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Consider the following optimization problem: \[ \begin{align} \max ~~ & \sum_{i=1}^{n} f(x_i) \\ \text{subject to~~} & \sum_{i=1}^{n} x_{i} = a \\ & x_{i} \in \{0,\dots,c_i\} \forall i \in \{1,\dots,n\}. \end{align} \] Here, $f(x)$ is a non-linear function of $x$ and is defined for each non-negative integer $x$. I am interested to know where this class of optimization problems can arise in practice. Sorry, but I tried my best to use MathJax. It seems that this tutorial does not work here. The case of $ c_i=1 \forall i \in \{1, \ldots, n \} $ resembles the best subset selection problem in regression, for a suitable loss-measuring objective function (so your maximization would need to be converted to minimization). If we change the objective function to $\min \, \sum_{i} f_i(x_i)$, there are several potential applications in supply chain management. One is splitting an order for $a$ units of a part across multiple suppliers with different price schedules (which could include quantity discounts). Another is splitting a production lot of $a$ identical items across multiple machines (minimizing the total cost, with possible economies or diseconomies of scale on each machine). A third is splitting a load of $a$ items across multiple carriers (whose costs depend on load quantity). Some of those examples might make sense if all cost functions were identical (drop the subscript on $f$), some maybe not so much. For maximizing, maybe something like allocating $a$ sales people or repair people to different regions, where the objective maximizes sales made, downed power lines restored, or whatever. answered Paul Rubin ♦♦
I am at a loss how to transform the following function: $\qquad \Bigg[\Bigg(\omega^{2}-\alpha+\betaz^{4}\Bigg)^{2}+ (\delta\omega)^{2}\Bigg]*z^{2} = \gamma$ into an explicit dependance of $z$ on $\omega$, i.e., $z = f(\omega)$. I have tried solving the equation using Matlab and Mathematica, but both programs return odd complex equations. Is there any way to get a plot using this implicit function? P.S. FYI this is the frequency response of a quintic Duffing oscillator.
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ... @EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics. Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They... @JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;) I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears. @ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int\|\nabla u\|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$ @BalarkaSen sorry if you were in our discord you would know @ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$. @Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication. @Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist. Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union. since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap) I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic?
5.1.1 Boundary value problems We have encountered several diﬀerent eigenvalue problems such as: \[ X''(x)+ \lambda X(x)=0\] with diﬀerent boundary conditions $X(0)=0~~~~~X(L)=0~~~~~{\rm{(Dirichlet)~or,}}$ $X'(0)=0~~~~~X'(L)=0~~~~~{\rm{(Neumann)~or,}}$ $X'(0)=0~~~~~X(L)=0~~~~~{\rm{(Mixed)~or,}}$ $X(0)=0~~~~~X'(L)=0~~~~~{\rm{(Mixed),...}}$ For example for the insulated wire, Dirichlet conditions correspond to applying a zero temperature at the ends, Neumann means insulating the ends, etc…. Other types of endpoint conditions also arise naturally, such as the Robin boundary conditions \[hX(0)-X'(0)=0~~~~~ hX(L)+X'(L)=0, \] for some constant $h$. These conditions come up when the ends are immersed in some medium. Boundary problems came up in the study of the heat equation $u_t=ku_{xx}$ when we were trying to solve the equation by the method of separation of variables. In the computation we encountered a certain eigenvalue problem and found the eigenfunctions $X_n(x)$. We then found the eigenfunction decomposition of the initial temperature $f(x)=u(x,0)$ in terms of the eigenfunctions \[f(x)= \sum_{n=1}^{\infty}c_nX_n(x).\] Once we had this decomposition and found suitable $T_n(t)$ such that $T_n(0)=1$ and $T_n(t)X(x)$ were solutions, the solution to the original problem including the initial condition could be written as \[u(x,t)= \sum_{n=1}^{\infty}c_nT_n(t)X_n(x).\] We will try to solve more general problems using this method. First, we will study second order linear equations of the form \[ \frac{d}{dx}\left( p(x)\frac{dy}{dx} \right)-q(x)y+\lambda r(x)y=0.\] Essentially any second order linear equation of the form $a(x)y''+b(x)y'+c(x)y+\lambda d(x)y=0$ can be written as (5.1.5) after multiplying by a proper factor. Example $\PageIndex{1}$: Sturm-Liouville Problem Put the following equation into the form (5.1.5): \[x^2y''+xy'+(\lambda x^2-n^2)y=0.\] Multiply both sides by $ \frac{1}{x}$ to obtain \[\frac{1}{x}(x^2y''+xy'+(\lambda x^2-n^2)y)=xy''+y'+ \left( \lambda x -\frac{n^2}{x}\right)y= \frac{d}{dx}\left( x \frac{dy}{dx} \right)-\frac{n^2}{x}y+\lambda xy=0.\] The so-called Sturm-Liouville problem 1 is to seek nontrivial solutions to \[ \frac{d}{dx}\left( p(x)\frac{dy}{dx} \right)-q(x)y+\lambda r(x)y=0,~~~~~a<x<b, \\ \alpha_1y(a)-\alpha_2y'(a)=0, \\ \beta_1y(b)+\beta_2y'(b)=0.\] In particular, we seek $\lambda$s that allow for nontrivial solutions. The $\lambda$s that admit nontrivial solutions are called the eigenvalues and the corresponding nontrivial solutions are called eigenfunctions. The constants $\alpha_1$ and $\alpha_2$ should not be both zero, same for $\beta_1$ and $\beta_2$. Theorem 5.1.1. Suppose $p(x), p'(x),q(x)$ and $r(x)$ are continuous on $[a,b]$ and suppose $p(x)>0$ and $r(x)>0$ for all $x$ in $[a,b]$. Then the Sturm-Liouville problem (5.1.8) has an increasing sequence of eigenvalues \[\lambda_1<\lambda_2<\lambda_3< \cdots\] such that \[\lim_{n \rightarrow \infty} \lambda_n= +\infty\] and such that to each $\lambda_n$ there is (up to a constant multiple) a single eigenfunction $y_n(x)$. Moreover, if $q(x) \geq 0$ and $\alpha_1, \alpha_2,\beta_1, \beta_2 \geq 0$, then $\lambda_n \geq 0$ for all $n$. Problems satisfying the hypothesis of the theorem are called regular Sturm-Liouville problems and we will only consider such problems here. That is, a regular problem is one where $p(x), p'(x),q(x)$ and $r(x)$ are continuous, $p(x)>0$, $r(x)>0$, $q(x) \geq 0$, and $\alpha_1, \alpha_2,\beta_1, \beta_2 \geq 0$. Note: Be careful about the signs. Also be careful about the inequalities for $r$ and $p$, they must be strict for all $x$! When zero is an eigenvalue, we usually start labeling the eigenvalues at $0$ rather than at $1$ for convenience. Example $\PageIndex{2}$: The problem $y''+ \lambda y, 0<x<L,y(0)=0$, and $y(L)=0$ is a regular Sturm-Liouville problem. $p(x)=1,q(x)=0,r(x)=1$, and we have $p(x)1>0$ and $r(x)1>0$. The eigenvalues are $\lambda_n=\frac{n^2 \pi^2}{L^2}$ and eigenfunctions are $y_n(x)=\sin(\frac{n \pi}{L}x)$. All eigenvalues are nonnegative as predicted by the theorem. Exercise $\PageIndex{1}$: Find eigenvalues and eigenfunctions for \[y''+\lambda y=0,~~~~~y'(0)=0,~~~~~y'(1)=0.\] Identify the $p,q,r,\alpha_j,\beta_j$. Can you use the theorem to make the search for eigenvalues easier? (Hint: Consider the condition $-y'(0)=0$) Example $\PageIndex{3}$: Find eigenvalues and eigenfunctions of the problem \[y''+\lambda y=0,~~~~~0<x<1, \\ hy(0)-y'(0)=0,~~~~~y'(1)=0,~~~~~h>0.\] These equations give a regular Sturm-Liouville problem. Exercise $\PageIndex{2}$: Identify $p,q,r,\alpha_j,\beta_j$ in the example above. First note that $\lambda \geq 0$ by Theorem 5.1.1. Therefore, the general solution (without boundary conditions) is \[ y(x)=A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x)~~~~~~\rm{if~}\lambda >0, \] \[ y(x)=Ax + B~~~~~~~~~~~~~~~\rm{if~}\lambda =0.\] Let us see if $\lambda=0$ is an eigenvalue: We must satisfy $0=hB-A$ and $A=0$, hence $B=0$ (as $h>0$), therefore, $0$ is not an eigenvalue (no nonzero solution, so no eigenfunction). Now let us try $h>0$. We plug in the boundary conditions. \[ 0=hA- \sqrt{\lambda}B, \\ 0=-A\sqrt{\lambda}\sin(\sqrt{\lambda})+B\sqrt{\lambda}\cos(\sqrt{\lambda}).\] If $A=0$, then $B=0$ and vice-versa, hence both are nonzero. So $B=\frac{hA}{\sqrt{\lambda}}$, and $0=-A \sqrt{\lambda}\sin(\sqrt{\lambda})+\frac{hA}{\sqrt{\lambda}}\sqrt{\lambda}\cos(\sqrt{\lambda})$. As $A \neq 0$ we get \[0=- \sqrt{\lambda}\sin(\sqrt{\lambda})+h\cos(\sqrt{\lambda}),\] or \[\frac{h}{\sqrt{\lambda}}= \tan \sqrt{\lambda}.\] Now use a computer to ﬁnd $\lambda_n$. There are tables available, though using a computer or a graphing calculator is far more convenient nowadays. Easiest method is to plot the functions $\frac{h}{x}$ and $\tan(x)$ and see for which they intersect. There is an inﬁnite number of intersections. Denote by $\sqrt{\lambda_1}$ the ﬁrst intersection, by $\sqrt{\lambda_2}$ the second intersection, etc…. For example, when $h=1$, we get that $\sqrt{\lambda_1}\approx 0.86, \sqrt{\lambda_2}\approx 3.43, ...$. That is $y_1 \approx 0.74,y_2 \approx 11.73,... $, …. A plot for $h=1$ is given in Figure 5.1. The appropriate eigenfunction (let $A=1$ for convenience, then $B= \frac{h}{\sqrt{\lambda}}$) is \[y_n(x)=\cos(\sqrt{\lambda_n}x)+\frac{h}{\sqrt{\lambda_n}}\sin(\sqrt{\lambda_n}x).\] When $h=1$ we get (approximately) \[y_1(x) \approx \cos(0.86x)+ \frac{1}{0.86} \sin(0.86x),~~~~~y_2(x) \approx \cos(3.43x)+ \frac{1}{3.43} \sin(3.43x),~~~~~....\] Figure 5.1: Plot of $\frac{1}{x}$ and $\tan x$. 5.1.2 Orthogonality We have seen the notion of orthogonality before. For example, we have shown that $\sin(nx)$ are orthogonal for distinct $n$ on $[0, \pi]$. For general Sturm-Liouville problems we will need a more general setup. Let $r(x)$ be a weight function(any function, though generally we will assume it is positive) on $[a, b]$. Two functions $f(x)$, $g(x)$ are said to be orthogonal with respect to the weight function $r(x)$ when \[\int_a^bf(x)g(x)r(x)dx=0.\] In this setting, we deﬁne the inner product as \[ \langle f,g \rangle \overset{\rm{def}}= \int_a^bf(x)g(x)r(x)dx,\] and then say $f$ and $g$ are orthogonal whenever $\langle f,g \rangle=0$. The results and concepts are again analogous to ﬁnite dimensional linear algebra. The idea of the given inner product is that those $x$ where $r(x)$ is greater have more weight. Nontrivial (nonconstant) $r(x)$ arise naturally, for example from a change of variables. Hence, you could think of a change of variables such that $d \xi =r(x)dx$. We have the following orthogonality property of eigenfunctions of a regular Sturm-Liouville problem. Theorem 5.1.2. Suppose we have a regular Sturm-Liouville problem \[\frac{d}{dx} \left( p(x) \frac{dy}{dx} \right) -q(x)y+\lambda r(x)y=0, \\ \alpha_1y(a)- \alpha_2y'(a)=0, \\ \beta_1y(b)+ \beta_2y'(b)=0.\] Let $y_j$ and $y_k$ be two distinct eigenfunctions for two distinct eigenvalues $\lambda_j$ and $\lambda_k$. Then \[\int_a^by_j(x)y_k(x)r(x)dx=0,\] that is, $y_j$ and $y_k$ are orthogonal with respect to the weight function $r$. 5.1.3 Fredholm alternative We also have the Fredholm alternative theorem we talked about before for all regular Sturm-Liouville problems. We state it here for completeness. Theorem 5.1.3 (Fredholm alternative). Suppose that we have a regular Sturm-Liouville problem. Then either \[\frac{d}{dx} \left( p(x) \frac{dy}{dx} \right) -q(x)y+\lambda r(x)y=0, \\ \alpha_1y(a)- \alpha_2y'(a)=0, \\ \beta_1y(b)+ \beta_2y'(b)=0,\] has a nonzero solution, or \[\frac{d}{dx} \left( p(x) \frac{dy}{dx} \right) -q(x)y+\lambda r(x)y=f(x), \\ \alpha_1y(a)- \alpha_2y'(a)=0, \\ \beta_1y(b)+ \beta_2y'(b)=0,\] has a unique solution for any $f(x)$ continuous on $[a,b]$. This theorem is used in much the same way as we did before in § 4.4. It is used when solving more general nonhomogeneous boundary value problems. The theorem does not help us solve the problem, but it tells us when a unique solution exists, so that we know when to spend time looking for it. To solve the problem we decompose $f(x)$ and $y(x)$ in terms of the eigenfunctions of the homogeneous problem, and then solve for the coeﬃcients of the series for $y(x)$. 5.1.4 Eigenfunction series What we want to do with the eigenfunctions once we have them is to compute the eigenfunction decomposition of an arbitrary function $f(x)$. That is, we wish to write \[f(x)= \sum_{n=1}^{\infty}c_ny_n(x),\] where $y_n(x)$ the eigenfunctions. We wish to ﬁnd out if we can represent any function $f(x)$ in this way, and if so, we wish to calculate (and of course we would want to know if the sum converges). OK, so imagine we could write $f(x)$ as (5.1.24). We will assume convergence and the ability to integrate the series term by term. Because of orthogonality we have \[ \langle f,y_m \rangle = \int_a^bf(x)y_m(x)r(x)dx \\ = \sum_{n=1}^{\infty}c_n \int_a^by_n(x)y_m(x)r(x)dx \\ =c_m \int_a^by_m(x)y_m(x)r(x)dx= c_m \langle y_m,y_m \rangle .\] Hence, \[c_m= \frac{\langle f,y_m \rangle}{\langle y_m,y_m \rangle}= \frac{\int_a^bf(x)y_m(x)r(x)dx}{\int_a^b(y_m(x))^2r(x)dx}.\] Note that $y_m$ are known up to a constant multiple, so we could have picked a scalar multiple of an eigenfunction such that $\langle y_m,y_m \rangle=1$ (if we had an arbitrary eigenfunction $\tilde{y}_m$, divide it by $\sqrt{\langle \tilde{y}_m,\tilde{y}_m \rangle}$). When $\langle y_m,y_m \rangle=1$ we have the simpler form $c_m=\langle f,y_m \rangle$ as we did for the Fourier series. The following theorem holds more generally, but the statement given is enough for our purposes. Theorem 5.1.4. Suppose $f$ is a piecewise smooth continuous function on . If $y_1,y_2, \ldots$ are the eigenfunctions of a regular Sturm-Liouville problem, then there exist real constants $c_1,c_2, \ldots$ given by (5.1.26) such that (5.1.24) converges and holds for $a<x<b$. Example $\PageIndex{4}$: Take the simple Sturm-Liouville problem \[ y''+ \lambda y=0, ~~~~~0<x<\frac{\pi}{2}, \\ y(0)=0,~~~~~ y' \left( \frac{\pi}{2}\right)=0.\] The above is a regular problem and furthermore we know by Theorem 5.1.1 that $\lambda \geq 0$. Suppose $\lambda = 0$, then the general solution is $y(x)Ax+B$, we plug in the initial conditions to get $0=y(0)=B$, and $0=y'(\pi/2)=A$, hence $\lambda=0$ is not an eigenvalue. The general solution, therefore, is \[y(x)=A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x).\] Plugging in the boundary conditions we get $0=y(0)=A$ and $0=y'(\pi/2)=\sqrt{\lambda}B\cos(\sqrt{\lambda}\frac{\pi}{2})$. $B$ cannot be zero and hence $\cos(\sqrt{\lambda}\frac{\pi}{2}=0)$. This means that $\sqrt{\lambda}\frac{\pi}{2}$ must be an odd integral multiple of $\frac{\pi}{2}$, i.e. $(2n-1)\frac{\pi}{2}=\sqrt{\lambda_n}\frac{\pi}{2}$. Hence \[\lambda_n=(2n-1)^2.\] We can take $B=1$. Hence our eigenfunctions are \[ y_n(x)= \sin((2n-1)x).\] Finally we compute \[ \int_0^{\frac{\pi}{2}}(\sin((2n-1)x))^2dx=\frac{\pi}{4}.\] So any piecewise smooth function on $[0, \pi/2]$ can be written as \[f(x)=\sum_{n=1}^{\infty}c_n\sin((2n-1)x),\] where \[c_n= \frac{\langle f,y_n \rangle}{\langle y_n,y_n \rangle}= \frac{\int_0^{\frac{\pi}{2}}\sin((2n-1)x)dx}{\int_0^{\frac{\pi}{2}}(\sin((2n-1)x))^2dx}= \frac{4}{\pi} \int f(x)\sin((2n-1)x)dx.\] Note that the series converges to an odd $2\pi$-periodic (not $\pi$-periodic!) extension of $f(x)$. Exercise $\PageIndex{3}$ In the above example, the function is deﬁned on $0<x< \pi/2$, yet the series converges to an odd $2\pi$-periodic extension of $f(x)$. Find out how is the extension deﬁned for $\pi/2 <x< \pi$.
The following script generates two plots in a row, both of them controlled by Manipulate. I would like to combine them together in one. The problem I don't know how to solve is that the plotting region in Q is different for the two plots, even though it is included in the range $Q \in [0,1]$. Manipulate[Row[{ Plot3D[{ (1 - 2 Q - Sqrt[1 - Δ^2] Cos[2 θ]) (1 + Sqrt[1 - Δ^2] Cos[2 θ]) }, {θ, 0.001, π/2 - 0.001}, {Q, 0, Min[((1 - Δ^2) Sin[2 θ]^2)/(2 (1 + (-1)^0 Sqrt[1 - Δ^2] Cos[2 θ])), -(-1)^0 Sqrt[1 - Δ^2] Cos[2 θ]]}, AxesLabel -> Automatic, PlotRange -> {Automatic, Automatic, {0, 1}}, ImageSize -> Large ], Plot3D[{ (1 - Sqrt[1 - Δ^2] Cos[2 θ]) (1 - 2 Q + Sqrt[1 - Δ^2] Cos[2 θ]) }, {θ, 0.001, π/2 - 0.001}, {Q, 0, Min[((1 - Δ^2) Sin[2 θ]^2)/(2 (1 + (-1)^1 Sqrt[1 - Δ^2] Cos[2 θ])), -(-1)^1 Sqrt[1 - Δ^2] Cos[2 θ]]}, AxesLabel -> Automatic, PlotRange -> {Automatic, Automatic, {0, 1}}, ImageSize -> Large ] }], {{Δ, 0}, -1, 1}] PS The range in $\theta$ is $[0.001, \pi/2 -0.001] $ to prevent errors.
I am reading Huber's Robust Statistics (2nd). On page 2 and 3 he gave an example. The basic facts are summarized here. Let $(X_n)$ be a sequence of random variables and define two measures of spread as follows. Mean Absolute Deviation: $d_n := \frac{1}{n}\sum\|x_i-\bar x\|$. Standard Deviation: $s_n := \sqrt{\frac{1}{n}\sum (x_i-\bar x)^2}$. Then he mentioned that Fisher claimed that for identically distributed normal observations $s_n$ is about 12% more efficient than $d_n$. In addition, $s_n$ converges to $\sigma$ while $d_n$ converges to $\sigma\sqrt{2/\pi}\doteq 0.8\sigma$. I have several questions about these statements. How to prove that $s_n$ is 12% more efficient, please? As least where to find the proof, please? How to prove that $d_n$ converges to $\sigma\sqrt{2/\pi}\doteq 0.8\sigma$, please? Again at least where to find the proof, please? I did some simulation to test all the above statements. Here are the codes and outcome. n <- 10000 # number of samples x <- array(list(), n) set.seed(2014) for(i in 1:n){ x[[i]] <- rnorm(10000) # the 10000 here is the size of each sample } dn <- rep(0, n) # mad sn <- rep(0, n) # sd for(i in 1:1000){ dn[i] <- mean(abs(x[[i]]-mean(x[[i]]))) # mad sn[i] <- sqrt(var(x[[i]])999/1000) # sd } mean(dn) # 0.07979068 check out* mean(sn) # 0.09995901 check out var(dn)/var(sn) # 0.6371817 As the above simulation shows, the 12% efficiency of $s_n$ does not check out. Why is this the case, please? Did I make errors in my simulation, please? Thank you!
Covers of Sets in a Topological Space Definition: Let $X$ be a topological space and let $A \subseteq X$. Then a Cover/Covering of $A$ is a collection of subsets $\mathcal F$ of $X$ such that $\displaystyle{A \subseteq \bigcup_{U \in \mathcal F} U}$. If $\mathcal F$ is a collection of open sets that satisfies the above inclusion then $\mathcal F$ is called an Open Cover/Covering of $A$. Similarly, if $\mathcal F$ is a collection of closed sets that satisfies the above inclusion then $\mathcal F$ is called a Closed Cover/Covering of $A$. If $\mathcal F^* \subseteq \mathcal F$ also covers $A$ then we said that $\mathcal F^*$ is a Subcover of $A$ from $\mathcal F$. Sometimes the term “cover” is used to denote an “open cover” when the context is clear and no ambiguity can arise. For example, consider the topological space $\mathbb{R}$ with the usual topology. Consider the set $A = \mathbb{N} \subset \mathbb{R}$. Every singleton set in $\mathbb{R}$ is closed and so the collection $\{ \{ n \} : n \in \mathbb{N} \}$ is a closed covering of $\mathbb{N}$ since:(1) On the otherhand, the collection $\{ \left ( n - \frac{1}{2}, n + \frac{1}{2} \right ) : n \in \mathbb{N} \}$ is an open covering of $\mathbb{N}$ as every set in this collection is an open interval centered at $n$ with radius $\frac{1}{2}$, and:(2) For another example, consider the set $A = [0, 1) \subset \mathbb{R}$. Then the collection $\mathcal F = \left \{ \left (-1, \frac{2}{3} \right ), \left ( \frac{1}{2}, 2 \right ) \right \}$ is an open cover of $[0, 1)$. In general, many open covers for subsets of a topological space may exist.
Example Problems Regarding Equivalent Binary Quadratic Forms Recall from the Equivalent Binary Quadratic Forms page that if $f$ and $g$ are binary quadratic forms then $f$ is said to be equivalent to $g$ if there exists $m_{11}, m_{12}, m_{21}, m_{22} \in \mathbb{Z}$ such that:(1) and $m_{11}m_{22} - m_{12}m_{21} = 1$. We will now look at some practice problems regarding equivalent binary quadratic forms. Example 1 On the page above, we proved that equivalence of binary quadratic forms is indeed an equivalence relation. Prove that $\sim$ is reflexive and symmetric directly. First we show that $\sim$ is reflexive. Let $m_{11} = 1$, $m_{12} = 0$, $m_{21} = 0$, and $m_{22} = 1$. Then:(2) and $m_{11}m_{22} - m_{12}m_{22} = 1$. So $f \sim f$ and $\sim$ is reflexive. Second we show that $\sim$ is symmetric. Suppose that $f \sim g$. Then $f(x, y) = g(m_{11}x + m_{12}y, m_{21}x + m_{22}y)$ where $m_{11}m_{22} - m_{12}m_{21} = 1$. Let $n_{11} = m_{22}$, $n_{12} = -m_{12}$, $n_{21} = -m_{21}$, and $n_{22} = m_{11}$. Then:(3) So $g \sim f$ and $\sim$ is symmetric. Example 2 Let $f(x, y) = x^2 + 3xy + 4y^2$. Use the matrix $M = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$ to obtain an equivalent binary quadratic form $g(x, y)$. We have that $m_{11} = 3$, $m_{12} = 2$, $m_{21} = 1$, and $m_{22} = 1$, so:(4)
T4 Normal Hausdorff Topological Spaces Recall from the T0 Kolmogorov Topological Spaces page that a topological space $X$ is said to be a T 0 space or a Kolmogorov space if for every pair of distinct points $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that either $x \not \in V$ or $y \not \in U$. On the T1 Fréchet Topological Spaces page we said that a topological space $X$ is said to be a T 1 space or a Fréchet space if for every pair of distinct points $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $x \not \in V$ and $y \not \in U$. On the T2 Hausdorff Topological Spaces page we said that a topological space $X$ is said to be a T 2 space or a Hausdorff space if for every pair of distinct points $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$. On the T3 Regular Hausdorff Topological Spaces page we saw that a topological space $X$ is said to be regular if for every $x \in X$ and for every closed set $F$ not containing $x$ there exists open sets $U$ and $V$ that separate $\{ x \}$ and $F$, i.e., $x \in U$ and $F \subseteq V$ where $U \cap V = \emptyset$. We then said that $X$ is a T 3 space or a regular Hausdorff space if $X$ is both regular and T 1 We will need to define another type of space before we can formally define a T 4 space. Definition: A topological space $X$ is said to be Normal if for every pair of disjoint closed sets $E$ and $F$ of $X$ there exists open sets $U$ and $V$ such that $U \subseteq E$, $V \subseteq F$, and $U \cap V = \emptyset$. We will now define what a T 4 space is. Definition: Let $X$ be a topological space. Then $X$ is said to be a T or a 4 Space Normal Hausdorff Space if $X$ is both a normal space and a T 1 space. As we will see in the following Theorem, every T 4 space is a T 3 space. Theorem 1: Let $X$ be a topological space. If $X$ is a T 4 space then $X$ is a T 3 space. Proof:Let $X$ be a T 4topological space. Then $X$ is both a normal space and a T 1space. We only need to show that $X$ is a regular space. Let $x \in X$ and let $F$ be a closed set in $X$ that does not contain $x$. Since $X$ is a T 1space, the singleton sets $\{ x \}$ are closed. So since $X$ is a regular space, there exists open sets $U$ and $V$ with $\{ x \} \subseteq U$ and $F \subseteq V$ such that $U \cap V = \emptyset$. Therefore $X$ is regular. Since $X$ is a regular space and is a T 1space we have that $X$ is a T 3space. $\blacksquare$
Hello one and all! Is anyone here familiar with planar prolate spheroidal coordinates? I am reading a book on dynamics and the author states If we introduce planar prolate spheroidal coordinates $(R, \sigma)$ based on the distance parameter $b$, then, in terms of the Cartesian coordinates $(x, z)$ and also of the plane polars $(r , \theta)$, we have the defining relations $$r\sin \theta=x=\pm R^2−b^2 \sin\sigma, r\cos\theta=z=R\cos\sigma$$ I am having a tough time visualising what this is? Consider the function $f(z) = Sin\left(\frac{1}{cos(1/z)}\right)$, the point $z = 0$a removale singularitya polean essesntial singularitya non isolated singularitySince $Cos(\frac{1}{z})$ = $1- \frac{1}{2z^2}+\frac{1}{4!z^4} - ..........$$$ = (1-y), where\ \ y=\frac{1}{2z^2}+\frac{1}{4!... I am having trouble understanding non-isolated singularity points. An isolated singularity point I do kind of understand, it is when: a point $z_0$ is said to be isolated if $z_0$ is a singular point and has a neighborhood throughout which $f$ is analytic except at $z_0$. For example, why would $... No worries. There's currently some kind of technical problem affecting the Stack Exchange chat network. It's been pretty flaky for several hours. Hopefully, it will be back to normal in the next hour or two, when business hours commence on the east coast of the USA... The absolute value of a complex number $z=x+iy$ is defined as $\sqrt{x^2+y^2}$. Hence, when evaluating the absolute value of $x+i$ I get the number $\sqrt{x^2 +1}$; but the answer to the problem says it's actually just $x^2 +1$. Why? mmh, I probably should ask this on the forum. The full problem asks me to show that we can choose $log(x+i)$ to be $$log(x+i)=log(1+x^2)+i(\frac{pi}{2} - arctanx)$$ So I'm trying to find the polar coordinates (absolute value and an argument $\theta$) of $x+i$ to then apply the $log$ function on it Let $X$ be any nonempty set and $\sim$ be any equivalence relation on $X$. Then are the following true: (1) If $x=y$ then $x\sim y$. (2) If $x=y$ then $y\sim x$. (3) If $x=y$ and $y=z$ then $x\sim z$. Basically, I think that all the three properties follows if we can prove (1) because if $x=y$ then since $y=x$, by (1) we would have $y\sim x$ proving (2). (3) will follow similarly. This question arised from an attempt to characterize equality on a set $X$ as the intersection of all equivalence relations on $X$. I don't know whether this question is too much trivial. But I have yet not seen any formal proof of the following statement : "Let $X$ be any nonempty set and $∼$ be any equivalence relation on $X$. If $x=y$ then $x\sim y$." That is definitely a new person, not going to classify as RHV yet as other users have already put the situation under control it seems... (comment on many many posts above) In other news: > C -2.5353672500000002 -1.9143250000000003 -0.5807385400000000 C -3.4331741299999998 -1.3244286800000000 -1.4594762299999999 C -3.6485676800000002 0.0734728100000000 -1.4738058999999999 C -2.9689624299999999 0.9078326800000001 -0.5942069900000000 C -2.0858929200000000 0.3286240400000000 0.3378783500000000 C -1.8445799400000003 -1.0963522200000000 0.3417561400000000 C -0.8438543100000000 -1.3752198200000001 1.3561451400000000 C -0.5670178500000000 -0.1418068400000000 2.0628359299999999 probably the weirdness bunch of data I ever seen with so many 000000 and 999999s But I think that to prove the implication for transitivity the inference rule an use of MP seems to be necessary. But that would mean that for logics for which MP fails we wouldn't be able to prove the result. Also in set theories without Axiom of Extensionality the desired result will not hold. Am I right @AlessandroCodenotti? @AlessandroCodenotti A precise formulation would help in this case because I am trying to understand whether a proof of the statement which I mentioned at the outset depends really on the equality axioms or the FOL axioms (without equality axioms). This would allow in some cases to define an "equality like" relation for set theories for which we don't have the Axiom of Extensionality. Can someone give an intuitive explanation why $\mathcal{O}(x^2)-\mathcal{O}(x^2)=\mathcal{O}(x^2)$. The context is Taylor polynomials, so when $x\to 0$. I've seen a proof of this, but intuitively I don't understand it. @schn: The minus is irrelevant (for example, the thing you are subtracting could be negative). When you add two things that are of the order of $x^2$, of course the sum is the same (or possibly smaller). For example, $3x^2-x^2=2x^2$. You could have $x^2+(x^3-x^2)=x^3$, which is still $\mathscr O(x^2)$. @GFauxPas: You only know $\|f(x)\|\le K_1 x^2$ and $\|g(x)\|\le K_2 x^2$, so that won't be a valid proof, of course. Let $f(z)=z^{n}+a_{n-1}z^{n-1}+\cdot\cdot\cdot+a_{0}$ be a complex polynomial such that $\|f(z)\|\leq 1$ for $\|z\|\leq 1.$ I have to prove that $f(z)=z^{n}.$I tried it asAs $\|f(z)\|\leq 1$ for $\|z\|\leq 1$ we must have coefficient $a_{0},a_{1}\cdot\cdot\cdot a_{n}$ to be zero because by triangul... @GFauxPas @TedShifrin Thanks for the replies. Now, why is it we're only interested when $x\to 0$? When we do a taylor approximation cantered at x=0, aren't we interested in all the values of our approximation, even those not near 0? Indeed, one thing a lot of texts don't emphasize is this: if $P$ is a polynomial of degree $\le n$ and $f(x)-P(x)=\mathscr O(x^{n+1})$, then $P$ is the (unique) Taylor polynomial of degree $n$ of $f$ at $0$.
Attention Conservation Notice:A tedious exercise in elementary probability theory showing that $R^{2}$ doesn’t behave the way you think. But you already knew that. It is a fact universally known that, when the simple linear regression model \[ Y = \beta_{0} + \beta_{1} X + \epsilon\] is correct, the value of $R^{2}$ depends on both: the variance of the noise term $\epsilon$ and the variance of the predictor $X$. The larger the value of $\operatorname{Var}(X)$, the larger the $R^{2}$, even for a fixed noise level. This, to me, is a pretty knock-down argument for why $R^{2}$ is a silly quantity to use for goodness-of-fit, since it only indirectly measures how good a model is, and does little-to-nothing to tell you how well the model fits the data. I even called $R^{2}$ “stupid” in my intro stats class last year, being careful to distinguish between $R^{2}$ being stupid, versus those who use $R^{2}$. I’m always looking for other examples of how $R^{2}$ behaves counterintuitively, and came across a nice example by John Myles White. He considers a logarithmic regression \[ Y = \log X + \epsilon,\] and demonstrates that for this model, the $R^{2}$ of a linear regression behaves non-monotonically as the variance of the $X$ values increase, first increasing and then decreasing. I wanted to use this to make a Shiny demo showing how $R^{2}$ depends on the variance of the predictor, but stumbled onto a weird “bug”: no matter what value of $\operatorname{Var}(X)$ I chose, I got the same value for $R^{2}$. I eventually figured out why my results differed from White’s, but I will save that punchline for the end. First, a great deal of tedious computations. Consider the following model for the predictors and the regression function. Let $X \sim U(0, b)$, $\epsilon \sim N(0, \sigma^{2})$, with $X$ and $\epsilon$ independent, and \[ Y = \log X + \epsilon.\] Then the optimal linear predictor of $Y$ is $\widehat{Y} = \beta_{0} + \beta_{1} X$, where \[\begin{align} \beta_{1} &= \operatorname{Cov}(X, Y)/\operatorname{Var}(X) \\ \beta_{0} &= E[Y] - \beta_{1} E[X] \end{align}\] The standard definitions of $R^{2}$ are either as the ratio of the variance of the predicted value of response to the variance of the response: \[ R^{2} = \frac{\operatorname{Var}(\widehat{Y})}{\operatorname{Var}(Y)} = \frac{\beta_{1}^{2} \operatorname{Var}(X)}{\operatorname{Var}(\log X) + \sigma^{2}}\] or as the squared correlation between the predicted value and the response: \[ R^{2} = (\operatorname{Cor}(\widehat{Y}, Y))^{2}.\] For a linear predictor, these two definitions are equivalent. Because $X$ is uniform on $(0, b)$, $\operatorname{Var}(X) = b^{2}/12$. We compute $\operatorname{Var}(\log X)$ in the usual way, using the “variance shortcut”, as \[ \operatorname{Var}(\log X) = E[(\log X)^{2}] - (E[\log X])^{2}.\] Direct computation shows that \[ E[\log X] = \frac{1}{b} \int_{0}^{b} \log x \, dx = \frac{1}{b} b (\log b - 1) = \log b - 1\] and \[ E[(\log X)^{2}] = \frac{1}{b} \int_{0}^{b} (\log x)^2 \, dx = \frac{1}{b} \cdot (b((\log (b)-2) \log (b)+2)).\] This gives the (perhaps suprising) result that $\operatorname{Var}(\log X) = 1,$ regardless of $b$. So far, we have that \[ R^{2} = \frac{\beta_{1}^{2} \frac{b^2}{12}}{1 + \sigma^{2}},\] so all that remains is computing $\beta_{1}$ from the covariance of $X$ and $Y$ and the variance of $X$. Using the “covariance shortcut,” \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y],\] and again via direct computation, we have that \[ E[Y] = E[\log X + \epsilon] = E[\log X] = \log b - 1\] \[ E[X] = \frac{b}{2}\] \[ E[XY] = E[X (\log X + \epsilon)] = E[X \log X] + E[X \epsilon] = E[X \log X] = \frac{2 \, b^{2} \log\left(b\right) - b^{2}}{4 \, b}.\] Overall, this gives us that \[ \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y] = \frac{1}{4} b \] and thus \[ \beta_{1} = \operatorname{Cov}(X,Y) / \operatorname{Var}(X) = \frac{\frac{1}{4} b}{\frac{b^2}{12}} = \frac{3}{b}\] Substituting this expression for $\beta_{1}$ into $R^{2}$, we get \[ R^{2} = \frac{3}{4} \frac{1}{1 + \sigma^{2}},\] and thus we see that the $R^{2}$ for the optimal linear predictor, for this model system, only depends on the variance of the noise term. My model differs from White’s in that his has a predictor that is uniform but bounded away from 0, i.e. $X \sim U(a, b)$ with $a > 0$. Making this choice prevents all of the nice cancellation from occurring (I will spare you the computations), and gives us the non-monotonic depence of $R^{2}$ on $\operatorname{Var}(X)$. So now we have at least three possible behaviors for $R^{2}$: $R^{2}$ can increase monotonically with $\operatorname{Var}{X}$, when the linear regression is correct. $R^{2}$ can increase and then decrease with $\operatorname{Var}{X}$, when White’s model is correct. $R^{2}$ can remain constant with $\operatorname{Var}{X}$, when the model I chose is correct. The zoo of possible behaviors for $R^{2}$ hopefully begins to convince you that “getting an $R^{2}$ close to 1 is good!” is far too simple of a story. For our model, we can get an $R^{2}$ of 3/4, in the limit as $\sigma \to 0$, which in some circles would be considered “good,” even though the linear model is clearly wrong for a logarithmic function. So I’ll stick with calling $R^{2}$ dumb. In fact, I generally avoid teaching it at all, following Wittgenstein’s maxim: If someone does not believe in fairies, he does not need to teach his children ‘There are no fairies’; he can omit to teach them the word ‘fairy.’ Though I suppose an opposing opinion would be that I should inoculate my students against $R^{2}$’s misuse. But there’s only so much time in a 13 week semester…
It is the half-twist. If you have an $2n$-gon, then rotate it by $n$. This is your centre (along with the trivial element). Note that for $G=\langle a, b, \ldots, c\rangle$ we have that $g\in Z(G)\Leftrightarrow ga=ag, gb=bg, \ldots, gc=cg$. That is, $g$ is in the centre if and only if it commutes with all the generators. You can prove that the half-twist is central by rotating the $2n$-gon in your head: Clearly the rotating generator $\alpha$ commutes with the half-twist (for the half-twist is $\alpha^n$), while the flip generator $\beta$ commutes with it too as $\alpha^n$ fixes the two vertices on the line of symmetry through the "initial" vertex. You can show $\alpha^n$ is the only non-trivial element of the centre by realising that every element has the form $\alpha^i$ or $\alpha^i\beta$ or $\beta$. Then, think about what these elements do to $\alpha$ and to $\beta$. The way I would do this is by using the fact that $ab=ba\Leftrightarrow aba^{-1}=b$, so you have to verify that $\alpha^i\beta\alpha^{-i}\neq \beta$ (unless $i=n$) and $\beta^{-1}\alpha^{-i}\alpha\alpha^i\beta=\beta\alpha\beta\neq\alpha$ (this last bit covers the last two cases - that $\alpha^i\beta$ and $\beta$ are not central). Do the flipping in your head - it is good practice! If your polygon has an odd number of vertices, then the centre is trivial. Use the ideas of the previous paragraph to prove this.
I am looking for a solution to a wave equation $\frac{\partial^2 u}{\partial \tau^2} = \frac{\partial^2 u}{\partial \xi^2}$ in which $t_c\tau = t$, $L\xi = x$, and $t_c = L/v_c$ is the characteristic time, $L$ is the sample thickness, and $v_c$ is the characteristic wave speed, with an IC of $\left [\frac{\partial u}{\partial \tau} \right]_{x,t=0} = \theta \left (x, t=0 \right)$ and a BC of $\left [\frac{\partial u}{\partial \xi} \right]_{x=0,t} = \phi \left (x=0, t \right)$ I have tried the D' Alembert solution, but I get a function $u\left(\xi, \tau \right)$ that is a function of the integral of phi which I don't know since it is not analytic, and it also introduces two new unknowns, $f\left (\tau_0 \right)$ and $g\left (\tau_0 \right)$ and I'm actually trying to find $\frac{\partial u}{\partial \tau}$ and $\frac{\partial u}{\partial \xi}$ not u. I haven't tried separation of variables, Sturm-Liouville or Fourier transform yet. This system is similar to Cauchy-Riemann equations.
Difference between revisions of "Literature on Carbon Nanotube Research" (→Ultra-high-yield growth of vertical single-walled carbon nanotubes: Hidden roles of hydrogen and oxygen) Line 86: Line 86: industry. industry. − == Ultra-high-yield growth of vertical single-walled carbon nanotubes + == Ultra-high-yield growth of vertical single-walled carbon nanotubes Hidden roles of hydrogen and oxygen == [http://www.pnas.org/content/102/45/16141.full.pdf+html G. Zhang et al.,PNAS, '''102''', no. 45, 16141-16145, 2005]<br> [http://www.pnas.org/content/102/45/16141.full.pdf+html G. Zhang et al.,PNAS, '''102''', no. 45, 16141-16145, 2005]<br> Revision as of 18:46, 20 March 2009 I have hijacked this page to write down my views on the literature on Carbon Nanotube (CNT) growths and processing, a procedure that should give us the cable/ribbon we desire for the space elevator. I will try to put as much information as possible here. If anyone has something to add, please do not hesitate! Contents 1 Direct mechanical measurement of the tensile strength and elastic modulus of multiwalled carbon nanotubes 2 Direct Spinning of Carbon Nanotube Fibers from Chemical Vapour Deposition Synthesis 3 Multifunctional Carbon Nanotube Yarns by Downsizing an Ancient Technology 4 Ultra-high-yield growth of vertical single-walled carbon nanotubes - Hidden roles of hydrogen and oxygen 5 Sustained Growth of Ultralong Carbon Nanotube Arrays for Fiber Spinning 6 In situ Observations of Catalyst Dynamics during Surface-Bound Carbon Nanotube Nucleation 7 High-Performance Carbon Nanotube Fiber Direct mechanical measurement of the tensile strength and elastic modulus of multiwalled carbon nanotubes B. G. Demczyk et al., Materials and Engineering, A334, 173-178, 2002 The paper by Demczyk et al. (2002) is the basic reference for the experimental determination of the tensile strengths of individual Multi-wall nanotube (MWNT) fibers. The experiments are performed with a microfabricated piezo-electric device. On this device CNTs in the length range of tens of microns are mounted. The tensile measurements are obseverd by transmission electron microscopy (TEM) and videotaped. Measurements of the tensile strength (tension vs. strain) were performed as well as Young modulus and bending stiffness. Breaking tension is reached for the SWNT at 150GP and between 3.5% and 5% of strain. During the measurements 'telescoping' extension of the MWNTs is observed, indicating that single-wall nanotubes (SWNT) could be even stronger. However, 150GPa remains the value for the tensile strength that was experimentally observed for carbon nanotubes. Direct Spinning of Carbon Nanotube Fibers from Chemical Vapour Deposition Synthesis Y.-L. Li, I. A. Kinloch, and A. H. Windle, Science, 304,276-278, 2004 The work described in the paper by Y.-L. Li et al. is a follow-on of the famous paper by Zhu et al. (2002), which was cited extensively in Brad's book. This article goes a little more into the details of the process. If you use a mixture of ethene (as the source of carbon), ferrocene, and theophene (both as catalysts, I suppose) into a furnace (1050 to 1200 deg C) using hydrogen as carrier gas, you apparently get an 'aerogel' or 'elastic smoke' forming in the furnace cavity, which comprises the CNTs. Here's an interesting excerpt: Under these synthesis conditions, the nanotubes in the hot zone formed an aerogel, which appeared rather like “elastic smoke,” because there was sufficient association between the nanotubes to give some degree of mechanical integrity. The aerogel, viewed with a mirror placed at the bottom of the furnace, appeared very soon after the introduction of the precursors (Fig. 2). Itwas then stretched by the gas flow into the form of a sock, elongating downwards along the furnace axis. The sock did not attach to the furnace walls in the hot zone, which accordingly remained clean throughout the process.... The aerogel could be continuously drawn from the hot zone by winding it onto a rotating rod. In this way, the material was concentrated near the furnace axis and kept clear of the cooler furnace walls,... The elasticity of the aerogel is interpreted to come from the forces between the individual CNTs. The authors describe the procedure to extract the aerogel and start spinning a yarn from it as it is continuously drawn out of the furnace. In terms of mechanical properties of the produced yarns, the authors found a wide range from 0.05 to 0.5 GPa/g/ccm. That's still not enough for the SE, but the process appears to be interesting as it allows to draw the yarn directly from the reaction chamber without mechanical contact and secondary processing, which could affect purity and alignment. Also, a discussion of the roles of the catalysts as well as hydrogen and oxygen is given, which can be compared to the discussion in G. Zhang et al. (2005, see below). Multifunctional Carbon Nanotube Yarns by Downsizing an Ancient Technology M. Zhang, K. R. Atkinson, and R. H. Baughman, Science, 306, 1358-1361, 2004 In the research article by M. Zhang et al. (2004) the procedure of spinning long yarns from forests of MWNTs is described in detail. The maximum breaking strength achieved is only 0.46 GPa based on the 30micron-long CNTs. The initial CNT forest is grown by chemical vapour deposition (CNT) on a catalytic substrate, as usual. A very intersting formula for the tensile strength of a yarn relative to the tensile strength of the fibers (in our case the MWNTs) is given: <math> \frac{\sigma_{\rm yarn}}{\sigma_{\rm fiber}} = \cos^2 \alpha \left(1 - \frac{k}{\sin \alpha} \right) </math> where <math>\alpha</math> is the helix angle of the spun yarn, i.e. fiber direction relative to yarn axis. The constant <math>k=\sqrt(dQ/\mu)/3L</math> is given by the fiber diameter d=1nm, the fiber migration length Q (distance along the yarn over which a fiber shifts from the yarn surface to the deep interior and back again), the quantity <math>\mu=0.13</math> is the friction coefficient of CNTs (the friction coefficent is the ratio of maximum along-fiber force divided by lateral force pressing the fibers together), <math>L=30{\rm \mu m}</math> is the fiber length. A critical review of this formula is given here. In the paper interesting transmission electron microscope (TEM) pictures are shown, which give insight into how the yarn is assembled from the CNT forest. The authors describe other characteristics of the yarn, like how knots can be introduced and how the yarn performs when knitted, apparently in preparation for application in the textile industry. Ultra-high-yield growth of vertical single-walled carbon nanotubes - Hidden roles of hydrogen and oxygen Important aspects of the production of CNTs that are suitable for the SE is the efficiency of the growth and the purity (i.e. lack of embedded amorphous carbon and imperfections in the Carbon bounds in the CNT walls). In their article G. Zhang et al. go into detail about the roles of oxygen and hydrogen during the chemical vapour deposition (CVD) growth of CNT forests from hydrocarbon sources on catalytic substrates. In earlier publications the role of oxygen was believed to be to remove amorphous carbon by oxidation into CO. The authors show, however, that, at least for this CNT growth technique, oxygen is important, because it removes hydrogen from the reaction. Hydrogen has apparently a very detrimental effect on the growth of CNTs, it even destroys existing CNTs as shown in the paper. Since hydrogen radicals are released during the dissociation of the hydrocarbon source compount, it is important to have a removal mechanism. Oxygen provides this mechanism, because its chemical affinity towards hydrogen is bigger than towards carbon. In summary, if you want to efficiently grow pure CNT forests on a catalyst substrate from a hydrocarbon CVD reaction, you need a few percent oxygen in the source gas mixture. An additional interesting information in the paper is that you can design the places on the substrate, on which CNTs grow by placing the the catalyst only in certain areas of the substrate using lithography. In this way you can grow grids and ribbons. Figures are shown in the paper. In the paper no information is given on the reason why the CNT growth stops at some point. The growth rate is given with 1 micron per minute. Of course for us it would be interesting to eliminate the mechanism that stops the growth so we could grow infinitely long CNTs. This article can be found in our archive. Sustained Growth of Ultralong Carbon Nanotube Arrays for Fiber Spinning Q. Li et al. have published a paper on a subject that is very close to our hearts: growing long CNTs. The longer the fibers, which we hope have a couple of 100GPa of tensile strength, can hopefully be spun into the yarns that will make our SE ribbon. In the paper the method of chemical vapour deposition (CVD) onto a catalyst-covered silicon substrate is described, which appears to be the leading method in the publications after 2004. This way a CNT "forest" is grown on top of the catalyst particles. The goal of the authors was to grow CNTs that are as long as possible. The found that the growth was terminated in earlier attempts by the iron catalyst particles interdiffusing with the substrate. This can apparently be avoided by putting an aluminium oxide layer of 10nm thickness between the catalyst and the substrate. With this method the CNTs grow to an impressive 4.7mm! Also, in a range from 0.5 to 1.5mm fiber length the forests grown with this method can be spun into yarns. The growth rate with this method was initially <math>60{\rm \mu m\ min.^{-1}}</math> and could be sustained for 90 minutes, This is very different from the <math>1{\rm \mu m\ min.^{-1}}</math> reported by G. Zhang et al. (2005), which shows that the growth is very dependent on the method and materials used. The growth was prolonged by the introduction of water vapour into the mixture, which achieved the 4.7mm after 2h of growth. By introducing periods of restricted carbon supply, the authors produced CNT forests with growth marks. This allowed to determine that the forest grew from the base. This is in line with the in situ observations by S. Hofmann et al. (2007). Overall the paper is somewhat short on the details of the process, but the results are very interesting. Perhaps the 5mm CNTs are long enough to be spun into a usable yarn. In situ Observations of Catalyst Dynamics during Surface-Bound Carbon Nanotube Nucleation The paper by S. Hofmann et al. (2007) is a key publication for understanding the microscropic processes of growing CNTs. The authors describe an experiment in which they observe in situ the growth of CNTs from chemical vapour deposition (CVD) onto metallic catalyst particles. The observations are made in time-lapse transmission electron microscopy (TEM) and in x-ray photo-electron spectroscopy. Since I am not an expert on spectroscopy, I stick to the images and movies produced by the time-lapse TEM. In the observations it can be seen that the catalysts are covered by a graphite sheet, which forms the initial cap of the CNT. The formation of that cap apparently deforms the catalyst particle due to its inherent shape as it tries to form a minimum-energy configuration. Since the graphite sheet does not extend under the catalyst particle, which is prevented by the catalyst sitting on the silicon substrate, the graphite sheet cannot close itself. The deformation of the catalyst due to the cap forming leads to a retoring force exerted by the crystaline stracture of the catalyst particle. As a consequence the carbon cap lifts off the catalyst particle. On the base of the catalyst particle more carbon atoms attach to the initial cap starting the formation of the tube. The process continues to grow a CNT as long as there is enough carbon supply to the base of the catalyst particle and as long as the particle cannot be enclosed by the carbon compounds. During the growth of the CNT the catalyst particle breathes so drives so the growth process mechanically. Of course for us SE community the most interesting part in this paper is the question: can we grow CNTs that are long enough so we can spin them in a yarn that would hold the 100GPa/g/ccm? In this regard the question is about the termination mechanism of the growth. The authors point to a very important player in CNT growth: the catalyst. If we can make a catalyst that does not break off from its substrate and does not wear off, the growth could be sustained as long as the catalyst/substrate interface is accessible to enough carbon from the feedstock. If you are interested, get the paper from our archive, including the supporting material, in which you'll find the movies of the CNTs growing. High-Performance Carbon Nanotube Fiber K. Koziol et al., Science, 318, 1892, 2007. The paper "High-Performance Carbon Nanotube Fiber" by K. Koziol et al. is a research paper on the production of macroscopic fibers out of an aerogel (low-density, porous, solid material) of SWNT and MWNT that has been formed by carbon vapor deposition. They present an analysis of the mechanical performance figures (tensile strength and stiffness) of their samples. The samples are fibers of 1, 2, and 20mm length and have been extracted from the aerogel with high winding rates (20 metres per minute). Indeed higher winding rates appear to be desirable, but the authors have not been able to achieve higher values as the limit of extraction speed from the aerogel was reached, and higher speeds led to breakage of the aerogel. They show in their results plot (Figure 3A) that typically the fibers split in two performance classes: low-performance fibers with a few GPa and high-performance fibers with around 6.5 GPa. It should be noted that all tensile strengths are given in the paper as GPa/SG, where SG is the specific gravity, which is the density of the material divided by the density of water. Normally SG was around 1 for most samples discussed in the paper. The two performance classes have been interpreted by the authors as the typical result of the process of producing high-strength fibers: since fibers break at the weakest point, you will find some fibers in the sample, which have no weak point, and some, which have one or more, provided the length of the fibers is in the order of the frequency of occurrence of weak points. This can be seen by the fact that for the 20mm fibers there are no high-performance fibers left, as the likelihood to encounter a weak point on a 20mm long fiber is 20 times higher than encountering one on a 1mm long fiber. As a conclusion the paper is bad news for the SE, since the difficulty of producing a flawless composite with a length of 100,000km and a tensile strength of better than 3GPa using the proposed method is enormous. This comes back to the ribbon design proposed on the Wiki: using just cm-long fibers and interconnect them with load-bearing structures (perhaps also CNT threads). Now we have shifted the problem from finding a strong enough material to finding a process that produces the required interwoven ribbon. In my opinion the race to come up with a fiber of better than Kevlar is still open.
Newform invariants Coefficients of the $q$-expansion are expressed in terms of a basis $1,\beta_1,\beta_2$ for the coefficient ring described below. We also show the integral $q$-expansion of the trace form. Basis of coefficient ring in terms of $\nu = \zeta_{14} + \zeta_{14}^{-1}$: $\beta_{0}$ $=$ $ 1 $ $\beta_{1}$ $=$ $ \nu $ $\beta_{2}$ $=$ $ \nu^{2} - 2 $ $1$ $=$ $\beta_0$ $\nu$ $=$ $\beta_{1}$ $\nu^{2}$ $=$ $\beta_{2}\mathstrut +\mathstrut $ $2$ For each embedding $\iota_m$ of the coefficient field, the values $\iota_m(a_n)$ are shown below. For more information on an embedded modular form you can click on its label. This newform does not admit any (nontrivial) inner twists. $ p $ Sign $71$ $1$ $113$ $1$ This newform can be constructed as the kernel of the linear operator $T_{2}^{3} $ $\mathstrut -\mathstrut 2 T_{2}^{2} $ $\mathstrut -\mathstrut T_{2} $ $\mathstrut +\mathstrut 1 $ acting on $S_{2}^{\mathrm{new}}(\Gamma_0(8023))$.
I believe this integral $$\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$$ can not be computed exactly. However is there a method or transformation to express this integral in terms of the cosine integral or similar? I am referring to the integrals here. $a$ is real number; with the change of variable this integral becomes $$ \int_0^a\cos(u\sin t) \ \mathrm dt $$ with $$ x=a\sin t, $$ So, the new integral is $$ \int_0^{\pi /2}\cos(ua\sin t) \ \mathrm dt $$
I recently saw the identity $$ \frac{1}{1} - {1 \over 2} +{1 \over 3} - {1 \over 4} + {1 \over 5} - {1 \over 6} \dotsb = \log(2) $$ which I found rather interesting. I was intrigued by the way a transcendental number like $\log(2)$ could be expressed by using the reciprocals of every integer once. I then wondered what other real numbers could be represented this way, by using the infinite harmonic series, but changing the $+$ or $-$ signs before each fraction. I soon realized that every real number could be represented this way, although not necessarily with such a clear pattern to the signs. Since + and - are binary options, I realized that the pattern of $+$ or $-$'s in the series could be represented by using a binary decimal number between 0 and 0.11111... I called this number the "harmonic sign number" (or "HSN" for short). For instance, an example of a harmonic sign number for $\log(2)$ would be $0.\overline{10}$, because it alternates between $+$ and $-$, with the 1's representing a + and the 0's representing a -. However, $\log(2)$ is probably a special case, as it has a harmonic sign number that is rational. It seems like most numbers could only be represented by irrational harmonic sign numbers, however that is still up for debate. Another example of a rational HSN would be $0.\overline{1100}$, which is the HSN for ${\pi \over 4} + {\log(2) \over 2}$ Also, every real number has an infinite amount of HSN's, but every HSN corresponds to only one real number (or not, if it doesn't converge). Some questions that I have thought about: Which numbers have rational harmonic sign numbers? Can only trancendental numbers have rational HSN's? If not, is there a quick way to tell whether a number has a rational HSN? Are there any rational numbers that have rational or at least algebraic HSN's? Does there exist a number which is it's own HSN? Are there more than one? Infinite? Which HSN's actually converge? I would suggest that they would have to be normal. Please note, this is done purely out of interest, so only respond if you are actually interested! If you would like clarification also please ask in comments and I will edit. Also, I'm not looking for full concrete answers. Feel free to answer with anything you have found or is of interest. It does not need to answer the questions posed above.
The Method of Integrating Factors Let $p$ and $g$ be functions of $t$ and consider the following first order differential equation:(1) If we multiply the both sides of the equation above by the function $\mu (t)$ we get that:(2) If we can guarantee that $\mu ' (t) = \mu (t) p(t)$, then notice that by applying the product rule for differentiation that we get: $\frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) \frac{dy}{dt} + \mu ' (t) y$ and substituting $\mu ' (t) = \mu (t) y$ and we get that $\frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) \frac{dy}{dt} + \mu (t) p(t) y$ which is exactly the lefthand side of the equation above. Thus we get that:(3) The above differential equation can be solved by integrating both sides of the equation with respect to $t$ and isolating $y$. The question now arises on how we can find such a function $\mu (t)$. Definition: If $\frac{dy}{dt} + p(t) y = g(t)$ is a first order differential equation, then $\mu (t)$ is called an Integrating Factor if for $\mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t)$ we have that $\mu ' (t) = \mu (t) p(t)$. The following proposition will give us a formula for obtaining the integrating factor for differential equations in the form $\frac{dy}{dt} + p(t) y = g(t)$. Proposition 1: If $\frac{dy}{dt} + p(t) y = g(t)$ is a differential equation, then an integrating factor $\mu (t)$ of this equation is given by the formula $\mu (t) = e^{\int p(t) \: dt}$. Proof:We want to find $\mu (t)$ such that $\mu ' (t) = \mu (t) p(t)$. We can rewrite this equation as as $\frac{\mu '(t)}{\mu (t)} = p(t)$ and then: Since we only need one integrating factor to solve differential equations in the form $\frac{dy}{dt} + p(t) y = g(t)$, we can more generally note that $\mu (t) = e^{\int p(t) \: dt}$ is an integrating factor of this differential equation. $\blacksquare$ Notice that from proposition 1 that integrating factors $\mu (t)$ are not unique. In fact, there are infinitely many integrating factors. This can be see when evaluating the indefinite integral in $\mu (t) = e^{\int p(t) \: dt}$ which will result in getting $\mu (t) = e^{P(t) + C}$ where $P$ is any antiderivative if $p$ and where $C$ is a constant. We will always use the simplest integrating factor in solving differential equations of this type. Let's now look at some examples of applying the method of integrating factors. Example 1 Find all solutions to the differential equation $\frac{dy}{dt} + \frac{2y}{t} = \frac{\sin t}{t^2}$. We first notice that our differential equation is in the appropriate form $\frac{dy}{dt} + p(t) y = g(t)$ where $p(t) = \frac{2}{t}$ and $g(t) = \frac{\sin t}{t^2}$. We compute our integrating factor as:(5) Thus we have that for $C$ as a constant:(6) Example 2 Find all solutions to the differential equation $\frac{dy}{dt} - \frac{y}{t} + te^{-t} = 0$. We first rewrite our differential equation as $\frac{dy}{dt} - \frac{y}{t} = - te^{-t}$. We note that in this form we have $p(t) = - \frac{1}{t}$ and $g(t) = -t e^{-t}$. We now find an integrating factor:(7) Thus we have that for $C$ as a constant:(8)
Four balls are transfered from an urn holding 3 white and 5 black balls. Two balls are drawn from the new urn and are white and are then replaced; which I presume means they were first drawn without replacement. Let random variable $X$ count the number of white balls transfered to the new urn, $E$ the event that two white balls are drawn from the new urn, and $F$ the event that a white ball will be drawn after the evidence is replaced. $E$ and $F$ are conditionally independent given a value for $X$. As you observed, there are two plausible values of $X$ given the evidence that at least two white balls are drawn. Yet you neglect to evaluate the probabilities, and for this Bates' Theorem is useful. $\mathsf P(X{=}2)=\left.\binom 32\binom 52\middle/\binom 84\right.$ is the probability for selecting two from three white balls and two from five black balls when selecting foutfrom eight balls in the first urn. $\mathsf P(E\mid X{=}2)= \left.\binom 22\binom 20\middle/\binom 42\right.$ is the probability for selecting two from the white balls and none from the black balls when selecting two from four balls given two of them are white. $\mathsf P(F\mid X{=}2)= \left.\binom 21\binom 20\middle/\binom 41\right.$ is the probability for selecting one from the white balls and none from the black balls when selecting two from four balls given two of them are white. And so forth. Then by the Law of Total Probability, the conditional independence, and Bayes' Theorem: $$\begin{align}\mathsf P(F\mid E)&=\mathsf P(F\mid X{=}2)\,\mathsf P(X{=}2\mid E)+\mathsf P(F\mid X{=}3)\,\mathsf P(X{=}3\mid E)\\[1ex]&=\dfrac{\mathsf P(F\mid X{=}2)\,\mathsf P(E\mid X{=}2)\,\mathsf P(X{=}2)+\mathsf P(F\mid X{=}3)\,\mathsf P(E\mid X{=}3)\,\mathsf P(X{=}3)}{\mathsf P(E\mid X{=}2)\,\mathsf P(X{=}2)+\mathsf P(E\mid X{=}3)\,\mathsf P(X{=}3)}\\[1ex]&~~\vdots\\[1ex]&=\dfrac{7}{12}\end{align}$$
I am trying to solve the following problem: $$\dfrac{\partial{\phi}}{\partial{t}} = \dfrac{\partial^2{\phi}}{\partial{x}^2} - \cos(x), \ x > 0, t > 0$$ $$\phi(x, 0) = 0, \ x > 0$$ $$\phi(0, t) = e^{-t}, \ t > 0$$ Taking the Laplace transform (in $t$, of course) gives $$\mathcal{L} \left\{ \dfrac{\partial{\phi}}{\partial{t}} \right\} = \mathcal{L} \left\{ \dfrac{\partial^2{\phi}}{\partial{x}^2} - \cos(x) \right\}$$ $$\therefore s \mathcal{L}\{\phi\} - \phi(0) = \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2} - \dfrac{\cos(x)}{s}$$ And since $\phi(x, 0) = 0$, we have $$s \mathcal{L} \{ \phi \} = \dfrac{d^2 \mathcal{L} \{ \phi \}}{dx^2} - \dfrac{\cos(x)}{s}$$ This is an ODE with constant coefficients. From now on, let $\mathcal{L}\{ \phi \} = \bar{\phi}$ (for the sake of conciseness). So our constant-coefficient ODE is $$\dfrac{d^2 \bar{\phi}}{dx^2} - s \bar{\phi} = \dfrac{\cos(x)}{s}$$ This is a inhomogeneous ODE and, judging by the inhomogeneous term $\dfrac{\cos(x)}{s}$, can be solved using the method of undetermined coefficients. The homogeneous version of the ODE is $$\dfrac{d^2 \bar{\phi}}{dx^2} - s \bar{\phi} = 0$$ The characteristic polynomial is $$m^2 - s = 0$$ $$\therefore m = \pm \sqrt{s}$$ Therefore, the complementary equation for this inhomogeneous ODE is $$y_c(x) = Ae^{x\sqrt{s}} + Be^{-x\sqrt{s}}$$ And our particular solution for this inhomogeneous ODE is $$y_p(x) = -\dfrac{1}{s^2 - s} \cos(x)$$ Therefore, our solution to the inhomogeneous ODE is $$y(x) = Ae^{x\sqrt{s}} + Be^{-x\sqrt{s}} - \dfrac{1}{s^2 - s} \cos(x)$$ Therefore, we have that $$\bar{\phi}(x, s) = Ae^{x\sqrt{s}} + Be^{-x\sqrt{s}} - \dfrac{1}{s^2 - s} \cos(x)$$ Here is where I am stumped. HOWEVER, one way that I have seen other problems proceed is by assuming another boundary condition: $$\lim\limits_{x \to \infty} \phi(x, t) = 0$$ This makes sense from a physical standpoint, but it was not specified in the problem statement, so I am unsure if it is valid for me to use it? Also, I don't have full solutions for these problems, so I have no way of checking the intermediate steps. The final solution should be $$\phi(x, t) = \text{erfc}\left( \dfrac{x}{2\sqrt{t}} \right) - \cos(1 - e^{-t})$$ I would greatly appreciate it if people could please help me solve this problem. EDIT: If we proceed with assuming the extra boundary condition (that is, if we proceed with assuming that $\lim\limits_{x \to \infty} \phi(x, t) = 0$), then we continue from above as follows: Now, since $\lim\limits_{x \to \infty} \phi(x, t) = 0$, we have that $Ae^{-x \sqrt{s}} = \dfrac{A}{e^{x \sqrt{s}}} \to 0$ as $x \to \infty$, and $- \dfrac{1}{s^2 - s} \cos(x)$ does not exist as $x \to \infty$, since $\cos(x)$ will not converge. So what do we do from here? It seems that the $- \dfrac{1}{s^2 - s} \cos(x)$ term is causing us problems? IGNORE EVERYTHING BELOW THIS POINT $$\bar{\phi}(x, s) = \int_0^\infty \phi(x, t) e^{-st} \ dt,$$ we have that $\bar{\phi}(x, s) \to 0$ as $x \to \infty$, and hence $B = 0$, since $Ae^{-x \sqrt{s}} = \dfrac{A}{e^{x \sqrt{s}}} \to 0$ as $x \to \infty$. Therefore, we now have $$\bar{\phi}(x, s) = Ae^{-x \sqrt{s}}.$$ We now use the given boundary conditions. Letting $x = 0$ in the Laplace transform of $\phi$ gives $$\bar{\phi}(0, s) = \int_0^\infty \phi(0, t)e^{-st} \ dt = \int_0^\infty e^{-t}e^{-st} \ dt = \int_0^\infty e^{-t(1 + s)} \ dt$$ $$= \dfrac{-1}{1 + s} \int_0^{-\infty} e^u \ du = \dfrac{1}{1 + s},$$ since $\phi(0, t) = e^{-t}$ for all $t > 0$.
Using a single integral we were able to compute the center of mass for a one-dimensional object with variable density, and a two dimensional object with constant density. With a double integral we can handle two dimensions and variable density. Just as before, the coordinates of the center of mass are $$\bar x={M_y\over M} \qquad \bar y={M_x\over M},$$ where $M$ is the total mass, $M_y$ is the moment around the $y$-axis, and $M_x$ is the moment around the $x$-axis. (You may want to review the concepts in Section 9.6.) The key to the computation, just as before, is the approximation of mass. In the two-dimensional case, we treat density $\sigma$ as mass per square area, so when density is constant, mass is $(\hbox{density})(\hbox{area})$. If we have a two-dimensional region with varying density given by $\sigma(x,y)$, and we divide the region into small subregions with area $\Delta A$, then the mass of one subregion is approximately $\sigma(x_i,y_j)\Delta A$, the total mass is approximately the sum of many of these, and as usual the sum turns into an integral in the limit: $$M=\int_{x_0}^{x_1}\int_{y_0}^{y_1} \sigma(x,y)\,dy\,dx,$$ and similarly for computations in cylindrical coordinates. Then as before $$\eqalign{ M_x &= \int_{x_0}^{x_1}\int_{y_0}^{y_1} y\sigma(x,y)\,dy\,dx\cr M_y &= \int_{x_0}^{x_1}\int_{y_0}^{y_1} x\sigma(x,y)\,dy\,dx.\cr }$$ Example $\PageIndex{1}$ Find the center of mass of a thin, uniform plate whose shape is the region between $y=\cos x$ and the $x$-axis between $x=-\pi/2$ and $x=\pi/2$. Since the density is constant, we may take $\sigma(x,y)=1$. It is clear that $\bar x=0$, but for practice let's compute it anyway. First we compute the mass: \[\begin{align} M&=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} 1\,dy\,dx \\[4pt]&=\int_{-\pi/2}^{\pi/2} \cos x\,dx \\[4pt]&=\left.\sin x\right\|_{-\pi/2}^{\pi/2}=2.\end{align}\] Next, \[\begin{align} M_x&=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} y\,dy\,dx \\[4pt]&=\int_{-\pi/2}^{\pi/2} {1\over2}\cos^2 x\,dx\\[4pt]&={\pi\over4}.\end{align}\] Finally, \[\begin{align} M_y&=\int_{-\pi/2}^{\pi/2} \int_0^{\cos x} x\,dy\,dx \\[4pt]&=\int_{-\pi/2}^{\pi/2} x\cos x\,dx\\[4pt]&=0.\end{align}\] So $\bar x=0$ as expected, and $\bar y=\pi/4/2=\pi/8$. This is the same problem as in example 9.6.4; it may be helpful to compare the two solutions. Example $\PageIndex{2}$ Find the center of mass of a two-dimensional plate that occupies the quarter circle $x^2+y^2\le1$ in the first quadrant and has density $k(x^2+y^2)$. It seems clear that because of the symmetry of both the region and the density function (both are important!), $\bar x=\bar y$. We'll do both to check our work. Jumping right in: \[\begin{align} M&=\int_0^1 \int_0^{\sqrt{1-x^2}} k(x^2+y^2)\,dy\,dx \\[4pt]&=k\int_0^1 x^2\sqrt{1-x^2}+{(1-x^2)^{3/2}\over3}\,dx. \end{align}\] This integral is something we can do, but it's a bit unpleasant. Since everything in sight is related to a circle, let's back up and try polar coordinates. Then $x^2+y^2=r^2$ and \[\begin{align} M&=\int_0^{\pi/2} \int_0^{1} k(r^2)\,r\,dr\,d\theta \\[4pt]&=k\int_0^{\pi/2}\left.{r^4\over4}\right\|_0^1\,d\theta \\[4pt]&=k\int_0^{\pi/2} {1\over4}\,d\theta \\[4pt]&=k{\pi\over8}.\end{align}\] Much better. Next, since $y=r\sin\theta$, \[\begin{align} M_x&=k\int_0^{\pi/2} \int_0^{1} r^4\sin\theta\,dr\,d\theta \\[4pt]&=k\int_0^{\pi/2} {1\over5}\sin\theta\,d\theta \\[4pt]&=k\left.-{1\over5}\cos\theta\right\|_0^{\pi/2}={k\over5}.\end{align}\] Similarly, \[\begin{align} M_y&=k\int_0^{\pi/2} \int_0^{1} r^4\cos\theta\,dr\,d\theta \\[4pt]&=k\int_0^{\pi/2} {1\over5}\cos\theta\,d\theta \\[4pt]&=k\left.{1\over5}\sin\theta\right\|_0^{\pi/2}={k\over5}.\end{align}\] Finally, $\bar x = \bar y = {8\over5\pi}$.
In Exercises $(2.4E.1)$ to $(2.4E.14)$, find a particular solution. Exercise $\PageIndex{1}$ $y''-3y'+2y=e^{3x}(1+x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{2}$ $y''-6y'+5y=e^{-3x}(35-8x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{3}$ $y''-2y'-3y=e^x(-8+3x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{4}$ $y''+2y'+y=e^{2x}(-7-15x+9x^2)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{5}$ $y''+4y=e^{-x}(7-4x+5x^2)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{6}$ $y''-y'-2y=e^x(9+2x-4x^2)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{7}$ $y''-4y'-5y=-6xe^{-x}$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{8}$ $y''-3y'+2y=e^x(3-4x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{9}$ $y''+y'-12y=e^{3x}(-6+7x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{10}$ $2y''-3y'-2y=e^{2x}(-6+10x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{11}$ $y''+2y'+y=e^{-x}(2+3x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{12}$ $y''-2y'+y=e^x(1-6x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{13}$ $y''-4y'+4y=e^{2x}(1-3x+6x^2)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{14}$ $9y''+6y'+y=e^{-x/3}(2-4x+4x^2)$ Answer Add texts here. Do not delete this text first. In Exercises $(2.4E.15)$ to $(2.4E.19)$, find the general solution. Exercise $\PageIndex{15}$ $y''-3y'+2y=e^{3x}(1+x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{16}$ $y''-6y'+8y=e^x(11-6x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{17}$ $ y''+6y'+9y=e^{2x}(3-5x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{18}$ $y''+2y'-3y=-16xe^x$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{19}$ $y''-2y'+y=e^x(2-12x)$ Answer Add texts here. Do not delete this text first. In Exercises $(2.4E.20)$ to $(2.4E.23)$, solve the initial value problem and plot the solution. Exercise $\PageIndex{20}$ $y''-4y'-5y=9e^{2x}(1+x), \quad y(0)=0,\quad y'(0)=-10$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{21}$ $y''+3y'-4y=e^{2x}(7+6x), \quad y(0)=2,\quad y'(0)=8$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{22}$ $y''+4y'+3y=-e^{-x}(2+8x), \quad y(0)=1,\quad y'(0)=2$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{23}$ $y''-3y'-10y=7e^{-2x}, \quad y(0)=1,\quad y'(0)=-17$ Answer Add texts here. Do not delete this text first. In Exercises $(2.4E.24)$ to $(2.4E.29)$, use the principle of superposition to find a particular solution. Exercise $\PageIndex{24}$ $y''+y'+y=xe^x+e^{-x}(1+2x)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{25}$ $y''-7y'+12y=-e^x(17-42x)-e^{3x}$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{26}$ $y''-8y'+16y=6xe^{4x}+2+16x+16x^2$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{27}$ $y''-3y'+2y=-e^{2x}(3+4x)-e^x$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{28}$ $y''-2y'+2y=e^x(1+x)+e^{-x}(2-8x+5x^2)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{29}$ $y''+y=e^{-x}(2-4x+2x^2)+e^{3x}(8-12x-10x^2)$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{30}$ (a) Prove that $y$ is a solution of the constant coefficient equation \begin{equation}\label{eq:2.4E.1} ay''+by'+cy=e^{\alpha x}G(x) \end{equation} if and only if $y=ue^{\alpha x}$, where $u$ satisfies \begin{equation}\label{eq:2.4E.2} au''+p'(\alpha)u'+p(\alpha)u=G(x) \end{equation} and $p(r)=ar^2+br+c$ is the characteristic polynomial of the complementary equation \begin{eqnarray} ay''+by'+cy=0. \end{eqnarray} For the rest of this exercise, let $G$ be a polynomial. Give the requested proofs for the case where \begin{eqnarray} G(x)=g_0+g_1x+g_2x^2+g_3x^3. \end{eqnarray} (b) Prove that if $e^{\alpha x}$ isn't a solution of the complementary equation then \eqref{eq:2.4E.2} has a particular solution of the form $u_p=A(x)$, where $A$ is a polynomial of the same degree as $G$, as in Example $(2.4.4)$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $y_p=e^{\alpha x}A(x)$. (c) Show that if $e^{\alpha x}$ is a solution of the complementary equation and $xe^{\alpha x}$ isn't, then \eqref{eq:2.4E.2} has a particular solution of the form $u_p=xA(x)$, where $A$ is a polynomial of the same degree as $G$, as in Example $(2.4.5)$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $y_p=xe^{\alpha x}A(x)$. (d) Show that if $e^{\alpha x}$ and $xe^{\alpha x}$ are both solutions of the complementary equation then \eqref{eq:2.4E.2} has a particular solution of the form $u_p=x^2A(x)$, where $A$ is a polynomial of the same degree as $G$, and $x^2A(x)$ can be obtained by integrating $G/a$ twice, taking the constants of integration to be zero, as in Example $(2.4.6)$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $y_p=x^2e^{\alpha x}A(x)$. Answer Add texts here. Do not delete this text first. Exercises $(2.4E.31)$ to $(2.4E.36)$ treat the equations considered in Examples $(2.4.1)$ to $(2.4.6)$. Substitute the suggested form of $y_p$ into the equation and equate the resulting coefficients of like functions on the two sides of the resulting equation to derive a set of simultaneous equations for the coefficients in $y_p$. Then solve for the coefficients to obtain $y_p$. Compare the work you've done with the work required to obtain the same results in Examples $(2.4.1)$ to $(2.4.6)$. Exercise $\PageIndex{31}$ Compare with Example $(2.4.1)$: \begin{eqnarray} y''-7y'+12y=4e^{2x};\quad y_p=Ae^{2x} \end{eqnarray} Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{32}$ Compare with Example $(2.4.2)$: \begin{eqnarray} y''-7y'+12y=5e^{4x};\quad y_p=Axe^{4x} \end{eqnarray} Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{33}$ Compare with Example $(2.4.3)$: \begin{eqnarray} y''-8y'+16y=2e^{4x};\quad y_p=Ax^2e^{4x} \end{eqnarray} Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{34}$ Compare with Example $(2.4.4)$: \begin{eqnarray} y''-3y'+2y=e^{3x}(-1+2x+x^2),\quad y_p=e^{3x}(A+Bx+Cx^2) \end{eqnarray} Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{35}$ Compare with Example $(2.4.5)$: \begin{eqnarray} y''-4y'+3y=e^{3x}(6+8x+12x^2),\quad y_p=e^{3x}(Ax+Bx^2+Cx^3) \end{eqnarray} Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{36}$ Compare with Example $(2.4.6)$: \begin{eqnarray} 4y''+4y'+y=e^{-x/2}(-8+48x+144x^2),\quad y_p=e^{-x/2}(Ax^2+Bx^3+Cx^4) \end{eqnarray} Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{37}$ Write $y=ue^{\alpha x}$ to find the general solution. (a) $y''+2y'+y=\displaystyle{e^{-x}\over\sqrt x}$ (b) $y''+6y'+9y=e^{-3x}\ln x$ (c) $y''-4y'+4y=\displaystyle{e^{2x}\over1+x}$ (d) $4y''+4y'+y=\displaystyle{4e^{-x/2}\left({1\over x}+x\right)}$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{38}$ Suppose $\alpha\ne0$ and $k$ is a positive integer. In most calculus books integrals like $\int x^k e^{\alpha x}\,dx$ are evaluated by integrating by parts $k$ times. This exercise presents another method. Let \begin{eqnarray} y=\int e^{\alpha x}P(x)\,dx \end{eqnarray} with \begin{eqnarray} P(x)=p_0+p_1x+\cdots+p_kx^k, \mbox{\quad (where $p_k\ne0$)}. \end{eqnarray} (a) Show that $y=e^{\alpha x}u$, where \begin{equation}\label{eq:2.4E.3} u'+\alpha u=P(x). \end{equation} (b) Show that \eqref{eq:2.4E.3} has a particular solution of the form \begin{eqnarray} u_p=A_0+A_1x+\cdots+A_kx^k, \end{eqnarray} where $A_k$, $A_{k-1}$, $\dots$, $A_0$ can be computed successively by equating coefficients of $x^k,x^{k-1}, \dots,1$ on both sides of the equation \begin{eqnarray} u_p'+\alpha u_p=P(x). \end{eqnarray} (c) Conclude that \begin{eqnarray} \int e^{\alpha x}P(x)\,dx=\left(A_0+A_1x+\cdots+A_kx^k\right)e^{\alpha x} +c, \end{eqnarray} where $c$ is a constant of integration. Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{39}$ Use the method of Exercise $(2.4E.38)$ to evaluate the integral. (a) $ \int e^x(4+x)\,dx$ (b) $\int e^{-x}(-1+x^2)\,dx$ (c) $\int x^3e^{-2x}\,dx$ (d) $\int e^x(1+x)^2\,dx$ (e) $\int e^{3x}(-14+30x+27x^2)\,dx$ (f) $\int e^{-x}(1+6x^2-14x^3+3x^4)\,dx$ Answer Add texts here. Do not delete this text first. Exercise $\PageIndex{40}$ Use the method suggested in Exercise $(2.4E.38)$ to evaluate $\int x^ke^{\alpha x}\,dx$, where $k$ is an arbitrary positive integer and $\alpha\ne0$. Answer Add texts here. Do not delete this text first.
I'm reading through some notes one locally convex spaces ("lcs" from now on) analysis and there the following version of the Banach-Steinhaus theorem is given Theorem (Banach-Steinhaus)$\quad$ The pointwise limit of a sequence of continuous, linear mappings from a barrelled lcs $U$ to a lcs $V$ is again a continuous, linear mapping. followed by the remark If we replace "sequence" with "net" this needn't be the case: For a discontinuous functional $f:U\rightarrow \mathbb{K}$ we can construct for each subspace $W\subseteq U$ a continuous, linear functional $F_W$ such that $f\big\|_W=F_W\big\|_W$. Can someone explain, or give me a hint, how to make this construction from the last sentence above explicit ? I'm also not sure how to use this to obtain a counterexample to the theorem above ? I somehow can't think of a way to make use of a point of discontinuity in $x_0\in U$ of $f$ to show that the net $(F_W)_W$ doesn't converge at all at $x_0$ (at least I intuitively think that this is the case - opposed to that the net indeed converges everywhere, but not to a continuous, linear functional).
Here I transfered the question from the comment The relationship between spin and spinor curvature How $\mathcal{R}_{ab} = \frac{1}{4}R_{abst}\gamma^s \gamma^t$ is from $\Psi \mapsto \Psi + \frac{1}{4} \epsilon_{\mu\nu}\gamma^\mu\gamma^\nu \Psi$ arise? When $\mathcal{R}_{ab}$ is defined by $\newcommand{\Rcal}{\mathcal{R}} \Rcal_{ab} \Psi = [D_a, D_b] \Psi$ and $R_{abst}$ is Riemann tensor constructed by metric. $D_a$ is the covariant derivative in the Dirac equation in the curved spacetimeDirac Equation in General RelativityThis post imported from StackExchange Physics at 2014-06-25 20:56 (UCT), posted by SE-user user48875
Joint work with Øystein Linnebo, University of Oslo. J. D. Hamkins and Ø. Linnebo, “The modal logic of set-theoretic potentialism and the potentialist maximality principles,” to appear in Review of Symbolic Logic, 2018. @ARTICLE{HamkinsLinnebo:Modal-logic-of-set-theoretic-potentialism, author = {Hamkins, Joel David and Linnebo, \O{}ystein}, title = {The modal logic of set-theoretic potentialism and the potentialist maximality principles}, journal = {to appear in Review of Symbolic Logic}, year = {2018}, volume = {}, number = {}, pages = {}, month = {}, note = {}, abstract = {}, keywords = {to-appear}, source = {}, eprint = {1708.01644}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://wp.me/p5M0LV-1zC}, doi = {}, } Abstract.We analyze the precise modal commitments of several natural varieties of set-theoretic potentialism, using tools we develop for a general model-theoretic account of potentialism, building on those of Hamkins, Leibman and Löwe (Structural connections between a forcing class and its modal logic), including the use of buttons, switches, dials and ratchets. Among the potentialist conceptions we consider are: rank potentialism (true in all larger $V_\beta$); Grothendieck-Zermelo potentialism (true in all larger $V_\kappa$ for inaccessible cardinals $\kappa$); transitive-set potentialism (true in all larger transitive sets); forcing potentialism (true in all forcing extensions); countable-transitive-model potentialism (true in all larger countable transitive models of ZFC); countable-model potentialism (true in all larger countable models of ZFC); and others. In each case, we identify lower bounds for the modal validities, which are generally either S4.2 or S4.3, and an upper bound of S5, proving in each case that these bounds are optimal. The validity of S5 in a world is a potentialist maximality principle, an interesting set-theoretic principle of its own. The results can be viewed as providing an analysis of the modal commitments of the various set-theoretic multiverse conceptions corresponding to each potentialist account. Set-theoretic potentialism is the view in the philosophy of mathematics that the universe of set theory is never fully completed, but rather unfolds gradually as parts of it increasingly come into existence or become accessible to us. On this view, the outer reaches of the set-theoretic universe have merely potential rather than actual existence, in the sense that one can imagine “forming” or discovering always more sets from that realm, as many as desired, but the task is never completed. For example, height potentialism is the view that the universe is never fully completed with respect to height: new ordinals come into existence as the known part of the universe grows ever taller. Width potentialism holds that the universe may grow outwards, as with forcing, so that already existing sets can potentially gain new subsets in a larger universe. One commonly held view amongst set theorists is height potentialism combined with width actualism, whereby the universe grows only upward rather than outward, and so at any moment the part of the universe currently known to us is a rank initial segment $V_\alpha$ of the potential yet-to-be-revealed higher parts of the universe. Such a perspective might even be attractive to a Platonistically inclined large-cardinal set theorist, who wants to hold that there are many large cardinals, but who also is willing at any moment to upgrade to a taller universe with even larger large cardinals than had previously been mentioned. Meanwhile, the width-potentialist height-actualist view may be attractive for those who wish to hold a potentialist account of forcing over the set-theoretic universe $V$. On the height-and-width-potentialist view, one views the universe as growing with respect to both height and width. A set-theoretic monist, in contrast, with an ontology having only a single fully existing universe, will be an actualist with respect to both width and height. The second author has described various potentialist views in previous work. Although we are motivated by the case of set-theoretic potentialism, the potentialist idea itself is far more general, and can be carried out in a general model-theoretic context. For example, the potentialist account of arithmetic is deeply connected with the classical debates surrounding potential as opposed to actual infinity, and indeed, perhaps it is in those classical debates where one finds the origin of potentialism. More generally, one can provide a potentialist account of truth in the context of essentially any kind of structure in any language or theory. Our project here is to analyze and understand more precisely the modal commitments of various set-theoretic potentialist views. After developing a general model-theoretic account of the semantics of potentialism and providing tools for establishing both lower and upper bounds on the modal validities for various kinds of potentialist contexts, we shall use those tools to settle exactly the propositional modal validities for several natural kinds of set-theoretic height and width potentialism. Here is a summary account of the modal logics for various flavors of set-theoretic potentialism. In each case, the indicated lower and upper bounds are realized in particular worlds, usually in the strongest possible way that is consistent with the stated inclusions, although in some cases, this is proved only under additional mild technical hypotheses. Indeed, some of the potentialist accounts are only undertaken with additional set-theoretic assumptions going beyond ZFC. For example, the Grothendieck-Zermelo account of potentialism is interesting mainly only under the assumption that there are a proper class of inaccessible cardinals, and countable-transitive-model potentialism is more robust under the assumption that every real is an element of a countable transitive model of set theory, which can be thought of as a mild large-cardinal assumption. The upper bound of S5, when it is realized, constitutes a potentialist maximality principle, for in such a case, any statement that could possibly become actually true in such a way that it remains actually true as the universe unfolds, is already actually true. We identify necessary and sufficient conditions for each of the concepts of potentialism for a world to fulfill this potentialist maximality principle. For example, in rank-potentialism, a world $V_\kappa$ satisfies S5 with respect to the language of set theory with arbitrary parameters if and only if $\kappa$ is $\Sigma_3$-correct. And it satisfies S5 with respect to the potentialist language of set theory with parameters if and only if it is $\Sigma_n$-correct for every $n$. Similar results hold for each of the potentialist concepts. Finally, let me mention the strong affinities between set-theoretic potentialism and set-theoretic pluralism, particularly with the various set-theoretic multiverse conceptions currently in the literature. Potentialists may regard themselves mainly as providing an account of truth ultimately for a single universe, gradually revealed, the limit of their potentialist system. Nevertheless, the universe fragments of their potentialist account can often naturally be taken as universes in their own right, connected by the potentialist modalities, and in this way, every potentialist system can be viewed as a multiverse. Indeed, the potentialist systems we analyze in this article—including rank potentialism, forcing potentialism, generic-multiverse potentialism, countable-transitive-model potentialism, countable-model potentialism—each align with corresponding natural multiverse conceptions. Because of this, we take the results of this article as providing not only an analysis of the modal commitments of set-theoretic potentialism, but also an analysis of the modal commitments of various particular set-theoretic multiverse conceptions. Indeed, one might say that it is possible ( ahem), in another world, for this article to have been entitled, “ The modal logic of various set-theoretic multiverse conceptions.” For more, please follow the link to the arxiv where you can find the full article. J. D. Hamkins and Ø. Linnebo, “The modal logic of set-theoretic potentialism and the potentialist maximality principles,” to appear in Review of Symbolic Logic, 2018. @ARTICLE{HamkinsLinnebo:Modal-logic-of-set-theoretic-potentialism, author = {Hamkins, Joel David and Linnebo, \O{}ystein}, title = {The modal logic of set-theoretic potentialism and the potentialist maximality principles}, journal = {to appear in Review of Symbolic Logic}, year = {2018}, volume = {}, number = {}, pages = {}, month = {}, note = {}, abstract = {}, keywords = {to-appear}, source = {}, eprint = {1708.01644}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://wp.me/p5M0LV-1zC}, doi = {}, }
Connectedness of Finite Topological Products Recall that if $X$ and $Y$ are topological spaces and $X \times Y$ is the Cartesian product of $X$ and $Y$, then we defined the product topology on $X \times Y$ to be the topology whose basis is:(1) We define the topological product of $X$ and $Y$ to then be $X \times Y$ with the product topology. More generally, if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces with $\displaystyle{X_1 \times X_2 \times ... \times X_n = \prod_{i=1}^{n} X_i}$ then the product topology on $\displaystyle{\prod_{i=1}^{n} X_i}$ has the basis:(2) Suppose that we have a finite collection of connected topological spaces, $\{ X_1, X_2, ..., X_n \}$. Then is the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ connected too? The answer is yes! Lemma 1: If $X$ and $Y$ are topological spaces such that $X$ is connected. Let $b \in Y$ and give the singleton $\{ b \}$ the subspace topology from $Y$. Then the product $X \times \{ b \}$ is connected. Proof:Suppose not, i.e., suppose that $X \times \{ b \}$ is disconnected. Then there exists open sets $U_1 \times V_1, U_2 \times V_2 \subseteq X \times \{ b \}$ such that $(U_1 \times V_1), (U_2 \times V_2) \neq \emptyset$, $(U_1 \times V_1) \cap (U_2 \times V_2) = \emptyset$ and: Notice that $V_1, V_2 \neq \emptyset$ since then $U_1 \times V_1 = \emptyset$ and/or $U_2 \times V_2 = \emptyset$ which is a contradiction. The only other open set in $\{ b \}$ with the subspace topology is the whole set $\{ b \}$, so $V_1 = \{ b \}$ and $V_2 = \{ b \}$. So $U_1 \times V_1 = U_1 \times \{ b \}$ and $U_2 \times V_2 = U_2 \times \{ b \}$ and do not contain any points in common we must have that $U_1, U_2 \subset X$, $U_1, U_2 = \emptyset$, and $U_1 \cap U_2 = \emptyset$. Note that: This implies that $X = U_1 \cup U_2$. Therefore $\{ U_1, U_2 \}$ is a separation of $X$ which contradicts $X$ being connected. So the assumption that $X \times \{ b \}$ was disconnected is false. $\blacksquare$ Theorem 1: Let $\{ X_1, X_2, ..., X_n \}$ be a finite collection of connected topological spaces. Then the topological product $\displaystyle{\prod_{i=1}^{n} X_i}$ is connected. Proof:Let $X$ and $Y$ be any two connected topological spaces. Let $(a, b)$ be any point in $X \times Y$. Since $X$ is connected, $X \times \{ b \}$ is also connected by Lemma 1. Similarly, for all $x \in X$ we have that $\{ x \} \times Y$ is also connected. Define $T_x$ by: Note that $T_x$ is connected by the theorem presented on the Common Point Criterion for Connectedness of Unions of Topological Subspaces page. This is because for each $x \in X$, $(x, b) \in (X \times \{ b \})$ and $(x, b) \in (\{ x \} \times Y)$. Then: Each $T_x$ contains the point $(a, b)$ since $(a, b) \in (X \times \{ b \}) \subset (X \times \{ b \}) \cup (\{ x \} \times Y) = T_x$, and by the theorem referenced above, $X \times Y$ is therefore connected. Inductively, we see that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces then $\displaystyle{\prod_{i=1}^{n} X_i}$ is also connected. $\blacksquare$ Theorem 1 can be generalized as true for an arbitrary collection of topological spaces. We state the result below but we will not prove it. Theorem 2: Let $\{ X_i \}_{i \in I}$ be an arbitrary collection of connected topological spaces. Then the topological product $\displaystyle{\prod_{i \in I} X_i}$ is also connected.
When $k=2$ and $n$ is even,$$D_{k,r} S^n \simeq BB_{r+} \wedge S^{rn}$$is the $rn$-fold unreduced suspension of the classifying space of the braid group $B_r$ on $r$ strings. I once cited Cohen-Mahowald-Milgram ``The stable decomposition for the double loop space of a sphere'' (1978) for this, and if I recall correctly they credited Arnold for the idea. Hence $rn$ is the maximal number of suspensions that can appear in this case. Edit: For $k=2$ and $n=2$ the map $F(\mathbb{R}^2,r) \times_{\Sigma_r} \mathbb{R}^{2r} \to \mathbb{R}^{2r}$ taking $[z,\xi] = [z_1, \dots, z_r, \xi_1, \dots, \xi_r]$ to$$(\sum_i \xi_i, \sum_i z_i \xi_i, \dots, \sum_i z_i^{r-1} \xi_i)$$trivializes the source as an $\mathbb{R}^{2r}$-bundle over $F(\mathbb{R}^2,r)/\Sigma_r$. Here the $z_i$ and $\xi_i$ in $\mathbb{R}^2$ are viewed as complex numbers, and the products in the displayed formula are formed in $\mathbb{C}$. A similar formula works for any even $n$, by taking the direct sum of $n/2$ copies of this $\mathbb{R}^{2r}$-bundle. Hence the Thom complex of this bundle, which is your $D_{2,r} S^n$, is the $rn$-th suspension of $(F(\mathbb{R}^2,r)/\Sigma_r)_+ \simeq BB_{r+}$. I recalled this in Proposition 9.10 of my paper "Topological logarithmic structures" GTM 16 (2009). For $k=2$ and $n=1$, or more generally for $n$ odd, the stable splitting of $\Omega^2 S^3$ is closely connected to the story of Brown-Gitler spectra. Regarding the case when $k$ is a higher power of $2$, you might first try $k=4$ and identify $\mathbb{R}^4$ with the quaternions. However, in that case the quaternionic Vandermonde matrix might well be singular. For instance, with $r=3$ and $z = (i,j,k)$ the matrix$\begin{pmatrix} 1 & 1 & 1 \\ i & j & k \\ -1 & -1 & -1 \end{pmatrix}$annihilates $\xi = (1,0,1)$, so Arnold's idea does not extend to this case.
Root-Finding Techniques for Polynomials We will now look at a few techniques to find roots of polynomials. The first theorem will tell us that if we have a rational root (in lowest terms) to a polynomial with integer coefficients, then the numerator/denominator of this rational root is restricted to be certain values. Theorem 1 (The Rational Root Theorem): If $r = \frac{p}{q} \in \mathbb{Q}$ is a rational root (in lowest terms) of the polynomial $f(x) = a_0 + a_1x + ... + a_nx^n$ with integer coefficients $a_0, a_1, ..., a_n \in \mathbb{Z}$ and $a_0 \neq 0$ then $p$ divides $a_0$ and $q$ divides $a_n$. Note that if a rational number $r = \frac{p}{q}$ is in "lowest terms", then the greatest common divisor of $p$ and $q$ is $1$, sometimes symbolically written as $\mathrm{gcd} (p, q) = 1$ or simply $(p, q) = 1$. Proof:Let $r = \frac{p}{q}$ be a rational root (in lowest terms) of $f(x) = a_0 + a_1x + ... + a_nx^n$ where $a_0 \neq 0$. Then $f(r) = f \left ( \frac{p}{q} \right ) = 0$ and so: If we multiply both sides of this equation by $q^n$ we get that: Therefore $p$ divides $a_0$ since $\mathrm{gcd} (p, q) = 1$. Furthermore, note that: Therefore $q$ divides $a_np^n$, and hence $q$ divides $a_n$ since $\mathrm{gcd} (p, q) = 1$. $\blacksquare$ Theorem 2 (Descarte's Rule of Signs): If $p(x)$ is a polynomial with real coefficients written in descending powers of $x$, that is $p(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ then: a) The number of positive real roots of $p(x)$ is equal to the number of sign changes in $p(x)$ or the number of sign changes in $p(x)$ minus an even positive integer. b) The number of negative real roots of $p(x)$ is equal to the number of sign changes in $p(-x)$ or the number of sign changes in $p(-x)$ minus an even positive integer. We will not prove Descarte's Rule of signs here. Theorem 3 (Bound's Theorem): Let $p(x) = a_0 + a_1x + ... + a_nx^n$ be a polynomial with complex coefficients, and let $\lambda$ be a root of $p$. Then $\mid \lambda \mid < \frac{\mathrm{max} \{ \mid a_0 \mid, \mid a_1 \mid, ..., \mid a_{n-1} \mid \}}{\mid a_n \mid} + 1$. Proof:Theorem 3 above is trivially true for $\mid \lambda \mid ≤ 1$. Suppose that $\mid \lambda \mid > 1$. Let $M = \mathrm{max} \{ \mid a_0 \mid, \mid a_1 \mid , ..., \mid a_{n-1} \mid \}$. Now since $\lambda$ is a root of $p$, we have that: The sum $1 + \mid \lambda \mid + ... + \mid \lambda \mid^{n-1}$ can be replaced as a geometric series $1 + \mid \lambda \mid + ... + \mid \lambda \mid^{n-1} = \frac{1 - \mid \lambda \mid^n}{1 - \mid \lambda \mid}$, and so: If we divide both sides of the equation above by $\mid \lambda \mid^n$ and since $\mid \lambda \mid > 1$ we have that:
I am trying to understand whether some non trivial explicit estimate is known about the following $$\sum_{n \leq x} \mu(n) \chi(n)$$ as a function of $x$. Here $\chi$ is a Dirichlet character modulo $q$ and $q > (\log x)^2$ (for example). MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.Sign up to join this community I am trying to understand whether some non trivial explicit estimate is known about the following $$\sum_{n \leq x} \mu(n) \chi(n)$$ as a function of $x$. Here $\chi$ is a Dirichlet character modulo $q$ and $q > (\log x)^2$ (for example). Estimating this sum is much the same as estimating the error term in the prime number theorem for arithmetic progressions. See Exercises 7-8 in Section 11.3 of Montgomery-Vaughan: Multiplicative number theory I (Cambridge University Press, 2006). (Your sum is denoted by $M(X,\chi)$ in this book, as introduced by (11.39) there.) In particular, there is a constant $c>0$ such that for any $A>0$ we have $$ \sum_{n \leq X} \mu(n) \chi(n)\ll_A x\exp(-c\sqrt{\log x})$$ as long as $q\leq(\log x)^A$ . Explicit estimates for the Moebius function are much harder than for the prime counting function. Ramare (From explicit estimates for the primes to explicit estimates for the Moebius function, Acta Arithmetica 157, (2013), 365-379) has shown how to translate estimates for primes to estimates for the Moebius function, as far as I know this approach is superior to a direct estimate using complex integration. It should be possible without too much work to change Sections 4, 5, 6 of that paper to the character case, as there is a lot of literature on various sums involving primes in arithmetic progressions (see the work of Rosser, Schoenfeld, McCurley, Rumely, Ramare). I would be worried about Lemma 9.1, but this is not necessary for your question. There is an equivalent of the Riemann explicit formula for those : If all the zeros of $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s}$ are simple, then $\displaystyle\frac{1}{L(s,\chi)} = \sum_{n=1}^\infty \mu(n)\chi(n) n^{-s}= s\int_1^\infty (\sum_{n < x}\mu(n) \chi(n))x^{-s-1}dx$ leads to $$\sum_{n < x} \mu(n) \chi(n) = \frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{1}{L(s,\chi)}\frac{x^s}{s}ds = \frac{1}{L(0,\chi)} + \sum_\beta \frac{x^{\beta}}{\beta L'(\beta,\chi)}$$ by Mellin inverse transform and the residue theorem, where $\sigma > 1$, $\beta$ are the zeros (trivial and non-trivial) of $L(s,\chi)$, and $\sum_\beta$ is convergent only when grouping the terms correctly. (note if some zeros are of order $k$, it is not so different except you'll get some additional terms $x^\beta (\ln x)^{m}, m < k$)
For a Hohmann transfer to Saturn, I get 15.7 km/s for both burns. The transfer time is also a simple formula. I obtain roughly 6 years. Compare to the lunar ice. It is roughly 2.8 km/s to get to, and the trip time would be a few days, even from Low Earth Orbit. As suggested by the other answer, you could compare to Earth's surface. If we're using some reference frame far from Earth's surface, then it's nearly 12.8 km/s going up (EDIT: corrected from newbie mistake, LEO gets Oberth). But this gets complicated. The first 9 km/s is constrained by atmospheric physics. Although, leaving the moon also has a similar thrust-to-weight requirement placed on the rockets by gravity drag. These numbers aren't totally fair to Saturn. If I use the edge of the rings (nearly the "F" ring), I get an orbital velocity of 16.6 km/s. In order to mine these rings, you'll need to follow a hyperbolic trajectory, and then do your final Hohmann burn when you're close to your destination, inside Saturn's gravity well. This will reduce the Delta V you'll need for the final burn. That'll reduce the final burn by a little bit. Breaking up the 15.7 km/s, the initial insertion burn is 10.3 km/s and the heliocentric circularization burn is 5.4 km/s. So we would need to correct this all for the Oberth effect, which is a multiplier of $\sqrt{1 + \frac{2 V_\text{esc}} {\Delta v}} $, sphere of influence assumptions, yada yada. That reduces the previous numbers to (respectively) 5.8 km/s and 1.4 km/s. Boy that was an unpleasant calculation, but the total comes to approximately 7.2 km/s. Commentary: It's true that mining the lunar ice would involve challenges that Saturn's rings don't have. But if the destination is close to Earth, those challenges would be more workable than the nuclear, ion, or whatever kind of drives that a Saturn hauler would involve. A similar tradeoff might present itself for other sources closer than Saturn, for instance, Jupiter trojan asteroids. I won't dispute the quality of Saturn's ice. It is likely more pure than the alternative sources. It might also be in convenient size chunks. Importantly (in its favor) it can be delivered by drives with a low thrust-to-weight ratio. Any value would do for this, although there might be some Delta V penalty due to reduced Oberth effect. Solar electric would be penalized heavily due to the distance from the sun. Nuclear power sources are vastly superior at that distance. 9.5 AU --> squared means 90 times less power per unit area than near Earth. The real problem is justifying that 6 year time frame. Who is going to wait 12 years for their return on investment? Now I grant you, because of the absence of a gravity drag penalty, it's possible that a ship of monstrous size could deliver huge quantities of ice. Because you're not dealing with any surface-to-orbit transition, your ship could be fragile, and your engine could have a low thrust. Given that, the only thing that scales directly with the payload is the propellant. If we can get something like 30 km/s reaction velocities, then it's believable that making the round trip would work... although only if you can cannibalize water for propellant, and at somewhat poor margins. So, sure, there is a business case in a very specific scenario where we have a mature market for water in space, someone could reduce costs of procurement by making a ridiculous-looking bloated spaceship that runs the Saturn to Earth orbit route. I'm still not convinced that it would be better than other alternatives beyond the frost line, but most of those have angular clustering. A business might want deliveries more than once every year, and for that, Saturn versus Jupiter trojans (for instance) provide some seasonal diversity. But the advantage is nil except for with extremely large quantities. That raises the question - how could the depreciation possibly compete with lunar ice? If you can run regular ferries from lunar poles to orbit, that seems more economic unless your bloated Saturn ship could be much lower cost (because the lunar scheme has much higher trip frequency). Maybe it could. We can't really say. The biggest variable will be your rate of return. If you demand 6% rate of return, then you're effectively cutting the value of the ice in half over 12 years. This is why it's a hard sell, but that's not really the rocket science part of it. Addition: Found the example of a "bloated spaceship" that I had in mind. Saw it on Atomic Rockets. Here is the original source. Apparently they had Phobos in mind. I have some problems with their off-the-bat assumptions. If Phobos has water, why doesn't Eros? I don't know these things. I'm doubtful that anyone else knows the bulk composition or the difficulty of refining the below-the-surface precious water of such inner system bodies. Myself, I would like to picture that "spaceship" with the bladder being 100x its above relatively size. These things will make a hot air balloon look lean.
The X-15 had a reaction control system for all three axes using thrusters with hydrogen-peroxide monopropellant. There was an automatic as well as a manual mode. The manual mode used a single three-axis control joystick.There were two completely independent systems. Each system used six RCS thrusters, two for each axis for both rotation directions. See ... The attitude thrusters and TCMs are mechanically identical, all Aerojet MR-103s. From the Voyager Press Kit:The 16 thrusters on the mission module each deliver 0.89 N (0.2-lb.) thrust. Four are used to execute trajectory correction maneuvers; the others in two redundant six-thruster branches, to stabilize the spacecraft on its three axes. Only one branch ... The vanes move to stabilse the spacecraft - although that may only be a first order stabilisation "Compensationforanunbalanceinsolarradiationpressureisprovidedbymoveablepaddleslocatedonthetipsofthesolarpanels" The thruster configuration can be seen better in this image:(cropped from this document ).The attitude control thrusters are not in plane with the direction control thrusters. This Raytheon patent on MKVs states that "attitude control system includes multiple thrusters offset from the center of gravity that provide yaw, pitch and roll control." All four of ... Gemini had a four-gimbal system as well:The IMU is the usual gimballed stable platform with accelerometers and angular resolvers as in Apollo, except for a key difference that the Gemini IMU had four gimbals rather than the three gimbals of Apollo. This means that it was not subject to the phenomenon of "gimbal lock", and hence the software used to ... The Shuttle Orbiter's Inertial Measurement Units (IMUs) had four gimbals.The IMU consists of a platform isolated from vehicle rotations by fourgimbals. Since the platform does not rotate with the vehicle, itsorientation remains fixed, or inertial, in space. The gimbal orderfrom outermost to innermost is outer roll, pitch, inner roll, and... This is not a full answer, but some numbers:What are the ISS moments of inertia around design axes?The total moment of inertia of the station is about $M = 55\cdot 10^6 \rm kg m^2$How often must the ISS desaturate its control moment gyros?The reaction wheels are desaturated when they reach $13000 \rm ft lbf sec$ which is $L = 17 \rm kJs = 17 \rm kgm^2s^... Without any time markings, it's impossible to tell if the ground track indicates a "hard left" or very gentle maneuvers over a long period of time.The annotated transcript gives us some hints, though:[Pete is descending very slowly as he flies along the north rim of Surveyor Crater, looking for a good spot to land.][Conrad, from the 1969 ... I don't think it's happened in a long time, but in the early stages of assembly, the ISS sometimes flew in a "XPH" attitude when the beta angle* was between 10 and 75 degrees.The only public info I could find on this is from a rather annoying flash animation NASA page; it shows some animations of the orbits, here are two frames from the XPH example showing ... The conversion of the attitude data as it appeared (undocumented) from Lightstreamer to roll, pitch, and yaw was the biggest pile of guesses and assumptions atop assumptions I made back when Matt and I were coding this. It is almost certainly wrong if it is varying from the ISS Live! website.We've been working on a complete rewrite of the code and it'd be ... Voyager's thrusters are in 3 groups:2 branches of 6 attitude control thrusters (which provide rotation around the major axes). The 2 branches provide redundancy1 set of 4 trajectory correction maneuver thrusters which were designed to provide translation. By firing them individually instead of in pairs, they can provide pitch and yaw (but not roll).... In this figure, ζ is the input axis, η is the output axis, and the spin axis is obvious.If the vehicle is rotated around the axis ζ (the input axis) with an angular velocity ωζ, the frame will rotate around the axis η (the output axis) to an angle β. The dependence of this angle on ωζ is given by the equation β = ωζH/c, where H is the moment of momentum of ... By integrating inertial measurements, initialized from the final star tracking about 20 or 30 minutes earlier.Also, from Spaceflight 101:With the separation of the Cruise Stage seven minutes before re-entry, the InSight spacecraft solely relies on its MIMU for attitude propagation, rate measurement and the deceleration trigger for the critical ... When you're trying to rotate something, there are two cases:1) The torque you're applying is large compared to the angular momentum the body has, i.e. when the body isn't rotating. Then it starts to rotate in the direction of the torque you're applying. This is the more intuitive case.2) The torque you're applying is small compared to the angular ... A hemispherical resonator is an example of a mechanical gyroscope that has no bearing parts and for practical purposes no moving parts either. See the Wiki page here.I actually find the description of the principal of operation there a little hard to follow so I've just taken the black box principle, that the vibration patterns in the surface respond to ... It boils down to: how much spacecraft resource is required by the attitude control method you propose using? And sometimes the mission's pointing requirements play a significant role.Thrusters use propellant. Reaction wheels (and momentum wheels) use electric power. Spin stabilization uses neither, as long as you don't need to repoint the spacecraft — but ... The equation you listed above is greatly simplified and fails to account for additional elements such as the spacecraft surface area exposed to the sun. Assuming you calculate solar radiation pressure as the following:$$ F_{s}=\frac{\phi }{c}$$Where:c = speed of lightϕ = Solar constant at your distance from the sunMy RecommendationI would ... I think you are right, the pressure of the gas does not inhibit the evaporation of lubricants. The same is true for air and water vapour. If we compress air from outdoors to 200 bar, a lot of water condenses, but the partial pressure of water is the same, it depends only on temperature, but not on ambient pressure. If the compressed air is expanded to 1 bar ... Most reaction wheel assemblies use ball bearings between the rotor and housing, with some kind of lubricant, typically liquid, coating the ball bearings. There's a problem with liquid lubricants: They evaporate in vacuum. For this reason, most reaction wheel assemblies are hermitically sealed with a low pressure inert gas inside the container.Alternatives ... There are 12 thrusters on InSight, mounted around the outer edge of the lander, all pointed in fixed generally downward orientation. The thrusters have electrically operated valves which can be operated in quite short pulses - the thruster can fire for a fraction of a second at a time.By controlling the rate and/or duration of the pulses differently on ... Passive damper booms dissipate energy internally so that oscillations around the stable orientation gradually damp down.There are some interesting, if slightly dated NASA monographs about these:Tubular Spacecraft Booms, Extendible, Reel Stored SP-8065 DODGE is mentioned 14 times.Passive Gravity-Gradient Libration Dampers, SP-8071How do they work? ... According to the Blue Origin publicity web site https://www.blueorigin.com/new-shepard/ , the aft fin hydraulics are effective up to mach 4 (altitude isn't specified, but actual max is [remember it's sub-orbital] mach 3), and those fins are also used as steering canards during descent.If you are interested in re-entry control, then you should also check ... The ISS Live page is using the ISSACS system (ISS Analysis frame) which is the normal coordinate frame for this data.Just doing the transforms in my head, it looks like the other web site is using a frame where the Y axis is the same as ISSACS, but the frame is pitched so that +X points to nadir and +Z points to ISS aft. Thus pitch is the same in both ... The torque will be the same - application of point torque to an object (in this case the point is the reaction wheel motor axis) results in the same torque exerted regardless of the point. But the moment of inertia won't be - because the layout of mass will differ. As the reaction wheels are not massles, they are a part of the system, and while mass won't ... I think the Int-Ball uses all fans in a "push-configuration" and only does have passive air inlets in it's chassis. The internals of the chassis are therefore "empty" enough to enable sufficient airflow to the fans.On this website if found a picture credited to JAXA:You can see those "Air inlets" in your first picture at the bottom of the drone. Gyroscopes used as sensors can be mechanical, ie with spinning parts, or electrical in which case lasers are often used. Movement is measured and the output is some sort of data which goes to a guidance system.Gyroscopes used to provide physical force to stabilize or orient a craft have to be mechanical as the electric signals or lasers used in electronic ... Some of the confusion may arise in that the gyros under discussion are not used to actually change the attitude of the HST...that is done by Reaction Wheel Assemblies, (separate spinning devices) and magnetic torquers. The gyros in question are sensors only.Knowing that, it makes sense that.Both the FHST and the FGS measure two degrees-of-freedom of ...
"The concept of a conditional probability with regard to an isolated hypothesis whose probability equals 0 is inadmissible." A. Kolmogorov For continuous random variables, $X$ and $Y$ say, conditional distributions are defined by the property that they recover the original probability measure, that is, for all measurable sets $A\in\mathcal{B}(\mathbf{X})$, $B\in\mathcal{B}(\mathbf{Y})$,$$\mathbb{P}(X\in A,Y\in B)=\int_B \text{d}P_Y(y) \int_B \text{d}P_{X\|Y}(x\|y)$$This implies that the conditional density is defined arbitrarily on sets of measure zero or, on other words, that the conditional density $p_{X\|Y}(x\|y)$ is defined almost everywhere. Since the set $\{5,6\}$ is of measure zero against the Lebesgue measure, this means that you can define both $p(5)$ and $p(6)$ in absolutely arbitrary manners and hence that the probability $$\mathbb{P}(U=5\|U\in\{5,6\})$$can take any value. This does not mean you cannot define a conditional density by the ratio formula $$f(y\|x)=f(x,y)\big/f(x)$$as in the bivariate normal case but simply that the density is only defined almost everywhere for both $x$ and $y$. "Many quite futile arguments have raged - between otherwise competent probabilists - over which of these results is 'correct'." E.T. Jaynes The fact that the limiting argument (when $\epsilon$ goes to zero) in the above answer seems to give a natural and intuitive answer is related with Borel's paradox. The choice of the parametrisation in the limit matters, as shown by the following example I use in my undergrad classes. Take the bivariate normal $$X,Y\stackrel{\text{i.i.d.}}{\sim}\mathcal{N}(0,1)$$ What is the conditional density of $X$ given that $X=Y$? If one starts from the joint density $\varphi(x)\varphi(y)$, the "intuitive" answer is [proportional to] $\varphi(x)^2$. This can be obtained by considering the change of variable $$(x,t)=(x,y-x) \sim \varphi(x)\varphi(t+x)$$ where $T=Y-X$ has the density $\varphi(t/\sqrt{2})/\sqrt{2}$. Hence $$f(x\|t)=\dfrac{\varphi(x)\varphi(t+x)}{\varphi(t/\sqrt{2})/\sqrt{2}}$$ and $$f(x\|t=0)=\dfrac{\varphi(x)\varphi(x)}{\varphi(0/\sqrt{2})/\sqrt{2}}=\varphi(x)^2\sqrt{2}$$ However, if one considers instead the change of variable $$(x,r)=(x,y/x) \sim \varphi(x)\varphi(rx)\|x\|$$ the marginal density of $R=Y/X$ is the Cauchy density $\psi(r)=1/\pi\{1+r^2\}$ and the conditional density of $X$ given $R$ is $$f(x\|r)=\varphi(x)\varphi(rx)\|x\| \times \pi \{1+r^2\}$$ Therefore, $$f(x\|r=1)= \pi\varphi(x)^2\|x\|/2\,.$$And here lies the "paradox": the events $R=1$ and $T=0$ are the same as $X=Y$, but they lead to different conditional densities on $X$.
In my post Trigonometry Yoga, I discussed how defining sine and cosine as lengths of segments in a unit circle helps develop intuition for these functions. I learned the circle definitions of sine and cosine in my junior year of high school, in the class that would now be called pre-calculus (it was called “Trig Senior Math”). Two years earlier, I’d learned the triangle definitions of sine, cosine, and tangent in geometry class. I don’t remember any of my teachers ever mentioning a circle definition of the tangent function. The geometric definition of the tangent function, which predates the triangle definition, is the length of a segment tangent to the unit circle. The tangent really is a tangent! Just as for sine and cosine, this one-variable definition helps develop intuition. Here is the definition, followed by an applet to help you get a feel for it: Let OA be a radius of the unit circle, let B = (1,0), and let $ \theta =\angle BOA$. Let C be the intersection of $\overrightarrow{OA}$ and the line x=1, i.e. the tangent to the unit circle at B. Then $\tan \theta$ is the y-coordinate of C, i.e. the signed length of segment BC. Move the blue point below; the tangent is the length of the red segment. (If a label is getting in the way, right click and toggle “show label” from the menu). The circle definition of the tangent function leads to geometric illustrations of many standard properties and identities. (If this were my class, I would stop here and tell you to explore on your own and with others). Some things to notice: $\left\| \tan \theta \right\|$ gets big as $\theta$ approaches $\pm 90{}^\circ $. $\tan (\pm 90{}^\circ)$ is undefined, because at these angles, $\overline{OA}$ is parallel to x=1, so the two lines don’t intersect, and point C doesn’t exist. $\tan 90{}^\circ$ tends toward $+\infty$, $\tan (-90{}^\circ)$ tends toward $-\infty$. $\tan \theta$ is positive in the first and third quadrants, negative in the second and fourth quadrants. $\tan \theta$=$tan (\theta+180{}^\circ)$ — the angles $\theta$ and $\theta +180{}^\circ$ form the same line. Thus the period of the tangent function is $180 {}^\circ = \pi$ radians. $\tan \theta$ = $- \tan (-\theta)$. Moving from $\theta$ to $-\theta$ reflects $OC$ about the x-axis. $\tan \theta$ is equal to the slope of OA (rise = $\tan \theta$ , run =1), which is also equal to $\dfrac{\sin\theta}{\cos\theta}$, as well as Opposite over Adjacent for angle $\theta$ in right triangle CBO. $\tan (45{}^\circ)=1$. When $\theta=45{}^\circ$, triangle CBO is a 45-45-90 triangle, and OB=1. Similarly, $\tan (-45{}^\circ)=-1$, etc. For small values of $\theta$, $\tan \theta$ is close to $\sin \theta$, which is close to the arc length of AB, i.e. the measure of $\theta$ in radians. If we define $\arctan \theta$ as the function whose input is the signed length of BC and whose output is the angle $\theta$ corresponding to that tangent length, then the domain of that function is the reals, and it makes sense to define the range as $-90 {}^\circ< \theta <90{}^\circ$ (in radians $-\pi/2<\theta < \pi/2$ and arctan’s output is an arc length). This range includes all the angles we need and avoids the discontinuity at $\theta= \pm 90{}^\circ =\pm \pi/2$ radians. For $\left\| \theta \right\|\leq 45{}^\circ$, $\left\| \tan \theta \right\|\leq 1$. Half of the input values of $\tan \theta$ give outputs with absolute values less than or equal to 1, and the other half give values on the rest of the number line. This mapping also occurs with fractions and slopes, but there’s something very compelling about seeing the lengths change dynamically. Applets like the one above could also help students develop intuition about slopes. $\tan (180{}^\circ-\theta) = -\tan \theta$. We reflect BC over the x-axis to form $B{C}’$. Then $\angle BO{C}’=\theta$ and $\angle BOD =(180{}^\circ-\theta)$. $B{C}’$ (the blue segment) is the tangent of $(180{}^\circ-\theta)$. $\tan (\theta \pm 90{}^\circ)$ = $-1/\tan \theta$. The picture below illustrates the geometry of this identity when $\theta$ is in the first quadrant. The line formed at $\theta + 90{}^\circ$ is perpendicular to OC and $\triangle COB\sim \triangle ODB$. Thus $\dfrac{BD}{OB}=\dfrac{OB}{BC}$, and with appropriate signs, $\tan (\theta + 90{}^\circ)$ = $-1/\tan \theta$. Since $\tan \theta$=$\tan (\theta+180{}^\circ)$, $\tan (\theta +90{}^\circ)=\tan(\theta-90{}^\circ)$. The applet below shows the geometry in all quadrants, and it gives a dynamic sense of the relationship between $\tan\theta$ and $\tan(-\theta)$. Again, move the blue point: Special Bonus: The Secant Function The signed length of the segment OC is called the secant function, $\sec\theta$. Using similar triangles, we see that $\sec \theta = \dfrac{1}{\cos \theta}$. The Pythagorean Theorem applied to $\triangle COB$ shows that $\tan^2\theta+1=\sec^2 \theta$. When the tangent function is big, so is the secant function, and when the tangent function is small, so is the secant function. Also $\sec \theta$ is close to $\pm 1$ when $\theta$ is close to the x-axis and when $\tan \theta$ is close to 0. The graphs of the two functions look nice together:
This question already has an answer here: Why is E(XY)=E(XE(Y\|X))? Is this using the properties of conditional expectation and is there a general formula that can be applied when you have E(...)=E(..(E(Y\|X))? Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.Sign up to join this community This question already has an answer here: Why is E(XY)=E(XE(Y\|X))? Is this using the properties of conditional expectation and is there a general formula that can be applied when you have E(...)=E(..(E(Y\|X))? By the expression $\mathbb{E}(XY),$ we mean the expectation of XY under their joint distribution. I.e., if these both are continuous, we have that $$ \mathbb{E}(XY) = \int\int xyf(x, y)dxdy, $$ where $f(x, y)$ is the joint pdf of X and Y. For this reason, we sometimes write $\mathbb{E}_{(X, Y)}(\cdot),$ or $\mathbb{E}_f(\cdot)$ in order to make it explicit which distribution the expectation is to be taken over. We can factorise distributions in a certain way, specifically we can write the pdf of a $d$-dimensional continuous random vector as $$f(x_1, \dots, x_d) = g_1(x_1)g_2(x_2\vert x_1)g_3(x_3\vert x_2, x_1)\cdots g_d(x_d\vert x_1, \dots, x_{d-1}).$$Applying this to $(X, Y),$ we can write $f(x, y) = g(x)h(y\vert x).$ We can then see that\begin{align}\mathbb{E}_{(X,Y)}(XY) &= \int\int xyf(x, y)dxdy\\&= \int\int xyg(x)h(y\vert x)dxdy\\&= \int x\int yh(y\vert x) dy f(x)dx\\&= \mathbb{E}_X\left(X\int y h(y\vert x) dy\right)\\&= \mathbb{E}_X\left(\mathbb{E}_{(Y\vert X)}\left(Y\right)\right).\end{align}This is also true for distributions that are not continuous, and can be generalised to a higher dimension analoguosly to the factorisation property of distributions. To show this in a very general context, you need some measure-theoretic arguments. The general formula that you request is often referred to as the law of iterated expectations,the tower rule, the smoothing theorem, or the law of total expectation. Another way to show it is directly using Law of iterated expectations: $$\begin{align}E[XY] &= E[E[XY\|X]]=E[XE[Y\|X]]\end{align}$$
I'm implementing a 2D physics engine where collision response is based on impulse computation. I'm going to first expose the context, then the problem, and finally the questions. Context Simple collision between two solids, the first one is static, the second one falls. The second solid has its colliding face parallel to the second solid. Two collisions point are generated (one at each side of the second solid colliding edge). Given two solids A and B who collided. \$\vec Va\$ is the vector from A's center of mass to one Colliding Point. \$\vec Vb\$ is the vector from B's center of mass to one Colliding Point. \$\vec N\$ is the collision normal. \$Ia\$ and \$Ib\$ are inertia tensor in world space of solids A and B. Here is the impulse formula if you only take linear speed into account: $$J = \frac{-(1 + bounceCoeff) * (\vec Va - \vec Vb) \cdot \vec N}{\frac{1}{mA} + \frac{1}{mB}}$$ Here is the impulse formula if you take linear and angular speed into account: $$J = \frac{-(1 + bounceCoeff) * (\vec Va - \vec Vb) \cdot \vec N}{\frac{1}{mA} + \frac{1}{mB} + ((\frac{\vec Va \times \vec N}{Ia}) + (\frac{\vec Vb \times \vec N}{Ib})) \cdot \vec N}$$ Here is the way I update velocities for n contact points: float \$pointCoeff = \frac{1}{\textbf{n}}\$ for(currentCollidingPoint : points) { \$\vec Va(t+1) = \vec Va(t) + \vec N * (J * \frac{1}{massA}) * pointCoeff\$ \$\vec Vb(t+1) = \vec Vb(t) - \vec N * (J * \frac{1}{massB}) * pointCoeff\$ } Problem Case 0: If I only take the linear speed into account, the second solid falls on the static one and bounce upward (the resulting velocity post collision is positive on the y axis). Case 1: If I take the linear and the angular speed into account, the second solid falls on the static one but doesn't bounce (the resulting velocity post collision is still negative on the y axis). Question Am I right to state this is not normal that, because colliding faces are parallel and only B is moving down, B should bounce no matter if we take angular speed into account (A only performs translation, no rotation)? I think that because if I only generate one collision point in the middle of B's colliding edge, B bounces If I'm right, I guess that can be due to the way I consider how the pointCoeff variable is used in the algorithm. Do you have any suggestion or piece of advice ?
κ(G) ≤ λ(G) ≤ δ(G) ≤ 2\|E(G)\|/\|V(G)\| ≤ Δ(G) Relation of Min Vertex Cutsets, Min Edge Cutsets, The Minimum Degree, and Maximum Degree We are going to prove the following inequality relating the minimum vertex cutset $\kappa (G)$, the minimum edge cutset $\lambda (G)$, the minimum degree in a graph $\delta (G)$, and the maximum degree in a graph $\Delta (G)$.(1) \begin{align} \kappa (G) ≤ \lambda (G) ≤ \delta (G) ≤ 2 \frac{\| \: E(G) \: \|}{\| \: V(G) \: \|} ≤ \Delta (G) \end{align} We will not prove all of these components, however, we will prove a few of them. Lemma 1: The minimum edge cutset of a graph $\lambda (G)$ is less than or equal to the minimum degree in a graph, $\delta (G)$, that is $\lambda (G) ≤ \delta (G)$. Proof: Suppose we have a graph G with a vertex $x \in V(G)$ where $\deg (x) = \sigma (G)$. If we take all of the edges incident with x and remove them, we obtain an edge cutset as vertex x will be isolated. We can always obtain an edge cutset this way. Hence, our minimum edge cutset will either be equal to our minimum degree or less. So $\lambda (G) ≤ \delta (G)$. Lemma 2: The minimum degree in a graph, $\delta (G)$ is less than or equal to twice the number of edges of $G$ divided by the number of vertices in $G$, that is $\delta (G) ≤ \frac{2 \mid E(G) \mid }{\mid V(G) \mid}$. (2) Proof: We acknowledge that $2 \frac{\| \: E(G) \: \|}{\| \: V(G) \: \|}$ represents the average degree in a graph G. Nevertheless we can prove this inequality algebraically. First we know by the Handshaking Lemma that: \begin{align} \sum_{x \in V(G)} \deg (x) = 2 \| \: E(G) \: \| \end{align} (3) Hence it follows that $\delta (G)$ is the minimum degree in the graph, then: \begin{align} \sum \delta (G) ≤ \sum_{x \in V(G)} \\ n \delta (G) ≤ 2 \| \: E(G) \: \| \\ \delta (G) ≤ 2 \frac{\| \: E(G) \: \|}{n} \end{align} (4) But we know that n represents the number of vertices in the graph. Hence: \begin{align} \delta (G) ≤ 2 \frac{\| \: E(G) \: \|}{\| \: V(G) \: \|} \end{align} Lemma 3: Twice the number of edges divided by the number of vertices $\frac{2 \mid E(G) \mid}{\mid V(G) \mid}$ is less than or equal to the maximum degree of $G$, $\Delta (G)$, that is $\frac{2 \mid E(G) \mid}{\mid V(G) \mid} ≤ \Delta (G)$. (5) Proof: We will complete this proof in the same manner as the last lemma. By the Handshaking lemma we know that: \begin{align} \sum_{x \in V(G)} \deg (x) = 2 \| \: E(G) \: \| \end{align} (6) And we also know that if $\Delta (G)$ is the maximum degree in our graph, then: \begin{align} \sum_{x \in V(G)} \deg (x) ≤ \sum \Delta (G) \\ \sum_{x \in V(G)} \deg (x) ≤ n \Delta (G) \\ 2 \| \: E(G) \: \| ≤ n \Delta (G) \\ 2 \frac{\| \: E(G) \: \|}{n} ≤ \Delta (G) \end{align} (7) But n represents the number of vertices in G, hence we are done the proof and: \begin{align} 2 \frac{\| \: E(G) \: \|}{\| \: V(G) \: \|} ≤ \Delta (G) \end{align}
The Cauchy-Schwarz Inequality We are about to look at one of the more significant inequalities in mathematics known as the Cauchy-Schwarz inequality. Before we do so though, we will first need to be able to write each vector $u$ in an inner product space as a sum of a scalar multiple of a vector $v$ and a vector $w$ that is orthogonal to $v$. Let $a \in \mathbb{F}$. Then $u = av + u - av = av + (u - av)$. We now want $(u - av)$ to be orthogonal to $v$. This happens if:(1) Thus we have that:(2) We are now ready to state and prove the Cauchy-Schwarz inequality. Theorem 1 (The Cauchy-Schwarz Inequality): Let $V$ be an inner product space. Then: a) $\mid <u, v> \mid ≤ \\| u \\| \\| v \\|$ for all $u, v \in V$. b) $\mid <u, v> \mid = \\| u \\| \\| v \\|$ if and only if one of $u$ or $v$ is a scalar multiple of the other. Proof of a)Let $V$ be an inner product space and let $u, v, w \in V$ and suppose that $v$ is orthogonal to $w$. Suppose that $v = 0$. Then $\mid <u, v> \mid = \mid <u, 0> \mid = 0$ and $\\| u \\| \\| v \\| = \\| u \\| \\| 0 \\| = 0$ so the inequality holds. Suppose instead that $v \neq 0$. Let $u$ be written as the sum of $v$ and $w$ such that: Now we can apply The Pythagorean Theorem for Inner Product Spaces since $u$ is the sum of two vectors $v$ and $w$ which are orthogonal. We now multiply both sides of this inequality by $\\| v \\|^2$ and then take the square root of both sides Proof of b)$\Rightarrow$ Suppose that $\mid <u, v> \mid = \\| u \\| \\| v \\|$. Then from the Cauchy-Schwarz inequality proof above we must have that: But then this implies that $\\| w \\|^2 = 0$ and so $w = 0$. But then $u = \frac{<u, v>}{\\| v \\|^2} v + w = \frac{<u, v>}{\\| v \\|^2} v$. Therefore $u$ is a scalar multiple of $v$. $\Leftarrow$ Suppose that $u$ is a scalar multiple of $v$. Then $u = kv$ for some $k \in \mathbb{F}$ ($\mathbb{R}$ or $\mathbb{C}$) and so:
amp-mathml Displays a MathML formula. Required Script <script async custom-element="amp-mathml" src="https://cdn.ampproject.org/v0/amp-mathml-0.1.js"></script> Supported Layouts container Examples amp-mathml.amp.html This extension creates an iframe and renders a MathML formula. <amp-mathml layout="container" data-formula="\[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\]"> </amp-mathml> <amp-mathml layout="container" data-formula="\[f(a) = \frac{1}{2\pi i} \oint\frac{f(z)}{z-a}dz\]"> </amp-mathml> <amp-mathml layout="container" data-formula="$$ \cos(θ+φ)=\cos(θ)\cos(φ)−\sin(θ)\sin(φ) $$"> </amp-mathml> This is an example of a formula of <amp-mathml layout="container" inline data-formula="`x`"></amp-mathml>, <amp-mathml layout="container" inline data-formula="$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$"></amp-mathml> placed inline in the middle of a block of text. <amp-mathml layout="container" inline data-formula="$ \cos(θ+φ) $"></amp-mathml> This shows how the formula will fit inside a block of text and can be styled with CSS. Specifies the formula to render. If specified, the component renders inline ( inline-block in CSS). See amp-mathml rules in the AMP validator specification. You've read this document a dozen times but it doesn't really cover all of your questions? Maybe other people felt the same: reach out to them on Stack Overflow.Go to Stack Overflow Found a bug or missing a feature? The AMP project strongly encourages your participation and contributions! We hope you'll become an ongoing participant in our open source community but we also welcome one-off contributions for the issues you're particularly passionate about.Go to GitHub
My question is the following: Suppose $M$ is an $n \times n$ symmetric real matrix. I want to find an $n \times n$ symmetric real matrix X such that $\|\| X -M\|\|_F$ is minimized with the constraint that $U^T X U \succeq 0$ (positive semidefinite), where $U$ is an $n \times d$ matrix, $U^TU = I_d$ and $n > d$. $\|\| \cdot \|\|_F$ is the frobenius norm of matrix. In short: \begin{equation} X^* = \displaystyle \text{argmin}_{X:U^TXU\succeq 0}\|\|X-M \|\|_F, X, M \in \mathcal{S_n} \end{equation} Some useful information may be: $U$ can be obtained from eigenvalue decomposition of $BB^T$ or QR decomposition of $B$ in the sense that there exists $Q = [U,V]$ such that $Q^TQ = QQ^T = I_n, U^TB = 0$. I would like to find a formula or construction based on eigenvalue decomposition or singular value decomposition for M. For example, if the constraint is $X \succeq 0$ instead of $U^T X U \succeq 0$, we can first apply eigenvalue decomposition to $M$ such that $M = Q \text{diag}(\lambda_1,\cdots, \lambda_n) Q^T$, then $X^* =Q \text{diag}(\max\{\lambda_1,0\},\cdots, \max\{\lambda_n,0\}) Q^T $ would minimize $\|\| X-M \|\|_F$. I used CVX package to compute $X^$ for some examples and I found some interesting facts: $ (X^-M) B = 0$ where $U^TB=0$ and $X^* -M$ is always positive semidefinite while $M$ and $X^*$ may not be positive semidefinite. Any help or suggestion of what method or tools I should use for these kind of problems would be much appreciated! Thanks~
We can use prime factorization to find the smallest common multiple of two positive integers. The least common multiple (l.c.m.) of two positive integers is the smallest positive integer that is a multiple of both. We denote the least common multiple of two positive integers $a$ an $b$ by $\langle a,b\rangle$. $\langle2,8\rangle=8$, $\langle5,8\rangle=40$ We can figure out $\langle a,b\rangle$ once we have the prime factorization of $a$ and $b$. To do that, let \[a=p_1^{a_1}p_2^{a_2}...p_m^{a_n}\] and \[b=p_1^{b_1}p_2^{b_2}...p_m^{b_n},\] where (as above) we exclude any prime with 0 power in both $a$ and $b$. Then $\langle a,b\rangle=p_1^{\max(a_1,b_1)}p_2^{\max(a_2,b_2)}...p_m^{\max(a_n,b_n)}$, where $\max(a,b)$ is the maximum of the two integers $a$ and $b$. We now prove a theorem that relates the least common multiple of two positive integers to their greatest common divisor. In some books, this theorem is adopted as the definition of the least common multiple. To prove the theorem we present a lemma If a and b are two real numbers, then \[\min(a,b)+\max(a,b)=a+b\] Assume without loss of generality that $a\geq b$. Then \[\max(a,b)=a \hspace{0.2cm}\mbox{and} \ \ \min(a,b)=b,\] and the result follows. Note Let $a$ and $b$ be two positive integers. Then $\langle a,b\rangle\geq 0$; $\langle a,b \rangle=ab/(a,b)$; If $a\mid m$ and $b \mid m$, then $\langle a,b \rangle \mid m$ Proof The proof of part 1 follows from the definition. As for part 2, let \[a=p_1^{a_1}p_2^{a_2}...p_m^{a_n} \mbox{and} \ \ b=p_1^{b_1}p_2^{b_2}...p_m^{b_n}.\] Notice that since \[(a,b)=p_1^{\min(a_1,b_2)}p_2^{\min(a_2,b_2)}...p_n^{\min(a_n,b_n)}\] and \[\langle a,b\rangle=p_1^{\max(a_1,b_1)}p_2^{\max(a_2,b_2)}...p_m^{\max(a_n,b_n)},\] then \[\begin{aligned} \langle a,b \rangle(a,b)&=&p_1^{\max(a_1,b_1)}p_2^{\max(a_2,b_2)}...p_m^{\max(a_n,b_n)} p_1^{\min(a_1,b_2)}p_2^{min(a_2,b_2)}...p_n^{\min(a_n,b_n)}\\ &=& p_1^{\max(a_1,b_1)+\min(a_1,b_1)}p_2^{\max(a_2,b_2)+\min(a_2,b_2)}...p_m^{\max(a_n,b_n)+\min(a_n,b_n)}\\&=& p_1^{a_1+b_1}p_2^{a_2+b_2}...p_n^{(a_n+b_n)}\\&=&p_1^{a_1}p_2^{a_2}...p_m^{a_n}p_1^{b_1}p_2^{b_2}...p_m^{b_n}=ab\end{aligned}\] Note also that we used Lemma 8 in the above equations. For part 3, it would be a nice exercise to show that $ab/(a,b) \mid m$ (Exercise 6). Thus $\langle a,b \rangle \mid m$. $\square$ Exercises Find the least common multiple of 14 and 15. Find the least common multiple of 240 and 610. Find the least common multiple and the greatest common divisor of $2^55^67^211$ and $2^35^87^213$. Show that every common multiple of two positive integers $a$ and $b$ is divisible by the least common multiple of $a$ and $b$. Show that if $a$ and $b$ are positive integers then the greatest common divisor of $a$ and $b$ divides their least common multiple. When are the least common multiple and the greatest common divisor equal to each other. Show that $ab/(a,b) \mid m$ where $m=<a,b>$.
The plane through the ray and the origin intersects your ellipsoid to form a planar ellipse. To find this planar ellipse one must establish a basis vector on the plane such that any point on the plane can be represented by two parameters ($\alpha$ and $\beta$). $$ \vec{r} = \alpha \vec{p} + \beta \vec{d} $$ The ellipse on the plane is now given by the equation $$ \alpha^2 C_{11} + 2 \alpha\beta C_{12} + \beta^2 C_{22} = 1 $$ $$ \pmatrix{\alpha \\ \beta \\ 1}^\top \pmatrix{C_{11} & C_{12} & \\ C_{12} & C_{22} & \\ & & -1} \pmatrix{\alpha \\ \beta \\ 1} =0 $$ where $$\begin{align} C_{11} &= \frac{x_p^2}{a^2} + \frac{y_p^2}{a^2} + \frac{z_p^2}{b^2} \\ C_{22} &= \frac{x_d^2}{a^2} + \frac{y_d^2}{a^2} + \frac{z_d^2}{b^2} \\ C_{12} &= \frac{x_p x_d}{a^2} + \frac{y_p y_d}{a^2} + \frac{z_p z_d}{b^2} \end{align} $$ Now for the fun part. To be perpendicular to the ray the point $\vec{r}$ must obey the projection $$ \vec{d}^\top (\vec{r}-\vec{p}) =0 $$ From the expansion of $\vec{r}$ in terms of the planar coordinates $\alpha$ and $\beta$ this projection is used to solve for $$ \alpha = 1- \frac{\vec{d}^\top \vec{d}}{\vec{d}^\top \vec{p}} \beta = 1-\frac{x_d^2+y_d^2+z_d^2}{x_d x_p + y_d y_p + z_d z_p} \beta = 1- \lambda \, \beta $$ So given any distance along the ray $\beta$ the 3D position of the point is $\vec{r} = (1-\lambda \,\beta) \vec{p} + \beta \vec{d} $ where $\lambda$ is the fixed ratio $\lambda = \frac{x_d^2+y_d^2+z_d^2}{x_d x_p + y_d y_p + z_d z_p}$. The points on the ellipse that are perpendicular to the ray solve the equation $$ (1-\lambda\,\beta)^2 C_{11} + 2 (1-\lambda\,\beta)\beta C_{12} + \beta^2 C_{22} = 1 \\ (C_{11} \lambda^2 -2 C_{12} \lambda + C_{22}) \beta^2 + 2 (C_{12}-C_{11} \lambda) \beta + (C_{11-1}) = 0 $$ This has two solutions (as expected) $$ \beta = \frac{C_{11} \lambda-C_{12} \pm \sqrt{ C_{11} (\lambda^2-C_{22}) + C_{12}^2 -2 C_{12} \lambda + C_{22}}}{C_11 \lambda^2 -2 C_{12} \lambda + C_{22}} $$ And the back substitution $$ \alpha = 1 - \lambda\, \beta \\ \vec{r} = \alpha \vec{p} + \beta \vec{d} $$ Edit 1To find the lines tangent to the planar ellipse, and passing through P you do the following: The planar coordinates of P are $(\alpha=1,\beta=0)$, or in homogeneous coordinates $$P=\pmatrix{1,& 0,& 1}$$ The homogeneous coordinates of the planar ellipse are $$C=\left\| \matrix{C_{11} & C_{12} & \\ C_{12} & C_{22} & \\ & & -1} \right\|$$ The polar line of P is $L=C\,P$ $$L = \left[ \matrix{C_{11}, & C_{12},& -1} \right]$$ The polar line intersects the ellipse at the tangent points $Q_1$ and $Q_2$. To find these points form the equation of the line $C_{11} \alpha + C_{12} \beta - 1 =0$ and plug it into the ellipse equation to find $$Q_1 = \pmatrix{ 1 + \tfrac{C_{12} \sqrt{C_{11}-1}}{K}, & - \tfrac{C_{11} \sqrt{ C_{11}-1 }}{K}, & C_{11} }$$ $$Q_2 = \pmatrix{ 1 - \tfrac{C_{12} \sqrt{C_{11}-1}}{K}, & \tfrac{C_{11} \sqrt{ C_{11}-1 }}{K}, & C_{11} }$$where $K=\sqrt{C_{11} C_{22}-C_{12}^2}$. The tangent line coordinates are $T_1 = C \,Q_1$ and $T_2 = C \, Q_2$$$ T_1 = \left[ \matrix{ C_{11}, & C_{12} -K \sqrt{C_{11}-1} , & -C_{11} } \right] $$ $$ T_2 = \left[ \matrix{ C_{11}, & C_{12} +K \sqrt{C_{11}-1} , & -C_{11} } \right] $$ In equation form the tangent lines are$$ (C_{11}) \alpha + (C_{12} - K \sqrt{C_{11}-1}) \beta - C_{11} = 0 $$$$ (C_{11}) \alpha + (C_{12} + K \sqrt{C_{11}-1}) \beta - C_{11} = 0 $$
This is a technical problem which must have been solved already. It won't be in beginners textbooks but there should be a solution somewhere. I welcome reading suggestions. Maybe someone with experience in solving Navier-Stokes equations numerically can help me. Here goes: An incompressible fluid flowing down a pipe obeys the Navier-Stokes equations $$\partial_t \mathbf{v} + (\mathbf{v} \cdot \nabla) \mathbf{v} = \nu \Delta \mathbf{v} - \nabla P \, ,$$ and the pressure is related to the velocity field by the incompressibility condition, $\nabla \cdot \mathbf{v} = 0$. When the flow is time dependent one usually specifies the velocity field at $t=t_0$, $\mathbf{v}(t_0,\mathbf{x}) = \mathbf{v}_0(\mathbf{x})$. The no-slip boundary condition requires the velocity to vanish on the edges of the pipe for all time $\mathbf{v}(t,r=R,\theta,z) = 0$. I use cylindrical coordinates $\mathbf{x}=(r \cos{\theta},r \sin{\theta},z)$. $R$ is the radius of the pipe. In practice however the pressure must be expressed in terms of the velocity field at all times. To the best of my knowledge this is achieved by taking the divergence of Navier-Stokes equations. Using the incompressibility condition one can write this as \begin{align} \nabla \cdot \left[(\mathbf{v} \cdot \nabla) \mathbf{v}\right] = (\partial_i v_j) (\partial_j v_i) = -\nabla^2 P \, . && () \end{align} Then the time evolution is computed as follows: Assume that instead of the no-slip boundary condition we specifiy the pressure on the boundary for all times, $P(t,R,\theta,z) = P_R(t,\theta,z)$. Then the initial velocity field $\mathbf{v}_0(\mathbf{x})$ is changed by a small time increment from $t=t_0$ in the following way, \begin{align}\mathbf{v}(t_0+dt,\mathbf{x}) = \mathbf{v}_0(\mathbf{x}) + dt \left[ - (\mathbf{v}_0 \cdot \nabla) \mathbf{v}_0 + \nu \Delta \mathbf{v}_0 - \nabla P(t_0) \right]\, . && (*) \end{align} The pressure at time $t_0$ is then the solution of an inhomogenous Laplace problem with the specified boundary condition given by $P_R(t_0,\theta,z)$. Specifying the pressure on the boundary $P_R(\theta,z)$ is mathematically fine. We can use this to integrate the flow equations for all times by iterating the procedure that I just outlined. It is however not very physical. We usually control the velocity field on the boundary and prefer to use the no-slip boundary condition on the velocity field. One then goes back to the no-slip boundary condition by tuning $P_R(t_0,\theta,z)$ in such a way that $\mathbf{v}_0(t_0+dt,R,\theta,z)=0$. If we choose initial conditions such that $v_0(R,\theta,z)=0$ we extract the pressure through $$ \nu \Delta \mathbf{v}_0(R,\theta,z) = \nabla P(t_0,R,\theta,z) \, .$$ My problem is the following: How can we be sure that this last step is possible? On the left-hand side we have an arbitrary (?) vector. It may not be possible to write it as the gradient of a scalar field. It looks like the no-slip boundary condition contains more information than the specification of $P(t,R,\theta,z) = P_R(t,\theta,z)$. Edit (18 sept 2015): An obvious solution to my problem would be to consider a potential velocity field, $$\mathbf{v} = \mathbf{\nabla} \phi \, .$$ Indeed, in this case we can simply identify the Laplacian of the potential with the pressure on the boundary, $$\nu \nabla^2 \phi_0(R,\theta,z) = P(t_0,R,\theta,z) \, . $$ This is however a strong restriction which I do not want since I am interested in the transition to turbulence. Edit (8 Oct 2015): I do not want to actually solve this problem numerically. I know that the scheme that I propose here is naive. What I want is to be convinced that the solution exists, is smooth and obeys the no-slip boundary conditions for all times.
PCTeX Talk Discussions on TeX, LaTeX, fonts, and typesetting Author Message zedler Joined: 03 Mar 2006 Posts: 15 Posted: Mon Mar 27, 2006 11:53 am Post subject: spacing of mathrm Hello, \documentclass{book} \usepackage{times,mtpro2} \begin{document} $(\mathrm j$ \end{document} gives touching glyphs. Michael Michael Spivak Joined: 10 Oct 2005 Posts: 52 Posted: Tue Mar 28, 2006 12:29 pm Post subject: Re: spacing of mathrm zedler wrote: Hello, \documentclass{book} \usepackage{times,mtpro2} \begin{document} $(\mathrm j$ \end{document} gives touching glyphs. Michael There's not much I can do about that---if you are using Times as the text font, then in text (j also touches! [though $(\mathrm j$ is worse, with more overlap]. I'm wondering how this arose. Assuming that you didn't really want \mathrm{(j ... I would guess that you are using roman letters as a set of variables, either in addition to, or in place of, the MTPro2 italic letters. In that case, you really would want a special font for this purpose, in the same way that MTPro's \mathbf font has different spacing than the Times-bold, so that subscripts and superscripts will work better. zedler Joined: 03 Mar 2006 Posts: 15 Posted: Wed Mar 29, 2006 5:23 am Post subject: Re: spacing of mathrm Quote: There's not much I can do about that---if you are using Times as the text font, then in text (j also touches! [though $(\mathrm j$ is worse, with more overlap]. I'm wondering how this arose. Assuming that you didn't really want \mathrm{(j ... I would guess that you are using roman letters as a set of variables, either in addition to, or in place of, the MTPro2 italic letters. In that case, you really would want a special font for this purpose, in the same way that MTPro's \mathbf font has different spacing than the Times-bold, so that subscripts and superscripts will work better. Yes, I really want to typeset $\exp(\mathrm j\omega\tau=$ ;-) I suppose this can only be corrected by increasing the bracket side bearings, but your approach was to have very tight bracket side bearings and adjust/increase the spacing using kerns. This of course fails for \mathrm... The tight bracket side bearings were also an issue in my previous example, $\[\]_{xy}$. CM, Fourier, Lucida and MnSymbol don't have this problem... Michael Michael Spivak Joined: 10 Oct 2005 Posts: 52 Posted: Wed Mar 29, 2006 6:39 am Post subject: Re: spacing of mathrm zedler wrote: Quote: There's not much I can do about that---if you are using Times as the text font, then in text (j also touches! [though $(\mathrm j$ is worse, with more overlap]. I'm wondering how this arose. Assuming that you didn't really want \mathrm{(j ... I would guess that you are using roman letters as a set of variables, either in addition to, or in place of, the MTPro2 italic letters. In that case, you really would want a special font for this purpose, in the same way that MTPro's \mathbf font has different spacing than the Times-bold, so that subscripts and superscripts will work better. Yes, I really want to typeset $\exp(\mathrm j\omega\tau=$ ;-) I suppose this can only be corrected by increasing the bracket side bearings, but your approach was to have very tight bracket side bearings and adjust/increase the spacing using kerns. This of course fails for \mathrm... The tight bracket side bearings were also an issue in my previous example, $\[\]_{xy}$. CM, Fourier, Lucida and MnSymbol don't have this problem... Michael Actually, I didn't, and one can't, adjust the spacing after a left parentheses or before a right parenthesis using kerns [I mentioned this on some posting somewhere once before]; even if you put kerns into the tfm file, they are ignored because the left parenthesis is an "opening", which determines its own spacing, and similarly the right parenthesis is a "closing". I chose side bearings for the parenthesis that worked well with the italic letters on the math italic font. Even if that were not the case, the real problem is that in the expression \exp(\mathrm j the ( comes from the math italic font, while the j is coming from an entirely different font, the Times-Roman font, and TeX has no way of kerning characters in different fonts. If you were to use some other roman font as your text font, then the problem might very well be less or much more---it would depend entirely on the left side bearing of j in that particular font. I suspect that j is being used here as a some special character (perhaps in electrical engineering, although I thought they preferred bold j); in that case, I would just define something like \myj to give a small kern followed by j---in fact, it's easier to type \myj than to type \mathrm j. Sorry that [] doesn't work out for you, but I've never seen something like that in any mathematics paper, and since I like the way brackets work with the math italic characters in general, I wouldn't want to change the side bearings just for this special case (once again, changes couldn't be overridden with kerns). zedler Joined: 03 Mar 2006 Posts: 15 Posted: Wed Mar 29, 2006 10:00 am Post subject: Re: spacing of mathrm Quote: Sorry that [] doesn't work out for you, but I've never seen something like that in any mathematics paper, and since I like the way brackets work with the math italic characters in general, I wouldn't want to change the side bearings just for this special case (once again, changes couldn't be overridden with kerns). I can apply manual spacings, the "\mathrm j" is stored in a macro anyway and the empty brackets I need only once. Perhaps you're interested, I've put together a collection showing how different math font setups behave in the above mentioned cases: http://www.hft.ei.tum.de/mz/mtpro2_sidebearings.pdf Michael Michael Spivak Joined: 10 Oct 2005 Posts: 52 Posted: Wed Mar 29, 2006 1:24 pm Post subject: Re: spacing of mathrm Quote: ="zedler I can apply manual spacings, the "\mathrm j" is stored in a macro anyway and the empty brackets I need only once. Perhaps you're interested, I've put together a collection showing how different math font setups behave in the above mentioned cases: http://www.hft.ei.tum.de/mz/mtpro2_sidebearings.pdf Michael Interesting. I'd say that CM looks the worst (especially the \omega and \tau, as well as being so thin). Lucida is somewhat "klunky", though definitely easy to read! (Is this Lucida or Lucida-Bright?) Some one mentioned that section headings are sometimes printed in sans-serif, so that a sans-serif math might be nice to have; I suspect that the Lucida greek letters would work well for that. If \mathrm j is in a macro, then probably there should also be some space on the right; certainly needed for CM, not really needed for Lucida or Minion, useful for Fourier and MTPro2. By the way, what is []_{\langle6\times6\rangle} ? zedler Joined: 03 Mar 2006 Posts: 15 Posted: Wed Mar 29, 2006 3:25 pm Post subject: Re: spacing of mathrm Quote: (Is this Lucida or Lucida-Bright?) Pctex's Lucida fonts. Quote: By the way, what is []_{\langle6\times6\rangle} ? Excerpt from a paper (Deadline tomorrow Mar 30, Hawai time ;-)) Code: \begin{document}\let\mathbf\mbf ... Next, the impedance matrix of the outer 12-port is obtained by inverting $\mathbf{Y}^{\langle 16\times16\rangle}$ and taking the upper left $\langle 12\times12\rangle$ submatrix \begin{equation} \mathbf{Z}^{\langle 12\times12\rangle}=\left[{\mathbf{Y}^{\langle 16\times16\rangle}}^{-1}\right]_{\langle 12\times12\rangle} \end{equation} where the operator $[\,]_{\langle 12\times12\rangle}$ denotes taking the submatrix. The $\mathbf Z^{\langle 6\times 6\rangle}=\mathbf Z$ matrix of the outer six-port is obtained by Perhaps not the best notation, do you have a better idea? BTW, quite funny that both you and my boss are aficionados of differential forms ;-) Wish you wedge and hodge, Michael Michael Spivak Joined: 10 Oct 2005 Posts: 52 Posted: Wed Mar 29, 2006 3:39 pm Post subject: Re: spacing of mathrm zedler wrote: Excerpt from a paper (Deadline tomorrow Mar 30, Hawai time ;-)) Code: \begin{document}\let\mathbf\mbf ... Next, the impedance matrix of the outer 12-port is obtained by inverting $\mathbf{Y}^{\langle 16\times16\rangle}$ and taking the upper left $\langle 12\times12\rangle$ submatrix \begin{equation} \mathbf{Z}^{\langle 12\times12\rangle}=\left[{\mathbf{Y}^{\langle 16\times16\rangle}}^{-1}\right]_{\langle 12\times12\rangle} \end{equation} where the operator $[\,]_{\langle 12\times12\rangle}$ denotes taking the submatrix. The $\mathbf Z^{\langle 6\times 6\rangle}=\mathbf Z$ matrix of the outer six-port is obtained by Perhaps not the best notation, do you have a better idea? BTW, quite funny that both you and my boss are aficionados of differential forms ;-) Wish you wedge and hodge, Michael Not really, but I would probably have used something like UL_{\langle 12\times\12\rangle}(...) with U and L roman (or perhaps bold). And I probably would actually have used something like UL_{[12]}, with the idea that for square matrices [12] would mean \langle12\times12\rangle. 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The Norm of a Vector Definition: If $\vec{u} \in \mathbb{R}^n$, then the Norm or Magnitude of $\vec{u}$ denoted $\\| \vec{u} \\|$ is defined as the length or magnitude of the vector and can be calculated using the formula: $\\| \vec{u} \\| = \sqrt{u_1^2 + u_2^2 + ... + u_n^2}$. We will note that the norm of a vector is sometimes denoted with single bars, that is $\mid \vec{u} \mid$ is a notation commonly used to denote what we have defined. We will not use this notation to prevent confusion with mistaking the norm of a vector and the absolute value of a scalar. Example 1 Calculate the norm of the vector $\vec{u} = (3, 4)$. We first note that $\vec{u} \in \mathbb{R}^2$, and we will thus use the formula $\\| \vec{u} \\| = \sqrt{u_1^2 + u_2^2}$. In fact, this formula should make sense geometrically as it analogous to the Pythagorean theorem as illustrated: When we substitute our values in, we obtain that $\\| \vec{u} \\| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Thus our vector $\vec{u}$ has length 5. Example 2 Find the norm of the vector $\vec{u} = (2, -2, 3, -4)$. Since $\vec{u} \in \mathbb{R}^4$, we will use the formula $\\| \vec{u} \\| = \sqrt{u_1^2 + u_2^2 + u_3^2 + u_4^2} = \sqrt{4 + 4 + 9 + 16} = \sqrt{33}$. So the norm of our vector $\vec{u}$ is the square root of 33. The Distance Between Two Points Remember, we can write a vector that starts at some initial point $P$, and some terminal point $Q$. Recall that we can calculate this vector in 3-space with the formula $\vec{PQ} = (x_{Q} - x_{P}, y_{Q} - y_{P}, z_{Q} - z_{P})$. We can thus apply this formula to obtain the norm of $\vec{PQ}$:(1) We note that the norm of $PQ$ is also equal to the distance between $P$ and $Q$, so an alternative way to write this formula is:(2) Where $d$ denotes the distance between the initial point $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$. Example 3 Find the distance between the point $P(2, 3, 4)$, and the point $Q(5, 4, 3)$: Applying the formula we just established we obtain that:(3) Therefore, the distance between point $P$ and $Q$ is $\sqrt{11}$. The Norm of a Scalar Multiple of a Vector Suppose that we have a vector $\vec{u}$. If we multiply this vector by a scalar $k$, then the norm of the vector $k\vec{u}$ will be k-times larger than $k$. There is a problem though. We define the norm to be the magnitude or length of the vector so the norm must be positive. Hence we draw the following relation:(4) We note that $\mid k \mid$ represents the absolute value of k, that is: Thus if $k$ is negative, then we take the positive of $k$, otherwise $k$ stays the same. For example, $\mid -3 \mid = 3$. Note that sometimes the notation $\mathrm{abs} (k)$ will be used to denote the absolute value. From this we can draw a relationship between a scalar multiple of a vector and its norm, that is:(6) Example 4 Given that $\\| \vec{u} \\| = 4$, find $\\| -3\vec{u} \\|$. From the formula given, we note that $\\| -3 \vec{u} \\| = \mid -3 \mid \\| \vec{u} \\| = (3)(4) = 12$.
Here’s a simple and intuitive way of looking at the geometry of a least squares regression: Take the bottom left point in the triangle below as the origin O. For the linear model: $$ Y=X\beta + \epsilon $$ Both $Y$ and $X\beta$ are vectors, and the residual vector $\epsilon$ is the difference. The standard least squares error technique uses $\epsilon^2$ or $(Y-X\beta)^T(Y-X\beta)$ as the error measure to be minimised, and this leads to the calculation of the $\beta$ coefficient vector. Geometrically, the beta coefficients calculated by the least squares regression minimise the squared length of the error vector. This turns out to be the projection of $Y$ on to $X\beta$ – i.e. the perpendicular vector that turns (O, $Y$, $X\beta$) into a right-angled triangle. The projection of $Y$ onto $X\beta$ is done using the projection matrix P, which is defined as \[ P = X\left(X^{T}X\right)^{-1}X^{T} \] So $ X\beta = \hat{Y} = PY $. Using the Pythagorean theorem: $ Y^TY = \hat{Y}^T\hat{Y} + (Y-X\beta)^T(Y-X\beta) $ In other words, the total sum of squares = sum of squares due to regression + residual sum of squares. This is a fundamental part of analysis of variance techniques.
General Background In this class we dealt with various crystals models, like the tableau models. However, many of these models only work for specific classes of Kac-Moody Lie algebras (eg finite dimensional), and often must be changed to fit with any particular Lie algebra. One of the ideas behind the model presented in this paper is that it can be applied to any complex, symmetrizable Kac-Moody algebra. We will just briefly go over the definitions, and important facts without proof. $ \lambda $ Chains (of roots) First we define a pair of counting functions to shorten our notation. Let $ \lbrace \beta_i \rbrace_{\lbrace i \in I\rbrace} $ be any set, indexed by a completed ordered set $ I $. Then: $ N(\alpha)= $ number of times $ \alpha $ appears in the sequence. $ N_i( \alpha )= $number of times $ \alpha $ appears in the sequence strictly before $ i $. Definition: Let $ \lambda \in P_+$ be a dominant weight. A $ \lambda-chain$ is a sequence of positive roots, $ \lbrace \beta_i \rbrace_{\lbrace i \in I\rbrace}$, such that the following two conditions hold: (1) $ N(\alpha) = \langle \lambda , \alpha^\vee \rangle$ for all positive roots $ \alpha \in \Phi^+$. (2) If $ \gamma^\vee = \alpha^\vee + k \beta^\vee$, and $ \beta_i=\beta$, then: $ N_i(\gamma)=N_i(\alpha)+kN_i(\beta)$ The idea is to pick any ambient $ \lambda $ chain, and that this will give us an equivalent crystal later on. However, in order to make sure that $ \lambda $-chains exist, we will construct one. Fix an ordering on the simple roots: $ \alpha_1,...\alpha_r $. We will order the set $ \lbrace ( \alpha,k) : \alpha \in \Phi^+,0 \leq k < \langle \lambda , \alpha^\vee \rangle \rbrace $. This will provide a $ \lambda $-chain as long as the second condition holds. Let $ \alpha^\vee = \sum\limits_p a_p \alpha_p^\vee $ , $ \beta^\vee = \sum\limits_p b_p \alpha_p^\vee $ . Then we let: $ (\alpha , k) < (\beta , l) $ if and only if $ \frac{1}{\langle \lambda , \alpha^\vee \rangle} (k,a_1,...,a_r) < \frac{1}{\langle \lambda , \beta^\vee \rangle} (l,b_1,...,b_r) $ in lexicographical order. The proof that this gives a $ \lambda $-chain, isn't particularly enlightening, but can be found in the paper. Folding Chains (of roots) Next we build a set of foldings recursively by first creating a basis element using the $ \lambda $-chain, and then defining folding operators which generate the rest of the foldings. $ \Gamma ( \varnothing ) = ( \lbrace (\beta_i , 1) \rbrace_{i \in I} , \rho ) $ Where $ \rho $ is the dominant weight that pairs to one with all simple roots. More generally, a folding will take the following form: $ \Gamma = ( \lbrace ( \gamma_i, \epsilon_i ) \rbrace_{i \in I} , \gamma_\infty ) $ Where $ \gamma_i $ are (possibly negative) roots, $ \epsilon_i \in \lbrace \pm 1 \rbrace $, and $ \gamma_\infty $ is just some weight (in the orbit of $ \rho $ under the Weyl group action) . Our folding operators are as follows: $ \phi_i ( \Gamma ) = ( \lbrace \delta_j , \zeta_j ) \rbrace_{j \in I} , t_i(\gamma_\infty )) $ Where: $ t_i = s_{\gamma_i} $, the reflection about the hyperplane perpendicular to $ \gamma_i $ , and $ \delta_j = \gamma_j $ if $ j \leq i $, $ \delta_j = t_i(\gamma_j) $ if $ j > i $, $ \zeta_j = \epsilon_j $ if $ j \neq i $ , $ \zeta_j = - \epsilon_j $ if $ j = i $ . Proposition: $ \phi_i , \phi_j $ commute for all $ i $ , $ j $ and so in particular, each folding is uniquely determined by which $ \epsilon_i $ are negative, and so is in bijection with finite subsets of $ I $ . For our arbitrary $ \Gamma $ this subset will be $ J = \lbrace j_1, ... , j_k \rbrace $. We will often use $ \Gamma $ and $ J $ interchangeably. Since this is supposed to give us a crystal, we want to define the parts that make it a crystal. First we will define the weight, but to do this we need to define the level sequence. Definition: The levels of a folding are: $ l_i = -\delta_{\gamma < 0} + \sum\limits_{j<i, \epsilon_j=1,\gamma_j = \pm \gamma_i} sgn(\gamma_j) $. We call the level sequence of folding is $ (l_i)_{i \in I} $ . In particular, for we write $ (l_i^\varnothing)_{i \in I} $ for the level sequence of $ \Gamma(\varnothing) $, Definition: $ \mu(\Gamma(J)) = \hat{r_{j_1}} \cdot \cdot \cdot \hat{r_{j_k}} $ which we call the Weight of $ \Gamma $. Where $ r_i = s_{\beta_i} $ , $ \hat{r_i} $ is the reflection about $ \langle \lambda , \beta_i \rangle = l^\varnothing_i $ . We also define $ \kappa(\Gamma) = r_{j_1} \cdot \cdot \cdot r_{j_k} $. Finally we define some particular subsets of interest, of both $ I $, and the level sequences. Definition: $ I_\alpha = \lbrace i \in I : \gamma_i = \pm \alpha \rbrace \subseteq I $ , $ \hat{I_\alpha} = I_\alpha \cup \lbrace \infty \rbrace $, Definition: $ L_\alpha = \lbrace l_i \in I : i \in I_\alpha \rbrace $ , $ \hat{I_\alpha} =L_\alpha \cup \lbrace l_\alpha^\infty = \langle \mu(\gamma), \alpha^\vee \rangle \rbrace $, Admissible Foldings We want to restrict to a subset of all possible foldings, which will give us our crystal. Definition: If $ (1) < r_{j_1} < r_{j_1}r_{j_2}<...<r_r_{j_1}r_{j_2} \cdot \cdot \cdot r_{j_k} = \kappa(J) $ are in fact coverings in the Bruhat order, we call $ \Gamma(J) $ (the folding associated the subset $ J $ ), an admissible folding. Now we will define the rest of the crystal. Definition: $ \epsilon_p = M= $ maximum value in $ \hat{L_\alpha} $ . ( $ \phi_p = \langle \mu(J) , \alpha_p^\vee \rangle - M $ ) Definition: $m= $ minimum index that gives $l_m = M $ . Let k be the predecessor of m in $ \hat{I_\alpha} $ . Proposition: $ l_k=l_m-1 $ Proposition: Such a k always exists. Definition: $ f_p(J) = \phi_k \phi_m (J) $. We define $ e_p $ similarly. Proposition: $ Proposition: \( e_p$, $f_p $ send admissible folds to admissible folds. Finally, here is the main theorem: Theorem: The collection of admissible subsets along with the above defined root operators form a (semi-perfect) crystal. Works Cited All information presented on this page is paraphrased or directly taken from the following paper: A Combinatorial Model for Crystals of Kac-Moody Algebras, Trans. of AMS, pg 4349-4381, August 2008 Contributors Colin Hagemeyer (UC Davis)
The Set of Accumulation Points under Homeomorphisms on Topological Spaces Recall from the Homeomorphisms on Topological Spaces page that if $X$ and $Y$ are topological spaces then a bijective map $f : X \to Y$ is said to be a homeomorphism if it is continuous and open. Furthermore, if such a homeomorphism exists then we say that $X$ and $Y$ are homeomorphic and write $X \simeq Y$. We will now look at a nice topological property which says that if $f$ is a homeomorphism from $X$ to $Y$ and if $A$ is a subset of $X$ then the image of the set of accumulation points of $A$ is equal to the set of accumulation points of the image of $A$. Theorem 1: Let $X$ and $Y$ be topological spaces, let $f : X \to Y$ be a homeomorphism, and let $A \subseteq X$. Then $f(A') = (f(A))'$. Proof:Let $x \in f(A)'$. Then $f^{-1}(x) \in A'$, so $f^{-1}(x)$ is an accumulation point of $A$. So, for all open neighbourhoods $U$ in $X$ of $f^{-1}(x)$ we have that: Then we have that: Since $f$ is a homeomorphism and $U$ is an open set in $X$ we have that then $f(U)$ is an open set in $Y$. Furthermore, every open set in $Y$ is of the form $f(U)$ since $f$ is a bijection. Therefore $x$ is an accumulation point of $f(A)$, so $x \in (f(A))'$. Therefore: Now let $x \in (f(A))'$. Then $x$ is an accumulation point of $f(A)$ and so for every open neighbourhoods $V$ in $Y$ of $x$ we have that: Therefore we have that: Since $f$ is a homeomorphism and $V$ is an open set we have that $f^{-1} (V)$ is an open set in $X$. Furthermore, every open set in $X$ is of this form since $f$ is a bijection. Therefore $f^{-1} (x)$ is an accumulation point of $A$, so $f^{-1}(x) \in A'$ and $x \in f(A')$. Therefore: We conclude that then $f(A') = (f(A))'$. $\blacksquare$
Answer $-\frac{2}{5}$ Work Step by Step 1.Reciprocal identity $\cos\theta=\frac{1}{\sec\theta}$ 2. Substitute the given value: $\cos\theta=\frac{1}{-\frac{5}{2}}=-\frac{2}{5}$ 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.
Euler's Method for Approximating Solutions to Differential Equations We are now going to look at a numerical method for approximating a solution of a first order differential equation initial value problem known as Euler's Method or The Tangent Line Method. Suppose that $\frac{dy}{dt} = f(t, y)$ is a first order differential equation with the initial condition that $y(t_0) = y_0$, and suppose that both $f$ and $\frac{\partial f}{\partial y}$ are continuous on an interval containing $t_0$. Then we know that a solution exists and that this solution is unique, so call it $y = \phi (t)$. Now consider the line that passes through the point $(t_0, y_0)$ (which is given by the initial condition of the initial value problem) and has the slope $f(t_0, y_0)$. Then this line is tangent to the unique solution $y = \phi (t)$ at $(t_0, y_0)$ (since the slope of the tangent line at $(t_0, y_0)$ on $y = \phi (t)$ is $f(t_0, y_0)$). Furthermore, the equation of this line in point-slope form is given by the following formula:(1) Note that this tangent line is a pretty accurate approximation of the solution $y = \phi (t)$ for values of $t$ that are very close to $t_0$. Now suppose that $t_1$ is close to $t_0$, and that we want to approximate the value of $\phi (t_1)$ using the line approximation $y = y_0 + f(t_0, y_0)(t - t_0)$. Then by plugging in $t = t_1$ to the equation given above, we will obtain an approximation, call it $y_1$, of $\phi (t_1)$.(2) Since $y_1$ is an approximation of $\phi(t_1)$, we have that $y_1 \approx \phi(t_1)$. We will now consider the line that passes through the point $(t_1, y_1)$ and has the slope $f(t_1, y_1)$. In point-slope form, the equation of this line is given by the following formula:(3) Once again, if $t_2$ is a value that is close to $t_1$, then we can obtain an approximation $y_2 = y(t_2)$ of the true value $\phi (t_2)$ using this new line, $y = y_1 + f(t_1, y_1)(t - t_1)$, by plugging in $t = t_2$:(4) If we repeat this process, then the equation for the tangent line that passes through the the point $(t_n, y_n)$ for $n = 0, 1, 2, ...$ is given by:(5) Furthermore, the approximation $y_{n+1}$ of the value $\phi(t_{n+1})$ is obtained by plugging $t = t_{n+1}$ into the equation of the tangent line that passes through the point $(t_n, y_n)$ whose formula is:(6) Now suppose that we want to approximate the solution $y = \phi (t)$ of our differential equation $\frac{dy}{dt} = f(t, y)$ using $t_0, t_1, t_2, ...$. If we connect the line segments between the points $(t_0, y_0)$, $(t_1, y_1)$, …, $(t_n, y_n)$, … then we obtained a piecewise linear function which approximates $y = \phi (t)$. Of course, for practical reasons, it's nice if the difference between consecutive values $t_j$ and $t_{j+1}$ are equal, that is the step size between $t_j$ and $t_{j+1}$ is say $h > 0$ for $j = 0, 1, 2, ...$. The formula for the approximation of $y_{n+1}$ of $\phi(t_{n+1})$ can thus be rewritten as:(7) Of course, the accuracy of our piecewise approximation of the unique solution $y = \phi (t)$ varies depending on how large are steps are, how large our interval we're approximating is, and the behaviour of the solution itself. Let's look at an example of using Euler's Method in order to approximate the solution to a first order differential equation. Example 1 Consider the first order differential equation $\frac{dy}{dt} = 0.5 - t + 2y$ with the initial condition $y(0) = 1$. Approximate the values of the unique solution $y = \phi (t)$ for this initial value problem using Euler's Method at the values $t = 0.1, 0.2, 0.3, 0.4$ with the step size $h = 0.1$. The tangent line that passes through the point $(0, 1)$ (which is given by the initial condition as $y(0) = 1$) has the slope $f(0, 1) = 0.5 - 0 + 2 = 2.5$. Therefore, the equation of this tangent line is:(8) We will now use this line in order to approximate the value of $\phi(0.1)$. Plugging in $0.1$ to the tangent line above and we get that:(9) We thus obtain the point $(0.1, 1.25)$ as an approximation of $\phi(0.1)$. The slope of the tangent line that passes through this point is $f(0.1, 1.25) = 0.5 - 0.1 + 2(1.25) = 2.9$. Therefore the equation of this tangent line is:(10) We will now use this line in order to approximate the value of $\phi(0.2)$. Plugging in $0.2$ to the tangent line above and we get that:(11) We thus obtain the point $(0.2, 1.54)$ as an approximation of $\phi (0.2)$. The slope of the tangent line that passes through this point is $f(0.2, 1.54) = 0.5 - 0.2 + 2(1.54) = 3.38$. Therefore the equation of this tangent line is:(12) We will now use this line in order to approximate the value of $\phi(0.3)$. Plugging in $0.3$ to the tangent line above and we get that:(13) We thus obtain the point $(0.3, 1.878)$ as an approximation of $\phi(0.3)$. The slope of the tangent line that passes through this point is $f(0.3, 1.878) = 0.5 - 0.3 + 2(1.878) = 4.056$. Therefore the equation of this tangent line is:(14) We will now use this line in order to approximate the value of $\phi(0.4)$. Plugging in $0.4$ to the tangent line above and we get that:(15) We thus obtain the point $(0.4, 2.2836)$ The following curve in blue represents the actual solution to this initial value problem, $y = 0.5t + e^{2t}$. The curve in green is the piecewise curve created by joining the tangent line segments to create a function that approximates the actual solution.
It's impossible to find explicit formulas for solutions of some differential equations. Even if there are such formulas, they may be so complicated that they're useless. In this case we may resort to graphical or numerical methods to get some idea of how the solutions of the given equation behave. In Section~3.4 we'll take up the question of existence of solutions of a first order equation \begin{equation} \label{eq:3.4.1} y'=f(x,y). \end{equation} In this section we'll simply assume that \eqref{eq:3.4.1} has solutions and discuss a graphical method for approximating them. In Chapter~3 we discuss numerical methods for obtaining approximate solutions of \eqref{eq:3.4.1}. Recall that a solution of \eqref{eq:3.4.1} is a function $y=y(x)$ such that $y'(x)=f(x,y(x))$ for all values of $x$ in some interval, and an integral curve is either the graph of a solution or is made up of segments that are graphs of solutions.Therefore, not being able to solve \eqref{eq:3.4.1} is equivalent to not knowing the equations of integral curves of \eqref{eq:3.4.1}. However, it's easy to calculate the slopes of these curves. To be specific, the slope of an integral curve of \eqref{eq:3.4.1} through a given point $(x_0,y_0)$ is given by the number $f(x_0,y_0)$. This is the basis of $ \textcolor{blue}{\mbox{the method of direction field}} $ If $f$ is defined on a set $R$, we can construct a $ \textcolor{blue}{\mbox{direction field}} $ for \eqref{eq:3.4.1} in $R$ by drawing a short line segment through each point $(x,y)$ in $R$ with slope $f(x,y)$. Of course, as a practical matter, we can't actually draw line segments through $ \textcolor{blue}{\mbox{every}} $ point in $R$; rather, we must select a finite set of points in $R$. For example, suppose $f$ is defined on the closed rectangular region $R:\{a\le x\le b, c\le y\le d\}.$ Let $a= x_0< x_1< \cdots< x_m=b$ be equally spaced points in $[a,b]$ and $c=y_0<y_1<\cdots<y_n=d$ be equally spaced points in $[c,d]$. We say that the points $(x_i,y_j),\quad 0\le i\le m,\quad 0\le j\le n,$ form a $ \textcolor{blue}{\mbox{rectangular grid}} $ Figure~\ref{figure:3.4.1}). Through each point in the grid we draw a short line segment with slope $f(x_i,y_j)$. The result is an approximation to a direction field for \eqref{eq:3.4.1} in $R$. If the grid points are sufficiently numerous and close together, we can draw approximate integral curves of \eqref{eq:3.4.1} by drawing curves through points in the grid tangent to the line segments associated with the points in the grid. \begin{figure}[H] \centering \includegraphics[width = 0.7\textwidth]{fig010301} \color{blue} \caption{\quad A rectangular grid} \label{figure:3.4.1} \end{figure} Unfortunately, approximating a direction field and graphing integral curves in this way is too tedious to be done effectively by hand. However, there is software for doing this. As you'll see, the combination of direction fields and integral curves gives useful insights into the behavior of the solutions of the differential equation even if we can't obtain exact solutions. We'll study numerical methods for solving a single first order equation \eqref{eq:3.4.1} in Chapter~3. These methods can be used to plot solution curves of \eqref{eq:3.4.1} in a rectangular region $R$ $ \textcolor{blue}{\mbox{if \(f$ is continuous on}} \) $R$. Figures~\ref{figure:3.4.2}, \ref{figure:3.4.3}, and \ref{figure:3.4.4} show direction fields and solution curves for the differential equations $y'=\frac{x^2-y^2}{1+x^2+y^2}, y'=1+xy^2, y'=\frac{x-y}{1+x^2}$ which are all of the form \eqref{eq:3.4.1} with $f$ continuous for all $(x,y)$. \begin{figure}[htbp] \color{blue} \begin{minipage}[b]{0.5\linewidth} \centering \scalebox{.6}{ \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010302}} \caption{\,A direction field and integral curves for $y =\dst{\frac{x^{2}-y^{2}}{1+x^{2}+y^{2}}}$} \label{figure:1.3.2} \end{minipage} \hspace{0.6cm} \begin{minipage}[b]{0.5\linewidth} \centering \scalebox{.7}{ \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010303}} \caption{A direction field and integral curves for $y'=1+xy^2$} \label{figure:1.3.3} \end{minipage} \end{figure} \begin{figure}[tbp] \centering \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010304} \color{blue} \caption{A direction and integral curves for $y'=\dst{\frac{x-y}{1+x^2}}$} \label{figure:1.3.4} \end{figure} The methods of Chapter~3 won't work for the equation \begin{equation} \label{eq:3.4.2} y'=-x/y \end{equation} if $R$ contains part of the $x$-axis, since $f(x,y)=-x/y$ is undefined when $y=0$. Similarly, they won't work for the equation \begin{equation} \label{eq:3.4.3} y'={x^2\over1-x^2-y^2} \end{equation} if $R$ contains any part of the unit circle $x^2+y^2=1$, because the right side of \eqref{eq:3.4.3} is undefined if $x^2+y^2=1$. However, \eqref{eq:3.4.2} and \eqref{eq:3.4.3} can written as \begin{equation} \label{eq:3.4.4} y'={A(x,y)\over B(x,y)} \end{equation} where $A$ and $B$ are continuous on any rectangle $R$. Because of this, some differential equation software is based on numerically solving pairs of equations of the form \begin{equation} \label{eq:1.3.5} {dx\over dt}=B(x,y),\quad {dy\over dt}=A(x,y) \end{equation} where $x$ and $y$ are regarded as functions of a parameter $t$. If $x=x(t)$ and $y=y(t)$ satisfy these equations, then $y'={dy\over dx}={dy\over dt}\left/{dx\over dt}\right.={A(x,y)\over B(x,y)},$ so $y=y(x)$ satisfies \eqref{eq:3.4.4}. Eqns.~\eqref{eq:3.4.2} and \eqref{eq:3.4.3} can be reformulated as in \eqref{eq:3.4.4} with ${dx\over dt}=-y,\quad {dy\over dt}=x$ and ${dx\over dt}=1-x^2-y^2,\quad {dy\over dt}=x^2$ respectively. Even if $f$ is continuous and otherwise "nice'' throughout $R$, your software may require you to reformulate the equation $y'=f(x,y)$ as ${dx\over dt}=1,\quad {dy\over dt}=f(x,y)$ which is of the form \eqref{eq:3.4.5} with $A(x,y)=f(x,y)$ and $B(x,y)=1$. Figure~\ref{figure:3.4.5} shows a direction field and some integral curves for \eqref{eq:3.4.2}. As we saw earlier, the integral curves of \eqref{eq:3.4.2} are circles centered at the origin. \begin{figure}[H] \centering \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010305} \color{blue} \caption{A direction field and integral curves for $y'=-\dst{x\over y}$} \label{figure:1.3.5} \end{figure} Figure~\ref{figure:3.4.6} shows a direction field and some integral curves for \eqref{eq:3.4.3}. The integral curves near the top and bottom are solution curves. However, the integral curves near the middle are more complicated. For example, Figure~\ref{figure:3.4.7} shows the integral curve through the origin. The vertices of the dashed rectangle are on the circle $x^2+y^2=1$ $(\approx.846, b\approx.533)$, where all integral curves of \eqref{eq:3.4.3} have infinite slope. There are three solution curves of \eqref{eq:3.4.3} on the integral curve in the figure: the segment above the level $y=b$ is the graph of a solution on $(-\infty,a)$, the segment below the level $y=-b$ is the graph of a solution on $(-a,\infty)$, and the segment between these two levels is the graph of a solution on $(-a,a)$. As you study from this book, you'll often be asked to use computer software and graphics. Exercises with this intent are marked as \Cex\, (computer or calculator required), \CGex\, (computer and/or graphics required), or \Lex\ (laboratory work requiring software and/or graphics). Often you may not completely understand how the software does what it does. This is similar to the situation most people are in when they drive automobiles or watch television, and it doesn't decrease the value of using modern technology as an aid to learning. Just be careful that you use the technology as a supplement to thought rather than a substitute for it. \begin{figure}[htbp] \color{blue} \begin{minipage}[b]{0.5\linewidth} \centering \scalebox{.7}{ \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010306} } \caption{A direction field and integral curves for $y'=\dst{x^2\over1-x^2-y^2}$} \label{figure:1.3.6} \end{minipage} \hspace{0.6cm} \begin{minipage}[b]{0.5\linewidth} \centering \scalebox{.7}{ \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010307} } \caption{} \label{figure:1.3.7} \end{minipage} \end{figure} Contributors Trench, William F., "Elementary Differential Equations" (2013). . 8. Faculty Authored and Edited Books & CDs https://digitalcommons.trinity.edu/mono/8
Covering Spaces Definition: Let $X$ be a topological space. A Covering Space of $X$ is a pair $(\tilde{X}, p)$ where: 1. $\tilde{X}$ is a path connected and locally path connected topological space. 2. $p : \tilde{X} \to X$ is a continuous surjective mapping such that for every $x \in X$ there is an path connected open neighbourhood $U$ of $x$ such that the restriction of $p$ on every path component of $p^{-1}(U)$ is a homeomorphism onto $U$. The space $\tilde{X}$ is called a Cover of $X$. The map $p : \tilde{X} \to X$ is called a Covering Map. The open neighbourhoods $U$ of $x$ described above are called Elementary Neighbourhoods. For example, let $X = S^1$ - the unit circle. Then $\tilde{X} = \mathbb{R}$ is a covering space of $X$. To view this, we wrap $\mathbb{R}$ into a coil and view $X$ as the "shadow" this coil makes. We will now describe special covering spaces. Definition: Let $X$ be a topological space. The Trivial Covering Space of $X$ is the covering space consisting of $\tilde{X} = X$ and $p = \mathrm{id}$. A covering space of $X$ is said to be a Universal Covering Space of $X$ if $\pi_1(\tilde{X}, x)$ is the trivial group. From above we see that the trivial covering space of the circle is the circle, and that $\mathbb{R}$ is a universal covering space of $X$ since $\pi_1(\mathbb{R}, x)$ is the trivial group.
I have solution that can guarantee the end goal being reached in a maximum of four five steps. (Many thanks for @dmg and @Taemyr for their comments to fix this solution.) The trick is to: Reduce the cups into the following configuration: $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $ where turning any two cups along the diagonal will satisfy the end goal For a constructive proof, we first consider the following two three CASES: CASE 1) the orientation of a single cup is different from the others, e.g. $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix} $ or $\begin{bmatrix} \bullet & \circ \\ \bullet & \bullet \end{bmatrix} $ CASE 2a) two cups are in each orientation along the diagonal, e.g. $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $ CASE 2b) two cups are in each orientation along two sides, e.g. $\begin{bmatrix} \circ & \bullet \\ \circ & \bullet \end{bmatrix} $ Step 1 Take two cups along the diagonal. - If they are different, flip $\circ$ to $\bullet$. If this doesn't sound the bell, we either have $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix}$ or $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$, with the majority orientation being $\circ$. In either case, proceed to step 2. - If they are the same, flip both. If this doesn't sound the bell, we either have $\begin{bmatrix} \bullet & \circ \\ \bullet & \bullet \end{bmatrix}$ or $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Either way, the majority orientation is known, and we can proceed to step 3. Step 2 Take two cups along the diagonal. - If they are different, we now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Flip $\bullet$ to $\circ$ to win. - If they are the same, flip both. If this doesn't sound the bell, we now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Proceed to Step 3. Step 3 At this point, we have already figured out what the majority orientation is, so we assume we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix} $ WLOG. Then: Take two cups along the diagonal. - If the cups are different, flip $\bullet$ to $\circ$ to win. - If the cups are both $\circ$, flip one of them. We now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \bullet \end{bmatrix}$. Step 4 Take two adjacent cups along an edge and flip both. - If the cups chosen are the same, we win. - If they are different, we now know we have $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $ Step 5 Take two cups along a diagonal and flip both. We win!
I can't seem to figure out how to do the following: compute the implied correlation $ρ_{imp}$ by using the observed market price $M_{quote}$ of a Margrabe option, and solving the non-linear equation shown below: $$M_{quote} = e^{−q_0T}\times S_0(0)\times N(d_+)−e^{−q_1T}\times S_1(0)\times N(d_−)$$ where: $$\begin{align} & d_\pm = \frac{\log\frac{S_0(0)}{S_1(0)}+(q_1 − q_0 ±σ^2/2)T}{\sigma\sqrt{T}} \\[4pt] & \sigma = \sqrt{\sigma^2_0 + \sigma^2_1 − 2\rho_{imp}\sigma_0 \sigma_1} \end{align}$$ Note that $d_− = d_+ − σ\sqrt{T}$.
GolfScript (23 chars) {:^((1${\.*2^?%}+}:f; The sentinel result for a non-existent inverse is 0. This is a simple application of Euler's theorem. \$x^{\varphi(2^n)} \equiv 1 \pmod {2^n}\$, so \$x^{-1} \equiv x^{2^{n-1}-1} \pmod {2^n}\$ Unfortunately that's rather too big an exponential to compute directly, so we have to use a loop and do modular reduction inside the loop. The iterative step is \$x^{2^k-1} = \left(x^{2^{k-1}-1}\right)^2 \times x\$ and we have a choice of base case: either k=1 with {1\:^(@{\.*2^?%}+}:f; or k=2 with {:^((1${\.*2^?%}+}:f; I'm working on another approach, but the sentinel is more difficult. The key observation is that we can build the inverse up bit by bit: if \$xy \equiv 1 \pmod{2^{k-1}}\$ then \$xy \in \{ 1, 1 + 2^{k-1} \} \pmod{2^k}\$, and if \$x\$ is odd we have \$x(y + xy-1) \equiv 1 \pmod{2^k}\$. (If you're not convinced, check the two cases separately). So we can start at any suitable base case and apply the transformation \$y' = (x+1)y - 1\$ a suitable number of times. Since \$0x \equiv 1 \pmod {2^0}\$ we get, by induction \$x\left(\frac{1 - (x+1)^n}{x}\right) \equiv 1 \pmod {2^n}\$ where the inverse is the sum of a geometric sequence. I've shown the derivation to avoid the rabbit-out-of-a-hat effect: given this expression, it's easy to see that (given that the bracketed value is an integer, which follows from its derivation as a sum of an integer sequence) the product on the left must be in the right equivalence class if \$x+1\$ is even. That gives the 19-char function {1$)1$?@/~)2@?%}:f; which gives correct answers for inputs which have an inverse. However, it's not so simple when \$x\$ is even. One potentially interesting option I've found is to add x&1 rather than 1. {1$.1&+1$?@/~)2@?%}:f; This seems to give sentinel values of either \$0\$ or \$2^{n-1}\$, but I haven't yet proved that. Taking that one step further, we can ensure a sentinel of \$0\$ for even numbers by changing the expression \$1 - (x+1)^n\$ into \$1 - 1^n\$: {1$.1&)1$?@/~)2@?%}:f; That ties with the direct application of Euler's theorem for code length, but is going to have worse performance for large \$n\$. If we take the arguments the other way round, as n x f, we can save one character and get to 22 chars: {..1&)2$?\/~)2@?%}:f;
Focus Questions The following questions are meant to guide our study of the material in this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions. What is a trigonometric equation? What does it mean to solve a trigonometric equation? How is a trigonometric equation different from a trigonometric identity? We have already learned how to solve certain types of trigonometric equations. In Section 2.6 we learned how to use inverse trigonometric functions to solve trigonometric equations. Beginning Activity Refer back to the method from Section 2.6 to find all solutions to the equation $\sin(x) = 0.4$. Trigonometric Equations When a light ray from a point $P$ reflects off a surface at a point $R$ to illuminate a point $Q$ as shown at left in Figure 4.1, the light makes two angles $\alpha$ and $\beta$ with a perpendicular to the surface. The angle $\alpha$ is called the angle of incidence and the angle $\beta$ is called the angle of reflection. The Law of Reflection states that when light is reflected off a surface, the angle of incidence equals the angle of reflection. What happens if the light travels through one medium (say air) from a point $P$, deflects into another medium (say water) to travel to a point $Q$? Think about what happens if you look at an object in a glass of water. See Figure 4.1 at right. Again the light makes two angles $\alpha$ and $\beta$ with a perpendicular to the surface. The angle $\alpha$ is called the angle of incidence and the angle $\beta$ is called the angle of refraction. If light travels from air into water, the Law of Refraction says that \[\dfrac{\sin(\alpha)}{\sin(\beta)} = \dfrac{c_{a}}{c_{w}}\] Figure $\PageIndex{1}$: Reflection and refraction. where $c_{a}$ is the speed of light in air and $c_{w}$ is the speed of light in water. The ratio $\dfrac{c_{a}}{c_{w}}$ of the speed of light in air to the speed of light in water can be calculated by experiment. In practice, the speed of light in each medium is compared to the speed of light in a vacuum. The ratio of the speed of light in a vacuum to the speed of light in water is around 1.33. This is called the index of refraction for water. The index of refraction for air is very close to 1, so the ratio $\dfrac{c_{a}}{c_{w}}$ is close to 1.33. We can usually measure the angle of incidence, so the Law of Refraction can tells us what the angle of refraction is by solving equation (6). Trigonometric equations arise in a variety of situations, like in the Law of Refraction, and in a variety of disciplines including physics, chemistry, and engineering. As we develop trigonometric identities in this chapter, we will also use them to solve trigonometric equations. Recall that Equation (6) is a conditional equation because it is not true for all allowable values of the variable. To solve a conditional equation means to find all of the values for the variables that make the two expressions on either side of the equation equal to each other. Equations of Linear Type Section 2.6 showed us how to solve trigonometric equations that are reducible to linear equations. We review that idea in our first example. Example $\PageIndex{1}$: (Solving an Equation of Linear Type) Consider the equation \[2\sin(x) = 1.\] We want to find all values of $x$ that satisfy this equation. Notice that this equation looks a lot like the linear equation $2y = 1$, with $\sin(x)$ in place of $y$. So this trigonometric equation is of linear type and we say that it is linear in $\sin(x)$. We know how to solve $2y = 1$, we simply divide both sides of the equation by 2 to obtain $y = \dfrac{1}{2}$. We can apply the same algebraic operation to $2\sin(x) = 1$ to obtain the equation \[\sin(x) = \dfrac{1}{2}.\] Now we could proceed in a couple of ways. From previous work we know that $\sin(x) = \dfrac{1}{2}$ when $x = \dfrac{\pi}{6}$. Alternatively, we could apply the inverse sine to both sides of our equation to see that one solution is $x = \sin^{-1}(\dfrac{1}{2}) = \dfrac{\pi}{6}$. Recall, however, this is not the only solution. The first task is to find all of the solutions in one complete period of the sine function. We can use the interval with $0 \leq x \leq 2\pi$ but we often use the interval $-\pi \leq x \leq \pi$. In this case, it makes no difference since the sine function is positive in the second quadrant. Using $\dfrac{\pi}{6}$ as a reference angle, we see that $x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ is another solution of this equation. (Use a calculator to check this.) We now use the fact that the sine function is period with a period of $2\pi$ to write formulas that can be used to generate all solutions of the equation $2\sin(x) = 1$. So the angles in the first quadrant are $\dfrac{\pi}{6} + k(2\pi)$. and the angles in the second quadrant are $\dfrac{5\pi}{6} + k(2\pi)$ where $k$ is an integer. So for the solutions of the equation $2\sin(x) = 1$, we write $x = \dfrac{\pi}{6} + k(2\pi)$ or $x = \dfrac{5\pi}{6} + k(2\pi)$, where $k$ is an integer. We can always check our solutions by graphing both sides of the equation to see where the two expressions intersect. Figure 4.2 shows that graphs of $y = 2\sin(x)$ and $y = 1$ on the interval $[-2\pi, 3\pi]$. We can see that the points of intersection of these two curves occur at exactly the solutions we found for this equation. Exercise $\PageIndex{1}$ Find the exact values of all solutions to the equation $4\cos(x) = 2\sqrt{2}$. Do this by first finding all solutions in one complete period of the cosine function and Figure $\PageIndex{2}$: The graphs of $y = 2\sin(x)$ and $y = 1$ then using the periodic property to write formulas that can be used to generate all solutions of the equation. Draw appropriate graphs to illustrate your solutions. Answer We divide both sides of the equation $4\cos(x) = 2\sqrt{2}$ to get $\cos(x) = \dfrac{\sqrt{2}}{2}$. So \[x = \dfrac{\pi}{4} + k(2\pi)\] or \[x = \dfrac{7\pi}{4} + k(2\pi)\] where $k$ is an integer. Solving an Equation Using an Inverse Function When we solved the equation $2\sin(x) = 1$, we used the fact that we know that $\sin(\dfrac{\pi}{6}) = \dfrac{1}{2}$. When we cannot use one of the common arcs, we use the more general method of using an inverse trigonometric function. This is what we did in Section 2.6. See “A Strategy for Solving a Trigonometric Function” on page 158. We will illustrate this strategy with the equation $\cos(x) = 0.7$. We start by applying the inverse cosine function to both sides of this equation to obtain \[\cos(x) = 0.7\] \[\cos^{-1}(\cos(x)) = \cos^{-1}(0.7)\] \[x = \cos^{-1}(0.7)\] This gives the one solution for the equation that is in interval $[0, \pi]$. Before we use the periodic property, we need to determine the other solutions for the equation in one complete period of the cosine function. We can use the interval $[0, 2\pi]$ but it is easier to use the interval $[-\pi, \pi]$. One reason for this is the following so-called “negative arc identity” stated on page 82. \[\cos(-x) = \cos(x)\] for every real number $x$. Hence, since one solution for the equation is $x = \cos^{-1}(0.7)$, another solution is $x = -\cos^{-1}(0.7)$. This means that the two solutions of the equation \[x = \cos(x)\] on the interval $[-\pi, \pi]$ are $x = \cos^{-1}(0.7)$ and $x = -\cos^{-1}(0.7)$ Since the period iof the cosine function is $2\pi$, we can say that any solution of the equation $\cos(x) = 0.7$ will be of the form $x = \cos^{-1}(0.7) + k(2\pi)$ and $x = -\cos^{-1}(0.7) + k(2\pi)$ where $k$ is some integer. Note Note: The beginning activity for this section had the equation $\sin(x) = 0.4$. The solutions for this equation are $x = \arcsin(0.4) + k(2\pi)$ and $x = (\pi - \arcsin(0.4)) + k(2\pi)$ where $k$ is some integer. We can write the solutions in approximate form as $x = 0.41152 + k(2\pi)$ and $x = 2.73008 + k(2\pi)$ where $k$ is an integer. Exercise $\PageIndex{2}$ Determine formulas that can be used to generate all solutions to the equation $5\sin(x) = 2$. Draw appropriate graphs to illustrate your solutions in one period of the sine function. Approximate, to two decimal places, the angle of refraction of light passing from air to water if the angle of incidence is $40^\circ$. Answer 1. We divide both sides of the equation $5\sin(x) = 2$ by $5$ to get $\sin(x) = 0.4$. So \[x = \sin^{-1}(0.4) + k(2\pi)\] or \[x = (\pi - \sin^{-1}(0.4)) + k(2\pi)\] where $k$ is an integer. 2. We use $\alpha = 40^\circ$ and $\dfrac{c_{a}}{c_{w}} = 1.33$ in the Law of Refraction. \[\dfrac{\sin(40^\circ)}{\sin(\beta)} = 1.33\] \[\sin(\beta) = \dfrac{\sin(40^\circ)}{1.33} \approx 0.483299\] \[\beta \approx 28.90^\circ\] The angle of refraction is approximately $28.90^\circ$. Solving Trigonometric Equations Using Identities We can use known trigonometric identities to help us solve certain types of trigonometric equations. Example $\PageIndex{1}$: (Using Identities to Solve Equations) Consider the trigonometric equation \[\cos^{2}(x) - \sin^{2}(x) = 1\] This equation is complicated by the fact that there are two different trigonometric functions involved. In this case we use the Pythagorean Identity \[\sin^{2}(x) + \cos^{2}(x) = 1\] by solving for $\cos^{2}(x)$ to obtain \[\cos^{2}(x) = 1- \sin^{2}(x).\] We can now substitute into equation (7) to get \[(1 - \sin^{2}(x)) - \sin^{2}(x) = 1\] Note that everything is in terms of just the sine function and we can proceed to solve the equation from here: \[(1 - \sin^{2}(x)) - \sin^{2}(x) = 1\] \[1 - 2\sin^{2}(x) = 1\] \[-2\sin^{2}(x) = 0\] \[\sin^{2}(x) = 0\] \[\sin(x) = 0\] We know that $\sin(x) = 0$ when $x = \pi k$ for any integer $k$, so the solutions to the equation $\cos^{2}(x) - \sin^{2}(x) = 1$ are $x = \pi k$ for any integer $k$. This is illustrated by Figure $\PageIndex{3}$. Exercise $\PageIndex{3}$ Find the exact values of all solutions to the equation $\sin^{2}(x) = 3\cos^{2}(x)$. Draw appropriate graphs to illustrate your solutions. Answer We will use the identity $\cos^{2}(x) = 1 - \sin(x)$. So we have \[\sin^{2}(x) = 3(1 - \sin^{2}(x))\] \[\sin^{2}(x) = \dfrac{3}{4}\] So we have $\sin(x) = \dfrac{\sqrt{3}}{2}$ or $\sin(x) = -\dfrac{\sqrt{3}}{2}$. For the first equation, we see that \[x = \dfrac{\pi}{3} + 2\pi k\] or \[x = -\dfrac{\pi}{3} + 2\pi k\] where $k$ is an integer, and for the second equation, we have \[x = \dfrac{\pi}{3} + 2\pi k\] or \[x = \dfrac{\pi}{3} + 2\pi k\] where $k$ is an integer. The graphs of $y^{2} = \sin^{2}(x)$ and $y^{2} = 3\cos^{2}(x)$ will show $4$ points of intersection on the interval $[0, 2\pi]$. Other Methods for Solving Trigonometric Equations Just like we did with linear equations, we can view some trigonometric equations as quadratic in nature and use tools from algebra to solve them. Figure $\PageIndex{3}$: The graphs of $y = \cos^{2}(x) - \sin^{2}(x)$ and $y = 1$ Example $\PageIndex{2}$: (Solving Trigonometric Equations of Quadratic Type) Consider the trigonometric equation \[\cos^{2}(x) - 2\cos(x) + 1 = 0\] This equation looks like a familiar quadratic equation $y^{2} - 2y + 1 = 0$. We can solve this quadratic equation by factoring to obtain $(y - 1)^{2} = 0$. So we can apply the same technique to the trigonometric equation $\cos^{2}(x) - 2\cos(x) + 1 = 0$. Factoring the left hand side yields \[(\cos(x) - 1)^{2} = 0\] The only way $(\cos(x) - 1)^{2} = 0$ can be 0 is if $(\cos(x) - 1)^{2}$ is 0. This reduces our quadratic trigonometric equation to a linear trigonometric equation. To summarize the process so far we have \[\cos^{2}(x) - 2\cos(x) + 1 = 0\] \[(\cos(x) - 1)^{2} = 0\] \[\cos(x) - 1 = 0\] \[\cos(x) = 1\] We know that $\cos(x) = 1$ when $x = 2\pi k$ for integer values of $k$. Therefore, the solutions to our original equation are \[x = 2\pi k\] where $k$ is any integer. As a check, the graph of $y = \cos^{2}(x) - 2\cos(x) + 1$ is shown in Figure $\PageIndex{4}$. The figure appears to show that the graph of $y = \cos^{2}(x) - 2\cos(x) + 1$ intersects the $x$-axis at exactly the points we found, so our solution is validated by graphical means. Figure $\PageIndex{4}$: the graph of $y = \cos^{2}(x) - 2\cos(x) + 1$ Exercise $\PageIndex{4}$ Find the exact values of all solutions to the equation $\sin^{2}(x - 4\sin(x) = -3)$. Draw appropriate graphs to illustrate your solutions. Answer We write the equation as $\sin(x) - 4\sin(x) + 3 = 0$ and factor the right side to get $(\sin(x) - 3)(\sin(x) - 1) = 0$. So we see that $\sin(x) - 3 = 0$ or $\sin(x) - 1 = 0$. However, the equation $\sin(x) - 3 = 0$ is equivalent to $\sin(x) = 3$, and this equation has no solution. We write $\sin(x) - 1 = 0$ as $\sin(x) = 1$ and so the solutions are \[x = \dfrac{\pi}{2} + 2\pi k\] where $k$ is an integer. Summary In this section, we studied the following important concepts and ideas: A trigonometric equation is a conditional equation that involves trigonometric functions. If it is possible to write the equation in the form $\text{“some trigonometric function of } x \text{"} = \text{a number}$ we can use the following strategy to solve the equation: Find all solutions of the equation within one period of the function. This is often done by using properties of the trigonometric function. Quite often, there will be two solutions within a single period. Use the period of the function to express formulas for all solutions by adding integer multiples of the period to each solution found in the first step. For example, if the function has a period of $2\pi$ and $x_{1}$ and $x_{2}$ are the only two solutions in a complete period, then we would write the solutions for the equation as \[x = x_{1} + k(2\pi), x = x_{2} + k(2\pi)\] where $k$ is an integer: We can sometimes use trigonometric identities to help rewrite a given equation in the form of equation (1).
I try to give an answer, welcome corrections! When we write a state, we must notice the reference frame which the state lies in, because the form of state is diffrent in diffrent frame. Now I give 3 frames $O,O', O''$. Their correlations are $t'=t-\tau'$, $t''=t-\tau''$, with $\tau'=-\infty$, $\tau''=+\infty$. Let's specify that the collision happens in $t=0$, in $O$ frame's view. When we say $\Psi^\pm$ are tranform as free particles, we actually mean $\Psi^+$ transforms as free particles just in frame $O'$, and $\Psi^-$ transforms as free particles just in frame $O''$. Turn now to the inner product between in and out states $(\Psi^-, \Psi^+)$. Noticing that the product must be calculated in the same frame, we specify the frame is $O$. Now, we do a transformation $T$, and to see how the inner product changes. Please note, in frame $O$, neither of $\Psi^\pm$ transforms as free particles. However, we can use time translation to take $\Psi^+$ to frame $O'$, and act on it with $U_0(T)$, then take back to frame $O$. So, under transformation $T$, $(\Psi^-, \Psi^+)$ changes into $$(exp(+iH\infty)U_0(T)exp(-iH\infty)\Psi^-, exp(-iH\infty)U_0(T)exp(+iH\infty)\Psi^+)$$ Obviously, they are not equal, unless there are some restrictions on $H$.
Difference between revisions of "Capillary waves" (Stuff added from my contribution to wikipedia -- still to be rewritten a bit) m (Tidy up (still work to do)) Line 36: Line 36: ===Derivation=== ===Derivation=== − This is a sketch of the derivation of the general dispersion relation, see Ref. + This is a sketch of the derivation of the general dispersion relation, see Ref. for a more detailed description. − + + Three contributions to the energy are involved: the [[surface tension]], gravity, and hydrodynamics. The part due to gravity is the simplest: integrating the potential energy density due to gravity, <math>\rho g z</math> from a reference height to the position of the surface, <math>z=h(x,y)</math>: Three contributions to the energy are involved: the [[surface tension]], gravity, and hydrodynamics. The part due to gravity is the simplest: integrating the potential energy density due to gravity, <math>\rho g z</math> from a reference height to the position of the surface, <math>z=h(x,y)</math>: − :<math>E_\mathrm{g}= \int dx\, dy\, \int_0^h dz \rho g z = \rho g + :<math>E_\mathrm{g}= \int dx\, dy\, \int_0^h dz \rho g z = \rho g2\int dx\, dy\, h^2.</math> (For simplicity, we are neglecting the density of the fluid above, which is often acceptable.) (For simplicity, we are neglecting the density of the fluid above, which is often acceptable.) An increase in area of the surface causes a proportional increase of energy: An increase in area of the surface causes a proportional increase of energy: − :<math>E_\mathrm{st}= \sigma \int dx\, dy\ \sqrt{(dh + :<math>E_\mathrm{st}= \sigma \int dx\, dy\ \sqrt{( dhdx)^2+( dhdy)^2} \approx \sigma2\int dx\, dy\ [ ( dhdx)^2+( dhdy)^2 ],</math> where the fist equality is the area in this ([[de Monge]]) representation, and the second where the fist equality is the area in this ([[de Monge]]) representation, and the second applies for small values of the derivatives (surfaces not too rough). applies for small values of the derivatives (surfaces not too rough). The last contribution involves the [[kinetic energy]] of the fluid: The last contribution involves the [[kinetic energy]] of the fluid: − :<math>T= \rho + :<math>T= \rho2 \int dx\, dy\, \int_{-\infty}^h dz v^2,</math> − where <math>v</math> is the module of the velocity field <math>\vec{v}</math>. If the fluid is + where <math>v</math> is the module of the velocity field <math>\vec{v}</math>. + + + + + + + + + + + + + If the fluid is incompressible and its flow is irrotational (often, sensible approximations), its incompressible and its flow is irrotational (often, sensible approximations), its flow will be [[potential flow\|potential]]: <math>\vec{v}=\nabla\psi</math>, and <math>\psi</math> flow will be [[potential flow\|potential]]: <math>\vec{v}=\nabla\psi</math>, and <math>\psi</math> Line 66: Line 80: #J. S. Rowlinson and B. Widom "Molecular Theory of Capillarity". Dover 2002 (originally: Oxford University Press 1982) ISBN 0486425444 #J. S. Rowlinson and B. Widom "Molecular Theory of Capillarity". Dover 2002 (originally: Oxford University Press 1982) ISBN 0486425444 #[http://dx.doi.org/10.1103/PhysRevLett.91.166103 E. Chacón and P. Tarazona "Intrinsic profiles beyond the capillary wave theory: A Monte Carlo study", Physical Review Letters '''91''' 166103 (2003)] #[http://dx.doi.org/10.1103/PhysRevLett.91.166103 E. Chacón and P. Tarazona "Intrinsic profiles beyond the capillary wave theory: A Monte Carlo study", Physical Review Letters '''91''' 166103 (2003)] + #[http://dx.doi.org/10.1103/PhysRevLett.99.196101 P. Tarazona, R. Checa, and E. Chacón "Critical Analysis of the Density Functional Theory Prediction of Enhanced Capillary Waves", Physical Review Letters '''99''' 196101 (2007)] #[http://dx.doi.org/10.1103/PhysRevLett.99.196101 P. Tarazona, R. Checa, and E. Chacón "Critical Analysis of the Density Functional Theory Prediction of Enhanced Capillary Waves", Physical Review Letters '''99''' 196101 (2007)] [[Category: Classical thermodynamics ]] [[Category: Classical thermodynamics ]] Revision as of 17:12, 10 April 2008 Contents Thermal capillary waves Thermal capillary waves are oscillations of an interfacewhich are thermal in origin. These take place at the molecular level, where only the surface tensioncontribution is relevant. Capillary wave theory (CWT) is a classic account of how thermalfluctuations distort an interface (Ref. 1). It starts from some intrinsic surfacethat is distorted. By performing a Fourier analysis treatment, normal modes are easily found.Each contributes a energy proportional to the square of its amplitude; therefore, according toclassical statistical mechanics, equipartition holds, and themean energy of each mode will be . Surprisingly, this result leads to a divergent surface (the width of the interface is bound to diverge with its area) (Ref 2).This divergence isnevertheless very mild: even for displacements on the order of meters the deviation of the surfaceis comparable to the size of the molecules. Moreover, the introduction of an external fieldremoves the divergence: the actionof gravity is sufficient to keep the width fluctuation on the orderof one molecular diameter for areas larger than about 1 mm 2 (Ref. 2). Recently, a procedure has been proposed to obtain a molecular intrinsicsurface from simulation data (Ref. 3). The density profiles obtainedfrom this surface are, in general, quite different from the usual mean density profiles. Gravity-capillary waves These are ordinary waves excited in an interface, such as ripples on a water surface. Their dispersion relation reads, for waves on the interface between two fluids of infinite depth: Derivation This is a sketch of the derivation of the general dispersion relation, see Ref. 4 for a more detailed description. Defining the problem Three contributions to the energy are involved: the surface tension, gravity, and hydrodynamics. The part due to gravity is the simplest: integrating the potential energy density due to gravity, from a reference height to the position of the surface, : (For simplicity, we are neglecting the density of the fluid above, which is often acceptable.) An increase in area of the surface causes a proportional increase of energy: where the fist equality is the area in this (de Monge) representation, and the second applies for small values of the derivatives (surfaces not too rough). The last contribution involves the kinetic energy of the fluid: where is the module of the velocity field . Wave solutions Let us try separation of variables: where is a two dimensional wave number vector, and the position. In this case, where a factor of that will appear every integration is dropped for convenience. If the fluid is incompressible and its flow is irrotational (often, sensible approximations), its flow will be potential: , and must satisfy Laplace equation . This equation can be solved with the proper boundary conditions: must vanish well below the surface (in the "deep water" case, which is the one we consider), and it vertical component must match the motion of the surface: at . Performing the integration one is left with a surface integral for the kinetic energy. One is then left with two contribution to the potential energy involving , , and , and one for the kinetic energy involving , all three being surface integrals. Constructing the Lagrangian of this system one readily finds a wave-like equation, whose oscillatory solutions satisfy the same dispersion as above if is neglected. References F. P. Buff, R. A. Lovett, and F. H. Stillinger, Jr. "Interfacial density profile for fluids in the critical region" Physical Review Letters 15pp. 621-623 (1965) J. S. Rowlinson and B. Widom "Molecular Theory of Capillarity". Dover 2002 (originally: Oxford University Press 1982) ISBN 0486425444 E. Chacón and P. Tarazona "Intrinsic profiles beyond the capillary wave theory: A Monte Carlo study", Physical Review Letters 91166103 (2003) Samuel Safran "Statistical thermodynamics of surfaces, interfaces, and membranes" Addison-Wesley 1994 P. Tarazona, R. Checa, and E. Chacón "Critical Analysis of the Density Functional Theory Prediction of Enhanced Capillary Waves", Physical Review Letters 99196101 (2007)
2019-10-09 06:01 HiRadMat: A facility beyond the realms of materials testing / Harden, Fiona (CERN) ; Bouvard, Aymeric (CERN) ; Charitonidis, Nikolaos (CERN) ; Kadi, Yacine (CERN)/HiRadMat experiments and facility support teams The ever-expanding requirements of high-power targets and accelerator equipment has highlighted the need for facilities capable of accommodating experiments with a diverse range of objectives. HiRadMat, a High Radiation to Materials testing facility at CERN has, throughout operation, established itself as a global user facility capable of going beyond its initial design goals. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPRB085 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPRB085 詳細記錄 - 相似記錄 2019-10-09 06:01 Commissioning results of the tertiary beam lines for the CERN neutrino platform project / Rosenthal, Marcel (CERN) ; Booth, Alexander (U. Sussex (main) ; Fermilab) ; Charitonidis, Nikolaos (CERN) ; Chatzidaki, Panagiota (Natl. Tech. U., Athens ; Kirchhoff Inst. Phys. ; CERN) ; Karyotakis, Yannis (Annecy, LAPP) ; Nowak, Elzbieta (CERN ; AGH-UST, Cracow) ; Ortega Ruiz, Inaki (CERN) ; Sala, Paola (INFN, Milan ; CERN) For many decades the CERN North Area facility at the Super Proton Synchrotron (SPS) has delivered secondary beams to various fixed target experiments and test beams. In 2018, two new tertiary extensions of the existing beam lines, designated “H2-VLE” and “H4-VLE”, have been constructed and successfully commissioned. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW064 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW064 詳細記錄 - 相似記錄 2019-10-09 06:00 The "Physics Beyond Colliders" projects for the CERN M2 beam / Banerjee, Dipanwita (CERN ; Illinois U., Urbana (main)) ; Bernhard, Johannes (CERN) ; Brugger, Markus (CERN) ; Charitonidis, Nikolaos (CERN) ; Cholak, Serhii (Taras Shevchenko U.) ; D'Alessandro, Gian Luigi (Royal Holloway, U. of London) ; Gatignon, Laurent (CERN) ; Gerbershagen, Alexander (CERN) ; Montbarbon, Eva (CERN) ; Rae, Bastien (CERN) et al. Physics Beyond Colliders is an exploratory study aimed at exploiting the full scientific potential of CERN’s accelerator complex up to 2040 and its scientific infrastructure through projects complementary to the existing and possible future colliders. Within the Conventional Beam Working Group (CBWG), several projects for the M2 beam line in the CERN North Area were proposed, such as a successor for the COMPASS experiment, a muon programme for NA64 dark sector physics, and the MuonE proposal aiming at investigating the hadronic contribution to the vacuum polarisation. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW063 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW063 詳細記錄 - 相似記錄 2019-10-09 06:00 The K12 beamline for the KLEVER experiment / Van Dijk, Maarten (CERN) ; Banerjee, Dipanwita (CERN) ; Bernhard, Johannes (CERN) ; Brugger, Markus (CERN) ; Charitonidis, Nikolaos (CERN) ; D'Alessandro, Gian Luigi (CERN) ; Doble, Niels (CERN) ; Gatignon, Laurent (CERN) ; Gerbershagen, Alexander (CERN) ; Montbarbon, Eva (CERN) et al. The KLEVER experiment is proposed to run in the CERN ECN3 underground cavern from 2026 onward. The goal of the experiment is to measure ${\rm{BR}}(K_L \rightarrow \pi^0v\bar{v})$, which could yield information about potential new physics, by itself and in combination with the measurement of ${\rm{BR}}(K^+ \rightarrow \pi^+v\bar{v})$ of NA62. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPGW061 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPGW061 詳細記錄 - 相似記錄 2019-09-21 06:01 Beam impact experiment of 440 GeV/p protons on superconducting wires and tapes in a cryogenic environment / Will, Andreas (KIT, Karlsruhe ; CERN) ; Bastian, Yan (CERN) ; Bernhard, Axel (KIT, Karlsruhe) ; Bonura, Marco (U. Geneva (main)) ; Bordini, Bernardo (CERN) ; Bortot, Lorenzo (CERN) ; Favre, Mathieu (CERN) ; Lindstrom, Bjorn (CERN) ; Mentink, Matthijs (CERN) ; Monteuuis, Arnaud (CERN) et al. The superconducting magnets used in high energy particle accelerators such as CERN’s LHC can be impacted by the circulating beam in case of specific failure cases. This leads to interaction of the beam particles with the magnet components, like the superconducting coils, directly or via secondary particle showers. [...] 2019 - 4 p. - Published in : 10.18429/JACoW-IPAC2019-THPTS066 Fulltext from publisher: PDF; In : 10th International Particle Accelerator Conference, Melbourne, Australia, 19 - 24 May 2019, pp.THPTS066 詳細記錄 - 相似記錄 2019-09-20 08:41 Shashlik calorimeters with embedded SiPMs for longitudinal segmentation / Berra, A (INFN, Milan Bicocca ; Insubria U., Varese) ; Brizzolari, C (INFN, Milan Bicocca ; Insubria U., Varese) ; Cecchini, S (INFN, Bologna) ; Chignoli, F (INFN, Milan Bicocca ; Milan Bicocca U.) ; Cindolo, F (INFN, Bologna) ; Collazuol, G (INFN, Padua) ; Delogu, C (INFN, Milan Bicocca ; Milan Bicocca U.) ; Gola, A (Fond. Bruno Kessler, Trento ; TIFPA-INFN, Trento) ; Jollet, C (Strasbourg, IPHC) ; Longhin, A (INFN, Padua) et al. Effective longitudinal segmentation of shashlik calorimeters can be achieved taking advantage of the compactness and reliability of silicon photomultipliers. These photosensors can be embedded in the bulk of the calorimeter and are employed to design very compact shashlik modules that sample electromagnetic and hadronic showers every few radiation lengths. [...] 2017 - 6 p. - Published in : IEEE Trans. Nucl. Sci. 64 (2017) 1056-1061 詳細記錄 - 相似記錄 2019-09-20 08:41 Performance study for the photon measurements of the upgraded LHCf calorimeters with Gd$_2$SiO$_5$ (GSO) scintillators / Makino, Y (Nagoya U., ISEE) ; Tiberio, A (INFN, Florence ; U. Florence (main)) ; Adriani, O (INFN, Florence ; U. Florence (main)) ; Berti, E (INFN, Florence ; U. Florence (main)) ; Bonechi, L (INFN, Florence) ; Bongi, M (INFN, Florence ; U. Florence (main)) ; Caccia, Z (INFN, Catania) ; D'Alessandro, R (INFN, Florence ; U. Florence (main)) ; Del Prete, M (INFN, Florence ; U. Florence (main)) ; Detti, S (INFN, Florence) et al. The Large Hadron Collider forward (LHCf) experiment was motivated to understand the hadronic interaction processes relevant to cosmic-ray air shower development. We have developed radiation-hard detectors with the use of Gd$_2$SiO$_5$ (GSO) scintillators for proton-proton $\sqrt{s} = 13$ TeV collisions. [...] 2017 - 22 p. - Published in : JINST 12 (2017) P03023 詳細記錄 - 相似記錄 2019-04-26 08:32 Baby MIND: A magnetised spectrometer for the WAGASCI experiment / Hallsjö, Sven-Patrik (Glasgow U.)/Baby MIND The WAGASCI experiment being built at the J-PARC neutrino beam line will measure the ratio of cross sections from neutrinos interacting with a water and scintillator targets, in order to constrain neutrino cross sections, essential for the T2K neutrino oscillation measurements. A prototype Magnetised Iron Neutrino Detector (MIND), called Baby MIND, has been constructed at CERN and will act as a magnetic spectrometer behind the main WAGASCI target. [...] SISSA, 2018 - 7 p. - Published in : PoS NuFact2017 (2018) 078 Fulltext: PDF; External link: PoS server In : 19th International Workshop on Neutrinos from Accelerators, Uppsala, Sweden, 25 - 30 Sep 2017, pp.078 詳細記錄 - 相似記錄 2019-04-26 08:32 詳細記錄 - 相似記錄 2019-04-26 08:32 ENUBET: High Precision Neutrino Flux Measurements in Conventional Neutrino Beams / Pupilli, Fabio (INFN, Padua) ; Ballerini, G (Insubria U., Como ; INFN, Milan Bicocca) ; Berra, A (Insubria U., Como ; INFN, Milan Bicocca) ; Boanta, R (INFN, Milan Bicocca ; Milan Bicocca U.) ; Bonesini, M (INFN, Milan Bicocca) ; Brizzolari, C (Insubria U., Como ; INFN, Milan Bicocca) ; Brunetti, G (INFN, Padua) ; Calviani, M (CERN) ; Carturan, S (INFN, Legnaro) ; Catanesi, M G (INFN, Bari) et al. The ENUBET project aims at demonstrating that the systematics in neutrino fluxes from conventional beams can be reduced to 1% by monitoring positrons from K$_{e3}$ decays in an instrumented decay tunnel, thus allowing a precise measurement of the $\nu_e$ (and $\overline{\nu}_e$) cross section. This contribution will report the results achieved in the first year of activities. [...] SISSA, 2018 - 8 p. - Published in : PoS NuFact2017 (2018) 087 Fulltext: PDF; External link: PoS server In : 19th International Workshop on Neutrinos from Accelerators, Uppsala, Sweden, 25 - 30 Sep 2017, pp.087 詳細記錄 - 相似記錄
I have a question where I need to find the general solution of the differential equation $y-3y^{7}=(y^{3}+6x)y'$, where the solution is in the form $F(x,y)=C$. I am only concerned about finding $F(x,y)$, and I have been advised to rewrite this equation in differential form, but I am still unsure where to go from here. Any help would be greatly appreciated with this problem! $$\frac{d\mu(y)}{dy}(-3y^7+y)(-21y^6+1)\mu(y)=-6\mu(y)$$ $$\frac{\frac{\partial \mu(y)}{\partial y}}{\mu(y)}=-\frac{7}{y}$$ $$\mu(y)=\frac{1}{y^7}$$ and then we have $$-3+\frac{1}{y(x)^6}-\frac{(6x+y(x)^3)y'(x)}{y(x)^7}=0$$ and let $$P(x,y)=\frac{1}{y^6}-3,Q(x,y)=-\frac{6x+y^3}{y^7}$$ and we get $$\frac{\partial P(x,y}{\partial y}=-\frac{6}{y^7}=\frac{\partial Q(x,y)}{\partial x}$$ and the solution is given by $$f(x,y)=C_1$$ where $C_1$ is a arbitrary constant Hint: Write it in terms of $x'(y)$. You will see its a linear differential equation! $$\frac{dx}{dy} + \frac{6x}{y-3y^7} = \frac{y^3}{y-3y^7}$$ Hint: Write $(y-3y^7)dx-(y^3+6x)dy=0$ then $$p(y)=\dfrac{M_y-N_x}{-M}=\dfrac{7}{-y}$$ then $$I=e^{\int p(y)dy}=\dfrac{1}{y^7}$$ is integrating factor.
You can do it by combining grammars. Notation: $A,B,C,...$ will be non-terminals $a,b,c,...$ will be terminals $\alpha,\beta, ...$ will be "whatever can be here" $x$ and $y$ will be regular expressions The grammar corresponding to a regular expression $x$ will be denoted $f(x)=(N_x,\Sigma,P_x,S_x)$ $f(a)=(\{S\},\Sigma, \{S\to a\}, S)$ $f(x\mid y)=(N, \Sigma, P, S)$ $N=N_x\cup N_y \cup \{S\}$ $P=P_x\cup P_y\cup\{S\to \alpha \mid S_x\to \alpha\in P_x\}\cup\{S\to \alpha \mid S_y\to \alpha\in P_y\} $ The idea is that you want to be able to go one way or the other but if you add $S\to S_a$ and $S\to S_b$, then your grammar isn't regular anymore. You could want to replace both $S_x$ and $S_y$ and their associated rules but if you do that, if the grammars use $S_x$ or $S_y$ somewhere on the right hand side, it might fail. For example if they represent $a^$ and $b^$ in such a way, then instead of $a^\mid b^$, your grammar would represent $(a\mid b)^$ which isn't the same thing. Here's the grammar for $a^$: $S\to \varepsilon$ $S\to aS$ $f(x\cdot y)=(N, \Sigma, P, S_x)$ $N=N_x\cup N_y$ $P=\{A\to bC\mid A\to bC\in P_x\} \cup \{A\to bS_y \mid A\to b \in P_x\}\cup \{A\to S_y \mid A\to \varepsilon \in P_x\} \cup P_y$ The idea is that since your grammar is regular, you'll always have a unique non-terminal until you finish your word with a rule not producing a non-terminal so you simply replace those rules in the first grammar by rules saying to continue on the second grammar. $f(a)=(N_a,\Sigma, P_a\cup \{A\to bS_a\mid A\to b\in P_a\}\cup \{A\to S_a\mid A\to \varepsilon\in P_a\}\cup \{S_a\to \varepsilon\}, S_a)$ $f(x^)=(N_x,\Sigma, P, S_x)$ $P=P_x \cup \{A\to bS_x \mid A\to b \in P_x\}\cup \{A\to S_x \mid A\to \varepsilon \in P_x\}\cup \{S_x\to \varepsilon\}$ Same thing as above but you also take the rules giving simply the non-terminal or $\varepsilon$ to allow to end right away.
Connected and Disconnected Topological Spaces One very important characteristic of a topological space is the concept of connectedness and disconnectedness. We define these terms below. Definition: Let $X$ be a topological space. Then $X$ is said to be Disconnected if there exists open sets $A, B \subset X$ such that $A, B \neq \emptyset$, $A \cap B = \emptyset$, and $X = A \cup B$ and the pair $\{ A, B \}$ is called a Separation of $X$. If $X$ is not disconnected then we say that $X$ is Connected. For example, consider the topological space $\mathbb{R}$ with the usual topology formed from open intervals. Consider the topological subspace $\mathbb{Q}$. We claim that $\mathbb{Q}$ is hence a disconnected topological space. Consider the irrational number $\sqrt{2}$ and let $A$ and $B$ be subsets of $\mathbb{Q}$ defined by:(1) We claim that $\{ A, B \}$ is a separation of $\mathbb{Q}$. First note that $A$ and $B$ are open in $\mathbb{Q}$ with the subspace topology since $A = \mathbb{Q} \cap (-\infty, \sqrt{2})$ and $B = \mathbb{Q} \cap (\sqrt{2}, \infty)$ and $(-\infty, \sqrt{2})$ and $(\sqrt{2}, \infty)$ are open sets in $\mathbb{R}$ with the usual topology. Furthermore, $A$ and $B$ are nonempty sets since $0 \in A$ and $2 \in B$, for example. We also see that $A \cap B = \emptyset$ since there exists no rational number that is simultaneously less than $\sqrt{2}$ and greater than $\sqrt{2}$. Also, it's not hard to see that:(3) Thus $\{ A, B \}$ is a separation of the topological space $\mathbb{Q}$ and so $\mathbb{Q}$ is a disconnected topological spaces. Note that in general it is much easier to show that a topological space is disconnected than it is to show that a topological space is connected. If we consider the topological space $\mathbb{R}$ with the usual topology as mentioned above then $\mathbb{R}$ is actually a connected topological space. Showing this (for the time being) is rather cumbersome though. We will develop a stronger type of connectedness later on and prove this as a very simple consequence later on.
The queen wishes to build a patio paved with of a circular center stone surrounded by circular rings of circular stones. All the stones in a ring will be the same size with the same number of stones in each ring. The stones in the innermost ring will be placed touching (tangent to) the adjacent stones in the ring and the central stone. The stones in the other rings will touch the two adjacent stones in the next inner ring and their neighbors in the same ring. The figures below depict a patio with one ring of three stones and a patio with $5$ rings of $11$ stones. The patio is to be surrounded by a fence that goes around the outermost stones and straight between them (the heavier line in the figures). The queen does not yet know how many stones there will be in each circle nor how many circles of stones there will be. To be prepared for whatever she decides, write a program to calculate the sizes of the stones in each circle and the length of the surrounding fence. The radius of the central stone is to be one queenly foot. The first line of input contains a single integer $P$, ($1 ≤ P ≤ 1000$), which is the number of data sets that follow. Each data set should be processed identically and independently. Each data set consists of a single line of input. It contains the data set number, $K$, the number, $N$ ($3 ≤ N ≤ 20$), of stsones in each circle and the number, $M$ ($1 ≤ M ≤ 15$), of circles of stones around the central stone. For each data set there is a single line of output. It contains the data set number, $K$, followed by a single space which is then followed by the radius (in queenly feet) of the stones in the outermost ring (to $3$ decimal places) which is followed by a single space which is then followed by the length (in queenly feet) of the fence (to $3$ decimal places). 3 1 3 1 2 7 3 3 11 5 1 6.464 79.400 2 3.834 77.760 3 2.916 82.481 $t$组数据,第一个是编号,第二个是一层有几个,第三个是有几层. 问你最外层的圆的半径$r$和围栏长度$l$. 首先,最外围的围栏长度$l$必定是最外围$1$个圆的周长$+n\times 2\times$最外围圆的半径$r$. 那么问题变为如何求最外围圆的半径$r$. 对于第一个图,很明显,最外层,即第一层的半径为$\frac{\sin\theta}{(1.0 - \sin\theta)}$,这里$\theta=\frac{\pi}{n}$ 就像这样,理解一下. 我们改一下图,让第二个图只有两层那么问题变为如何求$r_2$. 我们连接两球的圆心,外切圆和外切圆的外切圆由$r_1=R_1\times\sin\theta,r_2=R_2\times\sin\theta$我们可以得到一个公式: 一系列化简得到 $$r_2=R_2*\sin\theta$$ 之后每个$R_i=f(i-1)$递推求得. 1 int main()
Search Now showing items 1-2 of 2 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 ... 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 ...
Sometimes it is useful to compute the length of a curve in space; for example, if the curve represents the path of a moving object, the length of the curve between two points may be the distance traveled by the object between two times. Recall that if the curve is given by the vector function $\bf r$ then the vector $\Delta {\bf r}={\bf r}(t+\Delta t)-{\bf r}(t)$ points from one position on the curve to another, as depicted in figure . If the points are close together, the length of $\Delta {\bf r}$ is close to the length of the curve between the two points. If we add up the lengths of many such tiny vectors, placed head to tail along a segment of the curve, we get an approximation to the length of the curve over that segment. In the limit, as usual, this sum turns into an integral that computes precisely the length of the curve. First, note that $$\|\Delta {\bf r}\|={\|\Delta {\bf r}\|\over \Delta t}\,\Delta t\approx \|{\bf r}'(t)\|\,\Delta t,$$ when $\Delta t$ is small. Then the length of the curve between ${\bf r}(a)$ and ${\bf r}(b)$ is $$\lim_{n\to\infty} \sum_{i=0}^{n-1}\|\Delta {\bf r}\| =\lim_{n\to\infty} \sum_{i=0}^{n-1} {\|\Delta {\bf r}\|\over \Delta t}\,\Delta t =\lim_{n\to\infty} \sum_{i=0}^{n-1} \|{\bf r}'(t)\|\,\Delta t= \int_a^b\|{\bf r}'(t)\|\,dt.$$ (Well, sometimes. This works if between $a$ and $b$ the segment of curve is traced out exactly once.) Example 13.3.1 Let's find the length of one turn of the helix ${\bf r}=\langle \cos t, \sin t, t\rangle$ (see figure 13.1.1). We compute ${\bf r}'=\langle -\sin t, \cos t, 1\rangle$ and $\|{\bf r}'\|=\sqrt{\sin^2 t+\cos^2 t+1}=\sqrt2$, so the length is $$\int_0^{2\pi} \sqrt2\,dt = 2\sqrt2\pi.$$ Example 13.3.2 Suppose $y=\ln x$; what is the length of this curve between $x=1$ and $x=\sqrt3$? Although this problem does not appear to involve vectors or three dimensions, we can interpret it in those terms: let ${\bf r}(t)=\langle t,\ln t,0\rangle$. This vector function traces out precisely $y=\ln x$ in the $x$-$y$ plane. Then ${\bf r}'(t)=\langle 1,1/t,0\rangle$ and $\|{\bf r}'(t)\|=\sqrt{1+1/t^2}$ and the desired length is $$\int_1^{\sqrt3} \sqrt{1+{1\over t^2}}\,dt=2-\sqrt2+\ln(\sqrt2+1)-{1\over2}\ln3.$$ (This integral is a bit tricky, but requires only methods we have learned.) Notice that there is nothing special about $y=\ln x$, except that the resulting integral can be computed. In general, given any $y=f(x)$, we can think of this as the vector function ${\bf r}(t)=\langle t,f(t),0\rangle$. Then ${\bf r}'(t)=\langle 1,f'(t),0\rangle$ and $\|{\bf r}'(t)\|=\sqrt{1+(f')^2}$. The length of the curve $y=f(x)$ between $a$ and $b$ is thus $$\int_a^b \sqrt{1+(f'(x))^2}\,dx.$$ Unfortunately, such integrals are often impossible to do exactly and must be approximated. One useful application of arc length is the arc length parameterization. A vector function ${\bf r}(t)$ gives the position of a point in terms of the parameter $t$, which is often time, but need not be. Suppose $s$ is the distance along the curve from some fixed starting point; if we use $s$ for the variable, we get ${\bf r}(s)$, the position in space in terms of distance along the curve. We might still imagine that the curve represents the position of a moving object; now we get the position of the object as a function of how far the object has traveled. Example 13.3.3 Suppose ${\bf r}(t)=\langle \cos t,\sin t,0\rangle$. We know that this curve is a circle of radius 1. While $t$ might represent time, it can also in this case represent the usual angle between the positive $x$-axis and ${\bf r}(t)$. The distance along the circle from $(1,0,0)$ to $(\cos t,\sin t,0)$ is also $t$---this is the definition of radian measure. Thus, in this case $s=t$ and ${\bf r}(s)=\langle \cos s,\sin s,0\rangle$. Example 13.3.4 Suppose ${\bf r}(t)=\langle \cos t,\sin t,t\rangle$. We know that this curve is a helix. The distance along the helix from $(1,0,0)$ to $(\cos t,\sin t,t)$ is $$s=\int_0^t \|{\bf r}'(u)\|\,du=\int_0^t \sqrt{\cos^2u+\sin^2u+1}\,du= \int_0^t \sqrt{2}\,du=\sqrt2t.$$ Thus, the value of $t$ that gets us distance $s$ along the helix is $t=s/\sqrt2$, and so the same curve is given by $\hat{\bf r}(s)= \langle \cos(s/\sqrt2),\sin(s/\sqrt2),s/\sqrt2\rangle$. In general, if we have a vector function ${\bf r}(t)$, to convert it to a vector function in terms of arc length we compute $$s=\int_a^t \|{\bf r}'(u)\|\,du=f(t),$$ solve $s=f(t)$ for $t$, getting $t=g(s)$, and substitute this back into ${\bf r}(t)$ to get $\hat{\bf r}(s)={\bf r}(g(s))$. Suppose that $t$ is time. By the Fundamental Theorem of Calculus, if we start with arc length $$s(t)=\int_a^t \|{\bf r}'(u)\|\,du$$ and take the derivative, we get $$s'(t)=\|{\bf r}'(t)\|.$$ Here $s'(t)$ is the rate at which the arc length is changing, and we have seen that $\|{\bf r}'(t)\|$ is the speed of a moving object; these are of course the same. Suppose that ${\bf r}(s)$ is given in terms of arc length; what is $\|{\bf r}'(s)\|$? It is the rate at which arc length is changing relative to arc length; it must be 1! In the case of the helix, for example, the arc length parameterization is $\langle \cos(s/\sqrt2),\sin(s/\sqrt2),s/\sqrt2\rangle$, the derivative is $\langle -\sin(s/\sqrt2)/\sqrt2,\cos(s/\sqrt2)/\sqrt2,1/\sqrt2\rangle$, and the length of this is \[\sqrt{\dfrac{\sin^2 (s/\sqrt{2})}{2} + \dfrac{\cos^2(s/\sqrt{2})}{2} + \dfrac{1}{2}}=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}}=1. \] So in general, ${\bf r}'$ is a unit tangent vector. Given a curve ${\bf r}(t)$, we would like to be able to measure, at various points, how sharply curved it is. Clearly this is related to how "fast'' a tangent vector is changing direction, so a first guess might be that we can measure curvature with $\|{\bf r}''(t)\|$. A little thought shows that this is flawed; if we think of $t$ as time, for example, we could be tracing out the curve more or less quickly as time passes. The second derivative $\|{\bf r}''(t)\|$ incorporates this notion of time, so it depends not simply on the geometric properties of the curve but on how quickly we move along the curve. Example 13.3.5 Consider ${\bf r}(t)=\langle \cos t,\sin t,0\rangle$ and ${\bf s}(t)=\langle \cos 2t,\sin 2t,0\rangle$. Both of these vector functions represent the unit circle in the $x$-$y$ plane, but if $t$ is interpreted as time, the second describes an object moving twice as fast as the first. Computing the second derivatives, we find $\|{\bf r}''(t)\|=1$, $\|{\bf s}''(t)\|=4$. To remove the dependence on time, we use the arc length parameterization. If a curve is given by ${\bf r}(s)$, then the first derivative ${\bf r}'(s)$ is a unit vector, that is, ${\bf r}'(s)={\bf T}(s)$. We now compute the second derivative ${\bf r}''(s)={\bf T}'(s)$ and use $\|{\bf T}'(s)\|$ as the "official'' measure of {\dfont curvature}\index{curvature}, usually denoted $\kappa$. Example 13.3.6 We have seen that the arc length parameterization of aparticular helix is ${\bf r}(s)=\langle \cos(s/\sqrt2),\sin(s/\sqrt2),s/\sqrt2\rangle$. Computing the second derivative gives ${\bf r}''(s)= \langle -\cos(s/\sqrt2)/2,-\sin(s/\sqrt2)/2,0\rangle$ with length $1/2$. What if we are given a curve as a vector function ${\bf r}(t)$, where $t$ is not arc length? We have seen that arc length can be difficult to compute; fortunately, we do not need to convert to the arc length parameterization to compute curvature. Instead, let us imagine that we have done this, so we have found $t=g(s)$ and then formed $\hat{\bf r}(s)={\bf r}(g(s))$. The first derivative $\hat{\bf r}'(s)$ is a unit tangent vector, so it is the same as the unit tangent vector ${\bf T}(t)={\bf T}(g(s))$. Taking the derivative of this we get $${d\over ds}{\bf T}(g(s))= {\bf T}'(g(s)) g'(s)={\bf T}'(t){dt\over ds}.$$ The curvature\index{curvature formula} is the length of this vector: $$\kappa = \|{\bf T}'(t)\|\|{dt\over ds}\|={\|{\bf T}'(t)\|\over\|ds/dt\|}= {\|{\bf T}'(t)\|\over\|{\bf r}'(t)\|}.$$ (Recall that we have seen that $ds/dt=\|{\bf r}'(t)\|$.) Thus we can compute the curvature by computing only derivatives with respect to $t$; we do not need to do the conversion to arc length. Example 13.3.7 Returning to the helix, suppose we start with the parameterization ${\bf r}(t)=\langle \cos t,\sin t,t\rangle$. Then ${\bf r}'(t)=\langle -\sin t,\cos t,1\rangle$, $\|{\bf r}'(t)\|=\sqrt2$, and ${\bf T}(t)=\langle -\sin t,\cos t,1\rangle/\sqrt2$. Then ${\bf T}'(t)=\langle -\cos t,-\sin t,0\rangle/\sqrt2$ and $\|{\bf T}'(t)\|=1/\sqrt2$. Finally, $\kappa=1/\sqrt2/\sqrt2=1/2$, as before. Example 13.3.8 Consider this circle of radius $a$: ${\bf r}(t)=\langle a\cos t,a\sin t,1\rangle$. Then ${\bf r}'(t)=\langle -a\sin t,a\cos t,0\rangle$, $\|{\bf r}'(t)\|=a$, and ${\bf T}(t)=\langle -a\sin t,a\cos t,0\rangle/a$. Now ${\bf T}'(t)=\langle -a\cos t,-a\sin t,0\rangle/a$ and $\|{\bf T}'(t)\|=1$. Finally, $\kappa=1/a$: the curvature of a circle is everywhere the inverse of the radius. It is sometimes useful to think of curvature as describing what circle a curve most resembles at a point. The curvature of the helix in the previous example is $1/2$; this means that a small piece of the helix looks very much like a circle of radius $2$, as shown in figure. Figure 13.3.1. A circle with the same curvature as the helix. Example 13.3.9 Consider ${\bf r}(t)=\langle \cos t,\sin t,\cos 2t\rangle$, as shown in figure 13.2.4. ${\bf r}'(t)= \langle -\sin t,\cos t,-2\sin (2t)\rangle$ and $\|{\bf r}'(t)\|= \sqrt{1+4\sin^2(2t)}$, so $${\bf T}(t)=\left\langle {-\sin t\over \sqrt{1+4\sin^2(2t)}}, {\cos t\over \sqrt{1+4\sin^2(2t)}}, {-2\sin 2t\over \sqrt{1+4\sin^2(2t)}}\right\rangle.$$ Computing the derivative of this and then the length of the resulting vector is possible but unpleasant. Fortunately, there is an alternate formula for the curvature\index{curvature formula} that is often simpler than the one we have: $$\kappa = {\|{\bf r}'(t)\times{\bf r}''(t)\|\over\|{\bf r}'(t)\|^3}.$$ Example 13.3.10 Returning to the previous example, we compute the second derivative ${\bf r}''(t)= \langle -\cos t,-\sin t,-4\cos(2t)\rangle$. Then the cross product ${\bf r}'(t)\times{\bf r}''(t)$ is $$\langle -4\cos t\cos 2t-2\sin t\sin 2t, 2\cos t\sin 2t-4\sin t \cos2t,1\rangle.$$ Computing the length of this vector and dividing by $\|{\bf r}'(t)\|^3$ is still a bit tedious. With the aid of a computer we get $$\kappa = {\sqrt{48\cos^4 t - 48\cos^2 t + 17}\over (-16\cos^4 t +16\cos^2t+1)^{3/2}}.$$ Graphing this we get Compare this to figure 13.2.4. The highest curvature occurs where the curve has its highest and lowest points, and indeed in the picture these appear to be the most sharply curved portions of the curve, while the curve is almost a straight line midway between those points. Let's see why this alternate formula is correct. Starting with the definition of ${\bf T}$, ${\bf r}'=\|{\bf r}'\|{\bf T}$ so by the product rule ${\bf r}''=\|{\bf r}'\|'{\bf T}+\|{\bf r}'\|{\bf T}'$. Then by Theorem~\xrefn{thm:cross product properties} the cross product is $$\eqalign{ {\bf r}'\times{\bf r}''&=\|{\bf r}'\|{\bf T}\times\|{\bf r}'\|'{\bf T}+ \|{\bf r}'\|{\bf T}\times\|{\bf r}'\|{\bf T}'\cr &=\|{\bf r}'\|\|{\bf r}'\|'({\bf T}\times{\bf T})+\|{\bf r}'\|^2 ({\bf T}\times{\bf T}')\cr &=\|{\bf r}'\|^2({\bf T}\times{\bf T}')\cr }$$ because ${\bf T}\times{\bf T}={\bf 0}$, since ${\bf T}$ is parallel to itself. Then $$\eqalign{ \|{\bf r}'\times{\bf r}''\|&=\|{\bf r}'\|^2\|{\bf T}\times{\bf T}'\|\cr &=\|{\bf r}'\|^2\|{\bf T}\|\|{\bf T}'\|\sin\theta\cr &=\|{\bf r}'\|^2\|{\bf T}'\|\cr }$$ using exercise 8 in section 13.2 to see that $\theta=\pi/2$. Dividing both sides by $\|{\bf r}'\|^3$ then gives the desired formula. We used the fact here that ${\bf T}'$ is perpendicular to ${\bf T}$; the vector ${\bf N}={\bf T}'/\|{\bf T}'\|$ is thus a unit vector perpendicular to ${\bf T}$, called the {\dfont unit normal\/}\index{unit normal}\index{normal} to the curve. Occasionally of use is the {\dfont unit binormal\/}\index{unit binormal}\index{binormal} ${\bf B}={\bf T}\times{\bf N}$, a unit vector perpendicular to both ${\bf T}$ and ${\bf N}$.
I can provide an approach by finite elements and an application of the functional calculus of selfadjoint operators. Background The spectrum of the Laplacian on a bounded domain $\varOmega$ with sufficiently smooth boundary subject to homogeneous Dirichlet boundary conditions is discrete. By the spectral theorem, there are eigenfunctions $e_i \in H^1_0(\varOmega)$ and $\lambda_i \in \mathbb{R}$ with $-\Delta \, e_i = \lambda_i \, e_i$ and $0 < \lambda_1 \leq \lambda_2 \leq \dotsc$ and $$ \int_\varOmega e_i(x) \, e_j(x) \, \operatorname{d} x = \delta_{ij}.$$By functional calculus, the solution $u \in H^{2s}(\varOmega) \cap H^{s}_0(\varOmega)$ to $(-\Delta)^s \,u = f$ for $f \in L^2(\varOmega)$ can be represented by $$u = \sum_{i=1}^\infty \left( \lambda_i^{-s} \, e_i \cdot \int_\varOmega f(x) \, e_i(x) \, \operatorname{d} x \right).$$ Weyl's law states that the eigenvalues behave as $\lambda_i \sim i^\frac{2}{\operatorname{dim}(\varOmega)}$ for $i \to \infty$. (You can get an idea of the law by performing a $\sin$-Fourier transform on the eigenvalue equation $- \Delta v = \lambda \, v$ for functions on the square $[0, 2\pi] \times [0, 2\pi]$.) Thus, it is meaningful to use a truncated expansion in order to approximate $u$. Our aim is now to use finite elements to approximate the eigenfunctions $e_i$ and the eigenvalues $\lambda_i$. Implementation First, we create a nice domain. Disks are boring so I use the following starfish: R = DiscretizeRegion[ BoundaryMeshRegion[ Map[ t \[Function] (2 + Cos[5 t])/3 {Cos[t], Sin[t]}, Most@Subdivide[0., 2. Pi, 2000]], Line[Partition[Range[2000], 2, 1, 1]] ], MaxCellMeasure -> 0.001, MeshQualityGoal -> "Maximal" ] Next, we have to set up the finite element method. This is somewhat a mess but we cannot apply NDSolve directly; we need access to mass matrix and stiffness matrix of the system. Note that I use Sin[2 x] + Cos[x + 3 y] as right hand side $f$. Needs["NDSolve`FEM`"] (Initialization of Finite Element Method) Rdiscr = ToElementMesh[R, "MeshOrder" -> 1]; vd = NDSolve`VariableData[{"DependentVariables", "Space"} -> {{u}, {x, y}}]; sd = NDSolve`SolutionData[{"Space"} -> {Rdiscr}]; cdata = InitializePDECoefficients[vd, sd, "DiffusionCoefficients" -> {{-IdentityMatrix[2]}}, "MassCoefficients" -> {{1}}, "LoadCoefficients" -> {{Sin[2 x] + Cos[x + 3 y]}} ]; bcdata = InitializeBoundaryConditions[vd, sd, {{DirichletCondition[u[x, y] == 0., True]}}]; mdata = InitializePDEMethodData[vd, sd]; (Discretization) dpde = DiscretizePDE[cdata, mdata, sd]; dbc = DiscretizeBoundaryConditions[bcdata, mdata, sd]; {load, stiffness, damping, mass} = dpde["All"]; DeployBoundaryConditions[{load, stiffness}, dbc]; Having mass and stiffness matrix, we have to reduce them to the interior degrees of freedom. For FEM of order 1, the Dirichlet boundary conditions are deployed into the stiffness matrix by replacing rows that correspond to boundary vertices by the respective row of an identity matrix. Moreover, we exploit that Mathematica sorts boundary vertices in front. (Finding interior degrees of freedom) intplist = Min[UpperTriangularize[stiffness, 1]["NonzeroPositions"][[All, 1]]] ;;; stiffness2 = stiffness[[intplist, intplist]]; mass2 = mass[[intplist, intplist]]; load2 = load[[intplist]]; We need eigenvalues and eigenvectors. Since calculating them is expensive, we restrict our attention to the most relevant 5 percent. (Spectral decomposition and functional calculus) s = 3/4; reducedmodeldimension = Floor[Length[stiffness2] 0.05]; {Λ, U} = Reverse /@ Eigensystem[{stiffness2, mass2}, -reducedmodeldimension, Method -> {"Arnoldi", "MaxIterations" -> 4000 }]; // AbsoluteTiming U = Map[u \[Function] u/Sqrt[u.mass2.u], U]; The last step is necessary to ensure that the row vectors of U (representing the eigenfunctions of $-\Delta$) form an $L^2$-orthonormal system (the $L^2$-inner product being represented by the reduced mass matrix mass2). For example, we can draw the first 6 eigenfunction like this: GraphicsGrid[ Partition[ Table[ eigenvec = ConstantArray[0, Dimensions[stiffness][[2]]]; eigenvec[[intplist]] = Flatten[U[[i]]]; eigenfun = ElementMeshInterpolation[{Rdiscr}, eigenvec]; Plot3D[eigenfun[x, y], {x, y} ∈ R], {i, 1, 6}], 3], ImageSize -> Large] For the hidden operator L as defined in the comment below, we can easily optain its $L^2$-Moore-Penrose pseudoinverse Lpinv by (L=mass2.Transpose[Λ^s U].(U.mass2);) Lpinv = b \[Function] Transpose[U].(Λ^-s (U.b)); Next, we solve the reduced equations and write the result into the interior degrees of freedom of a zero vector. Finally, ElementMeshInterpolation provides us with an InterpolatingFunction that can be easily plotted. solution = ConstantArray[0, Dimensions[stiffness][[2]]]; solution[[intplist]] = Flatten[Lpinv[load2]]; // AbsoluteTiming solfun = ElementMeshInterpolation[{Rdiscr}, solution]; Plot3D[solfun[x, y], {x, y} ∈ R] As plausibility check: This is the result for $(-\Delta)^s u = 1$ (upon revisiting this post, I was puzzled by the solution being assymetric while OP asked for the solution of $(-\Delta)^s u = 1$ which ought to have the same symmetries as the domain): Discussion This method is however very slow: It has complexity $O(N^3)$, where $N$ is the number of interior vertices. Moreover, this method is not very accurate. Kernel based methods (e.g., convolution with fundamental solution) might be more accurate and might have complexity $O(N^2)$, but they might be limited to special domains where the fundamental solutions are known. Speaking about special domains: For the disk $B(0;1) \subset \mathbb{R}^2$, the surface of the sphere $S^2 \subset \mathbb{R}^3$, the unit ball $B(0;1) \subset \mathbb{R}^3$, the standard tori $\mathbb{T}^n = (S^1)^n$, and all cube-like domains like $Q = \coprod_{i=1}^n [a_i,b_i]$, the eigenvectors and eigenvalues are well studied. E.g., using FFT will speed up the process tremendously for $\mathbb{T}^n$ and $Q$. One could also circumvent the spectral decomposition step by directly assembling the Gagliardo bilinear form $$(u,v) \mapsto \pi^{-(2 s + n/2)}\frac{\Gamma(n/2+s)}{\Gamma(-s)}\int_\varOmega\int_\varOmega \frac{(u(x)-u(y))\,(v(x)-v(y))}{\| x-y \|^{n + 2 s}}\, \operatorname{d}x \, \operatorname{d}y$$ instead of the Laplacian stiffness matrix ($n=2$) (I am not exactly sure about the multiplicative constant in this formula; I got it from Hitchhiker's Guide to Fractional Sobolev Spaces, p. 8). This would require (partially singular) double integrals over $O(N^2)$ pairs of triangle elements. Moreover, that matrix is dense and needs $O(N^3)$ for factorization. Maybe one can apply FFT on a bounding box of the domain in order to construct a reasonable preconditioner for the conjugate gradient method.
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Ineffable cardinal Ineffable cardinals were introduced by Jensen and Kunen in [1] and arose out of their study of $\diamondsuit$ principles. An uncountable regular cardinal $\kappa$ is ineffable if for every sequence $\langle A_\alpha\mid \alpha<\kappa\rangle$ with $A_\alpha\subseteq \alpha$ there is $A\subseteq\kappa$ such that the set $S=\{\alpha<\kappa\mid A\cap \alpha=A_\alpha\}$ is stationary. Equivalently an uncountable regular $\kappa$ is ineffable if and only if for every function $F:[\kappa]^2\rightarrow 2$ there is a stationary $H\subseteq\kappa$ such that $F\upharpoonright [H]^2$ is constant [1]. This second characterization strengthens a characterization of weakly compact cardinals which requires that there exist such an $H$ of size $\kappa$. If $\kappa$ is ineffable, then $\diamondsuit_\kappa$ holds and there cannot be a slim $\kappa$-Kurepa tree [1] . A $\kappa$-Kurepa tree is a tree of height $\kappa$ having levels of size less than $\kappa$ and at least $\kappa^+$-many branches. A $\kappa$-Kurepa tree is slim if every infinite level $\alpha$ has size at most $\|\alpha\|$. Contents Ineffable cardinals and the constructible universe Ineffable cardinals are downward absolute to $L$. In $L$, an inaccessible cardinal $\kappa$ is ineffable if and only if there are no slim $\kappa$-Kurepa trees. Thus, for inaccessible cardinals, in $L$, ineffability is completely characterized using slim Kurepa trees. [1] Relations with other large cardinals Measurable cardinals are ineffable and stationary limits of ineffable cardinals. $\omega$-Erdős cardinals are stationary limits of ineffable cardinals, but not ineffable since they are $\Pi_1^1$-describable. [2] Ineffable cardinals are $\Pi^1_2$-indescribable [1]. Ineffable cardinals are limits of totally indescribable cardinals. [1] ([3] for proof) Weakly ineffable cardinal Weakly ineffable cardinals (also called almost ineffable) were introduced by Jensen and Kunen in [1] as a weakening of ineffable cardinals. An uncountable regular cardinal $\kappa$ is weakly ineffable if for every sequence $\langle A_\alpha\mid \alpha<\kappa\rangle$ with $A_\alpha\subseteq \alpha$ there is $A\subseteq\kappa$ such that the set $S=\{\alpha<\kappa\mid A\cap \alpha=A_\alpha\}$ has size $\kappa$. If $\kappa$ is weakly ineffable, then $\diamondsuit_\kappa$ holds. Weakly ineffable cardinals are downward absolute to $L$. [1] Weakly ineffable cardinals are $\Pi_1^1$-indescribable. [1] Ineffable cardinals are limits of weakly ineffable cardinals. Weakly ineffable cardinals are limits of totally indescribable cardinals. [1] ([3] for proof) Subtle cardinal Subtle cardinals were introduced by Jensen and Kunen in [1] as a weakening of weakly ineffable cardinals. A uncountable regular cardinal $\kappa$ is subtle if for every for every $\langle A_\alpha\mid \alpha<\kappa\rangle$ with $A_\alpha\subseteq \alpha$ and every closed unbounded $C\subseteq\kappa$ there are $\alpha<\beta$ in $C$ such that $A_\beta\cap\alpha=A_\alpha$. If $\kappa$ is subtle, then $\diamondsuit_\kappa$ holds. Subtle cardinals are downward absolute to $L$. [1] Weakly ineffable cardinals are limits of subtle cardinals. [1] Subtle cardinals are limits of totally indescribable cardinals. [1] The least subtle cardinal is not weakly compact as it is $\Pi_1^1$-describable. $\alpha$-Erdős cardinals are subtle. [1] Completely ineffable cardinal Completely ineffable cardinals were introduced in [3] as a strenthening of ineffable cardinals. Define that a collection $R\subseteq P(\kappa)$ is a stationary class if $R\neq\emptyset$, for all $A\in R$, $A$ is stationary in $\kappa$, if $A\in R$ and $B\supseteq A$, then $B\in R$. A cardinal $\kappa$ is completely ineffable if there is a stationary class $R$ such that for every $A\in R$ and $F:[A]^2\to 2$, there is $H\in R$ such that $F\upharpoonright [H]^2$ is constant. Completely ineffable cardinals are downward absolute to $L$. [3] Completely ineffable cardinals are limits of ineffable cardinals. [3] $n$-ineffable cardinal The $n$-ineffable cardinals for $2\leq n<\omega$ were introduced by Baumgartner in [4] as a strengthening of ineffable cardinals. A cardinal is $n$-ineffable if for every function $F:[\kappa]^n\rightarrow 2$ there is a stationary $H\subseteq\kappa$ such that $F\upharpoonright [H]^n$ is constant. A cardinal $\kappa$ is totally ineffable if it is $n$-ineffable for every $n$. $2$-ineffable cardinals are exactly the ineffable cardinals. an $n+1$-ineffable cardinal is a stationary limit of $n$-ineffable cardinals. [4] a $1$-iterable cardinal is a stationary limit of totally ineffable cardinals. (this follows from material in [5]) References Jensen, Ronald and Kunen, Kenneth. Some combinatorial properties of $L$ and $V$.Unpublished, 1969. www bibtex Jech, Thomas J. Third, Springer-Verlag, Berlin, 2003. (The third millennium edition, revised and expanded) www bibtex Set Theory. Abramson, Fred and Harrington, Leo and Kleinberg, Eugene and Zwicker, William. Flipping properties: a unifying thread in the theory of large cardinals.Ann Math Logic 12(1):25--58, 1977. MR bibtex Baumgartner, James. Ineffability properties of cardinals. I.Infinite and finite sets (Colloq., Keszthely, 1973; dedicated to P. Erdős on his 60th birthday), Vol. I, pp. 109--130. Colloq. Math. Soc. János Bolyai, Vol. 10, Amsterdam, 1975. MR bibtex Gitman, Victoria. Ramsey-like cardinals.The Journal of Symbolic Logic 76(2):519-540, 2011. www arχiv MR bibtex
x[1] := 1 x[n_] := n - 1 + Sum[j x[j], {j, 1, n - 1}] x /@ Range[10] {1, 2, 7, 29, 146, 877, 6140, 49121, 442090, 4420901} Edit $$x(1)=1$$ $$x(n+1)=n+\sum\limits_{j=1}^n j\cdot x(j) = 1+n-1+\sum\limits_{j=1}^{n-1}j\cdot x(j)+n\cdot x(n) = 1+x(n)+n\cdot x(n) \Rightarrow$$ $$x(n+1) = 1+(n+1)\cdot x(n)$$ but then $$x(2)=x(1+1)=1+(1+1)x(1)=1+2\cdot 1=3\neq 2$$ because in $$1+\left[n-1+\sum\limits_{j=1}^{n-1} j\cdot x(j)\right] + n\cdot x(n)$$ for $n=1$ the term in square brackets becomes $$1-1+\sum\limits_{j=1}^0 j\cdot x(j)$$ which doesn't make sense. So $x(2)$ has to be calculated from the first definition, and included in the RSolve as an initial condition: x[n] /. RSolve[{x[n + 1] == 1 + (n + 1) x[n], x[2] == 2}, x[n], n][[1]] // FullSimplify[#, n > 0 && n \[Element] Integers] & -((3 n!)/2) + E Gamma[1 + n, 1] as per Daniel's answer.
In practice, sound is a complex multitude of waves with different frequencies, phase, and amplitudes. When sound is recorded digitally, however, it becomes simply a collection of numbers. The properties of the many sound waves – frequency, phase, amplitude – cannot be discerned from the digitized data. In the world of audio then: Digital signal processing, or DSP, is the science of manipulating digital signals using nothing more than what is contained in the digital representation of the signal and with little knowledge of its precise properties. A digital audio signal is simply a set of numbers and the operations that modify these numbers to the needed end are mathematical in nature. DSP manipulates digitized signals typically with mathematical operations. It is a sizeable body of knowledge that has applications not only in music or mathematics, but also in physics, engineering, photography and video and many other fields. DSP, in fact, is such a large body of knowledge that the definition above is hardly precise. For the purposes of this site, however, it is appropriate. The following is a simple introduction to digital signal processing with examples of simple DSP operations. Properties of simple waves and the mathematical representation of complex signals The wave $$x(t)=A\,cos(2\pi\,f\,(t-\tau))$$ has initial phase equal to τ units of time, peak amplitude equal to A, frequency equal to f cycles per unit of time, and length of cycle equal to 1/f units of time. The three properties – amplitude, phase, and frequency (or length of cycle) – fully define a simple cosine wave as a function of time. When this wave is digitized, its value is recorded at various points of time. This process is called sampling. The number of samples taken in an interval of time is called the sampling rate. Commonly, sampling is at uniform time intervals and the sampling rate is constant. It is also common to record samples with comparable amplitude, called the sampling resolution. Sampling with a constant sampling rate and sampling resolution simplifies DSP and is called pulse code modulation or PCM. Suppose that we sample the simple wave A cos(2π f (t – τ)) over the time interval T with the sampling frequency f s. The times at which a sample is taken are t = 0, 1/f s, 2/f s, 3/f s, and so on. If these samples are numbered with k = 0, 1, 2, …, then the value of the wave at each sample will be $$x(k)=A\,cos(2\pi\,f\,(\frac{k}{f_s}-\tau))$$ In τ seconds, there would be m ≈ τ / T = τ f s samples taken and so we can, with some approximation, rewrite the above formula as follows. Given the sampling frequency f s, the wave $$x(k)=A\,cos(\frac{2\pi\,f\,(k-m)}{f_s})$$ has initial phase of m samples or m / f s units of time, peak amplitude of A, and frequency of f. Its cycle is 1/f units of time or f s / f samples. Again, the three properties – amplitude, phase, and frequency – fully define a simple cosine wave as a function of time. (Note here one of the most important theorems in DSP – the Nyquist-Shannon sampling theorem. In one of its many versions, the theorem states that the frequency content of a signal is fully represented by sampling at a certain frequency, if the signal does not contain frequencies higher than one-half of the sampling rate. That is, only waves with frequencies up to half of some sampling frequency can be recorded digitally with that sampling frequency.) Practical signals are often thought of as a sum of many simple waves. A continuous complex signal x(t) can be written as follows. $$x(t)=\sum_{n=0}^{N-1} A_n\,cos(2\pi\,f_n\,(t-\tau_n))$$ A n, f n, and τ n are the peak amplitude, frequency, and initial phase of each of the n simple waves in the signal. A discrete time complex signal x(t), sampled with the sampling frequency f s, would be $$x(k)=\sum_{n=0}^{N-1} A_n\,cos(\frac{2\pi\,f_n\,(k-m_n)}{f_s})$$ When processing digital signals, we will work with signals for which we do not know the precise values for N, A n, f n, and m n or τ n. As above, the task of DSP is to manipulate x(k) without knowing these values. The signal in the figure below, for example, consists of three simple waves. We can tell that the signal contains at least a higher frequency with smaller amplitude and some lower frequencies. It is impossible to tell, however, that there are exactly three frequencies and that they are exactly the simple waves cos(t), cos(0.5 t), and 0.4 cos(2π t). Practical signals One fundamental question remains: Whether all complex signals consist of simple waves. We do not actually know whether practical signals resemble sine or cosine waves in any way. As it turns out, whether practical signals are the sums of simple sine and cosine waves is not important. In 1807, Joseph Fourier showed that any periodic function with a period of 2π that is integrable over [-π, π] can be approximated with a (potentially infinite) linear sum of sine and cosine waves (see Fourier analysis). With some adjustments to Fourier's theory, similar approximations can be done for other periods and intervals. We can continue to treat complex signals as sums of simple waves, whether or not they are such in practice. Simple DSP operations The mixing of music – the combination of several audio tracks into one – is one of the simplest DSP operation. Take a recording that consists of four tracks that are yet not mixed: drums, bass, guitar, and vocals. Each track contains a sampled signal and there are four sampled signals, which we will label with x d(k), x b(k), x g(k), and x v(k) for the drums, bass, guitar and vocals respectively. We can mix the tracks into a single music piece if all tracks, when recorded digitally, are sampled with the same sampling frequency and that volumes, pans, envelopes, and various effects for each track have already been set and no other adjustments should be made. The output of the mixing process y(k) is the signal $$y(k)=x_d(k)+x_b(k)+x_g(k)+x_v(k)$$ The simplest delay DSP effect is one in which the signal is repeated once with some delay in time and some decay in amplitude. The output y(k) of the digital simple delay can be computed with the following equation. $$y(k)=x(k)+A\,x(k-m)$$ The output signal y(k) is the sum of the input signal x(k) and the same input signal x, but this time delayed by m samples and with amplitude scaled by A. Given the sampling frequency f s, the delay of m samples is equivalent to a delay of m / f s units of time. This operation is also known as a feedforward comb filter. Consider the following computation of the output signal y(k) from the input signal x(k). $$y(k)=\begin{cases} 0.75, & \|x(k)\| \ge 0.75 \\ x(k), & \|x(k)\| \lt 0.75 \end{cases}$$ This operation is known as a hard clip distortion. More complex DSP operations None of the simple DSP operations above make use of the fact that complex signals consist of simple sine waves. More complex DSP operations do. Most complex DSP operations rely on a couple of simple facts. First, the sumproduct (and convolution) of two simple waves at different frequencies is zero for appropriately chosen intervals (i.e., simple waves are orthogonal). Second, the sumproduct (and convolution) of two simple waves at the same frequency is constant for appropriately chosen intervals. This makes it easy to extract simple waves from complex signals and, in fact, discern their amplitude and phase. This is explained in the topic Fourier analysis. Relatively simple examples of the application of these facts is the alternative derivation of the low pass filter and the interpretation of the Fourier transform on this site. Of course, not all DSP relies on the facts above. See for example, data compression of digital signals with the Haar wavelets or the Daub4 wavelets. Note, however, that the wavelets themselves satisfy the same properties – they are orthonormal – and that they also may act as low pass and high pass filters on the signal.
Difference between revisions of "Vopenka" (bigger separate section) Line 7: Line 7: For any language $\mathcal{L}$ and any proper class $C$ of $\mathcal{L}$-structures, there are distinct structures $M, N\in C$ and an [[elementary embedding]] $j:M\to N$. For any language $\mathcal{L}$ and any proper class $C$ of $\mathcal{L}$-structures, there are distinct structures $M, N\in C$ and an [[elementary embedding]] $j:M\to N$. </blockquote> </blockquote> − For example, taking $\mathcal{L}$ to be the language with one unary and one binary predicate, we can consider for any ordinal $\eta$ the class of structures $\langle V_{\alpha+\eta},\{\alpha\},\in\rangle$, and conclude from Vopěnka's principle that a cardinal that is at least $\eta$-[[extendible]] exists. In fact if Vopěnka's principle holds then there + For example, taking $\mathcal{L}$ to be the language with one unary and one binary predicate, we can consider for any ordinal $\eta$ the class of structures $\langle V_{\alpha+\eta},\{\alpha\},\in\rangle$, and conclude from Vopěnka's principle that a cardinal that is at least $\eta$-[[extendible]] exists. In fact if Vopěnka's principle holds then there a stationary proper class of extendible cardinals; bounding the strength of the axiom from above, we have that if $\kappa$ is [[huge#Almost huge\|almost huge]], or even [[high-jump\|almost-high-jump]], then $V_\kappa$ satisfies Vopěnka's principle. ==Formalizations== ==Formalizations== Revision as of 04:50, 16 September 2019 Vopěnka's principle is a large cardinal axiom at the upper end of the large cardinal hierarchy that is particularly notable for its applications to category theory. In a set theoretic setting, the most common definition is the following: For any language $\mathcal{L}$ and any proper class $C$ of $\mathcal{L}$-structures, there are distinct structures $M, N\in C$ and an elementary embedding $j:M\to N$. For example, taking $\mathcal{L}$ to be the language with one unary and one binary predicate, we can consider for any ordinal $\eta$ the class of structures $\langle V_{\alpha+\eta},\{\alpha\},\in\rangle$, and conclude from Vopěnka's principle that a cardinal that is at least $\eta$-extendible exists. In fact if Vopěnka's principle holds then there is a stationary proper class of extendible cardinals; bounding the strength of the axiom from above, we have that if $\kappa$ is almost huge, or even almost-high-jump, then $V_\kappa$ satisfies Vopěnka's principle. Contents Formalizations As stated above and from the point of view of ZFC, this is actually an axiom schema, as we quantify over proper classes, which from a purely ZFC perspective means definable proper classes. A somewhat stronger alternative is to view Vopěnka's principle as an axiom in second-order set theory capable to dealing with proper classes, such as von Neumann-Gödel-Bernays set theory. This is a strictly stronger assertion. [1] Finally, one may relativize the principle to a particular cardinal, leading to the concept of a Vopěnka cardinal. Vopěnka's principle can be formalized in first-order set theory as a schema, where for each natural number $n$ in the meta-theory there is a formula expressing that Vopěnka’s Principle holds for all $Σ_n$-definable (with parameters) classes.[1] Vopěnka cardinals An inaccessible cardinal $\kappa$ is a Vopěnka cardinal if and only if $V_\kappa$ satisfies Vopěnka's principle, that is, where we interpret the proper classes of $V_\kappa$ as the subsets of $V_\kappa$ of cardinality $\kappa$. Because of a characterization of Vopěnka's principle in terms of graphs, a cardinal $\kappa$ is Vopěnka if and only if $\kappa$ is inaccessible and any set $\kappa$-sized set $G$ of $<\kappa$-sized nonisomorphic graphs has some $g_0$ and $g_1$ with $g_0$ a proper subgraph of $g_1$. (Need to cite sources) As we mentioned above, every almost huge cardinal is a Vopěnka cardinal. Equivalent statements $C^{(n)}$-extendible cardinals The schema form of Vopěnka's principle is equivalent to the existence of a proper class of $C^{(n)}$-extendible cardinals for every $n$; indeed there is a level-by-level stratification of Vopěnka's principle, with Vopěnka's principle for a $\Sigma_{n+2}$-definable class corresponds to the existence of a $C^{(n)}$-extendible cardinal greater than the ranks of the parameters (see section "Variants”). [3] Strong Compactness of Logics Vopěnka's principle is equivalent to the following statement about logics as well: For every logic $\mathcal{L}$, there is a cardinal $\mu_{\mathcal{L}}$ such that for any language $\tau$ and any $\mathcal{L}(\tau)$-theory $T$, $T$ is satisfiable if and only if every $t\subseteq T$ such that $\|t\|<\mu_{\mathcal{L}}$ is satisfiable. [4] This $\mu_{\mathcal{L}}$ is called the strong compactness cardinal of $\mathcal{L}$. Vopěnka's principle therefore is equivalent to every logic having a strong compactness cardinal. This is very similar in definition to the Löwenheim–Skolem number of $\mathcal{L}$, although it is not guaranteed to exist. Here are some examples of strong compactness cardinals of specific logics: If $\kappa\leq\lambda$ and $\lambda$ is strongly compact or $\aleph_0$, then the strong compactness cardinal of $\mathcal{L}_{\kappa,\kappa}$ is at most $\lambda$. Similarly, if $\kappa\leq\lambda$ and $\lambda$ is extendible, then for any natural number $n$, the strong compactness cardinal of $\mathcal{L}^n_{\kappa,\kappa}$ ($\mathcal{L}_{\kappa,\kappa}$ with $n+1$-th order logic) is at most $\lambda$. Therefore for any natural number $n$, the strong compactness cardinal of $n+1$-th order finitary logic is at most the least extendible cardinal. Locally Presentable Categories Vopěnka's principle is equivalent to the axiom stating "no large full subcategory $C$ of any locally presentable category is discrete." (Sources needed). Equivalently, no large full subcategory of Graph (the category of all graphs) is discrete; that is, for any proper class of simple directed graphs, there is at least one pair of nonequal graphs $G$ and $H$ in the class such that $G$ is a subgraph of $H$. This is a $\Pi^1_1$ statement, so the least Vopěnka cardinals are not even weakly compact (although the least weakly compact cardinal is much, much, much smaller than the least Vopěnka cardinal, if it exists). Intuitively, a "category" is just a class of mathematical objects with some notion of "morphism", "homomorphism", "isomorphism", (etc.). For example, in Set, the category of all sets, homomorphisms are just injections, and isomorphisms are bijections. In categories of groups and models, homomorphisms and isomorphisms share their actual names. A "locally small category" $C$ is one with only set-many morphisms between any two objects of $C$. This is one where the objects of $C$ behave "set-like" in the sense that, usually, the number of morphisms between two set-sized objects is at most the number of functions between their universes (like in groups and in graphs). A "locally presentable category" is a locally small category with a couple more really nice properties; you can "generate" all of the objects from set-many objects in the category. Vopěnka's principle intuitively states that if you have a locally presentable category $C$, then any proper class of objects of $C$ has some nonisomorphic objects $c$ and $d$ where $c$ has a morphism into $d$. Woodin cardinals There is a strange connection between the Woodin cardinals and the Vopěnka cardinals. In particular, Vopěnkaness is equivalent to two strengthening variants of Woodinness, namely the Woodin for Supercompactness cardinals and the $2$-fold Woodin cardinals. As a result, every Vopěnka cardinal is Woodin. Elementary Embeddings Between Ranks An equivalent statement to Vopěnka's principle is that for any proper class $C\subseteq ORD$, there are $\alpha\in C$, $\beta\in C$, and a nontrivial elementary embedding $j:\langle V_\alpha;\in,P\rangle\rightarrow\langle V_\beta;\in,P\rangle$. Vopěnka's principle quite obviously implies this. The reason the converse holds is because every elementary embedding can be "encoded" (in a sense) into one of these. For more information, see [5]. Other points to note Whilst Vopěnka cardinals are very strong in terms of consistency strength, a Vopěnka cardinal need not even be weakly compact. Indeed, the definition of a Vopěnka cardinal is a $\Pi^1_1$ statement over $V_\kappa$ (Vopěnka's principle itself is $\Pi^1_1$), and $\Pi^1_1$-indescribability is one of the equivalent definitions of weak compactness. Thus, the least weakly compact Vopěnka cardinal must have (many) other Vopěnka cardinals less than it. Variants (Boldface) $VP(\mathbf{Σ_n})$ denotes the fragment of Vopěnka’s Principle for $Σ_n$-definable classes and (lightface) $VP(Σ_n)$ is the weaker principle, where parameters are not allowed in the definition of the class (with analogous definitions for $Π_n$ and $∆_n$). Vopěnka-like principles $VP(κ, \mathbf{Σ_n})$ for cardinal $κ$ state that for every proper class $\mathcal{C}$ of structures of the same type that is $Σ_n$-definable with parameters in $H_κ$ (the collection of all sets of hereditary size less than $κ$), $\mathcal{C}$ reflects below $κ$, namely for every $A ∈ C$ there is $B ∈ H_κ ∩ C$ that elementarily embeds into $A$. Results: For every $Γ$, $VP(κ, Γ)$ for some $κ$ implies $VP(Γ)$. $VP(κ, \mathbf{Σ_1})$ holds for every uncountable cardinal $κ$. $VP(Π_1) \iff VP(κ, Σ_2)$ for some $κ \iff$ There exists a supercompact cardinal. $VP(\mathbf{Π_1}) \iff VP(κ, \mathbf{Σ_2})$ for a proper class of cardinals $κ \iff$ There is a proper class of supercompact cardinals. For $n ≥ 1$, the following are equivalent: $VP(Π_{n+1})$ $VP(κ, \mathbf{Σ_{n+2}})$ for some $κ$ There exists a $C(n)$-extendible cardinal. The following are equivalent: $VP(Π_n)$ for every n. $VP(κ, \mathbf{Σ_n})$ for a proper class of cardinals $κ$ and for every $n$. $VP$ For every $n$, there exists a $C(n)$-extendible cardinal. References Bagaria, Joan. $C^{(n)}$-cardinals.Archive for Mathematical Logic 51(3--4):213--240, 2012. www DOI bibtex Perlmutter, Norman. The large cardinals between supercompact and almost-huge., 2010. arχiv bibtex Bagaria, Joan and Casacuberta, Carles and Mathias, A R D and Rosický, Jiří. Definable orthogonality classes in accessible categories are small.Journal of the European Mathematical Society 17(3):549--589. arχiv bibtex Makowsky, Johann. Vopěnka's Principle and Compact Logics.J Symbol Logic www bibtex Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Bagaria, Joan and Gitman, Victoria and Schindler, Ralf. Generic {V}opěnka's {P}rinciple, remarkable cardinals, and the weak {P}roper {F}orcing {A}xiom.Arch Math Logic 56(1-2):1--20, 2017. www DOI MR bibtex
Search Now showing items 1-1 of 1 Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV (Elsevier, 2014-09) Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, \|y\| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ...
Charged-Lepton-Flavour Violation in the light of the Super-Kamiokande data 42 Downloads Citations Abstract. Motivated by the data from Super-Kamiokande and elsewhere indicating oscillations of atmospheric and solar neutrinos, we study charged-lepton-flavour violation, in particular the radiative decays $\mu \rightarrow e \gamma$ and $\tau \rightarrow \mu \gamma$, but also commenting on $\mu \to 3e$ and $\tau \to 3 \mu/e$ decays, as well as $\mu \to e$ conversion on nuclei. We first show how the renormalization group may be used to calculate flavour-violating soft supersymmetry-breaking masses for charged sleptons and sneutrinos in models with universal input parameters. Subsequently, we classify possible patterns of lepton-flavour violation in the context of phenomenological neutrino mass textures that accommodate the Super-Kamiokande data, giving examples based on Abelian flavour symmetries. Then we calculate in these examples rates for $\mu \rightarrow e \gamma$ and $\tau \rightarrow \mu \gamma$, which may be close to the present experimental upper limits, and show how they may distinguish between the different generic mixing patterns. The rates are promisingly large when the soft supersymmetry-breaking mass parameters are chosen to be consistent with the cosmological relic-density constraints. In addition, we discuss $\mu \rightarrow e$ conversion on Titanium, which may also be accessible to future experiments. KeywordsTitanium Input Parameter Renormalization Group Neutrino Mass Future Experiment Preview Unable to display preview. Download preview PDF.
I have a data set where many of the actual values are zero, so I can't use MAPE. It's not a time series, so I can't use MASE ala our very own Rob Hyndman. Is there another alternative to MAPE that I could use? Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.Sign up to join this community MASE is suitable for non-time series data also. See my textbook: https://www.otexts.org/fpp/2/5 In that case, you can scale the data using the mean as the base forecast. So if $e_j$ denotes a prediction error on the test data, then the scaled errors are $$ q_{j} = \frac{\displaystyle e_{j}}{\displaystyle\frac{1}{N}\sum_{i=1}^N \|y_i-\bar{y}\|}. $$ where $y_1,\dots,y_N$ denotes the training data.
Modelling Population Growth and Decay One important application of differential equations arises in modelling population growth or decay within a species. Let $y = y(t)$ denote the population size of a species at time $t$. If the rate of change of this population is proportional by some constant $r$ (where $r > 0$ implies that the population is growing and $r < 0$ implies that the population is decaying) then we can model the change in population with the following differential equation:(1) Suppose that the initial population at time $t = 0$ is $y_0$, that is $y(0) = y_0$ . We can solve the differential equation above by either using the method of integrating factors or as a separable equation. Let's use integrating factors. We first rewrite the differential equation above in the form:(2) Then $p(t) = -r$. Let $\mu (t) = e^{\int p(t) \: dt} = e^{\int -r \: dt} = e^{-rt}$ be our integrating factor. Multiplying both sides of this differential equation by $\mu (t)$ and we get that:(3) With the initial condition $y(0) = y_0$ we see that $C = y_0$ and so the solution to our differential equation is:(4) As we see, if $r > 0$ then the growth of the population exponentially increases and if $r < 0$ then the decay of the population exponentially decreases. Of course, if $r > 0$ then $\lim_{t \to \infty} y_0e^{rt} = \infty$. Realistically, as time increases, the population $y$ will not increase indefinitely as population dynamics are based off of many other variables that are bounded by limits. For example, a finite amount of space or food may impede on a population from growing indefinitely. Our original assumption that the population growth/decay rate $r$ was constant is not realistic. Instead, the growth/decay of a population is also dependent on the population itself. We replace the constant $r$ with the function $h(y)$ which depends on the population $y$. Therefore, we have the new differential equation:(5) The choice of function $h(y)$ can be determined based on various conditions that affect the rate of growth/decay in a population. A generic choice of this function is $h(y) = (r - ay)$, $a > 0$, and where $r$ is the growth/decay rate from earlier. Note that as $y$ gets very large, $h(y)$ does not get too large. Substituting this into the differential equation and letting $K = \frac{r}{a}$ gives us:(6) The constant $K$ is known as the Carrying Capacity for the population. Now, this differential equation is separable since it can be rewritten as: We we will now solve this separable differential equation by integrating both sides:(8) If we plug in our initial condition $y(0) = y_0$ we get that:(9) When we plug this into our differential equation and isolated for $y$ we get that:(10) The solution to this initial value problem models population growth in constraining the population growth as it approaches the carrying capacity.
Tabulating numbers Let $u(n,k)$ denote the number of upwards-pointing triangles of size $k$ included in a triangle of size $n$, where size is a short term for edge length. Let $d(n,k)$ likewise denote the number of down triangles. You can tabulate a few numbers to get a feeling for these. In the following table, row $n$ and column $k$ will contain two numbers separated by a comma, $u(n,k), d(n,k)$. $$\begin{array}{c\|cccccc\|c}n \backslash k &1 & 2 & 3 & 4 & 5 & 6 & \Sigma \\\hline1 & 1, 0 &&&&&& 1 \\2 & 3, 1 & 1,0 &&&&& 5 \\3 & 6, 3 & 3,0 & 1,0 &&&& 13 \\4 & 10, 6 & 6,1 & 3,0 & 1,0 &&& 27 \\5 & 15,10 & 10,3 & 6,0 & 3,0 & 1,0 && 48 \\6 & 21,15 & 15,6 & 10,1 & 6,0 & 3,0 & 1,0 & 78\end{array}$$ Finding a pattern Now look for patterns: $u(n, 1) = u(n - 1, 1) + n$ as the size change added $n$ upwards-pointing triangles $d(n, 1) = u(n - 1, 1)$ as the downward-pointing triangles are based on triangle grid of size one smaller $u(n, n) = 1$ as there is always exactly one triangle of maximal size $d(2k, k) = 1$ as you need at least twice its edge length to contain a downward triangle. $u(n, k) = u(n - 1, k - 1)$ by using the small $(k-1)$-sized triangle at the top as a representant of the larger $k$-sized triangle, excluding the bottom-most (i.e. $n$ th) row. $d(n, k) = u(n - k, k)$ as the grid continues to expand, adding one row at a time. Using these rules, you can extend the table above arbitrarily. The important fact to note is that you get the same sequence of $1,3,6,10,15,21,\ldots$ over and over again, in every column. It describes grids of triangles of same size and orientation, increasing the grid size by one in each step. For this reason, those numbers are also called triangular numbers. Once you know where the first triangle appears in a given column, the number of triangles in subsequent rows is easy. Looking up the sequence Now take that sum column to OEIS, and you'll find this to be sequence A002717 which comes with a nice formula: $$\left\lfloor\frac{n(n+2)(2n+1)}8\right\rfloor$$ There is also a comment stating that this sequence describes the Number of triangles in triangular matchstick arrangement of side $n$. Which sounds just like what you're asking. References If you want to know how to obtain that formula without looking it up, or how to check that formula without simply trusting an encyclopedia, then some of the references given at OEIS will likely help you out: J. H. Conway and R. K. Guy, The Book of Numbers, p. 83. F. Gerrish, How many triangles, Math. Gaz., 54 (1970), 241-246. J. Halsall, An interesting series, Math. Gaz., 46 (1962), 55-56. M. E. Larsen, The eternal triangle – a history of a counting problem, College Math. J., 20 (1989), 370-392. C. L. Hamberg and T. M. Green, An application of triangular numbers, Mathematics Teacher, 60 (1967), 339-342. (Referenced by Larsen) B. D. Mastrantone, Comment, Math. Gaz., 55 (1971), 438-440. Problem 889, Math. Mag., 47 (1974), 289-292. L. Smiley, A Quick Solution of Triangle Counting, Mathematics Magazine, 66, #1, Feb '93, p. 40.
Some programming principles help us make short work of this. The key principle is to encapsulate what's going on. First, the surface. It depends on some parameters, so let's be explicit about that, rather than letting those parameters run around loose as "global" variables. To illustrate, I begin by generating some (reproducible) random values for these parameters: n = 6; SeedRandom[17]; λ = RandomReal[GammaDistribution[4, 1/5], n]; ϵ = RandomReal[GammaDistribution[3, 1/4], n]; x0 = RandomVariate[BinormalDistribution[{2.5, 2}, {2, 1}, 1/2], n]; I am going to change the meaning of the $\epsilon$ a little, though: it is more natural that they be proportional to the widths of the "hills," so they will appear in the denominator below. Here is the "Gaussian radial" function: f[x_, {λ_, ϵ_, x0_}] := Sum[λ[[i]] Exp[-(Norm[x - x0[[i]]]/ϵ[[i]])^2], {i, 1, Length[λ]}]; Its first argument is an $(x,y)$ coordinate pair and its second is a list of the parameters. It computes the topographic elevation at $(x,y)$ by summing the values of the Gaussians at $(x,y)$. We can make a pretty pseudo-3D plot of the topography (and store it in a variable for later display): dem = Plot3D[ f[{x, y}, {λ, ϵ, x0}], {x, -1, 5}, {y, -1, 5}, PlotRange -> {Full, Full, Full}, PlotStyle -> Opacity[0.9], MeshFunctions -> {#3 &}, MeshStyle -> GrayLevel[0.7]] Second, the key Calculus idea is that the hikers' path is parameterized by their base curve $t\to \gamma(t) = (x(t), y(t))$; it merely needs to be "lifted" to the proper height given by $f(\gamma(t), \ldots)$. (The dots represent values of the parameters controlling the shape of $f$.) The fully parameterized hike therefore is $$t \to (x(t), y(t), f((x(t), y(t)), \ldots)).$$ Because we have encapsulated the key functionality, this is easy and straightforward to code: γ[t_] := {5/2, 2} + {9 Sin[4 t], -6 Cos[4 t]}/5; path[t_, {λ_, ϵ_, x0_}] := With[{x = γ[t]}, Append[x, f[x, {λ, ϵ, x0}]]]; (For a general-purpose solution you might want to make γ and f arguments of path, too, in keeping with the idea that everything a function like path depends on should be explicitly exhibited, within reason.) ParametricPlot3D takes care of the graphical display: hike = ParametricPlot3D[path[t, {λ, ϵ, x0}], {t, 0, π/2}, PlotStyle -> Directive[Thick, Red]]; Finally, use Show to assemble the plots of the surface and the path and control the final appearance: Show[dem, hike, Boxed -> False, BoxRatios -> Automatic, AxesLabel -> {x, y, z}] Edit If you wish to study the hike in more detail, you may introduce a second parameter to path to fill it in vertically: let it multiply the $z$ value and range from $0$ to $1$. I will leave the implementation as an exercise and just show the result: This also shows how the technique is capable of accurately drawing a hike which is not a closed loop.
In this paper the authors perform a high-temperature expansion of the correlation functions for a Heisenberg model on a lattice. Starting from $$\left<\mathbf{S}_i\cdot\mathbf{S}_j\right>_\beta = \mathrm{Tr}(\mathbf{S}_i\cdot\mathbf{S}_j\,e^{-\beta H})/\mathrm{Tr}\, (e^{-\beta H})$$ They write expanding in powers of $\beta$ and formally dividing out the denominator gives a cumulant expansion $$\left<\mathbf{S}_i\cdot\mathbf{S}_j\right>_\beta = \sum_{m=0}^\infty \frac{(-\beta)^m}{m!} \left<\mathbf{S}_i\cdot\mathbf{S}_j H^m\right>_c$$ where $\left<\cdots\right>_c$ is a cumulant average $$\left<ABC\cdots\right>_c = \left< ABC \cdots \right>_0 - \sum \left< \cdots \right>_0 \left< \cdots \right>_0 + 2! \sum \left< \cdots \right>_0\left< \cdots \right>_0\left< \cdots \right>_0 - 3! \left< \cdots \right>_0\left< \cdots \right>_0\left< \cdots \right>_0\left< \cdots \right>_0 + \cdots$$ and $\left< \cdots \right>_0$ is an average (at $\beta = 0$) over all orientations of the spin operators contained within the brackets. The sums, which are multiplied by $(-1)^{n-1}(n-1)!$ are over all ways that the product of the operators $\mathbf{S}_i\cdot\mathbf{S}_j$ and $H$ can be distributed into $n$ averages. I am familiar with the concept of cumulant expansions, such as done in the diffuse interacting gas model and path integrals, but I would like to derive for myself the above statements and understand their connection to both cumulants of probability distributions as well as the types of expansions performed when dealing with interacting systems. Those expansions generally split the Hamiltonian (or Lagrangian) into a non-interacting (Gaussian) part and an interacting part which can be treated as a perturbation to the non-interacting distribution. In this case, we are rather considering a sort of perturbation about the $\beta=0$ distribution (which is also that of non-interacting degrees of freedom), but I don't fully see how to flesh out that analogy. One resource I have looked at is the book "Series Expansion Methods" by Oitmaa, Hamer, and Zheng. They don't treat the high-T expansions in the way I am looking for, but in Appendix 6 they define a moment $\left<\,\,\right>$ as the average of a set of variables, and then define the cumulant $\left[\,\,\,\right]$ as follows $$\left<\alpha\cdots\zeta\right>=\sum_{P}\left[\alpha\cdots\beta\right]\left[\gamma\cdots\zeta\right]$$ where the sum goes over all possible partitions $P$ of the set of variables, or symbolically $$\left<\,\,\right> = \sum_{n=1}^\infty \frac{1}{n!}\left[\,\,\,\right]^n$$ For example, \begin{align} \left<\alpha\right>&=[\alpha]\\ \left<\alpha\beta\right>&=[\alpha\beta]+[\alpha][\beta] \\ \left<\alpha\beta\gamma\right>&=[\alpha\beta\gamma]+[\alpha\beta][\gamma]+[\beta\gamma][\alpha] + [\gamma\alpha][\beta]+[\alpha][\beta][\gamma] \end{align} [which] can be inverted to give symbolically $$\left[\,\,\,\right] = \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{n}\left<\,\,\right>^n$$ for example \begin{align} [\alpha]&=\left<\alpha\right>\\ [\alpha\beta]&=\left<\alpha\beta\right>-\left<\alpha\right>\left<\beta\right>\\ [\alpha\beta\gamma]&=\left<\alpha\beta\gamma\right>-\left<\alpha\beta\right>\left<\gamma\right>-\left<\beta\gamma\right>\left<\alpha\right> - \left<\gamma\alpha\right>\left<\beta\right>+2\left<\alpha\right>\left<\beta\right>\left<\gamma\right> \end{align} These "symbolic" equations are quite mystifying to me. I see how to make the inversion by identifying the first as $e^{[\,]}-1$ and obtaining the second by expanding $\ln(1+\left<\,\,\right>)$, but I don't understand how to interpret either of these equations to write down the corresponding expressions as given, since it is not clear what $[\,\,\,]^n$ and $\left<\,\,\right>^n$ mean. It is also unclear precisely how this relates to the $\beta=0$ expansion above (I see the mathematical relation in the definition of the cumulants, but not how to get from the expectation value to the cumulant expansion). I am looking to understand (1) How to obtain the high-temperature expansion of Harris, Kallin, and Berlinsky, (2) How to understand the "symbolic" equations of Oitmaa, Hamer, and Zheng, and (3) The relationship between cumulant, high-T, and perturbative expansions for interacting systems such as in QFT (preferably some literature resources where I can dig deeper into this)
The PhD thesis of Paul J. Werbos at Harvard in 1974 described backpropagation as a method of teaching feed-forward artificial neural networks (ANNs). In the words of Wikipedia, it lead to a "rennaisance" in the ANN research in 1980s. As we will see later, it is an extremely straightforward technique, yet most of the tutorials online seem to skip a fair amount of details. Here's a simple (yet still thorough and mathematical) tutorial of how backpropagation works from the ground-up; together with a couple of example applets. Feel free to play with them (and watch the videos) to get a better understanding of the methods described below! 1. Background To start with, imagine that you have gathered some empirical data relevant to the situation that you are trying to predict - be it fluctuations in the stock market, chances that a tumour is benign, likelihood that the picture that you are seeing is a face or (like in the applets above) the coordinates of red and blue points. We will call this data training examples and we will describe th training example as a tuple , where is a vector of inputs and is the observed output. Ideally, our neural network should output when given as an input. In case that does not always happen, let's define the error measure as a simple squared distance between the actual observed output and the prediction of the neural network: , where is the output of the network. 2. Perceptrons (building-blocks) The simplest classifiers out of which we will build our neural network are perceptrons (fancy name thanks to Frank Rosenblatt). In reality, a perceptron is a plain-vanilla linear classifier which takes a number of inputs , scales them using some weights , adds them all up (together with some bias ) and feeds everything through an activation function . A picture is worth a thousand equations: To slightly simplify the equations, define and . Then the behaviour of the perceptron can be described as , where and . To complete our definition, here are a few examples of typical activation functions: sigmoid:, hyperbolic tangent:, plain linearand so on. Now we can finally start building neural networks. The simplest kind of network that we can build is... exactly, one perceptron! Here's how we can train it to classify things! 3. Single-layer neural network We defined the error earlier as . Obviously, since we are using a single perceptron both our error and the output of the network () depend on the weights vector . Incorporating those observations into the updated error measure we obtain . Our goal is to find such a vector of weights that is minimised - that way our perceptron will correctly predict the output for all inputs of our training examples! We will do that by applying the gradient descent algorithm: in essence we will treat the error as a surface in n-dimensional space, then we will find a greatest downwards slope at the current point and will go in that direction to obtain . This way hopefully we will find a minimum point on the error surface and we will use the coordinates of that point as the final weight vector. By skipping a great deal of maths on whether the minimum point exists, is it unique and global, can we "overjump" it by accident, what are the conditions for the following partial derivatives to exist, etc, etc; we will dive straight in hoping for the best and will calculate the gradient of the error surface at . Then we will take a step in the opposite direction of the gradient (i.e. in the direction of the fastest decreasing slope on the error surface) to obtain . To express it in a slightly more mathematical way, we will start with some randomized (!) weight vector and will train our perceptron by updating the weights \begin{align} \vec{w}_{t+1} := \vec{w_t} - \eta \frac{\partial E(\vec{w})}{\partial \vec{w}} \bigg\|_{\vec{w_t}}, \end{align} where is known as a learning rate (a simple scaling factor that typically ranges between zero and one). Observe that \begin{align} \frac{\partial E(\vec{w})}{\partial \vec{w}} = \left( \frac{\partial E(\vec{w})}{\partial w_0},\frac{\partial E(\vec{w})}{\partial w_1}, ... ,\frac{\partial E(\vec{w})}{w_n} \right), \end{align} and we can calculate \begin{align} \frac{\partial E(\vec{w})}{\partial w_j} &= \frac{\partial}{\partial w_j} \sum_i (h_{\vec{w}}(\vec{x_i}) - y_i)^2 \\ &= \sum_i 2(h_{\vec{w}}(\vec{x_i}) - y_i) \frac{\partial}{\partial w_j} (h_{\vec{w}}(\vec{x_i}) - y_i) \\ &= \sum_i 2(h_{\vec{w}}(\vec{x_i}) - y_i) \frac{\partial}{\partial w_j} \sigma(\vec{x_i} \cdot \vec{w}) \\ &= \sum_i 2(h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (\vec{x_i} \cdot \vec{w}) \frac{d}{d w_j} \vec{x_i} \cdot \vec{w} \\ &= \sum_i 2(h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (\vec{x_i} \cdot \vec{w}) \frac{d}{d w_j} \sum_{k=1}^n x_{i,k} w_k \\ &= 2 \sum_i (h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (\vec{x_i} \cdot \vec{w}) x_{i,j} \end{align} for each . 3.1. Example single-layer neural network In this applet, a perceptron takes two inputs (normalized x and y coordinates and , i.e. , ) and uses sigmoid as an activation function with the learning rate . Then, using a previous general result \begin{align} \frac{\partial E(\vec{w})}{\partial w_j} &= 2 \sum_i (h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (\vec{x_i} \cdot \vec{w}) x_{i,j} \\ &= 2 \sum_i (\sigma(\vec{w} \cdot \vec{x_i}) - y_i) \sigma(\vec{x_i} \cdot \vec{w}) (1 - \sigma(\vec{x_i} \cdot \vec{w})) x_{i,j}, \end{align} (since for the sigmoid activation function ); and thus \begin{align} \frac{\partial E(\vec{w})}{\partial \vec{w}} = 2 \sum_i (\sigma(\vec{w} \cdot \vec{x_i}) - y_i) \sigma(\vec{x_i} \cdot \vec{w}) (1 - \sigma(\vec{x_i} \cdot \vec{w})) \vec{x_i}. \end{align} The final algorithm to update the weight vector (which is initially randomized) then is \begin{align} \vec{w}_{t+1} := \vec{w_t} - 0.2 \sum_i (h_{\vec{w}_t}(\vec{x_i}) - y_i) h_{\vec{w}_t}(\vec{x_i}) (1 - h_{\vec{w}_t}(\vec{x_i})) \vec{x_i}, \end{align} where . However, a single perceptron is extremely limited in the sense that different classes of examples must be separable with a hyperplane (hence the name, linear classifier), which is usually not the case in real-life applications. Time to bump things up a notch: let's connect a few of them together to obtain a multilayer feed-forward neural network! 4. Multilayer neural network Let's consider a general case first: a completely unrestricted feed-forward structure (with the only condition being that there are no loops between the perceptrons to avoid general madness and chaos). Since it is structurally more complex than just a single perceptron, take a look at the following figure that explains some more notation: Here the weight connects perceptrons and , the sum of the weighed inputs of perceptron is denoted by where iterates over all perceptrons connected to , and the output of is written as , where is 's activation function. We will use the same error measure , except now the weights vector will contain all the weights in the network, i.e. for all . To find that minimizes using gradient descent we have to calculate (again). However, this time it is (very slightly) more involved. First of all let's separate the contributions of individual training examples to the overall error using the following observation: \begin{align} \frac{\partial E(\vec{w})}{\partial \vec{w}} = \sum_i \frac{\partial E_i(\vec{w})}{\partial \vec{w}}, \end{align} where . Then \begin{align} \frac{\partial E_i(\vec{w})}{\partial w_{j \rightarrow k}} &= \frac{\partial}{\partial w_{j \rightarrow k}} (h_{\vec{w}}(\vec{x_i}) - y_i)^2 \\ &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial w_{j \rightarrow k}} \\ &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_k} \frac{\partial s_k}{\partial w_{j \rightarrow k}} \\ &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_k} z_j. \end{align} If is an output node, then \begin{align} \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_k} = \frac{d \sigma(s_k)}{d s_k} = \sigma' (s_k)\end{align} and thus \begin{align} \frac{\partial E_i(\vec{w})}{\partial w_{j \rightarrow k}} &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (s_k)\; z_j. \end{align} However, if is not an output node, then a change in can affect all the nodes which are connected to 's output, i.e. \begin{align} \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_k} &= \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial z_k} \frac{\partial z_k}{\partial s_k} \\ &= \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial z_k} \sigma ' (s_k) \\ &= \sum_{o \in \{ v \; \| \; v \text{ is connected to } k \}} \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_o} \frac{\partial s_o}{\partial z_k} \sigma ' (s_k) \\ &= \sum_{o \in \{ v \; \| \; v \text{ is connected to } k \}} \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_o} w_{k \rightarrow o} \; \sigma ' (s_k), \end{align} ... and we are almost done! All what is left to do is to place the th example at the inputs of our neural network, calculate and for all the nodes (the forward-propagation step) and work our way backwards from the output node calculating (hence the name, backpropagation). To summarize, if is an output node, then \begin{align} \frac{\partial E_i(\vec{w})}{\partial w_{j \rightarrow k}} &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (s_k)\; z_j, \end{align} otherwise \begin{align} \frac{\partial E_i(\vec{w})}{\partial w_{j \rightarrow k}} &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (s_k)\; z_j \sum_{o \in \{ v \; \| \; v \text{ conn. to } k \}} \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_o} w_{k \rightarrow o}. \end{align} Then after the following is obtained \begin{align} \frac{\partial E_i(\vec{w})}{\partial \vec{w}} = \left( \; \; \frac{\partial E_i(\vec{w})}{\partial w_{j \rightarrow k}} \; \; \right), \forall j, k \end{align} the weight vector can either be updated in one go ( batch update) \begin{align} \vec{w}_{t+1} := \vec{w_t} - \eta \frac{\partial E(\vec{w})}{\partial \vec{w}} \bigg\|_{\vec{w_t}} = \vec{w_t} - \eta \sum_i \frac{\partial E_i(\vec{w})}{\partial \vec{w}}\bigg\|_{\vec{w_t}}, \end{align} or it can be updated sequentially using one training example at a time: \begin{align} \vec{w}_{t+1} := \vec{w_t} - \eta \frac{\partial E_i(\vec{w})}{\partial \vec{w}} \bigg\|_{\vec{w_t}}.\end{align} 4.1. Example multilayer network If you launch and play with the applet above, you will see that it is able to separate classes non-linearly (indicating that it's using more than one perceptron). It is built using this two-layer neural network: The weights vector contains all the weights in the network, i.e. \begin{align} \vec{w} = ( w_{in_1 \rightarrow 1}, w_{in_x \rightarrow 1}, w_{in_y \rightarrow 1}, w_{in_1 \rightarrow 2}, ..., w_{in_y \rightarrow 5}, w_{in_1 \rightarrow 6}, w_{1 \rightarrow 6}, w_{2 \rightarrow 6}, ..., w_{5 \rightarrow 6}). \end{align} Each perceptron is using sigmoid as its activation function and the output of the perceptron is the output for the whole network, i.e. . Then an individual point i (with x and y coordinates normalized) is considered as an th training example and fed through the network. While it's being propagated, each and for are stored. Then the gradient of an th error surface is calculated as follows: \begin{align} \frac{\partial E_i(\vec{w})}{\partial \vec{w}} &= \left( \frac{\partial E_i(\vec{w})}{\partial w_{in_1 \rightarrow 1}},\frac{\partial E_i(\vec{w})}{\partial w_{in_x \rightarrow 1}}, ..., \frac{\partial E_i(\vec{w})}{\partial w_{in_y \rightarrow 5}},\frac{\partial E_i(\vec{w})}{\partial w_{in_1 \rightarrow 6}},\frac{\partial E_i(\vec{w})}{\partial w_{1 \rightarrow 6}},\frac{\partial E_i(\vec{w})}{\partial w_{2 \rightarrow 6}}, ..., \frac{\partial E_i(\vec{w})}{\partial w_{5 \rightarrow 6}} \right) , \end{align} where \begin{align} \frac{\partial E_i(\vec{w})}{\partial w_{in_1 \rightarrow 1}} &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (s_1)\; \frac{\partial h_{\vec{w}}(\vec{x_i})}{\partial s_6} w_{1 \rightarrow 6} \\ &= 2 (z_6 - y_i) \; \sigma (s_1) \; (1 - \sigma (s_1)) \; \sigma (s_6) \; (1 - \sigma (s_6)) \; w_{1 \rightarrow 6}, \\ \frac{\partial E_i(\vec{w})}{\partial w_{in_x \rightarrow 1}} &= 2 (z_6 - y_i) \; \sigma (s_1) \; (1 - \sigma (s_1)) \; {in}_x \; \sigma (s_6) \; (1 - \sigma (s_6)) \; w_{1 \rightarrow 6}, \\ & \vdots \\ \frac{\partial E_i(\vec{w})}{\partial w_{in_y \rightarrow 5}} &= 2 (z_6 - y_i) \; \sigma (s_5) \; (1 - \sigma (s_5)) \; {in}_y \; \sigma (s_6) \; (1 - \sigma (s_6)) \; w_{5 \rightarrow 6}, \\ \frac{\partial E_i(\vec{w})}{\partial w_{in_1 \rightarrow 6}} &= 2 (h_{\vec{w}}(\vec{x_i}) - y_i) \; \sigma ' (s_6) \\ &= 2 (z_6 - y_i) \; \sigma (s_6) \; (1 - \sigma (s_6)) , \\ \frac{\partial E_i(\vec{w})}{\partial w_{1 \rightarrow 6}} &= 2 (z_6 - y_i) \; \sigma (s_6) \; (1 - \sigma (s_6)) \; z_1, \\ \frac{\partial E_i(\vec{w})}{\partial w_{2 \rightarrow 6}} &= 2 (z_6 - y_i) \; \sigma (s_6) \; (1 - \sigma (s_6)) \; z_2, \\ & \vdots \\ \frac{\partial E_i(\vec{w})}{\partial w_{5 \rightarrow 6}} &= 2 (z_6 - y_i) \; \sigma (s_6) \; (1 - \sigma (s_6)) \; z_5. \end{align} Finally, the network is sequentially trained with the learning rate (starting with a random initial weight vector ) \begin{align} \vec{w}_{t+1} := \vec{w_t} - 0.5 \frac{\partial E_i(\vec{w})}{\partial \vec{w}} \bigg\|_{\vec{w_t}}.\end{align} That's it, I hope it sheds some light on the backpropagation!
▶▶ ◀◀ ▶ Announcements ▶ Schedule Room A208. Tue and Thu, 9-11. Teacher's office hours: Mon 9-11.. ▶ Course description Goal: Introduction to complex analysis. Contents: Complex numbers and geometry of the complex plane. Analytic functions, contour integrals and power series. Cauchy theory and applications. ▶ Books and lecture notes ▶ Student evaluation Intermediate exam 40%, final exam 60%. This remains the same for all further examination periods. ▶ Class diary [CB, Ch. 1, §1-4]: We saw what are complex numbers and how we to algebraic operations with them. We also saw what conjugate numbers are, the concept of the modulus of a complex number and several relations between them. We saw how to solve quadratic equations in complex numbers. Problems: [CB, Ch. 1, p. 13-14]: 1, 2, 4, 10, 13. ( 13-14 ) We went over several examples and problems covering the material we did last time. Towards the end of the lecture we spoke about the polar form of a complex number [CB, Ch. 1, § 5] and we proved the basic relation: $$ \arg(z w) = \arg z + \arg w. $$ [CB, Ch. 1, §5-7]: We defined what is the meaning of $e^z$, when $z \in \CC$: $$ e^z = e^{\Re{z}}(\cos{\Im{z}} + i \sin{\Im{z}}). $$ We saw that with this definition the important property of the exponential function $$ e^{z+w} = e^z e^w $$ is preserved. Then we used this form of representing a complex number $$ z = r e^{i\theta},\ \ \ \text{ where $r = \Abs{z}, \theta = \arg{z}$}, $$ to solve several problems. The last problem we solved was how to find the $n$-th roots (there are $n$ of them) of a given complex number $z_0$. Then we saw that a function of the form $$ z(t) = R e^{it},\ \ \ 0\le t \lt 2\pi, $$ parametrizes a circle of center 0 and radius $R\gt 0$. We also saw how to get parametrizations of different circles. We did several examples involving the polar form and roots of numbers. Then we talked about limits and continuity of sequences and functions, and we saw that nothing really changes from the case of real valued functions of a real variable. Changing the values of the function to be complex, instead of real, creates no problems at all. Limits and continuity can also be defined exactly as in the real case, for functions of a complex variable. What is different is the notion of a derivative when the variable, as well as the value, is complex. If $f:\CC\to\CC$ is a complex valued function of a complex variable then we say that $f$ is differentiable at $z_0 \in \CC$ if the limit $$ \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0} $$ exists. If it does we denote it by $f'(z_0)$ and call it the derivative of $f$ at $z_0$. Then we tried some simple functions: $f(z) = z, g(z) = z^2$ and $h(z) = \overline{z}$. We proved that the limit exists in the first two cases, and that $f'(z) = 1, g'(z) = 2z$, and that the limit does not exist in the case of $h(z)$. Today we did several problems from the first 2 problem sets. Then we talked about mapping properties of simple functions of the form $w = f(z)$. We started by discussing the function $f(z) = z^2$ and its mapping properties (how it transforms various regions in the complex plane). We also investigated on which domains this function is one-to-one and, therefore, on which domains we can define its inverse $z \to \sqrt{z}$. We did not give a complete answer to this last question but we saw that while it is possible to define the square root on the upper unit disk (via the formula $r e^{i\theta} \to \sqrt{r} e^{i \theta/2}$, where $0 \le \theta \le \pi$) it appears to be (and it is) impossible to define the square root on the entire unit disk as a continuous function. Next we proved that if a complex function $f$ is defined in a neighborhood of a point $z \in \CC$ and is differentiable at $z$ then, writing $$ f(z) = f(x+iy) = u(x, y) + i v(x, y),\ \ \ \text{ where $u, v \in \RR$}, $$ we have the Cauchy-Riemann (C-R) partial differential equations: $$ u_x = v_y,\ \ u_y = -v_x, $$ at the point $z = z+iy$. We then proved a converse to this: if the C-R equations hold and the partial derivatives of $u, v$ are assumed to not just exist but also to be continuous at $(x, y)$ then the function $f$ is differentiable at $z = x+iy$.
J. D. Hamkins and J. Reitz, “The set-theoretic universe $V$ is not necessarily a class-forcing extension of HOD,” ArXiv e-prints, 2017. (manuscript under review) @ARTICLE{HamkinsReitz:The-set-theoretic-universe-is-not-necessarily-a-forcing-extension-of-HOD, author = {Joel David Hamkins and Jonas Reitz}, title = {The set-theoretic universe {$V$} is not necessarily a class-forcing extension of {HOD}}, journal = {ArXiv e-prints}, year = {2017}, volume = {}, number = {}, pages = {}, month = {September}, note = {manuscript under review}, abstract = {}, keywords = {under-review}, source = {}, doi = {}, eprint = {1709.06062}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://jdh.hamkins.org/the-universe-need-not-be-a-class-forcing-extension-of-hod}, } Abstract.In light of the celebrated theorem of Vopěnka, proving in ZFC that every set is generic over $\newcommand\HOD{\text{HOD}}\HOD$, it is natural to inquire whether the set-theoretic universe $V$ must be a class-forcing extension of $\HOD$ by some possibly proper-class forcing notion in $\HOD$. We show, negatively, that if ZFC is consistent, then there is a model of ZFC that is not a class-forcing extension of its $\HOD$ for any class forcing notion definable in $\HOD$ and with definable forcing relations there (allowing parameters). Meanwhile, S. Friedman (2012) showed, positively, that if one augments $\HOD$ with a certain ZFC-amenable class $A$, definable in $V$, then the set-theoretic universe $V$ is a class-forcing extension of the expanded structure $\langle\HOD,\in,A\rangle$. Our result shows that this augmentation process can be necessary. The same example shows that $V$ is not necessarily a class-forcing extension of the mantle, and the method provides a counterexample to the intermediate model property, namely, a class-forcing extension $V\subseteq V[G]$ by a certain definable tame forcing and a transitive intermediate inner model $V\subseteq W\subseteq V[G]$ with $W\models\text{ZFC}$, such that $W$ is not a class-forcing extension of $V$ by any class forcing notion with definable forcing relations in $V$. This improves upon a previous example of Friedman (1999) by omitting the need for $0^\sharp$. In 1972, Vopěnka proved the following celebrated result. Theorem. (Vopěnka) If $V=L[A]$ where $A$ is a set of ordinals, then $V$ is a forcing extension of the inner model $\HOD$. The result is now standard, appearing in Jech (Set Theory 2003, p. 249) and elsewhere, and the usual proof establishes a stronger result, stated in ZFC simply as the assertion: every set is generic over $\HOD$. In other words, for every set $a$ there is a forcing notion $\mathbb{B}\in\HOD$ and a $\HOD$-generic filter $G\subseteq\mathbb{B}$ for which $a\in\HOD[G]\subseteq V$. The full set-theoretic universe $V$ is therefore the union of all these various set-forcing generic extensions $\HOD[G]$. It is natural to wonder whether these various forcing extensions $\HOD[G]$ can be unified or amalgamated to realize $V$ as a single class-forcing extension of $\HOD$ by a possibly proper class forcing notion in $\HOD$. We expect that it must be a very high proportion of set theorists and set-theory graduate students, who upon first learning of Vopěnka’s theorem, immediately ask this question. Main Question. Must the set-theoretic universe $V$ be a class-forcing extension of $\HOD$? We intend the question to be asking more specifically whether the universe $V$ arises as a bona-fide class-forcing extension of $\HOD$, in the sense that there is a class forcing notion $\mathbb{P}$, possibly a proper class, which is definable in $\HOD$ and which has definable forcing relation $p\Vdash\varphi(\tau)$ there for any desired first-order formula $\varphi$, such that $V$ arises as a forcing extension $V=\HOD[G]$ for some $\HOD$-generic filter $G\subseteq\mathbb{P}$, not necessarily definable. In this article, we shall answer the question negatively, by providing a model of ZFC that cannot be realized as such a class-forcing extension of its $\HOD$. Main Theorem. If ZFC is consistent, then there is a model of ZFC which is not a forcing extension of its $\HOD$ by any class forcing notion definable in that $\HOD$ and having a definable forcing relation there. Throughout this article, when we say that a class is definable, we mean that it is definable in the first-order language of set theory allowing set parameters. The main theorem should be placed in contrast to the following result of Sy Friedman. Theorem. (Friedman 2012) There is a definable class $A$, which is strongly amenable to $\HOD$, such that the set-theoretic universe $V$ is a generic extension of $\langle \HOD,\in,A\rangle$. This is a postive answer to the main question, if one is willing to augment $\HOD$ with a class $A$ that may not be definable in $\HOD$. Our main theorem shows that in general, this kind of augmentation process is necessary. It is natural to ask a variant of the main question in the context of set-theoretic geology. Question. Must the set-theoretic universe $V$ be a class-forcing extension of its mantle? The mantle is the intersection of all set-forcing grounds, and so the universe is close in a sense to the mantle, perhaps one might hope that it is close enough to be realized as a class-forcing extension of it. Nevertheless, the answer is negative. Theorem. If ZFC is consistent, then there is a model of ZFC that does not arise as a class-forcing extension of its mantle $M$ by any class forcing notion with definable forcing relations in $M$. We also use our results to provide some counterexamples to the intermediate-model property for forcing. In the case of set forcing, it is well known that every transitive model $W$ of ZFC set theory that is intermediate $V\subseteq W\subseteq V[G]$ a ground model $V$ and a forcing extension $V[G]$, arises itself as a forcing extension $W=V[G_0]$. In the case of class forcing, however, this can fail. Theorem. If ZFC is consistent, then there are models of ZFC set theory $V\subseteq W\subseteq V[G]$, where $V[G]$ is a class-forcing extension of $V$ and $W$ is a transitive inner model of $V[G]$, but $W$ is not a forcing extension of $V$ by any class forcing notion with definable forcing relations in $V$. Theorem. If ZFC + Ord is Mahlo is consistent, then one can form such a counterexample to the class-forcing intermediate model property $V\subseteq W\subseteq V[G]$, where $G\subset\mathbb{B}$ is $V$-generic for an Ord-c.c. tame definable complete class Boolean algebra $\mathbb{B}$, but nevertheless $W$ does not arise by class forcing over $V$ by any definable forcing notion with a definable forcing relation. More complete details, please go to the paper (click through to the arxiv for a pdf). J. D. Hamkins and J. Reitz, “The set-theoretic universe $V$ is not necessarily a class-forcing extension of HOD,” ArXiv e-prints, 2017. (manuscript under review) @ARTICLE{HamkinsReitz:The-set-theoretic-universe-is-not-necessarily-a-forcing-extension-of-HOD, author = {Joel David Hamkins and Jonas Reitz}, title = {The set-theoretic universe {$V$} is not necessarily a class-forcing extension of {HOD}}, journal = {ArXiv e-prints}, year = {2017}, volume = {}, number = {}, pages = {}, month = {September}, note = {manuscript under review}, abstract = {}, keywords = {under-review}, source = {}, doi = {}, eprint = {1709.06062}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://jdh.hamkins.org/the-universe-need-not-be-a-class-forcing-extension-of-hod}, }
The Hilbert Basis Theorem Recall from the Noetherian Rings page that a ring $R$ is said to be a Noetherian ring if it satisfies the ascending chain condition, that is, for all ascending chains of ideals $I_1 \subseteq I_2 \subseteq ... \subseteq I_n \subseteq ...$ there exists an $N \in \mathbb{N}$ such that for all $m \geq N$ we have that $I_m = I_N$. Equivalently, we proved that $R$ is Noetherian if and only if every ideal $I$ is finitely generated, that is, there exists $x_1, x_2, ..., x_n \in I$ such that $I = (x_1, x_2, ..., x_n)$. We about to prove a very important result known as the Hilbert basis theorem which tells us that if $R$ is a Noetherian ring then the corresponding ring of polynomials of a single variable $x$, $R[x]$, is a Noetherian ring. We first need the following lemma. We first need to get some notation out of the way. If $F \in R[x]$ then $F$ is of the form:(1) We define the function $\mathrm{cof} (F)$ to be the leading coefficient of $F$. That is, if $F$ has degree $n$ as above then:(2) For $m \geq 0$ and for an ideal $I$, we define:(3) Lemma 1: Let $R$ be a Noetherian ring and let $I$ be an ideal. Then for all $m \geq 0$, $J_m$ is an ideal. Proof:Let $a \in J_m$ and let $b \in R$. Then $a = \mathrm{cof}(F)$ for some polynomial $F \in I$ with $\deg F \leq m$. Consider the function $bF$. Then $bF \in I$ since $I$ is an ideal. Furthermore, $\deg (bF) = \deg (F) \leq m$. Therefore $\mathrm{cof} (bF) \in J_m$. But $\mathrm{cof} (bF) = ab$. So $ab \in J_m$. Now let $a, b \in J_m$. Then $a = \mathrm{cof} (F)$ and $b = \mathrm{cof} (G)$ for some polynomials $F, G \in I$ with $\deg F, \deg G \leq m$. Let $\deg F = s$ and let $\deg G = t$. Then $F$ and $G$ have the form: Without loss of generality, assume that $s \geq t$. Define a new polynomial $H$ by: Since $F, G \in I$ we have that $H \in I$. Furthermore observe that: Therefore $\mathrm{cof}(F + x^{s-t}G) = a + b$. Hence $(a + b) \in J_m$. Thus $J_m$ is an ideal. Theorem 1 (The Hilbert Basis Theorem): Let $R$ be a Noetherian ring. Then $R[x]$ is a Noetherian ring. Proof:Let $I \subseteq R[x]$ be an ideal and for each $m \geq 0$ let $J_m$ be defined in terms of $I$. Consider the following ascending chain of ideals: Since $R$ is Noetherian, there exists an $N \in \mathbb{N}$ such that for all $m \geq N$ we have that $J_m = J_N$. Also, since $R$ is a Noetherian ring every ideal in $R$ is finitely generated. So for each $m \geq 0$ there exists elements $a_{m_1}, a_{m_2}, ..., a_{m_k}$ such that: For each $1 \leq j \leq k$, choose a polynomial $F_{m_j} \in I$ such that $a_{m_j} = \mathrm{cof} (F_{m_j})$. Let $I'$ be defined as the ideal generated by the $F_{m_j}$s for all $m \leq N$. We claim that $I = I'$. By definition, we have that $I' \subseteq I$. Now suppose that $I' \not \supseteq I$. Then there exists a function $G \in I$ such that $G \not \in I'$. Let $G$ be chosen of minimal degree in $I'$ and let $\deg (G) = d$. Then $\mathrm{cof} (G) \in J_d$. So $\mathbf{cof} (G)$ has the form: Where $\alpha_j \in R$ and $F_{d_j}$ are in the list of generators for $I'$ and where $\deg (F_{d_j}) \leq d$. Let: Then $Q \in I$ and $\deg (Q) = d$. But $\mathrm{cof} (Q) = \mathbf{cof} (G)$. Then $G - Q \in I$. But $\deg (G - Q) \leq d - 1$. Since $G \not \in I'$ and $Q \in I'$ we hae that $G - Q \not \in I'$ and is such that $\deg (G - Q) < d$. But this contradicts $G$ having minimal degree in $I$. Therefore $I = I'$. Since $I'$ is finitely generated so is $I$. So every ideal in $R[x]$ is finite generated, i.e., $R[x]$ is Noetherian. $\blacksquare$
Difference between revisions of "Literature on Carbon Nanotube Research" m (→Tensile and Electrical Properties of Carbon Nanotube Yarns and Knitted Tubes in Pure or Composite Form) Line 16: Line 16: * Moderator: [[mailto:Markus.Landgraf@me.com Markus Landgraf]]<br /> * Moderator: [[mailto:Markus.Landgraf@me.com Markus Landgraf]]<br /> * Created: March 13th, 2009<br /> * Created: March 13th, 2009<br /> − * Modified: March + * Modified: March , 2009<br /> </div> </div> </td> </td> Revision as of 16:12, 23 March 2009 I have hijacked this page to write down my views on the literature on Carbon Nanotube (CNT) growths and processing, a procedure that should give us the cable/ribbon we desire for the space elevator. I will try to put as much information as possible here. If anyone has something to add, please do not hesitate! Contents 1 Direct mechanical measurement of the tensile strength and elastic modulus of multiwalled carbon nanotubes 2 Direct Spinning of Carbon Nanotube Fibers from Chemical Vapour Deposition Synthesis 3 Multifunctional Carbon Nanotube Yarns by Downsizing an Ancient Technology 4 Ultra-high-yield growth of vertical single-walled carbon nanotubes - Hidden roles of hydrogen and oxygen 5 Sustained Growth of Ultralong Carbon Nanotube Arrays for Fiber Spinning 6 In situ Observations of Catalyst Dynamics during Surface-Bound Carbon Nanotube Nucleation 7 High-Performance Carbon Nanotube Fiber 8 Tensile and Electrical Properties of Carbon Nanotube Yarns and Knitted Tubes in Pure or Composite Form Direct mechanical measurement of the tensile strength and elastic modulus of multiwalled carbon nanotubes B. G. Demczyk et al., Materials and Engineering, A334, 173-178, 2002 The paper by Demczyk et al. (2002) is the basic reference for the experimental determination of the tensile strengths of individual Multi-wall nanotube (MWNT) fibers. The experiments are performed with a microfabricated piezo-electric device. On this device CNTs in the length range of tens of microns are mounted. The tensile measurements are obseverd by transmission electron microscopy (TEM) and videotaped. Measurements of the tensile strength (tension vs. strain) were performed as well as Young modulus and bending stiffness. Breaking tension is reached for the SWNT at 150GP and between 3.5% and 5% of strain. During the measurements 'telescoping' extension of the MWNTs is observed, indicating that single-wall nanotubes (SWNT) could be even stronger. However, 150GPa remains the value for the tensile strength that was experimentally observed for carbon nanotubes. Direct Spinning of Carbon Nanotube Fibers from Chemical Vapour Deposition Synthesis Y.-L. Li, I. A. Kinloch, and A. H. Windle, Science, 304,276-278, 2004 The work described in the paper by Y.-L. Li et al. is a follow-on of the famous paper by Zhu et al. (2002), which was cited extensively in Brad's book. This article goes a little more into the details of the process. If you use a mixture of ethene (as the source of carbon), ferrocene, and theophene (both as catalysts, I suppose) into a furnace (1050 to 1200 deg C) using hydrogen as carrier gas, you apparently get an 'aerogel' or 'elastic smoke' forming in the furnace cavity, which comprises the CNTs. Here's an interesting excerpt: Under these synthesis conditions, the nanotubes in the hot zone formed an aerogel, which appeared rather like “elastic smoke,” because there was sufficient association between the nanotubes to give some degree of mechanical integrity. The aerogel, viewed with a mirror placed at the bottom of the furnace, appeared very soon after the introduction of the precursors (Fig. 2). Itwas then stretched by the gas flow into the form of a sock, elongating downwards along the furnace axis. The sock did not attach to the furnace walls in the hot zone, which accordingly remained clean throughout the process.... The aerogel could be continuously drawn from the hot zone by winding it onto a rotating rod. In this way, the material was concentrated near the furnace axis and kept clear of the cooler furnace walls,... The elasticity of the aerogel is interpreted to come from the forces between the individual CNTs. The authors describe the procedure to extract the aerogel and start spinning a yarn from it as it is continuously drawn out of the furnace. In terms of mechanical properties of the produced yarns, the authors found a wide range from 0.05 to 0.5 GPa/g/ccm. That's still not enough for the SE, but the process appears to be interesting as it allows to draw the yarn directly from the reaction chamber without mechanical contact and secondary processing, which could affect purity and alignment. Also, a discussion of the roles of the catalysts as well as hydrogen and oxygen is given, which can be compared to the discussion in G. Zhang et al. (2005, see below). Multifunctional Carbon Nanotube Yarns by Downsizing an Ancient Technology M. Zhang, K. R. Atkinson, and R. H. Baughman, Science, 306, 1358-1361, 2004 In the research article by M. Zhang et al. (2004) the procedure of spinning long yarns from forests of MWNTs is described in detail. The maximum breaking strength achieved is only 0.46 GPa based on the 30micron-long CNTs. The initial CNT forest is grown by chemical vapour deposition (CNT) on a catalytic substrate, as usual. A very intersting formula for the tensile strength of a yarn relative to the tensile strength of the fibers (in our case the MWNTs) is given: <math> \frac{\sigma_{\rm yarn}}{\sigma_{\rm fiber}} = \cos^2 \alpha \left(1 - \frac{k}{\sin \alpha} \right) </math> where <math>\alpha</math> is the helix angle of the spun yarn, i.e. fiber direction relative to yarn axis. The constant <math>k=\sqrt(dQ/\mu)/3L</math> is given by the fiber diameter d=1nm, the fiber migration length Q (distance along the yarn over which a fiber shifts from the yarn surface to the deep interior and back again), the quantity <math>\mu=0.13</math> is the friction coefficient of CNTs (the friction coefficent is the ratio of maximum along-fiber force divided by lateral force pressing the fibers together), <math>L=30{\rm \mu m}</math> is the fiber length. A critical review of this formula is given here. In the paper interesting transmission electron microscope (TEM) pictures are shown, which give insight into how the yarn is assembled from the CNT forest. The authors describe other characteristics of the yarn, like how knots can be introduced and how the yarn performs when knitted, apparently in preparation for application in the textile industry. Ultra-high-yield growth of vertical single-walled carbon nanotubes - Hidden roles of hydrogen and oxygen Important aspects of the production of CNTs that are suitable for the SE is the efficiency of the growth and the purity (i.e. lack of embedded amorphous carbon and imperfections in the Carbon bounds in the CNT walls). In their article G. Zhang et al. go into detail about the roles of oxygen and hydrogen during the chemical vapour deposition (CVD) growth of CNT forests from hydrocarbon sources on catalytic substrates. In earlier publications the role of oxygen was believed to be to remove amorphous carbon by oxidation into CO. The authors show, however, that, at least for this CNT growth technique, oxygen is important, because it removes hydrogen from the reaction. Hydrogen has apparently a very detrimental effect on the growth of CNTs, it even destroys existing CNTs as shown in the paper. Since hydrogen radicals are released during the dissociation of the hydrocarbon source compount, it is important to have a removal mechanism. Oxygen provides this mechanism, because its chemical affinity towards hydrogen is bigger than towards carbon. In summary, if you want to efficiently grow pure CNT forests on a catalyst substrate from a hydrocarbon CVD reaction, you need a few percent oxygen in the source gas mixture. An additional interesting information in the paper is that you can design the places on the substrate, on which CNTs grow by placing the the catalyst only in certain areas of the substrate using lithography. In this way you can grow grids and ribbons. Figures are shown in the paper. In the paper no information is given on the reason why the CNT growth stops at some point. The growth rate is given with 1 micron per minute. Of course for us it would be interesting to eliminate the mechanism that stops the growth so we could grow infinitely long CNTs. This article can be found in our archive. Sustained Growth of Ultralong Carbon Nanotube Arrays for Fiber Spinning Q. Li et al. have published a paper on a subject that is very close to our hearts: growing long CNTs. The longer the fibers, which we hope have a couple of 100GPa of tensile strength, can hopefully be spun into the yarns that will make our SE ribbon. In the paper the method of chemical vapour deposition (CVD) onto a catalyst-covered silicon substrate is described, which appears to be the leading method in the publications after 2004. This way a CNT "forest" is grown on top of the catalyst particles. The goal of the authors was to grow CNTs that are as long as possible. The found that the growth was terminated in earlier attempts by the iron catalyst particles interdiffusing with the substrate. This can apparently be avoided by putting an aluminium oxide layer of 10nm thickness between the catalyst and the substrate. With this method the CNTs grow to an impressive 4.7mm! Also, in a range from 0.5 to 1.5mm fiber length the forests grown with this method can be spun into yarns. The growth rate with this method was initially <math>60{\rm \mu m\ min.^{-1}}</math> and could be sustained for 90 minutes, This is very different from the <math>1{\rm \mu m\ min.^{-1}}</math> reported by G. Zhang et al. (2005), which shows that the growth is very dependent on the method and materials used. The growth was prolonged by the introduction of water vapour into the mixture, which achieved the 4.7mm after 2h of growth. By introducing periods of restricted carbon supply, the authors produced CNT forests with growth marks. This allowed to determine that the forest grew from the base. This is in line with the in situ observations by S. Hofmann et al. (2007). Overall the paper is somewhat short on the details of the process, but the results are very interesting. Perhaps the 5mm CNTs are long enough to be spun into a usable yarn. In situ Observations of Catalyst Dynamics during Surface-Bound Carbon Nanotube Nucleation The paper by S. Hofmann et al. (2007) is a key publication for understanding the microscropic processes of growing CNTs. The authors describe an experiment in which they observe in situ the growth of CNTs from chemical vapour deposition (CVD) onto metallic catalyst particles. The observations are made in time-lapse transmission electron microscopy (TEM) and in x-ray photo-electron spectroscopy. Since I am not an expert on spectroscopy, I stick to the images and movies produced by the time-lapse TEM. In the observations it can be seen that the catalysts are covered by a graphite sheet, which forms the initial cap of the CNT. The formation of that cap apparently deforms the catalyst particle due to its inherent shape as it tries to form a minimum-energy configuration. Since the graphite sheet does not extend under the catalyst particle, which is prevented by the catalyst sitting on the silicon substrate, the graphite sheet cannot close itself. The deformation of the catalyst due to the cap forming leads to a retoring force exerted by the crystaline stracture of the catalyst particle. As a consequence the carbon cap lifts off the catalyst particle. On the base of the catalyst particle more carbon atoms attach to the initial cap starting the formation of the tube. The process continues to grow a CNT as long as there is enough carbon supply to the base of the catalyst particle and as long as the particle cannot be enclosed by the carbon compounds. During the growth of the CNT the catalyst particle breathes so drives so the growth process mechanically. Of course for us SE community the most interesting part in this paper is the question: can we grow CNTs that are long enough so we can spin them in a yarn that would hold the 100GPa/g/ccm? In this regard the question is about the termination mechanism of the growth. The authors point to a very important player in CNT growth: the catalyst. If we can make a catalyst that does not break off from its substrate and does not wear off, the growth could be sustained as long as the catalyst/substrate interface is accessible to enough carbon from the feedstock. If you are interested, get the paper from our archive, including the supporting material, in which you'll find the movies of the CNTs growing. High-Performance Carbon Nanotube Fiber K. Koziol et al., Science, 318, 1892, 2007. The paper "High-Performance Carbon Nanotube Fiber" by K. Koziol et al. is a research paper on the production of macroscopic fibers out of an aerogel (low-density, porous, solid material) of SWNT and MWNT that has been formed by carbon vapor deposition. They present an analysis of the mechanical performance figures (tensile strength and stiffness) of their samples. The samples are fibers of 1, 2, and 20mm length and have been extracted from the aerogel with high winding rates (20 metres per minute). Indeed higher winding rates appear to be desirable, but the authors have not been able to achieve higher values as the limit of extraction speed from the aerogel was reached, and higher speeds led to breakage of the aerogel. They show in their results plot (Figure 3A) that typically the fibers split in two performance classes: low-performance fibers with a few GPa and high-performance fibers with around 6.5 GPa. It should be noted that all tensile strengths are given in the paper as GPa/SG, where SG is the specific gravity, which is the density of the material divided by the density of water. Normally SG was around 1 for most samples discussed in the paper. The two performance classes have been interpreted by the authors as the typical result of the process of producing high-strength fibers: since fibers break at the weakest point, you will find some fibers in the sample, which have no weak point, and some, which have one or more, provided the length of the fibers is in the order of the frequency of occurrence of weak points. This can be seen by the fact that for the 20mm fibers there are no high-performance fibers left, as the likelihood to encounter a weak point on a 20mm long fiber is 20 times higher than encountering one on a 1mm long fiber. As a conclusion the paper is bad news for the SE, since the difficulty of producing a flawless composite with a length of 100,000km and a tensile strength of better than 3GPa using the proposed method is enormous. This comes back to the ribbon design proposed on the Wiki: using just cm-long fibers and interconnect them with load-bearing structures (perhaps also CNT threads). Now we have shifted the problem from finding a strong enough material to finding a process that produces the required interwoven ribbon. In my opinion the race to come up with a fiber of better than Kevlar is still open. Tensile and Electrical Properties of Carbon Nanotube Yarns and Knitted Tubes in Pure or Composite Form The paper by S. Hutton et al. is the latest on yarns spun out of CNTs. The core of the paper is concerned with effect the different amounts of twist on the tensile strength and on the electrical conductivity of the yarn. The bad news for us is that they arrive only at 1GPa/ccm for the optimum tensile strength of the yarn. However, some insight is given in the spinning process and in the different methods of processing the CNTs. They use relatively short CNTs (0.2 to 0.3mm) grown into a MWNT forest by chemical vapour deposition (CVD) onto a silicon substrate covered with metal catalyst. The latter method appears to have become standard recently.
The category of simplicial presheaves on a small Grothendieck site $\mathcal{C}$ can be given a model structure by defining weak equivalences and cofibrations sectionwise. It's called the (global) injective model structure and has a mapping space functor $Hom$, given by $Hom(X, Y)_n = hom(X\times \Delta^n, Y)$. Using the given topology, one can actually define 'local' weak equivalences and Jardine showed that the left Bousfield localization at the class of local weak equivalence exists. This model category structure is called local injective model structure. Now, it's often mentioned that this is also the same as localizing the injective model structure at the class $S := \{X\rightarrow L^2X\}$ (where $L^2$ is the sheafification functor). So if $W$ denotes the class of local weak equivalences, this amounts to say that $S$-local is the same as $W$-local, that is for an injective fibrant simplicial presheaf $A$ the following are equivalent: 1) $Hom(L^2X, A) \cong Hom(X, A)$ for all $X$ 2) $Hom(Y, A) \cong Hom(X, Y)$ for all local weak equivalences $X\rightarrow Y$. Here, $\cong$ stands for weak equivalence of simplicial sets. 2) implies 1), of course. But how does 1) imply 2)? Or am I mistaken? Thanks!
For a sequence $X_1, X_2, \dots $, Let $F_n(x)$ denote the cdf of $X_n$. Suppose our sequence is $X_n \sim N(0,n) $ then for all $x$ the point-wise limit of $F_n(x)$ is $\frac{1}{2}$. How would one prove this? Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.Sign up to join this community More generally, let $X$ be any random variable with distribution $F$ having unit variance. Letting $(\mu_n)$ and $(\sigma_n)$ be any sequences of numbers, set $$X_n = \sigma_n X + \mu_n.$$ Define $F_n$ to be the distribution of $X_n$. Suppose that as $n\to\infty$, $$\sigma_n\to\infty$$ and $$\mu_n / \sigma_n\to z.$$ If $F$ is continuous in a neighborhood of $-z$, then the pointwise limit of $F_n(x)$ must be $F(-z)$. Intuitively, this is because The distributions of the $X_n$ are becoming more and more spread out but Relative to the spreads $\sigma_n$, any fixed number $x$ is becoming ever closer to $-z$. Thus the pointwise limit ought to be $F(-z)$. It remains to make this intuition rigorous. From the basic definitions and a little bit of algebra, observe that $$F_n(x) = \Pr(X_n \le x) = \Pr(\sigma_n X + \mu_n \le x) = \Pr\left(X \le \frac{x-\mu_n}{\sigma_n}\right) = F\left(\frac{x-\mu_n}{\sigma_n}\right).$$ The assumptions about the limiting values of $\mu_n/\sigma_n$ and $\sigma_n$ were made specifically to imply $\lim_{n\to \infty} (x-\mu_n)/\sigma_n = -z$. Applying the assumed continuity of $F$ in a neighborhood of $-z$ finishes the demonstration. (The details, which are straightforward, are left to the reader because this is a self-study question that asks for guidance and intuition rather than a complete answer.) Application of this result to the standard Normal distribution $F$, $(\mu_n) = (0)$, and $\sigma_n=(n)$ answers the question as stated, because $F(-0) = 1/2$.
Solution for $J_1$But what is J1? What is the expected number of steps for a frog that has only one leaf to go?The solution is $J_1 = 2(e-1)$ and other terms $J_n$ can be expressed as a sum.Rewriting the recurrence relation as a sumThe recurrence relation is not gonna solve the problem entirely (because one term in the initial conditions is not ... The solution is:$P(Y\in [a,b]\|X\in[a,b])=\frac{\int_{[a,b]}(F_{Y\|X}(b\|x)-F_{Y\|X}(a\|x))f_X(x)dx}{\int_{[a,b]}f_X(x)dx}$This can be seen, as:$P(Y\in[a,b]\cap X\in[a,b])= P(Y\leq b\cap X\in[a,b])-P(Y\leq a\cap X\in[a,b])= \int_{a}^b \left( \int_{-\infty}^bf_{Y\|X}(y\|x)dy \right)dx-\int_{a}^a \left( \int_{-\infty}^af_{Y\|X}(y\|x)dy \right)dx= \int_a^bF_{Y\|X}(... In order to calculate a parametric confidence intervall you need distributional assumptions. Simply using the L2 loss does not allow you to calculate confidence intervalls the "usual way". You need normally distributed errors in order for these confidence intervals to be legitimate. In that case the L2-optimizer is also the maximum likelihood estimate. So in ... It's simpler than it looks.To avoid writing lots of exponentials, let's work with the cumulant generating functions. These are the logarithms of the characteristic function: that is, the cgf of any random variable $X$ is$$\psi_X(s) = \log E\left[e^{isX}\right].$$For two variables $(X,Y)$ the cgf is$$\psi_{X,Y}(s,t) = \log E\left[e^{isX+itY}\right].$$... A graph $\Gamma$ is an ordered pair $(V,E)$ where $V$ (the vertices) is any set and $E\subset V\times V$ (the edges) is a collection of ordered pairs of vertices. We usually depict the vertices with point symbols at distinct locations in the plane and the edges $(v,w)$ as arrows running from the location of $v$ (the origin) to the location of $w$ (the ... Calculate the weighted mean. If you want to give sensor2 a 70% weight, you can calculate the respective weighted means via$\text{weighted mean} = 0.7 x_{\text{sensor_2}} + 0.3 x_{\text{sensor_1}}$Make sure the weights add up to 1!
Huge cardinal Huge cardinals (and their variants) were introduced by Kenneth Kunen in 1972 as a very large cardinal axiom. Kenneth Kunen first used them to prove that the consistency of the existence of a huge cardinal implies the consistency of $\text{ZFC}$+"there is a $\omega_2$-saturated $\sigma$-ideal on $\omega_1$". It is now known that only a Woodin cardinal is needed for this result. However, the consistency of the existence of an $\omega_2$-complete $\omega_3$-saturated $\sigma$-ideal on $\omega_2$, as far as the set theory world is concerned, still requires an almost huge cardinal. [1] Contents 1 Definitions 2 References 3 Consistency strength and size 4 Relative consistency results 5 In set theoretic geology 6 References Definitions Their formulation is similar to that of the formulation of superstrong cardinals. A huge cardinal is to a supercompact cardinal as a superstrong cardinal is to a strong cardinal, more precisely. The definition is part of a generalized phenomenon known as the "double helix", in which for some large cardinal properties n-$P_0$ and n-$P_1$, n-$P_0$ has less consistency strength than n-$P_1$, which has less consistency strength than (n+1)-$P_0$, and so on. This phenomenon is seen only around the n-fold variants as of modern set theoretic concerns. [2] Although they are very large, there is a first-order definition which is equivalent to n-hugeness, so the $\theta$-th n-huge cardinal is first-order definable whenever $\theta$ is first-order definable. This definition can be seen as a (very strong) strengthening of the first-order definition of measurability. Elementary embedding definitions $\kappa$ is almost n-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length less than $\lambda$ (that is, $M^{<\lambda}\subseteq M$). $\kappa$ is n-huge with target $\lambda$iff $\lambda=j^n(\kappa)$ and $M$ is closed under all of its sequences of length $\lambda$ ($M^\lambda\subseteq M$). $\kappa$ is almost n-hugeiff it is almost n-huge with target $\lambda$ for some $\lambda$. $\kappa$ is n-hugeiff it is n-huge with target $\lambda$ for some $\lambda$. $\kappa$ is super almost n-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is almost n-huge with target $\lambda$ (that is, the target can be made arbitrarily large). $\kappa$ is super n-hugeiff for every $\gamma$, there is some $\lambda>\gamma$ for which $\kappa$ is n-huge with target $\lambda$. $\kappa$ is almost huge, huge, super almost huge, and superhugeiff it is almost 1-huge, 1-huge, etc. respectively. Ultrahuge cardinals A cardinal $\kappa$ is $\lambda$-ultrahuge for $\lambda>\kappa$ if there exists a nontrivial elementary embedding $j:V\to M$ for some transitive class $M$ such that $j(\kappa)>\lambda$, $M^{j(\kappa)}\subseteq M$ and $V_{j(\lambda)}\subseteq M$. A cardinal is ultrahuge if it is $\lambda$-ultrahuge for all $\lambda\geq\kappa$. [1] Notice how similar this definition is to the alternative characterization of extendible cardinals. Furthermore, this definition can be extended in the obvious way to define $\lambda$-ultra n-hugeness and ultra n-hugeness, as well as the " almost" variants. Hyperhuge cardinals A cardinal $\kappa$ is $\lambda$-hyperhuge for $\lambda>\kappa$ if there exists a nontrivial elementary embedding $j:V\to M$ for some inner model $M$ such that $\mathrm{crit}(j) = \kappa$, $j(\kappa)>\lambda$ and $^{j(\lambda)}M\subseteq M$. A cardinal is hyperhuge if it is $\lambda$-hyperhuge for all $\lambda>\kappa$.[3, 4] Huge* cardinals A cardinal $κ$ is $n$-huge* if for some $α > κ$, $\kappa$ is the critical point of an elementary embedding $j : V_α → V_β$ such that $j^n (κ) < α$.[5] Hugeness* variant is formulated in a way allowing for a virtual variant consistent with $V=L$: A cardinal $κ$ is virtually $n$-huge* if for some $α > κ$, in a set-forcing extension, $\kappa$ is the critical point of an elementary embedding $j : V_α → V_β$ such that $j^n(κ) < α$.[5] Ultrafilter definition The first-order definition of n-huge is somewhat similar to measurability. Specifically, $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete ultrafilter, $U$, over $\kappa$. A cardinal $\kappa$ is n-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$, and cardinals $\kappa=\lambda_0<\lambda_1<\lambda_2...<\lambda_{n-1}<\lambda_n=\lambda$ such that: $$\forall i<n(\{x\subseteq\lambda:\text{order-type}(x\cap\lambda_{i+1})=\lambda_i\}\in U)$$ Where $\text{order-type}(X)$ is the order-type of the poset $(X,\in)$. [1] $\kappa$ is then super n-huge if for all ordinals $\theta$ there is a $\lambda>\theta$ such that $\kappa$ is n-huge with target $\lambda$, i.e. $\lambda_n$ can be made arbitrarily large. If $j:V\to M$ is such that $M^{j^n(\kappa)}\subseteq M$ (i.e. $j$ witnesses n-hugeness) then there is a ultrafilter $U$ as above such that, for all $k\leq n$, $\lambda_k = j^k(\kappa)$, i.e. it is not only $\lambda=\lambda_n$ that is an iterate of $\kappa$ by $j$; all members of the $\lambda_k$ sequence are. As an example, $\kappa$ is 1-huge with target $\lambda$ iff there is a normal $\kappa$-complete ultrafilter, $U$, over $\mathcal{P}(\lambda)$ such that $\{x\subseteq\lambda:\text{order-type}(x)=\kappa\}\in U$. The reason why this would be so surprising is that every set $x\subseteq\lambda$ with every set of order-type $\kappa$ would be in the ultrafilter; that is, every set containing $\{x\subseteq\lambda:\text{order-type}(x)=\kappa\}$ as a subset is considered a "large set." As for hyperhugeness, the following are equivalent:[4] $κ$ is $λ$-hyperhuge; $μ > λ$ and a normal, fine, κ-complete ultrafilter exists on $[μ]^λ_{∗κ} := \{s ⊂ μ : \|s\| = λ, \|s ∩ κ\| ∈ κ, \mathrm{otp}(s ∩ λ) < κ\}$; $\mathbb{L}_{κ,κ}$ is $[μ]^λ_{∗κ}$-$κ$-compact for type omission. Coherent sequence characterization of almost hugeness $C^{(n)}$-$m$-huge cardinals (this section from [6]) $κ$ is $C^{(n)}$-$m$-huge iff it is $m$-huge and $j(κ) ∈ C^{(n)}$ ($C^{(n)}$-huge if it is huge and $j(κ) ∈ C^{(n)}$). Equivalent definition in terms of normal measures: κ is $C^{(n)}$-$m$-huge iff it is uncountable and there is a $κ$-complete normal ultrafilter $U$ over some $P(λ)$ and cardinals $κ = λ_0 < λ_1 < . . . < λ_m = λ$, with $λ_1 ∈ C (n)$ and such that for each $i < m$, $\{x ∈ P(λ) : ot(x ∩ λ i+1 ) = λ i \} ∈ U$. It follows that “$κ$ is $C^{(n)}$-$m$-huge” is $Σ_{n+1}$ expressible. Every huge cardinal is $C^{(1)}$-huge. The first $C^{(n)}$-$m$-huge cardinal is not $C^{(n+1)}$-$m$-huge, for all $m$ and $n$ greater or equal than $1$. For suppose $κ$ is the least $C^{(n)}$-$m$-huge cardinal and $j : V → M$ witnesses that $κ$ is $C^{(n+1)}$-$m$-huge. Then since “x is $C^{(n)}$-$m$-huge” is $Σ_{n+1}$ expressible, we have $V_{j(κ)} \models$ “$κ$ is $C^{(n)}$-$m$-huge”. Hence, since $(V_{j(κ)})^M = V_{j(κ)}$, $M \models$ “$∃_{δ < j(κ)}(V_{j(κ)} \models$ “δ is huge”$)$”. By elementarity, there is a $C^{(n)}$-$m$-huge cardinal less than $κ$ in $V$ – contradiction. Assuming $\mathrm{I3}(κ, δ)$, if $δ$ is a limit cardinal (instead of a successor of a limit cardinal – Kunen’s Theorem excludes other cases), it is equal to $sup\{j^m(κ) : m ∈ ω\}$ where $j$ is the elementary embedding. Then $κ$ and $j^m(κ)$ are $C^{(n)}$-$m$-huge (inter alia) in $V_δ$, for all $n$ and $m$. If $κ$ is $C^{(n)}$-$\mathrm{I3}$, then it is $C^{(n)}$-$m$-huge, for all $m$, and there is a normal ultrafilter $\mathcal{U}$ over $κ$ such that $\{α < κ : α$ is $C^{(n)}$-$m$-huge for every $m\} ∈ \mathcal{U}$. References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Kentaro, Sato. Double helix in large large cardinals and iteration ofelementary embeddings., 2007. www bibtex Usuba, Toshimichi. The downward directed grounds hypothesis and very large cardinals.Journal of Mathematical Logic 17(02):1750009, 2017. arχiv DOI bibtex Boney, Will. Model Theoretic Characterizations of Large Cardinals.arχiv bibtex Gitman, Victoria and Shindler, Ralf. Virtual large cardinals.www bibtex Bagaria, Joan. $C^{(n)}$-cardinals.Archive for Mathematical Logic 51(3--4):213--240, 2012. www DOI bibtex Consistency strength and size Hugeness exhibits a phenomenon associated with similarly defined large cardinals (the n-fold variants) known as the double helix. This phenomenon is when for one n-fold variant, letting a cardinal be called n-$P_0$ iff it has the property, and another variant, n-$P_1$, n-$P_0$ is weaker than n-$P_1$, which is weaker than (n+1)-$P_0$. [2] In the consistency strength hierarchy, here is where these lay (top being weakest): measurable = 0-superstrong = 0-huge n-superstrong n-fold supercompact (n+1)-fold strong, n-fold extendible (n+1)-fold Woodin, n-fold Vopěnka (n+1)-fold Shelah almost n-huge super almost n-huge n-huge super n-huge ultra n-huge (n+1)-superstrong All huge variants lay at the top of the double helix restricted to some natural number n, although each are bested by I3 cardinals (the critical points of the I3 elementary embeddings). In fact, every I3 is preceeded by a stationary set of n-huge cardinals, for all n. [1] Similarly, every huge cardinal $\kappa$ is almost huge, and there is a normal measure over $\kappa$ which contains every almost huge cardinal $\lambda<\kappa$. Every superhuge cardinal $\kappa$ is extendible and there is a normal measure over $\kappa$ which contains every extendible cardinal $\lambda<\kappa$. Every (n+1)-huge cardinal $\kappa$ has a normal measure which contains every cardinal $\lambda$ such that $V_\kappa\models$"$\lambda$ is super n-huge" [1], in fact it contains every cardinal $\lambda$ such that $V_\kappa\models$"$\lambda$ is ultra n-huge". Every n-huge cardinal is m-huge for every m<n. Similarly with almost n-hugeness, super n-hugeness, and super almost n-hugeness. Every almost huge cardinal is Vopěnka (therefore the consistency of the existence of an almost-huge cardinal implies the consistency of Vopěnka's principle). [1] Every ultra n-huge is super n-huge and a stationary limit of super n-huge cardinals. Every super almost (n+1)-huge is ultra n-huge and a stationary limit of ultra n-huge cardinals. In terms of size, however, the least n-huge cardinal is smaller than the least supercompact cardinal (assuming both exist). [1] This is because n-huge cardinals have upward reflection properties, while supercompacts have downward reflection properties. Thus for any $\kappa$ which is supercompact and has an n-huge cardinal above it, $\kappa$ "reflects downward" that n-huge cardinal: there are $\kappa$-many n-huge cardinals below $\kappa$. On the other hand, the least super n-huge cardinals have both upward and downward reflection properties, and are all much larger than the least supercompact cardinal. It is notable that, while almost 2-huge cardinals have higher consistency strength than superhuge cardinals, the least almost 2-huge is much smaller than the least super almost huge. While not every $n$-huge cardinal is strong, if $\kappa$ is almost $n$-huge with targets $\lambda_1,\lambda_2...\lambda_n$, then $\kappa$ is $\lambda_n$-strong as witnessed by the generated $j:V\prec M$. This is because $j^n(\kappa)=\lambda_n$ is measurable and therefore $\beth_{\lambda_n}=\lambda_n$ and so $V_{\lambda_n}=H_{\lambda_n}$ and because $M^{<\lambda_n}\subset M$, $H_\theta\subset M$ for each $\theta<\lambda_n$ and so $\cup\{H_\theta:\theta<\lambda_n\} = \cup\{V_\theta:\theta<\lambda_n\} = V_{\lambda_n}\subset M$. Every almost $n$-huge cardinal with targets $\lambda_1,\lambda_2...\lambda_n$ is also $\theta$-supercompact for each $\theta<\lambda_n$, and every $n$-huge cardinal with targets $\lambda_1,\lambda_2...\lambda_n$ is also $\lambda_n$-supercompact. An $n$-huge* cardinal is an $n$-huge limit of $n$-huge cardinals. Every $n + 1$-huge cardinal is $n$-huge.[5] As for virtually $n$-huge:[5] If $κ$ is virtually huge, then $V_κ$ is a model of proper class many virtually extendible cardinals. A virtually $n+1$-huge cardinal is a limit of virtually $n$-huge* cardinals. A virtually $n$-huge* cardinal is an $n+1$-iterable limit of $n+1$-iterable cardinals. If $κ$ is $n+2$-iterable, then $V_κ$ is a model of proper class many virtually $n$-huge* cardinals. Every virtually rank-into-rank cardinal is a virtually $n$-huge* limit of virtually $n$-huge* cardinals for every $n < ω$. The $\omega$-huge cardinals A cardinal $\kappa$ is almost $\omega$-huge iff there is some transitive model $M$ and an elementary embedding $j:V\prec M$ with critical point $\kappa$ such that $M^{<\lambda}\subset M$ where $\lambda$ is the smallest cardinal above $\kappa$ such that $j(\lambda)=\lambda$. Similarly, $\kappa$ is $\omega$-huge iff the model $M$ can be required to have $M^\lambda\subset M$. Sadly, $\omega$-huge cardinals are inconsistent with ZFC by a version of Kunen's inconsistency theorem. Now, $\omega$-hugeness is used to describe critical points of I1 embeddings. Relative consistency results Hugeness of $\omega_1$ In [2] it is shown that if $\text{ZFC +}$ "there is a huge cardinal" is consistent then so is $\text{ZF +}$ "$\omega_1$ is a huge cardinal" (with the ultrafilter characterization of hugeness). Generalizations of Chang's conjecture Cardinal arithmetic in $\text{ZF}$ If there is an almost huge cardinal then there is a model of $\text{ZF+}\neg\text{AC}$ in which every successor cardinal is a Ramsey cardinal. It follows that (1) for all inner models $W$ of $\text{ZFC}$ and every singular cardinal $\kappa$, one has $\kappa^{+W} < \kappa^+$ and that (2) for all ordinal $\alpha$ there is no injection $\aleph_{\alpha+1}\to 2^{\aleph_\alpha}$. This in turn imply the failure of the square principle at every infinite cardinal (and consequently $\text{AD}^{L(\mathbb{R})}$, see determinacy). [3] In set theoretic geology If $\kappa$ is hyperhuge, then $V$ has $<\kappa$ many grounds (so the mantle is a ground itself).[3] This result has been strenghtened to extendible cardinals[7]. On the other hand, it s consistent that there is a supercompact cardinal and class many grounds of $V$ (because of the indestructibility properties of supercompactness).[3] References Kanamori, Akihiro. Second, Springer-Verlag, Berlin, 2009. (Large cardinals in set theory from their beginnings, Paperback reprint of the 2003 edition) www bibtex The higher infinite. Kentaro, Sato. Double helix in large large cardinals and iteration ofelementary embeddings., 2007. www bibtex Usuba, Toshimichi. The downward directed grounds hypothesis and very large cardinals.Journal of Mathematical Logic 17(02):1750009, 2017. arχiv DOI bibtex Boney, Will. Model Theoretic Characterizations of Large Cardinals.arχiv bibtex Gitman, Victoria and Shindler, Ralf. Virtual large cardinals.www bibtex Bagaria, Joan. $C^{(n)}$-cardinals.Archive for Mathematical Logic 51(3--4):213--240, 2012. www DOI bibtex
Compact T2 Spaces are T4 Spaces Recall from the T2 (Hausdorff) Topological Spaces page that a topological space $X$ is said to be T 2 or Hausdorff if for all $x, y \in X$, $x \neq y$ there exists open neighbourhoods $U$ of $x$ and $V$ of $y$ such that $U \cap V = \emptyset$. Also recall from the T4 (Normal Hausdorff) Topological Spaces page that a topological space $X$ is said to be T 4 or Normal Hausdorff if $X$ is both normal and T 1, that is, for every pair of disjoint closed sets $E$ and $F$ there exists open sets $U$ and $V$ with $E \subseteq U$, $F \subseteq V$ such that $U \cap V = \emptyset$ alongside with the T 1 propery. On the Compact T2 Spaces are T3 Spaces page we proved that if $X$ is a T 2 space that is compact then it is also a T 3 space and we noted that in fact every T 2 space that is compact is further a T 4 space. We will now prove this assertion. Theorem 1: Let $X$ be a topological space. If $X$ is T 2 and compact then $X$ is T 4. Proof:Let $X$ be a T 2space that is compact. Since $X$ is a T 2space, it is also a T 1space and so all that remains to show is that $X$ is normal. Let $E$ and $F$ be disjoint closed sets in $X$. We have already proven that $X$ is a T 3space, so, for every $x \in E$ there exists open sets $U_x$ and $V_x$ such that $x \in U_x$ and $F \subseteq V_x$. Consider the following collection: Then $\mathcal F$ is an open cover of $E$. Since $E$ is closed in $X$ and $X$ is compact we have that $E$ is also compact. So this open cover has a finite subcover, $\mathcal F^* = \{ U_{x_1} U_{x_2}, …, U_{x_n} \}$. Let $U = \bigcup_{i=1}^{n} U_{x_1}$. Then $A \subseteq U$. Now let $V = \bigcap_{i=1}^{n} V_{x_i}$. Then $V$ is open in $X$, $F \subseteq V$, and by construction, $U \cap V = \emptyset$. So $X$ is normal. Since $X$ is both T 1and normal we have that $X$ is a T 4space. $\blacksquare$
Here’s a simple and intuitive way of looking at the geometry of a least squares regression: Take the bottom left point in the triangle below as the origin O. For the linear model: $$ Y=X\beta + \epsilon $$ Both $Y$ and $X\beta$ are vectors, and the residual vector $\epsilon$ is the difference. The standard least squares error technique uses $\epsilon^2$ or $(Y-X\beta)^T(Y-X\beta)$ as the error measure to be minimised, and this leads to the calculation of the $\beta$ coefficient vector. Geometrically, the beta coefficients calculated by the least squares regression minimise the squared length of the error vector. This turns out to be the projection of $Y$ on to $X\beta$ – i.e. the perpendicular vector that turns (O, $Y$, $X\beta$) into a right-angled triangle. The projection of $Y$ onto $X\beta$ is done using the projection matrix P, which is defined as \[ P = X\left(X^{T}X\right)^{-1}X^{T} \] So $ X\beta = \hat{Y} = PY $. Using the Pythagorean theorem: $ Y^TY = \hat{Y}^T\hat{Y} + (Y-X\beta)^T(Y-X\beta) $ In other words, the total sum of squares = sum of squares due to regression + residual sum of squares. This is a fundamental part of analysis of variance techniques.
2019-07-18 17:03 Precision measurement of the $\Lambda_c^+$, $\Xi_c^+$ and $\Xi_c^0$ baryon lifetimes / LHCb Collaboration We report measurements of the lifetimes of the $\Lambda_c^+$, $\Xi_c^+$ and $\Xi_c^0$ charm baryons using proton-proton collision data at center-of-mass energies of 7 and 8 TeV, corresponding to an integrated luminosity of 3.0 fb$^{-1}$, collected by the LHCb experiment. The charm baryons are reconstructed through the decays $\Lambda_c^+\to pK^-\pi^+$, $\Xi_c^+\to pK^-\pi^+$ and $\Xi_c^0\to pK^-K^-\pi^+$, and originate from semimuonic decays of beauty baryons. [...] arXiv:1906.08350; LHCb-PAPER-2019-008; CERN-EP-2019-122; LHCB-PAPER-2019-008.- 2019-08-02 - 12 p. - Published in : Phys. Rev. D 100 (2019) 032001 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; Record dettagliato - Record simili 2019-07-02 10:45 Observation of the $\Lambda_b^0\rightarrow \chi_{c1}(3872)pK^-$ decay / LHCb Collaboration Using proton-proton collision data, collected with the LHCb detector and corresponding to 1.0, 2.0 and 1.9 fb$^{-1}$ of integrated luminosity at the centre-of-mass energies of 7, 8, and 13 TeV, respectively, the decay $\Lambda_b^0\to \chi_{c1}(3872)pK^-$ with $\chi_{c1}\to J/\psi\pi^+\pi^-$ is observed for the first time. The significance of the observed signal is in excess of seven standard deviations. [...] arXiv:1907.00954; CERN-EP-2019-131; LHCb-PAPER-2019-023; LHCB-PAPER-2019-023.- 2019-09-03 - 21 p. - Published in : JHEP 1909 (2019) 028 Article from SCOAP3: PDF; Fulltext: LHCb-PAPER-2019-023 - PDF; 1907.00954 - PDF; Related data file(s): ZIP; Supplementary information: ZIP; Record dettagliato - Record simili 2019-06-21 17:31 Updated measurement of time-dependent CP-violating observables in $B^0_s \to J/\psi K^+K^-$ decays / LHCb Collaboration The decay-time-dependent {\it CP} asymmetry in $B^{0}_{s}\to J/\psi K^{+} K^{-}$ decays is measured using proton-proton collision data, corresponding to an integrated luminosity of $1.9\,\mathrm{fb^{-1}}$, collected with the LHCb detector at a centre-of-mass energy of $13\,\mathrm{TeV}$ in 2015 and 2016. Using a sample of approximately 117\,000 signal decays with an invariant $K^{+} K^{-}$ mass in the vicinity of the $\phi(1020)$ resonance, the {\it CP}-violating phase $\phi_s$ is measured, along with the difference in decay widths of the light and heavy mass eigenstates of the $B^{0}_{s}$-$\overline{B}^{0}_{s}$ system, $\Delta\Gamma_s$. [...] arXiv:1906.08356; LHCb-PAPER-2019-013; CERN-EP-2019-108; LHCB-PAPER-2019-013.- Geneva : CERN, 2019-08-22 - 42 p. - Published in : Eur. Phys. J. C 79 (2019) 706 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; Record dettagliato - Record simili 2019-06-21 17:07 Measurement of $C\!P$ observables in the process $B^0 \to DK^{0}$ with two- and four-body $D$ decays / LHCb Collaboration Measurements of $C\!P$ observables in $B^0 \to DK^{0}$ decays are presented, where $D$ represents a superposition of $D^0$ and $\bar{D}^0$ states. The $D$ meson is reconstructed in the two-body final states $K^+\pi^-$, $\pi^+ K^-$, $K^+K^-$ and $\pi^+\pi^-$, and, for the first time, in the four-body final states $K^+\pi^-\pi^+\pi^-$, $\pi^+ K^-\pi^+\pi^-$ and $\pi^+\pi^-\pi^+\pi^-$. [...] arXiv:1906.08297; LHCb-PAPER-2019-021; CERN-EP-2019-111.- Geneva : CERN, 2019-08-07 - 30 p. - Published in : JHEP 1908 (2019) 041 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; Record dettagliato - Record simili 2019-05-16 14:53 Record dettagliato - Record simili 2019-05-16 14:31 Measurement of $CP$-violating and mixing-induced observables in $B_s^0 \to \phi\gamma$ decays / LHCb Collaboration A time-dependent analysis of the $B_s^0 \to \phi\gamma$ decay rate is performed to determine the $CP$-violating observables $S_{\phi\gamma}$ and $C_{\phi\gamma}$, and the mixing-induced observable $\mathcal{A}^{\Delta}_{\phi\gamma}$. The measurement is based on a sample of $pp$ collision data recorded with the LHCb detector, corresponding to an integrated luminosity of 3 fb$^{-1}$ at center-of-mass energies of 7 and 8 TeV. [...] arXiv:1905.06284; LHCb-PAPER-2019-015; CERN-EP-2019-077; LHCb-PAPER-2019-015; CERN-EP-2019-077; LHCB-PAPER-2019-015.- 2019-08-28 - 10 p. - Published in : Phys. Rev. Lett. 123 (2019) 081802 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Record dettagliato - Record simili 2019-04-10 11:16 Observation of a narrow pentaquark state, $P_c(4312)^+$, and of two-peak structure of the $P_c(4450)^+$ / LHCb Collaboration A narrow pentaquark state, $P_c(4312)^+$, decaying to $J/\psi p$ is discovered with a statistical significance of $7.3\sigma$ in a data sample of $\Lambda_b^0\to J/\psi p K^-$ decays which is an order of magnitude larger than that previously analyzed by the LHCb collaboration. The $P_c(4450)^+$ pentaquark structure formerly reported by LHCb is confirmed and observed to consist of two narrow overlapping peaks, $P_c(4440)^+$ and $P_c(4457)^+$, where the statistical significance of this two-peak interpretation is $5.4\sigma$. [...] arXiv:1904.03947; LHCb-PAPER-2019-014 CERN-EP-2019-058; LHCB-PAPER-2019-014.- Geneva : CERN, 2019-06-06 - 11 p. - Published in : Phys. Rev. Lett. 122 (2019) 222001 Article from SCOAP3: PDF; Fulltext: PDF; Fulltext from Publisher: PDF; Related data file(s): ZIP; Supplementary information: ZIP; External link: SYMMETRY Record dettagliato - Record simili 2019-04-03 11:16 Measurements of $CP$ asymmetries in charmless four-body $\Lambda^0_b$ and $\Xi_b^0$ decays / LHCb Collaboration A search for $CP$ violation in charmless four-body decays of $\Lambda^0_b$ and $\Xi_b^0$ baryons with a proton and three charged mesons in the final state is performed. To cancel out production and detection charge-asymmetry effects, the search is carried out by measuring the difference between the $CP$ asymmetries in a charmless decay and in a decay with an intermediate charmed baryon with the same particles in the final state. [...] arXiv:1903.06792; LHCb-PAPER-2018-044; CERN-EP-2019-13; LHCb-PAPER-2018-044 and CERN-EP-2019-13; LHCB-PAPER-2018-044.- 2019-09-07 - 30 p. - Published in : Eur. Phys. J. C 79 (2019) 745 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Record dettagliato - Record simili 2019-04-01 11:42 Observation of an excited $B_c^+$ state / LHCb Collaboration Using $pp$ collision data corresponding to an integrated luminosity of $8.5\,\mathrm{fb}^{-1}$ recorded by the LHCb experiment at centre-of-mass energies of $\sqrt{s} = 7$, $8$ and $13\mathrm{\,Te\kern -0.1em V}$, the observation of an excited $B_c^+$ state in the $B_c^+\pi^+\pi^-$ invariant-mass spectrum is reported. The state has a mass of $6841.2 \pm 0.6 {\,\rm (stat)\,} \pm 0.1 {\,\rm (syst)\,} \pm 0.8\,(B_c^+) \mathrm{\,MeV}/c^2$, where the last uncertainty is due to the limited knowledge of the $B_c^+$ mass. [...] arXiv:1904.00081; CERN-EP-2019-050; LHCb-PAPER-2019-007.- Geneva : CERN, 2019-06-11 - 10 p. - Published in : Phys. Rev. Lett. 122 (2019) 232001 Article from SCOAP3: PDF; Fulltext: PDF; Fulltext from Publisher: PDF; Related data file(s): ZIP; Supplementary information: ZIP; Record dettagliato - Record simili 2019-04-01 09:46 Near-threshold $D\bar{D}$ spectroscopy and observation of a new charmonium state / LHCb Collaboration Using proton-proton collisiondata, corresponding to an integrated luminosity of 9 fb$^{-1}$, collected with the~LHCb detector between 2011 and 2018, a new narrow charmonium state, the $X(3842)$ resonance, is observed in the decay modes $X(3842)\rightarrow D^0\bar{D}^0$ and $X(3842)\rightarrow D^+D^-$. The mass and the natural width of this state are measured to be \begin{eqnarray} m_{X(3842)} & = & 3842.71 \pm 0.16 \pm 0.12~ \text {MeV}/c^2\,, \\ \Gamma_{X(3842)} & = & 2.79 \pm 0.51 \pm 0.35 ~ \text {MeV}\,, \end{eqnarray} where the first uncertainty is statistical and the second is systematic. [...] arXiv:1903.12240; CERN-EP-2019-047; LHCb-PAPER-2019-005; LHCB-PAPER-2019-005.- Geneva : CERN, 2019-07-08 - 23 p. - Published in : JHEP 1907 (2019) 035 Article from SCOAP3: PDF; Fulltext: PDF; Related data file(s): ZIP; Supplementary information: ZIP; Record dettagliato - Record simili
I have an unweighted, undirected graph $G$ with $(v_1, ..., v_N) \in V$ vertices, where each vertex in the lattice represents a string of some length ($\leq 200$ characters), and I need to construct $G$ from an unordered list of edges between vertices (i.e. a list saying that vertex $v_i$ is connected to vertex $v_j$ for some $i \neq j$). Here, $\|\|V\|\|$ is very large, on order $10^{9}$ to $10^{12}$ or so, but the degree of each vertex $v_i$ is bounded by some small integer value s.t. the maximum vertex degree is $\Delta G\approx 16$. What is a good way to store or process $G$ in Mathematica (some sort of suffix tree or hash table?) s.t. I can efficiently retrieve a list of vertices in the $k$-hop neighborhood of some vertex $v_i$? By $k$-hop neighborhood, I mean the set of vertices that can be reached from $v_i$ by traveling along at most $k$ edges. To provide an example, I might start with a list of vertices corresponding to unique binary strings: $v_1 = 10101011110...$ $v_2 = 01110100101...$ ... $v_N = 00000011110...$ And an unordered list of (undirected) edges between vertices: $v_{36544} \to v_{3}$ $v_{740943} \to v_{92034}$ $v_{36544} \to v_{674}$ ... Is there a nice / efficient way to process the about lists in Mathematica s.t. I can retrieve the set of nearest-neighbors for, say, $v_{36544}$, which includes $(v_{3}, v_{674}, ...)$? The naive way to proceed is to simply specify a graph structure as: Graph[{...}], but is this efficient for very large instances of $G$? Update - OK, some data. Intel Xeon CPU (X5690) 3.47 GHz 64 bit system (No attempt at parallelization) Lookup time for the $k=1$ hop neighborhood of an arbitrary vertex (using NeighborhoodGraph[G]): G = 10 x 10 integer lattice: 27.4 milliseconds G = 50 x 50 integer lattice: 0.765 seconds G = 100 x 100 integer lattice: 3.08 seconds The scaling appears exactly linear with graph size: http://www.wolframalpha.com/input/?i=%7B0.0274%2C+10%5E2%7D%2C%7B0.765%2C2500%7D%2C%7B3.08%2C100%5E2%7D This is awful, especially considering that Mathematica can draw a spring minimized graph in relatively short order for the 100 x 100 integer lattice.
The set $V= \{ A\in M_n(\mathbb{C})\ \|\ A\bar A = I\}$ is a smooth submanifold of $M_n(\mathbb{C})$ with real dimension $n^2$. Proof: Consider the involution $\iota:\mathrm{GL}(n,\mathbb{C})\to \mathrm{GL}(n,\mathbb{C})$ defined by$$\iota(A) = (\bar A)^{-1}.$$This is an anti-holomorphic involution of the complex manifold $\mathrm{GL}(n,\mathbb{C})$, and its fixed locus is precisely $V$. Thus, $V$ is a totally real submanifold of $\mathrm{GL}(n,\mathbb{C})$ with (real) dimension $n^2$. Also: While it's true that the map $f:M_n(\mathbb{R})\to \mathrm{GL}(n,\mathbb{C})$ defined by $$f(a) = \exp(ia)$$has its image in $V$, it is not a 'parametrization' everywhere, i.e., $f$ is not a local diffeomorphism everywhere. While this is true on a neighborhood of $0\in M_n(\mathbb{R})$, at other places, the map $f$ definitely is not a local diffeomorphism. For example, if $a$ has $n$ eigenvalues of the form $2k_i\pi$ (where $k_1,\ldots, k_n$ are integers, not all zero), then $f(a) = I$, but the space of such real matrices $a$ has positive dimension for any $n$-tuple $(k_1,\ldots, k_n)$ for which not all of the $k_i$ are equal. It turns out that $f$ is surjective. The proof uses the fact that, for $A\in V$, we have $A\bar A = \bar A A = I$, so, in particular, $A$ and $\bar A$ commute and hence can be put simultaneously in Jordan normal form by a real conjugation. Then, breaking $\mathbb{C}^n$ into a sum of complexifications of real subspaces according to the eigenvalues of $A$ and using the semi-simple/nilpotent decomposition appropriately, one can reduce to dealing with the upper triangular case, and the result can then be proved using simple facts about power series in commuting nilpotent variables. Details upon request ( see below). Of course, all of this is probably a special case of known facts about affine symmetric spaces. I have now realized that $V$ is simply the Cartan embedding for the affine symmetric space $\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})$. This Cartan embedding$$\sigma:\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})\longrightarrow V\subset\mathrm{GL}(n,\mathbb{C})$$is given by $\sigma(B\cdot\mathrm{GL}(n,\mathbb{R})) = B\,(\,\overline B\,)^{-1}$. The exponential map we have been discussing is just the geodesic mapping of this affine symmetric space, so, probably all of this follows from general theory. Details: Let $A\in\mathrm{GL}(n,\mathbb{C})$ satisfy $A\overline{A} = I$. Let $\lambda\in\mathbb{C}$ be an eigenvalue of $A$ of multiplicity $m\le n$ and let$$V_\lambda = \{ v\in\mathbb{C}^n \ \|\ (A{-}\lambda)^mv = 0\ \}.$$be the associated generalized eigenspace. Of course, $\mathbb{C}^n$ is the direct sum of these generalized eigenspaces. Since $\overline{A}$ commutes with $A$, it preserves each of these subspaces. Moreover, one clearly has $$\overline{V_\lambda} = V_{1/\bar\lambda},$$so each of the spaces $V_\lambda+V_{1/\bar\lambda}$ is invariant under conjugation and hence is the complexification of a real subspace $W_\lambda\subset \mathbb{R}^n$. It follows that, by conjugating by a real matrix, we can assume that $A$ (and hence $\overline A$) is in block diagonal form, so it suffices to treat the case where either $A$ has a single eigenvalue $\lambda$ satisfying $\lambda\bar\lambda=1$, so that $\mathbb{C}^n=V_\lambda$, or else $A$ has an eigenvalue $\lambda$ satisfying $\lambda\bar\lambda > 1$ and $\mathbb{C}^n=V_\lambda\oplus V_{1/\bar\lambda}$. In either case, we can assume that $\lambda = r\ge 1$ is real, since, if $\lambda = r e^{i\theta}$, we can replace $A$ by $e^{-i\theta}A$ and show that $e^{-i\theta}A$ is in the image of $f$ (since $I$ commutes with everything and $e^{i\theta} I = \exp(i\theta I)$.) First, consider the case $\lambda = 1$. Then $A = C + i S$ where $C$ and $S$ are real matrices and $((C-I) + iS)^n = (A-I)^n = 0$. Moreover, $N=C-I$ and $S$ commute since $I = A\bar A = C^2+S^2 +i(SC-CS) = I + 2N + N^2 + S^2$. Note that $N$ and $S$ are nilpotent commuting matrices. Since they satisfy $$2N + N^2 = -S^2,$$it follows that $N = p(-S^2)$ where $p(t) = -\tfrac12 t^2 + \cdots $ is the (unique) power series (with real coefficients) that satisfies $2p(t) + p(t)^2 = -t^2$. Since $S$ is nilpotent, it follows that $N = p(-S^2)$ expresses $N$ canonically as a polynomial in $S$. Now let $q(t)$ be the (unique) power series with real coefficients that satisfies $\sin q(t) = t$. Note that we must have $p(-q(t)^2) = \cos(t)-1$. Now, because $S$ is nilpotent, we have $S = \sin q(S)$, where $q(S)$ is a real polynomial in $S$. Putting all of this together, we have$$A = I + p(-S^2) + iS = \cos(q(S)) + i\sin(q(S)) = \exp(iq(S)).$$Finally, before leaving this special case, let us note that, because $A$ satisfies its characteristic polynomial $(A-I)^n = 0$, it follows that $\overline A = A^{-1}$ is expressed as a universal polynomial in $A$ with real coefficients, so $iS = \tfrac12(A-\overline{A})$ is also expressed as a universal polynomial with real coefficients. Since $q(t)$ is an odd power series, it satisfies $iq(it) = f(t)$ where $f$ has real coefficients, so there is actually a formula of the form $A = \exp( i g_n(A))$ for all $A \in \mathrm{GL}(n,\mathbb{C})$ satisfying $A\bar A = I$ and $(A-I)^n=0$ for some universal polynomial $g_n(t)$ with real coefficients that also has the property that $g_n(A)$ is a real $n$-by-$n$ matrix for all such $A$. (This remark will be used below.) Now, finally, let us assume that $A$ has eigenvalues $\lambda = e^t$ and $1/\lambda = e^{-t}$ for some $t>0$. Then $V_\lambda$ and $\overline{V_\lambda} = V_{1/\lambda}$ are disjoint, complementary complex subspaces of $\mathbb{C}^n$ and so we must have that there exists a matrix $Q\in M_n(\mathbb{R})$ such that$$V_\lambda = \{ v + iQv \ \| \ v\in \mathbb{R}^n \}.$$Because $V_\lambda$ is closed under multiplication by $i$, it follows that $Q^2 = -I$. Setting $$S = \cosh t + i\sinh t\,Q = \exp(i t Q),$$we see that $Sw= e^t w$ for all $w \in V_\lambda$ and $Sw = e^{-t}w$ for all $w \in V_{1/\lambda}$, so $S$ is the semi-simple part of $A$. Hence $S$ can be written as $S = s_\lambda(A)$, where $s_\lambda(t)$ a polynomial in $t$ with real coefficients (that depend on $\lambda$). Thus, $\bar S = s_\lambda(\bar A)$ can be written as a universal polynomial in $A$, implying that $Q$ itself can be written as a universal polynomial in $A$ and hence, in particular, it commutes with $A$ and $\bar A$. Now, writing $$A = \exp(i t Q) B,$$we see that $B\bar B = I$ and that $B$ can be written as a polynomial in $A$. Moreover, the eigenvalues of $B$ are now all equal to $1$, so $(B-I)^n=0$, and so $B = \exp(i g_n(B))$ (as per above), where $g_n(B)$ is a polynomial in $A$, which, therefore, commutes with $Q$ (which is a polynomial in $A$). Finally, we have$$A = \exp(i t Q)\exp(i g_n(B)) = \exp(i (t Q+g_n(B))),$$as desired.
What is the difference between Base Correlation and Implied Correlation for a CDO tranche? An implied correlation $\rho_i(k_1,k_2)$ is a correlation that matches the $(k_1,k_2)$ tranche price $P_{k_1}^{k_2}$ (usually computed under a gaussian or student t copula) $$ C(k_1,k_2,\rho_i(k_1,k_2)) = P_{k_1}^{k_2} $$ For mezzanine tranches, there can sometimes be two different implied correlations matching the tranche price. A base correlation $b_i(k_2)$ is a correlation that matches the price of the tranche, plus all higher-risk tranches "beneath" it, so we can write it as $$ b_i(k_2) = \rho_i(0,k_2) $$ where we obtain $P_{0}^{k_2}$ as $$ P_{0}^{k_2} = \sum_{k_i\leq{k_2}}P_{k_{i-1}}^{k_i} $$ The pricing function $C(0,k_2,\rho)$ is monotonic in $\rho$, hence the base correlation is unique. This allows practitioners to think about correlations a bit more like they previously thought about implied volatility (and volatility skews) for options. The super-senior tranche has (trivially) a base correlation that matches the price of the entire underlying instrument, since it is $\rho_i(0,1)$.
This post was first published on 12/29/15, and has since been migrated to Blogger. This is a tutorial on implementing Ian Goodfellow's Generative Adversarial Nets paper in TensorFlow. Adversarial Nets are a fun little Deep Learning exercise that can be done in ~80 lines of Python code, and exposes you (the reader) to an active area of deep learning research (as of 2015): Generative Modeling! Code on Github Scenario: Counterfeit MoneyTo help explain the motivations behind the paper, here's a hypothetical scenario: Danielle is a teller at a bank, and one of her job responsibilities is to discriminate between real money and counterfeit money. George is a crook and is trying to make some counterfeit money, becase free money would be pretty radical. Let's simplify things a bit and assume the only distinguishing feature of currency is one unique number, $X$, printed on the each bill. These numbers are randomly sampled from a probability distribution, whose density function $p_{data}$ is only known to the Treasury (i.e. neither Danielle nor George know the function). For convenience, this tutorial uses $p_{data}$ to refer to both the distribution and the density function (though semantically a distribution and its density function are not the same). George's goal is to generate samples $x^\prime$ from $p_{data}$, so his counterfeit currency is indistinguishable from "real" currency. You might ask: how can George generate samples from $p_{data}$ if he doesn't know $p_{data}$ in the first place? We can create computationally indistinguishable samples without understanding the "true" underlying generative process [1]. The underlying generative process is the method that the Treasury itself is using to generate samples of $X$ - perhaps some efficient algorithm for sampling $p_{data}$ that relies on the analytical formula for the pdf. We can think of this algorithm as the "natural (function) basis", the straightforward method the Treasury would actually use to print our hypothetical currency. However, a (continuous) function can be represented as a combination of a different set of basis functions; George can express the same sampler algorithm in a "neural network basis" or "Fourier basis" or other basis that can be used to build a universal approximator. From an outsider's perspective, the samplers are computationally indistinguishable, and yet George's model doesn't reveal to him the structure of the "natural" sampler basis or the analytical formula of $p_{data}$. Background: Discriminative vs. Generative ModelsLet $X$, $Y$ be the "observed" and "target" random variables. The joint distribution for $X$ and $Y$ is $P(X,Y)$, which we can think of as a probability density over 2 (possibly dependent) variables. A Discriminative model allows us to evaluate the conditional probability $P(Y\|X)$. For instance, given a vector of pixel values $x$, what is the probability that $Y=6$? (where "6" corresponds to the categorical class label for "tabby cat"). MNIST LeNet, AlexNet, and other classifiers are examples of a discriminative models. On the other hand, a Generative model can allows us to evaluate the joint probability $P(X,Y)$. This means that we can propose value pairs $(X,Y)$ and do rejection-sampling to obtain samples $x$,$y$ from $P(X,Y)$. Another way to put this is that with the right generative model, we can convert a random number from $[0,1]$ into a picture of a rabbit. That's awesome. Of course, generative models are much harder to construct than discriminative models, and both are active areas of research in statistics and machine learning. Generative Adversarial NetworksGoodfellow's paper proposes a very elegant way to teach neural networks a generative model for any (continuous) probability density function. We build two neural networks $D$ (Danielle) and $G$ (George), and have them play an adversarial cat-and-mouse game: $G$ is a generator and attempts to counterfeit samples from $p_{data}$ and $D$ is a decider that tries to not get fooled. We train them simultaneously, so that they both improve by competing with each other. At convergence, we hope that $G$ has learned to sample exactly from $p_{data}$, at which point $D(x)=0.5$ (random guessing). Advesarial Nets have been used to great success to synthesize the following from thin air: Cat Faces Churches Anime Characters In this tutorial we won't be doing anything nearly as amazing, but hopefully you'll come away with a better fundamental understanding of adversarial nets. ImplementationWe'll be training a neural network to sample from the simple 1-D normal distribution $\mathcal{N}(-1,1)$ Let $D$,$G$ be small 3-layer perceptrons, each with a meager 11 hidden units in total. $G$ takes as input a single sample of a noise distribution: $z \sim \text{uniform}(0,1)$. We want $G$ to map points $z_1,z_2,...z_M$ to $x_1,x_2,...x_M$, in such a way that mapped points $x_i=G(z_i)$ cluster densely where $p_{data}(X)$ is dense. Thus, G takes in $z$ and generates fake data $x^\prime$. Meanwhile, the discriminator $D$, takes in input $x$ and outputs a likelihood of the input belonging to $p_{data}$. Let $D_1$ and $D_2$ be copies of $D$ (they share the same parameters so $D_1(x)=D_2(x)$). The input to $D_1$ is a single sample of the legitimate data distribution: $x \sim p_{data}$, so when optimizing the decider we want the quantity $D_1(x)$ to be maximized. $D_2$ takes as input $x^\prime$ (the fake data generated by $G$), so when optimizing $D$ we want to $D_2(x^\prime)$ to be minimized. The value function for $D$ is: $$ \log(D_1(x))+\log(1-D_2(G(z))) $$ Here's the Python code: The reason we go through the trouble of specifying copies $D_1$ and $D_2$ is that in TensorFlow, we need one copy of $D$ to process the input $x$ and another copy to process the input $G(z)$; the same section of the computational graph can't be re-used for different inputs. batch=tf.Variable(0)obj_d=tf.reduce_mean(tf.log(D1)+tf.log(1-D2))opt_d=tf.train.GradientDescentOptimizer(0.01) .minimize(1-obj_d,global_step=batch,var_list=theta_d) When optimizing $G$, we want the quantity $D_2(X^\prime)$ to be maximized (successfully fooling $D$). The value function for $G$ is: $$ \log(D_2(G(z))) $$ Instead of optimizing with one pair $(x,z)$ at a time, we update the gradient based on the average of $M$ loss gradients computed for $M$ different $(x,z)$ pairs. The stochastic gradient estimated from a minibatch is closer to the true gradient across the training data. batch=tf.Variable(0)obj_g=tf.reduce_mean(tf.log(D2))opt_g=tf.train.GradientDescentOptimizer(0.01) .minimize(1-obj_g,global_step=batch,var_list=theta_g) The training loop is straightforward: # Algorithm 1, GoodFellow et al. 2014for i in range(TRAIN_ITERS): x= np.random.normal(mu,sigma,M) # sample minibatch from p_data z= np.random.random(M) # sample minibatch from noise prior sess.run(opt_d, {x_node: x, z_node: z}) # update discriminator D z= np.random.random(M) # sample noise prior sess.run(opt_g, {z_node: z}) # update generator G Manifold AlignmentFollowing the above recipe naively will not lead to good results, because we are sampling $p_{data}$ and $\text{uniform}(0,1)$ independently each iteration. Nothing is enforcing that adjacent points in the $Z$ domain are being mapped to adjacent points in the $X$ domain; in one minibatch, we might train $G$ to map $0.501 \to -1.1$, $0.502 \to 0.01$, and $0.503 \to -1.11$. The mapping arrows cross each other too much, making the transformation very bumpy. What's worse, the next minibatch might map $0.5015 \to 1.1$, $0.5025 \to -1.1$, and $0.504 \to 1.01$. This implies a completely different mapping $G$ from the previous minibatch, so the optimizer will fail to converge. To remedy this, we want to minimize the total length of the arrows taking points from $Z$ to $X$, because this will make the transformation as smooth as possible and easier to learn. Another way of saying this is that the "vector bundles" carrying $Z$ to $X$ should be correlated between minibatches. First, we'll stretch the domain of $Z$ to the same size of $X$. The normal distribution centered at -1 has most of its probability mass lying between $[-5,5]$, so we should sample $Z$ from $\text{uniform}(-5,5)$. Doing this means $G$ no longer needs to learn how to "stretch" the domain $[0,1]$ by a factor of 10. The less $G$ has to learn, the better. Next, we'll align the samples of $Z$ and $X$ within a minibatch by sorting them both from lowest to highest. Instead of sampling $Z$ via np.random.random(M).sort(), we'll use via stratified sampling - we generate $M$ equally spaced points along the domain and then jitter the points randomly. This preserves sorted order and also increases the representativeness the entire training space. We then match our stratified, sorted $Z$ samples to our sorted $X$ samples. Of course, for higher dimensional problems it's not so straightforward to align the input space $Z$ with the target space $X$, since sorting points doesn't really make sense in 2D and higher. However, the notion of minimizing the transformation distance between the $Z$ and $X$ manifolds still holds [2]. The modified algorithm is as follows: for i in range(TRAIN_ITERS): x= np.random.normal(mu,sigma,M).sort() z= np.linspace(-5.,5.,M)+np.random.random(M).01 # stratified sess.run(opt_d, {x_node: x, z_node: z}) z= np.linspace(-5.,5.,M)+np.random.random(M).01 sess.run(opt_g, {z_node: z}) This step was crucial for me to get this example working: when dealing with random noise as input, failing to align the transformation map properly will give rise to a host of other problems, like massive gradients that kill ReLU units early on, plateaus in the objective function, or performance not scaling with minibatch size. Pretraining DThe original algorithm runs $k$ steps of gradient descent on $D$ for every step of gradient descent on $G$. I found it more helpful to pre-train $D$ a larger number of steps prior to running the adversarial net, using a mean-square error (MSE) loss function to fit $D$ to $p_{data}$. This loss function is nicer to optimize than the log-likelihood objective function for $D$ (since the latter has to deal with stuff from $G$). It's easy to see that $p_{data}$ is the optimal likelihood decision surface for its own distribution. Here is the decision boundary at initialization. After pretraining: Close enough, lol. Other Troubleshooting Comments Using too many parameters in the model often leads to overfitting, but in my case making the network too big failed to even converge under the minimax objective - the network units saturated too quickly from large gradients. Start with a shallow, tiny network and only add extra units/layers if you feel that the size increase is absolutely necessary. I started out using ReLU units but the units kept saturating (possibly due to manifold alignment issues). The Tanh activation function seemed to work better. I also had to tweak the learning rate a bit to get good results. ResultsBefore training, here is $p_{data}$, the pre-trained decision surface of $D$, and the generative distribution $p_g$. . Here's the loss function as a function of training iterations. After training, $p_g$ approximates $p_{data}$, and the discriminator is about uniformly confused ($D=0.5$) for all values of $X$. And there you have it! $G$ has learned how to sample (approximately) from $p_{data}$ to the extent that $D$ can no longer separate the "reals" from the forgeries. Footnotes Here's a more vivid example of computational indistinguishability: suppose we train a super-powerful neural network to sample from the distribution of cat faces. The underlying (generative) data distribution for "true" cat faces involves 1) a cat being born and 2) somebody eventually taking a photograph of the cat. Clearly, our neural network isn't learning this particular generative process, as no real cats are involved. However, if our neural network can generate pictures that we cannot distinguish from "authentic" cat photos (armed with polynomial-time computational resources), then in some ways these photos are "just as legitimate" as regular cat photos. This is worth pondering over, in the context of the Turing Test, Cryptography, and fake Rolexes. Excessive mapping crossover can be viewed from the perspective of "overfitting", where the learned regression/classification function has been distorted in "contradictory" ways by data samples (for instance, a picture of a cat classified as a dog). Regularization techniques implicitly discourage excessive "mapping crossover", but do not explicitly implement a sorting algorithm to ensure the learned transformation is continuous / aligned in $X$-space with respect to $Z$-space. Such a sorting mechanism might be very useful for improving training speeds...
101 people flip a fair coin. Everyone who tosses heads is on one team and everyone who tosses tails is on another other team. The team with more people on it wins. What are the odds that, given you are one of the 101 players, you will win? (101 players and coins eliminates ties but I am also interested the case where there are 100 players where you can win/lose/tie). If the 100 other players divide into two teams of size 50, your chances of winning are 100%. Otherwise, the winning team will not change with your choice hence your chances of winning are 50%. Thus your overall probability of winning is 50%+50%P(50-50 divide), that is, $$\frac12\left(1+\frac1{2^{100}}{100\choose50}\right)\approx53.98\%.$$ The expected number of winners is: $$2\sum\limits_{n=51}^{101}\frac{n\cdot\binom{101}{n}}{2^{101}}\approx54.5193$$ So the probability of being a winner is: $$\frac{54.5193}{101}\approx0.5397$$ You and 100 other people flip coins. You always win if exactly 50 of the others got heads. In that case, you are the tie-breaker and always end up on the winning side. If the split was not even, then you win if your flip matched the larger half ($50\%$). Thus, your odds of winning are: $${1\over 2} \times \left(1 + {{100 \choose 50}\over 2^{100}}\right) \approx 54\% $$ For a total of $100$ people, you win exactly $50\%$ of the time. The other side never splits evenly, and you win if your coin flip matches the larger half. It's a tie if the split (including yours) is even: $${{100 \choose 50}\over 2^{100}} \approx 8\%$$ And you lose about $42\%$ of the time when there are $100$ people.
Model A model of a theory $T$ is a set $M$ together with relations (eg. two: $a$ and $b$) satisfying all axioms of the theory $T$. Symbolically $\langle M, a, b \rangle \models T$. According to the Gödel completeness theorem, in $\mathrm{PA}$ (Peano arithmetic) (so also in $\mathrm{ZFC}$) a theory has models iff it is consistent. According to Löwenheim–Skolem theorem, in $\mathrm{ZFC}$ if a countable first-order theory has an infinite model, it has infinite models of all cardinalities. A model of a set theory (eg. $\mathrm{ZFC}$) is a set $M$ such that the structure $\langle M,\hat\in \rangle$ satisfies all axioms of the set theory. If $\hat \in$ is base theory's $\in$, the model is called a transitive model. Gödel completeness theorem and Löwenheim–Skolem theorem do not apply to transitive models. (But Löwenheim–Skolem theorem together with Mostowski collapsing lemma show that if there is a transitive model of ZFC, then there is a countable such model.) See Transitive ZFC model. Contents Class-sized transitive models One can also talk about class-sized transitive models. Inner model is a transitive class containing all ordinals. Forcing creates outer models, but it can also be used in relation with inner models.[1] Among them are canonical inner models like the core model the canonical model $L[\mu]$ of one measurable cardinal HOD and generic HOD (gHOD) mantle $\mathbb{M}$ (=generic mantle $g\mathbb{M}$) outer core the constructible universe $L$ Mantle $α$th inner mantle $\mathbb{M}^α$ is defined by $\mathbb{M}^0=V$, $\mathbb{M}^{α+1} = \mathbb{M}^{\mathbb{M}^α}$ (mantle of the previous inner mantle) and $\mathbb{M}^α = \bigcap_{β<α} \mathbb{M}^β$ for limit $α$. If there is uniform presentation of $\mathbb{M}^α$ for all ordinals $α$ as a single class, one can talk about $\mathbb{M}^\mathrm{Ord}$, $\mathbb{M}^{\mathrm{Ord}+1}$ etc. If an inner mantle is a ground, it is called the outer core.[1] It is conjenctured (unproved) that every model of ZFC is the $\mathbb{M}^α$ of another model of ZFC for any desired $α ≤ \mathrm{Ord}$, in which the sequence of inner mantles does not stabilise before $α$. It is probable that in the some time there are models of ZFC, for which inner mantle is undefined (Analogously, a 1974 result of Harrington appearing in (Zadrożny, 1983, section 7), with related work in (McAloon, 1974), shows that it is relatively consistent with Gödel-Bernays set theory that $\mathrm{HOD}^n$ exists for each $n < ω$ but the intersection $\mathrm{HOD}^ω = \bigcap_n \mathrm{HOD}^n$ is not a class.).[1] For a cardinal $κ$, we call a ground $W$ of $V$ a $κ$-ground if there is a poset $\mathbb{P} ∈ W$ of size $< κ$ and a $(W, \mathbb{P})$-generic $G$ such that $V = W[G]$. The $κ$-mantle is the intersection of all $κ$-grounds.[3] The $κ$-mantle is a definable, transitive, and extensional class. It is consistent that the $κ$-mantle is a model of ZFC (e.g. when there are no grounds), and if $κ$ is a strong limit, then the $κ$-mantle must be a model of ZF. However it is not known whether the $κ$-mantle is always a model of ZFC.[3] Mantle and large cardinals $\kappa$-model A weak $κ$-model is a transitive set $M$ of size $\kappa$ with $\kappa \in M$ and satisfying the theory $\mathrm{ZFC}^-$ ($\mathrm{ZFC}$ without the axiom of power set, with collection, not replacement). It is a $κ$-model if additionaly $M^{<\kappa} \subseteq M$.[4, 5] References Fuchs, Gunter and Hamkins, Joel David and Reitz, Jonas. Set-theoretic geology.Annals of Pure and Applied Logic 166(4):464 - 501, 2015. www arχiv DOI bibtex Usuba, Toshimichi. The downward directed grounds hypothesis and very large cardinals.Journal of Mathematical Logic 17(02):1750009, 2017. arχiv DOI bibtex Usuba, Toshimichi. Extendible cardinals and the mantle.Archive for Mathematical Logic 58(1-2):71-75, 2019. arχiv DOI bibtex Hamkins, Joel David and Johnstone, Thomas A. Strongly uplifting cardinals and the boldface resurrection axioms., 2014. arχiv bibtex Holy, Peter and Schlicht, Philipp. A hierarchy of Ramsey-like cardinals.Fundamenta Mathematicae 242:49-74, 2018. www arχiv DOI bibtex
Table of Contents Repeated Roots of The Characteristic Equation Examples 1 Recall from the Repeated Roots of The Characteristic Equation page that for a first order linear homogenous differential equation with constant coefficients, $a \frac{d^2 y}{dt^2} + b \frac{dy}{dt} + cy = 0$ where $a, b, c \in \mathbb{R}$, then if the roots $r_1$ and $r_2$ of the characteristic equation $ar^2 + br + c = 0$ are real and not distinct, that is $r_1 = r_2$, or simply just denote this root as $r$, then the general solution of this differential equation is given by:(1) We will now look at some examples of solving differential equations of this type. Example 1 Find the general solution of the differential equation $\frac{d^2y}{dt^2} + 4 \frac{dy}{dt} + 4 y = 0$. Describe the behaviour of the general solution as $t$ approaches infinity. We have that the characteristic equation for this differential equation is given by $r^2 + 4r + 4 = 0$. We can easily factor the characteristic equation as $(r + 2)(r + 2) = (r + 2)^2 = 0$. Therefore there is only one distinct root to our characteristic equation, that being $r = 2$. Therefore the general solution to this differential equation is:(2) As $t \to \infty$ we have that $e^{2t} \to \infty$ and so it should be relatively clear that $\lim_{t \to \infty} \left ( Ce^{2t} + Dte^{2t} \right ) = \infty$. The following image is a graph of a few of the functions to this differential equation. Example 2 Find the general solution of the differential equation $\frac{d^2y}{dt^2} + 6 \frac{dy}{dt} + 9 y = 0$. Describe the behaviour of the general solution as $t$ approaches infinity. The characteristic equation for this differential equation is $r^2 +6r + 9 = 0$. This characteristic equation can be easily factored as $(r + 3)(r + 3) = (r + 3)^2 = 0$. Therefore the repeated root in the characteristic equation is $r = -3$ and the general solution to our differential equation is:(3) Note that as $t \to \infty$, $e^{-3t} \to 0$. Exponential functions take "control" over linear functions, and so $\lim_{t \to \infty} \left ( Ce^{-3t} + Dte^{-3t} \right )= 0$. The following is a graph of a few solutions to this differential equation.
Finishing some long overdue work, I was recently benchmarking different ways to compute orbital closures of permutation groups. Orbital Closures, and what others call them Let’s first lay down what I am computing. I have referred to Orbital Graphs before. Suppose that \Omega = \{1..n\} for some n \in \mathbb{N} and G \leq S_{\Omega}. We consider all \mathcal{OC}(G) = \bigcap\limits_{(\alpha,\beta) \in \Omega^2} \operatorname{Aut}(\Gamma_{(\alpha,\beta)}) of GWhich necessarily contains G, but might be bigger. Naive computations In GAP using the package digraphs, and optionally ferret, one can naively compute the group \mathcal{OC}\left(G\right) as follows 1 2 3 4 5 6 OrbitalClosure := function(G) local ogs, ogauts; ogs := OrbitalGraphs(G); ogauts := List(ogs, AutomorphismGroup); return Intersection(ogauts); end; The code above becomes unbearably slow if we maliciously produce examples of groups with a decent number of orbital graphs, such as direct products of isomorphic groups or even a wreath product. Trying to compute the orbital closure of G = D_{32} \wr D_{32} makes the wait already quite painful. Exploring what takes so long in this computation yields some insight: We compute the automorphism group of each orbital graph separately, intersecting the results. Group intersection is a hard problem, though it is no harder than graph automorphism finding. Cleverer construction A slightly cleverer way of computing the orbital closure is not to compute a truckload of graphs, their automorphism groups, and intersecting them, but to just compute one graph with coloured edges, where the edges represent which orbital graph it belongs to, and handing that to a graph automorphism solver. This should of course yield the same result, but using a very efficient graph automorphism finder works a lot better.thanks go to Finn Smith for adding code to the GAP package digraphs to compute automorphisms of edge-coloured digraphs Benchmarks in GAP with Digraphs and ferret Putting this together I did some benchmark comparisons using GAP 4.11, digraphs, and ferret and found that the edge-coloured method is up to a factor of 1000 faster. Hence the new default method in the orbitalgraphs package is using the edge-coloured graph method as a default when one calls OrbitalClosure(G). The following table shows the computation time (in milliseconds) for the orbital closure of D_{32} \wr D_{2n} for n \in \{4 \ldots 11\}. The second column contains the number of distinct orbital graphs. n #orbital graphs intersect edge-coloured 4 10 588.818 7.810 5 10 876.719 9.932 6 11 2311.209 15.972 7 11 103966.205 21.423 8 12 301819.541 34.307 9 12 445608.635 34.193 10 13 824163.351 42.039 11 13 1657053.153 65.100 As a disclaimer it should be said that this comparison is to a degree probably unfair, as GAP spends a Even more sophistication? In a future post I will be looking at Johannes Hahn’s suggested pull request for the orbitalgraphs package to improve performance more by avoiding more graphs.
This sum may be evaluated by considering the following contour integral in the complex plane: $$\oint_C dz \frac{\pi \cot{\pi z}}{(m z^2-n)^2}$$ where $C$ is a rectangular contour that encompasses the poles of the integrand in the complex plane, up to $z=\pm \left ( N +\frac12\right)$, where we consider the limit as $N \to\infty$. We note here that we assume that the ratio $n/m$ is not the square of an integer. Now, the contour integral is zero because the individual integrals along each piece of the contour cancel. On the other hand, the contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand. Working this out, we find that $$\sum_{k=-\infty}^{\infty} \frac{1}{(m k^2-n)^2} = -\sum_{\pm}\operatorname{Res}_{z=\pm \sqrt{n/m}} \frac{\pi \cot{\pi z}}{(m z^2-n)^2}$$ Since the pole is a double pole, we have $$\sum_{\pm}\operatorname{Res}_{z=\pm\sqrt{n/m}} \frac{\pi \cot{\pi z}}{(m z^2-n)^2} =\sum_{\pm} \frac{\pi}{m^2} \left [\frac{d}{dz} \frac{\cot{\pi z}}{(z\pm \sqrt{n/m})^2} \right ]_{z=\pm\sqrt{n/m}} $$ I assume that the reader can take the derivatives and do the subsequent algebra. I get for the sum $$-\frac{\pi}{m^2} \frac{m}{2 n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$ So we now have $$\sum_{k=-\infty}^{\infty} \frac{1}{(m k^2-n)^2} = \frac{\pi}{2 m n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$ We now exploit the evenness of the summand; the result is $$\sum_{k=1}^{\infty} \frac{1}{(m k^2-n)^2} = -\frac{1}{2 n^2} + \frac{\pi}{4 m n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$ The result to be proven follows. For the general sum $$\sum_{k=1}^{\infty} \frac{1}{(m k^2-n)^{\ell}}$$ where $\ell \gt 2$ and $\ell \in \mathbb{Z}$, we take the same approach. The residue is an $\ell-1$ derivative of the integrand.
V. Gitman, J. D. Hamkins, and A. Karagila, “Kelley-Morse set theory does not prove the class Fodor theorem.” (manuscript under review) @ARTICLE{GitmanHamkinsKaragila:KM-set-theory-does-not-prove-the-class-Fodor-theorem, author = {Victoria Gitman and Joel David Hamkins and Asaf Karagila}, title = {Kelley-Morse set theory does not prove the class {F}odor theorem}, journal = {}, year = {}, volume = {}, number = {}, pages = {}, month = {}, note = {manuscript under review}, abstract = {}, keywords = {under-review}, eprint = {1904.04190}, archivePrefix = {arXiv}, primaryClass = {math.LO}, source = {}, doi = {}, url = {http://wp.me/p5M0LV-1RD}, } V. Gitman, J. D. Hamkins, and A. Karagila, “Kelley-Morse set theory does not prove the class Fodor theorem.” (manuscript under review) Abstract. We show that Kelley-Morse (KM) set theory does not prove the class Fodor principle, the assertion that every regressive class function $F:S\to\newcommand\Ord{\text{Ord}}\Ord$ defined on a stationary class $S$ is constant on a stationary subclass. Indeed, it is relatively consistent with KM for any infinite $\lambda$ with $\omega\leq\lambda\leq\Ord$ that there is a class function $F:\Ord\to\lambda$ that is not constant on any stationary class. Strikingly, it is consistent with KM that there is a class $A\subseteq\omega\times\Ord$, such that each section $A_n=\{\alpha\mid (n,\alpha)\in A\}$ contains a class club, but $\bigcap_n A_n$ is empty. Consequently, it is relatively consistent with KM that the class club filter is not $\sigma$-closed. The class Fodor principle is the assertion that every regressive class function $F:S\to\Ord$ defined on a stationary class $S$ is constant on a stationary subclass of $S$. This statement can be expressed in the usual second-order language of set theory, and the principle can therefore be sensibly considered in the context of any of the various second-order set-theoretic systems, such as Gödel-Bernays (GBC) set theory or Kelley-Morse (KM) set theory. Just as with the classical Fodor’s lemma in first-order set theory, the class Fodor principle is equivalent, over a weak base theory, to the assertion that the class club filter is normal. We shall investigate the strength of the class Fodor principle and try to find its place within the natural hierarchy of second-order set theories. We shall also define and study weaker versions of the class Fodor principle. If one tries to prove the class Fodor principle by adapting one of the classical proofs of the first-order Fodor’s lemma, then one inevitably finds oneself needing to appeal to a certain second-order class-choice principle, which goes beyond the axiom of choice and the global choice principle, but which is not available in Kelley-Morse set theory. For example, in one standard proof, we would want for a given $\Ord$-indexed sequence of non-stationary classes to be able to choose for each member of it a class club that it misses. This would be an instance of class-choice, since we seek to choose classes here, rather than sets. The class choice principle $\text{CC}(\Pi^0_1)$, it turns out, is sufficient for us to make these choices, for this principle states that if every ordinal $\alpha$ admits a class $A$ witnessing a $\Pi^0_1$-assertion $\varphi(\alpha,A)$, allowing class parameters, then there is a single class $B\subseteq \Ord\times V$, whose slices $B_\alpha$ witness $\varphi(\alpha,B_\alpha)$; and the property of being a class club avoiding a given class is $\Pi^0_1$ expressible. Thus, the class Fodor principle, and consequently also the normality of the class club filter, is provable in the relatively weak second-order set theory $\text{GBC}+\text{CC}(\Pi^0_1)$. This theory is known to be weaker in consistency strength than the theory $\text{GBC}+\Pi^1_1$-comprehension, which is itself strictly weaker in consistency strength than KM. But meanwhile, although the class choice principle is weak in consistency strength, it is not actually provable in KM; indeed, even the weak fragment $\text{CC}(\Pi^0_1)$ is not provable in KM. Those results were proved several years ago by the first two authors, but they can now be seen as consequences of the main result of this article (see corollary 15. In light of that result, however, one should perhaps not have expected to be able to prove the class Fodor principle in KM. Indeed, it follows similarly from arguments of the third author in his dissertation that if $\kappa$ is an inaccessible cardinal, then there is a forcing extension $V[G]$ with a symmetric submodel $M$ such that $V_\kappa^M=V_\kappa$, which implies that $\mathcal M=(V_\kappa,\in, V^M_{\kappa+1})$ is a model of Kelley-Morse, and in $\mathcal M$, the class Fodor principle fails in a very strong sense. In this article, adapting the ideas of Karagila to the second-order set-theoretic context and using similar methods as in Gitman and Hamkins’s previous work on KM, we shall prove that every model of KM has an extension in which the class Fodor principle fails in that strong sense: there can be a class function $F:\Ord\to\omega$, which is not constant on any stationary class. In particular, in these models, the class club filter is not $\sigma$-closed: there is a class $B\subseteq\omega\times\Ord$, each of whose vertical slices $B_n$ contains a class club, but $\bigcap B_n$ is empty. Main Theorem. Kelley-Morse set theory KM, if consistent, does not prove the class Fodor principle. Indeed, if there is a model of KM, then there is a model of KM with a class function $F:\Ord\to \omega$, which is not constant on any stationary class; in this model, therefore, the class club filter is not $\sigma$-closed. We shall also investigate various weak versions of the class Fodor principle. Definition. For a cardinal $\kappa$, the class $\kappa$-Fodor principleasserts that every class function $F:S\to\kappa$ defined on a stationary class $S\subseteq\Ord$ is constant on a stationary subclass of $S$. The class ${<}\Ord$-Fodor principleis the assertion that the $\kappa$-class Fodor principle holds for every cardinal $\kappa$. The bounded class Fodor principleasserts that every regressive class function $F:S\to\Ord$ on a stationary class $S\subseteq\Ord$ is bounded on a stationary subclass of $S$. The very weak class Fodor principleasserts that every regressive class function $F:S\to\Ord$ on a stationary class $S\subseteq\Ord$ is constant on an unbounded subclass of $S$. We shall separate these principles as follows. Theorem. Suppose KM is consistent. There is a model of KM in which the class Fodor principle fails, but the class ${<}\Ord$-Fodor principle holds. There is a model of KM in which the class $\omega$-Fodor principle fails, but the bounded class Fodor principle holds. There is a model of KM in which the class $\omega$-Fodor principle holds, but the bounded class Fodor principle fails. $\text{GB}^-$ proves the very weak class Fodor principle. Finally, we show that the class Fodor principle can neither be created nor destroyed by set forcing. Theorem. The class Fodor principle is invariant by set forcing over models of $\text{GBC}^-$. That is, it holds in an extension if and only if it holds in the ground model. Let us conclude this brief introduction by mentioning the following easy negative instance of the class Fodor principle for certain GBC models. This argument seems to be a part of set-theoretic folklore. Namely, consider an $\omega$-standard model of GBC set theory $M$ having no $V_\kappa^M$ that is a model of ZFC. A minimal transitive model of ZFC, for example, has this property. Inside $M$, let $F(\kappa)$ be the least $n$ such that $V_\kappa^M$ fails to satisfy $\Sigma_n$-collection. This is a definable class function $F:\Ord^M\to\omega$ in $M$, but it cannot be constant on any stationary class in $M$, because by the reflection theorem there is a class club of cardinals $\kappa$ such that $V_\kappa^M$ satisfies $\Sigma_n$-collection. Read more by going to the full article: V. Gitman, J. D. Hamkins, and A. Karagila, “Kelley-Morse set theory does not prove the class Fodor theorem.” (manuscript under review) @ARTICLE{GitmanHamkinsKaragila:KM-set-theory-does-not-prove-the-class-Fodor-theorem, author = {Victoria Gitman and Joel David Hamkins and Asaf Karagila}, title = {Kelley-Morse set theory does not prove the class {F}odor theorem}, journal = {}, year = {}, volume = {}, number = {}, pages = {}, month = {}, note = {manuscript under review}, abstract = {}, keywords = {under-review}, eprint = {1904.04190}, archivePrefix = {arXiv}, primaryClass = {math.LO}, source = {}, doi = {}, url = {http://wp.me/p5M0LV-1RD}, }
The set $\mathbb{Z}$ of all integers, which this book is all about, consists of all positive and negative integers as well as 0. Thus $\mathbb{Z}$ is the set given by \[\mathbb{Z}=\{...,-4,-3,-2,-1,0,1,2,3,4,...\}.\] While the set of all positive integers, denoted by $\mathbb{N}$, is defined by \[\mathbb{N}=\{1,2,3,4,...\}.\] On $\mathbb{Z}$, there are two basic binary operations, namely addition (denoted by $+$) and multiplication (denoted by $\cdot$), that satisfy some basic properties from which every other property for $\mathbb{Z}$ emerges. The Commutativity property for addition and multiplication\[\begin{aligned} a+b=b+a\\ a\cdot b=b\cdot a\end{aligned}\] Associativity property for addition and multiplication\[\begin{aligned} (a+b)+c&=&a+(b+c)\\ (a\cdot b)\cdot c&=& a\cdot (b\cdot c)\end{aligned}\] The distributivity property of multiplication over addition\[\begin{aligned} a\cdot (b+c)&=&a\cdot b+a\cdot c.\end{aligned}\] In the set $\mathbb{Z}$ there are "identity elements" for the two operations $+$ and $\cdot$, and these are the elements $0$ and $1$ respectively, that satisfy the basic properties \[\begin{aligned} a + 0 =0+a=a\\ a\cdot 1 = 1\cdot a=a\end{aligned}\] for every $a\in\mathbb{Z}$. The set $\mathbb{Z}$ allows additive inverses for its elements, in the sense that for every $a\in\mathbb{Z}$ there exists another integer in $\mathbb{Z}$, denoted by $-a$, such that \[a+(-a)=0.\] While for multiplication, only the integer 1 has a multiplicative inverse in the sense that 1 is the only integer $a$ such that there exists another integer, denoted by $a^{-1}$ or by $1/a$, (namely 1 itself in this case) such that \[a\cdot a^{-1}=1.\] From the operations of addition and multiplication one can define two other operations on $\mathbb{Z}$, namely subtraction (denoted by $-$) and division (denoted by $/$). Subtraction is a binary operation on $\mathbb{Z}$, i.e. defined for any two integers in $\mathbb{Z}$, while division is not a binary operation and thus is defined only for some specific couple of integers in $\mathbb{Z}$. Subtraction and division are defined as follows: $a-b$ is defined by $a+(-b)$, i.e. $a-b=a+(-b)$ for every $a,b\in\mathbb{Z}$ $a/b$ is defined by the integer $c$ if and only if $a=b\cdot c$.

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