Split (1)

train · ~4.74M rows (showing the first 1.13M)

text stringlengths 256 16.4k
It is essentially impossible to answer the general question of "how does multilinearity come up naturally in physics?" because of the myriad of possible examples that make up the total answer. Instead, let me describe a situation that very loudly cries out for the use of tensor products of two vectors. Consider the problem of conservation of momentum for a continuous distribution of electric charge and current, which interacts with an electromagnetic field, under the action of no other external force. I will describe it more or less along the lines of Jackson ( Classical Electrodynamics, 3 rd edition, §6.7) but depart from it towards the end. This will get very electromagneticky for a while, so if you want to skip to the tensors, you can go straight to equation (1). The rate of change of the total mechanical momentum of the system is the total Lorentz force, given by $$\frac{ d\mathbf{P}_\rm{mech}}{dt}=\int_V(\rho\mathbf{E}+\mathbf{J}\times \mathbf{B})d\mathbf{x}.$$To simplify this, one can take $\rho$ and $\mathbf{J}$ from Maxwell's equations:$$\rho=\epsilon_0\nabla\cdot\mathbf{E}\ \ \ \text{ and }\ \ \\mathbf{J}=\frac1{\mu_0}\nabla\times \mathbf{B}-\epsilon_0\frac{\partial \mathbf{E}}{\partial t}.$$(In particular, this means that what follows is only valid "on shell": momentum is only conserved if the equations of motion are obeyed. Of course!) One can then put these expressions back, to a nice vector calculus work-out, and come up with the following relation:$$\begin{align}{}\frac{ d\mathbf{P}_\rm{mech}}{dt}+&\frac{d}{dt}\int_V\epsilon_0\mathbf{E}\times \mathbf{B}d\mathbf{x} \\ &=\epsilon_0\int_V \left[\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E}) + c^2 \mathbf{B} (\nabla \cdot \mathbf{B})- c^2 \mathbf{B} \times (\nabla \times \mathbf{B})\right]d\mathbf{x}.\end{align}$$ The integral on the left-hand side can be identified as the total electromagnetic momentum, and differs from the integral of the Poynting vector by a factor of $1/c^2$. To get this in the proper form for a conservation law, though, such as the one for energy in this setting,$$\frac{dE_\rm{mech}}{dt}+\frac{d}{dt}\frac{\epsilon_0}{2}\int_V(\mathbf{E}^2+c^2\mathbf{B}^2)d\mathbf{x}=-\oint_S \mathbf{S}\cdot d\mathbf{a},$$we need to reduce the huge, ugly volume integral into a surface integral. The way to do this, is, of course, the divergence theorem. However, that theorem is for scalars, and what we have so far is a vector equation. To work further then, we need to (at least temporarily) work in some specific basis $\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$, and write $\mathbf{E}=\sum_i E_i \mathbf{e}_i$. Let's work with the electric field term first; after that the results also apply to the magnetic term. Thus, to start with,$$\begin{align}{}\int_V \left[\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E})\right]d\mathbf{x}=\sum_i \mathbf{e}_i\int_V \left[E_i(\nabla\cdot \mathbf{E})-\mathbf{e}_i\cdot\left(\mathbf{E} \times(\nabla \times \mathbf{E})\right)\right]d\mathbf{x}.\end{align}$$These terms should be simplified using the vector calculus identities$$E_i(\nabla\cdot \mathbf{E})=\nabla\cdot\left(E_i \mathbf{E}\right) - \mathbf{E}\cdot \nabla E_1$$and$$\mathbf{E} \times(\nabla \times \mathbf{E})=\frac12\nabla(\mathbf{E}\cdot\mathbf{E})-(\mathbf{E}\cdot\nabla)\mathbf{E},$$which mean that the whole combination can be simplified as$$\begin{align}{}\int_V \left[\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E})\right]d\mathbf{x}=\sum_i \mathbf{e}_i\int_V \left[\nabla\cdot\left(E_i \mathbf{E}\right) -\mathbf{e}_i\cdot\left(\frac12\nabla(\mathbf{E}\cdot\mathbf{E})\right)\right]d\mathbf{x},\end{align}$$since the terms in $\mathbf{E}\cdot \nabla E_i$ and $\mathbf{e}_i\cdot\left( (\mathbf{E}\cdot\nabla)\mathbf{E}\right)$ cancel. This means we can write the whole integrand as the divergence of some vector field, and use the divergence theorem:$$\begin{align}{}\int_V \left[\mathbf{E}(\nabla\cdot \mathbf{E})-\mathbf{E} \times(\nabla \times \mathbf{E})\right]d\mathbf{x}&=\sum_i \mathbf{e}_i\int_V \nabla\cdot\left[E_i \mathbf{E}-\frac12 \mathbf{e}_i E^2\right]d\mathbf{x}\\ & =\sum_i \mathbf{e}_i\oint_S\left[E_i \mathbf{E}-\frac12 \mathbf{e}_i E^2\right]\cdot d\mathbf{a}. \tag 1\end{align}$$ In terms of conservation law structure, we're essentially done, as we've reduced the rate of change of momentum to a surface term. However, it is crying out for some simplification. In particular, this expression is basis-dependent, but it is so close to being basis independent that it's worth a closer look. The first term, for instance, is simply crying out for a simplification that would look something like $$\sum_i \mathbf{e}_i\oint_SE_i \mathbf{E}\cdot d\mathbf{a}=\oint_S\mathbf{E}\, \mathbf{E}\cdot d\mathbf{a}$$if we could only make sense of an object like $\mathbf{E}\, \mathbf{E}$. Even better, if we could make sense of such a combination, then it turns out that the seemingly basis-dependent combination that would come up in the second term, $\sum_i \mathbf{e}_i\,\mathbf{e}_i$, turns out to be basis independent: one can prove that for any two orthonormal bases $\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$ and $\{\mathbf{e}_1', \mathbf{e}_2', \mathbf{e}_3'\}$, those combinations are the same:$$\sum_i \mathbf{e}_i\,\mathbf{e}_i = \sum_i \mathbf{e}_i'\,\mathbf{e}_i'$$as long as the product $\mathbf{u}\,\mathbf{v}$ of two vectors, whatever it ends up being, is linear on each component, which is definitely a reasonable assumption. So what, then, should this new vector multiplication be? One key to realizing what we really need is noticing the fact that we haven't yet assigned any real physical meaning to the combination $\mathbf{E}\,\mathbf{E}$; instead, we're only ever interacting with it by dotting "one of the vectors of the product" with the surface area element $d\mathbf{a}$, and that leaves a vector $\mathbf{E}\,\mathbf{E}\cdot d\mathbf{a}$ which we can integrate to get a vector, and that requires no new structure. Let's then write a list of how we want this new product to behave. To keep things clear, let's give it some fancy new symbol like $\otimes$, mostly to avoid unseemly combinations like $\mathbf{u}\,\mathbf{v}$. We want then, a function $\otimes:V\times V\to W$, which takes euclidean vectors in $V=\mathbb R^3$ into some vector space $W$ in which we'll keep our fancy new objects. Combinations of the form $\mathbf{u}\otimes \mathbf{v}$ should be linear in both $\mathbf{u}$ and $\mathbf{v}$. For all vectors $w$ in $V$, and all combinations $(\mathbf{u},\mathbf{v})\in V\times V$, we want the combination $(\mathbf{u}\otimes \mathbf{v})\cdot\mathbf{w}$ to be a vector in $V$. Even more, we want that to be the vector $(\mathbf{v}\cdot\mathbf{w})\mathbf{u}\in V$. That last one looks actually pretty strong, but there's evidently room for improvement. For one, it depends on the euclidean structure, which is not actually necessary: we can make an equivalent statement that uses the vector space's dual. For all $(\mathbf{u},\mathbf{v})\in V\times V$ and all $f\in V^\ast$, we want $f_\to(\mathbf{u}\otimes \mathbf{v})=f(\mathbf{v})\mathbf{u}\in V$ to hold, where $f_\to$ simply means that $f$ acts on the factor on the right. Finally, if we're doing stuff with the dual, we can reformulate that in a slightly prettier way. Since two vectors $\mathbf{u},\mathbf{v}\in V$ are equal if and only if $f(\mathbf{u})=f(\mathbf{v})$ for all $f\in V^\ast$, we can give another equivalent statement of the same statement: For all $(\mathbf{u},\mathbf{v})\in V\times V$ and all $f,g\in V^\ast$, we want $g_\leftarrow f_\to(\mathbf{u}\otimes \mathbf{v})=g(\mathbf{u})f(\mathbf{v})\in V$. [Note, here, that this last rephrasing isn't really that fancy. Essentially, it is saying that the vector equation (1) is really to be interpreted as a component-by-component equality, and that's not really off the mark of how we actually do things.] I could keep going, but it's clear that this requirement can be rephrased into the universal property of the tensor product, and that rephrasing is a job for the mathematicians. Thus, you can see the story like this: Upon hitting equation (1), we give to the mathematicians this list of requirements. They go off, think for a bit, and come back telling us that such a structure does exist (i.e. there exist rigorous constructions that obey those requirements) and that it is essentially unique, in the sense that multiple such constructions are possible, but they are canonically isomorphic. For a physicist, what that means is that it's OK to write down objects like $\mathbf{u}\otimes \mathbf{v}$ as long as one does keep within the rules of the game. As far as electromagnetism goes, this means that we can write our conservation law in the form$$\frac{ d\mathbf{P}_\rm{mech}}{dt}+\frac{d}{dt}\int_V\epsilon_0\mathbf{E}\times \mathbf{B}d\mathbf{x} =\oint_A \mathcal T\cdot d\mathbf{a}$$where$$\mathcal T=\epsilon_0\left[\mathbf{E}\otimes\mathbf{E}+c^2\mathbf{B}\otimes\mathbf{B}-\frac12\sum_i\mathbf{e}_i\otimes\mathbf{e}_i\left(E^2+c^2 B^2\right)\right]$$is, of course, the Maxwell stress tensor. I could go on and on about this, but I think this really captures the essence of how and where it happens in physics that a situation is really begging the use of a tensor product. There are other such situations, of course, but this is the clearest one I know.
I recently learned about irreducible and prime elements in a commutative ring. However, my professor was not quite sure who was the first to make this distinction, or give an example of an irreducible element which is not prime. Was it Dedekind? If so, where did he make this distinction? Could you also give some references regarding the history of commutative rings, UFD, and the evolution of these concepts? The phenomenon of nonunique factorization appears to have been first explicitly articulated in the setting of cyclotomic fields, by Kummer in the 1830s and 1840s: $\mathbf Z[\zeta_{23}]$ is the first cyclotomic ring that fails to have unique factorization. One of Kummer's first papers (in 1847) where he discusses how to save unique factorization in cyclotomic rings $\mathbf Z[\zeta_\lambda]$ for prime $\lambda$, using "ideal numbers," is http://gdz.sub.uni-goettingen.de/en/dms/loader/img/?PID=GDZPPN002146061&physid=PHYS_0340. (For readers who know algebraic number theory, on the last page of this paper at http://gdz.sub.uni-goettingen.de/en/dms/loader/img/?PID=PPN243919689_0035\|LOG_0003&physid=PHYS_0380 Kummer lists some examples of exponents $n$ of class groups of these cyclotomic rings, which are not always equal to the class number $h$ of these rings. For example, when $\lambda$ is $29$ and $41$ he finds $n$ is $2$ and $11$, but $h$ is $8$ and $121$, respectively.) See also Lenstra's article http://pub.math.leidenuniv.nl/~lenstrahw/PUBLICATIONS/1979a/art.pdf. I do not believe at this time that anyone created special terminology to distinguish primes from irreducibles like we do today. Moving a few decades later, in Dedekind's famous XI-th Supplement on algebraic number theory in Dirichlet's Lectures on Number Theory (Vorlesungen über Zahlentheorie), which is from the 1870s, there is the example $3 \cdot 7 = (1+2\sqrt{-5})(1-2\sqrt{-5})$ from $\mathbf Z[\sqrt{-5}]$ in Section 159, where Dedekind says each of the four factors is unzerlegbare Zahl, which means an irreducible number. Dedekind's goal was to rescue unique factorization by using ideals instead of elements, and he does not dwell on the phenomenon of non-UFDs long enough to introduce new vocabulary like prime vs. irreducible. Thanks to Franz Lemmermeyer, I can point you to a reference from 1903 that makes the distinction between primes and irreducibles. On the bottom of page 12 of the book Einleitung in die allgemeine Theorie der algebraischen Groessen, by Gyulya König, (see https://books.google.com/books?id=3LhUAAAAYAAJ&pg=PA12&lpg=PA12&dq=irreduzibel+holoiden&source=bl&ots=aVbTTDuTi6&sig=oOb2Gna_A0T_e8ONlmwsSYy7tEM&hl=en&sa=X&ved=0ahUKEwjGreu8_dLMAhWIQSYKHYlxCY0Q6AEIHDAA#v=onepage&q=irreduzibel%20holoiden&f=false) the terms Prim-größe and irreduzibel are introduced, where the first one means prime element and the second means irreducible irreducible element. At the top of page 13 he writes that each prime element is irreducible but not conversely, and a counterexample is presented over the course of pp. 20-23 in $\mathbf Z[\sqrt{-5}]$, where $2-\sqrt{-5}$ is shown to be irreducible but not prime. That it's not prime follows from $3 \cdot 3 = (2+\sqrt{-5})(2-\sqrt{-5})$. (König was writing before the general concept of a ring was created. He always works in an integral domain, which he calls a Bereich, and he limits his definitions of prime and irreducible to what he calls holoid domains, which are integral domains of characteristic $0$, even though we'd recognize a restriction on the characteristic as an irrelevant hypothesis today. His notation for $\mathbf Z$ is $[1]$, for $\mathbf Q$ is $(1)$, and for $\mathbf Z[\sqrt{d}]$ is $[\sqrt{d}]$.) Since König is not well-known for his work in algebra, I'll mention another early 20-th century reference: on p. 11 of Krull's 1937 book Idealtheorie (see https://books.google.com/books?id=jCvKBgAAQBAJ&pg=PA11&lpg=PA11&dq=unzerlegbare+element&source=bl&ots=y_Y9BXubOy&sig=ZVQsi9-A-xYKb1TE5UAHE4-p1wk&hl=en&sa=X&ved=0ahUKEwi71Z2NrdHMAhVDVD4KHfjsDUYQ6AEILTAD#v=onepage&q=unzerlegbare%20element&f=false) he defines unzerlegbar and Primelement the way you'd define irreducible and prime, and he also cautions that while every prime element is irreducible, not every irreducible element is prime.
Search Now showing items 1-5 of 5 Measurement of electrons from beauty hadron decays in pp collisions at root √s=7 TeV (Elsevier, 2013-04-10) The production cross section of electrons from semileptonic decays of beauty hadrons was measured at mid-rapidity (\|y\| < 0.8) in the transverse momentum range 1 < pT <8 GeV/c with the ALICE experiment at the CERN LHC in ... 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 ... Elliptic flow of muons from heavy-flavour hadron decays at forward rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (Elsevier, 2016-02) The elliptic flow, $v_{2}$, of muons from heavy-flavour hadron decays at forward rapidity ($2.5 < y < 4$) is measured in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The scalar ... Centrality dependence of the pseudorapidity density distribution for charged particles in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2013-11) We present the first wide-range measurement of the charged-particle pseudorapidity density distribution, for different centralities (the 0-5%, 5-10%, 10-20%, and 20-30% most central events) in Pb-Pb collisions at $\sqrt{s_{NN}}$ ... Beauty production in pp collisions at √s=2.76 TeV measured via semi-electronic decays (Elsevier, 2014-11) The ALICE Collaboration at the LHC reports measurement of the inclusive production cross section of electrons from semi-leptonic decays of beauty hadrons with rapidity \|y\|<0.8 and transverse momentum 1<pT<10 GeV/c, in pp ...
Abbreviation: InRL An is a structure $\mathbf{A}=\langle A, \vee, \wedge, \cdot, 1, \sim, -\rangle$ of type $\langle 2, 2, 2, 0, 1, 1\rangle$ such that involutive residuated lattice $\langle A, \vee, \wedge, \neg\rangle$ is an involutive lattice $\langle A, \cdot, 1\rangle$ is a monoid $xy\le z\iff x\le \neg(y(\neg z))\iff y\le \neg((\neg z)x)$ Let $\mathbf{A}$ and $\mathbf{B}$ be involutive residuated lattices. A morphism from $\mathbf{A}$ to $\mathbf{B}$ is a function $h:A\rightarrow B$ that is a homomorphism: $h(x \vee y)=h(x) \vee h(y)$, $h(x \cdot y)=h(x) \cdot h(y)$, $h({\sim}x)={\sim}h(x)$ and $h(1)=1$. An is a structure $\mathbf{A}=\langle A,...\rangle$ of type $\langle...\rangle$ such that … $...$ is …: $axiom$ $...$ is …: $axiom$ Example 1: Feel free to add or delete properties from this list. The list below may contain properties that are not relevant to the class that is being described. $\begin{array}{lr} f(1)= &1\\ f(2)= &\\ f(3)= &\\ f(4)= &\\ f(5)= &\\ \end{array}$ $\begin{array}{lr} f(6)= &\\ f(7)= &\\ f(8)= &\\ f(9)= &\\ f(10)= &\\ \end{array}$
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box.. There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$. What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation? Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach. Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line? Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$? Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?" @Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider. Although not the only route, can you tell me something contrary to what I expect? It's a formula. There's no question of well-definedness. I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer. It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time. Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated. You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system. @A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago. @Eric: If you go eastward, we'll never cook! :( I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous. @TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$) @TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite. @TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator
Forgot password? New user? Sign up Existing user? Log in I can't believe I saw Calvin (the math challenge master) walk into my class and talk to my teacher and us O.O Note by Victor Loh 5 years, 8 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: My English teacher was like, "Ok, we have a visitor today," and this man just walked in to our class. He had a name tag with the words "Visitor Calvin" on it. Our English teacher said, "This guy is a math genius. He works in San Francisco." We were interested to find out more about him, so he told us where he studied, which seemed vaguely familiar to me as I had been to his profile page before. Then, he told us, "I came back to Singapore to see my teachers. I will also be going to another school later, where I went for junior college. Also, I work at Brilliant.org". Immediately, those in my class who were active users of Brilliant couldn't believe their ears, and I asked, "Are you Calvin Lin, the Math Challenge Master?" Then, he said yes, and we asked him a few more questions. We wanted to take a photo with him but we had 51 English questions to do so we bade him farewell. After speaking to our teacher for a few more minutes, he left. By the way, he doesn't really look like his profile picture. He wears black spectacles. Also, we asked him to say out the square root of 2 truncated to 100 digits but he didn't answer. He was like, "What..." XD Log in to reply Black spectacles are cool! Lool! My favourite part is when he was like "What"? haha! People from Brilliant are amazing. Proof that this is true: http://math.stackexchange.com/users/54563/calvin-lin Vote this comment up if you want Calvin Sir to upload his latest picture! I'm desperate! What?!?!?! Explain in very great detail\textit{very great detail}very great detail please! He studied in singapore so he came back to see his teachers Y he only visit your class? Y not mine? Woah cool how coincidental I also saw him at our SIMO (Singapore International Math Olympiad) training party and yup I took a photo with him WHY NO COME MY CLASS!? Victor is my schoolmate.My age is wrong btw Yx, u dun need to keep on talking abt ur age LOL! Victor Loh is known for lying dun trust him XDDDD i know him v. well >:D I know right, Charlton Jowell, your class is right opposite mine, you didnt see calvin? @Victor Loh – He came. I'm serious. @Victor Loh – What time was it @Charlton Teo – I'm upstairs lol... @Victor Loh – I may be in the midst of a lesson. Anyway I am seating closest to the window so I can only see the people at the door. Are you serious? What? In Singapore? That's insane! Seriously. He studied there XD I NEED TO GO TO RAFFLES!! @Finn Hulse – Too bad :P @Joshua Ong – lol raffles rocks XD @Victor Loh – I'm definitely going to study abroad and go there. ;) @Finn Hulse – Try to come, it's totally awesome Really? WHAT? Victor you are really a lucky guy but please give details of his visit,PLEASE. Whoa! Details please... WHAATT?! it's near indonesia. Why he didnt visit my country xD Guys, its true. Hi Charlton (whoever you might be)I would say that I won't really care if you don't believe me. I posted this because I wanted to share my experience with others. You can even ask my teacher (According to you, you said that you knew me well, so you should know my classmates and teachers). You can even ask Calvin yourself. dun act like u dont know me vic -.- @Charlton Teo – Ok whatever XD o_o I wish he came to visit us. We had an exhibition in our school. He could have come here as the chief guest Problem Loading... Note Loading... Set Loading...
60J65 Brownian motion [See also 58J65] Refine We show that the intersection local times $\mu_p$ on the intersection of $p$ independent planar Brownian paths have an average density of order three with respect to the gauge function $r^2\pi\cdot (log(1/r)/\pi)^p$, more precisely, almost surely, \[ \lim\limits_{\varepsilon\downarrow 0} \frac{1}{log \|log\ \varepsilon\|} \int_\varepsilon^{1/e} \frac{\mu_p(B(x,r))}{r^2\pi\cdot (log(1/r)/\pi)^p} \frac{dr}{r\ log (1/r)} = 2^p \mbox{ at $\mu_p$-almost every $x$.} \] We also show that the lacunarity distributions of $\mu_p$, at $\mu_p$-almost every point, is given as the distribution of the product of $p$ independent gamma(2)-distributed random variables. The main tools of the proof are a Palm distribution associated with the intersection local time and an approximation theorem of Le Gall. We show that the occupation measure on the path of a planar Brownian motion run for an arbitrary finite time intervalhas an average density of order three with respect to thegauge function t^2 log(1/t). This is a surprising resultas it seems to be the first instance where gauge functions other than t^s and average densities of order higher than two appear naturally. We also show that the average densityof order two fails to exist and prove that the density distributions, or lacunarity distributions, of order threeof the occupation measure of a planar Brownian motion are gamma distributions with parameter 2. The Kallianpur-Robbins law describes the long term asymptotic behaviour of the distribution of the occupation measure of a Brownian motion in the plane. In this paper we show that this behaviour can be seen at every typical Brownian path by choosing either a random time or a random scale according to the logarithmic laws of order three. We also prove a ratio ergodic theorem for small scales outside an exceptional set of vanishing logarithmic density of order three.
When there are $n$ variables $X_1, \ldots, X_n,$ their mutual covariances can be assembled into a matrix $\Sigma = (\sigma_{ij})$ where $$\sigma_{ij} = \operatorname{Cov}(X_i,X_j).$$ Note that the diagonal elements $\sigma_{ii} = \operatorname{Cov}(X_i,X_i) = \operatorname{Var}(X_i)$ are the variances. It will be convenient to extract these elements into a diagonal matrix by setting all off-diagonal entries to zero; sometimes this is written $$\operatorname{Diag}(\Sigma)_{ij} = \left\{\matrix{\sigma_{ii} & i=j \\ 0 & \text{otherwise.}}\right.$$ Note, too, that a positive square root of this matrix is well-defined and unique because all its entries are positive: $$\sqrt{\operatorname{Diag}(\Sigma)}_{ij} = \left\{\matrix{\sqrt{\sigma_{ii}} & i=j \\ 0 & \text{otherwise.}}\right.$$ One formula for the corresponding correlation coefficient, as given in the question, is $$\rho_{ij} = \operatorname{Cor}(X_i,X_j) = \frac{\operatorname{Cov}(X_i,X_j)}{\sqrt{\operatorname{Var}(X_i)\operatorname{Var}(X_j)}} = \frac{\sigma_{ij}}{\sqrt{\sigma_{ii}\,\sigma_{jj}}}\ .$$ These can be assembled into a correlation matrix $P= (\rho_{ij}).$ It follows directly from the definitions of matrix multiplication and inversion that the preceding formula is the result of the following combination of operations: $$P = \sqrt{\operatorname{Diag}(\Sigma)}^{-1}\ \Sigma\ \sqrt{\operatorname{Diag}(\Sigma)}^{-1}.$$ By focusing on part of the matrices, this formula extends directly to the block matrix setup of the question, where the covariances of the variables in the $k$-vector $Y_{r_1}$ and those in the $m$-vector $Y_{r_2}$ appear in the $k\times m$ matrix $\Sigma_{12},$ $\Sigma_{11}$ is the $k\times k$ covariance matrix of $Y_{r_1},$ and $\Sigma_{22}$ is the $m\times m$ covariance matrix of $Y_{r_2}.$ As before, the correlations between the variables in $Y_{r_1}$ and those in $Y_{r_2}$ can be arranged into a $k\times m$ mutual correlation matrix $P_{12}$ and $$P_{12} = \sqrt{\operatorname{Diag}(\Sigma_{11})}^{-1}\ \Sigma_{12}\ \sqrt{\operatorname{Diag}(\Sigma_{22})}^{-1}.$$ For example, let $$\Sigma = \pmatrix{1 & -1 & 2 \\ -1 & 4 & -1 \\ 2 & -1 & 9}.$$ Then $$\operatorname{Diag}(\Sigma) = \pmatrix{1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 9},$$ $$\sqrt{\operatorname{Diag}(\Sigma)} = \pmatrix{1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3},$$ $$\sqrt{\operatorname{Diag}(\Sigma)}^{-1} = \pmatrix{1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}},$$ giving the correlation matrix $$P = \pmatrix{1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}}\ \pmatrix{1 & -1 & 2 \\ -1 & 4 & -1 \\ 2 & -1 & 9}\ \pmatrix{1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}} = \pmatrix{1 & -\frac{1}{2} & \frac{2}{3} \\ -\frac{1}{2} & 1 & -\frac{1}{6} \\ \frac{2}{3} & -\frac{1}{6} & 1}.$$ Taking $k=1$ and $m=2$ makes $\Sigma_{11} = \pmatrix{1},$ $\Sigma_{22} = \pmatrix{4 & -1 \\ -1 & 9},$ and $\Sigma_{12} = \pmatrix{-1 & 2}.$ Thus the correlations between the first variable ($Y_{r_1}$) and the remaining two variables ($Y_{r_2}$) are given by the $1\times 2$ matrix $$P_{12} = \pmatrix{\sqrt{1}}^{-1} \pmatrix{-1 & 2}\ \pmatrix{\sqrt{4} & 0 \\ 0 & \sqrt{9}}^{-1} = \pmatrix{-\frac{1}{2} & \frac{2}{3}},$$ which is precisely the upper right hand block of $P,$ as claimed.
skills to develop To learn what a standard normal random variable is. To learn how to compute probabilities related to a standard normal random variable. Definition: standard normal random variable A standard normal random variable is a normally distributed random variable with mean $\mu =0$ and standard deviation $\sigma =1$. It will always be denoted by the letter $Z$. The density function for a standard normal random variable is shown in Figure $\PageIndex{1}$. Figure $\PageIndex{1}$: Density Curve for a Standard Normal Random Variable To compute probabilities for $Z$ we will not work with its density function directly but instead read probabilities out of Figure $\PageIndex{2}$. The tables are tables of cumulative probabilities; their entries are probabilities of the form $P(Z< z)$. The use of the tables will be explained by the following series of examples. Figure $\PageIndex{2}$: Cumulative Normal Probability Example $\PageIndex{1}$ Find the probabilities indicated, where as always $Z$ denotes a standard normal random variable. $P(Z< 1.48)$. $P(Z< -0.25)$. Solution: Figure $\PageIndex{3}$ shows how this probability is read directly from the table without any computation required. The digits in the ones and tenths places of $1.48$, namely $1.4$, are used to select the appropriate row of the table; the hundredths part of $1.48$, namely $0.08$, is used to select the appropriate column of the table. The four decimal place number in the interior of the table that lies in the intersection of the row and column selected, $0.9306$, is the probability sought: \[P(Z< 1.48)=0.9306\] Figure $\PageIndex{3}$ : Computing Probabilities Using the Cumulative Table The minus sign in $-0.25$ makes no difference in the procedure; the table is used in exactly the same way as in part (a): the probability sought is the number that is in the intersection of the row with heading $-0.2$ and the column with heading $0.05$, the number $0.4013$. Thus $P(Z< -0.25)=0.4013$. Example $\PageIndex{2}$ Find the probabilities indicated. $P(Z> 1.60)$. $P(Z> -1.02)$. Solution: Because the events $Z> 1.60$ and $Z\leq 1.60$ are complements, the Probability Rule for Complementsimplies that \[P(Z> 1.60)=1-P(Z\leq 1.60)\] Since inclusion of the endpoint makes no difference for the continuous random variable $Z$, $P(Z\leq 1.60)=P(Z< 1.60)$, which we know how to find from the table in Figure $\PageIndex{2}$. The number in the row with heading $1.6$ and in the column with heading $0.00$ is $0.9452$. Thus $P(Z< 1.60)=0.9452$ so \[P(Z> 1.60)=1-P(Z\leq 1.60)=1-0.9452=0.0548\] Figure $\PageIndex{4}$ illustrates the ideas geometrically. Since the total area under the curve is $1$ and the area of the region to the left of $1.60$ is (from the table) $0.9452$, the area of the region to the right of $1.60$ must be $1-0.9452=0.0548$. Figure $\PageIndex{4}$: Computing a Probability for a Right Half-Line The minus sign in $-1.02$ makes no difference in the procedure; the table is used in exactly the same way as in part (a). The number in the intersection of the row with heading $-1.0$ and the column with heading $0.02$ is $0.1539$. This means that $P(Z<-1.02)=P(Z\leq -1.02)=0.1539$. Hence \[P(Z>-1.02)=P(Z\leq -1.02)=1-0.1539=0.8461\] Example $\PageIndex{3}$ Find the probabilities indicated. $P(0.5<Z<1.57)$. $P(-2.55<Z<0.09)$. Solution: Figure $\PageIndex{5}$ illustrates the ideas involved for intervals of this type. First look up the areas in the table that correspond to the numbers $0.5$ (which we think of as $0.50$ to use the table) and $1.57$. We obtain $0.6915$ and $0.9418$, respectively. From the figure it is apparent that we must take the difference of these two numbers to obtain the probability desired. In symbols, \[P(0.5<Z<1.57)=P(Z<1.57)-P(Z<0.50)=0.9418-0.6915=0.2503\] Figure $\PageIndex{5}$: Computing a Probability for an Interval of Finite Length The procedure for finding the probability that $Z$ takes a value in a finite interval whose endpoints have opposite signs is exactly the same procedure used in part (a), and is illustrated in Figure $\PageIndex{6}$ "Computing a Probability for an Interval of Finite Length". In symbols the computation is \[P(-2.55<Z<0.09)=P(Z<0.09)-P(Z<-2.55)=0.5359-0.0054=0.5305\] Figure $\PageIndex{6}$: Computing a Probability for an Interval of Finite Length The next example shows what to do if the value of $Z$ that we want to look up in the table is not present there. Example $\PageIndex{4}$ Find the probabilities indicated. $P(1.13<Z<4.16)$. $P(-5.22<Z<2.15)$. Solution: We attempt to compute the probability exactly as in Example $\PageIndex{3}$ by looking up the numbers $1.13$ and $4.16$ in the table. We obtain the value $0.8708$ for the area of the region under the density curve to left of $1.13$ without any problem, but when we go to look up the number $4.16$ in the table, it is not there. We can see from the last row of numbers in the table that the area to the left of $4.16$ must be so close to 1 that to four decimal places it rounds to $1.0000$. Therefore \[P(1.13<Z<4.16)=1.0000-0.8708=0.1292\] Similarly, here we can read directly from the table that the area under the density curve and to the left of $2.15$ is $0.9842$, but $-5.22$ is too far to the left on the number line to be in the table. We can see from the first line of the table that the area to the left of $-5.22$ must be so close to $0$ that to four decimal places it rounds to $0.0000$. Therefore \[P(-5.22<Z<2.15)=0.9842-0.0000=0.9842\] The final example of this section explains the origin of the proportions given in the Empirical Rule. Example $\PageIndex{5}$ Find the probabilities indicated. $P(-1<Z<1)$. $P(-2<Z<2)$. $P(-3<Z<3)$. Solution: Using the table as was done in Example $\PageIndex{3}$ we obtain \[P(-1<Z<1)=0.8413-0.1587=0.6826\] Since $Z$ has mean $0$ and standard deviation $1$, for $Z$ to take a value between $-1$ and $1$ means that $Z$ takes a value that is within one standard deviation of the mean. Our computation shows that the probability that this happens is about $0.68$, the proportion given by the Empirical Rule for histograms that are mound shaped and symmetrical, like the bell curve. Using the table in the same way, \[P(-2<Z<2)=0.9772-0.0228=0.9544\] This corresponds to the proportion 0.95 for data within two standard deviations of the mean. Similarly, \[P(-3<Z<3)=0.9987-0.0013=0.9974\] which corresponds to the proportion 0.997 for data within three standard deviations of the mean. Key takeaway A standard normal random variable $Z$ is a normally distributed random variable with mean $\mu =0$ and standard deviation $\sigma =1$. Probabilities for a standard normal random variable are computed using Figure $\PageIndex{2}$. Contributor Anonymous
Answer $90.2$ $ cm^{2}$ Work Step by Step Using the formula given in the previous exercise, the area of the parallelogram is: $a \times b \times \sin \gamma$ $8 cm \times 12 cm \times \sin 70^{\circ}$$= 90.2$ $ cm^{2}$ You can help us out by revising, improving and updating this answer.Update this answer After you claim an answer you’ll have 24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback.
Airspeed is measured with a pitot tube. A pitot tube has two pressure measurement ports. One that measures the total pressure $p_t$. This port is facing the incoming airflow. The other measures the static pressure $p$ and is placed perpendicular to the airflow. The difference between the two pressures is called impact pressure (pressure rise do to the airflow impacting the pitot tube) and is denoted $q_c$. The impact pressure is related to the speed of the airflow the pitot tube is exposed to. If the flow is considered incompressible (which is an acceptable approximation for speeds up to 200 knots) the impact pressure can be derived from Bernouilli's equation. $q_c = \frac{1}{2}\rho V^2$ $q_c$ is the impact pressure in Pa $\rho$ is the density in kg/m 3 $V$ is the true airspeed in m/s Equivalent airspeed The airspeed indicator is calibrated for standard sea level conditions, where $\rho$ is 1.225 kg/m 3. In reality the aircraft will fly at altitude and therefor the actual density of the air is lower. Therefor the airspeed as measured by impact pressure will be lower as well. For example if an aircraft flies 75 m/s (about 146 knots) at 6000 ft the density will be 1.02393 kg/m 3. $q_c = \frac{1}{2} 1.02393 \cdot 75 ^2 = 2879.8 \textrm{ Pa}$ The equivalent airspeed at sea level for the same $q_c$ is: $V_{EAS} = \sqrt{\frac{2 q_c}{\rho_0}} = \sqrt{\frac{2 \cdot 2879.8}{1.225}} = 68.6 \textrm{ m/s}$ Your airspeed indicator (assuming no errors) will read only 68.6 m/s (133 knots) despite the fact that you are moving with 75 m/s (146 knots) with respect to the air. Conversion of true airspeed to the equivalent airspeed can be done directly by: $V_{EAS} =V\cdot \sqrt{\frac{\rho}{\rho_0}}$ $V_{EAS}$ equivalent airspeed (m/s) $V$ true airspeed (m/s) $\rho$ actual air density (kg/m 3). $\rho_0$ density at standard sea level conditions (1.225 kg/m 3) Calibrated airspeed The effects of the lower density on your aispeed indicator become more pronounced the higher you go. Once you go faster than about 100 m/s true airspeed the effects of compressibility can no longer be ignored and the above no longer applies. Airspeed indicators are corrected for effects of compressibility and therefore don't use the equivalent airspeed but instead use calibrated airspeed for calibration. $V_{CAS}=a_{0}\sqrt{5\left[\left(\frac{q_c}{p_{0}}+1\right)^\frac{2}{7}-1\right]}$ $V_{CAS}$ is calibrated airspeed $a_{0}$ is the speed of sound under standard sea level conditions (340.3 m/s) $p_0$ is the static air pressure under standard sea level conditions (101325 Pa) ${q_c}$ is the impact pressure The impact pressure is also a bit more complex for compressible flow: $\;q_c = p\left[\left(1+0.2 M^2 \right)^\tfrac{7}{2}-1\right]$ $p$ the static pressure $M$ the Mach number Indicated airspeed The airspeed that is actually indicate on the airspeed indicator deviates from the calibrated airspeed because of several error factors: instrument error position error installation error Instrument error are errors within the airspeed indicator when converting the static pressure and the total pressure to a speed indication. In mechanical instruments these are often more pronounced that in digital systems. Position error are errors in the position of the static port (not measuring exactly static pressure, but also some effects of moving air) and position of the total pressure port (not exactly measuring the full ram rise). Finally there are installation errors, which include for example leaking tubes between the instrument and the pitot ports. Given the above, we can now derive the relation between Calibrated Airspeed and Equivalent airspeed Calibrated airspeed is dependent on the impact pressure, which is in turn depending on the Mach number. The Mach number is the ratio between the true air speed and the speed of sound $a = \sqrt{\gamma R T}$.We can now express the Mach number as a function of the equivalent airspeed: $M = \frac{V}{\sqrt{\gamma R T}} = \frac{V_{EAS}\sqrt{\frac{\rho_0}{\rho}}}{\sqrt{\gamma R T}}$ From the ideal gas law it follows that $p = \rho R T$ and so we can simplify the Mach number to : $ M = V_{EAS}\sqrt{\frac{\rho_0}{\gamma p}} $ From this it follows that impact pressure for compressible flow is: $\;q_c = p\left[\left(1+0.2 M^2 \right)^\tfrac{7}{2}-1\right] = p\left[\left(1+ \frac{\rho_0}{5\gamma p} V_{EAS}^2 \right)^\tfrac{7}{2}-1\right] $ That brings the relation between CAS and EAS to : $V_{CAS}=a_{0}\sqrt{5\left[\left(\frac{p}{p_{0}}\left[\left(1+ \frac{\rho_0}{5\gamma p} V_{EAS}^2 \right)^\tfrac{7}{2}-1\right]+1\right)^\frac{2}{7}-1\right]}$ Solving it for EAS = f(CAS) is left to the reader. Parts of this answer are taken from this answer.
I have the following statement whith an argumentation which I do not understand. Fix integers $k,l$ such that $0\leq l\leq k$. Then the identity map on $C^\infty(\mathbb{T}^d)$ extends to the injective continuous operator $$i_{k,l}:H^k(\mathbb{T}^d)\rightarrow H^l(\mathbb{T}^d)$$ of norm one. Moreover, for every $j\in\{1,...,d\}$ the partial derivative $$\partial_j:C^\infty(\mathbb{T}^d)\rightarrow C^\infty(\mathbb{T}^d)$$ extends to the continuous operator $$\partial'_j:H^k(\mathbb{T}^d)\rightarrow H^{k-1}(\mathbb{T}^d)$$ with norm less or equal then one. The proof consider the embedding of the $C^\infty$ space in $(L^2)^{K(k)}$ (The sobolev space is defined as the closure of the embedding). My problem is more for the norms. I can't see why should be so. And the proof states it is clear... Could someone explain? We defined the Sobolev space on $L^2(\mathbb{T}^d)$ as the closure of $C^\infty(\mathbb{T}^d)$ given an integer factor $k$. We use the embedding $f\mapsto (f_\alpha)_{\parallel \alpha\parallel_1\leq k}\in L^2(\mathbb{T}^d)^{K(k)}$ where $K(k)$ is the number of multi indices less or equal than $k$.
Skills to Develop Confidence limits tell you how accurate your estimate of the mean is likely to be. Introduction After you've calculated the mean of a set of observations, you should give some indication of how close your estimate is likely to be to the parametric ("true") mean. One way to do this is with confidence limits. Confidence limits are the numbers at the upper and lower end of a confidence interval; for example, if your mean is $7.4$ with confidence limits of $5.4$ and $9.4$, your confidence interval is $5.4$ to $9.4$. Most people use $95\%$ confidence limits, although you could use other values. Setting $95\%$ confidence limits means that if you took repeated random samples from a population and calculated the mean and confidence limits for each sample, the confidence interval for $95\%$ of your samples would include the parametric mean. To illustrate this, here are the means and confidence intervals for $100$ samples of $3$ observations from a population with a parametric mean of $5$. Of the $100$ samples, $94$ (shown with $X$ for the mean and a thin line for the confidence interval) have the parametric mean within their $95\%$ confidence interval, and $6$ (shown with circles and thick lines) have the parametric mean outside the confidence interval. Fig. 3.4.1 Mean and confidence intervals for 100 samples of 3 observations With larger sample sizes, the $95\%$ confidence intervals get smaller: Fig. 3.4.2 Mean and confidence intervals for 100 samples of 20 observations When you calculate the confidence interval for a single sample, it is tempting to say that "there is a $95\%$ probability that the confidence interval includes the parametric mean." This is technically incorrect, because it implies that if you collected samples with the same confidence interval, sometimes they would include the parametric mean and sometimes they wouldn't. For example, the first sample in the figure above has confidence limits of $4.59$ and $5.51$. It would be incorrect to say that $95\%$ of the time, the parametric mean for this population would lie between $4.59$ and $5.51$. If you took repeated samples from this same population and repeatedly got confidence limits of $4.59$ and $5.51$, the parametric mean (which is $5$, remember) would be in this interval $100\%$ of the time. Some statisticians don't care about this confusing, pedantic distinction, but others are very picky about it, so it's good to know. Confidence limits for measurement variables To calculate the confidence limits for a measurement variable, multiply the standard error of the mean times the appropriate t-value. The $t$-value is determined by the probability ($0.05$ for a $95\%$ confidence interval) and the degrees of freedom ($n-1$). In a spreadsheet, you could use =(STDEV(Ys)/SQRT(COUNT(Ys)))*TINV(0.05, COUNT(Ys)-1), where $Ys$ is the range of cells containing your data. You add this value to and subtract it from the mean to get the confidence limits. Thus if the mean is $87$ and the $t$-value times the standard error is $10.3$, the confidence limits would be $76.7$ and $97.3$. You could also report this as "$87\pm 10.3$ ($95\%$ confidence limits)." People report both confidence limits and standard errors as the "mean $\pm $ something," so always be sure to specify which you're talking about. All of the above applies only to normally distributed measurement variables. For measurement data from a highly non-normal distribution, bootstrap techniques, which I won't talk about here, might yield better estimates of the confidence limits. Confidence limits for nominal variables There is a different, more complicated formula, based on the binomial distribution, for calculating confidence limits of proportions (nominal data). Importantly, it yields confidence limits that are not symmetrical around the proportion, especially for proportions near zero or one. John Pezzullo has an easy-to-use web page for confidence intervals of a proportion. To see how it works, let's say that you've taken a sample of $20$ men and found $2$ colorblind and $18$ non-colorblind. Go to the web page and enter $2$ in the "Numerator" box and $20$ in the "Denominator" box," then hit "Compute." The results for this example would be a lower confidence limit of $0.0124$ and an upper confidence limit of $0.3170$. You can't report the proportion of colorblind men as "$0.10\pm something$," instead you'd have to say "$0.10$ with $95\%$ confidence limits of $0.0124$ and $0.3170$." An alternative technique for estimating the confidence limits of a proportion assumes that the sample proportions are normally distributed. This approximate technique yields symmetrical confidence limits, which for proportions near zero or one are obviously incorrect. For example, if you calculate the confidence limits using the normal approximation on $0.10$ with a sample size of $20$, you get $-0.03$ and $0.23$, which is ridiculous (you couldn't have less than $0\%$ of men being color-blind). It would also be incorrect to say that the confidence limits were $0$ and $0.23$, because you know the proportion of colorblind men in your population is greater than $0$ (your sample had two colorblind men, so you know the population has at least two colorblind men). I consider confidence limits for proportions that are based on the normal approximation to be obsolete for most purposes; you should use the confidence interval based on the binomial distribution, unless the sample size is so large that it is computationally impractical. Unfortunately, more people use the confidence limits based on the normal approximation than use the correct, binomial confidence limits. The formula for the $95\%$ confidence interval using the normal approximation is $p\pm 1.96\sqrt{\left [ \frac{p(1-p)}{n} \right ]}$, where $p$ is the proportion and $n$ is the sample size. Thus, for $P=0.20$ and $n=100$, the confidence interval would be $\pm 1.96\sqrt{\left [ \frac{0.20(1-0.20)}{100} \right ]}$, or $0.20\pm 0.078$. A common rule of thumb says that it is okay to use this approximation as long as $npq$ is greater than $5$; my rule of thumb is to only use the normal approximation when the sample size is so large that calculating the exact binomial confidence interval makes smoke come out of your computer. Statistical testing with confidence intervals This handbook mostly presents "classical" or "frequentist" statistics, in which hypotheses are tested by estimating the probability of getting the observed results by chance, if the null is true (the $P$ value). An alternative way of doing statistics is to put a confidence interval on a measure of the deviation from the null hypothesis. For example, rather than comparing two means with a two-sample t–test, some statisticians would calculate the confidence interval of the difference in the means. This approach is valuable if a small deviation from the null hypothesis would be uninteresting, when you're more interested in the size of the effect rather than whether it exists. For example, if you're doing final testing of a new drug that you're confident will have some effect, you'd be mainly interested in estimating how well it worked, and how confident you were in the size of that effect. You'd want your result to be "This drug reduced systolic blood pressure by $10.7 mm\; \; Hg$, with a confidence interval of $7.8$ to $13.6$," not "This drug significantly reduced systolic blood pressure ($P=0.0007$)." Using confidence limits this way, as an alternative to frequentist statistics, has many advocates, and it can be a useful approach. However, I often see people saying things like "The difference in mean blood pressure was $10.7 mm\; \; Hg$, with a confidence interval of $7.8$ to $13.6$; because the confidence interval on the difference does not include $0$, the means are significantly different." This is just a clumsy, roundabout way of doing hypothesis testing, and they should just admit it and do a frequentist statistical test. There is a myth that when two means have confidence intervals that overlap, the means are not significantly different (at the $P<0.05$ level). Another version of this myth is that if each mean is outside the confidence interval of the other mean, the means are significantly different. Neither of these is true (Schenker and Gentleman 2001, Payton et al. 2003); it is easy for two sets of numbers to have overlapping confidence intervals, yet still be significantly different by a two-sample t–test; conversely, each mean can be outside the confidence interval of the other, yet they're still not significantly different. Don't try compare two means by visually comparing their confidence intervals, just use the correct statistical test. Similar statistics Confidence limits and standard error of the mean serve the same purpose, to express the reliability of an estimate of the mean. When you look at scientific papers, sometimes the "error bars" on graphs or the ± number after means in tables represent the standard error of the mean, while in other papers they represent $95\%$ confidence intervals. I prefer $95\%$ confidence intervals. When I see a graph with a bunch of points and error bars representing means and confidence intervals, I know that most ($95\%$) of the error bars include the parametric means. When the error bars are standard errors of the mean, only about two-thirds of the bars are expected to include the parametric means; I have to mentally double the bars to get the approximate size of the $95\%$ confidence interval (because $t(0.05)$ is approximately $2$ for all but very small values of $n$). Whichever statistic you decide to use, be sure to make it clear what the error bars on your graphs represent. A surprising number of papers don't say what their error bars represent, which means that the only information the error bars convey to the reader is that the authors are careless and sloppy. Examples Measurement data The blacknose dace data from the central tendency web page has an arithmetic mean of $70.0$. The lower confidence limit is $45.3$ ($70.0-24.7$), and the upper confidence limit is $94.7$ ($70+24.7$). Nominal data If you work with a lot of proportions, it's good to have a rough idea of confidence limits for different sample sizes, so you have an idea of how much data you'll need for a particular comparison. For proportions near $50\%$, the confidence intervals are roughly $\pm 30\%,\; 10\%,\; 3\%$, and $1\%$ for $n=10,\; 100,\; 1000,$ and $10,000$, respectively. This is why the "margin of error" in political polls, which typically have a sample size of around $1,000$, is usually about $3\%$. Of course, this rough idea is no substitute for an actual power analysis. n proportion=0.10 proportion=0.50 10 0.0025, 0.4450 0.1871, 0.8129 100 0.0490, 0.1762 0.3983, 0.6017 1000 0.0821, 0.1203 0.4685, 0.5315 10,000 0.0942, 0.1060 0.4902, 0.5098 How to calculate confidence limits Spreadsheets The descriptive statistics spreadsheet descriptive.xls calculates $95\%$ confidence limits of the mean for up to $1000$ measurements. The confidence intervals for a binomial proportion spreadsheet confidence.xls calculates $95\%$ confidence limits for nominal variables, using both the exact binomial and the normal approximation. Web pages This web page calculates confidence intervals of the mean for up to $10,000$ measurement observations. The web page for confidence intervals of a proportion handles nominal variables. R Salvatore Mangiafico's $R$ Companion has sample R programs for confidence limits for both measurement and nominal variables. SAS To get confidence limits for a measurement variable, add CIBASIC to the PROC UNIVARIATE statement, like this: data fish; input location $ dacenumber; cards; Mill_Creek_1 76 Mill_Creek_2 102 North_Branch_Rock_Creek_1 12 North_Branch_Rock_Creek_2 39 Rock_Creek_1 55 Rock_Creek_2 93 Rock_Creek_3 98 Rock_Creek_4 53 Turkey_Branch 102 ; proc univariate data=fish cibasic; run; The output will include the $95\%$ confidence limits for the mean (and for the standard deviation and variance, which you would hardly ever need): Basic Confidence Limits Assuming Normality Parameter Estimate 95% Confidence Limits Mean 70.00000 45.33665 94.66335 Std Deviation 32.08582 21.67259 61.46908 Variance 1030 469.70135 3778 This shows that the blacknose dace data have a mean of $70$, with confidence limits of $45.3$ and $94.7$. You can get the confidence limits for a binomial proportion using PROC FREQ. Here's the sample program from the exact test of goodness-of-fit page: data gus; input paw $; cards; right left right right right right left right right right ; proc freq data=gus; tables paw / binomial( P=0.5); exact binomial; run; And here is part of the output: Binomial Proportion for paw = left -------------------------------- Proportion 0.2000 ASE 0.1265 95% Lower Conf Limit 0.0000 95% Upper Conf Limit 0.4479 Exact Conf Limits 95% Lower Conf Limit 0.0252 95% Upper Conf Limit 0.5561 The first pair of confidence limits shown is based on the normal approximation; the second pair is the better one, based on the exact binomial calculation. Note that if you have more than two values of the nominal variable, the confidence limits will only be calculated for the value whose name is first alphabetically. For example, if the Gus data set included "left," "right," and "both" as values, SAS would only calculate the confidence limits on the proportion of "both." One clumsy way to solve this would be to run the program three times, changing the name of "left" to "aleft," then changing the name of "right" to "aright," to make each one first in one run. References Payton, M. E., M. H. Greenstone, and N. Schenker. 2003. Overlapping confidence intervals or standard error intervals: what do they mean in terms of statistical significance? Journal of Insect Science 3: 34. Schenker, N., and J. F. Gentleman. 2001. On judging the significance of differences by examining overlap between confidence intervals. American Statistician 55: 182-186. Contributor John H. McDonald (University of Delaware)
Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box.. There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$. What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation? Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach. Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line? Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$? Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?" @Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider. Although not the only route, can you tell me something contrary to what I expect? It's a formula. There's no question of well-definedness. I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer. It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time. Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated. You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system. @A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago. @Eric: If you go eastward, we'll never cook! :( I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous. @TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$) @TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite. @TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator
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
There is no joint density function since the random variable $(U,V,W,Z)$ takes values on a subset $D=\{(f_1^{-1}(x),f_2^{-1}(x),f_3^{-1}(x),f_4^{-1}(x))\mid x\in\mathbb R\}$ of $\mathbb R^4$ which has Lebesgue measure zero. Informally, $D$ has co-dimension $3$, hence one can compare $D$ to a line in $\mathbb R^4$.Formally, for every measurable function $\varphi$ on $\mathbb R^4$,$$\mathrm E(\varphi(U,V,W,Z))=\int\varphi(f_1(x),f_2(x),f_3(x),f_4(x))\,g(x)\mathrm dx,$$where $g$ is the density of the distribution of $X$ hence $\mathrm E(\varphi(U,V,W,Z))$ is an integral on (a subset of) $\mathbb R$ instead of $\mathbb R^4$. The simplest analogue is when $U=V=X$ with $X$ uniformly distributed on $[0,1]$. Then $(U,V)$ is uniformly distributed on the diagonal $\Delta=\{(x,x)\mid x\in[0,1]\}$ hence the distribution of $(U,V)$ is$$\mathrm dP_{(U,V)}(u,v)=\mathbf 1_{u\in[0,1]}\,\delta_u(\mathrm dv)\,\mathrm du,$$where, for every $u$, $\delta_u$ is the Dirac distribution at $u$. One sees that $\mathrm dP_{(U,V)}(u,v)$ has no density with respect to Lebesgue measure $\mathrm du\mathrm dv$.
Let $N \subset \mathbb N$ denote the set of players and $v : 2^N \to \mathbb R$ , $v(\emptyset) = 0$, the characteristic function. We call $(N,v)$ a cooperative transferable utility (TU) game. Definition 1. A game $(N,v)$ is called convex, if for all $S,T \subseteq N$ it holds that\begin{align}v(S \cup T) + v(S \cap T) \geq v(S) + v(T) \tag{C.1}.\end{align}In Hafalir (2007, p. 255) it says that Def. (1) is equivalent to the following statement. Definition 2. A game $(N,v)$ is called convex, if for all $S,T \subseteq N$ with $\|S \setminus T\| = \|T \setminus S\| \leq 1$ it holds that \begin{align} v(S \cup T) + v(S \cap T) \geq v(S) + v(T) \tag{C.2}. \end{align} There is no proof of the statement, but the author refers to Moulin (1988, p. 112). But here it only gives an equivalent for Definition 1. Definition 1'. A game $(N,v)$ is called convex, if for all $i \in N$ and $S \subset T \subset N \setminus \{i\}$ it holds that \begin{align} v(S \cup \{i\}) - v(S) \leq v(T \cup \{i\}) - v(T) \tag{C.3}. \end{align} Definitions 1. and 1'. are standard. I never heard of 2. though. I have a problem, however, where I can show that Def. 2. is satisfied. And now I'm wondering how to show the equivalence of Def. 2. with either 1. or 1.'.
class Gudhi::cubical_complex::Bitmap_cubical_complex< T > Cubical complex represented as a bitmap. More... class Gudhi::cubical_complex::Bitmap_cubical_complex_base< T > Cubical complex represented as a bitmap, class with basic implementation. More... class Gudhi::cubical_complex::Bitmap_cubical_complex_periodic_boundary_conditions_base< T > Cubical complex with periodic boundary conditions represented as a bitmap. More... Bitmap_cubical_complex is an example of a structured complex useful in computational mathematics (specially rigorous numerics) and image analysis. The presented implementation of cubical complexes is based on the following definition. An elementary interval is an interval of a form $ [n,n+1] $, or $[n,n]$, for $ n \in \mathcal{Z} $. The first one is called non-degenerate, while the second one is degenerate interval. A boundary of a elementary interval is a chain $\partial [n,n+1] = [n+1,n+1]-[n,n] $ in case of non-degenerated elementary interval and $\partial [n,n] = 0 $ in case of degenerate elementary interval. An elementary cube $ C $ is a product of elementary intervals, $C=I_1 \times \ldots \times I_n$. Embedding dimension of a cube is n, the number of elementary intervals (degenerate or not) in the product. A dimension of a cube $C=I_1 \times ... \times I_n$ is the number of non degenerate elementary intervals in the product. A boundary of a cube $C=I_1 \times \ldots \times I_n$ is a chain obtained in the following way: \[\partial C = (\partial I_1 \times \ldots \times I_n) + (I_1 \times \partial I_2 \times \ldots \times I_n) + \ldots + (I_1 \times I_2 \times \ldots \times \partial I_n).\] A cubical complex $\mathcal{K}$ is a collection of cubes closed under operation of taking boundary (i.e. boundary of every cube from the collection is in the collection). A cube $C$ in cubical complex $\mathcal{K}$ is maximal if it is not in a boundary of any other cube in $\mathcal{K}$. A support of a cube $C$ is the set in $\mathbb{R}^n$ occupied by $C$ ( $n$ is the embedding dimension of $C$). Cubes may be equipped with a filtration values in which case we have filtered cubical complex. All the cubical complexes considered in this implementation are filtered cubical complexes (although, the range of a filtration may be a set of two elements). The implementation of Cubical complex provides a representation of complexes that occupy a rectangular region in $\mathbb{R}^n$. This extra assumption allows for a memory efficient way of storing cubical complexes in a form of so called bitmaps. Let $R = [b_1,e_1] \times \ldots \times [b_n,e_n]$, for $b_1,...b_n,e_1,...,e_n \in \mathbb{Z}$, $b_i \leq d_i$ be the considered rectangular region and let $\mathcal{K}$ be a filtered cubical complex having the rectangle $R$ as its support. Note that the structure of the coordinate system gives a way a lexicographical ordering of cells of $\mathcal{K}$. This ordering is a base of the presented bitmap-based implementation. In this implementation, the whole cubical complex is stored as a vector of the values of filtration. This, together with dimension of $\mathcal{K}$ and the sizes of $\mathcal{K}$ in all directions, allows to determine, dimension, neighborhood, boundary and coboundary of every cube $C \in \mathcal{K}$. Note that the cubical complex in the figure above is, in a natural way, a product of one dimensional cubical complexes in $\mathbb{R}$. The number of all cubes in each direction is equal $2n+1$, where $n$ is the number of maximal cubes in the considered direction. Let us consider a cube at the position $k$ in the bitmap. Knowing the sizes of the bitmap, by a series of modulo operation, we can determine which elementary intervals are present in the product that gives the cube $C$. In a similar way, we can compute boundary and the coboundary of each cube. Further details can be found in the literature. In the current implementation, filtration is given at the maximal cubes, and it is then extended by the lower star filtration to all cubes. There are a number of constructors that can be used to construct cubical complex by users who want to use the code directly. They can be found in the Bitmap_cubical_complex class. Currently one input from a text file is used. It uses a format inspired from the Perseus software (http://www.sas.upenn.edu/~vnanda/perseus/) by Vidit Nanda. -1's to indicate missing cubes. If you have missing cubes in your complex, please set their filtration to $+\infty$ (aka. inf in the file). The file format is described in details in Perseus file format section. Often one would like to impose periodic boundary conditions to the cubical complex. Let $ I_1\times ... \times I_n $ be a box that is decomposed with a cubical complex $ \mathcal{K} $. Imposing periodic boundary conditions in the direction i, means that the left and the right side of a complex $ \mathcal{K} $ are considered the same. In particular, if for a bitmap $ \mathcal{K} $ periodic boundary conditions are imposed in all directions, then complex $ \mathcal{K} $ became n-dimensional torus. One can use various constructors from the file Bitmap_cubical_complex_periodic_boundary_conditions_base.h to construct cubical complex with periodic boundary conditions. One can also use Perseus style input files (see Perseus). End user programs are available in example/Bitmap_cubical_complex and utilities/Bitmap_cubical_complex folders. GUDHI Version 3.0.0 - C++ library for Topological Data Analysis (TDA) and Higher Dimensional Geometry Understanding. - Copyright : MIT Generated on Tue Sep 24 2019 09:57:51 for GUDHI by Doxygen 1.8.13
Suppose that $R$ is a commutative ring (with unity) of characteristic $0$, and that the set $Z$ of zero divisors in $R$ forms an ideal. Does it follow that the characteristic of $R/Z$ is $0$? Equivalently (with less verbiage), suppose that the ring homomorphism $\iota:\mathbb{Z} \to R$ injects. Does it follows that $\iota(m)$ is not a zero divisor, for all $m$? This proposition holds when all zero divisors are nilpotent: Example: Suppose that $Z = \mathrm{nil}(R)$, the nilradical of $R$. Then $Z$ is an ideal, and if $\iota(m) \in Z$, then $\iota(m^n)=0$. Here $\mathrm{char}(R)=0$ forces $m^n=0$, so $m=0$ and $\mathrm{char}(R/Z)=0$. If this proposition does not hold under my given hypotheses, I would appreciate a counter-example or a new hypothesis on $Z$; something more restrictive than $Z$ forming an ideal and less restrictive than $Z =\mathrm{nil}(R)$.
The idea is to "back off the inf's" and use a "since $\epsilon$ is arbitrary" argument. More precisely, $1).\ A$ is bounded below, which implies that $A+k$ is also. This means that each has a finite greatest lower bound. $2).\ $ Set $\alpha=\inf A$ and $\beta=\inf (A+k)$. You want to show that $\beta=\alpha+k.$ As you rightly point out, if we can show $\beta\le\alpha+k$ and $\beta\ge\alpha+k,$ we will have proved the claim. Let $\epsilon>0$. $3).\ $ There is an $a\in A$ such that $a<\alpha+\epsilon.$ Then, $a+k\in A+k$ and so $\beta\le a+k<\alpha+k+\epsilon.$ Since $\epsilon>0$ is arbitrary, we get $\beta\le\alpha+k$. $4).$ There is a $b\in A+k$ such that $b\le \beta+\epsilon.$ But $b=a+k$ for some $a\in A$ (by definition of the set $A+k$) and of course $\alpha\le a.$ Therefore, $\alpha+k\le a+k=b< \beta+\epsilon$ and so by the same reasoning as in $(3)$, we get $\alpha +k\le \beta.$
Difference between revisions of "SageMath" m (Jupyter Notebook reference to sagemath kernel package added.) (Update outdated information) Line 23: Line 23: * {{Pkg\|sage-notebook}} includes the browser-based notebook interface. * {{Pkg\|sage-notebook}} includes the browser-based notebook interface. − {{Note\|Most + {{Note\|Most of the [http://doc.sagemath.org/html/en/installation/standard_packages.html standardpackages are available as [[pacman#Installing packages\|optional dependencies]] of the {{pkg\|sagemath}} package , 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 {{ic\|sage -i}}, in fact this command will not work if you installed SageMath with pacman.}} == Usage == == Usage == Line 41: Line 41: === Sage Notebook === === Sage Notebook === − {{Note\|The SageMath Flask notebook is + {{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 {{pkg\|sage-notebook-exporter}} application to convert your Flask notebooks to Jupyter}} A better suited interface for advanced usage in SageMath is the Notebook. A better suited interface for advanced usage in SageMath is the Notebook. Line 55: Line 55: === Jupyter Notebook === === Jupyter Notebook === − SageMath also provides a kernel for the [https://jupyter.org/ Jupyter] notebook in + SageMath also provides a kernel for the [https://jupyter.org/ Jupyter] notebook in {{Pkg\|sagemath-jupyter}} . To use it, launch the notebook with the command $ jupyter notebook $ jupyter notebook and choose "SageMath" in the drop-down "New..." menu. The SageMath Jupyter notebook supports [[LaTeX]] output via the {{ic\|%display latex}} command and 3D plots if {{Pkg\|jmol}} is installed. and choose "SageMath" in the drop-down "New..." menu. The SageMath Jupyter notebook supports [[LaTeX]] output via the {{ic\|%display latex}} command and 3D plots if {{Pkg\|jmol}} is installed. Revision as of 14:17, 29 September 2017 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.
Quadratic programming (QP) involves minimizing or maximizing an objective function subject to bounds, linear equality, and inequality constraints. Example problems include portfolio optimization in finance, power generation optimization for electrical utilities, and design optimization in engineering. Quadratic programming is the mathematical problem of finding a vector $x$ that minimizes a quadratic function: \[\min_{x} \left\{\frac{1}{2}x^{\mathsf{T}}Hx + f^{\mathsf{T}}x\right\}\] Subject to the constraints: \[\begin{eqnarray}Ax \leq b & \quad & \text{(inequality constraint)} \\A_{eq}x = b_{eq} & \quad & \text{(equality constraint)} \\lb \leq x \leq ub & \quad & \text{(bound constraint)}\end{eqnarray}\] The following algorithms are commonly used to solve quadratic programming problems: Interior-point-convex:solves convex problems with any combination of constraints Trust-region-reflective:solves bound constrained or linear equality constrained problems For more information about quadratic programming, see Optimization Toolbox™.
Heavenly Chips is a "new game plus" or "Ascension prestige" mechanic introduced in Cookie Clicker version 1.035. If you ascend by pressing the "Legacy" button, you will gain Heavenly Chips and be sent to the ascension screen. The amount of Heavenly Chips is dependent on your forfeited cookies of all games. Every reset, the amount of forfeited cookies is incremented by 'cookies baked (this game)', and will match 'cookies baked (all time)' until new cookies are baked. The easiest way to determine how many Heavenly Chips will be earned after a single reset is therefore by taking the difference of the amount of Heavenly Chips before and after the reset. Every Heavenly Chip you gain increases your prestige level. Heavenly Chips can be spent on various upgrades during an ascension. All upgrades are permanent when bought, and carry forward through ascensions. They include "Permanent Upgrade slots" to keep certain upgrades forever without having to buy them again. Other heavenly chip upgrades can unlock biscuits, the season switcher and season boosts, positive effects for golden cookies, the dragon, offline production of cookies and synergy upgrades, among other things. Wiping your save (formerly called hard reset) does not activate the Prestige System, and will also wipe your current amount of forfeited cookies and Heavenly Chips. Currently, the maximum amount of heavenly chips you can have is 564.38 Untrigintillion. (however this is impossible to reach without cheating) If you somehow earn enough cookies to get more than that, the "cookies baked (all time)" statistic will change to infinity, because the game doesn't support 2^1024 or higher numbers. "Your amount of prestige (heavenly chips, for now) at any time is dependent on all the cookies you've ever forfeited by soft-resetting. Each time you reset, your "cookies baked (all game [should be: this game])" are added to that pool." - Orteil Calculating Heavenly Chips (Version 2.0)Edit The number of cookies you have to bake to get to the next Heavenly chip is: $ \text{Cookies needed}=10^{12} \cdot ((K + 1)^3 - K^3) $ where Kis the number of heavenly chips you already have. The total number of cookies needed is 1 trillion times M cubed: $ \text{Cookies needed (total)}=10^{12} \cdot M^3 $ where Mis the number of heavenly chips desired. The total number of Heavenly Chips is calculated by the following formula (the outer brackets indicate a floor function). $ \text{Heavenly Chips}=\left\lfloor{\sqrt[3]{\frac{N}{10^{12}}}}\right\rfloor $ where Nis the number of cookies that have ever been made, including all previous resets. Also, you can use this script for displaying Heavenly Chips count after the next reset: var sucked = 0; for (var i = 0; i < Game.wrinklers.length; i++) { suck=Game.wrinklers[i].sucked; var toSuck=1.1; if (Game.Has('Sacrilegious corruption')) toSuck=1.05; if (Game.wrinklers[i].type==1) toSuck=3; suck=toSuck; if (Game.Has('Wrinklerspawn')) suck=1.05; sucked += suck; } var prestige=Game.HowMuchPrestige(Game.cookiesReset+Game.cookiesEarned+sucked), prestigeDifference=prestige-Game.prestige; Game.Notify('Wrinkler Cookies','Once Popped: '+Beautify(sucked),[19,7]); Game.Notify('Prestige','Additional PL and HC: '+Beautify(prestigeDifference),[19,7]); Game.Notify('Prestige levels','Total PL: '+Beautify(prestige),[19,7]); Game.Notify('Heavenly Chips','Total HC: '+Beautify(prestigeDifference+Game.heavenlyChips),[19,7]); For people who can't understand any of the calculations, the first HC "Costs" 1 3=1 trillion cookies. The second one costs 2 3=8 trillion. The third costs 3 3=27 trillion, etc. The following is a quick list of quantities of cookies baked (all-time), and how many Heavenly Chips they will yield after a reset (Keep in mind that this is not a complete table in the sense that you can receive Heavenly Chip quantities between displayed amounts, e.g. 42): If you don't want to proceed, here is a link to an online fanmade converter that can manage this for you if you don't want to deal with the intense math (outdated): https://cookie.riimu.net/help/ Short (A) Long (B) Scientific Cookies Baked (all time) Heavenly Chips 1T 1B 1x10 12 1,000,000,000,000 1 8T 8B 8×10 12 8,000,000,000,000 2 27T 27B 2.7×10 13 27,000,000,000,000 3 64T 64B 6.4×10 13 64,000,000,000,000 4 125T 125B 1.25×10 14 125,000,000,000,000 5 216T 216B 2.16×10 14 216,000,000,000,000 6 343T 343B 3.43×10 14 343,000,000,000,000 7 512T 512B 5.12×10 14 512,000,000,000,000 8 729T 729B 7.29×10 14 729,000,000,000,000 9 1Qa 1B×10 3 1×10 1 5 1,000,000,000,000,000 10 1Qi 1T 1×10 18 1,000,000,000,000,000,000 100 1Sx 1T×10 3 1×10 21 1,000,000,000,000,000,000,000 1,000 1Sp 1Qa 1×10 24 1,000,000,000,000,000,000,000,000 10,000 1Oc 1Qa×10 3 1×10 27 1,000,000,000,000,000,000,000,000,000 100,000 1No 1Qi 1×10 30 1,000,000,000,000,000,000,000,000,000,000 1,000,000 1Dc 1Qi×10 3 1×10 33 1,000,000,000,000,000,000,000,000,000,000,000 10,000,000 Tips and TricksEdit Before you reset, try to obtain all of the achievements (except those requiring 200 of most buildings, 100 Antimatter Condensers, 100 Prisms, Neverclick, Wholesome, and Leprechaun), so that getting back to where you were is as easy as possible. Note that the number of Golden Cookie clicks, as well as any achievements you may have obtained, is retained after resetting, so you do not need to worry about getting Leprechaun (and Black cat's paw) before resetting (but this doesn't apply only for popping wrinklers or reindeer). If you're already thinking about resetting, and you don't already have Neverclick, consider going for it to boost your milk for the next game (you have to be in the challenge mode to get this achievement). After you reset, click the big cookie no more than 15 times or grab Golden Cookies until you get Lucky! twice, then start buying cursors. You may hammer the big cookie again after baking 1 million. Heavenly chips are an additive bonus to your CpS multiplier; they do not multiply with any CpS upgrade other than kitten upgrades. Assuming you'll reset your first game with all cookie upgrades unlocked so far, and with all upgrades from the bingo facility, you'll have a pre-kitten multiplier of 610%. [As of v1.0393.] If you reset with 230 chips, you'll add 460% to that amount, and will unlock upgrades that together add for a new 150%. Adding everything up, you'll end up with 1220% multiplier + kittens, doubling your income. In order to double your income again, you'll need 598 MORE chips, for a total of 828 chips. This will add 1196% to your multiplier, and will unlock another 25% upgrade, for a total of 1220+(1196+25)% = 2441% To double it again, you'll need 1208 chips, for a total of 2036, and 2441+(2416 + upgrade unlocking 25)% = 4882% multiplier. The next double will require 2441 chips, adding up to 4477 chips, and 9764% multiplier. Permanent Upgrade slotsEdit When purchasing the Permanent Upgrade slots, you will get a chance to pick one upgrade you have purchased in the game session right before ascending, and apply it to all future games without needing to wait to unlock or purchase it. However, these choices are not locked, and can be changed at future ascensions for no cost. The fastest way to progress through the game is to put the highest "----illion Fingers" upgrade you have purchased in one of the permanent upgrade slots. Even a mere Quintillion Fingers and 101 (1 for Legacy (which enables further upgrades and 100 for a perm slot) heavenly cookies means that with 10 cursors and 5 grandmas you can produce 250k CpS instantly, increasing to 2.75M CpS as you unlock the power of your heavenly cookies. Bulk buy cursors and cheap buildings, and within a few minutes, you will have unlocked at least time machines. If you waited longer to ascend, you might be able to far exceed your previous cookies baked within less than an hour. Alternatively, some suggest Omelette (to speed up the search for Easter Eggs), the most expensive Kitten upgrade available and/or Golden Cookie-related upgrades to be put in the Permanent Upgrade slots. Heavenly Chips and Soft Resetting BugEdit If you have some HCs and your potential upgrades are done, there is a chance to have a small boost for the next early-game by unlocking cps achievement(s). It is because the game calculates the current cps based on 'delayed' HC count. Here is the scenario. 1. You now have 224B cps with 7854 % multiplier (192 HCs), and total cookie count of 3.642 x 10^18. If you reset the game, you will have 2698 HCs, boosting multiplier to 36563 % with all upgrades. 2. Export the save and reset. 3. Import the save, then and the game calculates the cps according to 2698 HCs, making the current cps over 1.04T and the achievement 'Let's never bake again' will be unlocked. 4. Of course, the HC count and cps are back to normal value instaneously, but this time due to 4% more milk, cps will be 227B 5. Reset again and start with increased milk. Don't worry about soft resetting!Edit Heavenly Chips' benefits far outweigh the costs of buying them. The CPS bonuses for having only a few hundred Heavenly Chips will allow one to trivially reach their previous status in only a day or so, almost always. If you're worried about resetting, simply export a save and email it to yourself (or simply save the code in a .txt file on your desktop. Nothing easier than that), just in case you're not satisfied with the Heavenly multipliers. Assume 6000 heavenly chips already present (last reset at 1810^18 chips = 18 Qi), current state 24.5 Qi chips produced = 7000 HC if you reset. A reasonable production rate for this state of progress is 1.9 T/s. Now, a daily passive production is 86400s 1.1 (wrinklers) * 1.9 T/s = 181 Qa (or about 29 HC on the first day). Calculate the current cookie multiplier: 100% base 695% flavoured cookies 15% bingo 12000% via 6000 HC = 12810% base multiplier *5.2273 (kitten upgrades at 388% milk) = 66962% Calculate the new cookie multiplier 810% other stuff as above 14000% via 7000 HC 5.2273 via kittens = 77416% which is 1.16 times the previous multiplier. Now, a 16% increment in production is an incredible amount in end-game. Consider buying one building of each type adds about 14 B = 0.014 T cookies per second in this scenario. Compared to the 1.8 T/s already reached, this is less than 0.8% improvement. To reach the 16% that is given by more HCs, one had to add 0.304 T/s or 22 buildings of each type. Remember, a level 127 condenser and corresponding other buildings already cost more than you earn per passive day. And at 6.5 Qi chips produced in a single game, you are likely to exceed these building levels by far, so a complete set of buildings are more like 10 days of passive cookie farming. Reaching previous cookie production rate after a reset however takes you usually less than a day when playing semi-active. But, as your multiplier does the job for you, your building levels can be much lower and therefore much cheaper (usually 10x!). On the other hand, a higher multiplier also means more CPS addition for the same amount spent in buildings... In the discussed case, a reset is badly needed. Doing it before this state of sluggish expansion is advised, perhaps at 6500 HC (21.1 Qi total) or a little later. Ideas for generating rules-of-thumb: Few HC: Calculate your new multiplier and effective gain (new multiplier/old multiplier), if it's larger than, say, 25% then reset! Hundreds to thousands of HC: If your antimatter condenser/prism takes more than one week of idling -> reset! Tens of thousands and more HC: Reset whenever you get time to rebuild everything from scratch to match at least your previous CPS General: Earn x% more cookies than in previous game. Don't compare to total cookie earnings, as HC generation slows down due to the calculation formula. Also, don't aim to add a fixed amount of HCs, like 100 per game or doubling it every time. Cookie Clicker game mechanics Cookies Clicking • Buildings General Achievements • CpS • Milk • Golden Cookies • News Ticker • Options • Cheating • Sugar Lump Upgrades Upgrades overview Multipliers: Flavored Cookies • Kittens Research: Grandmapocalypse • Wrath Cookies • Wrinklers • Shiny wrinklers Ascension Ascension • Heavenly Chips • Challenge Mode Seasons Seasons overview Valentine's Day • Business Day • Easter • Halloween • Christmas Minigames Minigames overview Garden • Pantheon • Grimoire Further reading Gameplay
I am trying to figure out the details on how to implement the structure tensor in Matlab and need some advice! For an image $\ I(x,y) $ the structure tensor S is given by: $$ S=\begin{pmatrix} W \ast I_x^2 & W \ast (I_xI_y)\\ W \ast (I_xI_y) & W \ast I_y^2 \end{pmatrix}$$ where W is a smoothing kernel, $\ \ast $ denotes convolution and subscript denotes partial derivative with respect to. I want to use a Gaussian as smoothing kernel. This is separable: $$ W = G_{\sigma}(G_{\sigma})^T$$ The constant $\ \sigma $ controls the outer/integration scale. E.g. if I truncate it at $\ +/- 2\sigma$ and use $\ \sigma = 1$ I get: $$ G_{\sigma} = [0.0103,0.2076,0.5642,0.2076,0.0103] $$ However the input to the Gaussian might have a mean different from 0. How do I compensate for this? I was thinking substracting the mean first. I could use central differences to estimate the partial derivatives: $$ I_x \approx [-0.5 \hspace{2 pt} 0 \hspace{2 pt} 0.5]I $$ However since the image is noisy it is probably better to use a more robust differentian operator. The Sobel operator is such an operator. However according to wikipedia it is not perfectly rotational symmetric. The Scharr operators are supposed to be perfectly rotational symmetric and robust approximations of the partial derivatives: $$ I_x \approx \frac{1}{32}\begin{pmatrix} -3 & 0 & 3\\ -10 & 0 & 10\\ -3 & 0 & 3 \end{pmatrix} * I$$ $$ I_y \approx \frac{1}{32}\begin{pmatrix} -3 & -10 & -3\\ 0 & 0 & 0\\ 3 & 10 & 3 \end{pmatrix} * I$$ The Scharr operators are separable: $$ \begin{pmatrix} -3 & 0 & 3\\ -10 & 0 & 10\\ -3 & 0 & 3 \end{pmatrix} = [-1,0,1][3,10,3]^T $$ $$ \begin{pmatrix} -3 & -10 & -3\\ 0 & 0 & 0\\ 3 & 10 & 3 \end{pmatrix} = [3,10,3][-1,0,1]^T$$ The elements of the structure tensor now becomes series of consecutive horizontal and vertical convolutions with the original image. EDIT: Another option (thanks to nikie!) is the Gaussian derivative. Introducing $$ GD_{\sigma}=\frac{-x}{\sigma}G{\sigma}$$ Is this then correct?$$ Ix = G{\sigma}^T(GD_{\sigma}I) $$$$ Iy = G{\sigma}(GD_{\sigma}^TI) $$ How do I Implement this easily in Matlab? How about in C++? Any comments regarding choice of smoothing kernel and differentiation operator? EDIT 2: My Matlab code so far: % I: image% si: inner (differetiation) scale,% so: outer (integration) scalefunction [Sxx, Sxy, Syy] = structureTensor(I,si,so)[m n] = size(I);Sxx = NaN(m,n);Sxy = NaN(m,n);Syy = NaN(m,n);% Robust differentiation by convolution with derivative of Gaussian:x = -2si:2si;g = exp(-0.5(x/si).^2);g = g/sum(g);gd = -x.g/si; % is this normalized?Ix = conv2( g',conv2(gd,I) );Iy = conv2( g,conv2(gd',I) );Ixx = Ix.^2;Ixy = Ix.Iy;Iyy = Iy.^2;% Smoothing:x = -2so:2so;g = exp(-0.5(x/so).^2);Sxx = conv2( g',conv2(g,Ixx) ); Sxy = conv2( g',conv2(g,Ixy) );Syy = conv2( g',conv2(g,Iyy) ); How do I normalize the Gaussian derivative?
Elasticity is the slope after taking a log transform. Throughout this post $\log$ refers to the natural logarithm. That is, the logarithm with base $e$, the ln function in Excel. The logarithm transforms multiplication into addition. Log differences are something akin to percent changes (and they're equivalent for small changes). Let's say we have: $$ Q_t = A P_t^b $$ Take the logarithm of both sides: $$ \log Q_t = \log A + b \log P_t$$ Log differences are conceptually similar to percent changes (for small percent changes). The slope of the line using the log transformed variables is $b$. This is the elasticity. $$ \frac{d \left( \log Q_t \right) }{\log P_t} = b $$ Why do people treat elasticities as percent changes? $$\log(Y) - \log(X) \approx \frac{Y-X}{X} \quad \quad \text{ for }\frac{y}{x} \approx 1$$ Log differences are approximately the percent change. Why? Take the first-order taylor expansion of the log to approximate it with a line near 1. \begin{align} \log(Y) &\approx \log(1) + \frac{1}{\log(1)}\left(Y - 1 \right) \\ &\approx Y - 1 \end{align}Hence $\log\left(\frac{Y}{X} \right) \approx \frac{Y-X}{X}$ for $Y \approx X$. This works well for small changes, eg. $\log(1.03) = .0296$, but it gets quite off for big changes $\log(1.5) = .4055$. This is probably where your stuff is getting off... Example where logs are nice... Let's say we have $1, 2, 4, 8$. Each is a 100% increase from the previous value. Taking logs we have $0, .693, .1386, 2.079$. The numbers increase linearly by $\log(2)$. Looking at percent increases though, it's less clean: $\frac{2-1}{1} = 100\%$ and $\frac{4-2}{2} = 100\%$ but $\frac{4-1}{1} = 300%$. $$ \left(1 + R_1 \right) \left(1 + R_2 \right) = 1 + R_1 + R_2 + R_1R_2$$Dealing with percent changes gets messy when $R_1R_2$ is big enough to matter. Taking logs though, let $r_1 = \log(1+R_1)$ etc... $$ \log\left((1+R_1)(1+R_2) \right)= r_1 + r_2$$ What I would do if I were you: Take a log and then it will be clean, and simple. Add a few columns to your spreadsheet. $q_t = \log Q_t$. $a = \log A$ $p_t = \log P_t$ Then the equation is:$$ q_t = a + b p_t$$ $$\frac{dq_t}{dp_t} = b $$ $$ \begin{array}{rrrrr} P & Q & \log P & \log Q & \Delta \log P & \Delta \log Q & \frac{\Delta \log P}{\Delta \log Q} \\1 & 100 & 0 & 4.605 & \\2 & 25 & .6931 & 3.2189 & .6931 & -1.3863 & -2.0\\3 & 11.11 & 1.0986 & 2.4079 & .4055 & -.8109 & -2.0\\4 & 6.25 & 1.3863 & 1.8326 & .2877 & -.5754 & -2.0\\5 & 4 & 1.6094 & 1.3863 & .2231 & -.4463 & -2.0\end{array}$$ $\Delta$ means "change in", hence $\Delta \log Q_i = \log Q_i - \log Q_{i-1}$.
We prove that the ideal $f^{-1}(P)$ is prime.Suppose that we have $ab\in f^{-1}(P)$ for $a, b\in R$. Then we have $f(ab) \in P$.Since $f$ is a ring homomorphism, we obtain\begin{align}f(a)f(b)=f(ab)\in P.\end{align} Since $P$ is a prime ideal, it follows that either $f(a)\in P$ or $f(b)\in P$.Hence we have either $a\in f^{-1}(P)$ or $b\in f^{-1}(P)$.This proves that the ideal $f^{-1}(P)$ is prime. Characteristic of an Integral Domain is 0 or a Prime NumberLet $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.Definition of the characteristic of a ring.The characteristic of a commutative ring $R$ with $1$ is defined as […] The Inverse Image of an Ideal by a Ring Homomorphism is an IdealLet $f:R\to R'$ be a ring homomorphism. Let $I'$ be an ideal of $R'$ and let $I=f^{-1}(I)$ be the preimage of $I$ by $f$. Prove that $I$ is an ideal of the ring $R$.Proof.To prove $I=f^{-1}(I')$ is an ideal of $R$, we need to check the following two […] Generators of the Augmentation Ideal in a Group RingLet $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring […] Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$Let\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2\|a\}\]be an ideal of the ring\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\]Then determine the quotient ring $\Z[\sqrt{10}]/P$.Is $P$ a prime ideal? Is $P$ a maximal ideal?Solution.We […]
so I found out that SR-71 Blackbird is using what's called "turboramjet" and I find the idea a bit appealing since they say with such engine, the Blackbird is more fuel efficient at it's top speed. I know that the mechanism for such engine is extremely complicated, but what's on my mind is not making an engine like Blackbird's, more to the point of having a direct supersonic fresh airflow to the afterburner to keep the fuel consumption slightly lower thanks to more oxygen to keep the afterburner effectively burn(because I heard somehwhere afterburners burns 3 times more fuel than dry thrust, I hope this can reduce it to only twice or even 1.5 times original dry thrust fuel consumption and have the same performance). Is it possible to do such thing? what would be the main problem to such engine/design? can it produce more thrust/more efficient in fuel consumption? and for the sake of calculations, let's say the turbofan engine in question is GE F-414-EPE, on afterburner, in supersonic speed. The J-58 took compressed air from the compressor at stage 4 and piped it directly into the flow aft of the turbine. This cooled the exhaust stream that entered the afterburner, so the starting temperature there was lower and the density higher, which increases efficiency and thrust. Note that the intake of the SR-71 slowed the flow down to Mach 0.4, so all internal flow was subsonic. Only when the hot exhaust gas in the afterburner expanded again would the flow speed increase to supersonic speeds again. This is only possible because the intake would already compress the air by a factor of nearly 40 when flying at Mach 3.2. This precompression scales with $$p_0 = p_{\infty}\cdot\frac{(1.2\cdot Ma^2)^{3.5}}{\left(1+\frac{5}{6}\cdot(Ma^2-1)\right)^{2.5}}$$ so the precompression is much lower (less than 6) for a typical F-414 maximum speed of Mach 1.8. ($Ma$ = Mach number, $p_0$ = ram pressure, $p_{\infty}$ = atmospheric pressure). The gains in efficiency scale with the ratio of starting and exhaust temperatures (measured from absolute zero), so the increase in efficiency is much lower than what you seem to hope for. Make sure that flow speed at the entry of the afterburner is decidedly subsonic. Supersonic flow at the start of the afterburner section will only delay ignition until the flow exits through the nozzle. All afterburners have rings inside, called flameholders, which cause local separated flow, so some burning gas is always present to ignite the newly arriving fuel-air mixture.
So we are in the midst of solving quadratic equations of the form:\[ {ax}^{2}\hspace{0.33em}{+}\hspace{0.33em}{bx}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0} \] where a, b, and c are some numbers. So far, we have been looking at equations like this that can be factored so that we can use the Null Factor Law. But most quadratics cannot be factored by hand and the solutions to the equations are frequently decimal numbers. So what to do? Well fortunately mathematicians have heard your concerns and way back in the year 628, an Indian mathematician Brahmagupta came up with a formula called the quadratic formula. This formula will solve any quadratic equation. So given a quadratic equation, if you identify the three coefficients a, b, and c, the solution can be obtained by using the following formula: {x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{b}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}}}{2a} \] This is a very powerful equation. It is not difficult to prove but I’d rather you research that yourself and for now, just accept my word for it truthfulness. Let’s use it in some examples.\[ {2}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{+}\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}{0} \] Here a = 2, b = 5, and c = 3. Let’s put these numbers in the quadratic formula. Notice that there is a symbol ± in the formula. That means that there are generally two solutions: one using the + symbol and the other using the – symbol. So the solutions to this equation are: {x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{5}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{2}{)(}{3}{)}}}{2(2)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}{1}}{4}\hspace{0.33em}{=}\hspace{0.33em}{-}{1}{,}\hspace{0.33em}{-}{1}{.}{5} \] So there are two solutions to this quadratic equation. Let’s do another one:\[ {3}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{0} \] Here a = 3, b = -4, and c = -7. It’s important to include the signs in the a, b, and c and include them in the quadratic formula. So the solutions are:\[ {x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{(}{-}{4}{)}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{(}{-}{4}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{3}{)(}{-}{7}{)}}}{2(3)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{4}\hspace{0.33em}\pm\hspace{0.33em}{10}}{6}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{33}{,}\hspace{0.33em}{-}{1} \] Now look at the \[ {b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac} \] part of the formula. If this is a positive number, like in the last two examples, then you can take the square root and get two solutions: one by using the + symbol and the other by using the – symbol. If this is zero, then you only get one solution as adding or subtracting a zero doesn’t give two solutions, just one. And if the expression is negative, well there is no solution in the real world because you can’t take the square root of a negative number. However, scientists and engineers work in a complex world where you can take the square root of a negative number, and this has a physical meaning. Maybe some day I will talk about these special numbers.
Relative efficiency between two unbiased estimators $\widehat{\theta }_{A}$and $\widehat{\theta }_{B}$ of an unknown parameter vector $\theta _{0}\in \mathbb{R}^{K}$ is usually defined as follow (see for instance Ruud, 2001).The estimator $\widehat{\theta }_{A}$ is said to be efficient relative to $\widehat{\theta }_{B}$ if we have:$$ \mathrm{V}\left[ \widehat{\theta }_{A}\right] \equiv \Omega _{A}<<\Omega_{B}\equiv \mathrm{V}\left[ \widehat{\theta }_{B}\right] .$$ If $\Omega _{B}-\Omega _{A}$ is positive definite, then the diagonal termsof $\Omega _{B}$ and $\Omega _{A}$ are necessarily such that $\sigma_{Bii}^{2}>\sigma _{Aii}^{2}.$ Indeed if $v^{T}\left( \Omega _{B}-\Omega_{A}\right) v>0$ for any $v\neq 0$ then for $v=e_{i}$ we catch the diagonalterms out of $\Omega _{B}-\Omega _{A}$ and find that $\sigma_{Bii}^{2}>\sigma _{Aii}^{2}.$ The converse, however, it is not true: themere conditions $\sigma _{Bii}^{2}>\sigma _{Aii}^{2}$ do not garantee that $\Omega _{B}-\Omega _{A}$ be positive definite. Covariances matter. The answer to your second question is: no it never happens that $\sigma _{Bii}^{2}<\sigma _{Aii}^{2}$ when $\Omega _{B}>>\Omega _{A}$. The answer to your first question is: it is important to consider all covariance terms. If we want that the confidence ellipse (at any threshold) of $\widehat{\theta }_{A}$ be nested within the confidence ellipse of $\widehat{\theta }_{B}$ then we need $\Omega _{A}<<\Omega _{B}$ and not just $\sigma _{Aii}^{2}<\sigma _{Bii}^{2}$. See Ruud, (2001, chapter 9) for a proof and detailed explanations. An example is provided here, illustrating that the confidence ellipses are nested when $\Omega _{B}-\Omega _{A}$ positive definite, and not nested if $\Omega _{B}-\Omega _{A}$ is not positive definite. Example: \begin{eqnarray}\widehat{\theta }_{A}\left( a\right) &\sim &\mathcal{N}\left( 0,\Omega_{A}\left( a\right) \right) ,\qquad \widehat{\theta }_{B}\sim \mathcal{N}%\left( 0,\Omega _{B}\right) \\\Omega _{A}\left( a\right) &=&\left( \begin{array}{cc}4 & a \\ a & 9%\end{array}%\right) ,\qquad \Omega _{B}=\left( \begin{array}{cc}5 & 0 \\ 0 & 10%\end{array}%\right).\end{eqnarray} For $a=0$ the matrix $\Omega _{B}-\Omega _{A}\left( 0\right) $ is positivedefinite and the $95\%$ threshold confidence ellipses are nested. The Figurebelow (left panel) represents the iso-curves centered on $\theta _{0}=0$ andwhose equations are given by $v^{T}\left( \Omega _{A}\left( 0\right) \right)^{-1}v=5.99$ and $x^{T}\Omega _{B}^{-1}x=5.99$ and illustrates that for $a=0$the later is nested within the former. This is no longer true for $a=-5$(right panel) in which case $\Omega _{B}-\Omega _{A}\left( -5\right) $ is nolonger positive definite. In this case the probability that $\widehat{\theta }_{A}$ is further away than $\widehat{\theta }_{B}$ from the true value $%\theta _{0}=0$ is positive, and $\widehat{\theta }_{A}$ is no longerefficient (case with $a=-5$) relatively to $\widehat{\theta }_{B}$. Notethat the variances always satisfy $\sigma _{Bii}^{2}>\sigma _{Aii}^{2}$ butthis is not sufficient for the ellipses to be nested, the covariances mustalso satisfy $\left( \sigma _{B11}^{2}-\sigma _{A11}^{2}\right) \left(\sigma _{B22}^{2}-\sigma _{A22}^{2}\right) -\left( \sigma _{B12}-\sigma_{A12}\right) ^{2}>0$, which is not the case in this example for $a=-5$.
ISSN: 1556-1801 eISSN: 1556-181X All Issues Networks & Heterogeneous Media March 2011 , Volume 6 , Issue 1 Select all articles Export/Reference: Abstract: We consider the Neumann spectral problem for a second order differential operator, with piecewise constants coefficients, in a domain $\Omega_\varepsilon$ of $R^2$. Here $\Omega_\varepsilon$ is $\Omega \cup \omega_\varepsilon \cup \Gamma$, where $\Omega$ is a fixed bounded domain with boundary $\Gamma$, $\omega_\varepsilon$ is a curvilinear band of variable width $O(\varepsilon)$, and $\Gamma=\overline{\Omega}\cap \overline {\omega_\varepsilon}$. The densityand stiffnessconstants are of order $O(\varepsilon^{-m-1})$ and $O(\varepsilon^{-1})$ respectively in this band, while they are of order $O(1)$ in $\Omega$; $m$ is a positive parameter and $\varepsilon \in (0,1)$, $\varepsilon\to 0$. Considering the range of the low, middle and high frequencies, we provide asymptotics for the eigenvalues and the corresponding eigenfunctions. For $m>2$, we highlight the middle frequencies for which the corresponding eigenfunctions may be localized asymptotically in small neighborhoods of certain points of the boundary. Abstract: We propose a mathematical model for the process of dry pasta cooking with specific reference to spaghetti. Pasta cooking is a two-stage process: water penetration followed by starch gelatinization. Differently from the approach adopted so far in the technical literature, our model includes free boundaries: the water penetration front and the gelatinization onset front representing a fast stage of the corresponding process. Behind the respective fronts water sorption and gelatinization proceed according to some kinetics. The outer boundary is also moving and unknown as a consequence of swelling. Existence and uniqueness are proved and numerical simulations are presented. Abstract: This paper concerns periodic multiscale homogenization for fully nonlinear equations of the form $u^\epsilon+H^\epsilon (x,\frac{x}{\epsilon},\ldots,\frac{x}{epsilon^k},Du^\epsilon,D^2u^\epsilon)=0$. The operators $H^\epsilon$ are a regular perturbations of some uniformly elliptic, convex operator $H$. As $\epsilon\to 0^+$, the solutions $u^\epsilon$ converge locally uniformly to the solution $u$ of a suitably defined effective problem. The purpose of this paper is to obtain an estimate of the corresponding rate of convergence. Finally, some examples are discussed. Abstract: We prove that for any $H: R^2 \to R$ which is $Z^2$-periodic, there exists $H_\varepsilon$, which is smooth, $\varepsilon$-close to $H$ in $L^1$, with $L^\infty$-norm controlled by the one of $H$, and with the same average of $H$, for which there exists a smooth closed curve $\gamma_\varepsilon$ whose curvature is $H_\varepsilon$. A pinning phenomenon for curvature driven flow with a periodic forcing term then follows. Namely, curves in fine periodic media may be moved only by small amounts, of the order of the period. Abstract: An elastic medium with a large number of small axially symmetric solid particles is considered. It is assumed that the particles are identically oriented and under the influence of elastic medium they move translationally or rotate around symmetry axis but the direction of their symmetry axes does not change. The asymptotic behavior of small oscillations of the system is studied, when the diameters of particles and distances between the nearest particles are decreased. The equations, describing the homogenized model of the system, are derived. It is shown that the homogenized equations correspond to a non-standard dynamics of elastic medium. Namely, the homogenized stress tensor linearly depends not only on the strain tensor but also on the rotation tensor. Abstract: The paper deals with a periodic homogenization problem for a non-stationary convection-diffusion equation stated in a thin infinite cylindrical domain with homogeneous Neumann boundary condition on the lateral boundary. It is shown that homogenization result holds in moving coordinates, and that the solution admits an asymptotic expansion which consists of the interior expansion being regular in time, and an initial layer. Abstract: We analyse the lower non trivial part of the spectrum of the generator of the Glauber dynamics for a $d$-dimensional nearest neighbour Ising model with a bounded random potential. We prove conjecture 1 in [1]: for sufficiently large values of the temperature, the first band of the spectrum of the generator of the process coincides with a closed non random segment of the real line. Abstract: This paper addresses a three-dimensional model for isothermal stress-induced transformation in shape memory polycrystalline materials in presence of permanent inelastic effects. The basic features of the model are recalled and the constitutive and the three-dimensional quasi-static evolution problem are proved to be well-posed. Finally, we discuss the convergence of the model to reduced/former ones by means of a rigorous $\Gamma$-convergence analysis. Readers Authors Editors Referees Librarians Email Alert Add your name and e-mail address to receive news of forthcoming issues of this journal: [Back to Top]
ISSN: 1556-1801 eISSN: 1556-181X All Issues Networks & Heterogeneous Media June 2011 , Volume 6 , Issue 2 Select all articles Export/Reference: Abstract: We consider solutions to a nonlinear reaction diffusion equation when the reaction term varies randomly with respect to the spatial coordinate. The nonlinearity is the KPP type nonlinearity. For a stationary and ergodic medium, and for certain initial condition, the solution develops a moving front that has a deterministic asymptotic speed in the large time limit. The main result of this article is a central limit theorem for the position of the front, in the supercritical regime, if the medium satisfies a mixing condition. Abstract: We prove convergence of discrete duality finite volume (DDFV) schemes on distorted meshes for a class of simplified macroscopic bidomain models of the electrical activity in the heart. Both time-implicit and linearised time-implicit schemes are treated. A short description is given of the 3D DDFV meshes and of some of the associated discrete calculus tools. Several numerical tests are presented. Abstract: We investigate the differentiability of minimal average energy associated to the functionals $S_\epsilon (u) = \int_{\mathbb{R}^d} \frac{1}{2}\|\nabla u\|^2 + \epsilon V(x,u)\, dx$, using numerical and perturbative methods. We use the Sobolev gradient descent method as a numerical tool to compute solutions of the Euler-Lagrange equations with some periodicity conditions; this is the cell problem in homogenization. We use these solutions to determine the average minimal energy as a function of the slope. We also obtain a representation of the solutions to the Euler-Lagrange equations as a Lindstedt series in the perturbation parameter $\epsilon$, and use this to confirm our numerical results. Additionally, we prove convergence of the Lindstedt series. Abstract: The global theory of transmission line networks with nonconservative junction conditions is developed from a spectral theoretic viewpoint. The rather general junction conditions lead to spectral problems for nonnormal operators. The theory of analytic functions which are almost periodic in a strip is used to establish the existence of an infinite sequence of eigenvalues and the completeness of generalized eigenfunctions. Simple eigenvalues are generic. The asymptotic behavior of an eigenvalue counting function is determined. Specialized results are developed for rational graphs. Abstract: We study a class of reaction-diffusion type equations on a finite network with continuity assumptions and a kind of Abstract: The dynamical stability of planar networks of non-uniform Timoshenko beams system is considered. Suppose that the displacement and rotational angle is continuous at the common vertex of this network and the bending moment and shear force satisfies Kirchhoff's laws, respectively. Time-delay terms exist in control inputs at exterior vertices. The feedback control laws are designed to stabilize this kind of networks system. Then it is proved that the corresponding closed loop system is well-posed. Under certain conditions, the asymptotic stability of this system is shown. By a complete spectral analysis, the spectrum-determined-growth condition is proved to be satisfied for this system. Finally, the exponential stability of this system is discussed for a special case and some simulations are given to support these results. Abstract: We analyze stability of consensus algorithms in networks of multi-agents with time-varying topologies and delays. The topology and delays are modeled as induced by an adapted process and are rather general, including i.i.d. topology processes, asynchronous consensus algorithms, and Markovian jumping switching. In case the self-links are instantaneous, we prove that the network reaches consensus for all bounded delays if the graph corresponding to the conditional expectation of the coupling matrix sum across a finite time interval has a spanning tree almost surely. Moreover, when self-links are also delayed and when the delays satisfy certain integer patterns, we observe and prove that the algorithm may not reach consensus but instead synchronize at a periodic trajectory, whose period depends on the delay pattern. We also give a brief discussion on the dynamics in the absence of self-links. Readers Authors Editors Referees Librarians Email Alert Add your name and e-mail address to receive news of forthcoming issues of this journal: [Back to Top]
A series of recent works has shown that Principal Ideal-SVP is not always as hard as finding short vectors in general lattices, and some schemes were broken using quantum algorithms --- the Soliloquy encryption scheme, Smart-Vercauteren fully homomorphic encryption scheme from PKC 2010, and Gentry-Garg-Halevi cryptographic multilinear-maps from Eurocrypt 2013. Those broken schemes were using a special class of principal ideals, but these works also showed how to solve SVP for principal ideals in the worst-case in quantum polynomial time for an approximation factor of $\exp(\tilde O(\sqrt n))$. This exposed an unexpected hardness gap between general lattices and some structured ones, and called into question the hardness of various problems over structured lattices, such as Ideal-SVP and Ring-LWE. In this work, we generalize the previous result to general ideals. Precisely, we show how to solve the close principal multiple problem (CPM) by exploiting the classical theorem that the class-group is annihilated by the (Galois-module action of) the so-called Stickelberger ideal. Under some plausible number-theoretical hypothesis, our approach provides a close principal multiple in quantum polynomial time. Combined with the previous results, this solves Ideal-SVP in the worst case in quantum polynomial time for an approximation factor of $\exp(\tilde O(\sqrt n))$. Although it does not seem that the security of Ring-LWE based cryptosystems is directly affected, we contribute novel ideas to the cryptanalysis of schemes based on structured lattices. Moreover, our result shows a deepening of the gap between general lattices and structured ones.
In QFT, a representation of the Lorentz group is specified as follows:$$U^\dagger(\Lambda)\phi(x) U(\Lambda)= R(\Lambda)~\phi(\Lambda^{-1}x)$$Where $\Lambda$ is an element of the Lorentz group, $\phi(x)$ is a quantum field with possibly many components, $U$ is unitary, and $R$ an element in a representation of the Lorentz group. We know that a representation is a map from a Lie group on to the group of linear operators on some vector space. My question is, for the representation specified as above, what is the vector space that the representation acts on? Naively it may look like this representation act on the set of field operators, for $R$ maps some operator $\phi(x)$ to some other field operator $\phi(\Lambda^{-1}x)$, and if we loosely define field operators as things you get from canonically quantizing classical fields, we can possibly convince ourselves that this is indeed a vector space. But then we recall that the dimension of a representation is simply the dimension of the space that it acts on. This means if we take $R$ to be in the $(1,1)$ singlet representation, this is a rep of dimension 1, hence its target space is of dimension 1. Then if we take the target space to the space of fields, this means $\phi(x)$ and $\phi(\Lambda x)$ are related by linear factors, which I am certainly not convinced of. EDIT: This can work if we view the set of all $\phi$ as a field over which we define the vector space, see the added section below. I guess another way to state the question is the following: we all know that scalar fields and vector fields in QFT get their names from the fact that under Lorentz transformations, scalars transform as scalars, and vectors transform like vectors. I would like to state the statement "a scalar field transforms like a scalar" by precisely describing the target vector space of a scalar representation of the Lorentz group, how can this be done? ADDED SECTION: Let me give an explicit example of what I'm trying to get at:Let's take the left handed spinor representation, $(2,1)$. This is a 2 dimensional representation. We know that acts on things like $(\phi_1,\phi_2)$. Let's call the space consisting of things of the form $(\phi_1,\phi_2)$ $V$. Is $V$ 2 dimensional? Viewed as a classical field theory, yes, because each $\phi_i$ is just a scalar. As a quantum field theory, each $\phi_i$ is an operator. We see that in order for $V$ space to be 2 dimensional after quantization, we need to able to view the scalar quantum fields as scalar multipliers of vectors in $V$. i.e. we need to view $V$ as a vector space defined over a (mathematical) field of (quantum) fields. We therefore have to check whether the set of (quantum) fields satisfy (mathematical) field axioms. Can someone check this? Commutativity seem to hold if we, as in quantum mechanics, take fields and their complex conjugates to live in adjoint vector spaces, rather than the same one. Checking for closure under multiplication would require some axiomatic definition of what a quantum field is. This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
2 1 ##cos(\omega)## is $$\frac{e^{j \omega } + e^{-j \omega }}{2}$$ ##sin(\omega)## is $$\frac{e^{j \omega } - e^{-j \omega }}{2 j}$$ I also know that ##cos(\omega - \pi / 2) = sin(\omega)##. I've been trying to show this using exponentials, but I can't seem to manipulate one form into the other. Specifically, I can't figure out how to manipulate $$\frac{e^{j (\omega - \pi / 2)} + e^{-j( \omega - \pi / 2)}}{2}$$ into $$\frac{e^{j \omega } - e^{-j \omega }}{2 j}$$ or the other way around. I seem to remember from back in school that there was some trick to this, but I can't remember it. I keep getting stuck on the fact that one has a plus between the exponential terms, and one has a minus. $$\frac{e^{j \omega } + e^{-j \omega }}{2}$$ ##sin(\omega)## is $$\frac{e^{j \omega } - e^{-j \omega }}{2 j}$$ I also know that ##cos(\omega - \pi / 2) = sin(\omega)##. I've been trying to show this using exponentials, but I can't seem to manipulate one form into the other. Specifically, I can't figure out how to manipulate $$\frac{e^{j (\omega - \pi / 2)} + e^{-j( \omega - \pi / 2)}}{2}$$ into $$\frac{e^{j \omega } - e^{-j \omega }}{2 j}$$ or the other way around. I seem to remember from back in school that there was some trick to this, but I can't remember it. I keep getting stuck on the fact that one has a plus between the exponential terms, and one has a minus.
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
In QFT, a representation of the Lorentz group is specified as follows:$$U^\dagger(\Lambda)\phi(x) U(\Lambda)= R(\Lambda)~\phi(\Lambda^{-1}x)$$Where $\Lambda$ is an element of the Lorentz group, $\phi(x)$ is a quantum field with possibly many components, $U$ is unitary, and $R$ an element in a representation of the Lorentz group. We know that a representation is a map from a Lie group on to the group of linear operators on some vector space. My question is, for the representation specified as above, what is the vector space that the representation acts on? Naively it may look like this representation act on the set of field operators, for $R$ maps some operator $\phi(x)$ to some other field operator $\phi(\Lambda^{-1}x)$, and if we loosely define field operators as things you get from canonically quantizing classical fields, we can possibly convince ourselves that this is indeed a vector space. But then we recall that the dimension of a representation is simply the dimension of the space that it acts on. This means if we take $R$ to be in the $(1,1)$ singlet representation, this is a rep of dimension 1, hence its target space is of dimension 1. Then if we take the target space to the space of fields, this means $\phi(x)$ and $\phi(\Lambda x)$ are related by linear factors, which I am certainly not convinced of. EDIT: This can work if we view the set of all $\phi$ as a field over which we define the vector space, see the added section below. I guess another way to state the question is the following: we all know that scalar fields and vector fields in QFT get their names from the fact that under Lorentz transformations, scalars transform as scalars, and vectors transform like vectors. I would like to state the statement "a scalar field transforms like a scalar" by precisely describing the target vector space of a scalar representation of the Lorentz group, how can this be done? ADDED SECTION: Let me give an explicit example of what I'm trying to get at:Let's take the left handed spinor representation, $(2,1)$. This is a 2 dimensional representation. We know that acts on things like $(\phi_1,\phi_2)$. Let's call the space consisting of things of the form $(\phi_1,\phi_2)$ $V$. Is $V$ 2 dimensional? Viewed as a classical field theory, yes, because each $\phi_i$ is just a scalar. As a quantum field theory, each $\phi_i$ is an operator. We see that in order for $V$ space to be 2 dimensional after quantization, we need to able to view the scalar quantum fields as scalar multipliers of vectors in $V$. i.e. we need to view $V$ as a vector space defined over a (mathematical) field of (quantum) fields. We therefore have to check whether the set of (quantum) fields satisfy (mathematical) field axioms. Can someone check this? Commutativity seem to hold if we, as in quantum mechanics, take fields and their complex conjugates to live in adjoint vector spaces, rather than the same one. Checking for closure under multiplication would require some axiomatic definition of what a quantum field is. This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
Google is constantly running experiments to test new search algorithms. For example, Google might test three algorithms using a sample of 10,000 google.com search queries. Table 6.15 shows an example of 10,000 queries split into three algorithm groups. 20 The group sizes were specified before the start of the experiment to be 5000 for the current algorithm and 2500 for each test algorithm. Search algorithm Counts current 5000 test 1 2500 test 2 2500 Total 10000 20Google regularly runs experiments in this manner to help improve their search engine. It is entirely possible that if you perform a search and so does your friend, that you will have different search results. While the data presented in this section resemble what might be encountered in a real experiment, these data are simulated. Example $\PageIndex{1}$ What is the ultimate goal of the Google experiment? What are the null and alternative hypotheses, in regular words? The ultimate goal is to see whether there is a difference in the performance of the algorithms. The hypotheses can be described as the following: H 0: The algorithms each perform equally well. H A: The algorithms do not perform equally well. In this experiment, the explanatory variable is the search algorithm. However, an outcome variable is also needed. This outcome variable should somehow reect whether the search results align with the user's interests. One possible way to quantify this is to determine whether (1) the user clicked one of the links provided and did not try a new search, or (2) the user performed a related search. Under scenario (1), we might think that the user was satis ed with the search results. Under scenario (2), the search results probably were not relevant, so the user tried a second search. Table 6.16 provides the results from the experiment. These data are very similar to the count data in Section 6.3. However, now the different combinations of two variables are binned in a two-way table. In examining these data, we want to evaluate whether there is strong evidence that at least one algorithm is performing better than the others. To do so, we apply a chi-square test to this two-way table. The ideas of this test are similar to those ideas in the one-way table case. However, degrees of freedom and expected counts are computed a little differently than before. Search algorithm current test 1 test 2 Total No new search New search 3511 1489 1749 751 1818 682 7078 2922 Total 5000 2500 2500 10000 What is so different about one-way tables and two-way tables? A one-way table describes counts for each outcome in a single variable. A two-way table describes counts for combinations of outcomes for two variables. When we consider a two-way table, we often would like to know, are these variables related in any way? That is, are they dependent (versus independent)? The hypothesis test for this Google experiment is really about assessing whether there is statistically significant evidence that the choice of the algorithm affects whether a user performs a second search. In other words, the goal is to check whether the search variable is independent of the algorithm variable. Expected Counts in Two-way Tables Example 6.35 From the experiment, we estimate the proportion of users who were satisfied with their initial search (no new search) as $\frac {7078}{10000} = 0.7078$. If there really is no difference among the algorithms and 70.78% of people are satisfied with the search results, how many of the 5000 people in the "current algorithm" group would be expected to not perform a new search? About 70.78% of the 5000 would be satis ed with the initial search: \[0.7078 \times 5000 = 3539 users\] That is, if there was no difference between the three groups, then we would expect 3539 of the current algorithm users not to perform a new search. Exercise $\PageIndex{1}$ Exercise 6.36 Using the same rationale described in Example 6.35, about how many users in each test group would not perform a new search if the algorithms were equally helpful? 21 21We would expect 0.7078 * 2500 = 1769.5. It is okay that this is a fraction. We can compute the expected number of users who would perform a new search for each group using the same strategy employed in Example 6.35 and Exercise 6.36. These expected counts were used to construct Table 6.17, which is the same as Table 6.16, except now the expected counts have been added in parentheses. Search algorithm current test 1 test 2 Total No new search New search 3511 (3539) 1489 (1461) 1749 (1769.5) 751 (730.5) 1818 (1769.5) 682 (730.5) 7078 2922 Total 5000 2500 2500 10000 The examples and exercises above provided some help in computing expected counts. In general, expected counts for a two-way table may be computed using the row totals, column totals, and the table total. For instance, if there was no difference between the groups, then about 70.78% of each column should be in the rst row: \[0.7078 \times \text {(column 1 total)} = 3539\] \[0.7078 \times \text {(column 2 total)} = 1769.5\] \[0.7078 \times \text {(column 3 total)} = 1769.5\] Looking back to how the fraction 0.7078 was computed - as the fraction of users who did not perform a new search (\frac {7078}{10000}\)) - these three expected counts could have been computed as \[ \frac {\text {row 1 total}}{\text {table total}} \text {(column 1 total)} = 3539\] \[ \frac {\text {row 1 total}}{\text {table total}} \text {(column 2 total)} = 1769.5\] \[ \frac {\text {row 1 total}}{\text {table total}} \text {(column 3 total)} = 1769.5\] This leads us to a general formula for computing expected counts in a two-way table when we would like to test whether there is strong evidence of an association between the column variable and row variable. Computing expected counts in a two-way table To identify the expected count for the ith row and jth column, compute \[ \text {Expected Count}_{\text{row i, col j}} = \frac {\text {(row i total)} \times \text {(column j total)}}{\text {table total}}\] The chi-square Test for Two-way Tables The chi-square test statistic for a two-way table is found the same way it is found for a one-way table. For each table count, compute \[ \text {General formula} \frac {\text {(observed count - expected count)}^2}{\text {expected count}}\] \[ \text {Row 1, Col 1} \frac {(3511 - 3539)^2}{3539} = 0.222\] \[ \text {Row 1, Col 2} \frac {(1749 - 1769.5)^2}{1769.5} = 0.237\] \[\vdots \vdots\] \[\text {Row 2, Col 3} \frac {(682 - 730.5)^2}{730.5} = 3.220\] Adding the computed value for each cell gives the chi-square test statistic X 2: \[ X^2 = 0.222 + 0.237 + \dots + 3.220 = 6.120\] Just like before, this test statistic follows a chi-square distribution. However, the degrees of freedom are computed a little differently for a two-way table. 22 For two way tables, the degrees of freedom is equal to \[df = \text {(number of rows minus 1)} \times \text {(number of columns minus 1)}\] In our example, the degrees of freedom parameter is \[df = (2 - 1) \times (3 - 1) = 2\] If the null hypothesis is true (i.e. the algorithms are equally useful), then the test statistic X 2 = 6.12 closely follows a chi-square distribution with 2 degrees of freedom. Using this information, we can compute the p-value for the test, which is depicted in Figure 6.18. Figure 6.18: Computing the p-value for the Google hypothesis test. Definition: degrees of freedom for a two-way table When applying the chi-square test to a two-way table, we use \[df = (R - 1) \times (C - 1)\] where R is the number of rows in the table and C is the number of columns. 22Recall: in the one-way table, the degrees of freedom was the number of cells minus 1. Congress Obama Democrats Republicans Total Approve Disapprove 842 616 736 646 541 842 2119 2104 Total 1458 1382 1383 4223 TIP: Use two-proportion methods for 2-by-2 contingency tables When analyzing 2-by-2 contingency tables, use the two-proportion methods introduced in Section 6.2. Example $\PageIndex{1}$ Compute the p-value and draw a conclusion about whether the search algorithms have different performances. Solution Looking in Appendix B.3 on page 412, we examine the row corresponding to 2 degrees of freedom. The test statistic, X 2 = 6.120, falls between the fourth and fth columns, which means the p-value is between 0.02 and 0.05. Because we typically test at a significance level of $\alpha$ = 0.05 and the p-value is less than 0.05, the null hypothesis is rejected. That is, the data provide convincing evidence that there is some difference in performance among the algorithms. Example $\PageIndex{1}$ Table 6.19 summarizes the results of a Pew Research poll. 23 We would like to determine if there are actually differences in the approval ratings of Barack Obama, Democrats in Congress, and Republicans in Congress. What are appropriate hypotheses for such a test? Solution H 0: There is no difference in approval ratings between the three groups. H A: There is some difference in approval ratings between the three groups, e.g. perhaps Obama's approval differs from Democrats in Congress. 23See the Pew Research website: www.people-press.org/2012/03/14/romney-leads-gop-contest-trails-in-matchup-with-obama. The counts in Table 6.19 are approximate. Exercise $\PageIndex{1}$ A chi-square test for a two-way table may be used to test the hypotheses in Example 6.38. As a rst step, compute the expected values for each of the six table cells. 24 24The expected count for row one / column one is found by multiplying the row one total (2119) and column one total (1458), then dividing by the table total (4223): $\frac {2119 \times 1458}{3902} = 731.6$. Similarly for the first column and the second row: \frac {2104 \times 1458}{4223} = 726.4\). Column 2: 693.5 and 688.5. Column 3: 694.0 and 689.0 Exercise $\PageIndex{1}$ Compute the chi-square test statistic. 25 25For each cell, compute $\frac {\text {(obs - exp)}^2}{exp}$ . For instance, the rst row and rst column: $\frac {(842-731.6)^2}{731.6} = 16.7$. Adding the results of each cell gives the chi-square test statistic: $X^2 = 16.7 + \dots + 34.0 = 106.4$. Exercise $\PageIndex{1}$ Because there are 2 rows and 3 columns, the degrees of freedom for the test is df = (2 - 1) (3 - 1) = 2. Use X 2 = 106.4, df = 2, and the chi-square table on page 412 to evaluate whether to reject the null hypothesis. 26 26The test statistic is larger than the right-most column of the df = 2 row of the chi-square table, meaning the p-value is less than 0.001. That is, we reject the null hypothesis because the p-value is less than 0.05, and we conclude that Americans' approval has differences among Democrats in Congress, Republicans in Congress, and the president. Contributors David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University) David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University)
A Linear Transformation Preserves Exactly Two Lines If and Only If There are Two Real Non-Zero Eigenvalues Problem 472 Let $T:\R^2 \to \R^2$ be a linear transformation and let $A$ be the matrix representation of $T$ with respect to the standard basis of $\R^2$. Prove that the following two statements are equivalent. (a) There are exactly two distinct lines $L_1, L_2$ in $\R^2$ passing through the origin that are mapped onto themselves: \[T(L_1)=L_1 \text{ and } T(L_2)=L_2.\] (b) The matrix $A$ has two distinct nonzero real eigenvalues. Proof. (a) $\implies$ (b). Suppose that the statement (a) holds. That is, there are lines $L_1, L_2$ in $\R^2$ passing through the origin that are mapped onto themselves, and no other lines passing through the origin are mapped onto themselves. A line passing through the origin in $\R^2$ is a one-dimensional subspace in $\R^2$. Thus, it is spanned by a single nonzero vector. We have \[L_1=\Span(\mathbf{v}_1) \text{ and } L_2=\Span(\mathbf{v}_2)\] for some nonzero vectors $\mathbf{v}_1, \mathbf{v}_2$. Since $T(L_1)=L_1$, we have $A\mathbf{v}_1\in L_1=\Span(\mathbf{v}_1)$. Hence there exists $\lambda_1 \in \R$ such that \[A\mathbf{v}_1=\lambda_1\mathbf{v}_1.\] Since $\mathbf{v}_1$ is a nonzero vector, this means that $\lambda_1$ is an eigenvalue of $A$. We claim that $\lambda_1\neq 0$. Otherwise, we have $A\mathbf{v}_1=\mathbf{0}$. Any vector in $L_1$ is of the form $\mathbf{v}=a\mathbf{v}_1$ for some $a\in \R$. So we have \[A\mathbf{v}=A(a\mathbf{v}_1)=aA\mathbf{v}_1=\mathbf{0},\] and this yields that $T(L_1)=\{\mathbf{0}\}$, a contradiction. Hence $\lambda_1$ is a nonzero real eigenvalue of $A$. By the same argument, there is a nonzero real eigenvalue $\lambda_2$ such that $A\mathbf{v}_2=\lambda_2 \mathbf{v}_2$. It remains to show that $\lambda_1 \neq \lambda_2$. Assume on the contrary that $\lambda:=\lambda_1=\lambda_2$. Since $\mathbf{v}_1, \mathbf{v}_2$ are a basis of two distinct lines, they form a basis of $\R^2$. Hence any vector $\mathbf{v}\in \R^2$ can be written as a linear combination \[\mathbf{v}=a\mathbf{v}_1+b\mathbf{v}_2\] for some $a, b \in \R$. Then we have \begin{align} A\mathbf{v}&=A(a\mathbf{v}_1+b\mathbf{v}_2)\\ &=aA\mathbf{v}_1+bA\mathbf{v}_2\\ &=a\lambda \mathbf{v}_1+b\lambda \mathbf{v}_2\\ &=\lambda (a\mathbf{v}_1+b\mathbf{v}_2)\\ &=\lambda \mathbf{v}. \end{align} This implies that any line spanned by a nonzero vector $\mathbf{v}$ is mapped onto itself by the linear transformation $T$. This contradicts our assumption that $L_1, L_2$ are the only such lines. Therefore, $\lambda_1 \neq \lambda_2$, and we have proved that $A$ has two distinct nonzero real eigenvalues. (b) $\implies$ (a). Now we suppose that the statement (b) holds. Namely, we suppose that there are two distinct nonzero real eigenvalues of $A$. Let us call them $\lambda_1, \lambda_2$ and let $\mathbf{v}_1, \mathbf{v}_2$ be eigenvectors corresponding to $\lambda_1, \lambda_2$, respectively. In general eigenvectors corresponding to distinct eigenvalues are linearly independent. Thus, $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent. Hence the lines $L_1, L_2$ spanned by $\mathbf{v}_1, \mathbf{v}_2$ are distinct. Each vector on $L_1$ is of the form $\mathbf{v}=a\mathbf{v}_1$ for some $a \in R$. Hence we have \begin{align} A\mathbf{v}=A(a\mathbf{v}_1)=aA\mathbf{v}_1=a\lambda_1\mathbf{v}_1=\lambda \mathbf{v}. \end{align} It follows that each vector $\mathbf{v}\in L_1$ is mapped onto $L_1$. (Since $\lambda \neq 0$, the image under $T$ is one-dimensional, hence $T(L_1)=L_1$.) Similarly, we have $T(L_2)=L_2$. Finally, we show that no other lines passing through the origin are mapped onto themselves by $T$. Assume that we have a line $L=\Span(\mathbf{w})$ such that $T(L)=L$. Then since we have $A\mathbf{w}\in L=\Span(\mathbf{w})$, there is $\mu\in \R$ such that \[A\mathbf{w}=\mu\mathbf{w}. \tag{}\] Since $\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis of $\R^2$, we have \[\mathbf{w}=a\mathbf{v}_1+b\mathbf{v}_2\] for some $a, b\in \R$. Then the left hand side of () becomes \begin{align} A\mathbf{w}&=A(a\mathbf{v}_1+b\mathbf{v}_2)\\ &=aA\mathbf{v}_1+bA\mathbf{v}_2\\ &=a\lambda_1 \mathbf{v}_1+b\lambda_2 \mathbf{v}_2. \end{align} Thus the relation () yields that \begin{align} a\lambda_1 \mathbf{v}_1+b\lambda_2 \mathbf{v}_2=\mu a\mathbf{v}_1+\mu b\mathbf{v}_2\\ a(\lambda_1-\mu )\mathbf{v}_1+b(\lambda_2-\mu)\mathbf{v}_2=\mathbf{0}. \end{align*} Since $\mathbf{v}_1, \mathbf{v}_2$ are linearly independent, we have \[a(\lambda_1-\mu )=0 \text{ and } b(\lambda_2-\mu)=0.\] The equality $a(\lambda_1-\mu )=0$ implies that either $a=0$ or $\lambda_1=\mu$. If $a=0$, then we have $\mathbf{w}=b \mathbf{v}_2$, and hence $L=L_2$. If $\mu=\lambda_1$, then the second equality $b(\lambda_2-\mu)=0$ implies that $b=0$ since $\lambda_1\neq \lambda_2$. Then we have $\mathbf{w}=a\mathbf{v}_1$, and hence $L=L_1$. In either case, $L$ must be $L_1$ or $L_2$. This completes the proof. Add to solve later
I know the BRDF of specular reflection is nonzero only in the reflection direction. But why it is infinite? A paragraph on page 36 of Advanced Global Illumination: Computer Graphics Stack Exchange is a question and answer site for computer graphics researchers and programmers. It only takes a minute to sign up.Sign up to join this community In the BRDF curve, the total area under the curve is the albedo: what fraction of incident light is reflected in total (as opposed to being absorbed or transmitted). For a perfect reflector, the area under the curve sums to 1, because it reflects all of the incident light. This area limits how high the curve can be. For example, a perfect diffuse reflector has a BRDF that's a flat line at $\frac{1}{2\pi}$, because it reflects the light evenly across the whole hemisphere, the solid angle of a hemisphere is $2\pi$, and the area has to be 1. In a perfect specular reflector, the area under the curve is still 1: it still reflects all of the incident light. But this time, it's reflecting all of that light in one direction. The BRDF curve is a single spike. If you draw a graph with an infinitely thin spike, and the area under the graph has to be 1, then the spike must also be infinitely high. It's the only way to make the area correct. Obviously a real material can't be a perfect specular reflector, so it will have two differences. First, the area under the curve will be less than 1, because some light is absorbed. This doesn't matter much to us. More importantly, the reflection peak can't be infinitely thin: the reflection will be blurred ever so slightly. This means the peak won't be infinitely high: the wider the peak, the less high it has to be to still have an area of 1. The remark about "$\delta$-functions" is because of a function (properly speaking, it's not a function but a distribution) called the Dirac $\delta$ (delta). This distribution is an infinitely thin, infinitely tall spike at $x = 0$, whose area is 1, just like our perfect specular BRDF.
You can take the way SNR is calculated in RGB images as a starting point. In RGB images, the image SNR is calculated using the familiar formula $$\frac{\mu}{\sigma}$$ Where $\mu$ is the "whole image" mean and $\sigma$ is the "whole image" standard deviation. To calculate these "whole image" values, each of the RGB channels is first converted to the common currency of luminosity by one of several formulas, generally weighted sums:$$L = W_{R}\mu_{R} + W_{G}\mu_{G} + W_{B}\mu_{G}$$ Where $\mu$ are means and $W$ are respective weights for the R, G, and B channels of the image (in the Adobe formula, for example, weights are 0.2974, 0.6273, and 0.0753 respectively), and L is luminosity. For whole image standard deviation, the variance includes the covariances of the channels, so the full formula is$$Var(L) = \sum_{i,j} (r_{i,j}^2Var_{R} + g_{i,j}^2Var_{G} + ...$$$$b_{i,j}^2Var_{B}+ 2r_{i,j}g_{i,j}Cov_{R, G} + ...$$$$2r_{i,j}b_{i,j}Cov_{R, B} + 2g_{i,j}b_{i,j}Cov_{G, B} )$$where $i$ and $j$ are pixel indices, which is a more expensive calculation and not as commonly used in industry. Oscar de Lama reports that the Adobe formula simply disregards the covariances:$$W_{R}^2Var_{R} + W_{G}^2Var_{G} + W_{B}^2Var_{B}$$and suggests a compromise formula:$$W_{R}Var_{R} + W_{G}Var_{G} + W_{B}Var_{B} ...$$$$+ (2/3)(W_{R}\mu_{R}^2 + W_{G}\mu_{G}^2 + W_{B}\mu_{B}^2) ...$$$$- (2/3)⋅(\hat{W}_{RG}⋅\mu{R}⋅\mu{G} + \hat{W}_{RG}⋅\mu{R}⋅\mu{B} + \hat{W}_{RB}⋅\mu{G}⋅\mu{B})$$Where $W$ are the earlier weights, and $\hat{W}(X,Y) = \sqrt{XY} / K$ where $K$ normalizes $\hat{W}$ such that $\sum \hat{W} = 1$. If the vector-valued image weights all components equally, then W reduces to 1, simplifying the formulas.
On p10 of these EFT lecture notes, the "relevance" of operators in a Lagrangian is determined by comparing their mass dimension to the spacetime "d" one considers such that an operator is Relevant if its dimension is $ < d$ Marginal if its dimension is $ = d$ And irrelevant if its dimension is $ > d$ This means for example for the action of e scalar field in $d=4$$S[\phi] = \int d^d x\left( \frac{1}{2} \partial_{\mu}\phi \partial^{\mu} \phi - \frac{1}{2} m^2\phi^2- \frac{\lambda}{4!} \phi^4 - \frac{\tau}{6!} \phi^6 + ...\right)$ that the mass term is relevant, the $\phi^4$ coupling is marginal, and the $\phi^6$ coupling is irrelevant, etc However, when analyzing the RG flow around a fixed point $S^{}[(\phi)]$, the (ir)rrelevance of an operator is determined by linearizing the RG flow equation around this fixed point ($M$ is the linearized right hand side of the for example Wilson RG flow equation, $t$ is the RG time), $\frac{\partial S}{\partial t} = M S^{}[\phi]$ solving the corresponding Eigenvalue problem $M O_i(\phi) = \lambda_i O_i(\phi)$ and looking at the sign of each Eigenvalue $\lambda_i$. The action around the fixed point can then be approximated as $S[\phi] = S^{*}[\phi] + \sum\limits_i \alpha_i e^{\lambda_i t} O_i(t)$ and the operator $O_i$ (or direction in coupling space) is said to be Relevant, if $\lambda_i > 0$ (leads away from the fixed point) Marginal, if $\lambda_i = 0$ Irrelevant, if $\lambda_i < 0$ (these operators describe the fixed point theory) So my question is: What is the relationship between these two notions / definitions of (ir)relevant and marginal operators in an effective field theory? Are they they equivalent, and if so how can I see this (mathematically) ?
I wonder if someone could provide me with a simple MWE for using TikZ together with \boxed to produce a colored equation background. If possible with rounded corners. I am using \boxed{} inside the split environment from amsmath. TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. It only takes a minute to sign up.Sign up to join this community There are some packages that can help you, for example: and along this site some answers could be a good starting point: Here, I provide you a MWE using the hf-tikz package. Actually \boxed{} is not used at all, thus I don't know if this will meet your requirements. Notice that, in the following example, it is shown how to highlight the whole equation or just a part of it, that is the major potentiality of the package. You should compile twice to get the right result. \documentclass{article}\usepackage{amsmath}\usepackage[customcolors]{hf-tikz}\begin{document}\begin{equation}\label{e:barwq}\begin{split}\tikzmarkin{a}(0.2,-0.5)(-0.2,0.65)H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{a}\end{split}\end{equation}\hfsetfillcolor{blue!10}\hfsetbordercolor{blue}\begin{equation}\label{e:barwq2}\begin{split}H_c&=\tikzmarkin{b}(0,-0.6)(0,0.65)\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\tikzmarkend{b}\\&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\end{split}\end{equation}\hfsetfillcolor{green!10}\hfsetbordercolor{green!50!black}\begin{equation}\label{e:barwq3}\begin{split}H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot\tikzmarkin{c}(0.05,-0.6)(-0.05,0.65)\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{c}\end{split}\end{equation}\end{document} Result: According to the request in the comments, here are the two possibilities to get rigid corners. It is needed the version 0.2 of the package. norndcorners to have always rigid corners: just loading the package with \usepackage[customcolors,norndcorners]{hf-tikz} the previous document becomes: shade option and then use the key disable rounded corners=true in the tikzmarkin command. In this example, the second equation is highlighted with rigid corners while the other two with rounded corners: \documentclass{article}\usepackage{amsmath}\usepackage[customcolors,shade]{hf-tikz}\begin{document}\begin{equation}\label{e:barwq}\begin{split}\tikzmarkin{a}(0.2,-0.5)(-0.2,0.65)H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{a}\end{split}\end{equation}\hfsetfillcolor{blue!10}\hfsetbordercolor{blue}\begin{equation}\label{e:barwq2}\begin{split}H_c&=\tikzmarkin[disable rounded corners=true]{b}(0,-0.6)(0,0.65)\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\tikzmarkend{b}\\&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\end{split}\end{equation}\hfsetfillcolor{green!10}\hfsetbordercolor{green!50!black}\begin{equation}\label{e:barwq3}\begin{split}H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot\tikzmarkin{c}(0.05,-0.6)(-0.05,0.65)\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].\tikzmarkend{c}\end{split}\end{equation}\end{document} The result: I think the best solution was already provided by Claudio. I am only putting this here to do something about that \boxed command. But I think this solution will not work with split environment so this does not really answer your question. As pointed out already, you may better use the solution by Claudio or use empheq package instead. Here is what I came up with. \documentclass[10pt]{article}\usepackage{amsmath}\usepackage{tikz}\usetikzlibrary{calc}% put color to \boxed math command\newcommand*{\boxcolor}{orange}\makeatletter\renewcommand{\boxed}[1]{\textcolor{\boxcolor}{%\tikz[baseline={([yshift=-1ex]current bounding box.center)}] \node [rectangle, minimum width=1ex,rounded corners,draw] {\normalcolor\m@th$\displaystyle#1$};}} \makeatother\begin{document}\begin{equation}f(x)=\boxed{ax^2+bx+c}\end{equation}\begin{equation}x_{1,2}=\boxed{\frac{-b\pm \sqrt{b^2-4ac}}{2a}}\end{equation}\end{document} You may tweak to your liking.
If the Quotient Ring is a Field, then the Ideal is MaximalLet $R$ be a ring with unit $1\neq 0$.Prove that if $M$ is an ideal of $R$ such that $R/M$ is a field, then $M$ is a maximal ideal of $R$.(Do not assume that the ring $R$ is commutative.)Proof.Let $I$ be an ideal of $R$ such that\[M \subset I \subset […] Is the Set of Nilpotent Element an Ideal?Is it true that a set of nilpotent elements in a ring $R$ is an ideal of $R$?If so, prove it. Otherwise give a counterexample.Proof.We give a counterexample.Let $R$ be the noncommutative ring of $2\times 2$ matrices with real […] Ideal Quotient (Colon Ideal) is an IdealLet $R$ be a commutative ring. Let $S$ be a subset of $R$ and let $I$ be an ideal of $I$.We define the subset\[(I:S):=\{ a \in R \mid aS\subset I\}.\]Prove that $(I:S)$ is an ideal of $R$. This ideal is called the ideal quotient, or colon ideal.Proof.Let $a, […] Equivalent Conditions For a Prime Ideal in a Commutative RingLet $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:(a) The ideal $P$ is a prime ideal.(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.Proof. […] Prime Ideal is Irreducible in a Commutative RingLet $R$ be a commutative ring. An ideal $I$ of $R$ is said to be irreducible if it cannot be written as an intersection of two ideals of $R$ which are strictly larger than $I$.Prove that if $\frakp$ is a prime ideal of the commutative ring $R$, then $\frakp$ is […] Nilpotent Element a in a Ring and Unit Element $1-ab$Let $R$ be a commutative ring with $1 \neq 0$.An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.We give two proofs.Proof 1.Since $a$ […] Characteristic of an Integral Domain is 0 or a Prime NumberLet $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.Definition of the characteristic of a ring.The characteristic of a commutative ring $R$ with $1$ is defined as […]
Exams can be easily created in LaTeX by means of the class exam.cls. This class makes it straightforward to typeset questions, and it sets a 1in margin in all paper sizes and provides special commands to write and compute grades. This article explains how to edit with exam.cls. Contents Let's see a simple working example of the exam class: \documentclass{exam} \usepackage[utf8]{inputenc} \begin{document} \begin{center} \fbox{\fbox{\parbox{5.5in}{\centering Answer the questions in the spaces provided. If you run out of room for an answer, continue on the back of the page.}}} \end{center} \vspace{5mm} \makebox[\textwidth]{Name and section:\enspace\hrulefill} \vspace{5mm} \makebox[\textwidth]{Instructor’s name:\enspace\hrulefill} \begin{questions} \question Is it true that $x^n + y^n = z^n$ if $x,y,z$ and $n$ are positive integers?. Explain. \question Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ \question Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \end{questions} To use the exam class you must put the line \documentclass{exam} on top of your .tex file. This will enable the package's exam-related commands, and set the page format to allow margins for corrections. The syntax of the questions environment is very similar to that of the itemize and enumerate environments. Each question is typed by putting the command \question before it. The other commands in this example are not specific to the exam class, but may be useful to create a quick header for your exam. In the previous section, a basic example showing how to create question was presented. Questions can be further customized, and this section explains how. If the students are required to answer the exam in the space provided, that space can be manually set or evenly distributed. See the example below: \begin{questions} \question Is it true that $x^n + y^n = z^n$ if $x,y,z$ and $n$ are positive integers?. Explain. \vspace{\stretch{1}} \question Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ \vspace{\stretch{1}} \question Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \vspace{\stretch{1}} \end{questions} \clearpage In this example the command \vspace{\stretch{1}} after each question equally distributes the available space. The command \clearpage inserts a page break point to continue typing questions in a new page. If you want to manually assign the space to each question, use the command \vspace{} and in between the braces write the units of space you need. For instance, \vspace{1in} inserts a 1-inch vertical space. Check the documentation about lenghts in LaTeX for a list of available units. If your questions have several parts focused on some subtopics you can use the environments parts, subparts, subsubparts and the corresponding commands \part, \subpart and \subsubpart. See the next example: \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \part What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \begin{subparts} \subpart Who actually proved the theorem? \vspace{\stretch{1}} \subpart How long did actually take to solve this problem? \vspace{\stretch{1}} \end{subparts} \end{parts} \question Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ ... \end{questions} The environments parts and subparts provide question-like nested lists. Jut like in questions you can set manually the vertical spacing. There are four environments to create multiple choice questions. \question Which of these famous physicists invented time? \begin{oneparchoices} \choice Stephen Hawking \choice Albert Einstein \choice Emmy Noether \choice This makes no sense \end{oneparchoices} \question Which of these famous physicists published a paper on Brownian Motion? \begin{checkboxes} \choice Stephen Hawking \choice Albert Einstein \choice Emmy Noether \choice I don't know \end{checkboxes} In this example, two different environments are used to list the possible choices for multiple-choice questions. oneparchoices labels the choices with upper case letters and prints them horizontally. If you want the choices to be printed in a list-like format, the environment choices is the right choice. checkboxes prints check boxes before each choice. If you need the choices to be printed horizontally use the environment oneparcheckboxes instead. Another important feature of the exam class is that it provides commands to make grading the exams easier. You can add a parameter to each \question or \part to print the number of points you attain by correctly answering it \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part[10] For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part[10] For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \part[10] What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \end{parts} \question[20] Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \vspace{\stretch{1}} \end{questions} The additional parameter inside brackets after a question or a part represents the number of points assigned to it. You can change the appearance and the place where the points are printed, see the reference guide for additional commands. Sometimes it's convenient to include half points as value for parts of a questions. You can do this and then print then the value of the whole question. See the example below: \documentclass[addpoints]{exam} \usepackage[utf8]{inputenc} \begin{document} \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part[5] For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part[2 \half] For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \part[2 \half] What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \end{parts} \droptotalpoints \question[20]... \end{questions} The command \half adds half points to a question. The command \droptotalpoints prints the total number of points for the last question. For this last command to work you must add the option [addpoints] to the document class statement. It is possible to add bonus questions, this extra points will later show up in the grading table. Adding bonus questions and parts is actually as simple as creating regular questions and parts. \begin{questions} \question Given the equation $x^n + y^n = z^n$ for $x,y,z$ and $n$ positive integers. \begin{parts} \part[5] For what values of $n$ is the statement in the previous question true? \vspace{\stretch{1}} \part[2 \half] For $n=2$ there's a theorem with a special name. What's that name? \vspace{\stretch{1}} \bonuspart[2 \half] What famous mathematician had an elegant proof for this theorem but there was not enough space in the margin to write it down? \vspace{\stretch{1}} \end{parts} \droptotalpoints \question[20] Compute \[\int_{0}^{\infty} \frac{\sin(x)}{x}\] \vspace{\stretch{1}} \bonusquestion[30] Prove that the real part of all non-trivial zeros of the function $\zeta(z)$ is $\frac{1}{2}$ \vspace{\stretch{1}} \end{questions} The commands \bonusquestion and \bonuspart print "(bonus)" next to the point value of the question. A table that show the points of each question can be printed with a special command. There are three commands to print a table of grades: \gradetable \bonusgradetable \combinedgradetable These commands take two extra parameters, each parameter inside brackets. [h] for a horizontal table or [v] for a vertical table. [questions] to index the points by question and [pages] to list the points by page number. There is no support for other languages than English in the exam class. Nevertheless, it's easy to translate the default words for those in your local language. The next snippet shows how to translate the example presented in the previous sections to Spanish. ss[addpoints]{exam} \usepackage[utf8]{inputenc} \usepackage[spanish]{babel} \pointpoints{punto}{puntos} \bonuspointpoints{punto extra}{puntos extra} \totalformat{Pregunta \thequestion: \totalpoints puntos} \chqword{Pregunta} \chpgword{Página} \chpword{Puntos} \chbpword{Puntos extra} \chsword{Puntos obtenidos} \chtword{Total} ... The rest of the document would be exactly the same shown in previous examples. The commands typed here change the default words in the exam class. \pointpoints{punto}{puntos} \bonuspointpoints{punto extra}{puntos extra} \totalformat{Pregunta \thequestion: \totalpoints puntos} \droptotalpoints. In the example it prints \chqword{Pregunta} chpgword for \chpword for \chbpword for \chsword for \chtword for Placing and formatting the points mark for questions These commands can be typed in the preamble to change the format of the whole document or right before a question to change the format from that question down to the next formatting command or the end of the document. \poinstmargin. The point values are printed in the left margin. Use \nopointsmargin to revert this command to the default format. \pointsmarginright. The point values are printed in the right margin. The command \nopoinstmarginright will revert to the normal behaviour. \bracketedpoints. Uses brackets instead of parentheses around the point values. \boxedpoints. Draws a box around the point values. Changing default names in Grade Tables The commands depend on the format and the information displayed on the table. The h and v within each command mean horizontal or vertical orientation. If the command is preceded by a b means it changes the format in a bonus table, if the command is preceded by a c means it works on combined tables. For instance, to change the word "Score" in a vertical oriented bonus table for the words "Points Awarded" you should use \bvsword{Points Awarded}. Below a table with the default values is shown. horizontal vertical grades table \hpgword{Page:} \hpword{Points:} \hsword{Score:} \htword{Total} \vpgword{Page} \vpword{Points} \vsword{Score} \vtword{Total:} bonus points table \bhpgword{Page:} \bhpword{Bonus Points:} \bhsword{Score:} \bhtword{Total} \bvpgword{Page} \bvpword{Bonus Points} \bvsword{Score} \bvtword{Total:} combined table \chpgword{Page:} \chpword{Points:} \chbpword{Bonus Points:} \chsword{Score:} \chtword{Total} \cvpgword{Page} \cvpword{Points} \cvbpword{Bonus Points} \cvsword{Score} \cvtword{Total:} For more information see:
Tagged: direct sum Problem 61 Let $V$ and $W$ be subspaces of $\R^n$ such that $V \cap W =\{\mathbf{0}\}$ and $\dim(V)+\dim(W)=n$. (a) If $\mathbf{v}+\mathbf{w}=\mathbf{0}$, where $\mathbf{v}\in V$ and $\mathbf{w}\in W$, then show that $\mathbf{v}=\mathbf{0}$ and $\mathbf{w}=\mathbf{0}$. (b) If $B_1$ is a basis for the subspace $V$ and $B_2$ is a basis for the subspace $W$, then show that the union $B_1\cup B_2$ is a basis for $R^n$. (c) If $\mathbf{x}$ is in $\R^n$, then show that $\mathbf{x}$ can be written in the form $\mathbf{x}=\mathbf{v}+\mathbf{w}$, where $\mathbf{v}\in V$ and $\mathbf{w} \in W$. Add to solve later (d) Show that the representation obtained in part (c) is unique.
Tagged: nilpotent matrix Problem 582 A square matrix $A$ is called nilpotent if some power of $A$ is the zero matrix. Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix. Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$. Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample. Problem 453 Let $P_n$ be the vector space of all polynomials with real coefficients of degree $n$ or less. Consider the differentiation linear transformation $T: P_n\to P_n$ defined by \[T\left(\, f(x) \,\right)=\frac{d}{dx}f(x).\] (a) Consider the case $n=2$. Let $B=\{1, x, x^2\}$ be a basis of $P_2$. Find the matrix representation $A$ of the linear transformation $T$ with respect to the basis $B$. (b) Compute $A^3$, where $A$ is the matrix obtained in part (a). (c) If you computed $A^3$ in part (b) directly, then is there any theoretical explanation of your result? Add to solve later (d) Now we consider the general case. Let $B$ be any basis of the vector space of $P_n$ and let $A$ be the matrix representation of the linear transformation $T$ with respect to the basis $B$. Prove that without any calculation that the matrix $A$ is nilpotent. Problem 336 A complex square ($n\times n$) matrix $A$ is called normal if \[A^* A=A A^,\] where $A^$ denotes the conjugate transpose of $A$, that is $A^*=\bar{A}^{\trans}$. A matrix $A$ is said to be nilpotent if there exists a positive integer $k$ such that $A^k$ is the zero matrix. (a) Prove that if $A$ is both normal and nilpotent, then $A$ is the zero matrix. You may use the fact that every normal matrix is diagonalizable. (b) Give a proof of (a) without referring to eigenvalues and diagonalization. Add to solve later (c) Let $A, B$ be $n\times n$ complex matrices. Prove that if $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$, where $I$ is the $n\times n$ identity matrix. Problem 146 Let $A$ be an $n \times n$ nilpotent matrix, that is, $A^m=O$ for some positive integer $m$, where $O$ is the $n \times n$ zero matrix. Prove that $A$ is a singular matrix and also prove that $I-A, I+A$ are both nonsingular matrices, where $I$ is the $n\times n$ identity matrix.Add to solve later Problem 77 A square matrix $A$ is called if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix. nilpotent (a) If $A$ is a nilpotent $n \times n$ matrix and $B$ is an $n\times n$ matrix such that $AB=BA$. Show that the product $AB$ is nilpotent. (b) Let $P$ be an invertible $n \times n$ matrix and let $N$ be a nilpotent $n\times n$ matrix. Is the product $PN$ nilpotent? If so, prove it. If not, give a counterexample. Add to solve later Problem 11 An $n\times n$ matrix $A$ is called nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix. Prove the followings. (a) The matrix $A$ is nilpotent if and only if all the eigenvalues of $A$ is zero. Add to solve later (b) The matrix $A$ is nilpotent if and only if $A^n=O$. Read solution
A One-Line Proof that there are Infinitely Many Prime Numbers Problem 446 Prove that there are infinitely many prime numbers in ONE-LINE. Contents Background There are several proofs of the fact that there are infinitely many prime numbers. Proofs by Euclid and Euler are very popular. In this post, I would like to introduce an elegant one-line proof published by Sam Northshield in 2015. Because the published paper really contains only one line, it is hard to cite only a part of it. So I decided to cite all of his proof word for word. Proof by Sam Northshield (2015). More details in this one-line proof. The above proof was given by Sam Northshield in 2015. Let us give more details hidden in this one line proof. Suppose that there are only finitely many prime numbers $p_1, p_2, \dots, p_n$. Since prime numbers must be greater than $1$, we have \[\sin\left(\, \frac{\pi}{p_i} \,\right) > 0\] for any $i=1, \dots, n$. Thus, the product \[\prod_{i=1}^n\sin\left(\, \frac{\pi}{p_i} \,\right)\] is still positive since the product of finitely many positive numbers is positive. This explains the first inequality in (). Recall the following basic property of the sine function. For any integer $m$, we have \[\sin(\theta+2\pi m)=\sin(\theta)\] for any $\theta \in \R$. We have \begin{align} \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} &=\frac{\pi}{p_i}+\frac{2\pi \prod_{j=1}^n p_j}{p_i}. \end{align} Note that since $\prod_{j=1}^n p_j$ is the product of all $p_1, \dots, p_n$, one factor is $p_i$. Hence \[m_i:=\frac{\prod_{j=1}^n p_j}{p_i}\] is just an integer, not a fraction. It follows from this observation that we have for each $i=1, \dots, n$ \begin{align} \sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} \,\right)&=\sin\left(\, \frac{\pi}{p_i}+2\pi m_i \,\right)\\ &=\sin\left(\, \frac{\pi}{p_i} \,\right) \end{align} by the property of the sine function mentioned above. Taking the product of these over all $i=1, \dots, n$, we obtain \begin{align} \prod_{i=1}^n \sin \left(\, \frac{\pi}{p_i} \,\right)=\prod_{i=1}^n \sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} \,\right). \end{align} This is the first equality in (). To see the last equality in (), we consider the number \[1+2\prod_{j=1}^n p_j\] in the numerator. If this is a prime number, then it must be one of $p_1, \dots, p_n$. If it is not a prime number, then it is divisible by some prime number $p_1, \dots, p_n$. In either case, there is a prime number $p_{i_0}$ among $p_1, \dots, p_n$ such that \[\frac{1+2\prod_{j=1}^n p_j}{p_{i_0}}\] is an integer. Therefore, we have \[\sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_{i_0}} \,\right)=0.\] Since one of the factors is zero, the product \[\prod_{i=1}^n \sin\left(\, \frac{\pi\cdot (1+2\prod_{j=1}^n p_j)}{p_i} \,\right)\] is also zero. This proves the last equality in (). The inequality obtained in (*) is clearly a contradiction. Hence there must be infinitely many prime numbers. This completes the proof. Comment. The one-line proof of Northshield is very clever and elegant. But to explain the proof to high-school students, I feel that I need to probably decipher the proof and give more explanations as I did in this post. Once you understand the details, the one-line proof is a very convenient and beautiful way to hide the details. Reference The one-line proof was published in the paper Northshield, Sam. A one-line proof of the infinitude of primes. Amer. Math. Monthly 122 (2015), no. 5, 466. Add to solve later
As the order of the polynomial (the highest power of x) increases, it usually gets harder to factor. In my last post on this topic, I will cover a way to reduce the order by one for each iteration of the process. If you can get the polynomial to a degree 2, there are many ways to factor these. A polynomial of degree 2 is called a quadratic. I covered factoring quadratics or solving quadratic equations (equations where the quadratic is set equal to 0) in several posts before. Please review these but I will just remind you of them here. A quadratic is a polynomial of the form \[{ax}^{2}\hspace{0.33em}{+}\hspace{0.33em}{bx}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}\] where a, b, and c are some numbers. Now for this set of posts, I am restricting a to be 1. So we would like to factor the quadratic to look like \[{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}\] Basically, the method is to do a reverse distributive property (please see my posts on this). Let’s do an example. Let’s factor \[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{24}\hspace{0.33em}\] We need to find two numbers, a and b, so that they multiply to equal -24 and add or subtract to equal +2, the coefficient in front of the x. 8 and 3 do not work as they do not add to equal 2. However, 6 and 4 look like contenders. The signs of 6 and 4 must be such that they add to equal +2 but multiply to equal -24. Looks like +6 and -4 work so \[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}{24}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}\] Another method you can use is to find the zeroes of the quadratic directly instead of factoring. This method is the quadratic formula. Please see my prior post on this. The quadratic formula is \[{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{b}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}}}{2a}\] where a, b, and c are the coefficients in the general form of a quadratic. From the example I just factored, we can see that x = -6 and x = 4 are zeroes of the quadratic. I could find these directly using the quadratic formula: {x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{2}\pm\sqrt{{2}^{2}{-}{4}{(}{1}{)(}{-}{24}{)}}}{2(1)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{2}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{100}}{2}\hspace{0.33em}{=}\hspace{0.33em}{4}{,}\hspace{0.33em}{-}{6} \] So looks like we have a few tools available to factor quadratics. But what can we do if the order of the polynomial is higher than 2? I will cover a method to do this in my next post.
I read your question a little differently than the other people who have answered. The role of the Fourier transform in physics is as a bi-directional one-to-one mapping between conjugate variables. The most famous example of a conjugate variable pair is position/momentum, but there are many others. Some examples of conjugate variable pairs: position and momentum over Planck's constant $[x,p/\hbar]$ time and energy $[t,E]$ aperture and the sine of the diffraction angle divided by the wavelength $\left[x,\frac{\sin\theta}{\lambda}\right]$ One of the consequences in quantum mechanics is that the operators (quantum mechanics uses differential operators which are related to the classical variables) do not commute. So, using the position/momentum example; $xp\neq px$. This leads to an uncertainty relation between conjugate variables. The usual statement of the Heisenberg uncertianty principle is that the more well one knows the position of a quantum particle, the less one can know about the momentum. Mathematically; $$\sigma_x\sigma_ p\geq\frac{\hbar}{2}.$$However, the more general statement of the uncertainty principle is that the uncertainty between any pair of conjugate variables is given by $$\sigma_1\sigma_2\geq\frac12.$$This uncertainty relation leads to some rather strange effects in quantum physics. For example, the uncertainty relation between energy and time allow for the violation of energy conservation on very brief timescales, a law which is sacred in classical physics. If you are looking for more information on the uses of the Fourier transform in physics and engineering with many worked examples, have a look at this free textbook by J. F. James.
closed as no longer relevant by Robin Chapman, Akhil Mathew, Yemon Choi, Qiaochu Yuan, Pete L. Clark Aug 22 '10 at 9:00 This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question. $e^{\pi i} + 1 = 0$ Stokes' Theorem Trivial as this is, it has amazed me for decades: $(1+2+3+...+n)^2=(1^3+2^3+3^3+...+n^3)$ $$ \frac{24}{7\sqrt{7}} \int_{\pi/3}^{\pi/2} \log \left\| \frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right\| dt\\ = \sum_{n\geq 1} \left(\frac n7\right)\frac{1}{n^2}, $$where $\left(\frac n7\right)$ denotes the Legendre symbol. Not really my favorite identity, but it has the interesting feature that it is aconjecture! It is a rare example of a conjectured explicit identitybetween real numbers that can be checked to arbitrary accuracy.This identity has been verified to over 20,000 decimal places.See J. M. Borwein and D. H. Bailey, Mathematics by Experiment: Plausible Reasoning in the 21st Century, A K Peters, Natick, MA,2004 (pages 90-91). There are many, but here is one. $d^2=0$ Mine is definitely $$1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}+\cdots=\frac{\pi^2}{6},$$ an amazing relation between integers and pi. There's lots to choose from. Riemann-Roch and various other formulas from cohomology are pretty neat. But I think I'll go with $$\sum\limits_{n=1}^{\infty} n^{-s} = \prod\limits_{p \text{ prime}} \left( 1 - p^{-s}\right)^{-1}$$ 1+2+3+4+5+... = -1/12 Once suitably regularised of course :-) $$\frac{1}{1-z} = (1+z)(1+z^2)(1+z^4)(1+z^8)...$$ Both sides as formal power series work out to $1 + z + z^2 + z^3 + ...$, where all the coefficients are 1. This is an analytic version of the fact that every positive integer can be written in exactly one way as a sum of distinct powers of two, i. e. that binary expansions are unique. $V - E + F = 2$ Euler's characteristic for connected planar graphs. I'm currently obsessed with the identity $\det (\mathbf{I} - \mathbf{A}t)^{-1} = \exp \text{tr } \log (\mathbf{I} - \mathbf{A}t)^{-1}$. It's straightforward to prove algebraically, but its combinatorial meaning is very interesting. $196884 = 196883 + 1$ For a triangle with angles a, b, c $$\tan a + \tan b + \tan c = (\tan a) (\tan b) (\tan c)$$ Given a square matrix $M \in SO_n$ decomposed as illustrated with square blocks $A,D$ and rectangular blocks $B,C,$ $$M = \left( \begin{array}{cc} A & B \\\ C & D \end{array} \right) ,$$ then $\det A = \det D.$ What this says is that, in Riemannian geometry with an orientable manifold, the Hodge star operator is an isometry, a fact that has relevance for Poincare duality. But the proof is a single line: $$ \left( \begin{array}{cc} A & B \\\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\\ B^t & D^t \end{array} \right). $$ It's too hard to pick just one formula, so here's another: the Cauchy-Schwarz inequality: \|\|x\|\| \|\|y\|\| >= \|(x.y)\|, with equality iff x&y are parallel. Simple, yet incredibly useful. It has many nice generalizations (like Holder's inequality), but here's a cute generalization to three vectors in a real inner product space: \|\|x\|\| 2\|\|y\|\| 2\|\|z\|\| 2+ 2(x.y)(y.z)(z.x) >= \|\|x\|\| 2(y.z) 2+ \|\|y\|\| 2(z.x) 2+ \|\|z\|\| 2(x.y) 2, with equality iff one of x,y,z is in the span of the others. There are corresponding inequalities for 4 vectors, 5 vectors, etc., but they get unwieldy after this one. All of the inequalities, including Cauchy-Schwarz, are actually just generalizations of the 1-dimensional inequality: \|\|x\|\| >= 0, with equality iff x = 0, or rather, instantiations of it in the 2 nd, 3 rd, etc. exterior powers of the vector space. I always thought this one was really funny: $1 = 0!$ I think that Weyl's character formula is pretty awesome! It's a generating function for the dimensions of the weight spaces in a finite dimensional irreducible highest weight module of a semisimple Lie algebra. $2^n>n $ It has to be the ergodic theorem, $$\frac{1}{n}\sum_{k=0}^{n-1}f(T^kx) \to \int f\:d\mu,\;\;\mu\text{-a.e.}\;x,$$ the central principle which holds together pretty much my entire research existence. Gauss-Bonnet, even though I am not a geometer. Ἐν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τετραγώνοις. That is, In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle. The formula $\displaystyle \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2+1} dx = \frac{\pi}{e}$. It is astounding in that we can retrieve $e$ from a formula involving the cosine. It is not surprising if we know the formula $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, yet this integral is of a purely real-valued function. It shows how complex analysis actually underlies even the real numbers. It may be trivial, but I've always found $\sqrt{\pi}=\int_{-\infty}^{\infty}e^{-x^{2}}dx$ to be particularly beautiful. For X a based smooth manifold, the category of finite covers over X is equivalent to the category of actions of the fundamental group of X on based finite sets: \pi-sets === et/X The same statement for number fields essentially describes the Galois theory. Now the ideathat those should be somehow unifiedwas one of the reasons in the development of abstract schemes, a very fruitful topic that is studied in the amazing area of mathematics called the abstract algebraic geometry. Also, note that "actions on sets" is very close to "representations on vector spaces" and this moves us in the direction of representation theory. Now you see, this simple line actually somehow relates number theory and representation theory. How exactly? Well, if I knew, I would write about that, but I'm just starting to learn about those things. (Of course, one of the specific relations hinted here should be the Langlands conjectures, since we're so close to having L-functions and representations here!) E[X+Y]=E[X]+E[Y] for any 2 random varibles X and Y $\prod_{n=1}^{\infty} (1-x^n) = \sum_{k=-\infty}^{\infty} (-1)^k x^{k(3k-1)/2}$ $ D_A\star F = 0 $ Yang-Mills $\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}$. My favorite is the Koike-Norton-Zagier product identity for the j-function (which classifies complex elliptic curves): j(p) - j(q) = p -1 \prod m>0,n>-1 (1-p mq n) c(mn), where j(q)-744 = \sum n >-2 c(n) q n = q -1 + 196884q + 21493760q 2 + ... The left side is a difference of power series pure in p and q, so all of the mixed terms on the right cancel out. This yields infinitely many identities relating the coefficients of j. It is also the Weyl denominator formula for the monster Lie algebra.
Today I am presenting the IPython notebook at the meeting of the Sage Paris group. This post gathers what I prepared. Installation First you can install the ipython notebook in Sage as explained in this previous blog post. If everything works, then you run: sage -ipython notebook and this will open a browser. Turn on Sage preparsing Create a new notebook and type: In [1]: 3 + 3 6 In [2]: 2 / 3 0 In [3]: matrix Traceback (most recent call last): ... NameError: name 'matrix' is not defined By default, Sage preparsing is turn off and Sage commands are not known. To turn on the Sage preparsing (thanks to a post of Jason on sage-devel): %load_ext sage.misc.sage_extension Since sage-6.2, according to sage-devel, the command is: %load_ext sage You now get Sage commands working in ipython: In [4]: 3 + 4 Out[4]: 7 In [5]: 2 / 3 Out[5]: 2/3 In [6]: type(_) Out[6]: <type 'sage.rings.rational.Rational'> In [7]: matrix(3, range(9)) Out[7]: [0 1 2] [3 4 5] [6 7 8] Scroll and hide output If the output is too big, click on Out to scrollor hide the output: In [8]: range(1000) Sage 3d Graphics 3D graphics works but open in a new Jmol window: In [9]: sphere() Sage 2d Graphics Similarly, 2D graphics works but open in a new window: In [10]: plot(sin(x), (x,0,10)) Inline Matplotlib graphics To create inline matplotlib graphics, the notebook must be started with this command: sage -ipython notebook --pylab=inline Then, a matplotlib plot can be drawn inline (example taken from this notebook): import matplotlib.pyplot as plt import numpy as np x = np.linspace(0, 3np.pi, 500) plt.plot(x, np.sin(x2)) plt.title('A simple chirp'); Or with: %load http://matplotlib.org/mpl_examples/showcase/integral_demo.py According to the previous cited notebook, it seems, that the inline mode can also be decided from the notebook using a magic command, but with my version of ipython (0.13.2), I get an error: In [11]: %matplotlib inline ERROR: Line magic function `%matplotlib` not found. Use latex in a markdown cell Change an input cell into a markdown cell and then you may use latex: Test $\alpha+\beta+\gamma$ Output in latex The output can be shown with latex and mathjax using the ipython display function: from IPython.display import display, Math def my_show(obj): return display(Math(latex(obj))) y = 1 / (x^2+1) my_show(y) ipynb format Create a new notebook with only one cell. Name it range_10and save: In [1]: range(10) Out[1]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] The file range_10.ipynb is saved in the directory. You can also download itfrom File > Download as > IPython (.ipynb). Here is the content of the file range_10.ipynb: { "metadata": { "name": "range_10" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "code", "collapsed": false, "input": [ "range(10)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "pyout", "prompt_number": 1, "text": [ "[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]" ] } ], "prompt_number": 1 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] } ipynb is just json A ipynb file is written in json format. Below, weuse json to open the file `range_10.ipynb as a Python dictionnary. sage: s = open('range_10.ipynb','r').read() sage: import json sage: D = json.loads(s) sage: type(D) dict sage: D.keys() [u'nbformat', u'nbformat_minor', u'worksheets', u'metadata'] sage: D {u'metadata': {u'name': u'range_10'}, u'nbformat': 3, u'nbformat_minor': 0, u'worksheets': [{u'cells': [{u'cell_type': u'code', u'collapsed': False, u'input': [u'range(10)'], u'language': u'python', u'metadata': {}, u'outputs': [{u'output_type': u'pyout', u'prompt_number': 1, u'text': [u'[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]']}], u'prompt_number': 1}, {u'cell_type': u'code', u'collapsed': False, u'input': [], u'language': u'python', u'metadata': {}, u'outputs': []}], u'metadata': {}}]} Load vaucanson.ipynb Download the file vaucanson.ipynb from the last meeting of Paris SageUsers. You can view the complete demo including pictures of automaton even ifyou are not able to install vaucanson on your machine. IPython notebook from a Python file In a Python file, separate your code with the following line to create cells: # <codecell> For example, create the following Python file. Then, import it in the notebook. It will get translated to ipynb format automatically. # -- coding: utf-8 -- # <nbformat>3.0</nbformat> # <codecell> %load_ext sage.misc.sage_extension # <codecell> matrix(4, range(16)) # <codecell> factor(2^40-1) (NEW: See also the demo I made at the Sage Paris group meeting in March 2014.) Ticket #12719 (Upgrade to IPython 0.13) was merged into sage-5.7.beta1. This took a lot of energy (see the number of comments in the ticket and especially the number of patches and dependencies). Big thanks to Volker Braun, Mike Hansen, Jason Grout, Jeroen Demeyer who worked on this since almost one year. Note that in December 2012, the IPython project has received a $1.15M grant from the Alfred P. Sloan foundation, that will support IPython development for the next two years. I really like this IPython sage command line interface so it is really good news! The IPython notebook Since version 0.12 (December 21 2011), IPython is released with its own notebook. The differences with the Sage Notebook are explained by Fernando Perez, leader of IPython project, in the blog post The IPython notebook: a historical retrospective he wrote in January 2012. One of the differences is that the IPython Notebook run in its own directory whereas each cell of the Sage Notebook lives in its directory. As William Stein says in the presentation Browser-based notebook interfaces for mathematical software - past, present and future he gave last December at ICERM, there are plenty of projects and directions these days for those interfaces. In May 2012, I tested the same ticket which was to upgrade to IPython 0.12 at that time. Today, I was curious to test it again. First, I installed sage-5.7.beta4: ./sage -version Sage Version 5.7.beta4, Release Date: 2013-02-09 Install tornado: ./sage -sh easy_install-2.7 tornado [update March 6th, 2014] Note that some linux user have to install libssl-dev before tornado: sudo apt-get install libssl-dev Install zeromq and pyzmq: ./sage -i zeromq ./sage -i pyzmq Start the ipython notebook: ./sage -ipython notebook [NotebookApp] Using existing profile dir: u'/Users/slabbe/.sage/ipython-0.12/profile_default' [NotebookApp] Serving notebooks from /Users/slabbe/Applications/sage-5.7.beta4 [NotebookApp] The IPython Notebook is running at: http://127.0.0.1:8888/ [NotebookApp] Use Control-C to stop this server and shut down all kernels. Create a new notebook. One may use sage commands by adding the line fromsage.all import in the first cell. The next things I want to look at are: Test the conversion of files from .pyto .pynb. Test the conversion of files from .rstto .pynb. Test the qtconsole. Test the parallel computing capacities of the IPython.
So now that you know about matrices, we can use them to add a third way to solve a system of equations. You will need to read my previous 3 posts on matrices if you are unfamiliar with how to multiply matrices. In the last post on System of Equations, I looked at the system: 2 x + 3 y = 51 3 x + 2 y = 49 And in my last post on Matrices, I showed you how a 2×2 matrix of numbers and a 2×1 matrix of unknowns can be multiplied together to get a 2×1 matrix that looks suspiciously like the left sides of a system of equations. This is , in fact, true. If I form a matrix using the coefficients on the left side of the above system, I get a matrix which I will call A: \[ {\textbf{A}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right] \] Let me now define a matrix x (which is different from the single variable x which is in italics and not bold): \[ {\textbf{x}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{x}\\{y}\end{array}}\right] \] Now I will define a matrix b: \[ {\textbf{b}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{51}\\{49}\end{array}}\right] \] Now see what happens if I multiply A by x: \[ {\textbf{A}}{\textbf{x}}\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right]\times\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{{2}{x}{+}{3}{y}}\\{{3}{x}{+}{2}{y}}\end{array}}\right] \] The rows of this result look just like the left side of our system of equations. And b is the right side. So the matrix equivalent of the system is \[ \begin{array}{c} {{\textbf{A}}{\textbf{x}}\hspace{0.33em}{=}\hspace{0.33em}{\textbf{b}}}\\ {\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right]\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{51}\\{49}\end{array}}\right]} \end{array} \] This is easy to form directly. You just form A as the matrix of coefficients (with the unknowns in the same order in each equation), x is the matrix of unknowns, and b is the matrix of the numbers on the right sides. So how do we solve this? From my last post, I defined the inverse of a matrix A as A -1. This is the matrix that if I multiply A by its inverse, I get the identity matrix which is the equivalent of “1” in scalar maths. The process of isolating (solving) for variables in a matrix equation is exactly the same as for scalar equations: you do the same thing to both sides with the goal of having the unknowns by themselves on one side. So if I pre-multiply (remember, order of multiplication in matrix maths is important) both sides of our matrix equation by A -1, the left side is the identity matrix times x which is equal to just x. The right side multiplies out to form the solution. As I said before, finding A -1 is beyond the scope of this set of posts. I will just tell you what it is. However, many modern calculators will do this for you, and you can also use the internet and search for “matrix inverse calculator”. It turns out that A -1 is: \[ \left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right] \] So taking the matrix equation and pre-multiply both sides by A -1 gives A -1 Ax = A -1 b ⟹ Ix = A -1 b ⟹ x = A -1 b \[ \left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right]\left[{\begin{array}{cc}{2}&{3}\\{3}&{2}\end{array}}\right]\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right]\left[{\begin{array}{c}{51}\\{49}\end{array}}\right] \]\[ \Longrightarrow\hspace{0.33em}\left[{\begin{array}{c}{x}\\{y}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{cc}{\frac{{-}{2}}{5}}&{\frac{3}{5}}\\{\frac{3}{5}}&{\frac{{-}{2}}{5}}\end{array}}\right]\left[{\begin{array}{c}{51}\\{49}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{9}\\{11}\end{array}}\right] \] Which is the same answer as before, x = 9 and y = 11. This is a very powerful method for large systems of equations. Next time I will solve a system of 4 equations with 4 unknowns. For those of you who have done this manually, you will appreciate the ease matrix algebra provides.
Well, it is indeed an abuse of notation though, by no means you could say (1) is an equality in $L^2(0,T;V')$, because (1) only holds in a finite dimensional subspace. Multiply (1) by the basis of the finite dimensional subspace$$(u_n',w_j) - (\Delta u_n,w_j) = (P_n f,w_j).$$Proper assumption about the boundary condition and continuity leads to:$$(u_n',w_j) - (\Delta u_n,w_j) = (u_n',w_j) + (\nabla u_n, \nabla w_j).$$Hence:$$(P_n f,w_j) = (f,w_j), \quad \text{for } j=1,\ldots,n,$$which is to say, $P_n f$ is the $L^2$-projection of $f$ onto this finite dimensional subspace. I didn't see the paper, but I guess the reason to multiply a function in $L^2(0,T;V)$ is to get some a priori error estimate for this approximation by exploiting the fact that$$(u'-u_n',w_j) + \big(\nabla (u-u_n), \nabla w_j\big) = (f-P_n f,w_j).$$ EDIT: question in the comment, instead of claiming (1) holds, rather what we can do is the following$$\begin{aligned}&(u_n',\phi) + a(u_n,\phi)\\=& (u_n',P_n\phi) + a(u_n,P_n\phi) + (u_n',\phi - P_n\phi) + a(u_n,\phi-P_n\phi) \\=& (P_n f,P_n \phi) + (u_n',\phi - P_n\phi) + a(u_n,\phi-P_n\phi) \\=& (P_n f,\phi) + \color{blue}{(u_n',\phi - P_n\phi)} + \color{red}{a(u_n,\phi-P_n\phi) }.\end{aligned}$$Blue term could vanish if we integrate w.r.t. time and integration by parts to move the derivative to $\phi-P_n \phi$. However for the red term to vanish, we must have something like $\Delta u_n = 0$... EDIT2: If we say $f = g$ in $L^2(0,T;V')$, I don't know what the author did, but according to the Evans' book (see Section 7.1), it is $$\newcommand{\lsub}[2]{{\vphantom{#2}}_{#1}{#2}}\lsub{V'}{\langle f(t,\cdot),v \rangle}_V = \lsub{V'}{\langle g(t,\cdot),v \rangle}_V,$$for any $v\in V$, at a.e. time $t\in [0,T]$. This is the same with what you wrote in the comment though.
203 12 Summary Is there any connection between the new definition of the kilogram based on the Planck constant and the mass-energy equivalence? Hi, today I stumbled upon a 2016 article in Scientific American about the (then) possibility of re-defining the kilogram through Planck's constant. The article is really a very quick review of the topic. At some point the author states the following "So for years, physicists have chased an elusive dream: replacing the physical kilogram with a standard inherent in properties of nature such as the speed of light, the wavelength of photons and the Planck constant (also called h-bar), which links the energy a wave carries with its frequency of oscillation. Scientists could use the Planck constant to compare the energy of a wave with Einstein's iconic ##E=m c^2## equation; in that way, they would determine mass solely through the physical constants." I really do not see the relevance of Einstein's equation in the redefinition of the kilogram. As far as I understand, the point here is to define a standard for mass. This is now done basically by choosing some fundamental constants of Nature and assuming them as units for their "physical dimension". In this approach, "everyday" physical dimensions (such as mass), which used to be fundamental, become composite physical dimensions that can be obtained from those defined by the constants. Thus,, maybe oversimplifying, if action is measured in units of ##h##, frequency is measured in units of ##\Delta \nu_{\rm Cs}## and speed is defined in units of ##c##, then the (now composite) unit of mass corresponds to $$ 1 kg = 1.4755214\cdot 10^{40} \frac{h\,\Delta \nu_{\rm Cs}}{c^2} $$ I might be wrong, but I'm under the impression that the author was a bit shoddy in referring to the mass-energy equivalence. By the way, she was not very precise about Planck constant either. It is not exactly true that the Planck constant is called "h-bar". That is the Does anyone know whether the mass-energy relation really played a role in the choice of ##h## for the definition of the mass unit? I looked into a previous discussion, but I was not able to find a connection. Thanks a lot Franz today I stumbled upon a 2016 article in Scientific American about the (then) possibility of re-defining the kilogram through Planck's constant. The article is really a very quick review of the topic. At some point the author states the following "So for years, physicists have chased an elusive dream: replacing the physical kilogram with a standard inherent in properties of nature such as the speed of light, the wavelength of photons and the Planck constant (also called h-bar), which links the energy a wave carries with its frequency of oscillation. Scientists could use the Planck constant to compare the energy of a wave with Einstein's iconic ##E=m c^2## equation; in that way, they would determine mass solely through the physical constants." I really do not see the relevance of Einstein's equation in the redefinition of the kilogram. As far as I understand, the point here is to define a standard for mass. This is now done basically by choosing some fundamental constants of Nature and assuming them as units for their "physical dimension". In this approach, "everyday" physical dimensions (such as mass), which used to be fundamental, become composite physical dimensions that can be obtained from those defined by the constants. Thus,, maybe oversimplifying, if action is measured in units of ##h##, frequency is measured in units of ##\Delta \nu_{\rm Cs}## and speed is defined in units of ##c##, then the (now composite) unit of mass corresponds to $$ 1 kg = 1.4755214\cdot 10^{40} \frac{h\,\Delta \nu_{\rm Cs}}{c^2} $$ I might be wrong, but I'm under the impression that the author was a bit shoddy in referring to the mass-energy equivalence. By the way, she was not very precise about Planck constant either. It is not exactly true that the Planck constant is called "h-bar". That is the reducedPlanck constant, whose value is ##h/2\pi##. Does anyone know whether the mass-energy relation really played a role in the choice of ##h## for the definition of the mass unit? I looked into a previous discussion, but I was not able to find a connection. Thanks a lot Franz
For periodic not-necessarily smooth $f$ and a range of $m$, say $0\ldots 31$, I want to compute $\int_{-\pi}^\pi f(t) \cos (mt) dt$ (and maybe the same integral with $\sin$ instead of $\cos$) to machine precision (or close-ish to it). The obvious thing to do is to evaluate $f$ at an equispaced grid and take a DFT, but this is only giving me linear convergence, and getting enough precision will take a very fine grid. The other obvious thing (which I haven't yet tried) would be to pre-compute $w_i \cos(m t_i)$ for $w_i$ $t_i$ the Legendre weights and points (scaled to $(-\pi,\pi)$), then at evaluation time compute $f(t_i)$ and find all the integrals with a matvec. (All else being equal, I'd prefer not to have to have a bunch of pre-computed data to keep track of, but it's better than re-computing it every time if that's the only other option) My only other idea (also not yet tried) was to figure out a closed-form solution for $\int_{-\pi}^\pi T_n(t/\pi)\cos(mt)$ for $T_n$ the Chebyshev basis, then at evaluation time do a fast Chebyshev transform to get $f \approx \sum_i c_i T_i $ and use that to compute the integrals. None of these seem particularly satisfying. So my question is: is there some neat technique I'm missing that will let me compute banks of these coefficients with spectral (or at least quadratic) accuracy and with good complexity?
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
Regression is a very fundamental concept in statistics, machine learning and in Neural Network. Imagine plotting the correlation between rainfall frequency and agriculture production in high school. Increase in rainfall generally increases agriculture production. Fitting a line to those points enables us to predict the production rate under different rain conditions. It was actually a very simplest form of linear regression. In simple words, regression is a study of how to best fit a curve to summarize a collection of data. It’s one of the most powerful and well-studied types of supervised learning algorithms. In regression, we try to understand the data points by discovering the curve that might have generated them. In doing so, we seek an explanation for why the given data is scattered the way it is. The best-fit curve gives us a model for explaining how the dataset might have been produced. There are many types of regression e.g. simple linear regression, polynomial regression, multivariate regression. In this post, we will discuss simple linear regression only and later we will discuss the rest. We will also provide the python code from scratch at the end of the post Simple regression, as the name implies, it’s just a very simple form of regression, where we assume that we just have one input and we’re just trying to fit a line. Data Set Consider a data set containing age and the number of homicide deaths in the US in the year 2015: age num_homicide_deaths 21 652 22 633 23 653 24 644 25 610 If we plot the dataset and the line best fit to it we see: When we are talking about regression, our goal is to predict a continuous variable output given some input variables. For simple regression, we only have one input variable x which is the age in our case and our desired output y which is num of homicide deaths for each age. Our dataset then consists of many examples of x and y, so: $$ D = \{(x_1,y_1), (x_2,y_2), …, (x_N,y_N)\} $$ where $N$ is the number of examples in the data set. So, our data set will look like: $$ D=\{(21,652),(22,633), …,(50,197)\} $$ Model So, how can we mathematically model single linear regression? Since the goal is to find the perfect line, let’s start by defining the model (the mathematical description of how predictions will be created) as a line. It’s very simple. We’re assuming we have just one input, which in this case is, age of people and one output which is the number of homicide deaths and we’re just gonna fit a line. And what’s the equation of a line? $$f(x) = w_0 + w_1x$$ what this regression model then specifies is that each one of our observations $ y_i$ is simply that function evaluated at $x_i$, so that’s $w_0$ plus $w_1x_1$ plus the error term which we call $\epsilon_i$. So this is our regression model $$y_i = w_0 + w_1x_i + \epsilon_i$$ and to be clear, this error, $\epsilon_i$, is the distance from our specific observation back down to the line. The parameters of this model are$w_0$and $w_1$ are intercept and slope and we call these the regression coefficients. Quality Metric We have chosen our model with two regression coefficients $w_0$ and $w_1$. For our data set, there can be infinitely many choices of these parameters. So our task is to find the best choice of parameters and we have to know how to measure the quality of the parameters or measure the quality of the fit. So in particular, we define a loss function (also called a cost function), which measures how good or bad a particular choice of $w_0$ and $w_1$ is. Values of $w_0$ and $w_1$ that seem poor should result in a large value of the loss function, whereas good values of $w_0$ and $w_1$ should result in small values of the loss function. So what’s the cost of using a specific line? It has many forms. But the one we’re gonna talk about here is Residual Sum of Squares (RSS): $$ RSS(w_0, w_1)= \sum_{i=1}^N(y_i-[ w_0 + w_1x_i])^2$$ what Residual sum of squares assumes is that we’re just gonna add up the errors we made between what we believe the relationship is or what we’ve estimated the relationship to be between $x$ and $y$ and what the actual observation $y$ is. And, we talked about the error as the $\epsilon_i$. Model Optimization Our cost was to find the residual sum of squares, and for any given line, we can compute the cost of that line. So for example, we have two different lines for two different residual sums of squares. How do we know which choice of parameters is better? Ones with the minimum RSS. Our goal is to minimize over all possible $w_0$ and $w_1$ intercepts and slopes respectively, but a question is, how are we going to do this? The mathematical notation for this minimization over all possible $w_0$ , $w_1$ is $$min_{w_0,w_1}\sum_{i=1}^N(y_i – [w_0 +w_1x_i])^2$$ So we want to find the specific value of $w_0$ and $w_1$ we’ll call that $\hat{w_0}$ and $\hat{w_1}$ respectively that minimize this residual sum of squares. The red dot marked below on the above plot shows where the desired minimum is. We need an algorithm to find this minimum. We will discuss two approaches e.g. Closed-form Solution and Gradient Descent. Closed-form Solution for Linear Regression From calculus, we know that at the minimum the derivatives will be $0$. So, if we compute the gradient of our RSS: $$ \begin{aligned} RSS(w_0, w_1) &= \sum_{i=1}^N(y_i-[ w_0 + w_1x_i])^2 \end{aligned}$$ $$\begin{aligned} &\nabla RSS(w_0, w_1) = \begin{bmatrix} \frac{\partial RSS}{\partial w_0} \\ \\ \frac{\partial RSS}{\partial w_1} \end{bmatrix} \\ &=\begin{bmatrix} -2\sum_{i=1}^N{[y_i – (w_0 + w_1 x_i)]} \\ \\ -2\sum_{i=1}^N[y_i – (w_0 + w_1 * x_i)] x_i \end{bmatrix} \end{aligned}$$ Take this gradient, set it equal to zero and find the estiamates for $w_0$ ,$w_1$. Those are gonna be the estimates of our two parameters of our model that define our fitted line. $$ \begin{aligned} &\nabla RSS(w_0, w_1) = 0 \end{aligned} $$ implies, $$\begin{aligned} &\begin{bmatrix} -2\sum_{i=1}^N{[y_i – (w_0 + w_1 x_i)]} \\ \\ -2\sum_{i=1}^N[y_i – (w_0 + w_1 * x_i)] x_i \end{bmatrix} = 0 \\ &-2\sum_{i=1}^N{[y_i – (w_0 + w_1 x_i)]} = 0, \\ &-2\sum_{i=1}^N{[y_i – (w_0 + w_1 * x_i)]} * x_i = 0 \end{aligned}$$ Solving for $w_0$ and $w_1$ we get, $$ \begin{aligned} \hat{w_0} = \frac{\sum_{i = 1}^N y_i}{N} – \hat{w_1}\frac{\sum_{i=1}^N x_i}{N} \\ \hat{w_1} = \frac{\sum y_i x_i – \frac{\sum y_i \sum x_i}{N}}{\sum x_i^2 – \frac{\sum x_i \sum x_i}{N}} \end{aligned} $$ Now that we have the solutions, we just have to compute $ \hat{w}_1$ and then plug that in and compute $\hat{w}_0$. To compute $ \hat{w}_1$ we need to compute a couple of terms e.g. sum over all of our observations $\sum y_i$ and sum over all of our inputs $\sum x_i$ and then a few other terms that are multipliers of our input and output $\sum y_i x_i$ and $\sum x_i^2$. Plug them into these equations and we get out what our optimal $\hat{w}_0$ and $\hat{w}_1$ are, that minimize our residual sum of squares. Gradient Descent for Linear Regression The other approach that we will discuss is Gradient descent where we’re walking down this surface of residual sum of squares trying to get to the minimum. Of course, we might overshoot it and go back and forth but that’s a general idea that we’re doing this iterative procedure. $$\begin{aligned} &\nabla RSS(w_0, w_1) \\ &= \begin{bmatrix} -2\sum_{i=1}^N{[y_i – (w_0 + w_1 * x_i)]} \\ \\ -2\sum_{i=1}^N[y_i – (w_0 + w_1 * x_i)] x_i \end{bmatrix} \\&= \begin{bmatrix} -2\sum_{i=1}^N{[y_i – \hat{y}_i (w_0, w_1)]} \\ \\ -2\sum_{i=1}^N[y_i – \hat{y}_i(w_0,w_1)] x_i \end{bmatrix} \end{aligned}$$ Then our gradient descent algorithm will be: $$\begin{aligned}&while \; not \; converged: \begin{bmatrix} w_0^{(t+1)} \\ w_1^{(t+1)} \end{bmatrix}\\ &= \begin{bmatrix}w_0^{(t)} \\ w_1^{(t)} \end{bmatrix} – \eta* \begin{bmatrix} -2\sum{[y_i – \hat{y}_i (w_0, w_1)]}\\ \\-2\sum[y_i – \hat{y}_i(w_0,w_1)] x_i \end{bmatrix}\\ &= \begin{bmatrix} w_0^{(t)} \\ w_1^{(t)} \end{bmatrix} +2\eta \begin{bmatrix} \sum{[y_i – \hat{y}_i (w_0, w_1)]} \\ \\ \sum[y_i – \hat{y}_i(w_0,w_1)] x_i \end{bmatrix} \end{aligned}$$ So gradient descent does this, we’re going to repeatedly update our weights. So set $W$ to $W$ minus $\eta$ times the derivative, where $W$ is the vector. We will repeatedly do that until the algorithm converges. $\eta$ here is the learning rate and controls how big a step we take on each iteration of gradient descent and the derivative quantity is basically the update or the change we want to make to the parameters $W$. Prediction for Linear Regression After all the hard work now we need to test our machine learning model. The dataset we work on, generally split into two parts. One part is called training data where we do all the training and another is called the test data where we test our network. We have developed equations for training and using them we have got a calibrated set of weights. We will then use this set of weights to predict the result for our new data using the equation $$ Prediction = \hat{w}_0 + \hat{w}_1 data $$ where $ \hat{w}_0$ and $\hat{w}_1$ are the optimized set weights. Now that we have finished the theoretical part of the tutorial now you can see the code and try to understand different blocks of the code. Also published on Medium.
Sorry for the basic question. But take for example: $$3x-8\leq 0$$ $$-2+3x-x^2\leq 0$$ If we sum these two inequalities we obtain: $$0\leq x^2-6x+10$$ The solution of this inequality is of course any $x \in \mathbb{R}$. However, we can also attempt to solve them separetly and obtain: $$x \leq 8/3$$ $$(x-2)(x-1)\geq 0$$ which of course implies that $x \in (-\infty,1]\cup[2,8/3]$.
I'm trying to find a function $u: (0, \infty) \to \mathbb{R}$ which satisfies these conditions: i) $u$ is bounded. ii) $u^2$ increases over $(0, \infty)$. iii) $\dot u(t)$ does not converge to $0$ as $t$ tends to infinity. I think it's easier to choose first the derivative of $u^2$ and then integrate it. For example, I chose $$\frac{d}{dt} u^2 = \frac{\sin^2 t}{(t+1)^2},$$ so the function $u$ is $$u(t) = \sqrt{\int_0^t \frac{\sin^2 x}{(1+x)^2}dx}.$$ This function is bounded, but its derivative converges to $0$ as $t$ tends to infinity, which does not satisfy the iii) condition. Those are my ideas for the problem. Thank you very much for any idea, hint or solution.
My question is: How can I integrate $\sqrt{1+\frac{1}{3x}} \, dx$ ? (I don't mind about negative Xs). I'm aware that there's a theorem which you can use that integrates the function using the points where the function can't be solved or something like that(improper integrals?) but I haven't learned anything but parts and substitution yet. I know that this integral can be solved using just those and that's what I am searching for =). Unfortunately, this is no homework (and my calculus teacher has no idea how to solve this... I asked =( ). I just saw this integral by accident. I've been struggling with it for like a week now. I just want the answer and preferably the steps. Possible starting points: The best approach I could find is: $$u^2=3x \implies 2u\ du = 3\ dx$$ $$\frac{2}{3}\int \sqrt{ 1+\frac{1}{u^2} } \, \, u\ du$$ $$\frac{2}{3}\int \sqrt{ u^2+1 } \, \, du$$ Now I know that I can use sinh but I have no clue how. That was one approach that a guy told me. If you don't like it you could start with the more traditional way: $$u=3x$$ $$ \frac{1}{3} \int\sqrt{1+\frac{1}{u}}\,du.$$ Now $$w^2=1+\dfrac{1}{u}$$ $$2w\,dw=-\frac{du}{u^2}=-(w^2-1)^2 \,du.$$ $$\frac{1}{3}\int-\frac{2w^2\,dw}{(w^2-1)^2}.$$ After that I've been told that you could integrate by parts but it's just too hard for me. I simply get nowhere. I won't type everything cause it's a waste of time. If anyone wants me to type in more work I'd be happy to if it helps in any way. I'm in a stage where I just want to see how this super-complex (at least for me) integral is solved. I've used wolframalpha but it's nowhere near a human approach. So well... thanks a lot for any help guys! And sorry for the long post! =)!
You said it yourself in the question: use L'Hospital's Rule applied to $$\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}$$The numerator and denominator each approach $\infty$, given your conditions on $f$. EDIT: The denominator approaches $\infty$ simply because $\alpha$ is positive. If $f(0)$ is nonzero, then $f$ is bounded below by some positive $\epsilon$ in a neighborhood of $0$. So the integral is bounded below by $\epsilon\int_x^\delta\frac{1}{t^{\alpha+1}}\,dt+\int_{\delta}^1\frac{f(t)}{t^{\alpha+1}}\,dt$ which diverges to infinity as $x$ approaches $0^+$, since the power of $t$ is greater than $1$. (And if $f(0)=0$ the numerator might not approach $\infty$. But it still approaches something since $f$ is integrable on $[0,1]$ and $t^{\alpha+1}$ is monotonic. Let's call it $L$. Then L'Hospital's Rule is not needed - the OP's limit is $0\cdot L$, or just $0$. This is consistent with the formula given below when $f(0)\neq0$.) The result is $$\begin{align}\lim_{x\to0^+}\frac{\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{1/x^{\alpha}}&=\lim_{x\to0^+}\frac{\frac{d}{dx}\int_x^1\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}1/x^{\alpha}}\\ &=\lim_{x\to0^+}\frac{-\frac{d}{dx}\int_1^x\frac{f(t)}{t^{\alpha+1}}\,dt}{\frac{d}{dx}x^{-\alpha}}\\ &=\lim_{x\to0^+}\frac{-\frac{f(x)}{x^{\alpha+1}}}{-\alpha x^{-\alpha-1}}\\ &= \lim_{x\to0^+}\frac{f(x)}{\alpha}\\ &=\frac{f(0)}{\alpha} \end{align}$$
In the game I am designing I have to simulate a gold market. A market is, essentially, a place where people buy and sell a specific good, in this case, gold. The prices change, ideally according to offer and demand, in a continuous way. The most well known market is the stock exchange. I am not an economist, and being exceptionally realistic is not a goal of this design. This primarily needs to resemble the behavior while still being slightly predictable and not so complex, in order to be a funny part of the game itself. I know that there is a lot of very complex models that I could apply, but this is a simple approach that I think it can work in a game. My approach will use a random walk. It is usually defined in a recursive way, like this: V(t+1) = V(t) + \Phi (t) where t \equiv \text{time period} \\ V(t) \equiv \text{value (current price)} \\ \phi (t) \equiv \text{random variable} This definition implies defining V(0) and also \phi (t). The first would be the initial price and the second one the “step”, a random value that is added to the initial price (it can be positive or negative) in order to get the next price (that can be bigger or smaller than the previous). At this point, as I don’t know really how the rest of the game will be, I establish that V(0) = 100. In order to get variability, I thought of the following definition: \phi (t) \sim U([ – \frac{V(0)}{X_{v}}, \frac{V(0)}{X_{v}} ])\\X_{v} \equiv \text{inverse of market variability} U is the uniform random distribution. The higher the variability, the smaller the possible change, thus the smaller its inverse, the bigger the possible change. So in this small model we have two free variables that we can control for the evolution of the market V(0) and X_{v}. It’s time for a simulation or two. I think that Excel will cut it. In this images you can see already several flaws that this small model has: once the tendency is set (raising or lowering price), it is difficult that such tendency is reverted and it goes below zero (which for obvious reasons should not be allowed). I will tackle this flaws and refine this model by adding more variables and trying to model tendencies and tendency changes.
There are several and almost similar inequalities in MSE that some of them can be proved in long page. some of these questions listed below: For $abc=1$ prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}.$ For positive $a$, $b$, $c$ with $abc=1$, show $\sum_{cyc} \left(\frac{a}{a^7+1}\right)^7\leq \sum_{cyc}\left(\frac{a}{a^{11}+1}\right)^7$ Inequality $\frac{x}{x^{10}+1}+\frac{y}{y^{10}+1}+\frac{z}{z^{10}+1}\leq \frac{3}{2}$ Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$ If $abc=1$ so $\sum\limits_{cyc}\frac{a}{a^2+b^2+4}\leq\frac{1}{2}$ For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$ If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$. and so on. One cane pose many many similar question in this way: Let $f(x)$ be a continuous (and maybe with a special property) then prove that $\sum_{x\in\{a,b,c\}}f(x)\leq 3f(1)$ whenever $abc=1$. or one can generalize this for arbitrary number of variables: $\sum_{cyc}f(x)\leq nf(1)$ whenever $\prod_{i=1}^n x_i=1$. My argument based on what I read in Problem-Solving Through Problems by Loren C. Larson that principle of insufficient reason, which can be stated briefly as follows: " Where there is no sufficient reason to distinguish, there can be no distinction." So my question is Is a (short and) beautiful proof for similar inequalities must exist always as the OPs want for desired answer?
Search Now showing items 1-10 of 27 Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider (American Physical Society, 2016-02) The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ... Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (Elsevier, 2016-02) Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < \|\eta\| < 4.0$) and associated particles in the central range ($\|\eta\| < 1.0$) are measured with the ALICE detector in ... Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV (Springer, 2016-08) The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ... Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Elsevier, 2016-03) The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ... Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV (Elsevier, 2016-03) Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ... Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV (Elsevier, 2016-07) The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ... $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV (Elsevier, 2016-03) The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ... Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV (Elsevier, 2016-09) The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ... Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV (Elsevier, 2016-12) We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ... Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV (Springer, 2016-05) Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
Search Now showing items 1-10 of 34 Search for top squark pair production in final states with one isolated lepton, jets, and missing transverse momentum in √s = 8 TeV pp collisions with the ATLAS detector (Springer, 2014-11) The results of a search for top squark (stop) pair production in final states with one isolated lepton, jets, and missing transverse momentum are reported. The analysis is performed with proton-proton collision data at s√ ... Search for supersymmetry in events with large missing transverse momentum, jets, and at least one tau lepton in 20 fb−1 of √s = 8 TeV proton-proton collision data with the ATLAS detector (Springer, 2014-09-18) A search for supersymmetry (SUSY) in events with large missing transverse momentum, jets, at least one hadronically decaying tau lepton and zero or one additional light leptons (electron/muon), has been performed using ... Measurement of the top quark pair production charge asymmetry in proton-proton collisions at √s = 7 TeV using the ATLAS detector (Springer, 2014-02) This paper presents a measurement of the top quark pair ( tt¯ ) production charge asymmetry A C using 4.7 fb−1 of proton-proton collisions at a centre-of-mass energy s√ = 7 TeV collected by the ATLAS detector at the LHC. ... Measurement of the low-mass Drell-Yan differential cross section at √s = 7 TeV using the ATLAS detector (Springer, 2014-06) The differential cross section for the process Z/γ ∗ → ℓℓ (ℓ = e, μ) as a function of dilepton invariant mass is measured in pp collisions at √s = 7 TeV at the LHC using the ATLAS detector. The measurement is performed in ... Measurements of fiducial and differential cross sections for Higgs boson production in the diphoton decay channel at √s=8 TeV with ATLAS (Springer, 2014-09-19) Measurements of fiducial and differential cross sections are presented for Higgs boson production in proton-proton collisions at a centre-of-mass energy of s√=8 TeV. The analysis is performed in the H → γγ decay channel ... Measurement of the inclusive jet cross-section in proton-proton collisions at $ \sqrt{s}=7 $ TeV using 4.5 fb−1 of data with the ATLAS detector (Springer, 2015-02-24) The inclusive jet cross-section is measured in proton-proton collisions at a centre-of-mass energy of 7 TeV using a data set corresponding to an integrated luminosity of 4.5 fb−1 collected with the ATLAS detector at the ... ATLAS search for new phenomena in dijet mass and angular distributions using pp collisions at $\sqrt{s}$=7 TeV (Springer, 2013-01) Mass and angular distributions of dijets produced in LHC proton-proton collisions at a centre-of-mass energy $\sqrt{s}$=7 TeV have been studied with the ATLAS detector using the full 2011 data set with an integrated ... Search for direct chargino production in anomaly-mediated supersymmetry breaking models based on a disappearing-track signature in pp collisions at $\sqrt{s}$=7 TeV with the ATLAS detector (Springer, 2013-01) A search for direct chargino production in anomaly-mediated supersymmetry breaking scenarios is performed in pp collisions at $\sqrt{s}$ = 7 TeV using 4.7 fb$^{-1}$ of data collected with the ATLAS experiment at the LHC. ... Search for heavy lepton resonances decaying to a $Z$ boson and a lepton in $pp$ collisions at $\sqrt{s}=8$ TeV with the ATLAS detector (Springer, 2015-09) A search for heavy leptons decaying to a $Z$ boson and an electron or a muon is presented. The search is based on $pp$ collision data taken at $\sqrt{s}=8$ TeV by the ATLAS experiment at the CERN Large Hadron Collider, ... Evidence for the Higgs-boson Yukawa coupling to tau leptons with the ATLAS detector (Springer, 2015-04-21) Results of a search for $H \to \tau \tau$ decays are presented, based on the full set of proton--proton collision data recorded by the ATLAS experiment at the LHC during 2011 and 2012. The data correspond to integrated ...
Search Now showing items 1-10 of 26 Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV (Elsevier, 2017-12-21) We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ... Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV (American Physical Society, 2017-09-08) The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ... Online data compression in the ALICE O$^2$ facility (IOP, 2017) The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ... Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV (American Physical Society, 2017-09-08) In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ... J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV (American Physical Society, 2017-12-15) We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($\|y\| < 0.9$) ... Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions (Nature Publishing Group, 2017) At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ... K$^{}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV (American Physical Society, 2017-06) The production of K$^{}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ... Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV (Springer, 2017-06) The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ... Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC (Springer, 2017-01) The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ... Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC (Springer, 2017-06) We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ...
What is the relationship, if any, between Kalman filtering and (repeated, if necessary) least squares polynomial regression? 1. There is a Difference in terms of optimality criteria Kalman filter is a Linear estimator. It is a linear optimal estimator - i.e. infers model parameters of interest from indirect, inaccurate and uncertain observations. But optimal in what sense? If all noise is Gaussian, the Kalman filter minimizes the mean square error of the estimated parameters. This means, that when underlying noise is NOT Gaussian the promise no longer holds. In case of nonlinear dynamics, it is well-known that the problem of state estimation becomes difficult. In this context, no filtering scheme clearly outperforms all other strategies. In such case, Non-linear estimators may be better if they can better model the system with additional information. [See Ref 1-2] Polynomial regression is a form of linear regression in which the relationship between the independent variable x and the dependent variable y is modeled as an nth order polynomial. $$ Y = a_0 + a_1x + a_2x^2 + \epsilon $$ Note that, while polynomial regression fits a nonlinear model to the data, these models are all linear from the point of view of estimation, since the regression function is linear in terms of the unknown parameters $a_0, a_1, a_2$. If we treat $x, x^2$ as different variables, polynomial regression can also be treated as multiple linear regression. Polynomial regression models are usually fit using the method of least squares. In the least squares method also, we minimize the mean squared error. The least-squares method minimizes the variance of the unbiased estimators of the coefficients, under the conditions of the Gauss–Markov theorem. This theorem, states that ordinary least squares (OLS) or linear least squares is the Best Linear Unbaised Estimator (BLUE) under following conditions: a. when errors have expectation zero i.e. $E(e_i) = 0 $ b. have equal variances i.e. $ Variance(e_i) = \sigma^2 < \infty $ c. and errors are uncorrelated i.e. $ cov(e_i,e_j) = 0 $ NOTE: that here, errors don't have to be Gaussian nor need to be IID. It only needs to be uncorrelated. 2. Kalman Filter is an evolution of estimators from least square In 1970, H. W. Sorenson published an IEEE Spectrum article titled "Least-squares estimation: from Gauss to Kalman." [See Ref 3.] This is a seminal paper that provides great insight about how Gauss' original idea of least squares to today's modern estimators like Kalman. Gauss' work not only introduced the least square framework but it was actually one of the earliest work that used a probabilistic view. While least squares evolved in the form of various regression methods, there was another critical work that brought filter theory to be used as an estimator. The theory of ﬁltering to be used for stationary time series estimation was constructed by Norbert Wiener during 1940s (during WW-II) and published in 1949 which is now known as Wiener filter. The work was done much earlier, but was classiﬁed until well after World War II). The discrete-time equivalent of Wiener's work was derived independently by Kolmogorov and published in 1941. Hence the theory is often called the Wiener-Kolmogorov filtering theory. Traditionally filters are designed for the desired frequency response. However, in case of Wiener filter, it reduces the amount of noise present in a signal by comparison with an estimation of the desired noiseless signal. Weiner filter is actually an estimator. In an important paper, however, Levinson (1947) [See Ref 6] showed that in discrete time, the entire theory could be reduced to least squares and so was mathematically very simple. See Ref 4 Thus, we can see that Weiner's work gave a new approach for estimation problem; an evolution from using least squares to another well-established filter theory. However, the critical limitation is that Wiener filter assumes the inputs are stationary. We can say that Kalman filter is a next step in the evolution which drops the stationary criteria. In Kalman filter, state space model can dynamically be adapted to deal with non-stationary nature of signal or system. The Kalman filters are based on linear dynamic systems in discrete time domain. Hence it is capable of dealing with potentially time varying signal as opposed to Wiener. As the Sorenson's paper draws parallel between Gauss' least squares and Kalman filter as ...therefore, one sees that the basic assumption of Gauss and Kalman are identical except that later allows the state to change from one time to next. The difference introduces a non-trivial modification to Gauss' problem but one that can be treated within the least squares framework. 3. They are same as far as causality direction of prediction is concerned; besides implementation efficiency Sometimes it is perceived that Kalman filter is used for prediction of future events based on past data where as regression or least squares does smoothing within end to end points. This is not really true. Readers should note that both the estimators (and almost all estimators you can think of) can do either job. You can apply Kalman filter to apply Kalman smoothing. Similarly, regression based models can also be used for prediction. Given the training vector, $X_t$ and you applied $Y_t$ and discovered the model parameters $α_0 ... a_K$ now for another sample $X_k$ we can extrapolate $Y_K$ based on the model. Hence, both methods can be used in the form of smoothing or fitting (non-causal) as well as for future predictions (causal case). However, the critical difference is the implementation which is significant. In case of polynomial regression - with entire process needs to get repeated and hence, while it may be possible to implement causal estimation but it might be computationally expensive. [While, I am sure there must be some research by now to make things iterative]. On the other hand, Kalman filter is inherently recursive. Hence, using it for prediction for future only using on past data will be very efficient. Here is another good presentation that compares several methods: Ref 5 References Best Introduction to Kalman Filter - Dan Simon Kalman FilteringEmbedded Systems Programming JUNE 2001 page 72 Levinson, N. (1947). "The Wiener RMS error criterion in filter design and prediction." J. Math. Phys., v. 25, pp. 261–278. The difference is quite huge, since they are two completely different models which can be used to tackle the same problem. Let's do a quick recap. Polynomial regression is a way of function approximation. We have a data set of the form $\lbrace x_i, z_i \rbrace$ and wish to determine the functional relationship, which is often expressed by estimating the probability density $p(z\|x)$. Under the assumption that this $p$ is a Gaussian, we get the least squares solution as a maximum likelihood estimator. Kalman filtering is a special way of inference in a linear dynamical system. LDSs are a special case of state space models, in which we assume that the data we observe is generated by a the application of a linear transform to the subsequent steps of a Markov chain over Gaussian random variables. Thus what we actually do is to model $p(x_{1:T})$, which is the probability of a time series. The process of Kalman Filtering is then to predict the next value of a time series, e.g. maximize $p(x_{t+1}\|x_{1:t})$. But the same model can be used to do inference on smoothing, interpolation and many more things. Thus: polynomial regression does function approximation, Kalman filtering does time series prediction. Two totally different things, but time series prediction is a special case of function approximation. Also, both models base quite different assumptions on the data they observe. Not an expert on kalman filters, however I believe traditional Kalman filtering presumes a linear relationship between the observable data, and data you wish to infer, in contrast to more intricate ones like the Extended Kalman filters that can assume non-linear relationships. With that in mind, I believe that for a traditional Kalman filter, on-line linear regression, would be similar to Kalman in performance. However, a polynomial regression can also be used that would assume a non-linear relationship that a traditional Kalman might not be able to capture. Kalman filtering gives multiple predictions for the next state, where an extrapolation of a regression would not. The Kalman filters are also focused on including noise factors (based on Gaussian distributions). A lot has been said already, allow me to add some comments: Kalman filters are an application of Bayesian probability theory, which means that "a priori information" or "prior uncertainty" can (and must) be specified. As I understand, this is not the case with traditional least-squares fitting. While observations (data) can be weighted with probabilities in LSQ fitting, prior knowledge of a solution cannot be readily taken into account. In summary, solutions found by a KF will depend on a) a model to provide 'predictions' b) measurements which are 'observations' c) uncertainty on predictions and observations d) a priori knowledge of the solution. "prior knowledge" is specified as a variance on the initial guess, but is not relevant or utilized to the same extent in every application. As mentioned before, a common use of the KF is to reduce noise in real-time observations. Comparing observations with model predictions can help to estimate the 'true measurement' devoid of noise. This common application is why the KF is called a filter. The initial guess in this example would be the assumed solution at time zero from which the KF starts, with associated "prior uncertainty". Often you will have some unknown parameters in the predictive model, but which can be constrained by the measurements i.e. are "observable". The KF will improve its estimates of both those parameters and the "true measurements" as it moves through the time series of data. In that case the initial state is often specified to simply result in consistent filtering performance: defined as the actual estimation errors being within the uncertainty bounds that the KF provides with its solution. In this example, the prior uncertainty on the initial state may be specified to be large, giving the KF opportunity to correct any errors it contains. Small values may also be specified, to nudge the filter towards future estimates that make sense (are consistent with observations). This area of KF design may involve trial and error, or engineering judgement, to come up with values of the initial state and its uncertainty that result in good performance. For this reason, this and other aspects of KF filter design which involve specifying uncertainties to result in good performance (be it numerical, estimation, prediction...) is often referred to as "filter tuning". But in other applications, a more rigorous and useful approach to prior uncertainties can be adopted. The previous example was about real-time estimation (to filter noise out of uncertain measurements). The initial state and its variance (prior uncertainty) are almost a necessary evil to initialize the filter at an early time, after which the initial state becomes increasingly unimportant as future observations are used to improve estimates. Consider now a Kalman filter applied to measurements and model predictions at a specific time t_s. We have uncertain observations, an uncertain model, but we also have some prior knowledge about the solution we are looking for. Let's say we know its Gaussian PDF: mean and variance. In this case the solution could depend very strongly on the prior uncertainty, meaning item d) above, where the hope is that the added information improves the KF solution (smaller error and less uncertainty). This feature, which is fundamental to Bayesian theory, allows the KF to solve stochastic problems while taking into account every kind of uncertainty/information that is typically available. Because the KF has been developed and applied for decades, its basic features are not always described in detail. In my experience, many papers and books focus on optimality and linearization (extended KF, unscented KF, and so on). But I've found great descriptions of the links between Bayesian theory and KF, by reading introductory papers and texts on "particle filters". Those are another and more recent implementation of Bayesian estimation, look them up if you're interested!
I have perhaps a silly question. I have a set of vectors $\xi_\ell$, with $\ell\in J_k\subset \{1,\ldots,k\}$. Now I want to create a long vector $\xi$ where these individual vectors are stacked on top of each other, and a matrix $\Xi$ of which the columns are these individual vectors. I am stuck with the proper notation. I cannot use the traditional way where, for example, $\ell\in\{1,\ldots,k\}$, I can define $\xi:=[\xi_1^\top,\ldots,\xi_k^\top]^\top$, since in my case the $\ell$ may not be contiguous or start at $1$. Similar problem for defining the matrix $\Xi$. Does anyone have a suggestion or is anyone aware of the proper way defining this vector and matrix? Thanks a lot!
Let us consider the subset\[Z:=\{z\in R \mid zr=rz \text{ for any } r\in R\}.\](This is called the center of the ring $R$.) This is a subgroup of the additive group $R$.In fact, if $z, z’\in Z$, then we have for any $r\in R$,\begin{align}(z-z’)r=zr-z’r=rz-rz’=r(z-z’).\end{align}It follows that $z-z’\in Z$, and thus $Z$ is a subgroup of $R$. Note that $0, 1 \in Z$, hence $Z$ is not a trivial subgroup.Thus, we have either $\|Z\|=p, p^2$ since $R$ is a group of order $p^2$. If $\|Z\|=p^2$, then we have $Z=R$.By definition of $Z$, this implies that $R$ is commutative. It remains to show that $\|Z\|\neq p$.Assume that $\|Z\|=p$.Then $R/Z$ is a cyclic group of order $p$.Let $\alpha$ be a generator of $R/Z$. Since $Z\neq R$, there exist $r, s\in R$ such that $rs\neq sr$.Write\[r=m\alpha+z \text{ and } s=n\alpha+z’\]for some $m, n\in \Z$, $z, z’\in Z$. Then we have\begin{align}rs&=(m\alpha+z)(n\alpha+z’)\\&=(m\alpha)(n\alpha)+m\alpha z’ + n z\alpha +z z’\\&=(n\alpha)(m\alpha)+m z’ \alpha +n \alpha z +z’ z\\&=(n\alpha+z’)(m\alpha+z)\\&=sr.\end{align} This contradicts $rs\neq sr$, and we conclude that $\|Z\|\neq p$. Generators of the Augmentation Ideal in a Group RingLet $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring […] Primary Ideals, Prime Ideals, and Radical IdealsLet $R$ be a commutative ring with unity. A proper ideal $I$ of $R$ is called primary if whenever $ab \in I$ for $a, b\in R$, then either $a\in I$ or $b^n\in I$ for some positive integer $n$.(a) Prove that a prime ideal $P$ of $R$ is primary.(b) If $P$ is a prime ideal and […] Three Equivalent Conditions for a Ring to be a FieldLet $R$ be a ring with $1$. Prove that the following three statements are equivalent.The ring $R$ is a field.The only ideals of $R$ are $(0)$ and $R$.Let $S$ be any ring with $1$. Then any ring homomorphism $f:R \to S$ is injective.Proof. […] Ideal Quotient (Colon Ideal) is an IdealLet $R$ be a commutative ring. Let $S$ be a subset of $R$ and let $I$ be an ideal of $I$.We define the subset\[(I:S):=\{ a \in R \mid aS\subset I\}.\]Prove that $(I:S)$ is an ideal of $R$. This ideal is called the ideal quotient, or colon ideal.Proof.Let $a, […] Every Integral Domain Artinian Ring is a FieldLet $R$ be a ring with $1$. Suppose that $R$ is an integral domain and an Artinian ring.Prove that $R$ is a field.Definition (Artinian ring).A ring $R$ is called Artinian if it satisfies the defending chain condition on ideals.That is, whenever we have […] Equivalent Conditions For a Prime Ideal in a Commutative RingLet $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:(a) The ideal $P$ is a prime ideal.(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.Proof. […] Nilpotent Element a in a Ring and Unit Element $1-ab$Let $R$ be a commutative ring with $1 \neq 0$.An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.We give two proofs.Proof 1.Since $a$ […]
According to a review by Legge & Bigelow the arc or degrees of visual angle ($\alpha$) is, $$\alpha = 57.3 \times S/D,$$ where S is height of object and D is distance to object. [1] $S/D$ is the small angles approximation of $2 \times arctan(S/2D)$ which follows form geometry. Image 1: the equation comes straight from trigonometric definitions. Since humans continuously scan the page with the fovea, small angles assumption is a good approximation for entire pages. The factor $57.3$ is simply $180/\pi$ or conversion form radians to angles. So given that we do not know the nature of $tg$, then its hard to say. Looks a bit fishy, but if tg is some sort of inversion of S then its plausible. [1] G. Legge & C Bigelow. 2011. Jornal of vision. Does print size matter for reading? A review of findings from vision science and typography.
It is also useful to be able to compare two means for small samples. For instance, a teacher might like to test the notion that two versions of an exam were equally difficult. She could do so by randomly assigning each version to students. If she found that the average scores on the exams were so different that we cannot write it off as chance, then she may want to award extra points to students who took the more difficult exam. In a medical context, we might investigate whether embryonic stem cells can improve heart pumping capacity in individuals who have suffered a heart attack. We could look for evidence of greater heart health in the stem cell group against a control group. In this section we use the t distribution for the difference in sample means. We will again drop the minimum sample size condition and instead impose a strong condition on the distribution of the data. Sampling Distributions for the Difference in two Means In the example of two exam versions, the teacher would like to evaluate whether there is convincing evidence that the difference in average scores between the two exams is not due to chance. It will be useful to extend the t distribution method from Section 5.3 to apply to a difference of means: \[\bar {x}_1 - \bar {x}_2\] as a point estimate for \[\mu _1 - \mu _2\] Our procedure for checking conditions mirrors what we did for large samples in Section 5.2. First, we verify the small sample conditions (independence and nearly normal data) for each sample separately, then we verify that the samples are also independent. For instance, if the teacher believes students in her class are independent, the exam scores are nearly normal, and the students taking each version of the exam were independent, then we can use the t distribution for inference on the point estimate $\bar {x}_1 - \bar {x}_2$. The formula for the standard error of $\bar {x}_1 - \bar {x}_2$., introduced in Section 5.2, also applies to small samples: \[ SE_{\bar {x}_1- \bar {x}_2} = \sqrt {SE^2_{\bar {x}_1} + SE^2_{\bar {x}_2}} = \sqrt { \dfrac {s^2_1}{n_1} +\dfrac {s^2_2}{n_2}} \tag {5.27}\] 19We use the row with 29 degrees of freedom. The value T = 2.39 falls between the third and fourth columns. Because we are looking for a single tail, this corresponds to a p-value between 0.01 and 0.025. The p-value is guaranteed to be less than 0.05 (the default signi cance level), so we reject the null hypothesis. The data provide convincing evidence to support the company's claim that student scores improve by more than 100 points following the class. 20This is an observational study, so we cannot make this causal conclusion. For instance, maybe SAT test takers tend to improve their score over time even if they don't take a special SAT class, or perhaps only the most motivated students take such SAT courses. Because we will use the t distribution, we will need to identify the appropriate degrees of freedom. This can be done using computer software. An alternative technique is to use the smaller of $n_1 - 1$ and $n_2 - 1$, which is the method we will apply in the examples and exercises. 21 Using the t distribution for a difference in means The t distribution can be used for inference when working with the standardized difference of two means if (1) each sample meets the conditions for using the t distribution and (2) the samples are independent. We estimate the standard error of the difference of two means using Equation \ref{5.27}. Two Sample t test Summary statistics for each exam version are shown in Table 5.19. The teacher would like to evaluate whether this difference is so large that it provides convincing evidence that Version B was more difficult (on average) than Version A. Version n $\bar {x}$ s min max A B 30 27 79.4 74.1 14 20 45 32 100 100 Exercise $\PageIndex{1}$ Construct a two-sided hypothesis test to evaluate whether the observed difference in sample means, $\bar {x}_A - \bar {x}_B = 5.3$, might be due to chance. Solution Because the teacher did not expect one exam to be more difficult prior to examining the test results, she should use a two-sided hypothesis test. H 0: the exams are equally difficult, on average. $\mu_A - \mu_B = 0$. HA: one exam was more difficult than the other, on average. $\mu_A - \mu_B \ne 0$. Exercise $\PageIndex{1}$ To evaluate the hypotheses in Exercise 5.28 using the t distribution, we must first verify assumptions. Does it seem reasonable that the scores are independent within each group? What about the normality condition for each group? Do you think scores from the two groups would be independent of each other (i.e. the two samples are independent)? 23 Solution (a) It is probably reasonable to conclude the scores are independent. (b) The summary statistics suggest the data are roughly symmetric about the mean, and it doesn't seem unreasonable to suggest the data might be normal. Note that since these samples are each nearing 30, moderate skew in the data would be acceptable. (c) It seems reasonable to suppose that the samples are independent since the exams were handed out randomly. After verifying the conditions for each sample and confirming the samples are independent of each other, we are ready to conduct the test using the t distribution. In this case, we are estimating the true difference in average test scores using the sample data, so the point estimate is $\bar {x}_A - \bar {x}_B = 5.3$. The standard error of the estimate can be calculated using Equation \ref{5.27}: \[ SE = \sqrt {\dfrac {s^2_A}{n_A} + \dfrac {s^2_B}{n_B}} = \sqrt {\dfrac {14^2}{30} + \dfrac {20^2}{27}} = 4.62\] 21This technique for degrees of freedom is conservative with respect to a Type 1 Error; it is more difficult to reject the null hypothesis using this df method. Figure 5.20: The t distribution with 26 degrees of freedom. The shaded right tail represents values with T $\ge$ 1.15. Because it is a two-sided test, we also shade the corresponding lower tail. Finally, we construct the test statistic: \[T = \dfrac {\text {point estimate - null value}}{SE} = \dfrac {(79.4 - 74.1) - 0}{4.62} = 1.15\] If we have a computer handy, we can identify the degrees of freedom as 45.97. Otherwise we use the smaller of $n_1 - 1 \text {and} n_2 - 1$: df = 26. Exercise $\PageIndex{1}$ Exercise 5.30 Identify the p-value, shown in Figure 5.20. Use df = 26. Solution We examine row df = 26 in the t table. Because this value is smaller than the value in the left column, the p-value is larger than 0.200 (two tails!). Because the p-value is so large, we do not reject the null hypothesis. That is, the data do not convincingly show that one exam version is more difficult than the other, and the teacher should not be convinced that she should add points to the Version B exam scores. In Exercise 5.30, we could have used df = 45.97. However, this value is not listed in the table. In such cases, we use the next lower degrees of freedom (unless the computer also provides the p-value). For example, we could have used df = 45 but not df = 46. Exercise $\PageIndex{1}$ Do embryonic stem cells (ESCs) help improve heart function following a heart attack? Table 5.21 contains summary statistics for an experiment to test ESCs in sheep that had a heart attack. Each of these sheep was randomly assigned to the ESC or control group, and the change in their hearts' pumping capacity was measured. A positive value generally corresponds to increased pumping capacity, which suggests a stronger recovery. Set up hypotheses that will be used to test whether there is convincing evidence that ESCs actually increase the amount of blood the heart pumps. Check conditions for using the t distribution for inference with the point estimate $\bar {x}_1 - \bar {x}_2$. To assist in this assessment, the data are presented in Figure 5.22. 25 Solution (a) We first setup the hypotheses: H 0: The stem cells do not improve heart pumping function. $\mu _{esc} - \mu _{control} = 0$. H A: The stem cells do improve heart pumping function. $\mu _{esc} - \mu _{control} > 0$. (b) Because the sheep were randomly assigned their treatment and, presumably, were kept separate from one another, the independence assumption is reasonable for each sample as well as for between samples. The data are very limited, so we can only check for obvious outliers in the raw data in Figure 5.22. Since the distributions are (very) roughly symmetric, we will assume the normality condition is acceptable. Because the conditions are satisfied, we can apply the t distribution. Figure 5.22: Histograms for both the embryonic stem cell group and the control group. Higher values are associated with greater improvement. We don't see any evidence of skew in these data; however, it is worth noting that skew would be difficult to detect with such a small sample. n $\bar {x}$ s ESCs control 9 9 3.50 -4.33 5.17 2.76 Figure 5.23: Distribution of the sample difference of the test statistic if the null hypothesis was true. The shaded area, hardly visible in the right tail, represents the p-value. Example $\PageIndex{1}$ Use the data from Table 5.21 and df = 8 to evaluate the hypotheses for the ESC experiment described in Exercise 5.31. Solution First, we compute the sample difference and the standard error for that point estimate: \[\bar {x}_{esc} - \bar {x}_{control} = 7.88\] \[SE = \dfrac {\dfrac {5.17^2}{9} + \dfrac {2.76^2}{9}} = 1.95\] The p-value is depicted as the shaded slim right tail in Figure 5.23, and the test statistic is computed as follows: \[T = \dfrac {7.88 - 0}{1.95} = 4.03\] We use the smaller of $n_1 - 1$ and $n_2 - 1$ (each are the same) for the degrees of freedom: df = 8. Finally, we look for T = 4.03 in the t table; it falls to the right of the last column, so the p-value is smaller than 0.005 (one tail!). Because the p-value is less than 0.005 and therefore also smaller than 0.05, we reject the null hypothesis. The data provide convincing evidence that embryonic stem cells improve the heart's pumping function in sheep that have suffered a heart attack. Two sample t confidence interval The results from the previous section provided evidence that ESCs actually help improve the pumping function of the heart. But how large is this improvement? To answer this question, we can use a confidence interval. Exercise $\PageIndex{1}$ In Exercise 5.31, you found that the point estimate, $\bar {x}_{esc} - \bar {x}_{control} = 7.88$, has a standard error of 1.95. Using df = 8, create a 99% confidence interval for the improvement due to ESCs. Solution We know the point estimate, 7.88, and the standard error, 1.95. We also veri ed the conditions for using the t distribution in Exercise 5.31. Thus, we only need identify $t_8$ to create a 99% con dence interval: $t_8 = 3.36$. The 99% con dence interval for the improvement from ESCs is given by \[\text {point estimate} \pm t*_8 SE \rightarrow 7.88 \pm 3.36 \times 1.95 \rightarrow (1.33, 14.43)\] That is, we are 99% con dent that the true improvement in heart pumping function is somewhere between 1.33% and 14.43%. Pooled Standard Deviation Estimate (special topic) Occasionally, two populations will have standard deviations that are so similar that they can be treated as identical. For example, historical data or a well-understood biological mechanism may justify this strong assumption. In such cases, we can make our t distribution approach slightly more precise by using a pooled standard deviation. The pooled standard deviation of two groups is a way to use data from both samples to better estimate the standard deviation and standard error. If s1 and s2 are the standard deviations of groups 1 and 2 and there are good reasons to believe that the population standard deviations are equal, then we can obtain an improved estimate of the group variances by pooling their data: \[s^2_{pooled} = \dfrac {s^2_1 \times (n_1 - 1) + s^2_2 \times (n_2 - 1)}{n_1 + n_2 - 2}\] where $n_1$ and $n_2$ are the sample sizes, as before. To use this new statistic, we substitute $s^2_{pooled}$ in place of $s^2_1$ and $s^2_2$ in the standard error formula, and we use an updated formula for the degrees of freedom: \[df = n_1 + n_2 - 2\] The bene ts of pooling the standard deviation are realized through obtaining a better estimate of the standard deviation for each group and using a larger degrees of freedom parameter for the t distribution. Both of these changes may permit a more accurate model of the sampling distribution of $\bar {x}^2_1 - \bar {x}^2_2$ Caution: Pooling standard deviations should be done only after careful research A pooled standard deviation is only appropriate when background research indicates the population standard deviations are nearly equal. When the sample size is large and the condition may be adequately checked with data, the benefits of pooling the standard deviations greatly diminishes. Contributors David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University) David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University)
Let $n$ be a positive integer and there be $p_1,p_2,p_3,........p_n$ prime numbers such that all of them are greater than $5$. If $6$ divides $p_1 ^2 + p_2 ^2 + p_3 ^2+\ldots +p_n ^2$, prove that $6$ divides $n$. NOTE:- This problem is the $2nd$ question of $1998$ $RMO$ (Regional Mathematical Olympiad). I tried using congruences with $2$ and $3$ on the first condition but it does not work. Let $n$ be a Notice that every prime$>3$ is of the form $6n\pm1$, So square of every such prime will be $36n^2\pm12n+1$, so when you are adding all those squares, you will get something like: $36(n_1^2+n_2^2+n_3^2......)+12(\pm n_1+\pm n_2.....)+ n(1)$.. The two terms containing $6$ and $12$ are divisible by $6$, the divisibility of whole by $6$ depends upon divisibilty of $n$ by $6$ and vice versa. The sum of squares of $n$ odd numbers is $\equiv n\pmod 8$ because a single odd square is $\equiv 1\pmod 8$; similarly the sum of squares of $n$ numbers not divisible by $3$ is $\equiv n\pmod 3$. Hence under the given conditions, we even have $p_1^2+\ldots +p_n^2\equiv n\pmod {24}$, but of course in particular $p_1^2+\ldots +p_n^2\equiv n\pmod {6}$. Using congruences on $2$ and $3$ works as follows: For all primes $p_i\ge 5$, we know that $2 \nmid p_i$ and $3\nmid p_i$. Therefore $p_i^2\equiv 1 \bmod 2$ and $p_i^2\equiv 1 \bmod 3$, which immediately gives us that $p_i^2 \equiv 1 \bmod 6$ by the Chinese Remainder Theorem. Clearly adding $n$ such prime squares to a total $s$ will only give us $s\equiv 0\bmod 6 $ (that is, $s\mid 6$) if $n\mid 6 $ also.
Z-Scores The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. Definition: Z-Score If $X$ is a normally distributed random variable and $X \sim N(\mu, \sigma)$, then the z-score is: \[z = \dfrac{x=\mu}{\sigma} \label{zscore}\] The z-score tells you how many standard deviations the value $x$ is above (to the right of) or below (to the left of) the mean, $\mu$. Values of $x$ that are larger than the mean have positive $z$-scores, and values of $x$ that are smaller than the mean have negative $z$-scores. If $x$ equals the mean, then $x$ has a $z$-score of zero. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows: \[ \begin{align} x &= \mu + (z)(\sigma) \\[5pt] &= 5 + (3)(2) = 11 \end{align}\] The z-score is three. Since the mean for the standard normal distribution is zero and the standard deviation is one, then the transformation in Equation \ref{zscore} produces the distribution $Z \sim N(0, 1)$. The value $x$ comes from a normal distribution with mean $\mu$ and standard deviation $\sigma$. A z-score is measured in units of the standard deviation. Example $\PageIndex{1}$ Suppose $X \sim N(5, 6)$. This says that $x$ is a normally distributed random variable with mean $\mu = 5$ and standard deviation $\sigma = 6$. Suppose $x = 17$. Then (via Equation \ref{zscore}): \[z = \dfrac{x-\mu}{\sigma} = \dfrac{17-5}{6} = 2 \nonumber\] This means that $x = 17$ is two standard deviations (2$\sigma$) above or to the right of the mean $\mu = 5$. The standard deviation is $\sigma = 6$. Notice that: $5 + (2)(6) = 17$ (The pattern is $\mu + z \sigma = x$) Now suppose $x = 1$. Then: \[z = \dfrac{x=\mu}{\sigma} = \dfrac{1-5}{6} = -0.67 \nonumber\] (rounded to two decimal places) This means that $x = 1$ is $0.67$ standard deviations ($–0.67\sigma$) below or to the left of the mean $\mu = 5$. Notice that: $5 + (–0.67)(6)$ is approximately equal to one (This has the pattern $\mu + (–0.67)\sigma = 1$) Summarizing, when $z$ is positive, $x$ is above or to the right of $\mu$ and when $z$ is negative, $x$ is to the left of or below $\mu$. Or, when $z$ is positive, $x$ is greater than $\mu$, and when $z$ is negative $x$ is less than $\mu$. Exercise $\PageIndex{1}$ What is the $z$-score of $x$, when $x = 1$ and $X \sim N(12, 3)$? Answer $z = \dfrac{1-12}{3} \approx -3.67$ Example $\PageIndex{2}$ Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let $X =$ the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. $X \sim N(5, 2)$. Fill in the blanks. Suppose a person lostten pounds in a month. The $z$-score when $x = 10$ pounds is $x = 2.5$ (verify). This $z$-score tells you that $x = 10$ is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Suppose a person gainedthree pounds (a negative weight loss). Then $z =$ __________. This $z$-score tells you that $x = -3$ is ________ standard deviations to the __________ (right or left) of the mean. Answers a. This $z$-score tells you that $x = 10$ is 2.5 standard deviations to the right of the mean five. b. Suppose the random variables $X$ and $Y$ have the following normal distributions: $X \sim N(5, 6)$ and $Y \sim N(2, 1)$. If $x = 17$, then $z = 2$. (This was previously shown.) If $y = 4$, what is $z$? \[z = \dfrac{y-\mu}{\sigma} = \dfrac{4-2}{1} = 2 \nonumber\] where $\mu = 2$ and $\sigma = 1$. The $z$-score for $y = 4$ is $z = 2$. This means that four is $z = 2$ standard deviations to the right of the mean. Therefore, $x = 17$ and $y = 4$ are both two (of their own) standard deviations to the right of their respective means. The z-score allows us to compare data that are scaled differently. To understand the concept, suppose $X \sim N(5, 6)$ represents weight gains for one group of people who are trying to gain weight in a six week period and $Y \sim N(2, 1)$ measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since $x = 17$ and $y = 4$ are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means. Exercise $\PageIndex{2}$ Fill in the blanks. Jerome averages 16 points a game with a standard deviation of four points. $X \sim N(16, 4)$. Suppose Jerome scores ten points in a game. The $z$–score when $x = 10$ is $-1.5$. This score tells you that $x = 10$ is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?). Answer 1.5, left, 16 The Empirical Rule If $X$ is a random variable and has a normal distribution with mean $\mu$ and standard deviation $\sigma$, then the Empirical Rule says the following: About 68% of the $x$ values lie between –1$\sigma$ and +1$\sigma$ of the mean $\mu$ (within one standard deviation of the mean). About 95% of the $x$ values lie between –2$\sigma$ and +2$\sigma$ of the mean $\mu$ (within two standard deviations of the mean). About 99.7% of the $x$ values lie between –3$\sigma$ and +3$\sigma$ of the mean $\mu$ (within three standard deviations of the mean). Notice that almost all the $x$ values lie within three standard deviations of the mean. The $z$-scores for +1$\sigma$ and –1$\sigma$ are +1 and –1, respectively. The $z$-scores for +2$\sigma$ and –2$\sigma$ are +2 and –2, respectively. The $z$-scores for +3$\sigma$ and –3$\sigma$ are +3 and –3 respectively. The empirical rule is also known as the 68-95-99.7 rule. Figure $\PageIndex{1}$ Example $\PageIndex{3}$ The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let $X =$ the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then $X \sim N(170, 6.28)$. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The $z$-score when $x = 168$ cm is $z =$ _______. This $z$-score tells you that $x = 168$ is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a $z$-score of $z = 1.27$. What is the male’s height? The $z$-score ($z = 1.27$) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. Answers –0.32, 0.32, left, 170 177.98, 1.27, right Exercise $\PageIndex{3}$ Use the information in Example $\PageIndex{3}$ to answer the following questions. Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The $z$-score when $x = 176$ cm is $z =$ _______. This $z$-score tells you that $x = 176$ cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a $z$-score of $z = –2$. What is the male’s height? The $z$-score ($z = –2$) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean. Answer Solve the equation $z = \dfrac{x-\mu}{\sigma}$ for $z$. $x = \mu+ (z)(\sigma)$ $z = \dfrac{176-170}{6.28}$, This z-score tells you that $x = 176$ cm is 0.96 standard deviations to the right of the mean 170 cm. Answer Solve the equation $z = \dfrac{x-\mu}{\sigma}$ for $z$. $x = \mu+ (z)(\sigma)$ $X = 157.44$ cm, The $z$-score($z = –2$) tells you that the male’s height is two standard deviations to the left of the mean. Example $\PageIndex{4}$ From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let $Y =$ the height of 15 to 18-year-old males from 1984 to 1985. Then $Y \sim N(172.36, 6.34)$. The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let $X =$ the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then $X \sim N(170, 6.28)$. Find the z-scores for $x = 160.58$ cm and $y = 162.85$ cm. Interpret each $z$-score. What can you say about $x = 160.58$ cm and $y = 162.85$ cm? Answer The $z$-score (Equation \ref{zscore}) for $x = 160.58$ is $z = –1.5$. The $z$-score for $y = 162.85$ is $z = –1.5$. Both $x = 160.58$ and $y = 162.85$ deviate the same number of standard deviations from their respective means and in the same direction. Exercise $\PageIndex{4}$ In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean $\mu = 496$ and a standard deviation $\sigma = 114$. Let $X =$ a SAT exam verbal section score in 2012. Then $X \sim N(496, 114)$. Find the $z$-scores for $x_{1} = 325$ and $x_{2} = 366.21$. Interpret each $z$-score. What can you say about $x_{1} = 325$ and $x_{2} = 366.21$? Answer The z-score (Equation \ref{zscore}) for $x_{1} = 325$ is $z_{1} = –1.14$. The z-score (Equation \ref{zscore}) for $x_{2} = 366.21$ is $z_{2} = –1.14$. Student 2 scored closer to the mean than Student 1 and, since they both had negative $z$-scores, Student 2 had the better score. Example $\PageIndex{5}$ Suppose $x$ has a normal distribution with mean 50 and standard deviation 6. About 68% of the $x$ values lie between $-1\sigma$ = (-1)(6) = -6\) and $1 \sigma = (1)(6) = 6$ of the mean 50. The values $50 - 6 = 44$ and $50 + 6 = 56$ are within one standard deviation of the mean 50. The $z$-scores are –1 and +1 for 44 and 56, respectively. About 95% of the $x$ values lie between $-2 \sigma = (–2)(6) = –12$ and $2 \sigma = (2)(6) = 12$. The values $50 – 12 = 38$ and $50 + 12 = 62$ are within two standard deviations of the mean 50. The $z$-scores are –2 and +2 for 38 and 62, respectively. About 99.7% of the $x$ values lie between $–3 \sigma = (–3)(6) = –18$ and $3 \sigma = (3)(6) = 18$ of the mean 50. The values $50 – 18 = 32$ and $50 + 18 = 68$ are within three standard deviations of the mean 50. The $z$-scores are –3 and +3 for 32 and 68, respectively. Exercise $\PageIndex{5}$ Suppose $X$ has a normal distribution with mean 25 and standard deviation five. Between what values of $x$ do 68% of the values lie? Answer between 20 and 30. Example $\PageIndex{6}$ From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let $Y =$ the height of 15 to 18-year-old males in 1984 to 1985. Then $Y \sim N(172.36, 6.34)$. About 68% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively. About 95% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________ respectively. About 99.7% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively. Answer About 68% of the values lie between 166.02 and 178.7. The $z$-scores are –1 and 1. About 95% of the values lie between 159.68 and 185.04. The $z$-scores are –2 and 2. About 99.7% of the values lie between 153.34 and 191.38. The $z$-scores are –3 and 3. Exercise $\PageIndex{6}$ The scores on a college entrance exam have an approximate normal distribution with mean, $\mu = 52$ points and a standard deviation, $\sigma = 11$ points. About 68% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively. About 95% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively. About 99.7% of the $y$ values lie between what two values? These values are ________________. The $z$-scores are ________________, respectively. Answer a About 68% of the values lie between the values 41 and 63. The $z$-scores are –1 and 1, respectively. Answer b About 95% of the values lie between the values 30 and 74. The $z$-scores are –2 and 2, respectively. Answer c About 99.7% of the values lie between the values 19 and 85. The $z$-scores are –3 and 3, respectively. Summary A $z$-score is a standardized value. Its distribution is the standard normal, $Z \sim N(0, 1)$. The mean of the $z$-scores is zero and the standard deviation is one. If $y$ is the z-score for a value $x$ from the normal distribution $N(\mu, \sigma)$ then $z$ tells you how many standard deviations $x$ is above (greater than) or below (less than) $\mu$. Formula Review $Z \sim N(0, 1)$ $z = a$ standardized value ($z$-score) mean = 0; standard deviation = 1 To find the $K$ th percentile of $X$ when the $z$-scores is known: $k = \mu + (z)\sigma$ $z$-score: $z = \dfrac{x-\mu}{\sigma}$ $Z =$ the random variable for z-scores $Z \sim N(0, 1)$ Glossary Standard Normal Distribution a continuous random variable (RV) $X \sim N(0, 1)$; when $X$ follows the standard normal distribution, it is often noted as $Z \sim N(0, 1)\. \(z$-score the linear transformation of the form $z = \dfrac{x-\mu}{\sigma}$; if this transformation is applied to any normal distribution $X \sim N(\mu, \sigma$ the result is the standard normal distribution $Z \sim N(0,1)$. If this transformation is applied to any specific value $x$ of the RV with mean $\mu$ and standard deviation $\sigma$, the result is called the $z$-score of $x$. The $z$-score allows us to compare data that are normally distributed but scaled differently. References “Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewre...reportid=11960 (accessed May 14, 2013). “The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013). “2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digita...Group-2012.pdf (accessed May 14, 2013). “Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d...s/dt09_147.asp (accessed May 14, 2013). Data from the San Jose Mercury News. Data from The World Almanac and Book of Facts. “List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_o...ms_by_capacity (accessed May 14, 2013). Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013). Contributors Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/30189442-699...b91b9de@18.114.
The statement is in general false. We give a counterexample. Let us consider the following $2\times 2$ matrix:\[A=\begin{bmatrix}1 & 2\\2& 1\end{bmatrix}.\]The matrix $A$ satisfies the required conditions, that is, $A$ is symmetric and its diagonal entries are positive. The determinant $\det(A)=(1)(1)-(2)(2)=-3$ and the inverse of $A$ is given by\[A^{-1}=\frac{1}{-3}\begin{bmatrix}1 & -2\\-2& 1\end{bmatrix}=\begin{bmatrix}-1/3 & 2/3\\2/3& -1/3\end{bmatrix}\]by the formula for the inverse matrix for $2\times 2$ matrices. This shows that the diagonal entries of the inverse matrix $A^{-1}$ are negative. Diagonalizable by an Orthogonal Matrix Implies a Symmetric MatrixLet $A$ be an $n\times n$ matrix with real number entries.Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix.Proof.Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$.The orthogonality of the […] Construction of a Symmetric Matrix whose Inverse Matrix is ItselfLet $\mathbf{v}$ be a nonzero vector in $\R^n$.Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$.Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by\[A=I-a\mathbf{v}\mathbf{v}^{\trans},\]where […] Find a Matrix that Maps Given Vectors to Given VectorsSuppose that a real matrix $A$ maps each of the following vectors\[\mathbf{x}_1=\begin{bmatrix}1 \\1 \\1\end{bmatrix}, \mathbf{x}_2=\begin{bmatrix}0 \\1 \\1\end{bmatrix}, \mathbf{x}_3=\begin{bmatrix}0 \\0 \\1\end{bmatrix} \]into the […] Questions About the Trace of a MatrixLet $A=(a_{i j})$ and $B=(b_{i j})$ be $n\times n$ real matrices for some $n \in \N$. Then answer the following questions about the trace of a matrix.(a) Express $\tr(AB^{\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\trans}$ is the transpose matrix of […] Symmetric Matrices and the Product of Two MatricesLet $A$ and $B$ be $n \times n$ real symmetric matrices. Prove the followings.(a) The product $AB$ is symmetric if and only if $AB=BA$.(b) If the product $AB$ is a diagonal matrix, then $AB=BA$.Hint.A matrix $A$ is called symmetric if $A=A^{\trans}$.In […] Quiz 13 (Part 1) Diagonalize a MatrixLet\[A=\begin{bmatrix}2 & -1 & -1 \\-1 &2 &-1 \\-1 & -1 & 2\end{bmatrix}.\]Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$.That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that […]
Search Now showing items 1-9 of 9 Production of $K(892)^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$ =7 TeV (Springer, 2012-10) The production of K(892)$^0$ and $\phi$(1020) in pp collisions at $\sqrt{s}$=7 TeV was measured by the ALICE experiment at the LHC. The yields and the transverse momentum spectra $d^2 N/dydp_T$ at midrapidity \|y\|<0.5 in ... Transverse sphericity of primary charged particles in minimum bias proton-proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV (Springer, 2012-09) Measurements of the sphericity of primary charged particles in minimum bias proton--proton collisions at $\sqrt{s}$=0.9, 2.76 and 7 TeV with the ALICE detector at the LHC are presented. The observable is linearized to be ... Pion, Kaon, and Proton Production in Central Pb--Pb Collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (American Physical Society, 2012-12) In this Letter we report the first results on $\pi^\pm$, K$^\pm$, p and pbar production at mid-rapidity (\|y\|<0.5) in central Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV, measured by the ALICE experiment at the LHC. The ... Measurement of prompt J/psi and beauty hadron production cross sections at mid-rapidity in pp collisions at root s=7 TeV (Springer-verlag, 2012-11) The ALICE experiment at the LHC has studied J/ψ production at mid-rapidity in pp collisions at s√=7 TeV through its electron pair decay on a data sample corresponding to an integrated luminosity Lint = 5.6 nb−1. The fraction ... Suppression of high transverse momentum D mesons in central Pb--Pb collisions at $\sqrt{s_{NN}}=2.76$ TeV (Springer, 2012-09) The production of the prompt charm mesons $D^0$, $D^+$, $D^{*+}$, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at the LHC, at a centre-of-mass energy $\sqrt{s_{NN}}=2.76$ TeV per ... J/$\psi$ suppression at forward rapidity in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (American Physical Society, 2012) The ALICE experiment has measured the inclusive J/ψ production in Pb-Pb collisions at √sNN = 2.76 TeV down to pt = 0 in the rapidity range 2.5 < y < 4. A suppression of the inclusive J/ψ yield in Pb-Pb is observed with ... Production of muons from heavy flavour decays at forward rapidity in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV (American Physical Society, 2012) The ALICE Collaboration has measured the inclusive production of muons from heavy flavour decays at forward rapidity, 2.5 < y < 4, in pp and Pb-Pb collisions at $\sqrt {s_{NN}}$ = 2.76 TeV. The pt-differential inclusive ... Particle-yield modification in jet-like azimuthal dihadron correlations in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (American Physical Society, 2012-03) The yield of charged particles associated with high-pT trigger particles (8 < pT < 15 GeV/c) is measured with the ALICE detector in Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV relative to proton-proton collisions at the ... Measurement of the Cross Section for Electromagnetic Dissociation with Neutron Emission in Pb-Pb Collisions at √sNN = 2.76 TeV (American Physical Society, 2012-12) The first measurement of neutron emission in electromagnetic dissociation of 208Pb nuclei at the LHC is presented. The measurement is performed using the neutron Zero Degree Calorimeters of the ALICE experiment, which ...
Measurements are used to measure something. It is the estimation of ratios of quantity. It is made by comparing a quantity with a standard unit. They are used to find the size, length or amount of something. Measurement Formulas are used to find the distance, area, surface area, volume, circumference etc. They also include some conversion formulas like conversion of inch to feet, meter to miles etc. Some of the Measurement Formulas are given below: Measurement Problems Some solved problems based on measurements are given below: Solved Examples Question 1:Fine the area of a triangle with base 3 units and height 8 units? Solution: Height = 8 units Area of a Triangle= $\frac{bh}{2}$ = $\frac{3 \times 8}{2}$ = $\frac{24}{2}$ = 12 sq.units Question 2:Find the distance between two points (2, 3) and (7, 5)? Solution: 1, y 1) = (2, 3) (x 2, y 2) = (7, 5) Distance between the points= $\sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}$ = $\sqrt{(7 – 2)^{2} + (5 – 3)^{2}}$ = $\sqrt{29}$ = 5.39 units More topics in Measurement Formulas Square Footage Formula Temperature Conversion Formula Unit Rate Formula Degree and Radian Measure Formula
Disclaimer: I admit that the question is not very clear. I think it cannot be helped because the question is very open-ended. First of all, I present the interested type of circuits. We only consider this structure of Boolean / Arithmetic circuits: a DAG consisting of multiple inputs, 1 output and multiple gates. Each gate has fan-in $\le 2$ and unlimited fan-out (in principle). Also, we only consider Boolean gates (NOT, AND, OR), and Arithmetic gates (ADD, MULTIPLY). Let $f: \mathbb{B}^n \rightarrow \mathbb{B}$ be a Boolean function and $G(f)$ be the set of its corresponding Boolean circuits with size poly$(n)$. For each circuit $C \in G(f)$, we can transform it into $C' \in G(f)$ only consisting of NOT and AND gates, by using De Morgan's laws. Then, we place weights on every edge of $C'$ as follows (called Rule 0): place indeterminate $X$ on every input edge of AND gates, and place $1$ on every input edge of NOT gates. Now we can construct $f_0(C) \in \mathbb{Z}[X]$ as follows: At each input node of $C'$, assign $1 \in \mathbb{Z}[X]$ as its value. At each of other nodes of $C'$, say node $u$, compute its value $\in \mathbb{Z}[X]$: $$\text{val}(u) = \sum_{\text{edge } (v,u)} \text{val}(v)\cdot\text{weight}(v, u)$$ $f_0(C)$ is the value of the output node of $C'$. With $f_0(C)$, we have an alternative definition of circuit depth: the depth of $C$ is $\deg f_0(C)$. Therefore, in a way, the best circuits to compute $f$ are the minimum circuits in $G(f)$ w.r.t $\deg f_0(\cdot)$. In a similar manner, we can change the weighting rule into Rule 1 to get $f_1(C)$. Particularly, Rule 1 is as follows: place indeterminate $X$ on only one of the input edges of each AND gate, and place $1$ on each of other edges. Now $f_1(C)$ is the minimum-degree value of the output node of $C'$, out of all weighting scenarios. So we have a variant definition of circuit depth: the depth of $C$ is $\deg f_1(C)$. Naturally, the research question has risen: what is the best circuits to compute $f$ w.r.t $\deg f_1(\cdot)$. In other words, given $f$, what are the minimum circuits in $G(f)$ w.r.t $\deg f_1(\cdot)$? Lastly, I quickly present why there is an Arithmetic aspect of all this. One can see easily that we can transform $C'$ into Arithmetic poly$(n)$-size circuits $D_0, D_1$, both of which have 1 input (indeterminate $X$), 1 output, and $f_0(C), f_1(C)$ as their outputs, respectively. In the transformations, the input nodes of $C'$ become the constant nodes of 1.
Before I get to the core of today’s post, I would like to show the details of calculating the standard deviation. From my last post, we have two sets of data: one from Nice David and the other from Evil David. Let’s look a t Nice David’s data which are the last five test scores of his students: 83, 83, 85, 87, and 90. Now in my last post, I gave you the formula for calculating the standard deviation:\[{\mathrm{standard}}\hspace{0.33em}{\mathrm{deviation}}\hspace{0.33em}\hspace{0.33em}{=}\hspace{0.33em}\sigma{=}\hspace{0.33em}\sqrt{\frac{\sum{{(}{x}\hspace{0.33em}{-}\hspace{0.33em}\overline{x}{)}^{2}}}{n}}\] This says (much more elegantly than english) that I need to subtract the mean of the data from each data point, square each difference, add these up, divide by the number of data points, then take the square root of the result. The mean of this data is 85.6. Taking the first data point of 83, subtracting 85.6 gives -2.6. Squaring this gives 6.76. I do this to each data point. After squaring each difference, I add these up and I get 35.2. Dividing this by the number of data points (5), I get 7.04. Then finally taking the square root, I get the standard deviation of 2.65. Doing the same thing to Evil David’s data gives standard deviation of 12.09. By the way, if you know how to use Excel, these calculations are very easy to do when you have lots of data. Now this post is about one of the ways you can use the standard deviation to make a decision as to which tutor you should use. Now there is a lot of development that I will be leaving out here and I will also be making several assumptions to simplify the presentation, but the final result is still valid. I will be assuming that the data from each tutor is normally distributed, which means we can make certain statements about the standard deviations. This is not an extreme assumption as this is usually assumed in statistics. For data that is normally distributed, an interval of the mean minus one standard deviation to the mean plus one standard deviation contains 68% of the data. So for Nice David’s data, the mean minus one standard deviation is 85.6 – 2.65 = 82.95. The mean plus one standard deviation is 85.6 + 2.65 = 88.25. Now your test score will be another data point in Nice David’s data. Though you do not know what your test score will be, based on Nice David’s historical data, you can be 68% confident that your test score will be between 82.95 and 88.25. The interval of all numbers between 82.95 and 88.25 is called a confidence interval. In this case, it is a 68% confidence interval. Let’s calculate Evil David’s 68% confidence interval. Evil David’s standard deviation is 12.09, so his interval is 85.6 – 12.09 = 73.51 to 85.6 + 12.09 = 97.69. So with Evil David, you can be 68% confident that your test score will be between 73.51 am 97.69. Now most calculations in statistics center around the 95% confidence interval. For normally distributed data that we are assuming here, that is an interval that is two standard deviations about the mean. So for Nice David, the 95% interval is 85.6 – 2×2.65 = 80.3 to 85.6 +2×2.65 = 90.9. So for Nice David as your tutor, you can be 95% confident that your test score will be between 80.3 to 90.9. What about Evil David? His 95% interval is 85.6 – 2×12.09 = 61.42 to 85.6 + 2*12.09 = 109.78. Now it’s not possible to get over 100 so you can be 95% confident that with Evil David as your tutor, your test score will be between 61.42 and 100. Who do you choose? If 65 is a passing score on your test, you would be risking a failing grade with Evil David. Not so much with nice David. Though you do have a chance of getting a very high score with Evil David (if you like lotteries), your test score is 95% guaranteed to be a passing one with Nice David. Being the unbiased person that I am, I would go with the tutor with more consistent results!
> Input > Input >> 1² >> (3] >> 1%L >> L=2 >> Each 5 4 >> Each 6 7 >> L⋅R >> Each 9 4 8 > {0} >> {10} >> 12∖11 >> Output 13 Try it online! Returns a set of all possible solutions, and the empty set (i.e. \$\emptyset\$) when no solution exists. How it works Unsurprisingly, it works almost identically to most other answers: it generates a list of numbers and checks each one for inverse modulus with the argument. If you're familiar with how Whispers' program structure works, feel free to skip ahead to the horizontal line. If not: essentially, Whispers works on a line-by-line reference system, starting on the final line. Each line is classed as one of two options. Either it is a nilad line, or it is a operator line. Nilad lines start with >, such as > Input or > {0} and return the exact value represented on that line i.e > {0} returns the set \$\{0\}\$. > Input returns the next line of STDIN, evaluated if possible. Operator lines start with >>, such as >> 1² or >> (3] and denote running an operator on one or more values. Here, the numbers used do not reference those explicit numbers, instead they reference the value on that line. For example, ² is the square command (\$n \to n^2\$), so >> 1² does not return the value \$1^2\$, instead it returns the square of line 1, which, in this case, is the first input. Usually, operator lines only work using numbers as references, yet you may have noticed the lines >> L=2 and >> L⋅R. These two values, L and R, are used in conjunction with Each statements. Each statements work by taking two or three arguments, again as numerical references. The first argument (e.g. 5) is a reference to an operator line used a function, and the rest of the arguments are arrays. We then iterate the function over the array, where the L and R in the function represent the current element(s) in the arrays being iterated over. As an example: Let \$A = [1, 2, 3, 4]\$, \$B = [4, 3, 2, 1]\$ and \$f(x, y) = x + y\$. Assuming we are running the following code: > [1, 2, 3, 4] > [4, 3, 2, 1] >> L+R >> Each 3 1 2 We then get a demonstration of how Each statements work. First, when working with two arrays, we zip them to form \$C = [(1, 4), (2, 3), (3, 2), (4, 1)]\$ then map \$f(x, y)\$ over each pair, forming our final array \$D = [f(1, 4), f(2, 3), f(3, 2), f(4, 1)] = [5, 5, 5, 5]\$ Try it online! How this code works Working counter-intuitively to how Whispers works, we start from the first two lines: > Input > Input This collects our two inputs, lets say \$x\$ and \$y\$, and stores them in lines 1 and 2 respectively. We then store \$x^2\$ on line 3 and create a range \$A := [1 ... x^2]\$ on line 4. Next, we jump to the section >> 1%L >> L=2 >> Each 5 4 >> Each 6 7 The first thing executed here is line 7, >> Each 5 4, which iterates line 5 over line 4. This yields the array \$B := [i \: \% \: x \: \| \: i \in A]\$, where \$a \: \% \: b\$ is defined as the modulus of \$a\$ and \$b\$. We then execute line 8, >> Each 6 7, which iterates line 6 over \$B\$, yielding an array \$C := [(i \: \% \: x) = y \: \| \: i \in A]\$. For the inputs \$x = 5, y = 2\$, we have \$A = [1, 2, 3, ..., 23, 24, 25]\$, \$B = [0, 1, 2, 1, 0, 5, 5, ..., 5, 5]\$ and \$C = [0, 0, 1, 0, 0, ..., 0, 0]\$ We then jump down to >> L⋅R >> Each 9 4 8 which is our example of a dyadic Each statement. Here, our function is line 9 i.e >> L⋅R and our two arrays are \$A\$ and \$C\$. We multiply each element in \$A\$ with it's corresponding element in \$C\$, which yields an array, \$E\$, where each element works from the following relationship: $$E_i =\begin{cases}0 & C_i = 0 \\A_i & C_i = 1\end{cases}$$ We then end up with an array consisting of \$0\$s and the inverse moduli of \$x\$ and \$y\$. In order to remove the \$0\$s, we convert this array to a set ( >> {10}), then take the set difference between this set and \$\{0\}\$, yielding, then outputting, our final result.
There are many examples where the notation of $\partial{U}$ is used to represent the surface boundary of some volume $U$. For example, the following representation of the Divergence Theorem: $$ \int_U{\nabla \cdot \mathbf{F}}\,dV = \oint_{\partial{U}}\,\mathbf{F}\cdot\mathbf{n}\,dS $$ What is the history and reasoning for the notation of using what appears to be a partial derivative of the volume for the surface. I am guessing that the derivative represents the tangent at the surface and thus refers to surface boundary but is there more then that? There seem to be two different rationales, one for how it first appeared, and the other for why it stuck, see MO thread. Francois Ziegler dug up the following remark in Michèle Audin's Henri Cartan & André Weil: du vingtième siècle et de la topologie $d$ is for differential, no doubt, but what is $\partial$ for? Neither boundary nor frontier, in neither French nor German (nor even English), so? well, Rand, in German, written in gothic and abbreviated $\mathfrak{Rd}$, then $\mathfrak d$ (two letters, that was too much), then $\partial$! He also found the corresponding use in §16 of Seifert and Threlfall's Lehrbuch der Topologie (1934):" Der Rand wird mit dem Symbole $\mathscr R\partial$ bezeichnet" [the boundary is denoted by the symbol $\mathscr R\partial$]. It seems unlikely that this convoluted justification would have impressed many in adopting the symbol. Fortunately, it has other things to recommend itself. E.g. it dovetails nicely with the dual use of $d$ in cohomology and calculus on manifolds, e.g. in the Stokes formula, $\int_M d\omega = \int_{\partial M} \omega$, of which all integral formulas of vector analysis, like the OP one, are particular cases. And it has some derivative like properties: $\partial(A\cap B)=(\partial A\cap B)\cup(A\cap\partial B)$ and $\partial(A\times B) =(\partial A\times B) \cup (A \times \partial B)$.
Is the electric field between two positively charged parallel infinite plates one with a charge density twice the other effect the electric field on the outside of the plates? I am thinking no, because the field lines cannot cross in the field between the plates since they have positive charges. What does the electric field in the middle look like? How would I draw electric field lines to geometrically interpret this situation? If I have a Gaussian cylinder between the plates will the field go out the walls of the cylinder like in the case of two positive point charges within a Gaussian cylinder? Yes the additional positive charge to one plate increases electric field behind the other one. To see this let's look at general Gauss's law: $$\oint_{S}\vec{E} \,d\vec{A}=Q/\epsilon_0$$ The $S$ is the Gaussian surface which we are free to choose therefore it could include both plates even if the surface goes to infinity with both plates. The charge Q however could be written as charge density $\sigma$ times surface area or in general: $$Q = \int_A \sigma \, dA $$ where $\sigma$ could be $\sigma_1$, $\sigma_2$ or $\sigma_1 + \sigma_2$. Now we are ready to use symmetry of configuration. Lets assume that field is homogeneous at the middle of of both infinities and rewrite the left side of Gauss law as: $$\oint_{S} \vec{E}\, d\vec{A} = \oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} $$ The right side with the same division: $$\int_A = \int_{in} +\int_{out}$$ Now we are ready for the main part: $$\oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} =\int_{in}\sigma \, dA + \int_{out}\sigma \, dA$$ And the Gauss law says that charges which is outside from closed surface does not affect the flux or $\vec{E} = \vec{E}_{in} + \vec{E}_{out}$ and the $\oint_{out} \vec{E}_{in} \, dA = 0 $ which makes to satisfy: $$\oint_{S_{out}} \vec{E}\, d\vec{A} = \int_{out}\sigma \, dA$$ And the only thing which is left is always satisfied: $$\oint_{S_{in}} \vec{E}\, d\vec{A} = \int_{in}\sigma \, dA$$ which for homogeneous field means: $E_{top} + E_{bottom} = \sigma / \epsilon_0$ where the signs for field is positive if it is directed out of surface.
ISSN: 1937-5093 eISSN: 1937-5077 Kinetic & Related Models September 2016 , Volume 9 , Issue 3 Issue in memory of Alfredo Lorenzi Select all articles Export/Reference: Abstract: In this paper, we establish a new blowup criterion for the strong solutions in a smooth bounded domain $\Omega\subset\mathbb{R}^3$. In [13], Wen, Yao, and Zhu prove that if the strong solutions blow up at finite time $T^$, the mass in $L^\infty(\Omega)$ norm must concentrate at $T^$. Here we extend Wen, Yao, and Zhu's work in the sense of the concentration of mass in $BMO(\Omega)$ norm at $T^*$. The method can be applied to study the blow-up criterion in terms of the concentration of density in $BMO(\Omega)$ norm for the strong solutions to compressible Navier-Stokes equations in smooth bounded domains. Therefore, as a byproduct, we can also improves the corresponding result about Navier-Stokes equations in [11]. Moreover, the appearance of vacuum is allowed in the paper. Abstract: In this paper we establish the uniform estimates of strong solutions with respect to the Mach number and the dielectric constant to the full compressible Navier-Stokes-Maxwell system in a bounded domain. Based on these uniform estimates, we obtain the convergence of the full compressible Navier-Stokes-Maxwell system to the incompressible magnetohydrodynamic equations for well-prepared data. Abstract: The motion of a collisionless plasma - a high-temperature, low-density, ionized gas - is described by the Vlasov-Maxwell (VM) system. These equations are considered in one space dimension and two momentum dimensions without the assumption of relativistic velocity corrections. The main results are bounds on the spatial and velocity supports of the particle distribution function and uniform estimates on derivatives of this function away from the critical velocity $\vert v_1 \vert = 1$. Additionally, for initial particle distributions that are even in the second velocity argument $v_2$, the global-in-time existence of solutions is shown. Abstract: We study the nonlinear stability of rarefaction waves to the Cauchy problem of one-dimensional compressible Navier-Stokes equations for a viscous and heat conducting ideal polytropic gas when the transport coefficients depend on both temperature and density. When the strength of the rarefaction waves is small or the rarefaction waves of different families are separated far enough initially, we show that rarefaction waves are nonlinear stable provided that $(\gamma- 1)\cdot H^3(\mathbb{R})$-norm of the initial perturbation is suitably small with $\gamma>1$ being the adiabatic gas constant. Abstract: We are concerned with the Cauchy problem of the relativistic Vlasov-Maxwell-Boltzmann system for short range interaction. For perturbative initial data with suitable regularity and integrability, we prove the large time stability of solutions to the relativistic Vlasov-Maxwell-Boltzmann system, and also obtain as a byproduct the convergence rates of solutions. Our proof is based on a new time-velocity weighted energy method and some optimal temporal decay estimates on the solution itself. The results also extend the case of ``hard ball" model considered by Guo and Strain [Comm. Math. Phys. 310: 49--673 (2012)] to the short range interactions. Abstract: This paper deals with the Cauchy problem for tropical climate model with the fractional velocity diffusion which was derived by Frierson-Majda-Pauluis in [16]. We establish the local well-posedness of strong solutions to this generalized model. Abstract: We propose a kinetic relaxation-model to describe a generation-recombination reaction of two species. The decay to equilibrium is studied by two recent methods [9,13] for proving hypocoercivity of the linearized equations. Exponential decay of small perturbations can be shown for the full nonlinear problem. The macroscopic/fast-reaction limit is derived rigorously employing entropy decay, resulting in a nonlinear diffusion equation for the difference of the position densities. Abstract: The main concern of this paper is to analyze a boundary layer called a sheath that occurs on the surface of materials when in contact with a multicomponent plasma. For the formation of a sheath, the generalized Bohm criterion demands that ions enter the sheath region with a high velocity. The motion of a multicomponent plasma is governed by the Euler--Poisson equations, and a sheath is understood as a monotone stationary solution to those equations. In this paper, we prove the unique existence of the monotone stationary solution by assuming the generalized Bohm criterion. Moreover, it is shown that the stationary solution is time asymptotically stable provided that an initial perturbation is sufficiently small in weighted Sobolev space. We also obtain the convergence rate, which is subject to the decay rate of the initial perturbation, of the time global solution toward the stationary solution. Abstract: The ellipsoidal BGK model (ES-BGK) is a generalized version of the original BGK model, designed to yield the correct Prandtl number in the Navier-Stokes limit. In this paper, we make two observations on the entropy production functional of the ES-BGK model. First, we show that the Cercignani type estimate holds for the ES-BGK model in the whole range of relaxation parameter $-1/2<\nu<1$. Secondly, we observe that the ellipsoidal relaxation operator satisfies an unexpected sign-definite property. Some implications of these observations are also discussed. Readers Authors Editors Referees Librarians Email Alert Add your name and e-mail address to receive news of forthcoming issues of this journal: [Back to Top]
Running your code, it seems like your pulse looks kinda like this: (sorry for not adding units to the plots, I used the same as you, i.e. t is in fs and w in rad/fs) So, the FWHM is not correct (should be 4 fs) and the angular carrier/central frequency is messed up (should be ca. 2.3 / ca. 0.37 per fs). I think there is two things not quite right here: 1) the definition of your w-axis 2) the conversion of frequency to time As by definition of the FFT, where there is already a $2\pi$ in the exponents, we get$$ \sigma_t = \frac{1}{2\sigma_w} $$where $\sigma_t$ and $\sigma_w$ are the standard deviations of the intensity in t- and w-space respectively. Now, according to wikipedia and with $\Delta t$ and $\Delta w$ as the FWHM in t- and w- space respectively$$ \text{fac} \cdot \sigma_t = \Delta t $$and respectively$$ \text{fac} \cdot \sigma_w = \Delta w $$with $\text{fac} = \sqrt{8\cdot \text{log}2}$. Plugging all these in you get$$ \Delta w = \text{fac}^2 \cdot \frac{1}{2\Delta t} \approx 2\pi \cdot 0.441 \cdot \frac{1}{\Delta t}$$as you did correctly. However, in laser physics, people use the FWHM of the intensity of the field, rather than the FWHM of the amplitude. This is why we use $\sigma_t \sigma_w = \frac{1}{2}$. The $\frac{1}{2}$ originates from the fact that the intensity is the amplitude squared. This is a small but important detail, as we have to pay attention to put factors of $2$ and $\sqrt{2}$ in the appropriate places. Now for 1), the FFT-implementation is going to consider the spectrum to start at w=0. The way you defined the w-axis, what the algorithm saw in E_w, was a Gaussian centered half the way up your w-axis, because you had the axis symmetrically spread around your peak. That means, after the FFT, your central frequency was significantly higher than it should have been. This is fixed by letting w start at zero and go to some high enough value (you chose 16 bandwidths, so I stuck with that too): #angular frequencies [rad/fs] w = np.linspace(0, delta_w16, 29) As above, for the Gaussian you need the $\sigma_w$ from the FWHM. Don't forget a factor $\sqrt{2}$ because of the whole intensity-amplitude-issue: sigma_w = delta_w/fac # standard deviation of intensity sigma_w_field = np.sqrt(2)sigma_w # standard deviation of amplitude #to make the spectrum centered around the carrier frequency diff_w = w-w0 I removed the phase function, as it didn't seem to do anything here and also make sure to use sigma_w_field: spectrum_w = np.exp(-(diff_w*2)/(2sigma_w_field*2)) #field in the frequency domain E_w = spectrum_w plt.figure() plt.xlabel("w") plt.ylabel("E_w") plt.plot(w, E_w) plt.grid() plt.show() Technically, the above plot is not 100% correct, because the left shoulder of the Gaussian might reappear in the high end of the spectrum. This is not an issue here, because at w=0 the Gaussian has decayed enough, but if you choose a wider band width or a lower central frequency, you should take care of that somehow (I have no idea how to do this elegantly though). This fixes Item 1). For Item 2), let's look at the variable timestep. timestep is supposed to be the inverse of the sampling frequency, however, I don't think 0.01 was the correct value for that. The inverse of the sampling frequency, i.e. the sampling time, is the length of your signal in f-space divided by the number of samples. Here this is 16 bandwidths over $2^9$ (here called w_s). This is however in w-space and fftfreq wants f-space, so one division by $2\pi$ is requiered. And again, delta_w was computed for the intensity, so we need another $\sqrt{2}$ to turn it into the band-width of the spectrum: #FT: n = len(E_w) w_s = np.sqrt(2)16delta_w/29 timestep = w_s/(2np.pi) #fa = 1.0/timestep t_1 = np.fft.fftfreq(n,d = timestep) t = np.fft.fftshift(t_1) field_ft = np.fft.ifft(E_w) new = np.fft.fftshift(field_ft) plt.figure() plt.xlabel("t") plt.ylabel("field") plt.xlim(-10,10) plt.plot(t,new, label="new") plt.plot(t,np.abs(new), label="abs(new)") plt.plot(t, np.ones(len(new))0.5np.max(np.abs(new)), label="half maximum") plt.grid() plt.legend() plt.show() The carrier-wave-plot can be improved by making the range of w larger.I hope this is the pulse you were looking for ;)
An $n\times n$ matrix $A$ is said to be idempotent if $A^2=A$.It is also called projective matrix. Proof. In general, an $n \times n$ matrix $B$ is diagonalizable if there are $n$ linearly independent eigenvectors. So if eigenvectors of $B$ span $\R^n$, then $B$ is diagonalizable. We prove that $\R^n$ is spanned by eigenspaces. Every vector $\mathbf{v}\in \R^n$ can be expresses as\[\mathbf{v}=(\mathbf{v}-A\mathbf{v})+A\mathbf{v}=\mathbf{v}_0+\mathbf{v}_1,\]where we put $\mathbf{v}_0=\mathbf{v}-A\mathbf{v}$ and $\mathbf{v}_1=A\mathbf{v}$. We claim that $\mathbf{v}_0$ and $\mathbf{v}_1$ are elements in the eigenspaces corresponding to (possible) eigenvalues $0$ and $1$, respectively.To see this, we compute\begin{align}A\mathbf{v}_0&=A(\mathbf{v}-A\mathbf{v})\\&=A\mathbf{v}-A^2\mathbf{v}\\&=A\mathbf{v}-A\mathbf{v} && \text{since $A$ is idempotent}\\&=O=0\mathbf{v}_0.\end{align} Thus, we have $A\mathbf{v}_0=0\mathbf{v}_0$, and this means that $\mathbf{v}_0$ is a vector in the eigenspace corresponding to the eigenvalue $0$.(If $0$ is not an eigenvalue of $A$, then $\mathbf{v}_0=\mathbf{0}$.) We also have\begin{align}A\mathbf{v}_1=A(A\mathbf{v})=A^2\mathbf{v}=A\mathbf{v}=\mathbf{v}_1,\end{align}where the third equality holds as $A$ is idempotent.This implies that $\mathbf{v}_1$ is a vector in the eigenspace corresponding to eigenvalue $1$. (If $1$ is not an eigenvalue of $A$, then $\mathbf{v}_1=\mathbf{0}$.) It follows that every vector $\mathbf{v}\in \R^n$ is a sum of eigenvectors (or the zero vector).That is, $\R^n$ is spanned by eigenvectors. By the general fact mentioned at the beginning of the proof, we conclude that the idempotent matrix $A$ is diagonalizable. Idempotent Matrices are DiagonalizableLet $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.We give three proofs of this problem. The first one proves that $\R^n$ is a direct sum of eigenspaces of $A$, hence $A$ is diagonalizable.The second proof proves […] Unit Vectors and Idempotent MatricesA square matrix $A$ is called idempotent if $A^2=A$.(a) Let $\mathbf{u}$ be a vector in $\R^n$ with length $1$.Define the matrix $P$ to be $P=\mathbf{u}\mathbf{u}^{\trans}$.Prove that $P$ is an idempotent matrix.(b) Suppose that $\mathbf{u}$ and $\mathbf{v}$ be […] Maximize the Dimension of the Null Space of $A-aI$Let\[ A=\begin{bmatrix}5 & 2 & -1 \\2 &2 &2 \\-1 & 2 & 5\end{bmatrix}.\]Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.Your score of this problem is equal to that […] Idempotent Matrix and its EigenvaluesLet $A$ be an $n \times n$ matrix. We say that $A$ is idempotent if $A^2=A$.(a) Find a nonzero, nonidentity idempotent matrix.(b) Show that eigenvalues of an idempotent matrix $A$ is either $0$ or $1$.(The Ohio State University, Linear Algebra Final Exam […] Quiz 13 (Part 1) Diagonalize a MatrixLet\[A=\begin{bmatrix}2 & -1 & -1 \\-1 &2 &-1 \\-1 & -1 & 2\end{bmatrix}.\]Determine whether the matrix $A$ is diagonalizable. If it is diagonalizable, then diagonalize $A$.That is, find a nonsingular matrix $S$ and a diagonal matrix $D$ such that […] How to Diagonalize a Matrix. Step by Step Explanation.In this post, we explain how to diagonalize a matrix if it is diagonalizable.As an example, we solve the following problem.Diagonalize the matrix\[A=\begin{bmatrix}4 & -3 & -3 \\3 &-2 &-3 \\-1 & 1 & 2\end{bmatrix}\]by finding a nonsingular […]
ISSN: 1937-5093 eISSN: 1937-5077 Kinetic & Related Models December 2016 , Volume 9 , Issue 4 Issue on Nonautonomous Dynamics Select all articles Export/Reference: Abstract: The mathematical description of the interaction between a plasma and a solid surface is a major issue that still remains challenging. In this paper, we model this interaction as a stationary and bi-kinetic Vlasov-Poisson-Ampère boundary value problem with boundary conditions that are consistent with the physics. In particular, we show that the wall potential can be determined from the ampibolarity of the particle flows as the unique solution of a non linear equation. Based on variational techniques, our analysis establishes the well-posedness of the model, provided that the incoming ion distribution satisfies a moment condition that generalizes the historical Bohm criterion of plasma physics. Quantitative estimates are also given, together with numerical illustrations that validate the robustness of our approach. Abstract: We study the time evolution of a single species positive plasma, confined in a cylinder and having infinite charge. We extend the result of a previous work by the same authors, for a plasma density having compact support in the velocities, to the case of a density having unbounded support and gaussian decay in the velocities. Abstract: This paper investigates the existence of a uniform in time $L^{\infty}$ bounded weak solution for the $p$-Laplacian Keller-Segel system with the supercritical diffusion exponent $1 < p < \frac{3d}{d+1}$ in the multi-dimensional space ${\mathbb{R}}^d$ under the condition that the $L^{\frac{d(3-p)}{p}}$ norm of initial data is smaller than a universal constant. We also prove the local existence of weak solutions and a blow-up criterion for general $L^1\cap L^{\infty}$ initial data. Abstract: This paper investigates the generalized Keller-Segel (KS) system with a nonlocal diffusion term $-\nu(-\Delta)^{\frac{\alpha}{2}}\rho~(1<\alpha<2)$. Firstly, the global existence of weak solutions is proved for the initial density $\rho_0\in L^1\cap L^{\frac{d}{\alpha}}(\mathbb{R}^d)~(d\geq2)$ with $\\|\rho_0\\|_{\frac {d}{\alpha}} < K$, where $K$ is a universal constant only depending on $d,\alpha,\nu$. Moreover, the conservation of mass holds true and the weak solution satisfies some hyper-contractive and decay estimates in $L^r$ for any $1< r<\infty$. Secondly, for the more general initial data $\rho_0\in L^1\cap L^2(\mathbb{R}^d)$$~(d=2,3)$, the local existence is obtained. Thirdly, for $\rho_0\in L^1\big(\mathbb{R}^d,(1+\|x\|)dx\big)\cap L^\infty(\mathbb{R}^d)(~d\geq2)$ with $\\|\rho_0\\|_{\frac{d}{\alpha}} < K$, we prove the uniqueness and stability of weak solutions under Wasserstein metric through the method of associating the KS equation with a self-consistent stochastic process driven by the rotationally invariant $\alpha$-stable Lévy process $L_{\alpha}(t)$. Also, we prove the weak solution is $L^\infty$ bounded uniformly in time. Lastly, we consider the $N$-particle interacting system with the Lévy process $L_{\alpha}(t)$ and the Newtonian potential aggregation and prove that the expectation of collision time between particles is below a universal constant if the moment $\int_{\mathbb{R}^d}\|x\|^\gamma\rho_0dx$ for some $1<\gamma<\alpha$ is below a universal constant $K_\gamma$ and $\nu$ is also below a universal constant. Meanwhile, we prove the propagation of chaos as $N\rightarrow\infty$ for the interacting particle system with a cut-off parameter $\varepsilon\sim(\ln N)^{-\frac{1}{d}}$, and show that the mean field limit equation is exactly the generalized KS equation. Abstract: We study a modified Kac model where the classical kinetic energy is replaced by an arbitrary energy function $\phi(v)$, $v \in \mathbb{R}$. The aim of this paper is to show that the uniform density with respect to the microcanonical measure is $Ce^{-z_0\phi(v)}$-chaotic, $C,z_0 \in \mathbb{R}_+$. The kinetic energy for relativistic particles is a special case. A generalization to the case $v\in \mathbb{R}^d$ which involves conservation momentum is also formally discussed. Abstract: We consider the global existence of the two-dimensional Navier-Stokes flow in the exterior of a moving or rotating obstacle. Bogovski$\check{i}$ operator on a subset of $\mathbb{R}^2$ is used in this paper. One important thing is to show that the solution of the equations does not blow up in finite time in the sense of some $L^2$ norm. We also obtain the global existence for the 2D Navier-Stokes equations with linearly growing initial velocity. Abstract: We establish a blow up criterion for the two-dimensional $k$-$\varepsilon$ model equations for turbulent flows in a bounded smooth domain $\Omega$. It is shown that for the initial-boundary value problem of the 2D $k$-$\varepsilon$ model equations in a bounded smooth domain, if $\\|\nabla u\\|_{L^{1}(0, T; L^{\infty})}+\\|\nabla\rho\\|_{L^{2}(0, T; L^{\infty})} +\\|\varepsilon\\|_{L^{2}(0, T; L^{\infty})}$ $<\infty$, then the strong solution $(\rho, u, h,k, \varepsilon)$ can be extended beyond $T$. Abstract: The article is retracted by a unanimous decision by the three Editors-in-chief of KRM on the ground that almost the same article was submitted and got quickly published in Boundary Value Problems (DOI 10.1186/s13661-015-0481-7) by the same authors right before their submission to KRM. Abstract: N/A Readers Authors Editors Referees Librarians Email Alert Add your name and e-mail address to receive news of forthcoming issues of this journal: [Back to Top]
I want to show by a double-counting argument that$${{n+m+1}\brace m}=\sum_{k=0}^mk{{n+k}\brace k}$$for $m,n\in\Bbb N$ (where $0\in\Bbb N$). Note that ${{n}\brace k}$ is a Stirling number of the second kind (i.e. the number of partitions of $[n]:=\{1,2,\ldots,n\}$ into $k$ blocks). I am just starting to do double-counting proofs, so I just want to make sure I am doing this right. I would like some feedback on my proof: whether it is correct or not, what are some improvements I could make, etc. Here is my proof: We know that ${{n+m+1}\brace m}$ is the number of ways to partition $[n+m+1]$ into $m$ blocks. Now we focus on the $n+k+1$ element of $[n+m+1]$ for some $0\leq k\leq m$. Every element past $n+k+1$ we put in its own block in the partition, which accounts for $(n+m+1)-(n+k+1)=m-k$ of the blocks. We partition the first $n+k$ elements into the remaining $k$ blocks in ${{n}\brace k}$ ways. Then we stick the element $n+k+1$ into one of these $k$ blocks (which can be done in $k$ ways). Hence, there are $k{{n+k}\brace k}$ ways to partition the set with $m$ blocks with our choice of $k$. We sum over all possibilities of $k$ to account for all choices of the $n+k+1$ element to obtain ${{n+m+1}\brace m}=\sum_{k=0}^mk{{n+k}\brace k}$ total partitions of $[n+m+1]$ into $m$ blocks. Thanks in advance for any feedback. If you have another way to prove this I'd love to see that as well.
Suppose that a wooden cube, whose edge is $3$ inch, is painted red, then cut into $27$ pieces of $1$ inch edge. Find total surface area of unpainted? First of all, I have tried to draw the cube using MS Paint, below is given picture: The surface area of cube is $6\cdot a^2$, where $a$ is the length of cube, when cube is painted, I'm trying to imagine it visually, how much surface would be unpainted? If it's painted,does not it mean that all pages are painted? EDIT: A $3\times3\times3$ cube gets painted red. Then it gets split into $27$, $1\times1\times1$ cubes. Find the number of painted faces: On the $3$ inch cube, there are $9$ $1$-inch faces on each face, and there are $6$ such faces, so the total red 1-inch faces is: $$ 9\cdot 6=54 $$ Find the total number of faces of all the cubes: There are $6$ faces per cube and $27$ cubes, so: $$ 27\cdot 6=162. $$ Of these, we know $54$ are painted: $$ 162-54=108 $$ There is a total of $108$ unpainted $1\times1$ squares each having an area of 1, so there are $108$ unpainted square inches. EDIT: I'd like to post also my approaches if the cube would be divided into two identical parts, actually if we have a cube with length $a$, divided into $2$ parts, then we get two cubes, which volume is $a^3/2$. So length is the cubic root of $a^3/2$, or length of this two cubes is $a$ divided by cubic root of $2$. Now we have length $3$, which means that it's length would be $3$ divided by cubic root of $2$, on each face with length $3$, we would have: $$ \frac{9}{3/\sqrt[3]{2}}=3\cdot \sqrt[3]{2} $$ in total $4$ such cubes, so $24$ of such faces, total face would be $26=12$,so unpainted would be $12-8=4$ is it correct?
Let $A \in Mat_{n,n}(\mathbb C)$. Let $g_A(t)$ denote $\exp(tA)$ and $\exp(A)$ denote the matrix $g_A(1)$. Let $\alpha \in \mathbb R$ and $B = \alpha A$. I've shown $\exp(B) = g_A(\alpha)$ by just substituting $B$ with $\alpha A$ (is this correct?). I've shown that $\exp(A)$ is invertibel by considering $\exp(-tA)$ and considering the derivative of product $\exp(-tA) \exp(tA)$ which is constant equal to $I_n$ (an easier way to do this?). I've shown that $\exp(0) = I_n$ by using $A = 0$ and $\alpha = 0$ and we know that $\exp(0A) = I_n$ by definition. Question: However I cannot show that $\exp(A)$ is a real matrix if $A$ is a real matrix ? (I should relate the question to solutions of linear differential equations). Now don't assume $A$ is real. Also if $C$ is a matrix that commutes with $A$, that is $AC = CA$, I can show that $C$ commutes with $\exp(A)$ by using Putzer's algorithm. But how can I show $\exp(A) \exp(B) = \exp(A+B)$ ? (Again I should relate the question to solutions of linear differential equations).
Tagged: determinant Problem 548 An $n\times n$ matrix $A$ is said to be invertible if there exists an $n\times n$ matrix $B$ such that $AB=I$, and $BA=I$, where $I$ is the $n\times n$ identity matrix. If such a matrix $B$ exists, then it is known to be unique and called the inverse matrix of $A$, denoted by $A^{-1}$. In this problem, we prove that if $B$ satisfies the first condition, then it automatically satisfies the second condition. So if we know $AB=I$, then we can conclude that $B=A^{-1}$. Let $A$ and $B$ be $n\times n$ matrices. Suppose that we have $AB=I$, where $I$ is the $n \times n$ identity matrix. Prove that $BA=I$, and hence $A^{-1}=B$.Add to solve later Problem 452 Let $A$ be an $n\times n$ complex matrix. Let $S$ be an invertible matrix. (a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix. (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$. Add to solve later (c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$. Problem 438 Determine whether each of the following statements is True or False. (a) If $A$ and $B$ are $n \times n$ matrices, and $P$ is an invertible $n \times n$ matrix such that $A=PBP^{-1}$, then $\det(A)=\det(B)$. (b) If the characteristic polynomial of an $n \times n$ matrix $A$ is \[p(\lambda)=(\lambda-1)^n+2,\] then $A$ is invertible. (c) If $A^2$ is an invertible $n\times n$ matrix, then $A^3$ is also invertible. (d) If $A$ is a $3\times 3$ matrix such that $\det(A)=7$, then $\det(2A^{\trans}A^{-1})=2$. (e) If $\mathbf{v}$ is an eigenvector of an $n \times n$ matrix $A$ with corresponding eigenvalue $\lambda_1$, and if $\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_2$, then $\mathbf{v}+\mathbf{w}$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda_1+\lambda_2$. ( Stanford University, Linear Algebra Exam Problem) Read solution Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue Problem 419 (a) Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$. Add to solve later (b) Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. Problem 391 (a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$? (b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}$? (c) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 0& 2 \end{bmatrix}$? Add to solve later (d) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ -1& 4 \end{bmatrix}$? Problem 389 (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$. (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$. (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$? ( Harvard University, Linear Algebra Exam Problem) Problem 374 Let \[A=\begin{bmatrix} a_0 & a_1 & \dots & a_{n-2} &a_{n-1} \\ a_{n-1} & a_0 & \dots & a_{n-3} & a_{n-2} \\ a_{n-2} & a_{n-1} & \dots & a_{n-4} & a_{n-3} \\ \vdots & \vdots & \dots & \vdots & \vdots \\ a_{2} & a_3 & \dots & a_{0} & a_{1}\\ a_{1} & a_2 & \dots & a_{n-1} & a_{0} \end{bmatrix}\] be a complex $n \times n$ matrix. Such a matrix is called circulant matrix. Then prove that the determinant of the circulant matrix $A$ is given by \[\det(A)=\prod_{k=0}^{n-1}(a_0+a_1\zeta^k+a_2 \zeta^{2k}+\cdots+a_{n-1}\zeta^{k(n-1)}),\] where $\zeta=e^{2 \pi i/n}$ is a primitive $n$-th root of unity. Problem 363 (a) Find all the eigenvalues and eigenvectors of the matrix \[A=\begin{bmatrix} 3 & -2\\ 6& -4 \end{bmatrix}.\] Add to solve later (b) Let \[A=\begin{bmatrix} 1 & 0 & 3 \\ 4 &5 &6 \\ 7 & 0 & 9 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 4 \end{bmatrix}.\] Then find the value of \[\det(A^2B^{-1}A^{-2}B^2).\] (For part (b) without computation, you may assume that $A$ and $B$ are invertible matrices.) Problem 338 Each of the following sets are not a subspace of the specified vector space. For each set, give a reason why it is not a subspace. (1) \[S_1=\left \{\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \R^3 \quad \middle \| \quad x_1\geq 0 \,\right \}\] in the vector space $\R^3$. (2)\[S_2=\left \{\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \in \R^3 \quad \middle \| \quad x_1-4x_2+5x_3=2 \,\right \}\] in the vector space $\R^3$. (3)\[S_3=\left \{\, \begin{bmatrix} x \\ y \end{bmatrix}\in \R^2 \quad \middle \| \quad y=x^2 \quad \,\right \}\] in the vector space $\R^2$. (4)Let $P_4$ be the vector space of all polynomials of degree $4$ or less with real coefficients. \[S_4=\{ f(x)\in P_4 \mid f(1) \text{ is an integer}\}\] in the vector space $P_4$. (5)\[S_5=\{ f(x)\in P_4 \mid f(1) \text{ is a rational number}\}\] in the vector space $P_4$. (6)Let $M_{2 \times 2}$ be the vector space of all $2\times 2$ real matrices. \[S_6=\{ A\in M_{2\times 2} \mid \det(A) \neq 0\} \] in the vector space $M_{2\times 2}$. (7)\[S_7=\{ A\in M_{2\times 2} \mid \det(A)=0\} \] in the vector space $M_{2\times 2}$. ( Linear Algebra Exam Problem, the Ohio State University) (8)Let $C[-1, 1]$ be the vector space of all real continuous functions defined on the interval $[a, b]$. \[S_8=\{ f(x)\in C[-2,2] \mid f(-1)f(1)=0\} \] in the vector space $C[-2, 2]$. (9)\[S_9=\{ f(x) \in C[-1, 1] \mid f(x)\geq 0 \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$. (10)Let $C^2[a, b]$ be the vector space of all real-valued functions $f(x)$ defined on $[a, b]$, where $f(x), f'(x)$, and $f^{\prime\prime}(x)$ are continuous on $[a, b]$. Here $f'(x), f^{\prime\prime}(x)$ are the first and second derivative of $f(x)$. \[S_{10}=\{ f(x) \in C^2[-1, 1] \mid f^{\prime\prime}(x)+f(x)=\sin(x) \text{ for all } -1\leq x \leq 1\}\] in the vector space $C[-1, 1]$. (11)Let $S_{11}$ be the set of real polynomials of degree exactly $k$, where $k \geq 1$ is an integer, in the vector space $P_k$. (12)Let $V$ be a vector space and $W \subset V$ a vector subspace. Define the subset $S_{12}$ to be the complementof $W$, \[ V \setminus W = \{ \mathbf{v} \in V \mid \mathbf{v} \not\in W \}.\] Add to solve later
Cryptology ePrint Archive: Report 2016/717 Comparison between Subfield and Straightforward Attacks on NTRU Paul Kirchner and Pierre-Alain Fouque Abstract: Recently in two independent papers, Albrecht, Bai and Ducas and Cheon, Jeong and Lee presented two very similar attacks, that allow to break NTRU with larger parameters and GGH Multinear Map without zero encodings. They proposed an algorithm for recovering the NTRU secret key given the public key which apply for large NTRU modulus, in particular to Fully Homomorphic Encryption schemes based on NTRU. Hopefully, these attacks do not endanger the security of the NTRUE NCRYPT scheme, but shed new light on the hardness of this problem. The basic idea of both attacks relies on decreasing the dimension of the NTRU lattice using the multiplication matrix by the norm (resp. trace) of the public key in some subfield instead of the public key itself. Since the dimension of the subfield is smaller, the dimension of the lattice decreases, and lattice reduction algorithm will perform better. Here, we revisit the attacks on NTRU and propose another variant that is simpler and outperforms both of these attacks in practice. It allows to break several concrete instances of YASHE, a NTRU-based FHE scheme, but it is not as efficient as the hybrid method of Howgrave-Graham on concrete parameters of NTRU. Instead of using the norm and trace, we propose to use the multiplication by the public key in some subring and show that this choice leads to better attacks. We √ can then show that for power of two cyclotomic fields, the time complexity is polynomialFinally, we show that, under heuristics, straightforward lattice reduction is even more efficient, allowing to extend this result to fields without non-trivial subfields, such as NTRU Prime. We insist that the improvement on the analysis applies even for relatively small modulus ; though if the secret is sparse, it may not be the fastest attack. We also derive a tight estimation of security for (Ring-)LWE and NTRU assumptions. when $q=2^{\Omega(\sqrt{n \log \log n})}$. Category / Keywords: public-key cryptography / cryptanalysis,lattice techniques,number theory ,post quantum cryptography,NTRU Date: received 19 Jul 2016 Contact author: paul kirchner at ens fr Available format(s): PDF \| BibTeX Citation Version: 20160721:150011 (All versions of this report) Short URL: ia.cr/2016/717 [ Cryptology ePrint archive ]
I've deduced simple relationships that satisfy each even perfecf number (even numbers $n$ for which $\sum_{d\mid n}d=2n$) and now I wondered about related conjectures. For each integer $m\geq 1$ we denote the sum of divisors function $\sum_{d\mid m}d$ as $\sigma(m)$, and the Euler's totient function as $\varphi(m)$. Claim 1. It's easy to prove that each even perfect number $n$ satisfies $$\sigma(n)=\frac{1}{2}\left(1+8\varphi(n)+\sqrt{1+8n}\right).\tag{1}$$ Claim 2. Thus each even perfect number satisfies also $$4n=1+8\varphi(n)+\sqrt{1+8n}.\tag{2}$$ Question.I would like to know what work can be done about the following conjectures (prove it or provide us what calculations/reasonings can be done, or refute these finding a counterexample): C1) If an integer$m\geq 1$ satifies$(1)$, then$m$ is an even perfect number. C2) Similarly to previous conjecture, each integer$l\geq 1$ satisfying$(2)$ is an even perfect number. Many thanks. I've tested that seems there are no counterexamples in my experiments, say us $\leq 10^5$. I don't know if these equations are in the literature, thus answer as a reference request this question adding the articles where there are information about these equations and I try to search and read those from the literature.
Let $f:\mathbb Z^3\to\mathbb Z^4$ be the group homomorphism given by $$f(a,b,c)=(a+b+c,a+3b+c,a+b+5c,4a+8b).$$ Let $H$ be the image of $f$. Find an element of infinite order in $\mathbb Z^4 /H$ and find the order of the torsion subgroup of that quotient. This problem certainly has to do with presentation matrices, but I'm too confused. Let $A$ denote the matrix with rows $(1,1,1),(1,3,1),(1,1,5),(4,8,0)$. Then $A\mathbb Z^3=H$. So $A$ must be a presentation matrix for $\mathbb Z^4/H$ according to the definition from here (or from here, both of them agree with Artin's definition). So in the quotient module the relations $$v_1+v_2+v_3=0,\\v_1+3v_2+v_3=0,\\v_1+v_2+5v_3=0,\\ 4v_1+8v_2=0$$ hold. But then according to Artin's text, the matrix should be $3\times 4$, not $4\times 3$ (see Example 14.5.2 in this question). So what is the size of a presentation matrix in this case? My size doesn't agree with Artin's notation, even though I use his definitions and conventions. Example 14.5.2The $\mathbb{Z}-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete set of relations $$ 3v_1+2v_2+v_3=0\\ 8v_1+4v_2+2v_3=0\\ 7v_1+6v_2+2v_3=0\\ 9v_1+6v_2+v_3=0 $$ is presented by the matrix $$ A=\begin{bmatrix} 3 & 8 & 7 & 9 \\ 2 & 4 & 6 & 6 \\ 1 & 2 & 2 & 1 \\ \end{bmatrix}. $$ Its columns are the coefficients of the (above) relations: $(v_1, v_2, v_3)A=(0, 0, 0, 0)$.
What is the result of $\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{\tan(n)}\right)^{\tan(n)}$ = ? The limit does not exist as stated by Adam Rubinson. Looks like the requested answer is e but then the question is wrong. What is the correct question to give the answer e? Did answered correctly. But I think this is some kind of typographical error. This question appeared on a high school test and the intended question should have a small error. What is the correct question to give the answer e changing only one value? What is the result of $\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{\tan(n)}\right)^{\tan(n)}$ = ? $$ \lim_{n\to\color{red}{\pi/2}}\left(1+\frac{1}{\tan(n)}\right)^{\tan(n)} = e$$ What is the correct question to give the answer $\mathrm e$? $$ \lim_{n\to \infty}\left(1+\tan(1/n)\right)^{1/\tan(1/n)} =\ ? $$
I've seen a number of 2D Poisson disc sampling algorithms online that use a grid to accelerate checking for existing points within the minimum radius [![r][r image]][r link] of a candidate point. For example: They use a grid of squares of side $\frac{r}{\sqrt2}$, which is the same side length that I intuitively came up with when implementing this myself. I can see the reason - that is the largest square that cannot contain more than 1 point (assuming the minimum is not attainable - the distance between two points must be strictly greater than $r$). However, having thought about it further, I adjusted the grid size to $\frac{r}2$ instead. This finer grid means 4 additional squares need to be checked (the 4 corner squares are now within the radius), but the total area covered by the required squares is less, so that on average fewer points will need to go through the Euclidean distance check. The difference can be visualized using the same style as the diagram in the first linked article. For a candidate new point, existing points must be checked in all squares that are within a radius $r$ of the corners of the candidate's square. Here the two grid sizes are shown side by side, to scale, for the same radius $r$. This shows clearly that a significantly smaller area is being checked. Each square is exactly half the area of the previous approach, and even if the 4 outer corner squares are excluded in the previous approach (left image), this still gives an area $2 \cdot \frac{21}{25} = 1.68$ times larger than in the new approach. My main question is this: Is this approach still correct, and does it give identical results? I'm also interested to know whether there is any reason to favor the $\frac{r}{\sqrt2}$ approach. Using $\frac{r}{2}$ seems more efficient in time, which seems worth the cost in space efficiency. Is there anything I'm missing? Images produced with this jsfiddle (in case I need to edit them later).
For discussion of specific patterns or specific families of patterns, both newly-discovered and well-known. gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm Kazyan wrote: Component found in a CatForce result: Code: Select all x = 42, y = 67, rule = LifeHistory A$.2A$2A5$7.A$8.2A$7.2A19$24.2A$24.2A2$39.A$37.A3.A$36.A$36.A4.A$36. 5A$14.2A.2D$13.A.AD.D$13.A$12.2A25$5.3A$7.A$6.A! That can be done with 4 gliders, although it's still interesting that it was found accidentally: Code: Select all x = 21, y = 30, rule = B3/S23 10b2o$11bo$11bobo$12b2o14$10bo4bo$10b2ob2o$9bobo2b2o8$2o17bo$b2o15b2o$ o17bobo! What were you looking for, exactly? A MWSS-to-herschel converter? Kazyan Posts: 864 Joined: February 6th, 2014, 11:02 pm gmc_nxtman wrote:What were you looking for, exactly? A MWSS-to-herschel converter? I'd settle for any signal, but yes. The current Orthogonoids have geometry challenges that pad their size, and the limiting factor in their repeat time is the syringe. Repeat time is more important for single-channel operations than probably any other constructor design, so I'm trying to give that fire some better fuel. Tanner Jacobi mniemiec Posts: 1055 Joined: June 1st, 2013, 12:00 am gmc_nxtman wrote:4-glider trans-boat with tail edgeshoot: ... Even though was already buildable from 4 gliders, this method improves syntheses of one still-life and 18 pseudo-objects. Kazyan Posts: 864 Joined: February 6th, 2014, 11:02 pm Potential component spotted in a failed eating reaction: Code: Select all x = 21, y = 17, rule = B3/S23 o$3o$3bo$2b2o2$6bo$5bobo2$5b3o$19bo$8bo9bo$18b3o3$15bo$14b2o$14bobo! Tanner Jacobi gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm Unusual still life in 8 gliders: Code: Select all x = 18, y = 26, rule = B3/S23 11bo$10bobo$10b2o2$10bo$9b2o$9bobo5$obo$b2o$bo2$3b2o$4b2o$3bo$9bo$9b2o $8bobo2$15b3o$7b2o6bo$6bobo7bo$8bo! EDIT: This also gives 21.41458 in 9 gliders. Kazyan Posts: 864 Joined: February 6th, 2014, 11:02 pm Potentially grow out a BTS into a structure like a snorkel loop: Code: Select all x = 15, y = 25, rule = B3/S23 2b2obo$3bob3o$bobo4bo$ob2ob2obo$o4bobo$b3obo$3bob2o3$10b3o2$8bo5bo$8bo 5bo$8bo5bo2$10b3o6$11b2o$10bo2bo$11b2o$11bo! I suspect that the drifter catalyst and its variants also have odd transformations, since both objects are robust. Tanner Jacobi gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm Haven't seen a component quite like this before: Code: Select all x = 27, y = 18, rule = B3/S23 20bobo$20b2o$21bo7$15bo$15bobo$15b2o2$3o9bobo$b3o9b2o$13bo10b3o$24bo$ 25bo! EDIT: Better version: Code: Select all x = 15, y = 11, rule = B3/S23 13bo$12bo$12b3o3$7bo$6bobo$6bobo2b2o$7bo2b2o$3o9bo$b3o! Gamedziner Posts: 796 Joined: May 30th, 2016, 8:47 pm Location: Milky Way Galaxy: Planet Earth p8 c/2 derived from blinker puffer 1 : Code: Select all 2bo$o3bo$5bo$o4bo$b5o5$b2o2b2o$bob2ob2o$2b5o$3b3o$4bo$2bo3bo$7bo$2bo4bo$3b5o! Code: Select all x = 81, y = 96, rule = LifeHistory 58.2A$58.2A3$59.2A17.2A$59.2A17.2A3$79.2A$79.2A2$57.A$56.A$56.3A4$27. A$27.A.A$27.2A21$3.2A$3.2A2.2A$7.2A18$7.2A$7.2A2.2A$11.2A11$2A$2A2.2A $4.2A18$4.2A$4.2A2.2A$8.2A! mniemiec Posts: 1055 Joined: June 1st, 2013, 12:00 am This is known. It can be easily synthesized from 10 gliders: Code: Select all x = 88, y = 26, rule = B3/S23 34bobo$35boo$35bo3$45bo$bo44boo$bbo42boo$3o$20boo18boo6bo$bbo17bobo17b obo4bo$boo18bo19bo5b3o$bobo$$43boo$44boo7b3o22bo4b3o$31b3o9bo9bobbo20b 3o3bobbo$33bo19bo16b3o3boobo3bo$32bo20bo3bo12bobbobb3o4bo3bo$53bo16bo 6boo4bo$54bobo13bo3bo3bo5bobo$70bo$71bobo$77bo$78bo$77bo! gameoflifemaniac Posts: 774 Joined: January 22nd, 2017, 11:17 am Location: There too Code: Select all x = 17, y = 17, rule = B3/S23 8bo$7bobo$4bo2bobo2bo$3bobobobobobo$2bo2bobobobo2bo$3b2o2bobo2b2o$7bob o$b6o3b6o$o15bo$b6o3b6o$7bobo$3b2o2bobo2b2o$2bo2bobobobo2bo$3bobobobob obo$4bo2bobo2bo$7bobo$8bo! Code: Select all x = 17, y = 17, rule = B3/S23 8bo$7bobo$4bo2bobo2bo$3bobobobobobo$2bo2bobobobo2bo$3b2o2bobo2b2o$7bob o$b6o3b6o$o7bo7bo$b6o3b6o$7bobo$3b2o2bobo2b2o$2bo2bobobobo2bo$3bobobob obobo$4bo2bobo2bo$7bobo$8bo! dvgrn Moderator Posts: 5874 Joined: May 17th, 2009, 11:00 pm Location: Madison, WI Contact: While incompetently welding a tremi-Snark this evening... Code: Select all x = 23, y = 31, rule = LifeHistory $3.4B$4.4B$5.4B5.2A$6.4B4.2A$7.9B$8.6B$8.4BA3B$6.7BA2B$6.5B3A2B$6.11B $4.2AB.10B$4.2AB3.B2A4B$9.2B2A5B$10.8B$10.6B$11.5B$5.2A5.3B$5.A.A3.5B $7.A2.B2AB2A$7.2A2.2A.AB2.2A$10.B3.A.A2.A$5.10A.2A$5.A$6.12A$17.A$8. 7A$8.A5.A$11.A$10.A.A$11.A! ... I ended up with a p3 that I didn't really want. Doesn't seem worth keeping it around until people are synthesizing all the 58-bit p3's, but it seemed mildly entertaining anyway. A for awesome Posts: 1901 Joined: September 13th, 2014, 5:36 pm Location: 0x-1 Contact: dvgrn wrote: Code: Select all x = 23, y = 31, rule = LifeHistory $3.4B$4.4B$5.4B5.2A$6.4B4.2A$7.9B$8.6B$8.4BA3B$6.7BA2B$6.5B3A2B$6.11B $4.2AB.10B$4.2AB3.B2A4B$9.2B2A5B$10.8B$10.6B$11.5B$5.2A5.3B$5.A.A3.5B $7.A2.B2AB2A$7.2A2.2A.AB2.2A$10.B3.A.A2.A$5.10A.2A$5.A$6.12A$17.A$8. 7A$8.A5.A$11.A$10.A.A$11.A! Pointless reduction: Code: Select all x = 17, y = 25, rule = LifeHistory 4B$.4B$2.4B5.2A$3.4B4.2A$4.9B$5.6B$5.4BA3B$3.7BA2B$3.5B3A2B$3.11B$.2A B.10B$.2AB3.B2A4B$6.2B2A5B$7.8B$7.6B$8.5B$9.3B$8.5B$7.B2AB2A$4.2A2.2A .AB2.2A$4.A2.B3.A.A2.A$5.7A.3A2$7.2A.4A$7.2A.A2.A! x₁=ηx V ⃰_η=c²√(Λη) K=(Λu²)/2 Pₐ=1−1/(∫^∞_t₀(p(t)ˡ⁽ᵗ⁾)dt) $$x_1=\eta x$$ $$V^*_\eta=c^2\sqrt{\Lambda\eta}$$ $$K=\frac{\Lambda u^2}2$$ $$P_a=1-\frac1{\int^\infty_{t_0}p(t)^{l(t)}dt}$$http://conwaylife.com/wiki/A_for_all Aidan F. Pierce gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm Can someone salvage this? (Look at T≈20) Code: Select all x = 16, y = 12, rule = B3/S23 bo$2bo$3o7$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo! BlinkerSpawn Posts: 1905 Joined: November 8th, 2014, 8:48 pm Location: Getting a snacker from R-Bee's gmc_nxtman wrote: Can someone salvage this? (Look at T≈20) Code: Select all x = 16, y = 12, rule = B3/S23 bo$2bo$3o7$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo! The red pattern inserted at gen 16 would do it: Code: Select all x = 17, y = 14, rule = LifeHistory 13.D$11.2D$.A14.D$2.A8.5D$3A7$3.2A5.2A2.2A$4.2A3.A.A.2A$3.A7.A3.A! AbhpzTa Posts: 475 Joined: April 13th, 2016, 9:40 am Location: Ishikawa Prefecture, Japan gmc_nxtman wrote: Can someone salvage this? (Look at T≈20) Code: Select all x = 16, y = 12, rule = B3/S23 bo$2bo$3o7$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo! Code: Select all x = 27, y = 19, rule = B3/S23 16bo$4bo9b2o$5bo9b2o$3b3o$22bo$20b2o$21b2o$bo$2bo$3o2$25b2o$24b2o$26bo 3$3b2o5b2o2b2o$4b2o3bobob2o$3bo7bo3bo! Iteration of sigma(n)+tau(n)-n [sigma(n)+tau(n)-n : OEIS A163163] (e.g. 16,20,28,34,24,44,46,30,50,49,11,3,3, ...) : 965808 is period 336 (max = 207085118608). gmc_nxtman Posts: 1147 Joined: May 26th, 2015, 7:20 pm Reduced an old synthesis from eleven (I think) down to eight gliders: Code: Select all x = 34, y = 34, rule = B3/S23 10bo$bobo7bo19bobo$2b2o5b3o19b2o$2bo29bo3$32bo$30b2o$31b2o3$12bo$12bob o$12b2o11$14b2o$14bobo$14bo12b2o$27bobo$27bo3$b2o$obo$2bo! Extrementhusiast Posts: 1796 Joined: June 16th, 2009, 11:24 pm Location: USA Glider + two-glider loaf/tub/block/blinker constellation lasts for over 10K gens: Code: Select all x = 16, y = 13, rule = B3/S23 3bobo$3b2o$4bo9bo$13bobo$4b2o8bo$4b2o3$6bo$5bobo$4bo2bo$5b2o$3o! I Like My Heisenburps! (and others) Entity Valkyrie Posts: 247 Joined: November 30th, 2017, 3:30 am A glider synthesis of Sawtooth 311 x = 193, y = 140, rule = B3/S23 40bo$41bo$39b3o$72bo$70b2o$71b2o19$32bobo$33b2o$33bo30bo$63bo$63b3o2$ 75bo$74b2o$24bo49bobo$25b2o$24b2o$49bo$47b2o$48b2o2$2bo$obo$b2o7$34b2o $35b2o$34bo3$67b2o$67bobo$53b2o12bo$53bobo$53bo6$27b2o93b2o$26bobo93bo bo$28bo93bo2$53b3o$53bo74bobo$54bo73b2o$4bo124bo$4b2o172b2o$3bobo171b 2o$174b2o3bo$31b2o140bobo$30b2o98b2o43bo$32bo97bobo36bobo$130bo3bobo 10bo22b2o$135b2o11b2o20bo$135bo4bobo4b2o34bobo$115b2o21bobobobo6b2o20b o9b2o$116b2o21b2ob2o6b2o22bo9bo$115bo36bo19b3o2$120bo23b2ob2o23b3o5bo$ 120b2o21bobobobo24bo4bo$119bobo13b2ob2o5bobo25bo5b3o$134bobobobo$115bo 20bobo13b2o25b3o$116b2o12b2o19b2o22bo3bo$115b2o14b2o20bo22bo3bo$130bo 35bo4bo2b3o$164bobo2b2o$113b3o49b2o3b2o2b3o4bobo$115bo60bo4b2o$114bo 60bo6bo$179bo$126bo25bo18bo7b2o$125b2o23b2o19b2o5bobo$125bobo23b2o17bo bo$146b2o$147b2o$121b2o23bo$120b2o$122bo2$156b2o$142bobo10bobo$134bobo 5b2o13bo$134b2o7bo$135bo$132bo6bo$132b2o4bo$131bobo4b3o2$138b3o43bo$ 134bo3bo22bo20b2o$135bo3bo22b2o19b2o$133b3o25b2o13bobo$174bobobobo7bo$ 133b3o5bo25bobo5b2ob2o6bobo$135bo4bo24bobobobo15b2o3bo$63b3o68bo5b3o 23b2ob2o19b2o$63bo127b2o$64bo75b3o19bo$130bo9bo22b2o6b2ob2o$130b2o9bo 20b2o6bobobobo$129bobo34b2o4bobo$144bo20b2o$143b2o22bo12b2o$143bobo33b 2o$181bo2$139bo$138bo$138b3o2$137bo$136b2o$136bobo! Entity Valkyrie Posts: 247 Joined: November 30th, 2017, 3:30 am This Simkin-Glider=Gun=like object actually produces two MWSS: Code: Select all x = 53, y = 17, rule = B3/S23 44b2o5b2o$44b2o5b2o2$47b2o$47b2o$12bo$12b3o$12bobo$14bo4$4b2o$4b2o2$2o 5b2o$2o5b2o! mniemiec Posts: 1055 Joined: June 1st, 2013, 12:00 am Entity Valkyrie wrote:A glider synthesis of Sawtooth 311 ... It's nice to have syntheses like this. Unfortunately, in this case, there are several pairs of gliders that would have had to pass through each other earlier (i.e. they would have already collided before this phase). To make sure this doesn't happen, it is usually a good idea to backtrack all the gilders a certain amount (e.g. far enough away that they are in four distinct clouds, one coming from each direction) and then run them to see if any unwanted interactions occur first. Rhombic Posts: 1056 Joined: June 1st, 2013, 5:41 pm This component (the reverse component would have been more useful). Found accidentally though. Code: Select all x = 12, y = 14, rule = B3/S23 11bo$9b3o$8bo$9bo$6b4o$6bo$2b2o3b3o$2b2o5bo$9bobo$2bo7b2o$bobo$bob2o$o $2bo! Code: Select all x = 13, y = 15, rule = B3/S23 7bo$7b3o$10bo$2b2ob3o2bo$o2bobo2bob2o$2o4b3o3bo$9bobo$3b2o3b2ob2o$3b2o 2$3bo$2bobo$2bob2o$bo$3bo! Extrementhusiast Posts: 1796 Joined: June 16th, 2009, 11:24 pm Location: USA Switch engine turns two rows of beehives into two rows of table on tables: Code: Select all x = 88, y = 96, rule = B3/S23 13b2o$12bo2bo$13b2o6$21b2o$8b3o9bo2bo$9bo2bo8b2o$13bo$10bobo4$29b2o$ 28bo2bo$29b2o2$bo$obo$obo$bo$37b2o$36bo2bo$37b2o2$9bo$8bobo$8bobo$9bo$ 45b2o$44bo2bo$45b2o2$17bo$16bobo$16bobo$17bo$53b2o$52bo2bo$53b2o2$25bo $24bobo$24bobo$25bo$61b2o$60bo2bo$61b2o2$33bo$32bobo$32bobo$33bo$69b2o $68bo2bo$69b2o2$41bo$40bobo$40bobo$41bo$77b2o$76bo2bo$77b2o2$49bo$48bo bo$48bobo$49bo$85b2o$84bo2bo$85b2o2$57bo$56bobo$56bobo$57bo5$65bo$64bo bo$64bobo$65bo5$73bo$72bobo$72bobo$73bo! I Like My Heisenburps! (and others) KittyTac Posts: 533 Joined: December 21st, 2017, 9:58 am Extrementhusiast wrote: Switch engine turns two rows of beehives into two rows of table on tables: Code: Select all x = 88, y = 96, rule = B3/S23 13b2o$12bo2bo$13b2o6$21b2o$8b3o9bo2bo$9bo2bo8b2o$13bo$10bobo4$29b2o$ 28bo2bo$29b2o2$bo$obo$obo$bo$37b2o$36bo2bo$37b2o2$9bo$8bobo$8bobo$9bo$ 45b2o$44bo2bo$45b2o2$17bo$16bobo$16bobo$17bo$53b2o$52bo2bo$53b2o2$25bo $24bobo$24bobo$25bo$61b2o$60bo2bo$61b2o2$33bo$32bobo$32bobo$33bo$69b2o $68bo2bo$69b2o2$41bo$40bobo$40bobo$41bo$77b2o$76bo2bo$77b2o2$49bo$48bo bo$48bobo$49bo$85b2o$84bo2bo$85b2o2$57bo$56bobo$56bobo$57bo5$65bo$64bo bo$64bobo$65bo5$73bo$72bobo$72bobo$73bo! And then explodes. I wonder if there's a way to eat it at the end. dvgrn Moderator Posts: 5874 Joined: May 17th, 2009, 11:00 pm Location: Madison, WI Contact: KittyTac wrote: Extrementhusiast wrote:Switch engine turns two rows of beehives into two rows of table on tables... And then explodes. I wonder if there's a way to eat it at the end. Yeah, switch engine/swimmer eaters definitely aren't a problem: Code: Select all x = 96, y = 98, rule = B3/S23 13b2o$12bo2bo$13b2o6$8b3o10b2o$20bo2bo$8bo3bo8b2o$9b4o$12bo4$29b2o$28b o2bo$29b2o2$bo$obo$obo$bo$37b2o$36bo2bo$37b2o2$9bo$8bobo$8bobo$9bo$45b 2o$44bo2bo$45b2o2$17bo$16bobo$16bobo$17bo$53b2o$52bo2bo$53b2o2$25bo$ 24bobo$24bobo$25bo$61b2o$60bo2bo$61b2o2$33bo$32bobo$32bobo$33bo$69b2o$ 68bo2bo$69b2o2$41bo$40bobo$40bobo$41bo$77b2o$76bo2bo$77b2o2$49bo$48bob o$48bobo$49bo$85b2o$84bo2bo$85b2o2$57bo$56bobo$56bobo$57bo5$65bo$64bob o$64bobo$65bo5$73bo$72bobo$72bobo17b2o$73bo18bo$93b3o$95bo! #C [[ AUTOSTART STEP 9 THEME 2 ]] kiho park Posts: 50 Joined: September 24th, 2010, 12:16 am I found this c/3 diagonal fuse while searching c/3 long barge crawler. Code: Select all x = 10, y = 11, rule = B3/S23:T40,27 8b2o$7bo2$6bobo$5bo2bo$4bobo$3bobo$2bobo$bobo$obo$bo!
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
Here we want to give an easy mathematical bootstrap argument why solutions to the time independent 1D Schrödinger equation (TISE) tend to be rather nice. First formally rewrite the differential form$$-\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) + V(x) \psi(x) ~=~ E \psi(x) \tag{1}$$into the int... [Some time travel comments] Since in the previous paragraph, we have explained how travelling to the future will not necessary result in you to arrive in the future that is resulted as if you have never time travelled (via twin paradox), what is the reason that the past you travelled back, has to be the past you learnt from historical records :? @0ßelö7 Well, I'd omit the explanation of the notation on the slide itself, and since there seems to be two pairs of formulae, I'd just put one of the two and then say that there's another one with suitable substitutions. I mean, "Hey, I bet you've always wondered how to prove X - here it is" is interesting. "Hey, you know that statement everyone knows how to prove but doesn't bother to write down? Here is the proof written down" significantly less so Sorry I have a quick question: For questions like this physics.stackexchange.com/questions/356260/… where the accepted answer clearly does not answer the original question what is the best thing to do; downvote, flag or just leave it? So this question says express $u^0$ in terms of $u^j$ where $u$ is the four-velocity and I get what $u^0$ and $u^j$ are but I'm a bit confused how to go about this one? I thought maybe using the space-time interval and evaluating for $\frac{dt}{d\tau}$ but it's not workin out for me... :/ Anyone give me a quickie starter please? :p Although a physics question, this is still important to chemistry. The delocalized electric field is related to the force (and therefore the repulsive potential) between two electrons. This in turn is what we need to solve the Schrödinger Equation to describe molecules. Short answer: You can calculate the expectation value of the corresponding operator, which comes close to the mentioned superposition. — Feodoran13 hours ago If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position? @0ßelö7 I just looked back at chat and noticed Phase's question, I wasn't purposefully ignoring you - do you want me to look over it? Because I don't think I'll gain much personally from reading the slides. Maybe it's just me having not really done much with Eigenbases but I don't recognise where I "put it in terms of M's eigenbasis". I just wrote it down for some vector v, rather than a space that contains all of the vectors v If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position? Honey, I Shrunk the Kids is a 1989 American comic science fiction film. The directorial debut of Joe Johnston and produced by Walt Disney Pictures, it tells the story of an inventor who accidentally shrinks his and his neighbor's kids to a quarter of an inch with his electromagnetic shrinking machine and throws them out into the backyard with the trash, where they must venture into their backyard to return home while fending off insects and other obstacles.Rick Moranis stars as Wayne Szalinski, the inventor who accidentally shrinks his children, Amy (Amy O'Neill) and Nick (Robert Oliveri). Marcia...
Suppose you have two equtaions: $$2xy + y^2 = 0$$ $$x^2 + 2xy + 1 = 0$$ Subtracting the second from the first yields $y^2 - x^2 - 1 = 0$. Isolating y, we discover that $y = \pm\sqrt{x^2 + 1}$. However, by inspection we can wee that the $2xy$ term in both equations must be negative, which means that a single value of x cannot have both a positive and negative corresponding y-value. (i.e. if x is positive, then y must be negative). It seems that the equation $y = \pm\sqrt{x^2 + 1}$ contains an "extraneous root", but I'm struggling to wrap my mind around how that could be. After all, I solved for y w.r.t. x without squaring both sides. Could anyone help me understand what is going on? The problem is that your reasoning isn't reversible. Your two equations together imply $y^2-x^2-1=0$, but the converse is not true: that equation does not imply your original system. Compare your question to the following argument. The system $$x+y=1$$ $$2x+y=1$$ yields, by subtraction, that $x=0$. But this second equation allows $y$ to be arbitrary! So we get a whole lot of "extraneous solutions": $(0,y)$, for any $y$. The problem, of course, is that the equation $x=0$ does not, in turn, imply the original system. I take it you do not find this situation puzzling. You generally lose information when you replace a system of equations with a linear combination of them. The combination will include the solutions of your original equation, but it likely will include non-solutions, as well. The form of the reasoning is: "any solution of the original equations will be a solution of this equation, too." Yes, but then you've only found a superset containing the solution set. To characterize the solution set exactly, you must worry about the converse. When you solve a system of equations (whether it be by elimination or other means), you cannot discard the individual equations you started with. The reason is that each one of these equations carry more information (restrictions for example) than the one equation you end up with. This goes for other example as well. The function $f(x) = \sqrt{x} - \sqrt{x}$ is not equal to $0$. The reason being any $x < 0 $ is not in the domain of $f$. By eliminating $\sqrt{x}$ , you have lost a key piece of information about this function (its domain). I think the issue here is at least partially that you're assuming the rest of the solution will go something like: We know that $y=\pm\sqrt{x^2+1}$. We'll use this to find the possible values for $x$, then substitute in for $y$. and this is missing something: When we go to solve for $x$, we're going to need to know whether $y$ is positive or negative. That is, we need to split into cases $y=\sqrt{x^2+1}$ or $y=-\sqrt{x^2+1}$ and then solve for $x$ by substituting in $y$. This is to say that when you write: After all, I solved for y w.r.t. x without squaring both sides. you're missing something. You didn't solve for $y$. You reduced one statement to saying that $y$ was among two possibilities. You have to proceed to the end of the argument with each possibility separate, since you can't substitute "either this or that" in for $y$. To be sure, what elimination does is we start with a system like $$2xy+y^2=0$$ $$x^2+2xy+1=0$$ And then you've arranged into an equivalent statement: $$y=\pm\sqrt{x^2+1}$$ $$x^2+2xy+1=0$$ - or whatever you take the second equation to be. This system is equivalent to the first since all of your steps are reversible (noting that you've correctly used $y=\pm\sqrt{z}$ as the equivalent to $y^2=z$). So, you aren't introducing extraneous roots - it's just that your formula for $x$ depends on the $\pm$, so you don't get to choose $x$, then choose the sign of $y$. Another way to say this is that your error is in this statement: A more simple example might be a system like: $$y^2=1$$ $$x+y=0$$ You get that $y=\pm 1$, but you can't really substitute that into the second equation itself since it depends on a variable $\pm$. You have taken a square root. Whenever you take a square root you can introduce an extraneous root. Even for the simple system $x^2=y^2$, $x=y$ if you took a square root of the first equation you'd get an extraneous root. You must always consider all the equations as they all contain information about the solution. If you had gone about solving it via elimination or substitution (rather than merging the two equations together randomly like you did) then you would have not encountered this. If you solve your equation fully (and use complex numbers) you get two pairs of roots: $$\left(\frac{1}{\sqrt3},-\frac{2}{\sqrt3}\right), \left(-\frac{1}{\sqrt3},\frac{2}{\sqrt3}\right),(i,0),(-i,0)$$ The second pair of roots does fit your reduced equation (with a small rearrangement), namely $x=\pm\sqrt{y^2-1}$.
skills to develop To learn how to find, for a normal random variable $X$ and an area $a$, the value $x^\ast$ of $X$ so that $P(X<x^\ast )=a$ or that $P(X>x^\ast )=a$ , whichever is required. Definition: Left and Right Tails The left tail of a density curve Figure $\PageIndex{1}$: Right and Left Tails of a Distribution The probabilities tabulated in Figure 5.3.1 are areas of left tails in the standard normal distribution. Tails of the Standard Normal Distribution At times it is important to be able to solve the kind of problem illustrated by Figure $\PageIndex{2}$. We have a certain specific area in mind, in this case the area $0.0125$ of the shaded region in the figure, and we want to find the value $z^\ast$ The idea for solving such a problem is fairly simple, although sometimes its implementation can be a bit complicated. In a nutshell, one reads the cumulative probability table for $Z$ in reverse, looking up the relevant area in the interior of the table and reading off the value of $Z$ from the margins. $Z$ Figure $\PageIndex{2}$: Value that Produces a Known Area Example $\PageIndex{1}$ Find the value $z^\ast$ Solution: The number that is known, $0.0125$, is the area of a left tail, and as already mentioned the probabilities tabulated in Figure 5.3.1 are areas of left tails. Thus to solve this problem we need only search in the interior of Figure 5.3.1 for the number $0.0125$. It lies in the row with the heading $-2.2$ and in the column with the heading $0.04$. This means that $P(Z < -2.24)= 0.0125$, hence $z^\ast=-2.24$ Example $\PageIndex{2}$ Find the value $z^\ast$ $Z$ Figure $\PageIndex{3}$: Value that Produces a Known Area Solution: The important distinction between this example and the previous one is that here it is the area of a right tail that is known. In order to be able to use Figure 5.3.1 we must first find that area of the left tail cut off by the unknown number $z^\ast$ Definition: standard normal random variable The value of the standard normal random variable $Z$ that cuts off a right tail of area $c$ is denoted $z_c$. By symmetry, value of $Z$ that cuts off a left tail of area $c$ is $-z_c$. See Figure $\PageIndex{4}$. Figure $\PageIndex{4}$: The Numbers $z_c$ and $-z_c$ The previous two examples were atypical because the areas we were looking for in the interior of Figure 5.3.1 were actually there. The following example illustrates the situation that is more common. Example $\PageIndex{3}$ Find $z_{.01}$ and $-z_{.01}$, the values of $Z$ that cut off right and left tails of area $0.01$ in the standard normal distribution. Solution: Since $-z_{.01}$ cuts off a left tail of area $0.01$ and Figure 5.3.1 is a table of left tails, we look for the number $0.0100$ in the interior of the table. It is not there, but falls between the two numbers $0.0102$ and $0.0099$ in the row with heading $-2.3$. The number $0.0099$ is closer to $0.0100$ than $0.0102$ is, so for the hundredths place in $-z_{.01}$ we use the heading of the column that contains $0.0099$, namely, $0.03$, and write $-z_{.01}\approx -2.33$. The answer to the second half of the problem is automatic: since $-z_{.01}=-2.33$, we conclude immediately that $z_{.01}=2.33$ We could just as well have solved this problem by looking for $z_{.01}$ first, and it is instructive to rework the problem this way. To begin with, we must first subtract $0.01$ from $1$ to find the area $ \(z_{.01}$ Figure $\PageIndex{5}$: Computation of the Number Tails of General Normal Distributions The problem of finding the value $x^\ast$ Suppose $X$ is a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$. To find the value $x^\ast$ of $X$ that cuts off a left or right tail of area $c$ in the distribution of $X$: find the value $z^\ast$ of $Z$ that cuts off a left or right tail of area $c$ in the standard normal distribution; $z^\ast$is the $z$-score of $x^\ast$ ; compute $x^\ast$ using the destandardization formula \[x^\ast =\mu +z^\ast \sigma\] In short, solve the corresponding problem for the standard normal distribution, thereby obtaining the $z$-score of $x^\ast$ Example $\PageIndex{4}$ Find $x^\ast$ Solution: All the ideas for the solution are illustrated in Figure $\PageIndex{6}$. Since $0.9332$ is the area of a left tail, we can find $z^\ast$ Figure $\PageIndex{6}$: Tail of a Normally Distributed Random Variable Example $\PageIndex{5}$ Find $x^\ast$ Solution: The situation is illustrated in Figure $\PageIndex{7}$. Since $0.65$ is the area of a right tail, we first subtract it from $1$ to obtain $1-0.65=0.35$, the area of the complementary left tail. We find $z^\ast$ Figure $\PageIndex{7}$: Tail of a Normally Distributed Random Variable Example $\PageIndex{6}$ Scores on a standardized college entrance examination (CEE) are normally distributed with mean $510$ and standard deviation $60$. A selective university decides to give serious consideration for admission to applicants whose CEE scores are in the top $5\%$ of all CEE scores. Find the minimum score that meets this criterion for serious consideration for admission. Solution: Let $X$ denote the score made on the CEE by a randomly selected individual. Then $X$ is normally distributed with mean $510$ and standard deviation $60$. The probability that $X$ lie in a particular interval is the same as the proportion of all exam scores that lie in that interval. Thus the minimum score that is in the top $5\%$ of all CEE is the score $x^\ast$ Figure $\PageIndex{8}$: Tail of a Normally Distributed Random Variable Since $0.0500$ is the area of a right tail, we first subtract it from $1$ to obtain $1-0.0500=0.9500$, the area of the complementary left tail. We find $z^\ast =z_{.05}$by looking for $0.9500$ in the interior of Figure 5.3.1. It is not present, and lies exactly half-way between the two nearest entries that are, $0.9495$ and $0.9505$. In the case of a tie like this, we will always average the values of $Z$ corresponding to the two table entries, obtaining here the value $z^\ast =1.645$ Example $\PageIndex{7}$ All boys at a military school must run a fixed course as fast as they can as part of a physical examination. Finishing times are normally distributed with mean $29$ minutes and standard deviation $2$ minutes. The middle $75\%$ of all finishing times are classified as “average.” Find the range of times that are average finishing times by this definition. Solution: Let $X$ denote the finish time of a randomly selected boy. Then $X$ is normally distributed with mean $29$ and standard deviation $2$. The probability that $X$ lie in a particular interval is the same as the proportion of all finish times that lie in that interval. Thus the situation is as shown in Figure $\PageIndex{9}$. Because the area in the middle corresponding to “average” times is $0.75$, the areas of the two tails add up to $1 - 0.75 = 0.25$ in all. By the symmetry of the density curve each tail must have half of this total, or area $0.125$ each. Thus the fastest time that is “average” has $z$-score $-z_{.125}$, which by Figure 5.3.1 is $-1.15$, and the slowest time that is “average” has $z$-score $z_{.125}=1.15$ and Figure $\PageIndex{9}$: A boy has an average finishing time if he runs the course with a time between $26.7$ and $31.3$ minutes, or equivalently between $26$ minutes $42$ seconds and $31$ minutes $18$ seconds. Key Takeaways The problem of finding the number $z^\ast$ so that the probability $P(Z<z^\ast )$ is a specified value $c$ is solved by looking for the number $c$ in the interior of Figure 5.3.1 and reading $z^\ast$ from the margins. The problem of finding the number $z^\ast$ so that the probability $P(Z>z^\ast )$ is a specified value $c$ is solved by looking for the complementary probability $1-c$ in the interior of Figure 5.3.1 and reading $z^\ast$ from the margins. For a normal random variable $X$ with mean $\mu$ and standard deviation $\sigma$, the problem of finding the number $x^\ast$ so that $P(X<x^\ast )$ is a specified value $c$ (or so that $P(X>x^\ast )$ is a specified value $c$) is solved in two steps: (1) solve the corresponding problem for $Z$ with the same value of $c$, thereby obtaining the $z$-score, $z^\ast$, of $x^\ast$; (2) find $x^\ast$ using $x^\ast =\mu +z^\ast \sigma$. The value of $Z$ that cuts off a right tail of area $c$ in the standard normal distribution is denoted $z_c$.
This is a subtle issue. First let's present a numerical example to see the head-scratching riddle.Assume $$\alpha =1/2 \implies Y = K^{1/2}L^{1/2}$$ and that the exogenously given input prices are $$r=1/8,\,\, w=4.$$ The f.o.c are $$\begin{cases} \frac {Y}{2K} = 1/8 \\\\\frac{Y}{2L} = 4 \end{cases} \implies K/4 = 8L \implies \left(L/K\right)^* = 1/32$$ This gives us the value of $L/K$ that satisfies both first-order conditions. But let's follow the OP's logic, and write/solve the f.o.c. as $$\begin{cases} \frac {K^{1/2}L^{1/2}}{2K} = 1/8 \\\\\frac{K^{1/2}L^{1/2}}{2L} = 4 \end{cases} \implies \begin{cases} \frac {L}{K} = 1/16 \\\\\frac{L}{K} = 1/64 \end{cases}$$ Oops. The system now seems impossible. But it appears that just above we have solved the exact same system. And we didn't do any illegal math operation, like dividing by zero or anything so how is it possible to obtain the above contradictory results?... But indeed we did something "illegal", although not in terms of mathematical operations in the narrow sense: by solving each f.o.c. with respect to $L/K$ we transformed a problem of optimizing a bivariate function that separately leads to a system with two unknowns and two equations, into a problem of directly solving a system with unknown ($L/K$), and two equations. one No surprise then that it will have a solution only for specific combinations of the exogenous parameters, which is expressed by eq. $(4)$ of the OP. But this is no longer the profit maximizing problem of the competitive firm. Another way to say it is that, given that what is treated as exogenous here is the input prices, not all input ratios satisfy the f.o.c.

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.