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A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ... So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$. Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$. Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow. Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$. Well, we do know what the eigenvalues are... The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$. Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker "a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers. I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd... Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work. @TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now) Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $\|\|v\|\| = \|\|(v)_S\|\|$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism @AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$. Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1) For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism
Let $A \in \mathfrak{U}$ where $\mathfrak{U}$ is a $C^$ algebra and let $A=A^$ with $\sigma(A)\subseteq \mathbb{R}^+$. Let $f := \sqrt{\cdot}\in \mathcal{C}(\sigma(A))$ where $\mathcal{C}(X)$ is the set of continuous functions from $X$ to $\mathbb{R}$ 1.Prove that $f(A)$ is well-defined using the continuous functional calculus Show that $\sqrt{A}^2 = A$ These are my questions, I'm studying elements of mathematical theory between quantum mechanics and classical mechanics. I don't quite understand how we can apply the continuous functional calculus usually I'm having trouble to show it. These are my goals: I really think that this is not something difficult. But still I could not use the correct line of reason here. Using the continuous functional calculus we know that for an $f$ like the one in the question there exists an Gelfand homomorphism $\phi: \mathcal{C}(\sigma(A)) \to \mathfrak{U}$ such that, by the Weierstrass theorem, $\phi(f)$ is the limit of the operator norms of $\phi(p)$ with $p$ the polynomials that approximate to $f$ in the sup norm. So, we define $\phi$ in this way. Then, we use for the second part of the problem that $\sqrt{A}^2 = \phi(f)\phi(f) = \phi(f\cdot f) = \phi(f^2) = A$ Can someone help me in showing some technicalities behind these types of questions? Thanks in advance. Another try: 1. Let's use the continuous functional calculus to say that it is properly defined. So, what we have is that, for any function $f \in \mathcal{C}(\sigma(A))$ there is a $\phi: \mathcal{C}(\sigma(A))\to \mathfrak{U}$ such that $\phi(f)$ is defined as the limit of $\phi(p)$ for the $p$ polynomials in Weierstrass approximation theorem. This makes $\phi(f)$ well-defined. Now, for $f := \sqrt{\cdot}$ we define $\phi(f):=\sqrt{A}$. This concludes the first part. 2.With this an the properties of $\phi$ as a $\star$-homomorphism we have that, (see that $f^2 = \text{id}_{\mathcal{C}(\sigma(A))}$) $$\tag{1}\sqrt{A}^2 = \phi(f^2) = \phi(\text{id}_{\mathcal{C}(\sigma(A))}) = \text{id}_{\mathfrak{U}}(A) = A$$ because we have that $\mathfrak{U}$ is a $C^*$ unital algebra. The $(a)$ line I think that there is a problem in here. Because how can I say the last equality? This seems something that I'm cheeting to get the correct answer.
The following is a FAQ that I sometimes get asked, and it occurred to me that I do not have an answer that I am completely satisfied with. In Rudin's Principles of Mathematical Analysis, following Theorem 3.29, he writes: One might thus be led to conjecture that there is a limiting situation of some sort, a “boundary” with all convergent series on one side, all divergent series on the other side—at least as far as series with monotonic coefficients are concerned. This notion of “boundary” is of course quite vague. The point we wish to make is this: No matter how we make this notion precise, the conjecture is false. Exercises 11(b) and 12(b) may serve as illustrations. Exercise 11(b) states that if $\sum_n a_n$ is a divergent series of positive reals, then $\sum_n a_n/s_n$ also diverges, where $s_n = \sum_{i=1}^n a_n$. Exercise 12(b) states that if $\sum_n a_n$ is a convergent series of positive reals, then $\sum_n a_n/\sqrt{r_n}$ converges, where $r_n = \sum_{i\ge n} a_i$. Although these two exercises are suggestive, they are not enough to convince me of Rudin’s strong claim that no matter how we make this notion precise, the conjecture is false. Are there any stronger theorems in this direction? Edit. For example, are there any theorems about the topology/geometry of the spaces of all convergent/divergent series, where a series is viewed as a point in $\mathbb{R}^\infty$ or $(\mathbb{R}^+)^\infty$ in the obvious way?
Forgot password? New user? Sign up Existing user? Log in If a=8−42a=8-4\sqrt{2}a=8−42 then find the value of : a+2a−2−a+22a−22 \dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}}a−2a+2−a−22a+22 Note by Nihar Mahajan 4 years, 4 months ago Easy Math Editor This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: italics _italics_ bold __bold__ - bulleted- list 1. numbered2. list paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org) > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines# 4 spaces, and now they show# up as a code block.print "hello world" \[ \] 2 \times 3 2^{34} a_{i-1} \frac{2}{3} \sqrt{2} \sum_{i=1}^3 \sin \theta \boxed{123} Sort by: ha ha componendo dividendo solved it for me isnt that way easier??? Log in to reply How did you solve this using componendo dividendo ? Also , post your solution. sorry i cant dont know a thing about latex. @Kaustubh Miglani – You can post a solution without latex too. I will understand it. Or at least post your approach.I am curious about this method. a+2a−2−a+22a−22=8−42+28−42−2−8−42+228−42−22=10−426−42−8−228−62=5−223−22−4−24−32=(5−22)(3+22)9−8−(4−2)(4+32)16−18=15+42−8+16+82−62=7+42+5+42=12+82 \begin{aligned} \dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}} & = \dfrac{8-4\sqrt{2}+2}{8-4\sqrt{2}-2} - \dfrac{8-4\sqrt{2}+2\sqrt{2}}{8-4\sqrt{2}-2\sqrt{2}} \\ & = \dfrac{10-4\sqrt{2}}{6-4\sqrt{2}} - \dfrac{8-2\sqrt{2}}{8-6\sqrt{2}} \\ & = \dfrac{5-2\sqrt{2}}{3-2\sqrt{2}} - \dfrac{4-\sqrt{2}}{4-3\sqrt{2}} \\ & = \dfrac{(5-2\sqrt{2})(3+2\sqrt{2})}{9-8} - \dfrac{(4-\sqrt{2})(4+3\sqrt{2})}{16-18} \\ & = 15+4\sqrt{2} - 8 + \dfrac{16 + 8\sqrt{2} - 6}{2} \\ & = 7 +4\sqrt{2} + 5 + 4\sqrt{2} \\ & = 12+8\sqrt{2} \end{aligned} a−2a+2−a−22a+22=8−42−28−42+2−8−42−228−42+22=6−4210−42−8−628−22=3−225−22−4−324−2=9−8(5−22)(3+22)−16−18(4−2)(4+32)=15+42−8+216+82−6=7+42+5+42=12+82 Err, sir, There is a typo in the line of the answer. It should be "+" Thanks. @Chew-Seong Cheong – ⌣¨\huge\ddot\smile⌣¨ The correct ans is 12+8212+8\sqrt{2}12+82. Yeah. Thank you sir. :) I needed the answer. I found out the procedure. Thanks everyone @Nihar Mahajan @Sravanth Chebrolu @Soumya Khurana and of course @Sandeep Bhardwaj sir. No need to @Mention!!! ⌣¨\huge \ddot \smile⌣¨ Yes , I am correct then. :P The answer I am getting is 12+8212 + 8\sqrt{2}12+82 , Is it correct? Simply put the value of aaa and calculate. :P 12+10\sqrt{2} @Soumya Khurana , I am afraid that there is no such option in the paper I got this from. Can u tell me the options @Mehul Arora @Soumya Khurana – Sure. a. 12+212+ \sqrt {2}12+2 b. 12−212- \sqrt {2}12−2 c. 2 d. -2 @Mehul Arora – Please see the explanation, According to the question, we have to find, 8−42+28−42−2−8−42+228−42−22=10−426−42×6+426+2−8−228−62×8+628+62=7+42−(−5−42)=12+82\huge \dfrac{8-4\sqrt{2}+2}{8-4\sqrt{2}-2}-\dfrac{8-4\sqrt{2}+2\sqrt{2}}{8-4\sqrt{2}-2\sqrt{2}} \\ \huge =\dfrac{10-4\sqrt{2}}{6-4\sqrt{2}} \times \dfrac{6+4\sqrt{2}}{6+\sqrt{2}} - \dfrac{8-2\sqrt{2}}{8-6\sqrt{2}} \times \dfrac{8+6\sqrt{2}}{8+6\sqrt{2}} \\ \huge = 7+4\sqrt{2}-(-5-4\sqrt{2}) \\ \huge = \boxed{12+8\sqrt{2}} 8−42−28−42+2−8−42−228−42+22=6−4210−42×6+26+42−8−628−22×8+628+62=7+42−(−5−42)=12+82 @Sravanth Chebrolu – You have made a mistake.Find it. @Mehul Arora – @Mehul Arora I checked for sure using a claculator...None of the options match. :( @Mehul Arora – @Mehul Arora , I am afraid that there is no such option in the paper which is the correct value of the expression above. @Nihar Mahajan – Thanks! I just got juggled up with the signs. cheers!cheers!cheers! @Nihar Mahajan – There should be one more option as "None of the given". :P @Sandeep Bhardwaj – Yeah... I calculated it manually. :P Problem Loading... Note Loading... Set Loading...
The generalized Gauss-Bonnet theorem says that: Let $M$ be a closed oriented Riemannian manifold with an even dimension $n$, then $$ \int_{M}\Omega=\chi(M). $$ In that formula, $\chi$ is the euler characteristic and $$ \Omega =\frac{(2\pi)^{-\frac{n}{2}}}{2^{n}(\frac{n}{2})!}\sum_{\sigma,\tau\in S_{n}}g^{-1}\mathrm{sgn}(\sigma)\mathrm{sgn}(\tau)R_{\sigma(1)\sigma(2)\tau(1)\tau(2)}\cdots R_{\sigma(n-1)\sigma(n)\tau(n-1)\tau(n)}, $$ where $g$ is the determinant of the matrix of the metric and, if $R$ is the curvature tensor $$ R_{i,j,k,l}=\langle R(e_{i},e_{j})e_{l},e_{k}\rangle. $$ First of all, I would like to know if the formula that I have written for $\Omega$ is the correct one. Now, does this coincide with the Gauss-Bonnet Theorem in the two dimensional case? I mean, if $n=2$, then, taking into account the properties of the curvature tensor $$ \Omega=\frac{1}{2\pi}\cdot\frac{R_{1212}}{g}. $$ Is this $\frac{1}{2\pi}\mathcal{K}$ if $M$ is a surface in $\mathbb {R}^{3}$?
For $t \in \mathbb{R}$ define $$ F(t) = \sum_{n=1}^{[t]} \frac{(-1)^{(n-1)}}{n^{\frac12 + it}}$$ Let $\operatorname{Arg}(t)$ be $\operatorname{atan2}(\Im t , \Re t)$ - basically this is $\arctan$, but the sign depends on the quadrant. Observation $\operatorname{Arg}(F(t))$ jumps (usually from negativeto positive) very near all nontrivial zeta zeros on the critical line and in seemingly rare occasions without zeros. Computing $F(t)$ thenaiive way is not efficient for me. $F(t)$ is truncated Dirichlet eta function on the critical line, but it is not $0$ at zeros of zeta, though $\|F(t)\|$ has local minima near zeros. Added Wolfram Alpha found closed form for $F(t)$ in terms of Hurwitz zeta and zeta: \begin{align} & \sum_{n=1}^k\frac{(-1)^{n-1}}{n^{\frac12 + i t}} = \\ & 2^{-1/2-i t}(-(-1)^k \zeta(i t+1/2, (k+1)/2)+ \\ & (-1)^k \zeta(i t+1/2, (k+2)/2)+2^{1/2+i t} \zeta(1/2 i (2 t-i))-2 \zeta(1/2 i (2 t-i))) \end{align} Setting $k=[t]$ gives $F(t)$. Numerical evidence supports the closed form. In comments Greg Martin suggested $F(t)$ might not be correlated to higher zeros, though numerical evidence suggests it is correlated at height $10^6$, including closely spaced zeros. Another observation is $\|F(t)\|$ appears to have local minima close to zeta zeros on the critical line. Setting $$ G(t) = \sum_{n=1}^{[t]} \frac{(-1)^{(n-1)}}{n^{1 + it}}$$ the jumps of $G(t)$ appear zeros of $\eta(1+i t)$ and looks like $\|G(t)\| \sim \|\eta(1 + i t)\|$ Can this be explained? Counterexamples? Plot:
Size and mass of very large stars: Most massive example, the blue Pistol Star (150 M ☉ ). Others are Rho Cassiopeiae (40 M ☉ ), Betelgeuse (20 M ☉ ), and VY Canis Majoris (17 M ☉ ). The Sun (1 M ☉ ) which is not visible in this thumbnail is included to illustrate the scale of example stars. Earth's orbit (grey), Jupiter's orbit (red), and Neptune's orbit (blue) are also given. The solar mass ( ) is a standard unit of mass in astronomy that is used to indicate the masses of other stars, as well as clusters, nebulae and galaxies. It is equal to the mass of the Sun, about two nonillion kilograms: M ☉ M ☉ = 55±0.00025)×10 30 kg(1.988 [1] [2] The above mass is about 946 times the mass of the 332Earth ( M ⊕), or times the mass of 1048Jupiter ( M J). Because the Earth follows an elliptical orbit around the Sun, its solar mass can be computed from the equation for the orbital period of a small body orbiting a central mass. [3] Based upon the length of the year, the distance from the Earth to the Sun (an astronomical unit or AU), and the gravitational constant ( G), the mass of the Sun is given by: M_\odot = \frac{4 \pi^2 \times (1\,\mathrm{AU})^3}{G \times (1\,\mathrm{yr})^2} The value of the gravitational constant was first derived from measurements that were made by Henry Cavendish in 1798 with a torsion balance. The value he obtained differs by only 1% from the modern value. [4] The diurnal parallax of the Sun was accurately measured during the transits of Venus in 1761 and 1769, [5] yielding a value of (9 9″arcseconds, compared to the present 1976 value of 148″). If we know the value of the diurnal parallax, we can determine the distance to the Sun from the geometry of the Earth. 8.794 [6] The first person to estimate the mass of the Sun was Isaac Newton. In his work Principia, he estimated that the ratio of the mass of the Earth to the Sun was about 1/28 700. Later he determined that his value was based upon a faulty value for the solar parallax, which he had used to estimate the distance to the Sun (1 AU). He corrected his estimated ratio to 1/169 282 in the third edition of the Principia. The current value for the solar parallax is smaller still, yielding an estimated mass ratio of 1/332 946. [7] As a unit of measurement, the solar mass came into use before the AU and the gravitational constant were precisely measured. This is because the relative mass of another planet in the Solar System or the combined mass of two binary stars can be calculated in units of Solar mass directly from the orbital radius and orbital period of the planet or stars using Kepler's third law, provided that orbital radius is measured in astronomical units and orbital period is measured in years. The mass of the Sun has decreased since the time it formed. This has occurred through two processes in nearly equal amounts. First, in the Sun's core hydrogen is converted into helium by nuclear fusion, in particular the pp chain, and this reaction converts some mass into energy in the form of gamma ray photons. Most of this energy eventually radiates away from the Sun. Second, high-energy protons and electrons in the atmosphere of the Sun are ejected directly into outer space as a solar wind. The original mass of the Sun at the time it reached the main sequence remains uncertain. The early Sun had much higher mass-loss rates than at present, so it may have lost anywhere from 1–7% of its natal mass over the course of its main-sequence lifetime. [8] The Sun gains a very small mass through the impact of asteroids and comets; however the Sun already holds 99.86% of the Solar System's total mass, so these impacts cannot offset the mass lost by radiation and ejection. Contents Related units 1 See also 2 References 3 Further reading 4 Related units One solar mass, M ☉, can be converted to related units: It is also frequently useful in general relativity to express mass in units of length or time. See also References ^ 2014 Astronomical Constants http://asa.usno.navy.mil/static/files/2014/Astronomical_Constants_2014.pdf ^ NIST CODATA http://physics.nist.gov/cgi-bin/cuu/Value?bg ^ Harwit, Martin (1998), Astrophysical concepts, Astronomy and astrophysics library (3rd ed.), Springer, pp. 72, 75, ^ Holton, Gerald James; Brush, Stephen G. (2001). Physics, the human adventure: from Copernicus to Einstein and beyond (3rd ed.). Rutgers University Press. p. 137. ^ Pecker, Jean Claude; Kaufman, Susan (2001). Understanding the heavens: thirty centuries of astronomical ideas from ancient thinking to modern cosmology. Springer. pp. 291–291. ^ Barbieri, Cesare (2007). Fundamentals of astronomy. CRC Press. pp. 132–140. ^ Leverington, David (2003). Babylon to Voyager and beyond: a history of planetary astronomy. Cambridge University Press. p. 126. ^ Sackmann, I.-Juliana; Boothroyd, Arnold I. (February 2003), "Our Sun. V. A Bright Young Sun Consistent with Helioseismology and Warm Temperatures on Ancient Earth and Mars", The Astrophysical Journal 583 (2): 1024–1039, Further reading I.-J. Sackmann; A. I. Boothroyd (2003). "Our Sun. V. A Bright Young Sun Consistent with Helioseismology and Warm Temperatures on Ancient Earth and Mars". The Astrophysical Journal 583 (2): 1024–1039. This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. 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TL;DR: Naphthalene anion radical won't tolerate a presence of unshielded alkali metal cation nor in solution, neither in crystal. Reduction of naphthalene is possible strictly under air- and moisture free conditions by alkali-metals in the presence of coordinating solvents ( cryptands). In this case solvent-separated cations don't form coordination bonds with naphthalenide radical. On the other hand, further reduced ( dianion) naphthalene readily forms various compounds not only with alkali metals, but also with lanthanides, typically $\eta^4$-complexes. Various studies on stabilization of radical anions, especially in solid state, show that complexation (preferably with polar aprotic solvent or polydentate ligands) of counter-cations is essential: $$\ce{C_nH_m + [M^0]_{\infty} + $x$ Solv + $n$ L <=> [M+(Solv_x\|L_n)](C_nH_m^{.-})},$$ where $\ce{[M^0]_{\infty}}$ - any metal in oxidation state $0$ as a bulk reactant, $\ce{Solv}$ and $\ce{L}$ - solvent and ligand, correspondingly. If cation is optimally solvated (complexated/sterically protected), radical counter anions $\ce{C_nH_m^{.-}}$ with extended $\pi$ systems are preferred because they lack extensively charged centers which favor contact ion formation (1). For example, crystals of bis(diglyme-O,O',O'')-sodium naphthalenide radical$\ce{[Na+(diglume)2](C10H8^{.-})}$ ( [Na]) (1) has been isolated based on a principle of increasing enthalpy (calcd. based on crystal structures): $$\Delta H (\ce{[Na+(THF)6]}) = \pu{-587 kJ mol^{-1}}$$ $$\Delta H (\ce{[Na+(DME)3]}) = \pu{-671 kJ mol^{-1}}$$ $$\Delta H (\ce{[Na+(diglume)2]}) = \pu{-677.3 kJ mol^{-1}}$$ Under certain conditions, the reduction of the naphthalene can proceed further to the naphthalene dianion, $\ce{C10H8^{2-}}$. For example, a complex with radical ion $\ce{[Li+(TMEDA)2](C10H8^{.-})}$ ( [Li]) can reversibly be transformed into naphthalene dianion-complex $\ce{[Li+(TMEDA)]2(C10H8^{2-})}$ (2): Of course, bigger cations require better shielding, commonly resulting in a usage of bulkier solvents or supramolecular assemblies, such as ([2,2,2]-cryptand)-potassium naphthalenide radical, $\ce{[K+(crypt-222)](C10H8^{.-})}$ ( [K]) (3), in comparison to previously mentioned [Li] and [Na] naphthalenides (in the following crystal structures sphere-packing model is applied to metal atoms, and the rest is represented with sticks for simplicity): It is also worth mentioning that naphthalene dianion forms numerous complexes with rare earth elements (4), e.g. with [Y] (see picture): $\ce{CpYC10H8(DME)}$, ($\eta^5$-Cyclopentadienyl)-(1,2-dimethoxyethane)-($2\eta^1:\eta^2(2\sigma,\pi)$-naphthalenide)-yttrium(III) with a $\ce{C6}$ ring being bent by $26.1^\circ$: $\ce{La}$ and $\ce{Eu}$ complexes with bridged $\mu-\eta^4:\eta^4$-naphthalene are also obtained ($\ce{[LaI2(THF)3]2(C10H8)}$, $\ce{[EuI(DME)]2(C10H8)}$). (1) Bock, H.; Arad, C.; Näther, C.; Havlas, Z. J. Chem. Soc., Chem. Commun. 1995, 23, 2393–2394. DOI: 10.1039/C39950002393 (2) Melero, C.; Guijarro, A.; Yus, M. Dalton Trans. 2009, 0 (8), 1286–1289. DOI: 10.1039/B821119C (3) Rosokha, S. V.; Kochi, J. K. J. Org. Chem. 2006, 71 (25), 9357–9365. DOI: 10.1021/jo061695a (4) Protchenko, A. V.; Zakharov, L. N.; Fukin, G. K.; Struchkov, V. T.; Bochkareva, M. N. Russ Chem Bull 1996, 45 (4), 950–953. DOI: 10.1007/BF01431330
Finite differences in infinite domains Because of my friend, Edward Villegas, I ended up thinking about using a change of variables when solving an eigenvalue problem with finite difference. The problem Let's say that we want to solve a differential equation over an infinite domain. A common case is to solve the Time independent Schrödinger equation subject to a potential $V(x)$. For example where we want to find the pairs of eigenvalues/eigenfunctions $(E_n, \psi_n(x))$. What I normally do when using finitedifferencesis to regularly divide the domain. Where I take a large enough domain,so the solution have decayed close to zero. What I do in this post is tomake a change of variable to render the interval finite first and thenregularly divide the transformed domain in finite intervals. My usual approach My usual approach is to approach the second derivative with a centered difference for the point $x_i$ like this with $\Delta x$ the separation between points. We can solve this in Python with the following snippet: import numpy as np from scipy.sparse import diags from scipy.sparse.linalg import eigs def regular_FD(pot, npts=101, x_max=10, nvals=6): """ Find eigenvalues/eigenfunctions for Schrodinger equation for the given potential `pot` using finite differences """ x = np.linspace(-x_max, x_max, npts) dx = x[1] - x[0] D2 = diags([1, -2, 1], [-1, 0, 1], shape=(npts, npts))/dx*2 V = diags(pot(x)) H = -0.5D2 + V vals, vecs = eigs(H, k=nvals, which="SM") return x, np.real(vals), vecs Let's setup the plotting details. # Jupyter notebook plotting setup & imports %matplotlib notebook import matplotlib.pyplot as plt gray = '#757575' plt.rcParams["figure.figsize"] = 6, 4 plt.rcParams["mathtext.fontset"] = "cm" plt.rcParams["text.color"] = gray fontsize = plt.rcParams["font.size"] = 12 plt.rcParams["xtick.color"] = gray plt.rcParams["ytick.color"] = gray plt.rcParams["axes.labelcolor"] = gray plt.rcParams["axes.edgecolor"] = gray plt.rcParams["axes.spines.right"] = False plt.rcParams["axes.spines.top"] = False Let's consider the Quantum harmonic oscillator, that has as eigenvalues Using the finite difference method we have values that are really close to the analytic ones. with output array([0.4996873 , 1.49843574, 2.49593063, 3.49216962, 4.48715031, 5.4808703 ]) With the analytic ones If we plot these two, we obtain the following. plt.figure() plt.plot(anal_vals) plt.plot(vals, "o") plt.xlabel(r"$n$", fontsize=16) plt.ylabel(r"$E_n$", fontsize=16) plt.legend(["Analytic", "Finite differences"]) plt.tight_layout(); Let's see the eigenfunctions. plt.figure() plt.plot(x, np.abs(vecs[:, :3])2) plt.xlim(-6, 6) plt.xlabel(r"$x$", fontsize=16) plt.ylabel(r"$\|\psi_n(x)\|^2$", fontsize=16) plt.yticks([]) plt.tight_layout(); One inconvenient with this method is the redundant sampling towards the extreme of the intervals, while under sample the middle part. Changing the domain Let's now consider the case where we transform the infinite domain to a finite one with a change of variable with $\xi \in (-1, 1)$. Two options for this transformation are: $\xi = \tanh x$; and $\xi = \frac{2}{\pi} \arctan x$. Making this change of variable the equation we need to solve the following equation The following snippet solve the eigenproblem for the mapped domain: def mapped_FD(pot, fun, dxdxi, dxdxi2, npts=101, nvals=6, xi_tol=1e-6): """ Find eigenvalues/eigenfunctions for Schrodinger equation for the given potential `pot` using finite differences over a mapped domain on (-1, 1) """ xi = np.linspace(-1 + xi_tol, 1 - xi_tol, npts) x = fun(xi) dxi = xi[1] - xi[0] D2 = diags([1, -2, 1], [-1, 0, 1], shape=(npts, npts))/dxi2 D1 = 0.5diags([-1, 1], [-1, 1], shape=(npts, npts))/dxi V = diags(pot(x)) fac1 = diags(dxdxi(xi)2) fac2 = diags(dxdxi2(xi)) H = -0.5fac1.dot(D2) - 0.5fac2.dot(D1) + V vals, vecs = eigs(H, k=nvals, which="SM") return x, np.real(vals), vecs First transformation: $\xi = \tanh(x)$ Let's consider first the transformation In this case and We need to define the functions pot = lambda x: 0.5x2 fun = lambda xi: np.arctanh(xi) dxdxi = lambda xi: 1 - xi2 dxdxi2 = lambda xi: -2xi(1 - xi*2) and run the function And we obtain the following array([0.49989989, 1.4984226 , 2.49003572, 3.46934257, 4.46935021, 5.59552989]) If we compare with the analytic values we get the following. plt.figure() plt.plot(anal_vals) plt.plot(vals, "o") plt.legend(["Analytic", "Finite differences"]) plt.xlabel(r"$n$", fontsize=16) plt.ylabel(r"$E_n$", fontsize=16) plt.tight_layout(); And the following are the eigenfunctions. Second transformation: $\xi = \frac{2}{\pi}\mathrm{atan}(x)$ Let's consider first the transformation In this case and Once, again, we define the functions. pot = lambda x: 0.5x*2 fun = lambda xi: np.tan(0.5np.pixi) dxdxi = lambda xi: 2/np.pi np.cos(0.5np.pixi)*2 dxdxi2 = lambda xi: -4/np.pi np.cos(0.5np.pixi)*4 np.tan(0.5np.pixi) and run the function to obtain the following eigenvalues array([0.49997815, 1.49979632, 2.49930872, 3.49824697, 4.49627555, 5.49295665]) with this plot plt.figure() plt.plot(anal_vals) plt.plot(vals, "o") plt.legend(["Analytic", "Finite differences"]) plt.xlabel(r"$n$", fontsize=16) plt.ylabel(r"$E_n$", fontsize=16) plt.tight_layout(); and the following eigenfunctions. Conclusion The method works fine, although the differential equation is more convoluted due to the change of variable. Although there are more elegant methods to consider infinite domains, this is simple enough to be solved in 10 lines of code. We can see that the mapping $\xi = \mathrm{atan}(x)$, covers better the domain than $\xi = \tanh(x)$, where most of the points are placed in the center of the interval. Thanks for reading! This post was written in the Jupyter notebook. You can download this notebook, or see a static view on nbviewer.
Let $M$ be a closed connected oriented smooth manifold and $\mathrm{Diff}_{+}(M)$ the group of orientation preserving diffeomorphisms of $M$ endowed with the compact-open topology. Pick a base point $x_{0} \in M$ and consider the evaluation map $$\mathrm{ev}\colon \mathrm{Diff}_{+}(M) \to M, \quad \mathrm{ev}(g) = g(x_{0}).$$ I would like to know when the evaluation map admits a global section, i.e. a continuous map $s\colon M \to \mathrm{Diff}_{+}(M)$ such that $\mathrm{ev} \circ s = \mathrm{id}_{M}$. Examples: If $M = G$ is a Lie group, then a section $s$ exists: for $x\in G$ put $s(x) := \text{left multiplication by }x\cdot x_{0}^{-1}$. However, for $M = S^{2}$, the 2-sphere, such a section $s$ cannot exist: it would induce an injectivemap on homotopy groups $s_{*}\colon \pi_{k}(S^{2}) \to \pi_{k}(\mathrm{Diff}_{+}(S^{2}))$, but $\pi_{2}(S^2) \cong \mathbb{Z}$ while $$\pi_{2}(\mathrm{Diff}_{+}(S^{2})) \cong \pi_{2}(\mathrm{SO}(3)) = 0.$$ It is probably hard to decide on the existence of a section for general $M$, so let us restrict to spheres $S^{n}$: Question: For which $n\in \mathbb{N}$ does the evaluation map $\mathrm{ev}\colon \mathrm{Diff}_{+}(S^{n}) \to S^{n}$ admit a continuous global section? Since $S^1$ and $S^3$ are naturally Lie groups, the section exists in thsese two cases by the first example above. We can analogously use octonionic multiplication to construct a global section for $S^{7}$. Are there any other cases than $n=1,3,7$?
Multi-armed bandit framework To start solving the problem of exploration, we are going to introduce the Multi-armed bandits framework. But what exactly does this solve? Just think that you are executing a clinical trial with 4 pills. You know that the pills have a survival rate but you don’t know what that survival rate is. Your goal: find the pill with the highest chance of survival in X trials. This is an example of a Multi-bandit framework, to put this more formal: Given $K$ actions (=arms) from a list of actions $A = {a_1, a_2, …, a_k}$ and knowing that the reward $r$ for a certain action is unknown $R_a = Pr(r \mid a)$ Note: for those who are new to probability theory, $\mid$ is a conditional, so $Pr(r \mid a)$ means the probability of getting reward $r$ if action $a$ happens. Now at each time step $t = 1, 2, …, T$ Choose an action: $a_t \in A$ See the reward received: $r_t ~ Pr(r mid a_t)$ Goal: Maximize $\sum_tr_t$ We call this kind of setup a stochastic setup, since $r_t$ is stochastic. Note: Stochastic means something that is randomly determined. Regret What is important in this framework, is to be able to quantify the “price of information”, or in other words: “If I choose path B instead of A, how hard will my reward decrease? how much will I regret taking path B?”. This means that we can say that for a successful algorithm, we want to to minimize our regret $L_t$, or maximize the reward $\sum_t{r_t}$ Our regret can be formulated as: $L_T$ = “The optimal way” - “what we managed to achieve without knowing the optimal route” = $L_T = T\mathbf{E}[r \mid a^{}] - \sum_T\mathbf{E}[r \mid a_t]$ . Exploration Algorithms The Naive Algoritm Let’s go back to the example of the pills, where we have T time steps (= population) and k actions. How would we divide the pills naively? Here we would just pick a round-robin approach, where we divide the pills equally. But we can do better! Greedy Algorithm A greedy algorithm is an algorithm that makes the choice that seems to be the best at the moment. Which means that we will make a local-optimal choice rather than a global-optimal solution. Now what happens if we try to apply this greedy algorithm to our Pill division problem? Then we would pick the pill that yields the best results, how you would say? By counting the amounts we were successful, and then keeping to pick this one. We thus want to estimate what the reward could be for any of the actions? or in mathematical terms: $\hat{r}_a \approx \mathbf{E} [ r \mid a ]$, but how can we calculate this? Note $\mathbf{E}$ represents the expected value, or what is our random variable expected to be if we repeat something $\infty$ times? Well let’s create a new variable that represents on how many times we chose a sample: $n_a$, then we say that $\hat{r}_a = \frac{r_t}{n_a}$. So our best action would be defined by $a_t = argmax \hat{r}_a$ The problem with this algorithm however, is that it could lock-on suboptimal forever due to the fact that we do not do any exploration. For example, if we take pills and we have a positive result the first time, then we would not go and try the other ones. Our Greedy Algorithm has linear regret /* * Use the greedy empirical estimate to choose an arm * * @param {Array} armCounts nArms * 2, representing the observed counts * @returns {Integer} the arm to be pulled at the next timestep (0 index based) /const greedyChoice = (armCounts) => { let pMax = -1; let choice = -1; for (let i = 0; i < armCounts.length; i++) { let nPull = (armCounts[i][0] + armCounts[i][1]); if (nPull < 0.5) { pHat = 0.5; } else { pHat = armCounts[i][0] / nPull; } if (pHat > pMax) { pMax = pHat; choice = i; } } return choice;} Optimistic Greedy Algorithm ($\epsilon$-Greedy) Since the greedy algorith locks-on a suboptimal choice, we want to be able to “explore” more to try to find the global minimum. Therefore we will have to introduce a probability to explore. This probability is what we can write as our $\epsilon$ variable. So formal this will be that with probability $\epsilon$ we will select a random action and with probability 1 - $\epsilon$ we will select our best action. /* * Use the greedy empirical estimate to choose an arm * * @param {Array} armCounts nArms * 2, representing the observed counts * @param {Integer} epsilon A value between 0 and 1 representing * how much chance there is to pick a random action * @returns {Integer} the arm to be pulled at the next timestep (0 index based) /const greedyEpsilonChoice = (armCounts, epsilon) => { let choice = -1; if (Math.random() < epsilon) { choice = Math.floor((Math.random() armCounts.length)); } else { choice = greedyChoice(armCounts); } return choice;} Upper Confidence Bound (UCB) Algorithm We just saw two algorithms (Greedy and $\epsilon$-greedy) that are able to achieve linear regret. But can we do better? Well as proven Lai and Robbins, it appears that the lower bound of Regret is Logarithmic. But how can we achieve this? Well we can achieve this through a principle known as optimism in the face of uncertainty. To achieve low regret, we only need to identify the optimal action. Therefore we need use to use the collected data to eliminate actions that are sub-optimal as much as possible. The algorithm is written as follows: For each action a, maintain the amount of times we took that action $n_a$ and the empirical average of pay-offs for that action $\hat{r}_a$ For the first $k$ rounds, play each action once At round t, play $a_t = argmax(\hat{r}_a + \sqrt{\frac{2logt}{n_a}})$ This algorithm achieves a Logarithmic regret, but only if the amount of rounds is larger than our amount of actions. It does this by adding an exp bonus to unexplored actions/less frequently called actions, represented by $\sqrt{\frac{2logt}{n_a}}$. This however decreases rapidly once $n_a$ increases. /** * @param {Array} armCounts nArms * 2, representing the observed counts * @param {Integer} timestep the number of timesteps elapsed * @returns {Integer} the arm to be pulled at the next timestep (0 index based) /const ucbChoice = (armCounts, timestep) => { let pMax = -1; let choice = -1; for (let i = 0; i < armCounts.length; i++) { let nPull == (armCounts[i][0] + armCounts[i][1]); if (nPull < 0.5) { nPull = 1; } let pHat = armCounts[i][0] / nPull; let pUpper = pHat + Math.sqrt(Math.log(timestep) / nPull); if (pUpper > pMax) { pMax = pUpper; choice = i; } } return choice;} Posterior Sampling Algorithm (=Thompson Sampling) If we have prior knowledge on what the payoffs could be, then posterior sampling says to take action a according to the probability that a is optimal. Resulting in a convergence to the a method. In our drug discovery example, the rewards were Bernoulli (live or die), which is a random Binary Variable). We can thus say that Rewards = Bernoulli($P_a$), assuming that $P_a \approx Beta(1, 1)$ with $Beta$ being a Beta distribution. Note: A beta distribution is a uniform chance distribution with 2 parameters. Algorithm: Maintain patients who lived or died At round T $\hat{P}_a \approx Beta(1 + live, 1 + die)$ (the posterior distribution for each arm) Play $a_t = argmax(P_a)$. (We will play the most optimistic looking sample) /** * @param {Array} armCounts nArms * 2, representing the observed counts * @returns {Integer} the arm to be pulled at the next timestep (0 index based) /const posteriorSamplingChoice = (armCounts, timestep) => { let pMax = -1; let choice = -1; let priorA = 1; let priorB = 1; for (let i = 0; i < armCounts.length; i++) { let pSample = rBeta(armCounts[i][0] + priorA, armCounts[i][1] + priorB); if (pSample > pMax) { pMax = pSample; choice = i; } } return choice;}function rbeta(alpha, beta) { var alpha_gamma = rgamma(alpha, 1); return alpha_gamma / (alpha_gamma + rgamma(beta, 1));}var SG_MAGICCONST = 1 + Math.log(4.5);var LOG4 = Math.log(4.0);function rgamma(alpha, beta) { // does not check that alpha > 0 && beta > 0 if (alpha > 1) { // Uses R.C.H. Cheng, "The generation of Gamma variables with non-integral // shape parameters", Applied Statistics, (1977), 26, No. 1, p71-74 var ainv = Math.sqrt(2.0 alpha - 1.0); var bbb = alpha - LOG4; var ccc = alpha + ainv; while (true) { var u1 = Math.random(); if (!((1e-7 < u1) && (u1 < 0.9999999))) { continue; } var u2 = 1.0 - Math.random(); v = Math.log(u1/(1.0-u1))/ainv; x = alphaMath.exp(v); var z = u1u1u2; var r = bbb+cccv-x; if (r + SG_MAGICCONST - 4.5z >= 0.0 \|\| r >= Math.log(z)) { return x beta; } } } else if (alpha == 1.0) { var u = Math.random(); while (u <= 1e-7) { u = Math.random(); } return -Math.log(u) * beta; } else { // 0 < alpha < 1 // Uses ALGORITHM GS of Statistical Computing - Kennedy & Gentle while (true) { var u3 = Math.random(); var b = (Math.E + alpha)/Math.E; var p = bu3; if (p <= 1.0) { x = Math.pow(p, (1.0/alpha)); } else { x = -Math.log((b-p)/alpha); } var u4 = Math.random(); if (p > 1.0) { if (u4 <= Math.pow(x, (alpha - 1.0))) { break; } } else if (u4 <= Math.exp(-x)) { break; } } return x beta; }} References https://www.edx.org/course/reinforcement-learning-explained http://iosband.github.io/2015/07/28/Beat-the-bandit.html Tze Leung Lai and Herbert Robbins. Asymptotically efﬁcient adaptive allocation rules. Advances in applied mathematics, 6(1):4–22, 1985 https://jeremykun.com/2013/10/28/optimism-in-the-face-of-uncertainty-the-ucb1-algorithm/ https://stackoverflow.com/questions/9590225/is-there-a-library-to-generate-random-numbers-according-to-a-beta-distribution-f
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions. Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not. (a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular? (b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular? (c) Let $A$ be a $4\times 4$ matrix and let\[\mathbf{v}=\begin{bmatrix}1 \\2 \\3 \\4\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}4 \\3 \\2 \\1\end{bmatrix}.\]Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular? Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible. Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$. Prove that the matrix\[A=\begin{bmatrix}0 & 1\\-1& 0\end{bmatrix}\]is diagonalizable.Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix. The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes. This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems. Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$. (a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$. (b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$. Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular. Problem 9.Determine whether each of the following sentences is true or false. (a) There is a $3\times 3$ homogeneous system that has exactly three solutions. (b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric. (c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$. (d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent. An $n\times n$ matrix $A$ is called nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.Using the definition of a nonsingular matrix, prove the following statements. (a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular. (b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then: The matrix $B$ is nonsingular. The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.) Let $A$ be an $n\times (n-1)$ matrix and let $\mathbf{b}$ be an $(n-1)$-dimensional vector.Then the product $A\mathbf{b}$ is an $n$-dimensional vector.Set the $n\times n$ matrix $B=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$. Prove that $B$ is a singular matrix for any choice of $\mathbf{b}$. For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A\|I]$, where $I$ is the $3\times 3$ identity matrix.
SageMath SageMath (formerly Sage) is a program for numerical and symbolic mathematical computation that uses Python as its main language. It is meant to provide an alternative for commercial programs such as Maple, Matlab, and Mathematica. SageMath provides support for the following: Calculus: using Maxima and SymPy. Linear Algebra: using the GSL, SciPy and NumPy. Statistics: using R (through RPy) and SciPy. Graphs: using matplotlib. An interactive shellusing IPython. Access to Python modulessuch as PIL, SQLAlchemy, etc. Contents Installation contains the command-line version; for HTML documentation and inline help from the command line. includes the browser-based notebook interface. Note:Most of the standard and optional Sage packages are available as optional dependencies of the package or in AUR, therefore they have to be installed additionally as normal Arch packages in order to take advantage of their features. Note that there is no need to install them with sage -i, in fact this command will not work if you installed SageMath with pacman. Usage SageMath mainly uses Python as a scripting language with a few modifications to make it better suited for mathematical computations. SageMath command-line SageMath can be started from the command-line: $ sage For information on the SageMath command-line see this page. Note, however, that it is not very comfortable for some uses such as plotting. When you try to plot something, for example: sage: plot(sin,(x,0,10)) SageMath opens the plot in an external application. Sage Notebook Note:The SageMath Flask notebook is deprecated in favour of the Jupyter notebook. The Jupyter notebook is recommended for all new worksheets. You can use the application to convert your Flask notebooks to Jupyter A better suited interface for advanced usage in SageMath is the Notebook. To start the Notebook server from the command-line, execute: $ sage -n The notebook will be accessible in the browser from http://localhost:8080 and will require you to login. However, if you only run the server for personal use, and not across the internet, the login will be an annoyance. You can instead start the Notebook without requiring login, and have it automatically pop up in a browser, with the following command: $ sage -c "notebook(automatic_login=True)" Jupyter Notebook SageMath also provides a kernel for the Jupyter notebook in the package. To use it, launch the notebook with the command $ jupyter notebook and choose "SageMath" in the drop-down "New..." menu. The SageMath Jupyter notebook supports LaTeX output via the %display latex command and 3D plots if is installed. Cantor Cantor is an application included in the KDE Edu Project. It acts as a front-end for various mathematical applications such as Maxima, SageMath, Octave, Scilab, etc. See the Cantor page on the Sage wiki for more information on how to use it with SageMath. Cantor can be installed with the official repositories.package or as part of the or groups, available in the Optional additions SageTeX If you have TeX Live installed on your system, you may be interested in using SageTeX, a package that makes the inclusion of SageMath code in LaTeX files possible. TeX Live is made aware of SageTeX automatically so you can start using it straight away. As a simple example, here is how you include a Sage 2D plot in your TEX document (assuming you use pdflatex): include the sagetexpackage in the preamble of your document with the usual \usepackage{sagetex} create a sagesilentenvironment in which you insert your code: \begin{sagesilent} dob(x) = sqrt(x^2 - 1) / (x * arctan(sqrt(x^2 - 1))) dpr(x) = sqrt(x^2 - 1) / (x * log( x + sqrt(x^2 - 1))) p1 = plot(dob,(x, 1, 10), color='blue') p2 = plot(dpr,(x, 1, 10), color='red') ptot = p1 + p2 ptot.axes_labels(['$\\xi$','$\\frac{R_h}{\\max(a,b)}$']) \end{sagesilent} create the plot, e.g. inside a floatenvironment: \begin{figure} \begin{center} \sageplot[width=\linewidth]{ptot} \end{center} \end{figure} compile your document with the following procedure: $ pdflatex <doc.tex> $ sage <doc.sage> $ pdflatex <doc.tex> you can have a look at your output document. The full documentation of SageTeX is available on CTAN. Troubleshooting TeX Live does not recognize SageTex If your TeX Live installation does not find the SageTex package, you can try the following procedure (as root or use a local folder): Copy the files to the texmf directory: # cp /opt/sage/local/share/texmf/tex/* /usr/share/texmf/tex/ Refresh TeX Live: # texhash /usr/share/texmf/ texhash: Updating /usr/share/texmf/.//ls-R... texhash: Done.
Definition: A renamable Horn formulais a Boolean formula that can be transformed into a Horn formula by flipping the polarity of every instance of one of more of its variables. Example: $\qquad (x_1 \lor x_2 \lor x_3)\land (\lnot x_1 \lor \lnot x_2 \lor x_3)$ This formula is renamable Horn because flipping the polarity of $x_2$ and $x_3$ produces the Horn formula $\qquad (x_1 \lor \lnot x_2 \lor \lnot x_3)\land (\lnot x_1 \lor x_2 \lor \lnot x_3)$ Can I extend the test procedure for identifying renamable Horn formulas, as described in the Harry Lewis paper "Renaming a Set of Clauses as a Horn Set", to quantified formulas? The paper states: Let $S$ be a set of clauses, say $S = (C_1 ..... C_m)$, where each $C_i = (L_{i1} .... , L_{il})$. Then define $S^$ to be the set of clauses $\qquad \bigcup_{i=1}^m \bigcup_{1\leq j <k\leq l} ((L_{ij}, L_{ik}))$. Then $S$ is renamable-Horn if and only if $S^$ is satisfiable. Is this procedure applicable to quantified Boolean formulas also?
I recently began browsing the 5e PHB when I noticed that there was no distance per round when falling under the Falling category. Is there a set fall speed and if so, what is it? [This answer superseded by the release of Xanathar's Guide to Everything, Nov 2017, as detailed in this answer.] The rules have no explicit guidance on falling kinematics. Mostly. Free-falling motion isn't tackled in the rules. Back to that in a moment. Feather Fall allows one to fall at 60 ft. per round (6 sec.), or at a speed of 10 fps without suffering damage. Free-fall, which is injurious, should be faster than that. A little high-school physics will tell us that a body falling freely (assuming g=32 ft/s 2) for 10 ft. will attain a final speed of ~25 fps. So this all makes sense: 10fps=no damage, 25fps=1d6 damage. Distance fallen: To me this means it's not inherently unreasonable to use the simple classical physics in this situation: assuming acceleration due to gravity similar to that experienced at sea level on Earth and ignoring air resistance at low speeds: starting from rest: \$ d_{\text{1 round}} = 576\text{ ft} \$ starting from rest: \$ d_{n\text{ rounds}} = 576 \times n^2\text{ ft} \$ Falling speed: your average velocity during the fall would be \$\sqrt{16d}\$, in feet per second. (Your final velocity is twice that.) For those who really want a refresher on simple kinematics, assuming uniform acceleration and starting velocities of zero: \$ \text{distance traveled} = \frac{1}{2} \times \text{acceleration} \times \text{time}^2 \$ \$ \text{final velocity} = \sqrt{2 \times \text{acceleration} \times \text{distance traveled}} \$ \$ \text{average velocity} = \dfrac{\text{final velocity}}{2} \$ \$ \text{time of fall} = \sqrt{\dfrac{2 \times \text{distance traveled}}{\text{acceleration due to gravity}}} \$ In non-SI units the acceleration due to gravity is approximately 32 feet per second 2. (up to) 500 feet, by rule. Xanathar's Guide to Everything p. 77 gives this optional rule for the rate of falling (particularly for long falls): When you fall from a great height you instantly descend up to 500 feet. If you're still falling on your next turn you descend up to 500 feet at the end of that turn. This process continues until the fall ends. The 500ft rule (c. 170m) is designed to simplify things. I also don't like the use of the word 'instantly' here. I think a more accurate representation, especially for sentient entities, would be 'before one can take an action', explaining why entities who can negate falling by taking an action, such as casting levitate or fly, cannot stop themselves in the first round of action. Why? Because they are surprised at suddenly falling and hence cannot take an action. But, it seems plausible that other entities might be able to act before one has fallen 500ft. I'm not suggesting that new rules applying the laws of physics (in our world) or dissecting a 6-second round into sub-components are required or even desired. Rather, I would remind everyone of the first rule of D&D: The GM is always right. The rules are meant to be guidelines, not constraints. 20d6 will kill a lot of characters, and for the sake of the story the GM may not want a character to die. On the other hand, 20d6 will not kill a lot of characters, meaning that a fall of tens of thousands of feet will not kill them, and the GM may decide that that is patently absurd and have death ensue. The angle here is that the GM and players are telling a story, not playing a video game.
Good day. But before I implement the algorithm, I have decided to test the DACE model on a simple 1-d function. I am receiving strange results; I believe I have located the problem, but am uncertain, whether my implementation is wrong, or if what I found is a weak point in the model. To initiate the algorithm, we need first to determine a set of parameters $\theta = \{\theta_h\}$ and $p = \{p_h\}$ from some sample data we measure. The way we do so is as follows: Assume we have $X = \{X^1, ..., X^n\}$ a set of points we have evaluated the function at, and $Y = \{y_1, ..., y_n\}$ the values at those points. Denote $X^i = (x^i_1, ... x^i_k)$. Then define $$ d_{ij} = \sum_{h=1}^k \theta_h \|x_h^i - x_h^j\|^{p_h}$$ $$R_{ij} = e^{-d_{ij}}$$ We estimate the values $\theta, p$ by maximizing the likelihood function $$\frac1{(2\pi)^{n/2}(\sigma^2)^{n/2}\|R\|^{1/2}} e^{-\frac{(...)R^{-1}(...)}{\sigma^2}}$$ (The exact expression in the exponent does not matter for what I am about to say, but is present in the files linked to in the beginning. Also, $\sigma^2 = \sigma^2(\theta,p)$). I carry the maximization of the likelihood by the Nelder-Meads algorithm (fminsearch in MATLAB). But here is my problem: in some cases, when the initial data set is too large (I am minimizing $-sin(x)$ on $[0,\pi]$, and 'too large' is ~10) what happens, is that the values the algorithm goes for are such, that the matrix $R$ becomes nearly singular ($cond(R)$ ~ $1e^{16}$). This can be explained, since we have a determinant of the matrix in negative power in the likelihood function. Thus, for a singular matrix the determinant becomes zero, and the likelihood function itself diverges to infinity. But because my calculations also feature $R^{-1}$, having a nearly-singular matrix is problematic for the algorithm (and that is indeed the case, the returned estimated function behaves nothing like the original). Note, that this does not happen always, and for some initial data of ~4 points, the aprroximation to the function behaves very well, so I have reason to believe that the mistake is not simply a coding mistake. Thus I ask again: is there another way to estimate the parameters? Am I doing something wrong? Is this a weak point in the algorithm?
Recall that in physical chemistry, the spontaneity of a reaction at constant pressure and a given temperature $T$ can be quantified by the Gibbs free energy $$\Delta G = \Delta H - T \Delta S$$ A reaction is spontaneous if $\Delta G < 0$. For an endothermic reaction to be spontaneous, $\Delta H < 0$, $\Delta S > 0$ and $\|\Delta H\|< \|T\Delta S\|$ Now suppose we monitor the rate of change of $\Delta G$ as the reaction progressed. This means \begin{align} \frac{\mathrm d\Delta G}{\mathrm dt} & = \frac{\mathrm d\Delta H}{\mathrm dt} - \frac{\mathrm d(T \Delta S)}{\mathrm dt}\\ \frac{\mathrm d\Delta G}{\mathrm dt} & = \frac{\mathrm dT}{\mathrm dt}\left(\frac{\partial \Delta H}{\partial T} - T\frac{\partial \Delta S}{\partial T}- \Delta S\right) \end{align} Now suppose for simplicity $\Delta H$ is roughly constant with temperature. Then: $$\frac{\mathrm d\Delta G}{\mathrm dt} = \frac{\mathrm dT}{\mathrm dt}\left( - T\frac{\partial \Delta S}{\partial T}- \Delta S\right)$$ Thus for $\frac{\mathrm dT}{\mathrm dt} < 0$, a reaction to be self sustaining indefinitely if the following holds for all $t > 0$ \begin{align} - T\frac{\partial \Delta S}{\partial T}- \Delta S & > 0\\ - T\frac{\partial \Delta S}{\partial T} & > \Delta S \end{align} In all closed systems, entropy will eventually maximises according to the second law, thus there will be a point where $\frac{\partial\Delta S}{\partial T} < 0$ (i.e. the increase in entropy as the reaction progressed will slow down over time) until the reaction stops (thus explaining why both exothermic and endothermic reactions will eventually go to completion) Therefore, my question on self sustaining endothermic reaction that last for hours is boiled down to finding a time $t>0$ such that the above inequality holds in the order of hours, given a $\Delta H$ as positive as physically realisable Any example of an endothermic reactions that has $\Delta H > 50{-}100\ \mathrm{kJ\ mol^{-1}}$ and can sustain itself in the order of hours?
okay, this is answer Part 2. doing it as a separate answer because, as soon as there are LaTeX equations put in, the simultaneous rendering and typing get very slow. so my questions (1) and (2) were meant to lead you to make a couple of basic conclusions that can help in understanding the source of the Octave Problem and, with such understanding, maybe craft code that can avoid some (maybe not all) of the octave errors. about (1), the issue is, we hear stuff with our ears and brain (and as such we hear a "pitch" of a tone that is most often associated with the fundamental frequency $f_0$) but the Pitch Detection Algorithm (PDA) is not hearing anything but is doing math and making logical decision that it is programmed to do. so, , if a tone is judged to be periodic with fundamental frequency of 440 Hz, it is just as well a tone of 220 Hz or 146.67 Hz or 110 Hz or 88 Hz or 55 Hz. it is actually just as periodic with those fundamentals (and the periods associated with those fundamentals) as it is at 440 Hz. mathematically so then, how do we choose which one. since these are all periods that are integer multiples of the same root period (1/440 second), we normally choose the smallest such period that satisfies the conditions of periodicity. so then, what's the problem? seems like we have a well-defined rule: "choose the smallest value of $T$ that satisfies $x(t) \approx x(t+T)$." problem 1: satisfying $x(t) \approx x(t+T)$ is the same as satisfying $x(t) - x(t+T) \approx 0$. how do you determine that? because this difference can be either positive or negative, then, to be consistent we try to pick $T$ to minimize something like $$Q(T) \triangleq \int_{-\infty}^{+\infty} \|x(t)-x(t+T)\|^p w(t-t_0) \ dt$$ $w(t-t_0)$ is a window function centered at $t_0$ (the portion of the tone of current interest) and $p$ is some power. if $p=1$, then we have the Average Magnitude Difference Function (AMDF) and if $p=2$, we have the Average Squared Difference Function (ASDF), which i prefer for a number of nice mathematical reasons. so when we say "$x(t) \approx x(t+T)$", we are also saying "$Q(T)$ is small". how small? and how small is necessary to say that $T$ is a potential period? that is a "thresholding problem" (the bane of many a DSP algorithm because of the ungraceful failures that occur when the threshold is not met). now, the way we sometimes get around threshold problems is to use a reasonably loose and inclusive threshold and get lots of pitch candidates (of which only one is the candidate you will ultimately choose). two criteria to pick one candidate over another is (a) which candidate has the shortest period (so we pick $T$ over $2T$) and (b) which candidate has the lowest $Q(T)$? but these two criteria do not always agree. so then which candidate do you choose? is what IP, patents, and trade secrets are made of so i will go no further with that. that problem 2: this has to do with question (2) of the previous answer. Adamski, you almost got it right, but i would say that "but with a very slightly deeper tone" is off-the-mark. what if the 220 Hz added to the 440 Hz is 80 dB down (instead of 60 dB as i originally posed the question)? no one would hear any difference at all. , mathematically, it is a periodic function with period 1/220 second and not quite one of period 1/440 second. if you yet always choose the mathematically truest and bestest period, a little bit of inaudible sub-harmonic will screw you. so, somehow, you need to bias or prefer period candidates that are shorter over the longer periods that may fit better, mathematically. how to do that is also stuff that secret sauce is made of, so i am going no further with that. problem 3: jitter or jumping around in pitch selection. usually, when you hear an octave error, it isn't that the pitch of whole note was judged to be an octave off (high or low) but that, while most of the note had the correct pitch assigned, as the note evolves in time (with changing timbre), some small snippet of that note jumps up or down an octave and, if that drives a synthesizer, will sound quite annoying. so, somehow, you need to put in a little hysteresis and make the pitch you select at earlier times be a little "sticky" and preferred, so that when there is suddenly a single frame that concludes that another candidate an octave high or low is the pitch, you can stick with the pitch you already have chosen for earlier frames. a clean way of doing that is also stuff of which commercial solutions are made, so i won't tell you exactly how to do that. so, even though i didn't spell out a solution (i will if you pay me), i have hopefully pointed you in the right direction for you to figure it out with a little creative thinking.
Say I have time-series data that is unevenly spaced, with anything between 4-50 hours of spacing in between. The data comes from a trading account history, which has captured the balance of the portfolio after each trade. I'd like to calculate the annualised daily volatility of this account in order to compute a sensible Sharpe ratio. As such, I am assuming volatility does not vary with time. I've read How do you estimate the volatility of a sample when points are irregularly spaced? but do not have access to the knowledge, experience or computational power required to solve for the maximum in the formula proposed. What would be a reasonably good way to approximate the daily volatility of this time series? I can think of a few solutions and would be interested to hear if you have comments on them or if you had other ideas. Solution #1:Pretend my data is, in fact, regularly spaced Solution #2:Cut all but the last data point of each day Solution #3:As #2 but fill any gaps with some short EMA estimate (perhaps making use of Eckner, 2015) Solution #4:Use a method akin to that oulined in pp.38 of Eckner, 2014, (if I've understood it correctly) and approximating the vol to the ATR scaled by $ \sqrt{ \frac{ \pi }{ 8 \rho } } $
Here is a couple of examples of the similarity from Wikipedia, in which the expressions differ only in signs. I encountered other analogies as well. $${\begin{aligned}\gamma &=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1-xy)\ln xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right).\end{aligned}}$$ $${\begin{aligned}\ln {\frac {4}{\pi }}&=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1+xy)\ln xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left((-1)^{n-1}\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right)\right).\end{aligned}}$$ $${\begin{aligned}\gamma &=\sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}\\\ln {\frac {4}{\pi }}&=\sum _{n=1}^{\infty }{\frac {N_{1}(n)-N_{0}(n)}{2n(2n+1)}},\end{aligned}}$$ I wonder whether is there any algebraic system where $4e^{-\gamma}$ would play a role similar to what $\pi$ plays, say in complex numbers, or a geometric system where $4e^{-\gamma}$ would play some special role, like $\pi$ in Euclidean and Riemannian geometries.
I am new to the beamerclass and I am currently trying to write a presentation containing lots of derivations. On one point I want to overlay a derivation by its conclusion and I use the overprint-environment. The problem is that the environment shifts the whole derivation downwards so that the last line is only partially visible. How can I avoid this? There is a workaround by making a second frame, copying the whole content and replacing the derivation with the conclusion, but I do not consider that "elegant" and it would falsify the frame counter. I would appreciate any help. A minimal example is attached: \documentclass[xcolor=dvipsnames]{beamer}\mode<presentation>\usetheme{Boadilla}\usepackage[ngerman]{babel}\usepackage[utf8]{inputenc}\usepackage{amsfonts,amsmath,amssymb,amsthm}\DeclareMathOperator{\Var}{Var}\DeclareMathOperator{\E}{\mathbb E} \title{Dummy-Titel}\subtitle{Dummy-Untertitel}\author{Autor}\date{\today}\begin{document}\begin{frame}{Erwartungswert} \begin{itemize} \item $\E(X_{t+1} \| X_t = i) = \sum_{j=0}^{2N} j \cdot P(X_{t+1}=j \| X_t = i)$ \item $X_{t+1}$ binomialverteilt mit Parametern $n=2N$ und $p=\frac{i}{2N}$: $\Rightarrow \E(X_{t+1} \| X_t=i) = n \cdot p = i$ \item Martingal-Eigenschaft: $\E(X_{t+1} \| X_t) = X_t$ \end{itemize} \begin{overprint} \onslide<1> \begin{align} \E(X_t) &= \E(\E(X_{t+1} \| X_t)) = \sum_{i=0}^{2N} E(X_{t+1} \| X_t = i) \cdot P(X_t = i) \\ &= \sum_{i=0}^{2N} \left( \sum_{j=0}^{2N} j \cdot P(X_{t+1}=j \| X_t = i) \right) \cdot P(X_t = i) \\ &= \sum_{j=0}^{2N} j \cdot \left( \sum_{i=0}^{2N} \cdot P(X_{t+1}=j , X_t = i) \right) \\ &= \sum_{j=0}^{2N} j \cdot P(X_{t+1}=j) = \E(X_{t+1}) \end{align} \onslide<2> \begin{equation} \Rightarrow \E(X_0) = \E(X_1) = \E(X_2) = \ldots \end{equation} \end{overprint}\end{frame}\end{document}
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions. Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not. (a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular? (b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular? (c) Let $A$ be a $4\times 4$ matrix and let\[\mathbf{v}=\begin{bmatrix}1 \\2 \\3 \\4\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}4 \\3 \\2 \\1\end{bmatrix}.\]Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular? Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible. Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$. Prove that the matrix\[A=\begin{bmatrix}0 & 1\\-1& 0\end{bmatrix}\]is diagonalizable.Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix. The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes. This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems. Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$. (a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$. (b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$. Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular. Problem 9.Determine whether each of the following sentences is true or false. (a) There is a $3\times 3$ homogeneous system that has exactly three solutions. (b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric. (c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$. (d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent. An $n\times n$ matrix $A$ is called nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.Using the definition of a nonsingular matrix, prove the following statements. (a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular. (b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then: The matrix $B$ is nonsingular. The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.) Let $A$ be an $n\times (n-1)$ matrix and let $\mathbf{b}$ be an $(n-1)$-dimensional vector.Then the product $A\mathbf{b}$ is an $n$-dimensional vector.Set the $n\times n$ matrix $B=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$. Prove that $B$ is a singular matrix for any choice of $\mathbf{b}$. For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A\|I]$, where $I$ is the $3\times 3$ identity matrix.
implicitly. Next, we postulate the existence of a conserved property, named energy, as a function of the state variables $ E = E(\mathbf{C}(t)) $. Differentiating the energy yields \[ \frac{\diff E}{\diff t} = \sum_j \frac{\partial E}{\partial C_j} \frac{\diff C_j}{\diff t} , \] which provides an exact expression of the rate of change of the generic coordinate $ C_i $ \[ \frac{\diff C_i}{\diff t} = \left( \frac{\partial E}{\partial C_i} \right)^{-2} \frac{\diff E}{\diff t} \frac{\partial E}{\partial C_i} - \sum_{j\ne i} \frac{\diff C_j}{\diff t} \left( \frac{\partial E}{\partial C_i} \right)^{-1} \frac{\partial E}{\partial C_j} = \sum_j L_{ij} \frac{\partial E}{\partial C_j} . \] We can write the above equation in vector-matrix form \[ \frac{\diff \mathbf{C}}{\diff t} = \mathbf{L} \frac{\partial E}{\partial \mathbf{C}} = \mathbf{K} \mathbf{C} . \] This is a general equation for the deterministic evolution of any system whose state is given by a non-stochastic vector $ \mathbf{C} $. The scope of this equation of evolution is beyond mechanics because $ \mathbf{C} $ is not limited to the positions and velocities (or momenta) of particles. Note that even if we restrict the vector to $ \mathbf{C} = (\mathbf{p}, \mathbf{q}) $ the description is still more general than Hamiltonian mechanics because the equation (3) can deal with dissipative systems. Uncertainty and stability The above expressions are deterministic. To introduce fluctuations and uncertainty we seek a generalized equation \[ \frac{\diff \mathbf{C}}{\diff t} = \mathbf{K} \mathbf{C} + \mathbf{f} , \] with $ \mathbf{f} $ measuring the difference between the actual rate and the deterministic prediction
On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about Resolve, the replacement for Vengeance. Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec. In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts. Mathemagic Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$: $$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$ where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds. The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this: My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve. Once you’re in combat and taking damage, the second term makes a contribution: I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage: $$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$ The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve. The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns. The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve. However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation. You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth. Comparing to Vengeance Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading. Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss. Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was. Celestalon’s post also outlines a few other significant differences: No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not. However, Resolve ramps up a lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds. To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason. As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented). The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having any ramp-up mechanism. Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists. This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage: You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly. One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown. The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction: It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve. I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference: Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords. If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta. Is It Fixed Yet? If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years. But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around. It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve. While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices. All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve.
Straight Lines Various Forms of the Equation of a Line Equation of horizontal line passing through (x 1y 1) is y = y 1 Equation of vertical line passing through (x 1y 1) is x = x 1 The equation of the line whose slope is ‘m’ and which cuts an intercept ‘c’ on the y-axis is y = mx + c The equation of the line whose slope is ‘m’ and which cuts an intercept ‘a’ on the x - axis is y = m(x – a) The equation of a straight line having x – intercept ‘a’ and ‘y’ intercept ‘b’ is \tt \frac{x}{a}+\frac{y}{b}=1 (a, b ≠0) The equation of the line with slope ‘m’ and passing through the point (x 1y 1) is y – y 1= m (x 1y 1) The equation of a line passing through ‘2’ points (x 1y 1) and (x 2y 2) is (y – y 1) (x 2– x 1) = (y 2– y 1) (x – x 1) The equation of the straight line upon which the length of the normal drawn from origin is ‘P’ and this perpendicular makes an angle ‘∝’ , (0 ≤ ∝ < 2π) with positive x–axis is x cos ∝ + y sin ∝ = P (P > 0) View the Topic in this video From 21:22 To 50:00 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers. 1. Equation of the horizontal line having distance a from the x-axis is either y = a or y = −a. 2. Equation of the vertical line having distance b from the y-axis is either x = b or x = −b. 3. The point (x, y) lies on the line with slope m and through the fixed point (x 0, y 0), if and only if its coordinates satisfy the equation y − y 0 = m (x − x 0). 4. Equation of the line passing through the points (x 1, y 1) and (x 2, y 2) is given by \tt y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right). 5. The point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c. 6. If a line with slope m makes x-intercept d. Then equation of the line is y = m (x − d). 7. Equation of a line making intercepts a and b on the x-and y-axis, respectively, is \tt \frac{x}{a}+\frac{y}{b}=1. 8. The equation of the line having normal distance from origin p and angle between normal and the positive x-axis ω is given by x cos ω + y sin ω = p 9. The equation of the straight line passing through (x 1 y 1) and makes an angle ‘θ’ with the positive direction of x – axis is \tt \frac{x-x_{1}}{\cos \theta}+\frac{y-y_{1}}{\sin \theta}
I'm trying to understand the proof of the Barr-Diaconescu theorem about Boolean covers for Grothendieck sites. Precisely, the versions you can find in Jardine's book "Local Homotopy Theory" or in Mac Lane - Moerdijk "Sheaves in Geometry and Logic", which are essentially the same. That is, Theorem. For every Grothendieck topos $\mathcal{E}$ there exists a complete Boolean algebra $\mathcal{B}$ and a topos morphism $p: \mathbf{Sh}(\mathcal{B}) \longrightarrow \mathcal{E}$ such that the inverse image functor $p^* : \mathcal{E} \longrightarrow \mathbf{Sh}(\mathcal{B}) $ is faithful. Particularly, I'm having troubles with something that must be definitively elementary. In both books the first step goes like this: you take a Grothendieck site $\mathcal{X}$ such that $\mathcal{E} = \mathbf{Sh}(\mathcal{X})$ and you consider the poset of strings of arrows $\mathbf{String}({\mathcal{X}})$ of $\mathcal{X}$, where a string is a sequence of composable arrows in $\mathcal{X}$, $$ s = \left( U_n \stackrel{\alpha_n}{\longrightarrow} U_{n-1} \stackrel{\alpha_{n-1}}{\longrightarrow} \dots \stackrel{\alpha_1}{\longrightarrow} U_0 \right) $$ These strings carry a natural partial order, both books say (and here, perhaps, begins my problem): For two strings $s$ and $t$ we write $t \leq s$ if $t$ prolongs $s$ to the left. That is, for a string $s$ like the one above, $t$ must be of the form $$ t = \left( U_{n+m} \longrightarrow \dots \longrightarrow U_{n+1} \longrightarrow U_n \stackrel{\alpha_n}{\longrightarrow} \dots \longrightarrow U_0 \right) \ . $$ So far so good, but, if I understand everything correctly, now this poset $\mathbf{String}({\mathcal{X}})$ must also be a frame. Because: (1) Mac Lane-Moerdijk says so in his theorem 1; (2) also because the second step of the Diaconescu-Barr theorem (theorem 3.39 in Jardine's book, lemma 3 in Mac Lane-Moerdijk's one) says: Proposition. For every frame $\mathcal{S}$ there exists a complete Boolean algebra $\mathcal{B}$ and a topos morphism $i: \mathbf{Sh}(\mathcal{B}) \longrightarrow \mathbf{Sh}(\mathcal{S})$ such that the inverse image functor $i^*: \mathbf{Sh}(\mathcal{S}) \longrightarrow \mathbf{Sh}(\mathcal{B}) $ is faithful. Hence, you put $\mathcal{S} = \mathbf{String}({\mathcal{X}}) $ in this proposition and you are done: you have the Diaconescu-Barr theorem proved for you. Notice, however, that in the proof of this proposition you do indeed need the frame structure of $\mathcal{S}$. Ok, so my question is the following: Question. Which are the meets $\wedge$ and joints $\vee$ of the frame $\mathbf{String}({\mathcal{X}})$? Neither book says anything about them, so I imagine this must be an easy exercise -that I'm unfortunately unable to solve. Let me explain my lack of understanding to you. (1) As for joints, I would say that, given strings $$ s = \left( U_n \stackrel{\alpha_n}{\longrightarrow} U_{n-1} \stackrel{\alpha_{n-1}}{\longrightarrow} \dots \stackrel{\alpha_1}{\longrightarrow} U_0 \right) $$ and $$ t = \left( V_m \stackrel{\beta_m}{\longrightarrow} V_{m-1} \stackrel{\beta_{n-1}}{\longrightarrow} \dots \stackrel{\beta_1}{\longrightarrow} V_0 \right) $$ their meet could be something like the $0$-string $$ s \vee t = U_0 \sqcup V_0 \ , $$ where $\sqcup$ is the coproduct in the category $\mathcal{X}$: by definition, we have morphisms $U_0 \longrightarrow U_0 \sqcup V_0 \longleftarrow V_0$, so both $s$ and $t$ prolong $s\vee t$ to the left. Right. But: the existence of coproducts is not included in the definition of the Grothendieck site $\mathcal{X}$, is it? (2) As for meets, my lack of understanding is even worse: if $s \wedge t$ should prolong both $s$ and $t$ to the left, then I cannot imagine how every pair of strings $s$ and $t$ could have a meet. Because, if there is a string $s\wedge t$ that prolongs both $s$ and $t$ to the left, then shouldn't $s$ and $t$ be essentially the same string (except for having, perhaps, some arrows deleted in one of them with respect to the other)? I guess the answers to my questions are pretty obvious, but at this moment I'm stuck with them. What am I missing? Thank you for your help, even if you embarras me a lot.
Let $L/K$ be a finite abelian extension of local fields. Although, there is no generic form for the image of the norm map, $N^{L}_K$, in practice one can follow the following procedure to determine its image. Choose a uniformizer $\pi_L$ in $\mathcal{O}_L.$ Then $L^{\times}$ is equal to the group generated by $\pi_L$ and $\mathcal{O}_L^{\times}.$ It follows that $N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L) \rangle N^{L}_K \mathcal{O}_L^{\times},$ Hence to determine $N^{L}_K L^{\times}$ it is enough to establish the image of $\mathcal{O}_L^{\times}$ under the norm mapping. The group $N^L_K \mathcal{O}_L^{\times}$ is a subgroup of $\mathcal{O}_K^{\times}$ and by a group cohomology argument it can be shown that $[\mathcal{O}_K^{\times}:N^{L}_K\mathcal{O}_L^{\times}] = e(L\|K) = [L^{\times} : K^{\times}\mathcal{O}_L^{\times}].$ In particular, if $L/K$ is unramified, $\mathcal{O}_K^{\times} =N^{L}_K\mathcal{O}_L^{\times}$ and hence $N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L) \rangle \mathcal{O}_K^{\times}.$ The norm group of a tamely ramified extension is similarly easy to deduce. Write $$\mathcal{O}_L^{\times} = \langle \zeta_{q_L - 1} \rangle U_L$$ where $q_L$ is the residue field characteristic of $L$ and $\zeta_n$ denotes a primitive n-th root of unity. Denote the residue field of $L$ by $l$ and that of $K$ by $k,$ then $$N^L_K(\zeta_{q_L -1}) = N^l_k(\zeta_{q_L -1})^{e(L\|K)} = \zeta_{q_K -1}^{e(L\|K)}.$$ As $U_K$ is a pro-$p$ group and contains the image of $U_L$ under $N^L_K,$ we have in the tamely ramified case that $U_K = N^L_K(U_L)$ else $\mathcal{O}_K^{\times}/N^{L}_K\mathcal{O}_L^{\times}$ would contain an element of $p$-power order contradicting the equality $$[\mathcal{O}_K^{\times}:N^{L}_K\mathcal{O}_L^{\times}] = e(L\|K)$$ and the fact that $L/K$ was assumed tamely ramfied. It follows in the tamely ramified case that $$N^{L}_K L^{\times} = \langle N^{L}_K(\pi_L), \zeta_{q_K -1}^{e(L\|K)}\rangle U_K$$ The case of wild ramification is more difficult. But two facts are helpful. First, in the case $K$ is a p-adic field $U_K^{ap^n} \supset 1 +\mathcal{M}_K^{2ne(k\|\mathbb{Q}_p) + 1}$ where $a$ and $p$ are relatively prime. Hence, it is enough to determine the image of the norm mapping in the units of $\mathcal{O}_K \mod \mathcal{M}_K^{2ne(k\|\mathbb{Q}_p) + 1},$ a finite set. Another technique is to determine the higher ramification filtration of $Gal(L/K)^{ab}.$ In practice this can be done by examining the derivatives of the irreducible polynomial of $\pi_L.$ These groups map under the inverse of the artin map to the higher unit groups. As the domain of the inverse of the artin map is $K^{\times}/N^{L}_K L^{\times}$ their sizes reveal norm indexes within the higher unit groups. For a good exposition see Serre's book Local Fields.
A Markov chain with transition matrix \begin{align} P = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{align} has a continuum of stationary distributions $$ \gamma \begin{bmatrix} 0 \\ 1 \end{bmatrix} + (1 - \gamma) \begin{bmatrix} 1 \\ 0 \end{bmatrix}; \; \; \forall \gamma \in [0,1] $$ And invariant functions $$ \begin{bmatrix} 0 \\ \alpha_1 \end{bmatrix} \; and \; \begin{bmatrix} \alpha_2 \\ 0 \end{bmatrix}; \; \; \alpha_1, \alpha_2 \in \mathbb{R} $$ However the book states the process is not ergodic when $\gamma \in (0,1)$, as neither invariant function is constant across states that receive positive probability according to a stationary distribution associated with $\gamma \in (0,1)$. Why is this? If $\gamma = .5$ then the Markov process has equal chance of starting in state $1$ or state $2$, but will stay there forever, hence the random variable is constant and should be invariant.
I want to convert an rgb color triplet to a quaternion w + xi + yj + zk.I thought of it as of rotation, and (using axis-angle representation and Euclidian length of a unit quaternion equals 1) I came to following equation system:\begin{align} w &= \sqrt(1-(x^2+y^2+z^2))\\ w &= \cos(\arcsin(r/x))\\ w &= \cos(\arcsin(g/y))\\ w &= \cos(\arcsin(b/z)).\end{align*} Here $r, g, b$ are constants and i need to obtain $x, y, z, w$. I am a bit stumped. On the one hand there are 4 variables and 4 equations. On the other hand... how can I solve this? Could this even be solved?
I wish to solve an eigenvalue problem: $$\nabla^{2}f=Ef $$ If I assume spherical symmetry $f(r,\theta,\phi)=f(r)$, I can reduce the problem to 1D: $$(\frac{2}{r}\frac{d}{dr}+\frac{d^{2}}{dr^{2}})f=Ef$$ My boundary conditions are $f(r=0)=0$ and $f(r=L)=0$, where $L$ is the size of the discrete grid. I am attempting to solve this problem using a orthogonal basis set expansion, but the issue I am experiencing is the divergence of $r$ at $r=0$. Is there a better method for solving this type of problem? How can I accommodate the singularity?
@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$ Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine" @LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$ I enjoyed developing the lebesgue integral though. I thought that was cool But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons? uh wait a sec... ... I need arbitrary intersection to produce singletons from open intervals... hmm... 0 does not even have a nbhd, since any set containing 0 is closed I have no idea how to deal with points having empty nbhd o wait a sec... the open set of any topology must contain the whole set itself so I guess the nbhd of 0 is $\Bbb{R}$ Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…) Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time So going from a to b directly is no more efficient than go from a to 0 and then 0 to b hmm... $d(A \to B \to C) = d(A,B)+d(B,C) = \|a\|+\|b\|+\|b\|+\|c\|$ $d(A \to 0 \to C) = d(A,0)+d(0,C)=\|a\|+\|c\|$ so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination But if the starting point is nonzero, then the distance increases quadratically Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius) only when travelling via the origin, will such qudratic penalty in travelling distance be not apply More interesting things can be said about slight generalisations of this metric: Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ... To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ... 1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d... Any alg geom guys on? I know zilch about alg geom to even start analysing this question Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — baluApr 13 at 18:24 grr, thought I can get some more intuition in SR by using open balls tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two. The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$. And all it has in the proof is the assertion that the above is clearly true. I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k... Consider $a,b$ both irrational and the interval $[a,b]$ Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$ It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.One way for an irrational number $\alpha$ to be in this new set is b... Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals.Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat... (For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs) Let $\lambda^$ be the Lebesgue outer measure We previously proved that $\lambda^(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero. We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting, or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi… W also knew that one way to compute $\lambda^(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^(\Bbb{I}\cap [-1,1])$ is to use $\lambda^(\Bbb{Q}\cap [-1,1])+\lambda^(\Bbb{I}\cap [-1,1]) = \lambda^([-1,1])$ and thus deducing $\lambda^(\Bbb{I}\cap [-1,1]) = 2$ However, what I am interested here is to compute $\lambda^(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below: We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages: First denote an enumeration of the rationals as follows: $\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short: Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $\|p-q\| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works... Let's say that for $N$ clients, Lotta will take $d_N$ days to retire. For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$ Where $d_1$ = 1. Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps.
If $A$ is a Skew-Symmetric Matrix, then $I+A$ is Nonsingular and $(I-A)(I+A)^{-1}$ is Orthogonal Problem 468 Let $A$ be an $n\times n$ real skew-symmetric matrix. (a) Prove that the matrices $I-A$ and $I+A$ are nonsingular. (b) Prove that \[B=(I-A)(I+A)^{-1}\] is an orthogonal matrix. Contents Proof. (a) Prove that the matrices $I-A$ and $I+A$ are nonsingular. The eigenvalues of a skew-symmetric matrix are either $0$ or purely imaginary numbers. (See the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even” for a proof of this fact.) Namely, the eigenvalues of $A$ are of the form $ib$, where $i=\sqrt{-1}$ and $b$ is a real number. The eigenvalues of the matrix $I\pm A$ are of the form $1\pm \lambda$, where $\lambda$ is an eigenvalue of $A$. Since $\lambda=ib$, we have $1\pm \lambda \neq 0$. Thus, $I\pm A$ do not have $0$ as an eigenvalue. Since the determinant is the product of all eigenvalues, it follows that the determinants of the matrices $I\pm A$ are not zero, hence they are nonsingular. (b) Prove that $B=(I-A)(I+A)^{-1}$ is an orthogonal matrix. Note that by part (a), the matrix $I+A$ is nonsingular, hence it is invertible. Thus the expression $B=(I-A)(I+A)^{-1}$ is well-defined. Our goal is to show that $B^{\trans}B=I$. Recall the following basic properties of transpose and inverse matrices. $(AB)^{\trans}=B^{\trans} A^{\trans}$ $(A^{-1})^{\trans}=(A^{\trans})^{-1}$ if $A$ is invertible. $(A+B)^{\trans}=A^{\trans}+B^{\trans}$. We have \begin{align} B^{\trans}&=\left(\, (I-A)(I+A)^{-1} \,\right)^{\trans}\\ &=\left(\, (I+A)^{-1} \,\right)^{\trans}(I-A)^{\trans} && \text{by property 1}\\ &=\left(\, (I+A)^{\trans} \,\right)^{-1}(I-A)^{\trans} && \text{by property 2}\\ &=(I^{\trans}+A^{\trans})^{-1}(I^{\trans}-A^{\trans}) && \text{by property 3}\\ &=(I+A^{\trans})^{-1}(I-A^{\trans}) && \text{since } I^{\trans}=I\\ &=(I-A)^{-1}(I+A), \end{align} where the last step follows since $A$ is skew-symmetric: $A^{\trans}=-A$. Hence we have \begin{align} B^{\trans} B&=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}. \end{align} We note that the middle two matrices $I+A$ and $I-A$ commutes. In fact we have \begin{align} (I+A)(I-A)&=(I+A)I-(I+A)A=I+A-A-A^2=I-A^2 \text{ and }\\ (I-A)(I+A)&=(I-A)I+(I-A)A=I-A+A-A^2=I-A^2. \end{align} It yields that \[(I+A)(I-A)=(I-A)(I+A) \tag{}.\] Hence we have \begin{align} &B^{\trans} B\\ &=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}\\ &=(I-A)^{-1}(I-A)(I+A)(I+A)^{-1} && \text{ by ()}\\ &=I\cdot I=I, \end{align} and we have obtained $B^{\trans}B=I$. Therefore, the matrix $B$ is an orthogonal matrix. Add to solve later
Examples of parents endowed with multiple realizations¶ class sage.categories.examples.with_realizations. SubsetAlgebra( R, S)¶ An example of parent endowed with several realizations We consider an algebra $A(S)$ whose bases are indexed by the subsets $s$ of a given set $S$. We consider three natural basis of this algebra: F, In, and Out. In the first basis, the product is given by the union of the indexing sets. That is, for any $s, t\subset S$\[F_s F_t = F_{s\cup t}\] The Inbasis and Outbasis are defined respectively by:\[In_s = \sum_{t\subset s} F_t \qquad\text{and}\qquad F_s = \sum_{t\supset s} Out_t\] Each such basis gives a realization of $A$, where the elements are represented by their expansion in this basis. This parent, and its code, demonstrate how to implement this algebra and its three realizations, with coercions and mixed arithmetic between them. EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: A.base_ring() Rational Field The three bases of A: sage: F = A.F() ; F The subset algebra of {1, 2, 3} over Rational Field in the Fundamental basis sage: In = A.In() ; In The subset algebra of {1, 2, 3} over Rational Field in the In basis sage: Out = A.Out(); Out The subset algebra of {1, 2, 3} over Rational Field in the Out basis One can quickly define all the bases using the following shortcut: sage: A.inject_shorthands() Defining F as shorthand for The subset algebra of {1, 2, 3} over Rational Field in the Fundamental basis Defining In as shorthand for The subset algebra of {1, 2, 3} over Rational Field in the In basis Defining Out as shorthand for The subset algebra of {1, 2, 3} over Rational Field in the Out basis Accessing the basis elements is done with basis()method: sage: F.basis().list() [F[{}], F[{1}], F[{2}], F[{3}], F[{1, 2}], F[{1, 3}], F[{2, 3}], F[{1, 2, 3}]] To access a particular basis element, you can use the from_set()method: sage: F.from_set(2,3) F[{2, 3}] sage: In.from_set(1,3) In[{1, 3}] or as a convenient shorthand, one can use the following notation: sage: F[2,3] F[{2, 3}] sage: In[1,3] In[{1, 3}] Some conversions: sage: F(In[2,3]) F[{}] + F[{2}] + F[{3}] + F[{2, 3}] sage: In(F[2,3]) In[{}] - In[{2}] - In[{3}] + In[{2, 3}] sage: Out(F[3]) Out[{3}] + Out[{1, 3}] + Out[{2, 3}] + Out[{1, 2, 3}] sage: F(Out[3]) F[{3}] - F[{1, 3}] - F[{2, 3}] + F[{1, 2, 3}] sage: Out(In[2,3]) Out[{}] + Out[{1}] + 2Out[{2}] + 2Out[{3}] + 2Out[{1, 2}] + 2Out[{1, 3}] + 4Out[{2, 3}] + 4Out[{1, 2, 3}] We can now mix expressions: sage: (1 + Out[1]) * In[2,3] Out[{}] + 2Out[{1}] + 2Out[{2}] + 2Out[{3}] + 2Out[{1, 2}] + 2Out[{1, 3}] + 4Out[{2, 3}] + 4Out[{1, 2, 3}] class Bases( parent_with_realization)¶ The category of the realizations of the subset algebra class ParentMethods¶ from_set( args)¶ Construct the monomial indexed by the set containing the elements passed as arguments. EXAMPLES: sage: In = Sets().WithRealizations().example().In(); In The subset algebra of {1, 2, 3} over Rational Field in the In basis sage: In.from_set(2,3) In[{2, 3}] As a shorthand, one can construct elements using the following notation: sage: In[2,3] In[{2, 3}] one()¶ Returns the unit of this algebra. This default implementation takes the unit in the fundamental basis, and coerces it in self. EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: In = A.In(); Out = A.Out() sage: In.one() In[{}] sage: Out.one() Out[{}] + Out[{1}] + Out[{2}] + Out[{3}] + Out[{1, 2}] + Out[{1, 3}] + Out[{2, 3}] + Out[{1, 2, 3}] super_categories()¶ EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: C = A.Bases(); C Category of bases of The subset algebra of {1, 2, 3} over Rational Field sage: C.super_categories() [Category of realizations of The subset algebra of {1, 2, 3} over Rational Field, Join of Category of algebras with basis over Rational Field and Category of commutative algebras over Rational Field and Category of realizations of unital magmas] class Fundamental( A)¶ The Subset algebra, in the fundamental basis INPUT: A– a parent with realization in SubsetAlgebra EXAMPLES: sage: A = Sets().WithRealizations().example() sage: A.F() The subset algebra of {1, 2, 3} over Rational Field in the Fundamental basis sage: A.Fundamental() The subset algebra of {1, 2, 3} over Rational Field in the Fundamental basis one()¶ Return the multiplicative unit element. EXAMPLES: sage: A = AlgebrasWithBasis(QQ).example() sage: A.one_basis() word: sage: A.one() B[word: ] one_basis()¶ Returns the index of the basis element which is equal to ‘1’. EXAMPLES: sage: F = Sets().WithRealizations().example().F(); F The subset algebra of {1, 2, 3} over Rational Field in the Fundamental basis sage: F.one_basis() {} sage: F.one() F[{}] product_on_basis( left, right)¶ Product of basis elements, as per AlgebrasWithBasis.ParentMethods.product_on_basis(). INPUT: left, right– sets indexing basis elements EXAMPLES: sage: F = Sets().WithRealizations().example().F(); F The subset algebra of {1, 2, 3} over Rational Field in the Fundamental basis sage: S = F.basis().keys(); S Subsets of {1, 2, 3} sage: F.product_on_basis(S([]), S([])) F[{}] sage: F.product_on_basis(S({1}), S({3})) F[{1, 3}] sage: F.product_on_basis(S({1,2}), S({2,3})) F[{1, 2, 3}] class In( A)¶ The Subset Algebra, in the Inbasis INPUT: A– a parent with realization in SubsetAlgebra EXAMPLES: sage: A = Sets().WithRealizations().example() sage: A.In() The subset algebra of {1, 2, 3} over Rational Field in the In basis class Out( A)¶ The Subset Algebra, in the $Out$ basis INPUT: A– a parent with realization in SubsetAlgebra EXAMPLES: sage: A = Sets().WithRealizations().example() sage: A.Out() The subset algebra of {1, 2, 3} over Rational Field in the Out basis a_realization()¶ Returns the default realization of self EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: A.a_realization() The subset algebra of {1, 2, 3} over Rational Field in the Fundamental basis base_set()¶ EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: A.base_set() {1, 2, 3} indices()¶ The objects that index the basis elements of this algebra. EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: A.indices() Subsets of {1, 2, 3} indices_key( x)¶ A key function on a set which gives a linear extension of the inclusion order. INPUT: x– set EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: sorted(A.indices(), key=A.indices_key) [{}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}] supsets( set)¶ Returns all the subsets of $S$ containing set INPUT: set– a subset of the base set $S$ of self EXAMPLES: sage: A = Sets().WithRealizations().example(); A The subset algebra of {1, 2, 3} over Rational Field sage: A.supsets(Set((2,))) [{1, 2, 3}, {2, 3}, {1, 2}, {2}]
On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about Resolve, the replacement for Vengeance. Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec. In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts. Mathemagic Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$: $$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$ where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds. The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this: My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve. Once you’re in combat and taking damage, the second term makes a contribution: I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage: $$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$ The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve. The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns. The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve. However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation. You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth. Comparing to Vengeance Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading. Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss. Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was. Celestalon’s post also outlines a few other significant differences: No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not. However, Resolve ramps up a lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds. To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason. As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented). The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having any ramp-up mechanism. Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists. This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage: You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly. One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown. The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction: It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve. I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference: Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords. If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta. Is It Fixed Yet? If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years. But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around. It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve. While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices. All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve.
For the reaction,$$\ce{M -> M+ + e-}$$the heat liberated is highest for lithium owing to its high negative $E^\circ$ value so one would think that the reaction must be most vigorous.The reason behind the more violent reactivity of potassium rather than lithium lies in kinetics and not in thermodynamics.No doubt, maximum energy is evolved with ... Can I predict the products of any chemical reaction?In theory, yes!Every substance has characteristic reactivity behavior. Likewise pairs and sets of substances have characteristic behavior. For example, the following combinations of substances only have one likely outcome each:$$\ce{HCl + NaOH -> NaCl + H2O} \\[2ex]\ce{CH3CH2CH2OH->[$1.$ (COCl)... Indeed, $\ce{Zn}$ is lower than $\ce{Na}$ in activity series of metals, so the following reaction won't take place$$\require{cancel}\ce{Zn + 2NaOH \cancel{→} Zn(OH)2 + 2Na}$$Remember, however, that $\ce{Zn}$ is amphoteric, so it can reacts with a strong base such as $\ce{NaOH}$ as an acid forming sodium zincate$$\ce{Zn + 2 H2O + 2 NaOH -> Na2Zn(OH)4 ... Interesting question. The documentary was almost certainly referring to the extremely hot lava generated by the nuclear meltdown at Chernobyl coming into contact with water. This lava is made of a substance colloquially known as 'corium' and is a combination of molten fuel rods, moderator, reactor walls and whatever else is melted by the incredible ... Like aniline, phenol too reacts to a very less extent during Friedel-Crafts reaction.The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid.In fact most substituents with lone pair would give poor yield.The two pathways involved in the reaction with phenol reduce the overall yeild:Phenols ... First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answerThe reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).A nice summary of some work by Jens K. ... In fact, both reagents you noted here have quite complex structure and are not nucleophiles at all: both are electrophiles, because metallic atom here has too little neighbors to draw electrons from.Let me explain, using Grignard reagent $\ce{EtMgCl}$ . In reality it has complex structure with $\ce{Mg}$ atom coordinated to alkyl fragment and two diethyl ... Absolutely yes.Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if ... Here is my "old school" explanation.Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left. As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well ... Relativistic effects account for gold's lack of reactivity. Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic ... Though the monomers in cyanoacrylate glues contain an ester, their polymerization doesn't rely on that ester group directly. The Wikipedia article for cyanoacrylates shows the polymerization more clearly than I can easily explain in words. The many hydroxyl groups in cellulose do start polymerization effectively, and the large surface area of cotton wool ... The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester. Look at the resonance structures drawn below. Both the ketone and ester have a resonance ... Normal dioxygen ($\ce{O2}$) exists as a ground state, triplet biradical. This is an example of a molecule that, thermodynamically should be quite reactive, yet is kinetically unreactive - once again, a case of kinetics vs. thermodynamics.Because spin must be conserved in a reaction, if ground state, triplet $\ce{O2}$ were to react with something, it ... You might want to look up some terms, such asnoble and less noble metalsreduction potentialgalvanic seriesHere, reaction means thathydrogen gas is formedthe metal is dissolvedIn order to form hydrogen, protons need to be reduced to hydrogen atoms which then combine to $\ce{H2}$.$$\ce{2 H+ + 2 e- -> H2}$$The metal serves as an electron ... First, let me point out that a rate difference of 500 is really not that large. There are solvolysis reactions with relative rate differences on the order of $\mathrm{10^{10}}$ or greater (1).The transition state for an $\mathrm{S_{N}2}$ reaction involves hypercoordinate (or hypervalent) bonding. The transition state is, more or less, a trigonal ... Looking at the partial charges of esters and ketones, unfortunately ron's answer is only half true. For a simple model I have chosen 3-pentanone and ethyl acetate. You can see, that the carbonyl carbon in the ketone has a smaller positive charge ($q=0.6$) than in the ester ($q=0.8$), so the latter should be more prone to nucleophilic attacks.This is not ... For this approach I am basically employing Frontier Molecular Orbital Theory (FMO) to predict the reactivity of carbonyl compounds towards nucleophiles.For the purpose of this explanation I have chosen water as nucleophile. In principle we are looking at the addition of an electron rich particle to an electron poor system. In this case, water will attack ... Another possible explanation: water can act as a neutron reflector, at least under some conditions. Therefore, it might be possible for a subcritical mass of uranium to become supercritical if immersed in water, as some of the emitted neutrons could be reflected back into the uranium. This could lead to a criticality accident, with large amount of ionizing ... Very interesting question… I would give two answers, at different levels:At undergrad level, I think the carbenium ion formed at the bridge head is less stabilized that “regular” ternary carbenium ions, because it cannot adopt the planar trigonal geometry which would be ideal for it.At higher level, this is much more complicated: the norboryl cation is an ... As you can see:Sulfur (16) is significantly smaller than Se (34). That size differencemeans that there is more room between the fluorine atoms on Se than onSulfur. That is what steric hindrance means. Think of steric hindranceas limited access. The smaller the atom the less access there is.Oxygen is so small that less than 6 fluorine atoms can ... You have to think about the whole process. When a metal loses electrons to make a metal ion the following happens:The metallic bonds holding the metal atoms together are broken.The metal atom loses the electrons.The resulting metal ion is hydrated.In your analysis you are only focusing on step 2. The enthalpy and entropy of the entire process factor ... To add to @user223679's answer. Phenol can react via two pathways with acyl chlorides to give either esters, via O-acylation, or hydroxyarylketones, via C-acylation.However, phenol esters also undergo a Fries rearrangement under Friedel-Crafts conditions to produce the C-alkylated, hydroxyarylketones. This reaction is promoted by having an excess of ... Dioxygen, $\ce{O2}$ is a very special molecule. A good majority of organisms on earth use dioxygen to breathe and survive. Oxygen is also the second most abundant element in the sky, as well as the most abundant elements in the earth's crust. What makes it so special? Its because of its abundance, which is obvious, and also because of it's strong oxidizing ... In principle, non-oxidizing acids cannot directly oxidize copper since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ cannot oxidize $\ce{Cu}$ to $\ce{Cu^2+}$ or to $\ce{Cu+}$:$$\begin{alignat}{2}\ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V}\\\ce{Cu^2+ + 2e- \;&<=> Cu}\quad &&E^\circ =... $\Delta G = \Delta H - T \Delta S$In the case of the $\ce{N2 + O2 -> 2NO}$ , $\Delta H$ and $\Delta S$ are both positive, so the reaction is thermodynamically favorable at high temperature (such as in lightning) but not at low temperature.If the temperature drops to room temperature after NO is formed, it is thermodynamically favorable for NO to ... The difference in electronegativity between copper (1.9) and magnesium (1.3) is the key difference.Since copper is more electronegative than magnesium its electronegativity is much closer to that of carbon (2.55). This results in carbon-copper bonds being less polarized and more covalent than carbon-magnesium bonds. The electrons in a carbon-copper ... You should clarify whether the iodide anion is a good nucleophile in a polar protic or polar aprotic solvent.Also let's note that nucleophilicity is a kinetic property, while acidity/basicity are thermodynamic properties. Iodide ion's lack of basicity in water reflects its conjugate acid's ($\ce{HI}$) lack of stability; it's relatively easy to ionize its ... There are a number of factors that can influence the rate of an $\mathrm{S_{N}2}$ reaction. Solvent, leaving group stability, attacking group nucleophilicity, steric factors and electronic factors.In the series of compounds you've presented, all of these parameters are held constant except for the steric and electronic factors. Considering only steric ... It has to do with the stability of the electrons that are ejected from the bond that breaks. In B-ketoacids, the electrons can resonate onto the oxygen and form an enol/enolate intermediate, which is much more stable than forming a carbanion on a gamma or delta ketoacid (because there is nowhere for the electrons to go other than go on the carbon atom that ... I would not expect boric acid to cause ignition of nitrocellulose. However, boric acid may react with un-nitrated hydroxyl groups on the cellulose backbone--likely a benign reaction. You might need more than a few drops of boric acid solution to see a green color - perhaps immersion in boric acid solution will be needed. Copper salts might give a good green ...
Colored Permutations¶ Todo Much of the colored permutations (and element) class can be generalized to $G \wr S_n$ class sage.combinat.colored_permutations. ColoredPermutation( parent, colors, perm)¶ A colored permutation. colors()¶ Return the colors of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: s1,s2,t = C.gens() sage: x = s1s2t sage: x.colors() [1, 0, 0] has_left_descent( i)¶ Return Trueif iis a left descent of self. Let $p = ((s_1, \ldots s_n), \sigma)$ be a colored permutation. We say $p$ has a left $n$-descent if $s_n > 0$. If $i < n$, then we say $p$ has a left $i$-descent if either $s_i \neq 0, s_{i+1} = 0$ and $\sigma_i < \sigma_{i+1}$ or $s_i = s_{i+1}$ and $\sigma_i > \sigma_{i+1}$. This notion of a left $i$-descent is done in order to recursively construct $w(p) = \sigma_i w(\sigma_i^{-1} p)$, where $w(p)$ denotes a reduced word of $p$. EXAMPLES: sage: C = ColoredPermutations(2, 4) sage: s1,s2,s3,s4 = C.gens() sage: x = s4s1s2s3s4 sage: [x.has_left_descent(i) for i in C.index_set()] [True, False, False, True] sage: C = ColoredPermutations(1, 5) sage: s1,s2,s3,s4 = C.gens() sage: x = s4s1s2s3s4 sage: [x.has_left_descent(i) for i in C.index_set()] [True, False, False, True] sage: C = ColoredPermutations(3, 3) sage: x = C([[2,1,0],[3,1,2]]) sage: [x.has_left_descent(i) for i in C.index_set()] [False, True, False] sage: C = ColoredPermutations(4, 4) sage: x = C([[2,1,0,1],[3,2,4,1]]) sage: [x.has_left_descent(i) for i in C.index_set()] [False, True, False, True] inverse()¶ Return the inverse of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: s1,s2,t = C.gens() sage: ~t [[0, 0, 3], [1, 2, 3]] sage: all(x * ~x == C.one() for x in C.gens()) True length()¶ Return the length of selfin generating reflections. This is the minimal numbers of generating reflections needed to obtain self. EXAMPLES: sage: C = ColoredPermutations(3, 3) sage: x = C([[2,1,0],[3,1,2]]) sage: x.length() 7 sage: C = ColoredPermutations(4, 4) sage: x = C([[2,1,0,1],[3,2,4,1]]) sage: x.length() 12 one_line_form()¶ Return the one line form of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: s1,s2,t = C.gens() sage: x = s1s2t sage: x [[1, 0, 0], [3, 1, 2]] sage: x.one_line_form() [(1, 3), (0, 1), (0, 2)] permutation()¶ Return the permutation of self. This is obtained by forgetting the colors. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: s1,s2,t = C.gens() sage: x = s1s2t sage: x.permutation() [3, 1, 2] reduced_word()¶ Return a word in the simple reflections to obtain self. EXAMPLES: sage: C = ColoredPermutations(3, 3) sage: x = C([[2,1,0],[3,1,2]]) sage: x.reduced_word() [2, 1, 3, 2, 1, 3, 3] sage: C = ColoredPermutations(4, 4) sage: x = C([[2,1,0,1],[3,2,4,1]]) sage: x.reduced_word() [2, 1, 4, 3, 2, 1, 4, 3, 2, 4, 4, 3] to_matrix()¶ Return a matrix of self. The colors are mapped to roots of unity. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: s1,s2,t = C.gens() sage: x = s1s2ts2; x.one_line_form() [(1, 2), (0, 1), (0, 3)] sage: M = x.to_matrix(); M [ 0 1 0] [zeta4 0 0] [ 0 0 1] The matrix multiplication is in the oppositeorder: sage: M == s2.to_matrix()t.to_matrix()s2.to_matrix()s1.to_matrix() True class sage.combinat.colored_permutations. ColoredPermutations( m, n)¶ The group of $m$-colored permutations on $\{1, 2, \ldots, n\}$. Let $S_n$ be the symmetric group on $n$ letters and $C_m$ be the cyclic group of order $m$. The $m$-colored permutation group on $n$ letters is given by $P_n^m = C_m \wr S_n$. This is also the complex reflection group $G(m, 1, n)$. We define our multiplication by\[((s_1, \ldots s_n), \sigma) \cdot ((t_1, \ldots, t_n), \tau) = ((s_1 t_{\sigma(1)}, \ldots, s_n t_{\sigma(n)}), \tau \sigma).\] EXAMPLES: sage: C = ColoredPermutations(4, 3); C 4-colored permutations of size 3 sage: s1,s2,t = C.gens() sage: (s1, s2, t) ([[0, 0, 0], [2, 1, 3]], [[0, 0, 0], [1, 3, 2]], [[0, 0, 1], [1, 2, 3]]) sage: s1s2 [[0, 0, 0], [3, 1, 2]] sage: s1s2s1 == s2s1s2 True sage: t^4 == C.one() True sage: s2ts2 [[0, 1, 0], [1, 2, 3]] We can also create a colored permutation by passing either a list of tuples consisting of (color, element): sage: x = C([(2,1), (3,3), (3,2)]); x [[2, 3, 3], [1, 3, 2]] or a list of colors and a permutation: sage: C([[3,3,1], [1,3,2]]) [[3, 3, 1], [1, 3, 2]] There is also the natural lift from permutations: sage: P = Permutations(3) sage: C(P.an_element()) [[0, 0, 0], [3, 1, 2]] REFERENCES: as_permutation_group()¶ Return the permutation group corresponding to self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.as_permutation_group() Complex reflection group G(4, 1, 3) as a permutation group cardinality()¶ Return the cardinality of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.cardinality() 384 sage: C.cardinality() == 43 factorial(3) True codegrees()¶ Return the codegrees of self. Let $G$ be a complex reflection group. The codegrees $d_1^* \leq d_2^* \leq \cdots \leq d_{\ell}^$ of $G$ can be defined by:\[\prod_{i=1}^{\ell} (q - d_i^ - 1) = \sum_{g \in G} \det(g) q^{\dim(V^g)},\] where $V$ is the natural complex vector space that $G$ acts on and $\ell$ is the rank(). If $m = 1$, then we are in the special case of the symmetric group and the codegrees are $(n-2, n-3, \ldots 1, 0)$. Otherwise the degrees are $((n-1)m, (n-2)m, \ldots, m, 0)$. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.codegrees() (8, 4, 0) sage: S = ColoredPermutations(1, 3) sage: S.codegrees() (1, 0) coxeter_matrix()¶ Return the Coxeter matrix of self. EXAMPLES: sage: C = ColoredPermutations(3, 4) sage: C.coxeter_matrix() [1 3 2 2] [3 1 3 2] [2 3 1 4] [2 2 4 1] sage: C = ColoredPermutations(1, 4) sage: C.coxeter_matrix() [1 3 2] [3 1 3] [2 3 1] degrees()¶ Return the degrees of self. The degrees of a complex reflection group are the degrees of the fundamental invariants of the ring of polynomial invariants. If $m = 1$, then we are in the special case of the symmetric group and the degrees are $(2, 3, \ldots, n, n+1)$. Otherwise the degrees are $(m, 2m, \ldots, nm)$. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.degrees() (4, 8, 12) sage: S = ColoredPermutations(1, 3) sage: S.degrees() (2, 3) We now check that the product of the degrees is equal to the cardinality of self: sage: prod(C.degrees()) == C.cardinality() True sage: prod(S.degrees()) == S.cardinality() True fixed_point_polynomial( q=None)¶ The fixed point polynomial of self. The fixed point polynomial $f_G$ of a complex reflection group $G$ is counting the dimensions of fixed points subspaces:\[f_G(q) = \sum_{w \in W} q^{\dim V^w}.\] Furthermore, let $d_1, d_2, \ldots, d_{\ell}$ be the degrees of $G$, where $\ell$ is the rank(). Then the fixed point polynomial is given by\[f_G(q) = \prod_{i=1}^{\ell} (q + d_i - 1).\] INPUT: q– (default: the generator of ZZ['q']) the parameter $q$ EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.fixed_point_polynomial() q^3 + 21q^2 + 131q + 231 sage: S = ColoredPermutations(1, 3) sage: S.fixed_point_polynomial() q^2 + 3q + 2 gens()¶ Return the generators of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.gens() ([[0, 0, 0], [2, 1, 3]], [[0, 0, 0], [1, 3, 2]], [[0, 0, 1], [1, 2, 3]]) sage: S = SignedPermutations(4) sage: S.gens() ([2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3], [1, 2, 3, -4]) index_set()¶ Return the index set of self. EXAMPLES: sage: C = ColoredPermutations(3, 4) sage: C.index_set() (1, 2, 3, 4) sage: C = ColoredPermutations(1, 4) sage: C.index_set() (1, 2, 3) is_well_generated()¶ Return if selfis a well-generated complex reflection group. A complex reflection group $G$ is well-generated if it is generated by $\ell$ reflections. Equivalently, $G$ is well-generated if $d_i + d_i^ = d_{\ell}$ for all $1 \leq i \leq \ell$. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.is_well_generated() True sage: C = ColoredPermutations(2, 8) sage: C.is_well_generated() True sage: C = ColoredPermutations(1, 4) sage: C.is_well_generated() True matrix_group()¶ Return the matrix group corresponding to self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.matrix_group() Matrix group over Cyclotomic Field of order 4 and degree 2 with 3 generators ( [0 1 0] [1 0 0] [ 1 0 0] [1 0 0] [0 0 1] [ 0 1 0] [0 0 1], [0 1 0], [ 0 0 zeta4] ) number_of_reflection_hyperplanes()¶ Return the number of reflection hyperplanes of self. The number of reflection hyperplanes of a complex reflection group is equal to the sum of the codegrees plus the rank. EXAMPLES: sage: C = ColoredPermutations(1, 2) sage: C.number_of_reflection_hyperplanes() 1 sage: C = ColoredPermutations(1, 3) sage: C.number_of_reflection_hyperplanes() 3 sage: C = ColoredPermutations(4, 12) sage: C.number_of_reflection_hyperplanes() 276 one()¶ Return the identity element of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.one() [[0, 0, 0], [1, 2, 3]] order()¶ Return the cardinality of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.cardinality() 384 sage: C.cardinality() == 4*3 factorial(3) True rank()¶ Return the rank of self. The rank of a complex reflection group is equal to the dimension of the complex vector space the group acts on. EXAMPLES: sage: C = ColoredPermutations(4, 12) sage: C.rank() 12 sage: C = ColoredPermutations(7, 4) sage: C.rank() 4 sage: C = ColoredPermutations(1, 4) sage: C.rank() 3 simple_reflection( i)¶ Return the i-th simple reflection of self. EXAMPLES: sage: C = ColoredPermutations(4, 3) sage: C.gens() ([[0, 0, 0], [2, 1, 3]], [[0, 0, 0], [1, 3, 2]], [[0, 0, 1], [1, 2, 3]]) sage: C.simple_reflection(2) [[0, 0, 0], [1, 3, 2]] sage: C.simple_reflection(3) [[0, 0, 1], [1, 2, 3]] sage: S = SignedPermutations(4) sage: S.simple_reflection(1) [2, 1, 3, 4] sage: S.simple_reflection(4) [1, 2, 3, -4] class sage.combinat.colored_permutations. SignedPermutation( parent, colors, perm)¶ A signed permutation. has_left_descent( i)¶ Return Trueif iis a left descent of self. EXAMPLES: sage: S = SignedPermutations(4) sage: s1,s2,s3,s4 = S.gens() sage: x = s4s1s2s3s4 sage: [x.has_left_descent(i) for i in S.index_set()] [True, False, False, True] inverse()¶ Return the inverse of self. EXAMPLES: sage: S = SignedPermutations(4) sage: s1,s2,s3,s4 = S.gens() sage: x = s4s1s2s3s4 sage: ~x [2, 3, -4, -1] sage: x * ~x == S.one() True to_matrix()¶ Return a matrix of self. EXAMPLES: sage: S = SignedPermutations(4) sage: s1,s2,s3,s4 = S.gens() sage: x = s4s1s2s3s4 sage: M = x.to_matrix(); M [ 0 1 0 0] [ 0 0 1 0] [ 0 0 0 -1] [-1 0 0 0] The matrix multiplication is in the oppositeorder: sage: m1,m2,m3,m4 = [g.to_matrix() for g in S.gens()] sage: M == m4 * m3 * m2 * m1 * m4 True class sage.combinat.colored_permutations. SignedPermutations( n)¶ Group of signed permutations. The group of signed permutations is also known as the hyperoctahedral group, the Coxeter group of type $B_n$, and the 2-colored permutation group. Thus it can be constructed as the wreath product $S_2 \wr S_n$. EXAMPLES: sage: S = SignedPermutations(4) sage: s1,s2,s3,s4 = S.group_generators() sage: x = s4s1s2s3s4; x [-4, 1, 2, -3] sage: x^4 == S.one() True This is a finite Coxeter group of type $B_n$: sage: S.canonical_representation() Finite Coxeter group over Number Field in a with defining polynomial x^2 - 2 with a = 1.414213562373095? with Coxeter matrix: [1 3 2 2] [3 1 3 2] [2 3 1 4] [2 2 4 1] sage: S.long_element() [-1, -2, -3, -4] sage: S.long_element().reduced_word() [1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 4, 3, 2, 4, 3, 4] We can also go between the 2-colored permutation group: sage: C = ColoredPermutations(2, 3) sage: S = SignedPermutations(3) sage: S.an_element() [-3, 1, 2] sage: C(S.an_element()) [[1, 0, 0], [3, 1, 2]] sage: S(C(S.an_element())) == S.an_element() True sage: S(C.an_element()) [-3, 1, 2] There is also the natural lift from permutations: sage: P = Permutations(3) sage: x = S(P.an_element()); x [3, 1, 2] sage: x.parent() Signed permutations of 3 REFERENCES: long_element( index_set=None)¶ Return the longest element of self, or of the parabolic subgroup corresponding to the given index_set. INPUT: index_set– (optional) a subset (as a list or iterable) of the nodes of the indexing set EXAMPLES: sage: S = SignedPermutations(4) sage: S.long_element() [-1, -2, -3, -4] one()¶ Return the identity element of self. EXAMPLES: sage: S = SignedPermutations(4) sage: S.one() [1, 2, 3, 4] simple_reflection( i)¶ Return the i-th simple reflection of self. EXAMPLES: sage: S = SignedPermutations(4) sage: S.simple_reflection(1) [2, 1, 3, 4] sage: S.simple_reflection(4) [1, 2, 3, -4]
In this paper, the authors make a simple model with: (1) A global bank, who is risk-neutral but has a Value-at-Risk constraint: $$\max_{x_t^B} E_t[x_t^B\prime R_{t+1}]$$ s.t. $$\alpha (Var(x_t^B\prime R_{t+1}))^{\frac{1}{2}} <= 1$$ where $R_{t+1}$ is a (n x 1) vector of returns, $x_t^B$ is a (n x 1) vector of weights $\alpha$ is a parameter, and $Var$ is the variance operator. I tried setting up a lagrangean which should yield: $$ \mathcal{L} = E_t[x_t^B\prime R_{t+1}] - \lambda_t (1 - \alpha (Var(x_t^B\prime R_{t+1}))^{\frac{1}{2}}) $$ The first order conditions w.r.t. $x_t^B$ are yielding me: $$ E_t(R_{t+1}) - \lambda_t \alpha Var(x_t^B\prime R_{t+1})^{-\frac{1}{2}} Var(R_{t+1}) x_t^B = 0 $$ In comparison with the solution given on the paper it seems that I have an extra term $ Var(x_t^B\prime R_{t+1})^{-\frac{1}{2}}$. Can anyone help me on this? Thanks.
Current browse context: cond-mat Change to browse by: References & Citations Bookmark(what is this?) Condensed Matter > Materials Science Title: Temperature and high fluence induced ripple rotation on Si(100) surface (Submitted on 7 Apr 2016) Abstract: Topography evolution of Si(100) surface due to oblique incidence low energy ion beam sputtering (IBS) is investigated. Experiments were carried out at different elevated temperatures from 20$^{\circ}$C to 450$^{\circ}$C and at each temperature, the ion fluence is systematically varied in a wide range from $\sim$ 1$\times$10$^{18}$cm$^{-2}$ to 1$\times$10$^{20}$cm$^{-2}$. The ion sputtered surface morphologies are characterized by atomic force microscopy and high-resolution cross-sectional transmission electron microscopy. At room temperature, the ion sputtered surfaces show periodic ripple nanopatterns and their wave-vector remains parallel to ion beam projection for entire fluence range. With increase of substrate temperature, these patterns tend to demolish and reduce into randomly ordered mound-like structures around 350$^{\circ}$C. Further rise in temperature above 400$^{\circ}$C leads, surprisingly, orthogonally rotated ripples beyond fluence 5$\times$10$^{19}$cm$^{-2}$. All the results are discussed combining the theoretical framework of linear, non-linear and recently developed mass redistribution continuum models of pattern formation by IBS. These results have technological importance regarding the control over ion induced pattern formation as well as it provides useful information for further progress in theoretical field. Submission historyFrom: Debasree Chowdhury [view email] [v1]Thu, 7 Apr 2016 16:35:38 GMT (1286kb,D)
I have this function $$ f(x,y) = \left\{ \begin{array}{ll} \frac{x^3}{x^2 + y^2} & \mbox{if } (x,y) \neq (0,0) \\ 0 & \mbox{if } (x,y) = (0,0) \end{array}\right.$$And I want to find the directional derivative in the $(1,1)$ direction at $(0,0)$, so using the limit definition this gives me$$\begin{align} D_vf(0,0) = \lim_{t\to 0}\frac{f((0,0) + t(1,1)) - f(0,0)}{t} = \lim_{t\to 0}\frac{t^3}{2t^2 t} = \frac{1}{2}\end{align}$$But wolfram alpha is giving me $0$ as a result Wolfram alpha result Is there something wrong with my procedure? Any help appreciated :) I have this function $$ f(x,y) = \left\{ \begin{array}{ll} \frac{x^3}{x^2 + y^2} & \mbox{if } (x,y) \neq (0,0) \\ 0 & \mbox{if } (x,y) = (0,0) \end{array}\right.$$And I want to find the directional derivative in the $(1,1)$ direction at $(0,0)$, so using the limit definition this gives me$$\begin{align} D_vf(0,0) = \lim_{t\to 0}\frac{f((0,0) + t(1,1)) - f(0,0)}{t} = \lim_{t\to 0}\frac{t^3}{2t^2 t} = \frac{1}{2}\end{align}$$But wolfram alpha is giving me $0$ as a result Wolfram alpha result You are right and Wolfram is wrong. What happens is that the computer is using the gradient to calculate the directional derivative. But the formula to calculate the directional derivative using the gradient uses the chain rule, which assumes the function to be differentiable. Your function is not differentiable at $(0,0)$, so the gradient cannot be used to find the directional derivatives there. It is worth mentioning that WA is making a second mistake: since the partial derivatives are not continuous at $(0,0)$, they cannot be calculated using differentiation rules (and I don't really know how they get the value zero). Calculating explicitly, $$ \frac{\partial f}{\partial x}f(0,0)=\lim_{h\to 0}\frac{h^3}{h(h^2+0)}=1, $$ $$ \frac{\partial f}{\partial y}f(0,0)=\lim_{h\to 0}\frac{0}{h(0+h^2)}=0. $$ Still, the formula $(a,b)\cdot\nabla f(0,0)=a$ gives the wrong result because $f$ is not differentiable.
again! I simply need help understanding this example. QUESTION: Show that the function $$ g(x) =\begin{cases} \frac{xy^2}{x^2 +y^4}, & \text{if $(x,y) \neq 0$}\\0, & \text{if $(x,y) = (0,0)$} \\\end{cases} $$ has directional derivatives at $0 = (0,0)$ but is not differentiable at $0 = (0,0)$ (What does this even mean?) In any case, the answer is given as: ANSWER Direct computation yields for every $v=(x,y) \in \Bbb R^2$, $$D_v(g) = \lim_{t\to0} \frac {g(0+tv) - g(0)}{t} = \lim_{t\to0} \frac {txy^2}{x^2 + t^2y^4} = 0$$ Thus $g$ has directional derivatives in all directions at $(0,0)$ On the other hand we notice that $g(x^2,x) =\displaystyle \frac{1}{2} \not\to 0 = g(0,0)$ as $g(x^2,x) \to 0$ thus g is not continuous at $0$ and is there not differentiable. Now I understand the first part. But my concern is with the second one. Why are we discussing $g(x^2,x)$ in the matter of continuity? I understand the idea is to prove indifferentiablity but why $g(x^2,x)$ to be exact? I know it's simply an example. But I may be thrown off by the use of x in both vector components...
Tagged: reduced row echelon form Problem 648 Determine whether the following augmented matrices are in reduced row echelon form, and calculate the solution sets of their associated systems of linear equations. (a) $\left[\begin{array}{rrr\|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 6 \end{array} \right]$. (b) $\left[\begin{array}{rrr\|r} 1 & 0 & 3 & -4 \\ 0 & 1 & 2 & 0 \end{array} \right]$. Add to solve later (c) $\left[\begin{array}{rr\|r} 1 & 2 & 0 \\ 1 & 1 & -1 \end{array} \right]$. Read solution Problem 643 For each of the following matrices, find a row-equivalent matrix which is in reduced row echelon form. Then determine the rank of each matrix. (a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$. (b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$. (c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$. (d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$. Add to solve later (e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$. Problem 569 For an $m\times n$ matrix $A$, we denote by $\mathrm{rref}(A)$ the matrix in reduced row echelon form that is row equivalent to $A$. For example, consider the matrix $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 &2 &2 \end{bmatrix}$ Then we have \[A=\begin{bmatrix} 1 & 1 & 1 \\ 0 &2 &2 \end{bmatrix} \xrightarrow{\frac{1}{2}R_2} \begin{bmatrix} 1 & 1 & 1 \\ 0 &1 & 1 \end{bmatrix} \xrightarrow{R_1-R_2} \begin{bmatrix} 1 & 0 & 0 \\ 0 &1 &1 \end{bmatrix}\] and the last matrix is in reduced row echelon form. Hence $\mathrm{rref}(A)=\begin{bmatrix} 1 & 0 & 0 \\ 0 &1 &1 \end{bmatrix}$. Find an example of matrices $A$ and $B$ such that \[\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B).\] Problem 272 Let \[A=\begin{bmatrix} 1 & 3\\ 2& 4 \end{bmatrix}.\] Then (a) Find all matrices \[B=\begin{bmatrix} x & y\\ z& w \end{bmatrix}\] such that $AB=BA$. Add to solve later (b) Use the results of part (a) to exhibit $2\times 2$ matrices $B$ and $C$ such that \[AB=BA \text{ and } AC \neq CA.\] Problem 267 Solve the following system of linear equations by transforming its augmented matrix to reduced echelon form (Gauss-Jordan elimination). Find the vector form for the general solution. \begin{align} x_1-x_3-3x_5&=1\\ 3x_1+x_2-x_3+x_4-9x_5&=3\\ x_1-x_3+x_4-2x_5&=1. \end{align} Problem 260 Let \[A=\begin{bmatrix} 1 & 1 & 2 \\ 2 &2 &4 \\ 2 & 3 & 5 \end{bmatrix}.\] (a) Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$. (b) Find a basis for the null space of $A$. (c) Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors. Add to solve later (d) Exhibit a basis for the row space of $A$. Problem 249 Suppose that the following matrix $A$ is the augmented matrix for a system of linear equations. \[A= \left[\begin{array}{rrr\|r} 1 & 2 & 3 & 4 \\ 2 &-1 & -2 & a^2 \\ -1 & -7 & -11 & a \end{array} \right],\] where $a$ is a real number. Determine all the values of $a$ so that the corresponding system is consistent. Problem 159 Let $T: \R^2 \to \R^2$ be a linear transformation. Let \[ \mathbf{u}=\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \mathbf{v}=\begin{bmatrix} 3 \\ 5 \end{bmatrix}\] be 2-dimensional vectors. Suppose that \begin{align} T(\mathbf{u})&=T\left( \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right)=\begin{bmatrix} -3 \\ 5 \end{bmatrix},\\ T(\mathbf{v})&=T\left(\begin{bmatrix} 3 \\ 5 \end{bmatrix}\right)=\begin{bmatrix} 7 \\ 1 \end{bmatrix}. \end{align} Let $\mathbf{w}=\begin{bmatrix} x \\ y \end{bmatrix}\in \R^2$. Find the formula for $T(\mathbf{w})$ in terms of $x$ and $y$. Problem 157 Let $P_2$ be the vector space of all polynomials of degree two or less. Consider the subset in $P_2$ \[Q=\{ p_1(x), p_2(x), p_3(x), p_4(x)\},\] where \begin{align} &p_1(x)=x^2+2x+1, &p_2(x)=2x^2+3x+1, \\ &p_3(x)=2x^2, &p_4(x)=2x^2+x+1. \end{align} (a) Use the basis $B=\{1, x, x^2\}$ of $P_2$, give the coordinate vectors of the vectors in $Q$. (b) Find a basis of the span $\Span(Q)$ consisting of vectors in $Q$. Add to solve later (c) For each vector in $Q$ which is not a basis vector you obtained in (b), express the vector as a linear combination of basis vectors. Problem 154 Define the map $T:\R^2 \to \R^3$ by $T \left ( \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\right )=\begin{bmatrix} x_1-x_2 \\ x_1+x_2 \\ x_2 \end{bmatrix}$. (a) Show that $T$ is a linear transformation. (b) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for each $\mathbf{x} \in \R^2$. Add to solve later (c) Describe the null space (kernel) and the range of $T$ and give the rank and the nullity of $T$. Problem 152 Let $V$ be the vector space of all $2\times 2$ matrices, and let the subset $S$ of $V$ be defined by $S=\{A_1, A_2, A_3, A_4\}$, where \begin{align} A_1=\begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}, \quad A_2=\begin{bmatrix} 0 & -1 \\ 1 & 4 \end{bmatrix}, \quad A_3=\begin{bmatrix} -1 & 0 \\ 1 & -10 \end{bmatrix}, \quad A_4=\begin{bmatrix} 3 & 7 \\ -2 & 6 \end{bmatrix}. \end{align} Find a basis of the span $\Span(S)$ consisting of vectors in $S$ and find the dimension of $\Span(S)$. Problem 115 Express the vector $\mathbf{b}=\begin{bmatrix} 2 \\ 13 \\ 6 \end{bmatrix}$ as a linear combination of the vectors \[\mathbf{v}_1=\begin{bmatrix} 1 \\ 5 \\ -1 \end{bmatrix}, \mathbf{v}_2= \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \mathbf{v}_3= \begin{bmatrix} 1 \\ 4 \\ 3 \end{bmatrix}.\] ( The Ohio State University, Linear Algebra Exam) Problem 102 Determine whether the following systems of equations (or matrix equations) described below has no solution, one unique solution or infinitely many solutions and justify your answer. (a)\[\left\{ \begin{array}{c} ax+by=c \\ dx+ey=f, \end{array} \right. \] where $a,b,c, d$ are scalars satisfying $a/d=b/e=c/f$. (b)$A \mathbf{x}=\mathbf{0}$, where $A$ is a singular matrix. (c)A homogeneous system of $3$ equations in $4$ unknowns. (d)$A\mathbf{x}=\mathbf{b}$, where the row-reduced echelon form of the augmented matrix $[A\|\mathbf{b}]$ looks as follows: \[\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 &1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}.\] ( The Ohio State University, Linear Algebra Exam) Read solution Add to solve later Problem 65 Consider the system of linear equations \begin{align} x_1&= 2, \\ -2x_1 + x_2 &= 3, \\ 5x_1-4x_2 +x_3 &= 2 \end{align} (a) Find the coefficient matrix and its inverse matrix. (b) Using the inverse matrix, solve the system of linear equations. ( The Ohio State University, Linear Algebra Exam)
On Monday, Celestalon kicked off the official Alpha Theorycrafting season by posting a Theorycrafting Discussion thread on the forums. And he was kind enough to toss a meaty chunk of information our way about Resolve, the replacement for Vengeance. Resolve: Increases your healing and absorption done to yourself, based on Stamina and damage taken (before avoidance and mitigation) in the last 10 sec. In today’s post, I want to go over the mathy details about how Resolve works, how it differs from Vengeance, and how it may (or may not) fix some of the problems we’ve discussed in previous blog posts. Mathemagic Celestalon broke the formula up into two components: one from stamina and one from damage taken. But for completeness, I’m going to bolt them together into one formula for resolve $R$: $$ R =\frac{\rm Stamina}{250~\alpha} + 0.25\sum_i \frac{D_i}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t_i )}{10} \right )$$ where $D_i$ is an individual damage event that occurred $\Delta t_i$ seconds ago, and $\alpha$ is a level-dependent constant, with $\alpha(100)=261$. The sum is carried out over all damaging events that have happened in the last 10 seconds. The first term in the equation is the stamina-based contribution, which is always active, even when out of combat. There’s a helpful buff in-game to alert you to this: My premade character has 1294 character sheet stamina, which after dividing by 250 and $\alpha(90)=67$, gives me 0.07725, or about 7.725% Resolve. It’s not clear at this point whether the tooltip is misleadingly rounding down to 7% (i.e. using floor instead of round) or whether Resolve is only affected by the stamina from gear. The Alpha servers went down as I was attempting to test this, so we’ll have to revisit it later. We’ve already been told that this will update dynamically with stamina buffs, so having Power Word: Fortitude buffed on you mid-combat will raise your Resolve. Once you’re in combat and taking damage, the second term makes a contribution: I’ve left this term in roughly the form Celestalon gave, even though it can obviously be simplified considerably by combining all of the constants, because this form does a better job of illustrating the behavior of the mechanic. Let’s ignore the sum for now, and just consider an isolated damage event that does $D$ damage: $$0.25\times\frac{D}{\rm MaxHealth}\left ( \frac{2 ( 10-\Delta t )}{10} \right )$$ The 0.25 just moderates the amount of Resolve you get from damaging attacks. It’s a constant multiplicative factor that they will likely tweak to achieve the desired balance between baseline (stamina-based) Resolve and dynamic (damage-based) Resolve. The factor of $D/{\rm MaxHealth}$ means that we’re normalizing the damage by our max health. So if we have 1000 health and take an attack that deals 1000 damage (remember, this is before mitigation), this term gives us a factor of 1. Avoided auto-attacks also count here, though instead of performing an actual damage roll the game just uses the mean value of the boss’s auto-attack damage. Again, nothing particularly complicated here, it just makes Resolve depend on the percentage of your health the attack would have removed rather than the raw damage amount. Also note that we’ve been told that dynamic health effects from temporary multipliers (e.g. Last Stand) aren’t included here, so we’re not punished for using temporary health-increasing cooldowns. The term in parentheses is the most important part, though. In the instant the attack lands, $\Delta t=0$ and the term in parentheses evaluates to $2(10-0)/10 = 2.$ So that attack dealing 1000 damage to our 1000-health tank would give $0.25\times 1 \times 2 = 0.5,$ or 50% Resolve. However, one second later, $\Delta t = 1$, so the term in parentheses is only $2(10-1)/10 = 1.8$, and the amount of resolve it grants is reduced to 45%. The amount of Resolve granted continues to linearly decrease as time passes, and by the time ten seconds have elapsed it’s reduced to zero. Each attack is treated independently, so to get our total Resolve from all damage taken we just have to add up the Resolve granted by every attack we’ve taken, hence the sum in my equation. You may note that the time-average of the term in parentheses is 1, which is how we get the advertised “averages to ~Damage/MaxHealth” that Celestalon mentioned in the post. In that regard, he’s specifically referring to just the part within the sum, not the constant factor of 0.25 outside of it. So in total, your average Resolve contribution from damage is 25% of Damage/MaxHealth. Comparing to Vengeance Mathematically speaking, there’s a world of difference between Resolve and Vengeance. First and foremost is the part we already knew: Resolve doesn’t grant any offensive benefit. We’ve talked about that a lot before, though, so it’s not territory worth re-treading. Even in the defensive component though, there are major differences. Vengeance’s difference equation, if solved analytically, gives solutions that are exponentials. In other words, provided you were continuously taking damage (such that it didn’t fall off entirely), Vengeance would decay and adjust to your new damage intake rather smoothly. It also meant that damage taken at the very beginning of an encounter was still contributing some amount of Vengeance at the very end, again, assuming there was no interruption. And since it was only recalculated on a damage event, you could play some tricks with it, like taking a giant attack that gave you millions of Vengeance and then riding that wave for 20 seconds while your co-tank takes the boss. Resolve does away with all of that. It flat-out says “look, the only thing that matters is the last 10 seconds.” The calculation doesn’t rely on a difference equation, meaning that when recalculating, it doesn’t care what your Resolve was at any time previously. And it forces a recalculation at fixed intervals, not just when you take damage. As a result, it’s much harder to game than Vengeance was. Celestalon’s post also outlines a few other significant differences: No more ramp-up mechanism No taunt-transfer mechanism Resolve persists through shapeshifts Resolve only affects self-healing and self-absorbs The lack of ramp-up and taunt-transfer mechanisms may at first seem like a problem. But in practice, I don’t think we’ll miss either of them. Both of these effects served offensive (i.e. threat) and defensive purposes, and it’s pretty clear that the offensive purposes are made irrelevant by definition here since Resolve won’t affect DPS/threat. The defensive purpose they served was to make sure you had some Vengeance to counter the boss’s first few hits, since Vengeance had a relatively slow ramp-up time but the boss’s attacks did not. However, Resolve ramps up a lot faster than Vengeance does. Again, this is in part thanks to the fact that it isn’t governed by a difference equation. The other part is because it only cares about the last ten seconds. To give you a visual representation of that, here’s a plot showing both Vengeance and Resolve for a player being attacked by a boss. The tank has 100 health and the boss swings for 30 raw damage every 1.5 seconds. Vengeance is shown in arbitrary units here since we’re not interested in the exact magnitude of the effect, just in its dynamic properties. I’ve also ignored the baseline (stamina-based) contribution to Resolve for the same reason. As a final note, while the blog post says that Resolve is recalculated every second, it seemed like it was updating closer to every half-second when I fooled with it on alpha, so these plots use 0.5-second update intervals. Changing to 1-second intervals doesn’t significantly change the results (they just look a little more fragmented). The plot very clearly shows the 50% ramp-up mechanism and slow decay-like behavior of Vengeance. Note that while the ramp-up mechanism gets you to 50% of Vengeance’s overall value at the first hit (at t=2.5 seconds), Resolve hits this mark as soon as the second hit lands (at 4.0 seconds) despite not having any ramp-up mechanism. Resolve also hits its steady-state value much more quickly than Vengeance does. By definition, Resolve gets there after about 10 seconds of combat (t=12.5 seconds). But with Vengeance, it takes upwards of 30-40 seconds to even approach the steady-state value thanks to the decay effect (again, a result of the difference equation used to calculate Vengeance). Since most fights involve tank swaps more frequently than this, it meant that you were consistently getting stronger the longer you tanked a boss. This in turn helped encourage the sort of “solo-tank things that should not be solo-tanked” behavior we saw in Mists. This plot assumes a boss who does exactly 30 damage per swing, but in real encounters the boss’s damage varies. Both Vengeance and Resolve adapt to mimic that change in the tank’s damage intake, but as you could guess, Resolve adapts much more quickly. If we allow the boss to hit for a random amount between 20 and 40 damage: You can certainly see the similar changes in both curves, but Resolve reacts quickly to each change while Vengeance changes rather slowly. One thing you’ve probably noticed by now is that the Resolve plot looks very jagged (in physics, we might call this a “sawtooth wave”). This happens because of the linear decay built into the formula. It peaks in the instant you take the attack – or more accurately, in the instant that Resolve is recalculated after that attack. But then every time it’s recalculated it linearly decreases by a fixed percent. If the boss swings in 1.5-second intervals, then Resolve will zig-zag between its max value and 85% of its max value in the manner shown. The more frequently the boss attacks, the smoother that zig-zag becomes; conversely, a boss with a long swing timer will cause a larger variation in Resolve. This is apparent if we adjust the boss’s swing timer in either direction: It’s worth noting that every plot here has a new randomly-generated sequence of attacks, so don’t be surprised that the plots don’t have the same profile as the original. The key difference is the size of the zig-zag on the Resolve curve. I’ve also run simulations where the boss’ base damage is 50 rather than 30, but apart from the y-axis having large numbers there’s no real difference: Note that even a raw damage of 50% is pretty conservative for a boss – heroic bosses in Siege have frequently had raw damages that were larger than the player’s health. But it’s not clear if that will still be the case with the new tanking and healing paradigm that’s been unveiled for Warlords. If we make the assumption that raw damage will be lower, then these rough estimates give us an idea of how large an effect Resolve will be. If we guess at a 5%-10% baseline value from stamina, these plots suggest that Resolve will end up being anywhere from a 50% to 200% modifier on our healing. In other words, it has the potential to double or triple our healing output with the current tuning numbers. Of course, it’s anyone’s guess as to whether those numbers are even remotely close to what they’ll end up being by the end of beta. Is It Fixed Yet? If you look back over our old blog posts, the vast majority of our criticisms of Vengeance had to do with its tie-in to damage output. Those have obviously been addressed, which leaves me worrying that I’ll have nothing to rant about for the next two or three years. But regarding everything else, I think Resolve stands a fair chance of addressing our concerns. One of the major issues with Vengeance was the sheer magnitude of the effect – you could go from having 50k AP to 600k AP on certain bosses, which meant your abilities got up to 10x more effective. Even though that’s an extreme case, I regularly noted having over 300k AP during progression bosses, a factor of around 6x improvement. Resolve looks like it’ll tamp down on that some. Reasonable bosses are unlikely to grant a multiplier larger than 2x, which will be easier to balance around. It hasn’t been mentioned specifically in Celestalon’s post, but I think it’s a reasonable guess that they will continue to disable Resolve gains from damage that could be avoided through better play (i.e. intentionally “standing in the bad”). If so, there will be little (if any) incentive to take excess damage to get more Resolve. Our sheer AP scaling on certain effects created situations where this was a net survivability gain with Vengeance, but the lower multiplier should make that impossible with Resolve. While I still don’t think it needs to affect anything other than active mitigation abilities, the fact that it’s a multiplier affecting everything equally rather than a flat AP boost should make it easier to keep talents with different AP coefficients balanced (Eternal Flame and Sacred Shield, specifically). And we already know that Eternal Flame is losing its Bastion of Glory interaction, another change which will facilitate making both talents acceptable choices. All in all, I think it’s a really good system, if slightly less transparent. It’s too soon to tell whether we’ll see any unexpected problems, of course, but the mechanic doesn’t have any glaring issues that stand out upon first examination (unlike Vengeance). I still have a few lingering concerns about steady-state threat stability between tanks (ironically, due to the removal of Vengeance), but that is the sort of thing which will become apparent fairly quickly during beta testing, and at any rate shouldn’t reflect on the performance of Resolve.
Statistics - (Univariate\|Simple\|Basic) Linear Regression Table of Contents 1 - About 2 - Articles Related 3 - Linear Equation In a standard notation: <math> Y = m + b.X + e </math> or in a general notation (that introduces the idea of multiple regression): <math> Y = \hat{Y} + e = (B_0 + B_1.{X_1}) + e </math> where: <math>e</math> is the error (residual) ie <math>Y - \hat{Y}</math> (<math>m</math> of <math>B_0</math>) is the intercept also known as the regression constant The intercept is the point in the Y axis when X is null. And the slope tells that for each X units (1 on the X scale), Y increases of the slope value. The intercept and the slope are also known as coefficients or parameters. 4 - Assumptions 5 - Estimation of the parameters The values of the regression coefficients are estimated such that the regression model yields optimal predictions. The regression coefficient is what we observed. In simple regression, the standardized regression coefficient will be the same as the correlation coefficient 5.1 - (Ordinary) Least Squares Method An optimal prediction for one variable is reached: by minimizing the residuals (ie the predictions errors), ie by minimizing the overall distance between the line and each individual dot <math>\displaystyle \sum_{i=1}^{N}(Y_i - \hat{Y}_i)</math>. ie by minimizing the sum of the squared residuals (RSS) This idea is called the Ordinary Least Squares estimation. The least squares approach chooses <math>B_0</math> and <math>B_1</math> to minimize the Residual sum of Squares (RSS). <MATH> \begin{array}{rrl} \hat{B}_1 & = & \frac { \displaystyle \sum_{i=1}^{\href{sample_size}{N}} (X_i - \href{mean}{\bar{X}})(Y_i - \href{mean}{\bar{Y}}) } { \displaystyle \sum_{i=1}^{\href{sample_size}{N}} (X_i - \href{mean}{\bar{X}})^2 } \\ \hat{B}_0 & = & \href{mean}{\bar{Y}} - \hat{B}_1 . \href{mean}{\bar{X}} \end{array} </MATH> 5.2 - Formula 5.2.1 - Unstandardized Formula for the unstandardized coefficient <MATH> \text{Regression Coefficient (Slope)} = \href{little_r}{r} . \frac{\href{Standard Deviation}{\text{Standard Deviation of Y}}}{\href{Standard Deviation}{\text{Standard Deviation of X}}} </MATH> where: The standard deviation division is needed in order to take into account the scale and variability of X and Y. 5.2.2 - Standardized In the z scale, the standard deviation is 1 and the mean is zero. The Regression Coefficient becomes there: <MATH> \begin{array}{rrl} \text{Standardized Regression Coefficient - } \beta & = & \href{little_r}{r} . \frac{\href{Standard Deviation}{\text{Standard Deviation of Y}}}{\href{Standard Deviation}{\text{Standard Deviation of X}}} \\ \beta & = & \href{little_r}{r}.\frac{1}{1} \\ \beta & = & \href{little_r}{r} \\ \end{array} </MATH> The coefficient are standardized to be able to compare them. For instance, in a multiple regression. The unstandardized coefficient are linked to the scale of each predictor and therefore can not be compared. 6 - Accuracy The Accuracy can be assessed for: the parameters the overall model 6.1 - Parameters The slope of the predictor can be assess , both in terms of: confidence intervals and hypothesis test. 6.1.1 - Standard Error <MATH> \begin{array}{rrl} SE(\hat{B}_1)^2 & = & \frac{ \displaystyle \sigma^2}{\displaystyle \sum_{i=1}^{N}(X_i - \bar{X})^2} \\ SE(\hat{B}_0)^2 & = & \sigma^2 \left [ \frac{1}{N} + \frac{ \displaystyle \bar{X}^2 } { \displaystyle \sum_{i=1}^{N}(X_i - \bar{X})^2 } \right ] \\ \end{array} </MATH> where sigma square is the noise (ie the variance of the square around the line). <math>\sigma^2 = \href{variance}{var}(\href{residual}{\epsilon})</math> 6.1.2 - Confidence Interval <MATH> \begin{array}{rrl} \hat{B}_1 & = & 2.SE(\hat{B}_1) \end{array} </MATH> As we assume that the errors are normally distributed, there is then approximately a 95% chance that the following interval will contain the true value of <math>B_1</math> (under a scenario where we got repeated samples ie repeated training set). If we calculate 100 confidence intervals on 100 sample data set, 95% of the time, they will contain the true value. <MATH> \begin{array}{rrl} [ \hat{B}_1 - 2.SE(\hat{B}_1) , \hat{B}_1 + 2.SE(\hat{B}_1) ] \end{array} </MATH> Visually, if we can't get a flat regression line (Slope = 0) in confidence interval region (the shadowed one), it means that it's a significant effect. 6.1.3 - t-statistic Standard errors can also be used to perform a null hypothesis tests on the coefficients. It corresponds to testing: <math>H_0 : B_1 = 0 </math> (Expected Slope, Null Hypothesis) versus <math>H_A : B_1 \neq 0 </math> (Observed Slope). To test the null hypothesis, a t-statistic is computed: <MATH> \begin{array}{rrl} \text{t-value} & = & \frac{(\text{Observed Slope} - \text{Expected Slope})}{\href{Standard Error}{\text{What we would expect just due to chance}}} & \\ \text{t-value} & = & \frac{\href{regression coefficient#unstandardised}{\text{Observed Slope}} - 0}{\href{Standard Error}{\text{What we would expect just due to chance}}}\\ \text{t-value} & = & \frac{\href{regression coefficient#unstandardised}{\text{Unstandardised regression coefficient}}}{\href{Standard Error}{\text{Standard Error}}}\\ \text{t-value} & = & \frac{\hat{B}_1}{SE(\hat{B}_1)} \\ \end{array} </MATH> where: The probability (p-value) is the chance of observing any value equal to <math>\|t\|</math> or larger. In a linear regression, when the p value is less than 0.5 we will reject the null hypothesis. With a p-value of for instance, 10 to the minus 4, the chance of seeing this data, under the assumption that the null hypothesis (There's no effect, ie no relationship, ie the slope is null ) is less than the p-value (10 to the minus 4). In this case, it's very unlikely to have seen this data. It's possible, but very unlikely under the assumption that the predictor X has no effect on the outcome Y. The conclusion, therefore, will be that the predictor X has an effect on the outcome Y. 6.2 - Model 6.2.1 - Residual Standard Error (RSE) <MATH> \begin{array}{rrl} \text{Residual Standard Error (RSE)} & = & \sqrt{\frac{1}{n-2}\href{rss}{\text{Residual Sum-of-Squares}}} \\ RSE & = & \sqrt{\frac{1}{n-2}\href{rss}{RSS}} \\ RSE & = & \sqrt{\frac{1}{n-2}\sum_{i=1}^{N}(Y_i -\hat{Y}_i)} \\ \end{array} </MATH> 6.2.2 - R (Big R) 6.2.3 - R-squared R-squared or fraction of variance explained measures how closely two variables are associated. <MATH> \begin{array}{rrl} R^2 & = & \frac{\href{tss}{TSS} - \href{rss}{RSS}}{\href{tss}{TSS}} \\ R^2 & = & 1 - \frac{\href{rss}{RSS}}{\href{tss}{TSS}} \\ \end{array} </MATH> where: RSS is the error after optimization with the least squares method. This statistic measures, how much was reduced the total sum of squares (TSS-RSS) relative to itself (TSS). So this is the fraction of variance explained. In this simple linear regression, <math>R^2 = r^2</math>, where r is the correlation coefficient between X and Y. The higher the correlation, the more that we'll explain the variance. A value <math>R^2 = 0.51</math> means that the variance was reduced by 51%. A <math>R^2 = 0.9 </math> indicates a fairly strong relationship between X and Y. The domain of application is really important to have a judgement of how good an R squared is. Example of good value by domain: In business, financial application and some kind of physical sciences: <math>R^2 \approx 0.51</math> In medicine, <math>R^2 \approx 0.05</math>
In this MathStackExchange post the question in the title was asked without much outcome, I feel. Edit: As Douglas Zare kindly observes, there is one more answer in MathStackExchange now. I am not used to basic Probability, and I am trying to prepare a class that I need to teach this year. I feel I am unable to motivate the introduction of random variables. After spending some time speaking about Kolmogoroff's axioms I can explain that they allow to make the following sentence true and meaningful: The probability that, tossing a coin $N$ times, I get $n\leq N$ tails equals $$\tag{$\ast$}{N \choose n}\cdot\Big(\frac{1}{2}\Big)^N.$$ But now people (i.e. books I can find) introduce the "random variable $X\colon \Omega\to\mathbb{R}$ which takes values $X(\text{tails})=1$ and $X(\text{heads})=0$" and say that it follows the binomial rule. To do this, they need a probability space $\Omega$: but once one has it, one can prove statement $(\ast)$ above. So, what is the usefulness of this $X$ (and of random variables, in general)? Added: So far my question was admittedly too vague and I try to emend. Given a discrete random variable $X\colon\Omega\to\mathbb{R}$ taking values $\{x_1,\dots,x_n\}$ I can define $A_k=X^{-1}(\{x_k\})$ for all $1\leq k\leq n$. The study of the random variable becomes then the study of the values $p(A_k)$, $p$ being the probability on $\Omega$. Therefore, it seems to me that we have not gone one step further in the understanding of $\Omega$ (or of the problem modelled by $\Omega$) thanks to the introduction of $X$. Often I read that there is the possibility of having a family $X_1,\dots,X_n$ of random variables on the same space $\Omega$ and some results (like the CLT) say something about them. But then I know no example—and would be happy to discover—of a problem truly modelled by this, whereas in most examples that I read there is either a single random variable; or the understanding of $n$ of them requires the understanding of the power $\Omega^n$ of some previously-introduced measure space $\Omega$. It seems to me (but admit to have no rigourous proof) that given the above $n$ random variables on $\Omega$ there should exist a $\Omega'$, probably much bigger, with a single $X\colon\Omega'\to\mathbb{R}$ "encoding" the same information as $\{X_1,\dots,X_n\}$. In this case, we are back to using "only" indicator functions. I understand that this process breaks down if we want to make $n\to \infty$, but I also suspect that there might be a deeper reason for studying random variables. All in all, my doubts come from the fact that random variables still look to me as being a poorer object than a measure (or, probably, of a $\sigma$-algebra $\mathcal{F}$ and a measure whose generated $\sigma$-algebra is finer than $\mathcal{F}$, or something like this); though, they are introduced, studied, and look central in the theory. I wonder where I am wrong. Caveat: For some reason, many people in comments below objected that "throwing random variables away is ridiculous" or that I "should try to come out with something more clever, then, if I think they are not good". That was not my point. I am sure they must be useful, lest all textbooks would not introduce them. But I was unable to understand why: many useful and kind answers below helped much.
The aim of this test case is to validate the following functions: The simulation results of SimScale were compared to the numerical results presented in [HPLA100]. The mesh used in (A) and (C) was created with the automatic-tetrahedralization-tool on the SimScale platform. The mesh used in (B) und (D) was created locally. A B C D x [m] 0.0195 0.0205 0.0205 0.0195 y [m] 0 0 0 0 z [m] 0.01 0.01 0 0 To obtain the solid body the face ABCD is rotated 45° around the z-axis. Because of the symmetry of the cylinder only one quarter was modelled. Tool Type : Code_Aster Analysis Type : Static Mesh and Element types : Case Mesh type Number of nodes Element type (A) quadratic tetrahedral 2893 3D isoparametric (B) quadratic hexahedral 2583 3D isoparametric (C) quadratic tetrahedral 17460 3D isoparametric (D) quadratic hexahedral 9285 3D isoparametric Material: Constraints: Loads: $$u(r) = \frac{-(1+\nu)(1-2\nu)}{(1-\nu)E} \rho \omega^2 \frac{r^3}{8} + Ar + \frac{B}{r}$$ $$\sigma_{zz}(r) = \frac{-\nu}{(1-\nu)} \rho \omega^2 \frac{r^2}{2} + \frac{2 \nu E}{(1+\nu)(1-2\nu)} A$$ $$A = \frac{(3-2\nu)(1+\nu)(1-2\nu)}{4(1-\nu)E} \rho \omega^2 R^2 (1-x^2) = 7.13588 \cdot 10^{-12}$$ $$B = \frac{(3-2\nu)(1+\nu)}{8(1-\nu)E} \rho \omega^2 R^4 (1-x^2)^2 = 3.561258 \cdot 10^{-15} m^2$$ $$x = \frac {h} {2R} = \frac {0.001m} {2 \cdot 0.02 m} = 0.025$$ All stated equations used to solve the problem are derived in [HPLA100]. The parameters h and R in (5) are the thickness of the cross section and the radius of the middle surface of the cylinder respectively. Important The values for the comparison of the displacement ur and the stresses σzz are averaged over on edge and an area respectively Comparison of the displacement U r and the stresses $\sigma_{zz}$ of the inner and the outter face obtained with SimScale and the results derived from [HPLA100]. Case Quantity [HPLA100] SimScale Error (A) $U{r}$ (r=0.0195m) [m] 2.9424E-013 2.83491E-013 3.653% (A) $U{r}$ (r=0.0205m) [m] 2.8801E-013 2.8972E-013 -0.594% (A) $\sigma_{zz}$ (r=0.0195m) [N/m 2] 0.99488 0.992259 0.263% (A) $\sigma_{zz}$ (r=0.0205m) [N/m 2] 0.92631 0.929711 -0.367% (B) $U{r}$ (r=0.0195m) [m] 2.9424E-013 2.94195E-013 0.015% (B) $U{r}$ (r=0.0205m) [m] 2.8801E-013 2.87966E-013 0.015% (B) $\sigma_{zz}$ (r=0.0195m) [N/m 2] 0.99488 1.00893 -1.412% (B) $\sigma_{zz}$ (r=0.0205m) [N/m 2] 0.92631 0.914645 1.259% (C) $U{r}$ (r=0.0195m) [m] 2.9424E-013 2.94234E-013 0.002% (C) $U{r}$ (r=0.0205m) [m] 2.8801E-013 2.88003E-013 0.002% (C) $\sigma_{zz}$ (r=0.0195m)[N/m 2] 0.99488 0.995217 -0.034% (C) $\sigma_{zz}$ (r=0.0205m) [N/m 2] 0.92631 0.926616 -0.033% (D) $U{r}$ (r=0.0195m) [m] 2.9424E-013 2.94237E-013 0.001% (D) $U{r}$ (r=0.0205m) [m] 2.8801E-013 2.88007E-013 0.001% (D) $\sigma_{zz}$ (r=0.0195m) [N/m 2] 0.99488 0.994995 -0.012% (D) $\sigma_{zz}$ (r=0.0205m) [N/m 2] 0.92631 0.926415 -0.011% [HPLA100] (1, 2, 3, 4) HPLA100 – Cylindre creux thermoélastique pesanten rotation uniforme
Alvin Jin Somehow, I'm finding myself reading a lot more homotopy theory than I thought I ever would. Currently diving into $\Omega^{\infty} \Sigma^{\infty}$-land and Goodwillie calculus (cf. Calculus III: Taylor Series https://arxiv.org/abs/math/0310481). Stockholm, Sweden Member for 4 years 1 profile view Last seen Sep 14 at 15:24 Communities (2) Top network posts 26 Prove that $(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ 20 Lecture Notes for Hatcher's Algebraic Topology 15 Renewing a US Passport without getting rid of the old one 11 Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal? 8 Many questions about a Klein bottle. 6 Why $\mathbb S^1\times \mathbb S^1$ is isomorphic to the torus? 6 Question on indefinite integrals View more network posts →
I am struggling to apply Freize et al. paper to break a truncated LCG. A truncated LCG is a pseudo random generator that outputs the $n$ leading bits $y_i$ of $x_i$, where $(x_i)$ is such that $x_{i+1} = \alpha \cdot x_i + \beta\mod M$, and $\alpha$ is an integer, coprime with $M$. The first state $x_1$ is unknown. The objective is to break the LCG, ie, from one or several consecutive outputs, find the current state $x_i$. We suppose that $\alpha$ and $M$ are known, and $\beta = 0$ (and is known too). The theory The paper states (pages 5 and 6) that their method solves systems of modular equations of the form $\displaystyle\sum_{j=1}^k a_{ij}x_j \equiv c_i \mod M$, for all $i \in \{1, \dots, k\}$. Deriving a lattice from the vectors $a_i$, the authors apply LLL to find a small basis, of vectors $w_1, \dots, w_k$. By multiplying the matrix $\begin{pmatrix} w_1\\ \vdots\\ w_k \end{pmatrix}$ on the right by $x$, they derive a new set of equations: $\displaystyle\sum_{j=1}^k w_{ij}x_j \equiv c_i' \mod M$, for all $i \in \{1, \dots, k\}$. Here, i guess that the $c_i'$ are just obtained from the $c_i$ by multiplying (on the left) by the same matrix that allowed us to go from our original basis to our reduced basis. Now, if the $c_i'$ are small enough, the equalities actually hold on $\mathbb{Z}$ and we can solve them: we have effectively solved the system. In practice Now, the authors talk a bit more about their method for a LCG, part 3 (pages 11 and 12). We have the relations $\alpha^{k-1} x_{1} - x_k = 0$ (edit: fixed typo), so I guess that it implies that all my $c_i$ are $0$. Now, if I want to obtain my $c_i'$, I should multiply $c$ on the left by the same matrix as the one used to change the basis of the lattice. But, given that I have $c = 0$, I should have $c' = 0$ as well, so we don't need to compute this matrix. Furthermore, my lattice is, as descibed by the paper, $$L = \begin{pmatrix} M & 0 && \dots & 0\\ \alpha & -1 &0& \dots & 0\\ \alpha^2 & 0 & -1 & \dots & 0\\ \vdots &&&\ddots & \vdots\\ \alpha^{k-1} & 0 & 0 & \dots & -1\\ \end{pmatrix}$$ which is trivially nonsingular, and as such, the only solution to $Lx = 0$ is $x = 0$, which is obviously wrong. Even worse, this method does not use the $y_i$, but it obviously should at some point, so I guess I'm missing something. My questions (Most probably related) In my case, what are the $c_i$, and do I really obtain the $c_i'$ from them by the basis change? How are the $y_i$ supposed to take part in this method?
Since you indicated that the power spectrum of your background noise is flat, I'll assume it is white. A major drawback with your current approach is that you are discarding a large amount of the signal power; even with the effect of front-end bandlimiting shown by in your diagram by the exponential-rise step response, a single ADC sample near the end of the rounded pulse provides a snapshot of the receiver input that is rather localized in time. You can take advantage of more of the signal power by sampling at a higher rate and applying a matched filter at the higher sample rate. Theory: You can look at this as a relatively simple problem in detection theory. In each symbol interval, your receiver needs to decide between two hypotheses: $$ \begin{eqnarray} H_0 &:& signal \ is \ not \ present \\ H_1 &:& signal \ is \ present \end{eqnarray} $$ This sort of problem is often solved using Bayesian decision rules, which attempt to make the optimum decision according to some specified measure of risk. This provides a framework where one can optimally make detection decisions based on a flexible set of criteria. For example, if there is a large penalty to your system for failing to detect the signal if it is in fact present (i.e. you pick $H_0$ when $H_1$ is true), then you can build that into your decision rule if needed. For a detection problem such as yours, where you're trying to decide between zeros and ones at the receiver output, the penalty is typically assumed to be equal (outputting a zero when a one was transmitted, and vice versa, "hurt equally"). The Bayesian approach in that case reduces to a maximum-likelihood estimator (also described here): you pick the hypothesis that is most likely, given the observation that your receiver makes. That is, if the quantity that your receiver observes is $x$, then it would generate a decision based on the hypothesis that has the largest likelihood function value. For the binary decision case, the likelihood ratio can be used instead: $$ \Lambda(x) = \frac{P(x\ \|\ H_0 \ is \ true)}{P(x\ \|\ H_1 \ is \ true)} = \frac{P(x\ \|\ signal \ is \ not \ present)}{P(x\ \|\ signal \ is \ present)} $$ Using the above model, for each observation of the channel $x$, the optimal receiver would decide that the signal was not present (therefore outputting a zero) if the likelihood ratio $\Lambda(x)$ is greater than one (and therefore the signal was most likely to be not present based on the observation), and vice versa. What remains is a model for the signal of interest and any other components in the receiver detection statistic $x$ that could affect its decisions. For a digital communications like this, it might be modeled as follows: $$ \begin{eqnarray} H_0 &:& x = N \\ H_1 &:& x = s + N \end{eqnarray} $$ where $n$ is a random variable taken from some distribution (often assumed to be zero-mean Gaussian) and $s$ is a deterministic component of the observation that is due to the signal that you're looking for. The distribution of the receiver observable $x$, therefore, varies depending upon whether hypothesis $H_0$ or $H_1$ is true. In order to evaluate the likelihood ratio, you need a model for what those distributions are. For the Gaussian case referenced above, the mathematics look like this: $$ \Lambda(x) = \frac{P(x\ \|\ H_0 \ is \ true)}{P(x\ \|\ H_1 \ is \ true)} = \frac{P(x\ \|\ x = N)}{P(x\ \|\ x = s + N)} $$ $$ \Lambda(x) = \frac{P(x\ \|\ H_0 \ is \ true)}{P(x\ \|\ H_1 \ is \ true)} = \frac{e^{\frac{-x^2}{2 \sigma^2}}}{e^{\frac{-(x - s)^2}{2 \sigma^2}}} $$ where $\sigma^2$ is the variance of the Gaussian noise term. Note that the additive signal component only has the function of shifting the mean of the resulting Gaussian distribution of $x$. The log-likelihood ratio can be used to get rid of the exponentials: $$ \ln(\Lambda(x)) = \ln\left(\frac{e^{\frac{-x^2}{2 \sigma^2}}}{e^{\frac{-(x - s)^2}{2 \sigma^2}}}\right) = \left(\frac{-x^2}{2 \sigma^2}\right) - \left(\frac{-(x - s)^2}{2 \sigma^2}\right) $$ Recall that our decision rule picked $H_0$ if the likelihood ratio was greater than one. The equivalent log-likelihood decision rule is to pick $H_0$ if the log-likelihood is greater than zero. Some algebra shows that the decision rule reduces to: $$ \begin{eqnarray} x < \frac{s}{2} \rightarrow choose\ H_0 \\ x > \frac{s}{2} \rightarrow choose\ H_1 \end{eqnarray} $$ Note that if $x = \frac{s}{2}$, then both hypotheses are equally likely, and you would need to just pick one; this isn't a practical concern for continuously-valued signals, however. So, given a known signal amplitude $s$, we can detect its presence against a background of Gaussian noise optimally by setting a threshold $T = \frac{s}{2}$; if the observed value $x$ is greater than $T$, we declare the signal present and emit a one, and vice versa. Practice: There are a few practical issues that creep into this simple, toy theoretical example. One: just mapping the scenario that you described into a deceptively simple-looking model might not seem straightforward. Secondly, it's very rare that you would know the amplitude $s$ of the signal that you're looking for, so threshold selection requires some thought. As I referenced before, noise is often assumed to be Gaussian because the normal distribution is so easy to work with: the sum of a bunch of independent Gaussians is still Gaussian, and their mean and variances just add also. Also, the first- and second-order statistics of the distribution are enough to completely characterize them (given the mean and variance of a Gaussian distribution, you can write its pdf). So, hopefully that's a decent approximation at least for your application. There are two ways to improve the performance of the detector given the model described above: you can increase $s$ (i.e. increase the signal power), making it stand out more against the noise. You could decrease $N$ (i.e. reduce the amount of noise), reducing the amount of interference that makes the presence of $s$ unclear. Or, equivalently, you can think of the signal to noise ratio instead. To see its importance, let's go back to the theory for a second. What is the probability of a bit error given our decision rule? $$ \begin{eqnarray} P_e &=& P(choose \ H_0 \ \|\ H_1\ true) P(H_1\ true) + P(choose \ H_1 \ \|\ H_0\ true) P(H_0\ true) \\ &=& \frac{1}{2} P(x < \frac{s}{2} \ \|\ x = s + N) + \frac{1}{2} P(x > \frac{s}{2} \ \|\ x = N) \\ &=& \frac{1}{2} F_{x\ \|\ x = s + N}\left(\frac{s}{2}\right) + \frac{1}{2} \left(1 - F_{x\ \|\ x = N}\left(\frac{s}{2}\right)\right) \end{eqnarray} $$ where $F_{x\ \|\ x = s + N}(z)$ is the cumulative distribution function of the distribution of the observation $x$, given that $x = s + N$ (and likewise for the other function). Substituting in the cdf for the Gaussian distribution, we get: $$ \begin{eqnarray} P_e &=& \frac{1}{2} \left(1 - Q\left(\frac{\frac{s}{2} - s}{\sigma}\right)\right) + \frac{1}{2} Q\left(\frac{\frac{s}{2}}{\sigma}\right) \\ &=& \frac{1}{2} + \frac{1}{2} \left(-Q\left(\frac{\frac{s}{2} - s}{\sigma}\right) + Q\left(\frac{\frac{s}{2}}{\sigma}\right)\right) \\ &=& \frac{1}{2} + \frac{1}{2} \left(-Q\left(\frac{-s}{2\sigma}\right) + Q\left(\frac{s}{2\sigma}\right)\right) \\ &=& \frac{1}{2} + \frac{1}{2} \left(-Q\left(\frac{-SNR}{2}\right) + Q\left(\frac{SNR}{2}\right)\right) \\ &=& Q\left(\frac{SNR}{2}\right) \end{eqnarray} $$ where $Q(x)$ is the Q function: $$ Q(x) = \frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} e^{\frac{-z^2}{2}} dz $$ (i.e. the tail integral of the standard normal distribution's pdf, or $1$ minus the distribution's cdf) and $SNR$ is the signal-to-noise ratio $\frac{s}{\sigma}$. The above function is a strictly decreasing function of $SNR$; as you increase the ratio of the signal amplitude $s$ to the noise standard deviation $\sigma$, the probability of making a bit decision error decreases. So, it behooves you to do whatever you can to increase this ratio. Remember our assumption that the noise was white and Gaussian? That can help us now. If the noise is white and Gaussian, then the noise components contained in each observation are jointly independent of one another. An important property of independent random variables is that when you sum them together, their means and variances sum. So, let's consider another simple case, where instead of taking one sample per symbol interval, you take two, then sum them together. I'll assume for simplicity that the pulse shape is rectangular (not an exponential rise), so the signal component $s$ in each observation $x_1$ and $x_2$ is the same. What is the difference in signal to noise ratio between just a single observation $x_1$ and the sum of two independent ones? $$ SNR_1 = \frac{s}{\sigma} $$ $$ SNR_2 = \frac{2s}{\sqrt{2 \sigma}} = \sqrt{2}SNR_1 $$ So, the signal to noise ratio in the combined observation is larger than using only a single sample (under the assumption of equal signal component and equal-variance white Gaussian noise in both samples that we took). This is a basic observation that points out the potential benefits of taking more than one sample per symbol interval and integrating them together (which, for a rectangular pulse, is a matched filter). In general, you want to cover the entire symbol interval with samples so that your receiver "ingests" as much of the transmitted energy for each symbol, thus maximizing the SNR in the combined output. The ratio of symbol energy to the background noise variance $\frac{E_s}{N_0}$ is often used as a figure of merit when evaluating digital communications system performance. More rigorously, it can be shown that a matched filter has an impulse response that is identical in shape (that is, "matched", with the only subtle exception being that the impulse response is reversed in time) to the pulse shape that the receiver sees (so it weights more strongly samples that have larger signal components). That shape is a function of the transmitted pulse shape as well as any effects induced by the channel or receiver front end, such as bandlimiting or multipath. To implement this sort of arrangement in practice, you would convolve the stream of samples taken by your ADC with the time-reversed expected pulse shape. This has the effect of calculating the cross-correlation between the pulse shape and the received signal for all possible time offsets. Your implementation is aided by the precise time synchronization that you have available, so you'll know exactly which matched filter output samples correspond to correct sampling instants. The filter outputs at those times are used as the detection statistic $x$ in the theoretical model above. I referred to threshold selection before, which can be a complicated topic, and there are many different ways that you can choose one, depending upon your system's structure. Selecting a threshold for an on-off-keyed system is complicated by the likely-unknown signal amplitude $s$; other signal constellations, like antipodal signaling (e.g. binary phase shift keying, or BPSK) have a more obvious threshold choice (for BPSK, the best threshold is zero for equally-likely data). One simple implementation of a threshold selector for OOK might calculate the mean of many observations. Assuming that zeros and ones are equally likely, the expected value of the resulting random variable is half of the signal amplitude, which is the threshold that you seek. Performing this operation over a sliding window can allow you to be somewhat adaptive to varying background conditions. Note that this is only intended to be a high-level introduction to the issues inherent in digital communications with respect to detection theory. It can be a very complicated topic, with a lot of statistics involved; I tried to make it somewhat easy to understand while keeping true to the underlying theory. For a better explanation, go get a good textbook, like Sklar's.
Statistical Learning - Simple Linear Discriminant Analysis (LDA) Table of Contents 1 - About Linear Discriminant Analysis with only one variable (p = 1). For a generalization, see Statistics - Fisher (Multiple Linear Discriminant Analysis\|multi-variant Gaussian) 2 - Articles Related 3 - Assumption The variance <math>\sigma_k</math> from the distribution of the value <math>X_i</math> when <math>Y_i = k</math> is the same in each of the classes k. It is an important convenience as it's going to determine whether the discriminant function that we get, the discriminant analysis, gives us linear functions or quadratic functions. 4 - Model Construction 4.1 - Gaussian density The Gaussian density has the form: <MATH> f_k(x) = \frac{1}{\sqrt{2\pi}\sigma_k} e^{\displaystyle -\frac{1}{2} \left (\frac{x-\mu_k}{\sigma_k} \right )^2} </MATH> where: <math>\mu_k</math> is the mean in class k <math>\sigma_k</math> is the variance in class k 4.2 - Bayes Formula 4.2.1 - Total Plugging the gaussian density into the Bayes formula, we get a rather complex expression. <MATH> \begin{array}{rrl} Pr(Y = k\|X = x) & = & \frac{\displaystyle Pr(X = x\|Y = k) Pr(Y = k)}{\displaystyle Pr(X = x)} \\ p_k(x) & = & \frac {\displaystyle \pi_k \frac{1}{\sqrt{2\pi}\sigma_k} e^{\displaystyle -\frac{1}{2} \left (\frac{x-\mu_k}{\sigma_k} \right )^2}} {\displaystyle \sum^K_{l=1} \pi_l \frac{1}{\sqrt{2\pi}\sigma_k} e^{\displaystyle -\frac{1}{2} \left (\frac{x-\mu_l}{\sigma_k} \right )^2}} \end{array} </MATH> 4.2.2 - Simplification Luckily, thanks to the assumptions, there's some simplifications and cancellations. To classify an observation to a class, we don't need to initially evaluate the probabilities. We just need to see which is the largest. Whenever you see exponentials the first thing you want to do is take the logs. And if you discard terms that do not depend on k, that amounts to doing a lot of cancellation of terms that don't count. This is equivalent to assigning to the class with the largest discriminant score. <MATH> \delta_k(x) = x.\frac{\mu_k}{\sigma^2}-\frac{\mu_k^2}{2\sigma^2}+log(\pi_k) </MATH> It involves: x, a single variable in this case. the mean <math>\mu_k</math> for the class k the variance <math>\sigma</math> of the distribution the prior <math>\pi_k</math> for the class k And importantly, <math>\delta_k(x)</math> is a linear function of x. There's: a constant <math>\frac{\mu_k}{\sigma^2}</math> a constant term <math>-\frac{\mu_k^2}{2\sigma^2}+log(\pi_k)</math> For each of the classes, we get one of those functions . 4.2.3 - Binary If: there are two classes (K = 2) <math>\pi_1 = \pi_2 = 0,5</math> , you can simplify even further and see that the decision boundary is at <MATH> x = \frac{\mu_1 + \mu_2}{2} </MATH> 5 - Parameters Estimation The priors are just the number in each class divided by the sample size <MATH> \hat{\pi_k} = \frac{\displaystyle N_k}{N} </MATH> The mean for the class k is the sum of all variable when the attribute Y is equal to the class divided by the number of case for this class <MATH> \hat{\mu_k} = \frac{1}{N_k}\sum_{i:y_i=k}x_i </MATH> The notation <math>\displaystyle \sum_{i:y_i=k}</math> will just sum the <math>x_i</math>'s that are in class k. As we're assuming that the variance is the same in each of the classes, this formula is called a pooled variance estimate. <MATH> \begin{array}{rrl} \hat{\sigma}^2 & = & \frac{1}{n-K}\sum_{k=1}^K\sum_{i:y_i=k}(x_i-\hat{\mu}_k)^2 \\ \end{array} </MATH> The formula: subtract from each <math>x_i</math> the mean for its class. (the same as when we compute the variance for the class k) sum all those square differences. sum them over all the classes and then divide it by n minus k. estimate the sample variance separately in each of the classes and then average them in order to weight each of them. The weight has to do with how many observations were in that class relative to the total number of observations. (minus 1 and the minus k is a detail that is to do with how many parameters we've estimated for each of these estimates) A simplified version is: <MATH> \begin{array}{rrl} \hat{\sigma}^2 & = & \sum_{k=1}^K \frac{n_k-1}{n-K}.\hat{\sigma}^2_k \end{array} </MATH> where <math>\hat{\sigma}^2_k</math> is the usual formula for the estimated variance in the kth class ie: <MATH> \begin{array}{rrl} \hat{\sigma}^2_k & = & \frac{1}{n_k-1} \sum_{i:y_i=k} (x_i-\hat{\mu_k})^2 \end{array} </MATH>
Recently it occurred to me that one could potentially design an interactive "compression" scheme which seems to allow for arbitrarily high compression ratios, at the sole expense of time. This already sounds pretty cranky, I realize that, but here is a formal description of the protocol which I hope makes sense: Define a data storeas a tuple $(\mathcal{O}, n, \delta)$ with $\mathcal{O} : \mathbb{N} \to \{ 0, 1\}^n$ a random oracle. We say that the data store "contains" the value $x$ at a (real) time $t \geq 0$ if $\mathcal{O}(\lfloor t \delta^{-1} \rfloor) = x$, such that the contents of the data store randomly change every $\delta$ units of time. A data store could be realized in software using the current time in some agreed-upon representation and any cryptographic hash function. Alice wants to sends a message to Bob encoded as a bit string as part of an interactive protocol (Alice and Bob are both present and active during the transmission). First they both agree on a suitable $\delta$ such that they can both resolve the current time to within $\delta$ units of time. Then, Alice chooses an integer $n \geq 2$, instantiates a data store $(\mathcal{O}, n, \delta)$, and sends these parameters to Bob, and then breaks up the message into $n$-bit blocks. For each block, Alice then monitorsthe data store until its contents equal the block, at which point Alice sends a single bit "1" to Bob. When Bob receives this signal, he looks up the contents of the data store and retrieves the corresponding $n$-bit block (possibly rewinding as needed to account for any transmission delay, this is feasible as long as the transmission delay is fairly constant and known to within $\delta$ units of time). When the last block is sent, Alice sends the bit "0", Bob computes a checksum of the message and sends it to Alice for verification. If the message contained $m$ $n$-bit blocks, the total amount of data actually sent (outside of protocol overhead) was $m$ bits, achieving a compression ratio of $1 - 1/n$, and it would take an expected $m \delta 2^{n}$ time to successfully send it (that is, $\delta 2^{n} / n$ time per bit). Also note that since this scheme doesn't actually use any structure in the message, it can be given already compressed datato compound any already attained compression ratios: the two are not mutually exclusive. Clearly this doesn't allow for amazingly high transmission rates in general. A realistic usecase could be communication with a space probe, where the light-time transmission delay may be accurately computed allowing for a very low $\delta$, say, $\delta = 10 ~ \text{ns}$. Then we can achieve decent compression ratio/transmission rate tradeoffs, e.g. 94% at 3 KB/s ($n = 16$) or 90% at 122 KB/s ($n = 10$). It gets exponentially more time-consuming to send data as $n$ increases, so 95% compression ratio is probably about the best one can practically achieve in reasonable time in the general case. Obviously this only works as an interactive protocol, not as an offline protocol. Also, it doesn't really work "as-is" in real life because you can't just send 1 bit on most channels, as each communication entails a nontrivial amount of overhead (TCP/IP packet, etc..) but in principle I don't see why it couldn't work. So, does anyone know if this kind of "temporal communication" has been studied before, and can point out any fundamental or practical problems or errors I overlooked that might explain why it isn't used?
I want to show that for every deterministic pushdown automaton with the language of the empty stack there is a deterministic pushdown automaton that always halts. my transition function of $P$ is $\delta: \Gamma \times Q \times \Sigma_\lambda \to \Gamma^* \times Q$ I tried to use look ahead pushdown method. I think I should create a $DFA$ for every $q \in Q$ that recognize if $P$ is in the lambda-cycle starting from $q$ with input string reversing of containing the stack of $P$. I define NFA with states of set $\Gamma \times Q$ and it has transition from $(\alpha,p)$ to $(\alpha',p')$ reading $\beta$ if there is $(\beta\alpha',p') \in \delta(\alpha,p,\lambda)$ and start state is $(Z_0,q)$ and I think every state should be final state. then reverse $NFA$ and convert it to $DFA$. I don't know this works or not. I after that I can create $DPDA$ named $P'$ such that its stack simulate all $DFAs$ and also original stack of $P$ and after that I can decide if $P$ will be in the loop in future so in $P'$ it halts. how can I edit creating $DFA$ that works well? Edit I changed the way using above automata, first of all, I design $DFA$ to check if in future there is a sequence of lambda-transitions that empties stack? if there is so I should let process continues because it will stop in future. $NFA$ is like above example with final states in $(\lambda,q)$ for each $q\in Q$ and simply we can convert $NFA$ to $DFA$, if there is not such sequence I ask another question, is it possible that we have some non-lambda transition? checking this is also simple we create exactly same $NFA$ as above $NFA$ and after that, if in a state we have non-lambda transition we go to a special final state. if there is no such transition it means we will stock on loop in future.
Structure of Atom Nature of electromagnetic radiation and Photoelectric effect Nature of Light : Different Theories Newton's Corpuscular Theory : Light comes out of the source as small particles called corpuscles. It traces in straight lines. Huygen's Wave Theory : According to Huygen, light has wave character. He compared light with mechanical waves. Maxwell's Electromagnetic Theory : A reaction which consists of both electric and magnetic components which are perpendicular to each other and perpendicular to direction of propagation of light Wave length (λ) : Units cm/s ; m/sec Frequency(ν) : \tt \nu = \frac{C}{\lambda} units Hz (or) C ps (or) S -1 eg : λ 400nm = 400 × 10 -9 = 4 × 10 -7 m wave number ( \overline{\nu}) : \overline{\nu} = \frac{1}{\lambda} units = cm -1 (or) m -1 Velocity of light (c) : C = 3 × 10 8 m/s (or) 3 × 10 10 cm/sec Amplitude (A) : I ∝ a 2 I = Intensity Electromagnetic spectrum : Photo electric effect : The emission of electrons by a metal surface when a radiation having same frequency h\nu = W + KE \ \ \ \ \ \ \ \ KE = \frac{1}{2}mv^{2} W → Work function (The amount of energy required to free the electron from the metal surface) W = hν 0 → The minimum frequency ν 0 = Threshold frequency → The minimum frequency required to free the electron. Part1: View the Topic in this Video from 33:09 to 56:12 Part2: View the Topic in this video From 0:40 to 43:40 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers. 1. Frequency = \tt V=\frac{C}{\lambda} 2. Wavelength = \tt \lambda=\frac{C}{V} 3. Wave number = \tt \overline{v}=\frac{1}{\lambda}, v=\frac{c}{\lambda}=c\overline{v} 4. Time period \tt T=\frac{1}{\upsilon} 5. No. of waves \tt n=\frac{2\pi r}{\lambda}\left(where \ \lambda=\frac{h}{m\upsilon}\right) 6. No. of revolutions of e - per second is =\tt \frac{\upsilon}{2\pi r}
Alcohols, Phenols and Ethers Methods for Preparation of Alcohols and Phenols Preparations of alcohols: From alkenes: (i) Acid catalysed hydration From alkynes: \tt CH\equiv CH\xrightarrow[kucherer(or)kucherov\ reaction]{H_2O/Hg^{2+},-H^+}CH_3CHO\xrightarrow[\Delta]{O_2/(CH_3COO)_2Mg}CH_3COOH From haloalkanes: Trick: Replacement of H by alkyl groups is related with the formation of 2° and 3°-alcohols. From 1° amines: \tt R-NH_2+O=N-O-H\xrightarrow{NaNO_{2}/HCl}R-OH+H_2O+N_2\uparrow Methanol (wood spirit) from water gas: \tt (CO+H_2)+H_2\xrightarrow[650K/200atm]{Zn/Cr_2O_3/CuO}CH_3OH Ethanol from sugar by fermentation: \tt C_{12}+H_{22}O_{11}(Molasses\ from\ sugar\ industry)\xrightarrow[Invertase]{H_2O/(NH_4)_2SO_4/yeast}C_6H_{12}O_6(Glucose)+C_6H_{12}O_6(Fructose)\xrightarrow{Zymase}C_2H_5OH(ethanol)+CO_2 Preparation of Phenols (Carbolic acid): (a) From benzene: Industrial process: (i) Rasching process: \tt 2C_6H_6(vapour)+2HCl(Gas)+O_2\xrightarrow[-2H_2O]{CuCl_2/FeCl_3/523K}2C_6H_5Cl\xrightarrow[700K]{steam}C_6H_5OH(phenol)+HCl (ii) Cumene process: (b) From chlorobenzene: Industrial process: Dow's method \tt C_6H_5Cl+2NaOH\xrightarrow[-NaCl,-H_2O]{200atm/250^0C}C_6H_5O^\ominus Na^\oplus \xrightarrow{H_2O/H^+}C_6H_5OH (c) Acidic character of some compounds is (a) HCOOH > CH 3COOH > H 2CO 3 > C 6H 5OH > CH 3OH > H 2O > ROH C 1° > 2° > 3° > NH 3 > Acetylene > CH 2=CH 2 > CH 3—CH 3 (b) Part1: View the Topic in this Video from 0:08 to 25:00 Part2: View the Topic in this Video from 0:07 to 24:09 Part3: View the Topic in this Video from 0:08 to 12:23 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers.
Problem statement: The search space A involves elements $\|0\rangle $, $\|2\rangle$... $\|d-1\rangle$. An oracle is provided for the function $f(x)$ where \begin{align} f(x)&=1 \quad x=x^{'}\in A \\ f(x)&=0 \quad x \neq x^{'} \end{align} The aim is to find $x^{'}$. We have an oracle U corresponding to $f(x)$ whose action is the following $$\|x\rangle \|y\rangle \rightarrow \|x\rangle \|y \oplus f(x)\rangle$$ We can consider the search space as the states of a d-level quantum system. If we input the superposition $$ \frac{1}{\sqrt{d}}\sum_{x=0}^{d-1} \|x\rangle \otimes \|\phi\rangle $$ where $\|\phi\rangle=\omega^d\|0\rangle+\omega^{d-1}\|1\rangle+\cdots + \omega\|d-1\rangle$, $\omega=e^{\frac{2\pi i}{d}}$ is the $d^{th}$ root of unity into the oracle corresponding to $f(x)$, using phase kickback we can create the superposition $$ \|\Psi\rangle=\frac{1}{\sqrt{d}}\sum_{x=0}^{d-1} \omega^{f(x)}\|x\rangle. $$ Let us write $\|\Psi\rangle$ as $$ \|\Psi\rangle=\sum_{x=0}^{d-1} c_x\|x\rangle$$ where $c_x=\frac{1}{\sqrt{d}}\omega^{f(x)}$. We can use quantum state tomography (for example using weak value measurements as described in this paper) to find the coefficients $c_x$. The $\|x\rangle$ corresponding to $c_x=\frac{\omega}{\sqrt{d}}$ is the required search result. But, quantum tomography requires an ensemble of the unknown quantum state so as to reduce the statistical error. Suppose $M_0$ is the required size of the ensemble such that the error in the measurement is small enough to determine the search result $x^{'}$. This requires $M_0$ identical copies of $\|\Psi\rangle$. But, as the state $\|\Psi\rangle$ is unknown according to No-Cloning theorem we cannot make copies of $\|\Psi\rangle$ using $\|\Psi\rangle$ alone. Hence, the ensemble has to be made by querying the oracle $M_0$ times with the uniform superposition (4). But, note that $M_0$ is decided on the basis of the acceptable amount of error in the tomography process and hence it does not scale with the size of the search space A. From this can I conclude that the query complexity of this algorithm is O(1)?
First of all, the axiom of countable choice says that given a countable family of non-empty sets, you can choose from each set simultaneously. If you want to choose from one, then from another, then from another, and so on you need a strictly stronger form of choice called Dependent Choice, abbreviated as $\sf DC$. To your questions, the definition of the $\aleph$ numbers uses absolutely no choice, although the proof that any of them is regular does use the axiom of choice (well, except $\aleph_0$). So it is consistent, for example, that $\aleph_1$ is singular. But assuming $\sf DC$ will prevent that. Whether or not you can have every cardinal $\geq\aleph_2$ singular with $\sf ZF+DC$ is still open. The definition of $\beth$ numbers, on the other hand, goes out the window. The axiom of choice is equivalent to saying that the power set of a well-ordered set is well-ordered. So if the axiom of choice fails, there will be some ordinal $\alpha$ whose power set cannot be well-ordered. This means that at some point $\beth$ cardinals will not be $\aleph$ numbers. You can still talk about $\beth$ numbers, of course, as iterated powers of $\omega$, but that gives you significantly less information in that sense. As for the continuum hypothesis, as noted by others it is still unprovable. But now you even get different forms of the continuum hypothesis. $2^{\aleph_0}=\aleph_1$ is no longer equivalent to "Every uncountable set of reals is equipollent with the reals themselves". Indeed, even without $\sf DC$, it is consistent that $2^{\aleph_0}$ and $\aleph_1$ are incomparable, and every uncountable set of reals has cardinality $2^{\aleph_0}$. Finally, for large cardinals, most properties which are reserved for very large cardinals can be made compatible with $\aleph_1$ and $\sf ZF+DC$. For a more complete survey, look at What sort of large cardinal can $\aleph_1$ be without the axiom of choice?.
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
In order to preform nearest neighbour search (NN) on large and high dimensional datasets you must overcome 2 big problems: ineffectiveness of distance measures in high dimensions computational intractability of exhaustive search on large datasets By employing hash tables w/ a novel hashing function that optimises for collisions we reduce the search space and mitigate the curse of dimensionality for NN search. Figure 1: The curse of dimensionality visualized: In high dimensions, every point is equally far away!. Hash Function Given a vector $x$ in $\mathbb{R^d}$ and a projection matrix $\mathrm{P}$ in $\mathbb{R^{r\times d}}$ we can formulate a hashing function $h(x)$: where $h(x)$ is a vector in the $\mathbb{R^r}$ space and $\mathrm{P}$ is random matrix parameterised by $r$ $(r\ll d)$. The key here is that $x$ is referenced in a smaller dimension than before $r$ vs $d$ . Hash Table Given a hashing function that maximises collisions and a dictionary data structure we can build a hash table $D$: where $D$ is a mapping of $h(x)$ to $[x_1,…,x_n]$ where each $x$ has a degree of similarity. Formally, we can find $x$ in $D$ in $O(1)$ compared to $O(n)$ for exhaustive search. Nearest Neighbour In order to preform NN in the pathological case, it would be $O(n)$ search w/ no upper bound on $n$. However, in the LSH case, we can put an upper bound on $n$ and guarantee that: $n_{LSH}\ll n_{NN}$ as we are guaranteed to never have to iterate the whole search space: Reduce search space by querying $\ell \leftarrow D : h(x)$ Preform classical NN search on the drastically reduced search space $\ell$ where the upper bound on $n$ is $\ell$ Conclusion In order to preform NN on large and high dimensional datasets you must overcome 2 big problems: ineffectiveness of distance measures in high dimensions computational intractability of exhaustive search on large datasets By employing hash tables w/ a novel hashing function that optimises for collisions we reduce the search space and mitigate the curse of dimensionality for NN search. Permalink: locality-sensitive-hashing-lsh Tags:
Moving charges and magnetism Magnetic Field on the Axis of a Circular Current Loop, Ampere’s Circuital Law The magnetic field at the centre of the coil is \tt B = \frac{\mu_{0} ni}{2r} The magnetic field at the centre when conductor is bent in the form of a coil. \tt B = \frac{\mu_{0} ni}{4 \pi r} (\theta) (θ in radius) If the wire is in the form of semi-circle \tt B = \frac{\mu_{0} i}{4 \pi r} (\pi) = \frac{\mu_{0}i}{4 r} For comparing magnetic fields \tt \frac{B_{1}}{B_{2}} = \left(\frac{n_{1}}{n_{2}}\right)^{2} = \left(\frac{r_{1}}{r_{2}}\right)^{2} If two circular coils are connected in series. \tt \frac{B_{1}}{B_{2}} = \left(\frac{n_{1}}{n_{2}}\right) \left(\frac{r_{1}}{r_{2}}\right) If two coils are connected in parallel. \tt \frac{B_{1}}{B_{2}} = \left(\frac{r_{2}}{r_{1}}\right)^{2} If two concentric circular loops carrying currents i 1, i 2and their planes are inclined at an angle θ. The resultant magnetic induction at the common centre is \tt BR = \sqrt{B_{1}^{2} + B_{2}^{2} + 2 B_{1}B_{2} \cos \theta } If θ = 0 ⇒ B = B 1+ B 2and If θ = 90° then \tt B = \sqrt{B_{1}^{2} + B_{2}^{2}} Null point in the following situation \tt x = \frac{d}{\left(\frac{i_{1}}{i_{2}}\right)^{1/3} \pm 1} Magnetic field due to straight conductor of finite length at a perpendicular distance “r” is \tt B = \frac{\mu_{0} i}{4 \pi r} (\sin \theta_{1} + \sin \theta_{2}) Magnetic induction at a point due to a straight conductor of infinite length at ‘r’ \tt B = \frac{\mu_{0} i}{2 \pi r} If the point is ‘φ’ along the length of the conductor at that point B = 0 “B” at a perpendicular distance r for infinite wire is \tt B = \frac{\mu_{0} i}{4 \pi r} ‘B’ at a perpendicular distance r for a finite conductor is \tt B = \frac{\mu_{0} i}{4 \pi r} \frac{l}{\sqrt{l^{2} + r^{2}}} Ampere’s circuital law used for calculating the magnetic field under highly symmetrical conditions and is given by \tt \oint B . dl = \mu_{0} i Force acting on a magnetic pole (m) kept at a distance “r” from a infinite current carrying conductor is \tt F = \frac{\mu_{0} i}{2 \pi r} m If magnetic pole ‘m’ is n times moved around the conductor were done \tt \oint \overline{f} . \overline{d} s = \frac{\mu_{0} i m}{2 \pi r} . 2 \pi r = \mu_{0} i m n The magnetic field induction inside the current carrying vary long solid cylinder at a distance ‘r’ is \tt B = \frac{\mu_{0} i r}{2 \pi R^{2}} (R = Radius of the conductor) (r < R) \tt B = \frac{\mu_{0} i }{2 \pi R} (r > R) Magnetic induction at the centre of current carrying wire bent in the form of a square “l” is \tt B = 8\sqrt{2} \frac{\mu_{0}}{4 \pi} \left(\frac{i}{l}\right) Magnetic induction at the centre of current carrying wire bent in the form of a triangle of side “l” is \tt B = 18 \frac{\mu_{0} i}{4 \pi l} Magnetic induction at the centre of current carrying wire bent is the form of hexagon of side ‘l’ is \tt B = 4\sqrt{3} \frac{\mu_{0}}{4 \pi} \left(\frac{i}{a}\right) \tt B_{P} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{1}} - \frac{i_{2}}{(r - r_{1})}\right] \tt B_{Q} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{2}} + \frac{i_{2}}{(r + r_{2})}\right] \tt B_{P} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{1}} + \frac{i_{2}}{(r - r_{1})}\right] \tt B_{Q} = \frac{\mu_{0}}{2 \pi} \left[\frac{i_{1}}{r_{2}} + \frac{i_{2}}{(r + r_{2})}\right] Null point position \tt S_{1} = \frac{r}{\frac{i_{2}}{i_{1}}+1} Null point position \tt S_{2} = \frac{r}{\frac{i_{2}}{i_{1}}-1} Ampere's law View the Topic in this video From 01:44 To 41:34 View the Topic in this video From 41:14 To 55:45 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers. 1. Ampere's Circuital Law It states that the line integral of magnetic field around any closed path in vacuum is equal to μ 0 times the total current passing the closed path. \oint \overrightarrow{B}.\overrightarrow{dl} = \mu_{0}l 2. Magnetic field at a point outside the wire i.e. ( r > a) is B = \frac{\mu_{0}I}{2 \pi r}
I want solve the next exercise. The author defined the experiment for the cryptosystem $\Pi$, the adversary $A$ and the security parameter $n$ as follows $\mathsf{PRIV_{EAV}}(\Pi,A,n)$ The adversary $A$ get the input $1^n$ and choose two messages $m^{(0)}$, $m^{(1)} \in M$ with same length; One key $k \in K$ is generated using $\mathsf{Gen}(1^n)$, and one bit is chosen uniformly at random. Then the message $m^{(b)}$ is encrypted and sent to $A$; Eventually $A$ outputs b'; If $b=b'$, the experiment result is 1. Otherwise 0. Definition A cryptosystem $\Pi$ has indistiguishable ciphertexts in the presence of an eavesdropper if for all adversaries $A$ there exist a negligible function $\mathsf{negl}$ such that $$\Pr[\mathsf{PRIV_{EAV}}(\Pi,A,n)=1]\leq 1/2 + \mathsf{negl}(n).$$ My question: How will I be able to prove that the cryptosystem described below is not secure under the definition above. Cryptosystem $\Pi$: $\mathsf{Gen}(1^n)$ chooses a bitstring $k \in \{0,1\}^n$ uniformly at random, $\mathsf{Enc}(m,k)$ works as follows: The message $m$ is split into blocks $m_0, m_1, \dots ,m_l$ of $n$ bits. Each block is encrypted separately, as $c_i = m_i \oplus k$. The ciphertext is then the concatenation of the encrypted blocks $c_0, c_1, \dots ,c_l$; $\mathsf{Dec}(c,k)$ is analogous to Enc, but permuting $c$ and $m$. Notation $K$ is the key space, $M$ the message space and $C$ the ciphertex space
Let $R$ be a ring with $1$ and let $M$ be an $R$-module. Let $I$ be an ideal of $R$.Let $M’$ be the subset of elements $a$ of $M$ that are annihilated by some power $I^k$ of the ideal $I$, where the power $k$ may depend on $a$.Prove that $M’$ is a submodule of $M$. Let $R$ be a ring with $1$. Let $M$ be an $R$-module. Consider an ascending chain\[N_1 \subset N_2 \subset \cdots\]of submodules of $M$.Prove that the union\[\cup_{i=1}^{\infty} N_i\]is a submodule of $M$. (a) Let $R$ be a commutative ring. If we regard $R$ as a left $R$-module, then prove that any two distinct elements of the module $R$ are linearly dependent. (b) Let $f: M\to M’$ be a left $R$-module homomorphism. Let $\{x_1, \dots, x_n\}$ be a subset in $M$. Prove that if the set $\{f(x_1), \dots, f(x_n)\}$ is linearly independent, then the set $\{x_1, \dots, x_n\}$ is also linearly independent. Let $R$ be a ring with $1$ and let $M$ be a left $R$-module.Let $S$ be a subset of $M$. The annihilator of $S$ in $R$ is the subset of the ring $R$ defined to be\[\Ann_R(S)=\{ r\in R\mid rx=0 \text{ for all } x\in S\}.\](If $rx=0, r\in R, x\in S$, then we say $r$ annihilates $x$.) Suppose that $N$ is a submodule of $M$. Then prove that the annihilator\[\Ann_R(N)=\{ r\in R\mid rn=0 \text{ for all } n\in N\}\]of $M$ in $R$ is a $2$-sided ideal of $R$. Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$.The set of torsion elements is denoted\[\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.\] (a) Prove that if $R$ is an integral domain, then $\Tor(M)$ is a submodule of $M$.(Remark: an integral domain is a commutative ring by definition.) In this case the submodule $\Tor(M)$ is called torsion submodule of $M$. (b) Find an example of a ring $R$ and an $R$-module $M$ such that $\Tor(M)$ is not a submodule. (c) If $R$ has nonzero zero divisors, then show that every nonzero $R$-module has nonzero torsion element. Let $R$ be a ring with $1$ and $M$ be a left $R$-module. (a) Prove that $0_Rm=0_M$ for all $m \in M$. Here $0_R$ is the zero element in the ring $R$ and $0_M$ is the zero element in the module $M$, that is, the identity element of the additive group $M$.To simplify the notations, we ignore the subscripts and simply write\[0m=0.\]You must be able to and must judge which zero elements are used from the context. (b) Prove that $r0=0$ for all $s\in R$. Here both zeros are $0_M$. (c) Prove that $(-1)m=-m$ for all $m \in M$. (d) Assume that $rm=0$ for some $r\in R$ and some nonzero element $m\in M$. Prove that $r$ does not have a left inverse. Let $A$ be an $n\times n$ matrix. Suppose that $\mathbf{y}$ is a nonzero row vector such that\[\mathbf{y}A=\mathbf{y}.\](Here a row vector means a $1\times n$ matrix.)Prove that there is a nonzero column vector $\mathbf{x}$ such that\[A\mathbf{x}=\mathbf{x}.\](Here a column vector means an $n \times 1$ matrix.) Recall that a complex matrix is called Hermitian if $A^=A$, where $A^=\bar{A}^{\trans}$.Prove that every Hermitian matrix $A$ can be written as the sum\[A=B+iC,\]where $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix. Let $A$ be an $n\times n$ matrix. Suppose that $A$ has real eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$ with corresponding eigenvectors $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$.Furthermore, suppose that\[\|\lambda_1\| > \|\lambda_2\| \geq \cdots \geq \|\lambda_n\|.\]Let\[\mathbf{x}_0=c_1\mathbf{u}_1+c_2\mathbf{u}_2+\cdots+c_n\mathbf{u}_n\]for some real numbers $c_1, c_2, \dots, c_n$ and $c_1\neq 0$. Define\[\mathbf{x}_{k+1}=A\mathbf{x}_k \text{ for } k=0, 1, 2,\dots\]and let\[\beta_k=\frac{\mathbf{x}_k\cdot \mathbf{x}_{k+1}}{\mathbf{x}_k \cdot \mathbf{x}_k}=\frac{\mathbf{x}_k^{\trans} \mathbf{x}_{k+1}}{\mathbf{x}_k^{\trans} \mathbf{x}_k}.\] Prove that\[\lim_{k\to \infty} \beta_k=\lambda_1.\]
Continuation 2 has been added, with seemingly ideal results. Ben Aveling’s solution.inspired three new solutions (“Continuations”)to this surprisingly playful puzzlethat is interestingly reminiscent ofBenford’sandZipf’slaws.$\require{begingroup} \begingroup \def\safeMathJax{\text{\endgroup error}} \def \p {{\kern 1mu p}} \def \r {{\large r}} \def \x {{x \kern1mu}} \def \P {{ \kern1mu p \kern2mu}} \def \s{{ \large s}} \def \= {~ = ~} \def \* {\kern2mu \cdot \kern1mu}$ Beginning with an untouched pizza, I will... ...make a cut from anywhere on the edge of the pizza to the center. With that taken care of, more than one way to continue have come to mind. Continuation 1 — Based on error correction, simplified to introduce analysis. Now to write a note to my fellow pizza partier/prisoners. Dear fellow pizza partier/prisoner, You are invited to... 1. Find the largest slice. If you see only one cut in the pizza, consider the whole thing to be the largest slice. 2. Measure that slice’s angle in degrees and call it $A \LARGE\raise-.3ex\strut$. 3. Cut an angle of $\dfrac {\raise-2mu 1} { \frac{2}{A} + \frac{{\large\ln}\,2}{720} \, }$ degrees into the slice. 4. By the way, $ \small \dfrac{\raise-2mu{\ln2}}{720} \approx \, 0.0009627044174443684853... $ With love, fellow pizza partier/prisoner. If my fellow pizza partier/prisoners make sense of that,this is how the first nine cuts should go. \| \| After my cut \| 360.0° \| \| \| After prisoner 2 \| 153.4° \| 206.6° \| \| \| \| After prisoner 3 \| \| 94.0° \| 112.6° \| \| \| \| \| After prisoner 4 \| 71.4° \| 82.0° \| \| \| \| \| \| \| \| After prisoner 5 \| \| \| \| 53.4° \| 59.2° \| \| \| \| \| \| \| After prisoner 6 \| \| \| 44.9° \| 49.0° \| \| \| \| \| \| \| \| \| \| After prisoner 7 \| \| 39.4° \| 42.5° \| \| \| \| \| \| \| \| \| \| \| \| \| After prisoner 8 \| 34.5° \| 36.9° \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| After prisoner 9 \| 34.5° \| 36.9° \| 39.4° \| 42.5° \| 44.9° \| 49.0° \| 53.4° \| 28.8° \| 30.4° \| . \| \| \| \| \| \| \| \| \| \| . \| . . . . . . . . \| . \| . . . . . . . . \| Now for some analysis while waiting for pizza, or execution.Begin with an ideal sequence of cuts where slices are re-cutin the same order as they were produced.The sequence above begins ideallybut hasn’t been examined thoroughly for monotonicity. .---. : :.------. 360° 360° : SLICE SIZES AS A LEAPFROGGING SEQUENCE \| \| .-'-. \| \| : : \| \| :.----:-------. Each prisoner p finds p-1 \| \| 207° : .-'-. slices and divides the slice \| \| \| : : : of size s_p into slices of \| \| \| :.----:-----:-------. size s_(2p-1) and s_2p . \| \| \| 153° : : : \| \| \| \| :.----:--------:----------. \| \| \| \| 113° : : : \| \| \| \| \| :.-------:-----------:-----------. \| \| \| \| \| 94° : : : \| \| \| \| \| \| .-'-. : : \| \| \| \| \| \| : : .-'-. : \| \| \| \| \| \| 82° : : : : \| \| \| \| \| \| \| 71° : : .-'-. \| \| \| \| \| \| \| \| 59° 53° : : \| \| \| \| \| \| \| \| \| \| 49° 45° . . . \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| s_1 s_2 s_3 s_4 s_5 s_6 s_7 s_8 s_9 s_10 s_11 s_12 . . . \| \| \| s_p = s_(2p-1) + s_2p For aesthetics as much as fairness, the sequence ofslice sizes (angles), $\s_p \kern2mu$, would ideally form a smooth curve.For convenience, how about a slice-size function, $\s(\p)$,that approximates $\s_p$ andsatisfies $\s(\p) \approx \s(2\P{-}1) + \s(2\p) \,$? As such, prisoner $p$ arrives around time = $p$and divides a slice of size $\s(\p)$ into two slices intended to be re-cutby prisoners $2\P{-}1$ and $2\p$ around time = $2\p$. $$ \s_p= \s(\p) = \dfrac {\large\tfrac{360}{\ln2}} {\, \P-\tfrac12 ~} $$ That’s just $\dfrac{\raise-2mu 1}{\, \raise4mu p \,}$, scaled and shifted so that all current slices add up to a full 360° of pizza. $$ \sum_{\P+1}^{2\p} \s_i ~\approx \int_{\P+\frac12}^{2\p+\frac12} \!\! \s(x)dx ~\approx~ 360^\circ $$ If at any stage the largest slice’s angle to be divided, $A$,happens to be larger or smaller than ideal,it may be treated as merely being out of place in the sequence.Thus every slice’s ideal position, $x$,is deduced such that $\s(x)=A$, before dividing $A$into slices with angles $\s(2x)$ and $A{-}\s(2x)$. \begin{array}{rrl}{} & A \kern-1em{} & \= \, \s(x) \,{} \= \dfrac{\large\tfrac{360}{\ln2}} {\, x-\tfrac12 ~}{} \\[1ex]{} \Longrightarrow & x \kern-1em{} & \= {\large\tfrac{360}{A \ln2}} + \tfrac12{} \\[2ex]{} \Longrightarrow & \s(2x) \kern-1em{} & \= \dfrac {\large\tfrac{360}{\ln2}} {\, 2x-\tfrac12 ~}{} \= \dfrac {\large\tfrac{360}{\ln2}} {\,{} {\large\tfrac{720}{A \ln2}} + \tfrac12 ~}{} \= \dfrac {1} {\, {\large\tfrac{2}{A}}{} + {\large\tfrac{\ln2}{720}} ~}{} \end{array} That last calculation is what was recommended in the note to other prisoners. This approach was designed to be relatively simple to calculate.It is also meant to be good at correcting mistakes in earlier slices,which will be tested in the section for Continuation 3. Continuation 2 — Each cut is based on the number of slices. Steps 2 through 4 of Continuation 2’s note to fellowpizza partier/prisoners are: 2. Count the number of slices so far and call that number $n \raise-2ex\strut$. 3. Cut the largest slice into two pieces whose angles have a ratio of $\ln(2n{+}1)-\ln 2n ~$ to $\ln(2n{+}2)-\ln(2n{+}1)$. 4. You can do it, pizza pal, just remember that $ \ln x \= \x{-}1 - \frac{(\x{-}1)^2}2 + \frac{(\x{-}1)^3}3 - \cdots $ And here is how the first nine cuts would go this time. \| \| After my cut \| 360.0° \| \| \| After prisoner 2 \| 210.6° \| 149.4° \| \| \| \| After prisoner 3 \| 115.9° \| 94.7° \| \| \| \| \| \| After prisoner 4 \| \| \| 80.1° \| 69.4° \| \| \| \| \| \| After prisoner 5 \| 61.2° \| 54.7° \| \| \| \| \| \| \| \| \| \| After prisoner 6 \| \| \| 49.5° \| 45.2° \| \| \| \| \| \| \| \| \| \| After prisoner 7 \| \| \| \| \| 41.6° \| 38.5° \| \| \| \| \| \| \| \| \| \| After prisoner 8 \| \| \| \| \| \| \| 35.8° \| 33.°5 \| \| \| \| \| \| \| \| \| \| After prisoner 9 \| 31.5° \| 29.7° \| 54.7° \| 49.5° \| 45.2° \| 41.6° \| 38.5° \| 35.8° \| 33.5° \| . \| \| \| \| \| \| \| \| \| \| . \| . . . . . . . . \| . \| . . . . . . . . \| This comes from howBen Aveling figured out a wayto obtain a very fair result by basing each cutsolely on the number of slices cut so far.Extending the terms used here, Ben Aveling’s solution is... $$ \r_p \= \frac{\s_{2\P-1}}{\s_{2\p}} \= \frac{n+1}{n} $$ ...prisoner $p$ divides slice $\s_p$into slices $\s_{2\P-1}$ and $\s_{2\p}$to have a size ratio, $\r_p$,based on the number of slices so far, $n$.Incidentally, $n = \P{-}1$. A plot of $\log\sum\kern-1mu\textsf{error}\kern1mu^2$shows how much smoother this is than Continuation 1or straightforward halving of the largest slice. Each $\sum\kern-1mu\textsf{error}\kern1mu^2$ reflectsthe unfairness when there are $n$ slicesand each of those slices has an $\textsf{error}$that equals the difference between its sizeand the $n$ slices’ average size. This plot also shows how fluctuations repeat at intervals that keep doubling. The fluctuations also reveal thatdifferent approaches take turns being better than the others. Along comes Continuation 2, a tweak on Ben Aveling’s approach,to produce a fluctuation-free(!) error graph. And along comes Continuation 2’s derivation of $\r_p$. The fluctuations in the first plotmake clear how any cut’s inaccuracy is echoedamong an infinite cascade of future cuts,which is part of the fun challenge in this puzzle.Writing out the consequences of $\s_3$ and $\s_4$is enough to establish a general formulafor $\s_p$ and hence for an ideally smooth $\r_p$. \begin{array}{rl}{} \s_3 \kern-1em{} & \= \s_5 + \s_6{} \= (\s_9 + \s_{10}) + (\s_{11} + \s_{12}){} \\[-1ex]{} & \= \s_{17}+\s_{18} + \s_{19}+\s_{20} + \s_{21}+\s_{22} + \s_{23}+\s_{24}{} \= \cdots{} \= {\displaystyle \lim_{i\to\infty} \int_{\large 2\2^i{+}1}^{\large 3\2^i} \!\! \s(x)dx}{} \\[1ex]{} & \= \frac{360}{\ln2} (\ln3-\ln2){} \\[4ex]{} \s_4 \kern-1em{} & \= \s_7 + \s_8{} \= (\s_{13}+\s_{14}) + (\s_{15}+\s_{16}){} \\[-1ex]{} & \= \s_{25}+\s_{26} + \s_{27}+\s_{28} + \s_{29}+\s_{30} + \s_{31}+\s_{32}{} \= \cdots{} \= {\displaystyle \lim_{i\to\infty} \int_{\large 3\2^i{+}1}^{\large 4\2^i} \!\! \s(x)dx}{} \\[1ex]{} & \= \frac{360}{\ln2} (\ln4-\ln3){} \\[5ex]{} \s_p \kern-1em{} & \= \frac{360}{\ln2} \big( \ln p - \ln(\P-1) \, \big){} \end{array} $ \begin{array}{rrl}{} \Longrightarrow & \r_p \kern-1em{} & \= \dfrac {\s_{2\P-1}} {\s_{2\p}}{} \= \dfrac {\ln(2\P{-}1)-\ln(2\P{-}2)} {\ln2\p-\ln(2\P{-}1)}{} \\[2ex]{} & & \= \dfrac {\ln(2n{+}1)-\ln 2n} {\ln (2n{+}2)-\ln(2n{+}1)}{} \end{array} $ Continuation 3 — More accurate version of error-correcting Continuation 1. Being refined. $\endgroup$
Prove that if $d_1$ and $d_2$ are metrics over $X$, $\tau_1$ and $\tau_2$ are the family of open subsets of their respective metric spaces then: (i) $\implies$ (ii) $\iff$ (iii) Where: (i) There exists a $c>0$ that for any $x,y \in X$ we have $c \cdot d_1(x,y) \geq d_2(x,y)$ (ii) For any $x \in X$ and $r > 0 $, there exists a $r'>0$ where $B_{d_1}(x,r') \subset B_{d_2}(x,r)$ (iii) $\tau_2 \subset \tau_1$ If (i) holds, then for given $x \in X$ and $r>0$, define $r'=\frac{r}{c}$. Then if $y \in B_{d_1}(x,r')$ then $d_1(x,y) < r'$. Also by (i): $$d_2(x,y) \le c\cdot d_1(x,y) < c \cdot r' = c \cdot \frac{r}{c}=r$$ so that $y \in B_{d_2}(x,r)$ , showing the inclusion. and so $\text{(i)} \implies \text{(ii)}$ holds. The equivalence of (ii) and (iii) is quite obvious from the definitions. What does $O \in \tau_2$ mean? Also recall that open balls are open sets in their induced topologies.
Problem 616 Suppose that $p$ is a prime number greater than $3$. Consider the multiplicative group $G=(\Zmod{p})^$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$. Add to solve later (c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. Problem 613 Let $m$ and $n$ be positive integers such that $m \mid n$. (a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective. Add to solve later (d) Determine the group structure of the kernel of $\phi$. Problem 612 Let $C[-2\pi, 2\pi]$ be the vector space of all real-valued continuous functions defined on the interval $[-2\pi, 2\pi]$. Consider the subspace $W=\Span\{\sin^2(x), \cos^2(x)\}$ spanned by functions $\sin^2(x)$ and $\cos^2(x)$. (a) Prove that the set $B=\{\sin^2(x), \cos^2(x)\}$ is a basis for $W$. Add to solve later (b) Prove that the set $\{\sin^2(x)-\cos^2(x), 1\}$ is a basis for $W$. Problem 611 An $n\times n$ matrix $A$ is called orthogonal if $A^{\trans}A=I$. Let $V$ be the vector space of all real $2\times 2$ matrices. Consider the subset \[W:=\{A\in V \mid \text{$A$ is an orthogonal matrix}\}.\] Prove or disprove that $W$ is a subspace of $V$. Problem 607 Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less. Let \[S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},\] where \begin{align} p_1(x)&=1+3x+2x^2-x^3 & p_2(x)&=x+x^3\\ p_3(x)&=x+x^2-x^3 & p_4(x)&=3+8x+8x^3. \end{align*} (a) Find a basis $Q$ of the span $\Span(S)$ consisting of polynomials in $S$. (b) For each polynomial in $S$ that is not in $Q$, find the coordinate vector with respect to the basis $Q$. Add to solve later (The Ohio State University, Linear Algebra Midterm) Read solution Problem 606 Let $V$ be a vector space and $B$ be a basis for $V$. Let $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ be vectors in $V$. Suppose that $A$ is the matrix whose columns are the coordinate vectors of $\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5$ with respect to the basis $B$. After applying the elementary row operations to $A$, we obtain the following matrix in reduced row echelon form \[\begin{bmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) What is the dimension of $V$? (b) What is the dimension of $\Span\{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \mathbf{w}_4, \mathbf{w}_5\}$? Add to solve later (The Ohio State University, Linear Algebra Midterm) Read solution Problem 605 Let $T:\R^2 \to \R^3$ be a linear transformation such that \[T\left(\, \begin{bmatrix} 3 \\ 2 \end{bmatrix} \,\right) =\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \text{ and } T\left(\, \begin{bmatrix} 4\\ 3 \end{bmatrix} \,\right) =\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}.\] (a) Find the matrix representation of $T$ (with respect to the standard basis for $\R^2$). (b) Determine the rank and nullity of $T$. Add to solve later (The Ohio State University, Linear Algebra Midterm) Read solution Problem 604 Let \[A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.\] (a) Find a basis for the null space $\calN(A)$. (b) Find a basis of the range $\calR(A)$. (c) Find a basis of the row space for $A$. Add to solve later (The Ohio State University, Linear Algebra Midterm) Read solution Problem 603 Let $C[-2\pi, 2\pi]$ be the vector space of all continuous functions defined on the interval $[-2\pi, 2\pi]$. Consider the functions \[f(x)=\sin^2(x) \text{ and } g(x)=\cos^2(x)\] in $C[-2\pi, 2\pi]$. Prove or disprove that the functions $f(x)$ and $g(x)$ are linearly independent. Add to solve later (The Ohio State University, Linear Algebra Midterm) Read solution Problem 601 Let $V$ be the vector space of all $2\times 2$ matrices whose entries are real numbers. Let \[W=\left\{\, A\in V \quad \middle \| \quad A=\begin{bmatrix} a & b\\ c& -a \end{bmatrix} \text{ for any } a, b, c\in \R \,\right\}.\] (a) Show that $W$ is a subspace of $V$. (b) Find a basis of $W$. (c) Find the dimension of $W$. Add to solve later (The Ohio State University, Linear Algebra Midterm) Read solution
Yesterday I came up with this problem and I just cannot get it out of my head: Problem: Consider a finite $I$ set of integer-valued intervals $I_i=(a_i,b_i)$ $i = 1,\ldots,n; a_i \in \mathbb{N_0}, k \in \mathbb{N}, I=( I_1,\ldots,I_n\ )$ Assume there is a map $\phi: I \longrightarrow \{1,\ldots,k\}$ such that $$(): \{l, m\} \subset \{1,\ldots,n\}, l \neq m, \phi(I_l)=\phi(I_m) \Rightarrow I_l \cap I_m=\emptyset$$ Now if $M=(x,y)$ is an inteval such that $(): M\cap\left(\bigcap_{i=1}^k\left(\bigcup\{I_j\|\phi(I_j) = i\}\right)\right)=\emptyset$. then $M$ can be always appended to $I$ with a new map $\phi'$ such that property $()$ holds. Example: $k=3$, $I=( (0,2), (3,5), (2,4), (5,7), (0,3), (4,6) )$, $\phi(I)=( 1,1,2,2,3,3)$ This corresponds to the following image (edit: the $k$ on the left side is a bit misleading, better ignore it): As you can see the property $()$ is fullfilled (no intervals are overlapping in their time line). We can also see that for the interval $(2,4)$ he have the property $(*)$ i.e. we find spaces to put it into the image: Now the theorem tells us that there is a new image where the green interval does not have to be broken. And indeed we find $\phi(I \cup (2,4))=(1,2,3,2,2,2,1,1)$. Or in the image: I hope I could explain the problem well. I am not sure if the theorem is correct but I tried quite a few examples and it always worked out.
I'm doing some classical field theory exercises with the Lagrangian $$\mathscr{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$$ where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. To find the conjugate momenta $\pi^\mu_{\ \ \ \nu} = \partial \mathscr{L} / \partial(\partial_\mu A^\nu)$, I can use two methods. First method: directly apply this to $\mathscr{L}$. We get a a factor of $2$ since there are two $F$'s, and another factor of $2$ since each $F$ contains two $\partial_\mu A_\nu$ terms, giving $$\pi^\mu_{\ \ \ \nu} = -F^\mu_{\ \ \ \nu}.$$ Second method: get $\mathscr{L}$ in terms of $A$ by expanding and integrating by parts, yielding $$\mathscr{L} = \frac{1}{2}(\partial_\mu A^\mu)^2 - \frac{1}{2}(\partial_\mu A^\nu)^2.$$ Differentiating this gets factors of $2$ and gives $$\pi^\mu_{\ \ \ \nu} = \partial_\rho A^\rho \delta^\mu_\nu - \partial^\mu A_\nu.$$ These two answers are different! (They do give the same equations of motion, at least.) I guess that means doing the integration by parts changed the canonical momenta. Is this something I should be worried about? In particular, I have another exercise that wants me to show that one of the canonical momenta vanishes -- this isn't true for the ones I get from the second method! Plus, my stress-energy tensor is changed too. When a problem asks for "the" canonical momenta, am I forbidden from integrating by parts?
Chemical Bonding and Molecular Structure Fajan's rule and Dipole moment Fajan's Rule : (For covalent bond formation) smaller is the cation and larger is the anion covalent character ↑ Larger cation, smaller anion ionic character ↑ Charge on both cation and anion ↑ covalent character ↑ Cations with pseudo inert gas configuration covalent character ↑ Dipole moment : used to express the polarity of molecule \mu = \delta \times l , H^{\delta+} \xrightarrow[]{l} Cl^{\delta-} δ = magnitude of charge on poles \mu_{R} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_{1}\mu_{2} \ cos\theta} Part1: View the Topic in this Video from 0:07 to 4:25 Part2: View the Topic in this Video from 0:12 to 10:50 Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers. 1.Formal charge = \tt V-\begin{bmatrix}L+\frac{1}{2}S \end{bmatrix} V= Number of valence electrons in free state L= Number of lone pair of electrons that is total no. of e − of lone pair S=No. of shared electrons 2. Pauling Equation:%Ionic character = 18 (X A - X B) 1.4 E of A > B in A-B bond 3. Hannay and Smith Equation : % Ionic character = 16 (X A - X B) + 3.5 (X A - X B) 2 in A-B bond 4. μ=δ × d CGS units SI units δ 10 -10 esu 10 -19 Coulomb d 10 -8 cm 10 -10 m μ 1D= 10 -18 esu cm 1D = 3.33 × 10 -30 Cm 5. Percentage ionic character = \tt \frac{observed \ dipole \ moment}{dipole \ moment \ for \ 100\% \ ionic \ bond}\times 100 \ = \frac{\mu_{obs}}{\mu_{ionic}} \times 100
This formula can more insightfully be written as $$ p = \frac{1}{e^{\dfrac{\Delta E}{T}}}$$ where, in the context of SA, $\Delta E$ is the difference between the "energy" (or quality) of two possible solutions and $T$ is the temperature (which is either a constant or a function of the input size). So, the higher the energy difference $\Delta E$, the smaller the probability $p$. In the context of SA, if the cost of the current solution is greater than the cost of the new solution, then you replace the current solution with the new solution. However, if the cost of the current solution is smaller than the new solution, then you do not automatically discard the new solution. More concretely, with probability $p$, you accept the new worse solution. This the stochastic action taken by SA. So, the higher the diffence between the costs of the current and new solutions, the smaller the probability of accepting the new worse solution. The Boltzman distribution only happens to capture this idea of accepting new worse solutions, provided that they are not extremely worse than the current solution. In theory, you can use other (similar) distributions.
I am reading through the Computer Vision: Models, Learning, and Inference book to get an understanding of computer vision. The author describes the high-level steps taken to arrive at one of the derivations, but I'm stuck in working through the details. An excerpt from the book (available at above link), describing the problem: "Problem 4.3 Taking equation 4.29 as a starting point, show that the maximum likelihood parameters for the categorical distribution are given by$$\hat{\lambda_k} = \frac{N_k}{\sum_{m=1}^{6}N_m}$$where $N_k$ is the number of times that category $K$ was observed in the training data. Equation 4.29$$L = \sum_{k=1}^{6} N_k \log[\lambda_k] + v\Bigg(\sum_{k=1}^{6}\lambda_k - 1 \Bigg)$$where the second term uses the Lagrange multiplier $v$ to enforce the constraint on the parameters $\sum_{k=1}^{6}\lambda_k=1$". The high-level description of how to solve this is given: "We differentiate $L$ with respect to $\lambda_k$ and $v$, set the derivatives to zero, and solve for $\lambda_k$" I think some of my confusion comes in how to take the partial derivative when it comes to within a summation. I have found some questions which seem similar, but I'm not sure how to apply them to this equation: 1, 2. I'll now list out the steps I took so far and hope to be pointed in the right direction. Partial derivative of $L$ with respect to $\lambda_k$: $$ \frac{\partial L}{\partial \lambda_k} = \frac{\partial}{\partial \lambda_k} \Bigg[ \sum_{k=1}^{6} N_k \log[\lambda_k] + v\Bigg(\sum_{k=1}^{6}\lambda_k - 1 \Bigg) \Bigg] $$ $$ = \sum_{k=1}^{6} N_k \frac{1}{\lambda_k} + \frac{\partial}{\partial \lambda_k} \Bigg[ v\Bigg(\sum_{k=1}^{6}\lambda_k - 1 \Bigg) \Bigg] $$ $$ = \sum_{k=1}^{6} N_k \frac{1}{\lambda_k} + \frac{\partial}{\partial \lambda_k} \Bigg[ v \sum_{k=1}^{6}\lambda_k - v \Bigg] $$ $$ = \sum_{k=1}^{6} N_k \frac{1}{\lambda_k} + v \sum_{k=1}^{6}1 $$ $$ = \sum_{k=1}^{6} N_k \frac{1}{\lambda_k} + 6v $$ Partial derivative of $L$ with respect to $v$: $$ \frac{\partial L}{\partial v} = \frac{\partial}{\partial v} \Bigg[ \sum_{k=1}^{6} N_k \log[\lambda_k] + v\Bigg(\sum_{k=1}^{6}\lambda_k - 1 \Bigg) \Bigg] $$ $$ = \frac{\partial}{\partial v} \Bigg[ v\Bigg(\sum_{k=1}^{6}\lambda_k - 1 \Bigg) \Bigg] $$ $$ = \frac{\partial}{\partial v} v\sum_{k=1}^{6}\lambda_k - v $$ $$ = \sum_{k=1}^{6}\lambda_k - 1 $$ At this point, I imagine I would set both equations equal to zero and solve for $\lambda_k$. However, I don't know that I did the differentiation correctly and if so, I'm not sure how to proceed with solving for $\lambda_k$ (removing it from the summation?).
I am following some lecture notes on Ricci flow and linearizing an operator to obtain its principal symbol. We have $T \in \: \Gamma(Sym^2 T^{*}M)$ smooth, fixed and positive definite and then compute the time derivative for the divergence of $G(T)$: $\bigg(\frac{\partial}{\partial t}\delta G(T) \bigg)Z = -T \bigg((\delta G(h))^{\#},Z\bigg) + \bigg<h,\nabla T(.,.,Z) - \frac{1}{2}\nabla_{Z}T \bigg>, $ where $\frac{\partial g}{\partial t}=h$, $Z$ an arbitrary vector field and $G(T)=T-\frac{1}{2}(tr \: T)h$. The notes then say that this implies $\frac{\partial}{\partial t}T^{-1}\delta G(T) = -\delta G(h)\: + \:\: ...$ where the dots indicate terms which don't have derivatives of $h$. I see that you can apply $T^{-1}$ as it doesn't depend on $t$ and then lose the $Z$ as it is arbitrary, but not sure exactly what happens to the right-hand side such that we end up with $-\delta G(h)$ or where the sharp goes.
In many areas of application, one needs to solve a nonlinear system of equations $$ F(x) = 0. $$ Sometimes, the formulation $$ \\|F(x)\\|^2 \to\min $$ is used. Clearly, every solution $\hat{x}$ of $F(x)=0$ is also a solution of the second problem; the converse is also true (if a solution exists). The question is if one can tell a-priori which formulation is better suited for a given problem. Have people worked on this before? One example Consider the function $$ F(x, y) = \begin{pmatrix} x^3 - 3x y^2 - 1\\ 3 x^2 y - y^3 \end{pmatrix}. $$ It has the three roots $x_1=(1,0)$ (green in the figure below), $x_2=(-0.5,\sqrt{3}/2)$ (blue), $x_3=(-0.5,-\sqrt{3}/2)$ (red). When applying Newton's method to $F$, the starting point will determine to which of the three solutions we converge. The darker the color, the more Newton iterations were required. The typical Newton fractals appear. When finding critial points $\nabla (\\|F(x)\\|^2) = 0$, again with Newton's method, the picture is a little different. Note that the point $(0,0)$ is a critical point of $\\|F(x)\\|^2$, but no solution of $F(x)=0$. This highlights one possible problem with the $\min$-formulation.
Transition state theory or activated complex theory is a very advanced topic and deals with chemical reaction dynamics. It goes beyond the scope of my knowledge, however, I will attempt my best to explain with what I know. As stated, the activated complex is a temporary unstable product. As two particles with sufficient kinetic energy collide; along the reaction path, they distort the bonds and rearrange into new atoms. The location of the transition state for the activated complex is at the highest point; or peak, of an potential energy diagram. As stated in my general chem textbook Principles of Modern Chemistry by Oxtoby et. al, p. 685: The activated complex is assumed to exist as if it were in equilibrium with the reactants and the theory focuses on calculating the rate at which the activated complex passes through the transition state to form products. Statistical thermodynamics is required to calculate all motions of particles, including both rotational and vibrational energy. In the transition state, change from products is mediated when bonds between particles or molecules become out of phase (or antisymmetrically stretched) by vibrational energies. The frequency of the molecules becoming out of phase with one another decreases as the reaction proceeds forward and the new product is transformed. The energy transferred from vibrational energy to achieve the activated complex is turned into kinetic energy. The Eyring formula makes it possible to calculate reaction rates using transition state theory: $$ k_r = \kappa \frac{K_\mathrm{B}T}{h}K^†$$ The kappa term is referred to as the transmission coefficent and refers to the probability that the particles will create the transition state. $K_\mathrm{B}T/h$ measures the rate at which the activated complex dissociates to form into it's products. This equation is unwieldy to us; as you stated previously, it is extremely hard to determine equilibrium constants for the activated complex since the instability achieved at the height of the potential energy for the reaction exists for only a brief period of time. Instead, we can relate the Eyring formula to thermodynamics and substitute $\exp (-\Delta G^†/RT)$ for $K^†$. $$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta G^†/RT)$$$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta H^†/RT)\exp(\Delta S^†/R) $$ The equation that Arrhenius derived is: $$k_r = A\exp (-E_a/RT)$$ Activation energy is expressed as activation enthalpy and activation entropy. The pre-exponential factor $A$ is substituted with $K_\mathrm{B}T/h$. This equation and the use of enthalpy and entropy as driving forces is used extensively in enzyme catalyzed reactions.
I am a bit confused. Somehow I have a problem connecting two problems together. The Closest String problem and the problem of matching with mismatches. They seam to be related but, I fail to see the connection. The Closest String problem is defined as : Instance:Strings $S_{1},S_{2}...S_{n}$ over alphabet $\Sigma$ of length $L$ each and a non-negative integers $d$ and $n$. Parameters:$n,d$ Question:Is there a string $s$ of length $L$ such that $\delta(s,S_{i})\leq d$ for all $i=1..n$? Note: $\delta(x,y)\leq d$ is the Hamming distance between $x$ and $y$. This problem is proven to be NP-complete. On the other hand we have a problem of matching with mismatches, which is described as: The problem of string matching with $d$ mismatches consists of finding all occurrences of a pattern of length $m$ in a text of length $n$ such that in at most $d$ positions the text and the pattern have different symbols. In the following, we assume that $0 < d < m$ and $m\leq n$. Landau and Vishkin gave the first (to my knowledge) efficient algorithm to solve this problem in $O(kn)$ time. Now my question is: Is matching with mismatches, or can it be seen, as a special parametrized case of the Closest String problem and how is this connection made?
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem. Yeah it does seem unreasonable to expect a finite presentation Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections. How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th... Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ... Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ... The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place. Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$ Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$ So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$ Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$ But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$ For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor. Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$ You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices). Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)... @Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$. This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra. You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost. I'll use the latter notation consistently if that's what you're comfortable with (Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$) @Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$) Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$. Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms. That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$ Voila, Riemann curvature tensor Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean? Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$. Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$. Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$? Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle. You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form (The cotangent bundle is naturally a symplectic manifold) Yeah So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$. But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!! So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ? Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty @Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job. My only quibble with this solution is that it doesn't seen very elegant. Is there a better way? In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}. Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group Everything about $S_4$ is encoded in the cube, in a way The same can be said of $A_5$ and the dodecahedron, say
Show that $$\int_0^\pi{d\theta\over(a-b\cos\theta)^2}={a\pi\over (a^2-b^2)^{3\over 2}}$$ where $a>b>0$. I'm not sure how to simplify this integral or evaluate it. Any solutions or hints are greatly appreciated. closed as off-topic by Travis Willse, C. Dubussy, imranfat, heropup, Martin R Apr 21 '16 at 21:25 This question appears to be off-topic. The users who voted to close gave this specific reason: " This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis Willse, C. Dubussy, imranfat, heropup, Martin R HINTS: Note that $$\int_0^\pi \frac{1}{(a-b\cos(\theta))^2}\,d\theta=-\frac{\partial}{\partial a}\int_0^\pi \frac{1}{a-b\cos(\theta)}\,d\theta$$ Then, evaluate the integral on the right-hand side using either the classical Tangent Half-Angle Substitution or use contour integration. Under the assumption $a>b>0$, let us start with: $$ I(a,b) = \int_{0}^{\pi}\frac{dt}{a-b\cos t} = 2\int_{0}^{\pi/2}\frac{dt}{(a+b)-2b\cos^2 t} $$ that through the substitution $t=\arctan u$ becomes: $$ I(a,b) = 2\int_{0}^{+\infty}\frac{du}{(a+b)(1+u^2)-2b}=\frac{\pi}{\sqrt{a^2-b^2}}. $$ Our integral is just $-\frac{\partial}{\partial a} I(a,b)$, hence it equals $\frac{\pi a}{(a^2-b^2)^{3/2}}$.
5.10 Exercises Daily electricity demand for Victoria, Australia, during 2014 is contained in elecdaily. The data for the first 20 days can be obtained as follows. Plot the data and find the regression model for Demand with temperature as an explanatory variable. Why is there a positive relationship? Produce a residual plot. Is the model adequate? Are there any outliers or influential observations? Use the model to forecast the electricity demand that you would expect for the next day if the maximum temperature was $15^\circ$ and compare it with the forecast if the with maximum temperature was $35^\circ$. Do you believe these forecasts? Give prediction intervals for your forecasts. The following R code will get you started: Plot Demand vs Temperature for all of the available data in elecdaily. What does this say about your model? Data set mens400contains the winning times (in seconds) for the men’s 400 meters final in each Olympic Games from 1896 to 2016. Plot the winning time against the year. Describe the main features of the plot. Fit a regression line to the data. Obviously the winning times have been decreasing, but at what averagerate per year? Plot the residuals against the year. What does this indicate about the suitability of the fitted line? Predict the winning time for the men’s 400 meters final in the 2020 Olympics. Give a prediction interval for your forecasts. What assumptions have you made in these calculations? Type easter(ausbeer)and interpret what you see. An elasticity coefficient is the ratio of the percentage change in the forecast variable ($y$) to the percentage change in the predictor variable ($x$). Mathematically, the elasticity is defined as $(dy/dx)\times(x/y)$. Consider the log-log model, \[ \log y=\beta_0+\beta_1 \log x + \varepsilon. \] Express $y$ as a function of $x$ and show that the coefficient $\beta_1$ is the elasticity coefficient. The data set fancyconcerns the monthly sales figures of a shop which opened in January 1987 and sells gifts, souvenirs, and novelties. The shop is situated on the wharf at a beach resort town in Queensland, Australia. The sales volume varies with the seasonal population of tourists. There is a large influx of visitors to the town at Christmas and for the local surfing festival, held every March since 1988. Over time, the shop has expanded its premises, range of products, and staff. Produce a time plot of the data and describe the patterns in the graph. Identify any unusual or unexpected fluctuations in the time series. Explain why it is necessary to take logarithms of these data before fitting a model. Use R to fit a regression model to the logarithms of these sales data with a linear trend, seasonal dummies and a “surfing festival” dummy variable. Plot the residuals against time and against the fitted values. Do these plots reveal any problems with the model? Do boxplots of the residuals for each month. Does this reveal any problems with the model? What do the values of the coefficients tell you about each variable? What does the Breusch-Godfrey test tell you about your model? Regardless of your answers to the above questions, use your regression model to predict the monthly sales for 1994, 1995, and 1996. Produce prediction intervals for each of your forecasts. Transform your predictions and intervals to obtain predictions and intervals for the raw data. How could you improve these predictions by modifying the model? The gasolineseries consists of weekly data for supplies of US finished motor gasoline product, from 2 February 1991 to 20 January 2017. The units are in “million barrels per day”. Consider only the data to the end of 2004. Fit a harmonic regression with trend to the data. Experiment with changing the number Fourier terms. Plot the observed gasoline and fitted values and comment on what you see. Select the appropriate number of Fourier terms to include by minimising the AICc or CV value. Check the residuals of the final model using the checkresiduals()function. Even though the residuals fail the correlation tests, the results are probably not severe enough to make much difference to the forecasts and prediction intervals. (Note that the correlations are relatively small, even though they are significant.) To forecast using harmonic regression, you will need to generate the future values of the Fourier terms. This can be done as follows. where fitis the fitted model using tslm(), Kis the number of Fourier terms used in creating fit, and his the forecast horizon required. Forecast the next year of data. Plot the forecasts along with the actual data for 2005. What do you find? Data set hurongives the water level of Lake Huron in feet from 1875 to 1972. Plot the data and comment on its features. Fit a linear regression and compare this to a piecewise linear trend model with a knot at 1915. Generate forecasts from these two models for the period up to 1980 and comment on these. (For advanced readers following on from Section 5.7). Using matrix notation it was shown that if $\bm{y}=\bm{X}\bm{\beta}+\bm{\varepsilon}$, where $\bm{e}$ has mean $\bm{0}$ and variance matrix $\sigma^2\bm{I}$, the estimated coefficients are given by $\hat{\bm{\beta}}=(\bm{X}'\bm{X})^{-1}\bm{X}'\bm{y}$ and a forecast is given by $\hat{y}=\bm{x}^\hat{\bm{\beta}}=\bm{x}^(\bm{X}'\bm{X})^{-1}\bm{X}'\bm{y}$ where $\bm{x}^$ is a row vector containing the values of the regressors for the forecast (in the same format as $\bm{X}$), and the forecast variance is given by $var(\hat{y})=\sigma^2 \left[1+\bm{x}^(\bm{X}'\bm{X})^{-1}(\bm{x}^*)'\right].$ Consider the simple time trend model where $y_t = \beta_0 + \beta_1t$. Using the following results, \[ \sum^{T}_{t=1}{t}=\frac{1}{2}T(T+1),\quad \sum^{T}_{t=1}{t^2}=\frac{1}{6}T(T+1)(2T+1) \] derive the following expressions: $\displaystyle\bm{X}'\bm{X}=\frac{1}{6}\left[ \begin{array}{cc} 6T & 3T(T+1) \\ 3T(T+1) & T(T+1)(2T+1) \\ \end{array} \right]$ $\displaystyle(\bm{X}'\bm{X})^{-1}=\frac{2}{T(T^2-1)}\left[ \begin{array}{cc} (T+1)(2T+1) & -3(T+1) \\ -3(T+1) & 6 \\ \end{array} \right]$ $\displaystyle\hat{\beta}_0=\frac{2}{T(T-1)}\left[(2T+1)\sum^T_{t=1}y_t-3\sum^T_{t=1}ty_t \right]$ $\displaystyle\hat{\beta}_1=\frac{6}{T(T^2-1)}\left[2\sum^T_{t=1}ty_t-(T+1)\sum^T_{t=1}y_t \right]$ $\displaystyle\text{Var}(\hat{y}_{t})=\hat{\sigma}^2\left[1+\frac{2}{T(T-1)}\left(1-4T-6h+6\frac{(T+h)^2}{T+1}\right)\right]$
Tagged: ideal Problem 624 Let $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism. Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively. (a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$. (b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$ Add to solve later (c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$ Problem 526 A ring is called local if it has a unique maximal ideal. (a) Prove that a ring $R$ with $1$ is local if and only if the set of non-unit elements of $R$ is an ideal of $R$. Add to solve later (b) Let $R$ be a ring with $1$ and suppose that $M$ is a maximal ideal of $R$. Prove that if every element of $1+M$ is a unit, then $R$ is a local ring. Problem 525 Let \[R=\left\{\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \quad \middle \| \quad a, b\in \Q \,\right\}.\] Then the usual matrix addition and multiplication make $R$ an ring. Let \[J=\left\{\, \begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix} \quad \middle \| \quad b \in \Q \,\right\}\] be a subset of the ring $R$. (a) Prove that the subset $J$ is an ideal of the ring $R$. Add to solve later (b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$. Problem 524 Let $R$ be the ring of all $2\times 2$ matrices with integer coefficients: \[R=\left\{\, \begin{bmatrix} a & b\\ c& d \end{bmatrix} \quad \middle\| \quad a, b, c, d\in \Z \,\right\}.\] Let $S$ be the subset of $R$ given by \[S=\left\{\, \begin{bmatrix} s & 0\\ 0& s \end{bmatrix} \quad \middle \| \quad s\in \Z \,\right\}.\] (a) True or False: $S$ is a subring of $R$. Add to solve later (b) True or False: $S$ is an ideal of $R$. Problem 432 (a) Let $R$ be an integral domain and let $M$ be a finitely generated torsion $R$-module. Prove that the module $M$ has a nonzero annihilator. In other words, show that there is a nonzero element $r\in R$ such that $rm=0$ for all $m\in M$. Here $r$ does not depend on $m$. Add to solve later (b) Find an example of an integral domain $R$ and a torsion $R$-module $M$ whose annihilator is the zero ideal. Problem 431 Let $R$ be a commutative ring and let $I$ be a nilpotent ideal of $R$. Let $M$ and $N$ be $R$-modules and let $\phi:M\to N$ be an $R$-module homomorphism. Prove that if the induced homomorphism $\bar{\phi}: M/IM \to N/IN$ is surjective, then $\phi$ is surjective.Add to solve later Problem 417 Let $R$ be a ring with $1$ and let $M$ be an $R$-module. Let $I$ be an ideal of $R$. Let $M’$ be the subset of elements $a$ of $M$ that are annihilated by some power $I^k$ of the ideal $I$, where the power $k$ may depend on $a$. Prove that $M’$ is a submodule of $M$.
Recently, I am reading the book [1]. On pages 16-17, the author wrote: Theorem 12 A non-trivial connected graph has an Euler circuit iff each vertex has even degree. A connected graph has an Euler trail from a vertex $x$ to a vertex $y \neq x$ iff $x$ and $y$ are the only vertices of odd degree. Proof.The conditions are clearly necessary. For example, if $G$ has an Euler circuit $x_1 x_2 \cdots x_m$, and $x$ occurs $k$ times in the sequence $x_1, x_2, ..., x_m$, then $d(x) = 2 k$. We prove the sufficiency of the first condition by induction on the number of edges. If there are no edges, there is nothing to prove, so we proceed to the induction step. Let $G$ be a non-trivial conneted graph in which each vertex has even degree. Since $e(G) \geq 1$, we find that $\delta(G) \geq 2$, so by Corollary 9, $G$ contains a cycle. Let $C$ be a circuitin $G$ with the maximal number of edges. Suppose $C$ is not Eulerian. As $G$ is connected, $C$ contains a vertex $x$ that is in a non-trivial component $H$ of $G - E(C)$. ... Corollary 9 is: Corollary 9 A tree of order at least 2 contains at least 2 vertices of degree 1. My question is: Why does $C$ contain a vertex $x$ that is in a non-trivial component $H$ of $G - E(C)$? The following is my attempt to answer the above question: First, according to page 6 of [1], A maximal connected subgraphis a componentof the graph. Then I search for the definitions of "connected" and "subgraph". On page 6 of [1], I find: A graph is connectedif for every pair $\{x, y\}$ of distinct vertices there is a path from $x$ to $y$. On page 2 of [1], I find: We say that $G' = (V', E')$ is a subgraphof $G = (V, E)$ if $V' \subset V$ and $E' \subset E$. Then I don't know how to continue. Could anyone please give me some comments or an answer? Thanks in advance. Reference [1] B. Bollobas, Modern Graph Theory, Springer, 1998.
a) $(\lnot(P \land Q)) \lor (Q \land R)$ b) $(P \lor Q) \land \lnot(Q)$ How do I simplify these 2 expression? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community You can often simplify boolean propositions by fiddling around with the algebra. Sometimes it helps to push in all negations to bring the formula to negation normal form. This means using de Morgan's law repeatedly until all negation signs are directly in front of literals: $\neg (P\wedge Q) \vee (Q\wedge R) = \neg P\vee \neg Q \vee (Q\wedge R)$ [de Morgan's] The next step you should do is either to bring the formula to conjunctive normal form or disjunctive normal form. Both of these forms only have to "layers" of nesting in terms of bracketing, provided you write $A \wedge (B \wedge C)$ as $A \wedge B \wedge C$, and similarly also don't write brackets in chains of disjunctions (i.e. "$\vee$"s) . To bring a formula to disjunctive normal form, you have to push in $\wedge$ using the law $A \wedge ( B \vee C) = (A \wedge B) \vee (A \wedge C)$. As it happens, we've already reached disjunctive normal form in the above formula. From here, there are many ways to continue. In the above example, the formula is clearly true if $\neg P$ or $\neg Q$ are true. Then what happens if $P$ and $Q$ are true? Well, in that case, you must have $Q \wedge R$ true, and we're already assuming $Q$ is true, it's enough that $R$ is true. So, a simplified version of your formula is $\neg P\vee \neg Q \vee R$. In your second formula, considering the different possible values of $Q$ will also help you out. If you want to be more systematic in your simplifications, then a good approach is to use Karnaugh maps. These are a very human-friendly way of simplifying formulas with up to four variables, but after that the maps get very unwieldy. If you need to cope with more variables in a systematic way, you can use the Quine-McCluskey algorithm. You don't have to put your formula in disjunctive normal form to use Karnaugh maps or the Quine-McCluskey algorithm, you just have to compute the truth table for your formula. a) 1. $(\lnot(P \land Q)) \lor (Q \land R)$ 2. $(\lnot P \lor \lnot Q) \lor (Q\land R)$ 3. $((\lnot P \lor \lnot Q) \lor Q) \land ((\lnot P \lor \lnot Q) \lor R)$ 4. $(\lnot P \lor (\lnot Q \lor Q)) \land (\lnot P \lor \lnot Q \lor R)$ 5. $(\lnot P \lor True) \land (\lnot P \lor \lnot Q \lor R)$ 6. $(True ) \land (\lnot P \lor \lnot Q \lor R)$ 7. $\lnot P \lor \lnot Q \lor R\qquad$ (Simplified normal disjunctive form) Note: By DeMorgan's, (7) is equivalent to $\lnot(P \land Q) \lor R$ b) 1. $(P \lor Q) \land \lnot(Q)$ 2. $(P \land \lnot Q) \lor (Q \land \lnot Q)$ 3. $(P \land \lnot Q) \lor (False)$ 4. $(P \land \lnot Q)\qquad$ (simplified in normal disjunctive form) Can you understand how to move from one step to the next? Essentially, I am simply applying "distribution": e.g. $(a \lor b) \land c \equiv (a \land c) \lor (b \land c)$ also $a \lor (b \land c) \equiv (a \lor b) \land (a \lor c)$ and I've used DeMorgan's in the first simplifications. For the first, try writing $\lnot (P \land Q)$ using an "or" instead of an "and" - you should know a rule to convert between the two. For the second, what can you say about P, given you know that Q is false?
Any element of the ring $\Z[\sqrt{-5}]$ is of the form $a+b\sqrt{-5}$ for some integers $a, b$.The associated (field) norm $N$ is given by\[N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^2+5b^2.\] Consider the case when $a=2, b=1$.Then we have\begin{align}(2+\sqrt{-5})(2-\sqrt{-5})=9=3\cdot 3. \tag{}\end{align} We claim that the numbers $3, 2\pm \sqrt{-5}$ are irreducible elements in the ring $\Z[\sqrt{-5}]$. To prove the claim at once, we show that any element in $\Z[\sqrt{-5}]$ of norm $9$ is irreducible. Let $\alpha$ be an element in $\Z[\sqrt{-5}]$ such that $N(\alpha)=9$.Suppose that $\alpha=\beta \gamma$ for some $\beta, \gamma \in \Z[\sqrt{-5}]$.Out goal is to show that either $\beta$ or $\gamma$ is a unit. We have\begin{align}9&=N(\alpha)=N(\beta)N(\gamma).\end{align}Since the norms are nonnegative integers, $N(\beta)$ is one of $1, 3, 9$. If $N(\beta)=1$, then it yields that $\beta$ is a unit. If $N(\beta)=3$, then we write $\beta=a+b\sqrt{-5}$ for some integers $a, b$, and we obtain\[3=N(\beta)=a^2+5b^2.\]A quick inspection yields that there are no integers $a, b$ satisfying this equality.Thus $N(\beta)=3$ is impossible. If $N(\beta)=9$, then $N(\gamma)=1$ and thus $\gamma$ is a unit. Therefore, we have shown that either $\beta$ or $\gamma$ is a unit. Note that the elements $3, 2\pm \sqrt{-5}$ have norm $9$, and hence they are irreducible by what we have just proved. It follows from the equalities in () that the factorization of the element $9$ into irreducible elements are not unique.Thus, the ring $\Z[\sqrt{-5}]$ is not a UFD. Related Question. Problem.Prove that the quadratic integer ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD). Ring of Gaussian Integers and Determine its Unit ElementsDenote by $i$ the square root of $-1$.Let\[R=\Z[i]=\{a+ib \mid a, b \in \Z \}\]be the ring of Gaussian integers.We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to\[N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.\]Here $\bar{\alpha}$ is the complex conjugate of […] The Ring $\Z[\sqrt{2}]$ is a Euclidean DomainProve that the ring of integers\[\Z[\sqrt{2}]=\{a+b\sqrt{2} \mid a, b \in \Z\}\]of the field $\Q(\sqrt{2})$ is a Euclidean Domain.Proof.First of all, it is clear that $\Z[\sqrt{2}]$ is an integral domain since it is contained in $\R$.We use the […] A ring is Local if and only if the set of Non-Units is an IdealA ring is called local if it has a unique maximal ideal.(a) Prove that a ring $R$ with $1$ is local if and only if the set of non-unit elements of $R$ is an ideal of $R$.(b) Let $R$ be a ring with $1$ and suppose that $M$ is a maximal ideal of $R$.Prove that if every […] 5 is Prime But 7 is Not Prime in the Ring $\Z[\sqrt{2}]$In the ring\[\Z[\sqrt{2}]=\{a+\sqrt{2}b \mid a, b \in \Z\},\]show that $5$ is a prime element but $7$ is not a prime element.Hint.An element $p$ in a ring $R$ is prime if $p$ is non zero, non unit element and whenever $p$ divide $ab$ for $a, b \in R$, then $p$ […] Three Equivalent Conditions for an Ideal is Prime in a PIDLet $R$ be a principal ideal domain. Let $a\in R$ be a nonzero, non-unit element. Show that the following are equivalent.(1) The ideal $(a)$ generated by $a$ is maximal.(2) The ideal $(a)$ is prime.(3) The element $a$ is irreducible.Proof.(1) $\implies$ […]
We receive a stream of $n-1$ pairwise different numbers from the set $\left\{1,\dots,n\right\}$. How can I determine the missing number with an algorithm that reads the stream once and uses a memory of only $O(\log_2 n)$ bits? Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It only takes a minute to sign up.Sign up to join this community You know $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, and because $S = \frac{n(n+1)}{2}$ could be coded in $O(\log(n))$ bits this can be done in $O(\log n)$ memory and in one path (just find $S - \mathrm{currentSum}$, this is missing number). But this problem could be solved in general case (for constant $k$): we have $k$ missing numbers, find out all of them. In this case instead of calculating just sum of $y_i$, calculate sum of j'st power of $x_i$ for all $1\le j \le k$ (I assumed $x_i$ is missing numbers and $y_i$ is input numbers): $\qquad \displaystyle \begin{align} \sum_{i=1}^k x_i &= S_1,\\ \sum_{i=1}^k x_i^2 &= S_2,\\ &\vdots \\ \sum_{i=1}^k x_i^k &= S_k \end{align}$ $\qquad (1)$ Remember that you can calculate $S_1,...S_k$ simply, because $S_1 = S - \sum y_i$, $S_2 = \sum i^2 - \sum y_i^2$, ... Now for finding missing numbers you should solve $(1)$ to find all $x_i$. You can compute: $P_1 = \sum x_i$, $P_2 = \sum x_i\cdot x_j$, ... , $P_k = \prod x_i$ $(2)$. For this remember that $P_1 = S_1$, $P_2 = \frac{S_1^2 - S_2}{2}$, ... But $P_i$ is coefficients of $P=(x-x_1)\cdot (x-x_2) \cdots (x-x_k)$ but $P$ could be factored uniquely, so you can find missing numbers. These are not my thoughts; read this. From the comment above: Before processing the stream, allocate $\lceil \log_2 n \rceil$ bits, in which you write $x:= \bigoplus_{i=1}^n \mathrm{bin}(i)$ ($\mathrm{bin}(i)$ is the binary representation of $i$ and $\oplus$ is pointwise exclusive-or). Naively, this takes $\mathcal{O}(n)$ time. Upon processing the stream, whenever one reads a number $j$, compute $x := x \oplus \mathrm{bin}(j)$. Let $k$ be the single number from $\{ 1, ... n\}$ that is not included in the stream. After having read the whole stream, we have $$ x = \left(\bigoplus_{i=1}^n \mathrm{bin}(i)\right) \oplus \left(\bigoplus_{i \neq k } \mathrm{bin}(i)\right) = \mathrm{bin}(k) \oplus \bigoplus_{i \neq k } (\mathrm{bin}(i) \oplus \mathrm{bin}(i)) = \mathrm{bin}(k), $$ yielding the desired result. Hence, we used $\mathcal{O}(\log n)$ space, and have an overall runtime of $\mathcal{O}(n)$. HdM's solution works. I coded it in C++ to test it. I can't limit the value to $O(\log_2 n)$ bits, but I'm sure you can easily show how only that number of bits is actually set. For those that want pseudo code, using a simple $\text{fold}$ operation with exclusive or ($\oplus$): $$\text{Missing} = \text{fold}(\oplus, \{1,\ldots,N\} \cup \text{InputStream})$$ Hand-wavey proof: A $\oplus$ never requires more bits than its input, so it follows that no intermediate result in the above requires more than the maximum bits of the input (so $O(\log_2 n)$ bits). $\oplus$ is commutative, and $x \oplus x = 0$, thus if you expand the above and pair off all data present in the stream you'll be left only with a single un-matched value, the missing number. #include <iostream>#include <vector>#include <cstdlib>#include <algorithm>using namespace std;void find_missing( int const * stream, int len );int main( int argc, char ** argv ){ if( argc < 2 ) { cerr << "Syntax: " << argv[0] << " N" << endl; return 1; } int n = atoi( argv[1] ); //construct sequence vector<int> seq; for( int i=1; i <= n; ++i ) seq.push_back( i ); //remove a number and remember it srand( unsigned(time(0)) ); int remove = (rand() % n) + 1; seq.erase( seq.begin() + (remove - 1) ); cout << "Removed: " << remove << endl; //give the stream a random order std::random_shuffle( seq.begin(), seq.end() ); find_missing( &seq[0], int(seq.size()) );}//HdM's solutionvoid find_missing( int const * stream, int len ){ //create initial value of n sequence xor'ed (n == len+1) int value = 0; for( int i=0; i < (len+1); ++i ) value = value ^ (i+1); //xor all items in stream for( int i=0; i < len; ++i, ++stream ) value = value ^ *stream; //what's left is the missing number cout << "Found: " << value << endl;}
This example uses systune to generate smooth gain schedules for a three-loop autopilot. This example uses a three-degree-of-freedom model of the pitch axis dynamics of an airframe. The states are the Earth coordinates , the body coordinates , the pitch angle , and the pitch rate . Figure 1 summarizes the relationship between the inertial and body frames, the flight path angle , the incidence angle , and the pitch angle . Figure 1: Airframe dynamics. We use a classic three-loop autopilot structure to control the flight path angle . This autopilot adjusts the flight path by delivering adequate bursts of normal acceleration (acceleration along ). In turn, normal acceleration is produced by adjusting the elevator deflection to cause pitching and vary the amount of lift. The autopilot uses Proportional-Integral (PI) control in the pitch rate loop and proportional control in the and loops. The closed-loop system (airframe and autopilot) are modeled in Simulink. addpath(fullfile(matlabroot,'examples','control','main')) % add example data open_system('rct_airframeGS') The airframe dynamics are nonlinear and the aerodynamic forces and moments depend on speed and incidence . To obtain suitable performance throughout the flight envelope, the autopilot gains must be adjusted as a function of and to compensate for changes in plant dynamics. This adjustment process is called "gain scheduling" and are called the scheduling variables. In the Simulink model, gain schedules are implemented as look-up tables driven by measurements of and . Gain scheduling is a linear technique for controlling nonlinear or time-varying plants. The idea is to compute linear approximations of the plant at various operating conditions, tune the controller gains at each operating condition, and swap gains as a function of operating condition during operation. Conventional gain scheduling involves three major steps: Trim and linearize the plant at each operating condition Tune the controller gains for the linearized dynamics at each operating condition Reconcile the gain values to provide smooth transition between operating conditions. In this example, we combine Steps 2. and 3. by parameterizing the autopilot gains as first-order polynomials in and directly tuning the polynomial coefficients for the entire flight envelope. This approach eliminates Step 3. and guarantees smooth gain variations as a function of and . Moreover, the gain schedule coefficients can be automatically tuned with systune. Assume that the incidence varies between -20 and 20 degrees and that the speed varies between 700 and 1400 m/s. When neglecting gravity, the airframe dynamics are symmetric in so consider only positive values of . Use a 5-by-9 grid of linearly spaced pairs to cover the flight envelope: nA = 5; % number of alpha values nV = 9; % number of V values [alpha,V] = ndgrid(linspace(0,20,nA)pi/180,linspace(700,1400,nV)); For each flight condition , linearize the airframe dynamics at trim (zero normal acceleration and pitching moment). This requires computing the elevator deflection and pitch rate that result in steady and . To do this, first isolate the airframe model in a separate Simulink model. open_system('rct_airframeTRIM') Use operspec to specify the trim condition, use findop to compute the trim values of and , and linearize the airframe dynamics for the resulting operating point. See the "Trimming and Linearizing an Airframe" example in Simulink Control Design for details. Repeat these steps for the 45 flight conditions . % Compute trim condition for each (alpha,V) pair clear op for ct=1:nAnV alpha_ini = alpha(ct); % Incidence [rad] v_ini = V(ct); % Speed [m/s] % Specify trim condition opspec = operspec('rct_airframeTRIM'); % Xe,Ze: known, not steady opspec.States(1).Known = [1;1]; opspec.States(1).SteadyState = [0;0]; % u,w: known, w steady opspec.States(3).Known = [1 1]; opspec.States(3).SteadyState = [0 1]; % theta: known, not steady opspec.States(2).Known = 1; opspec.States(2).SteadyState = 0; % q: unknown, steady opspec.States(4).Known = 0; opspec.States(4).SteadyState = 1; % TRIM Options = findopOptions('DisplayReport','off'); op(ct) = findop('rct_airframeTRIM',opspec,Options); end % Linearize at trim conditions G = linearize('rct_airframeTRIM',op); G = reshape(G,[nA nV]); G.u = 'delta'; G.y = {'alpha' 'V' 'q' 'az' 'gamma' 'h'}; This produces a 5-by-9 array of linearized plant models at the 45 flight conditions . The plant dynamics vary substantially across the flight envelope. sigma(G), title('Variations in airframe dynamics') The autopilot consists of four gains to be "scheduled" (adjusted) as a function of and . Practically, this means tuning 88 values in each of the corresponding four look-up tables. Rather than tuning each table entry separately, parameterize the gains as a two-dimensional gain surfaces, for example, surfaces with a simple multi-linear dependence on and : . This cuts the number of variables from 88 down to 4 for each lookup table. Use the tunableSurface object to parameterize each gain surface. Note that: TuningGrid specifies the "tuning grid" (design points). This grid should match the one used for linearization but needs not match the loop-up table breakpoints ShapeFcn specifies the basis functions for the surface parameterization (, , and ) Each surface is initialized to a constant gain using the tuning results for = 10 deg and = 1050 m/s (mid-range design). TuningGrid = struct('alpha',alpha,'V',V); ShapeFcn = @(alpha,V) [alpha,V,alphaV]; Kp = tunableSurface('Kp', 0.1, TuningGrid, ShapeFcn); Ki = tunableSurface('Ki', 2, TuningGrid, ShapeFcn); Ka = tunableSurface('Ka', 0.001, TuningGrid, ShapeFcn); Kg = tunableSurface('Kg', -1000, TuningGrid, ShapeFcn); Next create an slTuner interface for tuning the gain surfaces. Use block substitution to replace the nonlinear plant model by the linearized models over the tuning grid. Use setBlockParam to associate the tunable gain surfaces Kp, Ki, Ka, Kg with the Interpolation blocks of the same name. BlockSubs = struct('Name','rct_airframeGS/Airframe Model','Value',G); ST0 = slTuner('rct_airframeGS',{'Kp','Ki','Ka','Kg'},BlockSubs); % Register points of interest ST0.addPoint({'az_ref','az','gamma_ref','gamma','delta'}) % Parameterize look-up table blocks ST0.setBlockParam('Kp',Kp,'Ki',Ki,'Ka',Ka,'Kg',Kg); systune can automatically tune the gain surface coefficients for the entire flight envelope. Use TuningGoal objects to specify the performance objectives: loop: Track the setpoint with a 1 second response time, less than 2% steady-state error, and less than 30% peak error. Req1 = TuningGoal.Tracking('gamma_ref','gamma',1,0.02,1.3); viewGoal(Req1) loop: Ensure good disturbance rejection at low frequency (to track acceleration demands) and past 10 rad/s (to be insensitive to measurement noise). % Note: The disturbance is injected at the az_ref location RejectionProfile = frd([0.02 0.02 1.2 1.2 0.1],[0 0.02 2 15 150]); Req2 = TuningGoal.Gain('az_ref','az',RejectionProfile); viewGoal(Req2) loop: Ensure good disturbance rejection up to 10 rad/s. The disturbance is injected at the plant input delta. Req3 = TuningGoal.Gain('delta','az',600tf([0.25 0],[0.25 1])); viewGoal(Req3) Transients: Ensure a minimum damping ratio of 0.35 for oscillation-free transients MinDamping = 0.35; Req4 = TuningGoal.Poles(0,MinDamping); Using systune, tune the 16 gain surface coefficients to best meet these performance requirements at all 45 flight conditions. ST = systune(ST0,[Req1 Req2 Req3 Req4]); Final: Soft = 1.13, Hard = -Inf, Iterations = 57 The final value of the combined objective is close to 1, indicating that all requirements are nearly met. Visualize the resulting gain surfaces. % Get tuned gain surfaces TGS = getBlockParam(ST); % Plot gain surfaces clf subplot(221), viewSurf(TGS.Kp), title('Kp') subplot(222), viewSurf(TGS.Ki), title('Ki') subplot(223), viewSurf(TGS.Ka), title('Ka') subplot(224), viewSurf(TGS.Kg), title('Kg') First validate the tuned autopilot at the 45 flight conditions considered above. Plot the response to a step change in flight path angle and the response to a step disturbance in elevator deflection. clf subplot(211), step(getIOTransfer(ST,'gamma_ref','gamma'),5), grid title('Tracking of step change in flight path angle') subplot(212), step(getIOTransfer(ST,'delta','az'),3), grid title('Rejection of step disturbance at plant input') The responses are satisfactory at all flight conditions. Next validate the autopilot against the nonlinear airframe model. First use writeBlockValue to apply the tuning results to the Simulink model. This evaluates each gain surface formula at the breakpoints specified in the two Prelookup blocks and writes the result in the corresponding Interpolation block. writeBlockValue(ST) Now simulate the autopilot performance for a maneuver that takes the airframe through a large portion of its flight envelope. The code below is equivalent to pressing the Play button in the Simulink model and inspecting the responses in the Scope blocks. % Initial conditions h_ini = 1000; alpha_ini = 0; v_ini = 700; % Simulate SimOut = sim('rct_airframeGS', 'ReturnWorkspaceOutputs', 'on'); % Extract simulation data SimData = get(SimOut,'sigsOut'); Sim_gamma = getElement(SimData,'gamma'); Sim_alpha = getElement(SimData,'alpha'); Sim_V = getElement(SimData,'V'); Sim_delta = getElement(SimData,'delta'); Sim_h = getElement(SimData,'h'); Sim_az = getElement(SimData,'az'); t = Sim_gamma.Values.Time; % Plot the main flight variables clf subplot(211) plot(t,Sim_gamma.Values.Data(:,1),'r--',t,Sim_gamma.Values.Data(:,2),'b'), grid legend('Commanded','Actual','location','SouthEast') title('Flight path angle \gamma in degrees') subplot(212) plot(t,Sim_delta.Values.Data), grid title('Elevator deflection \delta in degrees') subplot(211) plot(t,Sim_alpha.Values.Data), grid title('Incidence \alpha in degrees') subplot(212) plot(t,Sim_V.Values.Data), grid title('Speed V in m/s') subplot(211) plot(t,Sim_h.Values.Data), grid title('Altitude h in meters') subplot(212) plot(t,Sim_az.Values.Data), grid title('Normal acceleration a_z in g''s') Tracking of the flight path angle profile remains good throughout the maneuver. Note that the variations in incidence and speed cover most of the flight envelope considered here ([-20,20] degrees for and [700,1400] for ). And while the autopilot was tuned for a nominal altitude of 3000 m, it fares well for altitude changing from 1,000 to 10,000 m. The nonlinear simulation results confirm that the gain-scheduled autopilot delivers consistently high performance throughout the flight envelope. The "gain surface tuning" procedure provides simple explicit formulas for the gain dependence on the scheduling variables. Instead of using look-up tables, you can use these formulas directly for an more memory-efficient hardware implementation. rmpath(fullfile(matlabroot,'examples','control','main')) % remove example data
This MathOverflow question seems to indicate that the state of the art in computing $$ M(x)=\sum_{n\le x}\mu(n) $$ takes time $\Theta(n^{2/3}(\log\log n)^{1/3}),$ which matches my understanding. Recently I came across a paper [1] which gives, almost as an afterthought, an algorithm for computing $M(x)$ in $O(\sqrt x)$ time (in section 3.2). The algorithm itself seems to be correct, being derived in a straightforward way from the identity $$ M(x)=1-\sum_{d\ge2}M(x/d). $$ But the claim that is runs in time $O(\sqrt x),$ or even $O(x^{1/2+\varepsilon}),$ seems unusual enough for me to ask for verification. First, this would be a major breakthrough in computing $M(x),$ enough that I would think it would merit more mention than a substep of another algorithm. At the least, I would expect a mention that previous algorithms were slower. Second, the algorithm is very simple, so that if the time is as indicated the implied constant should be low and the algorithm should be practical. In any case it is not hard to program. Third, on coding the algorithm I found it to be very slow. In fact, it was slower for those values tested than sum(n=1, 10^7, moebius(n)) in GP which involves factoring each number up to $10^7.$ The time needed to factor numbers of those sizes is about $\sqrt x/\log x$ on average, so that's a $\Theta(n^{3/2}/\log n)$ algorithm (admittedly, well-optimized) beating a $\Theta(n^{1/2})$ algorithm. The constants would have to be worse by a factor of $5\cdot10^6$ for that to happen, and there's nothing in the algorithm to suggest anything that bad. Of course I may have miscoded it (though I obtained the correct answer) or there may be reasons why the constant factors would be so high for this algorithm. But in any case this seemed suspicious enough to bring up here that I might be enlightened in any case. Of course even if the result claimed is not correct this is no mark against the author, as the paper is only a preprint and the claim is peripheral in any case. [1]: Jakub Pawlewicz, Counting square-free numbers (2011), arXiv:1107.4890.
The force ${\bf F}_e$ experienced by a particle at location ${\bf r}$ bearing charge $q$ in an electric field intensity ${\bf E}$ is \[{\bf F}_e = q {\bf E}({\bf r}) \label{m0061_eFeqE}\] If left alone in free space, this particle would immediately begin to move. The resulting displacement represents a loss of potential energy. This loss can quantified using the concept of work, $W$. The incremental work $\Delta W$ done by moving the particle a short distance $\Delta l$, over which we assume the change in ${\bf F}_e$ is negligible, is \[\Delta W \approx -{\bf F}_e\cdot\hat{\bf l}\Delta l \label{m0061_WeFdl}\] where in this case $\hat{\bf l}$ is the unit vector in the direction of the motion; i.e., the direction of ${\bf F}_e$. The minus sign indicates that potential energy of the system consisting of the electric field and the particle is being reduced. Like a spring that was previously compressed and is now released, the system is “relaxing.” To confirm that work defined in this way is an expression of energy, consider the units. The product of force (units of N) and distance (units of m) has units of N$\cdot$m, and 1 N$\cdot$m is 1 J of energy. Now, what if the motion of the particle is due to factors other than the force associated with the electric field? For example, we might consider “resetting” the system to it’s original condition by applying an external force to overcome ${\bf F}_e$. Equation \ref{m0061_WeFdl} still represents the change in potential energy of the system, but now $\hat{\bf l}$ changes sign. The same magnitude of work is done, but now this work is positive. In other words, positive work requires the application of an external force that opposes and overcomes the force associated with the electric field, thereby increasing the potential energy of the system. With respect to the analogy of a mechanical spring used above, positive work is achieved by compressing the spring. It is also worth noting that the purpose of the dot product in Equation \ref{m0061_WeFdl} is to ensure that only the component of motion parallel to the direction of the electric field is included in the energy tally. This is simply because motion in any other direction cannot be due to ${\bf E}$, and therefore does not increase or decrease the associated potential energy. We can make the relationship between work and the electric field explicit by substituting Equation \ref{m0061_eFeqE} into Equation \ref{m0061_WeFdl}, yielding \[\Delta W \approx -q {\bf E}({\bf r})\cdot\hat{\bf l}\Delta l \label{m0061_WqEdl}\] Equation \ref{m0061_WqEdl} gives the work only for a short distance around ${\bf r}$. Now let us try to generalize this result. If we wish to know the work done over a larger distance, then we must account for the possibility that ${\bf E}$ varies along the path taken. To do this, we may sum contributions from points along the path traced out by the particle, i.e., \[W \approx \sum_{n=1}^N \Delta W ({\bf r}_n)\] where ${\bf r}_n$ are positions defining the path. Substituting Equation \ref{m0061_WqEdl}, we have \[W \approx -q \sum_{n=1}^N {\bf E}({\bf r}_n)\cdot\hat{\bf l}({\bf r}_n)\Delta l\] Taking the limit as $\Delta l\to 0$ we obtain \[W = -q \int_{\mathcal C} {\bf E}({\bf r})\cdot\hat{\bf l}({\bf r}) dl\] where $\mathcal{C}$ is the path (previously, the sequence of ${\bf r}_n$’s) followed. Now omitting the explicit dependence on ${\bf r}$ in the integrand for clarity: \[W = -q \int_{\mathcal C} {\bf E}\cdot d{\bf l} \label{m0061_eWqint}\] where $d{\bf l} = \hat{\bf l}dl$ as usual. Now, we are able to determine the change in potential energy for a charged particle moving along any path in space, given the electric field. At this point, it is convenient to formally define the electric potential difference $V_{21}$ between the start point (1) and end point (2) of ${\mathcal C}$. $V_{21}$ is defined as the work done by traversing ${\mathcal C}$, per unit of charge: \[V_{21} \triangleq \frac{W}{q}\] This has units of J/C, which is volts (V). Substituting Equation \ref{m0061_eWqint}, we obtain: \[\boxed{ V_{21} = - \int_{\mathcal C} {\bf E} \cdot d{\bf l} } \label{m0062_eVAB}\] An advantage of analysis in terms of electrical potential as opposed to energy is that we will no longer have to explicitly state the value of the charge involved. The potential difference $V_{21}$ between two points in space, given by Equation \ref{m0062_eVAB}, is the change in potential energy of a charged particle divided by the charge of the particle. Potential energy is also commonly known as “voltage” and has units of V. Example $\PageIndex{1}$: Potential difference in a uniform electric field Consider an electric field ${\bf E}({\bf r}) = \hat{\bf z}E_0$, which is constant in both magnitude and direction throughout the domain of the problem. The path of interest is a line beginning at $\hat{\bf z}z_1$ and ending at $\hat{\bf z}z_2$. What is $V_{21}$? (It’s worth noting that the answer to this problem is a building block for a vast number of problems in electromagnetic analysis.) Solution From Equation \ref{m0062_eVAB} we have \[V_{21} = - \int_{z_1}^{z_2} \left( \hat{\bf z}E_0 \right) \cdot \hat{\bf z}dz = -E_0 (z_2 - z_1)\] Note $V_{21}$ is simply the electric field intensity times the distance between the points. This may seem familiar. For example, compare this to the findings of the battery-charged capacitor experiment described in Section 2.2. There too we find that potential difference equals electric field intensity times distance, and the signs agree. The solution to the preceding example is simple because the direct path between the two points is parallel to the electric field. If the path between the points had been perpendicular to ${\bf E}$, then the solution is even easier – $V_{21}$ is simply zero. In all other cases, $V_{21}$ is proportional to the component of the direct path between the start and end points that is parallel to ${\bf E}$, as determined by the dot product. Contributors Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 Licensed with CC BY-SA 4.0 https://creativecommons.org/licenses/by-sa/4.0. Report adoption of this book here. If you are a professor reviewing, adopting, or adapting this textbook please help us understand a little more about your use by filling out this form.
I have this problem, and I don't really understand how am I supposed to do this. Could someone please help me with this? I know what left quotient is. I also know about regular, irregular languages. But still I don't know how to show this... Any help would be appreciated, thanks in advance! Also sorry if something is not correctly in English, not my first language. Problem: Show that for any language $L ⊆ Σ^*$ and any DFA $A = \langle \Sigma, Q, q_0, \delta, F \rangle$, the left quotient $L \diagdown L (A)$ is a union of languages $L_q = \{v \| \delta(q,v) \in F\}$ for selected states $q \in Q$, and explain what are those selected states. Could someone please also demonstrate on a small (but non-trivial) example, in which the $L$ would be irregular?
Recently, I have been learning about path integrals and I was wondering, can the probability of a certain path be weighted more in a path integral? Said in another way, can certain paths have more probability in a path integral? closed as unclear what you're asking by ACuriousMind♦, CuriousOne, Kyle Kanos, BMS, Brandon Enright Dec 15 '14 at 7:21 Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question. In general the "weighting" of each path $q$ in a path integral is given by $e^{\frac{i}{\hbar}S[q]}$. Then paths for which the action $S$ is stationary with respect to small deviations from the path are the only ones which really contribute because the contributions from those with non stationary $S$ get averaged out as the phase changes very rapidly (because $\hbar$ is very small). The number $S[q]$ is defined as: $$S[q] = \int_{t_0}^{t_1}L(q(t),\dot{q}(t), t)dt$$ Where $t$ is some parameter that varies along the path and $L$ is the lagrangian. The Langrangian will depend on the details of your system but for a free particle it looks like the classical kinetic energy $L = \frac{1}{2}\dot{q}^2$.
You start by writing down the probability to find a particle at $y$ at time $t$ when it was at $x$ at time $0$, denoted as $K(y,t;x,0)$. You get this by solving the Schrödinger equation with the initial condition $\psi(y,0) = \delta(y-x)$. Then, $K(y,t;x,0) = \psi(y,t)$.Thus, to solve this, we need to know the time development of the initial condition $\psi(y,0)$. Let us start with the simple example of a free particle. This is easiest solved in momentum-representation, obtained by Fourier-transforming $\psi(y,t)$: $$\psi(y,t) = \frac{1}{\sqrt{2\pi\hbar}} \int dp \exp(ipy/\hbar) \tilde \psi(p,t)$$For $\tilde \psi$, the Schrödinger equation gives $$\tilde \psi(p,t) = \frac{1}{\sqrt{2\pi\hbar}} \exp \left(-\frac{i}{\hbar} \left[\frac{p^2 t}{2m} - px\right]\right)$$This can be inserted back into the equation for $\psi(y,t)$. The integral over $p$ can be solved exactly. The final result is $$K_\text{free}(y,t;x,0) = \sqrt{\frac{m}{2\pi i\hbar t}} \exp\left(\frac{im(x-y)^2}{2\hbar t}\right)$$ Next step: The solution of the Schrödinger equation can generally be written as $$\|\psi, t\rangle = \exp\left(-\frac{iHt}{\hbar}\right) \|\psi,0\rangle$$with $H$ being the Hamiltonian of your system. Writing $H = T+V$, the general formula for $K$ becomes$$K(y,t;x,0) = \langle y \mid \exp(-\frac{i(T+V)t}{\hbar}) \mid x \rangle$$We use the Trotter-Kato Formula (which holds under certain conditions which I won't go into detail at this point. It allows us to write $$K(y,t;x,0) = \lim_{N\rightarrow \infty} \langle y \mid \left[ \exp(-\frac{iTt}{N\hbar})\exp(-\frac{iVt}{N\hbar})\right]^N \mid x\rangle$$We insert the unity operator, decomposed as $1 = \int dx \| x \rangle \langle x \|$ $N-1$ times, which gives us $$K(y,t;x,0) = \int dx_1 dx_2 \dots dx_{N-1} \prod_{j=0}{N-1}\langle x_{j+1} \mid \exp(-iTt/N\hbar) \exp(-iVt/N\hbar) \mid x_j \rangle$$Note that $V$ as an operator acting on $\|x\rangle$ gives just $V(x) \|x\rangle$. And $\langle x_{j+1} \| \exp(-iTt/N\hbar) \| x_j \rangle$ gives us just the contribution of a free particle, i.e. $$\sqrt{\frac{mN}{2\pi i\hbar t}} \exp\left(\frac{imN}{2\hbar t}(x_{j+1} - x_j)\right)^2$$.If we abbreviate $\tau = t/N$, we can write:$$K(y,t;x,0) = \lim_{N\rightarrow \infty} \int dx_1 dx_2 \dots dx_{N-1} \left( \frac{m}{2\pi i\hbar \tau}\right)^{N/2} \times$$$$\exp \left(\frac{i\tau}{\hbar} \sum_{j=0}^{N-1} \left[ \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\tau}\right)^2 - V(x_j)\right]\right)$$ The next step is to see the values $x_j$ as points of a certain path $x(t')$ evaluated at points $t' = t_j = j\tau = jt/N$. If $\tau$ is small, we write $$\sum_{j=0}^{N-1} \tau f(t_j) \rightarrow \int f(t') dt'$$$$\frac{x_{j+1} - x_j}{\tau} \rightarrow \dot x(t')$$where the dot denotes the time-derivative. The argument of the exponential then becomes $$\frac{i}{\hbar} \int_0^t dt' \left( \frac{m\dot x(t')^2}{2} - V(x(t'))\right)$$You will have no trouble identifying the integrand as the Lagrangian $L = T-V$. The integral itself, therefore, is the classical action. Thus, the formula we have for $K$ can be interpreted as the sum over all possible paths from $(x,0)$ to $(y,t)$ of the function $\exp\left(\frac{i}{\hbar} S(t,0)\right)$ of the classical action. The interpretation of this was given in other answers: The classical path is that which minimizes the action, i.e. the action is stationary for the classical path. In your path-integral formula, this path will have a large contribution, as all paths that vary only slightly from the classical path will still have pretty much the same phase factor as the classical one, leading to constructive interference of those paths. For paths far from the classical path, the action will vary greater among the paths, so that there all possible phases occur, which will ultimately cancel each outer out. Reference A lecture on advanced quantum mechanics given by Prof. Crispin Gardiner. Lecture notes are, unfortunately, not freely available. It was a good lecture :)
“ ... a knotted springy wire cannot rest in stable equilibrium without points of self-contact—an experimentally observable fact. This fact leads to a rather curious “topologically constrained” variational problem; what actually happens if one forms a knot in a piece of springy wire? Experiments yield some beautiful curves with impressive symmetry ... —J. Langer & D. A. Singer Indeed, it is an interesting experiment to form a knot in a piece of springy wire, stick the endpoints together, and release the configuration. Can we predict the resulting shape? Langer and Singer's question [LS] leads to a free obstacle problem that involves techniques at the interface of geometric analysis, low-dimensional topology, modeling, numerical analysis, and nonlinear optimization. Elastic energy Neglecting all effects of twist and shear, we consider a simplistic model by assuming that the shape of the wire is only influenced by the bending energy of its centerline $\gamma:\mathbb{R}/\mathbb{Z}\to\mathbb{R}^3$. The latter amounts to \[ E_{\mathrm{bend}}(\gamma) = \int\limits_{\mathbb{R}/\mathbb{Z}}\kappa_\gamma(s)^2\;\mathrm d s \] where $\kappa_\gamma(s)$ denotes the scalar curvature at $\gamma(s)$ and $s$ is an arc-length parameter. The Model In order to model the impermeability of the curve, we regularize the bending energy by a functional $\mathcal{R}$ that models self-avoidance. This means that its values blow up on sequences of embedded curves converging to a curve with a self-intersection. There are several candidates for this purpose. As proposed in [vdM], we study the problem of minimizing the total energy \[ E_\vartheta = E_{\mathrm{bend}} + \vartheta \mathcal{R} \] within a given knot class. We expect that the limit of $E_\vartheta$ minimizers as $\vartheta\searrow0$ does not depend on the choice of $\mathcal{R}$. Employing the ropelength functional, i.e., the quotient of length over thickness, reflects the idea of a tube with a uniform radius. Thickness of a curve can be defined as the infimum over the radii of all circles passing through three distinct points of the curve [GM]. Two-bridge torus knots In case of the trefoil knot class, minimizers of $E_\vartheta$ tend to the doubly covered circle as $\vartheta\searrow0$ [GRvdM]. This statement also applies to any $(2,b)$ torus knot class where $b\ge3$ is an odd number. This proves a conjecture of Gallotti and Pierre-Louis [GPL] for two-bridge knots. The proof relies on a generalization of the Fáry–Milnor theorem [M] to the $C^1$ closure of a knot class which in turn bases on the existence of alternating quadrisecants for knotted curves due to Denne [D]. References [D] Elizabeth Denne. Alternating quadrisecants of knots. Doctoral dissertation, University of Illinois at Urbana-Champaign, 2004. [ arXiv ] [GPL] Riccardo Gallotti, Olivier Pierre-Louis. Stiff knots. Phys. Rev. E (3), 75(3):031801, 2007. [ arXiv \| doi ] [GRvdM] Henryk Gerlach, Philipp Reiter, and Heiko von der Mosel. The elastic trefoil is the doubly covered circle. Arch. Rat. Mech. Anal., 225(1):89–139, 2017. [ arXiv \| doi ] [LS] Joel Langer, David A. Singer. Curve straightening and a minimax argument for closed elastic curves. Topology, 24(1):75–88, 1985. [ open archive ] [M] John W. Milnor. On the total curvature of knots. Ann. of Math. (2), 52:248–257, 1950. [ doi ] [vdM] Heiko von der Mosel. Minimizing the elastic energy of knots. Asymptot. Anal., 18(1-2):49–65, 1998. [ preprint \| journal ]
Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing Abstract We cross-correlate galaxy weak lensing measurements from the Dark Energy Survey (DES) year-one (Y1) data with a cosmic microwave background (CMB) weak lensing map derived from South Pole Telescope (SPT) and Planck data, with an effective overlapping area of 1289 deg$$^{2}$$. With the combined measurements from four source galaxy redshift bins, we reject the hypothesis of no lensing with a significance of $$10.8\sigma$$. When employing angular scale cuts, this significance is reduced to $$6.8\sigma$$, which remains the highest signal-to-noise measurement of its kind to date. We fit the amplitude of the correlation functions while fixing the cosmological parameters to a fiducial $$\Lambda$$CDM model, finding $$A = 0.99 \pm 0.17$$. We additionally use the correlation function measurements to constrain shear calibration bias, obtaining constraints that are consistent with previous DES analyses. Finally, when performing a cosmological analysis under the $$\Lambda$$CDM model, we obtain the marginalized constraints of $$\Omega_{\rm m}=0.261^{+0.070}_{-0.051}$$ and $$S_{8}\equiv \sigma_{8}\sqrt{\Omega_{\rm m}/0.3} = 0.660^{+0.085}_{-0.100}$$. These measurements are used in a companion work that presents cosmological constraints from the joint analysis of two-point functions among galaxies, galaxy shears, and CMB lensing using DES, SPT and Planck data. Authors: Publication Date: Research Org.: Argonne National Lab. (ANL), Argonne, IL (United States); Oak Ridge National Lab. (ORNL), Oak Ridge, TN (United States); Lawrence Berkeley National Lab. (LBNL), Berkeley, CA (United States); Brookhaven National Lab. (BNL), Upton, NY (United States); SLAC National Accelerator Lab., Menlo Park, CA (United States); Fermi National Accelerator Lab. (FNAL), Batavia, IL (United States) Sponsoring Org.: USDOE Office of Science (SC), High Energy Physics (HEP) (SC-25) Contributing Org.: DES; SPT OSTI Identifier: 1487048 Report Number(s): arXiv:1810.02441; FERMILAB-PUB-18-513-AE 1697154 DOE Contract Number: AC02-07CH11359 Resource Type: Journal Article Journal Name: TBD Additional Journal Information: Journal Name: TBD Country of Publication: United States Language: English Subject: 79 ASTRONOMY AND ASTROPHYSICS Citation Formats Omori, Y., and et al. Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing. United States: N. p., 2018. Web. Omori, Y., & et al. Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing. United States. Omori, Y., and et al. Thu . "Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing". United States. https://www.osti.gov/servlets/purl/1487048. @article{osti_1487048, title = {Dark Energy Survey Year 1 Results: Cross-correlation between DES Y1 galaxy weak lensing and SPT+Planck CMB weak lensing}, author = {Omori, Y. and et al.}, abstractNote = {We cross-correlate galaxy weak lensing measurements from the Dark Energy Survey (DES) year-one (Y1) data with a cosmic microwave background (CMB) weak lensing map derived from South Pole Telescope (SPT) and Planck data, with an effective overlapping area of 1289 deg$^{2}$. With the combined measurements from four source galaxy redshift bins, we reject the hypothesis of no lensing with a significance of $10.8\sigma$. When employing angular scale cuts, this significance is reduced to $6.8\sigma$, which remains the highest signal-to-noise measurement of its kind to date. We fit the amplitude of the correlation functions while fixing the cosmological parameters to a fiducial $\Lambda$CDM model, finding $A = 0.99 \pm 0.17$. We additionally use the correlation function measurements to constrain shear calibration bias, obtaining constraints that are consistent with previous DES analyses. Finally, when performing a cosmological analysis under the $\Lambda$CDM model, we obtain the marginalized constraints of $\Omega_{\rm m}=0.261^{+0.070}_{-0.051}$ and $S_{8}\equiv \sigma_{8}\sqrt{\Omega_{\rm m}/0.3} = 0.660^{+0.085}_{-0.100}$. These measurements are used in a companion work that presents cosmological constraints from the joint analysis of two-point functions among galaxies, galaxy shears, and CMB lensing using DES, SPT and Planck data.}, doi = {}, journal = {TBD}, number = , volume = , place = {United States}, year = {2018}, month = {10} } Figures / Tables: i s(z) for the 4 tomographic bins for Metacalibration. The black line shows the CMB lensing kernel.
Root systems¶ See Root Systems for an overview. class sage.combinat.root_system.root_system. RootSystem( cartan_type, as_dual_of=None)¶ A class for root systems. EXAMPLES: We construct the root system for type $B_3$: sage: R=RootSystem(['B',3]); R Root system of type ['B', 3] Rmodels the root system abstractly. It comes equipped with various realizations of the root and weight lattices, where all computations take place. Let us play first with the root lattice: sage: space = R.root_lattice() sage: space Root lattice of the Root system of type ['B', 3] This is the free $\ZZ$-module $\bigoplus_i \ZZ.\alpha_i$ spanned by the simple roots: sage: space.base_ring() Integer Ring sage: list(space.basis()) [alpha[1], alpha[2], alpha[3]] Let us do some computations with the simple roots: sage: alpha = space.simple_roots() sage: alpha[1] + alpha[2] alpha[1] + alpha[2] There is a canonical pairing between the root lattice and the coroot lattice: sage: R.coroot_lattice() Coroot lattice of the Root system of type ['B', 3] We construct the simple coroots, and do some computations (see comments about duality below for some caveat): sage: alphacheck = space.simple_coroots() sage: list(alphacheck) [alphacheck[1], alphacheck[2], alphacheck[3]] We can carry over the same computations in any of the other realizations of the root lattice, like the root space $\bigoplus_i \QQ.\alpha_i$, the weight lattice $\bigoplus_i \ZZ.\Lambda_i$, the weight space $\bigoplus_i \QQ.\Lambda_i$. For example: sage: space = R.weight_space() sage: space Weight space over the Rational Field of the Root system of type ['B', 3] sage: space.base_ring() Rational Field sage: list(space.basis()) [Lambda[1], Lambda[2], Lambda[3]] sage: alpha = space.simple_roots() sage: alpha[1] + alpha[2] Lambda[1] + Lambda[2] - 2Lambda[3] The fundamental weights are the dual basis of the coroots: sage: Lambda = space.fundamental_weights() sage: Lambda[1] Lambda[1] sage: alphacheck = space.simple_coroots() sage: list(alphacheck) [alphacheck[1], alphacheck[2], alphacheck[3]] sage: [Lambda[i].scalar(alphacheck[1]) for i in space.index_set()] [1, 0, 0] sage: [Lambda[i].scalar(alphacheck[2]) for i in space.index_set()] [0, 1, 0] sage: [Lambda[i].scalar(alphacheck[3]) for i in space.index_set()] [0, 0, 1] Let us use the simple reflections. In the weight space, they work as in the number game: firing the node $i$ on an element $x$ adds $c$ times the simple root $\alpha_i$, where $c$ is the coefficient of $i$ in $x$: sage: s = space.simple_reflections() sage: Lambda[1].simple_reflection(1) -Lambda[1] + Lambda[2] sage: Lambda[2].simple_reflection(1) Lambda[2] sage: Lambda[3].simple_reflection(1) Lambda[3] sage: (-2Lambda[1] + Lambda[2] + Lambda[3]).simple_reflection(1) 2Lambda[1] - Lambda[2] + Lambda[3] It can be convenient to manipulate the simple reflections themselves: sage: s = space.simple_reflections() sage: s[1](Lambda[1]) -Lambda[1] + Lambda[2] sage: s[1](Lambda[2]) Lambda[2] sage: s[1](Lambda[3]) Lambda[3] Ambient spaces The root system may also come equipped with an ambient space. This is a $\QQ$-module, endowed with its canonical Euclidean scalar product, which admits simultaneous embeddings of the (extended) weight and the (extended) coweight lattice, and therefore the root and the coroot lattice. This is implemented on a type by type basis for the finite crystallographic root systems following Bourbaki’s conventions and is extended to the affine cases. Coefficients permitting, this is also available as an ambient lattice. In finite type $A$, we recover the natural representation of the symmetric group as group of permutation matrices: sage: RootSystem(["A",2]).ambient_space().weyl_group().simple_reflections() Finite family {1: [0 1 0] [1 0 0] [0 0 1], 2: [1 0 0] [0 0 1] [0 1 0]} In type $B$, $C$, and $D$, we recover the natural representation of the Weyl group as groups of signed permutation matrices: sage: RootSystem(["B",3]).ambient_space().weyl_group().simple_reflections() Finite family {1: [0 1 0] [1 0 0] [0 0 1], 2: [1 0 0] [0 0 1] [0 1 0], 3: [ 1 0 0] [ 0 1 0] [ 0 0 -1]} In (untwisted) affine types $A$, …, $D$, one can recover from the ambient space the affine permutation representation, in window notation. Let us consider the ambient space for affine type $A$: sage: L = RootSystem(["A",2,1]).ambient_space(); L Ambient space of the Root system of type ['A', 2, 1] Define the “identity” by an appropriate vector at level $-3$: sage: e = L.basis(); Lambda = L.fundamental_weights() sage: id = e[0] + 2e[1] + 3e[2] - 3Lambda[0] The corresponding permutation is obtained by projecting it onto the classical ambient space: sage: L.classical() Ambient space of the Root system of type ['A', 2] sage: L.classical()(id) (1, 2, 3) Here is the orbit of the identity under the action of the finite group: sage: W = L.weyl_group() sage: S3 = [ w.action(id) for w in W.classical() ] sage: [L.classical()(x) for x in S3] [(1, 2, 3), (3, 1, 2), (2, 3, 1), (2, 1, 3), (1, 3, 2), (3, 2, 1)] And the action of $s_0$ on these yields: sage: s = W.simple_reflections() sage: [L.classical()(s[0].action(x)) for x in S3] [(0, 2, 4), (-1, 1, 6), (-2, 3, 5), (0, 1, 5), (-1, 3, 4), (-2, 2, 6)] We can also plot various components of the ambient spaces: sage: L = RootSystem(['A',2]).ambient_space() sage: L.plot() Graphics object consisting of 13 graphics primitives For more on plotting, see Tutorial: visualizing root systems. Dual root systems The root system is aware of its dual root system: sage: R.dual Dual of root system of type ['B', 3] R.dualis really the root system of type $C_3$: sage: R.dual.cartan_type() ['C', 3] And the coroot lattice that we have been manipulating before is really implemented as the root lattice of the dual root system: sage: R.dual.root_lattice() Coroot lattice of the Root system of type ['B', 3] In particular, the coroots for the root lattice are in fact the roots of the coroot lattice: sage: list(R.root_lattice().simple_coroots()) [alphacheck[1], alphacheck[2], alphacheck[3]] sage: list(R.coroot_lattice().simple_roots()) [alphacheck[1], alphacheck[2], alphacheck[3]] sage: list(R.dual.root_lattice().simple_roots()) [alphacheck[1], alphacheck[2], alphacheck[3]] The coweight lattice and space are defined similarly. Note that, to limit confusion, all the output have been tweaked appropriately. See also ambient_lattice()¶ Return the ambient lattice for this root_system. This is the ambient space, over $\ZZ$. EXAMPLES: sage: RootSystem(['A',4]).ambient_lattice() Ambient lattice of the Root system of type ['A', 4] sage: RootSystem(['A',4,1]).ambient_lattice() Ambient lattice of the Root system of type ['A', 4, 1] Except in type A, only an ambient space can be realized: sage: RootSystem(['B',4]).ambient_lattice() sage: RootSystem(['C',4]).ambient_lattice() sage: RootSystem(['D',4]).ambient_lattice() sage: RootSystem(['E',6]).ambient_lattice() sage: RootSystem(['F',4]).ambient_lattice() sage: RootSystem(['G',2]).ambient_lattice() ambient_space( base_ring=Rational Field)¶ Return the usual ambient space for this root_system. INPUT: base_ring– a base ring (default: $\QQ$) This is a base_ring-module, endowed with its canonical Euclidean scalar product, which admits simultaneous embeddings into the weight and the coweight lattice, and therefore the root and the coroot lattice, and preserves scalar products between elements of the coroot lattice and elements of the root or weight lattice (and dually). There is no mechanical way to define the ambient space just from the Cartan matrix. Instead it is constructed from hard coded type by type data, according to the usual Bourbaki conventions. Such data is provided for all the finite (crystallographic) types. From this data, ambient spaces can be built as well for dual types, reducible types and affine types. When no data is available, or if the base ring is not large enough, None is returned. Warning for affine types See also EXAMPLES: sage: RootSystem(['A',4]).ambient_space() Ambient space of the Root system of type ['A', 4] sage: RootSystem(['B',4]).ambient_space() Ambient space of the Root system of type ['B', 4] sage: RootSystem(['C',4]).ambient_space() Ambient space of the Root system of type ['C', 4] sage: RootSystem(['D',4]).ambient_space() Ambient space of the Root system of type ['D', 4] sage: RootSystem(['E',6]).ambient_space() Ambient space of the Root system of type ['E', 6] sage: RootSystem(['F',4]).ambient_space() Ambient space of the Root system of type ['F', 4] sage: RootSystem(['G',2]).ambient_space() Ambient space of the Root system of type ['G', 2] An alternative base ring can be provided as an option: sage: e = RootSystem(['B',3]).ambient_space(RR) sage: TestSuite(e).run() It should contain the smallest ring over which the ambient space can be defined ($\ZZ$ in type $A$ or $\QQ$ otherwise). Otherwise Noneis returned: sage: RootSystem(['B',2]).ambient_space(ZZ) The base ring should also be totally ordered. In practice, only $\ZZ$ and $\QQ$ are really supported at this point, but you are welcome to experiment: sage: e = RootSystem(['G',2]).ambient_space(RR) sage: TestSuite(e).run() Failure in _test_root_lattice_realization: Traceback (most recent call last): ... AssertionError: 2.00000000000000 != 2.00000000000000 ------------------------------------------------------------ The following tests failed: _test_root_lattice_realization cartan_matrix()¶ EXAMPLES: sage: RootSystem(['A',3]).cartan_matrix() [ 2 -1 0] [-1 2 -1] [ 0 -1 2] cartan_type()¶ Returns the Cartan type of the root system. EXAMPLES: sage: R = RootSystem(['A',3]) sage: R.cartan_type() ['A', 3] coambient_space( base_ring=Rational Field)¶ Return the coambient space for this root system. This is the ambient space of the dual root system. See also EXAMPLES: sage: L = RootSystem(["B",2]).ambient_space(); L Ambient space of the Root system of type ['B', 2] sage: coL = RootSystem(["B",2]).coambient_space(); coL Coambient space of the Root system of type ['B', 2] The roots and coroots are interchanged: sage: coL.simple_roots() Finite family {1: (1, -1), 2: (0, 2)} sage: L.simple_coroots() Finite family {1: (1, -1), 2: (0, 2)} sage: coL.simple_coroots() Finite family {1: (1, -1), 2: (0, 1)} sage: L.simple_roots() Finite family {1: (1, -1), 2: (0, 1)} coroot_lattice()¶ Returns the coroot lattice associated to self. EXAMPLES: sage: RootSystem(['A',3]).coroot_lattice() Coroot lattice of the Root system of type ['A', 3] coroot_space( base_ring=Rational Field)¶ Returns the coroot space associated to self. EXAMPLES: sage: RootSystem(['A',3]).coroot_space() Coroot space over the Rational Field of the Root system of type ['A', 3] coweight_lattice( extended=False)¶ Returns the coweight lattice associated to self. This is the weight lattice of the dual root system. EXAMPLES: sage: RootSystem(['A',3]).coweight_lattice() Coweight lattice of the Root system of type ['A', 3] sage: RootSystem(['A',3,1]).coweight_lattice(extended = True) Extended coweight lattice of the Root system of type ['A', 3, 1] coweight_space( base_ring=Rational Field, extended=False)¶ Returns the coweight space associated to self. This is the weight space of the dual root system. EXAMPLES: sage: RootSystem(['A',3]).coweight_space() Coweight space over the Rational Field of the Root system of type ['A', 3] sage: RootSystem(['A',3,1]).coweight_space(extended=True) Extended coweight space over the Rational Field of the Root system of type ['A', 3, 1] dynkin_diagram()¶ Returns the Dynkin diagram of the root system. EXAMPLES: sage: R = RootSystem(['A',3]) sage: R.dynkin_diagram() O---O---O 1 2 3 A3 index_set()¶ EXAMPLES: sage: RootSystem(['A',3]).index_set() (1, 2, 3) is_finite()¶ Returns True if self is a finite root system. EXAMPLES: sage: RootSystem(["A",3]).is_finite() True sage: RootSystem(["A",3,1]).is_finite() False is_irreducible()¶ Returns True if self is an irreducible root system. EXAMPLES: sage: RootSystem(['A', 3]).is_irreducible() True sage: RootSystem("A2xB2").is_irreducible() False root_lattice()¶ Returns the root lattice associated to self. EXAMPLES: sage: RootSystem(['A',3]).root_lattice() Root lattice of the Root system of type ['A', 3] root_poset( restricted=False, facade=False)¶ Returns the (restricted) root poset associated to self. The elements are given by the positive roots (resp. non-simple, positive roots), and $\alpha \leq \beta$ iff $\beta - \alpha$ is a non-negative linear combination of simple roots. INPUT: restricted– (default:False) if True, only non-simple roots are considered. facade– (default:False) passes facade option to the poset generator. EXAMPLES: sage: Phi = RootSystem(['A',2]).root_poset(); Phi Finite poset containing 3 elements sage: sorted(Phi.cover_relations(), key=str) [[alpha[1], alpha[1] + alpha[2]], [alpha[2], alpha[1] + alpha[2]]] sage: Phi = RootSystem(['A',3]).root_poset(restricted=True); Phi Finite poset containing 3 elements sage: sorted(Phi.cover_relations(), key=str) [[alpha[1] + alpha[2], alpha[1] + alpha[2] + alpha[3]], [alpha[2] + alpha[3], alpha[1] + alpha[2] + alpha[3]]] sage: Phi = RootSystem(['B',2]).root_poset(); Phi Finite poset containing 4 elements sage: Phi.cover_relations() [[alpha[2], alpha[1] + alpha[2]], [alpha[1], alpha[1] + alpha[2]], [alpha[1] + alpha[2], alpha[1] + 2*alpha[2]]] root_space( base_ring=Rational Field)¶ Returns the root space associated to self. EXAMPLES: sage: RootSystem(['A',3]).root_space() Root space over the Rational Field of the Root system of type ['A', 3] weight_lattice( extended=False)¶ Returns the weight lattice associated to self. EXAMPLES: sage: RootSystem(['A',3]).weight_lattice() Weight lattice of the Root system of type ['A', 3] sage: RootSystem(['A',3,1]).weight_space(extended = True) Extended weight space over the Rational Field of the Root system of type ['A', 3, 1] weight_space( base_ring=Rational Field, extended=False)¶ Returns the weight space associated to self. EXAMPLES: sage: RootSystem(['A',3]).weight_space() Weight space over the Rational Field of the Root system of type ['A', 3] sage: RootSystem(['A',3,1]).weight_space(extended = True) Extended weight space over the Rational Field of the Root system of type ['A', 3, 1] sage.combinat.root_system.root_system. WeylDim( ct, coeffs)¶ The Weyl Dimension Formula. INPUT: type- a Cartan type coeffs- a list of nonnegative integers The length of the list must equal the rank type[1]. A dominant weight hwv is constructed by summing the fundamental weights with coefficients from this list. The dimension of the irreducible representation of the semisimple complex Lie algebra with highest weight vector hwv is returned. EXAMPLES: For $SO(7)$, the Cartan type is $B_3$, so: sage: WeylDim(['B',3],[1,0,0]) # standard representation of SO(7) 7 sage: WeylDim(['B',3],[0,1,0]) # exterior square 21 sage: WeylDim(['B',3],[0,0,1]) # spin representation of spin(7) 8 sage: WeylDim(['B',3],[1,0,1]) # sum of the first and third fundamental weights 48 sage: [WeylDim(['F',4],x) for x in ([1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1])] [52, 1274, 273, 26] sage: [WeylDim(['E', 6], x) for x in ([0, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 2], [0, 0, 0, 0, 1, 0], [0, 0, 1, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 1], [2, 0, 0, 0, 0, 0])] [1, 78, 27, 351, 351, 351, 27, 650, 351]
Books by Independent Authors Advanced Real Analysis 2017, 179-211 Chapter V. Distributions Abstract This chapter makes a detailed study of distributions, which are continuous linear functionals on vector spaces of smooth scalar-valued functions. The three spaces of smooth functions that are studied are the space $C^{\infty}_{\mathrm{com}}(U)$ of smooth functions with compact support in an open set $U$, the space $C^{\infty}(U)$ of all smooth functions on $U$, and the space of Schwartz functions $\mathcal{S}(\mathbb{R}^{N})$ on $\mathbb{R}^{N}$. The corresponding spaces of continuous linear functionals are denoted by $\mathcal{D}^{\prime}(U)$, $\mathcal{E}^{\prime}(U)$, and $\mathcal{S}^{\prime}(\mathbb{R}^{N})$. Section 1 examines the inclusions among the spaces of smooth functions and obtains the conclusion that the corresponding restriction mappings on distributions are one-one. It extends from $\mathcal{E}^{\prime}(U)$ to $\mathcal{D}^{\prime}(U)$ the definition given earlier for support, it shows that the only distributions of compact support in $U$ are the ones that act continuously on $C^{\infty}(U)$, it gives a formula for these in terms of derivatives and compactly supported complex Borel measures, and it concludes with a discussion of operations on smooth functions. Sections 2-3 introduce operations on distributions and study properties of these operations. Section 2 briefly discusses distributions given by functions, and it goes on to work with multiplications by smooth functions, iterated partial derivatives, linear partial differential operators with smooth coefficients, and the operation $(\cdot)^{\vee}$ corresponding to $x \mapsto -x$. Section 3 discusses convolution at length. Three techniques are used—the realization of distributions of compact support in terms of derivatives of complex measures, an interchange-of-limits result for differentiation in one variable and integration in another, and a device for localizing general distributions to distributions of compact support. Section 4 reviews the operation of the Fourier transform on tempered distributions; this was introduced in Chapter III. The two main results are that the Fourier transform of a distribution of compact support is a smooth function whose derivatives have at most polynomial growth and that the convolution of a distribution of compact support and a tempered distribution is a tempered distribution whose Fourier transform is the product of the two Fourier transforms. Section 5 establishes a fundamental solution for the Laplacian in $\mathbb{R}^{N}$ for $N > 2$ and concludes with an existence theorem for distribution solutions to $\Delta u = f$ when $f$ is any distribution of compact support. Chapter information Source Advanced Real Analysis, Digital Second Edition, Corrected version (East Setauket, NY: Anthony W. Knapp, 2017) Dates First available in Project Euclid: 21 May 2018 Permanent link to this document https://projecteuclid.org/euclid.bia/1526871319 Digital Object Identifier doi:10.3792/euclid/9781429799911-5 Rights Copyright © 2017, Anthony W. Knapp Citation Knapp, Anthony W. Chapter V. Distributions. Advanced Real Analysis, 179--211, Anthony W. Knapp, East Setauket, New York, 2017. doi:10.3792/euclid/9781429799911-5. https://projecteuclid.org/euclid.bia/1526871319
Next year is the 70th anniversary of the publication of the book 1984 by George Orwell. Here is a puzzle to start the anniversary celebrations off a bit early ... Warm up Can you assemble a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $246$. You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt{x}$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$, as long as all operands are either $1$, $9$, $8$, or $4$. Operands may of course also be derived from calculations e.g. $198(\sqrt{4})$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $8$ and $4$ to make the number $84$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations will get plus one from me. Main Event If you used concatenation above then make a formula using the numbers $1$, $9$, $8$, and $4$ in any order so that the results equals $246$ without using any concatenation. so, for example, you cannot put $8$ and $4$ together to make the number $84$. The rest of the rules above apply, but concatenation is not allowed. If you didn't use any concatenation above then you have solved the puzzle, but you could try to solve it with concatenation, that is concatenation of the initial numbers only. Note ( and perhaps hint): For this second part any finite number of functions can be used, though ingenious solutions with infinite numbers of functions will get plus one from me. Note that in all the puzzles above Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 \times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get the required answers, but will not mark them as correct - particularly because a general method was developed by @Carl Schildkraut to solve these puzzles. many thanks to the authors of the similar questions below for inspiring this question.
Answer The values of the components of the vector are $22\sqrt{3}\text{i}+22\text{j}$. Work Step by Step The speed of the football $=44$ feet per second. The angle that the football makes with the horizontal $=30{}^\circ $. The vector component along the x-axis or horizontal is i and the vector component along they-axis or the vertical direction is j. Therefore, the equation becomes $\begin{align} & v=\left\| \left\| v \right\| \right\|\cos \theta \text{i}\,+\left\| \left\| v \right\| \right\|\sin \theta \text{j} \\ & =44\cos 30{}^\circ \text{i}+\,44\sin 30{}^\circ \text{j} \\ & =44\left( \frac{\sqrt{3}}{2} \right)\text{i}+44\left( \frac{1}{2} \right)\text{j} \\ & =22\sqrt{3}\text{i}\,+22\text{j} \end{align}$
Consider two $\ce{H}$ atoms. Since the proton in one attracts the electron in another, they attract each other, and form a covalent bond ($\ce{H-H}$). Bond forming requires energy (436 kJ/mol for an $\ce{H-H}$ bond) – where does that energy come from? If I put two $\ce{H}$ atoms in a container, would they form an $\ce{H-H}$ bond by themselves, or would I need to add energy to the system in order for them to bond? Consider two $\ce{H}$ atoms. Since the proton in one attracts the electron in another, they attract each other, and form a covalent bond ($\ce{H-H}$). Bond forming requires energy (436 kJ/mol for an $\ce{H-H}$ bond) – where does that energy come from? Your view of bonding is a little simplified and confused. For the details it is worth reading the great answer referred to in some comments already: Fundamental forces behind covalent bonding . I'll stick with a simpler view that focusses just on the energy involved and not the mechanism of chemical bonding. Firstly, you are wrong that boding two hydrogen atoms needs an input of energy. It actually releases energy. This is a thermodynamic view of the reaction. This tells us whether there is more or less energy in the product of the reaction being described than in the chemicals being combined to create the product. But a lot of chemical reactions are thermodynamically favourable (that is the reaction releases energy) don't happen easily (this is probably not true for atomic hydrogen which should react very easily to form H 2). But there are plenty of other reactions that release energy according to thermodynamics but that don't happen easily. For example, elemental carbon plus oxygen will release energy if converted to CO 2. But neither the diamond in jewellery or the coal in your fireplace (which is basically impure carbon) reacts easily to give carbon dioxide despite being in an environment with a lot of Oxygen gas in it. Here the problem is that room temperature oxygen molecules and carbon just don't have enough energy to kick off the process that creates the product. This is where we have to look at the kinetic view of the reaction rather than just the thermodynamics. Although the overall reaction is favourable according to thermodynamics, we need to add energy (firelighters for coal fires, a blowtorch or diamond rings) to the mixture to give the atoms in the reactants enough energy to overcome the kinetic barriers to the reaction. Think of having to climb a tall mountain to get to a lower valley on the other side. Or think of giving oxygen molecules enough energy to break carbon atoms apart from their normal solid structure. Once coal or diamond are warm enough, the reaction will happen spontaneously and you have either destroyed your engagement ring or you have a nice coal fire. In this case you can feel the energy released from the reaction as a pleasant warming of your room. Back to the question. The energy involved in a reaction is an accounting of the amount of energy used in the bonding of the products versus the reactants. The energy released in reactions such as burning carbon goes into the environment as heat. Other reactions happen spontaneously (try dropping a small lump of sodium into water). Some reactions require a tiny amount of encouragement (oxygen and hydrogen in certain concentrations will react explosively to form water when encouraged with the tiniest spark). We can drive some reactions the other way using inputs of energy. For example we can electrolyse water into a mixture of hydrogen and oxygen. This requires us to add energy to the system (about 286kJ/mol) because the products are in a higher energy state then the reactants. If you just focus on the thermodynamics of a reaction the numbers (kcal or kJ/mol) tell you how much the energy difference between the products and reactants is. But many reactions that should release energy require some input to kick off the reaction. This is confusing, but chemistry wouldn't be interesting and we wouldn't exist if thermodynamics were the only thing driving reactions. Also energy is not created or destroyed here, just moved around. You can put energy in to a system to drive a reaction or the environment gets energy back from reactions that release it. So, some reactions require energy to drive them because the products have higher energy, some release energy because the products have lower energy but many release energy but, confusingly, require some energy input to kick off the actual reaction process. As others have mentioned energy is released on bonding, which is why $H_2$ is more stable that two H atoms. To answer your last question first, H atoms in a box would eventually form $H_2$ but only if some third body (another H atom or wall of vessel) can take away some of the energy released before the H atoms fly apart again. This leaves $H_2$ in a highly vibrational state, further collisions will remove the remaining energy until it is at equilibrium with its surroundings. Now as to bond formation, its easier to examine $H_2^+$, this has two protons and one electron; the simplest possible bond. The calculation involves evaluating the interaction between the two protons and either proton and the electron. These are of the coulomb type, that is product of charges divided by separation. (No complications from Pauli principle here as there is only one electron.) Solving Schroedinger's equation involves terms with kinetic and potential energy (this is the Hamiltonian $H_0$). The terms for electron proton interaction are negative ($-e^2/r_1 - e^2/r_2$, where e is the electronic charge and $r_1$ the distance of proton 1 to the electron and similarly for $r_2$)) and the potential energy between protons is $+e^2/R$ where R is the internuclear separation). The calculation proceeds by technical steps, say by using the variational theorem, and integrating over all distances and angles and distances using a basis set of atomic wavefunctions, call them $\phi_1$ and $\phi_2$.(for details see Cohen-Tannoudji, Diu & Laloe, Quantum Mechanics vol II for a full calculation). We find that there are Coulomb integrals of the form $\int \phi_1\frac{e^2}{r_2} \phi_1 d\tau$, and $\int \phi_2\frac{e^2}{r_1} \phi_2 d\tau$. ($d \tau$ simply means integrate over all coordinates). These give the energy of the electrostatic interaction between proton 2 and charge distribution of the electron were it in the 1s orbital of proton 1 and vice versa. In addition there are integrals of the form $A=\int \phi_1\frac{e^2}{r_1} \phi_2 d\tau$ which have no classical equivalent, i.e. they would not arise in a classical calculation. These are sometimes called 'resonance integrals' or 'exchange integrals' although this latter term can mean something else to physicists. The integral A is difficult to explain, other than to say that it arises out of the mathematics used when making a linear combination of atomic orbitals. The fact that integral A is not zero expresses the possibility of the electron ' jumping' as it were from the neighbourhood of one proton to that of the other and hence the possibility of 'quantum resonance' (The term 'resonance' arises by analogy with oscillating pendulums, see Coulson, Valence chapter 4 but does not imply actual motion). The combination of these integrals (and the overlap integral $S=\int \phi_1\phi_2 d\tau$) produces a minimum in the potential energy profile as the calculation is repeated at different internuclear separations R and hence describes the energy of the $H_2^+$ molecule vs. separation of the nuclei. The total energy comprises kinetic and potential energy terms, so is it the decrease in kinetic energy or potential energy that is responsible for the chemical bond? The overwhelming effect is due to the decrease in potential energy. This is also confirmed by Virial Theorem calculations. In the $H_2$ molecule we can carry over what has been done for $H_2^+$. Now we have two valence electrons instead of one, they will both occupy an orbital similar to that in $H_2^+$ and the electrons will have opposite spins by Pauli Principle. The main difference is that there are now two electron contributing to binding and the bond if stronger. The essential principles outlined here can be carried through to forming bonds between other atoms although the calculations become tremendously more involved as there are many more electrons.
This question already has an answer here: I am trying to solve the following problem: For each Turing machine $M_k$ and each string $x$ in $\{$0,1$\}$$^\ast$ let $time_k(x)$ = $\{$the number of steps executed by $M_k(x)$ if $M_k(x)$$\downarrow$ (halts), and $\infty$ if $M_k(x)$$\uparrow$ (does not halt)$\}$ Prove that the function $T$: $\mathbb{N}$ $\rightarrow$ $\mathbb{N}$ defined by $T(n)$ = max$\{$$time_k(x)$ \| $0$ $\leq$ $k$ $\leq$ $n$, $x$ $\in$ $\{$0,1$\}$$^\ast$, and $M_k(x)$$\downarrow$ (halts)$\}$ is uncomputable. So far, I have begun my proof by assuming that $T$ is computable. Thus, there exists a Turing machine $M$ such that for all $n$$\in$$\mathbb{N}$, $M$ produces $T(n)$ on its tape. Thus, we must show that we can decide the Halting Problem if $T$ is computable, which in turn lets us know that $T$ is uncomputable since the Halting Problem is uncomputable. I do not know where to go from there however. Any help would be greatly appreciated. Thanks in advance.
We define complex power functions using the exponential and logarithm functions:$$z^\alpha = e^{\alpha\log z}.$$The problem with this definition is that the complex logarithm function cannot be defined continuously on the entire punctured complex plane. (The punctured plane is the plane with zero removed. Of course you cannot define the logarithm to be continuous at zero, but there is a deeper problem. By contrast, the exponential function is defined continuously on the entire complex plane, so that part of the formula doesn't give a problem.) To explain why, switching to polar coordinates and writing $z=r e^{i\theta}$ gives$$\log z = \log r + i\theta.$$But in general, there is no continuous way to assign a complex argument (the angle $\theta$ above) to every complex number in the punctured plane. More specifically, it is clear from the formula above that all possible values of $\log z$ differ by an integer multiple of $2\pi i$. So, we can say informally that the logarithm function is multi-valued:$$\log z = \log r + i\theta + 2ni\pi.$$Moving back to the definition above of power function, this means that the power function $z^\alpha$ is multi-valued:$$z^\alpha = e^{\alpha\log z} = e^{2n\pi i\alpha}e^{\alpha\log z} = e^{2n\pi i\alpha}z^\alpha.$$So in other words, $z^\alpha$ is only defined up to a factor of powers of $e^{2\pi i\alpha}$. For example, the possible values of $i^i$ are$$\dots,e^{-3\pi},e^{-\pi},e^\pi,e^{3\pi},\dots.$$If $\alpha$ is an integer then $2n\pi i\alpha$ is an integer multiple of $2\pi i$ (since $e^{2\pi i}=1$; this is related to Euler's formula) and so $z^\alpha$ has only one possible value; this is why we can define e.g. $z^{45}$ without any ambiguity. So to finally answer your question, the formula $z^{a+bi} = z^a\cdot z^{bi}$ is valid as long as we choose the right values for $z^{a+bi}$, $z^a$, and $z^{bi}$, thinking of these as multi-valued functions.
In ZFC the domain of discourse is all sets. So when we say 'for all $X$' we mean for all sets $X$. So the power set axiom can be written in more detail as "for every set $X$ there is a set $P(X),$ whose elements are all sets that are subsets of $X.$" (What 'all' sets means is a different can of worms and varies from model to model. But in any event, proper classes are not sets by definition, so are not included. In fact they can't be directly expressed as objects in ZFC's language. Rather, they are talked about indirectly, via formulas. So a class is just a formula, whose extension may or may not correspond to a set.) Edit (In response to your comments) It is wrong to think of the ZFC axioms as 'defining what a set is' although I can see why you might be under that impression since most of the axioms essentially tell you how to define new sets from previously defined ones. Rather, the ZFC axioms define what a set theoretical universe is, in that a set theoretical universe is a collection of objects and a membership relation $\in$ on those objects for which the axioms hold. Although we certainly want such a theory to be consistent, and paradox avoidance is what motivated the restricted form of comprehension that ZFC has, it's also important to realize that ZFC makes some 'arbitrary' choices that have nothing to do with paradox avoidance. For instance, the axiom of extensionality implies that everything in our universe is a set and, for instance, there are no 'pure elements' (usually called urelements) that are not set-like in nature. (It does so by implying that the empty set is the only thing that doesn't contain anything else.) Also, the axiom of foundation guarantees that there are no strange sets that contain themselves, e.g. so-called Quine atoms obeying an equation like $x=\{x\},$ which are not inherently paradoxical despite not being as natural as urelements and superficially smelling like Russell's paradox. It is useful since it implies that the universe of sets is arranged in a Von Neumann hierarchy The other axioms (including perhaps choice) are less arbitrary, in the sense that they formalize our intuition from naive set theory of what kinds of sets we should be able to define based on others. It is also important to note that ZFC is not a full characterization of a unique universe of sets. For instance, you have probably heard that the continuum hypothesis is independent of ZFC and that the axiom of choice is independent of ZF. This means that there are models of ZF/C that differ on these questions, and thus we haven't uniquely specified a universe of sets. (In fact we could have never hoped to do anything close to this in a first order theory, with results like Lowenheim-Skolem. And yet if we try to move to a second order theory we run into another problem that we need set theory in our metatheory so set theoretical questions our theory 'answers' may well depend on what the answer is in the metatheory.) As other have noted in the comments, this is not a lot different from how we usually proceed in axiomatizing math. The group axioms don't 'define what a group element is', they tell you the properties all groups have (and note there are many groups just as there are many models of ZFC). Same goes for the axiomatization of any type of mathematical object. Sometimes, as in the second order Peano axioms, there is only one model (up to isomorphism) obeying them, but that is a rare case. This issue of circularity that you bring up is called impredicativity and a lot has been written about it. Essentially, it is the problem of circularity when the definition of an object involves quantifying over a universe containing that object. So for instance, when you define the empty set as 'the set that doesn't contain anything,' this definition includes an implicit clause that the empty set doesn't contain the empty set, and thus seems to require we know what the empty set is prior to our defining the empty set. Of course this is a silly example since it's reasonably clear that the notion of the empty set is self-consistent (see also the example of 'the tallest person in the room' from the Wikipedia page), so not all impredicative definitions are problematic. Russell originally believed impredicativity was the ultimate source of his paradox and thus something that should be eliminated from formal foundations mathematics. However he was unable to develop mathematics satisfactorily on a predicative basis. There have been more successful attempts since then, but in general, much of classical mathematics is either impossible or intolerably difficult to develop only using predicative definitions. (Compare to the situation with constructive mathematics.) Still, like constructivity, predicativity has its proponents, and consensus seems to be that it is at least something worth paying attention to as an organizing principle. For instance, in a second order theory, once can restrict comprehension principles to only apply to predicative formulas (i.e. ones that do not quantify over properties). ZFC is "hopelessly" impredicative, in the sense that if you think predicativity is a requirement in some sense, you probably reject ZFC foundations. You are correct to highlight the power set axiom as a weak point here, and there is also essential impredicativity in the separation and replacement axioms. (Kripke-Platek set theory (which is often viewed as a predicative fragment of ZFC) eliminates power set and choice and restricts both separation and replacement. Notice, however, according to the the view I expressed at the outset (which is the view Jech and the majority of modern set theorists hold), impredicativity is not a problem. To be a little more specific, way to think about it is we have a (perhaps hypothetical) collection $V$ of objects related by the membership relation $\in$ that obey the ZFC axioms. This universe is fixed (though we don't necessarily specify it fully other than that it obeys the ZFC axioms). Then it makes sense that the axioms say 'for all sets, yada yada'. They are true statements about this universe of sets. They do tell us some that sets with certain properties have to exist and for certain properties (like $\forall x (x=x)$ by Cantor's paradox) there can't be a set with that extension, but again, this is a description of the universe. If the axioms simply describe a universe of sets rather than 'building one from the ground up', then impredicative definitions pose no risk. It's just like the example with the tallest person in the room: if you really have a room with some people in it and aren't defining the people in the room into existence, there can't be any problems brought on by the circularity here. This isn't philosophically bulletproof since of course who says these set theoretical universes even exist, but still, given the difficulties with predicative mathematics and the general belief amongst set theorists that ZFC is consistent and that set theoretical universes either really exist in some sense, or are at least worth reasoning about as hypothetical objects, it is the prevailing attitude. The idea that taking a Platonist view of the domain makes predicativity less of a concern leads readily to the idea of relative predicativity, where certain objects are taken to 'exist at the outset' and everything else is defined predicatively. An example that came up in the comments was the constructible universe $L$. If you are given the ordinals, $L$ is a completely predicative construction where every set (except the ordinals you start with) can be regarded as 'defined into existence' from the bottom up. Edit 2This is already overlong and a bit unfocused (though I think the commenters-- including a few professional logicians in case authority makes a difference-- have done a really good job of hammering home the main take-aways) but I thought it would be useful to actually spell out how the definition of the power set should be thought of here. First of all, there is the subset property $y\subseteq x$ which is an abbreviation for $\forall z (z\in y \to z\in x),$ which hopefully already seemed sharp and unproblematic here. We don't even really quantify over 'all sets', just the elements of $y.$ Then, the power set axiom says $$\forall x \exists!z \forall u (u\in z \leftrightarrow u \subseteq x)$$ which informs us we are assuming that a set theoretical universe has, for any set $x,$ a unique power set $z$ whose elements are exactly those sets $u$ that obey the relation $u\subseteq x$ (just as we assume a group has an identity element with certain properties). (Note we don't actually have to put all this in the axiom... for instance extensionality already guarantees the uniqueness part, and if we wish, we can weaken the $\leftrightarrow$ to a $\leftarrow$ and use subset comprehension to pick up the slack.) Now, since we know there is a unique set with this property, we can define an operation $\mathcal P:V\to V$ such that given a set $x,$ $\mathcal P(x)$ is the unique set with this property. So we have argued that within any set-theortical universe, that this operation is well-defined and thus we have a power set for every set. If you are a bit unsatisfied and think it's tautoglogical and we rigged that to happen, well, that's the point of the power set axiom. It's a thing that guarantees the existence of a power set for every set.
Positive ground state solutions of a quadratically coupled schrödinger system School of Mathematical Sciences, Shanxi University, Taiyuan 030006, Shanxi, China $\begin{equation}\left\{\begin{array}{ll}-\Delta u+\lambda_1u=\mu_1u^2+2\alpha uv+\gamma v^2, & \mbox{in }\Omega,\\-\Delta v+\lambda_2v=\mu_2v^2+2\gamma uv+\alpha u^2, & \mbox{in }\Omega,\\u=v=0, & \mbox{on }\partial\Omega,\end{array}\right.\end{equation}$ J. Math. Anal. Appl., 430(2):950-970, 2015], the existence of positive ground state solutions of the system is established with more ingenious hypotheses. Keywords:Schrödinger system, quadratic nonlinearities, critical growth, ground state, ingenious hypotheses. Mathematics Subject Classification:Primary: 35J47; Secondary: 35J50. Citation:Zhanping Liang, Yuanmin Song, Fuyi Li. Positive ground state solutions of a quadratically coupled schrödinger system. Communications on Pure & Applied Analysis, 2017, 16 (3) : 999-1012. doi: 10.3934/cpaa.2017048 References: [1] [2] T. Bartsch, N. Dancer and Z.-Q. Wang, A Liouville theorem, a-priori bounds and bifurcating branches of positive solutions for a nonlinear elliptic system, [3] [4] H. Brézis and L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, [5] A. V. Buryak, P. Di Trapani, D. V. Skryabin and S. Trillo, Optical solitons due to quadratic nonlinearities: from basic physics to futuristic applications, [6] [7] [8] Z. Chen and W. Zou, Positive least energy solutions and phase separation for coupled Schrödinger equations with critical exponent, [9] Z. Chen and W. Zou, An optimal constant for the existence of least energy solutions of a coupled Schrödinger system, [10] Z. Chen and W. Zou, Standing waves for linearly coupled Schrödinger equations with critical exponent, [11] E. N. Dancer, J. Wei and T. Weth, A priori bounds versus multiple existence of positive solutions for a nonlinear Schrödinger system, [12] D. Gilbarg and N. S. Trudinger, [13] Z. Guo, Positive ground state solutions for a nonlinearly coupled Schrödinger system with critical exponents in ℝ [14] [15] [16] [17] A. C. Yew, A. R. Champneys and P. J. McKenna, Multiple solitary waves due to secondharmonic generation in quadratic media, [18] [19] L. Zhao, F. Zhao and J. Shi, Higher dimensional solitary waves generated by second-harmonic generation in quadratic media, show all references References: [1] [2] T. Bartsch, N. Dancer and Z.-Q. Wang, A Liouville theorem, a-priori bounds and bifurcating branches of positive solutions for a nonlinear elliptic system, [3] [4] H. Brézis and L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, [5] A. V. Buryak, P. Di Trapani, D. V. Skryabin and S. Trillo, Optical solitons due to quadratic nonlinearities: from basic physics to futuristic applications, [6] [7] [8] Z. Chen and W. Zou, Positive least energy solutions and phase separation for coupled Schrödinger equations with critical exponent, [9] Z. Chen and W. Zou, An optimal constant for the existence of least energy solutions of a coupled Schrödinger system, [10] Z. Chen and W. Zou, Standing waves for linearly coupled Schrödinger equations with critical exponent, [11] E. N. Dancer, J. Wei and T. Weth, A priori bounds versus multiple existence of positive solutions for a nonlinear Schrödinger system, [12] D. Gilbarg and N. S. Trudinger, [13] Z. Guo, Positive ground state solutions for a nonlinearly coupled Schrödinger system with critical exponents in ℝ [14] [15] [16] [17] A. C. Yew, A. R. Champneys and P. J. McKenna, Multiple solitary waves due to secondharmonic generation in quadratic media, [18] [19] L. Zhao, F. Zhao and J. Shi, Higher dimensional solitary waves generated by second-harmonic generation in quadratic media, [1] Marco A. S. Souto, Sérgio H. M. Soares. Ground state solutions for quasilinear stationary Schrödinger equations with critical growth. [2] Yanfang Xue, Chunlei Tang. Ground state solutions for asymptotically periodic quasilinear Schrödinger equations with critical growth. [3] Kaimin Teng, Xiumei He. Ground state solutions for fractional Schrödinger equations with critical Sobolev exponent. [4] Yongpeng Chen, Yuxia Guo, Zhongwei Tang. Concentration of ground state solutions for quasilinear Schrödinger systems with critical exponents. [5] Lun Guo, Wentao Huang, Huifang Jia. Ground state solutions for the fractional Schrödinger-Poisson systems involving critical growth in $ \mathbb{R} ^{3} $. [6] Kenji Nakanishi, Tristan Roy. Global dynamics above the ground state for the energy-critical Schrödinger equation with radial data. [7] Jianhua Chen, Xianhua Tang, Bitao Cheng. Existence of ground state solutions for a class of quasilinear Schrödinger equations with general critical nonlinearity. [8] Xu Zhang, Shiwang Ma, Qilin Xie. Bound state solutions of Schrödinger-Poisson system with critical exponent. [9] Yong-Yong Li, Yan-Fang Xue, Chun-Lei Tang. Ground state solutions for asymptotically periodic modified Schr$ \ddot{\mbox{o}} $dinger-Poisson system involving critical exponent. [10] Sitong Chen, Xianhua Tang. Existence of ground state solutions for the planar axially symmetric Schrödinger-Poisson system. [11] Sitong Chen, Junping Shi, Xianhua Tang. Ground state solutions of Nehari-Pohozaev type for the planar Schrödinger-Poisson system with general nonlinearity. [12] Claudianor Oliveira Alves, M. A.S. Souto. On existence and concentration behavior of ground state solutions for a class of problems with critical growth. [13] Miao-Miao Li, Chun-Lei Tang. Multiple positive solutions for Schrödinger-Poisson system in $\mathbb{R}^{3}$ involving concave-convex nonlinearities with critical exponent. [14] Chuangye Liu, Zhi-Qiang Wang. A complete classification of ground-states for a coupled nonlinear Schrödinger system. [15] Daniele Garrisi, Vladimir Georgiev. Orbital stability and uniqueness of the ground state for the non-linear Schrödinger equation in dimension one. [16] Zhitao Zhang, Haijun Luo. Symmetry and asymptotic behavior of ground state solutions for schrödinger systems with linear interaction. [17] Xianhua Tang, Sitong Chen. Ground state solutions of Nehari-Pohozaev type for Schrödinger-Poisson problems with general potentials. [18] Chenmin Sun, Hua Wang, Xiaohua Yao, Jiqiang Zheng. Scattering below ground state of focusing fractional nonlinear Schrödinger equation with radial data. [19] Xiaoyan Lin, Yubo He, Xianhua Tang. Existence and asymptotic behavior of ground state solutions for asymptotically linear Schrödinger equation with inverse square potential. [20] Chao Ji. Ground state solutions of fractional Schrödinger equations with potentials and weak monotonicity condition on the nonlinear term. 2018 Impact Factor: 0.925 Tools Metrics Other articles by authors [Back to Top]
The GBJ package implements the Generalized Berk-Jones (GBJ) test for set-based inference in genetic association studies. Also included in this package are routines to perform the Generalized Higher Criticism (GHC), Higher Criticism (HC), Berk-Jones (BJ), and Minimum p-value (minP) tests. Some guidance on choosing between these methods (which, in principal, test the same null hypothesis and could be used interchangeably) is also given in the above paper. The remainder of this vignette provides: We find it instructional to begin with a short anecdote from the class notes of John Tukey (Donoho and Jin, 2004): “A young psychologist administers many hypothesis tests as part of a research project, and finds that, of 250 tests 11 were significant at the 5% level. The young researcher feels very proud of this fact and is ready to make a big deal about it, until a senior research suggests that one would expect 12.5 significant tests even in the purely null case, merely by chance. In that sense, finding only 11 significant results is actually somewhat disappointing! … [Tukey] then proposed a sort of second-level significance testing, … [to] indicate a kind of significance of the overall body of tests.” All the tests performed in this package are designed to carry out the sort of second-level significance testing suggested in the story. That is, they are designed to test if there is at least one non-null hypothesis in the entire group of hypotheses. In statistical terms, assume that the researcher above had calculated 250 Z-scores, where $Z_{i}$ has mean $\mu_{i}$ and variance 1. Then the GBJ null hypothesis is that $\mu_{i}=0$ for all $i$, and the GBJ alternative is that $\mu_{i} \neq 0$ for at least one $i$. GBJ can be used to test either an entire collection of hypotheses, or it may make more sense to partition the group into smaller, predefined sets and then apply GBJ multiple times. For example, in the case of GWAS, we can group the individual SNP test statistics into genes/pathways and apply GBJ to each gene or pathway. In this case, GBJ is testing if the entire gene has any association with the outcome. Notable advantages to using GBJ are: GBJ is a generalization of the Berk-Jones test, which is known to be optimal - in a certain sense - for detecting rare and weak signals when factors in a set are independent. This is clearly a very relevant guarantee for the genetics setting. GBJ modifies Berk-Jones to provide better finite sample rejection regions when factors in a set are correlated. Analytic calculation of p-values (no need for permutation). Suppose we are interested in testing whether a specific gene is associated with pancreatic cancer. We have 1000 patients in our study, half with pancreatic cancer and half without. Our dataset consists of 50 SNPs in the gene of interest, and for each patient we have their minor allele count (0,1,2) at each of the 50 SNPs. Additionally we have information on each patient’s age and gender: library(GBJ)set.seed(0)cancer_status <- c(rep(1,500), rep(0,500))# All of our SNPs have minor allele frequency of 0.3 in this examplegenotype_data <- matrix(data=rbinom(n=100050, size=2, prob=0.3), nrow=1000)age <- round( runif(n=1000, min=30, max=80) )gender <- rbinom(n=1000, size=1, prob=0.5) # Let 1 denote a female and 0 a male Under the null hypothesis of no association between gene and pancreatic cancer, we can assume the true logistic model to be: \[\text{logit}(\mu_{i}) = \beta_{0} + \beta_{1}Age_{i} + \beta_{2}Gender_{i}\] (actually since we generated the data, we know $\beta_{1}=\beta_{2}=0$, but this is just for illustration) The function calc_score_stats() can be used to calculate score statistics for each of the 50 SNPs. Under the null, these statistics will have an asymptotic N(0,1) distribution. null_mod <- glm(cancer_status~age+gender, family=binomial(link="logit"))log_reg_stats <- calc_score_stats(null_model=null_mod, factor_matrix=genotype_data, link_function="logit")log_reg_stats$test_stats ## [1] -0.68383391 -1.52282289 -0.04324001 1.02900792 -2.08636456## [6] -0.47750816 1.02134595 0.57157512 0.09543809 1.88826217## [11] -0.50314642 -0.39237040 -1.00903095 0.70567669 -0.74408222## [16] 1.17911250 0.63650614 -1.45052034 -0.04092308 1.06692697## [21] -1.39140523 -0.15765491 0.11381805 0.44208014 -0.62752319## [26] 0.77393856 -0.06840144 -0.27651351 0.03332305 0.96798877## [31] 0.65627672 0.18681278 -1.11695309 -1.58261616 1.44547339## [36] 1.78314038 0.88944355 0.44694854 -1.34907210 1.02299625## [41] -0.59392262 -0.87602299 0.55980215 -1.39571015 -0.16295461## [46] -0.86944840 -0.21030648 -1.61408239 0.64425823 1.71642921 log_reg_stats$cor_mat[1:5,1:5] ## [,1] [,2] [,3] [,4] [,5]## [1,] NA 0.02020415 -0.01580513 0.03906158 0.005420753## [2,] 0.020204152 NA -0.03967446 0.03321533 0.028975255## [3,] -0.015805130 -0.03967446 NA -0.02172816 0.034011177## [4,] 0.039061583 0.03321533 -0.02172816 NA -0.043455163## [5,] 0.005420753 0.02897526 0.03401118 -0.04345516 NA If our outcome was continuous and we wanted to assume a linear regression model, then we could still use calc_score_stats() with link_function='linear', or if the outcome was non-negative count data and we assume a Poisson regression model, then use link_function='log'. Now we have both the test statistics and their correlation matrix, we can apply GBJ. cor_Z <- log_reg_stats$cor_matscore_stats = log_reg_stats$test_statsGBJ(test_stats=score_stats, cor_mat=cor_Z) ## $GBJ## [1] 0.6109885## ## $GBJ_pvalue## [1] 0.6091488## ## $err_code## [1] 0 And that’s it! Now you are not convinced that GBJ is the correct test for your application, you can also apply GHC, HC, BJ, or minP, as demonstrated below: GHC(test_stats=score_stats, cor_mat=cor_Z) ## $GHC## [1] 1.505614## ## $GHC_pvalue## [1] 0.6328255## ## $err_code## [1] 0 HC(test_stats=score_stats, cor_mat=cor_Z) ## $HC## [1] 1.514436## ## $HC_pvalue## [1] 0.633645 BJ(test_stats=score_stats, cor_mat=cor_Z) ## $BJ## [1] 0.6208709## ## $BJ_pvalue## [1] 0.607419 minP(test_stats=score_stats, cor_mat=cor_Z) ## $minP## [1] 0.03694561## ## $minP_pvalue## [1] 0.8439997 Suppose now that we only have GWAS summary statistics, not individual-level data, but we still want to perform a set-level test using GBJ. We then need to estimate the correlations between these summary statistics using genotypes from a reference panel. We have provided a function, estimate_ss_cor() to perform the estimation. estimate_ss_cor() requires as input (1) a matrix of $m$ PCs calculated from a reference panel (of the same ethnicity) and (2) a matrix of the genotypes at the summary statistic SNPs from the same reference panel. Here $m$ is approximately the number of PCs used in the original analysis that produced the summary statistics. # Load the genotype data at FGFR2 SNPs for 91 Great Britain (GBR) subjects from the 1000 Genomes Project (publically available).data(FGFR2)# Load PCs for these same 91 subjects (calculated, for example, with EIGENSTRAT)data(gbr_pcs)# Suppose we were given the following 64 test statistics for the FGFR2 SNPs (must be the same SNPs that# are in our genotype matrix!)FGFR2_stats <- rnorm(n=64)# Estimate correlation matrix for summary statisticsFGFR2_cor_mat <- estimate_ss_cor(ref_pcs=gbr_pcs, ref_genotypes=FGFR2, link_function='logit')# Run GBJGBJ(test_stats=FGFR2_stats, cor_mat=FGFR2_cor_mat) ## $GBJ## [1] 0.3366795## ## $GBJ_pvalue## [1] 0.6390038## ## $err_code## [1] 0 Note that in the first example (individual-level data) we know each of our factors (SNPs) are independent, so another option would be to just input a correlation matrix where all the off-diagonal elements are zero. This should not drastically change the results since our estimated correlations are so close to 0 anyway. Limits on the mathematical precision of R and C++ currently limit us somewhat for very large sets and very small p-values. There are plans to remove these limitations in future iterations of the software by using arbitrary precision libraries, however this has not yet been tested. At the moment, we limit sets to 2000 factors at most. Also p-values less than $110^{(-14)}$ are generally rounded to this value. Questions or novel applications? Please let me know! Contact information can be found in the package description.
There are $n$ bins and $m$ type of balls. The $i$th bin has labels $a_{i,j}$ for $1\leq j\leq m$, it is the expected number of balls of type $j$. You start with $b_j$ balls of type $j$. Each ball of type $j$ has weight $w_j$, and want to put the balls into the bins such that bin $i$ has weight $c_i$. A distribution of balls such that previous condition holds is called a feasible solution. Consider a feasible solution with $x_{i,j}$ balls of type $j$ in bin $i$, then the cost is $\sum_{i=1}^n \sum_{j=1}^m \|a_{i,j}-x_{i,j}\|$. We want to find a minimum cost feasible solution. This problem is clearly NP-hard if there is no restriction on $\{w_j\}$. The subset sum problem reduces to the existence of a feasible solution. However, if we add the condition that $w_j$ divides $w_{j+1}$ for every $j$, then the subset sum reduction no longer works, so it's not clear whether the resulting problem remains NP-hard. Checking for the existence of a feasible solution takes only $O(n\,m)$ time (attached at the end of the question), but this does not give us the minimum-cost feasible solution. The problem has an equivalent integer program formulation. Given $a_{i,j},c_i,b_j,w_j$ for $1\leq i\leq n,1\leq j\leq m$: \begin{align} \text{Minimize:} & \sum_{i=1}^n \sum_{j=1}^m \|a_{i,j}-x_{i,j}\| \\ \text{subject to:} & \sum_{j=1}^m x_{i,j}w_j = c_i \text{ for all } 1\leq i\leq n\\ & \sum_{i=1}^n x_{i,j} \leq b_j \text{ for all } 1\leq j \leq m\\ & x_{i,j}\geq 0 \text{ for all } 1 \leq i\leq n, 1\leq j \leq m\\ \end{align} My question is, Is the above integer program NP-hard when $w_j$ divides $w_{j+1}$ for all $j$? An algorithm to decide if there is a feasible solution in $O(n\,m)$ time: Define $w_{m+1}=w_m(\max_{j} c_j + 1)$ and $d_j = w_{j+1}/w_j$. Let $a\%b$ be the remainer when $a$ is divided by $b$. If there exists a $c_i$ that's not divisible by $w_1$, return "no feasible solution". (the invariant $c_i$ divides $w_j$ will always be maintained in the following loop) for $j$ from $1$ to $m$: $k \gets \sum_{i=1}^n (c_i/w_j)\%d_j$. (the minimum of balls of weight $w_j$ required) If $b_j<k$, return "no feasible solution". $c_i \gets c_i - ((c_i/w_j)\% d_j)$ for all $i$. (remove the minimum number of required balls of weight $w_j$) $b_{j+1} \gets \lfloor (b_j-k)/d_j \rfloor$. (group smaller balls into a larger ball) return "there is a feasible solution". A polynomial time solution to the special case where $n=1$, and $w_j$ are power of $2$s It is known that if $n=1$ and all $w_j$ are powers of $2$, then this special case can be solved in polynomial time. \begin{align} \text{Minimize:} & \sum_{j=1}^m \|a_j-x_j\| \\ \text{subject to:} & \sum_{j=1}^m w_j x_j = c\\ & 0 \leq x_j \leq b_j \text{ for all } 1\leq j \leq m\\ \end{align}
I would like to pose questions occasionally (perhaps I can make a series if this idea is well-received) of the type of the title. In particular I was thinking: Proposed Question: What are some of the immediate consequences of the Lindelmann Weierstraß Theorem? I am interested in collecting a list of some of the things that an undergraduate could prove in 4 or 5 lines after learning LW and knowing some pretty basic results about algebraic numbers. Let's assume we know that $\alpha +\beta i$ is algebraic iff $\alpha$ and $\beta$ are algebraic. Then as soon as one learns LW we can show that entire slews of numbers are transcendental. Knowing how powerful a workhorse LW is I am sure there are many things we can see after just a few sentences. I would almost prefer to see them unproven. Just leave the conclusion and statement like : No more than 3 sentences are required. Another user may elect to put comments below asking for clarification of the proof or may elect to update your answer with the proof. So the question: What is the fallout after achieving the following Lindemann–Weierstrass Theorem (Baker's reformulation) For algebraic numbers $a_1, \dots a_n$ not all equal to zero and distinct algebraic numbers $\alpha_1, \dots \alpha_n$ we have $a_1e^{\alpha_1}+a_2e^\alpha_2+ \dots a_n e^\alpha_n \neq 0$? This question is intended to be a repository for questions that can be quickly defeated by the LW theorem. It is not for a general discussion on the theorem. End of Proposed Question Some Cons to asking questions like this: It fails to meet many of the standards of this site. It is open ended. There are multiple correct answers. Some Pros: It would be nice to have a place to direct users when they ask questions that are the immediate consequence of the ________-Weierstraß theorem (which we can safely say is like $\epsilon>0$ of all mathematics. I swear $\epsilon \approx .02$. Weierstraß ate up everything and has left us only the crumbs... ) . like this one or this one or this one. We can use something like: "Hi! Welcome to MSE you have asked question that belongs in our "Is an immediate consequence of a Weierstraß theorem."-series. Here's a link to a collection of those. I am going to include an answer to this particular case on that post. Another pro would be that I think people would enjoy it. But democracy may prove that false. Vote your feels!

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