years. That is, we want to solve
\[ \exp{\left(\frac{-70}{\mu_\mathrm{lo}}\right)} = 0.025\]
That’s easily solved by taking natural logs of both sides, giving
\[ \mu_\mathrm{lo} = \frac{-70}{\ln{\left(0.025\right)}}= 19.0\text{ years}\]
A similar argument leads to an upper confidence limit, $\mu_\mathrm{hi}$, for the mean interval between bombs, by solving
\[ \exp{\left(\frac{-70}{\mu_\mathrm{hi}}\right)} = 0.975\]
so \[ \mu_\mathrm{hi} = \frac{-70}{\ln{\left(0.975\right)}}= 2765\text{ years}\]
If the worst case were true, and the mean interval between bombs was 19 years. then the distribution of the time to the next bomb would have an exponential probability density function, $f(t)$,
\[ f(t) = \frac{1}{19} \exp{\left(\frac{-70}{19}\right)} \]
There would be a 50% chance that the waiting time until the next bomb would be less than the median of this distribution, =19 ln(0.5) = 13.2 years.
In summary, the observation that there has been no explosion for 70 years implies that the mean time until the next explosion lies (with 95% confidence) between 19 years and 2765 years. If it were 19 years, there would be a 50% chance that the waiting time to the next bomb could be less than 13.2 years. Thus there is no reason at all to think that nuclear deterrence works well enough to protect the world from incineration.
Another approach
My statistical colleague, the ace probabilist Alan Hawkes, suggested a slightly different approach to the problem,
via likelihood. The likelihood of a particular value of the interval between bombs is defined as the probability of making the observation(s), given a particular value of $\mu$. In this case, there is one observation, that the interval between bombs is more than 70 years. The likelihood, $L\left(\mu\right)$, of any specified value of $\mu$ is thus
\[L\left(\mu\right)=\text{Prob}\left( \text{interval > 70 | }\mu\right) = \exp{\left(\frac{-70}{\mu}\right)} \]
If we plot this function (graph on right) shows that it increases with $\mu$ continuously, so the maximum likelihood estimate of $\mu$ is infinity. An infinite wait until the next bomb is perfect deterrence.
But again we need confidence limits for this. Since the upper limit is infinite, the appropriate thing to calculate is a one-sided lower 95% confidence limit. This is found by solving
\[ \exp{\left(\frac{-70}{\mu_\mathrm{lo}}\right)} = 0.05\]
which gives
\[ \mu_\mathrm{lo} = \frac{-70}{\ln{\left(0.05\right)}}= 23.4\text{ years}\]
Summary
The first approach gives 95% confidence limits for the average time until we get incinerated as 19 years to 2765 years. The second approach gives the lower limit as 23.4 years. There is no important difference between the two methods of calculation. This shows that the bland assurances of politicians that “nuclear deterrence works” is not justified.
It is not the purpose of this post to predict when the next bomb will explode, but rather to point out that the available information tells us very little about that question. This seems important to me because it contradicts directly the frequent assurances that deterrence works.
The only consolation is that, since I’m now 79, it’s unlikely that I’ll live long enough to see the conflagration.
Anyone younger than me would be advised to get off their backsides and do something about it, before you are destroyed by innumerate politicians.
Postscript
While talking about politicians and war it seems relevant to reproduce Peter Kennard’s powerful image of the Iraq war.
and with that, to quote the comment made by Tony Blair’s aide, Lance Price
It’s a bit like my feeling about priests doing the twelve stations of the cross. Politicians and priests masturbating at the expense of kids getting slaughtered (at a safe distance, of course). |
Let $X$ be a connected, based CW complex. Then the James splitting of $\Sigma\Omega\Sigma X$ gives, in particular, a weak equivalence of spectra $$ \Sigma^{\infty} \Omega\Sigma X_+ \quad \simeq \quad \Sigma^{\infty} (S^0 \vee X \vee X^{[2]} \vee X^{[3]} \vee \cdots ) , $$ where $X^{[n]}$ denotes the $n$-fold smash product of $X$ with itself. Each side of this equivalence has the structure of $A_\infty$-ring spectrum (the structure on the right side is induced by concatenation $X^{[n]} \wedge X^{[m]} \cong X^{[m+n]}$, and the right side can be seen as the tensor algebra over the sphere spectrum on the points of $X$).
Now, my understanding$^\dagger$ is that the Cartan formula for Hopf invariants implies that the above splitting is multiplicative up to homotopy.
Question: Can the above splitting be enriched to an equivalence of $A_\infty$-rings?
If so, can anyone provide me with a reference?
$^\dagger\tiny \text{From being once upon a time in Bill Richter's orbit.}$ |
The electric circuits are closed loop or path which forms a network of electrical components, where electrons are able to flow. This path is made using electrical wires and is powered by a source, like a battery. The start of the point from where the electrons start flowing is called the source whereas the point where electrons leave the electrical circuit is called the return.Let’s conduct a small experiment, you would need the following,
Electric bulb Wire Electrical tape A battery
Connect the wires to the bulb and connect one end of the wire to the battery, notice what happens, there won’t be any change in the bulb, but when you connect the free wire to the battery and complete the circuit, you will notice that the bulb starts glowing, therefore the circuit must be completed in order for the current to flow.
An electrical circuit complete only when there is at least one closed loop from positive to the negative end. This is the simplest form of electric circuit, the circuit found inside a television is more complicated and has different components.
You might have seen these danger signs on electric poles, transformers and sometimes even on some of the electrical equipment at home, it is to warn you about the dangers of electricity, and electricity, if not handled properly, could be lethal and can even cause deaths. Electric wires sockets and live wire should be properly insulated and kept away from the reach of children, the voltage in a small battery does not exceed more than 12 volts but the voltage in a transformer may reach up to 11000 volts.
Faulty electrical circuits because fire, for fires caused due to electrical failure use of water to extinguish must be avoided, there are special extinguishers for the same.
We have come to a conclusion that electric current flows only when there is a complete uninterrupted connection from a battery through different components back to the battery, any interruption will stop the flow of current
Electric batteries should also be handled carefully, never connect two terminals of the battery without a bulb or a load, because the chemicals inside the battery react so quickly that they generate an immense amount of energy, which can even cause them to burst.
Electric Circuit Symbols
Every component and product of electric circuit contain a symbol. The symbols represent parts of the circuit in a circuit diagram. Beneath are the basic set of symbols that are present in a circuit diagram.
Simple Circuit
A simple circuit comprises of the power source, conductors, switch and load.
Cell:It is the power source. Load:It is also termed as the resistor. It is a light bulb that lights when the circuit is turned on. Conductors:They are made of copper wires with no insulation. One end of the wire is connected the load to the power source and the other end of the wire connects the power source back to the load. Switch:It is a small gap in the circuit. There are various types of switches. A switch can be used to open or close a circuit. Electric Circuit Formula
Following are the list of formulas that are used in electric circuits are:
Quantity Formula Notations Electric current \(I=\frac{Q}{t}\) Resistance \(R=\rho .\frac{L}{A}\) Voltage \(\Delta V=I.R\) Power \(P=\frac{\Delta E}{t}\) Series circuit \(R_{eq}=R_{1}+R_{2}+R_{3}+…\) Parallel circuit \(\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+…\)
Stay tuned with BYJU’S to learn more about Electric Circuit like domestic electric circuits, etc. with the help of interactive video lessons. |
@ACuriousMind Something I was working on: Let $B$ be $\tilde g$-bounded. Fix $x\in B$ and let $(y_n)$ be a sequence. Connect $x$ to each $y_n$ with a curve $\gamma_n$. Each curve can be made to have finite length, in fact, one can bound them above. Then the sequence $L_{\tilde g}(\gamma_n)$ has a convergent subsequence.
Now the goal is to get $f$ to blow up along this sequence, and somehow get a contradiction
Esh (majuscule: Ʃ Unicode U+01A9, minuscule: ʃ Unicode U+0283) is a character used in conjunction with the Latin script. Its lowercase form ʃ is similar to a long s ſ or an integral sign ∫; in 1928 the Africa Alphabet borrowed the Greek letter Sigma for the uppercase form Ʃ, but more recently the African reference alphabet discontinued it, using the lowercase esh only. The lowercase form was introduced by Isaac Pitman in his 1847 Phonotypic Alphabet to represent the voiceless postalveolar fricative (English sh). It is today used in the International Phonetic Alphabet, as well as in the alphabets...
Uh, that will differ depending on the person typesetting the thing. That the "German" and "Russian" typographic traditions would have it that way doesn't mean it looks that way "in Germany" or "in Russia" |
I realise, that this question is a stretch, but I was wondering, how would a bonding orbital be called if it was formed from two $f_{x(x^2−3y^2)}$ or $f_{y(3x^2−y^2)}$ orbitals. Have there been any suggestions on this, was it anywhere proposed or discussed? I am not arguing about the existence of such a thing, but as a thought experiment, how would it be called? Phi, $\varphi$?
I am also throwing in a reference request here, since I am interested in any literature related to the topic.
Some background:
A $\sigma$ orbital is an orbital, that has no nodal plane between the bonding partners. It is $C_\infty$ symmetric with respect to the bonding axis (in first approximation - of course $d_{xy}$ orbitals can align in a $\sigma$ fashion, too, but for the sake of the argument, let's not consider this.) The point group of this orbital can further be assumed as $D_\mathrm{\infty h}$. It shares therefore most features of an $s$ orbital, hence $s$igma, $\sigma$. The most prominent example for this is probably the dihydrogen, $\ce{H2}$, molecule. A $\pi$ orbital has one nodal plane and the bonding axis of the involved atoms is part of this plane. It is therefore asymmetric, $C_1$, with respect to this plane. However, it is $C_\mathrm{s}$ symmetric with respect to the plane that is perpendicular, also sharing the bonding axis. The orbital has further the point group $C_\mathrm{2v}$. It has therefore most features of a $p$ orbital, hence $p$i, $\pi$. The most prominent examples for $\pi$ bonds are probably ethylene, $\ce{C2H4}$, and acetylene, $\ce{C2H2}$. A $\delta$ orbital has two nodal planes between the bonding partners. These are perpendicular to each other. The bond orbital is asymmetric with respect to both of them. As a result, it is $C_\mathrm{s}$ symmetric with respect to the bisecting mirror planes. The point group of this orbital is $D_\mathrm{2h}$. Therefore it has shares features with a $d$ orbital, hence $d$elta, $\delta$. One example is the $\ce{[Re2Cl8]^{2-}}$ ion, see Wikipedia's quadruple bond. I imagine two $f$ orbitals aligning with the bonding axis being simultaneously a $C_3$ symmetry axis, giving it an overall $D_\mathrm{3h}$ point group. Since there is no greek letter starting with an $f$, I think the logical thing would be to choose $\phi$ as it sounds a like. (I cannot be the only one thinking that.)
The following graphic should somewhat demonstrate what I mean:
I have to admit, that I have not done much research about this. It was more like an impulsive question for me to ask. |
$\displaystyle \sum_{n=0}^{\infty}\frac{7^n}{n!}x^n$
I'm still trying to get the hang of these and feel like I've done something wrong here. After applying the ratio test I end up with:
$\left|7x\right|\lim \limits_{n \to \infty}\left|\frac{1}{n+1}\right|$
That limit is $0$, so does this mean my radius of convergence is $\infty$ and my interval of convergence is $(-\infty, \infty)$? |
In this post, I discuss our recent paper, Categorical Reparameterization with Gumbel-Softmax, which introduces a simple technique for training neural networks with discrete latent variables. I'm really excited to share this because (1) I believe it will be quite useful for a variety of Machine Learning research problems, (2) this is my first published paper ever (on Arxiv, and submitted to a NIPS workshop and ICLR as well).
The TLDR;if you want categorical features in your neural nets, just let
sample = softmax((logits+gumbel noise)/temperature), and then backprop as usual using your favorite automatic differentiation software (e.g. TensorFlow, Torch, Theano).
You can find the code for this article here
IntroductionOne of the main themes in Deep Learning is to “let the neural net figure out all the intermediate features”. For example: training convolutional neural networks results in the self-organization of a feature detector hierarchy, while Neural Turing Machines automatically “discover” copying and sorting algorithms.
The workhorse of Deep Learning is the backpropagation algorithm, which uses dynamic programming to compute parameter gradients of the network. These gradients are then used to minimize the optimization objective via gradient descent. In order for this to work, all of the layers in our neural network — i.e. our learned intermediate features — must be continuous-valued functions.
What happens if we want to learn intermediate representations that are discrete? Many "codes" we want to learn are fundamentally discrete - musical notes on a keyboard, object classes (“kitten”, “balloon”, “truck”), and quantized addresses (“index 423 in memory”).
Gumbel-Softmax DistributionThe problem of backpropagating through stochastic nodes can be circumvented if we can re-express the sample $z \sim p_\theta(z)$, such that gradients can flow from $f(z)$ to $\theta$ without encountering stochastic nodes. For example, samples from the normal distribution $z \sim \mathcal{N}(\mu,\sigma)$ can be re-written as $z = \mu + \sigma \cdot \epsilon$, where $\epsilon \sim \mathcal{N}(0,1)$. This is also known as the “reparameterization trick”, and is commonly used to train variational autoencoders with Gaussian latent variables. The main contribution of this work is a “reparameterization trick” for the categorical distribution.Well, not quite – it’s actually a re-parameterization trick for a distribution that we can smoothly deforminto the categorical distribution. We use the Gumbel-Max trick, which provides an efficient way to draw samples $z$ from the Categorical distribution with class probabilities $\pi_i$:
$$ \DeclareMathOperator*{\argmax}{arg\,max} z = \verb|one_hot|\left(\argmax_{i}{\left[ g_i + \log \pi_i \right]}\right) $$
argmax is not differentiable, so we simply use the softmax function as a continuous approximation of argmax:
$$ y_i = \frac{\text{exp}((\log(\pi_i)+g_i)/\tau)}{\sum_{j=1}^k \text{exp}((\log(\pi_j)+g_j)/\tau)} \qquad \text{for } i=1, ..., k. $$
Hence, we call this the Gumbel-
SoftMax distribution*. $\tau$ is a temperature parameter that allows us to control how closely samples from the Gumbel-Softmax distribution approximate those from the categorical distribution. As $\tau \to 0$, the softmax becomes an argmax and the Gumbel-Softmax distribution becomes the categorical distribution. During training, we let $\tau > 0$ to allow gradients past the sample, then gradually anneal the temperature $\tau$ (but not completely to 0, as the gradients would blow up).
Below is an interactive widget that draws samples from the Gumbel-Softmax distribution. Keep in mind that samples are vectors, and a one-hot vector (i.e. one of the elements is 1.0 and the others are 0.0) corresponds to a discrete category. Click "re-sample" to generate a new sample, and try dragging the slider and see what samples look like when the temperature $\tau$ is small!
1.0
TensorFlow ImplementationUsing this technique is extremely simple, and only requires 12 lines of Python code:
Despite its simplicity, Gumbel-Softmax works surprisingly well - we benchmarked it against other stochastic gradient estimators for a couple tasks and Gumbel-Softmax outperformed them for both Bernoulli (K=2) and Categorical (K=10) latent variables. We can also use it to train semi-supervised classification models much faster than previous approaches. See our paper for more details.
Categorical VAE with Gumbel-SoftmaxTo demonstrate this technique in practice, here's a categorical variational autoencoder for MNIST, implemented in less than 100 lines of Python + TensorFlow code.
In standard Variational Autoencoders, we learn an encoding function that maps the data manifold to an isotropic Gaussian, and a decoding function that transforms it back to the sample. The data manifold is projected into a Gaussian ball; this can be hard to interpret if you are trying to learn the categorical structure within your data.
First, we declare the encoding network:
Next, we sample from the Gumbel-Softmax posterior and decode it back into our MNIST image.
Variational autoencoders minimizes reconstruction error of the data by maximizing an expectedlower bound (ELBO) on the likelihood of the data, under a generative model $p_\theta(x)$. For a derivation, see this tutorial on variational methods.
$$\log p_\theta(x) \geq \mathbb{E}_{q_\phi(y|x)}[\log p_\theta(x|y)] - KL[q_\phi(y|x)||p_\theta(y)]$$
Finally, we run train our VAE:
...and, that's it! Now we can sample randomly from our latent categorical code and decode it back into MNIST images:
here. Thank you for reading, and let me know if you find this technique useful!
AcknowledgementsI'm sincerely grateful to my co-authors, Shane Gu and Ben Poole for collaborating with me on this work. Shane introduced me to the Gumbel-Max trick back in August, and supplied the framework for comparing Gumbel-Softmax with existing stochastic gradient estimators. Ben suggested and implemented the semi-supervised learning aspect of the paper, did the math derivations in the Appendix, and helped me a lot with editing the paper. Finally, thanks to Vincent Vanhoucke and the Google Brain team for encouraging me to pursue this idea.
*Chris J. Maddison, Andriy Mnih, and Yee Whye Teh at Deepmind have discovered this technique independently and published their own paper on it - they call it the “Concrete Distribution”. We only found out about each other’s work right as we were submitting our papers to conferences (oops!). If you use this technique in your work, please cite both of our papers! They deserve just as much credit.
23 Apr 2017: Update - Chris Maddison and I integrated these distributions into TensorFlow's Distributions sub-framework. Here's a code example of how to implement a categorical VAE using the distributions API. |
Consider a multiset of natural numbers:
$$A_n=[a_j]_{j=1..n}$$
in the cases:
$a_j \not=a_k$ for all $j,k \in {\{1,2,3,...,n}\} \land n \gt 1\quad\quad\quad\quad\,\,\,\,\,\,\,\,\quad\quad\quad\quad\quad\quad\quad(\operatorname{i})$
$a_j \not=a_k$ for
some $j,k \in {\{1,2,3,...,n}\} \land n \gt 1$ and $a_j \not= 1$ $\quad\quad\quad\quad\quad\,\,\,\,\,(\operatorname{ii})$
We define the product of all elements of $A_n$ as a function of $n$ as follows:
$$\mathcal P(n)=\prod_{j=1}^{n}a_j$$
If we define the function $f_k$ as a function of some fixed $x$ to be the raising to a power of $x$ $k-1$ times so for example :
$$f_1(x)=1$$ $$f_2(x)=x$$ $$f_3(x)=x^x$$ $$f_4(x)=(x^x)^x$$
Then we define a rounding function $\mathcal R$ as:
$$\mathcal R(x)=\cases{\lfloor x \rfloor&${\{x}\} <\frac{1}{2}$\cr \lceil x \rceil&${\{x}\}\geq\frac{1}{2} $\cr}$$
where ${\{x}\}$ is the fractional part of $x$
We have that for any $A_n$ as previously defined,
$$\mathcal R\Biggl(f_j\Bigl(\frac{1}{\mathcal P(n)}\Bigr)\Biggr) \in {\{0,1}\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,\,(1)$$
$$\sum_{j=1}^{n-1}\frac{\mathcal R\Biggl(f_j\Bigl(\frac{1}{\mathcal P(n)}\Bigr)\Biggr)}{n} \in {\Biggl\{1,\frac{1}{2},\frac{2}{3},\frac{3}{5},\frac{4}{7},\frac{5}{9}}\Biggr\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)$$
$(1)$ is clearly true seeing that the sequence in $j$ is always a binary alternating sequence, But this does not explain the result of $(2)$, however it is supported by the results like:
$$\lim _{N\rightarrow \infty}\Biggr(\sum^{N}_{n=1}\frac{(1+(-1)^n)}{2N}\Biggl)=\frac{1}{2}$$
$$\lim _{N\rightarrow \infty}\Biggr(\sum^{N}_{n=0}\frac{(1+(-1)^n)}{2N}\Biggl)=\frac{1}{2}$$
So the proof will of course require the infinite, because we need to establish that it is true for any such $A_n$, which I think I will probably not be capable enough to do, so a counter example is actually a happy thing here I think
EDIT:
The counterexamples should occur as an example for $n=12$, as we expect sums like $\frac{6}{11},\frac{8}{13},\frac{9}{15}$ and so on to occur. |
Group Presentations
We defined a lot of terms on the Words Over a Set page. Recall that if $A$ is a nonempty set and $\mathcal R$ is a set of words over $A \cup A^{-}$ then we defined an equivalence relation $\sim$ on the set of words over $A \cup A^{-}$ for words $u$ and $v$ as follows. We say that $u \sim v$ if there exists a finite sequence of words $w_1, w_2, ..., w_n$ with $w_1 = u$ and $w_n = v$ such that for each $i \in \{ 1, 2, ..., n - 1\}$:
1)$w_{i+1}$ is obtained from $w_i$ by inserting an occurrence of $xx^{-1}$ in the word $w_i$. 2)$w_{i+1}$ is obtained from $w_i$ by deleting an occurrence $xx^{-1}$ in the word $w_i$. 3)$w_{i+1}$ is obtained from $w_i$ by inserting a word in $\mathcal R$ or $\mathcal R^{-}$ in the word $w_i$. 4)$w_{i+1}$ is obtained from $w_i$ by deleting a word in $\mathcal R$ or $\mathcal R^{-}$ in the word $w_i$.
Let $\langle A, \mathcal R \rangle$ denote the set of equivalence classes of words over $A \cup A^{-}$. Define an operation on these equivalence classes for all $[u], [v] \in \langle A, \mathcal R \rangle$ by:(1)
It can be shown that this operation is well-defined. Furthermore, for all equivalence classes $[u], [v], [w] \in \langle A, \mathcal R \rangle$ we have that:
1)$[u]([v][w]) = ([u][v])[w]$. (The Associativity Property) 2)$[u][()] = [u]$ and $[()][u] = [u]$ where $[()]$ is the equivalence class of the empty word $()$. (The Existence of an Identity) 3)$[u][u^{-1}] = [()]$ and $[u^{-1}[u] = [()]$. (The Existence of Inverses)
Therefore $\langle A, \mathcal R \rangle$ with this operation forms a group.
Definition: Let $A$ be a nonempty set and let $\mathcal R$ be a set of words over $A \cup A^{-}$. Let $G$ be a group. Let $f : A \to G$ be any function. We define a similar function $f : \{ \mathrm{words \:over \:} A \cup A^{-} \} \to G$ for all $u = a_1^{\epsilon_1}a_2^{\epsilon_2}...a_n^{\epsilon_n}$ where $a_1, a_2, ..., a_n \in A$ and $\epsilon_1, \epsilon_2, ..., \epsilon_n \in \{ +, - \}$ by:
$f(u) = f(a_1^{\epsilon_1}a_2^{\epsilon_2}...a_n^{\epsilon_n}) := f(a_1)^{\epsilon_1}f(a_2)^{\epsilon_2}...f(a_n)^{\epsilon_n}$.
And further define another similar function $f : \langle A, \mathcal R \rangle \to G$ by:
$f([u]) = f([a_1^{\epsilon_1}a_2^{\epsilon_2}...a_n^{\epsilon_n}]) = f([a_1])^{\epsilon_1}f([a_2])^{\epsilon_2}...f([a_n])^{\epsilon_n}$.
If the function $f : \langle A, \mathcal R \rangle \to G$ is well-defined then $f$ is a group homomorphism. However, this is not always the case. The following theorem gives us a sufficient and necessary condition for determining when $f$ is well-defined.
Theorem 1: Let $A$ be a set and let $\mathcal R$ be a set of words over $A \cup A^{-}$. Then the function $f : \langle A, \mathcal R \rangle \to G$ above is well-defined if and only if for every word $R \in \mathcal R$ we have that $f([R]) = 1$ where $1$ is the identity in $G$. Proof:$\Rightarrow$ If $f$ is well-defined then $f$ is a group homomorphism. Then for all $R \in \mathcal R$ we have that $[R] = [()]$. But group homomorphisms send the identity of one group to the identity of the other group. So indeed, $f([R]) = 1$. $\Leftarrow$ Let $[u], [v] \in \langle A, \mathcal R \rangle$ and that $[u] = [v]$. Then $f([u]) = f([v])$ if $v$ can be obtained from $u$ from either of the first two steps mentioned at the top of this page. For the steps $3$ and $4$, if $u = w_1w_2$ and $v = w_1Rw_2$ then: So indeed, $f$ is well-defined regardless of the representative chosen for the equivalence class $[u]$. $\blacksquare$
Definition: Let $A$ be a nonempty set and let $\mathcal R$ be a set of words over $A \cup A^{-}$. Let $G$ be a group. If $f : \langle A, \mathcal R \rangle \to G$ is well-defined (and hence a homomorphism) and is further a group isomorphism, then $\langle A, \mathcal R \rangle$ is said to be a Group Presentation of $G$. If $A$ and $\mathcal R$ are both finite sets then this group is said to be Finitely Presented. |
To do it for a particular number of variables is very easy to follow. Consider what you do when you integrate a function of x and y over some region. Basically, you chop up the region into boxes of area ${\rm d}x{~\rm d} y$, evaluate the function at a point in each box, multiply it by the area of the box. This can be notated a bit sloppily as:
$$\sum_{b \in \text{Boxes}} f(x,y) \cdot \text{Area}(b)$$
What you do when changing variables is to chop the region into boxes that are not rectangular, but instead chop it along lines that are defined by some function, call it $u(x,y)$, being constant. So say $u=x+y^2$, this would be all the parabolas $x+y^2=c$. You then do the same thing for another function, $v$, say $v=y+3$. Now in order to evaluate the expression above, you need to find "area of box" for the new boxes - it's not ${\rm d}x~{\rm d}y$ anymore.
As the boxes are infinitesimal, the edges cannot be curved, so they must be parallelograms (adjacent lines of constant $u$ or constant $v$ are parallel.) The parallelograms are defined by two vectors - the vector resulting from a small change in $u$, and the one resulting from a small change in $v$. In component form, these vectors are ${\rm d}u\left\langle\frac{\partial x}{\partial u}, ~\frac{\partial y}{\partial u}\right\rangle $ and ${\rm d}v\left\langle\frac{\partial x}{\partial v}, ~\frac{\partial y}{\partial v}\right\rangle $. To see this, imagine moving a small distance ${\rm d}u$ along a line of constant $v$. What's the change in $x$ when you change $u$ but hold $v$ constant? The partial of $x$ with respect to $u$, times ${\rm d}u$. Same with the change in $y$. (Notice that this involves writing $x$ and $y$ as functions of $u$, $v$, rather than the other way round. The main condition of a change in variables is that both ways round are possible.)
The area of a paralellogram bounded by $\langle x_0,~ y_0\rangle $ and $\langle x_1,~ y_1\rangle $ is $\vert y_0x_1-y_1x_0 \vert$, (or the abs value of the determinant of a 2 by 2 matrix formed by writing the two column vectors next to each other.)* So the area of each box is
$$\left\vert\frac{\partial x}{\partial u}{\rm d}u\frac{\partial y}{\partial v}{\rm d}v - \frac{\partial y}{\partial u}{\rm d}u\frac{\partial x}{\partial v}dv\right\vert$$
or
$$\left\vert \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\right\vert~{\rm d}u~{\rm d}v$$
which you will recognise as being $\mathbf J~{\rm d}u~{\rm d}v$, where $\mathbf J$ is the Jacobian.
So, to go back to our original expression
$$\sum_{b \in \text{Boxes}} f(x,y) \cdot \text{Area}(b)$$
becomes
$$\sum_{b \in \text{Boxes}} f(u, v) \cdot \mathbf J \cdot {\rm d}u{\rm d}v$$
where $f(u, v)$ is exactly equivalent to $f(x, y)$ because $u$ and $v$ can be written in terms of $x$ and $y$, and vice versa. As the number of boxes goes to infinity, this becomes an integral in the $uv$ plane.
To generalize to $n$ variables, all you need is that the area/volume/equivalent of the $n$ dimensional box that you integrate over equals the absolute value of the determinant of an n by n matrix of partial derivatives. This is hard to prove, but easy to intuit.
*to prove this, take two vectors of magnitudes $A$ and $B$, with angle $\theta$ between them. Then write them in a basis such that one of them points along a specific direction, e.g.:
$$A\left\langle \frac{1}{\sqrt 2}, \frac{1}{\sqrt 2}\right\rangle \text{ and } B\left\langle \frac{1}{\sqrt 2}(\cos(\theta)+\sin(\theta)),~ \frac{1}{\sqrt 2} (\cos(\theta)-\sin(\theta))\right\rangle $$
Now perform the operation described above and you get $$\begin{align} & AB\cdot \frac12 \cdot (\cos(\theta) - \sin(\theta)) - AB \cdot 0 \cdot (\cos(\theta) + \sin(\theta)) \\ = & \frac 12 AB(\cos(\theta)-\sin(\theta)-\cos(\theta)-\sin(\theta)) \\ = & -AB\sin(\theta) \end{align}$$
The absolute value of this, $AB\sin(\theta)$, is how you find the area of a parallelogram - the products of the lengths of the sides times the sine of the angle between them. |
It's clearly a norm: If $\|\nabla u\|_{L^2}=0$ then $u$ is constant, and the only constant in $L^2$ is $0$. On the other hand, if $H^1=W^{1,2}$ were complete under this norm we would have, by the bounded inverse theorem, that both norms are equivalent, which we know to be false (the scaling below should give you an idea of how to build explicit counterexamples), in fact $\| \nabla \cdot \|$
the norm of a Hilbert space, which is very important in applications. is
The completion of $C_c^\infty$ with respect to this norm I've seen called $\mathcal{D}^{1,2}=\dot{H}^1=L^{1,2}$. In any case, call it $X$, it is the space$$X=\{ u\in L^{2^*}: \nabla u\in L^2\}, \qquad 2^*=\frac{2n}{n-2}.$$This is a consequence of the Gagliardo-Nirenberg-Sobolev inequality that says,
$$\| u\|_{L^{2^*}} \leq C \| \nabla u\|_{L^2}, \qquad u\in X.$$Although usually this inequality is stated for functions in $H^1$, it's true in the more general $X$, and it turns out that this latter space is the appropriate one if we are looking for minimizers, i.e. functions $u$ that give an equality in the preceding (this functions statisfy $u\in X\setminus H^1$). The existence of such functions has very important implications, see for example this work of Kenig and Merle.
The fact that this inequality is invariant under the scaling $u\mapsto u_\lambda=\lambda^{(n-2)/2}u(\lambda \cdot)$, i.e. both norms are the same for $u$ and $u_\lambda$ provides a criterion for 'energy criticality' in dispersive PDE. In this setting a google search for "Energy critical Schrödinger" should give you plenty of examples where $X$ is used. |
Suppose that I want to check how good OLS works in some specific environment using Monte Carlo. I can simulate $Y=X\beta+\epsilon$. What should I do in Monte Carlo simulations, do I simulate the whole model on each replication, or do I simulate only $\epsilon$ in each replication, while $X$ is the same across all replications.
Generally you would have some distributional specification for $X$ and re-simulate $X$ on each monte carlo iteration as apposed to using the same $X$ on each iteration. This way the monte carlo simulation would apply to the entire population of $(y,X)$ as apposed to just one finite sample. Also, you have to have some specification for $\epsilon$. Lets say, $\epsilon \sim N(0,\sigma^2)$. Here is how I would do it in R (pseudo code).
b.estimates=matrix(NA,nrow=b,ncol=k) #k is the number of variables for(i in 1:b){ #Simulate X here e = rnorm(n,0,sd=sigma) y= X%*%Beta +e b.estimates[i,]=solve(t(X)%*%X)%*%t(X)%*%y }
Hope that helps |
Search
Now showing items 1-10 of 24
Production of Σ(1385)± and Ξ(1530)0 in proton–proton collisions at √s = 7 TeV
(Springer, 2015-01-10)
The production of the strange and double-strange baryon resonances ((1385)±, Ξ(1530)0) has been measured at mid-rapidity (|y|< 0.5) in proton–proton collisions at √s = 7 TeV with the ALICE detector at the LHC. Transverse ...
Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Inclusive photon production at forward rapidities in proton-proton collisions at $\sqrt{s}$ = 0.9, 2.76 and 7 TeV
(Springer Berlin Heidelberg, 2015-04-09)
The multiplicity and pseudorapidity distributions of inclusive photons have been measured at forward rapidities ($2.3 < \eta < 3.9$) in proton-proton collisions at three center-of-mass energies, $\sqrt{s}=0.9$, 2.76 and 7 ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of pion, kaon and proton production in proton–proton collisions at √s = 7 TeV
(Springer, 2015-05-27)
The measurement of primary π±, K±, p and p¯¯¯ production at mid-rapidity (|y|< 0.5) in proton–proton collisions at s√ = 7 TeV performed with a large ion collider experiment at the large hadron collider (LHC) is reported. ...
Two-pion femtoscopy in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(American Physical Society, 2015-03)
We report the results of the femtoscopic analysis of pairs of identical pions measured in p-Pb collisions at $\sqrt{s_{\mathrm{NN}}}=5.02$ TeV. Femtoscopic radii are determined as a function of event multiplicity and pair ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Charged jet cross sections and properties in proton-proton collisions at $\sqrt{s}=7$ TeV
(American Physical Society, 2015-06)
The differential charged jet cross sections, jet fragmentation distributions, and jet shapes are measured in minimum bias proton-proton collisions at centre-of-mass energy $\sqrt{s}=7$ TeV using the ALICE detector at the ...
Centrality dependence of high-$p_{\rm T}$ D meson suppression in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Springer, 2015-11)
The nuclear modification factor, $R_{\rm AA}$, of the prompt charmed mesons ${\rm D^0}$, ${\rm D^+}$ and ${\rm D^{*+}}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at a centre-of-mass ...
K*(892)$^0$ and $\Phi$(1020) production in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(American Physical Society, 2015-02)
The yields of the K*(892)$^0$ and $\Phi$(1020) resonances are measured in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV through their hadronic decays using the ALICE detector. The measurements are performed in multiple ... |
Inaccessible
Inaccessible cardinals are the traditional entry-point to the large cardinal hierarchy (although there are some weaker large cardinal notions, such as universe cardinals).
If $\kappa$ is inaccessible, then $V_\kappa$ is a model of ZFC, but this is not an equivalence, since the weaker notion of universe cardinal also have this feature, and are not all regular when they exist. Every inaccessible cardinal $\kappa$ is a beth fixed point, and consequently $V_\kappa=H_\kappa$. (Zermelo) The models of second-order ZFC are precisely the models $\langle V_\kappa,\in\rangle$ for an inaccessible cardinal $\kappa$. Solovay proved that if there is an inaccessible cardinal, then there is an inner model of a forcing extension satisfying ZF+DC in which every set of reals is Lebesgue measurable there. (citation) Shelah proved that Solovay's use of the inaccessible cardinal is necessary, in the sense that in any model of ZF+DC in which every set of reals is Lebesgue measurable, there is an inner model of ZFC with an inaccessible cardinal. Consequently, the consistency of the existence of an inaccessible cardinal with ZFC is equivalent to the impossibility of our constructing a non-measurable set of reals using only ZF+DC. The uncountable Grothedieck universes are precisely the sets of the form $V_\kappa$ for an inaccessible cardinal $\kappa$. The universe axiom is equivalent to the assertion that there is a proper class of inaccessible cardinals. Contents Weakly inaccessible
A cardinal $\kappa$ is
weakly inaccessible if it is an uncountable regular limit cardinal. Under the GCH, this is equivalent to inaccessibility, since under GCH every limit cardinal is a strong limit cardinal. So the difference between weak and strong inaccessibility only arises when GCH fails badly. Every inaccessible cardinal is weakly inaccessible, but forcing arguments show that any inaccessible cardinal can become a non-inaccessible weakly inaccessible cardinal in a forcing extension, such as after adding an enormous number of Cohen reals (this forcing is c.c.c. and hence preserves all cardinals and cofinalities and hence also all regular limit cardinals). Meanwhile, every weakly inaccessible cardinal is fully inaccessible in any inner model of GCH, since it will remain a regular limit cardinal in that model and hence also be a strong limit there. In particular, every weakly inaccessible cardinal is inaccessible in the constructible universe $L$. Consequently, although the two large cardinal notions are not provably equivalent, they are equiconsistent. Levy collapse
The Levy collapse of an inaccessible cardinal $\kappa$ is the $\lt\kappa$-support product of $\text{Coll}(\omega,\gamma)$ for all $\gamma\lt\kappa$. This forcing collapses all cardinals below $\kappa$ to $\omega$, but since it is $\kappa$-c.c., it preserves $\kappa$ itself, and hence ensures $\kappa=\omega_1$ in the forcing extension.
Inaccessible to reals
A cardinal $\kappa$ is
inaccessible to reals if it is inaccessible in $L[x]$ for every real $x$. For example, after the Levy collapse of an inaccessible cardinal $\kappa$, which forces $\kappa=\omega_1$ in the extension, the cardinal $\kappa$ is of course no longer inaccessible, but it remains inaccessible to reals. Grothendieck universe
The concept of Grothendieck universes arose in category theory out of the desire to create a hierarchy of notions of smallness, so that one may form such categories as the category of all small groups, or small rings or small categories, without running into the difficulties of Russell's paradox.
A
Grothendieck universe is a transitive set $W$ that is closed under pairing, power set and unions. That is, (transitivity) If $b\in a\in W$, then $b\in W$. (pairing) If $a,b\in W$, then $\{a,b\}\in W$. (power set) If $a\in W$, then $P(a)\in W$. (union) If $a\in W$, then $\cup a\in W$. Universe axiom
The
Grothendieck universe axiom is the assertion that every set is an element of a Grothendieck universe. This is equivalent to the assertion that the inaccessible cardinals form a proper class. |
Given the logarithmic spiral $$\alpha(t) = e^{-t}(\cos(t),\sin(t))$$ I take a ray from the origin given by $\lambda(\cos \theta, \sin \theta)$ and I have to prove that in $\alpha(\mathbb{R}) \cap R_{\theta}$ the tangents form a constant angle with the vector $(\cos \theta,\sin \theta)$ (constant in the sense that it does not depend on the point nor the angle $\theta$).
My approach
I compute the tangent line as having director vector $-e^{-t}(\cos(t)+\sin(t),\sin(t)-\cos(t))$ and then the intersection of the two lines is given by the equation $\lambda_1(\cos \theta,\sin \theta) = e^{-t}((1-\lambda_2)\cos(t)\sin(t),(1-\lambda_2)\sin(t)+\lambda_2\cos(t))$.
Solving this gives me $$t = -\frac{1}{2}log\left(\frac{\lambda_1^2}{(1-\lambda_2)^2+\lambda_2^2}\right)$$ But then the angle is given by $$cos \alpha = \frac{-e^{-t_0}((\cos t_0+\sin t_0) \cos \theta + (\sin t_0 - \cos t_0) \sin \theta}{\sqrt{t}e^{-t_0}}$$ which depends on $t_0$ (the point) and $\theta$.
Perhaps I should change to polar coordinates? |
The Robin's inequality says - If the Riemann hypothesis is true then - $$\sigma(n) < e^{\gamma}n \log(\log(n))$$ holds true for all $n \in \mathbb{N}$
Now it is proved for all $5-$ free integers .And thus for infinely many integers. So my question is does this proof also prove That there are infinitely many no - trivial zeros of zeta function zeros on real line $1/2$ ?
The Robin's inequality says - If the Riemann hypothesis is true then - $$\sigma(n) < e^{\gamma}n \log(\log(n))$$ holds true for all $n \in \mathbb{N}$
This result is quite nice; I should say that I am properly thought of as an amateur enthusiast about Colossally Abundant Numbers and Ramanujan's Superior Highly Composite Numbers; I do not know any of the methods of basic analytic number theory.
First things first. Robin's condition is that $\sigma(n) < e^\gamma \log \log n $ for $n \geq 5041 = 71^2 = 1 + 7! . $
Theorem 1.6 in Choie, Lichiardopol, Moree, and Sole is this: RH is true if and only if Robin's inequality is true for all natural numbers divisible by the fifth power of some prime.
There is an excellent reason that this does not change very much: The numbers that give very large values of $\sigma(n)$ relative to their size, are the Colossally Abundant Numbers, see Alaoglu and Erdos. These are almost all divisible by 32 and 243, for example, only a small finite set of them in the beginning are not so divisible. As a result, RH is true, roughly speaking, if and only if it is true for all CA numbers above 5040; note that the first counterexample is not guaranteed to be CA, only superabundant.
Oh, this is all-or-nothing. No conclusions about zeroes of the zeta function in the critical strip may be drawn.
For experimentation, I recommend the original condition of Nicolas, the adviser of Robin: RH is true if and only if, for every
primorial $P,$ $$ P > \, \varphi(P) \; e^\gamma \, \log \log P, $$ where $\varphi$ is the Euler totient function. I mention this because making a computer list of consecutive CA numbers is actually a bit difficult, while computing primorials (the product of consecutive primes beginning with $2$) is relatively easy. |
It is said that $p \left( \theta | y _ { 1 : N } \right) \propto _ { \theta } p \left( y _ { 1 : N } | \theta \right) p ( \theta )$.
And $p \left( \theta | y _ { 1 : N } \right)$ is the posterior, $ p \left( y _ { 1 : N } | \theta \right)$ is the likelihood, and $p ( \theta )$ is the prior.
Suppose we have a model M, and its weights are W. To my understanding:
As for $p \left( \theta | y _ { 1 : N } \right) \propto _ { \theta } p \left( y _ { 1 : N } | \theta \right) p ( \theta )$, I think $\theta$ is $W=\{w_1, w_2, ..., w_n\}$, $p(\theta)$ is a distribution for $w_i$, $ p \left( y _ { 1 : N } | \theta \right)$ is the accuracy, and $p \left( \theta | y _ { 1 : N } \right)$ is the ideal weights distribution. Are my understandings correct?
As for bayesian inference for $p \left( \theta | y _ { 1 : N } \right)$, first we select a distribution $q(.)$ for $w_i$, and the goal is to find the parameters(mean, variance, ...) for this distribution so that it is very close to the ideal weights distribution $p \left( \theta | y _ { 1 : N } \right)$. Using KL-divergence
$\mathrm { KL } ( q ( \cdot ) \| p ( \cdot | y ) ) := ... := \log p ( y ) - \int q ( \theta ) \log \frac { p ( \theta , y ) } { q ( \theta ) } d \theta$ , it is down to maximum $ \int q ( \theta ) \log \frac { p ( \theta , y ) } { q ( \theta ) } d \theta$. But how to solve this? I see some materials talking about SVI, but still do not understand this well. Where can I find some codes for SVI? |......| If me, I will first use Adam to optimize all the $w_i$, and then recompute the mean and variance for the distribution $q(.)$, after resampling, use Adam again to optimize these new $w_i$, ... . Is this a feasible method? |
Why does a 95% CI not imply a 95% chance of containing the mean?
There are many issues to be clarified in this question and in the majority of the given responses. I shall confine myself only to two of them.
a.
What is a population mean? Does exist a true population mean?
The concept of population mean is model-dependent. As all models are wrong, but some are useful, this population mean is a fiction that is defined just to provide useful interpretations. The fiction begins with a probability model.
The probability model is defined by the triplet$$(\mathcal{X}, \mathcal{F}, P),$$where $\mathcal{X}$ is the sample space (a non-empty set), $\mathcal{F}$ is a family of subsets of $\mathcal{X}$ and $P$ is a well-defined probability measure defined over $\mathcal{F}$ (it governs the data behavior). Without loss of generality, consider only the discrete case. The population mean is defined by$$\mu = \sum_{x \in \mathcal{X}} xP(X=x),$$ that is, it represents the central tendency under $P$ and it can also be interpreted as the center of mass of all points in $\mathcal{X}$, where the weight of each $x \in \mathcal{X}$ is given by $P(X=x)$.
In the probability theory, the measure $P$ is considered known, therefore the population mean is accessible through the above simple operation. However, in practice, the probability $P$ is hardly known. Without a probability $P$, one cannot describe the probabilistic behavior of the data. As we cannot set a precise probability $P$ to explain the data behavior, we set a family $\mathcal{M}$ containing probability measures that possibly govern (or explain) the data behavior. Then, the classical statistical model emerges$$(\mathcal{X}, \mathcal{F}, \mathcal{M}).$$The above model is said to be a parametric model if there exists $\Theta \subseteq \mathbb{R}^p$ with $p< \infty$ such that $\mathcal{M} \equiv \{P_\theta: \ \theta \in \Theta\}$. Let us consider just the parametric model in this post.
Notice that, for each probability measure $P_\theta \in \mathcal{M}$, there is a respective mean definition$$\mu_\theta = \sum_{x \in \mathcal{X}} x P_\theta(X=x).$$That is, there is a family of population means $\{\mu_\theta: \ \theta \in \Theta\}$ that depends tightly on the definition of $\mathcal{M}$. The family $\mathcal{M}$ is defined by limited humans and therefore it may not contain the true probability measure that governs the data behavior. Actually, the chosen family will hardly contain the true measure, moreover this true measure may not even exist. As the concept of a population mean depends on the probability measures in $\mathcal{M}$, the population mean is model-dependent.
The Bayesian approach considers a prior probability over the subsets of $\mathcal{M}$ (or, equivalently, $\Theta$), but in this post I will concentrated only on the classical version.
b.
What is the definition and the purpose of a confidence interval?
As aforementioned, the population mean is model-dependent and provides useful interpretations. However, we have a family of population means, because the statistical model is defined by a family of probability measures (each probability measure generates a population mean). Therefore, based on an experiment, inferential procedures should be employed in order to estimate a small set (interval) containing good candidates of population means. One well-known procedure is the ($1-\alpha$) confidence region, which is defined by a set $C_\alpha$ such that, for all $\theta \in \Theta$,$$P_\theta(C_\alpha(X) \ni \mu_\theta) \geq 1-\alpha \ \ \ \mbox{and} \ \ \ \inf_{\theta\in \Theta} P_\theta(C_\alpha(X) \ni \mu_\theta) = 1-\alpha,$$ where $P_\theta(C_\alpha(X) = \varnothing) = 0$ (see Schervish, 1995). This is a very general definition and encompasses virtually any type of confidence intervals. Here, $P_\theta(C_\alpha(X) \ni \mu_\theta)$ is the probability that $C_\alpha(X)$ contains $\mu_\theta$ under the measure $P_\theta$. This probability should be always greater than (or equal to) $1-\alpha$, the equality occurs at the worst case.
Remark: The readers should notice that it is not necessary to make assumptions on the state of reality, the confidence region is defined for a well-defined statistical model without making reference to any "true" mean. Even if the "true" probability measure does not exist or it is not in $\mathcal{M}$, the confidence region definition will work, since the assumptions are about statistical modelling rather than the states of reality.
On the one hand,
before observing the data, $C_\alpha(X)$ is a random set (or random interval) and the probability that "$C_\alpha(X)$ contains the mean $\mu_\theta$" is, at least, $(1-\alpha)$ for all $\theta \in \Theta$. This is a very desirable feature for the frequentist paradigm.
On the other hand,
after observing the data $x$, $C_\alpha(x)$ is just a fixed set and the probability that "$C_\alpha(x)$ contains the mean $\mu_\theta$" should be in {0,1} for all $\theta \in \Theta$.
That is,
after observing the data $x$, we cannot employ the probabilistic reasoning anymore. As far as I know, there is no theory to treat confidence sets for an observed sample (I am working on it and I am getting some nice results). For a while, the frequentist must believe that the observed set (or interval) $C_\alpha(x)$ is one of the $(1-\alpha)100\%$ sets that contains $\mu_\theta$ for all $\theta\in \Theta$.
PS: I invite any comments, reviews, critiques, or even objections to my post. Let's discuss it in depth. As I am not a native English speaker, my post surely contains typos and grammar mistakes.
Reference:
Schervish, M. (1995), Theory of Statistics, Second ed, Springer. |
23 people. In a room of just 23 people there’s a 50-50 chance of at least two people having the same birthday. In a room of 75 there’s a 99.9% chance of at least two people matching.
Put down the calculator and pitchfork, I don’t speak heresy. The birthday paradox is strange, counter-intuitive, and
completely true. It’s only a “paradox” because our brains can’t handle the compounding power of exponents. We expect probabilities to be linear and only consider the scenarios we’re involved in (both faulty assumptions, by the way).
Let’s see why the paradox happens and how it works.
Problem 1: Exponents aren’t intuitive
We’ve taught ourselves mathematics and statistics, but let’s not kid ourselves: it’s not natural.
Here’s an example: What’s the chance of getting 10 heads in a row when flipping coins? The untrained brain might think like this:
“Well, getting one head is a 50% chance. Getting two heads is twice as hard, so a 25% chance. Getting
ten heads is probably 10 times harder… so about 50%/10 or a 5% chance.”
And there we sit, smug as a bug on a rug. No dice bub.
After pounding your head with statistics, you know not to divide, but use exponents. The chance of 10 heads is not .5/10 but .5 10, or about .001.
But even after training, we get caught again. At 5% interest we’ll double our money in 14 years, rather than the “expected” 20. Did you naturally infer the Rule of 72 when learning about interest rates? Probably not. Understanding compound exponential growth with our linear brains is hard.
Problem 2: Humans are a tad bit selfish
Take a look at the news. Notice how much of the negative news is the result of acting without considering others. I’m an optimist and
do have hope for mankind, but that’s a separate discussion :).
In a room of 23, do you think of the 22 comparisons where
your birthday is being compared against someone else’s? Probably.
Do you think of the
231 comparisons where someone who is not you is being checked against someone else who is not you? Do you realize there are so many? Probably not.
The fact that we neglect the
10 times as many comparisons that don’t include us helps us see why the “paradox” can happen. Ok, fine, humans are awful: Show me the math!
The question: What are the chances that two people share a birthday in a group of 23?
Sure, we could list the pairs and count all the ways they could match. But that’s hard: there could be 1, 2, 3 or even 23 matches!
It’s like asking “What’s the chance of getting one or more heads in 23 coin flips?” There are so many possibilities: heads on the first throw, or the 3rd, or the last, or the 1st and 3rd, the 2nd and 21st, and so on.
How do we solve the coin problem? Flip it around (Get it? Get it?). Rather than counting every way to get heads,
find the chance of getting all tails, our “problem scenario”.
If there’s a 1% chance of getting all tails (more like .5^23 but work with me here), there’s a 99% chance of having
at least one head. I don’t know if it’s 1 head, or 2, or 15 or 23: we got heads, and that’s what matters. If we subtract the chance of a problem scenario from 1 we are left with the probability of a good scenario.
The same principle applies for birthdays. Instead of finding all the ways we match,
find the chance that everyone is different, the “problem scenario”. We then take the opposite probability and get the chance of a match. It may be 1 match, or 2, or 20, but somebody matched, which is what we need to find. Explanation: Counting Pairs
With 23 people we have 253 pairs:
(Brush up on combinations and permutations if you like).
The chance of 2 people having different birthdays is:
Makes sense, right? When comparing one person's birthday to another, in 364 out of 365 scenarios they won't match. Fine.
But making
253 comparisons and having them all be different is like getting heads 253 times in a row -- you had to dodge "tails" each time. (For now, let's pretend birthday comparisons are like coin flips -- see Appendix A for the exact calculation.)
We use exponents to find the probability:
Our chance of getting a single miss is pretty high (99.7260%), but when you take that chance hundreds of times, the odds of keeping up that streak drop. Fast.
The chance we find a match is: 1 – 49.95% = 50.05%, or just over half! If you want to find the probability of a match for any number of people n the formula is:
Interactive Example
I didn’t believe we needed only 23 people. The math works out, but is it real?
You bet. Try the example below: Pick a number of items (365), a number of people (23) and run a few trials. You’ll see the theoretical match and your actual match as you run your trials. Go ahead, click the button (or see the full page).
As you run more and more trials (keep clicking!) the actual probability should approach the theoretical one.
Examples and Takeaways
Here are a few lessons from the birthday paradox:
$\sqrt{n}$is roughly the number you need to have a 50% chance of a match with n items. $\sqrt{365}$ is about 20. This comes into play in cryptography for the birthday attack. Even though there are 2 128(1e38) GUIDs, we only have 2 64(1e19) to use up before a 50% chance of collision. And 50% is really, really high. You only need 13 people picking letters of the alphabet to have 95% chance of a match. Try it above (people = 13, items = 26). Exponential growth rapidly decreases the chance of picking unique items (aka it increases the chances of a match). Remember: exponents are non-intuitive and humans are selfish!
After thinking about it a lot, the birthday paradox finally clicks with me. But I still check out the interactive example just to make sure.
Appendix A: Repeated Multiplication Explanation (Geeky Math Alert!)
Remember how we assumed birthdays are independent? Well, they aren’t.
If Person A and Person B match, and Person B and C match, we know that A and C must match also. The outcome of matching A and C depends on their results with B, so the probabilities aren’t independent. (If truly independent, A and C would have a 1/365 chance of matching, but we know it's a 100% guaranteed match.)
When counting pairs, we treated birthday matches like coin flips, multiplying the same probability over and over. This assumption isn’t strictly true but it’s good enough for a small number of people (23) compared to the sample size (365). It’s unlikely to have multiple people match and screw up the independence, so it’s a good approximation.
It’s unlikely, but it can happen. Let’s figure out the real chances of each person picking a different number:
The first person has a 100% chance of a unique number (of course) The second has a (1 – 1/365) chance (all but 1 number from the 365) The third has a (1 – 2/365) chance (all but 2 numbers) The 23rd has a (1 – 22/365) (all but 22 numbers)
The multiplication looks pretty ugly:
But there’s a shortcut we can take. When x is close to 0, a coarse first-order Taylor approximation for $e^x$ is:
so
Using our handy shortcut we can rewrite the big equation to:
But we remember that adding the numbers 1 to n = n(n + 1)/2. Don’t confuse this with n(n-1)/2, which is C(n,2) or the number of pairs of n items. They look almost the same!
Adding 1 to 22 is (22 * 23)/2 so we get:
Phew. This approximation is very close, plug in your own numbers below:
Good enough for government work, as they say. If you simplify the formula a bit and swap in
n for 23 you get:
and
With the exact formula, 366 people has a guaranteed collision: we multiply by $1 - 365/365 = 0$, which eliminates $p(\text{different})$ and makes $p(\text{match}) = 1$. With the approximation formula, 366 has a near-guarantee, but is not exactly 1: $1 - e^{-365^2 / (2 \cdot 365)} \approx 1$ .
Appendix B: The General Birthday Formula
Let’s generalize the formula to picking
n people from T total items (instead of 365):
If we choose a probability (like 50% chance of a match) and solve for
n:
Voila! If you take $\sqrt{T}$ items (17% more if you want to be picky) then you have about a 50-50 chance of getting a match. If you plug in other numbers you can solve for other probabilities:
Remember that m is the
desired chance of a match (it’s easy to get confused, I did it myself). If you want a 90% chance of matching birthdays, plug m=90% and T=365 into the equation and see that you need 41 people.
Wikipedia has even more details to satisfy your inner nerd. Go forth and enjoy. |
I'm trying to understand the proof of the zero-one-law for first order logic as provided in (Ebbinghaus-Flum, 1995). It goes as follows:
Let $\tau$ be a relational signature. Let $r\in\mathbb{N}$, then $\Delta_{r+1}=\{\phi(v_1,\ldots,v_r,v_{r+1})\mid\phi\text{ has the form }R\bar x\text{, where }R\in\tau\text{ and where }v_{r+1}\text{ occurs in }\bar x\}$. For $\Phi\subseteq\Delta_{r+1}$, we say the extension axiom of $\Phi$ is: $$\chi_\Phi=\forall v_1\ldots\forall v_r(\bigwedge_{1\leq i<j\leq r}v_i\neq v_j\rightarrow \exists v_{r+1}(\bigwedge_{1\leq i\leq r} v_i\neq v_{r+1}\wedge\bigwedge_{\phi\in\Phi}\phi\wedge\bigwedge_{\phi\in\Phi^c}\neg\phi)).$$ Now they prove
Lemma. Any extension axiom $\chi_{\Phi}$ holds in almost all finite structures, i.e. $l(\chi_\Phi)=\lim_{n\rightarrow\infty} \frac{L_n(\chi_\Phi)}{L_n(\tau)}=\lim_{n\rightarrow\infty}\frac{|\{\mathcal{A}\mid\mathcal A\models\chi_\Phi\wedge|A|=\{1,\ldots,n\}\}|}{|\{\mathcal A\mid|A|=\{1,\ldots,n\}\}|}=1$. Proof. Given $\Phi$. For any tuple $a_1,\ldots,a_r$ of distinct elements in a structure $\mathcal A$ and any further object $a$ let $\delta$ be the probability that $a_1,\ldots,a_r,a$ satisfies $\Phi\cup\{\neg\phi\mid \phi\in\Phi^c\}$, when adding $a$ to $\mathcal A$ as a new element and randomly fixing the truth values of $R\bar b$ for any $R\in\tau$ and any sequence $\bar b$ in $A\cup\{a\}$ containing $a$. Clearly, if $c$ is the number of subsets of $\Delta_{r+1}$, $\delta=\frac1c$.
Until here I understand everything, but I don't understand the following derivation:
$$l_n(\neg\chi_\Phi)=l_n(\exists v_1\ldots\exists v_r(\ldots))\leq n^r(\tfrac{c-1}c)^{n-r}=n^r(1-\delta)^{n-r}.$$
I don't understand where the expression just after the inequality comes from, could someone explain this?
Update. I have found this article, in which they seem to prove a special case of the lemma above (see the proof of Lemma 0.3, (b) $\Rightarrow$ (c)). I'm still not fully able to make the generalisation to this case, could someone explain this? Update II. By replacing $\frac1{2^2k}$ by $\frac1c$ in the above article, and thinking some more about the arguments in 1. and 2., I have been able to prove the lemma. Update III. The replacement part in my last update wasn't really clear to me at that moment. I now have been able to fully prove it: Since we can choose our relations arbitrarily, for a relation $R\in\tau$ and a sequence of variables $\bar x$ of the right size, the probability of $R\bar x$ to hold equals $\tfrac12$. So in total there are$$\left(\dfrac12\right)^{\|\Delta_{r+1}\|}$$sequences $v_1,\ldots,v_r,v_{r+1}$ in $\{1,\ldots,n\}$ that realise $\psi_\Phi$. If now $c$ is the amount of subsets of $\Delta_{r+1}$, then $c=2^{\|\Delta_{r+1}\|}$, and filling that in in my original proof, finally proves the lemma. |
I'm looking for help translate these posts into different languages! Please email me at <myfirstname><mylastname>2004<at>gmail.com if you are interested. Xiaoyi Yin (尹肖贻) has kindly translated this post into Chinese (中文).
In my previous blog post, I described how simple distributions like Gaussians can be “deformed” to fit complex data distributions using normalizing flows. We implemented a simple flow by chaining 2D Affine Bijectors with PreLU nonlinearities to build a small invertible neural net.
However, this MLP flow is pretty weak: there are only 2 units per hidden layer. Furthermore, the non-linearity is monotonic and piecewise linear, so all it does is slightly warp the data manifold around the origin. This flow completely fails to implement more complex transformations like separating an isotropic Gaussian into two modes when trying to learn the “Two Moons” dataset below:
Fortunately, there are several more powerful normalizing flows that have been introduced in recent Machine Learning literature. We will explore several of these techniques in this tutorial.
Autoregressive Models are Normalizing Flows
$$p(x) = \prod_i{p(x_i \,|\, x_{1:i-1})}$$
The conditional densities usually have learnable parameters. For example, a common choice is an autoregressive density $p(x_{1:D})$ whose conditional density is a univariate Gaussian, whose mean and standard deviations are computed by neural networks that depend on the previous $x_{1:i-1}$.
$$p(x_i \,|\, x_{1:i-1}) = \mathcal{N}(x_i \,|\,\mu_i, (\exp\alpha_i)^2)$$
$$p(x_i \,|\, x_{1:i-1}) = \mathcal{N}(x_i \,|\,\mu_i, (\exp\alpha_i)^2)$$
$$\mu_i = f_{\mu_i}(x_{1:i-1})$$
$$\alpha_i = f_{\alpha_i}(x_{1:i-1})$$
Learning data with autoregressive density estimation makes the rather bold inductive bias that the ordering of variables are such that your earlier variables don’t depend on later variables. Intuitively, this shouldn’t be true at all for natural data (the top row of pixels in an image does have a causal, conditional dependency on the bottom of the image). However it’s still possible to generate plausible images in this manner (to the surprise of many researchers!).
To sample from this distribution, we compute $D$ “noise variates” $u_{1:D}$ from the standard Normal, $N(0,1)$, then apply the following recursion to get $x_{1:D}$.
$$x_i = u_i\exp{\alpha_i} + \mu_i$$
$$u_i \sim \mathcal{N}(0, 1)$$
The procedure of autoregressive sampling is a deterministic transformation of the underlying noise variates (sampled from $\mathcal{N}(0, \mathbb{I})$) into a new distribution, so autoregressive samples can actually be interpreted as a TransformedDistribution of the standard Normal!
Armed with this insight, we can stack multiple autoregressive transformations into a normalizing flow. The advantage of doing this is that we can change the ordering of variables $x_1,...x_D$ for each bijector in the flow, so that if one autoregressive factorization cannot model a distribution well (due to a poor choice of variable ordering), a subsequent layer might be able to do it.
The Masked Autoregressive Flow (MAF) bijector implements such a conditional-Gaussian autoregressive model. Here is a schematic of the forward pass for a single entry in a sample of the transformed distribution, $x_i$:
The gray unit $x_i$ is the unit we are trying to compute, and the blue units are the values it depends on. $\alpha_i$ and $\mu_i$ are scalars that are computed by passing $x_{1:i-1}$ through neural networks (magenta, orange circles). Even though the transformation is a mere scale-and-shift, the scale and shift can have complex dependencies on previous variables. For the first unit $x_1$, $\mu$ and $\alpha$ are usually set to learnable scalar variables that don’t depend on any $x$ or $u$.
More importantly, the transformation is designed this way so that computing the inverse $u = f^{-1}(x)$ does not require us to invert $f_\alpha$ or $f_\mu$. Because the transformation is parameterized as a scale-and-shift, we can recover the original noise variates by reversing the shift and scale: $u = (x-f_\mu(x))/\exp(f_\alpha(x))$. The forward and inverse pass of the bijector only depend on the forward evaluation of $f_\alpha(x)$ and $f_\mu(x)$, allowing us to use non-invertible functions like ReLU and non-square matrix multiplication in the neural networks $f_\mu$ and $f_\alpha$.
The inverse pass of the MAF model is used to evaluate density:
distribution.log_prob(bijector.inverse(x)) + bijector.inverse_log_det_jacobian(x))
Runtime Complexity and MADE
Autoregressive models and MAF can be trained “quickly” because all conditional likelihoods $p(x_1), p(x_2\,|\, x_1), ... p(x_D\,|\, x_{1:D-1}))$ can be evaluated simultaneously in a single pass of D threads, leveraging the batch parallelism of modern GPUs. We are operating under the assumption that parallelism, such as SIMD vectorization on CPUs/GPUs, has zero runtime overhead.
On the other hand, sampling autoregressive models is slow because you must wait for all previous $x_{1:i-1}$ to be computed before computing new $x_i$. The runtime complexity of generating a single sample is D sequential passes of a single thread, which fails to exploit processor parallelism.
Another issue: in the parallelizable inverse pass, should we use separate neural nets (with differently-sized inputs) for computing each $\alpha_i$ and $\mu_i$? That's inefficient, especially if we consider that learned representations between these D networks should be shared (as long as the autoregressive dependency is not violated). In the Masked Autoencoder for Distribution Estimation (MADE) paper, the authors propose a very nice solution: use a single neural net to output all values of $\alpha$ and $\mu$ simultaneously, but mask the weights so that the autoregressive property is preserved.
This trick makes it possible to recover all values of $u$ from all values of $x$ with a single pass through a single neural network (D inputs, D outputs). This is far more efficient than processing D neural networks simultaneously (D(D+1)/2 inputs, D outputs).
To summarize, MAF uses the MADE architecture as an efficiency trick for computing nonlinear parameters of shift-and-scale autoregressive transformations, and casts these efficient autoregressive models into the normalizing flows framework.
Inverse Autoregressive Flow (IAF)
In Inverse Autoregressive Flow, the nonlinear shift/scale statistics are computed using the previous noise variates $u_{1:i-1}$, instead of the data samples:
$$x_i = u_i\exp{\alpha_i} + \mu_i$$
$$\mu_i = f_{\mu_i}(u_{1:i-1})$$
$$\alpha_i = f_{\alpha_i}(u_{1:i-1})$$
The forward (sampling) pass of IAF is fast: all the $x_i$ can be computed in a single pass of $D$ threads working in parallel. IAF also uses MADE networks to implement this parallelism efficiently.
However, if we are given a new data point and asked to evaluate the density, we need to recover $u$ and this process is slow: first we recover $u_1 = (x-\mu_1) * \exp(-\alpha_1)$, then $u_i = (x-\mu_i(u_{1:i-1})) * \exp(-\alpha_i(u_{1:i-1}))$ sequentially. On the other hand, it’s trivial to track the (log) probability of samples generated by IAF, since we already know all of the $u$ values to begin with without having to invert from $x$.
The astute reader will notice that if you re-label the bottom row as x_1, .. x_D, and the top row as u_1, … u_D, this is exactly equivalent to the Inverse Pass of the MAF bijector! Likewise, the inverse of IAF is nothing more than the forward pass of MAF (with $x$ and $u$ swapped). Therefore in TensorFlow Distributions, MAF and IAF are actually implemented using the exact same Bijector class, and there is a convenient “Invert” feature for inverting Bijectors to swap their inverse and forward passes.
iaf_bijector = tfb.Invert(maf_bijector)
IAF and MAF make opposite computational tradeoffs - MAF trains quickly but samples slowly, while IAF trains slowly but samples quickly. For training neural networks, we usually demand way more throughput with density evaluation than sampling, so MAF is usually a more appropriate choice when learning distributions.
Parallel Wavenet
An obvious follow-up question is whether these two approaches can be combined to get the best of both worlds, i.e. fast training
andsampling.
The answer is yes! The much-publicized Parallel Wavenet by DeepMind does exactly this: an autoregressive model (MAF) is used to train a generative model efficiently, then an IAF model is trained to maximize the likelihood of its own samples under this teacher. Recall that with IAF, it is costly to compute density of external data points (such as those from the training set), but it can cheaply compute density of its
ownsamples by caching the noise variates $u_{1:D}$, thereby circumventing the need to call the inverse pass. Thus, we can train the “student” IAF model by minimizing the divergence between the student and teacher distributions.
This is an incredibly impactful application of normalizing flows research - the end result is a real-time audio synthesis model that is 20 times faster to sample, and is already deployed in real-world products like the Google Assistant.
NICE and Real-NVP
Finally, we consider is Real-NVP, which can be thought of as a special case of the IAF bijector.
In a NVP “coupling layer”, we fix an integer $0 < d < D$. Like IAF, $x_{d+1}$ is a shift-and-scale that depends on previous $u_{d}$ values. The difference is that we also force $x_{d+2}, x_{d+3}, … x_{D}$ to only depend on these $u_{d}$ values, so a single network pass can be used to produce $\alpha_{d+1:D}$ and $\mu_{d+1:D}$.
As for $x_1:d$ they are “pass-through” units that are set equivalently to $u_{1:d}$. Therefore, Real-NVP is also a special case of the MAF bijector (since $\alpha(u_{1:d}) = \alpha(x_{1:d})$).
Because the shift-and-scale statistics for the whole layer can be computed from either $x_{1:d}$ or $u_{1:d}$ in a single pass, NVP can perform forward and inverse computations in a single parallel pass (sampling and estimation are both fast). MADE is also not needed.
However, empirical studies suggest that Real-NVP tends to underperform MAF and IAF and my experience has been that NVP tends to fit my toy 2D datasets (e.g. SIGGRAPH dataset) more poorly when using the same number of layers. Real-NVP and IAF are nearly equivalent in the 2D case, except the first unit of IAF is still transformed via a scale-and-shift that does not depend on $u_1$, while Real-NVP leaves the first unit unmodified.
Real-NVP was a follow-up work to the NICE bijector, which is a shift-only variant that assumes $\alpha=0$. Because NICE does not scale the distribution, the ILDJ is actually constant!
Batch Normalization Bijector
In normalizing flows, batch norm is used in bijector.inverse during training, and the accumulated statistics are used to de-normalize data at “test time” (bijector.forward). Concretely, BatchNorm Bijectors are typically implemented as follows:
Inverse pass:
Compute the current mean and standard deviation of the data distribution $x$. Update running mean and standard deviation Batch normalize the data using current mean/std Use running mean and standard deviation to un-normalize the data distribution.
Thanks to TF Bijectors, this can be implemented with only a few lines of code:
The ILDJ can be derived easily by simply taking the log derivative of inverse function (consider the univariate case).
Code Example
Thanks to the efforts of Josh Dillon and the Google Bayesflow team, there is already a flexible implementation of MaskedAutoregressiveFlow Bijector that uses MADE networks to implement efficient recovery of $u$ for training.
I’ve created a complex 2D distribution, which is a point cloud in the shape of the letters “SIGGRAPH” using this blender script. We construct our dataset, bijector, and transformed distribution in a very similar fashion to the first tutorial, so I won’t repeat the code snippets here - you can find the Jupyter notebook here. This notebook can train a normalizing flow using MAF, IAF, Real-NVP with/without BatchNorm, for both the "Two Moons" and "SIGGRAPH" datasets.
One detail that’s easy to miss / introduce bugs on is that this doesn’t work at all unless you permute the ordering of variable at each flow. Otherwise, none of the layers’ autoregressive factorization will be learn structure of $p(x1 | x2)$. Fortunately, TensorFlow has a Permute bijector specially made for doing this.
Here’s the learned flow, along with the final result. It reminds me a lot of a taffy pulling machine.
Discussion
TensorFlow distributions makes normalizing flows easy to implement, and automatically accumulate all the Jacobians determinants in a chain for us in a way that is clean and highly readable. When deciding which Normalizing Flow to use, consider the design tradeoff between a fast forward pass and a fast inverse pass, as well as between an expressive flow and a speedy ILJD.
Although explicit-density models like normalizing flows are amenable to training via maximum likelihood, this is not the only way they can be used and are complementary to VAEs and GANs. It’s possible to use normalizing flow as a drop-in replacement for anywhere you would use a Gaussian, such as VAE priors and latent codes in GANs. For example, this paper use normalizing flows as flexible variational priors, and the TensorFlow distributions paper presents a VAE that uses a normalizing flow as a prior along with a PixelCNN decoder. Parallel Wavenet trains an IAF "student" model via KL divergence.
One of the most intriguing properties of normalizing flows is that they implement reversible computation (i.e. have a defined inverse of an expressive function). This means that if we want to perform a backprop pass, we can re-compute the forward activation values without having to store them in memory during the forward pass (potentially expensive for large graphs). In a setting where credit assignment may take place over very long time scales, we can use reversible computation to “recover” past decision states while keeping memory usage bounded. In fact, this idea was utilized in the RevNets paper, and was actually inspired by the invertibility of the NICE bijector. I’m reminded of the main character from the film Memento who is unable to store memories, so he uses invertible compute to remember things.
Thank you for reading.
Acknowledgements
I’m grateful to Dustin Tran, Luke Metz, Jonathan Shen, Katherine Lee, and Samy Bengio for proofreading this post.
References and Further Reading The content and outline of this blog post was heavily influenced by the Masked Autoregressive Flow for Density Estimation paper, which is very well-written and is more or less my primary source for understanding this topic. Give it a read! Some earlier work on NFs: https://math.nyu.edu/faculty/tabak/publications/Tabak-Turner.pdf and https://arxiv.org/pdf/1302.5125.pdf and https://arxiv.org/abs/1505.05770 Talk by Laurent Dinh & discussion by Twitter Cortex researchers. Some neat ideas and discussion here. Tutorial on Normalizing Flows using PyMC. There’s a body of work that I don’t fully understand yet, bridging Normalizing Flows to Langevin Flow and Hamiltonian Flow. As the number of Bijectors in a normalizing flow goes to infinity, one arrives at a Continuous-Time Flow, which apparently can express even richer transformations. |
If you have a list, e.g.
{1, 2, 3}
then you can extract the $k$th part using
Part (
list[[k]]):
In[1]:= {1, 2, 3}[[2]]Out[1]= 2
The problem is that if you provide a symbolic expression in the place of the list,
Part will try to decompose it:
In[1]:= list[[2]]Part::partd: Part specification list[[2]] is longer than depth of object.Out[1]= list[[2]]
That is,
the issue occurs when the overall expression is evaluated before the value of list is known. For example:
In[1]:= list[[2]] /. list -> {1, 2, 3}Part::partd: Part specification list[[2]] is longer than depth of object.Out[1]= 2
Although the final output is correct, this produces an annoying spurious error message.
It's even worse if
list is a compound expression, because then
Part will decompose it, unexpectedly changing its value. For example,
In[1]:= SinCos[x][[1]] /. SinCos -> (θ \[Function] {Sin@θ, Cos@θ})Out[1]= x
Example
I commonly run into this when plotting one dimension of nested functions and data structures.
Suppose I have a function that generates a list of functions according to some parameters:
functions[x_Real, a_, b_, c_] := {Sin@x, Cos@x, Sin[a*x], Cos[b*x], Sin[c*x]};
Now I want to plot the first, third, and fourth functions in the list of functions with parameters $1,2,3$; in this case, $y = \sin(x), \sin(1 \cdot x), \cos(2 \cdot x)$. The obvious way is:
Plot[functions[x, 1, 2, 3][[#]]& /@ {1,3,4} // Evaluate, {x, 0, 5}]
(The
Evaluate is needed to get
Plot to treat the values as separate functions, so that they will be styled differently.) And Mathematica will dutifully plot... $y = x, 2, 3$:
{Plot[functions[x,1,2,3][[#]]&/@{1,3,4}//Unevaluated,{x,0,5}], Plot[functions[x,1,2,3][[#]]&/@{1,3,4}//Evaluate,{x,0,5}]}
Question
How do I prevent
Part[] from trying to decompose symbolic expressions when it is evaluated?
I have a workaround below, but I'm interested in whether there's a better way to do it. Is there a standard built-in function that does what I'm looking for? |
Exercise \(\PageIndex{1}\)
Write each of the following as a complex number in standard form.
\[(4 + i) + (3 - 3i)\] \[5(2 - i) + i(3 - 2i)\] \[(4 + 2i)(5 - 3i)\] \[(2 + 3i)(1 + i) + (4 - 3i)\] Answer \((4 + i) + (3 - 3i) = 7 - 2i\) \(5(2 - i) + i(3 -2i) = 12 - 2i\) \((4 + 2i)(5 - 3i) = 26 - 2i\) \((2 + 3i)(1 + i) + (4 - 3i) = 3 + 2i\)
Exercise \(\PageIndex{2}\)
Use the quadratic formula to write the two solutions of each of the following quadratic equations in standard form.
\[x^{2} - 3x + 5 = 0\] \[2x^{2} = x - 7\] Answer
(a) \(x = \dfrac{3}{2} + \dfrac{\sqrt{11}}{2}i, x = \dfrac{3}{2} - \dfrac{\sqrt{11}}{2}i\).
Exercise \(\PageIndex{3}\)
For each of the following pairs of complex numbers \(w\) and \(z\), determine the sum \(w + z\) and illustrate the sum with a diagram.
\[w = 3 + 2i, z = 5 - 4i.\] \[w = 4i, z = -3 + 2i.\] \[w = 5, z = -7 + 2i.\] \[w = 6 - 3i, z = -1 + 7i.\] Answer \(w + z = 8 - 2i\) \(w + z = -3 + 6i\)
Exercise \(\PageIndex{4}\)
For each of the following complex numbers \(z\), determine \(\bar{z}\), \(|z|\), and \(z\bar{z}\).
\[z = 5 + 2i\] \[z = 3i\] \[z = 3 - 4i\] \[z = 7 + i\] Answer \(\bar{z} = 5 + 2i, |z| = \sqrt{29}, z\bar{z} = 29\) \(\bar{z} = -3i, |z| = 3, z\bar{z} = 9\)
Exercise \(\PageIndex{5}\)
Write each of the following quotients in standard form.
\[\dfrac{5 + i}{3 + 2i}\] \[\dfrac{3 + 3i}{i}\] \[\dfrac{i}{2 - i}\] \[\dfrac{4 + 2i}{1 - i}\] Answer \(\dfrac{5 + i}{3 + 2i} = \dfrac{17}{13} - \dfrac{7}{13}i\) \(\dfrac{3 + 3i}{i} = 3 - 3i\)
Exercise \(\PageIndex{6}\)
We know that \(i^{1} = i\) and \(i^{2} = -1\). We can then see that \[i^{3} = i^{2}\cdot i = (-1)i = -i.\] Show that \(i^{4} = 1\). Now determine \(i^{5}, i^{6}, i^{7}\), and \(i^{8}\). Note: Each power of \(i\) will equal \(1, -1, i\), or \(-i\). Notice that \(12 = 4\cdot 3 + 1\). We will use this to determine \(i^{13}\). \[i^{13} = i^{4\cdot 3 + 1} = i^{4\cdot 3}i^{1} = (i^{4})^{3}\cdot i\] So what is \(i^{13}\)? Using \(39 = 4\cdot 9 + 3\), determine \(i^{39}\). Determine \(i^{54}\).
Exercise \(\PageIndex{7}\)
Write the complex number \(i(2 + 2i)\) in standard form. Plot the complex numbers \(2 + 2i\) and \(i(2 + 2i)\) in the complex plane. What appears to be the angle between these two complex numbers? Repeat part (a) for the complex numbers \(2 - 3i\) and \(i(2 - 3i)\). Repeat part (a) for the complex numbers \(3i\) and \(i(3i)\). Describe what happens when the complex number \(a + bi\) is multiplied by the complex number \(i\). To make sense of solutions of quadratic equations that are not real, we introduce complex numbers. Although complex numbers arise naturally when solving quadratic equations, their introduction into mathematics came about from the problem of solving cubic equations.
Exercise \(\PageIndex{8}\)
Determine the polar (trigonometric) form of each of the following complex numbers.
\[3 +3i\] \[3 -3i\] \[-3 +3i\] \[5i\] \[4\sqrt{3} + 4i\] \[-4\sqrt{3} - 4i\] Answer \(3 + 3i = \sqrt{18}(\cos(\dfrac{\pi}{4}) + i\sin(\dfrac{\pi}{4}))\) \(4\sqrt{3} + 4i = 8(\cos(\dfrac{\pi}{6}) + i\sin(\dfrac{\pi}{6}))\)
Exercise \(\PageIndex{9}\)
In each of the following a complex number \(z\) is given. In each case, determine real numbers \(a\) and \(b\) so that \(z = a + bi\). If it is not possible to determine exact values for \(a\) and \(b\), determine the values of \(a\) and \(b\) correct to four decimal places.
\[z = 5(\cos(\dfrac{\pi}{2}) + i\sin(\dfrac{\pi}{2}))\] \[z = 2.5(\cos(\dfrac{\pi}{4}) + i\sin(\dfrac{\pi}{4}))\] \[z = 2.5(\cos(\dfrac{3\pi}{4}) + i\sin(\dfrac{3\pi}{4}))\] \[z = 3(\cos(\dfrac{7\pi}{6}) + i\sin(\dfrac{7\pi}{6}))\] \[z = 8(\cos(\dfrac{7\pi}{10}) + i\sin(\dfrac{7\pi}{10}))\] Answer \(5\cos(\dfrac{\pi}{2}) + i\sin(\dfrac{\pi}{2}) = 5i\) \(2.5(\cos(\dfrac{\pi}{4}) + i\sin(\dfrac{\pi}{4})) = 1.25\sqrt{2} + 1.25\sqrt{2}i\)
Exercise \(\PageIndex{10}\)
For each of the following, write the product \(wz\) in polar (trigonometric form). When it is possible, write the product in form \(a + bi\), where \(a\) and \(b\) are real numbers and do not involve a trigonometric function.
\[w = 5(\cos(\dfrac{\pi}{12}) + i\sin(\dfrac{\pi}{12})), z = 2(\cos(\dfrac{5\pi}{12}) + i\sin(\dfrac{5\pi}{12}))\] \[w = 2.3(\cos(\dfrac{\pi}{3}) + i\sin(\dfrac{\pi}{3})), z = 2(\cos(\dfrac{5\pi}{4}) + i\sin(\dfrac{5\pi}{4}))\] \[w = 2(\cos(\dfrac{7\pi}{10}) + i\sin(\dfrac{7\pi}{10})), z = 2(\cos(\dfrac{2\pi}{5}) + i\sin(\dfrac{2\pi}{5}))\] \[w = (\cos(24^\circ) + i\sin(24^\circ)), z = 2(\cos(33^\circ) + i\sin(33^\circ))\] \[w = 2(\cos(72^\circ) + i\sin(72^\circ)), z = 2(\cos(78^\circ) + i\sin(78^\circ))\] Answer \(wz = 10(\cos(\dfrac{6\pi}{12}) + i\sin(\dfrac{6\pi}{12})) = 10i\) \(wz = 6.9(\cos(\dfrac{19\pi}{12}) + i\sin(\dfrac{19\pi}{12}))\)
Exercise \(\PageIndex{11}\)
For the complex numbers in Exercise \(\PageIndex{10}\), write the quotient \(\dfrac{w}{z}\) in polar (trigonometric form). When it is possible, write the quotient in form \(a + bi\) , where \(a\) and \(b\) are real numbers and do not involve a trigonometric function.Add text here. For the automatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page.
Answer
(a) \(\dfrac{w}{z} = \dfrac{5}{2}(\cos(\dfrac{-4\pi}{12}) + i\sin(\dfrac{-4\pi}{12})) = \dfrac{5}{4} - \dfrac{5\sqrt{3}}{4}i\)
(b) \(\dfrac{w}{z} = \dfrac{23}{30}(\cos(\dfrac{-11\pi}{12}) + i\sin(\dfrac{-11\pi}{12}))\)
Exercise \(\PageIndex{12}\)
Write the complex number \(i\) in polar form. Let \(z = r(\cos(\theta) + i\sin(\theta))\). Determine the product \(i\cdot z\) in polar form. Use this to explain why the complex number \(i\cdot z\) and \(z\) will be perpendicular when both are plotted in the complex plane. Multiplication of complex numbers is more complicated than addition of complex numbers. To better understand the product of complex numbers, we first investigate the trigonometric (or polar) form of a complex number. This trigonometric form connects algebra to trigonometry and will be useful for quickly and easily finding powers and roots of complex numbers.
Exercise \(\PageIndex{13}\)
Use DeMoivre’s Theorem to determine each of the following powers of a complex number. Write the answer in the form \(a + bi\), where \(a\) and \(b\) are real numbers and do not involve the use of a trigonometric function.
\[(2 + 2i)^{6}\] \[(\sqrt{3} + i)^{8}\] \[(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i)^{3}\] \[2(\cos(\dfrac{\pi}{15}) + i\sin(\dfrac{\pi}{15}))^{10}\] \[(1 + i\sqrt{3})^{-4}\] \[(-3 + 3i)^{-3}\] Answer
(a) \((2 + 2i)^{6} = [\sqrt{8}(\cos(\dfrac{\pi}{4}) + i\sin(\dfrac{\pi}{4}))]^{6} = 512i\)
(b) \((\sqrt{3} + i)^{8} = [2(\cos(\dfrac{\pi}{6}) + i\sin(\dfrac{\pi}{6}))]^{8} = -128 - 128\sqrt{3}i\)
Exercise \(\PageIndex{14}\)
In each of the following, determine the indicated roots of the given complex number. When it is possible, write the roots in the form a C bi , where a andb are real numbers and do not involve the use of a trigonometric function. Otherwise, leave the roots in polar form.
The two square roots of \(16i\). The two square roots of \(2 + 2i\sqrt{3}\). The three cube roots of \(5(\cos(\dfrac{3\pi}{4}) + i\sin(\dfrac{3\pi}{4}))\). The five fifth roots of unity. The four fourth roots of \((\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i)\). The three cube roots of \(1 + \sqrt{3}i\). Answer
(a) Write \(16i = 16(\cos(\dfrac{\pi}{2}) + i\sin(\dfrac{\pi}{2}))\). The two square roots of \(16i\) are \[4(\cos(\dfrac{\pi}{4}) + i\sin(\dfrac{\pi}{4})) = 2\sqrt{2} + 2i\sqrt{2}\]
\[4(\cos(\dfrac{5\pi}{4}) + i\sin(\dfrac{5\pi}{4})) = -2\sqrt{2} - 2i\sqrt{2}\]
(c) The three cube roots of \(5(\cos(\dfrac{3\pi}{4}) + i\sin(\dfrac{3\pi}{4}))\) are
\[\sqrt[3]{5}(\cos(\dfrac{\pi}{4}) + i\sin(\dfrac{\pi}{4})) = \sqrt[3]{5}(\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i)\]
\[\sqrt[3]{5}(\cos(\dfrac{11\pi}{12}) + i\sin(\dfrac{11\pi}{12}))\] \[\sqrt[3]{5}(\cos(\dfrac{19\pi}{12}) + i\sin(\dfrac{19\pi}{12}))\] The trigonometric form of a complex number provides a relatively quick and easy way to compute products of complex numbers. As a consequence, we will be able to quickly calculate powers of complex numbers, and even roots of complex numbers.
Exercise \(\PageIndex{15}\)
Plot the following points with the specified polar coordinates.
Figure \(\PageIndex{1}\)
\[(7, \dfrac{\pi}{6})\] \[(3, \dfrac{3\pi}{4})\] \[(2, \dfrac{-\pi}{3})\]
\[(3, \dfrac{7\pi}{4})\] \[(5, \dfrac{-\pi}{4})\] \[(4, \dfrac{11\pi}{4})\]
\[(6, \dfrac{11\pi}{6})\] \[(-3, \dfrac{2\pi}{3})\] \[(-5, \dfrac{5\pi}{6})\]
Answer
Figure \(\PageIndex{2}\)
Exercise \(\PageIndex{16}\)
For each of the following points in polar coordinates, determine three different representations in polar coordinates for the point. Use a positive value for the radial distance \(r\) for two of the representations and a negative value for the radial distance \(r\) for the other representation.
\[(5, 30^\circ)\] \[(4, 100^\circ)\] \[(-2, 50^\circ)\] \[(7, -60^\circ)\] Answer
(a) Some correct answers are: \((5, 390^\circ)\), \((5, -330^\circ)\), and \((-5, 210^\circ)\).
(b) Some correct answers are: \((4, 460^\circ)\), \((4, -260^\circ)\), and \((-4, 280^\circ)\).
Exercise \(\PageIndex{17}\)
For each of the following points in polar coordinates, determine three different representations in polar coordinates for the point. Use a positive value for the radial distance \(r\) for two of the representations and a negative value for the radial distance \(r\) for the other representation.
Note: The angles are measured in radians. \[(5, \dfrac{\pi}{6})\] \[(4, \dfrac{5\pi}{9})\] \[(-2, \dfrac{5\pi}{18})\] \[(7, -\dfrac{\pi}{3})\] Answer Some correct answers are: \((5, \dfrac{13\pi}{6})\), \((5, -\dfrac{11\pi}{6})\), and \((-5, \dfrac{7\pi}{6})\). Some correct answers are: \((4, \dfrac{23\pi}{9})\), \((4, -\dfrac{13\pi}{9})\), and \((-4, \dfrac{14\pi}{9})\).
Exercise \(\PageIndex{18}\)
Determine rectangular coordinates for each of the following points in polar coordinates:
\[(10, \dfrac{2\pi}{3})\] \[(8, \dfrac{7\pi}{6})\] \[(-5, \dfrac{5\pi}{4})\] \[(10, -\dfrac{2\pi}{3})\] \[(3, \dfrac{5\pi}{3})\] \[(6, -\dfrac{\pi}{6})\] Answer
1. \((-5, 5\sqrt{3})\)
3. \((\dfrac{5\sqrt{2}}{2}, \dfrac{5\sqrt{2}}{2})\)
Exercise \(\PageIndex{19}\)
Determine polar coordinates for each of the following points in rectangular coordinates. Use a positive radial distance \(r\) and a polar angle with \(0 \leq \theta < 2\pi\). When necessary, use an inverse trigonometric function and round the angle (in radians) to the nearest thousandth.
\[(\dfrac{-5\sqrt{3}}{2}, \dfrac{5}{2})\] \[(3, 5)\] \[(\sqrt{2}, -\sqrt{2})\] \[(-3, -4)\] Answer
1. \((5, \dfrac{5\pi}{6})\)
2. \((\sqrt{34}, \tan^{-1}(\dfrac{5}{3})) \approx (\sqrt{34}, 1.030)\)
Exercise \(\PageIndex{20}\)
Convert each of the following polar equations into a rectangular equation. If possible, write the rectangular equation with \(y\) as a function of \(x\).
\[r = 5\] \[\theta = \dfrac{\pi}{3}\] \[r = 8\cos(\theta)\] \[r = 1 - \sin(\theta)\] \[r^{2}\sin(2\theta) = 1\] \[r = 1 - 2\cos(\theta)\] \[r = \dfrac{3}{\sin(\theta) + 4\cos(\theta)}\] Answer
1. \(x^{2} + y^{2} = 25\)
2. \(y = \dfrac{\sqrt{3}}{3}x\)
4. \(x^{2} + y^{2} = \sqrt{x^{2} + y^{2}} - y\)
Exercise \(\PageIndex{21}\)
Convert each of the following rectangular equations into a polar equation. If possible, write the polar equation with \(r\) as a function of \(\theta\).
\[x^{2} + y^{2} = 36\] \[y = 4\] \[x = 7\] \[x^{2} - 6x + y^{2} = 0\] \[x + y = 4\] \[y = x^{2}\] Answer
2. \(r\sin(\theta) = 4\) or \(r = \dfrac{4}{\sin(\theta)}\)
5. \(r\cos(\theta) + r\sin(\theta) = 4\) or \(r = \dfrac{4}{\cos(\theta) + \sin(\theta)}\)
Exercise \(\PageIndex{8}\)
Let a be a positive real number.
Convert the polar equation \(r = 2a\sin(\theta)\) to rectangular and then explain why the graph of this equation is a circle. What is the radius of the circle and what is the center of the circle in rectangular coordinates?
Convert the polar equation \(r = 2a\cos(\theta)\) to rectangular and then explain why the graph of this equation is a circle. What is the radius of the circle and what is the center of the circle in rectangular coordinates?
In our study of trigonometry so far, whenever we graphed an equation or located a point in the plane, we have used rectangular (or Cartesian) coordinates. The use of this type of coordinate system revolutionized mathematics since it provided the first systematic link between geometry and algebra. Even though the rectangular coordinate system is very important, there are other methods of locating points in the plane. We will study one such system in this section. |
In school, I have learnt to plot simple graphs such as $y=x^2$ followed by $y=x^3$.
A grade or two later, I learnt to plot other interesting graphs such as $y=1/x$, $y=\ln x$, $y=e^x$. I have also recently learnt about trigonometric graphs and circle equations. In the internet, I have seen users posting graphs of different shapes like a heart-shaped graph and a Batman logo-shaped graph. I am sure there are numerous more graphs that I have yet to see. Seeing that the graphs can be shaped into shapes like the Batman logo and a heart brings me to my question: Is it possible to plot a graph of any shape regardless of its complexity? Perhaps, shaped into an outline of a person or a landmark? Why or why not?
In school, I have learnt to plot simple graphs such as $y=x^2$ followed by $y=x^3$.
Simple answer: Yes--simply draw your person or landmark and then superimpose the $xy$-plane on top and suddenly you have all of the points (i.e., coordinates) that need to be filled in to create a plot of your graph.
Now,
how you come up with a good (read: not complex) mathematical description of these coordinates (e.g., using a function) is another issue entirely. Depending on the complexity of what you are drawing, you most likely won't get something pretty. For example, consider drawing the fictional character Donkey Kong:
The picture above was generated by Wolfram|Alpha. How complicated is the curve? Well, here you go:
That's pretty horrible. So yes, you can certainly plot whatever you want, but describing your plot effectively using whatever kind of function, parametric equations, etc., may not be very easy or nice in the end.
Added: Given the unexpected popularity of this post (both question and answer(s)), I thought I might add something that some may find helpful or useful. In 2012, I wrote an article entitled Bézier Curves with a Romantic Twist that appeared in the Math Horizons periodical. This piece largely dealt with using lower order Bézier curves (linear and cubic) to construct letters for a person's name on a graphing calculator; in the context of this post, the problem was to plot a graph of letters in the alphabet (along with a heart and parametrically-defined sequence). If you read the article, you will see that the math behind constructing such letters is not all too complicated--my reason for providing the Donkey Kong example was largely to show just how complicated it can be to effectively sketch something with equations.
But sketching letters and the like (as opposed to much more complicated representations like Captain Falcon, Pikachu, Sonic, etc.) is quite manageable. In fact, the avatar for my username even uses a simple construction to spell the word MATH:
For those interested, I will provide the equations I used for the M, the sequence, and the heart (as entered on a TI-89 calculator):
$$ \mathrm{M}= \begin{cases} xt1 & = & (1-t)10+11.25t\\ yt1 & = & (1-t)5+12.75t\\ xt2 & = & (1-t)11.25+12.5t\\ yt2 & = & (1-t)12.75+8.875t\\ xt3 & = & (1-t)12.5+13.75t\\ yt3 & = & (1-t)8.875+12.75t\\ xt4 & = & (1-t)13.75+15t\\ yt4 & = & (1-t)12.75+5t\\ \end{cases} $$
$$ \mathrm{Heart} = \begin{cases} xt5 & = & 4\sin(t)^3\\ yt5 & = & \frac{1}{2}\bigl(13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)\bigr)+34.2 \end{cases} $$
$$ \mathrm{Sequence}= \begin{cases} xt6 & = & t\\ yt6 & = & (3^t+5^t)^{1/t} \end{cases} $$
Of course, the A, T, and H are all similar to the M in that they are drawn using linear Bézier curves. A more interesting letter is something like C or S or even D or B (these will all use at least cubic Bézier curves).
Is it possible to plot a graph of any shape regardless of its complexity?
Definitely not, simply because it is possible to define shapes that are so complex they cannot be computed. More precisely, there exist uncomputable functions $f$, such that no program can compute their graphs $\{(x,y)\mid y=f(x)\}.$
In fact,
most functions that map $N \to N$ are uncomputable, in the sense that uncountably many are uncomputable, whereas only countably many are computable (there being uncountably many such functions altogether).
An example would be $f(x)$ defined as the number of $x$-state Busy-Beaver-class Turing machines.
NB: This is contary to all $5$ of the other answers -- perhaps because they assume the graph is supposed to be finite (and hence computable).
Yes. These graphs are often not functions, but can be written as several piecewise functions (which I imagine you have covered) with domain restrictions.
When I was a junior in high school, I graphed the words "Homecoming Date" and asked a girl to be my $f(x)$. I defined several functions as $f(x)$ to ensure that my wording was acceptable. She didn't seem to notice and accepted, haha.
But yes, your graphing possibilities are endless. Try some out yourself!
The answer to your question is yes, you can plot any graph like the one you're interested in. A way to do this (which I'm sure won't be completely satisfying) is to simply have the complete list of all co-ordinate points that you want to graph, and define your mapping m so that $m(x) = y$ whenever $(x,y)$ is a co-ordinate point. (Also note here that $x$ might map to more than one $y$ - for example in your heart shaped example).
What it seems like you'd like however is a 'nice looking' map like $y = e^x-4x+1$, or something like that. In order to do something like this, many methods exist. I would suggest looking up interpolation for a start.
It would be a good idea to make clear what a graph
is.
First, a
function is simply an “answering machine” where you put something in, and get something else out. It's not necessarily something you could write as a mathematical formula in the usual way. Only, it so happens that this is possible for many useful functions which simply map numbers to numbers – but that's just a very specific special case. For instance, a function could map manifolds to manifolds: it could map a sphere to a torus, a torus to a double torus, etc.. Or, perhaps an even better example would be the function that maps people to their parents. To define a function you only need to define, for each input argument, what the result should be.
In particular, given any shape, you can
define a function that takes an x-coordinate and yields the highest corresponding y-coordinate which is present in the shape †. The graph of this function is then, again by definition, the upper edge of your original shape.
It's relatively easy to generalise this to give not only the upper edge but a whole drawing:
if you can draw the shape with a pen, then this defines a function from time $t$ to position-of-the-pen-tip $p$. This can be used for a parametric plot, and by the implicit function theorem that is equivalent to a couple of graphs.
So, the answer to your question is trivially
yes, because drawing is nothing else but plotting a function.
What's a more interesting question is whether any shape can be the plot of a function that's reasonably simple to write down / store as a definition. Daniel W. Farlow gave an example for how the usual maths-formula way is not really well-suited for this in general; however there are more optimised ways to do it. In particular, computer graphics file formats are essentially nothing but clever conventions of how to efficiently write (in binary form) a mathematical description of a picture/shape. Some such formats, the vector graphics kind, can actually give an exact description of shapes, however for more complicated stuff like photographs an exact description is not feasible
‡; instead you approximate the image by something that looks almost the same. And finding such approximations is a pretty interesting topic mathematically, mainly in the branch of functional analysis. † That would actually be a partial function, because not for all $x$ can any point in the shape at all be found. ‡ In fact its not possible, physically: a picture is a collection of measured brightnesses. No physical measurement is exact, it always has some uncertainty. |
I'm trying to calculate confidence intervals for a neural network (rather than prediction intervals). I'm following this paper, which treats them in the same framework as any parametric (parameter-involving?) nonlinear model (which a neural net basically is). Calculating these CI's involves computing the Jacobian -- the matrix of partial derivatives of $\hat y$ with respect to the parameters.
In the one-outcome regression case, a neural net with two hidden layers (and no biases, WLOG) is $$ y = \sigma(\sigma(\mathbf{X\alpha)\beta)\gamma} + \epsilon $$ where
$y$ is the $N\times 1$ outcome $\mathbf{X}$ is the $N\times P_x$ data $\alpha$ is the $P_x \times P_\beta$ first set of weights $\beta$ is the $P_\beta \times P_\gamma$ second set of weights $\gamma$ is the $P_\gamma \times 1$ final set of weights, linking to $y$ $\epsilon$ is the mean-zero error $\sigma()$ is the activation function
The chain rule gives me the following partial derivatives of the fitted model: $$ \begin{array}{rcl} \displaystyle\frac{\partial\hat y}{\partial \hat\gamma} & = & \sigma(\sigma(\mathbf{X\alpha})\beta) \\ \displaystyle\frac{\partial\hat y}{\partial \hat\beta} & = & \sigma'(\sigma(\mathbf{X\alpha})\beta)\gamma\sigma(\mathbf{X}\alpha) \\ \displaystyle\frac{\partial\hat y}{\partial \hat\alpha} & = & \sigma'(\sigma(\mathbf{X\alpha})\beta)\gamma\sigma'(\mathbf{X}\alpha)\beta \mathbf{X} \end{array} $$
Now, the matrix dimensions on the above expressions are not conformable, which suggests that vectorizing the Jacobian calculation is not possible. Is this true?
Next, if I need to calculate the partial derivatives parameter-by-parameter, what is the appropriate way to deal with the network structure when evaluating each derivative?
For example, $\alpha_1$ links the $\mathbf{X}_1$ to $\sigma(\mathbf{X\alpha})_1$. That node in turn links to several different upper nodes. Let's say that there are $g$ hidden units in the topmost layer. Should the Jacobian column for $\alpha_1$ include $\displaystyle\sum_g \sigma(\mathbf{X\alpha})_1 \beta_g$ (i.e.: row-wise sum)? Or would the combination of the $\beta$'s take some other form than a sum?
Likewise, $\hat\alpha_1$ influences $\hat y$ through all of the top layer hidden-units. So $\hat\alpha_1$'s Jacobian column should also include $\sigma'(\sigma(\mathbf{X\alpha})\beta)\gamma$, treated as a matrix, which evaluates to a $N \times P_\gamma \times P_\gamma \times 1 = N \times 1$ matrix?
$\hat\alpha_1$'s Jacobian column would thus be $$ \frac{\partial\hat y}{\partial \hat\alpha_1} = \sigma'(\sigma(\mathbf{X\alpha})\beta)\gamma\circ\displaystyle\sum_g\sigma'(\mathbf{X}_1\alpha_1)\beta_g \circ\mathbf{X}_1 $$
Is this correct?
Is there a more efficient way to do this? Can a general statement be made about how much better the analytical Jacobian would be than one calculated by numerical approximation, for example in the
numDeriv package in R? |
After the excellent post by JD Long in this thread, I looked for a simple example, and the R code necessary to produce the PCA and then go back to the original data. It gave me some first-hand geometric intuition, and I want to share what I got. The dataset and code can be directly copied and pasted into R form Github.
I used a data set that I found online on semiconductors here, and I trimmed it to just two dimensions - "atomic number" and "melting point" - to facilitate plotting.
As a caveat the idea is purely illustrative of the computational process: PCA is used to reduce more than two variables to a few derived principal components, or to identify collinearity also in the case of multiple features. So it wouldn't find much application in the case of two variables, nor would there be a need to calculate eigenvectors of correlation matrices as pointed out by @amoeba.
Further, I truncated the observations from 44 to 15 to ease the task of tracking individual points. The ultimate result was a skeleton data frame (
dat1):
compounds atomic.no melting.point
AIN 10 498.0
AIP 14 625.0
AIAs 23 1011.5
... ... ...
The "compounds" column indicate the chemical constitution of the semiconductor, and plays the role of row name.
This can be reproduced as follows (ready to copy and paste on R console):
dat <- read.csv(url("http://rinterested.github.io/datasets/semiconductors"))
colnames(dat)[2] <- "atomic.no"
dat1 <- subset(dat[1:15,1:3])
row.names(dat1) <- dat1$compounds
dat1 <- dat1[,-1]
The data were then scaled:
X <- apply(dat1, 2, function(x) (x - mean(x)) / sd(x))
# This centers data points around the mean and standardizes by dividing by SD.
# It is the equivalent to `X <- scale(dat1, center = T, scale = T)`
The linear algebra steps followed:
C <- cov(X) # Covariance matrix (centered data)
$\begin{bmatrix} &\text{at_no}&\text{melt_p}\\\text{at_no}&1&0.296\\\text{melt_p}&0.296&1\end{bmatrix}$
The correlation function
cor(dat1) gives the same output on the non-scaled data as the function
cov(X) on the scaled data.
lambda <- eigen(C)$values # Eigenvalues
lambda_matrix <- diag(2)*eigen(C)$values # Eigenvalues matrix
$\begin{bmatrix} &\color{purple}{\lambda_{\text{PC1}}}&\color{orange}{\lambda_{\text{PC2}}}\\&1.296422& 0\\&0&0.7035783\end{bmatrix}$
e_vectors <- eigen(C)$vectors # Eigenvectors
$\frac{1}{\sqrt{2}}\begin{bmatrix} &\color{purple}{\text{PC1}}&\color{orange}{\text{PC2}}\\&1&\,\,\,\,\,1\\&1&-1\end{bmatrix}$
Since the first eigenvector initially returns as $\sim \small [-0.7,-0.7]$ we choose to change it to $\small [0.7, 0.7]$ to make it consistent with built-in formulas through:
e_vectors[,1] = - e_vectors[,1]; colnames(e_vectors) <- c("PC1","PC2")
The resultant eigenvalues were $\small 1.2964217$ and $\small 0.7035783$. Under less minimalistic conditions, this result would have helped decide which eigenvectors to include (largest eigenvalues). For instance, the relative contribution of the first eigenvalue is $\small 64.8\%$:
eigen(C)$values[1]/sum(eigen(C)$values) * 100, meaning that it accounts for $\sim\small 65\%$ of the variability in the data. The variability in the direction of the second eigenvector is $35.2\%$. This is typically shown on a scree plot depicting the value of the eigenvalues:
We'll include both eigenvectors given the small size of this toy data set example, understanding that excluding one of the eigenvectors would result in dimensionality reduction - the idea behind PCA.
The
score matrix was determined as the matrix multiplication of the scaled data (
X) by the
matrix of eigenvectors (or "rotations"):
score_matrix <- X %*% e_vectors
# Identical to the often found operation: t(t(e_vectors) %*% t(X))
The concept entails a linear
combination of each entry (row / subject / observation / superconductor in this case) of the centered (and in this case scaled) data weighted by the rows of each eigenvector, so that in each of the final columns of the score matrix, we'll find a contribution from each variable (column) of the data (the entire
X), BUT only the corresponding eigenvector will have taken part in the computation (i.e. the first eigenvector $[0.7, 0.7]^{T}$ will contribute to $\text{PC}\,1$ (Principal Component 1) and $[0.7, -0.7]^{T}$ to $\text{PC}\,2$, as in:
Therefore each eigenvector will influence each variable differently, and this will be reflected in the "loadings" of the PCA. In our case, the negative sign in the second component of the second eigenvector $[0.7, - 0.7]$ will change the sign of the melting point values in the linear combinations that produce PC2, whereas the effect of the first eigenvector will be consistently positive:
The eigenvectors are scaled to $1$:
> apply(e_vectors, 2, function(x) sum(x^2))
PC1 PC2
1 1
whereas the (
loadings) are the eigenvectors scaled by the eigenvalues (despite the confusing terminology in the in-built R functions displayed below). Consequently, the loadings can be calculated as:
> e_vectors %*% lambda_matrix
[,1] [,2]
[1,] 0.9167086 0.497505
[2,] 0.9167086 -0.497505
> prcomp(X)$rotation %*% diag(princomp(covmat = C)$sd^2)
[,1] [,2]
atomic.no 0.9167086 0.497505
melting.point 0.9167086 -0.497505
It is interesting to note that the rotated data cloud (the score plot) will have variance along each component (PC) equal to the eigenvalues:
> apply(score_matrix, 2, function(x) var(x))
PC1 PC2
53829.7896 110.8414
> lambda
[1] 53829.7896 110.8414
Utilizing the built-in functions the results can be replicated:
# For the SCORE MATRIX:
prcomp(X)$x
# or...
princomp(X)$scores # The signs of the PC 1 column will be reversed.
# and for EIGENVECTOR MATRIX:
prcomp(X)$rotation
# or...
princomp(X)$loadings
# and for EIGENVALUES:
prcomp(X)$sdev^2
# or...
princomp(covmat = C)$sd^2
Alternatively, the singular value decomposition ($\text{U}\Sigma \text{V}^\text{T}$) method can be applied to manually calculate PCA; in fact, this is the method used in
prcomp(). The steps can be spelled out as:
svd_scaled_dat <-svd(scale(dat1))
eigen_vectors <- svd_scaled_dat$v
eigen_values <- (svd_scaled_dat$d/sqrt(nrow(dat1) - 1))^2
scores<-scale(dat1) %*% eigen_vectors
The result is shown below, with first, the distances from the individual points to the first eigenvector, and on a second plot, the orthogonal distances to the second eigenvector:
If instead we plotted the values of the score matrix (PC1 and PC2) - no longer "melting.point" and "atomic.no", but really a change of basis of the point coordinates with the eigenvectors as basis, these distances would be preserved, but would naturally become perpendicular to the xy axis:
The trick was now to
recover the original data. The points had been transformed through a simple matrix multiplication by the eigenvectors. Now the data was rotated back by multiplying by the inverse of the matrix of eigenvectors with a resultant marked change in the location of the data points. For instance, notice the change in pink dot "GaN" in the left upper quadrant (black circle in the left plot, below), returning to its initial position in the left lower quadrant (black circle in the right plot, below).
Now we finally had the original data restored in this "de-rotated" matrix:
Beyond the change of coordinates of rotation of the data in PCA, the results must be interpreted, and this process tends to involve a
biplot, on which the data points are plotted with respect to the new eigenvector coordinates, and the original variables are now superimposed as vectors. It is interesting to note the equivalence in the position of the points between the plots in the second row of rotation graphs above ("Scores with xy Axis = Eigenvectors") (to the left in the plots that follow), and the
biplot (to the right):
The superimposition of the original variables as red arrows offers a path to the interpretation of
PC1 as a vector in the direction (or with a positive correlation) with both
atomic no and
melting point; and of
PC2 as a component along increasing values of
atomic no but negatively correlated with
melting point, consistent with the values of the eigenvectors:
PCA$rotation
PC1 PC2
atomic.no 0.7071068 0.7071068
melting.point 0.7071068 -0.7071068
This interactive tutorial by Victor Powell gives immediate feedback as to the changes in the eigenvectors as the data cloud is modified. |
It is true that K-means clustering and PCA appear to have very different goals and at first sight do not seem to be related. However, as explained in the Ding & He 2004 paper K-means Clustering via Principal Component Analysis, there is a deep connection between them.
The intuition is that PCA seeks to represent all $n$ data vectors as linear combinations of a small number of eigenvectors, and does it to minimize the mean-squared reconstruction error. In contrast, K-means seeks to represent all $n$ data vectors via small number of cluster centroids, i.e. to represent them as linear combinations of a small number of cluster centroid vectors where linear combination weights must be all zero except for the single $1$. This is also done to minimize the mean-squared reconstruction error.
So K-means can be seen as a super-sparse PCA.
What Ding & He paper does, it to make this connection more precise.
Unfortunately, the Ding & He paper contains some sloppy formulations (at best) and can easily be misunderstood. E.g. it might seem that Ding & He claim to have proved that cluster centroids of K-means clustering solution lie in the $(K-1)$-dimensional PCA subspace:
Theorem 3.3. Cluster centroid subspace is spanned by the first
$K-1$ principal directions [...].
For $K=2$ this would imply that projections on PC1 axis will necessarily be negative for one cluster and positive for another cluster, i.e. PC2 axis will separate clusters perfectly.
This is either a mistake or some sloppy writing; in any case, taken literally, this particular claim is false.
Let's start with looking at some toy examples in 2D for $K=2$. I generated some samples from the two normal distributions with the same covariance matrix but varying means. I then ran both K-means and PCA. The following figure shows the scatter plot of the data above, and the same data colored according to the K-means solution below. I also show the first principal direction as a black line and class centroids found by K-means with black crosses. PC2 axis is shown with the dashed black line. K-means was repeated $100$ times with random seeds to ensure convergence to the global optimum.
One can clearly see that even though the class centroids tend to be pretty close to the first PC direction, they do
not fall on it exactly. Moreover, even though PC2 axis separates clusters perfectly in subplots 1 and 4, there is a couple of points on the wrong side of it in subplots 2 and 3.
So the agreement between K-means and PCA is quite good, but it is not exact.
So what did Ding & He prove? For simplicity, I will consider only $K=2$ case. Let the number of points assigned to each cluster be $n_1$ and $n_2$ and the total number of points $n=n_1+n_2$. Following Ding & He, let's define
cluster indicator vector $\mathbf q\in\mathbb R^n$ as follows: $q_i = \sqrt{n_2/nn_1}$ if $i$-th points belongs to cluster 1 and $q_i = -\sqrt{n_1/nn_2}$ if it belongs to cluster 2. Cluster indicator vector has unit length $\|\mathbf q\| = 1$ and is "centered", i.e. its elements sum to zero $\sum q_i = 0$.
Ding & He show that K-means loss function $\sum_k \sum_i (\mathbf x_i - \boldsymbol \mu_k)^2$ (that K-means algorithm minimizes) can be equivalently rewritten as $-\mathbf q^\top \mathbf G \mathbf q$, where $\mathbf G$ is the $n\times n$ Gram matrix of scalar products between all points: $\mathbf G = \mathbf X_c^\top \mathbf X_c$, where $\mathbf X$ is the $n\times 2$ data matrix and $\mathbf X_c$ is the centered data matrix.
(Note: I am using notation and terminology that slightly differs from their paper but that I find clearer).
So the K-means solution $\mathbf q$ is a centered unit vector maximizing $\mathbf q^\top \mathbf G \mathbf q$. It is easy to show that the first principal component (when normalized to have unit sum of squares) is the leading eigenvector of the Gram matrix, i.e. it is also a centered unit vector $\mathbf p$ maximizing $\mathbf p^\top \mathbf G \mathbf p$. The only difference is that $\mathbf q$ is additionally constrained to have only two different values whereas $\mathbf p$ does not have this constraint.
In other words, K-means and PCA maximize the same objective function, with the only difference being that K-means has additional "categorical" constraint.
It stands to reason that most of the times the K-means (constrained) and PCA (unconstrained) solutions will be pretty to close to each other, as we saw above in the simulation, but one should not expect them to be identical. Taking $\mathbf p$ and setting all its negative elements to be equal to $-\sqrt{n_1/nn_2}$ and all its positive elements to $\sqrt{n_2/nn_1}$ will generally
not give exactly $\mathbf q$.
Ding & He seem to understand this well because they formulate their theorem as follows:
Theorem 2.2. For K-means clustering where $K= 2$, the continuous solution of the cluster indicator vector is the [first] principal component
Note that words "continuous solution". After proving this theorem they additionally comment that PCA can be used to initialize K-means iterations which makes total sense given that we expect $\mathbf q$ to be close to $\mathbf p$. But one still needs to perform the iterations, because they are not identical.
However, Ding & He then go on to develop a more general treatment for $K>2$ and end up formulating Theorem 3.3 as
Theorem 3.3. Cluster centroid subspace is spanned by the first
$K-1$ principal directions [...].
I did not go through the math of Section 3, but I believe that this theorem in fact also refers to the "continuous solution" of K-means, i.e. its statement should read "cluster centroid space of the continuous solution of K-means is spanned [...]".
Ding & He, however, do not make this important qualification, and moreover write in their abstract that
Here we prove
that principal components are the continuous
solutions to the discrete cluster membership
indicators for
K-means clustering. Equivalently, we show that the subspace spanned
by the cluster centroids are given by spectral expansion of the data covariance matrix truncated at $K-1$ terms.
The first sentence is absolutely correct, but the second one is not.
It is not clear to me if this is a (very) sloppy writing or a genuine mistake. I have very politely emailed both authors asking for clarification. (Update two months later: I have never heard back from them.) Matlab simulation code
figure('Position', [100 100 1200 600])
n = 50;
Sigma = [2 1.8; 1.8 2];
for i=1:4
means = [0 0; i*2 0];
rng(42)
X = [bsxfun(@plus, means(1,:), randn(n,2) * chol(Sigma)); ...
bsxfun(@plus, means(2,:), randn(n,2) * chol(Sigma))];
X = bsxfun(@minus, X, mean(X));
[U,S,V] = svd(X,0);
[ind, centroids] = kmeans(X,2, 'Replicates', 100);
subplot(2,4,i)
scatter(X(:,1), X(:,2), [], [0 0 0])
subplot(2,4,i+4)
hold on
scatter(X(ind==1,1), X(ind==1,2), [], [1 0 0])
scatter(X(ind==2,1), X(ind==2,2), [], [0 0 1])
plot([-1 1]*10*V(1,1), [-1 1]*10*V(2,1), 'k', 'LineWidth', 2)
plot(centroids(1,1), centroids(1,2), 'w+', 'MarkerSize', 15, 'LineWidth', 4)
plot(centroids(1,1), centroids(1,2), 'k+', 'MarkerSize', 10, 'LineWidth', 2)
plot(centroids(2,1), centroids(2,2), 'w+', 'MarkerSize', 15, 'LineWidth', 4)
plot(centroids(2,1), centroids(2,2), 'k+', 'MarkerSize', 10, 'LineWidth', 2)
plot([-1 1]*5*V(1,2), [-1 1]*5*V(2,2), 'k--')
end
for i=1:8
subplot(2,4,i)
axis([-8 8 -8 8])
axis square
set(gca,'xtick',[],'ytick',[])
end |
Consider this question,
Let $(X_1,Y_1),(X_2,Y_2),...,(X_n,Y_n)$ be independent and identically distributed pairs of random variables with $E(X_1) = E(Y_1), Var(X_1)= Var(Y_1) = 1$, and $Cov(X_1,Y_1) = \rho \in (-1,1).$
Given $\alpha \in (0,1)$, obtain a statistic $L_n$ which is a function of $(X_1,Y_1),(X_2,Y_2),...,(X_n,Y_n)$ such that \begin{align} \displaystyle \lim_{n\to\infty}P(L_n < \rho < 1) = \alpha \end{align}
My concern is that since it is given that $\rho \in (-1,1)$, can I drop the $1$ in the equality to be proven. I mean, will the statistic $L_n$ calculated such that it satisfies \begin{align} \displaystyle \lim_{n\to\infty}P(L_n < \rho) = \alpha \end{align} would also satisfy the given equation. And if not, how should I go about this problem? |
Hints will display for most wrong answers; explanations for most right answers. You can attempt a question multiple times; it will only be scored correct if you get it right the first time.
I used the official objectives and sample test to construct these questions, but cannot promise that they accurately reflect what’s on the real test. Some of the sample questions were more convoluted than I could bear to write. See terms of use. See the MTEL Practice Test main page to view random questions on a variety of topics or to download paper practice tests.
MTEL General Curriculum Mathematics Practice
Question 1
The picture below represents a board with pegs on it, where the closest distance between two pegs is 1 cm. What is the area of the pentagon shown?
Question 2
Each individual cube that makes up the rectangular solid depicted below has 6 inch sides. What is the surface area of the solid in square feet?
\( \large 11\text{ f}{{\text{t}}^{2}}\)
Hint:
Check your units and make sure you're using feet and inches consistently.
\( \large 16.5\text{ f}{{\text{t}}^{2}}\)
Hint:
Each square has surface area \(\dfrac{1}{2} \times \dfrac {1}{2}=\dfrac {1}{4}\) sq feet. There are 9 squares on the top and bottom, and 12 on each of 4 sides, for a total of 66 squares. 66 squares \(\times \dfrac {1}{4}\) sq feet/square =16.5 sq feet.
\( \large 66\text{ f}{{\text{t}}^{2}}\)
Hint:
The area of each square is not 1.
\( \large 2376\text{ f}{{\text{t}}^{2}}\)
Hint:
Read the question more carefully -- the answer is supposed to be in sq feet, not sq inches.
Question 3
What is the length of side \(\overline{BD}\) in the triangle below, where \(\angle DBA\) is a right angle?
\( \large 1\)
Hint:
Use the Pythagorean Theorem.
\( \large \sqrt{5}\)
Hint:
\(2^2+e^2=3^2\) or \(4+e^2=9;e^2=5; e=\sqrt{5}\).
\( \large \sqrt{13}\)
Hint:
e is not the hypotenuse.
\( \large 5\)
Hint:
Use the Pythagorean Theorem.
Question 4
An above-ground swimming pool is in the shape of a regular hexagonal prism, is one meter high, and holds 65 cubic meters of water. A second pool has a base that is also a regular hexagon, but with sides twice as long as the sides in the first pool. This second pool is also one meter high. How much water will the second pool hold?
\( \large 65\text{ }{{\text{m}}^{3}}\)
Hint:
A bigger pool would hold more water.
\( \large 65\cdot 2\text{ }{{\text{m}}^{3}}\)
Hint:
Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube.
\( \large 65\cdot 4\text{ }{{\text{m}}^{3}}\)
Hint:
If we think of the pool as filled with 1 x 1 x 1 cubes (and some fractions of cubes), then scaling to the larger pool changes each 1 x 1 x 1 cube to a 2 x 2 x 1 prism, or multiplies volume by 4.
\( \large 65\cdot 8\text{ }{{\text{m}}^{3}}\)
Hint:
Try a simpler example, say doubling the sides of the base of a 1 x 1 x 1 cube.
Question 5
The column below consists of two cubes and a cylinder. The cylinder has diameter y, which is also the length of the sides of each cube. The total height of the column is 5y. Which of the formulas below gives the volume of the column?
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{4}\)
Hint:
The cubes each have volume \(y^3\). The cylinder has radius \(\dfrac{y}{2}\) and height \(3y\). The volume of a cylinder is \(\pi r^2 h=\pi ({\dfrac{y}{2}})^2(3y)=\dfrac{3\pi {{y}^{3}}}{4}\). Note that the volume of a cylinder is analogous to that of a prism -- area of the base times height.
\( \large 2{{y}^{3}}+3\pi {{y}^{3}}\)
Hint:
y is the diameter of the circle, not the radius.
\( \large {{y}^{3}}+5\pi {{y}^{3}}\)
Hint:
Don't forget to count both cubes.
\( \large 2{{y}^{3}}+\dfrac{3\pi {{y}^{3}}}{8}\)
Hint:
Make sure you know how to find the volume of a cylinder.
Question 6
A car is traveling at 60 miles per hour. Which of the expressions below could be used to compute how many feet the car travels in 1 second? Note that 1 mile = 5,280 feet.
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot 60\dfrac{\text{seconds}}{\text{minute}} \)
Hint:
This answer is not in feet/second.
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot 5280\dfrac{\text{feet}}{\text{mile}}\cdot \dfrac{1}{60}\dfrac{\text{hour}}{\text{minutes}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}} \)
Hint:
This is the only choice where the answer is in feet per second and the unit conversions are correct.
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{foot}}{\text{miles}}\cdot 60\dfrac{\text{hours}}{\text{minute}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\)
Hint:
Are there really 60 hours in a minute?
\( \large 60\dfrac{\text{miles}}{\text{hour}}\cdot \dfrac{1}{5280}\dfrac{\text{mile}}{\text{feet}}\cdot 60\dfrac{\text{minutes}}{\text{hour}}\cdot \dfrac{1}{60}\dfrac{\text{minute}}{\text{seconds}}\)
Hint:
This answer is not in feet/second.
Question 7
The speed of sound in dry air at 68 degrees F is 343.2 meters per second. Which of the expressions below could be used to compute the number of kilometers that a sound wave travels in 10 minutes (in dry air at 68 degrees F)?
\( \large 343.2\times 60\times 10\)
Hint:
In kilometers, not meters.
\( \large 343.2\times 60\times 10\times \dfrac{1}{1000}\)
Hint:
Units are meters/sec \(\times\) seconds/minute \(\times\) minutes \(\times\) kilometers/meter, and the answer is in kilometers.
\( \large 343.2\times \dfrac{1}{60}\times 10\)
Hint:
Include units and make sure answer is in kilometers.
\( \large 343.2\times \dfrac{1}{60}\times 10\times \dfrac{1}{1000}\)
Hint:
Include units and make sure answer is in kilometers.
Question 8
A homeowner is planning to tile the kitchen floor with tiles that measure 6 inches by 8 inches. The kitchen floor is a rectangle that measures 10 ft by 12 ft, and there are no gaps between the tiles. How many tiles does the homeowner need?
30
Hint:
The floor is 120 sq feet, and the tiles are smaller than 1 sq foot. Also, remember that 1 sq foot is 12 \(\times\) 12=144 sq inches.
120
Hint:
The floor is 120 sq feet, and the tiles are smaller than 1 sq foot.
300
Hint:
Recheck your calculations.
360
Hint:
One way to do this is to note that 6 inches = 1/2 foot and 8 inches = 2/3 foot, so the area of each tile is 1/2 \(\times\) 2/3=1/3 sq foot, or each square foot of floor requires 3 tiles. The area of the floor is 120 square feet. Note that the tiles would fit evenly oriented in either direction, parallel to the walls.
Question 9
A cylindrical soup can has diameter 7 cm and height 11 cm. The can holds g grams of soup. How many grams of the same soup could a cylindrical can with diameter 14 cm and height 33 cm hold?
\( \large 6g\)
Hint:
You must scale in all three dimensions.
\( \large 12g\)
Hint:
Height is multiplied by 3, and diameter and radius are multiplied by 2. Since the radius is squared, final result is multiplied by \(2^2\times 3=12\).
\( \large 18g\)
Hint:
Don't square the height scale factor.
\( \large 36g\)
Hint:
Don't square the height scale factor.
Question 10
Which of the following is closest to the height of a college student in centimeters?
1.6 cm
Hint:
This is more the height of a Lego toy college student -- less than an inch!
16 cm
Hint:
Less than knee high on most college students.
160 cm
Hint:
Remember, a meter stick (a little bigger than a yard stick) is 100 cm. Also good to know is that 4 inches is approximately 10 cm.
1600 cm
Hint:
This college student might be taller than some campus buildings!
Question 11
A family on vacation drove the first 200 miles in 4 hours and the second 200 miles in 5 hours. Which expression below gives their average speed for the entire trip?
\( \large \dfrac{200+200}{4+5}\)
Hint:
Average speed is total distance divided by total time.
\( \large \left( \dfrac{200}{4}+\dfrac{200}{5} \right)\div 2\)
Hint:
This seems logical, but the problem is that it weights the first 4 hours and the second 5 hours equally, when each hour should get the same weight in computing the average speed.
\( \large \dfrac{200}{4}+\dfrac{200}{5} \)
Hint:
This would be an average of 90 miles per hour!
\( \large \dfrac{400}{4}+\dfrac{400}{5} \)
Hint:
This would be an average of 180 miles per hour! Even a family of race car drivers probably doesn't have that average speed on a vacation!
Question 12
The window glass below has the shape of a semi-circle on top of a square, where the side of the square has length x. It was cut from one piece of glass.
What is the perimeter of the window glass?
\( \large 3x+\dfrac{\pi x}{2}\)
Hint:
By definition, \(\pi\) is the ratio of the circumference of a circle to its diameter; thus the circumference is \(\pi d\). Since we have a semi-circle, its perimeter is \( \dfrac{1}{2} \pi x\). Only 3 sides of the square contribute to the perimeter.
\( \large 3x+2\pi x\)
Hint:
Make sure you know how to find the circumference of a circle.
\( \large 3x+\pi x\)
Hint:
Remember it's a semi-circle, not a circle.
\( \large 4x+2\pi x\)
Hint:
Only 3 sides of the square contribute to the perimeter.
If you found a mistake or have comments on a particular question, please contact me (please copy and paste at least part of the question into the form, as the numbers change depending on how quizzes are displayed). General comments can be left here. |
I am trying to decompose a time series of $n$ observations $\bf{\mathrm{v_c}}$ into the $n \times n$ variance-covariance structure $\sum$ and a random series $\bf{\mathrm{v}}$.
So, I can derive the variance-covariance matrix $\sum$ from the autocorrelation function of $\bf{\mathrm{v_c}}$. This will be a Toeplitz matrix, which is positive semidefinite. Therefore, I am able to compute a suitable matrix $\sum^{-\frac{1}{2}}$ to transform my correlated series into a random signal.
$\bf{\mathrm{v}} = \sum^{-\frac{1}{2}}\bf{\mathrm{v_c}}$
I am able to do this using the
sqrt(m) function in MATLAB, but can also find a Cholesky factorisation of the variance-covariance matrix and use this to induce the correlations. However, I get different (but somewhat similar) results for the random series using the sqrtm and Cholesky methods.
I have read through several texts to determine how I might ascertain the square root of various matrices, and have looked at eigenvalue decomposition methods and so on. I see there are only unique solutions under certain prescribed conditions - but I presume that these unique solutions are still only one of many roots?
My question is this: is there any way to argue that one particular square root is preferable over another. If not, is there a way to extract all possible solutions, such that all possible random functions may be obtained? |
Answer
A=$\frac{8}{3}$$\pi$-4$\sqrt 3$
Work Step by Step
height of the triangle=.5s$\sqrt 3$ h=2$\sqrt 3$ base of the triangle=r=4 A=$\frac{1}{6}$$\pi$4$^2$-$\frac{1}{2}$(4)(2$\sqrt 3$) A=$\frac{8}{3}$$\pi$-4$\sqrt 3$
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In practice, claiming that $x$ being of type $T$
usually is used to describe syntax, while claiming that $x$ is in set $S$ is usually used to indicate a semantic property. I will give some examples to clarify this difference in usage of types and sets. For the difference in what types and sets actually are, I refer to Andrej Bauer's answer. An example
To clarify this distinction, I will use the example given in Herman Geuvers' lecture notes. First, we look at an example of inhabiting a type:
$$3+(7*8)^5:\mathrm{Nat},$$ and an example of being member of a set:$$3\in \{n\in\mathbb{N}\mid \forall x,y,z\in\mathbb{N}^+ (x^n+y^n\neq z^n)\}$$
The main difference here is that to test whether the first expression is a natural number, we don't have to compute some semantic meaning, we merely have to 'read off' the fact that all literals are of type Nat and that all operators are closed on the type Nat.
However, for the second example of the set, we have to determine the semantic meaning of the $3$ in the context of the set. For this particular set, this is quite hard: the membership of $3$ for this set is equivalent to proving Fermat's last theorem! Do note that, as stated in the notes, the distinction between syntax and semantics cannot always be drawn that clearly. (and you might even argue that even
this example is unclear, as Programmer2134 mentions in the comments) Algorithms vs Proofs
To summarize, types are often used for 'simple' claims on the syntax of some expression, such that membership of a type can be checked by an
algorithm, while to test membership of a set, we would in usually require a proof.
To see why this distinction is useful, consider a compiler of a typed programming language. If this compiler has to create a formal proof to 'check types', the compiler is asked to do an almost impossible task (automated theorem proving is, in general, hard). If on the other hand the compiler can simply run an (efficient) algorithm to check the types, then it can realistically perform the task.
A motivation for a strict(er) interpretation
There are multiple interpretations of the semantic meaning of sets and types. While under the distinction made here extensional types and types with undecidable type-checking (such as those used in NuPRL, as mentioned in the comments) would not be 'types', others are of course free to call them as such (just as free as they are as to call them something else, as long as their definitions fit).
However, we (Herman Geuvers and I), prefer to not throw this interpretation out of the window, for which I (not Herman, although he might agree) have the following motivation:
First of all, the intention of this interpretation isn't
that far from that of Andrej Bauer. The intention of a syntax is usually to describe how to construct something and having an algorithm to actually construct it is generally useful. Furthermore, the features of a set are usually only needed when we want a semantic description, for which undecidability is allowed.
So, the advantage of our more stricter description is to keep the separation
simpler, to get a distinction more directly related to common practical usage. This works well, as long as you don't need or want to loosen your usage, as you would for, e.g. NuPRL. |
Equivalences:
The non-orthogonal vectors problem (as defined above) for a set $S$ of $n$ Boolean
vectors each of length $d$ and a positive integer $k$ is equivalent
the following:
Finding a $2$ by $k$ submatrix of 1's in a given $n$ by $d$ Boolean matrix.
Finding a $\mathrm{K}_{2,k}$ complete subgraph in a given bipartite graph where the first vertex set has size $n$ and the second vertex
set has size $d$.
Naive Algorithm:
The naive approach for the non-orthogonal vectors problem runs in $O(d \cdot n^2)$ time because it takes $O(d \cdot n^2)$ time to naively compute the dot product of every pair of vectors.
Answer to questions (2) & (3):
Yes, there are several algorithms that are more efficient in different
cases.
First approach:
We can solve the non-orthogonal vectors problem in $O(d \cdot n + k \cdot n^2)$ time.
Note: Since the dot product of two length $d$ Boolean vectors must be bounded by $d$, the problem only makes sense when $k \leq d$.
Proof. Let a set $S$ of $n$ Boolean vectors each of length $d$ and a positive integer $k$ be given. Consider an enumeration $\{s_i\}_{i\in[n]}$of the elements of $S$.
Create a hashmap $m$ from pairs $(a,b) \in [n] \times [n]$ to $\mathbb{N}$. Initially, $m$ maps each input to the value 0.
For each $i \in [d]$, we do the following. Enumerate through pairs of vectors $s_a$, $s_b$ such that $a < b$, the $i$th bit of $s_a$ is 1, and the $i$th bit of $s_b$ is 1. For each such $s_a$ and $s_b$ if $m(a,b) = k - 1$, then $s_a$ and $s_b$ are non-orthogonal i.e. $s_a \cdot s_b \geq k$. Otherwise, increment $m(a,b)$ and continue.
If we finish the enumeration, then no pair of vectors are non-orthogonal.
It takes $O(n \cdot d)$ time to scan through every bit of every vector. Then, it takes additional time for enumerating pairs of vectors. Because there are at most ${n \choose 2}$ pairs of vectors and each pair can show up at most $k-1$ times before they've been shown to be non-orthogonal, enumerating pairs takes at most $O(k \cdot n^2)$ time. Therefore, the total runtime is $O(d \cdot n + k \cdot n^2)$.
Note: When $k = 2$, we can improve this approach to $O(n \cdot d)$ time.
This is because when $k = 2$, we can reduce finding a pair of non-orthogonal vectors among $n$ Boolean vectors of length $d$ to finding a pair of non-orthogonal vectors among $d$ Boolean vectors of length $n$.
Second approach:
We can solve the non-orthogonal vectors problem in $O(k \cdot {d \choose k} \cdot n)$ time.
Proof. Let a set $S$ of $n$ Boolean vectors each of length $d$ and a positive integer $k$ be given.
Enumerate through sets $P \subseteq [d]$ such that $P$ has size $k$. For every vector $v \in S$, check if $v$ has all 1's at the positions in $P$. It there are two vectors that have all 1's at the positions in $P$, then we've found two non-orthogonal vectors.
In total, there are ${d \choose k}$ possible choices for $P$. And, for each choice, we scan through $k \cdot n$ bits from the vectors. Therefore, in total, the runtime is $O(k \cdot {d \choose k} \cdot n)$.
Third approach:
When $d \leq n$, we can solve the non-orthongal vectors problem in $O(d^{\omega - 2} \cdot n^2)$ time where $\omega$ is the exponent for integer matrix multiplication. When $d > n$, we can solve the non-orthongal vectors problem in $O(d \cdot n^{\omega - 1})$ time.
Note: As pointed out by @Rasmus Pagh, we can improve this algorithm to $O(n^{2 + o(1)})$ time when $d \leq n^{0.3}$. See here for more info: https://arxiv.org/abs/1204.1111
Proof. Let a set $S$ of $n$ Boolean vectors each of length $d$ and a positive integer $k$ be given.
Consider matrices $A$ and $B$. The first matrix $A$ has dimensions $n$ by $d$ where each row of $A$ is a vector from $S$. The second matrix $B$ has dimensions $d$ by $n$ where each column of $B$ is a vector from $S$.
We can compute the dot product of every pair of vectors in $S$ by computing $A \cdot B$ using algorithms for fast integer matrix multiplication.
When $d \leq n$, one approach is to convert the rectangular matrix multiplication into $(\frac{n}{d})^2$ multiplications of square $d$ by $d$ matrices. By using fast square matrix multiplication, we can compute all of the multiplications in $O((\frac{n}{d})^2 \cdot d^{\omega}) = O(d^{\omega - 2} \cdot n^2)$ time.
When $d > n$, one approach is to convert the rectangular matrix multiplication into $\frac{d}{n}$ multiplications of square $n$ by $n$ matrices. By using fast square matrix multiplication, we can compute all of the multiplications in $O((\frac{d}{n}) \cdot n^{\omega}) = O(d \cdot n^{\omega - 1})$ time. |
I am currently doing a problem in *Introduction to Quantum Mechanics, 2nd edition Griffiths *
QUESTION
In question 2.22 part D, we are asked to calculate $\langle p^2 \rangle$ for the Gaussian Wave packet $$ \Psi(x,0) = A e^{-ax^2} $$ In an earlier problem we calculate the normalization constant $ A = (\frac{2a}{\pi})^\frac{1}{4}$
PROBLEM
I tried calculating $\langle p^2 \rangle$ by letting the operator $p$ be $$ p = -i \hbar \frac{\partial}{\partial x} \Rightarrow p^2 = -\hbar^2 \frac{\partial^2}{\partial x^2} $$ Hence I get \begin{eqnarray} \langle p^2 \rangle &=& \int_{-\infty}^{\infty} A e^{-ax^2}-\hbar \frac{\partial^2}{\partial x^2}Ae^{-ax^2}dx\\ &=& A^2 -\hbar^2 \int_{-\infty}^{\infty} e^{-ax^2} 4a^2e^{-ax^2}dx\\ &=&A^2 -\hbar^2 4a^2 \int_{-\infty}^{\infty} e^{-ax^2} e^{-ax^2} dx\\ &=&A^2 -\hbar^2 4a^2 \int_{-\infty}^{\infty} e^{-2ax^2} dx \\ &=&A^2 -\hbar^2 4a^2 2\int_{0}^{\infty} e^{-2ax^2}dx \\ &=&A^2 -\hbar^2 4a^2 2 \frac{1}{2} (\frac{\pi}{a})^\frac {1}{2} \\ &=&(\frac{2a}{\pi})^\frac{1}{2} -\hbar^2 4a^2 (\frac{\pi}{a})^\frac {1}{2} \\ &=&(2)^\frac{1}{2} -\hbar^2 4a^2 \end{eqnarray} However this answer is wrong and the correct answer is $$ \langle p^2 \rangle = a \hbar^2 $$ I looked at the solutions manual and they way they did it was completely different and instead they substituted for a bunch of different variables. My question is why wouldn't the method I tried to do work with this wave function? |
I'm looking for an upright Greek font for single Greek characters (like "β-decay" or "µ-metal") which fits to the default CM/latin style, i.e. the upright version of the default italic math mode Greek letters (
\beta,
\mu).
The "default" upright Greek font (should be
cbgreek), which is used when writing with
babel or with the
textgreek package (
\textbeta,
\textmu), doesn't quite fit to the CM/latin font, especially the µ symbol. There are packages replacing the entire font families, e.g.
mathdesign or
kpfonts, providing a complete set of Greek, but none of them was made for CM/latin.
upgreek uses another font for math mode Greek (
euler I guess, also accessible through
textgreek with the respective option) which also doesn't quite fit with CM/latin.
I would already give up if it wasn't for the µ used by the
siunitx package. Its default "micro" prefix (
\si{\micro}) fits perfectly with CM/latin (see comparison below) and I wanted to know whether that µ is just a unique character of the CM font or whether there's a chance to also get an upright β and other letters in exactly that font. Also, how does the
siunitx package get that µ? The characters µ (micro symbol) and μ (greek letter mu) print by default with the same character in
cbgreek with UTF-8 encoding (see below)...
Here's a MWE (pdflatex):
\documentclass{minimal}\usepackage[utf8]{inputenc}\usepackage{siunitx}\usepackage{textgreek}\usepackage{upgreek}\begin{document}m\si{\micro}\textmu $\mu$\textbeta $\beta$$\upmu \upbeta$µμβ % U+00B5 (micro symbol) U+03BC (greek lower case mu) U+03B2 (greek lower case beta)\end{document}
Result: |
Exercise \(\PageIndex{1}\)
In the following exercises, state whether each statement is true, or give an example to show that it is false.
1. If \(\displaystyle \sum_{n=1}^∞a_nx^n\) converges, then \(\displaystyle a_nx^n→0\) as \(\displaystyle n→∞.\)
Answer
True. If a series converges then its terms tend to zero.
2. \(\displaystyle \sum_{n=1}^∞a_nx^n\) converges at \(\displaystyle x=0\) for any real numbers \(\displaystyle a_n\).
3. Given any sequence \(\displaystyle a_n\), there is always some \(\displaystyle R>0\), possibly very small, such that \(\displaystyle \sum_{n=1}^∞a_nx^n\) converges on \(\displaystyle (−R,R)\).
Answer
False. It would imply that \(\displaystyle a_nx^n→0\) for \(\displaystyle |x|<R\). If \(\displaystyle a_n=n^n\), then \(\displaystyle a_nx^n=(nx)^n\) does not tend to zero for any \(\displaystyle x≠0\).
4. If \(\displaystyle \sum_{n=1}^∞a_nx^n\) has radius of convergence \(\displaystyle R>0\) and if \(\displaystyle |b_n|≤|a_n|\) for all \(\displaystyle n\), then the radius of convergence of \(\displaystyle \sum_{n=1}^∞b_nx^n\) is greater than or equal to \(\displaystyle R\)
Exercise \(\PageIndex{2}\)
1. Suppose that \(\displaystyle \sum_{n=0}^∞a_n(x−3)^n\) converges at \(\displaystyle x=6\). At which of the following points must the series also converge? Use the fact that if \(\displaystyle \sum a_n(x−c)^n\) converges at \(\displaystyle x\), then it converges at any point closer to \(\displaystyle c\) than \(\displaystyle x\).
a. \(\displaystyle x=1\)
b. \(\displaystyle x=2\)
c. \(\displaystyle x=3\)
d. \(\displaystyle x=0\)
e. \(\displaystyle x=5.99\)
f. \(\displaystyle x=0.000001\)
Answer
It must converge on \(\displaystyle (0,6]\) and hence at: a. \(\displaystyle x=1\); b. \(\displaystyle x=2\); c. \(\displaystyle x=3\); d. \(\displaystyle x=0\); e. \(\displaystyle x=5.99\); and f. \(\displaystyle x=0.000001\).
2. Suppose that \(\displaystyle \sum_{n=0}^∞a_n(x+1)^n\) converges at \(\displaystyle x=−2\). At which of the following points must the series also converge? Use the fact that if \(\displaystyle \sum a_n(x−c)^n\) converges at \(\displaystyle x\), then it converges at any point closer to \(\displaystyle c\) than \(\displaystyle x\).
a. \(\displaystyle x=2\)
b. \(\displaystyle x=−1\)
c. \(\displaystyle x=−3\)
d. \(\displaystyle x=0\)
e. \(\displaystyle x=0.99\)
f. \(\displaystyle x=0.000001\)
Exercise \(\PageIndex{3}\)
In the following exercises, suppose that \(\displaystyle ∣\frac{a_{n+1}}{a_n}∣→1\) as \(\displaystyle n→∞.\) Find the radius of convergence for each series.
1. \(\displaystyle \sum_{n=0}^∞a_n2^nx^n\)
Answer
\(\displaystyle ∣\frac{a_{n+1}2^{n+1}x^{n+1}}{a_n2^nx^n}∣ =2|x|∣ \frac{a_{n+1}}{a_n}∣→2|x|\) so \(\displaystyle R=\frac{1}{2}\)
2. \(\displaystyle \sum_{n=0}^∞\frac{a_nx^n}{2^n}\)
Answer
\(\displaystyle|\frac{a_{n+1}x^{n+1}}{2^{n+1}} \frac{2^n}{a_nx^n}| = \frac{|x|}{2}| \frac{a_{n+1}}{a_n}∣→\frac{|x|}{2}\) so \(R=2\)
3. \(\displaystyle \sum_{n=0}^∞\frac{a_nπ^nx^n}{e^n}\)
Answer
\(\displaystyle ∣\frac{a_{n+1}(\frac{π}{e})^{n+1}x^{n+1}}{a_n(\frac{π}{e})^nx^n}∣ =\frac{π|x|}{e}∣\frac{a_{n+1}}{a_n}∣→\frac{π|x|}{e}\) so \(\displaystyle R=\frac{e}{π}\)
4. \(\displaystyle \sum_{n=0}^∞\frac{a_n(−1)^nx^n}{10^n}\)
Answer
\(\displaystyle|\frac{a_{n+1}(-1)^{n+1}x^{n+1}}{10^{n+1}} \frac{10^n}{a_n(-1)^nx^n}| = \frac{|x|}{10}| \frac{a_{n+1}}{a_n}∣→\frac{|x|}{10}\) so \(R=10\)
5. \(\displaystyle \sum_{n=0}^∞a_n(−1)^nx^{2n}\)
Answer
\(\displaystyle ∣\frac{a_{n+1}(−1)^{n+1}x^{2n+2}}{a_n(−1)^nx^{2n}}∣ =∣x^2∣∣\frac{a_{n+1}}{a_n}∣→∣x^2∣\) so \(\displaystyle R=1\)
6. \(\displaystyle \sum_{n=0}^∞a_n(−4)^nx^{2n}\)
Answer
\(\displaystyle ∣\frac{a_{n+1}(-4)^{n+1}x^{2(n+1)}}{a_n(-4)^nx^{2n}}∣ =4|x^2|∣ \frac{a_{n+1}}{a_n}∣→4|x^2|\) so \(\displaystyle R=\frac{1}{2}\)
Exercise \(\PageIndex{4}\)
In the following exercises, find the radius of convergence \(\displaystyle R\) and interval of convergence for \(\displaystyle \sum a_nx^n\) with the given coefficients \(\displaystyle a_n\).
1. \(\displaystyle \sum_{n=1}^∞\frac{(2x)^n}{n}\)
Answer
\(\displaystyle a_n=\frac{2^n}{n}\) so \(\displaystyle \frac{a_{n+1}x}{a_n}→2x\). so \(\displaystyle R=\frac{1}{2}\). When \(\displaystyle x=\frac{1}{2}\) the series is harmonic and diverges. When \(\displaystyle x=−\frac{1}{2}\) the series is alternating harmonic and converges. The interval of convergence is \(\displaystyle I=[−\frac{1}{2},\frac{1}{2})\).
2. \(\displaystyle \sum_{n=1}^∞(−1)^n\frac{x^n}{\sqrt{n}}\)
Answer
R=1
Interval of convergence (-1,1)
3. \(\displaystyle \sum_{n=1}^∞\frac{nx^n}{2^n}\)
Answer
\(\displaystyle a_n=\frac{n}{2^n}\) so \(\displaystyle \frac{a_{n+1}x}{a_n}→\frac{x}{2}\) so \(\displaystyle R=2\). When \(\displaystyle x=±2\) the series diverges by the divergence test. The interval of convergence is \(\displaystyle I=(−2,2)\).
4. \(\displaystyle \sum_{n=1}^∞\frac{nx^n}{e^n}\)
5. \(\displaystyle \sum_{n=1}^∞\frac{n^2x^n}{2^n}\)
Answer
\(\displaystyle a_n=\frac{n^2}{2^n}\) so \(\displaystyle R=2\). When \(\displaystyle x=±2\) the series diverges by the divergence test. The interval of convergence is \(\displaystyle I=(−2,2).\)
6. \(\displaystyle \sum_{k=1}^∞\frac{k^ex^k}{e^k}\)
7. \(\displaystyle \sum_{k=1}^∞\frac{π^kx^k}{k^π}\)
Answer:
\(\displaystyle a_k=\frac{π^k}{k^π}\) so \(\displaystyle R=\frac{1}{π}\). When \(\displaystyle x=±\frac{1}{π}\) the series is an absolutely convergent p-series. The interval of convergence is \(\displaystyle I=[−\frac{1}{π},\frac{1}{π}].\)
8. \(\displaystyle \sum_{n=1}^∞\frac{x^n}{n!}\)
9. \(\displaystyle \sum_{n=1}^∞\frac{10^nx^n}{n!}\)
Answer:
\(\displaystyle a_n=\frac{10^n}{n!},\frac{a_{n+1}x}{a_n}=\frac{10x}{n+1}→0<1\) so the series converges for all \(\displaystyle x\) by the ratio test and \(\displaystyle I=(−∞,∞)\).
10. \(\displaystyle \sum_{n=1}^∞(−1)^n\frac{x^n}{ln(2n)}\)
Exercise \(\PageIndex{5}\)
In the following exercises, find the radius of convergence of each series.
1. \(\displaystyle \sum_{k=1}^∞\frac{(k!)^2x^k}{(2k)!}\)
Answer
\(\displaystyle a_k=\frac{(k!)^2}{(2k)!}\) so \(\displaystyle \frac{a_{k+1}}{a_k}=\frac{(k+1)^2}{(2k+2)(2k+1)}→\frac{1}{4}\) so \(\displaystyle R=4\)
2. \(\displaystyle \sum_{n=1}^∞\frac{(2n)!x^n}{n^{2n}}\)
3. \(\displaystyle \sum_{k=1}^∞\frac{k!}{1⋅3⋅5⋯(2k−1)}x^k\)
Answer
\(\displaystyle a_k=\frac{k!}{1⋅3⋅5⋯(2k−1)}\) so \(\displaystyle \frac{a_{k+1}}{a_k}=\frac{k+1}{2k+1}→\frac{1}{2}\) so \(\displaystyle R=2\)
4. \(\displaystyle \sum_{k=1}^∞\frac{2⋅4⋅6⋯2k}{(2k)!}x^k\)
5. \(\displaystyle \sum_{n=1}^∞\frac{x^n}{(^{2n}_n)}\) where \(\displaystyle (^n_k)=\frac{n!}{k!(n−k)!}\)
Answer
\(\displaystyle a_n=\frac{1}{(^{2n}_n)}\) so \(\displaystyle \frac{a_{n+1}}{a_n}=\frac{((n+1)!)^2}{(2n+2)!}\frac{2n!}{(n!)^2}=\frac{(n+1)^2}{(2n+2)(2n+1)}→\frac{1}{4}\) so \(\displaystyle R=4\)
6. \(\displaystyle \sum_{n=1}^∞sin^2nx^n\)
Exercise \(\PageIndex{6}\)
In the following exercises, use the ratio test to determine the radius of convergence of each series.
1. \(\displaystyle \sum_{n=1}^∞\frac{(n!)^3}{(3n)!}x^n\)
Answer
\(\displaystyle \frac{a_{n+1}}{a_n}=\frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}→\frac{1}{27}\) so \(\displaystyle R=27\)
2. \(\displaystyle \sum_{n=1}^∞\frac{2^{3n}(n!)^3}{(3n)!}x^n\)
3. \(\displaystyle \sum_{n=1}^∞\frac{n!}{n^n}x^n\)
Answer
\(\displaystyle a_n=\frac{n!}{n^n}\) so \(\displaystyle \frac{a_{n+1}}{a_n}=\frac{(n+1)!}{n!}\frac{n^n}{(n+1)^{n+1}}=(\frac{n}{n+1})^n→\frac{1}{e}\) so \(\displaystyle R=e\)
4. \(\displaystyle \sum_{n=1}^∞\frac{(2n)!}{n^{2n}}x^n\)
Exercise \(\PageIndex{7}\)
In the following exercises, given that \(\displaystyle \frac{1}{1−x}=\sum_{n=0}^∞x^n\) with convergence in \(\displaystyle (−1,1)\), find the power series for each function with the given center a, and identify its interval of convergence.
1. \(\displaystyle f(x)=\frac{1}{x};a=1\) (Hint: \(\displaystyle \frac{1}{x}=\frac{1}{1−(1−x)})\)
Answer
\(\displaystyle f(x)=\sum_{n=0}^∞(1−x)^n\) on \(\displaystyle I=(0,2)\)
2. \(\displaystyle f(x)=\frac{1}{1−x^2};a=0\)
3. \(\displaystyle f(x)=\frac{x}{1−x^2};a=0\)
Answer
\(\displaystyle \sum_{n=0}^∞x^{2n+1}\) on \(\displaystyle I=(−1,1)\)
4. \(\displaystyle f(x)=\frac{1}{1+x^2};a=0\)
5. \(\displaystyle f(x)=\frac{x^2}{1+x^2};a=0\)
Answer
\(\displaystyle \sum_{n=0}^∞(−1)^nx^{2n+2}\) on \(\displaystyle I=(−1,1)\)
6. \(\displaystyle f(x)=\frac{1}{2−x};a=1\)
7. \(\displaystyle f(x)=\frac{1}{1−2x};a=0.\)
Answer
\(\displaystyle \sum_{n=0}^∞2^nx^n\) on \(\displaystyle (−\frac{1}{2},\frac{1}{2})\)
8. \(\displaystyle f(x)=\frac{1}{1−4x^2};a=0\)
9. \(\displaystyle f(x)=\frac{x^2}{1−4x^2};a=0\)
Answer
\(\displaystyle \sum_{n=0}^∞4^nx^{2n+2}\) on \(\displaystyle (−\frac{1}{2},\frac{1}{2})\)
10. \(\displaystyle f(x)=\frac{x^2}{5−4x+x^2};a=2\)
Exercise \(\PageIndex{8}\)
Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.
1. Explain why, if \(\displaystyle |a_n|^{1/n}→r>0,\) then \(\displaystyle |a_nx^n|^{1/n}→|x|r<1\) whenever \(\displaystyle |x|<\frac{1}{r}\) and, therefore, the radius of convergence of \(\displaystyle \sum_{n=1}^∞a_nx^n\) is \(\displaystyle R=\frac{1}{r}\).
Answer
\(\displaystyle |a_nx^n|^{1/n}=|a_n|^{1/n}|x|→|x|r\) as \(\displaystyle n→∞\) and \(\displaystyle |x|r<1\) when \(\displaystyle |x|<\frac{1}{r}\). Therefore, \(\displaystyle \sum_{n=1}^∞a_nx^n\) converges when \(\displaystyle |x|<\frac{1}{r}\) by the
nth root test.
2. \(\displaystyle \sum_{n=1}^∞\frac{x^n}{n^n}\)
3. \(\displaystyle \sum_{k=1}^∞(\frac{k−1}{2k+3})^kx^k\)
Answer
\(\displaystyle a_k=(\frac{k−1}{2k+3})^k\) so \(\displaystyle (a_k)^{1/k}→\frac{1}{2}<1\) so \(\displaystyle R=2\)
4. \(\displaystyle \sum_{k=1}^∞(\frac{2k^2−1}{k^2+3})^kx^k\)
5. \(\displaystyle \sum_{n=1}^∞a_n=(n^{1/n}−1)^nx^n\)
Answer
\(\displaystyle a_n=(n^{1/n}−1)^n\) so \(\displaystyle (a_n)^{1/n}→0\) so \(\displaystyle R=∞\)
6. Suppose that \(\displaystyle p(x)=\sum_{n=0}^∞a_nx^n\) such that \(\displaystyle a_n=0\) if \(\displaystyle n\) is even. Explain why \(\displaystyle p(x)=p(−x).\)
7. Suppose that \(\displaystyle p(x)=\sum_{n=0}^∞a_nx^n\) such that \(\displaystyle a_n=0\) if \(\displaystyle n\) is odd. Explain why \(\displaystyle p(x)=−p(−x).\)
Answer
We can rewrite \(\displaystyle p(x)=\sum_{n=0}^∞a_{2n+1}x^{2n+1}\) and \(\displaystyle p(x)=p(−x)\) since \(\displaystyle x^{2n+1}=−(−x)^{2n+1}\).
8. Suppose that \(\displaystyle p(x)=\sum_{n=0}^∞a_nx^n\) converges on \(\displaystyle (−1,1]\). Find the interval of convergence of \(\displaystyle p(Ax)\).
9. Suppose that \(\displaystyle p(x)=\sum_{n=0}^∞a_nx^n\) converges on \(\displaystyle (−1,1]\). Find the interval of convergence of \(\displaystyle p(2x−1)\).
Answer
If \(\displaystyle x∈[0,1],\) then \(\displaystyle y=2x−1∈[−1,1]\) so \(\displaystyle p(2x−1)=p(y)=\sum_{n=0}^∞a_ny^n\) converges.
Exercise \(\PageIndex{9}\)
In the following exercises, suppose that \(\displaystyle p(x)=\sum_{n=0}^∞a_nx^n\) satisfies \(\displaystyle \lim_{n→∞}\frac{a_{n+1}}{a_n}=1\) where \(\displaystyle a_n≥0\) for each \(\displaystyle n\). State whether each series converges on the full interval \(\displaystyle (−1,1)\), or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.
1. \(\displaystyle \sum_{n=0}^∞a_nx^{2n}\)
2. \(\displaystyle \sum_{n=0}^∞a_{2n}x^{2n}\)
Answer
Converges on \(\displaystyle (−1,1)\) by the ratio test
3. \(\displaystyle \sum_{n=0}^∞a_{2n}x^n\) (Hint:\(\displaystyle x=±\sqrt{x^2}\))
4. \(\displaystyle \sum_{n=0}^∞a_{n^2}x^{n^2}\) (Hint: Let \(\displaystyle b_k=a_k\) if \(\displaystyle k=n^2\) for some \(\displaystyle n\), otherwise \(\displaystyle b_k=0\).
Answer
Consider the series \(\displaystyle \sum b_kx^k\) where \(\displaystyle b_k=a_k\) if \(\displaystyle k=n^2\) and \(\displaystyle b_k=0\) otherwise. Then \(\displaystyle b_k≤a_k\) and so the series converges on \(\displaystyle (−1,1)\) by the comparison test.
5. Suppose that \(\displaystyle p(x)\) is a polynomial of degree \(\displaystyle N\). Find the radius and interval of convergence of \(\displaystyle \sum_{n=1}^∞p(n)x^n\).
Exercise \(\PageIndex{10}\)
1. Plot the graphs of \(\displaystyle \frac{1}{1−x}\) and of the partial sums \(\displaystyle S_N=\sum_{n=0}^Nx^n\) for \(\displaystyle n=10,20,30\) on the interval \(\displaystyle [−0.99,0.99]\). Comment on the approximation of \(\displaystyle \frac{1}{1−x}\) by \(\displaystyle S_N\) near \(\displaystyle x=−1\) and near \(\displaystyle x=1\) as \(\displaystyle N\) increases.
Answer
The approximation is more accurate near \(\displaystyle x=−1\). The partial sums follow \(\displaystyle \frac{1}{1−x}\) more closely as \(\displaystyle N\) increases but are never accurate near \(\displaystyle x=1\) since the series diverges there.
2. Plot the graphs of \(\displaystyle −ln(1−x)\) and of the partial sums \(\displaystyle S_N=\sum_{n=1}^N\frac{x^n}{n}\) for \(\displaystyle n=10,50,100\) on the interval \(\displaystyle [−0.99,0.99]\). Comment on the behavior of the sums near \(\displaystyle x=−1\) and near \(\displaystyle x=1\) as \(\displaystyle N\) increases.
3. Plot the graphs of the partial sums \(\displaystyle S_n=\sum_{n=1}^N\frac{x^n}{n^2}\) for \(\displaystyle n=10,50,100\) on the interval \(\displaystyle [−0.99,0.99]\). Comment on the behavior of the sums near \(\displaystyle x=−1\) and near \(\displaystyle x=1\) as \(\displaystyle N\) increases.
Answer
The approximation appears to stabilize quickly near both \(\displaystyle x=±1\).
4. Plot the graphs of the partial sums \(\displaystyle S_N=\sum_{n=1}^Nsinnx^n\) for \(\displaystyle n=10,50,100\) on the interval \(\displaystyle [−0.99,0.99]\). Comment on the behavior of the sums near \(\displaystyle x=−1\) and near \(\displaystyle x=1\) as \(\displaystyle N\) increases.
5. Plot the graphs of the partial sums \(\displaystyle S_N=\sum_{n=0}^N(−1)^n\frac{x^{2n+1}}{(2n+1)!}\) for \(\displaystyle n=3,5,10\) on the interval \(\displaystyle [−2π,2π]\). Comment on how these plots approximate \(\displaystyle sinx\) as \(\displaystyle N\) increases.
Answer
The polynomial curves have roots close to those of \(\displaystyle sinx\) up to their degree and then the polynomials diverge from \(\displaystyle sinx\).
6. Plot the graphs of the partial sums \(\displaystyle S_N=\sum_{n=0}^N(−1)^n\frac{x^{2n}}{(2n)!}\) for \(\displaystyle n=3 ,5,10\) on the interval \(\displaystyle [−2π,2π]\). Comment on how these plots approximate \(\displaystyle cosx\) as \(\displaystyle N\) increases. |
If you are an NLP person, chances are you know PCFGs (probabilistic context-free grammars) pretty well. These are just generative models for generating phrase-structure trees.
A CFG includes a set of nonterminals (\(N\)) with a designed start symbol \(\mathrm{S} \in N\), a set of terminal words (the “words”) and a set of context-free rules such as \(\mathrm{S \rightarrow NP\, VP}\) or \(\mathrm{NP \rightarrow DET\, NN}\), or \(\mathrm{NN \rightarrow\, dog}\). The left handside of a rule is a nonterminal, and the right handside is a string over the union of the nonterminals and the vocabulary.
A PCFG simply associates each rule with a weight, such that the sum of all weights for all rules with the same nonterminal on the left hand-side is 1. The weights, naturally, need to be positive.
Now, a PCFG defines a generative process for generating a phrase-structure tree which is followed by starting with the starting symbol \(\mathrm{S}\), and then repeatedly expanding the partially-rewritten tree until the yield of the tree contains only terminals.
Here is an example of a phrase-structure tree, which could be generated using a PCFG:
Each node (or its symbol, to be more precise) in this tree and its immediate children correspond to a context-free rule. The tree could have been generated probabilistically by starting with the root symbol, and then recursively expanding each node, each time probabilistically selecting a rule from the list of rules with the left handside being the symbol of the node being expanded. I am being succinct here on purpose, since I am assuming most of you know PCFGs to some extent. If not, then just searching for “probabilistic context-free grammars” will yield many results which are useful to learn the basics about PCFGs. (Here is a place to start, taken from Michael Collins’ notes.)
What is the end result, and what is the probability distribution that a PCFG induces? This is where we have to be careful. First, a PCFG defines a measure over phrase-structure trees. The measure of a tree is just the product of all rules that appear in that tree \(t\), so that:
$$\mu(t) = \prod_{r \in t} p_r.$$
Here, \(r \in t\) denotes that the rule \( r\) appears in \(t \) somewhere. \( t \) is treated as a list of rules (not a set! a rule could appear several times in \( t \), so we slightly abuse the \( \in \) notation). \( p_r \) is the rule probability associated with rule \( r \).
You might be wondering why I am using the word “measure,” instead of just saying that the above equation defines a distribution over trees. After all, aren’t we used to assuming that the probability of a tree is just the product of all rule probabilities that appear in the tree?
And that’s where the catch is. In order for the product of all rule probabilities to be justifiably called “a probability distribution”, it has to sum to 1 over all possible trees. But that’s not necessarily the case for any assignment of rule probabilities, even if they sum to 1 and are positive. The sum of all measures of finite trees could sum to *less* than 1, simply because some probability “leaks” to infinite trees.
Take for example the simple grammar with rules \(\mathrm{S \rightarrow S\, S}\) and \( \mathrm{S \rightarrow a} \). If the rule probability \(\mathrm{S \rightarrow S\, S}\) is larger than 0.5, then if we start generating a tree using this PCFG, there is a non-zero probability that we will never stop generating that tree! The rule \(\mathrm{S \rightarrow S\, S}\) has such large probability, that the tree we generate in a step-by-step derivation through the PCFG may potentially grow too fast and the derivation will never terminate.
Fortunately, we know that in a frequentist setting, if we estimate a PCFG using frequency count (with trees being observed) or EM (when trees are not being observed, but only strings are being observed), the resulting measure we get is actually a probability distribution. This means that this kind of estimation always leads to a *tight* PCFG. That’s the term used to denote a PCFG for which the sum of measures over finite trees is 1.
So where is the issue? The issue starts when we do not follow a vanilla maximum likelihood estimation. If we reguarlize, or if we smooth the estimated parameters or do something similar to that, we may end up with a non-tight PCFG.
This is especially true in the Bayesian setting, where typically we put a Dirichlet prior on each of the set of probabilities associated with a single non-terminal on the left handside. The support of the Dirichlet prior is all possible probability distributions over the \(K\) rules for nonterminal \(A\), and therefore, it assigns a non-zero probability to non-tight grammars.
What can we do to handle this issue? This is where we identified the following three alternatives (the paper can be found here).
Just ignore the problem. That means that you let the Bayesian prior, say a factorized Dirichlet, have non-zero probability on non-tight grammars. This is what people have done until now, perhaps without realizing that non-tight grammars are considered as well. Re-normalize the prior. Here, we take any prior that we start with (such as a Dirichlet), remove all the non-tight grammars, and re-normalize to get back a probability distribution over all possible tight grammars (and only the tight grammars). Re-normalize the likelihood. It has been shown that any non-tight PCFG can be converted to a tight PCFG which induces a distribution over finite trees which is identical to the distribution that we would get from the non-tight PCFG, after re-normalizing the non-tight PCFG to become a distribution over the finite trees by dividing the probability of all finite trees by the total probability mass that the finite trees get (Chi, 1999). So, one can basically map all non-tight distributions in the prior to tight PCFG distributions according to Chi’s procedure.
Notice the difference between 2 and 3, in where we “do the re-normalization.” In 2, we completely ignore all non-tight grammars, and assign them probability 0. In 3, we map any non-tight point to a tight point according to the procedure described in Chi (1999).
The three alternatives 1, 2 and 3 are not mathematically equivalent, as is shown in this Mathematica log. But we do believe they are equivalent in the following sense: any prior which is defined using one of the approaches can be transformed into a different prior that can be used with one of the other approaches and both would yield the same posterior over trees, conditioned on a string and marginalizing out the parameters. That’s an open problem, so you are more than welcome sending us ideas!
But the real question is, why would we want to handle non-tight grammars? Do we really care? This might be more of a matter of aesthetics, as we show in the paper, empirically speaking. Still, it is worth pointing out this issue, especially since people have worked hard to show what happens with non-tightness in the non-Bayesian setting. In the Bayesian setting, this problem has been completely swept under the rug until now!
Notes:
1. For those of you who are less knowledgable about Bayesian statisitics, here is an excellent blog post about it from Bob Carpenter. Non-tight PCFGs are also referred to as “inconsistent PCFGs.” |
The Fundamental Theorem of Arithmetic is one of the most important results in this chapter. It simply says that every positive integer can be written uniquely as a product of primes. The unique factorization is needed to establish much of what comes later. There are systems where unique factorization fails to hold. Many of these examples come from algebraic number theory. We can actually list an easy example where unique factorization fails.
Consider the class \(C\) of positive even integers. Note that \(C\) is closed under multiplication, which means that the product of any two elements in \(C\) is again in \(C\). Suppose now that the only number we know are the members of \(C\). Then we have \(12=2.6\) is composite where as \(14\) is prime since it is not the product of two numbers in \(C\). Now notice that \(60=2.30=6.10\) and thus the factorization is not unique.
We now give examples of the unique factorization of integers.
\(99=3\cdot 3\cdot 11=3^2\cdot 11\),
\(32=2\cdot 2\cdot 2\cdot 2\cdot 2=2^5\)
The Fundamental Theorem of Arithmetic
To prove the fundamental theorem of arithmetic, we need to prove some lemmas about divisibility.
Lemma 4
If a,b,c are positive integers such that \((a,b)=1\) and \(a \mid bc\), then \(a\mid c\).
Since \((a,b)=1\), then there exists integers \(x,y\) such that \(ax+by=1\). As a result, \(cax+cby=c\). Notice that since \(a \mid bc\), then by Theorem 4, \(a\) divides \(cax+cby\) and hence \(a\) divides \(c\).
We can generalize the above lemma as such: If \((a_,n_i)=1\) for every \(i=1,2,\cdots,n\) and \(a\mid n_1n_2\cdots n_{k+1}\), then \(a\mid n_{k+1}\). We next prove a case of this generalization and use this to prove the fundamental theorem of arithmetic.
lemma5
If \(p\) divides \(n_1n_2n_3...n_k\), where p is a prime and \(n_i >0\) for all \(1\leq i\leq k\), then there is an integer \(j\) with \(1\leq j\leq k\) such that \(p \mid n_j\).
We present the proof of this result by induction. For \(k=1\), the result is trivial. Assume now that the result is true for \(k\). Consider \(n_1n_2...n_{k+1}\) that is divisible by \(p\). Notice that either
\[(p,n_1n_2...n_k)=1\ \ \mbox{or} \ \ (p,n_1n_2...n_{k})=p.\]
Now if \((p,n_1n_2...n_k)=1\) then by Lemma 4, \(p \mid n_{k+1}\). Now if \(p\mid n_1n_2...n_k\), then by the induction hypothesis, there exists an integer \(i\) such that \(p\mid n_i\).
We now state the fundamental theorem of arithmetic and present the proof using Lemma 5.
Theorem: The Fundamental Theorem of Arithmetic
Every positive integer different from 1 can be written uniquely as a product of primes.
If \(n\) is a prime integer, then \(n\) itself stands as a product of primes with a single factor. If \(n\) is composite, we use proof by contradiction. Suppose now that there is some positive integer that cannot be written as the product of primes. Let \(n\) be the smallest such integer. Let \(n=ab\), with \(1<a<n\) and \(1<b<n\). As a result \(a\) and \(b\) are products of primes since both integers are less than \(n\). As a result, \(n=ab\) is a product of primes, contradicting that it is not. This shows that every integer can be written as product of primes. We now prove that the representation of a positive integer as a product of primes is unique. Suppose now that there is an integer \(n\) with two different factorizations say
\[n=p_1p_2...p_s=q_1q_2...q_r\]
where \(p_1,p_2,...p_s,q_1,q_2,...q_r\) are primes, \[p_1\leq p_2 \leq p_3\leq ...\leq p_s \mbox{and} \ \ q_1 \leq q_2 \leq q_3 \leq ... \leq q_r.\] Cancel out all common primes from the factorizations above to get
\[p_{j_1}p_{j_2}...p_{j_u}=q_{i_1}q_{i_2}...q_{i_v}\]
Thus all the primes on the left side are different from the primes on the right side. Since any \(p_{j_l}\) \((l=1,\cdots,n)\) divides \(p_{j_1}p_{j_2}...p_{j_u}\), then \(p_{j_l}\) must divide \(q_{i_1}q_{i_2}...q_{i_v}\), and hence by Lemma 5, \(p_{j_1}\) must divide \(q_{j_k}\) for some \(1\leq k \leq v\) which is impossible. Hence the representation is unique.
[remark1] The unique representation of a positive integer \(n\) as a product of primes can be written in several ways. We will present the most common representations. For example, \(n=p_1p_2p_3...p_k\) where \(p_i\) for \(1\leq i\leq k\) are not necessarily distinct. Another example would be \[n=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_j^{a_j}\] where all the \(p_i\) are distinct for \(1\leq i\leq j\). One can also write a formal product \[n=\prod_{all\hspace{0.1cm} primes\hspace{0.1cm} p_i}p_i^{\alpha_i},\] where all but finitely many of the \(\alpha_i's\) are 0.
The prime factorization of 120 is given by
\(120=2\cdot 2\cdot 2\cdot 3\cdot 5=2^3\cdot 3\cdot 5.\)
Notice that 120 is written in the two ways described in [remark1].
We know describe in general how prime factorization can be used to determine the greatest common divisor of two integers. Let
\[a=p_1^{a_1}p_2^{a_2}...p_n^{a_n} \hspace{0.3cm}\mbox{and} \ \ b=p_1^{b_1}p_2^{b_2}...p_n^{b_n},\]
where we exclude in these expansions any prime \(p\) with power 0 in both \(a\) and \(b\) (and thus some of the powers above may be 0 in one expansion but not the other). Of course, if one prime \(p_i\) appears in \(a\) but not in \(b\), then \(a_i\neq 0\) while \(b_i=0\), and vise versa. Then the greatest common divisor is given by
\[(a,b)=p_1^{\min(a_1,b_2)}p_2^{min(a_2,b_2)}...p_n^{\min(a_n,b_n)}\]
where \(\min(n,m)\) is the minimum of \(m\) and \(n\).
The following lemma is a consequence of the Fundamental Theorem of Arithmetic.
Lemma
Let \(a\) and \(b\) be relatively prime positive integers. Then if \(d\) divides \(ab\), there exists \(d_1\) and \(d_2\) such that \(d=d_1d_2\) where \(d_1\) is a divisor of \(a\) and \(d_2\) is a divisor of \(b\). Conversely, if \(d_1\) and \(d_2\) are positive divisors of \(a\) and \(b\), respectively, then \(d=d_1d_2\) is a positive divisor of \(ab\).
Let \(d_1=(a,d)\) and \(d_2=(b,d)\). Since \((a,b)=1\) and writing \(a\) and \(b\) in terms of their prime decomposition, it is clear that \(d=d_1d_2\) and \((d_1,d_2)=1\). Note that every prime power in the factorization of \(d\) must appear in either \(d_1\) or \(d_2\). Also the prime powers in the factorization of \(d\) that are prime powers dividing \(a\) must appear in \(d_1\) and that prime powers in the factorization of \(d\) that are prime powers dividing \(b\) must appear in \(d_2\).
Now conversely, let \(d_1\) and \(d_2\) be positive divisors of \(a\) and \(b\), respectively. Then \[d=d_1d_2\] is a divisor of \(ab\).
More on the Infinitude of Primes
There are also other theorems that discuss the infinitude of primes in a given arithmetic progression. The most famous theorem about primes in arithmetic progression is Dirichlet’s theorem
Dirichlet’s Theorem
Given an arithmetic progression of terms \(an+b\) , for \(n=1, 2, ...\) ,the series contains an infinite number of primes if \(a\) and \(b\) are relatively prime,
This result had been conjectured by Gauss but was first proved by Dirichlet. Dirichlet proved this theorem using complex analysis, but the proof is so challenging. As a result, we will present a special case of this theorem and prove that there are infinitely many primes in a given arithmetic progression. Before stating the theorem about the special case of Dirichlet’s theorem, we prove a lemma that will be used in the proof of the mentioned theorem.
If \(a\) and \(b\) are integers both of the form \(4n+1\), then their product \(ab\) is of the form \(4n+1\)
Let \(a=4n_1+1\) and \(b=4n_2+1\), then \[ab=16n_1n_2+4n_1+4n_2+1=4(4n_1n_2+n_1+n_2)+1=4n_3+1,\] where \(n_3=4n_1n_2+n_1+n_2\).
There are infinitely many primes of the form \(4n+3\), where \(n\) is a positive integer.
Suppose that there are finitely many primes of the form \(4n+3\), say \(p_0=3,p_1,p_2,...,p_n\). Let \[N=4p_1p_2...p_n+3.\] Notice that any odd prime is of the form \(4n+1\) or \(4n+3\). Then there is at least one prime in the prime factorization of \(N\) of the form \(4n+3\), as otherwise, by Lemma 7, \(N\) will be in the form \(4n+1\). We wish to prove that this prime in the factorization of \(N\) is none of \(p_0=3,p_1,p_2,...,p_n\). Notice that if \[3\mid N,\] then \(3 \mid (N-3)\) and hence \[3 \mid 4p_1p_2...p_n\] which is impossible since \(p_i\neq 3\) for every \(i\). Hence 3 doesn’t divide \(N\). Also, the other primes \(p_1,p_2,...,p_n\) don’t divide \(N\) because if \(p_i \mid N\), then \[p_i\mid (N-4p_1p_2...p_n)=3.\] Hence none of the primes \(p_0,p_1,p_2,...,p_n\) divides N. Thus there are infinitely many primes of the form \(4n+3\).
Exercises
Find the prime factorization of 32, of 800 and of 289.
Find the prime factorization of 221122 and of 9!.
Show that all the powers of in the prime factorization of an integer \(a\) are even if and only if a is a perfect square.
Show that there are infinitely many primes of the form \(6n+5\). |
The definition of incompressible is often unclear and changes depending on which community uses it. So let's look at some common definitions:
Constant density
This means the density is constant everywhere in space and time. So:$$\frac{D\rho}{Dt} = \frac{\partial \rho}{\partial t} + \vec{u}\cdot\nabla{\rho} = 0$$Because density is constant everywhere in space and time, the temporal derivative is zero, and the spatial gradient is zero.
Low Mach Number
This shows up when the flow velocity is relatively low and so all pressure changes are hydrodynamic (due to velocity motion) rather than thermodynamic. The effect of this is that $\partial \rho / \partial p = 0$. In other words, the small changes in pressure due to flow velocity changes do not change the density. This has a secondary effect -- the speed of sound in the fluid is $\partial p/\partial \rho = \infty$ in this instance. So there is an infinite speed of sound, which makes the equations elliptic in nature.
Although we assume density is independent of pressure, it is possible for density to change due to changes in temperature or composition if the flow is chemically reacting. This means:
$$\frac{D\rho}{Dt} \neq 0$$because $\rho$ is a function of temperature and composition. If, however, the flow is not reacting or multi-component, you will also get the same equation as the constant density case:
$$\frac{D\rho}{Dt} = 0$$
Therefore, incompressible can mean constant density, or it can mean low Mach number, depending on the community and the application. I prefer to be explicit in the difference because I work in the reacting flow world where it matters. But many in the non-reacting flow communities just use incompressible to mean constant density.
Example of non-constant density
Since it was asked for an example where the material derivative is zero but density is not constant, here goes:
$$\frac{D\rho}{Dt} = \frac{\partial \rho}{\partial t} + \vec{u}\cdot\nabla \rho = 0$$
Rearrange this:
$$\frac{\partial \rho}{\partial t} = -\vec{u}\cdot\nabla\rho$$
gives a flow where $\rho \neq \text{const.}$ yet $D\rho/Dt = 0$. It has to be an unsteady flow.
Is there another example of steady flow? In steady flow, the time derivative is zero, so you have:
$$\vec{u}\cdot\nabla\rho = 0$$
If velocity is not zero, $\vec{u} \neq 0$, then we have $\nabla \rho = 0$ and so any moving, steady flow without body forces (gravity) or temperature/composition differences must have constant density.
If velocity is zero, you can have a gradient in density without any issues. Think of a column of the atmosphere for example -- density is higher at the bottom than the top due to gravity, and there is no velocity. So again, $D\rho/Dt = 0$ but density is not constant everywhere. The challenge here of course is that the continuity equation is not sufficient to describe the situation since it becomes $0 = 0$. You would have to include the momentum equation to incorporate the gravity forces. |
Inner Product Spaces Review
Inner Product Spaces Review
We will now review some of the recent content regarding inner product spaces.
Recall from the Inner Product Spaces page that if $V$ is a vector space over $\mathbb{R}$ or $\mathbb{C}$ then an Inner Producton $V$ is a function which takes every pair of vectors $u, v \in V$ and maps it to a number $<u, v>$ that satisfies the following properties for all $u, v, w \in V$ and $a \in \mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$):
Positivity Property Definiteness Property Additivity in First Slot Homogeneity in First Slot Conjugate Symmetry Property $<u, u> ≥ 0$ $<u, u> = 0$ if and only $u = 0$. $<u + v, w> = <u, w> + <v, w>$ $<au, v> = a<u, v>$ $<u, v> = \overline{<v, u>}$ We also saw that inner product spaces also had additivity in the second slot, that is $<u, v + w> = <u, v> + <u, w>$ for all $u, v, w \in V$ and conjugate homogeneity in the second slot, that is $<u, av> = \overline{a}<u, v>$ for all $u, v \in V$ and $a \in \mathbb{F}$ ( $\mathbb{R}$ or $\mathbb{C}$). Perhaps the simplest inner product space in the generic dot product defined on $\mathbb{R}^n$. For $u = (u_1, u_2, ..., u_n)$ and $v = (v_1, v_2, ..., v_n)$ we have that $<u, v> = u_1v_1 + u_2v_2 + ... + u_nv_n$. Another example of an inner product space is defined on $\wp (\mathbb{R})$ as $<p(x), q(x)> = \int_0^1 p(x)q(x) \: dx$ for all $p(x), q(x) \in \wp(\mathbb{R})$. If $V$ is a vector space with an inner product then $V$ is called an Inner Product Space. Furthermore if $<u, v> = 0$ for $u, v \in V$ then $u$ and $v$ are said to be Orthogonalto each other. On the Formulas for The Inner Product we proved the following formulas for inner products. If $V$ is an inner product space over $\mathbb{R}$ then:
\begin{align} \quad <u, v> = \frac{ \| u + v \|^2 - \| u - v \|^2}{4} \end{align}
If $V$ is an inner product space over $\mathbb{C}$ then:
\begin{align} \quad <u, v> = \frac{ \| u + v \|^2 - \| u - v \|^2 + \| u + iv \|^2i - \| u - iv \|^2 i}{4} \end{align}
On The Pythagorean Theorem for Inner Product Spaces we proved the famous Pythagorean Theorem(for inner product spaces) which says that if $V$ is an inner product space and if $u, v \in V$ are orthogonal then:
\begin{align} \quad \| u + v \|^2 = \| u \|^2 + \| v \|^2 \end{align}
On The Cauchy-Schwarz Inequality page we proved one of the most famous inequalities in mathematics known as the Cauchy-Schwarz Inequalitywhich says that if $V$ is an inner product space then for all $u, v \in V$ with equality holding if and only if $u$ is a multiple of $v$:
\begin{align} \quad \mid <u, v> \mid ≤ \| u \| \| v \| \end{align}
We then proved the Triangle Inequalityon The Triangle Inequality for Inner Product Spaces page which says that if $V$ is an inner product space then for all $u, v \in V$ with equality holding if and only if $u$ is a nonnegative scalar multiple of $v$, we have that:
\begin{align} \quad \| u + v \| ≤ \| u \| + \| v \| \end{align}
Then we proved The Parallelogram Identityon The Parallelogram Identity for Inner Product Spaces page which says that if $u, v \in V$ then:
\begin{align} \quad \| u + v \|^2 + \| u - v \|^2 = 2 \| u \|^2 + 2 \| v \|^2 \end{align} |
Character conditionEdit
The most common character conditions found in
Fallout 3, Fallout: New Vegas, and Fallout 4 are poisons from Chems or from the environment, and damage from combat. Fallout 4 is currently the only one in the Fallout series to include illnesses. PoisonedEdit InjuriesEdit OtherEdit Fallout 4Edit IllnessEdit
All illness can be cured with antibiotics, being well rested for several days, or by any doctor. Specific illnesses can be prevented by taking the appropriate herbal remedy prior to the disease check.
Untreated Illness (if not fatal) generally persists for around a week of rest (well fed, hydrated and rested). However, there is a chance it will persist for an additional week.
Illness chance in general is increased by all forms of Fatigue but does not directly impact Action Points. Regardless of risk exposures, the bed slept in and the duration of sleep, there is always a slight chance of contracting an illness when going to sleep (but announced upon waking); sleeping for short periods of time (to save the game, for example) increases this chance overall, due to an increased number of (shorter) sleep sessions, although all sleep sessions after the first will be at the minimum probability, assuming no further risk disease exposures between sleeps.
ChecksEdit
An immediate check for contracting a disease is made 30 real time seconds after a high risk exposure. A check is also made when going to sleep. Apart from these two situations, risk exposure simply increases the probability (disease risk pool) of getting a disease at the next check. Checks are not "rolled for" unless the chance is 25%+.
All checks resulting in a new disease, clear the risk pool down to its minimum value (which depends on current Fatigue level). All checks that don't produce a new disease cause a small drain down of the risk pool. All checks upon sleeping also clear the risk pool down to the minimum, after the check.
Risk factorsEdit
In addition to poor sleep quality, causes of disease risk include: using chems, eating anything (but especially bad or uncooked meat), drinking or swimming in irradiated water, being in rain, drinking Nuka-Cola and taking damage from feral ghouls, bugs or other infected creatures. Each event puts the player character at increased risk of contracting an illness. High risk events trigger a near immediate disease check. Even if a single event doesn't trigger an illness to start, each event will increase the probability of contracting an illness with each event, up until the character sleeps (when the final check is made and the disease risk pool is cleared) or gets a new disease (which also clears down the risk pool).
List of diseasesEdit Fatigue - Require more frequent sleep to avoid debuffs. Infection - Periodic damage. Similar to Severely Dehydrated and Starving. Insomnia - Sleep provides half as much reduction in tiredness. Lethargy - 50% slower AP regeneration. Parasites - Must eat x2 as much food to avoid hunger or to eliminate any given hunger level. Weakness - Take +20% more damage. Herbal remediesEdit
Various herbal remedies become available for crafting, purchase and as a loot in Survival mode:
Each type of remedy will act (one time only per dose) to prevent one of the two specific diseases that it prevents against, but only if the herbal remedy is taken before the disease check. This means taking the herbal remedy prior to sleeping, or within 30 seconds (real time) of a high risk event.
Fallout 76Edit Equipment conditionEdit
Note that any equations are not to be taken as literal representations of what happens in-game, but as accurate, general representations of the logic.
If the condition of an item degrades all the way to 0%, the item is broken and cannot be used until repaired. In the case of equipped armor and clothing, bonuses to skills or attributes will still apply even if the item is broken, but once removed it will have to be repaired before it can be re-equipped.
Another consequence of poor condition when using weapons is the reload animation. The lower the condition of a weapon, the more likely it is to jam, causing the player to adjust the magazine, thus taking longer to complete the animation. A weapon at 100% condition (or at least 75% in
New Vegas) will never jam, while weapons in a poor state of repair will jam quite frequently. ValueEdit
The value of an item is calculated by
[1]: $ \text{Value} = \text{Base value} \times \text{Condition} ^ {1.5} $ Conditionis a number ranging from 0 to 1.
Also, knowing the current value of an item it is possible to find the condition:
$ \text{Condition} = \left(\frac{\text{Value}}{\text{Base value}}\right)^\frac{2}{3} $
For example, the 10mm pistol has a maximum value of 225 caps when in perfect condition. At 50% (or 0.5) condition, the pistol's value is 225 * (0.5 ^ 1.5) = 225 * 0.3535, or 35.3% of its base value (225): 79 caps.
Fallout 3Edit DamageEdit
As condition degrades, non-melee weapon damage will scale linearly from full damage at 100% condition to roughly two-thirds damage at 0% condition for single-shot ranged weapons and roughly half damage for fully-automatic ranged weapons and melee weapons.
[2] $ \text{Dam}_{\text{Skill}=100}= \text{Base} \times (\text{Minimum} + \text{Condition} \times (1 - \text{Minimum})) $ Conditionis a number ranging from 0 to 1. Minimumis .66 for single-shot weapons, .54 for fully-automatic weapons, .5 for melee weapons. $ \text{Dam}_{\text{Guns}=100}=9 \times (.66 + .25 \times .34) = 6.705 \approx 7 $ $ \text{Dam}_{\text{Guns}=100}=38 \times (.54 + .25 \times .46) = 24.89 \approx 25 $ Armor protectionEdit $ \text{DR or DT}=\text{Base} \times (.62 + \text{Condition} \times .38) $ Equipment degradationEdit
All equipable items in Fallout 3 and Fallout: New Vegas have
item health, which helps determine how quickly an item condition degrades from use and from being hit in combat. At full item health, an item is in 100% condition. Similarly, at close to 0 item health, an item is about to break.
In Fallout 3, item health decreases per use by a percentage of the base damage of the weapon.
Small Guns degrade at 3% of base weapon damage per shot. Energy Weapons degrade at 4% of base weapon damage per shot. Unarmed and Melee Weapons degrade at 5% of base weapon damage per attack. Big Guns degrade at 6% of base weapon damage per shot.
For example, the Plasma rifle has a base damage of 45 and an item health of 900. Each shot will decay the health by 1.8 health, which means that it takes 500 shots to take this plasma rifle from full condition to broken.
Fallout: New VegasEdit DamageEdit
As condition degrades, weapon damage will scale linearly from full damage at 75% condition to half the damage at 0% condition.
$ \text{Dam}_{\text{Skill}=100}= \text{Base} \times \left( 0.5 + \min \left( \frac{0.5\times \text{Condition}}{0.75}, 0.5 \right) \right) $ Armor protectionEdit
As condition degrades, armor protection will scale linearly from full protection at 50% condition to about 2/3 of full protection at 0% condition.
$ \text{DR or DT}=\text{Base} \times \left( 0.66 + \min \left( \frac{0.34\times \text{Condition}}{0.5}, 0.34 \right) \right) $ Equipment degradationEdit
Weapon condition degrades at a flat rate of 0.2 item health per shot. V.A.T.S. does not impose an additional penalty, but various types of ammunition can increase or decrease the degradation rate (indicated by a "CND x 1.5" or similar in the Pip-Boy 3000, which means a 1.5 multiplier for decay). Having Raul Tejada as a companion can make a weapon degrade slower (a .5 multiplier for Regular Maintenance, a .25 multiplier for Full Maintenance).
Armor condition degrades if an incoming attack power exceeds DT of the armor. |
Difference between revisions of "Inaccessible"
(Organized a bit)
(→Hyper-inaccessible: Meta-ordinal)
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Therefore $2$-inaccessibility is weaker than $3$-inaccessibility, which is weaker than $4$-inaccessibility... all of which are weaker than $\omega$-inaccessibility, which is weaker than $\omega+1$-inaccessibility, which is weaker than $\omega+2$-inaccessibility...... all of which are weaker than hyperinaccessibility, etc.
Therefore $2$-inaccessibility is weaker than $3$-inaccessibility, which is weaker than $4$-inaccessibility... all of which are weaker than $\omega$-inaccessibility, which is weaker than $\omega+1$-inaccessibility, which is weaker than $\omega+2$-inaccessibility...... all of which are weaker than hyperinaccessibility, etc.
−
==Hyper-inaccessible==
+
==Hyper-inaccessible ==
A cardinal $\kappa$ is ''hyperinaccessible'' if it is $\kappa$-inaccessible. One may similarly define that $\kappa$ is $\alpha$-hyperinaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$, it is a limit of $\beta$-hyperinaccessible cardinals. Continuing, $\kappa$ is ''hyperhyperinaccessible'' if $\kappa$ is $\kappa$-hyperinaccessible.
A cardinal $\kappa$ is ''hyperinaccessible'' if it is $\kappa$-inaccessible. One may similarly define that $\kappa$ is $\alpha$-hyperinaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$, it is a limit of $\beta$-hyperinaccessible cardinals. Continuing, $\kappa$ is ''hyperhyperinaccessible'' if $\kappa$ is $\kappa$-hyperinaccessible.
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More generally, $\kappa$ is ''hyper${}^\alpha$-inaccessible'' if it is hyperinaccessible and for every $\beta\lt\alpha$ it is $\kappa$-hyper${}^\beta$-inaccessible, where $\kappa$ is ''$\alpha$-hyper${}^\beta$-inaccessible'' if it is hyper${}^\beta$-inaccessible and for every $\gamma<\alpha$, it is a limit of $\gamma$-hyper${}^\beta$-inaccessible cardinals.
More generally, $\kappa$ is ''hyper${}^\alpha$-inaccessible'' if it is hyperinaccessible and for every $\beta\lt\alpha$ it is $\kappa$-hyper${}^\beta$-inaccessible, where $\kappa$ is ''$\alpha$-hyper${}^\beta$-inaccessible'' if it is hyper${}^\beta$-inaccessible and for every $\gamma<\alpha$, it is a limit of $\gamma$-hyper${}^\beta$-inaccessible cardinals.
−
Every [[Mahlo]] cardinal $\kappa$ is
+
Every [[Mahlo]] cardinal $\kappa$ is $\$-inaccessible .
Revision as of 14:05, 29 April 2019 Inaccessible cardinals are the traditional entry-point to the large cardinal hierarchy, although weaker notions such as the worldly cardinals can still be viewed as large cardinals.
A cardinal $\kappa$ being inaccessible implies the following:
$V_\kappa$ is a model of ZFC and so inaccessible cardinals are worldly. The worldly cardinals are unbounded in $\kappa$, so $V_\kappa$ satisfies the existence of a proper class of worldly cardinals. $\kappa$ is an aleph fixed point and a beth fixed point, and consequently $V_\kappa=H_\kappa$. (Solovay)there is an inner model of a forcing extension satisfying ZF+DC in which every set of reals is Lebesgue measurable; in fact, this is equiconsistent to the existence of an inaccessible cardinal. For any $A\subseteq V_\kappa$, the set of all $\alpha<\kappa$ such that $\langle V_\alpha;\in,A\cap V_\alpha\rangle\prec\langle V_\kappa;\in,A\rangle$ is club in $\kappa$.
An ordinal $\alpha$ being inaccessible is equivalent to the following:
$V_{\alpha+1}$ satisfies $\mathrm{KM}$. $\alpha>\omega$ and $V_\alpha$ is a Grothendiek universe. $\alpha$ is $\Pi_0^1$-Indescribable. $\alpha$ is $\Sigma_1^1$-Indescribable. $\alpha$ is $\Pi_2^0$-Indescribable. $\alpha$ is $0$-Indescribable. $\alpha$ is a nonzero limit ordinal and $\beth_\alpha=R_\alpha$ where $R_\beta$ is the $\beta$-th regular cardinal, i.e. the least regular $\gamma$ such that $\{\kappa\in\gamma:\mathrm{cf}(\kappa)=\kappa\}$ has order-type $\beta$. $\alpha = \beth_{R_\alpha}$. $\alpha = R_{\beth_\alpha}$. $\alpha$ is a weakly inaccessible strong limit cardinal (see weakly inaccessible below). Contents Weakly inaccessible cardinal
A cardinal $\kappa$ is
weakly inaccessible if it is an uncountable regular limit cardinal. Under GCH, this is equivalent to inaccessibility, since under GCH every limit cardinal is a strong limit cardinal. So the difference between weak and strong inaccessibility only arises when GCH fails badly. Every inaccessible cardinal is weakly inaccessible, but forcing arguments show that any inaccessible cardinal can become a non-inaccessible weakly inaccessible cardinal in a forcing extension, such as after adding an enormous number of Cohen reals (this forcing is c.c.c. and hence preserves all cardinals and cofinalities and hence also all regular limit cardinals). Meanwhile, every weakly inaccessible cardinal is fully inaccessible in any inner model of GCH, since it will remain a regular limit cardinal in that model and hence also be a strong limit there. In particular, every weakly inaccessible cardinal is inaccessible in the constructible universe $L$. Consequently, although the two large cardinal notions are not provably equivalent, they are equiconsistent.
There are a few equivalent definitions of weakly inaccessible cardinals. In particular:
Letting $R$ be the transfinite enumeration of regular cardinals, a limit ordinal $\alpha$ is weakly inaccessible if and only if $R_\alpha=\aleph_\alpha$ A nonzero cardinal $\kappa$ is weakly inaccessible if and only if $\kappa$ is regular and there are $\kappa$-many regular cardinals below $\kappa$; that is, $\kappa=R_\kappa$. A regular cardinal $\kappa$ is weakly inaccessible if and only if $\mathrm{REG}$ is unbounded in $\kappa$ (showing the correlation between weakly Mahlo cardinals and weakly inaccessible cardinals, as stationary in $\kappa$ is replaced with unbounded in $\kappa$) Levy collapse
The Levy collapse of an inaccessible cardinal $\kappa$ is the $\lt\kappa$-support product of $\text{Coll}(\omega,\gamma)$ for all $\gamma\lt\kappa$. This forcing collapses all cardinals below $\kappa$ to $\omega$, but since it is $\kappa$-c.c., it preserves $\kappa$ itself, and hence ensures $\kappa=\omega_1$ in the forcing extension.
Inaccessible to reals
A cardinal $\kappa$ is
inaccessible to reals if it is inaccessible in $L[x]$ for every real $x$. For example, after the Levy collapse of an inaccessible cardinal $\kappa$, which forces $\kappa=\omega_1$ in the extension, the cardinal $\kappa$ is of course no longer inaccessible, but it remains inaccessible to reals. Universes
When $\kappa$ is inaccessible, then $V_\kappa$ provides a highly natural transitive model of set theory, a universe in which one can view a large part of classical mathematics as taking place. In what appears to be an instance of convergent evolution, the same universe concept arose in category theory out of the desire to provide a hierarchy of notions of smallness, so that one may form such categories as the category of all small groups, or small rings or small categories, without running into the difficulties of Russell's paradox. Namely, a
Grothendieck universe is a transitive set $W$ that is closed under pairing, power set and unions. That is, (transitivity) If $b\in a\in W$, then $b\in W$. (pairing) If $a,b\in W$, then $\{a,b\}\in W$. (power set) If $a\in W$, then $P(a)\in W$. (union) If $a\in W$, then $\cup a\in W$.
The
Grothendieck universe axiom is the assertion that every set is an element of a Grothendieck universe. This is equivalent to the assertion that the inaccessible cardinals form a proper class. Degrees of inaccessibility
A cardinal $\kappa$ is
$1$-inaccessible if it is inaccessible and a limit of inaccessible cardinals. In other words, $\kappa$ is $1$-inaccessible if $\kappa$ is the $\kappa^{\rm th}$ inaccessible cardinal, that is, if $\kappa$ is a fixed point in the enumeration of all inaccessible cardinals. Equivalently, $\kappa$ is $1$-inaccessible if $V_\kappa$ is a universe and satisfies the universe axiom.
More generally, $\kappa$ is $\alpha$-inaccessible if it is inaccessible and for every $\beta\lt\alpha$ it is a limit of $\beta$-inaccessible cardinals.
$1$-inaccessibility is already consistency-wise stronger than the existence of a proper class of inaccessible cardinals, and $2$-inaccessibility is stronger than the existence of a proper class of $1$-inaccessible cardinals. More specifically, a cardinal $\kappa$ is $\alpha$-inaccessible if and only if for every $\beta<\alpha$: $$V_{\kappa+1}\models\mathrm{KM}+\text{There is a proper class of }\beta\text{-inaccessible cardinals}$$
As a result, if $\kappa$ is $\alpha$-inaccessible then for every $\beta<\alpha$: $$V_\kappa\models\mathrm{ZFC}+\text{There exists a }\beta\text{-inaccessible cardinal}$$
Therefore $2$-inaccessibility is weaker than $3$-inaccessibility, which is weaker than $4$-inaccessibility... all of which are weaker than $\omega$-inaccessibility, which is weaker than $\omega+1$-inaccessibility, which is weaker than $\omega+2$-inaccessibility...... all of which are weaker than hyperinaccessibility, etc.
Hyper-inaccessible and more
A cardinal $\kappa$ is
hyperinaccessible if it is $\kappa$-inaccessible. One may similarly define that $\kappa$ is $\alpha$-hyperinaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$, it is a limit of $\beta$-hyperinaccessible cardinals. Continuing, $\kappa$ is hyperhyperinaccessible if $\kappa$ is $\kappa$-hyperinaccessible.
More generally, $\kappa$ is
hyper${}^\alpha$-inaccessible if it is hyperinaccessible and for every $\beta\lt\alpha$ it is $\kappa$-hyper${}^\beta$-inaccessible, where $\kappa$ is $\alpha$-hyper${}^\beta$-inaccessible if it is hyper${}^\beta$-inaccessible and for every $\gamma<\alpha$, it is a limit of $\gamma$-hyper${}^\beta$-inaccessible cardinals.
Meta-ordinal terms are terms like $Ω^α · β + Ω^γ · δ +· · ·+Ω^\epsilon · \zeta + \theta$ where $α, β...$ are ordinals. They are ordered as if $Ω$ were an ordinal greater then all the others. $(Ω · α + β)$-inaccessible denotes $β$-hyper${}^α$-inaccessible, $Ω^2$-inaccessible denotes hyper${}^\kappa$-inaccessible $\kappa$ etc. Every Mahlo cardinal $\kappa$ is $\Omega^α$-inaccessible for all $α<\kappa$. Similar hierarchy exists for Mahlo cardinals below weakly compact. All such properties can be killed softly be forcing to make them any weaker properties from this family.[1] |
Table of Contents
Products of Paths Relative to {0, 1} in a Topological Space
Definition: Let $X$ be a topological space. A Path in $X$ is a continuous function $\alpha : [0, 1] \to X$.
Definition: Let $X$ be a topological space and let $\alpha, \beta : [0, 1] \to X$ be paths such that $\alpha(1) = \beta (0)$. The Product of the paths $\alpha$ and $\beta$ is the new path $\alpha \beta : [0, 1] \to X$ defined for all $t \in [0, 1]$ by $\alpha \beta = \left\{\begin{matrix} \alpha(2t) & \mathrm{if} \: 0 \leq t \leq \frac{1}{2}\\ \beta (2t - 1) & \mathrm{if} \: \frac{1}{2} \leq t \leq 1 \end{matrix}\right.$. Observe that the order of the notation for denoting the path product $\alpha \beta$, is important. We traverse the pathes in the product from left to right. We first traverse $\alpha$ at twice the speed, and hence $\beta$ at twice the speed. Since the terminal point of $\alpha$ is equal to the initial point of $\beta$, $\alpha \beta$ is indeed a continuous function.
Let $X$ be a topological space and let $\alpha : I \to X$ be a path. Observe that $I = [0, 1]$ is a topological space and that $A = \{0, 1 \} \subset I$. From the Homotopic Mappings Relative to a Subset of a Topological Space, we proved that the relation that two continuous functions are homotopic relative to a subset $A$ of the domain is an equivalence relation. We let:(1)
In the following proposition we show that multiplication of such equivalence classes is a well-defined operation.
Theorem 1: Let $X$ be a topological space and let $\alpha, \beta : I \to X$ be paths such that $\alpha (1) = \beta (0)$. Then $[\alpha][\beta] := [\alpha\beta]$ is well-defined. Proof:Let $\alpha, \beta : I \to X$ be paths and let $\alpha', \beta' : I \to X$ also be paths such that: We want to then show that $[\alpha][\beta] = [\alpha'][\beta']$, i.e., $[\alpha\beta] = [\alpha'\beta']$ which shows that the product in the theorem is well-defined and independent of the choice of representative path for the equivalence class. Since $\alpha \simeq_{\{0, 1\}} \alpha'$ there exists a continuous function $H : I \times I \to X$ such that $H_0 = \alpha$, $H_1 = \alpha'$, $H_t(0) = \alpha(0) = \alpha'(0)$ and $H_t(1) = \alpha(1) = \alpha'(1)$ for all $t \in I$. Similarly, since $\beta \simeq_{\{0, 1\}} \beta'$ there exists a continuous function $H' : I \times I \to X$ such that $H'_0 = \beta$ $H'_1 = \beta'$, $H'_t(0) = \beta(0) = \beta'(0)$ and $H'_t(1) = \beta(1) = \beta'(1)$ for all $t \in I$. Define a new function $H^* : I \times I \to X$ by: Then $H^*$ is continuous since $H$ and $H'$ are continuous and by The Gluing Lemma. Also: Therefore $\alpha \beta \simeq_{\{0, 1 \}} \alpha ' \beta '$. So $[\alpha\beta] = [\alpha'\beta']$. So indeed, the product $[\alpha][\beta] := [\alpha\beta]$ is well-defined. $\blacksquare$ |
I would like to ask that at what distance from the Earth's surface the curvature of the Earth is visible. What layer of the atmosphere is this?
I've noticed that at the height of 9-12 Km (the view from from aeroplanes) it is not visible.
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Depends on your eye. You can realise the curvature of the Earth by just going to the beach. Last summer I was on a scientific cruise in the Mediterranean. I took two pictures of a distant boat, within an interval of a few seconds: one from the lowest deck of the ship (left image), the other one from our highest observation platform (about 16 m higher; picture on the right):
A distant boat seen from 6 m (left) and from 22 m (right) above the sea surface. This boat was about 30 km apart. My pictures, taken with a 30x optical zoom camera.
The part of the boat that is missing in the left image is hidden by the quasi-spherical shape of the Earth. In fact, if you would know the size of the boat and its distance, we could infer the radius of the Earth. But since we already know this, let's do it the other way around and deduce the distance to which we can see the full boat:
The distance $d$ from an observer $O$ at an elevation $h$ to the visible horizon follows the equation (adopting a spherical Earth):
$$ d=R\times\arctan\left(\frac{\sqrt{2\times{R}\times{h}}}{R}\right) $$
where $d$ and $h$ are in meters and $R=6370*10^3m$ is the radius of the Earth. The plot is like this:
Distance of visibility d (vertical axis, in km), as a function of the elevation h of the observer above the sea level (horizontal axis, in m).
From just 3 m above the surface, you can see the horizon 6.2 km apart. If you are 30 m high, then you can see up to 20 km far away. This is one of the reasons why the ancient cultures, at least since the sixth century BC, knew that the Earth was curved, not flat. They just needed good eyes. You can read first-hand Pliny (1st century) on the unquestionable spherical shape of our planet in his
Historia Naturalis.
But addressing more precisely the question. Realising that the horizon is lower than normal (lower than the perpendicular to gravity) means realising the angle ($gamma$) that the horizon lowers below the flat horizon (angle between $OH$ and the tangent to the circle at
O, see cartoon below; this is equivalent to gamma in that cartoon). This angle depends on the altitude $h$ of the observer, following the equation:
$$ \gamma=\frac{180}{\pi}\times\arctan\left(\frac{\sqrt{2\times{R}\times{h}}}{R}\right) $$
where
gamma is in degrees, see the cartoon below. Angle of the horizon below the flat-Earth horizon ( gamma, in degrees, on the vertical axis of this plot) as a function of the observer's elevation h above the surface (meters). Note that the apparent angular size of the Sun or the Moon is around 0.5 degrees..
So, at an altitude of only 290 m above the sea level you can already see 60 km far and the horizon will be lower than normal by the same angular size of the sun (half a degree). While normally we are no capable of feeling this small lowering of the horizon, there is a cheap telescopic device called levelmeter that allows you to point in the direction perpendicular to gravity, revealing how lowered is the horizon when you are only a few meters high.
When you are on a plane ca. 10,000 m above the sea level, you see the horizon 3.2 degrees below the astronomical horizon (O-H), this is, around 6 times the angular size of the Sun or the Moon. And you can see (under ideal meteorological conditions) to a distance of 357 km.
Felix Baumgartner roughly doubled this number but the pictures circulated in the news were taken with very wide angle, so the ostensible curvature of the Earth they suggest is mostly an artifact of the camera, not what Felix actually saw. This ostensible curvature of the Earth is mostly an artifact of the camera's wide-angle objective, not what Felix Baumgartner actually saw.
A quick Google turned up a published article answering precisely this question (Lynch, 2008). The abstract states:
Reports and photographs claiming that visual observers can detect the curvature of the Earth from high mountains or high-flying commercial aircraft are investigated. Visual daytime observations show that the minimum altitude at which curvature of the horizon can be detected is at or slightly below 35,000 ft, providing that the field of view is wide (60°) and nearly cloud free. The high-elevation horizon is almost as sharp as the sea-level horizon, but its contrast is less than 10% that of the sea-level horizon. Photographs purporting to show the curvature of the Earth are always suspect because virtually all camera lenses project an image that suffers from barrel distortion. To accurately assess curvature from a photograph, the horizon must be placed precisely in the center of the image, i.e., on the optical axis.
Note that the given minimum of 35,000 feet (10.7 km) is a plausible cruise altitude for a commercial airliner, but you probably shouldn't expect to see the curvature on a typical commercial flight, because:
Lynch, D. K. (2008). Visually discerning the curvature of the Earth.
Applied Optics, 47(34), H39-H43.
It's hard to see the curvature of the earth from an altitude of 7 miles or 37,000 ft (typical cruising altitude of a jetliner) but easy to see from 250 miles (typical altitude of the ISS).
The line of sight from an aircraft at 37,000 feet = 235 miles. That's only about 3.4 degrees of the earth's surface. From the ISS at 250 miles, the line of sight is 1,435 miles, which covers about 19.8 degrees of the earth's surface - much easier to see the curve from this altitude.
Most people don't realize how large the earth is compared to the altitude of a passenger aircraft. It's easy to think we're really high up, but comparatively we're just skimming the surface.
The attached drawing is to scale, but the images of the jetliner and ISS are NOT to scale (much, much larger than their actual sizes).
Further to DrGC's excellent answer, a subjective assessment of visibility of the Earth's curvature can be gleaned from pilot's experioence over many decades. These can be summarized as:
High up on a peak in Hawaii surrounded by nothing but water in every direction, seeing the curvature can be really quite humbling. As far as the boat theory goes, its not something I'd be able to use considering Im aware of the unnerving sizes of deep sea swells and counting rogue waves, naturally between those the boat is at a low point. Having parents that use to go deep sea fishing often, spending over a week out at sea, the swells are...huge.
Is not the amount of curvature available to see reduced by looking at it from a very flat angle - ie multiplied by the sine of that small angle? At 35000 feet the horizon is 229 miles away and 440 miles long, with the maximum field of vision of the human eye of 110 degrees (not attainable in practice) so the curvature depth is 78 miles, but because of the flatness of view, it foreshortens to about 2.4 miles (and much less with a narrower field of view). To resolve 2.4 miles at 229 distance over 440 miles miles is going some, or perhaps about 1 mile or less in practice through a window. Using a telescope does not help as all it does is reduce the angle of the field of view proportionately.
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Linear Maps Examples 4
Recall from the Linear Maps page that a linear map or linear transformation from the vector space $V$ to the vector space $W$ is a function $T : V \to W$ such that for all $u, v \in V$ and for all $a \in \mathbb{F}$ we have that $T(u + v) = T(u) + T(v)$ (additivity property) and $T(av) = aT(v)$ (homogeneity property).
We will now look at some more example questions regarding linear maps.
Example 1 Let $V$ be a vector space and let $U$ be a subspace of $V$ such that $U \neq V$. Let $S \in \mathcal L (U, W)$ where $S$ is not the zero map, and define a function $T : V \to W$ by $T(v) = \left\{\begin{matrix} S(v) \quad \mathrm{if} \: v \in U \\ 0 \quad \mathrm{if} \: v \not \in U \end{matrix}\right.$. Prove that $T$ is NOT a linear map.
To show that $T$ is not a linear map, we must show that either the additivity or the homogeneity property does not hold for some vectors/scalars in $V$/$\mathbb{F}$.
Choose a vector $v \in U$ such that $S(v) \neq 0$. This is possible since $S$ is not identically equal to the zero map. Also choose a vector $w \not \in U$. Then we have that $(v + w) \not \in U$ (since if $(v + w) \in U$ then since $v \in U$ we have by the closure under addition of $U$ that $(v + w) - (v) = w \in U$ which is a contradiction). Therefore since $(v + w) \not \in U$ we have that:(1)
However, we note that:(2)
Therefore $T(v + w) = 0 \neq T(v) + T(w)$ and so the additivity property does not hold, and so $T$ is not a linear map. |
This story actually starts with Einstein's paper on the photoelectric effect. Einstein proposed that for light waves, $E \propto f$, with a proportionality constant that eventually became known as $h$. Using the relation $E = pc$ from special relativity, you can derive that $pc = hf$, and with $\lambda f = c$ you get $\lambda = \frac{h}{p}$. Remember, though, so far this only applies to
light. de Broglie's insight was to use the same relation to define the wavelength of a particle as a function of its momentum.
So where does your derivation go wrong? The key step is $v = \lambda f$, which applies to a
wave, not a particle. As Qmechanic says, the wave velocity is not the same as the particle velocity. (The former is the phase velocity and the latter is the group velocity.)
Even though $\lambda = \frac{h}{p}$ was originally taken as an assumption, you can work backwards (or forwards, depending on your view) and derive it from a more general quantum theory. For example, suppose you start with the Schroedinger equation in free space,
$$i\hbar\frac{\partial \Psi(t,x)}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi(t,x)}{\partial x^2}$$
Solutions to this equation take the form
$$\Psi(t,x) = \sum_n C_n\exp\biggl(-\frac{i}{\hbar}\bigl(E_nt \pm x\sqrt{2mE_n}\bigr)\biggr) = \sum_n C_n\exp\biggl[-i\biggl(\omega_nt \pm k_nx\biggr)\biggr]$$
This is a wave with multiple individual components, each having angular frequency $$\omega_n = E_n/\hbar$$ and wavenumber $$k_n = \frac{\sqrt{2mE_n}}{\hbar}$$ or equivalently, frequency $$f_n = E_n/h$$ and wavelength $$\lambda_n = \frac{h}{\sqrt{2mE_n}}$$ To come up with de Broglie's relation, you need to find an expression for the momentum carried by the wave. This is done using the momentum operator $\hat{p} = -i\hbar\frac{\partial}{\partial x}$ in $\hat{p}\Psi = p\Psi$. The thing is, it only works for a wavefunction with one component. So if (and only if) all the $C_n$ are zero except one, you can get
$$p_n = \mp\hbar k_n = \mp\sqrt{2m E_n}$$
and if you put that together with the definition of $\lambda_n$, you get $\lambda_n = \frac{h}{p_n}$.
It may seem like a problem that this procedure only works for single-component waves. It's okay, though, because the wave doesn't actually have a single well-defined wavelength anyway unless it consists of only one component. This is a key point: whenever you talk about the wavelength of a particle, or more precisely the wavelength of the matter wave associated with a particle, you're implicitly assuming that the matter wave has only a single frequency component. This is generally a useful approximation for real particles, but it's never
exactly true. |
Preservation of Connectivity under Continuous Maps
Table of Contents
Preservation of Connectivity under Continuous Maps
One particularly nice property of connected topological spaces $X$ is that if $f : X \to Y$ is a continuous map then the range $f(X)$ is also connected. We prove this result below.
Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a continuous map. If $X$ is connected then $f(X)$ is connected. Proof:Let $X$ be connected and suppose that $f(X)$ is instead disconnected. We will show that a contradiction arises. Let $\{ A, B \}$ be a separation of $f(X)$, i.e., $A, B \subset f(X)$ are open in $f(X)$ with the subspace topology, $A, B \neq \emptyset$, $A \cap B = \emptyset$, and:
\begin{align} \quad f(X) = A \cup B \end{align}
Since $f$ is continuous, we see that $f^{-1}(A)$ and $f^{-1}(B)$ are open in $X$. We claim that $\{ f^{-1}(A), f^{-1}(B) \}$ is a separation of $X$. We have already established that $f^{-1}(A)$ and $f^{-1}(B)$ are open in $X$. Furthermore, $f^{-1}(A) \neq \emptyset$ and $f^{-1}(B) \neq \emptyset$. Clearly since $f(X) = A \cup B$ we see that $X = f^{-1}(A) \cup f^{-1}(B)$. All we need to check is that $f^{-1}(A) \cap f^{-1}(B) = \emptyset$. Suppose not. Then there exists an $x \in f^{-1}(A) \cap f^{-1}(B)$, so $f(x) \in A$ and $f(x) \in B$. But this implies that $A \cap B \neq \emptyset$ which is a contradiction. Thus, $\{ f^{-1}(A), f^{-1}(B) \}$ is a separation of $X$ which shows that $X$ is disconnected. But this is a contradiction. Therefore the assumption that $f(X)$ is disconnected is false. So if $X$ is a connected topological space and $f : X \to Y$ is continuous then $f(X)$ is also connected. $\blacksquare$
Corollary 1: If $X$ and $Y$ are topological spaces and $f : X \to Y$ is a homeomorphism between $X$ and $Y$ then if $X$ is connected then $Y$ is connected. Proof:If $f : X \to Y$ is a homeomorphism between $X$ and $Y$ then by definition $f$ is a continuous function. Furthermore, since $f$ is bijective, $f(X) = Y$. By Theorem 1 we immediately get that $f(X) = Y$ is connected. $\blacksquare$ |
Just curious: This monday, I had an exam in Knowledge Processing. They asked what's the problem with FOL (compared to propositional), and I gave the textbook answer that iterating functions gives infinitely many ground terms and makes it undecidable. And since I'm a bigmouth, lavedida, I added that I strongly suspect throwing just that feature out won't suffice. (Cf. e.g. planners like STRIPS, which are far weaker than FOL.)
We simply let that one go since it was irrelevant for the exam (and neither of us had the math knowledge) but I still want to know: Can you restrict FOL so that you still have quantors (which are the whole point) but the resulting axiom system is decidable?
Just curious: This monday, I had an exam in Knowledge Processing. They asked what's the problem with FOL (compared to propositional), and I gave the textbook answer that iterating functions gives infinitely many ground terms and makes it undecidable. And since I'm a bigmouth, lavedida, I added that I strongly suspect throwing just that feature out won't suffice. (Cf. e.g. planners like STRIPS, which are far weaker than FOL.)
closed as off-topic by Andrej Bauer, Pace Nielsen, YCor, Piotr Hajlasz, Joseph Van Name Feb 25 at 13:14
This question does not appear to be about research level mathematics within the scope defined in the help center.
There are various decidable fragments of FOL. Beyond prefix classes which Emil Jeřábek mentioned, other common fragments are:
Two variable logics. The fragment of FOL in which only two variables x and y are allowed is decidable. There are extensions of this logic which are decidable, e.g. with counting quantifiers $\exists ^{\leq 5} x$. Note that allowing even three variables leads to undecidability.
Guarded logics. Here all quantification is guarded by a predicate from the signature, e.g. $\forall x,y,z\, (R(x,y,z) \to S(z,z,y,x))$. This remains true with various extensions including that go beyond FOL such as fixed point logic.
If you want to know more, have a look at: Decidable Fragments of First-Order and Fixed-Point Logic - From prefix vocabulary classes to guarded logics (2003) by Erich Grädel. I believe it is unpublished, but you can google a manuscript and slides. |
Here are the conditions for making a substitution:
$\int_{y_0}^{y_1}f(y) dy = \int_{x_0}^{x_1}f(h(x))h'(x) dx$
if:1. $f : [y_0, y_1] → R$ is continuous on $[y_0, y_1]$,
$h : [x_0, x_1] → [y_0, y_1]$ is differentiable on $[x_0, x_1]$, with $h′$ continuous on $[x_0, x_1]$, $h(x_0) = y_0$ and $h(x_1) = y_1$
note that this doesn't actually mean that h has to be bijective.
By taking limits you can extend this to improper integrals with infinities in the limits.
Also if you can satisfy the conditions 1 and 2 above but $h(x_0)=y_1$ and $h(x_1)=y_0$ ($y_1 > y_0$) you can fiddle about and show that you still $\int_{y_0}^{y_1}f(y) dy = \int_{x_0}^{x_1}f(h(x))h'(x) dx$ (do this by subtituting with $h_2(x) = h(x_0 + x_1 - x)$ which does satisfy the conditions above).
So you can actually get it to work both ways,
In your case:
$t(x) = x^2$
$t((- \infty ,0]) = [0, \infty)$
$t'(x) = 2x$
$\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t} dt = \int_{0}^{-\infty}|x^{-1}|e^{-x^2}*2x dx = \int_{0}^{-\infty}-2e^{-x^2} dx$ (x<=0)
$=\int_{-\infty}^{0}2e^{-x^2} dx$
And you can see that this is going to be the same as $\int_{0}^{\infty}2e^{-x^2} dx$ which is what you get if you do it with the other choice of limits. |
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Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ... |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
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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 ... |
Take $f:[0,1]\to [0,1]^n$ a continuous tour around $[0,1]^n,$ say, some iteration of a Hilbert curve. For $\varepsilon \in (0,1)$ what is the following thing called and are there any nontrivial upper bounds?
\begin{equation} \max_{|a-b|<\varepsilon} \|f(a)-f(b)\|. \end{equation}
Or if not a maximum, then the typical value for such $a,b$.
It seems that most research focuses on the opposite, more impressive direction. That is, for $\varepsilon>0$ characterizing how often $p,q\in [0,1]^n$ have their closest tour points $f(a)\approx p,f(b)\approx q$ such that $|a-b|<\varepsilon.$
For a k-th approximation to a Hilbert curve over $[0,1]^n$ is it true that for any $\varepsilon$-length interval roughly traverses not much more than a cube of ($\mathbb{R}^n$-)volume $\varepsilon$? |
Instrumental Cutsets Efficient Identification in Linear SCM
We’re going to be presenting our paper, “Efficient Identification in Linear Structural Causal Models with Instrumental Cutsets”, at NeurIPS 2019. The work is extremely technical, requiring a huge amount of specialized background knowledge, so I’m offering this post as a lightweight introduction to one of our core results that does not require you to dedicate a year of your life towards the nitty-gritty of linear identification.
Let’s start simple. What do the words in the title even mean?
SCM and Identification
Suppose you’re trying to decide on the effectiveness of a drug by observing all patients who took it, and seeing if they were cured. You are comparing this with patients who took an older drug, to make sure that the new medicine is actually superior. We can draw a “causal graph”, which basically says that the medicine can
cause patients to be cured, and not the other way around. This is denoted with an arrow from the medicine to the patient’s status:
However, that’s not the full picture. You see, this new drug is very expensive, so only rich people can afford it (this story is based in the USA). It turns out that wealthy people are also more likely to be cured even without treatment, simply because they do not need to keep working while sick, and can relax. This is called
latent confounding - the amount of money a patient has, which was not gathered as part of the dataset, affects both whether the patient gets medicine, and whether the patient is cured. We denote such an effect with a dashed bidirected arrow between the two nodes in our graph:
The problem of identification is therefore the question:
Given this causal graph, and the dataset, is it possible to find out the causal effect of taking the drug?
In the graph with latent confounding, the answer is
no [1] - it is not possible to find out how much of the effect on health was due to taking the drug, and how much was due to the fact that those who were taking it were rich, and therefore more likely to get better anyways!
The goal of randomized clinical trials is therefore to get rid of this confounding - if people get the drug through the flip of a coin rather than the contents of their wallet, we have the situation in the first graph, which is
identifiable, meaning that we can compute the effect of the drug on cure rates.
Unfortunately, a clinical trial is not always possible, whether due to ethical, monetary, or even feasibility issues. By creating algorithms that determine identifiability in arbitrary graphs, we allow researchers to simply gather as much data as they can, as well as the hypothesized causal relations between the data, and then stuff it all in a computer. The computer will either compute the causal effect, or tell the researchers that they’re out of luck [2]!
As an example, the graph above represents the situation where a doctor makes a treatment recommendation to the patient, sometimes recommending the new drug, and other times the old drug, independently of the patient’s financial status. It turns out that this model is not identifiable without further assumptions - if all the variables are binary, the effect can only be
bounded, but cannot be uniquely determined [3]. That is, we might be able to say that the actual causal effect lies somewhere between a 50% increase in cure rate and a 10% decrease, meaning that we can’t even conclude that the drug isn’t harmful! Linear SCM
The model shown above is called the “Instrumental Variable”, or IV, and becomes identifiable once we add a few assumptions about the mechanisms underlying the graph [4]. To demonstrate, we will modify the mechanisms of the example.
Previously, the doctor recommended one drug or the other, the patient likewise took a binary dose (drug 1 or 2), and was either cured or not cured. Instead, suppose the doctor’s recommendation is continuous: the doctor can advocate the drug either lightly, strongly, or anywhere in between. Let’s assign a number to the strength of this advocacy:
Similarly, now the patient chooses how much of the treatment regimen to take:
Next, instead of “cured” and “not cured”, let’s make the result be the count of a biomarker in the patient’s blood:
Finally, let’s assume that the effects are linear. That is, the amount of medicine the patient takes is linearly related to how strongly the doctor believed the medicine would work. Similarly, the biomarkers in the patient’s blood are linearly related to the amount of medicine taken. Our job, then, is to solve for the amount that the blood marker will decrease per day of the treatment regimen. This is represented by $\lambda_{mb}$ in the following system of equations:
Here, $\epsilon$ represent latent error terms, and have different values for each patient (they are random variables), and $\lambda$ are the causal effects, which are constant for all patients. Of course, the only data we have is samples of $(D,M,B)$ - we don’t actually know what any of the $\epsilon$ or $\lambda$ are! Can we solve for $\lambda_{mb}$? Yes we can! We can use the dataset to find the covariances between the variables:
To simplify the math (without loss of generality), let’s assume that $D,M,B$ are normalized, meaning that the data has mean 0. This makes $\sigma_{ab} = \mathbb{E}[AB]$.
In effect, we have turned a dataset of samples $(D,M,B)$ into known values for the covariances $\sigma_{dm},\sigma_{db},\sigma_{mb}$. Now, we can do a bit of mathematical magic:
We can use the fact that $D$ is uncorrelated with $\epsilon_b$, and that the variables have mean 0 to conclude:
As you can see, despite the patients’ choice of drug and their biomarkers being confounded by wealth, we were able to extract the drug’s causal effect
anyways. This property has made the instrumental variable a cornerstone of observational studies since their introduction, all the way back in 1928 [5].
In general, when you want to find the causal effect between $M$ and $B$, you look for a node ($D$) called the “instrument”, which has all of its paths to $B$ pass through $M$.
Unconditioned Instrumental Sets
Of course, if a researcher does not find an instrument in their causal graph, it does not mean that their target effect is not identifiable! An example of this is the instrumental set [6]:
Notice that $x_1\rightarrow y$ has no instruments - both $z_1$ and $z_2$ have paths to $y$ through $x_2$. Nevertheless, using the same math as in the previous example, we can get:
This system of equations is solvable for $\lambda_{x_1y}$ if $\det\begin{pmatrix} \sigma_{z_1x_1} & \sigma_{z_1x_2}\\ \sigma_{z_2x_1} & \sigma_{z_2x_2} \end{pmatrix} \ne 0$. An example where this requirement is not satisfied is:
In this example, we can create an identical system of equations, but this system will now be degenerate, meaning that $(\sigma_{z_1x_1},\sigma_{z_1x_2}) = \lambda_{z_1z_2}(\sigma_{z_2x_1},\sigma_{z_2x_2})$. In fact, $\lambda_{x_1y}$ is not identifiable here.
Thankfully, there is a simple condition that is sufficient to determine whether the system is generically solvable with unconditioned instrumental sets. We choose a set of $K$ candidate instruments, as well as $K$ parents of $Y$. If all paths from each of the instruments cross one of the $K$ chosen parents, and we also have a set of $K$ non-intersecting paths from instruments to parents, then we know that all $K$ chosen edges incoming to $Y$ are solvable. To demonstrate, let’s look at a couple examples:
In the first example, $z_1$ and $z_2$ have non-intersecting paths ($z_1\rightarrow x_1$ and $z_2\rightarrow x_2$), and $z_1,z_2$ have all their paths to $y$ cross either $x_1$ or $x_2$. This means that $z_1,z_2$ can be used as an instrumental set to solve for $\lambda_{x_1y}$ and $\lambda_{x_2y}$.
In the second example, $z_2$ has a path to $x_1$, but it can still get to $y$ through $x_2$. $z_1$ cannot help here, because all paths from $z_1$ to $x_2$ would cross $z_2$! This means that we don’t have a instrumental set.
In the final example, both $z_1$ and $z_2$ have nonintersecting paths to $x_1$ and $x_2$, but $z_2$ can still get to $y$ by crossing through $x_3$. Once again, there is no instrumental set here.
The Match-Block
This is where our paper starts: it isn’t clear how to
efficiently find an unconditioned instrumental set. If we know which parents of the target node ($y$) are part of the subset, then it is possible to find associated instruments using a max-flow [7]. Unfortunately, we don’t generally have this knowledge. Look at the graph here:
In this graph,
only $\{x_1,x_2,x_5\}$ has an instrumental subset $\{z_1,z_2,z_4\}$. How can we make a computer find it?
One way of approaching this is enumerating all subsets of edges incoming into the target node (i.e. all subsets of $\{x_1,x_2,x_3,x_4,x_5\}$), and then checking if an instrumental set exists to those nodes. This enumeration takes $2^N$ time, where $N$ is the number of edges incoming into that node. In other words, this requires
exponential-time, and that simply won’t do.
We developed a very simple algorithm that quickly finds the set. Here’s how it works. First, we find a max-flow from
all the candidate instruments $z$ to all the $x$:
Now, we realize that $x_4$ had no flow to it. This means that $x_4$ is not part of
any instrumental set. Furthermore, if $x_4$ isn’t part of an instrumental set, then none of its ancestors can be instruments (all of them would have a path to $x_4$). We therefore remove $x_4$ and $z_3,z_5$ from candidacy, and run another max-flow:
Once again, $x_3$ had no flow to it, so we disable it, and all of its ancestors (here, its ancestors are already disabled). Finally, we do one more max-flow:
All the remaining $x$ values are still active, meaning that we have found the desired instrumental set!
Instrumental Cutsets
It turns out that we can get a lot of mileage out of this simple algorithm. You see, most state-of-the-art identification algorithms have a similar structure to the instrumental set, where they needed to enumerate all subsets of edges incoming into a target node. We were able to adapt the match-block to these algorithms, and even extend it into a new method, which can solve for certain edges that no other efficient method we know of can solve! We called our method the “Instrumental Cutset”, and you can read about the details in our full paper on arXiv (forthcoming).
References P. G. Wright, Tariff on Animal and Vegetable Oils. Macmillan Company, New York, 1928. J. Pearl, Aspects Of Graphical Models Connected With Causality. 1993. J. Pearl, Causality: Models, Reasoning and Inference. 2000. C. Brito and J. Pearl, “Generalized Instrumental Variables,” in Proceedings of the Eighteenth Conference on Uncertainty in Artificial Intelligence, 2002, pp. 85–93. J. D. Angrist, G. W. Imbens, and D. B. Rubin, “Identification of Causal Effects Using Instrumental Variables,” Journal of the American Statistical Association, vol. 91, no. 434, pp. 444–455, 1996. R. Foygel, J. Draisma, and M. Drton, “Half-Trek Criterion for Generic Identifiability of Linear Structural Equation Models,” The Annals of Statistics, pp. 1682–1713, 2012. J. Tian and J. Pearl, “A General Identification Condition for Causal Effects,” p. 7, 2002. |
Focus Questions
The following questions are meant to guide our study of the material in this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions. What is a complex number? What does it mean for two complex numbers to be equal? How do we add two complex numbers together? How do we multiply two complex numbers together? What is the conjugate of a complex number? What is the modulus of a complex number? How are the conjugate and modulus of a complex number related? How do we divide one complex number by another?
The quadratic formula \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) allows us to find solutions to the quadratic equation \(a^x + bx + c = 0\). For example, the solutions to the equation \(x^{2} + x + 1 = 0\) are
\[x = \dfrac{-1 \pm \sqrt{1 - 4}}{2} = \dfrac{-1 \pm \sqrt{-3}}{2}. \nonumber\]
A problem arises immediately with this solution since there is no real number \(t\) with the property that \(t^{2} = -3\) or \(t = \sqrt{-3}\). To make sense of solutions like this we introduce
complex numbers. Although complex numbers arise naturally when solving quadratic equations, their introduction into mathematics came about from the problem of solving cubic equations.
If we use the quadratic formula to solve an equation such as \(x^{2} + x + 1 = 0\),
we obtain the solutions \(x = \dfrac{-1 + \sqrt{-3}}{2}\) and \(x = \dfrac{-1 - \sqrt{-3}}{2}\). These numbers are complex numbers and we have a special form for writing these numbers. We write them in a way that that isolates the square root of \(-1\). To illustrate, the number
\[\dfrac{-1 + \sqrt{-3}}{2} \nonumber\]
can be written as follows:
\[\dfrac{-1 + \sqrt{-3}}{2} = -\dfrac{1}{2} + \dfrac{\sqrt{-3}}{2} = -\dfrac{1}{2} + \dfrac{\sqrt{3}\sqrt{-1}}{2} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\sqrt{-1} \nonumber\]
Since there is no real number \(t\) satisfying \(t^{2} = -1\), the number \(\sqrt{-1}\) is not a real number. We call \(\sqrt{-1}\) an
imaginary number and give is a special label \(i\). Thus, \(i = \sqrt{-1}\) or \(i^{2} = -1\). With this in mind we can write
\[\dfrac{-1 + \sqrt{-3}}{2} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i \nonumber\]
and every complex number has this special form.
Definition: Complex Numbers
A
complex number is an object of the form
\[a + bi\]
where \(a\) and \(b\) are real numbers and \(i^{2} = -1\).
The form \(a + bi\), where a and b are real numbers is called the
standard form for a complex number. When we have a complex number of the form \(z = a + bi\), the number \(a\) is called the real part of the complex number \(z\) and the number \(b\) is called the imaginary part of \(z\). Since i is not a real number, two complex numbers \(a + bi\) and \(c + di\) are equal if and only if \(a = c\) and \(b = d\).
There is an arithmetic of complex numbers that is determined by an addition and multiplication of complex numbers. Adding and subtracting complex numbers is natural:
\[(a + bi) + (c + di) = (a + c) + (b + d)i\]
\[(a + bi) + (c + di) = (a + c) + (b + d)i\]
That is, to add (or subtract) two complex numbers we add (subtract)their real parts and add (subtract) their imaginary parts. Multiplication is also done in a natural way – to multiply two complex numbers, we simply expand the product as usual and exploit the fact that \(i^{2} = -1\). So the product of two complex number is
\[(a + bi) + (c + di) = ac + (ad)i + (bc)i + (bd)i^{2} = (ac - bd) + (ad + bc)i\]
Complex Number Properties
It can be shown that the complex numbers satisfy many useful and familiar properties, which are similar to properties of the real numbers. If \(u\), \(w\), and \(z\), are complex numbers, then
\(w + z = z + w\) \(u + (w + z) = (u + w) + z\) The complex number \(0 = 0 + 0i\) is an additive identity, that is \(z + 0 = z\). If \(z = a + bi\), then the additive inverse of \(z\) is \(-z = (-a) + (-b)i\). That is, \(z + (-z) = 0\). \(wz = zw\) \(u(wz) = (uw)z\) \(u(w + z) = uw + uz\) If \(wz = 0\), then \(w = 0\) or \(z = 0\).
We will use these properties as needed. For example, to write the complex product \((1 + i)i\) in the form \(a + bi\) with \(a\) and \(b\) real numbers, we distribute multiplication over addition and use the fact that \(i^{2} = -1\) to see that
\[(1 + i)i = i + i^{2} = i + (-1) = (-1) + i.\]
For another example, if \(w = 2 + i\) and \(z = 3 - 2i\), we can use these properties to write \(wz\) in the standard \(a + bi\) form as follows:
\[wz = (2 + i)z = 2z + iz = 2(3 - 2i) + i(3 - 2i) = (6 - 4i) + (3i - 2i^{2}) = 6 - 4i + 3i - 2(-1) = 8 - i\]
Exercise \(\PageIndex{1}\)
1. Write each of the sums or products as a complex number in standard form.
\((2 + 3i) + (7 - 4i)\) \((4 - 2i)(3 + i)\) \((2 + i)i - (3 + 4i)\)
2. Use the quadratic formula to write the two solutions to the quadratic equation \(x^{2} - x +2 = 0\) as complex numbers of the form \(r + si\) and \(u + vi\) for some real numbers \(r\), \(s\), \(u\), and \(v\). (
Hint: Remember: \(i = \sqrt{-1}\). So we can rewrite something like \(\sqrt{-4}\) as \(\sqrt{4} = \sqrt{4}\sqrt{-1} = 2i\).) Answer
1. (a) \((2 + 3i) + (7 - 4i) = 9 - i\)
(b) \((4 - 3i)(3 + i) = (4 - 2i)3 + (4 - 2i)i = 14 - 2i\)
(c) \((2 + i)i - (3 + 4i) = (2i - 1) - 3 - 4i = -4 - 2i\)
2. We use the quadratic formula to solve the equation and obtain \(x = \dfrac{1 \pm \sqrt{-7}}{2}\).
We can then write \(\sqrt{-7} = i\sqrt{7}\). So the two solutions of the quadratic equation are:
\[x = \dfrac{1 + i\sqrt{7}}{2}\]
\[x = \dfrac{1}{2} + \dfrac{\sqrt{7}}{2}i \nonumber\]
\[x = \dfrac{1 - i\sqrt{7}}{2}\]
\[x = \dfrac{1}{2} - \dfrac{\sqrt{7}}{2}i \nonumber\]
Division of Complex Numbers
We can add, subtract, and multiply complex numbers, so it is natural to ask if we can divide complex numbers. We illustrate with an example.
Example \(\PageIndex{1}\): Dividing by a Complex Number
Write the quotient \(\dfrac{2 + i}{3 + i}\) as a complex number in the form \(a + bi\).
Solution
This problem is rationalizing a denominator since \(i = \sqrt{-1}\). So in this case we need to “remove” the imaginary part from the denominator. Recall that the product of a complex number with its conjugate is a real number, so if we multiply the numerator and denominator of \(\dfrac{2 + i}{3 + i}\) by the complex conjugate of the denominator, we can rewrite the denominator as a real number. The steps are as follows. Multiplying the numerator and denominator by the conjugate \(3 - i\) or \(3 + i\) gives us
\[\dfrac{2 + i}{3 + i} = \left(\dfrac{2 + i}{3 + i}\right)\left(\dfrac{3 - i}{3 - i}\right) = \dfrac{(2 + i)(3 - i)}{(3 + i)(3 - i)} = \dfrac{(6 - i^{2}) + (-2 + 3)i}{9 - i^{2}} = \dfrac{7 + i}{10} \nonumber\]
Now we can write the final result in standard form as
\[\dfrac{7 + i}{10} = \dfrac{7}{10} + \dfrac{1}{10}i. \nonumber\]
Example \(\PageIndex{1}\) illustrates the general process for dividing one complex number by another. In general, we can write the quotient \(\dfrac{a + bi}{c + di}\) in the form \(r + si\) by multiplying numerator and denominator of our fraction by the conjugate \(c - di\) of \(c + di\) to see that
\[\dfrac{a + bi}{c + di} = \left(\dfrac{a + bi}{c + di}\right)\left(\dfrac{c - di}{c - di}\right) = \dfrac{(ac + bd) + (bc - ad)i}{c^{2} + d^{2}} = \dfrac{ac + bd}{c^{2} + d^{2}} + \dfrac{bc - ad}{c^{2} + d^{2}}i\]
Therefore, we have the formula for the quotient of two complex numbers.
Definition: quotient of Complex NUmbers
The
quotient \(\dfrac{a + bi}{c + di}\) of the complex numbers \(a + bi\) and \(c + di\) is the complex number
\[\dfrac{a + bi}{c + di} = \dfrac{ac + bd}{c^{2} + d^{2}} + \dfrac{bc - ad}{c^{2} + d^{2}}i\]
provided \(c + di \neq 0\).
Exercise \(\PageIndex{2}\)
Let \(z = 3 + 4i\) and \(w = 5 - i\).
Write \(\dfrac{w}{z} = \dfrac{5 - i}{3 + 4i}\) as a complex number in the form \(r + si\) where \(r\) and \(s\) are some real numbers. Check the result by multiplying the quotient by \(3 + 4i\). Is this product equal to \(5 - i\)? Find the solution to the equation \((3 + 4i)x = 5 - i\) as a complex number in the form \(x = u + vi\) where \(u\) and \(v\) are some real numbers. Answer
1. Using our formula with \(a = 5, b = -1, c = 3\) and \(d = 4\) gives us
\[\dfrac{5 - i}{3 + 4i} = \dfrac{15 - 4}{15} + \dfrac{-3 -20}{25}i = \dfrac{11}{25} - \dfrac{23}{25}i\]
As a check, we see that
\[\left(\dfrac{11}{25} - \dfrac{23}{25}i\right)\left(3 + 4i\right) = \left(\dfrac{23}{25} - \dfrac{69}{25}i\right) + \dfrac{44}{25}i - \dfrac{92}{25}i^{2} = \left(\dfrac{23}{25} + \dfrac{69}{25}\right) + \left(-\dfrac{69}{25}i + \dfrac{44}{25}i\right) = 5 - i\]
2. We can solve for \(x\) by dividing both sides of the equation by \(3 + 4i\) to see that
\[x = \dfrac{5 - i}{3 + 4i} = \dfrac{11}{25} - \dfrac{23}{25}i\]
Geometric Representations of Complex Numbers
Each ordered pair \((a , b)\) of real numbers determines:
A point in the coordinate plane with coordinates \((a , b)\). A complex number \(a + bi\) A vector \(a\textbf{i} + b\textbf{j} = \langle a, b \rangle\)
This means that we can geometrically represent the complex number \(a + bi\) with a vector in standard position with terminal point \((a , b)\). Therefore, we can draw pictures of complex numbers in the plane. When we do this, the horizontal axis is called
the real axis, and the vertical axis is called the imaginary axis. In addition, the coordinate plane is then referred to as the complex plane. That is, if \(\) we can think of z as a directed line segment from the origin to the point .a; b/, where the terminal point of the segment is a units from the imaginary axis and \(b\) units from the real axis. For example, the complex numbers \(3 + 4i\) and \(-8 + 3i\) are shown in Figure 5.1.
Figure \(\PageIndex{1}\): Two complex numbers.
In addition, the sum of two complex numbers can be represented geometrically using the vector forms of the complex numbers. Draw the parallelogram defined by \(w = a + bi\) and \(z = c + di\). The sum of w and z is the complex number represented by the vector from the origin to the vertex on the parallelogram opposite the origin as illustrated with the vectors \(w = 3 + 4i\) and \(z = -8 + 3i\) in Figure \(\PageIndex{2}\).
Exercise \(\PageIndex{3}\)
Let \(w = 3 + 4i\) and \(z = -8 + 3i\).
Write the complex sum \(w + z\) in standard form. Draw a picture to illustrate the sum using vectors to represent \(w\) and \(z\). Answer
1. The sum is \(w + z = (2 - 1) + (3 + 5)i = 1 + 8i\).
2. A representation of the complex sum using vectors is shown in the figure below.
We now extend our use of the representation of a complex number as a vector in standard position to include the notion of the length of a vector. Recall from Section3.6 (page234) that the length of a vector \(\textbf{v} = a\textbf{i} + b\textbf{j}\) is \(|\textbf{v}| = \sqrt{a^{2} + b^{2}}\).
Figure \(\PageIndex{2}\): The Sum of Two Complex Numbers.
When we use this idea with complex numbers, we call it the
norm or modulus of the complex number.
Definition: norm
The norm (or modulus) of the complex number \(z = a + bi\) is the distance from the origin to the point \((a, b)\) and is denoted by \(|z|\). We see
that \[|z| = |a + bi| = \sqrt{a^{2} + b^{2}}.\]
There is another concept related to complex number that is based on the following bit of algebra.
\[(a + bi)(a -bi) = a^{2} - (bi)^{2} = a^{2} - b^{2}i^{2} = a^{2} + b^{2}\]
The complex number \(a - bi\) is called the complex conjugate of \(a + bi\). If we let \(z = a + bi\), we denote the
complex conjugate of \(z\) as \(\bar{z}\). So \[\bar{z} = \bar{a + bi} = a - bi.\]
We also notice that
\[z\bar{z} = (a + bi)(a - bi) = a^{2} + b^{2},\]
and so the product of a complex number with its conjugate is a real number. In fact,
\[z\bar{z} = a^{2} + b^{2} = |z|^{2},\], and so \[|z| = \sqrt{z\bar{z}}\]
Exercise \(\PageIndex{4}\)
Let \(w = 2 + 3i\) and \(z = -1 + 5i\)
Find \(\bar{w}\) and \(\bar{z}\). Compute \(|w|\) and \(|z|\). Compute \(w\bar{w}\) and \(z\bar{z}\). What is \(\bar{z}\) if \(z\) is a real number? Answer
1. Using the definition of the conjugate of a complex number we find that \(\bar{w} = 2 - 3i\) and \(\bar{z} = -1 - 5i\).
2. Using the definition of the norm of a complex number we find that \(|w| = \sqrt{2^{2} + 3^{2}} = \sqrt{13}\) and \(|z| = \sqrt{(-1)^{2} + 5^{2}} = \sqrt{26}\). 3. Using the definition of the product of complex numbers we find that
\[w\bar{w} = (2 + 3i)(2 - 3i) = 4 + 9 = 13\]
\[z\bar{z} = (-1 + 5i)(-1 - 5i) = 1 + 25 = 26\]
4. Let \(z = a + 0i\) for some \(a \in \mathbb{R}\). Then \(\bar{z} = a - 0i\). Thus, \(\bar{z} = z\) when \(z \in \mathbb{R}\).
Summary
In this section, we studied the following important concepts and ideas: A complex numberis an object of the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i^{2} = -1\). When we have a complex number of the form \(z = a + bi\), the number \(a\) is called the real partof the complex number \(z\) and the number \(b\) is called the imaginary partof \(z\). We can add, subtract, multiply, and divide complex numbers as follows:
\[(a + bi) + (c + di) = (a + c) + (b + d)i \nonumber \]
\[(a + bi) + (c + di) = (a + c) + (b + d)i\nonumber\]
\[(a + bi)(c + di) = (ac - bd) + (ad + bc)i\nonumber\]
\[\dfrac{a + bi}{c + di} = \dfrac{ac + bd}{c^{2} + d^{2}} + \dfrac{bc - ad}{c^{2} + d^{2}}i\nonumber\] provided \(c + di \neq 0\)
A complex number \(a + bi\) can be represented geometrically with a vector in standard position with terminal point \((a, b)\). When we do this, the horizontal axis is called the real axis, and the vertical axis is called the imaginary axis. In addition, the coordinate plane is then referred to as the complex plane. That is, if \(z = a + bi\) we can think of \(z\) as a directed line segment from the origin to the point \((a, b)\), where the terminal point of the segment is a units from the imaginary axis and \(b\) units from the real axis. The norm(or modulus) of the complex number \(z = a + bi\) is the distance from the origin to the point \((a, b)\) and is denoted by \(|z|\). We see that \[|z| = |a + bi| = \sqrt{a^{2} + b^{2}} \nonumber\] The complex number \(a - bi\) is called the complex conjugate of \(a + bi\). Note that \[(a + bi)(a - bi) = a^{2} + b^{2} = |a + bi|^{2} \nonumber\] |
I am trying to prove or disprove the following statement:
$$\sum_{r|n} d(r^2) = d^2(n),$$
where $d$ is the number of divisors function.
Computing it for small numbers yields equality, so I at least think it's true. But I can't see how to prove it formally. I have tried working with prime factorisations, but finding equality between terms seems near-impossible at best. My best guess is there is a way to manipulate
$$\sum_{r\mid n} \sum_{k\mid r^2} 1$$
into a form which becomes the RHS, although my attempts at writing $n=ar$ and $r^2=bk$ for $a,b\in\mathbb{Z}$ become very messy when I attempt substitution.
Any assistance or hints would be greatly appreciated! Thank you! |
Answer
$S=2$
Work Step by Step
$\displaystyle \sum_{i=0}^{∞}(0.5)^i=(0.5)^0+(0.5)^1+(0.5)^2+...$ $a_1=(0.5)^0=1$ $r=\frac{a_2}{a_1}=\frac{(0.5)^1}{(0.5)^0}=0.5$ $S=\frac{a_1}{1-r}=\frac{1}{1-0.5}=\frac{1}{0.5}=2$
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The Neighbourhood of a Vertex
Definition: The Neighbourhood of the vertex $x \in V(G)$ is the set of all vertices which are adjacent to $x$. For the graph $G = (V(G), E(G))$ we have that the neighbourhood of $x$ denoted $N_{G} (x)$ is $\displaystyle{N_{G}(x) = \left\{ {y \in V(G) : {x, y} \in E(G)}\right\}}$. When understood, the notation $N_G(x)$ can be replaced with $N(x)$.
For example, the following graph:
The neighbourhood of vertex $x$ is $N_G(x) = \{a, b, c, d \}$ since the vertex $x$ forms an edge with $a, b, c$ and $d$. The neighbourhood of vertex $a$ is $N_G(a) = \{x, b\}$.
The Degree of a Vertex
Definition: The Degree of a vertex $x$ in a graph $G = (V(G), E(G))$, denoted $\mathrm{deg} (x)$ is the number of vertices adjacent to $x$. Equivalently, the degree of $x$ is the size of the neighbourhood of $x$, that is, $\mathrm{deg}(x) = \mid N_G(x) \mid$.
For example, from our graph above, the degree of vertex $x$ is $\mathrm{deg}(x) = 4$. The degree of vertex $a$ is $\mathrm{deg}(a) = 2$.
The Degree Sequence of a Graph
Definition: The Degree Sequence of a graph $G = (V(G), E(G))$ is the ordered sequence of degrees of vertices $x \in V(G)$ from smallest degree to greatest degree.
From the graph from earlier we have that the degrees of each vertex are given in the following table:
Vertex Degree $x$ $\mathrm{deg}(x) = 4$ $a$ $\mathrm{deg}(a) = 2$ $b$ $\mathrm{deg}(b) = 2$ $c$ $\mathrm{deg}(c) = 2$ $d$ $\mathrm{deg}(d) = 2$
So the degree sequence for our graph is $(2, 2, 2, 2, 4)$. Notice that the degree sequence allows for repetition as some vertices may have the same degree. |
The Handshaking Lemma
We will now look at a very important and well known lemma in graph theory.
Lemma 1 (The Handshaking Lemma): In any graph $G = (V(G), E(G))$, the sum of the degrees in the degree sequence of $G$ is equal to one half the number of edges in the graph, that is $\displaystyle{\sum_{v \in V(G)} \deg (v) = 2 \mid E(G) \mid}$. Proof:In any graph, each edge in $E(G)$ is attached to two vertices. Therefore each edge contributes $1$ to each of the two vertices it is connected to. Therefore $\sum_{v \in V(G)} \deg (v) = 2 \mid E(G) \mid$. $\blacksquare$
For example, let's look at the following graph where we label the degrees of each vertex in pink and each edge in blue.
Notice that the sum of the degrees in the degree sequence $(2 + 2 + 2 + 4 + 2 + 3 + 3)$ is $18$ and the number of edges is $9$. Hence we have verified that:(1) |
Bedforms and roughness
Bed forms are relief features initiated by the fluid motions generated downstream of small local obstacles at the bottom consisting of movable (alluvial) sediment materials.
Contents Introduction
Many types of bed forms can be observed in nature. The bed form regimes for steady flow over a sand bed can be classified into (see Figure 1):
lower transport regime with flat bed, ribbons and ridges, ripples, dunes and bars, transitional regime with washed-out dunes and sand waves, upper transport regime with flat mobile bed and sand waves (anti-dunes).
When the bed form crest is perpendicular (transverse) to the main flow direction, the bed forms are called transverse bed forms, such as ripples, dunes and anti-dunes. Ripples have a length scale much smaller than the water depth, whereas dunes have a length scale much larger than the water depth. The crest lines of the bed forms may be straight, sinuous, linguoid or lunate. Ripples and dunes travel downstream by erosion at the upstream face (stoss-side) and deposition at the downstream face (lee-side). Antidunes travel upstream by lee-side erosion and stoss-side deposition. Bed forms with their crest parallel to the flow are called longitudinal bed forms such as ribbons and ridges.
In the literature, various bed-form classification methods for sand beds are presented. The types of bed forms are described in terms of basic parameters (Froude number, suspension parameter, particle mobility parameter; dimensionless particle diameter).
A flat immobile bed may be observed just before the onset of particle motion, while a flat mobile bed will be present just beyond the onset of motion. The bed surface before the onset of motion may also be covered with relict bed forms generated during stages with larger velocities.
Ripples
Small-scale ribbon and ridge type bed forms parallel to the main flow direction have been observed in laboratory flumes and small natural channels, especially in case of fine sediments (grainsize [math]d_{50}[/math] typically smaller than 0.1 mm). They are probably generated by secondary flow phenomena and near-bed turbulence effects (burst-sweep cycle) in the lower and transitional flow regime. These bed forms are also known as parting lineations because of the streamwise ridges and hollows with a vertical scale equal to about 10 grain diameters. These bed forms are mostly found in fine sediments ([math]0.05 \lt d_{50} \lt 0.25 \; mm[/math]).
When the velocities are somewhat larger (10%-20%) than the critical velocity for initiation of motion and the median particle size is smaller than about 0.5 mm, small (mini) ripples are generated at the bed surface. Ripples that are developed during this stage remain small with a ripple length much smaller than the water depth. The characteristics of mini ripples are commonly assumed to be related to the turbulence characteristics near the bed (burst-sweep cycle). Current ripples have an asymmetric profile with a relatively steep downstream face (lee-side) and a relatively gentle upstream face (stoss-side). As the velocities near the bed become larger, the ripples become more irregular in shape, height and spacing yielding strongly three-dimensional ripples. In this case the variance of the ripple length and height becomes rather large. These ripples are known as lunate ripples when the ripple front has a concave shape in the current direction (crest is moving slower than wing tips) and are called linguoid ripples when the ripple front has a convex shape (crest is moving faster than wing tips). The largest ripples may have a length up to the water depth and are commonly called mega-ripples.
Dunes
Another typical bed form type of the lower regime is the dune-type bed form. Dunes have an asymmetrical (triangular) profile with a rather steep lee-side and a gentle stoss-side. A general feature of dune type bed forms is lee-side flow separation resulting in strong eddy motions downstream of the dune crest. The length of the dunes is strongly related to the water depth ([math]h[/math]) with values in the range of [math](3 - 15) h[/math]. Extremely large dunes with heights ([math]\Delta[/math]) of the order of 7 m and lengths ([math]\lambda[/math]) of the order of 500 m have been observed in the Rio Parana River (Argentina) at water depths of about 25 m, velocities of about 2 m/s and bed material sizes of about 0.3 mm. The formation of dunes may be caused by large-scale fluid velocity oscillations generating regions at regular intervals with decreased and increased bed-shear stresses, resulting in the local deposition and erosion of sediment particles.
Sand bars
The largest bed forms in the lower regime are sand bars (such as alternate bars, side bars, point bars, braid bars and transverse bars), which usually are generated in areas with relatively large transverse flow components (bends, confluences, expansions). Alternate bars are features with their crests near alternate banks of the river. Braid bars actually are alluvial "islands" which separate the anabranches of braided streams. Numerous bars can be observed distributed over the cross-sections. These bars have a marked streamwise elongation. Transverse bars are diagonal shoals of triangular-shaped plan along the bed. One side may be attached to the channel bank. These type of bars generally are generated in steep slope channels with a large width-depth ratio. The flow over transverse bars is sinuous (wavy) in plan. Side bars are bars connected to river banks in a meandering channel. There is no flow over the bar. The planform is roughly triangular. Special examples of side bars are point bars and scroll bars.
Transitional regime
It is a well-known phenomenon that the bed forms generated at low velocities are washed out at high velocities. It is not clear, however, whether the disappearance of the bed forms is accomplished by a decrease of the bed form height, by an increase of the bed form length or both. Flume experiments with sediment material of about 0.45 mm show that the transition from the lower to the upper regime is effectuated by an increase of the bed form length and a simultaneous decrease of the bed form height. Ultimately, relatively long and smooth sand waves with a roughness equal to the grain roughness were generated
[1].
In the transition regime the sediment particles will be transported mainly in suspension. This will have a strong effect on the bed form shape. The bed forms will become more symmetrical with relatively gentle lee-side slopes. Flow separation will occur less frequently and the effective bed roughness will approach to that of a plane bed. Large-scale bed forms with a relative height ([math]\Delta / h[/math]) of 0.1 to 0.2 and a relative length ([math]\lambda / h[/math]) of 5 to 15 were present in the Mississippi river at high velocities in the upper regime.
Antidunes
In the supercritical upper regime the bed form types will be plane bed and/or anti-dunes. The latter type of bed forms are sand waves with a nearly symmetrical shape in phase with the water surface waves. The anti-dunes do not exist as a continuous train of bed waves, but they gradually build up locally from a flat bed. Anti-dunes move upstream due to strong lee-side erosion and stoss-side deposition. Anti-dunes are bed forms with a length scale of less than 10 times the water depth. When the flow velocity further increases, finally a stage with chute and pools may be generated.
Bed roughness
Nikuradse
[2] introduced the concept of an equivalent or effective sand roughness height, [math]k_s[/math], to simulate the roughness of arbitrary roughness elements of the bottom boundary. In case of a movable bed consisting of sediments the effective bed roughness [math]k_s[/math] mainly consists of grain roughness ([math]k'_s[/math]) generated by skin friction forces and of form roughness ([math]k''_s[/math]) generated by pressure forces acting on the bed forms. Similarly, a grain-related bed-shear stress ([math]\tau'_b[/math]) and a form-related bed-shear stress ([math]\tau''_b[/math]) can be defined. The effective bed roughness for a given bed material size is not constant but depends on the flow conditions. Analysis results of [math]k_s[/math]-values computed from Mississippi River data (USA) show that [math]k_s[/math] strongly decreases from about 0.5 m at low velocities (0.5 m/s) to about 0.001 m at high velocities (2 m/s), probably because the bed forms become more rounded or are washed out at high velocities.The fundamental problem of bed roughness prediction is that the bed characteristics (bed forms) and hence the bed roughness depend on the main flow variables (depth, velocity) and sediment transport rate (sediment size). These hydraulic variables are, however, in turn strongly dependent on the bed configuration and its roughness.Another problem is the almost continuous variation of the discharge during rising and falling stages. Under these conditions the bed form dimensions and hence the Chézy-coefficient are not constant but vary with the flow conditions. See further Bed roughness and friction factors in estuaries. Related articles References Van Rijn, L.C., 1993, 2012. Principles of sediment transport in rivers, estuaries and coastal seas. Aqua Publications, Amsterdam, The Netherlands (WWW.AQUAPUBLICATIONS.NL) Nikuradse, J., 1932. Gesetzmässigkeiten der turbulente Strömung in glatten Rohren. Ver. Deut. Ing. Forschungsheft 356
Please note that others may also have edited the contents of this article.
Leo van Rijn (2019): Bedforms and roughness. Available from http://www.coastalwiki.org/wiki/Bedforms_and_roughness [accessed on 14-10-2019] |
Section 7.4 Exercises
Find a possible formula for the trigonometric function whose values are given in the following tables.
1.
\(x\) 0 3 6 9 12 15 18 \(y\) -4 -1 2 -1 -4 -1 2
2.
\(x\) 0 2 4 6 8 10 12 \(y\) 5 1 -3 1 5 1 -3
3. The displacement \(h(t)\), in centimeters, of a mass suspended by a spring is modeled by the function \(h\left(t\right)=8{\rm sin}(6\pi t)\), where t is measured in seconds. Find the amplitude, period, and frequency of this displacement.
4. The displacement \(h(t)\), in centimeters, of a mass suspended by a spring is modeled by the function \(h\left(t\right)=11{\rm sin}(12\pi t)\), where \(t\) is measured in seconds. Find the amplitude, period, and frequency of this displacement.
5. A population of rabbits oscillates 19 above and below average during the year, reaching the lowest value in January. The average population starts at 650 rabbits and increases by 160 each year. Find a function that models the population, \(P\), in terms of the months since January, \(t\).
6. A population of deer oscillates 15 above and below average during the year, reaching the lowest value in January. The average population starts at 800 deer and increases by 110 each year. Find a function that models the population, \(P\), in terms of the months since January, \(t\).
7. A population of muskrats oscillates 33 above and below average during the year, reaching the lowest value in January. The average population starts at 900 muskrats and increases by 7% each month. Find a function that models the population, \(P\), in terms of the months since January, \(t\).
8. A population of fish oscillates 40 above and below average during the year, reaching the lowest value in January. The average population starts at 800 fish and increases by 4% each month. Find a function that models the population, \(P\), in terms of the months since January, \(t\).
9. A spring is attached to the ceiling and pulled 10 cm down from equilibrium and released. The amplitude decreases by 15% each second. The spring oscillates 18 times each second. Find a function that models the distance, \(D\), the end of the spring is below equilibrium in terms of seconds, \(t\), since the spring was released.
10. A spring is attached to the ceiling and pulled 7 cm down from equilibrium and released. The amplitude decreases by 11% each second. The spring oscillates 20 times each second. Find a function that models the distance, \(D\), the end of the spring is below equilibrium in terms of seconds, \(t\), since the spring was released.
11. A spring is attached to the ceiling and pulled 17 cm down from equilibrium and released. After 3 seconds the amplitude has decreased to 13 cm. The spring oscillates 14 times each second. Find a function that models the distance, \(D\) the end of the spring is below equilibrium in terms of seconds, \(t\), since the spring was released.
12. A spring is attached to the ceiling and pulled 19 cm down from equilibrium and released. After 4 seconds the amplitude has decreased to 14 cm. The spring oscillates 13 times each second. Find a function that models the distance, \(D\) the end of the spring is below equilibrium in terms of seconds, \(t\), since the spring was released.
Match each equation form with one of the graphs.
13. a. \(ab^{x} +\sin \left(5x\right)\)
b. \(\sin \left(5x\right)+mx+b\)
14. a. \(ab^{x} \sin \left(5x\right)\)
b. \(\left(mx+b\right){\rm sin}(5x)\)
I II III IV
Find a function of the form \(y=ab^{x} +c\sin \left(\frac{\pi }{2} x\right)\) that fits the data given.
15.
\(x\) 0 1 2 3 \(y\) 6 29 96 379
16.
\(x\) 0 1 2 3 \(y\) 6 34 150 746
Find a function of the form \(y=a\sin \left(\frac{\pi }{2} x\right)+m+bx\) that fits the data given.
17.
\(x\) 0 1 2 3 \(y\) 7 6 11 16
18.
\(x\) 0 1 2 3 \(y\) -2 6 4 2
Find a function of the form \(y=ab^{x} \cos \left(\frac{\pi }{2} x\right)+c\) that fits the data given.
19.
\(x\) 0 1 2 3 \(y\) 11 3 1 3
20.
\(x\) 0 1 2 3 \(y\) 4 1 -11 1 Answer
1. \(y = 3\sin(\dfrac{\pi}{6} (x - 3)) - 1\)
3. Amplitude: 8, Period: \(\dfrac{1}{3}\) second, Frequency: 3 Hz (cycles per second)
5. \(P(t) = -19\cos(\dfrac{\pi}{6}t) + \dfrac{40}{3} t + 650\)
7. \(P(t) = -33\cos(\dfrac{\pi}{6}t) + 900(1.07)^t\)
9. \(D(t) = 10 (0.85)^t \cos(36 \pi t)\)
11. \(D(t) = 17(0.9145)^t \cos(28\pi t)\)
13. a. IV
b. III
15. \(y = 6(4)^x + 5 \sin(\dfrac{\pi}{2} x)\)
17. \(y = -3\sin(\dfrac{\pi}{2}) + 2x + 7\)
19. \(y = 8(\dfrac{1}{2})^x \cos(\dfrac{\pi}{2}x) + 3\) |
The suggestion in the comments to use brute force is a little unsatisfactory; what if $15$ were $15000$? Here is a general method.
Suppose$$m = \sigma(n) = \prod_{p^k ||\: n} \frac{p^{k+1}-1}{p-1}.\tag{1}$$Each factor $d$ in the product
is an integer (a finite geometric series). The prime factorization of $m$ will severely restrict the possible multiplicands on the r.h.s. of (1), and then their unusual shape (viz. $\frac{p^{k+1}-1}{p-1}$) will reveal much about the prime factorization of $n$.
To wit, if $d$ divides $m$ and $\frac{p^{k+1} - 1}{p-1} = d$ then $p^{k+1} = pd - (d - 1)$ which means
$p$ divides $d-1$. This yields one equation to solve per prime $p$ dividing $d-1$, namely the following equation: $$p^k = d - \frac{d-1}{p}.\tag{2}$$
Already we can see that $d = 1$ is forbidden, since it implies $p^k = 1$, which is absurd (contradicts the fundamental theorem of arithmetic). Thus the general procedure is to consider a factor $d \ne 1$ of $m$ and check, for each prime $p$ under $d-1$, whether the r.h.s. of (2) is a power of $p$; if so, repeat the process replacing $m$ by $m/d$ (and excluding the prime $p$ from further consideration). This can get complicated; luckily we are dealing here only with $m = 15 = 3 \cdot 5$, a semiprime.
If $d = 3$ then $p \mid 2$ so $p = 2$ and (2) becomes $2^k = 2$ which means $k = 1$. Thus, possibly, $n$ is singly even. However, moving on to the complementary divisor $15/3 = 5$, if $d = 5$ then $p \mid 4$ so $p = 2$ and this is a problem: the prime $2$ was supposed to contribute only one term in the r.h.s. of (1). [If this is unconvincing, use (2) again: $2^k = 3$ which can't happen ($k \in \mathbf{Z}$).] Thus the appearance of $5$ and $3$ together in the r.h.s. of (1) is inconsistent.
That leaves us with $d = 15$. This implies $p \mid 14$ so $p = 2$ or $7$. But $7^k = 13$ can't happen. On the other hand, $2^k = 8$ has the solution $k = 3$, and this is the only consistent description of $n$ we can glean.
It remains to check that $n = 2^3 = 8$ works: $\sigma(8) = 1 + 2 + 4 + 8 = 15$ ($ = 1111_2$ in binary!) so, yes! |
A first order ODE is an equation of the form
\[\dfrac{dy}{dx}=f(x,y)\]
or just
\[y'=f(x,y)\]
In general, there is no simple formula or procedure one can follow to find solutions. In the next few lectures we will look at special cases where solutions are not difficult to obtain. In this section, let us assume that \(f\) is a function of \(x\) alone, that is, the equation is
\[y'=f(x) \label{1.1.1}\]
We could just integrate (antidifferentiate) both sides with respect to \(x\).
\[\int y' (x) dx = \int f(x) dx + C\]
that is
\[y(x)=\int f(x) dx + C\]
This \(y(x)\) is actually the general solution. So to solve Equation \(\ref{1.1.1}\), we find some antiderivative of \(f(x)\) and then we add an arbitrary constant to get the general solution.
Now is a good time to discuss a point about calculus notation and terminology. Calculus textbooks muddy the waters by talking about the integral as primarily the so-called indefinite integral. The indefinite integral is really the antiderivative (in fact the whole one-parameter family of antiderivatives). There really exists only one integral and that is the definite integral. The only reason for the indefinite integral notation is that we can always write an antiderivative as a (definite) integral. That is, by the fundamental theorem of calculus we can always write \(\int f(x) dx + C\) as
\[\int_{x_0}^x f(t) dt + C\]
Hence the terminology to integrate when we may really mean to antidifferentiate. Integration is just one way to compute the antiderivative (and it is a way that always works, see the following examples). Integration is defined as the area under the graph, it only happens to also compute antiderivatives. For sake of consistency, we will keep using the indefinite integral notation when we want an antiderivative, and you should always think of the definite integral.
Example \(\PageIndex{1}\):
Find the general solution of \(y' = 3x^2\).
Solution
Elementary calculus tells us that the general solution must be \(y = x^3 + C\). Let us check: \(y' = 3x^2\). We have gotten precisely our equation back.
Normally, we also have an initial condition such as \(y(x_0) = y_0\) for some two numbers\({x_0}\) and \({y_0}\) \({x_0}\) is usually 0, but not always). We can then write the solution as a definite integral in a nice way. Suppose our problem is \(y' = f(x), \, y(x_0) = y_0\). Then the solution is
\[y(x) = \int_{x_0}^x f(s) ds + y_0 \label{1.1.2})\]
Let us check! We compute \(y' = f(x)\), via the fundamental theorem of calculus, and by Jupiter,\(y\) is a solution. Is it the one satisfying the initial condition? Well, \(y(x_0) = \int_{x_0}^{x_0} f(x) dx + y_0 = y_0\). It is!
Do note that the definite integral and the indefinite integral (antidifferentiation) are completely different beasts. The definite integral always evaluates to a number. Therefore, Equation \(\ref{1.1.2}\) is a formula we can plug into the calculator or a computer, and it will be happy to calculate specific values for us. We will easily be able to plot the solution and work with it just like with any other function. It is not so crucial to always find a closed form for the antiderivative.
Example \(\PageIndex{2}\):
Solve
\[y' = e^{-x^2}, ~~ y(0) = 1.\]
By the preceding discussion, the solution must be
\[y(x) = \int_0^x e^{-s^2} ds + 1.\]
Solution
Here is a good way to make fun of your friends taking second semester calculus. Tell them to find the closed form solution. Ha ha ha (bad math joke). It is not possible (in closed form). There is absolutely nothing wrong with writing the solution as a definite integral. This particular integral is in fact very important in statistics.
Using this method, we can also solve equations of the form
\[y' = f(y)\]
Let us write the equation in Leibniz notation.
\[\dfrac{dy}{dx} = f(y)\]
Now we use the inverse function theorem from calculus to switch the roles of \(x\) and \(y\) to obtain
\[\dfrac{dy}{dx} = \dfrac{1}{f(y)}\]
What we are doing seems like algebra with \(dx\) and \(dy\). It is tempting to just do algebra with \(dx\) and \(dy\) as if they were numbers. And in this case it does work. Be careful, however, as this sort of hand-waving calculation can lead to trouble, especially when more than one independent variable is involved. At this point we can simply integrate,
\[x(y) = \int \dfrac{1}{f(y)} dy + C\]
Finally, we try to solve for \(y\).
Example \(\PageIndex{3}\):
Previously, we guessed \(y' = ky\) (for some \(k > 0\)) has the solution \(y = Ce^{kx}\). We can now find the solution without guessing. First we note that \(y = 0\) is a solution. Henceforth, we assume \(y \ne 0\). We write
\[\dfrac{dx}{dy} = \dfrac{1}{ky}\]
We integrate to obtain
\[x(y) = x = \dfrac{1}{k} \ln \left\vert y \right\vert + D\]
where \(D\) is an arbitrary constant. Now we solve for\(y\) (actually for \(\left\vert y \right\vert\) ).
\[\left\vert y \right\vert = e^{kx-kD} = e^{-kD}e^{kx}\]
If we replace \(e^{-kD}\) with an arbitrary constant \(C\) we can get rid of the absolute value bars (which we can do as \(D\) was arbitrary). In this way, we also incorporate the solution \(y = 0\). We get the same general solution as we guessed before, \(y = Ce^{kx}\).
Example \(\PageIndex{4}\):
Find the general solution of \(y' = y^2\).
Solution
First we note that \(y = 0\) is a solution. We can now assume that \(y \ne 0\) . Write
\[\dfrac{dx}{dy} = \dfrac{1}{y^2}\]
We integrate to get
\[x = \dfrac{-1}{y} + C\]
We solve for\(y = \dfrac{1}{C-x}\). So the general solution is
\[y = \dfrac{1}{C-x} \,\, or\,\, y=0\]
Note the singularities of the solution. If for example \(C = 1\), then the solution “blows up” as we approach \(x = 1\). Generally, it is hard to tell from just looking at the equation itself how the solution is going to behave. The equation \(y' = y^2\) is very nice and defined everywhere, but the solution is only defined on some interval \(\left(-\infty, C \right)\) or \(\left(C, \infty \right)\) .
Classical problems leading to differential equations solvable by integration are problems dealing with velocity, acceleration and distance. You have surely seen these problems before in your calculus class.
Example \(\PageIndex{5}\):
Suppose a car drives at a speed \(e^{t/2}\) meters per second, where \(t\) is time in seconds. How far did the car get in 2 seconds (starting at \(t = 0\))? How far in 10 seconds?
Solution
Let \(x\) denote the distance the car traveled. The equation is
\[x' = e^{t/2}\]
We can just integrate this equation to get that
\[x(t) = 2e^{t/2} + C\]
We still need to figure out \(C\). We know that when \(t = 0\), then \(x = 0\). That is, \(x(0) = 0\). So
\[ 0 = x(0) = 2 e^{0/2} + C = 2 + C\]
Thus \(C = -2\) and
\[x(t) = 2 e^{t/2} - 2\]
Now we just plug in to get where the car is at 2 and at 10 seconds. We obtain
\[x(2) = 2e^{2/2} - 2 \approx 3.44 \text{~meters},~~~ x(10) = 2e^{10/2} - 2 \approx 294\text{~meters}\]
Example \(\PageIndex{6}\):
Suppose that the car accelerates at a rate of \(t^2 \frac{m}{s^2}\). At time \(t = 0\) the car is at the 1 meter mark and is traveling at 10
m/ s. Where is the car at time \(t = 10\). Solution
Well this is actually a second order problem. If \(x\) is the distance traveled, then \(x'\) is the velocity, and \(x''\) is the acceleration. The equation with initial conditions is
\[x'' = t^2, ~~~~ x(0) = 1, ~~~~x'(0) = 10\]
What if we say \(x' = v\). Then we have the problem
\[v' = t^2, ~~~~ v(0) = 10\]
Once we solve for \(v\), we can integrate and find \(x\). |
If $1_K \in K$ and $1_F \in F$ are the multiplicative neutral elements of $K$ and $F$, notice that $\Bbb Q$ naturally embeds in $K$ by the map $\Bbb Q \ni \frac p q \mapsto (p \cdot 1_K) \cdot (q \cdot 1_K)^{-1} \in K$. This is where we use the fact that the characteristic is $0$: this guarantees that $q \cdot 1_K \ne 0_K$, so we may invert it. Similar considerations hold for $F$.
Let now $\phi : F \to K$ be an injective field homomorphism. In general, $\phi (1_F) = \phi (1_F ^2) = \phi (1_F)^2$, whence it follows that either $\phi (1_F) = 0_K$ or that $\phi (1_F) = 1_K$. Injectivity eliminates the first possibility, whence it follows that $\phi (1_F) = 1_K$. We have, for every $m \in \Bbb N \setminus \{0\}$ that
$$\phi (m \cdot 1_F) = \phi ( \underbrace {1_F + \dots + 1_F} _{m \text{ times}}) = \underbrace {\phi (1_F) + \dots + \phi (1_F)} _{m \text{ times}} = \underbrace{ 1_K + \dots + 1_K} _{m \text{ times}} = m \cdot 1_K$$
It is obvious that $\phi (0 \cdot 1_F) = \phi (0_F) = 0_K = 0 \cdot 1_K$, so we deduce that $\phi (m \cdot 1_F) = m \cdot 1_K$ for all $m \in \Bbb N$.
Mow, if $m \in \Bbb N$ then $0_K = \phi (0_F) = \phi(m \cdot 1_F - m \cdot 1_F) = \phi (m \cdot 1_F) + \phi (- m \cdot 1_F)$, whence we deduce that $\phi (-m \cdot 1_F) = - \phi (m \cdot 1_F) = - m \cdot 1_K$, which finally allows us to say that $\phi (m \cdot 1_F) = m \cdot 1_K$ for all $m \in \Bbb Z$.
For $p \in \Bbb Z$ and $q \in \Bbb Z \setminus \{0\}$ we may now write
$$\phi \big( (p \cdot 1_F) (q \cdot 1_F)^{-1} \big) = \phi \big( (p \cdot 1_F) \big) \cdot \phi \big( (q \cdot 1_F)^{-1} \big) = (p \cdot 1_K) \cdot (q \cdot 1_K) ^{-1} ,$$
which shows that the copy of $\Bbb Q$ inside $F$ is taken isomorphically onto the copy of $\Bbb Q$ inside $K$. |
In the Hamilton-Jacobi equation, we take the partial time derivative of the action. But the action comes from integrating the Lagrangian over time, so time seems to just be a dummy variable here and hence I do not understand how we can partial differentiate $S$ with respect to time? A simple example would also be helpful.
The action functional and Hamilton's principal function are two different mathematical objects related to the same physical quantity.
The action along a trajectory $\gamma:[t_1,t_2]\rightarrow Q$ is given by $$ S[\gamma] = \int_{t_1}^{t_2}L(\gamma(t'),\dot\gamma(t'),t')dt' $$ whereas the principal function is the solution of the Hamilton-Jacobi equation $$ H(q,\nabla S(q,t),t) + \frac{\partial S}{\partial t}(q,t) = 0 $$
If you denote by $\gamma_{q,t}$ the solution of the Euler-Lagrange equations with $$ \gamma_{q,t}(t_0)=q_0\\ \gamma_{q,t}(t)=q $$ then $$ S(q,t):=S[\gamma_{q,t}]=\int_{t_0}^{t}L(\gamma_{q,t}(t'),\dot\gamma_{q,t}(t'),t')dt' $$ will solve the Hamilton-Jacobi equation.
On the flip side, for the principal function we have the following $$ \frac{d}{dt}S(\gamma(t),t)=L(\gamma(t),\dot\gamma(t),t) $$ and thus $$ S[\gamma]=S(\gamma(t_2),t_2)-S(\gamma(t_1),t_1) $$
Note that the last two equations only hold for trajectories with $$ \frac{\partial L}{\partial\dot q}(\gamma(t),\dot\gamma(t),t) = \nabla S(\gamma(t),t) $$
Geometrically, the choice of integration constants of the principal function selects a leave of a foliation of phase space, which corresponds to the choice of initial condition $\gamma_q(t_0)=q_0$ from above.
I) At least three different quantities in physics are customary called an action and denoted with the letter $S$.
The (off-shell) action $$\tag{1}S[q]~:=~ \int_{t_i}^{t_f}\! dt \ L(q(t),\dot{q}(t),t) $$ is a
functionalof the full position curve/path $q^i:[t_i,t_f] \to \mathbb{R}$ for alltimes $t$ in the interval $[t_i,t_f]$. See also this question. (Here the words on-shelland off-shellrefer to whether the equations of motion (eom) are satisfied or not.)
If the variational problem $(1)$ with well-posed boundary conditions, e.g. Dirichlet boundary conditions $$\tag{2} q(t_i)~=~q_i\quad\text{and}\quad q(t_f)~=~q_i,$$ has a unique extremal/classical path $q_{\rm cl}^i:[t_i,t_f] \to \mathbb{R}$, it makes sense to define an on-shell action $$ \tag{3} S(q_f;t_f;q_i,t_i) ~:=~ S[q_{\rm cl}], $$ which is a
functionof the boundary values. See e.g. MTW Section 21.1.
The Hamilton's principal function $S(q,\alpha, t)$ in Hamilton-Jacobi equation is a
functionof the position coordinates $q^i$, integration constants $\alpha_i$, and time $t$, see e.g. H. Goldstein, Classical Mechanics,chapter 10. The total time derivative $$\tag{4} \frac{dS}{dt}~=~ \dot{q}^i \frac{\partial S}{\partial q^i}+ \frac{\partial S}{\partial t}$$ is equal to the Lagrangian $L$ on-shell, as explained here. As a consequence, the Hamilton's principal function $S(q,\alpha, t)$ can be interpreted as an action on-shell.
II)
Example: A non-relativistic free particle in 1 dimension.
The off-shell action is $$\tag{5} S[q]~=~ \frac{m}{2}\int_{t_i}^{t_f}\! dt \ \dot{q}(t)^2. $$
If we assume Dirichlet boundary conditions (2), the unique classical trajectory $q_{\rm cl}$ has constant velocity $$\tag{6}\dot{q}_{\rm cl}~=~\frac{q_f-q_i}{t_f-t_i}.$$ The Dirichlet on-shell action (3) is $$ \tag{7} S(q_f,t_f;q_i,t_i) ~=~ \frac{m}{2} \cdot \frac{(q_f-q_i)^2}{t_f-t_i}. $$
The Hamilton's principal function, i.e. a solution to Hamilton-Jacobi equation, is $$\tag{8} S(q,E,t)~=~\pm\sqrt{2m E} q - Et, $$ where $E$ is an integration constant (=the total energy). Due to the interpretation of Hamilton's principal function as a type 2 generator of canonical transformations, the partial derivative $$\tag{9}Q~:=~ \frac{\partial S}{ \partial E}~\stackrel{(8)}{=}~\pm\sqrt{\frac{m}{2E}}q -t $$ must be a constant of motion. In other words, the position $q(t)$ is, as expected, an affine function of time $t$. This implies that the velocity is constant $$\tag{10} \dot{q} ~\approx~\pm\sqrt{\frac{2E}{m}}, $$ where the "$\approx$" symbol means equality modulo eom. The total time derivative of the Hamilton's principal function (8) is equal to the Lagrangian (=the kinetic energy) on-shell $$\tag{11} \frac{dS}{dt}~\stackrel{(8)}{=}~ \pm\sqrt{2m E} \dot{q} -E ~\stackrel{(10)}{\approx}~E.$$
Let us now compare point 2 and 3. With the Dirichlet boundary conditions (2), the energy becomes $$ \tag{12} E~=~ \frac{m}{2} \cdot \left(\frac{q_f-q_i}{t_f-t_i}\right)^2. $$ A comparison of eqs. (6) and (10) shows that we should use the plus (minus) branch of the solution (8) if $q_f\geq q_i$ ($q_f\leq q_i$), respectively. It is straightforward to check that the difference in the Hamilton's principal function becomes the on-shell action (7), $$ \tag{13} S(q_f,E,t_f)-S(q_i,E,t_i)~\stackrel{(8)+(12)}{=}~\frac{m}{2} \cdot \frac{(q_f-q_i)^2}{t_f-t_i}~\stackrel{(7)}{=}~ S(q_f,t_f;q_i,t_i).\qquad $$
I think the other two answers are overkill. The simpler answer is that time $t$ is not a dummy variable. The integration of $L$ over time here is an indefinite integration, so if we have $L(q,\dot{q},t)$, and we want to integrate it over time $t$, the result is $$\int_{t_i}^tL(q,\dot{q},\tau)d\tau$$ here $\tau$ is the dummy variable but $t$ is not, and the result is a function of $t$. |
Question: As we know, (1) the macroscopic spatial dimension of our universe is 3 dimension, and (2) gravity attracts massive objects together and the gravitational force is isotropic without directional preferences. Why do we have the spiral 2D plane-like Galaxy(galaxies), instead of spherical or elliptic-like galaxies? Input:Gravity is (at least, seems to be) isotropic from its force law (Newtonian gravity). It should show no directional preferences from the form of force vector $\vec{F}=\frac{GM(r_1)m(r_2)}{(\vec{r_1}-\vec{r_2})^2} \hat{r_{12}}$. The Einstein gravity also does not show directional dependence at least microscopically.
If the gravity attracts massive objects together
isotropically, and the macroscopic space dimension is 3-dimensional, it seems to be natural to have a spherical shape of massive objects gather together. Such as the globular clusters, or GC, are roughly spherical groupings Star cluster, as shown in the Wiki picture:
However, my impression is that, even if we have observed some more spherical or more-ball-like Elliptical galaxy, it is more common to find more-planar Spiral galaxy such as our Milky Way? (Is this statement correct? Let me know if I am wrong.)
Also, such have a look at this more-planar
Spiral "galaxy" as this NGC 4414 galaxy:
Is there some physics or math theory explains why the Galaxy turns out to be planar-like (or spiral-like) instead of spherical-like?
p.s. Other than the classical stability of a 2D plane perpendicular to a classical angular momentum, is there an interpretation in terms of the quantum theory of vortices in a macroscopic manner (just my personal speculation)?
Thank you for your comments/answers! |
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