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Say I have a market-making strategy that trades intraday. I start with a flat position and finish flat too. I end up with a daily P&L $p_{today}$. Over a year of trading I get $\vec{p} = (p_1,\dots,p_{252})$. There is no way to calculate returns here. As such I calculate $$Sharpe = S(\vec{p}) = \sqrt{252} \cdot \frac{\mathbb{E}[\vec{p}]}{\sqrt{\mathbb{V}[\vec{p}]}} = \sqrt{252} \cdot \frac{mean(p)}{sd(p)}$$ My questions are : Am I right to do it like this? Do you usually bootstrap your Sharpe? (I do not but I am interested in your view of it.)
I am reading the book "Numerical Recipes in Fortran 77: The Art of Scientific Computing" (Second Edition) and I came across some methods for numerical integration of 1D functions. More specifically the Gauss-Laguerre, Gauss Hermite, and Gauss Jacobi weights and abscissas appealed to me. I need them because I have some ill-behaved integrands that I need to integrate numerically. As far as I can tell there is no python implementation of these routines, the closest to these methods was this http://docs.scipy.org/doc/numpy/reference/generated/numpy.polynomial.hermite.hermgauss.html but I feel that documentation is a bit incomplete. I'm not sure what to do with this method and slightly puzzled why is there no standard method for these types of quadrature in python. EDIT: I have implemented and compared scipy quadrature with Gauss-Hermite Quadrature on the example problem: \begin{align} I = \int_{-\infty}^{\infty} \left( i\zeta(t) + \overline{i\zeta(t)} \right) \exp(i\omega t) \, \mathrm{d}t \end{align} here $i=\sqrt{-1}$ and $\zeta(t)$ is a complex function of $t$. The overbear denotes complex conjugate. $\zeta(t)$ is not a standard function, it is something I calculated and the samples can be found at http://speedy.sh/SX6JU/seData.npy as a numpy array. Each row in the array is of form [t real(zeta) imag(zeta)] i.e. the sampled real and imaginary values of the function at time t. import numpy as npimport matplotlib.pyplot as pltfrom scipy import integrateT1 = np.load('seData.npy') #: [Time, x1, x2] from downloaded filedef integrand(t, omega): x1 = np.interp(t, T1[:,0], T1[:,1]) x2 = np.interp(t, T1[:,0], T1[:,2]) z0 = x1+x21j zeta = z0/np.sqrt(z02+1) return (1jzeta+np.conjugate(1jzeta))np.exp(1jomegat) def integral(omega): def real_func(x,omega): return np.real(integrand(x,omega)) def imag_func(x,t): return np.imag(integrand(x,omega)) real_integral = integrate.quad(real_func, -80, 80, args=(omega)) imag_integral = integrate.quad(imag_func, -80, 80, args=(omega)) return real_integral[0] + 1jimag_integral[0]vintegral = np.vectorize(integral)omega = np.linspace(-30, 30, 601) I = vintegral(omega) Im not sure what the correct result should be but the "noisiness" present in the result should not be there I believe. Instead I tried a Gauss-Hermite routine: ti_her, wi_her, = np.polynomial.hermite.hermgauss(90)vintegrand = np.vectorize(integrand)def integral_hermite(omega): real_integral = np.dot(vintegrand(ti_her, omega).real, wi_her) imag_integral = np.dot(vintegrand(ti_her, omega).imag, wi_her) return real_integral + 1jimag_integralvintegral_hermite = np.vectorize(integral_hermite)I2 = vintegral_hermite(omega) I believe this is more in line with what I should be getting , however I am not sure I have done the correct thing. It seems the order of Gauss-hermite is very high and at low orders (90) it looks like my result shows some type of periodicity/aliasing.
I'm trying to simulate the growth of a brain tumor using a 3d reaction-diffusion model $ \partial_{t}u = \nabla.{(D\nabla{u}) } + ku.(1-u)$ , knowing the initial distribution of tumor $u^0$, the non-homogeneous diffusion coefficient $D$ and the non-homogeneous net-proliferation map $k$. So I first tried to proceed to simulation simply by dividing it into small time steps and directly simulate the process by simply replacing the derivatives what is appearing to me now as forward Euler finite difference. I got serious instability problems which, if I understand well, are due to the difference scheme I'm using. So I'm trying to learn how to correctly code finite difference algorithms. At this point there's something I don't understand and I hope someone will be able to explain me where I'm wrong. Let forget the proliferation term and do simply a 1D diffusion ( $ \partial_{t}u = \partial_{x}(D\partial_{x}u) $ ) using a Crank-Nicolson scheme for the time difference and a central difference for the spatial derivatives. So, if $D$ depends on $x$, we have $ u^{n+1}_{i} = u^{n}_{i} + \frac{\Delta{t}}{2} \{ \partial_{x}(D\partial_{x}u^{n}) + \partial_{x}(D\partial_{x}u^{n+1}) \} $ which develop to $ \Rightarrow ( 1 - \frac{\Delta t}{2}\partial_{x}(D\partial_{x}\cdot) ) u^{n+1}_{i} = ( 1 + \frac{\Delta t}{2}\partial_{x}(D\partial_{x}\cdot) ) u^{n}_{i}$ and using $ \partial_{x}(D\partial_{x}V) = \partial_{x}D.\partial_{x}V + D.\partial_{xx}V $, $\partial_{x}V = \frac{V_{i+1}-V_{i-1}}{2\Delta x}$ and $\partial_{xx}V = \frac{V_{i+1}+V_{i-1}-2V_{i}}{{\Delta x}^2}$ one get $ \partial_{x}(D\partial_{x}u_{i}) = \frac{(D_{i+1}-D_{i-1}).(u_{i+1}-u_{i-1})}{4{\Delta x}^2} + \frac{D_{i}.(u_{i+1}+u_{i-1}-2u_{i})}{{\Delta x}^2}$ ordering it according to the spatial indice of $u$ , one get $ \partial_{x}(D\partial_{x}u_{i}) = \frac{1}{4{\Delta x}^2}.\{ (D_{i+1}-D_{i-1}+4D_{i})u_{i+1} + 8D_{i}u_{i} - (D_{i+1}-D_{i-1}-4D_{i})u_{i-1} \} $ In matrix form, this can be written $ \partial_{x}(D\partial_{x}u) = \tilde{D}u $ with $ \tilde{D} \equiv \frac{1}{4{\Delta x}^2}\begin{bmatrix} -8D_1 & (D_2+4D_1) & 0 & \cdots \\ -(D_3 - D_1 -4D_2) & -8D_2 & (D_3-D_1+4D_2) & \cdots \\ \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & -(- D_N -4D_{N-1}) & -8D_N \end{bmatrix} $ Therefore, one have to solve, at each time step, the equation $ (1 - \frac{\Delta t}{2}\tilde{D})u^{n+1} = (1 + \frac{\Delta t}{2}\tilde{D})u^{n} $ my problem is that on every course/paper I read on this subject, the authors are explaining that one have to solve the equation at each time step which is the time consuming step. However, as it is written here at least, the matrix $\tilde{D}$ is completely determined from the (fixed in time) $D$ distribution and it looks to me that one can just simply generate it, then calculate the total matrix $ M = (1 - \frac{\Delta t}{2}\tilde{D})^{-1} \cdot (1 + \frac{\Delta t}{2}\tilde{D})$ and then simply calculate $u^n = M^nu^0 $ which is very fast. I tried to code it in MATLAB, and indeed, doing it step by step, or just at once gives me the same results ( and which are coherent with a simple Gaussian blurring with $\sigma = \sqrt{2D.T} $ in the case of an homogeneous D, just to check the validity of the simulation). So i must have missed something but what ? Maybe finite difference is not the tool to use in case of diffusion-only and are actually used because of the reaction term, but still, if I understand well, one can separate the simulation into smaller sub-time steps to perform only diffusion or only proliferation (solved analytically for the later) with Douglas-Gunn or equivalent approaches. In such a case, the diffusion step can still be solved using a simple matrix multiplication at each step which do not require a long processing ? So I would be very happy if someone can tell me if I'm going on the right track or if (as I suspect) I missed any fundamental issue... Many thanks ! D. Guez
This question has two parts, as I do not understand whether my problem is theoretical (identification of the parameters) or practical (insufficient R skills). Econometrics Most "probit" style models are identified through a normalization of the standard error to one. In my case, I would argue that this is not necessary as the order of magnitude is already set by fixing one coefficient to 1. More specifically, for each observation there is a dummy variable equal to one if the latent index is above a (observation-specific and observed) threshold : $$d_i=1 \text{ iff } K_i > X_i \beta + \epsilon_i$$ The error term is assumed $\epsilon_i \thicksim \mathcal{N}(0, \sigma)$. In my naive understanding, the likelihood of this problem should somehow look like (where $\Phi$ is the standard normal cdf): $$L= \prod_{i=1}^N \Phi \left(\frac{K_i - X_i \beta}{\sigma} \right)^{d_i} \Phi \left(\frac{X_i \beta -K_i}{\sigma} \right)^{1-d_i} $$ Is it possible to estimate $\sigma$ without further normalization? Estimation If the answer to previous part is "yes" -- then why does my R implementation not work? ### simulate dataset.seed(5849)N <- 2000b.cons <- 8 b.x <- 10sig <- 2x <- cbind(rep(1, N), runif(N)) #"observed variables"e <- rnorm(N, sig) # "unobserved error"k <- runif(N)10+8 # threshold: something random, but high enough to guarantee some variation in it <- x%%c(b.cons, b.x)+ei <- 1(k>t) #participation dummy### likelihood functionprobit.sim <- function(params, I, K, X) { params[1:2] -> b params[3] -> s z= (K-X%%b)/s pr.1 = pnorm(z) pr.1[pr.1==0] <- 0.001 #seems somehow weird to me, but how is this problem usually treated?? pr.1[pr.1==1] <- 0.999 pr.0 = 1-pr.1 llik = t(I)%%log(pr.1) + t(1-I)%%log(pr.0) -llik}### maximize likelihoodoptim(c(1,1,1), probit.sim, I = i, K = k, X = x) #using a random starting vectorst <- coef(lm(k*(1-i) ~ x-1)) #searching for better standard valuesoptim(c(st, 1), probit.sim, I = i, K = k, X = x) The estimated parameters are clearly not c(8, 10, 2) as they should be. I asked a related question on already on stack overflow, and the answer was "take better starting values", but this does not seem to do the trick here. Or maybe I don't know how to do it right. Any ideas? Alternative approach My alternative was to use standard statistical software and estimate a probit (needs a bit of twisting but should be possible to make it equivalent). This estimates a coefficient for K, which should be equal to $-1/\sigma$; how about taking this $\sigma$ and computing the "non-normalized"/"true" values of the other coefficients? Many thanks in advance for any suggestion on any of these 3 parts.
There are $$$n$$$ persons who initially don't know each other. On each morning, two of them, who were not friends before, become friends. We want to plan a trip for every evening of $$$m$$$ days. On each trip, you have to select a group of people that will go on the trip. For every person, one of the following should hold: Note that the friendship is not transitive. That is, if $$$a$$$ and $$$b$$$ are friends and $$$b$$$ and $$$c$$$ are friends, it does not necessarily imply that $$$a$$$ and $$$c$$$ are friends. For each day, find the maximum number of people that can go on the trip on that day. The first line contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$2 \leq n \leq 2 \cdot 10^5, 1 \leq m \leq 2 \cdot 10^5$$$, $$$1 \le k < n$$$) — the number of people, the number of days and the number of friends each person on the trip should have in the group. The $$$i$$$-th ($$$1 \leq i \leq m$$$) of the next $$$m$$$ lines contains two integers $$$x$$$ and $$$y$$$ ($$$1\leq x, y\leq n$$$, $$$x\ne y$$$), meaning that persons $$$x$$$ and $$$y$$$ become friends on the morning of day $$$i$$$. It is guaranteed that $$$x$$$ and $$$y$$$ were not friends before. Print exactly $$$m$$$ lines, where the $$$i$$$-th of them ($$$1\leq i\leq m$$$) contains the maximum number of people that can go on the trip on the evening of the day $$$i$$$. 4 4 2 2 3 1 2 1 3 1 4 0 0 3 3 5 8 2 2 1 4 2 5 4 5 2 4 3 5 1 4 1 3 2 0 0 0 3 3 4 4 5 5 7 2 1 5 3 2 2 5 3 4 1 2 5 3 1 3 0 0 0 0 3 4 4 In the first example, In the second example, In the third example, Name
2fFCS is an abbreviation 2 Focus Fluorescence Correlation Spectroscopy. Dual focus FCS might also be used. Correlation analysis that is applied to the fluctuations of the fluorescence intensity. The cross-correlation between the occurrence of events in two different detection volumes is used to introduce an absolute diffusion length into the analysis. see FCS for a introduction on FCS. The conjugated pinhole size should be chosen slightly larger than the excitation spot diameter. The conjugated pinhole size is the pinhole diameter divided by the magnification of the objective. e.g: Usually the size of the confocal volume or the diameter of the excitation spot is given as its FWHM. However the airy disc diameter is bigger and the 1/e2 diameter must be used. $2ln(2)=\frac{FWHM^2}{w_0^2}$ $w_0=0.849 \cdot FWHM$ The diameter of the excitation spot therefore is: $ d_{exc}=2 \cdot w_0 = 594 nm $ Together with the 60x magnification of the objective the airy disc at the pinhole location is $ d_{PH}=60\cdot 594nm = 35.6\mu m $ This equals 1 airy unit (AU) Occasionally, pinhole size can be used to adjust amount of photons to change the signal intensity and increase SNR. In addition to the “optimal” 1 AU, Pinhole 1-3 AU is the range of choice. Bigger pinhole give you stronger signal but with the compromised confocal effects. PicoQuant usually uses a 50 micron pinhole (1.4 AU) to get maximum detection efficiency, somewhat sacrificing background rejection. The distance between the foci in two-focus FCS is determined by the Nomarski prism. Usually the distance is around 400 nm. Therefore the 2 foci (separated by 400 nm) will occupy: $ d_{exc, 2fFCS} = 2 \cdot w_0 + 400 nm = 994 nm $ Together with the 60x magnification of the objective both airy discs together, at the the pinhole location, have a diameter $ d_{PH, 2fFCS}=60\cdot 994nm =59.6\mu m $
Current browse context: astro-ph.EP Change to browse by: References & Citations Bookmark(what is this?) Astrophysics > Earth and Planetary Astrophysics Title: Planet Hunters TESS I: TOI 813, a subgiant hosting a transiting Saturn-sized planet on an 84-day orbit (Submitted on 19 Sep 2019) Abstract: We report on the discovery and validation of TOI-813b (TIC 55525572 b), a transiting exoplanet identified by citizen scientists in data from NASA's Transiting Exoplanet Survey Satellite (TESS) and the first planet discovered by the Planet Hunters TESS project. The host star is a bright (V = 10.3 mag) subgiant ($R_\star=1.94\,R_\odot$, $M_\star=1.32\,M_\odot$). It was observed almost continuously by TESS during its first year of operations, during which time four individual transit events were detected. The candidate passed all the standard light curve-based vetting checks, and ground-based follow-up spectroscopy and speckle imaging enabled us to statistically validate the planetary nature of the companion. Detailed modelling of the transits yields a period of $83.8911_{ - 0.0031 } ^ { + 0.0027 }$ days, a planet radius of $6.71 \pm 0.38$ $R_{\oplus}$, and a semi major axis of $0.423_{ - 0.037 } ^ { + 0.031 }$ AU. The planet's orbital period combined with the evolved nature of the host star places this object in a relatively under-explored region of parameter space. We estimate that TOI-813b induces a reflex motion in its host star with a semi-amplitude of $\sim6$ ms$^{-1}$, making this system a promising target to measure the mass of a relatively long-period transiting planet. Submission historyFrom: Nora L. Eisner Miss [view email] [v1]Thu, 19 Sep 2019 17:00:03 GMT (2880kb,D)
Generally speaking, your $\delta x_i$ will be a manual estimate of the error in measurement $x_i$ - e.g. this could be the smallest grid size on your ruler. This is a rather rough estimate since it presumes very little about your measurement methodology and is reliant on the experimenter's best judgement. $S$ (usually we use the symbol $\sigma$ here) on the other hand is measuring how consistent your measurements of $x$ are. The presumption here is that the less consistently you can measure $x$, the larger $S$ is and the larger the error is. Maybe a concrete example will help here. So let's say you are measuring the length of a table while standing on a ship that rocks back and forth randomly. You have a ruler that is accurate down to the millimeter. One way an experimenter might report $\delta x_i$ is then $\delta x_i=1\text{mm}~\forall x_i$. The rms of this would again be $\bar{\delta x}=1\text{mm}$ . However, this is probably an underestimate of the error. How else should the experimenter report his error though? Eyeballing how far the ruler moves around during his measurements? Such efforts are difficult. This is when we introduce $S$. We let the experimenter measure the length of the table 100 times, we take the average of those measurements as the length of the table and then we take the measure of how consistently that experimenter will get some value - this is exactly what $S$ measures - as the error in that measurement. Perhaps even though the ruler the experimenter has has guides at every $1\text{mm}$, he gets values that are spread out such that $S=10\text{mm}$. Hopefully it is clear now that this latter value is a better estimate of the error. Now, why isn't $\delta x_i$ even considered when computing $S$? Well, presumably $S$ includes contributions that $\delta x_i$ would give. If you have a ruler that is only accurate to $1\text{mm}$, presumably over many measurements, your measurements will be spread out so that $S\geq 1\text{mm}$. Of course, one can conceive of instances where this might fail. It is of course the experimenter's perogative to report the best estimate of error he can. And if in some instances $\delta x_i$ is a better measure, then he should report those. But, generally speaking, $S$ is just a better measure overall.
Nemtsov function The Nemtsov function $y=\mathrm{Nem}_q(x)= x+x^3+q x^4$ is shown in figure at right versis $x$ for various $q\!\ge\!0$. Complex map of $\mathrm{Nem}_q$ is shown in figure 2 for $q\!=\!0$ and in figure 3 for $q\!=\!2$ with lines $u\!=\!\mathrm{const}$ and lines $v\!=\!\mathrm{const}$ assuming that $u\!+\!\mathrm i v=\mathrm{Nem}_0(x\!+\!\mathrm i y)$ . Contents Definition Let $q$ be non–negative real parameter. Then, Nemtsov Function $\mathrm{Nem}$ is defined for complex argument $z$ as follows: (1)$~ ~ ~ ~ ~ ~ \mathrm{Nem}_q(z)= z+z^3+q z^4$ Notations For Function $~ \mathrm{Nem}_q~$ by equation (1), at $q\!>\!0$, the algorithms, described in the first edition of the Russian version of the book Суперфункции cannot be applied "as is", and some modification, generalisation is required. The need of this modification had been revealed 2015.02.27, in the day, when Putin killed Nemtsov. Since February to July of 2015, no other scientific concept of that murder hat been suggested. Apparently, the total corruption in Russia does not allow the professional to investigate that case, and that terroristic act caused many publications. This makes name Nemtsov to be a good mark, label on the timeline of Hunan History. For this reason, function $\mathrm{Nem}_q$ by equation (1) is called after the name of Boris Nemtsov. In this article, properties of the Nemtsov function are considered, and also some properties of the related functions: Inverse function, denoted with ArqNem, $\mathrm{Nem}_q(\mathrm{ArqNem}_q(z))=z$ $\mathrm{SuNem}_q(z\!+\!1)=\mathrm{Nem}_q(\mathrm{SuNem}_q(z))$ $\mathrm{SuNem}_q(\mathrm{Nem}_q(z))=\mathrm{SuNem}_q(z)+1$ and the corresponding iterates $\mathrm{Nem}_q^n(z)=\mathrm{SuNem}_q\big(n+ \mathrm{SuNem}_q(z)\big)$ The inverse function is called ArqNem. This name allows to distinguish it from other inverse functions of the Nemtsov function. Two other inverse functions are called ArcNem and ArkNem. A priori, it had been difficult to guess, that namely ArqNem happens to be suitable for construction of the corresponding Abel function; so, all the three versions had been assigned (designated) the different names. These inverse functions have different positions of the cuts of the range of holomorphism. The Abel function for the Nemtov function is called AuNem, to indicate, that it is constructed by the exotic iterates at the fixed point zero, that is maximal (Upper) among the fixed points of the Nemtsov function. While no other iterates of the Nemtsov function are presented no special mark is used to denote the iterates $\mathrm{Nem}_q^n$. Later, perhaps, one additional subscript $_{\mathrm u}$ will be added to the notation, in order to distinguish this iterate from other iterates, constructed, for example, using the asymptotic behaviour of the Nemtsov function (and its iterates) at infinity. General properties For real $q$, the Nemtsov function is real holomorphic in the whole complex plane; $\mathrm{Nem}_q(z^)=\mathrm{Nem}_q(z)^$ At least for positive values of the argument, the Nemtsov function grows monotonously. The monotonous growth has also the inverse function ArqNem, the Abel function AuNem and the real iterates of the Nemtsov function. As for any non–trivial entire function, the inverse function $\mathrm{ArqNem}_q$ has the branch points. Two of them are complex; and one of them is expressed with function NemBran; at $z=\mathrm{NemBran}(q)$, function $\mathrm{ArqNem}_q(z)$ has infinite derivative. For construction of the inverse function, important are the complex soluitons $A$ of equation $\mathrm{Nem}_q^{\prime}=0$. One of these solutions is expressed with function NemBra, id est, $A=\mathrm{NemBra}(q)$. Parametric plot of this function is shown in figure 4: $x=\mathrm{Re}\big(\mathrm{NemBra}(q)\big)~$ $y=\mathrm{Im}\big(\mathrm{NemBra}(q)\big)~$ For positive $q$, both real and imaginary parts of the branchpoint are significantly smaller than unity. Inverse function The Nemtsov function has various inverse functions, as equation $\mathrm{Nem}_q(A)=z$ has many solutions $A$ ; there are three solutions for $q\!=\!0$ and four for $q\!\ne\!0$. The holomorphic inverse function unavoidably has the cut lines. The appropriate choice of these cut lines happens to be non–trivial problem. For construction of the Abel function, denoted as AuNem, the special inverse function ArqNem of the Nemtsov function is chosen. For function ArqNem, the cut lines are chosen in the following way. One cut line goes along the negative part of the real axis from $-\infty$ to zero. Two other cut lines go straightly from zero to the complex branchpoints $Z_o$ of the inverse function. These brach points can be expressed in the following way: $Z_{\mathrm o}=\mathrm{Nem}_q(z_{\mathrm o})$ where $z_{\mathrm o}=\mathrm{NemBran}(q)$ is solution of equation $\mathrm{Nem}_q^{\prime}( z_{\mathrm o})=0$ The following relation takes place: $\mathrm{NemBran}(q)=\mathrm{Nem}_q\big(\mathrm{NemBra}(q)\big)$ $x=\mathrm{Re}\big(\mathrm{NemBran}(q)\big)~$, $y=\mathrm{Im}\big(\mathrm{NemBran}(q)\big)~$ At small values of the argument, the Nemtsov function looks similar to the identity function; so, the parametric plot of function NemBran in figure 5 looks similar to that of function NemBra in figure 4. $u\!+\!\mathrm i v= \mathrm{ArqNem}_q(x\!+\!\mathrm i y)$ The complex double implementation of function ArqNem is loaded as arqnem.cin ; parameter $q$ is storem in the global variable $Q$. Before evaluation of $\mathrm{ArqNem}_q$ of complex argument, the complex branch point should be evaluated with routine nembran.cin and stored in the global variables tr and ti; in the version from year 2015, the real and imaginary parts of the branch point are stored as two global variables. At any change of parameter $q$, these values should be recalculated. Superfunction $F(z\!+\!1)=\mathrm{Nem}_q\big( F(z)\big)$ with specific asymptotic behaviour at infinity, namely, $\displaystyle F(z)=\frac{1}{\sqrt{-2 z}}\left( 1-\frac{q}{\sqrt{-2 z}} + O\big( \ln(-z)/z\big) \right)$ In order to specify function SuNem, the additional condition is assumed: $\mathrm{SuNem}_q(0)=1$ Explicit plot of function SuNem of the real argument is shown in figure at left, $y\!=\!\mathrm{SuNem}_q(x)$ is plotted versus $x$ for various values of $q$. Function SuNem grows monotonously from zero at $-\infty$, takes value unity at zero and then grows quickly to infinity for positive values of the argument. The larger is parameter $q$, the faster is the growth at $+\infty$. This behaviour corresponds to the intuitive expectations about this function. Complex maps of function SuNem are shown in figures at right for $q\!=\!0$, $q\!=\!1$ and for $q\!=\!2$; $u\!+\!\mathrm i v=\mathrm{SuNem}_q(x\!+\!\mathrm i y)$ Maps for different values of $q$ look similar; however, the greater is $q$, the faster is the growth of $\mathrm{SuNem}_q$ along the real axis. This is seen also at the explicit plot in figure at left. In order to construct function SuNem, first, any superfunction $F$ with appropriate asymptotic behaviour is constructed; then, $\mathrm{SuNem}_q(z)=F(x_1+z)$ where $x_1$ is real solution of equation $F(x_1)=1$. Abel function Explicit plot $y\!=\!\mathrm{AuNem}_q(x)$ versus $x$ is shown in figure at left for $q\!=\!0$, $q\!=\!1$ and $q\!=\!2$. The same plot can be obtained, reflecting the curves for SuNem from the bisektris of the First quadrant of the coordinate plane. This property can be used as a mumerical test of relation $\mathrm{SuNem}_q(\mathrm{АuNem}_q(x))=x$; it should be so, as $\mathrm{AuNem}_q=\mathrm{SuNem}_q^{-1}$ $\mathrm{AuNem}_q\big( \mathrm{Nem}_q(z)\big)=\mathrm{AuNem}_q(z)+1$ Function AuNem satisfies also the additional condition $\mathrm{AuNem}_q(1)=0$ that is determined by the corresponding property of the SuNem, namely, that $\mathrm{SuNem}_q(0)\!=\!1~$. The asymptotic expansion of function AuNem at zero can be found, inverting the asymptotic expansion of function SuNem at $-\infty$; the coeffcieints of this expansion can be found also from the Abel equation. The expansion can be written as follows: Iterates of the Nemtsov function $\mathrm{Nem}_q^{\,n}(z)=\mathrm{AuNem}_q\big( n+\mathrm{AuNem}_q(z)\big)$ For $q\!=\!0$, $q\!=\!1$ and for $q\!=\!2$, these iterates are shown in figures at right. These iterates look similar to other iterates of the fast growing functions. At positive number $n$ of iterate, the iterate shows the fast growth; at negative values of $n$, the growth is slow. The zeroth iterate ($n\!=\!0$), the iterates apperas as identity function. In the figures, the corresponding graphic is marked with green line. Due to the singularity of function ArqNem at zero, the non–integer iterates are not defined at zero and the negative part of the real axis, although they approach zero as the positive argument of the iterate becomes small. The curves of the iterates show the symmetry with respect to reflections from the bisectris of the First quadrant of the coordinate plane, as $\mathrm{Nem}_q^{-n}(z)=\mathrm{ArqNem}_q^{\,n}(z)$ Applivation and the Discussion The Nemtsov function is considered as a candidate of the transfer function, for which the superfunction and the Abel function are diffucult to construct with the exotic iterate at its fixed point zero. Indeed, the construction required certain efforts, but they sere related mainly with construction of the inverse function ArqNem, the efficient implementation and, especially, with guessing, that namely ArqNem shouls be used to get the Abel function and non–integer iterates, that are holomorphic in the most of the complex plane. For the Nemtsov function, two other inverse functions have been constructed, they are denoted with hames ArcNem and ArkNem. They have different positions of the cut lines, and, at the iterates, do not provide the holomorphic Abel function. Once the iterates of the inverse function leads to the fixed point of the transfer function, the exotic iterates are straightforward. Other transfer functions can be considered in the similar way, assuming, that their expansion at the fixed point begins with the identity function and the cubic term. Without loss of generality, the coefficient at the cubic term can be treated as unity; the corresponding transform to this case is shown in the last row of the Table of superfunctions. References http://mizugadro.mydns.jp/2016NEMTSOV/TRY00/23.pdf Dmitrii Kouznetsov. Nemtsov function and its iterates. 2016, in preparation.
As you say yourself, indeed every connection on a bundle is locally given by a Lie algebra valued 1-form and in general only locally. Let's say this more in detail: for $X$ any manifold, a $G$-principal connection on it is (in "Cech data"): a choice of good open cover $\{U_i \to X\}$; on each patch a 1-form $A_i \in \Omega^1(U_i)\otimes \mathfrak{g}$; on each double intersection of patches a gauge transformation function $g_{i j} \in C^\infty(U_i \cap U_j, G)$ such that on each double intersectin $U_i \cap U_j$ we have the equation $A_j = g_{i j}^{-1} A g_{i j} + g_{i j}^{-1} \mathbf{d} g_{i j}$ on each triple intersection $U_i \cap U_j \cap U_k$ we have the equation $g_{i j} g_{j k} = g_{i k}$. Okay, now you would like to form a Chern-Simons 3-form... something out of this. What you immediately get from the above data is a bunch of local differential 3-forms, one on each patch: $CS(A_i) \in \Omega^3(U_i)$. To make these 3-forms globally glue together to what is called a 3-form connection we need the evident data of higher gauge transformation on each patch we have the local 3-form $CS(A_i)$; on each double intersection there should be a 2-form $B_{i j} \in \Omega^2(U_i \cap U_j)$ which gauge transforms the respective CS-3-forms into each other, by $CS(A_j) = CS(A_i) + \mathbf{d} B_{i j}$; on each triple intersection there should be a 1-form $\alpha_{i j k} \in \Omega^1(U_i \cap U_j \cap U_k)$ which exhibits a second-order gauge transformation ("ghosts of ghosts"!) between the first order gauge trasformations, in that $B_{i j} + B_{j k} = B_{i k} + \mathbf{d} \alpha_{ i j k}$ finally on each quadruple intersection there should be a smooth function $h_{i j k l} \in C^\infty(U_i \cap U_j \cap U_k \cap U_l, U(1))$ which gauge-of-gauge-of-gauge-transforms the gauge-of-gauge-transforms into each other, in that $\alpha_{i j k} + \alpha_{i k l} = \alpha_{j k l} + \alpha_{i j l} + h_{i j k l}^{-1}\mathbf{d}h_{i j k l}$. That's the data that makes the local Chern-Simons 3-form into a globally well-defined 3-form field. (For instance the supergravity C-field is of this form, with some further twists and bells and whistles added, as we have discussed here). In mathematical language one says that this kind of local gauge-of-gauge-of-gauge gluing data for global definition of higher form fields is a "degree-4 cocycle in Cech-Deligne cohomology". This is precisely the right data needed to have a well-defined 3-dimensional higher holonomy as is needed here for the definition, because the Chern-Simons action functional is nothing but the 3-dimensional higher holonomy of this 3-form connection. If you can build it, that is. From the above it is not entirely obvious how to build the 3-form cocycle data $\{CS(A_i), B_{i j}, \alpha_{i j k}, h_{i j k}\}$ from the given gauge field data $\{A_i, g_{i j}\}$. But this can be done. This is what Cheeger-Simons differential characters were discovered for. An explicit construction that is very natural for the application to Chern-Simons theory we have given in Fiorenza, Schreiber, Stasheff, Cech cocycles for differential characteristic classes, Advances in Theoretical and Mathematical Phyiscs, Volume 16 Issue 1 (2012) (arXiv:1011.4735, web) Based on this we give a detailed introduction to and discussion of Chern-Simons action functionals for globally non-trivial situations like above in Fiorenza, Sati, Schreiber, A higher stacky perspective on Chern-Simons theory (arXiv:1301.2580, web) That article gives the local formulas that apply generally, discusses the simplifications that occur when the 3-manifold can be assumed to be bounding, discusses what happens if not, and then explores various other properties of globally defined Chern-Simons theory, such as how to couple Wilson lines to the above story. If you just look at the first part, I think you should find what you need. edit: In the comments below came up the question why a similar discussion is not also needed when writing down the Yang-Mills action functional, whose Lagrangian is the 4-form $\langle F_A \wedge \star F_A \rangle$ (where $\star$ is the Hodge star of a given metric (gravity) and $\langle -,-\rangle$ is an invariant polynomial, the "Killing form" or trace), or similarly the topological Yang-Mills action functional, whose Lagrangian is the 4-form $\langle F_A \wedge F_A \rangle$. The reason is that these Lagrangians are built from curvatures evaluated in an invariant polynomial. The very invariance of these invariant polynomials under the adjoint action of the gauge group on its arguments ensures that if $\{U_i \to X\}$ is a good open cover of 4-dimensional space(-time) and if the gauge field is given by the Cech-cocycle data $\{A_i, g_{i j}\}$ with respect to these local patches, that then on double overlaps the two (topological or not) Yang-Mills Lagrangians coming from two patches are already equal $$ \langle F_{A_i} \wedge F_{A_i}\rangle = \langle F_{A_j}\wedge F_{A_j}\rangle \,.$$ Hence if we write $\nabla = \{A_i, g_{i j}\}$ for the gauge field connection abstractly and denote the (topological) Yang-Mills Lagrangian globally by $\langle F_\nabla \wedge F_\nabla\rangle$, then this is already a globally defined 4-form. Mathematically, this statement is what is at the core of Chern-Weil theory. Notice that there is nevertheless an intricate relation to the story of the Chern-Simons functional. Namely the local Chern-Simons form $CS(A_i)$ has the special property (essentially by definition) that its differential is the topological Yang-Mills Lagrangian: $$ \mathbf{d}CS(A_i) = \langle F_{A_i} \wedge F_{A_i}\rangle \,.$$ This means that with the Chern-Simons Lagrangian regarded as a 3-form connection then the topological Yang-Mills Lagrangian is its curvature 4-form. Therefore the relation between the topological Yang-Mills Lagrangian 4-form and the Chern-Simons 3-form is precisely an analogue in higher gauge theory of the familiar relation two degrees down of how the electromagnetic potential 1-form -- which is not globally defined in general -- has a curvature 2-form that is globally well defined. Mathematically this is why Chern-Simons functionals are called "secondary invariants" Indeed, this is a bit more than just an analogy: the Chern-Simons 3-form is precisely a doubly higher analog of the electromagnetic field as we pass from the point, via the string, to the membrane. I have some lecture notes with more along these lines at nLab:twisted smooth cohomology in string theory.This post imported from StackExchange Physics at 2014-04-05 03:54 (UCT), posted by SE-user Urs Schreiber
I want to numerically solve the diffusion equation $\partial_t u = D \partial_x^2 u$ in Fourier space, and can think of multiple ways to do it. Setup Option 1 Differentiating $u$ twice in Fourier space brings down two factors of $ik$ from the exponential $e^{i k x}$. $$\partial_t \tilde{u}_k = D (ik)^2 \tilde{u}_k$$ which can be discretised, for example, like: $$\frac{\tilde{u}^{n+1} - \tilde{u}^n}{\Delta t} = -Dk^2\left(\frac{\tilde{u}^{n+1} + \tilde{u}^{n}}{2} \right)$$ Option 2 Taking the second order finite difference equation in real space: $$\partial_x^2 u \approx \frac{u_{i+1} - 2 u_i + u_{i-1}}{\Delta x^2}$$ where $u_{i+1}$ and $u_i$ are $\Delta x $ apart. Fourier transforming this, remembering that phase shift of $\Delta x$ is given by $e^{\Delta x k}$ we get: $$\partial_x^2 \tilde{u}_k \approx \frac{e^{+\Delta x k} -2 + e^{-\Delta x k} }{\Delta x^2} \tilde{u}_k = 2 \frac{\cos(\Delta x k) - 1}{\Delta x^2} \tilde{u}_k\qquad(1)$$ Applying again to the diffusion equation: $$\frac{\tilde{u}^{n+1}_k - \tilde{u}^n_k}{\Delta t} = D\frac{2 (\cos(\Delta x k)-1)}{\Delta x^2} \left(\frac{\tilde{u}^{n+1}_k + \tilde{u}^{n}_k}{2} \right)$$ I came across Option 2 in a few places, for example, in this paper. Question Which of these (if either) should I be using, and why? I'm guessing the discrete nature of the FFT is important, as the approaches are equivalent as $\Delta x \to 0$, seen by Taylor expanding Eq. (1): $$2 \frac{\cos(\Delta x k) - 1}{\Delta x^2} = 2 \frac{1 - \frac{(\Delta x k)^2}{2} + \mathcal{O}(\Delta x^4) - 1}{\Delta x^2} = -k^2 + \mathcal{O}(\Delta x^2)$$ Any help is greatly appreciated! Update Thinking more about this, the properties of the DFT are actually important. The maximum $k$ is $k_{max} =2\pi (N/2)$ where $\Delta x = L / N$. So as $\Delta x \to 0$, $k_{max} \to \infty$. Therefore the two methods are different for large $k$ at any $\Delta x$: This looks like the latter option is applying some smoothing at high $k$.
While the analytic definition of a continuous function involves open sets, we can define continuous functions in a completely equivalent way with closed sets. We can define a function between metric (or topological) spaces $f:X \rightarrow Y$ to be continuous if $f^{-1}(C)$ is a closed set if $C$ is. Typically we avoid closed sets when talking about differentiable functions. Open sets are nice because points are relatively homogenous in them. Locally about any point, the behavior of the set is similar. We can always find an open disk lying in our set about any point. However, on closed sets, we cannot always do this. Any point on the boundary has a different local picture, as any disk containing the point cannot lie solely in this set. But why is this a problem? Well, in the definition of a real differentiable function of a real variable, we consider the quotient$$\lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x-x_0}.$$Ideally, if we are working in the set $C$, we would want each $x$ that we are limiting over to be in $C$. Otherwise we would have to refer to the ambient space that $C$ lies in, and it would be a hassle. If $C$ is open, this is fine due to the discussion above. However, if $C$ is not open, it may be rather strange. What if $x_0$ is an isolated point of $C$ for example? I.e. there is no sequence of points in $C - \{x_0\}$ that approach $x_0$? How do we define the limit here? We cannot simply plug in $x_0$ into the difference quotient as it is not defined there. All in all, this makes it tedious and difficult to talk about differentiability on such sets. Instead, we can just ignore all of these pathologies by concerning ourselves with only open sets.
[Scroll to graphic below if math doesn’t render for you] Thanks to Mark Andrews for correcting some crucial typos (I hope I got it right this time!). Thanks also to Andrew Gelman for pointing out that the proof below holds only when the null hypothesis is a point null $H_0: \mu = 0$, and the dependent measure is continuous, such as reading time in milliseconds, or EEG responses. Someone asked this question in my linear modeling class: why is it that the p-value has a uniform distribution when the null hypothesis is true? The proof is remarkably simple (and is called the probability integral transform). First, notice that when a random variable Z comes from a $Uniform(0,1)$ distribution, then the probability that $Z$ is less than (or equal to) some value $z$ is exactly $z$: $P(Z\leq z)=z$. Next, we prove the following proposition: Proposition: If a random variable $Z=F(T)$, then $Z \sim Uniform(0,1)$. Note here that the p-value is a random variable, call it $Z$. The p-value is computed by calculating the probability of seeing a t-statistic or something more extreme under the null hypothesis. The t-statistic comes from a random variable $T$ that is a transformation of the random variable $\bar{X}$: $T=(\bar{X}-\mu)/(\sigma/\sqrt{n})$. This random variable T has a CDF $F$. So, if we can prove the above proposition, we have shown that the p-value’s distribution under the null hypothesis is $Uniform(0,1)$. Proof: Let $Z=F(T)$. $P(Z\leq z) = P(F(T)\leq z) = P(F^{-1} F(T) \leq F^{-1}(z) ) = P(T \leq F^{-1} (z) ) = F(F^{-1}(z))= z$. Since $P(Z\leq z)=z$, Z is uniformly distributed, that is, Uniform(0,1). A screengrab in case the above doesn’t render:
How do I calculate the distance to touchdown when the glideslope angle is different from 3°? For example, you are 670ft AGL on a 3.2° glideslope approach; what do I change in the formula to calculate the distance? Just in case you don't have a calculator at hand (yeah, I know...): WolframAlpha returns (unsurprisingly): 1.97nm (and lot of other entertaining info). Sidenote: Because $\tan(\alpha)\approx\alpha~for~small~\alpha$, you can use a simple linear approximation in this case. Your new glide distance will be more or less $\frac{3}{3.2}=\frac{15}{16}=0.9375$ times the original one. The $300ft/NM$ used to calculate your height above the runway at a particular distance is just an approximation that is easy to remember, and is close enough for our purposes. If you want a more exact number, you can use some trigonometry: $$\tan(3)=\frac{xft}{6,076ft/NM}=318ft/NM$$ (the "real" number). For 3.2 degrees, it would be: $$\tan(3.2)=\frac{xft}{6,076 ft/NM}=340ft/NM$$ So if you want to know the appropriate altitude for, say a 3 mile final on a 3.2 degree glideslope, it would be: $$Runway~Touchdown~Elevation+340ft/NM\times3NM=TDE+1,020ft$$ In your specific example where you already know the altitude and want to know when to start the descent, the answer would be $2NM$: $$\frac{670ft}{340ft/NM}=2.0NM$$ ($1.97$ if you want to be REALLY exact.) Since the 1 in 60 rule of thumb yields a solution of 300 feet for every NM from the touchdown point for a 3 degree slope, use 320 feet for a 3.2 degree slope. So 670ft, gives you a nadgers whisker over 2 nm. To be a little more precise, 2.09375 nm ($\frac{670}{320}$) $$tan(3.2)=\frac{670}{a}$$ $$\frac{670}{\tan(3.2)}=a$$ $$a=11,983ft$$ $$11,983ft=1.97NM$$ Since we all agree that a 3.2°GS give us 340'/nm, I suggest that you take your FAA, subtract by the TCH elevatin and divide by 340, the result is the distance to touchdown Thanks to all for answers. The $\frac{1.97}{96}$ was not correct. This is how you should do it: $$3.2=\frac{670ft\times60}{3.2}=12,562ft=2.066NM$$
I'm trying to understand the original of quark and lepton mass. Here's a paragraph from the book: "Massive neutrinos in physics and astrophysics" by R. N. Mohapatra and P. B. Pal : The gauge invariance prevent adding bare masses fro them in the Lagrangian. They arise from the following Yukawa interactions allowed by gauge symmetry: $$ -\mathcal{L}_{Y}=\sum_{a,b}[h^{(u)}_{ab}\bar{q}_{aL}\hat{\phi}u_{bR}+h^{(d)}_{ab}\bar{q}_{aL}\phi d_{bR}+h^{(l)}_{ab}\bar{\psi}_{aL}\phi l_{bR}]+\mathrm{h.c.}\ .\tag{2.14} $$ Here, $a,b$ stand for generation indices and $$\hat{\phi}=i\tau_{2}\phi^{*}\ .\tag{2.15}$$ On substituting the non-zero vacuum expectation values for $\phi_{0}=\sqrt{\frac{\mu^{2}}{2\lambda}}=v/\sqrt{2}$, the following mass terms for up and down quarks as well as the charge leptons are generated: $$-\mathcal{L}_{mass}= \sum_{a,b}[\bar{u}_{aL}M^{(u)}_{ab}u_{bR}+\bar{d}_{aL}M^{(d)}_{ab} d_{bR}+\bar{l}_{Lb} M^{(l)}_{ab}l_{bR}]+\mathrm{h.c.}\ ,\tag{2.16}$$ where $$ M^{(f)}_{ab}=h^{(f)}_{ab}v/\sqrt{2} \tag{2.17}$$ with $f=u,d,l$. By an appropriate choice of the quark and lepton basis, the coupling matrices $h^{(u)}$ and $h^{(l)}$ can be chosen diagonal so that we have $u^{0}_{a}=u_{a}$ and $l^{0}_{a}=l_{a}$. The $M^{(d)}$ is however, a complex non-diagonal matrix in this basis and can be diagonalized by the following biunitary transformation: $$V_{L}M^{(d)}V_{R}^{\dagger}=D^{(d)}. \tag{2.18}$$ Here's what I don't understand: Why does the Yukawa interaction Lagrangian look like that? I first thought that it has something to do with $$\mathcal{L}_{Yukawa}=\mathcal{L}_{Dirac}+\mathcal{L}_{Klein-Gordon}-g\bar{\psi}\psi\phi\ ,$$ but I cannot see any obvious connection between these two Lagrangians mathematically. I also don't understand what those h-matrices $h^{(u)}_{ab},h^{(d)}_{ab},h^{(l)}_{ab} $in equation (2.14) are. In equation (2.15), why does $\hat{\phi}$ look like that? In equation (2.16), how does $\bar{q}_{aL}$ become $\bar{u}_{aL}$ and so on. Also, how can I know that the matrices $M^{(u)}_{ab}$ and $M^{(l)}_{ab}$ are diagonal and $M^{(d)}_{ab}$ are not? At last, why do I need to put subscripts for the matrix $V$? I'd be very grateful for any one who finish reading and answer my lengthy questions.
Let's note $L(t,T_i,T_{i+1})$ the libor rate observed at $t$, fixing at $T_i$ with delivery at $T_{i+1}$. The natural delivery date for this rate is $T_{i+1}$, so a vanilla swap with no pay lag would be priced as : $Swap(t) = \sum_{i=1}^{n} \tau (T_i,T_{i+1}) P(t,T_{i+1}) \mathop{\mathbb{E}} ^{i+1}\left[ (L(T_i,T_i,T_{i+1})-K) \right]$ with $\mathop{\mathbb{E}} ^{i+1}$ the expectation under the forward measure associated with the bond $P(t,T_{i+1}) $ $\tau (T_i,T_{i+1}) = T_{i+1} - T_i$ for simplification Under the sum we use $P(t,T_{i+1}) $ for dicounting as the payments occurs at $T_{i+1} $. To price a CMS swap, we have the know expression for the swap rate : $$s_{m,n}(t)=\frac{ P(t,T_m) - P(t,T_n)}{ A_{m,n}(t) }$$ with $A_{m,n}$ the annuity given by : $$A_{m,n}(t)= \sum_{i=1}^{n} \tau (T_i,T_{i+1}) P(t,T_{i+1})$$ now a CMS swap can be priced as (Simply replacing the libor rate with the constant maturity swap rate) : $SwapCMS(t) = \sum_{i=1}^{n} \tau (T_i,T_{?}) P(t,T_{?}) \mathop{\mathbb{E}} ^{?}\left[ (s_{m,n}(T_i)-K) \right]$ I put questions marks in the formula because this is where i'm confused. When is the delivery of the swap rate? I mean the natural pay date so that I can choose the the bond for discounting and the forward measure needed? Or what is the discount I need to use and the forward measure in this case and why? Thank you
I have the following question: Suppose I have two matrices $X, Y$ both of size $m\times p$ and a random i.i.d Gaussian matrix $G$ of size $m \times k$, $m\gg p>k$. Is there a fast way to compute $\exp(-XY^T)G$? Perhaps by using the fact that both $X$ and $Y$ are much smaller than $XY^T$? Here, $\exp(\cdot)$ means entry-wise exponent (i.e. of $\textbf{each}$ entry of the matrix). Clearly, without the exponent, it's easy and can be done simply by $X(Y^TG)$, but the problem is that the element-wise exponent is no longer low rank. The motivation for the question is multiplying a kernel of the form $D_{ij}=\exp(-d_{ij})$ or $D_{ij}=\exp(-d_{ij}^2)$ with $d_{ij}=\Vert x_i-y_j \Vert$ and $x_i, y_i \in \mathbb{R}^p$ by a random matrix efficiently. An approximated solution is also fine.
Want to clarify few things. It is said that two $\lambda$-terms are equal up to renaming of bound variables, such as $\lambda x.x$ equals $\lambda y.y$, so I think it is a relation actually, about how two terms are equal. Is that (such equality up to renaming) is same as equivalence relation? To show a relation is an equivalence relation on a set, have to prove it is reflexive, symmetric, or transitive. I know, but have not done any practice on it. Let's say, I want to prove following $\frac{}{x = x}$ \two variables $\frac{ t_1 = t_3 \,\,\, t_2=t_4 } {t_1 \, t_2 \, = \,t_3 \, t_4}$ \applications $\frac{ t_2=[y / x]t_2 }{\lambda x.t_1 \, = \, \lambda y.t_2}$ \abstractions where $[y / x]t_2$ replace $x$ by $y$ in $t_2$ and $y$ is not occur in $t_2$. I think above 3 rules are true. How can I show proof of they holds? does that proof related to the proof of $\alpha$-equivalence is an equivalence relation on $\lambda$-terms? I am bit fonfused, it would be great if anyone help me clear things for me.
I was wondering if the following is a viable method using machine learning and neural networks to get to the ground states of the toric code and also understand the phase transition in the presence of perturbation that is $H_{p} = H_{TC} + h_{x(z)}\sum_{i}\sigma_{x(z)}^{i}$, where $H_{TC}$ is the toric code hamiltonian given by $H_{TC} = -\sum_{v}A_{v} - \sum_{p}B_{p}$, where $A_{v}, B_{p}$ are the vertex and face operators respectively Consider toric code with periodic boundary conditions, we know that the ground state is given by $$ \psi_{gs} = \prod_{v}(1+A_{v})\lvert\bf{0}\rangle,$$ given we are in the $\lvert0\rangle, \lvert1\rangle$ basis. For system sizes upto $N=24$ (on a system with decent memory) we can compute the ground state using the above expression and express it as a superposition of bit strings (as a superposition of 2^24 basis states, given by combination of the 24 bit strings of 0's and 1's, $\lvert000.....0\rangle, \lvert000.....1\rangle, \lvert111.....1\rangle$). We see that, if we represent the ground state as a superposition, not all 2^24 states are present in the superposition. Now we can use the number of spins (here 24) as an input to CNN (may be other architectures, I do not know much on this) to classify the toric code ground states as follows, if a bit string (a particular combination appearing in the basis) is present in the superposition we can label it as 1, and if it not we label it as 0. As we have labelled the entire data set, it should be possible to train a neural network. Given we do the above there are two points that arise If we look at the basis states which appear in the superposition, there are very few appearing in the superposition compared to the entire basis space (for $N=12$, we have 4096 bit strings (basis states) out of which only 36 appear in the ground state, even if we include the other ground states arising out of the application of the non-trivial loop operators, it is around 144 which is still small compared to 4096), there is a cornering of the hilbert space and how can this be captured in the neural network analysis (because it makes the learning completely biased) ? Also, if we can somehow get past the bias mentioned in the above point it means we have a neural network which has learnt about the states, now given we have neural network for $N=24$, how can this be extended to higher lattice sizes. In sense we have trained on a system with inputs $N=24$ and now we want to extend the knowledge of the trained network on lesser input and apply to a higher input, say $N=32$? In the sense the network trained on $N=24$ would predict whether a particular 32 bit string appears in the superposition or not, thereby providing a classification for higher system sizes and thereby getting to the ground states in higher system sizes. Now consider the toric code in the presence of perturbation, performing a similar analysis of representing the ground state in a superposition of bit-strings (here the ground state is obtained by ED for similar system sizes). The aim here is not to predict the ground state for higher system sizes but to predict the phase transition as the strength parameter is varied. For this the input to the neural network would be a bit string along with the strength parameter $h_{x(z)}$ and the output would be a 11 bit string output (here we see that the ground state is no longer a equal superposition of bit strings, to perform a binary classification, therefore we need more strings in the output to label the input bit string). For example if the probability of a bit-string, at say strength $h_{x}=0.3$ to appear in the ground state is given by 0.23, we assign the output 00100000000, if the probability is 0.i we assign the output which has 1 at lower bound of i and so on so forth. Here we do not have issue of cornering of Hilbert space, or heavy bias as for each strength the same basis state appear in the superposition but with a different probability. Therefore we can train the neural network at high strengths and low strengths of perturbation and then predict for all strengths whether a particular basis state appears or not and thereby from the superposition of these states conclude if it is in the ordered phase or trivial phase. The problem here is I do not see the classification happening using CNN, so it would nice if someone could suggest a better method and if the method itself is viable ?
Triangular Norms and Conorms Mirko Navara (2007), Scholarpedia, 2(3):2398. doi:10.4249/scholarpedia.2398 revision #137537 [link to/cite this article] Triangular norms and conorms are operations which generalize the logical conjunction and logical disjunction to fuzzy logic. They are a natural interpretation of the conjunction and disjunction in the semantics of mathematical fuzzy logics [Hájek (1998)] and they are used to combine criteria in multi-criteria decision making. Contents Triangular Norms Definition A triangular norm (abbreviation t-norm) is a binary operation $T$ on the interval [0,1] satisfying the following conditions: $T(x,y)=T(y,x)$ (commutativity) $T(x,T(y,z))=T(T(x,y),z)$ (associativity) $y\le z\Longrightarrow T(x,y)\le T(x,z)$ (monotonicity) $T(x,1)=x$ (neutral element 1) Examples $T_M(x,y)=\min(x,y) $ ( minimum or Gödel t-norm) $T_P(x,y)=x\cdot y$ ( product t-norm) $T_L(x,y)=\max(x+y-1,0)$ ( Lukasiewicz t-norm) No t-norm can attain greater values than $T_M\ .$There are many parametrized families of t-norms [Klement et al. (2000)].The Frank t-norms are defined for all $r>0, r \ne 1$ by\[T_{F_{r}}(x,y)=\log_r\left(1+\frac{(r^x-1)(r^y-1)}{r-1}\right)\,.\]The limit elements of this family are the above t-norms\[T_{F_0}=T_M\ ,\]$T_{F_1}=T_P\ ,$ and $T_{F_{\infty}}=T_L\ .$The only t-norms which are rational functions are the Hamacher t-norms defined for all $r>0$ by\[T_{H_r}(x,y) = \frac{xy}{r + (1-r)(x+y-xy)}\]and for $r=0$ by\[T_{H_0}(x,y)= \frac{x y}{x+y-xy}\]($T_{H_0}(0,0)=0$). Classification and representations The idempotents of a t-norm $T$ are those $x$ satisfying$T(x,x)=x\ .$ The bounds 0 and 1 are trivial idempotents. A t-norm is called Archimedean if each sequence $x_n, n\in\mathbb{N},$where $x_1<1$ and$x_{n+1}=T(x_n,x_n),$converges to 0.A continuous t-norm is Archimedean iff it has no idempotents between 0 and 1.A continuous Archimedean t-norm is called strict if $T(x,x)>0$ for all $x>0\ .$Continuous Archimedean t-norms which are not strict are called nilpotent.The product t-norm is strict, the Lukasiewicz t-norm is nilpotent. If $T^$ is a t-norm and $h\colon [0,1]\to[0,1]$ is an increasing bijection, then \[\tag{1} T(x,y)=h^{-1}(T^(h(x),h(y)))\] is a t-norm. This way,all strict t-norms can be obtained from the product t-norm and all nilpotent t-norms from the Lukasiewicz t-norm.(These t-norms serve as universal examples of these classes.) More generally, each continuous Archimedean t-norm can be obtained from the product t-norm using the formula\[T(x,y)=\left\{\begin{array}{ll}\theta^{-1}(\theta(x)\cdot\theta(y)) & \mbox{if }\theta(x)\cdot\theta(y)\ge b,\\0 & \mbox{otherwise.}\end{array}\right.\]where $\theta\colon [0,1]\to[b,1]\ (b\in\left[0,1\right[)$ is an increasing bijection called a multiplicative generator of $T\ .$ (It is not uniquely determined by $T\ .$) Each continuous Archimedean t-norm has also a (non-unique) additive generator, which is a decreasing bijection$t\colon [0,1]\to[0,B]\ (B\in\left]0,\infty\right])$ such that\[T(x,y)=\left\{\begin{array}{ll}t^{-1}(t(x)+t(y)) & \mbox{if }t(x)+t(y)\le B,\\0 & \mbox{otherwise.}\end{array}\right.\] Generalizations More generally, triangular norms can be defined (exactly the same way) on any ordered set with an upper bound (serving as the neutral element). They can be also restricted to (possibly finite) subsets of the unit interval. The term triangular norm is usually used for these operations, too. In particular, a t-norm $T$ on an interval [a,b] can be defined by (1), where $h\colon [a,b]\to[0,1]$ is an increasing bijection and $T^*$ is a t-norm on [0,1]. For a family of disjoint subintervals $\left]a_j,b_j\right[\subseteq[0,1],\ j\in J,$we may define a t-norm $T$ called an ordinal sum:\[T(x,y)=\left\{\begin{array}{ll}h_j^{-1}(T_j(h_j(x),h_j(y))) & \mbox{if } x,y \in \left]a_j,b_j\right[\mbox{ for some } j\in J,\\\min(x,y) & \mbox{otherwise,}\end{array}\right.\]where $h_j\colon [a_j,b_j]\to[0,1]$ are increasing bijections and $T_j$ are t-norms on $[0,1],\,j\in J.$All continuous t-norms are ordinal sums of Archimedean t-norms, we may choose $T_j\in \{T_L,T_P\}.$ There are t-norms which are not continuous or even not measurable. Triangular conorms Definition The dual notion to a triangular norm is a triangular conorm (abbreviation t-conorm, also s-norm), $S\ .$ Its neutral element is 0 instead of 1, all other conditions remain unchanged: $S(x,y)=S(y,x)$ (commutativity) $S(x,S(y,z))=S(S(x,y),z)$ (associativity) $y\le z\Longrightarrow S(x,y)\le S(x,z)$ (monotonicity) $S(x,0)=x$ (neutral element 0) Examples of t-conorms $S_M(x,y)=\max(x,y)$ ( maximum or Gödel t-conorm) $S_P(x,y)=x+y-x\cdot y$ ( product t-conorm, probabilistic sum) $S_L(x,y)=\min(x+y,1)$ ( Lukasiewicz t-conorm, bounded sum)) No t-conorm can attain smaller values than $S_M\ .$ If $T$ is a t-norm, then $S(x,y)=1-T(1-x,1-y)$is a t-conorm, and vice versa. We obtain a dual pair $(T,S)$ of a t-norm and a t-conorm. (Instead of the standard fuzzy negation, $x\mapsto 1-x\ ,$ another strong fuzzy negation can be used in the duality formula.) The classification and representations of t-conorms are dual to those of t-norms. Each continuous Archimedean t-conorm $S$ has a (non-unique) additive generator, which is an increasing bijection $s\colon [0,1]\to[0,B]\ (B\in\left]0,\infty\right])$ such that \[S(x,y)= \left\{\begin{array}{ll} s^{-1}(s(x)+s(y)) & \mbox{if }s(x)+s(y)\le B, \\1 & \mbox{otherwise.} \end{array}\right.\] Derived operations Fuzzy intersections and unions If $A,B$ are fuzzy sets and $\mu_A,\mu_B$ their membership functions, then the fuzzy intersection $C$ of $A$ and $B$ has the membership function$\mu_C(u)=T(\mu_A(u),\mu_B(u))\ .$ Thus a t-norm is sometimes called a fuzzy intersection.Depending on the choice of a t-norm, we obtain different fuzzy intersections.Dually, a t-conorm corresponds to a fuzzy union. Residua (fuzzy implications) Originally t-norms appeared in the context of probabilistic metric spaces [Schweizer and Sklar (1983)].Then they were used as a natural interpretation of the conjunction in the semantics of mathematical fuzzy logics [Hájek (1998)] and they are used to combine criteria in multi-criteria decision making.T-norms and t-conorms allow to evaluate the truth degrees of compound formulas.They are applied in fuzzy control to formulate assumptions of rules as conjunctions (fuzzy intersections) of fuzzy sets called antecedents or premises. (In such applications, the minimum or product t-norm are usually used because of a lack of motivation for other t-norms [Driankov et al. (1993)].)The Lukasiewicz t-conorm is closely related to the basic binary operation of MV-algebras.T-norms and t-conorms form also examples of aggregation operators.They play a crucial role in the axiomatic definition of the concept of triangularnorm based measure and, in particular, of a concept of probability of fuzzy events; the Frank family of t-norms and t-conorms plays a particular role here [Butnariu and Klement (1993)]. T-norms overlap with copulas [Nelsen (1999), Alsina et al. (2006)]: commutative associative copulas are t-norms;t-norms which satisfy the 1- Lipschitz condition are copulas.Some families of t-norms are known as families of copulas under different names. References Alsina, Claudi; Frank, Maurice J.; and Schweizer, Berthold: Associative Functions: Triangular Norms and Copulas.World Scientific, 2006. ISBN 981-256-671-6. doi:10.1142/6036. Butnariu, Dan and Klement, Erich Peter: Triangular Norm-Based Measures and Games wih Fuzzy Coalitions.Kluwer, Dordrecht, Netherlands, 1993. ISBN 0-7923-2369-6. doi:10.1007/978-94-017-3602-2_4. Driankov, Dimiter; Hellendoorn, Hans; and Reinfrank, Michael: An Introduction to Fuzzy Control.Springer, Berlin/Heidelberg, 1993. ISBN 3-540-56362-8. doi:10.1007/978-3-662-11131-4_1. Hájek, Petr: Metamathematics of Fuzzy Logic.Kluwer, Dordrecht, 1998. ISBN 0-7923-5238-6 Klement, Erich Peter; Mesiar, Radko; and Pap, Endre: Triangular Norms.Kluwer, Dordrecht, 2000. ISBN 0-7923-6416-3. doi:10.1007/978-94-015-9540-7. Nelsen, Roger B.: An Introduction to Copulas.Lecture Notes in Statistics 139, Springer, New York, 1999. ISBN 0-387-98623-5. doi:10.1007/978-1-4757-3076-0. Nguyen, Hung T. and Walker, Elbert A.: A First Course in Fuzzy Logic.2nd ed., Chapman & Hall/CRC, Boca Raton/London/New York/Washington, 2000. ISBN 0-8493-1659-6. Schweizer, Berthold and Sklar, Abe: Probabilistic Metric Spaces.North-Holland, New York, 1983. ISBN 0-444-00666-4. Internal references
Step 1 Let me formulate the problem to convey my notation. I have a matrix $A$ which is hermitian - and is diagonalisable by a transformation $$ U_A A\,\,U_A^{-1} = A_{diag}$$ Now the matrix is changed, using the small parameter $\lambda$. Therefore, $$ A \rightarrow A-\lambda B $$ where $B$ can be not Hermitian, in general. I tried to calculate the first order changes in eigenvalues and eigenvectors the following way - Step 2 The matrix $A-\lambda B $ has to be rotated by $U^\prime = e^{i\lambda \alpha}U_A$ to be diagonalized, $\alpha$ being a generator of rotation - and it makes sense that the "extra" rotation has to be proportional to $\lambda$. $$ A^\prime_{diag} = A_{diag} + C_1\lambda + C_2\lambda^2 + \mathcal{O}(\lambda^3) $$ I wrote this down using the BCH formula and set the linear coefficient, $C_1= 0$ and got the constraint $U_AB\;U_A^{-1} = \left[i\alpha,A_{diag}\right]$. The final form I obtained for $A^\prime_{diag}$ was $$ A^\prime_{diag} = A_{diag} -\frac{1}{2}\left[i\alpha,U_AB\;U_A^{-1}\right]\lambda^2 + \mathcal{O}(\lambda^3)$$ Question Can anyone tell me if - I am on the right track - and how to proceed from here to solve for $\alpha$ in terms of the known matrices. I need to approach this differently. it would help if $B$ were Hermitian? EDIT This is exactly the problem of doing perturbation theory using the SW transformation. Can anybody cite a reference?
I know that the set of all rational numbers is countable, and can be enumerated by a sequence, say $\{a_n\}$. But can we construct a monotonic $\{a_n\}_{n=1}^{\infty}$, e.g. with $a_k<a_{k+1}$? It doesn't seem plausible to me, because then $a_1$ would be the smallest rational number, which clearly can't be any finite number. Am I mistaken? if $a_k< a_{k+1}$ then $x := \frac{a_{k+1}+a_k}{2}$ is a rational number between those two, so no. You are not mistaken. Even assuming you want just the non-negative numbers, so that $a_0=0$, you cannot pick $a_1$ correctly because you will have skipped $a_1/2$. Alternatively, you could allow negative indices (and have $\ldots, a_{-1}, a_0, a_1,\ldots$), which would also solve your "there is no smallest rational number" problem, but it still has the problem that there are rational numbers between any two numbers. Specifically, if the list is complete, we must have some $n$ such that $a_n=0$. But then we necessarily miss $a_{n+1}/2$. As Arthur already stated, such a numbering cannot start with a finite index, i.e. either your sequence will count from $-\infty$ to $\infty$ or you only count the non-negative rational numbers (i.e. $\mathbb Q^+_0$). And as Thomas showed, a "normal" sequence also won't work since you'll always find a number in between. However, you can define a sequence of sequences $\{\{a_{nm}\}_{n=-\infty}^\infty\}_{m=1}^\infty$ such that its $\lim_{m\to\infty}$ yields a sequence counting all rational numbers. As an example, consider the typical enumeration sequence of $\mathbb Q$ (see e.g. here) and let $\{a_{nm}\}_n$ be the ordered sequence of the first m rational numbers obtained that way. The thing is, though, you will just end up with $\mathbb R$... As stated, the answer is no, because the question uses the symbol $<$ which has the implied meaning: The usual ordering of $\mathbb{Q}$ where $\frac{a}{b}<\frac{c}{d}$ iff $ad < bc$ in $\mathbb{Z}$. But. As mentioned in another answer, $\mathbb{Q}$ can be well-ordered, i.e. one can define a different order $\prec$ with the property that every nonempty subset of $\mathbb{Q}$ contains a least element with respect to $\prec$. For this ordering, a monotone sequence containing all of the rationals is easy to construct: let $x_1$ be the smallest rational, let $x_2$ be the smallest element of $\mathbb{Q} \setminus \{x_1\}$, etc. To parallel Thomas's answer with a more obscure one: If $a_k < a_{k+1}$ and $a_k = \frac{b_k}{c_k}$ then the mediant $x:= \frac{b_k+b_{k+1}}{c_k+c_{k+1}}$ is a rational number between those two, so no. ($b_k$ and $c_k$ are relatively prime/are in lowest terms) It is elementary but not obvious that the mediant $x$ is between the two (exercise for the reader). The reason this answer is even a thing (Thomas gave a much simpler and easier to understand answer) is because using the mediant we can construct an ordering on the rationals that is countable, and in addition gives all and only the rationals. The usual proof that the (positive) rationals have equal cardinality to the naturals is by 'dovetailing'. counting along antidiagonals of pairs and then ignoring rationals that have already been seen (ignoring if gcd $\neq 1$). This is somewhat unsatisfying because it doesn't give an explicit bijection with the naturals. To get the implied rational-natural bijection from gcd, see the Stern-Brocot tree for details.
Recently I had to solve a little problem related with Fibonacci numbers. I was using the Binet's formula for that: $$F_{n} = \frac{\Phi^n - \phi^n}{\sqrt{5}}$$ The effective solution I got by using computation by rounding formulas, related to Binet's formula that arises from simple inequality: $$\left\| \frac{\phi^n}{\sqrt{5}} \right\| < \frac{1}{2}, \quad n \leq 0.$$ $$F_{n} =\left\lfloor \frac{\Phi^n}{\sqrt{5}} + \frac{1}{2} \right\rfloor$$ I understand how it can be derived from above mentioned inequality, but I don't get it how I can derive that inequality rigorously in the first place. Any explanations would help me a lot! Thank you in advance.
I'm solving a nonlinear constrained optimization with constraint of following form: $$\mathbf{A}^T\mathbf{A}-\mathbf{I}=\mathbf{0}, \mathbf{B}^T\mathbf{B}-\mathbf{I}=\mathbf{0}$$ where $\mathbf{A}$ and $\mathbf{B}$ are matrices that I'm solving for. Usually the initial guess of $\mathbf{A}$ and $\mathbf{B}$ are diagonal matrices with entries of 1 or -1. I flatted the entries of $\mathbf{A}$ and $\mathbf{B}$ in column order into a vector as the variable $\mathbf{x}$ to solve for. I tried NLopt and Opt++ to solve my problem, both of them failed. While debugging I found that the gradient of constraint is a singular matrix. $\mathbf{A}$ and $\mathbf{B}$ are not necessarilly square matrices, thus neither is gradient of constraint. My question is: is there any algorithm to solve this nonlinear constrained optimization, which does not require gradient of the constraint? =======Edit========= The objective function is: $$\textrm{off}(\mathbf{A^T\Lambda_XA})+\textrm{off}(\mathbf{B^T\Lambda_YB})+\mu\left\\|\mathbf{F^T\Phi{}A}-\mathbf{G^T\Psi{}B}\right\\|_{\mathbf{F}}^2$$ where off() means sum of the squared off-diagonal elements, $\Lambda_X$ and $\Lambda_Y$ are diagonal matrices, $\mathbf{F}$ and $\mathbf{G}$, $\mathbf{\Phi}$ and $\mathbf{\Psi}$ are all known matrices.
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I believe it can be useful to define the following concepts (I won't be very formal here for pedagogical reasons): Any event can be described through four real numbers, which we take to be: the moment in time it happens, and the position in space where it takes place. We call this four numbers the coordinates of the event. We collect these numbers in a tuple, which we call $x\equiv (t,\boldsymbol r)$. These numbers depend, of course, on which reference frame we are using: we could, for example, use a different origin for $t$ or a different orientation for $\boldsymbol r$. This means: for $x$ to make sense, we must pick a certain reference frame. Call it $S$ for example. Had we chosen a different frame, say $S'$, the components of the same event would be $x'$, i.e., four real numbers, in principle different from those before. We declare that the new reference frame is inertial if and only if $x'$ and $x$ are related through$$x'=\Lambda x \tag{1}$$for a certain matrix $\Lambda$, that depends, for example, on the relative orientations of both reference frames. There are certain conditions $\Lambda$ must fulfill, which will be discussed in a moment. We define a vector to be any set of four real numbers such that, if its components in $S$ are $v=(v^0,\boldsymbol v)$, then in $S'$ its components must be$$v'=\Lambda v \tag{2}$$ For example, the coordinates $x$ of an event are, by definition, a vector, because of $(1)$. There are more examples of vectors in physics, for example, the electromagnetic potential, or the current density, the momentum of a particle, etc. It turns out that it is really useful to define the following operation for vectors: if $u,v$ are two vectors, then we define$$u\cdot v\equiv u^0 v^0-\boldsymbol u\cdot\boldsymbol v\tag{3}$$The reason this operation is useful is that it is quite ubiquitous in physics: there are many formulas that use this. For example, any conservation law, the wave equation, the Dirac equation, the energy-momentum relation, etc. We define the operation $\cdot$ through the components of the vectors, but we know these components are frame-dependent, so if $\cdot$ is to be a well-defined operation, we must have$$u\cdot v=u'\cdot v' \tag{4}$$because otherwise $\cdot$ would be pretty useless. This relation $(4)$ won't be true in general, but only for some matrices $\Lambda$. Thus, we declare that the matrices $\Lambda$ can only be those which make $(4)$ to be true. This is a restriction on $\Lambda$: only some matrices will represent changes of reference frames. Note that in pure mathematics, any invertible matrix defines a change of basis. In physics only a subset of matrices are acceptable changes of basis. So, what are the possible $\Lambda$'s that satisfy $(4)$? Well, the easier way to study this is to rewrite $(3)$ using a different notation: define $$\eta=\begin{pmatrix} 1 &&&\\&-1&&\\&&-1&\\&&&-1\end{pmatrix} \tag{5}$$ This is just a matrix that will simplify our discussion. We should not try to find a deep meaning for $\eta$ (it turns out there is a lot of geometry behind $\eta$, but this is not important right now). Using $\eta$, its easy to check that $(3)$ can be written as$$u\cdot v=u^\mathrm{T}\eta v \tag{6}$$where in the r.h.s. we use the standard matrix product. If we plug $v'=\Lambda v$ and $u'=\Lambda v$ here, and set $u\cdot v=u'\cdot v'$, we find that we must have$$\Lambda^\mathrm{T} \eta \Lambda=\eta \tag{7}$$ This is a relation that defines $\Lambda$: any possible change of reference frame must be such that $(7)$ is satisfied. If it is not, the $\Lambda$ cannot relate two different frames. This relation is not in fact a statement of how $\eta$ transforms (as you say in the OP), but actually a restriction of $\Lambda$. It is customary to say that $\eta$ transforms as $(7)$, which will be explained in a moment. For now, just think of $(7)$ as what are the possible matrices $\Lambda$. At this point, it is useful to introduce index notation. If $v$ is a vector, we call its components $v^\mu$, with $\mu=0,1,2,3$. On the other hand, we write the components of changes of frames $\Lambda^\mu{}_\nu$. With this notation, $(2)$ can be written as$$v'^\mu=\Lambda^\mu{}_\nu v^\nu \tag{8}$$ Also, using index notation, the product of two vectors can be written as$$u\cdot v=\eta_{\mu\nu}u^\mu v^\nu \tag{9}$$where $\eta_{\mu\nu}$ are the components of $\eta$. Index notation is useful because it allows us to define the following concept: a tensor is an object with several indices, e.g. $A^{\mu\nu}$. But not any object with indices is a tensor: the components of a tensor must change in different frames of reference, such that they are related through$$\begin{align}&A'^{\mu\nu}=\Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma\ A^{\rho\sigma} \\&B'^\mu{}_\nu=\Lambda^\mu{}_\rho(\Lambda^\mathrm{T})_\nu{}^\sigma\ B^\rho{}_\sigma\\&C'^{\mu\nu}{}_\pi{}^\tau=\Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma (\Lambda^\mathrm{T})_\pi{}^\psi \Lambda^\tau{}_\omega\ C^{\rho\sigma}{}_\psi{}^\omega\end{align}\tag{10}$$and the obvious generalisation for more indices: for every upper index, there is a factor of $\Lambda$, and for every lower index, a factor of $\Lambda^\mathrm{T}$. If the components of an object with indices don't satisfy $(10)$ then that object is not a tensor. According to this definition, any vector is a tensor (with just one index). I don't like to use index notation too much: $v'=\Lambda v$ is easier that $v'^\mu=\Lambda^\mu{}_\nu v^\nu$, don't you think?. But sometimes we have to use index notation, because matrix notation is not possible: when using tensors with three or more indices, matrices cannot be used. Tensors with one index are just vectors. You'll hear sometimes that matrices are tensors with two indices, which is not quite true: if you remember from your course on linear algebra, you know that when you make a change of basis, matrices transform like $M\to C^\mathrm{T} M C$, which is like $(10)$ in the case of one upper/one lower index. Therefore, matrices are like tensors with one uppe/one lower index. This is the reason we wrote $\Lambda$ as $\Lambda^\mu{}_\nu$. This is a matrix, but it is also a tensor. Also, $(7)$ pretty much looks like $(10)$, right? This is the reason people say $(7)$ expresses the transformation properties of $\eta$. While not false, I you recommend not to take this too seriously: formally, it is right, but in principle $\eta$ is just a set of numbers that simplifies our notation for scalar products. It turns out you can think of it as a tensor, but only a-posteriori. In principle, it is not defined as a tensor, but it turns out it is. Actually, it is a trivial tensor (the only one!) whose components are the same in every frame of reference (by definition). If you were to calculate what are the components of $\eta$ in another frame of reference using $(10)$, you'll find out that they are the same. This is stated as the metric is invariant. We actually define it to be invariant. We define what a change of reference frame through the restriction of $\eta$ being invariant. It doesn't make sense to try to prove $\eta$ is invariant, as this is a definition. $(7)$ doesn't really prove $\eta$ is invariant, but actually defines what a change of reference is. For completeness I'd like to make the following definitions: We say an object is invariant if it takes the same value on any frame of reference. You can check that if $v$ is a vector, then $v\cdot v$ takes the same value on any frame, i.e., $v^2$ is invariant. We say an object is covariant if it doesn't take the same value on every frame of reference, but the different values are related in a well defined way: the components of a covariant object must satisfy $(10)$. This means tensors are covariant by definition. For example, a vector is not invariant because its components are frame-dependent. But as vectors are tensors, they are covariant. We really like invariant objects because they simplify a lot of problems. We also like covariant objects because, even though these objects are frame-dependent, they transform in a well-defined way, making them easy to work with. You'll understand this better after you solve many problems in SR and GR: in the end you will be thankful for covariant objects. So, what does it mean for $\eta$ to be invariant? It means its components are the same in every (inertial) frame of reference. How do we prove this? we actually can't, because we define this to be true. How can we prove $\eta$ is the only invariant tensor? We can't, because it is not actually true. The most general invariant tensor is proportional to the metric. Proof: let $N^\mu{}_\nu$ be an invariant tensor by definition. Then, as it is a tensor, we have$$N'=\Lambda^\mathrm{T}N\Lambda \tag{11}$$ But we also must have $N'=N$ for it to be invariant. This means $\Lambda^\mathrm T N\Lambda=N$. Multiply on the right by $\eta \Lambda^\mathrm{T} \eta$ and use $(7)$ to get $[N,\Lambda^\mathrm{T}]=0$. By Shur's Lemma, $N$ must be proportional to the identity. QED. And what about the Levi-Civita symbol? we are usually told that it is also an invariant tensor, which is not actually true: it is invariant, but it is not a tensor, it is a pseudo-tensor. In SR it doesn't satisfy $(10)$ for any $\Lambda$, but only for a certain subset of matrices $\Lambda$ (check Proper Orthochronus Lorentz Group), and in GR it is a tensor density (discussed in many posts on SE). The proof of the covariance of the LC symbol is usually stated as follows (you'll have to fill in the details): the definition of the determinant of a matrix is can be stated as $\text{det}(A)\varepsilon^{\mu\nu\sigma\rho}=\varepsilon^{abcd}A^\mu{}_a A^\nu{}_b A^\rho{}_c A^\sigma{}_d$. The proper Orthochronus Lorentz Group consists of the subset of matrices with unit determinant, i.e., $\text{det}(\Lambda)=1$. If you use this together with the definition of $\text{det}$, you get $\varepsilon^{\mu\nu\rho\sigma}=\varepsilon^{abcd}\Lambda^\mu{}_a\Lambda^\nu{}_b\Lambda^\rho{}_c\Lambda^\sigma{}_d$, which is the same as $(10)$ for the object $\varepsilon^{\mu\nu\rho\sigma}$. This proves that, when restricted to this subset of the Lorentz Group, the Levi-Civita symbol is a tensor. Raising and Lowering indices: this is something that is usually made more important that it really is. IMHO, we can fully formulate SR and GR without even mentioning raising and lowering indices. If you define an object with its indices raised, you should keep its indices where they are. In general there is no good reason as why would someone want to move an index. That being said, I'll explain what these are, just for completeness. The first step is to define the inverse of the metric. Using matrix notation, the metric is its own inverse: $\eta \eta=1$. But we want to use index notation, so we define another object, call it $\zeta$, with components $\zeta^{\mu\nu}=\eta_{\mu\nu}$. With this, you can check that $\eta\eta=1$ can be writen as $\eta_{\mu\nu}\zeta^{\nu\rho}=\delta^\mu_\rho$, where $\delta$ is the Kronecker symbol. For now, $\delta$ is just a symbol that simplifies the notation. Note that $\zeta$ is not standard notation, but we will keep it for the next few paragraphs. (People usually use the same letter for both $\eta$ and $\zeta$, and write $\eta_{\mu\nu}=\eta^{\mu\nu}$; we'll discuss why in a moment. For now, note that these are different objects, with different index structure: $\eta$ has lower indices and $\zeta$ has upper indices) We can use $\eta$ and $\zeta$ to raise and lower indices, which we now define. Let's say you have a certain tensor $A^{\mu\nu}{}_\rho$. We want to define what it means to raise the index $\rho$: it means to define a new object $\bar A$ with components$$\bar A^{\mu\nu\rho}\equiv \zeta^{\rho\sigma}A^{\mu\nu}{}_\sigma \tag{12}$$(this is called to raise the index $\rho$ for obvious reasons) Using $(10)$ you can prove that this new object is actually a tensor. We usually drop the bar $\bar{\phantom{A}}$ and write $A^{\mu\nu\rho}$. We actually shouldn't do this: these objects are different. We can tell them apart from the index placement, so we relax the notation by not writing the bar. In this post, we'll keep the bar for pedagogical reasons. In an analogous way, we can lower an index, for example the $\mu$ index: we define another object $\tilde A$, with components$$\tilde A_\mu{}^\nu{}_\rho\equiv \eta_{\mu\sigma} A^{\sigma\nu}{}_\rho \tag{13}$$(we lowered $\mu$) This new object is also a tensor. The three objects $A,\bar A,\tilde A$ are actually different, but we can tell them apart through the indices placement, so we can drop the tildes and bars. For now, we won't. We'll discuss the usefulness of these operations in a moment. For now, note that if you raise both indices of the metric, you get$$\bar{\bar{\eta}}^{\mu\nu}\equiv\zeta^{\mu\rho}\zeta^{\nu\sigma} \eta_{\rho\sigma}=\zeta^{\mu\rho}\delta^\nu_\rho=\zeta^{\mu\nu} \tag{14}$$which means that $\bar{\bar{\eta}}=\zeta$. As we usually drop the bars, this means that we can use the same letter $\eta$ for both objects. In principle, they are different: $\eta_{\mu\nu}$ is the metric, and $\zeta^{\mu\nu}$ is its inverse. In practice, we use $\eta_{\mu\nu}$ and $\eta^{\mu\nu}$ for both these objects, and even call them both metric. From now on, we will use $\eta$ both for the metric and its inverse, but we keep the bars for other objects. With this in mind, we get the following important result:$$\eta_{\mu\nu}\eta^{\nu\rho}=\delta_\mu^\rho \tag{15}$$which is actually a tautology: it is the definition of the inverse of the metric. So, what is the use of these operations? for example, what do we get if we lower the index of a vector $v$? Well, we get a new tensor, but it is not a vector (you can check that $(2)$ is not satisfied), so we call it a covector. This is not really important in SR, but in other branches of physics vectors and covectors are really really different. So, what is the covector associated to $v$? Call this covector $\bar v$. Its components will be $\bar v_\mu=\eta_{\mu\nu} v^\nu$ by definition. Why is this useful? Well, one reason is that by lowering an index, the scalar product $\cdot$ turns into standard matrix product:$$u\cdot v=\bar u v \tag{16}$$as you can check (compare this to $(3)$ or $(6)$). So in principle, raising and lowering indices is supposed to simplify notation. Actually, in the end, you'll see that people write $uv$ instead of $u\cdot v$ or $u_\mu v^\mu$. So you see that the notation is simplified without the need of raising/lowering any index. The following fact is rather interesting: we know that if we raise both indices of the metric we get the metric again. But what do we get if we raise only one index to the metric? that is, what is $\bar \eta$?, or, put it another way, what is $\eta^\mu{}_\nu$? Well, according to the definition, it is$$\eta^\mu{}_\nu=\eta_{\nu\rho}\eta^{\mu\rho}=\delta^\mu_\nu \tag{17}$$where I used $(15)$. This means that $\bar \eta=\delta$: the metric is the same object as the Kronecker symbol, which is a cool result. As we know that raising and lowering indices from a tensor results in a new tensor, we find that the Kronecker symbol is actually a tensor! We can even prove this from the definition of tensors, i.e., we can check that $(10)$ is satisfied for $\delta$. But we don't need to: we know that it must be true (check it if you want to). As a side note: you (as many people) write prime marks on the indices, while I (as many others) write the primes on the tensors. IMHO the latter convention is the best, because it is the tensor what is changing, not the indices. For example, what you wrote $\eta_{\mu'\nu'}=\eta_{\mu\nu}$ looks better when written $\eta'_{\mu\nu}=\eta_{\mu\nu}$, because the $\mu\nu$ component of both objects are equal, and not the $\mu'$ is equal to the $\mu$ component (which actually makes no sense and makes the indices mismatched).
September 12th, 2016, 11:01 PM # 1 Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 Velocity and time A particle is projected vertically upwards with velocity u m/s. Its height is h after t1 and t2 seconds. Prove that t1 × t2 = 2h/g. Last edited by skipjack; September 13th, 2016 at 02:33 AM. September 12th, 2016, 11:32 PM # 2 Senior Member Joined: Sep 2015 From: USA Posts: 2,575 Thanks: 1423 Set up the quadratic equation describing the motion at height h. You know that t1, t2 are roots of this quadratic equation. Using the general quadratic formula, find out the expression for the product of two roots of a quadratic equation. These will be in terms of the $a,b,c$ in the quadratic formula. Using your equation you set up in step 1, note the values of these coefficients and plug them into the result you got above. profit $h(t) = -\dfrac 1 2 g t^2 + u t$ solve for $t_1,~t_2$ $h = -\dfrac {g t^2}{2}+u t$ $-\dfrac {g t^2}{2}+u t-h=0$ you'll get two roots in the form of $t_1,t_2 = \dfrac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$ where here $a=-\dfrac{g}{2},~b=u,~c=-h$ Now if you you multiply those two roots out you'll find that $t_1 \cdot t_2 = \dfrac c a = \dfrac{-h}{-g/2} = \dfrac {2h}{g}$ Last edited by skipjack; September 13th, 2016 at 02:34 AM. Tags time, velocity Thread Tools Display Modes Similar Threads Thread Thread Starter Forum Replies Last Post Velocity, speed, time problem themagicstick1 Physics 2 September 14th, 2015 02:01 AM velocity and time kunz398 Calculus 2 May 14th, 2015 07:20 AM Obtaining angular velocity given linear velocity and center of rotation 3D quarkz Calculus 0 April 18th, 2014 05:34 AM Velocity-time graph Chikis Algebra 7 October 5th, 2012 10:03 PM Derivative - position, velocity and time condemath Calculus 1 October 25th, 2011 02:53 PM
Consider $m$ linearly independent vectors in $n$-dimensional Euclidean space, $v_1,...,v_m \in \mathbb R^n$ where $1\leq m<n$, and let $U := {\rm span}(v_1,...,v_m)$ denote the $m$-dimensional subspace of $\mathbb R^n$ spanned by these vectors. Informally, I want to count the "integer lattice points" (i.e., elements of $\mathbb Z^n$) whose Euclidean distance from $U$ is smaller or equal to some given number $a\geq0$. More formally, if we consider the orthogonal projections onto $U$ of all elements of $\mathbb Z^n$ whose distance from $U$ is $\leq a$, then what I am looking for is an upper bound on the global density $\rho_U(a)$ in $U$ of these projection images ("density" is used here in the sense of number per volume in $U$). Note that if we assume in addition that the vectors $v_1,...,v_m$ are themselves elements of $\mathbb Z^n$, then the distribution of projection images in $U$ will be repetitive (more precisely, $m$-fold periodic with periodicities $v_1,...,v_m$), so that in this case one only needs to count images contained in a "unit cell" with finite volume in $U$. Now, since the density in $\mathbb R^n$ of lattice points $\mathbb Z^n$ is unity, I would heuristically expect (at least for large $a$) that the above density of projection images in $U$ behaves like the volume (in $\mathbb R^n$) of a "layer of thickness $a$ around $U$", so that\begin{align}\rho_U(a) \approx V_{n-m}\ a^{n-m}\ , \tag{(1)}\end{align}where $V_{k} := \frac{\pi^{\frac{k}2}}{\Gamma(\frac{k}2+1)}$ denotes the volume of the $k$-dimensional unit ball. Using Stirling's approximation for the $\Gamma$ function, we can write (1) as \begin{align}\rho_U(a) & \approx \frac1{\sqrt{\pi(n{-}m)}} \left( \!\frac{2\pi{\rm e}}{n{-}m} \!\right)^{\!\!\frac{n-m}2} \ a^{n-m}\\& \leq \left( \!\frac{2\pi{\rm e}\, a^2}{n{-}m} \!\right)^{\!\!\frac{n-m}2}\ . \tag{(2)}\end{align}Clearly, inequality (2) cannot be valid in general for arbitrarily small $a$; for instance, if we choose $v_1,...,v_m \in \mathbb Z^n$ then the set of lattice points lying directly on $U$ (i.e. at zero distance) has finite density $\rho_U(0) > 0$ since it contains at least all linear combinations of $v_1,...,v_m$ with integer coefficients. However, I believe that if we exclude all lattice points lying directly on $U$ from counting, an inequality of the form (2) may hold. More explicitly, let me try the following Conjecture.For an arbitrary but fixed $m$-dimensional subspace $U$ of $\mathbb R^n$ ($1\leq m<n$) and for any $a \geq 0$, let $\rho_U(a)$ denote the density in $U$ of orthogonal projection images of all lattice points (i.e., elements of $\mathbb Z^n$) that lie at distances $\leq a$ from $U$. There is a universal constant $C$ such that for any choice of $m$, $n$, and $U$, and for all $a$: \begin{align} \rho_U(a) - \rho_U(0) & \leq \left( \!\frac{C\, a^2}{n{-}m} \!\right)^{\!\!\frac{n-m}2} \ . \tag{(3)} \end{align} My questions, of course, are: Is it known whether some inequality similar to (3) holds? Perhaps with restrictions on the choice of the subspace $U$ or the range of $a$? If the conjecture (or a similar one) is true, how can we prove it? If is is false, can we find a counterexample? My feeling is that this problem ought to be solvable by relatively elementary arguments, but so far I have no clue how to approach it. Addendum: For the sake of clarity I will give a more formal definition of the quantity $\rho_U(a)$ here, assuming that $v_1,...,v_m \in \mathbb Z^n$, i.e. the vectors have only integer components. Consider the orthogonal projection (in $\mathbb R^n$) of all elements of $\mathbb Z^n$ which lie at distances $\leq a$ from $U = {\rm span}(v_1,...,v_m)$ onto $U$. Let $S_a \subset U$ be the set of these projection points. The distribution of $S_a$ within $U$ is $m$-fold periodic with periodicities $v_1,...,v_m$, hence we can define the mean density of these points in $U$ as their mean density in a fundamental domain ("unit cell") $P \subset U$. An obvious fundamental domain is the $m$-parallelotope spanned by $v_1,...,v_m$, defined as \begin{align}P := \big\{ \sum_{i=1}^m \alpha_i v_i : \forall_i\ 0 \leq \alpha_i \leq 1 \big\}\ . \tag{(4)}\end{align}The number of elements in the intersection of $S_a$ and $P$ is finite, so we can define the mean density in $P$ as this number divided by the volume of $P$:\begin{align}\rho_U(a) := \frac{\|S_a \cap P\|}{{\rm vol}(P) }\ . \tag{(5)}\end{align*}Note that if all components of $v_1,...,v_m$ are rational numbers instead of integers, one can proceed in exactly the same way, after multiplying them all with their common denominator (which doesn't change $U$). In contrast, if the components are arbitrary real numbers, generally there will be no periodicities and one has to resort to some limiting procedure to define $\rho_U(a)$. However, currently I am mostly interested in the integer case $v_1,...,v_m \in \mathbb Z^n$ where definition (5) is sufficient.
Counter Counting objects/people forms the basis for a range of high-tech solutions, including retail analytics, queue management, building management and security applications. By using the counter heuristic you can count incoming and outgoing objects (e.g. people). By defining two lines, the counter heuristics can count the incoming (green line) and outgoing (red line) objects. The quality of the Counter heuristic depends heavily on the accurancy of the chosen algorithm (Differential, Background subtraction, etc). Therefore we recommend to use the Background Subtraction algorithm as this is the most accurate one; the algorithm is used for segmentation and it will distinguish background and foreground. After the algorithm did its magic, the segmented image is used by the Counter heuristic to calculate some features: the center of mass is calculated for each foreground segment, and is stored in memory together with the height and width of the segment. By using the coordinates of the center of mass a trajectory is calculated. $$ \texttt{mu} _{ji}= \sum _{x,y} \left ( \texttt{array} (x,y) \cdot (x - \bar{x} )^j \cdot (y - \bar{y} )^i \right ) $$ $$ \bar{x} = \frac{\texttt{m}_{10}}{\texttt{m}_{00}} , ; \bar{y} = \frac{\texttt{m}_{01}}{\texttt{m}_{00}} $$ While the capture device is taking images, the counter heuristic will calculate the features (as mentioned before) for every subsequent frame. When calculated the heuristic will search for the best match: the closest, in terms of distance, center of mass is searched from the previous calculation. When the best match is found the center of mass (with its features) is added to the feature list of the best match; or if not found, a new feature list is created. The idea of the feature list is that it contains the trajectory of an object which is moving from one point to another (a person walking by). A feature list exists for every moving object in the viewport of the camera and each feature list contains the sequence of center of masses (the x- and y-coordinates of the object). After the center of masses are calculated, existing feature lists are expanded or new feature lists are created (the start of a new object). By using the feature lists, the direction of the moving objects can be calculated, as it contains the initial center of mass and the last found center of mass. As we know the trajectory of an object (and thus its direction), the intersection is calculated with the incoming (green line) and outgoing (red line). If the trajectory intersects both incoming and outgoing lines, the direction of the trajectory is used to determine which line was interesected first (and this is an incoming or outgoing object). Parameters The parameters of the Counter heuristic can be found in the config/heuristic.xml file, but you can also use the web to modify the parameters. Below you see a default configuration file. <heuristics> <Counter> <appearance type="number">5</appearance> <maxDistance type="number">90</maxDistance> <minArea type="number">1400</minArea> <onlyTrueWhenCounted type="bool">true</onlyTrueWhenCounted> <minimumChanges type="number">20</minimumChanges> <noMotionDelayTime type="number">1000</noMotionDelayTime> <markers type="twolines">100,100\|100,200\|200,100\|200,200</markers> </Counter></heuristics> Appearance The appearance parameters works as a time-out value. If a feature lists hasn't been updated for x (appearance) times in a row, it's removed. The idea is that when an object has moved outside the viewport of the camera, it can't be tracked anymore. Max Distance While calculating the center of masses, the best match is searched for each one. As the best match is calculated in function of the distance, a maximum distance is used to limit the matching process. Min Area The center of mass is calculated for each segment which has an area which is larger than the minimum area. Only true when counted By default the heuristic will only return true when one or more objects are marked as incoming or outgoing. However by unchecking this option you can make the heuristic return true when something changed (same behaviour as the sequence heuristic). The idea is that one wants to track the incoming and outgoing objects, but also want to have images for every change. Minimum changes The heuristic will only procede true if enough changes have been detected. No motion delay time When the heuristic is not valid, the heuristic will idle for some time. Markers The start- and en-coordinates of both the incoming and outgoing lines (delimited by the pipe symbol). Output The whole point of the counter heuristic is counting objects. Therefore when one or more incoming or outgoing objects are detected the results are added to the JSON object which is passed along the four passway. The JSON object can be retrieved by using the Webhook IO device. [ 'regionCoordinates' : [618, 317, 703, 493], 'numberOfChanges' : 5446, 'incoming' : 0, 'outgoing' : 1, 'timestamp' : '1465894497', 'microseconds' : '5-97451', 'token' : 695, 'pathToImage' : '1465894497_5-97451_frontdoor_618-317-703-493_5446_695.jpg', 'instanceName' : 'frontdoor',]
What Does a Circle Look Like?It is in the very beginning of Apollonius' treatise the Conics that he deals with the question we are concerned with, and I'll try to explain, roughly, what he does. ... Bill Casselman Introduction The obvious answer is certainly correct. Nonetheless this is a subtle question. You say you see a circle when you are looking at a circle, but the reasons for this response are not straightforward. The data that lead to it are what your eyes perceive, no more and no less. But what your eyes perceive is entirely coded in images formed on your retina, and they are almost certainly not circles. Instead they are some kind of oval, and to say that a circle looks like an oval would in some sense be a more precise description of what is going on. Your interpretation of ovals as circles is based on the activity of your brain in interpreting the world around you - your 3D perception, which tells you that parts of the oval are further away than others; and your experience with seeing other circles in a wide range of situations. In fact, the ovals are ellipses, except for some slight distortion caused by curvature in your retina, and this is an even more subtle fact. The reason it is subtle is that ellipses have a four-fold symmetry, whereas the only obvious symmetry in your perception of a circle is likely bilateral. Part of the circle is closer than other parts, and you might expect that part to bulge a bit, so the actual shape you perceive would look more like an egg than an ellipse. Famous and talented people have indeed asserted that this is the way things are. As the figure above suggests, the image recorded on your retina is the intersection of (a) the cone whose base is the given circle and whose apex is the lens of your eye and (b) the (approximate) plane spanned by your retina. It is, to a very close approximation, a It is in the very beginning of Apollonius' treatise the The equation of an ellipse Apollonius' principal tool in analyzing conic sections is a kind of coordinate geometry, along with equations for the sections in terms of these coordinates. The one for a circle is fundamental and simple. As the figure on the left shows, a triangle inscribed in a semicircle with a diameter as one side is right-angled. A simple argument about similar triangles then shows that $$y^{2} = x(d-x) $$ with $x$, $y$, $d$ indicated in the right-hand figure. An ellipse is obtained from a circle by compressing it, say vertically by a factor $f$. If we do this in the direction perpendicular to the horizontal diameter we get an equation $$ y^{2} = f^{2} x(d-x) $$ for the ellipse, as on the left below. But if we project onto another diameter, we still get an ellipse together with a diameter and conjugate chords. In this skew coordinate system we also get an equation of the same form. Any line through the center of an ellipse cuts out one of its diameters. Unless the ellipse is in fact a circle, there will be a unique family of parallel chords cutting this in turn that are bisected by the original diameter. The one through the center is called the Therefore, in order to verify that a plane section of a cone is an ellipse, we must (a) find a diameter, say of length $d$; (b) find the conjugate diameter; (c) assign the ellipse the corresponding coordinate system and verify that an equation of the form $y^{2} = f^{2} x(d-x)$ holds. In doing this, I follow Heath's translation of the Finding a diameter of a section In this section I lead up to Proposition I.7 of the Find the intersection $\Lambda$ of $\Pi$ with the base plane. Choose $XX'$ a diameter of the base circle perpendicular to $\Lambda$. Then the intersection of the plane $OXX'$ and $E$ is a diameter $D=PP'$ of $E$. Its conjugate chords are the segments in $E$ parallel to $\Lambda$. In general, the diameter $D$ and these will be skewed - i.e. not mutually perpendicular. Apollonius' derivation of the equation In this section I conclude the proof that a finite conic section is an ellipse. This is I.13 of the Suppose $V$ to be a point on the diameter $D$. To it are associated two points $Q$, $Q'$ at the ends of the chord passing through $V$. We wish to know that the ratio of $QV^{2}$ to $PV\cdot VP'$ is a constant, which will imply that $E$ is an ellipse. There is one thing of the sort that we do know: If we replace $E$ by the circle $C_{*}$ parallel to the base circle $C$ intersecting $\Pi$ in the chord $QVQ'$, and let $HH'$ be the intersection of this disk with the triangular slice, then $QV^{2} = HV\cdot VH'$. The constant can be evaluated explicitly from the figure to be $XY\cdot YX'/OY^{2}$. Dürer's egg It has probably often been thought that a conic section can be somewhat wider at one end than the other, but one case is famous. The Renaissance artist Albrecht Dürer was very well trained in mathematics. He bought his copy of Euclid in Venice in 1507 (one of the earliest of all printed editions), and it is now in the possession of the Herzog August Bibliothek in Wolfenbüttel. (You can see a photograph of its title page in this article.) But when he came to explaining how to draw conic sections in (Page 35 of the Dresden digital copy) Final remarksApollonius' book of Conics is notoriously difficult to read, and not always rewarding. It is not totally unfair to regard much of it as a first step towards algebraic treatment of geometry. But it is impressive, certainly ranking in effort and ingenuity with other major works of mathematics such as Euclid's Elements or Newton's Principia mathematica. For the Conics above all, one wonders, how did Apollonius arrive at his results? As in what we have seen earlier, one of his major tricks is to reduce three-dimensional geometry to two dimensions, but even so it required an almost unbelievable skill in mentally manipulating objects in three dimensions. Did the Greeks have 3D models? Where to read more Bill Casselman The AMS encourages your comments, and hopes you will join the discussions. We review comments before they're posted, and those that are offensive, abusive, off-topic or promoting a commercial product, person or website will not be posted. Expressing disagreement is fine, but mutual respect is required. Welcome to the These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics. Search Feature Column Feature Column at a glance
Just out of curiosity I started looking into how to get jupyter to display text and math formatted as markdown and latex in output cells. This got into my head when I was looking at symbolic integration and differentiation of functions and how to format the output in a more civilized manner. I don’t have the full solution to the question but formatting output as markdown turned out to be ridiculously easy since the heavy lifting had already been done. # Displaying markdown and latex in jupyter output cells from IPython.display import display, Markdown, Latex markdown = Markdown("### Displaying markdown and Latex in jupyter output cells") display(markdown) latex = Latex("$\phi$") display(latex) Displaying markdown and Latex in jupyter output cells¶ $$\phi$$ And of course the new WordPress editor is helpfully mangling both the code highlighting and output. Trust me, it works. I’ll have to sort out how to get “Gutenberg” to do the right thing later. Thrashing around a bit now get’s the code highlighted but now the crayon tag is visible. Still no love for the actual Latex. Rolling back to the old editor and fixing these posts. I’ll deal with the gutenberg editor issues when I’m dragged kicking and screaming to it So far this morning I’ve Gotten jupyter notebooks running natively on my workstation rather than using the linux subsystem as a workaround Learned how to integrate functions in python using scipy Got exporting of jupyter notebooks to html working properly Built a scale model of the terrain around my home town in Unity Gained three levels on my warlock Now I’m going to put my pants on. Integrating Functions Quick example on integrating functions with jupyter, matplotlib, numpy, and scipy Import the required libraries %matplotlib inline import numpy as np from scipy.integrate import quad from matplotlib import pyplot as plt Define your function For this example we’ll be using $f(x) = e^{-x}sin(3x)$ as our function. def f(x): """Sample equation""" return(np.exp(-x) * np.sin(3.0x)) Set the range for your plot and integration To plot $f(x)$ we get an array of linearly spaced values from numpy, pass that array and our function to the plot function, and show the plot. Evalating the integral $\int_{0}^{2\pi}e^{-x}sin(3x)$ simply requires passing our function and the range of the intgral to scipy’s quad function which will return a tuple of the integral and the estimated error. x = np.linspace(0, 2np.pi, 100) plt.plot(x, f(x)) plt.show() i, err = quad(f, 0, 2*np.pi) print("Integral:",i) print("Estimated error", err) Integral: 0.29943976718048754 Estimated error 5.05015300411582e-13 One of the benefits of coming back to an existing wordpress blog rather than continuing to try to roll my own in github.io is that I can easily piggyback on other people writing plugins to solve the same problems. So far, the Crayon and MathJax-Latex plugins seem to fit the bill. Testing syntax highlighting # hello_world.py class HelloWorld(): """An unnecessarily complex hello world""" def __init__(self): print("Hello world") if __name__ == "__main__": hello = HelloWorld() Testing Latex / Math $$ \begin{align} J(\theta) &= \frac{1}{m}\sum_{i=1}^{m}Cost(h_\theta(x)^{(i)},y^{(i)}) \\ Cost(h_\theta(x),y) &= -log(h_\theta(x))& y=1 \\ Cost(h_\theta(x),y) &= -log(1-h_\theta(x))& y=0 \end{align} $$
Before I state the questions I have in mind, let me give some background. If one considers $S^2$ then it is known due to Kneser that $\textrm{Homeo}^{+}(S^2)$ has the homotopy type of $SO(3)$. By Smale's work we also know that $\textrm{Diff}^{+}(S^2)$ is homotopic to $SO(3)$. However, when we work in the homotopy category this changes. Later Hansen considered $\textrm{Aut}_0(S^2)$, the connected component of identity in the space of all self homotopy equivalences. He showed that its homotopy type is that of $SO(3)\times \mathbf{\Omega}$, where $\mathbf{\Omega}$ is the universal cover of the connected component of the constant loop in the double loop space $\Omega^2 S^2$. Although this is a very nice and interesting fact, it's turns out to be hard to generalize his method of proof for higher spheres. For one, the homotopy type of $\textrm{Diff}(S^n)$ is not really known for $n\geq 7$ if I'm not mistaken. For another, Hansen crucially uses the fact that $S^2$ is the base of the usual fibration $SO(2)\to SO(3)\to S^2$ and the fact $SO(2)=S^1$ has no higher homotopy. This indubitably fails for higher spheres! Question 1 What is known about the homotopy type of $\textrm{Aut}_0(S^n)$ for $n\geq 3$? I should say that the rational homotopy type is fairly easily calculable via Sullivan's minimal models. So, I'm looking for a bit more here. Question 2 What is the homotopy type of the identity component of self homotopy equivalences of $\vee_k S^2$, a bouquet of $2$-spheres? Of course, one can ask this question for higher spheres but bearing in mind question 1, I decided I would be happy with an answer for $S^2$. If it helps, the homotopy groups of $\vee_k S^2$ can be calculated by Hilton-Milnor theorem. This fits in a long exact sequence of groups associated to $\textrm{Aut}_0^\ast(\vee_k S^2)\to \textrm{Aut}_0(\vee_k S^2)\to \vee_k S^2$ where the last map is evaluation of an automorphism at the common point of the bouquet and $\textrm{Aut}^\ast_0(\vee_k S^2)$ consists of based maps. But this doesn't seem to lead anywhere!
It is known that nucleons (proton, neutron) are composed of partons (quarks, etc.). How was this identified experimentally? In particular, how it has been identified that nucleons comprise of more than one constituent? Matt Strassler goes into detail with LHC data here: As you may know, when particles are scattered off a target, what actually gets measured is the differential cross section $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}$. This can basically be thought of as being related to the fraction of particles that come out of the collision in a particular direction. It's possible to calculate this quantity using quantum field theory, and we can then compare the results to the assumptions that go into that calculation. For the proton specifically, we express the differential cross section in terms of parton distribution functions (PDFs). Each of these functions, written $f_{i/\text{p}}(x, Q^2)$, gives something akin to the probability that a collision will involve a parton (quark or gluon) of type $i$ with a fraction $x$ of the proton's momentum, when the colliding particle has squared transverse momentum $Q^2$. So the overall cross section for, say, an electron hitting a proton has the basic structure $$\frac{\mathrm{d}\sigma_{\mathrm{ep}}}{\mathrm{d}\Omega} \sim \sum_{i}\int\mathrm{d}x\,f_{i/\text{p}}(x, Q^2)\frac{\mathrm{d}\hat\sigma_{\mathrm{e}i}}{\mathrm{d}\Omega}$$ In words, for each type of parton, you multiply the probability of finding that particular kind of parton with the cross section $\frac{\mathrm{d}\hat\sigma}{\mathrm{d}\Omega}$ for that particular interaction (which can be calculated from QFT). Then integrate the contributions from all possible values of $x$, and add that up for all possible parton types. That gets you the total cross section. (There's some other stuff involved which I'll leave out for simplicity.) Now, since $\frac{\mathrm{d}\hat\sigma_{\mathrm{e}i}}{\mathrm{d}\Omega}$ can be calculated, and $\frac{\mathrm{d}\sigma_{\mathrm{ep}}}{\mathrm{d}\Omega}$ can be measured, if you do enough different kinds of collisions at different energies, you can reconstruct the PDFs. And since each PDF is related to the density of its particular type of parton within the proton, if you know them all you basically know the composition of the proton. (Specifically, the quantity $xf_{i/\mathrm{p}}$ basically tells you the momentum distribution of parton type $i$.) For example: Suppose $xf_{\mathrm{u/p}}$ was measured and turned out to be a sharp spike at $x=1$. That means that when something collides with a proton, it only ever interacts with an up quark that carries all the proton's momentum, and that in turn tells you that the proton contains just an up quark and nothing else. Suppose $xf_{\mathrm{u/p}}$ instead turned out to be a spike at $x = \frac{1}{3}$. That would mean that when something collides with a proton, the only up quarks it ever interacts with carry $\frac{1}{3}$ of the proton's momentum. This would be the case if, for example, the proton was a rigidly bound collection of 3 quarks, each carrying $\frac{1}{3}$ of the total momentum, in which case the binding particles (gluons) would carry very little of the proton's momentum. You would, of course, have to measure the other PDFs (for down quarks, strange quarks, etc.) to determine what, if anything else, was in the proton. This is how you'd tell the difference between compositions of, say, $\mathrm{uuu}$ and $\mathrm{uud}$: in the latter case $xf_{\mathrm{d/p}}$ would have a spike of half the size, whereas in the former case $xf_{\mathrm{d/p}}$ would be zero everywhere. On the other hand, suppose you find that $xf_{\mathrm{u/p}}$ is a bump. That tells you that the up quarks in the proton don't have a definite fixed momentum, but rather that they move around with varying velocities. This would be a strong indication that the proton acts kind of like a fluid, with multiple constituent particles bouncing off each other and sharing their momentum around in the process. The peak of the bump would tell you (roughly) what is the most likely momentum fraction for the up quarks to have. Again, you would have to measure the other PDFs to figure out what other kinds of particles are in this "fluid," and in what proportions - for example, if there are twice as many up quarks as down quarks, you would find that the bump in $xf_{\mathrm{u/p}}$ is twice the height of $xf_{\mathrm{d/p}}$. It turns out (perhaps unsurprisingly) that this last case is most relevant for a real proton. Now, I couldn't find the picture I really wanted to illustrate this, but I did get to make the following graph using polarized structure function data from SLAC-143, and it makes a similar point. You can see that there's a wide bump centered around $x = \frac{1}{3}$. This indicates that the partons in the proton are most likely to be carrying a third of the proton's momentum each, and this is the sense in which one might say that there are 3 constituent particles. (Technically, because this is a polarized structure function, it's talking about the difference between particles with opposite spins. But the unpolarized equivalent would look roughly similar.) Taking into account all the data that has been collected over the last 40 years or so, the current PDFs for the proton look like this: As you can see, this combines some of the different features I discussed above. Most notably, there's a bump around $x = \frac{1}{3}$, and it's twice as large for up quarks as for down quarks, which means that there 2 valence up quarks and 1 valence down quark. But there is also a sharp rise in all the PDFs as you go to very small $x$, which is kind of like a spike. It indicates that there are a very large number of partons carrying very small fractions of the total momentum.
Difference between revisions of "Vertex equations of a quadratic function and it's inverse" From JSXGraph Wiki Line 1: Line 1: A parabola can be uniquely defined by its vertex ''V'' and one more point ''P''. A parabola can be uniquely defined by its vertex ''V'' and one more point ''P''. The function term of the parabola then has the form The function term of the parabola then has the form − :<math> + :<math>= R_0\cdot \gamma</math> Revision as of 12:27, 16 December 2014 A parabola can be uniquely defined by its vertex V and one more point P.The function term of the parabola then has the form [math]\alpha = R_0\cdot \gamma[/math] JavaScript code var b = JXG.JSXGraph.initBoard('box1', {boundingbox: [-5, 5, 5, -5], grid:true});var v = b.create('point', [0,0], {name:'V'}), p = b.create('point', [3,3], {name:'P'}), f = b.create('functiongraph', [ function(x) { var den = p.X()- v.X(), a = (p.Y() - v.Y()) / (den * den); return a * (x - v.X()) * (x - v.X()) + v.Y(); }]);})(); JavaScript code var b = JXG.JSXGraph.initBoard('box2', {boundingbox: [-5, 5, 5, -5], grid:true});var v = b.create('point', [0,0], {name:'V'}), p = b.create('point', [3,3], {name:'P'}), f = b.create('functiongraph', [ function(x) { var den = p.Y()- v.Y(), a = (p.X() - v.X()) / (den * den); return Math.sqrt((x - v.X()) / a) + v.Y(); }]);
Let us define the following cumulative distribution: \begin{align} \Pr (Y(t)\geq a)=\sum_{n=0}^\infty \frac{e^{-kt}(kt)^n }{n!}\int_a^\infty[\circledast_{f}\circ H_a ]^n\delta (x) dx \end{align} where the operator $[\circledast_{f_J}\circ H_a ]$ denotes the multiplication by a Heaviside function with a cut off at $a$ followed by the convolution by a pdf $f$. This is a cumulative distribution function in the sense that $\Pr (Y(t)> 0)=0$ and $\Pr (Y(t)\geq -\infty)=1$ (or we could say that $1-\Pr (Y(t)\geq a)$ is the CDF). A global question would be: what can I know about this distribution, given the pdf $f$? An intuitive property would be $\langle Y(t) \rangle\xrightarrow[t\rightarrow +\infty]{}-\infty \quad \forall f \;\|\; \mu_f<0$ A more specific one is: how to compute $\langle Y(+\infty) \rangle$ given $f$? My first idea was to notice that for $t\rightarrow + \infty$, we can just consider terms with a large $n$ in the series. Then I wanted to take inspiration from the central limit theorem. Given that the pdf $f$ has definite first and second moments: \begin{align} [\circledast_{f}]^n\delta (x)=ff...*f\sim\mathcal{N}\Big(\frac{x-\mu_f n}{\sigma_f \sqrt{n}}\Big)\quad \text{as }\quad n\rightarrow \infty \end{align} However we have here \begin{align} \int_a^\infty[\circledast_{f}\circ H_a]^n\delta (x)dx\sim \quad ? \quad <\int_a^\infty \mathcal{N}\Big(\frac{x-\mu_f n}{\sigma_f \sqrt{n}}\Big)dx\quad \text{as }\quad n\rightarrow \infty \end{align} Here are other complementary informations: For the special case $f(x)=p_-\delta(x+1)+p_+\delta(x-1)$, I have found, with a random walk approach, that $\langle Y(+\infty) \rangle = \frac{1}{1-\frac{p_+}{p_-}}$ for $p_+>p_-$. following this kind of approximation along with the previous result, we end up with the following approximation: $\langle Y(+\infty) \rangle \approx \frac{1}{2}\Big(\sqrt{\mu_f^2+\sigma_f^2}-\frac{\mu_f^2+\sigma_f^2}{\mu_f}\Big)$ for $\mu_f>0$, which is not exact according to numerical simulations but which gives the order of magnitude. This post is related to this one.
Branch misprediction is expensive: an exampleThe SciMark 2.0 Monte Carlo benchmark is calculating $\pi$ by generating random points $\{(x,y) \mid x,y \in [0,1]\}$ and calculating the ratio of points that are located within the quarter circle $\sqrt{x^2 + y^2} \le 1$. The square root can be avoided by squaring both sides, and the benchmark is implemented as double MonteCarlo_integrate(int Num_samples) { Random R = new_Random_seed(SEED); int under_curve = 0; for (int count = 0; count < Num_samples; count++) { double x = Random_nextDouble(R); double y = Random_nextDouble(R); if (xx + yy <= 1.0) under_curve++; } Random_delete(R); return ((double) under_curve / Num_samples) * 4.0; }GCC used to generate a conditional move for this if-statement, but a recent change made this generate a normal branch which caused a 30% performance reduction for the benchmark due to the branch being mispredicted (bug 79389). The randomization function is not inlined as it is compiled in a separate file, and it contains a non-trivial amount of loads, stores, and branches typedef struct { int m[17]; int seed, i, j, haveRange; double left, right, width; } Random_struct, Random; #define MDIG 32 #define ONE 1 static const int m1 = (ONE << (MDIG-2)) + ((ONE << (MDIG-2)) - ONE); static const int m2 = ONE << MDIG/2; static double dm1; double Random_nextDouble(Random R) { int I = R->i; int J = R->j; int m = R->m; int k = m[I] - m[J]; if (k < 0) k += m1; R->m[J] = k; if (I == 0) I = 16; else I--; R->i = I; if (J == 0) J = 16; else J--; R->j = J; if (R->haveRange) return R->left + dm1 * (double) k * R->width; else return dm1 * (double) k; }so I had expected the two calls to this function to dominate the running time, and that the cost of the branch would not affect the benchmark too much. But I should have known better — x86 CPUs can have more than 100 instructions in flight (192 micro-ops for Broadwell), and a mispredict need to throw away all that work and restart from the actual branch target. Branch overhead and branch predictionThe cost of branch instructions differ between different CPU implementations, and the compiler needs to take that into account when optimizing and generating branches. Simple processors with a 3-stage pipeline fetch the next instruction when previous two instructions are decoded and executed, but branches introduce a problem: the next instruction cannot be fetched before the address is calculated by executing the branch instruction. This makes branches expensive as they introduce bubbles in the pipeline. The cost can be reduced for conditional branches by speculatively fetching and decoding the instructions after the branch — this improves performance if the branch was not taken, but taken branches need to discard the speculated work, and restart from the actual branch target. Some CPUs have instructions that can execute conditionally depending on a condition, and this can be used to avoid branches. For example if (c) a += 3; else b -= 2;can be compiled to the following straight line code on ARM (assuming that a, b, and c are placed in r1, r2, and r0 respectively) cmp r0, #0 addne r1, r1, #3 subeq r2, r2, #2The cmp instruction sets the Z flag in the status register, and the addne instruction is treated as an addition if Z is 0, and as a nop instruction if Z is 1. subeq is similarly treated as a subtraction if Z is 1 and as a nop if Z is 0. The instruction takes time to execute, even when treated as a nop, but this is still much faster than executing branches. This means that the compiler should structure the generated code so that branches are minimized (using conditional execution when possible), and conditional branches should be generated so that the most common case is not taking the branch. Taken branches become more expensive as the CPUs get deeper pipelines, and this is especially annoying as loops must branch to the top of the loop for each iteration. This can be solved by adding more hardware to let the fetch unit calculate the target address of the conditional branch, and the taken branch can now be the cheap case. It is, however, nice to have the “not taken” case be the cheap case, as the alternative often introduce contrived control flow that fragments the instruction cache and need to insert extra “useless” (and expensive) unconditional branches. The way most CPUs solve this is to predict that forward branches are unlikely (and thus speculatively fetch from following instructions), and that backward branches are likely (and thus speculatively fetch from the branch target). The compiler should do similar work as for the simpler CPU, but structure the code so that conditional branches branching forward are not taken in the common case, and conditional branches branching backward are taken in the common case. There are many branches that the compiler cannot predict, so the next step up in complexity is adding branch prediction to the CPU. The basic idea is that the CPU keeps a cache of previous branch decisions and use this to predict the current branch. High-end branch predictors look at the history of code flow, and can correctly predict repetitive patterns in how the branch behaved. Hardware vendors do not publish detailed information about how the prediction work, but Agner Fog’s optimization manuals contain lots of information (especially part 3, “The microarchitecture of Intel, AMD and VIA CPUs”, that also have a good overview of different ways branch prediction can be done). Branch prediction in high-end CPUs is really good, so branches are essentially free, while conditional execution adds extra dependencies between instructions which constrain the out-of-order execution engine, so conditional execution should be avoided. This is essentially the opposite from how the simple CPUs should be handled. 😃 There is one exception — branches that cannot be predicted (such as the one in SciMark) should be generated using conditional instructions, as the branch will incur the misprediction cost of restarting the pipeline each time it is mispredicted. The compiler should not use conditional execution unless the condition is unpredictable. The code should be structured as for the static prediction (this is not strictly necessary, but most CPUs use the static prediction first time a branch is encountered. And it is also slightly more efficient for the instruction cache). So branches are free, except when they cannot be predicted. I find it amusing that many algorithms (balanced search trees, etc.) have the aim to make the branches as random as possible. I do not know how much this is a problem in reality, but Clang has a built-in function, __builtin_unpredictable, that can be used to tell the compiler that the condition is unpredictable. Heuristics for estimating branch probabilities The compiler estimates branch probabilities in order to generate the efficient form of the branch (there are more optimizations that need to know if a code segment is likely executed or not, such as inlining and loop unrolling). The general idea, as described in the PLDI ’93 paper “Branch Prediction for Free”, is to look at how the branches are used. For example, code such as GCC has a number of such predictors, for example The predictors are defined in if (p == NULL) return -1;comparing a pointer with NULL and returning a constant, is most likely error handling, and thus unlikely to execute the return statement. GCC has a number of such predictors, for example Branch ending with returning a constant is probably not taken. Branch from comparison using !=is probably taken, ==is probably not taken. Branch to a basic block calling a cold function is probably not taken. __builtin_expect often does not make any difference — the heuristics are already coming to the same conclusion! The predictors are defined in predict.def (some of the definitions seem reversed due to how the rules are implemented, e.g. PROB_VERY_LIKELY may mean “very unlikely”, but the comments describing each heuristic are correct). You can see how GCC is estimating the branch probabilities by passing -fdump-tree-profile_estimate to the compiler, which writes a file containing the output from the predictors for each basic block Predictions for bb 2 DS theory heuristics: 1.7% combined heuristics: 1.7% pointer (on trees) heuristics of edge 2->4: 30.0% call heuristics of edge 2->3: 33.0% negative return heuristics of edge 2->4: 2.0%as well as (when using GCC 7.x) the IR annotated with the estimated probabilities.
S H Kulkarni Articles written in Proceedings – Mathematical Sciences Volume 95 Issue 1 September 1986 pp 37-40 The relationship between the harmonicity and analyticity of a continuous map from the open unit disc to the underlying space of a real algebra is investigated. Volume 118 Issue 4 November 2008 pp 613-625 Let $H_1, H_2$ be Hilbert spaces and 𝑇 be a closed linear operator defined on a dense subspace $D(T)$ in $H_1$ and taking values in $H_2$. In this article we prove the following results: (i) Range of 𝑇 is closed if and only if 0 is not an accumulation point of the spectrum $\sigma(T^\ast T)$ of $T^\ast T$, In addition, if $H_1=H_2$ and 𝑇 is self-adjoint, then (ii) $\inf \{\\| Tx\\|:x\in D(T)\cap N(T)^\perp \\| x\\|=1\}=\inf\{\| \lambda\|:0\neq\lambda\in\sigma(T)\}$, (iii) Every isolated spectral value of 𝑇 is an eigenvalue of 𝑇, (iv) Range of 𝑇 is closed if and only if 0 is not an accumulation point of the spectrum $\sigma(T)$ of 𝑇, (v) $\sigma(T)$ bounded implies 𝑇 is bounded. We prove all the above results without using the spectral theorem. Also, we give examples to illustrate all the above results. Current Issue Volume 129 \| Issue 5 November 2019 Click here for Editorial Note on CAP Mode
This is a classic exercise of functional analysis, but I do not fully understand it after reading many answers in textbooks. So I am trying to reorganize the proof step by step in details. I am hoping that someone may review my proof very carefully and give comments or corrections. Then I will revise the proof and hopefully it could be helpful to the beginners of functional analysis. Let $c_0(\mathbb{N})$ be the space of sequences converging to $0$. Show that there is a well-defined, isometric isomorphism \begin{align} T: l^1(\mathbb{N}) \to \left(c_0(\mathbb{N})\right)^, \qquad T(g)(f) := \sum_{n\in\mathbb{N}}f(n)g(n). \end{align} That is, show that $T(g)$ is a bounded linear functional $c_0(\mathbb{N}) \to \mathbb{C}$ with $\\|T(g)\\| = \\|g\\|$ and that any bounded linear functional on $c_0(\mathbb{N})$ is of this form for a unique $g \in l^1(\mathbb{N})$. My proof: First of all, we denote the sequences $f \in c_0(\mathbb{N})$ and $g \in l^1(\mathbb{N})$, thus $T(g) \in c_0^: c_0(\mathbb{N}) \to \mathbb{C}$. By the way, can we say $T(f): l^1(\mathbb{N}) \to \mathbb{C}$? I think it is not well-defined. Boundedness: To show $T(g)$ is a bounded linear functional $c_0(\mathbb{N}) \to \mathbb{C}$ with $\\|T(g)\\| = \\|g\\|_{l^1}$, we first show the boundedness. \begin{align} \|T(g)(f)\| = \left\|\sum_{n\in\mathbb{N}}f(n)g(n)\right\| \le \sum_{n\in\mathbb{N}}\|f(n)\|\|g(n)\| \le \sup_{n\in\mathbb{N}}\|f(n)\|\sum_{n\in\mathbb{N}}\|g(n)\| = \\|f\\|_{l^\infty}\\|g\\|_{l^1} \end{align} Therefore, $T(g)$ is bounded by \begin{align} \\|T(g)\\| = \sup\{\|T(g)(f)\|: \forall f \in c_0(\mathbb{N}), \\|f\\|_{l^\infty}\le 1\} \le \\|g\\|_{l^1} \end{align} Linearity: To show the linearity, we define $f_1, f_2 \in c_0(\mathbb{N})$ and $a_1, a_2 \in \mathbb{C}$. Then \begin{align} T(g)(a_1 f_1 + a_2 f_2) &= \sum_{n\in\mathbb{N}} \left(a_1 f_1(n) + a_2 f_2(n)\right) g(n) \\ &= \sum_{n\in\mathbb{N}} \left(a_1 f_1(n) g(n) + a_2 f_2(n) g(n)\right) \\ &= a_1 \sum_{n\in\mathbb{N}} f_1(n) g(n) + a_2 \sum_{n\in\mathbb{N}} f_2(n) g(n) \\ &= a_1 T(g)(f_1) + a_2 T(g)(f_2) \end{align} implies that $T(g)$ is linear. Isometry: We already proved $T(g): c_0(\mathbb{N}) \to \mathbb{C}$ a bounded linear functional for all $g \in l^1(\mathbb{N})$, now can we say the operator $T: l^1(\mathbb{N}) \to \left(c_0(\mathbb{N})\right)^$ is therefore well-defined? Next we need to show that $T$ is an isometry for which $\\|T(g)\\| = \\|g\\|_{l^1}$. Since we already have $\\|T(g)\\| \le \\|g\\|_{l^1}$ from the boundedness, if we are able to show that there exists some $f \in c_0(\mathbb{N})$ for which $\\|T(g)\\| \ge \\|g\\|_{l^1}$, then $\\|T(g)\\| = \\|g\\|_{l^1}$. Let $g$ be a sequence in $l^1(\mathbb{N})$. If $g = 0$, then $\\|T(g)\\| = \\|g\\|_{l^1}$ holds trivially. Assuming $g \ne 0$, we define \begin{align} f(n) := \begin{cases} \frac{\|g(n)\|}{g(n)} &\qquad n \le N \\ 0 &\qquad n > N \end{cases} \end{align} which is a sequence in $c_0(\mathbb{N})$, with $T(g)(f) := \sum_{n\in\mathbb{N}}f(n)g(n) = \sum_{n\in\mathbb{N}}\|g(n)\| =: \\|g\\|_{l^1}$ and $\\|f\\|_{l^\infty} = 1$ by definition. Therefore, we have \begin{align} \\|g\\|_{l^1} = \|T(g)(f)\| \le \\|T(g)\\|\\|f\\|_{l^\infty} = \\|T(g)\\| \end{align} in addition to $\\|T(g)\\| \le \\|g\\|_{l^1}$, which implies $\\|T(g)\\| = \\|g\\|_{l^1}$, i.e., $T$ is an isometry and thus injective (one-to-one). Surjectivity: To prove that $T: l^1(\mathbb{N}) \to \left(c_0(\mathbb{N})\right)^$ is surjective (onto), we need to show for any bounded linear functional $\forall S \in \left(c_0(\mathbb{N})\right)^$ there exists $T(g) = S$ that has some preimage $g \in l^1(\mathbb{N})$. Let us build a basis $\{e_n: n = 1, 2,...\}$ of sequences for $c_0(\mathbb{N})$, where $e_n = (\delta_n)_{n \in \mathbb{N}} \in c_0(\mathbb{N})$. Any $f \in c_0(\mathbb{N})$ can be written coordinate-wisely by the linear combination of the sequences of the basis $f = \sum_{n\in\mathbb{N}} f(n) e_n$. Then by the linearity of $S$ we have\begin{align}S(f) = \sum_{n\in\mathbb{N}} f(n) S(e_n)\end{align}If we define $g(n) := S(e_n)$, then we find\begin{align}S(f) = \sum_{n\in\mathbb{N}} f(n) S(e_n) = \sum_{n\in\mathbb{N}} f(n) g(n) =: T(g)(f)\end{align}that implies $S = T(g)$. By definition, $g(n)$ maps each sequence of basis to $T(g)(e_n)$. In addition, we need to show this $g \in l^1(\mathbb{N})$ by $\\|g\\|_{l^1} \le \\|S\\|$ which I am not able to finish. In conclusion, we find $\exists g \in l^1(\mathbb{N})$ for $T(g)$ corresponding to $\forall S \in \left(c_0(\mathbb{N})\right)^$, therefore $T$ is surjective. Finally, $T$ is a well-defined isometric isomorphism. Please show me that $\\|g\\|_{l^1} \le \\|S\\|$ How to show this $g$ is unique?
@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$ Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine" @LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$ I enjoyed developing the lebesgue integral though. I thought that was cool But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons? uh wait a sec... ... I need arbitrary intersection to produce singletons from open intervals... hmm... 0 does not even have a nbhd, since any set containing 0 is closed I have no idea how to deal with points having empty nbhd o wait a sec... the open set of any topology must contain the whole set itself so I guess the nbhd of 0 is $\Bbb{R}$ Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…) Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time So going from a to b directly is no more efficient than go from a to 0 and then 0 to b hmm... $d(A \to B \to C) = d(A,B)+d(B,C) = \|a\|+\|b\|+\|b\|+\|c\|$ $d(A \to 0 \to C) = d(A,0)+d(0,C)=\|a\|+\|c\|$ so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination But if the starting point is nonzero, then the distance increases quadratically Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius) only when travelling via the origin, will such qudratic penalty in travelling distance be not apply More interesting things can be said about slight generalisations of this metric: Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ... To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ... 1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d... Any alg geom guys on? I know zilch about alg geom to even start analysing this question Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — baluApr 13 at 18:24 grr, thought I can get some more intuition in SR by using open balls tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two. The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$. And all it has in the proof is the assertion that the above is clearly true. I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k... Consider $a,b$ both irrational and the interval $[a,b]$ Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$ It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.One way for an irrational number $\alpha$ to be in this new set is b... Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals.Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat... (For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs) Let $\lambda^$ be the Lebesgue outer measure We previously proved that $\lambda^(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero. We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting, or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi… W also knew that one way to compute $\lambda^(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^(\Bbb{I}\cap [-1,1])$ is to use $\lambda^(\Bbb{Q}\cap [-1,1])+\lambda^(\Bbb{I}\cap [-1,1]) = \lambda^([-1,1])$ and thus deducing $\lambda^(\Bbb{I}\cap [-1,1]) = 2$ However, what I am interested here is to compute $\lambda^(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below: We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages: First denote an enumeration of the rationals as follows: $\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short: Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $\|p-q\| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works... Let's say that for $N$ clients, Lotta will take $d_N$ days to retire. For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$ Where $d_1$ = 1. Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps.
There are two ways to think of the Hilbert space as the space of sections of a line bundle. First, the exponentiated Chern-Simons action on a manifold $\Sigma\times[0,1]$ is a section of the determinant line bundle $\mathcal{L}_\Sigma$ on the space of flat connections on $\Sigma$. Moreover, Wilson loops (which can be thought of as a 1d TFT) contribute $R_i$ each. So, the Hilbert space (before remembering gauge invariance) is $\Gamma(\mathcal{L}_\Sigma^k)\otimes\bigotimes_i R_i$. Now, if $\Sigma = S^2$, which is simply-connected, the space of flat connections is a point, so $\Gamma(\mathcal{L}_\Sigma^k)=\mathbf{C}$. Finally, gauge invariance picks out the $G$-invariants in $\bigotimes_i R_i$. Note, that the Hilbert space for a non-simply-connected $\Sigma$ is nontrivial even without the punctures. Another way to think of this Hilbert space is to recall the 2d CFT <-> 3d TFT correspondence. The idea here is the following. Correlation functions of a 2d CFT live in a certain bundle over the moduli space of complex curves $M_{g,n}$ called the bundle of conformal blocks. The Knizhnik-Zamolodchikov equations on correlation functions correspond to a (projectively) flat connection on this bundle. So, a 2d CFT associates global sections of this bundle to a topological surface $\Sigma$, this is the Hilbert space in a 3d TFT. In the case of the Chern-Simons theory, the associated 2d CFT is the Wess-Zumino-Witten model. A down-to-earth description can be found in S. Elitzur, G. Moore, A. Schwimmer, N. Seiberg, Remarks on the Canonical Quantization of the Chern-Simons-Witten Theory, Nucl Phys B326 (1989), 108. Mathematically, this correspondence is an equivalence between modular functors (as defined by Segal in The definition of conformal field theory) and modular tensor categories which give rise to 3d TFTs (due to Reshetikhin and Turaev). All of that is discussed in an excellent book Lectures on tensor categories and modular functors by Bakalov and Kirillov.This post imported from StackExchange Physics at 2014-04-04 16:44 (UCT), posted by SE-user Pavel Safronov
2019-09-04 12:06 Soft QCD and Central Exclusive Production at LHCb / Kucharczyk, Marcin (Polish Academy of Sciences (PL)) The LHCb detector, owing to its unique acceptance coverage $(2 < \eta < 5)$ and a precise track and vertex reconstruction, is a universal tool allowing the study of various aspects of electroweak and QCD processes, such as particle correlations or Central Exclusive Production. The recent results on the measurement of the inelastic cross section at $ \sqrt s = 13 \ \rm{TeV}$ as well as the Bose-Einstein correlations of same-sign pions and kinematic correlations for pairs of beauty hadrons performed using large samples of proton-proton collision data accumulated with the LHCb detector at $\sqrt s = 7\ \rm{and} \ 8 \ \rm{TeV}$, are summarized in the present proceedings, together with the studies of Central Exclusive Production at $ \sqrt s = 13 \ \rm{TeV}$ exploiting new forward shower counters installed upstream and downstream of the LHCb detector. [...] LHCb-PROC-2019-008; CERN-LHCb-PROC-2019-008.- Geneva : CERN, 2019 - 6. Fulltext: PDF; In : The XXVII International Workshop on Deep Inelastic Scattering and Related Subjects, Turin, Italy, 8 - 12 Apr 2019 Подробен запис - Подобни записи 2019-08-15 17:39 LHCb Upgrades / Steinkamp, Olaf (Universitaet Zuerich (CH)) During the LHC long shutdown 2, in 2019/2020, the LHCb collaboration is going to perform a major upgrade of the experiment. The upgraded detector is designed to operate at a five times higher instantaneous luminosity than in Run II and can be read out at the full bunch-crossing frequency of the LHC, abolishing the need for a hardware trigger [...] LHCb-PROC-2019-007; CERN-LHCb-PROC-2019-007.- Geneva : CERN, 2019 - mult.p. In : Kruger2018, Hazyview, South Africa, 3 - 7 Dec 2018 Подробен запис - Подобни записи 2019-08-15 17:36 Tests of Lepton Flavour Universality at LHCb / Mueller, Katharina (Universitaet Zuerich (CH)) In the Standard Model of particle physics the three charged leptons are identical copies of each other, apart from mass differences, and the electroweak coupling of the gauge bosons to leptons is independent of the lepton flavour. This prediction is called lepton flavour universality (LFU) and is well tested. [...] LHCb-PROC-2019-006; CERN-LHCb-PROC-2019-006.- Geneva : CERN, 2019 - mult.p. In : Kruger2018, Hazyview, South Africa, 3 - 7 Dec 2018 Подробен запис - Подобни записи 2019-05-15 16:57 Подробен запис - Подобни записи 2019-02-12 14:01 XYZ states at LHCb / Kucharczyk, Marcin (Polish Academy of Sciences (PL)) The latest years have observed a resurrection of interest in searches for exotic states motivated by precision spectroscopy studies of beauty and charm hadrons providing the observation of several exotic states. The latest results on spectroscopy of exotic hadrons are reviewed, using the proton-proton collision data collected by the LHCb experiment. [...] LHCb-PROC-2019-004; CERN-LHCb-PROC-2019-004.- Geneva : CERN, 2019 - 6. Fulltext: PDF; In : 15th International Workshop on Meson Physics, Kraków, Poland, 7 - 12 Jun 2018 Подробен запис - Подобни записи 2019-01-21 09:59 Mixing and indirect $CP$ violation in two-body Charm decays at LHCb / Pajero, Tommaso (Universita & INFN Pisa (IT)) The copious number of $D^0$ decays collected by the LHCb experiment during 2011--2016 allows the test of the violation of the $CP$ symmetry in the decay of charm quarks with unprecedented precision, approaching for the first time the expectations of the Standard Model. We present the latest measurements of LHCb of mixing and indirect $CP$ violation in the decay of $D^0$ mesons into two charged hadrons [...] LHCb-PROC-2019-003; CERN-LHCb-PROC-2019-003.- Geneva : CERN, 2019 - 10. Fulltext: PDF; In : 10th International Workshop on the CKM Unitarity Triangle, Heidelberg, Germany, 17 - 21 Sep 2018 Подробен запис - Подобни записи 2019-01-15 14:22 Experimental status of LNU in B decays in LHCb / Benson, Sean (Nikhef National institute for subatomic physics (NL)) In the Standard Model, the three charged leptons are identical copies of each other, apart from mass differences. Experimental tests of this feature in semileptonic decays of b-hadrons are highly sensitive to New Physics particles which preferentially couple to the 2nd and 3rd generations of leptons. [...] LHCb-PROC-2019-002; CERN-LHCb-PROC-2019-002.- Geneva : CERN, 2019 - 7. Fulltext: PDF; In : The 15th International Workshop on Tau Lepton Physics, Amsterdam, Netherlands, 24 - 28 Sep 2018 Подробен запис - Подобни записи 2019-01-10 15:54 Подробен запис - Подобни записи 2018-12-20 16:31 Simultaneous usage of the LHCb HLT farm for Online and Offline processing workflows LHCb is one of the 4 LHC experiments and continues to revolutionise data acquisition and analysis techniques. Already two years ago the concepts of “online” and “offline” analysis were unified: the calibration and alignment processes take place automatically in real time and are used in the triggering process such that Online data are immediately available offline for physics analysis (Turbo analysis), the computing capacity of the HLT farm has been used simultaneously for different workflows : synchronous first level trigger, asynchronous second level trigger, and Monte-Carlo simulation. [...] LHCb-PROC-2018-031; CERN-LHCb-PROC-2018-031.- Geneva : CERN, 2018 - 7. Fulltext: PDF; In : 23rd International Conference on Computing in High Energy and Nuclear Physics, CHEP 2018, Sofia, Bulgaria, 9 - 13 Jul 2018 Подробен запис - Подобни записи 2018-12-14 16:02 The Timepix3 Telescope andSensor R&D for the LHCb VELO Upgrade / Dall'Occo, Elena (Nikhef National institute for subatomic physics (NL)) The VErtex LOcator (VELO) of the LHCb detector is going to be replaced in the context of a major upgrade of the experiment planned for 2019-2020. The upgraded VELO is a silicon pixel detector, designed to with stand a radiation dose up to $8 \times 10^{15} 1 ~\text {MeV} ~\eta_{eq} ~ \text{cm}^{−2}$, with the additional challenge of a highly non uniform radiation exposure. [...] LHCb-PROC-2018-030; CERN-LHCb-PROC-2018-030.- Geneva : CERN, 2018 - 8. Подробен запис - Подобни записи
Suppose that $f : (a, b) → \mathbb{R}$ is differentiable at $x ∈ (a, b)$. Prove that $\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}$ exists and equals $f'(x)$ Give an example of a function where the limit exists but the function is not differentiable. Proof: $\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}$ =$\lim_{h→0}\frac{f((x-h) + 2h) − f(x − h)}{2h}$ $= f'(x)$ by definition which exists as $f:(a,b) \to \mathbb{R}$ is differentiable $x \in (a,b)$. Can anyone verify what I have done? I think I'm going wrong somewhere. Can anyone please help? Also, I can't think of a function whose limit exists but it is not differentiable. I was thinking of $f(x) =\|x\|$ but don't know if its correct. Thank you. EDIT (After the comments and hints received): $\lim_{h→0}\frac{f(x + h) − f(x − h)}{2h}$=$\lim_{h→0}\frac{[f(x + h)-f(x)]+[f(x) − f(x − h)]}{2h}$=$\frac{1}{2}\lim_{h→0}\frac{[f(x + h)-f(x)]+[f(x) − f(x− h)]}{h}$ =$\frac{1}{2}[\lim_{h→0}\frac{f(x + h)-f(x)}{h}$ + $\lim_{h→0}\frac{f(x) − f(x − h)}{h}]=\frac{1}{2}[f'(x)+f'(x)]=f'(x)$ Is this ok? I'm particularly concerned about breaking the limits into two. Is that allowed and can anyone give me a proper theorem or a result/justification that allows me to do that? Thank you!!
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
Let ${\cal A}$ be a set of $k\times n $ matrices over ${\mathbb F}_2$. We call ${\cal A}$ a (n,k)-covering, if for every subset of columns$I=(i_1,\ldots,i_k)\subseteq [n]$, there is a matrix $A \in {\cal A}$such that the columns $(i_1,\ldots,i_k)$ in $A$ are linearly independent. It is known that when choosing a random $k\times k$ matrix over ${\mathbb F}_2$ the probability that it is non-singular is at least half. This means that if we pick at random a set of $r$ such matrices, the probability that for some $I=(i_1,\ldots,i_k)\subseteq [n]$ won't be covered (by any matrix $A \in {\cal A}$) is at most $2^{-r}$. Since there are $n \choose k$ such $I$'s, the probability that one of them will not be covered is at most (union bound): $${n \choose k} \cdot 2^{-r}.$$ If we force this probability to be strictly lower than 1 (which ensures such family of size $r$ exists), we get: $${n \choose k} \cdot 2^{-r}<1 \Rightarrow r > k \log n.$$ Is there a deterministic build of such family of size $O(k \log n)$, or at least of size $\text{poly}(k,\log n)$? Motivation for this problem: I have an algorithm that searches for subgraphs of a given graph $G$ which are similar to a smaller graph $H$ on $k$ vertices. As part of the algorithm it gets a $k\times n$ matrix (each vertex in the graph is associated with a binary vector of length k) and then it checks that all of the vertices which are matched to $H$ are different by checking that the induced $k \times k$ matrix is not singular. On the one hand, if two vertices of $H$ were mapped to the same vertex, their vectors are the same and the matrix's rank will surely be less than $k$. On the other, I'm not guaranteed that if the vertices were distinct the resulting matrix will be of full rank, so I run the algorithm for every matrix in my $(n,k)$-covering family. Currently I can draw random matrices and get the result with high probability, but an explicit build will allow me to derandomize the algorithm.
So we try by applying the definition and see how it goes. So we deal with 2 cases, either $\sup S=\infty$ or $\sup S=M<\infty$. If it is the second case, this means that for all $x\in S$, $x\leq M$, so naturally $ax\leq aM$. By definition of the least upper bound, we have $\sup aS\leq aM=a\sup S$. For the reverse, let $\epsilon>0$ be given. If $a>0$, then $\frac{\epsilon}{a}>0$. By the definition of sup, there exists $x\in S$ such that $M-\frac{\epsilon}{a}\leq x\leq M$. Hence $aM-\epsilon\leq x\leq aM$. Since this $\epsilon$ is arbitrary, so $\sup S=M$. If $a=0$ then we have nothing to say. And as for the infinity case, it should be quite evident, by choosing a sequence in $S$ such that it goes to infinity.
Root system data for affine Cartan types¶ class sage.combinat.root_system.type_affine. AmbientSpace( root_system, base_ring)¶ Ambient space for affine types. This is constructed from the data in the corresponding classical ambient space. Namely, this space is obtained by adding two elements $\delta$ and $\delta^\vee$ to the basis of the classical ambient space, and by endowing it with the canonical scalar product. The coefficient of an element in $\delta^\vee$, thus its scalar product with $\delta^\vee$ gives its level, and dually for the colevel. The canonical projection onto the classical ambient space (by killing $\delta$ and $\delta^\vee$) maps the simple roots (except $\alpha_0$) onto the corresponding classical simple roots, and similarly for the coroots, fundamental weights, … Altogether, this uniquely determines the embedding of the root, coroot, weight, and coweight lattices. See simple_root()and fundamental_weight()for the details. Warning In type $BC$, the null root is in fact: sage: R = RootSystem(["BC",3,2]).ambient_space() sage: R.null_root() 2e['delta'] Warning In the literature one often considers a larger affine ambient space obtained from the classical ambient space by adding four dimensions, namely for the fundamental weight $\Lambda_0$ the fundamental coweight $\Lambda^\vee_0$, the null root $\delta$, and the null coroot $c$ (aka central element). In this larger ambient space, the scalar product is degenerate: $\langle \delta,\delta\rangle=0$ and similarly for the null coroot. In the current implementation, $\Lambda_0$ and the null coroot are identified:sage: L = RootSystem([“A”,3,1]).ambient_space() sage: Lambda = L.fundamental_weights() sage: Lambda[0] e[‘deltacheck’] sage: L.null_coroot() e[‘deltacheck’] Therefore the scalar product of the null coroot with itself differs from the larger ambient space: sage: L.null_coroot().scalar(L.null_coroot()) 1 In general, scalar products between two elements that do not live on “opposite sides” won’t necessarily match. EXAMPLES: sage: R = RootSystem(["A",3,1]) sage: e = R.ambient_space(); e Ambient space of the Root system of type ['A', 3, 1] sage: TestSuite(e).run() Systematic checks on all affine types: sage: for ct in CartanType.samples(affine=True, crystallographic=True): ....: if ct.classical().root_system().ambient_space() is not None: ....: print(ct) ....: L = ct.root_system().ambient_space() ....: assert L ....: TestSuite(L).run() ['A', 1, 1] ['A', 5, 1] ['B', 1, 1] ['B', 5, 1] ['C', 1, 1] ['C', 5, 1] ['D', 3, 1] ['D', 5, 1] ['E', 6, 1] ['E', 7, 1] ['E', 8, 1] ['F', 4, 1] ['G', 2, 1] ['BC', 1, 2] ['BC', 5, 2] ['B', 5, 1]^ ['C', 4, 1]^* ['F', 4, 1]^* ['G', 2, 1]^* ['BC', 1, 2]^* ['BC', 5, 2]^* class Element¶ Bases: sage.modules.with_basis.indexed_element.IndexedFreeModuleElement associated_coroot()¶ Return the coroot associated to self. INPUT: self– a root EXAMPLES: sage: alpha = RootSystem(['C',2,1]).ambient_space().simple_roots() sage: alpha Finite family {0: -2e[0] + e['delta'], 1: e[0] - e[1], 2: 2e[1]} sage: alpha[0].associated_coroot() -e[0] + e['deltacheck'] sage: alpha[1].associated_coroot() e[0] - e[1] sage: alpha[2].associated_coroot() e[1] inner_product( other)¶ Implement the canonical inner product of selfwith other. EXAMPLES: sage: e = RootSystem(['B',3,1]).ambient_space() sage: B = e.basis() sage: matrix([[x.inner_product(y) for x in B] for y in B]) [1 0 0 0 0] [0 1 0 0 0] [0 0 1 0 0] [0 0 0 1 0] [0 0 0 0 1] sage: x = e.an_element(); x 2e[0] + 2e[1] + 3e[2] sage: x.inner_product(x) 17 scalar()is an alias for this method: sage: x.scalar(x) 17 Todo Lift to CombinatorialFreeModule.Element as canonical_inner_product scalar( other)¶ Implement the canonical inner product of selfwith other. EXAMPLES: sage: e = RootSystem(['B',3,1]).ambient_space() sage: B = e.basis() sage: matrix([[x.inner_product(y) for x in B] for y in B]) [1 0 0 0 0] [0 1 0 0 0] [0 0 1 0 0] [0 0 0 1 0] [0 0 0 0 1] sage: x = e.an_element(); x 2e[0] + 2e[1] + 3e[2] sage: x.inner_product(x) 17 scalar()is an alias for this method: sage: x.scalar(x) 17 Todo Lift to CombinatorialFreeModule.Element as canonical_inner_product coroot_lattice()¶ EXAMPLES: sage: RootSystem(["A",3,1]).ambient_lattice().coroot_lattice() Ambient lattice of the Root system of type ['A', 3, 1] Todo Factor out this code with the classical ambient space. fundamental_weight( i)¶ Return the fundamental weight $\Lambda_i$ in this ambient space. It is constructed by taking the corresponding fundamental weight of the classical ambient space (or $0$ for $\Lambda_0$) and raising it to the appropriate level by adding a suitable multiple of $\delta^\vee$. EXAMPLES: sage: RootSystem(['A',3,1]).ambient_space().fundamental_weight(2) e[0] + e[1] + e['deltacheck'] sage: RootSystem(['A',3,1]).ambient_space().fundamental_weights() Finite family {0: e['deltacheck'], 1: e[0] + e['deltacheck'], 2: e[0] + e[1] + e['deltacheck'], 3: e[0] + e[1] + e[2] + e['deltacheck']} sage: RootSystem(['A',3]).ambient_space().fundamental_weights() Finite family {1: (1, 0, 0, 0), 2: (1, 1, 0, 0), 3: (1, 1, 1, 0)} sage: RootSystem(['A',3,1]).weight_lattice().fundamental_weights().map(attrcall("level")) Finite family {0: 1, 1: 1, 2: 1, 3: 1} sage: RootSystem(['B',3,1]).ambient_space().fundamental_weights() Finite family {0: e['deltacheck'], 1: e[0] + e['deltacheck'], 2: e[0] + e[1] + 2e['deltacheck'], 3: 1/2e[0] + 1/2e[1] + 1/2e[2] + e['deltacheck']} sage: RootSystem(['B',3]).ambient_space().fundamental_weights() Finite family {1: (1, 0, 0), 2: (1, 1, 0), 3: (1/2, 1/2, 1/2)} sage: RootSystem(['B',3,1]).weight_lattice().fundamental_weights().map(attrcall("level")) Finite family {0: 1, 1: 1, 2: 2, 3: 1} In type $BC$ dual, the coefficient of ‘delta^vee’ is the level divided by $2$ to take into account that the null coroot is $2\delta^\vee$: sage: R = CartanType(['BC',3,2]).dual().root_system() sage: R.ambient_space().fundamental_weights() Finite family {0: e['deltacheck'], 1: e[0] + e['deltacheck'], 2: e[0] + e[1] + e['deltacheck'], 3: 1/2e[0] + 1/2e[1] + 1/2e[2] + 1/2e['deltacheck']} sage: R.weight_lattice().fundamental_weights().map(attrcall("level")) Finite family {0: 2, 1: 2, 2: 2, 3: 1} sage: R.ambient_space().null_coroot() 2e['deltacheck'] By a slight naming abuse this function also accepts "delta" as input so that it can be used to implement the embedding from the extended weight lattice:: sage: RootSystem(['A',3,1]).ambient_space().fundamental_weight("delta") e['delta'] is_extended()¶ Return whether this is a realization of the extended weight lattice: yes! See also EXAMPLES: sage: RootSystem(['A',3,1]).ambient_space().is_extended() True simple_coroot( i)¶ Return the $i$-th simple coroot $\alpha_i^\vee$ of this affine ambient space. EXAMPLES: sage: RootSystem(["A",3,1]).ambient_space().simple_coroot(1) e[0] - e[1] It is built as the coroot associated to the simple root $\alpha_i$: sage: RootSystem(["B",3,1]).ambient_space().simple_roots() Finite family {0: -e[0] - e[1] + e['delta'], 1: e[0] - e[1], 2: e[1] - e[2], 3: e[2]} sage: RootSystem(["B",3,1]).ambient_space().simple_coroots() Finite family {0: -e[0] - e[1] + e['deltacheck'], 1: e[0] - e[1], 2: e[1] - e[2], 3: 2e[2]} Todo Factor out this code with the classical ambient space. simple_root( i)¶ Return the $i$-th simple root of this affine ambient space. EXAMPLES: It is built straightforwardly from the corresponding simple root $\alpha_i$ in the classical ambient space: sage: RootSystem(["A",3,1]).ambient_space().simple_root(1) e[0] - e[1] For the special node (typically $i=0$), $\alpha_0$ is built from the other simple roots using the column annihilator of the Cartan matrix and adding $\delta$, where $\delta$ is the null root: sage: RootSystem(["A",3]).ambient_space().simple_roots() Finite family {1: (1, -1, 0, 0), 2: (0, 1, -1, 0), 3: (0, 0, 1, -1)} sage: RootSystem(["A",3,1]).ambient_space().simple_roots() Finite family {0: -e[0] + e[3] + e['delta'], 1: e[0] - e[1], 2: e[1] - e[2], 3: e[2] - e[3]} Here is a twisted affine example: sage: RootSystem(CartanType(["B",3,1]).dual()).ambient_space().simple_roots() Finite family {0: -e[0] - e[1] + e['delta'], 1: e[0] - e[1], 2: e[1] - e[2], 3: 2e[2]} In fact $\delta$ is really $1/a_0$ times the null root (see the discussion in WeightSpace) but this only makes a difference in type $BC$: sage: L = RootSystem(CartanType(["BC",3,2])).ambient_space() sage: L.simple_roots() Finite family {0: -e[0] + e['delta'], 1: e[0] - e[1], 2: e[1] - e[2], 3: 2e[2]} sage: L.null_root() 2*e['delta'] Note An alternative would have been to use the default implementation of the simple roots as linear combinations of the fundamental weights. However, as in type $A_n$ it is preferable to take a slight variant to avoid rational coefficient (the usual $GL_n$ vs $SL_n$ issue). classmethod smallest_base_ring( cartan_type)¶ Return the smallest base ring the ambient space can be defined on. This is the smallest base ring for the associated classical ambient space. See also EXAMPLES: sage: cartan_type = CartanType(["A",3,1]) sage: cartan_type.AmbientSpace.smallest_base_ring(cartan_type) Integer Ring sage: cartan_type = CartanType(["B",3,1]) sage: cartan_type.AmbientSpace.smallest_base_ring(cartan_type) Rational Field
The series $\sum_{n=1}^\infty\frac1{4n^2-1}$ telescopes. Find a simple formula for the $k$th partial sum $S_k$, and use it to determine whether the series converges. If it converges, find its value. I've mostly tried to intuit my way through. Calculating the partial sums to several degrees by varying $n$, I got $$\frac13+\frac1{15}+\frac1{35}+\frac1{63}+\frac1{99}+\cdots$$ and I can see that it is converging to $\frac12$. I computed the first few multiplicative differences between the values and saw that $\frac13$ was multiplied by $\frac15$ to get $\frac1{15}$, $\frac1{15}$ by $\frac37$, $\frac1{35}$ by $\frac7{11}$. I see that both the denominator and the numerator of the multiplier is increasing by two each time. So I see that it is certainly converging. But I cannot quite put the pieces together on the formula. Any help?
If $\exists a, b, m_1, m_2 \in \mathbb{Z}, a\equiv b\pmod {m_1}$ and $a\equiv b\pmod {m_2}$, then $\exists k \in \mathbb{Z}, a=b+{m_1m_2k\over gcd(m_1,m_2)} \implies a\equiv b\pmod L$, where $L$ is the l.c.m. of $m_1$ and $m_2$. I am trying to derive, starting with the first part, with no success $\exists l \in \mathbb{Z}, a -b = l\cdot m_1 \implies a =b +l\cdot m_1$ $\exists m \in \mathbb{Z}, a-b = m\cdot m_2 \implies a =b +m\cdot m_2$ Equating both, $b +l\cdot m_1 = b +m\cdot m_2$ Unable to start the proof Based on comment by @saulspatz, the new attempt to proof is: $\exists r_1, r_2 \in \mathbb{Z},$ s.t. (i) $m_1 = r_1\gcd(m_1, m_2),$ (ii) $m_2 = r_2\gcd(m_1, m_2)$ Multiplying, (i) by (ii), we get: $m_1m_2 = r_1r_2(m_1, m_2)^2$ if for suitable integer $k$, take $r_1r_2={1\over k}$, then not possible as $r_1, r_2$ are themselves integers. So, anyway attempt something: Let, $r_1r_2=k'$, $m_1m_2 = k'(m_1, m_2)^2 \implies (m_1, m_2) = {m_1m_2 \over {k'\cdot (m_1, m_2)}}$ But, $k'$ is an integer, and the final form has an integer $k$ in numerator.
I have mentioned this elsewhere, but it bears repeating because it is such an important concept: Sufficiency pertains to data reduction, not parameter estimation per se. Sufficiency only requires that one does not "lose information" about the parameter(s) that was present in the original sample. Students of mathematical statistics have a tendency to conflate sufficient statistics with estimators, because "good" estimators in general need to be sufficient statistics: after all, if an estimator discards information about the parameter(s) it estimates, it should not perform as well as an estimator that does not do so. So the concept of sufficiency is one way in which we characterize estimators, but that clearly does not mean that sufficiency is about estimation. It is vitally important to understand and remember this. That said, the Factorization theorem is easily applied to solve (a); e.g., for a sample $\boldsymbol x = (x_1, \ldots, x_n)$, the joint density is $$f(\boldsymbol x \mid \theta) = \prod_{i=1}^n \frac{\theta}{x_i^2} \mathbb 1 (x_i \ge \theta) \mathbb 1 (\theta > 0) = \mathbb 1 (x_{(1)} \ge \theta > 0) \, \theta^n \prod_{i=1}^n x_i^{-2},$$ where $x_{(1)} = \min_i x_i$ is the minimum order statistic. This is because the product of the indicator functions $\mathbb 1 (x_i \ge \theta)$ is $1$ if and only if all of the $x_i$ are at least as large as $\theta$, which occurs if and only if the smallest observation in the sample, $x_{(1)}$, is at least $\theta$. We see that we cannot separate $x_{(1)}$ from $\theta$, so this factor must be part of $g(\boldsymbol T(\boldsymbol x) \mid \theta)$, where $\boldsymbol T(\boldsymbol x) = T(\boldsymbol x) = x_{(1)}$. Note that in this case, our sufficient statistic is a function of the sample that reduces a vector of dimension $n$ to a scalar $x_{(1)}$, so we may write $T$ instead of $\boldsymbol T$. The rest is easy: $$f(\boldsymbol x \mid \theta) = h(\boldsymbol x) g(T(\boldsymbol x) \mid \theta),$$ where $$h(\boldsymbol x) = \prod_{i=1}^n x_i^{-2}, \quad g(T \mid \theta) = \mathbb 1 (T \ge \theta > 0) \theta^n,$$ and $T$, defined as above, is our sufficient statistic. You may think that $T$ estimates $\theta$--and in this case, it happens to--but just because we found a sufficient statistic via the Factorization theorem, this doesn't mean it estimates anything. This is because any one-to-one function of a sufficient statistic is also sufficient (you can simply invert the mapping). $T^2 = x_{(1)}^2$ is also sufficient (note while $m : \mathbb R \to \mathbb R$, $m(x) = x^2$ is not one-to-one in general, in this case it is because the support of $X$ is $X \ge \theta > 0$). Regarding (b), MLE estimation, we express the joint likelihood as proportional to $$\mathcal L(\theta \mid \boldsymbol x) \propto \theta^n \mathbb 1(0 < \theta \le x_{(1)}).$$ We simply discard any factors of the joint density that are constant with respect to $\theta$. Since this likelihood is nonzero if and only if $\theta$ is positive but not exceeding the smallest observation in the sample, we seek to maximize $\theta^n$ subject to this constraint. Since $n > 0$, $\theta^n$ is a monotonically increasing function on $\theta > 0$, hence $\mathcal L$ is greatest when $\theta = x_{(1)}$; i.e., $$\hat \theta = x_{(1)}$$ is the MLE. It is trivially biased because the random variable $X_{(1)}$ is never smaller than $\theta$ and is almost surely strictly greater than $\theta$; hence its expectation is almost surely greater than $\theta$. Finally, we can explicitly compute the density of the order statistic as requested in (c): $$\Pr[X_{(1)} > x] = \prod_{i=1}^n \Pr[X_i > x],$$ because the least observation is greater than $x$ if and only if all of the observations are greater than $x$, and the observations are IID. Then $$1 - F_{X_{(1)}}(x) = \left(1 - F_X(x)\right)^n,$$ and the rest of the computation is left to you as a straightforward exercise. We can then take this and compute the expectation $\operatorname{E}[X_{(1)}]$ to ascertain the precise amount of bias of the MLE, which is necessary to answer whether there is a scalar value $c$ (which may depend on the sample size $n$ but not on $\theta$ or the sample $\boldsymbol x$) such that $c\hat \theta$ is unbiased.
I'm having trouble with the Ho-Lee model for short rates and differentiating between how to find the values for the free parameter λ versus using the model to predict future rates. The Ho-Lee model for each step in a binomial tree: $$ \lambda_tdt + \sigma \sqrt dt $$ I've read that to set the free parameter at each step in a recombining binomial tree, you set the rate at state 0 to the current spot rate (ie: 1 month spot rate) and find a value for lambda that when plugged into the model will result in the current spot rate for the next time step (eg: starting with 1 month spot rate at state 0 and using a 1 month time step, the correct value for lambda when plugged into the model will produce the current 2 month spot rate etc). This confuses me. Once I've determined the value of lambda for each step in my tree, what inputs do I change to use the model with my binomial tree to predict futures rates .. ie: one month rate in one month, in two months etc? In case my description isn't clear, here is an except from Bruce Tuckman's book on the subject. ... find λ1 such that the model produces a two-month spot rate equal to that in the market. Then find λ2 such that the model produces a three-month spot rate equal to that in the market. Continue in this fashion until the tree ends.
I recently came across this in a textbook (NCERT class 12 , chapter: wave optics , pg:367 , example 10.4(d)) of mine while studying the Young's double slit experiment. It says a condition for the formation of interference pattern is$$\frac{s}{S} < \frac{\lambda}{d}$$Where $s$ is the size of ... The accepted answer is clearly wrong. The OP's textbook referes to 's' as "size of source" and then gives a relation involving it. But the accepted answer conveniently assumes 's' to be "fringe-width" and proves the relation. One of the unaccepted answers is the correct one. I have flagged the answer for mod attention. This answer wastes time, because I naturally looked at it first ( it being an accepted answer) only to realise it proved something entirely different and trivial. This question was considered a duplicate because of a previous question titled "Height of Water 'Splashing'". However, the previous question only considers the height of the splash, whereas answers to the later question may consider a lot of different effects on the body of water, such as height ... I was trying to figure out the cross section $\frac{d\sigma}{d\Omega}$ for spinless $e^{-}\gamma\rightarrow e^{-}$ scattering. First I wrote the terms associated with each component.Vertex:$$ie(P_A+P_B)^{\mu}$$External Boson: $1$Photon: $\epsilon_{\mu}$Multiplying these will give the inv... As I am now studying on the history of discovery of electricity so I am searching on each scientists on Google but I am not getting a good answers on some scientists.So I want to ask you to provide a good app for studying on the history of scientists? I am working on correlation in quantum systems.Consider for an arbitrary finite dimensional bipartite system $A$ with elements $A_{1}$ and $A_{2}$ and a bipartite system $B$ with elements $B_{1}$ and $B_{2}$ under the assumption which fulfilled continuity.My question is that would it be possib... @EmilioPisanty Sup. I finished Part I of Q is for Quantum. I'm a little confused why a black ball turns into a misty of white and minus black, and not into white and black? Is it like a little trick so the second PETE box can cancel out the contrary states? Also I really like that the book avoids words like quantum, superposition, etc. Is this correct? "The closer you get hovering (as opposed to falling) to a black hole, the further away you see the black hole from you. You would need an impossible rope of an infinite length to reach the event horizon from a hovering ship". From physics.stackexchange.com/questions/480767/… You can't make a system go to a lower state than its zero point, so you can't do work with ZPE. Similarly, to run a hydroelectric generator you not only need water, you need a height difference so you can make the water run downhill. — PM 2Ring3 hours ago So in Q is for Quantum there's a box called PETE that has 50% chance of changing the color of a black or white ball. When two PETE boxes are connected, an input white ball will always come out white and the same with a black ball. @ACuriousMind There is also a NOT box that changes the color of the ball. In the book it's described that each ball has a misty (possible outcomes I suppose). For example a white ball coming into a PETE box will have output misty of WB (it can come out as white or black). But the misty of a black ball is W-B or -WB. (the black ball comes out with a minus). I understand that with the minus the math works out, but what is that minus and why? @AbhasKumarSinha intriguing/ impressive! would like to hear more! :) am very interested in using physics simulation systems for fluid dynamics vs particle dynamics experiments, alas very few in the world are thinking along the same lines right now, even as the technology improves substantially... @vzn for physics/simulation, you may use Blender, that is very accurate. If you want to experiment lens and optics, the you may use Mistibushi Renderer, those are made for accurate scientific purposes. @RyanUnger physics.stackexchange.com/q/27700/50583 is about QFT for mathematicians, which overlaps in the sense that you can't really do string theory without first doing QFT. I think the canonical recommendation is indeed Deligne et al's Quantum Fields and Strings: A Course For Mathematicians , but I haven't read it myself @AbhasKumarSinha when you say you were there, did you work at some kind of Godot facilities/ headquarters? where? dont see something relevant on google yet on "mitsubishi renderer" do you have a link for that? @ACuriousMind thats exactly how DZA presents it. understand the idea of "not tying it to any particular physical implementation" but that kind of gets stretched thin because the point is that there are "devices from our reality" that match the description and theyre all part of the mystery/ complexity/ inscrutability of QM. actually its QM experts that dont fully grasp the idea because (on deep research) it seems possible classical components exist that fulfill the descriptions... When I say "the basics of string theory haven't changed", I basically mean the story of string theory up to (but excluding) compactifications, branes and what not. It is the latter that has rapidly evolved, not the former. @RyanUnger Yes, it's where the actual model building happens. But there's a lot of things to work out independently of that And that is what I mean by "the basics". Yes, with mirror symmetry and all that jazz, there's been a lot of things happening in string theory, but I think that's still comparatively "fresh" research where the best you'll find are some survey papers @RyanUnger trying to think of an adjective for it... nihilistic? :P ps have you seen this? think youll like it, thought of you when found it... Kurzgesagt optimistic nihilismyoutube.com/watch?v=MBRqu0YOH14 The knuckle mnemonic is a mnemonic device for remembering the number of days in the months of the Julian and Gregorian calendars.== Method ===== One handed ===One form of the mnemonic is done by counting on the knuckles of one's hand to remember the numbers of days of the months.Count knuckles as 31 days, depressions between knuckles as 30 (or 28/29) days. Start with the little finger knuckle as January, and count one finger or depression at a time towards the index finger knuckle (July), saying the months while doing so. Then return to the little finger knuckle (now August) and continue for... @vzn I dont want to go to uni nor college. I prefer to dive into the depths of life early. I'm 16 (2 more years and I graduate). I'm interested in business, physics, neuroscience, philosophy, biology, engineering and other stuff and technologies. I just have constant hunger to widen my view on the world. @Slereah It's like the brain has a limited capacity on math skills it can store. @NovaliumCompany btw think either way is acceptable, relate to the feeling of low enthusiasm to submitting to "the higher establishment," but for many, universities are indeed "diving into the depths of life" I think you should go if you want to learn, but I'd also argue that waiting a couple years could be a sensible option. I know a number of people who went to college because they were told that it was what they should do and ended up wasting a bunch of time/money It does give you more of a sense of who actually knows what they're talking about and who doesn't though. While there's a lot of information available these days, it isn't all good information and it can be a very difficult thing to judge without some background knowledge Hello people, does anyone have a suggestion for some good lecture notes on what surface codes are and how are they used for quantum error correction? I just want to have an overview as I might have the possibility of doing a master thesis on the subject. I looked around a bit and it sounds cool but "it sounds cool" doesn't sound like a good enough motivation for devoting 6 months of my life to it
156 10 Hi, I'm aware of a typical example of injective immersion that is not a topological embedding: figure 8 ##\beta: (-\pi, \pi) \to \mathbb R^2##, with ##\beta(t)=(\sin 2t,\sin t)## As explained here an-injective-immersion-that-is-not-a-topological-embedding the image of ##\beta## is compact in ##\mathbb R^2## subspace topology while the domain open interval is not, thus ##\beta## is not a smooth embedding. Consider it from the point of view of "homeomorphism onto its image" definition, I was trying to find out an instance of image subset open in the subspace topology that actually is not open in the domain topology or the other way around. Can you help me ? Thanks. ##\beta: (-\pi, \pi) \to \mathbb R^2##, with ##\beta(t)=(\sin 2t,\sin t)## As explained here an-injective-immersion-that-is-not-a-topological-embedding the image of ##\beta## is compact in ##\mathbb R^2## subspace topology while the domain open interval is not, thus ##\beta## is not a smooth embedding. Consider it from the point of view of "homeomorphism onto its image" definition, I was trying to find out an instance of image subset open in the subspace topology that actually is not open in the domain topology or the other way around. Can you help me ? Thanks.
I'm not sure if it is possible to estimate the flux linkage of a PMSM motor (star connection) by the controller itself, where I can measure the input voltage, all the phase currents and the velocity. Measurements of R and L are also available. The flux linkage is defined as: $$ \lambda_m = \sqrt{2/3} \frac{2}{P} K_b $$ Where P=Number of Poles and Kb theback EMF constant. So I need to come up with Kb first, but I don't know how I could achieve this. When I turn on the power supply to speed up the motor to a constant speed, I cannot measure the back EMF, because I can only measure the input voltage. Is it even possible to measure the flux linkage with this setup?
We define $G = \GSp^+(2g,\Q)$ by $$G = \{\gamma\in \GL_{2g}(\Z):\gamma^t J\gamma=r(\gamma)J,\text{ for all }\gamma\in\Bbb{Q}_+\}.$$ Let $L(\Gamma,G)$ be the free $\C$-module generated by the right cosets $\Gamma\alpha$ where $\alpha\in\Gamma \backslash G$. Note $\Gamma$ acts on $L(\Gamma,G)$ by right multiplication and we set $\mathcal{H}_g(\Gamma,G)=L(\Gamma,G)^\Gamma$. Let $T_1,T_2\in \mathcal{H}_g(\Gamma,G)$ and $T_i = \sum_{\alpha_i \in \Gamma \setminus G} c_i(\alpha) \Gamma\alpha.$ Then $T_1 T_2 = \sum_{\alpha,\alpha'\in \Gamma \setminus G} c_1(\alpha)c_2(\alpha')\Gamma\alpha\alpha'$. As in the classical case, we pay most attention to the Hecke operators at a prime $p$. It is known that $\mathcal{H}_g = \bigotimes_{p\text{ prime}} \mathcal{H}_{g,p}$ where the construction of the local Hecke algebra $\mathcal{H}_{g,p}$ is the same as before but with $G$ replaced with $G_p = G\cap \text{GL}_{2g}(\Z[p^{-1}])$. The generators of this local algrebra $\mathcal{H}_{g,p}$ are the double cosets $T(p)=\Gamma\text{diag}(I_g;pI_g)\Gamma$ and $T_i(p^2)=\Gamma \text{diag}(I_i,pI_{g-i};p^2I_i,pI_{g-i})\Gamma$ for $1\leq i \leq g$. Some authors also define $T_0(p^2)$, too. The operator $T(p^2)=\sum_{i=1}^g T_i(p^2)$. The space $\mathcal{H}_g$ acts on Siegel modular forms of degree $g$ and weight $k$ by $F\|_k\left(\sum c_i\Gamma\alpha_i\right)=\sum c_i F\|_k\alpha_i$ where $\left(F\|_k \alpha\right)(Z)=r(\alpha)^{gk-\frac{g(g+1)}{2}}\det(CZ+D)^{-k}F\left(\alpha\cdot Z \right)$. Some authors use a different normalization in this definition. A Hecke eigenform is a form in $M_k(\Gamma)$ which is a simultaneous eigenform for all the operators $T(p)$, $T(p^2)$,...,$T(p^g)$. Knowl status: Review status: beta Last edited by John Voight on 2018-06-28 01:01:31 Referred to by: Not referenced anywhere at the moment. History:(expand/hide all)
Finally, after a lot of hesitation, I rooted my phone (Samsung Galaxy On7, Android 6.0.1) using Magisk systemless root. The main aim was to remove pre-installed useless software (often termed "bloatware").I didn't want to flash Magisk using TWRP, because installing TWRP would remove the stock... Summary: I want to basically multipartition my pc and use MY android OS copied from my phone as the dual boot OS.Hi all, it's been awhile since I've posted. Sorry for AWOL.What I want to do:Partition my hard drive and run my exact copy of android on the other partition.Why:I want to... Because it holds that##\displaystyle\int_{1}^{x}\frac{dt}{t} = \log x##, and##\displaystyle\int_{1}^{x}\frac{dt}{t^a} = \frac{1}{a-1}\left(1-\frac{1}{x^{a-1}}\right)\hspace{20pt}##when ##a>1##it could be expected that##\displaystyle\frac{1}{a-1}\left(1-\frac{1}{x^{a-1}}\right)... I've recently started my new RA position, and I've been given the task of analyzing a root data file. I'm not completely lost, but I don't exactly know what I'm doing. The point of my post is not to ask for answers, merely advice. A place I could go for info on data analytics. Pointers on how to... I would like to know if there is an official name for the class of integers that are (not) perfect powers. A perfect power is a number that can be expressed as xn, where x and n are both integers > 1. I have been calling these integers "roots" - since they do not have any integer roots of their... 1. Homework StatementShow, using the axiom of completeness of ##\mathbb{R}##, that every positive real number has a unique n'th root that is a positive real number.Or in symbols:##n \in \mathbb{N_0}, a \in \mathbb{R^{+}} \Rightarrow \exists! x \in \mathbb{R^{+}}: x^n = a##2. Homework... I am using ROOT to calculate the Fourier transform of a digital signal. I can extract the individual parts of the transform, the magnitude and phase in the form of a 1D histogram. I am attempting to reconstruct the transforms from the phase and magnitude but cannot seem to figure it out. Any... 1. Homework StatementHi, I'm new here. I'm really rusty, I resume my career this year, and I'm reading 'the spivak book', (for Calculus 1).Making some exercises, I get curious about how to solve this: x+e^x=4I would love if someone could give me any trick2. Homework Equations3. The... I was going through some important points give in my textbook and I saw this:##\log_e x < \sqrt x##How did they get this?I know calculus so you can show this using differentiation, etc.One possible way is that they took##f(x)=\sqrt x-\log_e x##And tried to prove it is always greater than zero.
In the early 2000's, I wrote software code to process and recover telemetry data from archived raw telemetry files obtained from the Pioneer 10 and 11 spacecraft between 1972 and 2003. This led to my participation in the investigation of the Pioneer Anomaly, an as-yet unexplained small, anomalous acceleration of these two spacecraft that was measured as the spacecraft were leaving the Solar System. Over the years, I developed a precision orbit determination program (ODP) that can model spacecraft orbits and radio-metric tracking data with the necessary accuracy. I also developed some thermal modeling code, as part of our investigation of the possibility that much (all?) of the anomaly is due to the recoil force caused by heat radiated by the spacecraft in an anisotropic pattern. Pioneer 10 and 11 were launched in 1972 and 1973, respectively. These were humanity's first two spacecraft to leave the inner solar system, cross the asteroid belt, and make close-up observations of the gas giang Jupiter. The missions were tremendously successful. Both spacecraft reached Jupiter safely, providing the first ever close-up observations of the planet, its moons, and its immense radiation fields. Pioneer 11 used Jupiter's gravity for a maneuver that took it across the solar system, for an eventual encounter with Saturn several years later. The two spacecraft continued to operate well beyond their original design lifetime: Pioneer 11 was last contacted in 1995, whereas Pioneer 10's last transmission was received in 2003. The Pioneer 10 and 11 spacecraft were spin-stabilized. The spacecraft's axis of rotation coincided with their antenna axis, and was oriented in the direction of the Earth. Infrequent precession maneuvers were needed to ensure that the Earth remained within the antenna beamwidth. This meant that much of the time, the spacecraft flew completely undisturbed. As a result, Pioneer 10 and 11 remains the most precise large-scale gravitational experiment to date and for the foreseeable future. In the 1990s, it became apparent, however, that in order to achieve maximum agreement between theory and data, the theory needed to be modified. This small correction was in the form of a constant sunward acceleration, with an approximate magnitude of $a_P=(8.74\pm 1.33)\times 10^{-10}~{\rm m}/{\rm s}^2$. My first contribution to researching the Pioneer anomaly was a C++ software library that could process Master Data Records, a format used by NASA's DSN (Deep Space Network) to store raw Pioneer telemetry. Subsequently, I developed a full-blown orbit determination program that could utilize DSN radio-metric Doppler measurements of the Pioneer radio signal to model the Pioneer spacecraft orbits with high precision. The program can be run from the command line, but I also developed a Windows front-end using Visual C++ that made it easier to investigate test cases and monitor the calculation. The program solves the relativistic equations of motion \[\frac{d^2\vec{r}}{dt^2}=\frac{\mu_i}{\|\vec{r}_i-\vec{r}\|^3}(A_i(\vec{r}_i-\vec{r})+\vec{B}_i),\] where $\vec{r}$ is the spacecraft's position, $\vec{r}_i$ is the position, $\mu_i$ is the mass of the $i$-th solar system body, $t$ is the time and the post-Newtonian correction terms $A_i$ and $\vec{B}_i$ are given by \begin{align}A_i&=1-\frac{1}{c^2}\left\{2(\beta+\gamma)\sum\limits_j\frac{\mu_j}{\|\vec{r}_j-\vec{r}\|}+\gamma v^2+(1+\gamma)v_i^2-2(1+\gamma)\vec{v}\cdot\vec{v}_i-\frac{3}{2}\left[\frac{(\vec{r}-\vec{r}_i)\cdot\vec{v}_i}{\|\vec{r}_i-\vec{r}\|}\right]^2\right\},\\ \vec{B}_i&=\frac{1}{c^2}\left\{(\vec{r}-\vec{r}_i)\cdot\left[(2+2\gamma)\vec{v}-(1+2\gamma)\vec{v}_i\right]\right\}(\vec{v}-\vec{v}_i).\end{align} The parameters $\beta$ and $\gamma$ are the so-called Eddington-parameters; they are both 1 for general relativity, but may have different values for alternate theories of gravity. The software also models the effects of the oblateness of Jupiter and Saturn (these become important when the spacecraft is near these giant planets) as well as the propagation of the radio signal through the solar system, affected by gravity, charged particles from the Sun, and the Earth's atmosphere. The positions of solar system bodies and the precise locations of DSN ground stations are obtained from NASA data sets. Nongravitational forces, notably solar pressure, are also accounted for by the code. Uniquely, my program can also model the recoil force due to on-board generated heat that is radiated in an anisotropic pattern. Most of the heat on-board is due to two sources: waste heat from the spacecraft's radioisotope thermoelectric generators (RTGs) and electrical heat. It turns out that the thermal recoil force $\vec{F}_\mathrm{recoil}$ is proportional to a linear combination of the power $Q_\mathrm{rtg}$ and $Q_\mathrm{elec}$ of these two heat sources: \[\vec{F}_\mathrm{recoil}\propto \eta_\mathrm{rtg}Q_\mathrm{rtg}+\eta_\mathrm{elec}Q_\mathrm{elec},\] where the two coefficients $\eta_\mathrm{rtg}$ and $\eta_\mathrm{elec}$ are yet to be determined. To estimate the magnitude of these coefficients, and to verify that it is indeed legitimate to model the thermal recoil force this way, I also developed a finite element model of the spacecraft that, in combination with a raytracing algorithm, estimated $\eta_\mathrm{rtg}$ and $\eta_\mathrm{elec}$. I first developed a finite element model of the spacecraft using Maxima. In turn, the Maxima code that I wrote generated C++ code which was then utilized by a generic raytracing algorithm that I developed. The final program ran from the command line as it iteratively estimated the thermal output of the spacecraft. One version of the program could also estimate the resulting change in angular momentum (due to the way heat is reflected off some asymmetric spacecraft components, thermal radiation could also affect the spacecraft's rotation, and indeed, an anomalous change in the rotation rate of both spacecraft was detected.)
The Annals of Mathematical Statistics Ann. Math. Statist. Volume 37, Number 2 (1966), 458-463. All Admissible Linear Estimates of the Mean Vector Abstract Let $y$ be a single observation on a $p \times 1$ random vector which is distributed according to the multivariate normal distribution with mean vector $\theta$ and covariance matrix $I$. Consider the problem of estimating $\theta$ when the loss function is the sum of the squared errors in estimating the individual components of $\theta$. Let $G$ be a $p \times p$ real matrix. Then we will prove that the estimate $Gy$ is admissible if and only if $G$ is symmetric, and the characteristic roots of $G, g_i$ say, $i = 1, 2, \cdots, p$, satisfy $0 \leqq g_i \leqq 1$, with equality at one for at most two of the roots. The proof concerning the characteristic roots uses the results of Karlin [4] and James and Stein [3]. We give two proofs of the fact that $Gy$ is inadmissible when $G$ is asymmetrical. In the first proof we give an estimate $G^\ast y$ that is better than $Gy$, where $G^\ast$ is a symmetric matrix. This not only adds to the practicality of the result, but also enables us to resolve the question of which estimates are admissible in the restricted class of estimates of the form $Gy$. The method of the second proof, which utilizes a theorem of Sacks [6], leads to the following finding: If $Gy$ is admissible, then $Gy$ must be a generalized Bayes procedure, where the unique generalized prior distribution must be either a multivariate normal distribution with mean vector zero and a specified covariance matrix determined by $G$, or the product of a distribution which is multivariate normal over a subspace of the parameter space and a distribution which is uniformly distributed over a subspace of the parameter space. This latter finding generalizes the well known one dimensional case. In Section 2 then the main results are proved. In Section 3 some remarks concerned with generalizing the main results are given. We remark here that the decision theory terminology used is more or less that of Blackwell and Girshick [1]. Article information Source Ann. Math. Statist., Volume 37, Number 2 (1966), 458-463. Dates First available in Project Euclid: 27 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aoms/1177699528 Digital Object Identifier doi:10.1214/aoms/1177699528 Mathematical Reviews number (MathSciNet) MR189164 Zentralblatt MATH identifier 0146.40201 JSTOR links.jstor.org Citation Cohen, Arthur. All Admissible Linear Estimates of the Mean Vector. Ann. Math. Statist. 37 (1966), no. 2, 458--463. doi:10.1214/aoms/1177699528. https://projecteuclid.org/euclid.aoms/1177699528
By the Modularity Theorem (a.k.a. the Shimura--Taniyama--Weil Conjecture), if $E$ is an elliptic curve over $\textbf{Q}$ with conductor $N$, then there exists a “modular parametrization” $\psi: X_0(N) \to E$, a non-constant surjective morphism defined over $\textbf{Q}$. Let me call this “ geometric” modularity. It is known that this is equivalent to the existence of a holomorphic modular form of weight 2 and level $\Gamma_0(N)$ such that $L(E, \chi, s) = L(f, \chi, s)$ for all Dirichlet characters $\chi$. Indeed, over $\textbf{Q}$, such a result follows from the weaker result that $L(E, s) = L(f,s)$. Let me call this “ analytic” modularity. According to the Langlands philosophy, one expects all motivic L-functions to be automorphic. In particular, for Galois representations coming from the etale cohomology of any variety over a number field, one expects to be able to attach an automorphic representation (or some set of them, depending on the context); so we expect “analytic” modularity in some general sense. It seems like most modularity/automorphy results aim to prove this kind of statement. My question is the following: Does one also expect a general form of “geometric” modularity to hold? I have not seen any such parametrizations outside of the well-known case of modular parametrizations of elliptic curves over $\textbf{Q}$, and occasionally by Shimura curves (but am less familiar with the implications, compared to the modular curve case). Since many of the nice “coincidences” in this classical modularity setting seem to break down as one passes to more general settings, and since the connection between geometric and analytic modularity seems weaker in the function field setting (e.g. the relevant automorphic forms are not functions or sections of line bundles on the Drinfeld modular curve), does one expect a general "geometric" modularity statement to hold? For example, given a higher genus curve over $\textbf{Q}$, or an elliptic curve over a number field, or a surface (say, abelian) over $\textbf{Q}$, do you expect some form of modularity to correspond to the existence of a map from some space (like a Shimura variety) to the variety in question? I am aware that one can sometimes, say, find the appropriate motives in the cohomology of Shimura varieties (e.g. if V is a variety such that $L(V,s) = L(\pi, s)$ for some automorphic representation $\pi$, and then you may have a Galois representation attached to $\pi$ lying in the cohomology of some Shimura variety), but let me call this a case of "analytic" modularity instead of a weak form of "geometric" modularity, for the sake of the question. In other words, by "geometric" modularity, I am specifically referring to the existence of something like a morphism from, say, a Shimura variety to a variety over a number field. When I asked this question (IRL), some suggested that such a statement may be better formulated in the language of correspondences (or purely in the language of motives), but then that such an answer may be essentially the case of the variety $V$ above. I would be interested in any such answers, or in any corrections to the formulation of the question.
I have asked this question in cs.stackeschange, and it was recommended to me that I asked it here. Is there any literature on approximate maximum-weight perfect matchings where the approximation criterion is not a factor between the weight sum of the approximate and the optimal solutions, but instead the cardinality of the intersection of the edge sets of the approximate and the optimal solutions to the original maximum-weight perfect matching? The relevant graphs $G(V_1\cup V_2,E)$, $V_1\cap V_2=\varnothing$, are bipartite with non-negative possibly non-uniform edge weights and assumed to have perfect matchings. In particular, $\|V_1\| = \|V_2\|$. To be clear: Let $\mathcal{G}$ the the class of graphs $G$ as above, and let $\mathcal{E}_{G}$ be the set of maximum-weight perfect matchings of a $G\in\mathcal{G}$ with $\|V_1\| = N$. I seek an approximation algorithm which outputs a perfect matching $E''$ of $G$ satisfying $\max_{E'\in\mathcal{E}_{G}}\|E''\cap E'\|\geq(1-\epsilon)N$. If it helps, I am willing to restrict the problem to the class of graphs for which $\|\mathcal{E}_{G}\|=1$, and therefore the potentially complicating term "$\max_{E'\in\mathcal{E}_{G}}$" can be removed from the problem definition. Randomized algorithms, in the sense that their output satisfies the required inequality with probability $1 - \delta$, for some $\delta > 0$, are also welcome. I am familiar with Duan and Pettie's work, and most references therein.
Start by considering each coin in isolation, and the question becomes easier. Let $Y$ denote the number of heads for the first coin, and $Z$ denote the number of heads for the second coin, so $X=Y+Z$. $Y$ and $Z$ are identically distributed so let's just consider $Y$. First, we know that if $Y=y$, then the first coin must have come up heads exactly $y$ times, and then tails once. The probability of $y$ heads in a row is $\left ( \frac{1}{2} \right )^{y}$, and the probability of getting tails after that is $\frac{1}{2} $. Thus: $P(Y=y) = \left ( \frac{1}{2} \right )^{(y+1)}$ To calculate the expected value of $Y$, we sum $y \cdot P(Y=y)$ over all values of $Y$, from zero to infinity: \begin{align} E[Y] &= \sum_{y=0}^{\infty} y \cdot P(Y = y)\\&= \sum_{y=0}^{\infty} y \cdot \left ( \frac{1}{2} \right )^{(y+1)} \\&= 1\end{align} The expectation of the sum of two random variables is the sum of their expectations, so $E[X] = E[Y+Z] = E[Y] + E[Z] = 2$. How do we use $P(Y)$ to get $P(X)$? Here's an example: Say $X=2$. Then we know there are three possibilities: (1) $Y=2$ and $Z=0$, (2) $Y=1$ and $Z=1$, or (3) $Y=0$ and $Z=1$. Since $Y$ and $Z$ are independent, we have: \begin{align}P(X=2) &= \left( \left( \frac{1}{2} \right)^{3}\cdot \frac{1}{2} \right ) + \left( \left( \frac{1}{2} \right)^{2}\cdot \left(\frac{1}{2} \right)^2 \right) + \left( \frac{1}{2} \cdot \left( \frac{1}{2} \right)^{3} \right )\\&= 3 \cdot \left( \frac{1}{2} \right)^{4}\end{align} This example gives the intuition that maybe $P(X=x) = (x+1) \cdot \left( \frac{1}{2} \right)^{(x+2)}$. It is true for $X=0$: both heads have to come up tails on the first flip, and the probability of that occurring is $\frac{1}{4} = (0+1) \cdot \left( \frac{1}{2} \right)^{(0+2)}$. It should be simple to show by induction that this is true for all values of $X$. Here is a sketch. First note that if $X=x$ there are $x+1$ possible combinations of $Y$ and $Z$ values that can produce $y + z = x$. Each value of $Y$ corresponds to a unique series of heads and a tail (and likewise for $Z$). If we iterate, and ask what values of $Y$ and $Z$ could give $y' + z' = x+1$, we can start with our original set of possible combinations of $Y$ and $Z$ values, and just add an extra head to the start of each run for the first coin, which would multiply the probability of each combination by $\frac{1}{2}$. That is, we set $y'= y+1$ and $z'=z$. Then we need to add one new term to the sum, to account for the case where $y=0$, and $z'=x+1$.
Getting all (complex) solutions of a non polynomial equation Hi ! I was used to solve the following equation with Mathematica. \begin{equation} \alpha_1 + \alpha_2x + \alpha_3x^2 + x^4 + \frac{\alpha_4}{x^2-\alpha_0} + \frac{\alpha_5 x^2}{x^2-\alpha_0}=0 \end{equation} where $\alpha_i$ are constants. The Mathematica function "Solve" gives me all the numerical roots of this non polynomial equation very easily. These roots can be real or complex. I'm a very beginner at Sage. I have tried several methods to solve this equation automatically but it seems that all methods I've used work only for polynomial equations. Here they are : alpha0 = 0.25alpha1 = -2.5alpha2 = 6.9282alpha3 = -5.5alpha4 = 0.5alpha5 = -0.5x = var('x')eq = alpha1 + alpha2x + alpha3x2 + x4 + alpha4/(x*2 - alpha0) + alpha5x2/(x2 - alpha0) == 0.# test 1# solve(eq, x, ring=CC)# ==> [0 == 20000x^6 - 115000x^4 + 138564x^3 - 32500x^2 - 34641x + 22500]# test 2# solve(eq, x, ring=CC, solution_dict=True)# ==> [{0: 20000x^6 - 115000x^4 + 138564x^3 - 32500x^2 - 34641x + 22500}]# test 3# eq.roots(x, ring=CC,multiplicities=False)# ==> TypeError: fraction must have unit denominator Do you have any idea of a method to get the approximated roots of the equation ? Thanks in advance EDIT : correction of an error in the equation ; add few tests
Energy Density Formula - Electric Energy Density Formulas with Examples VIEW MORE Energy Density Formula Energy Density refers to the total amount of energy in a system per unit volume. (Eventhough generally energy per unit mass is also mentioned as energy density, the proper term for the same is specific energy. The term density usually measures the amount per unit spatial extension). Energy density is denoted by letter U. Magnetic and electric fields can also store the energy. In the case of electric field or capacitor, the energy density is given by \[U = \frac{1}{2}{\varepsilon _0}{E^2}\] The energy density in the case of magnetic field or inductor is given by, \[U = \frac{1}{{2{\mu _0}}}{B^2}\] For electromagnetic waves, both magnetic and electric fields are equally involved in contributing to energy density. Therefore, the energy density is the sum of the energy density of electric and magnetic fields. i.e.,\[U = \frac{1}{2}{\varepsilon _0}{E^2} + \frac{1}{{2{\mu _0}}}{B^2}\] Example 1 In a certain region of space, the magnetic field has a value of 1.0 × 10–2 T, and the electric field has a value of 2.0 ×106 Vm–1. Find the combined energy density of the electric and magnetic fields. Solution: E = 2.0 ×106 Vm–1; B = 1.0 × 10–2 T For the electric field, the energy density is \[{U_E} = \frac{1}{2}{\varepsilon _0}{E^2} = \frac{1}{2} \times 8.85 \times {10^{--12}}{(2.0 \times {10^6})^2} = 18\,J{m^{--3}}\] For the magnetic field, the energy density is \[{U_B} = \frac{1}{2}\frac{{{B^2}}}{{{\mu _0}}} = \frac{1}{2} \times \frac{{{{(1.0 \times {{10}^{--2}})}^2}}}{{4\pi \times {{10}^{--7}}}} = 40\,J{m^{--3}}\] The net energy density is the sum of the energy density due to the electric field and the energy density due to the magnetic field: U = UE + UB = 18 + 40 = 58 Jm–3 Practice question The total energy density associated with an electromagnetic wave is (assuming the same to be equally shared across the electric and magnetic fields): (a) \[\frac{1}{2}{\varepsilon _0}{E^2}\] (b) \[\frac{1}{{2{\mu _0}}}{B^2}\] (c) \[{\varepsilon _0}{E^2}\] (d) \[\frac{{2{B^2}}}{{{\mu _0}}}\] Ans (c).
Learning With Large Datasets Draw a learning curve and determine if more data needs to be collected. Stochastic Gradient Descent Cost Function in SGD : $cost(\theta, (x^{(i)}, y^{(i)})) = \frac {1}{2}(h_{\theta}(x^{(i)})-y^{(i)})^2$ randomly shuffle the data set a little gradient descent step using just one single training example maybe head in a bad direction, generally move the parameters in the direction of the global minimum, but not always it ends up doing is wandering around continuously in some region that’s in some region close to the global minimum Mini-Batch Gradient Descent In Batch gradient descent we will use all m examples in each generation. Whereas in Stochastic gradient descent we will use a single example in each generation. What Mini-batch gradient descent does is somewhere in between. Stochastic Gradient Descent Convergence$\alpha = \frac {const1}{iterationNumber + const2}$ We can compute the cost function on the last 1000 examples or so. And we can use this method both to make sure the stochastic gradient descent is okay and is converging or to use it to tune the learning rate alpha. Online Learning The online learning setting allows us to model problems where we have a continuous flood or a continuous stream of data coming in and we would like an algorithm to learn from that. We learn using that example like so and then we throw that example away. If you really have a continuous stream of data, then an online learning algorithm can be very effective. If you have a changing pool of users, or if the things you’re trying to predict are slowly changing like your user taste is slowly changing, the online learning algorithm can slowly adapt your learned hypothesis to whatever the latest sets of user behaviors are like as well. Map Reduce and Data Parallelism In the MapReduce idea, one way to do, is split this training set in to different subsets and use many different machines. multi-core machine multiple machines numerical linear algebra libraries like Hadoop
So what are spin-networks? Briefly, they are graphs with representations ("spins") of some gauge group (generally SU(2) or SL(2,C) in LQG) living on each edge. At each non-trivial vertex, one has three or more edges meeting up. What is the simplest purpose of the intertwiner? It is to ensure that angular momentum is conserved at each vertex. For the case of four-valent edge we have four spins: $(j_1,j_2,j_3,j_4)$. There is a simple visual picture of the intertwiner in this case. Picture a tetrahedron enclosing the given vertex, such that each edge pierces precisely one face of the tetrahedron. Now, the natural prescription for what happens when a surface is punctured by a spin is to associate the Casimir of that spin $ \mathbf{J}^2 $ with the puncture. The Casimir for spin $j$ has eigenvalues $ j (j+1) $. You can also see these as energy eigenvalues for the quantum rotor model. These eigenvalues are identified with the area associated with a puncture. In order for the said edges and vertices to correspond to a consistent geometry it is important that certain constraints be satisfied. For instance, for a triangle we require that the edge lengths satisfy the triangle inequality $ a + b \lt c $ and the angles should add up to $ \angle a + \angle b + \angle c = \kappa \pi$, with $\kappa = 1$ if the triangle is embedded in a flat space and $\kappa \ne 1$ denoting the deviation of the space from zero curvature (positively or negatively curved). In a similar manner, for a classical tetrahedron, now it is the sums of the areas of the faces which should satisfy "closure" constraints. For a quantum tetrahedron these constraints translate into relations between the operators $j_i$ which endow the faces with area. Now for a triangle giving its three edge lengths $(a,b,c)$ completely fixes the angles and there is no more freedom. However, specifying all four areas of a tetrahedron does not fix all the freedom. The tetrahedron can still be bent and distorted in ways that preserve the closure constraints (not so for a triangle!). These are the physical degrees of freedom that an intertwiner possesses - the various shapes that are consistent with a tetrahedron with face areas given by the spins, or more generally a polyhedron for n-valent edges. Some of the key players in this arena include, among others, Laurent Friedel, Eugenio Bianchi, E. Magliaro, C. Perini, F. Conrady, J. Engle, Rovelli, R. Pereira, K. Krasnov and Etera Livine. I hope this provides some intuition for these structures. Also, I should add, that at present I am working on a review article on LQG for and by "the bewildered". This post imported from StackExchange Physics at 2014-04-01 16:52 (UCT), posted by SE-user user346 I reserve the right to use any or all of the contents of my answers to this and other questions on physics.se in said work, with proper acknowledgements to all who contribute with questions and comments. This legalese is necessary so nobody comes after me with a bullsh*t plagiarism charge when my article does appear :P
This post is meant to teach new folks how to use MathJax and mhchem formatting on chem.SE Getting started On chemistry.SE, we use MathJax to format mathematical as well as chemical equations and similar expressions in questions, answers, and comments. MathJax allows us to typeset expressions using $\LaTeX$ notation. To use MathJax, enclose your math within single ( $...$) or double ( $$...$$) dollar signs. Single dollar signs make the math inline, for example, Let $x$ be a variable gives: Let $x$ be a variable. On the other hand, double dollar signs make the math a block element. It gets its own line, and is slightly larger. For example, The equation of motion is as follows: $$v=u+at$$ It is a SUVAT equation gives: The equation of motion is as follows: $$v=u+at$$ It is a SUVAT equation Note that, in math mode, MathJax ignores the spaces you type, e.g. $a b$ yields $a b$. MathJax formats expressions the way it is common in mathematics texts. However, the printing rules for signs and symbols used in the natural sciences and technology may require additional spaces (in particular, between the numerical value and the unit symbol). In math mode, use \ (backslash space) or ~ (tilde) if you want the equivalent of space in normal text. Where separation of numbers into groups of three digits is used, the groups shall be separated by a thin space \, (backslash comma); e.g. $299\,792\,458$ yields $299\,792\,458$. Basic chem We use the mhchem package for chemistry. It lets us easily format chemical formulas and reactions without typing too much. There really is only one command you need to know here: \ce{...}. \ce{...} takes its parameters and automatically formats it. For example, $\ce{HCl}$ dissociates in water as follows:$$\ce{H2O + HCl <=> H3O+ + Cl-}$$ Renders as $\ce{HCl}$ dissociates in water as follows: $$\ce{H2O + HCl <=> H3O+ + Cl-}$$ Note that spaces are very important for mhchem to separate super/subscripts from normal text. \ce{H3O+} will display $\ce{H3O+}$, but \ce{H2O +} will display $\ce{H2O +}$. When typesetting ions with more than a single charge, the argument has to be raised using the caret (^; also known as the circumflex accent) character, e.g. $\ce{Cu^2+}$ renders as $\ce{Cu^2+}$, while $\ce{Cu2+}$ would incorrectly render to $\ce{Cu2+}$. (Also see basic math below.) Various types of reaction arrows are supported, including ->, <=>, <==>>, etc. It also supports various types of bonds, via the \bond{..} command (to be called inside \ce{...}). You need not call \bond for normal bonds. Eg: \ce{H\bond{->}A-B=C#D\bond{~}E\bond{~-}F\bond{...}G\bond{<-}E} displays: $$\ce{H\bond{->}A-B=C#D\bond{~}E\bond{~-}F\bond{...}G\bond{<-}E}$$ Basic math Superscripts and subscripts You can denote superscripts via the ^ character, and subscripts via _. For example, x^2 renders as $x^2$, x_1 renders as $x_1$, and x_1^3 renders as $x_1^3$. If you want to include more than one character in the super/sub script, enclose it in curly braces ( {...}). For example, x^10 renders as $x^10$, but x^{10} renders as $x^{10}$ Fractions and square roots Fractions can be easily displayed using \frac{..}{..}. For example, $$\frac{a+b^c}{de+f}$$ renders as $$\frac{a+b^c}{de+f}$$ However, where it is necessary to include fractions in the body text, they shall, where possible, be reduced to a single level by using a solidus (/) or, where applicable, the negative index. For example, instead of $\frac{1}{\sqrt 2}$ write $1/\sqrt 2$ or $2^{-1/2}$; instead of $C_{\mathrm m,p} = 33.58\ \mathrm{\frac{J}{K \cdot mol}}$ write $C_{\mathrm m,p} = 33.58\ \mathrm{J/(K \cdot mol)}$ or $C_{\mathrm m,p} = 33.58\ \mathrm{J\cdot K^{-1} \cdot mol^{-1}}$. Protip: You can exclude the braces for single-character numerators/denominators (if the first character is a letter, you need to use a space after \frac, though). For example \frac12 renders as $\frac12$, and \frac ab renders as $\frac ab$. Square roots can be added in a similar manner, via \sqrt{....}. For example, \sqrt{x+y} renders as $\sqrt{x+y}$. Greek letters Greek letters can be added using a backslash ('\'), followed by the name of the letter. Captialise the first letter of the name for greek capital letters. Eg \alpha \beta \gamma \Omega renders as $\alpha \beta \gamma \Omega$. Make sure that you put spaces after these if you are typing normal alphabet characters. Eg e^{\pii} gives an error, you need to use e^{\pi i}. Note that there are special commands \varepsilon , \varsigma , \varrho , and \varpi to distinguish between the lunate Greek letters. MathJax extensions MathJax has available a variety of extensions enabling other features: strikeout lines, enclosures, text/background coloration, interactive equations, etc. Information about these extensions and how to enable them can be found at this meta post. Further reading Wikipedia TeX help page (extremely useful as a reference, useless as a tutorial) TeX/LaTeX Stack Exchange site Harvard intro to TeX LaTeX wikibook, Math section LaTeX wikibook, Advanced Math section If you would like to know more, you can continue reading about which symbols are written in italic (sloping) type and which are printed in roman (upright) type.
Is the following claim known? Claim: For any graph $G$ with $n$ vertices there exists a coloring of $G$ such that every independent set is colored by at most $O(\sqrt{n})$ colors. Theoretical Computer Science Stack Exchange is a question and answer site for theoretical computer scientists and researchers in related fields. It only takes a minute to sign up.Sign up to join this community The following claim is known to me, but may not count because it is unpublished: Any graph on $n$ vertices can be colored so that any induced subgraph $H$ of chromatic number at most $k$ uses at most $\chi(H)+B$ colors, where $B(B+1)\leq 2kn$. This is a proof by induction; the motivation was to consider colorings which use few colors not only on the graph but also on all induced subgraphs. I am not aware of any published results, though. Not quite what you ask for, but here's a lower bound - a graph for which any coloring will result in an independent set colored by $\sqrt{n}$ colors: Take $\sqrt{n}$ copies of $K_{\sqrt{n}}$, and connect all vertices to a single vertex $s$. Obviously, every set of $\sqrt{n}$ vertices from different $K$'s is independent, and in every copy of $K_{\sqrt{n}}$ you can find at least one "new" color. This lower bound can easily be improved to $\sqrt{2n}$ or so if we connect $K_1,K_2,..$ to a single vertex, but it remains only $\Omega(\sqrt{n})$ colors. What about the following proof? If $\alpha(G) \leq \sqrt{n}$, then the claim holds obviously. Suppose the contrary, and let $I$ be an independent set of $G$ with maximum cardinality $\alpha$. Color $I$ with color 1, and recursively color the graph $G - I$ with colors $2,...,c$. Now, if $K$ is an independent set of $G$, consider $K' = K - I$. By induction hypothesis, $K'$ is colored with at most $\sqrt{n-\alpha}$ colors, and thus $K$ is colored with at most $1+\sqrt{n-\alpha} \leq \sqrt{n}$ colors; the inequality holds by the assumption that $\alpha \geq \sqrt{n}$.
Matching pursuit Piotr J. Durka (2007), Scholarpedia, 2(11):2288. doi:10.4249/scholarpedia.2288 revision #140500 [link to/cite this article] Matching pursuit (MP) algorithm finds a sub-optimal solution to the problem of an adaptive approximation of a signal in a redundant set (dictionary) of functions. Commonly used with dictionaries of Gabor functions, it offers several advantages in time-frequency analysis of signals, in particular EEG/MEG. Contents Dictionaries and linear expansions Many signal analysis methods look for a linear expansion of the unknown signal x in terms of functions $g_n$ \[\tag{1} x = \sum_{n=1}^{N} a_n g_n\] We may say that in such a way the unknown signal x is explained using words (functions $g_n$) from the dictionary D, used for decomposition. If the dictionary D is an orthonormal basis (like orthogonal wavelets or Fourier bases), then the coefficients $a_n$ are given simply by the inner products of the dictionary's functions $g_n$ with the signal $\langle x, g_n \rangle\ .$ We would like to use a dictionary $D=\lbrace g_n\rbrace_{n=1..N} $ that would properly reveal the intrinsic properties of an unknown signal, or, almost equivalently, would give low entropy of the $\lbrace a_n\rbrace $ and possibilities of good lossy compression. Unfortunately, there is no universal recipe for such a prior choice. Adaptive approximations We may relax the requirement of exact signal representation (1), and try to automatically choose the functions $g_{\gamma_n}\ ,$ optimal for the representation of a given signal x, from a redundant dictionary D. The expansion becomes an approximation, and uses only the functions $g_{\gamma_n}$ chosen from the redundant dictionary D. In practice, the dictionary contains orders of magnitude more candidate functions $g_\gamma$ than the number M of functions chosen for the representation. \[\tag{2} x\approx \sum_{n=0}^{M-1} a_n g_{\gamma_n}\] A criterion of optimality of a given solution (for a fixed dictionary D, signal x, and number of used functions M) can be formulated as minimization of the error of representation \[\tag{3} \epsilon=\left\\| x-\sum_{n=0}^{M-1} a_n g_{\gamma_n} \right\\| = \min\] Finding the minimum requires checking all the possible combinations (subsets) of M functions from the dictionary, which leads to a combinatorial explosion. Therefore, the problem is intractable even for moderate dictionary sizes.Matching pursuit algorithm, proposed in (Mallat and Zhang 1993), finds a sub-optimal solution by means of an iterative procedure. Matching pursuit algorithm In the first step, the waveform $g_{\gamma_0}$ which gives the largest product with the signal x is chosen from the dictionary D, composed of normalized functions ($\|\|g_{\gamma_n}\|\|=1$).In each of the consecutive steps, the waveform $g_{\gamma_n}$ is matched to the signal $R^n x$ which is the residual left after subtracting results of previous iterations:\[\tag{4}\begin{cases} R^0 x = x \\ R^n x = \langle R^n x,g_{\gamma_n}\rangle g_{\gamma_n}+R^{n+1}x\\ g_{\gamma_n} = \arg \max_{g_{\gamma_i} \in D } \|\langle R^n x, g_{\gamma_i}\rangle \| \end{cases} \] Orthogonality of $R^{n+1} x$ and $g_{\gamma_n}$ in each step implies energy conservation \[\|\|x\|\|^2 =\sum_{n=0}^{M-1} {\|\langle R^n x, \;g_{\gamma_n}\rangle \|^2} + \|\|R^M x\|\|^2\] For a complete dictionary the procedure converges to x with $M\rightarrow\infty$ (Mallat and Zhang 1993). In practice we use finite expansions \[\tag{5} x\approx\sum_{n=0}^{M-1} {\langle R^n x,\; g_{\gamma_n}\rangle g_{\gamma_n} }\] Multivariate matching pursuit (MMP) Multivariate extensions of the matching pursuit algorithm can be achieved in many significantly different ways, possibly corresponding to the aims of analysis and/or to the model of generation of the signal. For example, multivariate time series may stem from simultaneous measurements at different sensors (channels), as in the case of multichannel EEG/MEG, or from a joint analysis of subsequent repetitions of evoked fields or potentials. Variety of multivariate MP algorithms (MMPs) can be mostly attributed to: (A) the structure of the multichannel dictionary (i.e., which parameters of the time-frequency atoms are allowed to vary across the jointly analyzed signals), (B) to the criterion used for choosing (in each iteration) the atom, best fitting the residua in all the signals simultaneously. The most straightforward multichannel extension of the MP can be defined by the following conditions (Gribonval 2003): (A) only the amplitude varies across channels, (B) we maximize the sum of squared products (energies) in all the channels: \[\tag{6} \max_{g_{\gamma} \in D } \sum_{i=1}^{N_s} \|\langle R^n\mathbf{x}^i, g_{\gamma}\rangle \|^2 \ ,\] where $N_s$ stands for the number of simultaneously analyzed signals $x^i\ .$ A sub-optimal approach, proposed in (Durka et al. 2005) for preprocessing for EEG Inverse Solutions, relies on maximizing the sum of products \[\tag{7} \max_{g_{\gamma} \in D } \left\| \sum_{i=1}^{N_s} \langle R^n\mathbf{x}^i, g_{\gamma}\rangle \right\| \ .\] Owing to the linearity of the residuum $R\ ,$ this assumption allows for implementation operating in each step on the average of signals. Similarly, we can allow for different phases in functions fitted to each channel/repetition $x^i\ .$ In the analysis of multichannel EEG/MEG, that would relate to a different model of propagation. If we take $x^i$ as repetitions of evoked potentials or fields, variable phase can partly account e.g. for the jitter of latencies. As exemplified above, other modifications of the above mentioned criteria (A) and (B) may lead to other variants of MMP. Dictionary of time-frequency atoms Dictionaries ( D), commonly used for time-frequency analysis, are composed of Gabor functions, that is Gaussian envelopes modulated by sinusoidal oscillations. Real-valued Gabor function can be expressed as $\tag{8} g_\gamma (t) = K(\gamma)e^{-\pi{ \left( \frac{t-u}{s} \right) }^2} \cos\left(\omega (t-u)+\phi\right)$ where $\gamma=\{u, s, \omega, \phi\}\ ,$ and $K(\gamma)$ is a normalization factor ensuring $\|\|g_{\gamma}\|\|=1\ .$ These functions provide a general and compact model for transient oscillations; the above equation can describe parametrically a wide variety of shapes ( Figure 1). Apart from that, Gabor functions exhibit compact time-frequency localization. Gabor dictionaries used for matching pursuit decomposition usually contain also complete Dirac and Fourier bases. Other functions can be also used with the matching pursuit algorithm. In practical implementations of matching pursuit with Gabor dictionaries we must restrict the search to a discrete and finite subset of the continuous 3-dimensional space of parameters $\{u, s, \omega\}\ ;$ phase ($\phi$) is optimized separately in numerical implementations. Time-frequency energy density Wigner-Ville transform It can be simply demonstrated that the squared modulus of the Fourier transform of the signal x, that is the power spectrum of x, can be computed as the Fourier transform of the autocorrelation function of x (Wiener-Khinchine theorem). That is, for a real-valued signal x\[\tag{9}\|\mathcal{F}(x)\|^2 = \mathcal{F} \left(\int_{-\infty}^{\infty} x(t+\frac{\tau}{2}) x(t-\frac{\tau}{2}) d\tau \right)\] where the Fourier transform of the signal x is given by\[\tag{10}\mathcal F(x)=\int_{-\infty}^{\infty} e^{-i\omega t} x(t) dt \] Substituting explicitly (10) into (9) we get the following formula for the power spectral density of x:\[\tag{11}\|\mathcal{F}(x)\|^2 = \int_{-\infty}^{\infty} e^{-i\omega \tau} \left(\int_{-\infty}^{\infty} x(t+\frac{\tau}{2}) x(t-\frac{\tau}{2}) dt \right) d \tau\] If we remove the middle integral, corresponding to the integration over time, we get a time-dependent spectral density as a 2-dimensional function \[\tag{12} \mathcal{W}(x) = \int_{-\infty}^{\infty} e^{-i\omega \tau} x(t+\frac{\tau}{2}) x(t-\frac{\tau}{2}) d \tau \] which is the Wigner-Ville transform of x. This transform exhibits several elegant and desirable mathematical properties, hence it is sometimes considered a 'fundamental' time-frequency representation. Unfortunately, it has also one major drawback which is the presence of cross-terms (aka cross-components) in the time-frequency plane, as illustrated in Figure 2. Minimization of the presence of cross-terms in time-frequency estimates can be achieved by applying a smoothing kernel, designed for a particular signal as in the upper panel of Figure 6. Unfortunately, there is no general recipe for such smoothing which would give satisfactory results for an unknown signal. Similarly to the Wigner-Ville transform of x, we can define a cross-Wigner transform of signals x and y:\[\tag{Wigner1:label exists!}\mathcal{W}(x, y) = \int_{-\infty}^{\infty} e^{-i\omega \tau} x(t+\frac{\tau}{2}) y(t-\frac{\tau}{2}) d \tau\] MP-based estimate of energy density Wigner distribution computed directly from (2) gives \[\tag{13} \mathcal{W}(x) \approx \mathcal{W}\left( \sum_{n=0}^{M-1} a_n g_{\gamma_n} \right) = \] $ \sum_{n=0}^{M-1} a_n^2\, \mathcal{W}(g_{\gamma_n}) + \, \sum_{n=0}^{M-1} \sum_{k=1, k\not=n}^{M-1} a_n a_k \mathcal{W}\left( g_{\gamma_n}, g_{\gamma_k} \right) $ where $a_n=\langle R^n x,\; g_{\gamma_n}\rangle\ .$ The double sum in (13) gathers cross-terms. Using linear expansion (5), we can omit them explicitly and construct the time-frequency representation of signal's energy density from the first sum, containing only auto-terms: \[\tag{14} \mathcal{E} x = \sum_{n=0}^{M-1} \|\langle R^n x,\; g_{\gamma_n}\rangle\|^2 \mathcal{W}g_{\gamma_n} \] Technical issues Matching pursuit provides a simple and intuitive decomposition of a signal, yet "under the hood" the procedure is highly nonlinear and complicated. First problem relates to the already mentioned need to select a discrete subset of the possible dictionary's functions for practical implementations. In general, it is expected, the larger (denser) the dictionary, the better the resulting matching pursuit decomposition at a higher computational cost. However, fixed schemes of subsampling the potentially continuous space of dictionary's parameters introduce a statistical bias, which becomes relevant when pooling together results of many matching pursuit decompositions (Durka et al. 2001). As for the number of waveforms in expansion (5), equal to the number of iterations of (4), there are several possible stopping criteria, but if we stop the iterations "too late" we may simply neglect the waveforms fitted in the latter iterations in further analysis. Reasonable settings for the dictionary size (density) and the number of iterations are important, because the computational cost of the matching pursuit, approximately proportional to these two factors, is still relatively high (Durka 2007). Not all the signal's structures can be efficiently modeled by Gabor functions. Some waveforms, as for example the chirp in Figure 3, may be expressed by several functions from the dictionary D, in which case the advantage of explicit parametrization is impaired. Nevertheless, we still get a robust and universal estimate of the time-frequency energy density, which is relatively independent of arbitrary factors. Influence of the prior choices, biasing other estimates of signals time-frequency energy density, is partly illustrated in Figure 6. Matching pursuit algorithm can be also used with different types of dictionaries, but up to now these attempts were mostly theoretical. Finally, there is a price to pay for using the sub-optimal solution of (3). Greedy matching pursuit algorithm can fail also in cases when the signal is constructed entirely from the structures present in the dictionary. An example is given in Figure 4: a perfect solution to (3) would contain two Gabor functions, but, according to the greedy strategy of explaining maximum energy in a single step, in the first iteration matching pursuit chooses an intermediate waveform embracing both the structures. However, such an error can occur only in case when both the structures are perfectly in phase; in some cases one could argue that explaining them in term of one oscillation may be not so erratic. Advantages in EEG/MEG analysis Major advantages of the matching pursuit in analysis of biomedical signals stem from the explicit parametrization of transients, robust estimate of the time-frequency energy density, and combinations of these two. Extensive summary of the matching pursuit-based frameworks successfully applied to the EEG analysis is given in (Durka 2007). Explicit parametrization of transients Gabor functions (8) fitted to the signal in the matching pursuit procedure (4) are defined by their amplitude, occurrence in time u, frequency $\omega\ ,$ time span s and phase $\phi\ .$ If we assume that Gabor functions model appropriately the relevant components of the signal, then we can treat the fitted Gabors as representing the investigated phenomena. For example, EEG sleep spindles known from the visual analysis are usually defined as waveforms of frequencies in the 11-15 Hz range, with more or less generally assumed restrictions also on the amplitude and time span ( Figure 5). These restrictions can be implemented directly in the space of parameters of functions from the signal's expansion (5). A robust and universal time-frequency estimate The major problem in estimating the time-frequency energy density of signals is the trade-off between the time and frequency resolutions, known as the uncertainty principle in signal analysis. Different estimates offer different solutions to this issue, based upon different prior settings regulating this trade-off ( Figure 6). Spectrogram (windowed Fourier transform) gives uniform time-frequency resolution based upon the prior choice of the length of the analyzing window, scalogram (wavelet transform) gives higher time resolution for higher frequencies, while a variety of Wigner-derived distributions from the Cohen's class allow for almost any setting of these constrains. Nevertheless, these settings have to be decided a priori, before the analysis. Their improper choice may severely bias the results. On the contrary, time-frequency estimate (14) derived from the matching pursuit decomposition (5) is based upon a representation, which adopts the time-frequency trade-off to the local signal structures in each iteration of the algorithm, yielding an estimate which is basically free from arbitrary settings--apart from choosing the Gabor functions. References S. Mallat and Z. Zhang (1993) Matching pursuit with time-frequency dictionaries. IEEE Transactions on Signal Processing, 41:3397-3415. P.J. Durka, D. Ircha, and K.J. Blinowska (2001) Stochastic Time-Frequency Dictionaries for Matching Pursuit, IEEE Transactions on Signal Processing, vol. 49, no. 3, pp. 507–510. R. Gribonval (2003) Piecewise Linear Source Separation, Proc. SPIE 03, San Diego, CA. P.J. Durka (2007) Matching Pursuit and Unification in EEG Analysis, Artech House, ISBN 978-1-58053-304-1. P.J. Durka, A. Matysiak, E. Martinez Montes, P. Valdes Sosa, K.J. Blinowska (2005) Multichannel matching pursuit and EEG inverse solutions. Journal of Neuroscience Methods, vol. 148/1, pp. 49-59. Internal references Valentino Braitenberg (2007) Brain. Scholarpedia, 2(11):2918. Paul L. Nunez and Ramesh Srinivasan (2007) Electroencephalogram. Scholarpedia, 2(2):1348. Tomasz Downarowicz (2007) Entropy. Scholarpedia, 2(11):3901.
I am looking at Trefethen and Bau Exercise 37.1: I have two normalizations of the Legendre polynomials with corresponding recurrence relations: $$P_n(1)=1$$ which follows $$P_n(x) = \frac{2n-1}{n} x P_{n-1}(x) -\frac{n-1}{n}P_{n-2}(x)$$ and $$\\|q_n\\|=1$$ which follows $$xq_n(x)=\beta_{n-1}q_{n-1}(x)+\alpha_n q_n(x) + \beta_n q_{n+1}(x)$$ where $$\beta_n = \tfrac12 (1-(2n)^{-2})^{-1/2}$$ Apparently, $q_{n+1}$ is proportional to $P_n$. I know that the entries of the Jacobi matrix $t_{ij} = (q_i,xq_j)$ in the $L^2([-1,1])$ inner-product space, and I managed to get the right answer for the $q_j$ with $0$ along the main diagonal, $\beta_{j-1}$ on the upper diagonal and $\beta_j$ along the lower diagonal. However, it doesn't seem that I am getting the right answer for the 1st normalization. Also, I am asked to find the relationship between the two Jacobi matrices--my intuition is that they are equal but I can't quite justify it--I think it has to do with the fact that the characteristic polynomials are the same monic polynomial. Furthermore, I am asked to find a formula for $q_{n+1}(1)=\\|P_n\\|$ using the Jacobi matrices. I've tried to solve this by equating the entries for the Jacobi matrices (where one includes a term of $\|\|P_n\|\|^2$) and solving. But I can't get the answer $\\|P_n\\|=\sqrt{\frac{2}{2n+1}}$.
Optical fibers are one of the most widely used form of optical waveguide. Unlike other form of waveguides, optical fibers can be manufactured long and bendable, and thus are primarily used for a transmission media for optical fiber communication. The history of optical fibers is believed to date back to mid 1800s, when an Irish physicist John Tyndall demonstrated light guidance in a fountain (see Figure 1). Light is guided in a stream of water by total internal reflection (TIR) even though the water stream is bent by gravity. This phenomenon highlights two very important features of optical fiber – long and bendable. Figure 1: Light guidance in fountain by TIR. Light guidance in optical fiber TIR takes place at any interface between two materials with different refractive indices, not only at the interface of water and air. Figure 2 shows schematically the most basic form of a modern optical fiber. A core is embedded in a cladding and modifies the refractive index in the transverse (x and y) direction, and the modification is uniform in the longitudinal (z) direction. The refractive-index modification enables the transverse confinement of light in the core, and light is guided longitudinally in the core. Figure 2: Schematic of optical fiber, and a ray confined in the core by TIR. Figure 2 shows schematically, using ray optics, the light guided in the core by TIR. The refractive indices of the core and cladding are n core and n clad, and a ray is travelling in the core with an angle θ. If n core > n clad and θ is larger than the critical angle θ c which is provided by $ \theta_{\mathrm{c}}=\sin^{-1}\left( \frac{\mathrm{n_{clad}}}{\mathrm{n_{core}}} \right), $ TIR takes place at the border between the core and cladding, and the ray is confined in the core. On the other hand, the power of a ray in the core rapidly decreases if θ is smaller than θ c, because the ray is only partially reflected back at the border. The refractive index of the core has to be higher than that of the cladding, in order to guide light by total internal reflection. The equation also suggests that a larger refractive-index difference between the core and cladding provides a smaller θ c. A small θ c allows for launching light into the fiber with a large angle ( $\pi/2−\theta$ in Figure 1), enabling the launch of higher power density into the core. As a measure of the ability to accept light with a large launch angle, the numerical aperture (NA) of a fiber is defined by $ \mathrm{NA}=\mathrm{n_{core}}\sin \left( \frac{\pi}{2}-\theta_{\mathrm{c}} \right)=\sqrt{\mathrm{n_{core}^2-n_{clad}^2}}. $ This equation suggests an optical fiber with a higher NA allows the launch of light with a larger launch angle, therefore enables the launch of higher power density into the core. Step-index and graded-index fiber There are two major types of refractive-index profiles – step index (SI) and graded index (GI) – and they are schematically shown in Figure 3. The core of a SI fiber has a uniformly raised refractive index profile, and that of a GI fiber has a continuously raised refractive index profile. The largest difference between these two types of optical fibers lies in the ability to maintain the pulse shape after propagation. Figure 3: Light guidance in (1) step-index and (2) graded-index fiber. When an optical pulse is launched into a SI fiber, the propagation velocity of the pulse is highly dependent on the incident angle. Figure 3(a) intuitively explains that the light launched at a larger incident angle (blue ray in the figure) propagates at a slower velocity, as the path length is longer. This is problematic for high-speed optical communication as the incident pulse shape is degraded at the output. GI fiber was introduced in order to overcome this limitation of SI fiber. Figure 3(b) shows that, light launched at a small angle (red ray) mainly travels near the center of the core, and light launched at a large angle (blue ray) travels at the outer side of the core. The path length is still longer for the blue ray; however the blue ray travels at a faster speed than the red ray as the refractive index of the outer side of the core is lower than that of the center. The difference in the propagation velocity, therefore, becomes much smaller and degradation in the pulse shape is reduced. Multimode and Single-mode fiber The number of allowed optical paths in an optical fiber, called modes, depends on the refractive-index profile. If more than one mode is allowed to propagate in a fiber at a certain wavelength, the fiber is called a multimode fiber (at that wavelength). If only one mode is allowed to propagate, the fiber is called a single-mode fiber. A single-mode fiber has a relatively smaller core size compared to a multimode fiber, and only one propagation mode is allowed (see Figure 4). It is theoretically possible to eliminate the modal delay in a multimode fiber by carefully tuning the graded index profile. In reality, however, small amount of modal delay always remains due to small variations in fiber fabrication. The small amount of modal delay is critical in high-speed and long-distance optical transmission, as even a small delay causes signal overlap between neighboring bits – the original signals can no longer be recovered at the receiver side. The use of single-mode fibers in an optical transmission line fundamentally eliminates signal distortions due to modal delay, and thus single-mode fibers have now become the most widely used type of optical fibers in optical communications. Figure 4: Multimode and single-mode fiber. Key characteristics of optical fiber Phase velocity: The phase of a wave travels at a certain speed in a medium, called the phase velocity. In an optical fiber, the phase velocity of a mode v p is provided by ω/β. Effective index: The effective index of a mode n eff is defined by β/k 0 (k 0 = 2π/λ). It is called “effective” index because the phase velocity of a mode with effective index n eff is the same as that of a plane wave travelling in a uniform medium with refractive index n eff. Group velocity: Each mode class travels at a different speed when it travels in an optical fiber as an optical pulse. The speed of the pulse is called the group velocity of a mode, and is given by v g = dβ/dω. Modal dispersion (or modal delay): When an optical pulse contains several different mode classes, the pulse spreads as it travels in an optical fiber because each mode class travels at a different group velocity. The different group velocity between mode classes is called modal dispersion, and is the dominant effect causing pulse spreading in a multimode fiber. Chromatic dispersion (or group velocity dispersion): Group velocity in an optical fiber is wavelength dependent thus optical pulses at different wavelengths travel at different speeds. The speed difference between different wavelengths is called chromatic dispersion. Any optical pulse spreads due to the chromatic dispersion, because an optical pulse always has a spectral bandwidth. This effect is the dominant effect accounting for pulse spreading in a single-mode fiber, and the spreading is pronounced for a short pulse, because a short pulse has a wider spectral bandwidth. Chromatic dispersion is induced both by the refractive-index profile and the wavelength dependency of the refractive index of the material; the former is called waveguide dispersion, and the latter is called material dispersion. Modal birefringence: One mode class is generally subdivided into several modes depending on the polarization (polarization modes), because an electromagnetic wave is a vector wave. When polarization modes in one mode class have the same β, the polarization modes are not maintained due to polarization coupling. On the other hand, when modes in the same mode class have slightly different β, the polarization coupling is reduced thus polarization modes are maintained. The difference in β within the same mode class Δβ is called modal birefringence; modal birefringence is usually normalized such that it has no unit, and provided as Bm = Δβ/k0. Large modal birefringence reduces polarization coupling, thus enables better ability to maintain polarization modes. These fibers are called Polarization-Maintaining Fiber (PMF). Group birefringence: In a PMF, the group velocity is generally polarization dependent within one mode class. This is the dominant effect accounting for pulse spreading in a single-mode fiber in which the pulse spreading due to chromatic dispersion is controlled and suppressed. Polarization Mode Dispersion (PMD) originates in Group birefringence. Mode area: A mode area represents the tightness of a guided mode in the core, and is a measure of nonlinearity of the fiber. An optical fiber with a small mode area generally displays high nonlinearity and is one way to produce a high-nonlinear fiber. On the other hand, an optical fiber with a large mode area shows low nonlinearity, and is called a large-mode-area fiber (LMA fiber). Cutoff: Modes are fundamentally wavelength dependent, and one mode guided at one wavelength may not be guided at another wavelength. If a guided mode loses guidance as the wavelength changes, it is called a cutoff. The wavelength limit, at which a guided mode loses the guidance, is called a cutoff wavelength. Fiber manufacturing and material Low-attenuation silica fiber for optical transmission Another important feature of optical fiber is its low attenuation. In 1966, Charles Kao predicted that the loss of optical fiber can be reduced significantly if it is made by high-purity glass . The prediction was proven in 1970 by three scientists in Corning – Donald Keck, Robert Maurer, and Peter Schultz (link). Now the loss of standard telecom optical fiber is well below 0.20 dB/km; light is lost by less than five percent per kilometer. In 2009, Charles Kao was awarded the Novel prize for his contribution to fiber optics. Today low-loss fibers for telecommunication are all silica-based fibers, and the low losses are enabled by vapor-phase manufacturing methods, for example, MCVD , VAD , and OVD . In manufacturing silica fiber by vapor-phase methods, SiCl 4 and GeCl 4 are typically used as starting raw materials. The vapor pressures of these two materials are much higher than those of undesirable contaminants (e.g. FeCl 3, PbCl 2) remained in the raw material, therefore high-purity raw materials can be delivered to a reaction chamber by vapor-phase delivery. The raw materials delivered to the chamber is then heated and turns into oxide glass, either by oxidation or hydrolysis: $ \text{SiCl}_{4}/\text{GeCl}_{4} + \text{O}_{2} \longrightarrow \text{SiO}_{2}/\text{GeO}_{2} + \text{2Cl}_{2}, $ (Oxidation) $ \text{SiCl}_{4}/\text{GeCl}_{4} + \text{4H}_{2} + \text{2O}_{2} \longrightarrow \text{SiO}_{2}/\text{GeO}_{2} + \text{4HCl} + \text{2H}_{2}\text{O}. $ (Hydrolysis) Silica optical fibers for telecommunication nowadays have negligible amount of contaminants, and the two factors that determine the losses are Rayleigh scattering and infrared absorption. Non-silica specialty optical fibers Low loss is arguably the most important characteristic as a transmission media for optical communication, and silica-based glass is by far the most transparent optical fiber material. There are, however, many other areas where fibers are used in a short piece (up to several tens of meters) and other characteristics have more importance. These application-specific fibers are often called as specialty fibers, and non-silica optical fiber materials may be used due to their unique natures. Fluoride fibers are used for fiber lasers/amplifiers and IR laser power delivery (see our fluoride fiber technology for detail). Chalcogenide fibers are used for optical signal processing due to its high nonliniarity. Plastic fibers are used for cars and consumer electronics, as handling of plastic fibers does not require special skill or training. FiberLabs’ optical fiber products FiberLabs manufactures ZBLAN and AlF3-based fluoride fibers in house. Fluoride fibers are mainly used as specialty fibers, and have been used in a variety of industries ranging from telecom to medical. We have been supplying various fluoride fibers, both standard and custom made, to meet customers’ needs. So please feel free to contact us if you would like to use fluoride fibers for your projects. References 113(7), 1151–1158 (1966) [http://doi.org/10.1049/piee.1966.0189]. 62(9), 1280–1281 (1974) [http://doi.org/10.1109/PROC.1974.9608]. 68(10), 1187–1190 (1980) [http://doi.org/10.1109/PROC.1980.11828]. 68(10), 1181–1184 (1980) [http://doi.org/10.1109/PROC.1980.11826]. 68(10), 1184–1187 (1980) [http://doi.org/10.1109/PROC.1980.11827].
Consider to following decision problem: Input: Undirected graph $G=(V,E)$ Question: Is the minimum numbers of colors needed to color the vertices (such that every two adjacent vertices aren't colored the same) exactly2015? Note that 2015 could be any fixed number. This number is not part of the input. I need to understand what is the minimal complexity class that contain this problem, between $\sf P,NP,coNP,NP\cup coNP,\Sigma _{2} \cap \Pi _{2}, \Sigma _{2} \cup \Pi _{2}$. I don't it's $\sf NP$ or $\sf coNP$ since I can't think of a verifier for both the language or it's complement. The problem is maybe in $\sf \Sigma _{2} \cap \Pi {2}$ since we can think of it as exists$f$ some 2009-coloring function, for-all$g$, 2008-coloring function, $f$ is a correct coloring and $g$ isn't a correct coloring. So by the above, it seems that the language is in $\sf \Sigma _{2}$, but the quantifiers order doesn't really matter here, so the language is also in $\sf \Pi _{2}$, I think. And maybe the there is some deterministic polynomial-time algorithm?
I have to solve this equation $$\ddot{x}x = -\frac{3}{2}\dot{x}^2 + \frac{\dot{x}}{h} + \sin(t)$$ where $h$ is defined by $$h = \left(\frac{\dot{x}}{h}\right)^{1/3} + \frac{\dot{x}}{h}$$ My idea is to initialize $h=1$, solve the first equation, calculate a new value for h: $$h = \frac{\dot{x}}{x\ddot{x} + (3/2)\dot{x}^2 - \sin(t)}$$ and do this many times because it guarantees that h has converged.In the following code I do not know how to express $\ddot{x}$, in fact MATLAB says that xp ($\dot{x}$) is undefined. Plesset.m file function xp = Plesset(t, x)xp = zeros(2, 1);xp(1) = x(2);h = 1;h = (x(2)/h)^(1/3) + x(2)/h;xp(2) = 1/x(1)(-1.5(x(2))^2 + x(2)/h+sin(t)); run.m file for i = 1:100 [t, x] = ode45('Plesset', [0,5], [0,0]); h = x(2)/(x(1)xp(2) + 3/2(x(2))^2 - sin(t));end[t, x(:,1)]plot(t, x(:,1))
According to remark 6.14 in Shigeru Mukai's An introduction to invariants and moduli (unfortunately, the page is not available on Google Books, so I explain it below), the GIT-quotient of an affine variety $X$ by a reductive group $G$ with respect to a nontrivial character $\chi: G \to \mathbb G_m$ may be considered as first taking an affine quotient $\operatorname{Spec} (\mathbb k[X]^{G_\chi})$ by $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then taking a quotient by $\mathbb G_m=G / G_\chi$ via replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings. As a result, I wonder why it is so much better then just taking an affine quotient $\operatorname{Spec}(\mathbb k[X]^G)$ if $G$ has much more characters like $\mathrm{GL}_{k_1} \times \ldots \times \mathrm{GL}_{k_r}$. It seems that replacing $\operatorname{Spec}$ to $\operatorname{Proj}$ of non-negative gradings is a correct way to take a quotient by $\mathbb G_m$, but the first step looks fairly stupid. Update: Yes, as Will Sawin said, the question is approximately why one does not iterate it (why there is no need/why nothing better happens)? Details of Mukai's description. For simplicity, work over an algebraically closed field $\mathbb k$ of characteristic $0$. Let $X$ be an affine space or any other affine irreducible algebraic variety with $\operatorname{Pic} X=0$, and let $G$ be a connected reductive group acting on $X$. Then a $G$-linearized line bundle $\mathcal L$ is equivalent to a character $\chi: G \to \mathbb G_m$, so Mumford's GIT quotient is$$X //_{\mathcal L} G:=\operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \Gamma(X, \mathcal L^{\otimes n}) \right) = \operatorname{Proj} \left( \bigoplus_{n \geqslant 0} \mathbb k[X]^G_{\chi^n} \right).$$ Here $$\mathbb k[X]^G_{\chi^n}=\{ f \in \mathbb k[X]: f(gx)=\chi(g)^nf(x) \}$$ are $\chi^n$-semi-invariants. If $G_\chi:=\operatorname{Ker}(\chi: G \to \mathbb G_m)$, then $\mathbb k[X]^{G_\chi}$ has an action of $\mathbb G_m=G / G_\chi$, so is graded -- exactly by $(\mathbb k[X]^{G_\chi})_n=\mathbb k[X]^G_{\chi^n}$. From this follows Mukai's description of $X //_{\mathcal L} G$. Generality of Mumford's GIT-quotient. By the way, I believe that I have seen somewhere that any geometric quotient arises as an open subset in Mumford's GIT-quotient for some line bundle or something like this. Could you give me a reference?
Answer $$x=-1.53, -10.47$$ Work Step by Step Using the quadratic formula, we obtain: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ $$x=\frac{12\pm \sqrt{(12)^2-4(1)(16)}}{2(1)}$$ $$x=-1.53, -10.47$$ 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.
Gain medium The active laser medium (also called gain medium or lasing medium) is the source of optical gain within a laser. The gain results from the stimulated emission of nuclear, electronic or molecular transitions to a lower energy state from a higher energy statepreviously populated by an electron current, chemical reaction or thepump source. Warning: This article is copypasted from http://en.wikipedia.org/wiki/Active_medium for technical needs, the most of wiki–articles cited are not yet loaded; consider to use the wikipedia article instead. Contents Examples of active laser media Certain crystals, typically doped with rare-earth ions (e.g. neodymium, ytterbium, or erbium) or transition metal ions (titanium or chromium); most often yttrium aluminium garnet (YAG), yttrium orthovanadate (YVO 4), or sapphire (Al 2O 3); [1] Glasses, e.g. silicate or phosphate glasses, doped with laser-active ions; [2] Gases, e.g. mixtures of helium and neon (HeNe), nitrogen, argon, carbon monoxide, carbon dioxide, or metal vapors; [3] Semiconductors, e.g. gallium arsenide (GaAs), indium gallium arsenide (InGaAs), or gallium nitride (GaN). [4] Liquids, in the form of dye solutions as used in dye lasers. [5] [6] In order to lase, the active gain medium must be in a nonthermal energy distribution known as a population inversion. The preparation of this state requires an external energy source and is known as laser pumping. Pumping may be achieved with electrical currents (e.g. semiconductors, or gases via high-voltage discharges) or with light, generated by discharge lamps or by other lasers (semiconductor lasers). More exotic gain media can be pumped by chemical reactions, nuclear fission, or with high-energy electron beams. [7] Example of a model of gain medium A universal model valid for all laser types does not exist. [8] The simplest model includes two systems of sub-levels: upper and lower. Within each level, the fast transitions lead to the Boltzmann distribution of excitations among sub-levels (fig.1). The upper level is assumed to be metastable.Also, gain and refractive index are assumed independent of a particular way of excitation. For good performance of the gain medium, the separation between sub-levels should be larger than working temperature; then, at pump frequency $~\omega_{\rm p}~$, the absorption dominates. In the case of amplification of optical signals, the lasing frequency is called signal frequency. However, the same term is used even in the laser oscillators, when amplified radiation is used to transfer energy rather than information. The model below seems to work well for most optically-pumped solid-state lasers. Cross-sections Let $~N~$ be concentration of active centers in the solid-state lasers. Let $~N_1~$ be concentration of active centers in the ground state. Let $~N_2~$ be concentration of excited centers. Let $~N_1+N_2=N~$. The relative concentrations can be defined as $~n_1=N_1/N~$ and $~n_2=N_2/N~$. The rate of transitions of an active center from ground state to the excited state can be expressed with $~ W_{\rm u}= \frac{I_{\rm p}\sigma_{\rm ap}}{ \hbar \omega_{\rm p} }+\frac{I_{\rm s}\sigma_{\rm as}}{ \hbar \omega_{\rm s} } ~$ and The rate of transitions back to the ground state can be expressed with $~W_{\rm d}=\frac{ I_{\rm p} \sigma_{\rm as}}{ \hbar \omega_{\rm p} }+\frac{I_{\rm s}\sigma_{\rm es}}{ \hbar \omega_{\rm s} } +\frac{1}{\tau}~$, where $~\sigma_{\rm as} ~$ and $~\sigma_{\rm ap} ~$ are effective cross-sections of absorption at the frequencies of the pump and the signal. $~\sigma_{\rm es} ~$ and $~\sigma_{\rm ep} ~$ are the same for stimulated emission; $~\frac{1}{\tau}~$ is rate of the spontaneous decay of the upper level. Then, the kinetic equation for relative populations can be written as follows\[~ \frac {{\rm d}n_2} {{\rm d}t} = W_{\rm u} n_1 - W_{\rm d} n_2 ~\], $~ \frac{{\rm d}n_1}{{\rm d}t}=-W_{\rm u} n_1 + W_{\rm d} n_2 ~$ However, these equations keep $~ n_1+n_2=1 ~$. The absorption $~ A ~$ at the pump frequency and the gain $~ G ~$ at the signal frequency can be written as follows\[~ A = N_1\sigma_{\rm pa} -N_2\sigma_{\rm pe} ~\], $~ G = N_2\sigma_{\rm se} -N_1\sigma_{\rm se} ~$. Steady-state solution The steady-state solution can be written\[~ n_2=\frac{W_{\rm u}}{W_{\rm u}+W_{\rm d}} ~\] , $~ n_1=\frac{W_{\rm d}}{W_{\rm u}+W_{\rm d}}.$ The dynamic saturation intensities can be defined\[~ I_{\rm po}=\frac{\hbar \omega_{\rm p}}{(\sigma_{\rm ap}+\sigma_{\rm ep})\tau} ~\], $~ I_{\rm so}=\frac{\hbar \omega_{\rm s}}{(\sigma_{\rm as}+\sigma_{\rm es})\tau} ~$. The absorption at strong signal\[~ A_0=\frac{ND}{\sigma_{\rm as}+\sigma_{\rm es}}~\]. The gain at strong pump\[~ G_0=\frac{ND}{\sigma_{\rm ap}+\sigma_{\rm ep}}~\], where $~ D= \sigma_{\rm pa} \sigma_{\rm se} - \sigma_{\rm pe} \sigma_{\rm sa} ~$ is determinant of cross-section. Gain never exceeds value $~G_0~$, and absorption never exceeds value $~A_0 U~$. At given intensities $~I_{\rm p}~$, $~I_{\rm s}~$ of pump and signal, the gain and absorption can be expressed as follows\[~A=A_0\frac{U+s}{1+p+s}~\], $~G=G_0\frac{p-V}{1+p+s}~$, where $~p=I_{\rm p}/I_{\rm po}~$ , $~s=I_{\rm s}/I_{\rm so}~$ , $~U=\frac{(\sigma_{\rm as}+\sigma_{\rm es})\sigma_{\rm ap}}{D}~$ , $~V=\frac{(\sigma_{\rm ap}+\sigma_{\rm ep})\sigma_{\rm as}}{D}~$ . Identities The following identities [9] take place\[U-V=1 ~ \] , $~ A/A_0 +G/G_0=1~.\ $ The state of gain medium can be characterized with a single parameter, such as population of the upper level, gain or absorption. Efficiency of the gain medium The efficiency of a gain medium can be defined as$~ E =\frac{I_{\rm s} G}{I_{\rm p}A}~$. Within the same model, the efficiency can be expressed as follows\[~E =\frac{\omega_{\rm s}}{\omega_{\rm p}} \frac{1-V/p}{1+U/s}~\]. For the efficient operation both intensities, pump and signal should exceed their saturation intensities; $~\frac{p}{V}\gg 1~$, and $~\frac{s}{U}\gg 1~$. The estimates above are valid for a medium uniformly filled with pump and signal light. The spatial hole burning may slightly reduce the efficiency because some regions are pumped well, but the pump is not efficiently withdrawn by the signal in the nodes of the interference of counter-propagating waves. Keywords The articles from the list below are not yet loaded. The list is here for technical reasons. While, use the wikipedia source http://en.wikipedia.org/wiki/Active_medium . References and notes Hecht, Jeff. The Laser Guidebook: Second Edition.McGraw-Hill, 1992. (Chapter 22) Hecht, Chapter 22 Hecht, Chapters 7-15 Hecht, Chapters 18-21 F. J. Duarte and L. W. Hillman (Eds.), Dye Laser Principles(Academic, New York, 1990). F. P. Schäfer (Ed.), Dye Lasers, 2nd Edition (Springer-Verlag, Berlin, 1990). Encyclopedia of laser physics and technology http://www.uscibooks.com/siegman.htm A.E.Siegman. Lasers. University Science Books (1986) sbn0-935702-11-3 http://josab.osa.org/abstract.cfm?id=84730D.Kouznetsov,J.F.Bisson, K.Takaichi, K.Ueda.Single-mode solid-state laser with short wide unstable cavityJOSAB 22p.1605–1619 (2005) http://en.wikipedia.org/wiki/Active_medium (source of this article) http://www.rp-photonics.com/gain_media.html Gain media, Encyclopedia of Laser Physics and Technology
Calculators¶ ClusterExpansionCalculator¶ class mchammer.calculators. ClusterExpansionCalculator( structure, cluster_expansion, name='Cluster Expansion Calculator', scaling=None, use_local_energy_calculator=True)¶ A ClusterExpansionCalculator object enables the efficient calculation of properties described by a cluster expansion. It is specific for a particular (supercell) structure and commonly employed when setting up a Monte Carlo simulation, see Ensembles. Cluster expansions, e.g., of the energy, typically yield property values per site. When running a Monte Carlo simulation one, however, considers changes in the totalenergy of the system. The default behavior is therefore to multiply the output of the cluster expansion by the number of sites. This behavior can be changed via the scalingkeyword parameter. Parameters: structure( ase.Atoms) – structure for which to set up the calculator cluster_expansion( ClusterExpansion) – cluster expansion from which to build calculator name( str) – human-readable identifier for this calculator scaling( Union[ int, float, None]) – scaling factor applied to the property value predicted by the cluster expansion use_local_energy_calculator( bool) – evaluate energy changes using only the local environment; this method is generally muchfaster; unless you know what you are doing do notset this option to False calculate_local_contribution( , local_indices, occupations)¶ Calculates and returns the sum of the contributions to the property due to the sites specified in local_indices Parameters: local_indices( List[ int]) – sites over which to sum up the local contribution occupations( List[ int]) – entire occupation vector Return type: float calculate_total( , occupations)¶ Calculates and returns the total property value of the current configuration. Parameters: occupations( List[ int]) – the entire occupation vector (i.e. list of atomic species) Return type: float cluster_expansion¶ cluster expansion from which calculator was constructed Return type: ClusterExpansion structure¶ atomic structure associated with calculator Return type: Atoms sublattices¶ Sublattices of the calculators structure. Return type: Sublattices update_occupations( indices, species)¶ Updates the occupation (species) of the associated atomic structure. Parameters: indices( List[ int]) – sites to update species( List[ int]) – new occupations (species) by atomic number TargetVectorCalculator¶ class mchammer.calculators.target_vector_calculator. TargetVectorCalculator( structure, cluster_space, target_vector, weights=None, optimality_weight=1.0, optimality_tol=1e-05, name='Target vector calculator')¶ A TargetVectorCalculatorenables evaluation of the similarity between a structure and a target cluster vector. Such a comparison can be carried out in many ways, and this implementation follows the measure proposed by van de Walle et al.in Calphad 42, 13 (2013) [WalTiwJon13]. Specifically, the objective function $Q$ is calculated as\[Q = - \omega L + \sum_{\alpha} \left\| \Gamma_{\alpha} - \Gamma^{\text{target}}_{\alpha} \right\|.\] Here, $\Gamma_{\alpha}$ are components in the cluster vector and $\Gamma^\text{target}_{\alpha}$ the corresponding target values. The factor $\omega$ is the radius of the largest pair cluster such that all clusters with the same or smaller radii have $\Gamma_{\alpha} - \Gamma^\text{target}_{\alpha} = 0$. Parameters: structure( Atoms) – structure for which to set up calculator cluster_space( ClusterSpace) – cluster space from which to build calculator target_vector( List[ float]) – vector to which any vector will be compared weights( Optional[ List[ float]]) – weighting of each component in cluster vector comparison, by default 1.0 for all components optimality_weight( float) – factor $L$, a high value of which effectively favors a complete series of optimal cluster correlations for the smallest pairs (see above) optimality_tol( float) – tolerance for determining whether a perfect match has been achieved (used in conjunction with $L$) name( str) – human-readable identifier for this calculator calculate_local_contribution( local_indices, occupations)¶ Not yet implemented, forwards calculation to calculate_total. Return type: float calculate_total( occupations)¶ Calculates and returns the similarity value $Q$ of the current configuration. Parameters: occupations( List[ int]) – the entire occupation vector (i.e. list of atomic species) Return type: float structure¶ atomic structure associated with calculator Return type: Atoms sublattices¶ Sublattices of the calculators structure. Return type: Sublattices update_occupations( indices, species)¶ Updates the occupation (species) of the associated atomic structure. Parameters: indices( List[ int]) – sites to update species( List[ int]) – new occupations (species) by atomic number mchammer.calculators.target_vector_calculator. compare_cluster_vectors( cv_1, cv_2, orbit_data, weights=None, optimality_weight=1.0, tol=1e-05)¶ Calculate a quantity that measures similarity between two cluster vecors. Parameters: cv_1( ndarray) – cluster vector 1 cv_2( ndarray) – cluster vector 2 orbit_data( OrderedDict) – orbit data as obtained by ClusterSpace.orbit_data weights( Optional[ List[ float]]) – Weight assigned to each cluster vector element optimality_weight( float) – quantity $L$ in [WalTiwJon13] (see mchammer.calculators.TargetVectorCalculator) tol( float) – numerical tolerance for determining whether two elements are exactly equal Return type: float
Current browse context: astro-ph.GA Change to browse by: Bookmark(what is this?) Astrophysics > Astrophysics of Galaxies Title: SEAGLE--II: Constraints on feedback models in galaxy formation from massive early type strong lens galaxies (Submitted on 4 Jan 2019) Abstract: We use ten different galaxy formation scenarios from the EAGLE suite of {\Lambda}CDM hydrodynamical simulations to assess the impact of feedback mechanisms in galaxy formation and compare these to observed strong gravitational lenses. To compare observations with simulations, we create strong lenses with $M_\star$ > $10^{11}$ $M_\odot$ with the appropriate resolution and noise level, and model them with an elliptical power-law mass model to constrain their total mass density slope. We also obtain the mass-size relation of the simulated lens-galaxy sample. We find significant variation in the total mass density slope at the Einstein radius and in the projected stellar mass-size relation, mainly due to different implementations of stellar and AGN feedback. We find that for lens selected galaxies, models with either too weak or too strong stellar and/or AGN feedback fail to explain the distribution of observed mass-density slopes, with the counter-intuitive trend that increasing the feedback steepens the mass density slope around the Einstein radius ($\approx$ 3-10 kpc). Models in which stellar feedback becomes inefficient at high gas densities, or weaker AGN feedback with a higher duty cycle, produce strong lenses with total mass density slopes close to isothermal (i.e. -d log({\rho})/d log(r) $\approx$ 2.0) and slope distributions statistically agreeing with observed strong lens galaxies in SLACS and BELLS. Agreement is only slightly worse with the more heterogeneous SL2S lens galaxy sample. Observations of strong-lens selected galaxies thus appear to favor models with relatively weak feedback in massive galaxies. Submission historyFrom: Sampath Mukherjee [view email] [v1]Fri, 4 Jan 2019 13:57:14 GMT (5688kb,D)
The database contains data for Maass forms on $\Gamma_0(N)$ with trivial character and Laplace eigenvalue $\lambda = \frac14 + R^2$ in three ranges: $1\le N \le 10$ and $0<R\le 10$, prime $100 < N < 250$ and $0<R\le 2$, and prime $250 < N < 1000$ and $0<R\le 1$, as well as many additional examples for small $N$. It is believed that each database entry corresponds to an actual Maass form and the given decimal numbers are reasonable approximations to the true value. However, some eigenvalues may be missing. Using current methods it is not feasible to prove that the data for an individual Maass form is correct, nor that the list of Maass eigenvalues in a given range is compete. You can read about the reasons for this difficulty.
Is it (always) true that $$\mathrm{Var}\left(\sum\limits_{i=1}^m{X_i}\right) = \sum\limits_{i=1}^m{\mathrm{Var}(X_i)} \>?$$ The answer to your question is "Sometimes, but not in general". To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then, $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) - \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2$$ Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore, $$ E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) = E \left( \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$ similarly, $$ \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \sum_{i=1}^{n} E(X_i) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$ so $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$ by the definition of covariance. Now regarding Does the variance of a sum equal the sum of the variances?: If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$ If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails. but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & 0.4 &-0.6 \\ 0.4 & 1 & 0.2 \\ -0.6 & 0.2 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$ Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general. $$\text{Var}\bigg(\sum_{i=1}^m X_i\bigg) = \sum_{i=1}^m \text{Var}(X_i) + 2\sum_{i\lt j} \text{Cov}(X_i,X_j).$$ So, if the covariances average to $0$, which would be a consequence if the variables are pairwise uncorrelated or if they are independent, then the variance of the sum is the sum of the variances. An example where this is not true: Let $\text{Var}(X_1)=1$. Let $X_2 = X_1$. Then $\text{Var}(X_1 + X_2) = \text{Var}(2X_1)=4$. I just wanted to add a more succinct version of the proof given by Macro, so it's easier to see what's going on. $\newcommand{\Cov}{\text{Cov}}\newcommand{\Var}{\text{Var}}$ Notice that since $\Var(X) = \Cov(X,X)$ For any two random variables $X,Y$ we have: \begin{align} \Var(X+Y) &= \Cov(X+Y,X+Y) \\ &= E((X+Y)^2)-E(X+Y)E(X+Y) \\ &\text{by expanding,} \\ &= E(X^2) - (E(X))^2 + E(Y^2) - (E(Y))^2 + 2(E(XY) - E(X)E(Y)) \\ &= \Var(X) + \Var(Y) + 2(E(XY)) - E(X)E(Y)) \\ \end{align} Therefore in general, the variance of the sum of two random variables is not the sum of the variances. However, if $X,Y$ are independent, then $E(XY) = E(X)E(Y)$, and we have $\Var(X+Y) = \Var(X) + \Var(Y)$. Notice that we can produce the result for the sum of $n$ random variables by a simple induction. Yes, if each pair of the $X_i$'s are uncorrelated, this is true. See the explanation on Wikipedia
There are a wide variety of notions of dimension of a measure. Your basic intuition is completely correct: for a dynamical system, the dimension of a natural invariant measure provides more relevant information than the dimension of the invariant set, since the system may spend more time in certain parts of the space. For sufficiently homogeneous measures, all reasonable notions of dimension will agree. By ``sufficiently homogeneous'' I mean something very precise: that$$C^{-1} \, r^s \le \mu(B(x,r)) \le C\, r^s$$for some constant $C\ge 1$, some $s\ge 0$ and all points $x$ in the support of $\mu$. Of course the dimension in this case is $s$. Such measures are often called Ahlfors-regular, and an example is the natural measure on the middle-thirds Cantor set. For more general measures, the local dimension is one of the most important concepts and has already been mentioned:$$\dim(\mu,x)=\lim_{r\to 0}\frac{\log \mu(B(x,r))}{\log r}.$$But this is really a function of the point $x$ (and not even, as the limit in the definition may not exist, although one can always speak of upper and lower local dimensions). There are several ways to globalize the information given by the local dimensions. Perhaps the easiest is to take the essential supremum/infimum of the upper/lower local dimensions. This results in four global concepts of dimensions, known as upper/lower packing/Hausdorff dimensions of the measure. They turn out (somewhat surprisingly) to be closely connected to the dimensions of the sets the measure ``sees''. For example, the upper Hausdorff dimension of a probability measure $\mu$ (that is, the essential supremum of the lower local dimensions), is the same as the infimum of the Hausdorff dimension of $A$ over all Borel sets $A$ of full measure. A finer study is provided by the multifractal spectrum of a measure $\mu$: for each $\alpha$, we form the level set $E_\alpha$ of all points $x$ where $\dim(\mu,x)=\alpha$. Then we try to understand how the size of $E_\alpha$ depends on $\alpha$, for example by studying the function $\alpha\to \dim_H(E_\alpha)$. There are (many!) other useful concepts of dimension which are not directly related to local dimension. In computing lower bounds for the Hausdorff dimension, the potential method is widely applicable: if a measure $\mu$ satisfies that the energy integral$$I_s(\mu) = \int \frac{d\mu(x)\, d\mu(y)}{\|x-y\|^s}$$is finite, then the support of $\mu$ has Hausdorff dimension at least $s$. So it makes sense to think of $\sup\{s: I_s(\mu)<\infty\}$ as a notion of dimension of $\mu$. This is often called the (lower) correlation dimension, and is one instance of a more general family of dimensions indexed by a real number $q$ (correlation dimension corresponds to $q=2$, and has several alternative definitions, perhaps pointing to its importance). Yet another notion of dimension has a dynamical underpinning. Given a probability measure $\mu$ say on the unit cube $[0,1]^d$, we may consider the entropy $H_k(\mu)$ of $\mu$ with respect to the partition into dyadic cubes of side length $2^{-k}$. We then define the entropy (also called information) dimension of $\mu$ as$$\lim_{k\to\infty} \frac{H_k(\mu)}{k\log 2}.$$ This is just a sample of the diverse zoo of dimensions of a measure. Which ones to use depends on the context and what you are able to compute/prove. Coming back to invariant measures, it is very often the case that the local dimension exists and takes a constant value at almost every point. Such measures are called exact dimensional, and have the property that lower and upper Hausdorff dimension, as well as entropy dimension, are all equal to this almost sure value. (But correlation dimension may be strictly smaller, and the multifractal spectrum may still be very rich; in other words, even though attained on a set of measure zero, other local dimensions may still be relevant). Proving that measures invariant under certain class of dynamics are exact dimensional may be very challenging. Eckmann and Ruelle conjectured in 1984 that hyperbolic measures ergodic a $C^{1+\delta}$ diffeomorphism are exact dimensional. This was proved by Barreira, Pesin and Schmeling in 1999; the paper appeared in Annals. For invariant measures, there is often a strong connection between their dimension and other dynamical characteristics (at least generically). The conformal expanding case is the easiest: in this case one has the well-known formula ``dimension=entropy/Lyapunov exponent". The nonconformal situation is much harder, but still a lot of deep research has been done, for example Ledrappier-Young theory.
Given a positive definite quadratic $m\times m$ positive definite even quadratic form $Q$, and given a degree $g$, define $$\vartheta_Q^{(g)}(\Omega) = \sum_{N\in\Bbb Z^{m \times g}} \exp(\pi i N' Q N \Omega ) $$ If $m$ is a multiple of 4, this is a Siegel modular form of degree $g$ of weight $m/2$ with respect to the subgroup $\Gamma_0(\ell)$ where $\ell$ is such that $\ell Q^{-1}$ is an even quadratic form. We say $P$ is pluriharmonic (spherical function) in the matrix variable $X=(x_{r,s})$ for $1\le r\le m$, $1\le s\le g$, if for each $1\le i,j\le n$, we have $\displaystyle{\sum_{t=1}^g\frac{\partial^2 P}{\partial_{i,t}\partial_{j,t}}=0}$ If $m$ is a multiple of 4, this is a Siegel cusp form of weight $m/2+\nu$ in degree $g$ with respect to the subgroup $\Gamma_0(\ell)$. When $m$ is not a multiple of $4$, one gets a Siegel modular form with a character. Knowl status: Review status: beta Last edited by Jerry Shurman on 2016-03-28 12:39:28 History:(expand/hide all)
Does the solution of the Schrodinger equation always have to be normalizable? By normalizable I mean, given a wavefunction $\psi(x)$ $$\int_{-\infty}^{\infty}\|\psi(x)\|^2 dx<\infty \qquad \text{or}\qquad \int_{0}^{\infty}\|\psi(x)\|^2 dx <\infty$$ What would be the physical implications if one (or both) of those integrals diverges. From the viewpoint that the Copenhagen interpretation is one of the most popular and the wave function is interpretated as a probability distribution in this case; will the wavefunction diverging be valid for any other interpretation of quantum mechanics. Does anyone know any wavefunctions which are not normalizable? What if there was a singularity at 0 that made it diverge at all times. For instance $$\int_{0}^{\infty}\|\psi(x)\|^2 dx \to\infty\quad \text{but}\quad \int_{a}^{\infty}\|\psi(x)\|^2 dx <\infty $$ where $a>0.$ Would the second integral count as a valid pdf?
Can someone please help me understand how the expressions after the equals (marked in red) are arrived at? I don't quite understand the very last one where the c.d.f. $\phi$ comes into the picture. My guess for the first expression is that the term added to $c$ is always going to be negative and hence its added in order to get to the next expression involving a c.d.f. Can someone please help me understand how the expressions after the equals (marked in red) are arrived at? I don't quite understand the very last one where the c.d.f. $\phi$ comes into the picture. My guess for the first expression is that the term added to $c$ is always going to be They are calculating the power function $\beta(\theta)$ of the test. Note that $\beta(\theta)$, a function of $\theta$, is the probability of rejecting the null hypothesis $H_0$ under $\theta$. Formally, for testing $H_0:\theta\in\Theta_0$ vs $H_1:\theta\in\Theta_1$, where $\Theta_0\subset\Theta$ and $\Theta_1\subset \Theta-\Theta_0$, we define the power function of the test as $$\beta(\theta)=P_{\theta}(\text{reject }H_0)\quad\forall\,\theta\in\Theta$$ Distribution of the sample mean $\overline X_n$ is given by $$\overline X_n\sim N(\theta,\sigma^2/n)$$ So they standardize $\overline X_n$ after the second equal sign: \begin{align} \beta(\theta)&=P_{\theta}\left(\overline X_n>c\sigma/\sqrt{n}+\theta_0\right) \\&=P_{\theta}\left(\frac{\overline X_n-\theta}{\sigma/\sqrt{n}}>c+\frac{\theta_0-\theta}{\sigma/\sqrt{n}}\right) \end{align} Hence the cdf $\Phi$ comes into play after a missing fourth equal sign: $$\beta(\theta)=1-\Phi\left(c+\frac{\theta_0-\theta}{\sigma/\sqrt{n}}\right)$$ For the first $=$, in the numerator of the term on the LHS of the equality, add and subtract $\theta$, rearrange terms to get $\bar{X}_n-\theta-(\theta_0-\theta)$ on the numerator and then move the second part to the RHS. For the second $=$, the LHS of the inequality inside the $P()$ is just a standardized $\bar{X}_n$, so the probability it exceeds the RHS is the probability a standard normal ($Z$) exceeds the RHS. Note that there should be another $=$ sign just before the $1-\Phi()$ term at the end.
I'll follow your lead and represent adipic acid as $\ce{H2A}$ for the sake of simplicity. Equations we know: (I) o.100 = $\ce{[H2A] + [HA^{-}] + [A^{2-}]}$ (II) $\dfrac{\ce{[H^+] [HA^{-}]}}{\ce{[H2A]}} = 3.9 * 10^{−5}$ (III) $\dfrac{\ce{[H^+] [A^{-2}]}}{\ce{[HA^{-}]}} = 3.9 * 10^{−6}$ (IV) $\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]}$ There are four equations in four unknowns so the system does have a solution. I could find any way to simply all the equations into anything which seems reasonable for an exact solution. The way to solve this problem is to solve for the first ionization and assume that the second ionization has an insignificant contribution. A check can be done with equation (IV) to verify that $\ce{[HA^{-}] \gg 2[A^{2-}]}$. I am going to show an excessive number of significant figures in the answers so that various methods of calculation can be compared. The acid constants are only given to 2 significant figures so that is the limiting number of significant figures that can justified. Quick and dirty Let $\ce{[H^+] = [HA^{-}]}$ and let $\ce{[HA]} = 0.100$ then from (II) $\ce{[H^{+}]^2 = [HA^{-}]^2 =} \sqrt{(0.100)(3.9E-5)}$ $\ce{[H^{+}] = [HA^{-}] =} 1.97E-3$ thus $[\ce{HA}] \approx 0.1 - 0.00197 = 0.09803$ from equation (III) (V) $\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$ So indeed $\ce{[HA^{-}] \gg [A^{-2}]}$ This method can be made somewhat better by noting x*(0.1/0.10197) HA 0.10000 0.09807 HA- 0.00197 0.00193 ------- ------- 0.10197 0.10000 Cleaner solution The Quick and dirty is a bit dodgy since $\ce{[H2A]}$ was assigned a value of 0.100 to calculate, then reassigned a value of 0.09803. Since the pK constant is only given to two significant figures this isn't really wrong, but a more consistent approach can be easily calculated. Let $\ce{[H^+] = [HA^{-}]} = x$ and let $\ce{[HA]} = 0.100 - x$, and also let's use $k_2$. Substituting into equation (III): $x^2 = (0.100 -x)k_2$ $x^2 + k_2x -0.1k_2$ This is a quadratic in x so roots are: $x = \dfrac{-k_2 \pm \sqrt{k_2^2 - 4(1)(-0.1k_2)}}{2(1)}$ The negative root doesn't work so the positive one is the only one. Also $k_2^2$ is so small that it can be neglected so; $x = \dfrac{-k_2 + \sqrt{0.4k_2}}{2} =0.00186$ $\ce{[H^+] = [HA^{-}]} = 0.00186$ $\ce{[HA]} = 0.100 - 0.00186 = 0.09814 $ Again from equation (V) $\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$ Exact Solution Solve equation (II) for $\ce{[HA^{-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get: $\ce{[HA^{-}]} = \dfrac{k_1\ce{[H2A]}}{\ce{[H^+]}}$ and multiply (II) and (III) to solve for $\ce{[HA^{2-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get: $\ce{[A^{2-}]} = \dfrac{k_1k_2\ce{[H2A]}}{\ce{[H^+]^2}}$ Both of the above can be substituted into the left hand side of the following three equations to get the values shown on the right. $\dfrac{\ce{[H2A]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}}$ $\dfrac{\ce{[HA^{-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k1[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $ $\dfrac{\ce{[A^{2-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k_1k_2[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $ Being a glutton for punishment I created a spreadsheet with the calculations. After some manual manipulation I was able to determine the "exact solution" to a totally ridiculous number of decimal places. $\ce{[\ce{H+}]} \text{ }= 0.00195932639309$ $\ce{[\ce{H2A}]}~~= 0.09804455814272$ $\ce{[\ce{HA-}]} ~= 0.00195155732146$ $\ce{[\ce{A^{2-}}]}~~= 0.00000388453582$ $\text{sum} \quad = 0.10000000000000$ $\ce{2[\ce{A^{2-}}] = [\ce{H+}] - [\ce{HA-}]} = 0.00000776907164$ $\ce{K_{a1}}$ from data = $3.9E-5$ $\ce{K_{a2}}$ from data = $3.9E-5$ $\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]} = 0.00195932639309$
This is a bit of an X-Y problem. You cannot accurately measure the resistances involved in the voltage dividers when the circuit is not powered, because the constituent resistors wind up in parallel. For instance, when you attempt to measure the resistance across the LDR, you wind up measuring the resistance of this circuit: simulate this circuit – Schematic created using CircuitLab Which, if you do the math gives you: \$R_{measured} = \frac{1}{ \frac{1}{R_{LDR}} +\frac{1}{ R_{1} + R_{2}} }\$ \$R_{measured} = \frac{1}{ \frac{1}{82k} +\frac{1}{ 51k + 10k} } = 34.97k\Omega\$ That said, this does not mean that the two dividers will interfere with each other in circuit. In that case, as long as the power supply is stable and has low impedance, the voltage across the whole divider can be considered fixed. The typical way that surrounding circuitry interferes with the output of a voltage divider is when enough current is drawn out of the voltage divider to create a meaningful error in the divider. simulate this circuit Ideally, the load current on the output of the voltage divider would be zero, and if the voltage divider were connected to the input of an op amp, that would be close enough to true. If, however, we draw a substantial amount of current \$I\$ out of the divider, then we get an error: If \$I\$ is zero: \$V_{out} = \frac{V}{R_{1}+R_{2}}\times R_{2}\$ If \$I\$ is non-zero: \$V_{out} = (\frac{V}{R_{1}+R_{2}}-I)\times R_{2}\$ (note that the sign of \$I\$ depends on the direction of current flow, so the error could increase or decrease our actual reading) This shows that the amount of current that flows out of the voltage divider \$(I)\$ needs to be substantially smaller than the current that flows through the divider \$ (\frac{V}{R_{1}+R_{2}})\$ to minimize this error. Now, you haven't mentioned whether your ADC readings are coming out as you expect, but let's discuss the main potential source of error there anyway. To start with, lets look at what a typical ADC channel looks like, electrically. In this case, it's going to have a sampling capacitor that is temporarily connected to the input for some period of time before the ADC performs the conversion. Something like this: simulate this circuit What this means is that during the sampling time, the input mux will have connected the appropriate input channel to the sampling capacitor, and the capacitor will charge (or discharge) towards the input voltage at a rate determined by the capacitance and the impedance of whatever's connected to the selected ADC input (in this case, one of your voltage dividers). In order to get an accurate reading, your sampling time must be some multiple of the time constant of the sampling capacitor and your total input impedance (check the datasheet for details). If your sampling time is inadequate, your ADC readings will be off, because the sampling capacitor does not have adequate time to charge or discharge to the correct voltage. For this reason, the sampling time on MCU ADCs is typically configurable, as the required sampling time will depend on the nature of the circuitry you're attempting to measure. If you have some really high impedance circuitry, you would probably want to use an op-amp buffer as another answer suggests. However, in your case, this is entirely unnecessary as long as your sampling time is set correctly.
I am surprisingly having a bit of difficulty with an indefinite integral which is interesting since the integral I solved before is $$ \int \frac{1+\cos2x}{\sin^2(2x)} dx$$ The integral I am currently working on is $$ \int \frac{\sin^2(2x)}{1+\cos2x} dx$$ I first divided out giving: ( this is the mistake I made early on, in this case you use the pythagorean identity for sin, and then cancel out $1+\cos2x$) $$ \int \sin^2(2x)+ \sin^2(2x)\cdot \sec(2x) \,dx$$ Then I factored out the $\sin^2(2x)$ resulting in: $$ \int \sin^2(2x)(1+ \sec(2x)) \,dx$$ Substituting for $u=2x$ and splitting the integral into two parts: $$\frac{1}{2} \int \sin^2(u) du + \frac{1}{2} \int \sin^2(u)\sec(u) \,du$$ lets call this eq. 1. Now, this is where I am having difficulty as 1.) dealing with even powers of sin and 2.) the $\sec(u)$ term is proving to be troublesome. Another form of the above equation is: $$ \frac{1}{2} \int \sin^2(u) \,du + \frac{1}{2} \int \sin(u)\tan(u) \,du$$ Some approaches I have tried are using different trigonometric identities e.g. $$\sin^2(u) = \frac{1}{2} (1-\cos(2u))$$ however, this results in for eq. 1 $$ \frac{1}{4} \int 1-\cos(2u) \,du + \frac{1}{4} \int \frac{(1-\cos(2u)}{\cos(u)} du $$ Then I would have to use the cosine angle addition formula which quickly gets out of hand. I understand there are different approaches to solving different indefinite integrals. The purpose of this problem is to only use substitutions. Questions that I have are as follows, 1.) is it possible to continue along with the steps I have taken?, 2.) or must I do an entirely different substitution at the beginning. Sorry for the long post and thank you for your time.
The Ornstein–Uhlenbeck process is defined as the stochastic process that solves the following SDE: $dx_t = \theta (\mu-x_t)\,dt + \sigma\, dW_t$ where $\theta>0$, $\mu$ and $\sigma>0$ are parameters and $W_t$ is Brownian motion. It is well known the solution to this equation. In particular, it is known that $E(x_t)=x_0 e^{-\theta t}+\mu(1-e^{-\theta t})$ and $\operatorname{cov}(x_s,x_t) = \frac{\sigma^2}{2\theta}\left( e^{-\theta(t-s)} - e^{-\theta(t+s)} \right).$ It can be easily seen that $\lim_{t\to+\infty}E(x_t)=\mu$ and that $\lim_{t\to+\infty}Var(x_t)=\frac{\sigma^2}{2\theta}$. Assume that $f(t)$ is a well behaved function. What is it known about the process $dx_t = \theta (f(t)-x_t)\,dt + \sigma\, dW_t$? In particular, assume that $f(t)$ is periodic with certain period $\tau$. What is the limit of $E(x_t)$?
Center Manifold Jack Carr (2006), Scholarpedia, 1(12):1826. doi:10.4249/scholarpedia.1826 revision #126955 [link to/cite this article] One of the main methods of simplifying dynamical systems is to reducethe dimension of the system. Centre manifold theory is a rigorous mathematical technique that makes this reduction possible, at least near equilibria. Contents An Example We first look at a simple example. Consider \[\tag{1} x' =ax^3 \,, \qquad y' =-y + y^2 \] where $a$ is a constant. Since the equations are uncoupled, we see that the stationary solution $ x=y =0$ of (1) is asymptotically stable if and only if $a < 0\ .$ Suppose now that \[\tag{2} x' =ax^3 + xy - xy^2\,, \qquad y' =-y + bx^2 +x^2 y \] Since the equations are coupled we cannot immediately decide if the stationary solution $ x=y =0$ of (2) is asymptotically stable. The key is an abstraction of the idea of uncoupled equations. A curve $y =h(x)\ ,$ defined for $\|x\|$ small, is said to be an invariant manifold for the system \[\tag{3}x' =f(x,y)\,, \qquad y' = g(x,y)\] if the solution of (3) with $x(0) =x_0\ ,$ $y(0) = h(x_0)$ lies on the curve $y =h(x)$ as long as $x(t)$ remains small. For the system (1), $y=0$ is an invariant manifold. Note that in deciding upon the stability of the stationary solution of (1), the only important equation is $x' = ax^3\ ,$ that is, we need only study a first order equation on a particular invariant manifold. Center manifold theory tells us that (2) has an invariant manifold $y =h(x) = \mbox{O}(x^2)$ for small $x\ .$ Furthermore, the local behaviour of solutions of the two dimensional system (2) can be determined by studying the scalar equation \[\tag{4} u' = au^3 + uh(u) -uh^2(u) \] The theory also tells us how to compute approximations to the invariant manifold $y = h(x)\ .$ For (2) we have that $h(x) = bx^2 + \mbox{O}(x^4)$ and using this information in (4) gives \[\tag{5} u' =(a+b)u^3 + \mbox{O}(u^5) \] Hence the stationary solution of (2) is asymptotically stable if $a+b < 0$ and unstable if $a+b>0\ .$ If $a+b = 0$ we need a better approximation to the invariant manifold in order to decide on the stability. Centre Manifolds Consider the system \[\tag{6} x' =Ax + f(x,y)\,, \qquad y' = By+g(x,y)\,, \qquad (x,y ) \in \R^n \times \R^m \] where all the eigenvalues of the matrix $A$ have zero real parts and all the eigenvalues of the matrix $B$ have negative real parts. The functions $f$ and $g$ are sufficiently smooth and \[ f(0,0) =0\,, \qquad Df(0,0) =0\,, \qquad g(0,0) =0\,, \qquad Dg(0,0) = 0 \] where $Df$ is the Jacobian matrix of $f\ .$ If $f$ and $g$ are identically zero then (6) has the two obvious invariant manifolds $x=0$ and $y=0\ .$ The invariant manifold $x=0$ is called the stable manifold, and on the stable manifold all solutions decay to zero exponentially fast. The invariant manifold $y=0$ is called the centre manifold. In general, an invariant manifold $y = h(x)$ for (6) defined for small $\|x\|$ with $h(0)=0$ and $Dh(0)=0$ is called a centre manifold. In more physical terms, the dynamics of y follows the dynamics of x and one may say that x enslaves the variable y. This interpretation has been called slaving principle. Main Results The general theory states that there exists a centre manifold $y =h(x)$ for (6) and that the equation on the centre manifold \[\tag{7} u' =Au + f(x,h(u))\,, \qquad u \in R^n \] determines the dynamics of (6) near $(x, y) =(0,0)\ .$ In particular, if the stationary solution $u=0$ of (7) is stable, we can represent small solutions of (6) as $t \rightarrow \infty$ by \[ x(t) =u(t) + \mbox{O}(e^{-\gamma t} )\,, \qquad y(t) =h(u(t)) + \mbox{O}(e^{-\gamma t}) \] where $\gamma > 0$ is a constant. To use the above theory, we need to have enough information about the centre manifold $y = h(x)$ in order to determine the local dynamics of (7). If we substitute $y(t) = h(x(t))$ into the second equation in (6) we obtain \[\tag{8} N(h(x)) =h'(x)\left[ Ax +f(x,h(x)) \right] - Bh(x) -g(x,h(x)) = 0 \] The general theory tells us that the solution $h$ of (8) can be approximated by a polynomial in $x\ ,$ that is, if $N(\phi(x)) = \mbox{O}(\|x\|^q)$ as $x \rightarrow 0$ then $h(x) =\phi (x) + \mbox{O}(\|x\|^q)\ .$ There is also an $m$ dimensional invariant manifold $W^s$ tangential to the y-axis called the stable manifold. On the stable manifold all solutions decay to zero exponentially fast. Figure 1 illustrates the local dynamics for equation (6). The details of the flow on the centre manifold $y = h(x)$ depend on the higher order terms in equation (7) and we cannot assign directions to the flow without further information. We have assumed that all of the eigenvalues of the matrix B in (6) have negative real parts. The theory can be extended to the case in which the matrix B has in addition some eigenvalues with positive real parts. In this case the stationary solution $x=0, y=0$ of (6) is unstable due to the unstable eigenvalues. There exists a centre manifold for (6) which captures the behaviour of small bounded solutions. In particular, this gives a method of studying all sufficiently small equilibria, periodic orbits and heteroclinic orbits. Local Bifurcations Centre manifold reduction is central to the development of bifurcation theory. We illustrate this by means of a simple example. Consider \[\tag{9} x' =\epsilon x -x^3 +xy\,, \qquad y' =-y + y^2 -x^2 \] where $\epsilon$ is a small scalar parameter. The goal is to study small solutions of (9). The linearised problem about the zero equilibrium has eigenvalues $-1$ and $\epsilon$ so the theory does not directly apply. We can write the equations in the equivalent form \[\tag{10} x' =\epsilon x -x^3 +xy\,, \qquad y' = -y + y^2 -x^2 \,, \qquad \epsilon' = 0 \ .\] When considered as an equation on $\R^3$ the $\epsilon x$ term in (10) is nonlinear and the system has an equilibrium at $(x,y,\epsilon) = (0,0,0)\ .$ The linearisation about this equilibrium has eigenvalues $-1, 0 ,0\ ,$ that is, it has two zero eigenvalues and one negative eigenvalue. . The theory now applies so that the extended system (10) has a two dimensional centre manifold $y =h(x,\epsilon)$ that can be approximated by a polynomial in $x$ and $\epsilon\ .$ The equation on the centre manifold is two dimensional and may be written in terms of the scalar variables $u$ and $\epsilon$ as \[ u' =\epsilon u - 2u^3 + \mbox{higher order terms} \,, \qquad \epsilon' = 0 \] and the local dynamics of (10) can be deduced from this equation. Notes and Further Reading The ideas for centre manifolds in finite dimensions have been around for a long time and have been developed by Carr (1981), Guckenheimer and Holmes (1983), Kelly (1967), Vanderbauwhede (1989) and others. For recent developments in the approximation of centre manifolds see Jolly and Rosa (2005). Pages 1-5 of the book by Li and Wiggins (1997) give an extensive list of the applications of centre manifold theory to infinite dimensional problems. Mielke (1996) has developed centre manifold theory for elliptic partial differential equations and has applied the theory to elasticity and hydrodynamical problems. Applications to phase transitions in biological, chemical and physical systems have been investigated by Haken (2004). In addition, it is interesting to note that there is a stochastic extension of the center manifold theorem, which has been introduced by Boxler (1989). In this case, for instance the center and stable manifolds may fluctuate randomly. References J. Carr (1981), Applications of Centre Manifold Theory, Springer-Verlag. J. Guckenheimer and P. Holmes (1983), Nonlinear Oscillations, Dynamical systems and Bifurcations of Vector Fields. Springer-Verlag. M. S. Jolly and R. Rosa (2005), Computation of non-smooth local centre manifolds, IMA Journal of Numerical Analysis , 25, no. 4, 698-725. A. Kelly (1967), The stable, center-stable, center, center-unstable and unstable manifolds. J. Diff. Eqns, 3, 546-570. Li and S. Wiggins (1997), Invariant manifolds and fibrations for perturbed nonlinear Schrödinger equations. Springer-Verlag. A. Mielke (1996), Dynamics of nonlinear waves in dissipative systems: reduction, bifurcation and stability. In Pitman Research Notes in Mathematics Series, 352. Longman. A. Vanderbauwhede (1989). Center Manifolds, Normal Forms and Elementary Bifurcations, In Dynamics Reported, Vol. 2. Wiley. H. Haken (2004), Synergetics: Introduction and Advanced topics, Springer Berlin P. Boxler (1989), A stochastic version of center manifold theory, Probability Theory and Related Fields, 83(4), 509-545 Internal references John W. Milnor (2006) Attractor. Scholarpedia, 1(11):1815. Jeff Moehlis, Kresimir Josic, Eric T. Shea-Brown (2006) Periodic orbit. Scholarpedia, 1(7):1358. Philip Holmes and Eric T. Shea-Brown (2006) Stability. Scholarpedia, 1(10):1838.
Do not passively use Yahoo where you need reliable historical data; it will just fail at one point (from what I have seen due to corporate actions/dividends not properly implemented). Paying for a single alternative data source will not save you either (Bloomberg sometimes reports crazy intraday prices); the only way is to write some data cleaning routines (... Let us denote $\delta$, the Libor's tenor (e.g. 3M), $P(t, T)$ the price of a zero coupon bond price paying 1 unit of currency at $T$, and $L_t(T, T + \delta)$ the forward 3M Libor starting at $T$ and ending at $T+\delta$, seen from $t$:$$L_0(T, T + \delta) = \frac{1}{\delta} \left(\frac{P(0, T)}{P(0, T + \delta)} - 1 \right)$$The vanilla case: payment ... As you pointed out there are many ways to adjust for the roll overs. Hence, I guess you would agree that there is no one-size-fits-all answer to this. It really depends on the usage of the data:First think about how the trades in your back test are structured. If they are longer-term trades and you hold over roll overs then think what you would do if you ... There is a major bug when you are getting information from exchanges outside USA. If you get the adjusted prices for BOVESPA (Brazilian Stock Exchange) for example, it will only consider the events that happened using the US Calendar and not the Brazilian calendar of working days, this leads to a lack of information on other exchanges.Be aware of this if ... Consider a date sequence\begin{align}0 \leq t_0 \leq T_s < T_e < T_p,\end{align}where $t_0$ is the valuation date, $T_s$ is the Libor start date, $T_e$ is the Libor end date, and $T_p$ is the payment date. Let $\Delta_s^e = T_e-T_s$.For $0\le t \le T_s$, define\begin{align*}L^e(t, T_s, T_e) = \frac{1}{\Delta_s^e}\bigg(\frac{P(t, T_s)}{P(t, T_e)... An oldie but goodie from 2000. Bob Fulks, Back-Adjusting Futures Contracts, http://www.nuclearphynance.com/User%20Files/7228/cntcontr.pdfThe article is a great summary of the most popular adjustment methods. I agree with and would like to stress the author's observation: "There is no "best" method in an absolute sense. All the methods have advantages and ... Errors in some data can cause the calculation to go awry. For EPD, I have reported that they believe the stock had a 2:1 split on August 21, 2014 and on August 22, 2014. Only one of these splits occurred, so all the split adjusted data is off by a factor of 2 before the split that did not happen. I reported this error in August, but in November I noticed ... Yahoo data is good enough, but it has its quirks. As people have mentioned, sometimes it does miss out on corporate actions.I remember a while back I was looking at price for Ford (F) around 1999 , and computing my own adjusted close using yahoo's methodology and noticed that yahoo was missing a dividend payment in 1999(which I verified from bloomberg). ... I'm going to guess that you might be getting the timing mismatched when computing value weights. (When I was a TA for a first year finance PhD class, I was surprised at how common this error was.)Let $s_{it}$ be the share price of firm $i$ at the end of month $t$.Let $n_{it}$ be the number of shares outstanding of firm $i$ at the end of month $t$.Let $r_{... This seems to be the established convention and I'm not aware of other approaches being commonly (or at all) used. It's specified for example in ISDA Definitions:Section 4.11. FRN Convention; Eurodollar Convention. “FRN Convention” or “Eurodollar Convention” means, in respect of either Payment Dates or Period End Dates ... that the Payment Dates or ... Actually, because IB hasTRADES - data adjusted for splits but not dividends, andADJUSTED_LAST - data adjusted for splits and dividendsI can check what I have saved from earlier that day or the previous day against their data, and compute an adjustment factor for only splits, only dividends, and both.Using that, I have an accurate daily adjustment ... The other two answers do a good job of explaining, within the context of mathematical financial models, why a convexity adjustment is necessary, but I think a more tangible perspective can also be useful.Consider two forward rate agreements (FRA) to receive fixed and pay floating, with the same fixing date $T_s$ and end date $T_e$. The first pays on the ... This article has some interesting descriptions about the different methods used and when you would them:https://www.quandl.com/collections/futures/continuousThought it might be somewhat relevant to your question.
Testing Time Dilation in Special Relativity with Spectroscopy of Fast Ions in Storage Rings: A New, Improved Limit x x The relativistic Doppler effect was already proposed as an experimental test of relativity by Einstein in 1907 [1]. Time dilation leads to the ether-independentrelativistic Doppler formula \nu_0=\nu_{\rm l}\gamma(1-\beta\cos\theta), where \nu_{\rm l} and \nu_0 denotethe frequencies in the laboratory reference frame of the observerand the particles' rest frame moving at velocity v=\beta c with respect to the observer, respectively; \theta is the angle of observation with respect to the particles' movement as measured in the lab frame, and \gamma=1/\sqrt{1-v^2/c^2}. We are using a modern version of the experiment by Ives and Stilwell [2], where two laser beams, parallel and anti-parallel with the atomic motion, excite the same transition of rest-frame frequency \nu_0. Within Special Relativity (SR), the lasers will have laboratory frequencies of \nu_p = \nu_0/\gamma (1-\beta) and \nu_a = \nu_0/\gamma (1+\beta), respectively, and the product of the two expressions yields \nu_0^2 = \nu_a \nu_p. It is common to parameterize possible deviations from SR using the Robertson-Mansouri-Sexl (RMS) test theory, where deviations in the time dilation sector are quantified by the test parameter \hat{\alpha}, and we get where c \vec{\beta}_{\rm lab} is the velocity of the lab against a preferred cosmic frame, which is generally taken to be the cosmic microwave background rest frame.The \beta^2 term used in our measurement allows to determine\hat\alpha absolutely without having to rely on the preciseknowledge of \beta_{\rm lab} (for \beta \gg \beta_{\rm lab}). Fig. 1 Schematic layout of the TSR. Li^+ ions circulate in the 55 m circumference ring. In the electron cooler, cold electrons are overlapped with the ions and provide cooling. The measurements at the two different velocities are carried out sequentially. In the experiment, the two lasers are coupled into the ring from the same side and are retro-reflected. In our experiment at the Max Planck Institute for Nuclear Physics, ^7Li^+ ions are accelerated by a tandem Van-de-Graaff accelerator and injected into the Test Storage Ring (TSR) shown in Fig.~1. The helium-like ^7$Li$^+ exhibits the strong 2s~^3S_{1}\rightarrow 2p~^3P_{2} transition at 548~nm in its metastable triplet spectrum. To extract time dilation from a measurement of the Doppler shifts at one ion velocity, the rest frame transition frequency needs to be known accurately. Since the best available measurement has an uncertainty of 400~kHz, which was the limiting factor in our previous time dilation measurement [3], The Doppler-shifted frequencies \nu^(1,2)_{\rm a}, \nu^(1,2)_{\rm p} measured at \beta_1 and \beta_2 can be combined(neglecting the sidereal term) to independent of the rest frame frequency. As \beta_2^2 - \beta_1^2 \approx 0.8 \beta_2^2, the sensitivity is not significantly diminished. The moving clocks are read using laser saturation spectroscopy. The laboratory frequencies \nu_{\rm p} and \nu_{\rm a} of the parallel and anti-parallel laser beams (with respect to the ion beam) must obey the above relation for resonance, which is indicated by a dip in the fluorescence spectrum. Through permanent cooling of the ions by a cold electron beam, the ion beam's width shrinks to \approx 250~ \mu m, the divergence to \approx 50~\mu rad, and the longitudinal momentum spread to \delta p/p=3.5\times 10^{-5}, leading to a Doppler width of the transition of about 2.5 GHz full-width half maximum. This broadening is overcome in saturation spectroscopy by selecting a narrow velocity class of the order of the natural linewidth; two lasers are overlapped parallel and anti-parallel with the ion beam, respectively, and excite the clock transition. The co-propagating laser (a Nd:YAG laser at 532 nm for \beta_1 and an argon-ion laser at 514 nm for \beta_2) is fixed in frequency by locking it to a well-known iodine (I_2) line, whereas the counter-propagating light is generated by a tuneable dye laser (at 565 nm and 585 nm for \beta_1 and \beta_2, respectively). The dye laser frequency is referenced to a second, I_2-stabilised dye laser by a tuneable frequency-offset lock. The iodine lines for the dye laser are calibrated using an optical frequency comb [4,5]. All laser frequencies are known absolutely to 70~kHz during the whole experiment. The mean velocity of the ion beam is adjusted for thefixed-frequency laser at \nu^(1,2)_{\rm p} to select ions in the centreof the velocity distribution.The dye laseris scanned around \nu^(1,2)_{\rm a} and the Lamb dip in the fluorescence is recorded with photomultipliers (PMT) from the side; itsfrequency is measured with respect to the I_2 clock in thelaboratory frame.The observed resonance widths are in accordance with the natural linewidth of the 2\,^3S \rightarrow 2\,^3P transition of 3.7~MHz, once the broadening mechanisms present in our experiment are accounted for. Taking all systematic errors into account, the transition frequencies \nu_{\rm a} and \nu_{\rm p} measured at \beta_1=0.030 and \beta_2=0.064 yield SR values for the rest frame frequency of respectively. From above follows a test parameter which is consistent with SR. In late 2007 this experiment came to an end with the final analysis and publication [6]. This measurement is the most stringent test of relativistic time dilation, and puts a more than one order of magnitude tighter limit on deviations from special relativity than any non-storage ring bases method. To make further progress at the TSR, the Lamb-dip frequency would have to be measured to significantly better than 100 kHz absolutely. At this level, distortions of the resonance due to laser forces on the ions and other effects start to become intractable. The obvious alternative is to use faster ion beams. A new experimental initiative has been started at the ESR storage ring at GSI in Darmstadt, Germany, where the Li^+ beam can be stored at much higher speeds. First data was taken at \beta = 0.338, where the co-propagating laser beam at 386 nm has essentially twice the frequency of the counter-propagating one at 780 nm. A first-order Doppler-free signal has been observed (Lambda-spectrosocopy rather than saturation), and the prospects for a further-improved test of time dilation appear to be very good. References: [1] A. Einstein, Ann. Phys. 328, 197(1907). [2] H. E. Ives and G. R. Stilwell. J. Opt. Soc. Am. 28, 215 (1938). [3] G. Saathoff, S. Karpuk, U. Eisenbarth, G. Huber, S. Krohn, R. M. Horta, S. Reinhardt, D. Schwalm, A. Wolf, and G. Gwinner. Phys. Rev. Lett. 91, 190403 (2003). [4] T. Udem, R. Holzwarth, and T. W. H\"ansch. Nature 416, 233 (2002). [5] S. Reinhardt, G. Saathoff, S. Karpuk, C. Novotny, G. Huber, M. Zimmermann, R. Holzwarth, T. Udem, T. H\"ansch, and G. Gwinner. Optics Communications 261, 282 (2006). [6] S. Reinhardt, G. Saathoff, H. Buhr, L. A. Carlson, A. Wolf, D. Schwalm, S. Karpuk, C. Novotny, G. Huber, M. Zimmermann, R. Holzwarth, Th. Udem, T. W. H\"ansch, and G. Gwinner. Nature Physics 3, 861 (2007).
Let $\Gamma\le SL(2,\mathbb{Z})$ be a congruence subgroup of level $N$. Let $R$ be a $\mathbb{Z}[1/N]$-algebra. Let $\mathcal{Y}(\Gamma)_R$ denote the moduli stack over $R$ of elliptic curves equipped with a $\Gamma$-structure. For an integer $k$, let $\mathcal{Y}(\Gamma,\Omega^k)_R$ denote the moduli stack of triples $(E/S,\alpha,\omega)$, where $f : E\rightarrow S$ is an elliptic curve over an $R$-scheme $S$, $\alpha$ a $\Gamma$-structure on $E/S$, and $\omega$ is a basis of the $\mathcal{O}_S$-module $(f_\Omega^1_{E/S})^{\otimes k}$. There is a natural map $p_k : \mathcal{Y}(\Gamma,\Omega^k)_R\rightarrow\mathcal{Y}(\Gamma)_R$ given by forgetting the differential $\omega$. My understanding is that a Katz modular form of weight $k$ is just a section of $p_k$ (perhaps up to 2-isomorphism?). Is this correct? (Here I want to consider weakly holomorphic modular forms - that is, modular forms "holomorphic on the upper half plane, meromorphic at the cusps") The set of such sections ("weakly holomorphic" Katz modular forms of weight $k$) form an $R$-module $KM^k(\Gamma,R)$. My question is, if we pick an embedding $R\hookrightarrow\mathbb{C}$, any section of $p_k$ can be pulled back to a section of $p_{k,\mathbb{C}} : \mathcal{Y}(\Gamma,\Omega^k)_\mathbb{C}\rightarrow\mathcal{Y}(\Gamma)_{\mathbb{C}}$, and over $\mathbb{C}$, I can more or less see that the $\mathbb{C}$-vector space $KM^k(\Gamma,\mathbb{C})$ is the "same" as the space of weakly holomorphic modular forms for $\Gamma$ defined classically. My question is - is $KM^k(\Gamma,R)\otimes_R\mathbb{C}\cong KM^k(\Gamma,\mathbb{C})$? If this is not always true, then under what circumstances will it be true? (perhaps some condition like the Tate curve over $\mathbb{Z}((q))\otimes_\mathbb{Z} R$ admitting "all $\Gamma$-structures"?) Also, if $\Gamma$ is torsion free, so that $\mathcal{Y}(\Gamma)_R$ is a scheme, then does $p_k$ (over $R$) always admit a section? Equivalently, if $f : E\rightarrow S$ is an elliptic curve equipped with a "fine level structure" (e.g. a $\Gamma(N)$-structure for $N\ge 3$, a $\Gamma_1(N)$-structure for $N\ge 4$,...etc), then is $f_\Omega^1_{E/S}\cong \mathcal{O}_S$? I apologize if this is obvious.
I need to optimize affine transformations for of a set of triangles using energy function based on the connectivity. The energy of an edge $e_j$ between triangles $T_a, T_b$ is given by $$ E_j = \left[(A_a v_{a,1} + t_a) - (A_b v_{1,b} + t_b) \right]+ \left[(A_a v_{a,2} + t_a) - (A_b v_{2,b} + t_b) \right] $$ where $(A_a, t_a)$ and $(A_b, t_b)$ are affine transformations on the triangles adjacent to the edge and $v_1$ and $v_2$ are the vertices of the edge. The solution $v_i \equiv 0$ is prevented by an additional deformation energy on the triangles, like the conformal energy. After the optimization, we want to have $$v_{a,1}=v_{b,1}, \quad v_{a,2}=v_{b,2} \quad \Rightarrow E_j = 0$$ My problem is, that this is a non-linear problem, which is hard for solvers. I am using Google Ceres with line-search, but it often fails to find a valid line-search stepsize when optimizing $$ \begin{pmatrix} a_1 & a_2 \\ a_3 & a_4 \end{pmatrix}, \begin{pmatrix}t_1 \\ t_2\end{pmatrix} $$ as a vector $(a_1, a_2, a_3, a_4, \dots, t_1, t_2, \dots)$. For some transformations, there are more favorable parameterizations, e.g. Rotation can be parameterizized as $$ \begin{pmatrix} \cos \alpha & -\sin\alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}, \begin{pmatrix}t_1 \\ t_2\end{pmatrix} $$ optimizing the vector $(\alpha, \dots, t_1, t_2, \dots)$. Shearing can be optimized using $$ \begin{pmatrix} 1 & s_1 \\ s_2 & 1 \end{pmatrix}, \begin{pmatrix}t_1 \\ t_2\end{pmatrix} $$ optimizing the vector $(s_1, s_2, \dots, t_1, t_2, \dots)$. But for general deformations, I do not see which parameterization of the transformation matrix would fit and optimizing the coefficients directly probably lets the solver fail (or not converge) because it yields no good gradient. So what is the best way to optimize affine transformations to find triangle deformations? The final optimization objective is: Given a set of Triangles $$T=\{t_1=[v_1, v_2, v_3], t_2=[v_4, v_5, v_6], ßdots\}$$ without shared vertices in the set and the information about the shared vertices of common edges as a separate set $$E=\{[a,b,c,d], \dots\|a\sim c, b\sim d\}$$ where $\overline{ab}$ and $\overline{cd}$ are the same edges, when all triangles are re-assembled to a mesh. Find affine transformations $A_i+t_i$ for all triangles, such that for two adjacent triangles $t_1=[a,b,x]$ and $t_2=[c,d,y]$ holds $$ A_1 a + t_1 = A_2 c + t_2 \quad\wedge\quad A_1 c + t_1 = A_2 d + t_2 $$ and for all transformations $A_i$: $$ \text{trace} \frac{(A_i^T+A_i)}{2 \det A_i} \to \min $$ The first condition ensures that common edges are merged and the second condition minimizes the conformal deformation, what e.g. prevents the mesh form collapsing to a single vertex. When I for example optimize the triangles of a cut cube, then the only transformations I need are rotation and translation to arrange the triangles in the plane. Optimzing $(\alpha, t_1, t_2, \dots)$ using line-search finds a good solution without problem.s But for example a sphere requires quite a bit of deformation to be embedded in the plane. And to optimize the deformation needed to assemble the triangles into a planar patch, the optimization of the raw values $(a_1, a_2, a_3, a_4, t_1, t_2)$ seems not to work very well, especially line-search with Google ceres fails to find a valid stepsize for which the wolfe condition holds. I suspect that in contrast to e.g. the gradient of the rotation, the gradient of this vector is not well-suited for finding the optimal transformation and it would need another local parameterization, which better suits the search space, just like the rotation/translation for rigid embeddings.
September 25th, 2017, 12:22 AM # 1 Newbie Joined: Jan 2017 From: India Posts: 9 Thanks: 0 Math Focus: Calculus Double Integration-Finding Volume Hello, Pic:- So I was solving some questions of double integration.And I come across this que. which I didn't get. I mean I get that it is asked to find out Volume but I don't get the given data.I mean Limits explained in a tricky manner plus the answer given directly without any explaination. So can you please simplify the question for me and tell me how to solve it? Last edited by LeitHunt; September 25th, 2017 at 12:40 AM. September 25th, 2017, 10:05 AM # 2 Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The area over which you want to integrate is the region bounded by the two parabola y= x^2 and x= y^2. There are a number of different ways to do this. I might choose to integrate with respect to y first, then with respect to x. The result must, of course, be a number so the limits of the "outer integral" must be numbers, not functions of x and y. I can see that, in order to cover that region, x must go from 0, on the right, to 1 on the left. The inner integral, with respect to y, must go from the lower parabola, y= x^2, to the upper parabola, x= y^2 or y= sqrt(x). That is, for each x, y goes from x^2 to sqrt(x): $\displaystyle \int_0^1 \int_{x^2}^{\sqrt{x}} f(x, y) dydx$. Or you could decide to integrate with respect to x first, then y. Then the "outer" integral will be with respect to y and, to cover the given region, must go from y= 0 at the bottom to y= 1 at the top. And, for each y, x must go from the left parabola, y^2= x or x= sqrt{y} to the upper parabola, x= y^2. The integral is $\displaystyle \int_0^1\int_{\sqrt{y}}^{y^2} f(x,y)dxdy$. Those are so similar because of the symmetry of the problem. Also, in going from x= y^2 to y= sqrt(x) and from y= x^2 to x= sqrt(y), we use the fact that the region of integration is in the first quadrant, so both x and y are positive to choose "sqrt(x)" and "sqrt(y)" rather than "-sqrt(x)" and "sqrt(x)". Last edited by greg1313; September 25th, 2017 at 10:37 AM. September 25th, 2017, 10:20 PM # 3 Newbie Joined: Jan 2017 From: India Posts: 9 Thanks: 0 Math Focus: Calculus Quote: But I have one confusion in this:- the question, it says constraints "x>y^2" & "y>x^2", how this is giving us the parabola? I mean I know equation of parabola but how x>y^2 can be said x=y^2 & y>x^2 can be said y=x^2. How these Limiting region is obtained from these constraints? Last edited by LeitHunt; September 25th, 2017 at 10:26 PM. Tags double, integrationfinding, volume Thread Tools Display Modes Similar Threads Thread Thread Starter Forum Replies Last Post Double integration shashank dwivedi Calculus 5 April 14th, 2017 12:35 PM Double/Triple Integral for volume between surfaces zollen Calculus 8 April 9th, 2017 09:51 AM Polar coordinates for volume using the double integral Aftermath Calculus 1 May 11th, 2015 01:59 AM Volume with Double Integral PedroMinsk Calculus 3 December 9th, 2010 03:59 AM Evaluate volume with double integral in polar coordinates jim626 Calculus 0 May 4th, 2008 05:19 PM
It is well known that the generating function of a regular language $L$, i.e. $\sum n_kz^k$ where $n_k$ is the number of words of length $k$ in $L$, is rational, i.e. a quotient of two polynomials $P(z)/Q(z)$. Suppose that $L$ is the language accepted by some finite automaton $\mathcal{A}$. How to find the polynomials $P, Q$ given $\mathcal{A}$? Is there a simple procedure and proof? Cannon does more than just refer to result that you ask for, he sketches the proof out. His proof is couched in the notation that he has set up for his application to hyperbolic groups. But it is easy enough to unravel the notation and express the proof in general. Label the state set of the automaton as $0,\ldots,N$ where $0$ is the start state. Consider the transition matrix whose $i,j$ entry is the number of directed edges from $i$ to $j$ in the automaton. The growth function we want is the power series $f(x) = \sum n_k x^k$ where $n_k$ is the number of directed paths starting at $0$ of length $k$ and stopping at the terminal states of the automaton. For simplicity I'll assume every state is an terminal state; otherwise one just has to change the notation. With this assumption, $f(x) = f_0(x) + \ldots + f_N(x)$ where $f_i(x)$ is the growth function whose $k^{th}$ coefficient is the number of directed paths from state $0$ to state $i$ of length $k$. Cannon then writes a linear recursion for these functions: $f_0(x) = 1$ (the interpretation is that there are not actually any directed edges ending at the start state); and $$f_j(x) = x \cdot \sum_{i=0}^N b_{ij} \cdot f_i(x), j=0,\ldots,N $$ He explains how to express the coefficients $b_{ij}$ as functions of the entries of the transition matrix. Then he writes "It is a routine problem in linear algebra to solve (these equations) for $f_0, f_1, \ldots, f_N$ and $f$", which I interpret as rewriting the equations in vector form $F = x B F$ where $F$ is the column vector whose entries are the functions $f_0(x),\ldots,f_N(x)$ and $B=(b_{ij})$. So we get $(I - xB) F = (1;0;...;0)$ and this can be solved for $F$. Start by making a deterministic finite automaton $M$. Now $n_k$ is the number of walks of length $k$ from the starting state to an accepting state, so $\sum n_k z^k$ is the sum of some entries of $(I-zA)^{-1}$, where $A=(a_{ij})$ is the integer matrix in which $a_{ij}$ is the number of transitions from state $i$ to state $j$. The entries you need to add are the $(k,\ell)$ entries where $k$ is the starting state and $\ell$ is an accepting state. To get this as a rational function, write $(I-zA)^{-1}$ using Cramer's rule. The denominator (before cancelling of any common factors) is the determinant $\|I-zA\|$. The numerator is the adjugate of $I-zA$, whose entries are cofactors, which are also determinants. So in total, if there are $m$ accepting states, you get a sum of $m$ determinants divided by one determinant, and all these determinants are polynomials in $z$. This is of course a special case of the Chomsky-Schutzenberger theorem that unambiguous context-free languages have algebraic generating functions. Restricted to a regular language it is like this. Assume the automata has state set $1,...,n$. Let $1$ be the initial state for convenience. Let A be the adjacency matrix of the automaton, let $e_1$ be the standard unit row vector and let $c$ be the column vector which is the characteristic vector of the terminal states. Then it is easy to see that the generating function is $$f(t)=\sum_{n=0}^{\infty}e_1A^nct^n = e_1\left[\sum_{n=0}^{\infty}A^nt^n\right]c= e_1(I-tA)^{-1}c.$$ Now using the classical adjoint formula for the inverse, you get that the denominator is $\det(I-tA)$ and the numerator is what it is. This is basically equivalent for finding an unambiguous regular expression for the language.This MO answer explains how to do it, given an DFA $\mathcal A$. The rest is easy: replace $\emptyset$ by $0$ replace $\epsilon$ by $1$ replace any symbol with $x$ replace concatenation with multiplication replace $\cup$ with $+$ replace Kleene star with $1/1-f(x)$. Source: this paper. Please see http://algo.inria.fr/flajolet/Publications/books.html ,the book Analytic combinatorics's first several chapters.
The first order necessary optimality conditions for a minimization problem with inequality constraints are not that the gradient vanishes, since the minimum can be attained at the boundary of the feasible set (where only the directional derivatives into the interior of the set have to be non-negative, i.e., going away from the boundary leads to an increase in function value). Specifically, for a problem of the form$$ \min_{x\in C} f(x)$$for a convex set $C$, the necessary optimality conditions for a minimizer $\bar x \in C$ are$$ \langle \nabla f(\bar x), \bar x-x \rangle \leq 0\qquad \text{for all }x\in C.$$For this condition, there is no sensible way to define a residual. But one way to reformulate this condition is using the metric projection $P_C$ onto $C$ as$$ \bar x = P_C(\bar x - \nabla f(\bar x)),$$so you could monitor $$ r(x^k) := \\| x^k - P_C(x^k - \nabla f(x^k))\\|,$$which should go to zero. That being said, if your minimizer $\bar x$ lies in the interior of $C$, either of these conditions can only hold if the gradient vanishes. If this is the case in your problem, your gradient norm should go to zero -- and $10^{-2}$ is definitely not zero. If your derivatives are correct, then the norm should go to zero up to machine precision (say, $10^{-16}$). Otherwise the method is limited by the accuracy of your gradient and Hessian evaluation, so I'd suspect that there's an error there (which you can check by comparing directional derivatives via inner products of the gradient with a direction and via finite difference approximations). EDIT: To test whether your gradient and Hessian are suitably accurate approximations of derivatives of your functional, you can do the following: Gradient: Pick $x$ and direction $h$ (e.g., $h=x$ or $h$ random -- always test multiple vectors!) and compare for $\varepsilon \to 0$ a) $(f(x+\varepsilon h)-f(x))/\varepsilon$ b) $\langle \nabla f(x),h\rangle$ Action of Hessian: $x,h$ as above and compare for $\varepsilon \to 0$ a) $(f(x+\varepsilon h)-2f(x) +f(x-\varepsilon h))/\varepsilon^2$ b) $\langle [\nabla^2 f(x)h],h\rangle$
Consider a thin uniform rod AB of mass $m$ and length $l$ sliding down a frictionless wall and on a frictionless floor. Point O is its instantaneous center of rotation, and C its center of mass (and geometric center). We need to calculate the angle $\theta$ after which the rod loses contact with the wall. The initial value of $\theta$ is given to be $\pi/3$ radians. I've marked the directions of motion of the end points and the center of mass of the rod (pardon my MS paint skills). The forces on the rod are its weight through C, $N_{wall}$ is the reaction force from the wall and $N_{ground}$ is the reaction force from the wall. The problem for me is that calculating the angular acceleration $\alpha$ about O gives me $ (ml^2/3) \alpha = mgl(cos{\theta})/2$ And when we do the same from A, we get $ (ml^2/3) \alpha = mgl (cos{\theta})/2 - N_{wall}lcos{\theta}$ And from B, we get $ (ml^2/3) \alpha = N_{ground}lcos{\theta} - mgl (cos{\theta})/2 $ I'm having trouble figuring out where I'm going wrong. Supposedly, the angular acceleration for a rigid body about all points should be the same. Which of the above equations is wrong, then? Is it because the points in question are non-inertial/accelerating? As far as the question itself is concerned, I think I can do it based on making a differential equation based on the given equation(s) and horizontal acceleration being zero when it does lose contact. I just think that there's some fundamentally wrong with the way I've approached the problem above, and would really appreciate some help. Thanks!
Let $R$ be a Noetherian ring, and let $I$ be an ideal of $R$. Fix a rational number $a=\frac{p}{q}$ with $p, q\in \mathbb{Z_\geq 0}$ $q\neq 0$. We deﬁne $I_a = \{x \in R: x^q\in \overline{I^p}\}$, where overline denotes integral closure. Huneke-Swanson show, (in their book on Integral Closure), that $I_a$ is well defined, is indeed an ideal and is integrally closed. If $a\leq b$, $I_b\subseteq I_a$ and if $n$ is a positive integer, then, $I_n=\overline{I^n}$. I was curious to know what else is known about these rational powers. Are there any computations done in special cases (e.g. monomial ideals)? Do these ideals find any use other than computing the integral closure of the extended Rees ring in the Laurent polynomial ring? I would appreciate if anyone has any references on this subject other than Huneke-Swanson book. So far I have only found a paper by David Rush "Rees valuations and asymptotic primes of rational powers in Noetherian rings and lattices" PS: I am not sure why, but I am unable to typeset curly braces (I am using backslash followed by brace sign, but it does not show up) and the "less than" (causes text after the symbol to vanish) symbol.
I think Higham's Accuracy and Stability of Numerical Algorithms addresses how one can analyze these types of problems. See Chapter 2, especially exercise 2.8. In this answer I'd like to point out something that isn't really addressed in Higham's book (it doesn't seem to be very widely known, for that matter). If you are interested in proving properties of simple numerical algorithms such as these, you can use the power of modern SMT solvers (Satisfiability Modulo Theories), such as z3, using a package such as sbv in Haskell. This is somewhat easier than using pencil and paper. Suppose I'm given that $0\leq x\leq y$, and I'd like to know if $z=(x+y)/2$ satisfies $x\leq z\leq y$. The following Haskell code import Data.SBV test1 :: (SFloat -> SFloat -> SFloat) -> Symbolic SBool test1 fun = do [x, y] <- sFloats ["x", "y"] constrain $ bnot (isInfiniteFP x) &&& bnot (isInfiniteFP y) constrain $ 0 .<= x &&& x .<= y let z = fun x y return $ x .<= z &&& z .<= y test2 :: (SFloat -> SFloat -> SFloat) -> Symbolic SBool test2 fun = do [x, y] <- sFloats ["x", "y"] constrain $ bnot (isInfiniteFP x) &&& bnot (isInfiniteFP y) constrain $ x .<= y let z = fun x y return $ x .<= z &&& z .<= y will let me do this automatically. Here test1 fun is the proposition that $x \leq \mathit{fun}(x,y) \leq y$ for all finite floats $x,y$ with $0\leq x\leq y$. λ> prove $ test1 (\x y -> (x + y) / 2) Falsifiable. Counter-example: x = 2.3089316e36 :: Float y = 3.379786e38 :: Float It overflows. Suppose I now take your other formula: $z=x/2+y/2$ λ> prove $ test1 (\x y -> x/2 + y/2) Falsifiable. Counter-example: x = 2.3509886e-38 :: Float y = 2.3509886e-38 :: Float Doesn't work (due to gradual underflow: $(x/2)\times2 \neq x$, which might be unintuitive due to all arithmetic being base-2). Now try $z=x + (y-x)/2$: λ> prove $ test1 (\x y -> x + (y-x)/2) Q.E.D. Works! The Q.E.D. is a proof that the test1 property holds for all floats as defined above. What about the same, but restricted to $x\leq y$ (instead of $0\leq x\leq y$)? λ> prove $ test2 (\x y -> x + (y-x)/2) Falsifiable. Counter-example: x = -3.1300826e34 :: Float y = 3.402721e38 :: Float Okay, so if $y-x$ overflows, how about $z = x + (y/2-x/2)$? λ> prove $ test2 (\x y -> x + (y/2 - x/2)) Q.E.D. So it seems that among the formulas I've tried here, $x + (y/2 - x/2)$ seems to work (with a proof, too). The SMT solver approach seems to me a much quicker way of answering suspicions about simple floating-point formulas than going through floating-point error analysis with pencil and paper. Finally, the goal of accuracy and stability is often at odds with the goal of performance. For performance, I don't really see how you can do better than $(x+y)/2$, especially since the compiler will still do the heavy lifting of translating this into machine instructions for you. P.S. This is all with single-precision IEEE754 floating-point arithmetic. I checked $x \leq x + (y/2-x/2) \leq y$ with double-precision arithmetic (replace SFloat with SDouble), and it works too. P.P.S. One thing to bear in mind when implementing this in code is that compiler flags like -ffast-math (some forms of such flags are sometimes turned on by default in some common compilers) will not result in IEEE754 arithmetic, which will invalidate the above proofs. If you do use flags that enable, e.g., associative addition optimizations, then there's no point in doing anything other than $(x+y)/2$. P.P.P.S. I got carried away a little looking only at simple algebraic expressions without conditionals. Don Hatch's formula is strictly better.
Scale-free networks Cesar A. Hidalgo and Albert-Laszlo Barabasi (2008), Scholarpedia, 3(1):1716. doi:10.4249/scholarpedia.1716 revision #140135 [link to/cite this article] A network that has a power-law degree distribution, regardless of any other structure, is called a scale-free network. Power-Law degree distribution The degree of a node is the number of links adjacent to it. If we call the degree of a node $k\ ,$a scale-free network is defined by a power-law degree distribution, which can be expressed mathematically as$P(k)\sim k^{-\gamma}$From the form of the distribution it is clear that when: $\gamma<2$ the average degree diverges. $\gamma<3$ the standard deviation of the degree diverges. It has been found that most scale-free networks have exponents between 2 and 3. Thus, they lack a characteristic degree or scale, and therefore their name. High degree nodes are called hubs. Scale-free network models There are several models that are able to create a scale-free network. But most ofthem introduce in one way or another two main ingredients, growth and preferential attachment. By growth we mean that the number of nodes in the network increases in time. Preferential attachment refers to the fact that new nodes tend to connect to nodes with large degree. One can naively argue that for large networks this is nonsense becauseit requires a global knowledge of the network, i.e., knowing which are the high degree nodes, but this is not the case.There are several local mechanisms that introduce preferential attachment (see below). In mathematical terms, preferential attachment means that the probability that a node with degree $k_i$ acquires a link goes as $P(k_i)=\frac{k_i}{\sum_{i}k_i}$ The Barabasi-Albert model The Barabasi-Albert model (a.k.a. BA model) introduced in 1998 explains the power-law degree distribution of networks by considering two main ingredients: growth and preferential attachment (Barabasi and Albert 1999). The algorithm used in the BA model goes as follows. Growth: Starting with a small number ($m_0$) of connected nodes, at every time step, we add a new node with $m(<m_0)$ edges that link the new node to $m$ different nodes already present in the network. Preferential attachment: When choosing the nodes to which the new node connects, we assume that the probability $P$ that a new node will be connected to node $i$ depends on the degree $k_i$ of node $i\ ,$ such that \[P\sim \frac{k_i}{\sum_{i}k_{i}}\] Numerical simulations and analytic results indicate that this network evolves into a scale-invariant state with the probability that a node has $k$ edges following a power law with an exponent $\gamma=3\ .$ The scaling exponent is independent of $m\ ,$ the only parameter in the model. Analytical solution for the BA model This model can be solved analytically by setting up a differential equation in which the rate at which a node acquires links is equal to the number of links added times the probability of acquiring a link \[ \frac{dk_i}{dt}= m\frac{k_i}{\sum_{j}k_j} \] This equation can be simplified by realizing that at each time step $m$ links are added, thus \[\sum_j k_j = 2mt\] and \[ \frac{dk_i}{dt}=\frac{k_i}{2t} \] \[ \ln{k_i}=\frac{1}{2}\ln{t} + C \] where $C$ is an integration constant that can be determined by using the fact that the $i^{th}$ node arrived to the network at time $t^i$ having degree $m\ .$ Thus \[k_i = m(\frac{t}{t_i})^{1/2}\ .\] The degree distribution can be calculated by finding the probability that a node has a degree smaller than $k$ \[ P(k_i(t) > k) = P(t_i < \frac{mt}{k^2}) = 1 - P(t_i > \frac{mt}{k^2}) \] Without loss of generality we can assume that nodes are added at a constant rate thus \[ P(t_i) =\frac{1}{m_0 + t} \] where $m_0$ is the total degree of the nodes that got the network started. Using this distribution \[ P(k_{i} - k) = 1 -\frac{mt}{k^2}\frac{1}{m_0 + t} \] Finally we get the degree distribution by differentiating and conclude that \[ \frac{d}{dk}P(k_i < k) = P(ki = k) = \frac{2m^2 t}{k^3}\frac{1}{m_0 + t} \] This mechanism was first introduced by Yule in the early 20th century to explain the distribution of different Taxa and was later generalized by Price in the 70s and coined as cumulative advantage. The example shown here is not the most general version of the Price model that can be found in the original paper as well as in Newman (2005). In any case, the lesson that should be learned from this is that whenever we found a system in which the probability of increasing is proportional to the actual value, we should expect its distribution to follow a power-law. Local alternatives for preferential attachment We could imagine that when nodes join the network they follow a link. For example, you move to a new town and an old friend tells you that you should visit a friend of him. If we consider that link to be a randomly chosen one, the probability $\pi$ that you were referred to a person with degree $k$ is \[\pi(k)= k P(k)\] which is the precise definition of preferential attachment. This is because $k$ links end up in an agent of degree $k\ .$ Duplication and divergence model In a biological context the "duplication and divergence" model has been proposed as an explanation for the scale-free nature of protein-protein interaction networks. Let us consider a protein that interacts with $k_p$ other proteins. Eventually some of the genes that encode a protein get duplicated and now 2 copies become available. This redundancy allows one copy of the gene to mutate without changing the fitness of the organism. This process ends up with different proteins which are likely to share some interactions. If these duplication and divergence process occurs at random, proteins that have a high degree are more likely to have one of its neighbors change, and again, the probability of winning a neighbor through this process is proportional to the actual number of neighbors. This is another example of linear preferential attachment. Limited information A different scenario that one can imagine is the one in which a node incorporates to the network with a limited or local information about it. If linear preferential attachment is used as the rule to create links in the local information context, a power-law degree distribution is also recovered, regardless of having limited information. Properties of scale-free networks Scale-free networks have qualitatively different properties from strictly random, Erdos and Renyi, networks. These are: Scale-free networks are more robust against failure. By this we mean that the network is more likely to stay connected than a random network after the removal of randomly chosen nodes. Scale-free networks are more vulnerable against non-random attacks. This means that the network quickly disintegrates when nodes are removed according to their degree. Scale-free networks have short average path lengths. In fact the average path length goes as $L\sim \log{N}/\log{\log{k}} $ Scale-free networks in nature Scale-free networks have been observed in social, technological and biological systems. These include the citation and co-author scientific networks, the internet and world-wide web, and protein-protein interaction and gene regulatory networks (Albert and Barabasi 2002). References R. Albert, A.-L. Barabási (2002) Statistical mechanics of complex networks. Reviews of Modern Physics 74, 47—97. A.-L. Barabási and R. Albert (1999) Emergence of scaling in random networks. Science 286, 509—512 M. E. J. Newman (2005) Contemporary Physics 46, 323—351. Internal references Wikipedia contributors. Scale-free network. Wikipedia, The Free Encyclopedia.January 20, 2012, 13:35 UTC. Available online. Accessed January 30, 2012.
Both answers are "No." There are well-known obstructions to the existence of an equivariant momentum mapping arising from the action by symplectomorphisms of a group $G$ on a symplectic manifold. They can be phrased in many ways, but if $G$ is connected and its Lie algebra is semisimple, for example, the obstructions vanish. A nice treatment can be found in the classic paper by Atiyah and Bott: "The moment map and equivariant cohomology" in Topology 1984. After-dinner update: Let me try to add some more details, since it seems I was a little too quick both reading the question and also the comments! (In my defense, I was getting really hungry and wanted to get home!) Let $(M,\omega)$ be a symplectic manifold and let $G$ be a connected Lie group acting on $M$ via symplectomorphisms. Let $\mathfrak{g}$ be the Lie algebra of $G$ and for every $X \in \mathfrak{g}$ let $\xi_X$ denote the corresponding vector field on $M$. Since $G$ acts symplectomorphically, $\xi_X$ is a symplectic vector field; that is, $\mathcal{L}_{\xi_X} \omega = 0$, which, as Ben points out, implies that $d i_{\xi_X}\omega = 0$. Let $\mathrm{Sym}(M)$ denote the Lie algebra of symplectic vector fields on $M$. We have a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$. A symplectic vector field $\xi$ is said to be hamiltonian if $i_\xi\omega$ is not merely closed, but also exact. The $G$-action is hamiltonian if the $\xi_X$ are hamiltonian for all $X \in \mathfrak{g}$. Let $\mathrm{Ham}(M)$ denote the Lie algebra of hamiltonian vector fields. It is not hard to show that the Lie bracket of two symplectic vector fields is hamiltonian, so $\mathrm{Ham}(M)$ is an ideal in $\mathrm{Sym}(M)$ with abelian quotient. This gives rise to a short exact sequence of Lie algebras$$0 \longrightarrow \mathrm{Ham}(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0$$where $H^1(M)$ is the first de Rham cohomology group thought of as an abelian Lie algebra. If $\xi \in \mathrm{Ham}(M)$, then $i_\xi \omega = df$ for some $f \in C^\infty(M)$. This defines a map $C^\infty(M) \to \mathrm{Ham}(M)$ which is also a Lie algebra homomorphism if we make $C^\infty(M)$ into a Lie algebra via the Poisson bracket. The kernel of this map consists of the locally constant functions, whence we have another short exact sequence of Lie algebras$$0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Ham}(M) \longrightarrow 0$$ Putting these two sequences together we have a four-term exact sequence starting and ending at the first two de Rham cohomology groups:$$0 \longrightarrow H^0(M) \longrightarrow C^\infty(M) \longrightarrow \mathrm{Sym}(M) \longrightarrow H^1(M) \longrightarrow 0$$ Now the symplectic $G$-action defines a Lie algebra homomorphism $\mathfrak{g} \to \mathrm{Sym}(M)$ and for the existence of a momentum map, we want this to lift to a Lie algebra morphism $\mathfrak{g} \to C^\infty(M)$. There is an immediate obstruction for the map to lift at the level of vector spaces, namely the map $\mathfrak{g} \to \mathrm{Sym}(M) \to H^1(M)$ is a Lie algebra cocycle with values in the trivial module $H^1(M)$, and so defines a class in $H^1(\mathfrak{g};H^1(M))$. If this class vanishes, we do get a map $\mathfrak{g} \to C^\infty(M)$ which may only be a homomorphism modulo $H^0(M)$. In other words, it defines a Lie algebra homomorphism to a central extension of $C^\infty(M)$ defined by a 2-cocycle with values in the trivial module $H^0(M)$. The moment map will exist if the class of this cocycle in $H^2(\mathfrak{g}; H^0(M))$ vanishes. For example, if $\mathfrak{g}$ is semisimple, then both $H^1$ and $H^2$ vanish and the momentum map exists. The beautiful observation of Atiyah and Bott is that the obstruction can be reinterpreted in terms of the Cartan model for the equivariant de Rham cohomology of $M$, where it becomes simply the obstruction to extending $\omega$ to an equivariant cocycle.
Getting all (complex) solutions of a non polynomial equation Hi ! I was used to solve the following equation with Mathematica. \begin{equation} \alpha_1 + \alpha_2x + \alpha_3x^2 + x^4 + \frac{\alpha_4}{x^2-\alpha_0} + \frac{\alpha_5 x^2}{x^2-\alpha_0}=0 \end{equation} where $\alpha_i$ are constants. The Mathematica function "Solve" gives me all the numerical roots of this non polynomial equation very easily. These roots can be real or complex. I'm a very beginner at Sage. I have tried several methods to solve this equation automatically but it seems that all methods I've used work only for polynomial equations. Here they are : alpha0 = 0.25alpha1 = -2.5alpha2 = 6.9282alpha3 = -5.5alpha4 = 0.5alpha5 = -0.5x = var('x')eq = alpha1 + alpha2x + alpha3x2 + x4 + alpha4/(x*2 - alpha0) + alpha5x2/(x2 - alpha0) == 0.# test 1# solve(eq, x, ring=CC)# ==> [0 == 20000x^6 - 115000x^4 + 138564x^3 - 32500x^2 - 34641x + 22500]# test 2# solve(eq, x, ring=CC, solution_dict=True)# ==> [{0: 20000x^6 - 115000x^4 + 138564x^3 - 32500x^2 - 34641x + 22500}]# test 3# eq.roots(x, ring=CC,multiplicities=False)# ==> TypeError: fraction must have unit denominator Do you have any idea of a method to get the approximated roots of the equation ? Thanks in advance EDIT : correction of an error in the equation ; add few tests

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