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# Non Homogeneous Partial Differential Equation With Constant Coefficient

Having a non-zero value for the constant c is what makes this equation non-homogeneous, and that adds a step to the process of solution. As with a general PDE, elliptic PDE may have non-constant coefficients and be non-linear. The Euler scheme is one of the standard schemes to obtain numerical approximations of solutions of stochastic differential equations (SDEs). Linear non-homogeneous ordinary differential equations and links to common methods for particular solutions, including method of undetermined coefficients, method of variation of parameters, method of reduction of order, and method of inverse operators. Methods for Solution. If the general solution $${y_0}$$ of the associated homogeneous equation is known, then the general solution for the nonhomogeneous equation can be found by using the method of variation of constants. This fact can often be used to solve constant coeﬃcient partial diﬀerential equation. An example of a first order linear non-homogeneous differential equation is. This inﬁnite series formula for u(x,t)is your solution to the entire partial differential equation problem. ? Damped harmonic oscillator: second order differential equations w/ constant coefficients? Non-homogeneous Second Order Differential Equation?. Section 4-6 : Nonconstant Coefficient IVP's. Homogeneous linear equations of order 2 with non constant coefficients We will show a method for solving more general ODEs of 2n order, and now we will allow non constant coefficients. How to Solve Linear Differential Equations Using the Method of Undetermined Coefficients. BSc Mathematics in hindi Linear Differential Equation with constant coefficient in hindi Partial Differential Equation - Solution of one dimensional heat flow. Homogeneous Equations where the Auxiliary Equation contains: 1. Hadamard Counterexample. However, there are some simple cases that can be done. Separation of variables. We'll look at two simple examples of ordinary differential equations below, solve them in. heterogeneous first-order linear constant coefficient ordinary differential equation:. Equation [4] is non-homogeneous. Homogeneous second order differential equation with constant coefficients. 6) (vi) Nonlinear Differential Equations and Stability (Ch. A partial differential equation for which the Cauchy problem is uniquely solvable for initial data specified in a neighbourhood of on any non-characteristic surface (cf. Computations in MATLAB are done in floating point arithmetic by default. ISBN 9789385676161 from SChand Publications. A second order differential equation is one containing the second derivative. 1 Homogeneous Linear Equations with Constant Coefficients127 13. An equation that contains partial derivatives is called a partial differential equation (PDE). • Partial Differential Equation: At least 2 independent variables. Partial differential equation, in mathematics, equation relating a function of several variables to its partial derivatives. A homogeneous $$n$$th-order ordinary differential equation with constant coefficients admits exactly $$n$$ linearly-independent solutions. Video of lectures given by Arthur Mattuck and Haynes Miller, mathlets by Huber Hohn, at Massachussette Institute of Technology. Differential Equations Calculators; Math Problem Solver (all calculators) Differential Equation Calculator. How to Solve Linear Differential Equations Using the Method of Undetermined Coefficients. Second, it is generally only useful for constant coefficient differential equations. Procedure for solving non-homogeneous second order differential equations: y" p(x)y' q(x)y g(x) 1. Advanced Math Solutions – Ordinary Differential Equations Calculator, Exact Differential Equations In the previous posts, we have covered three types of ordinary differential equations, (ODE). o Identify whether or not a differential equation is exact. The general form of a homogeneous differential equation is. The equation is said to be linear non-homogeneous difference equation if R (n) ≠ 0. A differential equation is an equation that involves a function and its derivatives. An Introduction to Partial Diﬀerential Equations in the Undergraduate Curriculum Andrew J. In the above six examples eqn 6. This is a classroom-tested and developed textbook designed for use in either one-or two term courses in Partial Differential Equations taught at the advanced undergraduate and beginning graduate levels of instruction. (A) First Order and First Degree Equations of Bernoulli (B) Differential Equations of First Order and Higher Degree Second Order Differential Equation Margham Publications - Differential Equations & Laplace Transforms III Semester - P. Solve linear second order equations with constant coefficients (both homogenous and non-homogeneous) using the method of undetermined coefficients, variation of parameters, and Laplace transforms. Merged partial differential equation with constant coefficients. Unit 2: Higher Order Differential Equations and Applications Level 2. Euler-Cauchy Equation. the function G(x) = 3e x + sin x. 1 A first order homogeneous linear differential equation is one of the form $\ds \dot y + p(t)y=0$ or equivalently $\ds \dot y = -p(t)y$. [Lars Hörmander] -- Vol. Classification of second order PDEs, General solution of higher order PDEs with constant coefficients, method of separation of variables. This fact can often be used to solve constant coeﬃcient partial diﬀerential equation. 4, 1631--1666. is any non -trivial linear function of x and/or y (except any multiple of yx O). We'll look at two simple examples of ordinary differential equations below, solve them in. How can I solve a 2nd order differential equation with non-constant coefficients like the following? Second order homogeneous differential equation with non. In this paper, we develop an auxiliary equation method for solving fractional partial differential equations, where the fractional derivative is defined in the sense of the conformable fractional derivative. Section 4-6 : Nonconstant Coefficient IVP's. 6) m' = 1, i. The term y 3 is not linear. The fourth order partial differential equation with variable coefficients governing vibrations of the non-prismatic prestressed Rayleigh beam is solved using. Since a homogeneous equation is easier to solve compares to its. Chapter 3: Higher Order Differential Equations. To solve the equation, use the substitution. Finally, we model the process of diffusion by the linear differential equation d dt u=(Aux)x in Q=(0,1)×[0,T] (1. F(x) = cxkeax, 2. Because g is a solution. When g(t) = 0 we call the Differential Equation Homogeneous and when we call the Differential Equation Non- Homogeneous. Computations in MATLAB are done in floating point arithmetic by default. Partial differential equations are useful for modelling waves, heat flow, fluid dispersion, and. ) 62-69 [131] Kuchment P A 1981 On Floquet theory for parabolic and elliptic boundary-value problems in a cylinder Dokl. Section 4-6 : Nonconstant Coefficient IVP's. The fourth order partial differential equation with variable coefficients governing vibrations of the non-prismatic prestressed Rayleigh beam is solved using. But am unsure of how to implement the complex roots into my DE solution. 1 Homogeneous Linear Equations with Constant Coefficients127 13. 2 we will learn how to reduce the order of homogeneous linear differential equations if one solution is known. They typically cannot be solved as written, and require the use of a substitution. The equation 2 2 2 2 2 x u c t u ∂ ∂ = ∂ ∂ [4] is an example of a partial differential equation. Solution of non-homogeneous PDE by direct integration. Bessel and Legendre equations. JANSSEN Koninklijke Militaire School, Leerstoel Wiskunde, Renaissancelaan 30, 1040 Brussel, Belgium Received 22 July 1986 Abstract: For second order homogeneous partial difference equations with constant coefficients in n variables, it is. We can write the general equation as ax double dot, plus bx dot plus cx equals zero. Damped driven harmonic oscillator in a steady state Equation of motion, functions solving differential equations. Here also, the complete solution = C. Laplace, Heat and wave equations. Homogeneous Linear Systems with Constant Coefficients: Solution via Matrix Eigenvalues (Real and Distinct Case) -- Lecture 25. This method does not obtain a solution of the partial differential equation but results in the evaluation of integrals over the domain which are evaluated by use of the physical data. (A) First Order and First Degree Equations of Bernoulli (B) Differential Equations of First Order and Higher Degree Second Order Differential Equation Margham Publications - Differential Equations & Laplace Transforms III Semester - P. A PDE is homogeneous if each term in the equation contains either the dependent variable or one of its derivatives. equations Finding annihilators Functions that can be annihilated by polynomial di erential operators are exactly those that can arise as solutions to constant-coe cient homogeneous linear di erential equations. Partial differential equation, in mathematics, equation relating a function of several variables to its partial derivatives. In the most general form, an Nth order ordinary differential equation (ODE) of a single-variable function can be expressed as. which is the equation. A third way of classifying differential equations, a DFQ is considered homogeneous if & only if all terms separated by an addition or a subtraction operator include the dependent variable; otherwise, it's non-homogeneous. To solve a homogeneous linear differential equation of second order with constant coefficients: {eq}a\frac {\partial^2 y. \) Therefore, we will look for a particular solution in the form Read moreSecond Order Linear Nonhomogeneous Differential Equations with Constant Coefficients - Page 2. How to Solve Linear Differential Equations Using the Method of Undetermined Coefficients. Higher order differential equations 1. A diﬀerential equation (de) is an equation involving a function and its deriva-tives. However, if you know one nonzero solution of the homogeneous equation you can find the general solution (both of the homogeneous and non-homogeneous equations). The determinant of these two algebraic equations, Coefficient[eq1, u0] Coefficient[eq2, v0] - Coefficient[eq1, v0] Coefficient[eq2, u0]) // Simplify; provides a relationship between the two coefficients d1 and d2. Studying it will pave the way for studying higher order constant coefficient equations in later sessions. Enrique Mateus NievesPhD in Mathematics Education. (2002) Asymptotical stability of partial difference equations with variable coefficients. The problem consists ofa linear homogeneous partial differential equation with lin ear homogeneous boundary conditions. the differential equation, we conclude that A=1/20. Since a homogeneous equation is easier to solve compares to its. A second order differential equation is one containing the second derivative. Separating the Variables. The price that we have to pay is that we have to know one solution. 4 Differential Equations with Variable Coefficients / 480 25 Partial Differential Equations Transform Methods / 481 25. Now consider a Cauchy problem for the variable coefficient equation tu x,t xt xu x,t 0, u x,0 sin x. Linear Homogeneous Recurrence Relations with Constant Coefficients: The equation is said to be linear homogeneous difference equation if and only if R (n) = 0 and it will be of order n. Power series solutions. BibTeX @MISC{Hu05particularsolutions, author = {Hsin-yun Hu and Heng-shuing Tsai and Zi-cai Li and Song Wang}, title = {Particular Solutions of Singularly Perturbed Partial Differential Equations with Constant Coefficients in Rectangular Domains, Part II: Computational Aspects}, year = {2005}}. Classical Partial Differential Equations The Coefficient Form of Partial Differential Equations. Solution space. To provide a framework for this. Characteristic surface). 7 Electrical. Determine the. In homogeneous media, A may be treated as a global constant. Linear Second Order Homogeneous Differential Equations - (two distict real roots) Ex: Linear Second Order Homogeneous Differential Equations - (two distict real roots) Ex: Solve and Verify the Solution of a Linear Second Order Homogeneous Differential Equation. Linear Non-homogeneous Differential Equations with Constant Coefficients 262 12. Differential Equations presents the basics of differential equations, adhering to the UGC curriculum for undergraduate courses on differential equations offered by all Indian universities. Welcome to Differential Equations. Homogeneous Equations with Constant Coefficients; Non-Homogeneous Equations with Constant Coefficients. That is, the equation y' + ky = f(t), where k is a constant. Partial Differential Equations Fall 2019 Linear Independence, Constant Coefficient Homogeneous Problems with Non-Homogeneous Boundary Conditions. 8) for a, b, and c constants with a2 +b2 > 0. The price that we have to pay is that we have to know one solution. The Relevance of the Use of the Method of Undetermined Coefficient for Solving Differential Equations. A PDE is homogeneous if each term in the equation contains either the dependent variable or one of its derivatives. Multiplying through by dx, dividing through by a(x)y, and re-arranging the terms gives. Variations of parameters. In particular, a partial differential equation for which the normal cone has no imaginary zones is a hyperbolic partial differential equation. They can be solved by the following approach, known as an integrating factor method. 2 Solution to Case with 4 Non-homogeneous is called a separation constant because the solution to the. a system of ordinary differential equations. Linear Equations of Higher Order 136 3. Since a homogeneous equation is easier to solve compares to its. In mathematics, an ordinary differential equation (ODE) is a differential equation containing one or more functions of one independent variable and the derivatives of those functions. If the general solution $${y_0}$$ of the associated homogeneous equation is known, then the general solution for the nonhomogeneous equation can be found by using the method of variation of constants. An equation that contains partial derivatives is called a partial differential equation (PDE). The term ln y is not linear. This is not always an easy thing to do. - A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. If one speaks, as a rule, of a vectorial non-linear partial differential equation or of a system of non-linear partial differential equations. Example 3: General form of the first order linear. There are six types of non-linear partial differential equations of first order as given below. We're now ready to solve non-homogeneous second-order linear differential equations with constant coefficients. 3 Classification of Linear PDEs in Two Independent Variables In addition to the distinction between linear and nonlinear PDEs, it is important for the computational scientist to know that there are different classes of PDEs. For our better understanding we all should know what homogeneous equation is. they contain ordinary derivatives as opposed to partial derivatives. Nonhomogeneous differential equations are the same as homogeneous differential equations, except. , 43 (2011), no. Often, systems described by differential equations are so complex, or the systems that they describe are so large, that a purely analytical solution to the equations is not tractable. Equations [1], [2], and [3] above are homogeneous equations. If the function in question has only one independent variable, the equation is known as an ordinary differential equation; if the function is of multiple variables, it is called a partial differential equation. Let's start working on a very fundamental equation in differential equations, that's the homogeneous second-order ODE with constant coefficients. Systems of Linear Partial Differential Equations with Constant Coefficients: Bounds on Solutions C. Laplace, Heat and wave equations. ? Damped harmonic oscillator: second order differential equations w/ constant coefficients? Non-homogeneous Second Order Differential Equation?. [1] Aït-Sahalia, Telling from discrete data whether the underlying continuous-time model is a diffusion,, Journal of Finance, 57 (2002), 2075. (*) Each such nonhomogeneous equation has a corresponding homogeneous equation: y″ + p(t. o Use integrating factors to convert a differential equation to an exact equation and then solve. In heterogeneous media, A varies with spatial location. If an ODE can be written in the form $$\frac{\partial y}{\partial t}=\frac{g(t)}{h(y)},$$ then the ODE is said to be separable. hi Torsten Thank u very much for your help :) Yesterday I tried to simplify the problem, so I started with a very simple sinusoidal signal of the following form: b = A sin (2 pi f t), I calculated the solution of this equation analytically, I found this expression : y(x,t) = -A x pi f cos (2 pi f t), It is clear that the solution has a sinusoidal shape. 336 course at MIT in Spring 2006, where the syllabus, lecture materials, problem sets, and other miscellanea are posted. If f(D) is a polynomial in the Differential operator, i. But since it's a partial differential equation, we know that is an arbitrary function. So this is going to be equal to 0. In the case where we assume constant coefficients we will use the following differential equation. Undetermined coefficients. Solving first order linear general PDE's with constant coefficients #2109. It is in these complex systems where computer simulations and numerical methods are useful. The analytic solution to a differential equation is generally viewed as the sum of a homogeneous solution and a particular solution. The Second Order linear refers to the equation having the setup formula of y”+p(t)y’ + q(t)y = g(t). Nevertheless, there are some particular cases that we will be able to solve: Homogeneous systems of ode's with constant coefficients, Non homogeneous systems of linear ode's with constant coefficients, and Triangular systems of differential equations. Power series solutions. The equation is said to be linear non-homogeneous difference equation if R (n) ≠ 0. has to be measured. They are the n roots of the associated. equation is given in closed form, has a detailed description. The general solution of the differential equation is then. Constant Coefficients and Maple Verification Method of Undermined coefficients Second Order Linear Differential Equations Solving non homogeneous differential equations LC Circuit With A Battery Differential equations. More precisely, the eigenfunctions must have homogeneous boundary conditions. Its convergence properties are well known in the case of globally Lipschitz continuous coefficients. A solution of a PDE in some region R of the space of independent variables is a. - A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow. For instance, an ordinary differential equation in x(t) might involve x, t, dx/dt, d 2 x/dt 2 and perhaps other derivatives. II, Differential operators with constant coefficients. In this presentation, we look at linear, nth-order autonomic and homogeneous differential equations with constant coefficients. Knowledge beyond the boundaries. If f (D,D ') is not homogeneous, then (1) is a non–homogeneous linear partial differential equation. They typically cannot be solved as written, and require the use of a substitution. This is another way of classifying differential equations. k (a constant) C linear in x Cx+D quadratic in x Cx2 +Dx+E ksinpx or kcospx C cospx+Dsinpx ke pxCe sum of the above sum of the above product of the above product of the above (where p is a constant) Note: If the suggested form of y PS already appears in the complementary function then multiply this suggested form by x. The Adobe Flash plugin is needed to view this content. The first step in the procedure is to find that homogeneous linear differential equation with constant coefficients which has as a particular solution the right-hand side of 2) i. Now consider a Cauchy problem for the variable coefficient equation tu x,t xt xu x,t 0, u x,0 sin x. Homogeneous and non-homogeneous equations Typically, differential equations are arranged so that all the terms involving the dependent variable are placed on the left-hand side of the equation leaving only constant terms or terms. The first method of solving linear ordinary differential equations with constant coefficients is due to Euler, who realized that solutions have the form ezx, for possibly-complex values of z. Orthogonal functions and Fourier expansions. Series solutions to differential equations can be grubby or elegant, depending on your perspective. The following sections are devoted to Laplace and Helmholtz equations as typical representatives of the elliptic partial differential equations. A first‐order differential equation is said to be homogeneous if M( x,y) and N( x,y) are both homogeneous functions of the same degree. This note covers the following topics: First Order Equations and Conservative Systems, Second Order Linear Equations, Difference Equations, Matrix Differential Equations, Weighted String, Quantum Harmonic Oscillator, Heat Equation and Laplace Transform. Set up the differential equation for simple harmonic motion. Differential Calculus Structural Analysis; Differential Equations of Order One. , determine what function or functions satisfy the equation. Lagrange's and Clairaut's Equations 257 12. Differential equations can either be solved analytically, or they can be solved numerically. Other topics include the following: solutions to non-linear equations, systems of linear differential equations, the construction of differential equations as mathematical models. Identify homogeneous equations, homogeneous equations with constant coefficients, and exact and linear differential equations. Next, substitute the eigenvalues found above into the second equation to find T(t). A differential equation is an equation which relates a function to at least one of its derivatives. Elliptic partial differential equations have applications in almost all areas of mathematics, from harmonic analysis to geometry to Lie theory, as well as numerous applications in physics. Determine the general solution y h C 1 y(x) C 2 y(x) to a homogeneous second order differential equation: y" p(x)y' q(x)y 0 2. 6 ans = 1 A couple of remarks about the above examples: MATLAB knows the number , which is called pi. Undetermined coefficients. You have a linear equation in y, so if you find y c which is the general solution of the homogenous equation, you only need to find one y p which is a solution to your non-homogenous equation. With Variable Coefficients. Separation of Variables Homogeneous Functions Equations with Homogeneous Coefficients. fD()J α isa linear differential operator 0 1<α <. Introduction to partial differential equations. So the problem we are concerned for the time being is the constant coefficients second order homogeneous differential equation. PARTIAL DIFFERENTIAL EQUATIONS Math 124A { Fall 2010 « Viktor Grigoryan [email protected] Free tutoring at the Teaching Center, SW Broward Hall. Second order homogeneous. Nonhomogeneous systems of first-order linear differential equations Nonhomogeneous linear system: y¢ = Ay + B(x), ( ) 2 1 b x b x b x B x n (8) The general solution y = yh + yp where yh is the general solution of the homogeneous system (6) and yp is a particular. by program, a standard approach to solving a nasty di erential equation is to convert it to an approximately equivalent di erence equation. 4) to ﬁnd the values for the c k's. Yuxia Tong, Shuang Liang, Shenzhou Zheng; Integrability of very weak solution to the Dirichlet problem of nonlinear elliptic system, Vol. 3 we will solve all homogeneous linear differential equations with constant coefficients. Buy Ordinary and Partial Differential Equations by M D Raisinghania PDF Online. Systems of linear differential equations with constant coefficients. The differential equation is not linear. CHOICE RULES :METHOD OF UNDETERMINED COEFFICIENTS odification Rule. Wu, Existence of global solutions and invariant measures for stochastic differential equations driven by Poisson type noise with non-Lipschitz coefficients,, Journal of Mathematical Analysis and. The first three worksheets practise methods for solving first order differential equations which are taught in MATH108. 3rd order homogeneous differential equation with non-constant coefficients Differential Equation — Non-constant Coefficients. [5 marks] (ii) By looking for a series solution of the form y= X1 n=0 a nx n+; show that = 0 satis es the indicial equation, and nd the second value of. Methods for Solution. Hadamard Counterexample. However, before we proceed to solve the Non-homogeneous equation, with method of undetermined Coefficients, we must look for some key factors into our differential equation. But for finding the C. Homogeneous PDE: If all the terms of a PDE contains the dependent variable or its partial derivatives then such a PDE is called non-homogeneous partial differential equation or homogeneous otherwise. For curiosity: Navier-Stokes Equations and Minimal Surface Equation and A rapid survey of the modern theory of PDEs: 01. The determinant of these two algebraic equations, Coefficient[eq1, u0] Coefficient[eq2, v0] - Coefficient[eq1, v0] Coefficient[eq2, u0]) // Simplify; provides a relationship between the two coefficients d1 and d2. An example of a first order linear non-homogeneous differential equation is. Procedure for solving non-homogeneous second order differential equations: y" p(x)y' q(x)y g(x) 1. This method does not obtain a solution of the partial differential equation but results in the evaluation of integrals over the domain which are evaluated by use of the physical data. Unit 4Unit 4Unit 4( (((iiii). 5 we will solve the non-homogeneous case. Studying it will pave the way for studying higher order constant coefficient equations in later sessions. An Introduction to Partial Diﬀerential Equations in the Undergraduate Curriculum Andrew J. Homogeneous equation is a differential equation, which is equal to zero. The Second Order linear refers to the equation having the setup formula of y”+p(t)y’ + q(t)y = g(t). Diﬀerential equations are called partial diﬀerential equations (pde) or or-dinary diﬀerential equations (ode) according to whether or not they contain partial derivatives. In this method, an operator is employed which transforms the original equation into a homogeneous Nth-order (N¿n) differential equation with constant coefficients; this can then be solved using one of several elementary procedures. But am unsure of how to implement the complex roots into my DE solution. o Solve second order linear differential equations with constant coefficients that have a characteristic equation with real and distinct roots. Linear non-homogeneous ordinary differential equations and links to common methods for particular solutions, including method of undetermined coefficients, method of variation of parameters, method of reduction of order, and method of inverse operators. Partial Differential Equations & Beyond Stanley J. , one will be a constant multiple of the other. Differential equations involve the differential of a quantity: how rapidly that quantity changes with respect to change in another. heterogeneous first-order linear constant coefficient ordinary differential equation:. Priority B. Solve linear second order equations with constant coefficients (both homogenous and non-homogeneous) using the method of undetermined coefficients, variation of parameters, and Laplace transforms. Formation of partial differential equations – Singul ar integrals -- Solutions of standard types of first order partial differential equations – Lagrange’s linear equation -- Linear partial differential equations of second and higher order with constant coefficients of both homogeneous and non-homogeneous types. o Identify whether or not a differential equation is exact. BSc Mathematics in hindi Linear Differential Equation with constant coefficient in hindi Partial Differential Equation - Solution of one dimensional heat flow. The general solution of an Nth-order, constant coefficient, homogeneous differential equation is a linear combination of N exponential terms. 1 Eigenfunction Expansions of Solutions Let us complicate our problems a little bit by replacing the homogeneous partial differential equation, X jk a jk ∂2u ∂xk∂xj + X l b l ∂u ∂xl + cu = 0 , with a corresponding nonhomogeneous partial differential equation, X jk a jk ∂2u ∂xk∂xj + X l b l ∂u. Second Order Linear Homogeneous Differential Equations with Constant Coefficients For the most part, we will only learn how to solve second order linear equation with constant coefficients (that is, when p(t) and q(t) are constants). Orthogonal functions and Fourier expansions. Complex Roots relate to the topic of Second order Linear Homogeneous equations with constant coefficients. Old Dominion University ODU Digital Commons Mathematics & Statistics Faculty Publications Mathematics & Statistics 2009 Gas-Kinetic Schemes for Direct Numerical Simulations of Com. txt) or view presentation slides online. As in the case of ordinary linear equations with constant coefficients the complete solution of. Solution of Linear Constant-Coefficient Difference Equations Two methods Direct method Indirect Method (z-transform) Direct solution Method: The total solution is the sum of two parts Part 1 homogeneous solution Part 2 particular solution The Homogeneous solution Assuming that the input. Linear differential equations with constant coefficient. Since we already know how to solve the general first order linear DE this will be a special case. same reason, the temperature equation takes the form • For a constant pressure it is very similar to the equation for the mass fraction Y i with an equal diffusion coefficient D=λ/ρ/c p for all reactive species and a spatially constant Lewis number may be written as. Knowledge beyond the boundaries. Because g is a solution. In fact, it is a formula that. 5 Nonhomogeneous Equations and Undetermined Coefﬁcients 184 3. Unit 4Unit 4Unit 4( (((iiii). We mostly deal with the plane case when the number of independent variables is restricted to be two. Second-order linear equations with non-constant coefficients don't always have solutions that can be expressed in closed form'' using the functions we are familiar with. Elimination of Arbitrary. Studying it will pave the way for studying higher order constant coefficient equations in later sessions. In section 4. Homogenization for stochastic partial differential equations derived from nonlinear filterings with feedback ICHIHARA, Naoyuki, Journal of the Mathematical Society of Japan, 2005; Symbolic Solution to Complete Ordinary Differential Equations with Constant Coefficients Navarro, Juan F. Next, substitute the eigenvalues found above into the second equation to find T(t). Lecture 12: Introduction to Partial Differential Equations. Step-by-step solutions for differential equations: separable equations, Bernoulli equations, general first-order equations, Euler-Cauchy equations, higher-order equations, first-order linear equations, first-order substitutions, second-order constant-coefficient linear equations, first-order exact equations, Chini-type equations, reduction of order, general second-order equations. Recently there has been an ongoing research effort to develop data-driven methods for discovering unknown physical laws. Series solutions to differential equations can be grubby or elegant, depending on your perspective. 4 Fundamental set of lin ear ODE with constant coefficients 2. A simple way of checking this property is by shifting all of the terms that include the. 57–69 June 30, 2013. made to extend these methods of identification to partial differential equations. the function G(x) = 3e x + sin x. In the most general form, an Nth order ordinary differential equation (ODE) of a single-variable function can be expressed as. This book has been Designed for the use of undergraduate (Honours) and postgraduate Students of various Indian Universities. If in differential equation y'=f(t,y), function f(t,y) has the property that f(at,ay)=f(t,y) then such differential equation is called homogeneous. differential equations with constant coefficients. 1 BACKGROUND OF THE STUDY. A second order homogeneous equation with constant coefficients is written as where a, b and c are constant. Recently there has been an ongoing research effort to develop data-driven methods for discovering unknown physical laws. 4 Fundamental set of lin ear ODE with constant coefficients 2. The general solution of the differential equation is then. (2002) On ordinary difference equations with variable coefficients. Namely, the two values of λ have been selected so that in each case the coeﬃcient determinant of the system will be zero, which means the equations will be dependent. Students with disabilities requesting accommodations should first register with the Disability Resource Center (352-392-8565) by providing appropriate documentation. Non-homogeneous equations. For our better understanding we all should know what homogeneous equation is. 6 Summary for Particular Integral137. First Order Non-homogeneous Differential Equation. The problem consists ofa linear homogeneous partial differential equation with lin ear homogeneous boundary conditions. 6) m' = 1, i. In the preceding section, we learned how to solve homogeneous equations with constant coefficients. would be classified as a second-order, linear PDE with variable coefficients. If f (D,D ') is not homogeneous, then (1) is a non–homogeneous linear partial differential equation. Using a calculator, you will be able to solve differential equations of any complexity and types: homogeneous and non-homogeneous, linear or non-linear, first-order or second-and higher-order equations with separable and non-separable variables, etc. For curiosity: Navier-Stokes Equations and Minimal Surface Equation and A rapid survey of the modern theory of PDEs: 01. UNIT II FOURIER SERIES. I of Lars Hormander's 4-volume treatise was an exposition of the theory of distributions and Fourier analysis preparing for the study of linear partial differential operators. PARTIAL DIFFERENTIAL EQUATIONS(PDEs): Lagrange method, Charpit method and Cauchy’s method for solving first order PDEs. Bernoﬀ LECTURE 1 What is a Partial Diﬀerential Equation? 1. A homogeneous linear partial differential equation of the n th order is of the form. The first method of solving linear ordinary differential equations with constant coefficients is due to Euler, who realized that solutions have the form ezx, for possibly-complex values of z. Welcome to Differential Equations. It also includes methods of solving higher- order differential equations: the methods of undetermined coefficients, variation of parameters, and inverse operators. 2012 (2012), No. Lagrange's and Clairaut's Equations 257 12. has to be measured. is an arbitrary constant. Therefore, for nonhomogeneous equations of the form $$ay″+by′+cy=r(x)$$, we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. Differential Equations Calculators; Math Problem Solver (all calculators) Differential Equation Calculator. In this work, we will assume that the searched equation is a linear evolution equation, with non-constant coefficients. Unit 2: Higher Order Differential Equations and Applications Level 2. In order to identify a nonhomogeneous differential equation, you first need to know what a homogeneous differential equation looks like. How can I solve a 2nd order differential equation with non-constant coefficients like the following? Second order homogeneous differential equation with non.


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# The invariance for time-reversal: some clarifications

On the Physics.SE I haven't found a simple definition of the invariance for time-reversal. Is it possible to have a simple explanation (also using differential equations) of the invariance for time-reversal?

• have you seen en.wikipedia.org/wiki/T-symmetry#Invariance – Adrian Howard Aug 8 '19 at 15:16
• @AdrianHoward In all honesty, no. But I have seen your link but for me it is all complicated. I asked for something very simple. – Sebastiano Aug 8 '19 at 15:20
• It simply means that an equations of motion are the same if you replace t by -t – Wolphram jonny Aug 8 '19 at 15:56
• @Wolphramjonny If $t\geq 0$, what is $-t$? If the law is $md^2\mathbf {r}/dt^2=\mathbf F$ what happen replacing $t$ by $-t$ if it has a significance? – Sebastiano Aug 8 '19 at 16:03
• it is better to call it $-t_p$ instead of just -t. $t_p$ grows into the past. the equation above will stay the same, with $t_p$ instead of t. which means that the system evolves in the same way into the past than into the future. It means that if I show you a movie of two planets orbiting you cannot tell if the movie is running forwards or backwards, there is nothing in the motion that lets you tell. – Wolphram jonny Aug 8 '19 at 16:22

Imagine any physical system (try a pendulum, without friction), on a given time (let's call this instant in time '$$t$$') the system is in a state you can know things about and make a note of the state, if you let advance time for a while ant take a note of the state of the system again, you'll notice it has changed, and label the new entry $$t+1$$.
A law of physics is said to be symmetric with respect of something if it has a parameter you can change in the functions that model the behavior and not change the result, for example if you label $$t$$ as $$12$$:$$03\,$$ and $$t+1$$ as $$12$$:$$05\,$$ but you discover your watch is two minutes fast and relabel them as $$t' = 12$$:$$01$$ and $$t'+1 = 12$$:$$03$$ nothing else in your experiment will change that's because the laws of physics are symmetric with respect of time translation.
Now if laws of physics are time reversible that would mean that they can be 'played backwards' so that you can relabel $$t' = 12$$:$$05\,$$ and $$t'+1 = 12$$:$$03$$ and the only change is that you get the notes on the states backwards, that would imply that you can always know what the past of the system by knowing its current state; we still don't know if this is true (but physicists would love that it was).

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[Solutions] International Mathematics Competition for University Students 2018

1. Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be two sequences of positive numbers. Show that the following statements are equivalent
a) There is a sequence $(c_n)_{n=1}^{\infty}$ of positive numbers such that $\displaystyle\sum_{n=1}^{\infty}{\frac{a_n}{c_n}}$ and $\displaystyle\sum_{n=1}^{\infty}{\frac{c_n}{b_n}}$ both converge;
b) $\displaystyle\sum_{n=1}^{\infty}{\sqrt{\frac{a_n}{b_n}}}$ converges.
2. Does there exist a field such that its multiplicative group is isomorphism to its additive group?
3. Determine all rational numbers $a$ for which the matrix $$\begin{pmatrix}a & -a & -1 & 0 \\a & -a & 0 & -1 \\1 & 0 & a & -a\\0 & 1 & a & -a\end{pmatrix}$$ is the square of a matrix with all rational entries.
4. Find all differentiable functions $f:(0,\infty) \to \mathbb{R}$ such that $$f(b)-f(a)=(b-a)f’(\sqrt{ab}),\,\forall a,b>0.$$
5. Let $p$ and $q$ be prime numbers with $p<q$. Suppose that in a convex polygon $P_1,P_2,…,P_{pq}$ all angles are equal and the side lengths are distinct positive integers. Prove that $$P_1P_2+P_2P_3+\cdots +P_kP_{k+1}\geqslant \frac{k^3+k}{2}$$holds for every integer $k$ with $1\leqslant k\leqslant p$.
6. Let $k$ be a positive integer. Find the smallest positive integer $n$ for which there exists $k$ nonzero vectors $v_1,v_2,…,v_k$ in $\mathbb{R}^n$ such that for every pair $i,j$ of indices with $|i-j|>1$ the vectors $v_i$ and $v_j$ are orthogonal.
7. Let $(a_n)_{n=0}^{\infty}$ be a sequence of real numbers such that $a_0=0$ and $$a_{n+1}^3=a_n^2-8,\, n=0,1,2,…$$ Prove that the following series is convergent $$\sum_{n=0}^{\infty}{|a_{n+1}-a_n|}.$$
8. Let $\Omega =\{ (x,y,z)\in \mathbb{Z}^3:y+1\geqslant x\geqslant y\geqslant z\geqslant 0\}$. A frog moves along the points of $\Omega$ by jumps of length $1$. For every positive integer $n$, determine the number of paths the frog can take to reach $(n,n,n)$ starting from $(0,0,0)$ in exactly $3n$ jumps.
9. Determine all pairs $P(x),Q(x)$ of complex polynomials with leading coefficient $1$ such that $P(x)$ divides $Q(x)^2+1$ and $Q(x)$ divides $P(x)^2+1$.
10. For $R>1$ let $\mathcal{D}_R =\{ (a,b)\in \mathbb{Z}^2: 0<a^2+b^2<R\}$. Compute $$\lim_{R\rightarrow \infty}{\sum_{(a,b)\in \mathcal{D}_R}{\frac{(-1)^{a+b}}{a^2+b^2}}}.$$
 MOlympiad.NET là dự án thu thập và phát hành các đề thi tuyển sinh và học sinh giỏi toán. Quý bạn đọc muốn giúp chúng tôi chỉnh sửa đề thi này, xin hãy để lại bình luận facebook (có thể đính kèm hình ảnh) hoặc google (có thể sử dụng $\LaTeX$) bên dưới. BBT rất mong bạn đọc ủng hộ UPLOAD đề thi và đáp án mới hoặc liên hệbbt.molympiad@gmail.comChúng tôi nhận tất cả các định dạng của tài liệu: $\TeX$, PDF, WORD, IMG,...


text:
# ON A CONSTANT ARISING IN MANIN'S CONJECTURE FOR DEL PEZZO SURFACES

@article{Derenthal2007ONAC,
title={ON A CONSTANT ARISING IN MANIN'S CONJECTURE FOR DEL PEZZO SURFACES},
author={U. Derenthal},
journal={Mathematical Research Letters},
year={2007},
volume={14},
pages={481-489}
}
• U. Derenthal
• Published 2007
• Mathematics
• Mathematical Research Letters
• For split smooth Del Pezzo surfaces, we analyse the structure of the effective cone and prove a recursive formula for the value of $\alpha$, appearing in the leading constant as predicted by Peyre of Manin's conjecture on the number of rational points of bounded height on the surface. Furthermore, we calculate $\alpha$ for all singular Del Pezzo surfaces of degree $\ge 3$.
26 Citations


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# If Karishma spends 53 seconds folding H napkins, how many seconds

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If Karishma spends 53 seconds folding H napkins, how many seconds  [#permalink]

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29 Sep 2018, 03:28
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If Karishma spends 53 seconds folding H napkins, how many seconds will it take her to fold B napkins at the same rate?

A) 53B/H

B) HB/53

C) 53/HB

D) 53H/B

E) 53HB

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29 Sep 2018, 04:36

Solution

Given:
• Karishma spends 53 seconds folding H napkins

To find:
• The time Karishma will take (in secs) to fold B napkins, working at the same rate

Approach and Working:

To fold H napkins, Karishma takes 53 seconds
• Hence, to fold 1 napkin, Karishma takes $$\frac{53}{H}$$ seconds
Therefore, to fold B napkins, Karishma takes $$\frac{53}{H} * B$$ seconds = $$\frac{53B}{H}$$ seconds

Hence, the correct answer is option A.

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29 Sep 2018, 21:39
Harshgmat wrote:
If Karishma spends 53 seconds folding H napkins, how many seconds will it take her to fold B napkins at the same rate?

A) 53B/H

B) HB/53

C) 53/HB

D) 53H/B

E) 53HB

The trap in this question: $$B$$ napkins = Work
Work needs to be divided by rate in order to get time

How many seconds to fold $$B$$ napkins?

Rate: $$\frac{H.napkins}{53secs}=\frac{H}{53}$$

Work: $$B$$ napkins

$$r*t=W$$
$$\frac{H}{53}*t=B$$
Time, $$t=\frac{B}{(\frac{H}{53})}=(B*\frac{53}{H})=\frac{53B}{H}$$

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30 Sep 2018, 02:18
Hi Everyone,

I solved the question in a different way. I feel that it is a quicker approach, so thought of sharing the approach.

There are 2 variables (H,B) in the question. Lets assume some values to both H and B , so that we can calculate it more easily.

Let H = 10 and B =20. (We can assume any number , it will lead to the same answer. Here I am taking 10 and 20 for making the calculations easy)

Now we will workout the problem with the information given in the question stem.
The question stem says Karishma takes 53 second to fold H napkins . Substituting H we get :-

10 napkins =53 secs
1 napkin = 53/10
B napkins = (53/10)*B
we have assumed B = 20, substituting the value of B in the above equation, we will get :-
20 napkins = (53/10)*20=106

Now we can substitute the values of B and H in the answer choices and the answer choice which provides the same result(106) as above will be the correct answer.

(A) - > 53B/H= (53*20)/10=106 .

which is same as the answer we got above.
Therefore Option A is correct answer.
If Karishma spends 53 seconds folding H napkins, how many seconds   [#permalink] 30 Sep 2018, 02:18
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# If Karishma spends 53 seconds folding H napkins, how many seconds

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# Feature extraction for time series classification

I have electric signals represented as 2(internal)d data. Each data point is in the form (<timestamp>, <power_value>). I want to be able to differentiate 4 different states that a window of the signal can belong to.

To give you a better understanding here is an example of how the signals look like

Here we can see all 4 different states represented.

• Off - it's when the signal is completely stationary and its amplitude is bellow the standard deviation in this case. You can see a small part a little bit before the end of the plot

• Idling - when the signal is stationary but its amplitude is close to the mean level. Like in the beginning of the plot.

• Stand-by - Areas where the signal has some noticeable variation like right after the second spike.

• On - Areas with a lot of variation and high frequency

I want to be able to determine in which of those states a specified length window (the length is something that must be discussed too) belongs to.

What I have done so far is quite simple and a bit amateurish. I trained a Random Forest using just two features, the mean of a window and the variance of a window. In order to run the classification I apply the classification on a sliding window and each time I label the first data point.

I started off with that because at first the classification had to be able to differentiate just 3 states (off, idling, on) which even with those simple statistical features is very easy. I could even as well right a simple if-then-else instead of training decision trees and the results would be as accurate because those three states have very distinct values for those features. But with adding a fourth state the classifier misfires a lot. And that's normal. It might not be very well depicted in that picture above but Stand-By and On, in particular areas, have a bit common variance and averaged values.

I am not interested in going for a pattern matching approach, although that could solve all those problems robustly but that would also mean that I would have to create sets of reference patterns which gives rise to many different implications.

What I am looking for is a set a features that will enable me to efficiently differentiate those states. First do you think that is even possible?

My understanding tells me that maybe a set of frequency domain features could improve the accuracy. I have only used such features (MFCC) once though, for audio classification, but that doesn't apply to this case.

If you think that this is possible, can you recommend a set of features that it would make sense to try out, or point me to a study that you think is well suited for this case?

update

After MikeP's suggestion I created a spectogram using Octave's specgram with the default arguments

Here is an instance of the signal

and here is the spectogram

update2

After some days working on the problem I managed to make some progress. I experimented with different sets of features and added a few new ones.

What I did was to extend my statistical features by adding the skewness and the kurtosis. I also moved towards a more pattern matching approach by adding dtw distances from reference patters (one of each state). That last step improved the results on the sliding window classification quite a bit.

My problem still lies on the sliding technique I use for the classification, since every time the window falls into intermediate (also unlabeled) cases the classifier has a hard time concluding on what is that it is seeing.

What I am trying to do now is to use SVM instead of RandomForests with a polynomial kernel (degree 4). I am using the scikit-learn implementation. The problem is that the classifier is always categorizing the data as state_3. I think I am imposing some kind of bias, but I don't know why and how. Do you have any ideas what might be going wrong?

To get an idea how the labeling looks, here are some labeled instances of the signal

update3

Instead of polynomial kernel, I used a linear kernel for the SVM, and the classifier worked correctly. Actually I got the best results so far. I might have to work extra on making the whole classification pipeline more efficient because svm's and the extra feature extraction is making everything very slow

• Can you create a spectrogram of the data? If it's in matlab, that should be very easy. By visualizing the amplitude over frequency, the features might become much clearer. – MikeP Mar 4 '16 at 15:32
• I updated the question accordingly – LetsPlayYahtzee Mar 4 '16 at 18:36
• Thanks!!! You might actually find fewer frequency bins (smaller fft size) better for classification. It 'looks' like the yellow vs blue is a little more intense midway up the frequency range during on, versus standby - and way more dense than off. – MikeP Mar 4 '16 at 19:11
• A numerical approach such as Fast Fourier Transforms (FFT) may help. FFTs are widely used to decompose signals like this. There are a variety of features that would be included such as frequency, amplitude, density, etc. Here's a Wiki link: en.wikipedia.org/wiki/Fast_Fourier_transform – Mike Hunter Mar 9 '16 at 14:24
• I tried to use the spectral centroid but it didn't increase the accuracy. But again maybe it did. As you can see from the last image basically all four states can be predicted just by the mean and var. With using the sliding window for applying on-line classification I am looking for features that will make the classifier be less prune to rapid changes. Adding features that increase that accuracy made the on-line classifer misfire a lot, probably due to over-fitting. But again, I didn't try out freq amp and density so I will do, thanks – LetsPlayYahtzee Mar 9 '16 at 14:31


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# The SErial Killer (Part 2)

This puzzle is part of a series - part 1 is here. Any comments I make on the OP are canonical and out of character (OOC). Any comments I post on answers are for flavour only.

Update: As it turns out, this was a very easy puzzle, and I'm actually having to rewrite the plot of the series thanks to SKV being far too clever for his own bleeding good (I love you really). The glory's gone, but try to solve it without checking out the answers. I enjoyed making it, anyway :-) Part 3 will be tougher.

You're staring at the list of Stacks when out of nowhere, you hear a voice going rik tika tik tik tik tik and before you wonder where it came from you realise something - the capital letters in the titles of all the dead sites can be re-arranged to spell "Moderators"!

You cast your eyes over the the list of murdered sites to find the Moderators Stack, and in your best Scottish-detective voice you announce "Och! There's bin a murrrrrrder, and ah nae where ahm going tae look!" (OOC: Sorry, Scotland). Tentatively, you click through to the Moderators stack and right at the top, you see a post entitled "We Will Conquer". You recognise the title from one of the dead Stacks, so you click on to read it...

You gasp in disbelief and check back to the Stacks list - sure enough, Ask Ubuntu is dead. You're about to click on it to make sure, but then remember the ominous warning about entering a new Stack, and think better of it. This "Handel" character doesn't seem like someone to mess around with.

You realise you're suddenly on a countdown, trying to minimise the number of stacks that die, and every false step you make could bring a flood of confused newbies asking inane questions about anything on to your beloved Puzzling.SE... or even worse, Puzzling.SE might get killed off!

Baffled and starting to panic a little, you sit back to mull things over. What's your next move? Where should you look?

• Goodness. Where do you learn to make the graphic? This is madly cool. – Realdeo Nov 12 '14 at 15:10
• @Realdeo it's a combination of Chrome's inspector and Photoshop. My day job is web development :-) – Joe Nov 12 '14 at 15:10
• What's your next move? flag it for OT...? – Bolu Nov 12 '14 at 15:18
• @Joe I guess the name Handel has nothing to do with DoorKnob :P – skv Nov 12 '14 at 15:37
• @skv Man, everything about this part has been a flop lol. I thought that was a nice disguise that I might be able to play out for a big reveal. Good news DoorKnob - you're no longer unwittingly playing the main villain in the series. Back to the drawing board... – Joe Nov 12 '14 at 15:41

English language & usage. The ampersand in the logo has hex color 90383b, which is the roman numeral in the tag.

• Solving a puzzle in 18 minutes. Impressive. – Realdeo Nov 12 '14 at 15:36
• You take a deep breath and settle yourself. In a moment of brilliance you realise that $\overline{XC}CCCLXXXIII$ is the Roman numeral for 90383, and the b on the end forms a hexadecimal colour. You quickly map this to the EL&U Stack - it's an exact match! You click on it... – Joe Nov 12 '14 at 15:37
• (OOC) Well darn it. I actually had an easier version of this part originally and decided it needed to be a bit more challenging. However, 8 minutes isn't much of a challenge. Expect part 3 to test you a bit more. Well done, though – Joe Nov 12 '14 at 15:38
• More luck than anything else really... A few good hits on google and I happened to have paint.net open. Also wanted to go for the reddish background of seasoned advice first - then decided to double-check the color. Then kept color-picking everything on the screenshot that was remotely red... – EagleV_Attnam Nov 12 '14 at 15:48
• wow! This would have taken me weeks to find out – markasoftware Nov 13 '14 at 1:37

Obviously the answer is already posted; but this solution seemed valid to me as well:

Server Fault is the next target. No self respecting serial killer would leave the clues from Part 1 without having the post for Part 2 already up, and yet the timestamp on the post is from 11/12! Clearly, the incorrect NTP configuration is the indication of the next target.

• (OOC) Real world time is different to puzzle time :-) I'm trying to have the puzzle feel like it's moving at a reasonably fast pace, but obviously it might take one or more days for each part. Glad people are enjoying it though :-) – Joe Nov 12 '14 at 18:41

programmers. purely because of "look sharp" which is reference to the type of coding used c#

• You decide that the Programmers Stack is a good place to look, but wary of killing Stacks unnecessarily you decide to Google a little bit first. You find that there's also an F# programming language and that musical notation uses sharp signs. You don't think you can justify risking a Stack on such a flimsy basis – Joe Nov 12 '14 at 15:29
• Took a stab and got stabbed :) – Oblongamous Nov 12 '14 at 16:04

What roman numerals, Julius Caesar (We see, We saw, god conquered) and George Friderich is known for?

Italian

So I visit Italian Language.

• You move your mouse towards the Italian Language stack, but just before you click you suddenly remember that actually, Handel was German, not Italian. Doubting yourself, you decide that "We will conquer" is a bit too tentative to link it with Caesar's Veni Vidi Vici which is actually Latin. You move the mouse away, and go back to thinking. – Joe Nov 12 '14 at 15:32
• He's a composer of Italian Baroque! – Realdeo Nov 12 '14 at 15:32
• (OOC) That's far too tenuous :P If I set that as a clue, I'd get so many complaints about being vague and obscure. That said, I may need to be a touch more vague and obscure in future parts of the series... – Joe Nov 12 '14 at 15:34
• I know. I know =p. That was a troll comment. – Realdeo Nov 12 '14 at 15:35


text:
# Euler-Mascheroni constant

I proved that $$\int_{1}^\infty \left(\frac{1}{\lfloor x\rfloor}-\frac{1}{x}\right)\mathrm dx=\gamma.$$

I now have to deduce that $$1-\int_1^\infty \frac{x-\lfloor x\rfloor}{x^2}\mathrm dx=\gamma.$$

So what I want to prove is that $$\int_1^\infty \frac{x-\lfloor x\rfloor}{x^2}\mathrm dx=1-\gamma=1-\int_1^\infty \left(\frac{1}{\lfloor x\rfloor}-\frac{1}{x}\right)\mathrm dx=1-\int_1^\infty \frac{x-\lfloor x\rfloor}{x\lfloor x\rfloor}\mathrm dx.$$

To do this, I use the fact that $$1=\int_1^\infty \frac{1}{x^2}\mathrm dx$$ and thus, I get that $$1-\int_1^\infty \frac{x-\lfloor x\rfloor}{x\lfloor x\rfloor}\mathrm dx=\int_1^\infty \left(\frac{1}{x^2}+\frac{\lfloor x\rfloor-x}{x\lfloor x\rfloor}\right)\mathrm dx.$$

But I don't see how to deduce that $$\int_1^\infty \left(\frac{1}{x^2}+\frac{\lfloor x\rfloor-x}{x\lfloor x\rfloor}\right)\mathrm dx=\int_1^\infty \frac{x-\lfloor x\rfloor}{x^2}\mathrm dx.$$

I tried many manipulation, but it wasn't conclusif.

The $1/x$ terms cancel, so what you want to prove is
$$\int_1^\infty\left(\frac1{\lfloor x\rfloor}-\frac{\lfloor x\rfloor}{x^2}\right)\mathrm dx=1\;.$$
$$\left[\frac{\lfloor x\rfloor}x\right]_{1\uparrow}^\infty+\int_{1\uparrow}^\infty\left(\frac1{\lfloor x\rfloor}-\frac{\lfloor x\rfloor'}x\right)\mathrm dx=1\;.$$
The first term is $1$, and in the integral both terms run through the positive integers, so they cancel.


text:
## Precalculus (6th Edition) Blitzer

The required value of the $\sin \frac{x}{2}$ is $\pm \sqrt{\frac{1-\cos x}{2}}$ .
In order to find the value of $\sin \frac{x}{2}$, the half angle formula is used that is as given below: ${{\sin }^{2}}\alpha =\frac{1-\cos 2\alpha }{2}$ Now, replace $\alpha$ with $\frac{x}{2}$ \begin{align} & {{\sin }^{2}}\frac{x}{2}=\frac{1-\cos 2\left( \frac{x}{2} \right)}{2} \\ & =\frac{1-\cos x}{2} \\ & =\pm \sqrt{\frac{1-\cos x}{2}} \end{align}


text:
# How to find a confidence interval for the total number of events

I have detector which will detect an event with some probability p. If the detector says that an event occured, then that is always the case, so there are not false-positives. After I run it for some time, I get k events detected. I would like to calculate what the total number of events that occured was, detected or otherwise, with some confidence, say 95%.

So for example, let's say I get 13 events detected. I would like to be able to calculate that there were between 13 and 19 events with 95% confidence based on p.

Here's what I've tried so far:

The probability of detecting k events if there were n total is:

binomial(n, k) * p^k * (1 - p)^(n - k)

The sum of that over n from k to infinity is:

1/p

Which means, that the probability of there being n events total is:

f(n) = binomial(n, k) * p^(k + 1) * (1 - p)^(n - k)

So if I want to be 95% sure I should find the first partial sum f(k) + f(k+1) + f(k+2) ... + f(k+m) which is at least 0.95 and the answer is [k, k+m]. Is this the correct approach? Also is there a closed formula for the answer?

-

## 4 Answers

I would choose to use the negative binomial distribution, which returns the probability that there will be X failures before the k_th success, when the constant probability of a success is p.

Using an example

k=17 # number of successes
p=.6 # constant probability of success

the mean and sd for the failures are given by

mean.X <- k*(1-p)/p
sd.X <- sqrt(k*(1-p)/p^2)

The distribution of the failures X, will have approximately that shape

plot(dnbinom(0:(mean.X + 3 * sd.X),k,p),type='l')

So, the number of failures will be (with 95% confidence) approximately between

qnbinom(.025,k,p)
[1] 4

and

qnbinom(.975,k,p)
[1] 21

So you inerval would be [k+qnbinom(.025,k,p),k+qnbinom(.975,k,p)] (using the example's numbers [21,38] )

-

I think you misunderstood the purpose of confidence intervals. Confidence intervals allow you to assess where the true value of the parameter is located. So, in your case, you can construct a confidence interval for $p$. It does not make sense to construct an interval for the data.

Having said that, once you have an estimate of $p$ you can calculate the probability that you will observe different realizations such as 14, 15 etc using the binomial pdf.

-
 Well I already know p. I also know the amount of events detected: k. So the total events is somewhere around k/p. I would like to find out an interval around k/p so I can be say 95% sure that the total number of events is inside it. Does that make more sense? – Statec Aug 5 '10 at 12:24 I believe the OP is trying to compute an interval for N in binomial sampling, where p is known. It makes sense to try to do that. – Glen_b May 24 '12 at 9:11

If you measure $k$ events and know your detection efficiency is $p$ you can automatically correct your measured result up to the "true" count $k_\mathrm{true} = k/p$.

Your question is then about finding the range of $k_\mathrm{true}$ where 95% of the observations will fall. You can use the Feldman-Cousins method to estimate this interval. If you have access to ROOT there is a class to do this calculation for you.

You would calculate the upper and lower limits with Feldman-Cousins from the uncorrected number of events $k$ and then scale them up to 100% with $1/p$. This way the actual number of measurements determines your uncertainty, not some scaled number that wasn't measured.

{
gSystem->Load("libPhysics");

const double lvl = 0.95;
TFeldmanCousins f(lvl);

const double p = 0.95;
const double k = 13;
const double k_true = k/p;

const double k_bg = 0;

const double upper = f.CalculateUperLimit(k, k_bg) / p;
const double lower = f.GetLowerLimit() / p;

std::cout << "["
lower <<"..."<<
k_true <<"..."<<
upper <<
"]" << std::endl;
}

-
 Thanks, that looks great. I think this is the answer I was looking for. – Statec Aug 6 '10 at 6:18 +1 thanks for the link to the Feldman/Cousins paper. – ars Aug 12 '10 at 0:08

Assuming you want to pick a distribution for n, p(n) you can apply Bayes law.

You know that the probability of k events occuring given that n have actually occured is governed by a binomial distribtion

$p(k|n) = {n \choose k} p^k (1-p)^{(n-k)}$

The thing you really want to know is the probability of n events having actually occured, given that you observed k. By Bayes lay:

$p(n|k) = \frac{p(k|n)p(n)}{p(k)}$

By applying the theorem of total probability, we can write:

$p(n|k) = \frac{p(k|n)p(n)}{\sum_{n'} p(k|n')p(n')}$

So without further information, about the distribution of $p(n)$ you can't really go any further.

However, if you want to pick a distribution for $p(n)$ for which there is a value $n$ greater than which $p(n) = 0$, or sufficiently close to zero, then you can do a bit better. For example, assume that the distribution of $n$ is uniform in the range $[0,n_{max}]$. this case:

$p(n) = \frac{1}{n_{max}}$

The Bayesian formulation simplifies to:

$p(n|k) = \frac{p(k|n)}{\sum_{n'} p(k|n')}$

As for the final part of the problem, I agree that the best approach is to perform a cumulative summation over $p(n|k)$, to generate the cummulative probability distribution function, and iterate until the 0.95 limit is reached.

Given that this question migrated from SO, toy sample code in python is attached below

import numpy.random

p = 0.8
nmax = 200

def factorial(n):
if n == 0:
return 1
return reduce( lambda a,b : a*b, xrange(1,n+1), 1 )

def ncr(n,r):
return factorial(n) / (factorial(r) * factorial(n-r))

def binomProbability(n, k, p):
p1 = ncr(n,k)
p2 = p**k
p3 = (1-p)**(n-k)
return p1*p2*p3

def posterior( n, k, p ):
def p_k_given_n( n, k ):
return binomProbability(n, k, p)
def p_n( n ):
return 1./nmax
def p_k( k ):
return sum( [ p_n(nd)*p_k_given_n(nd,k) for nd in range(k,nmax) ] )
return (p_k_given_n(n,k) * p_n(n)) / p_k(k)

observed_k   = 80
p_n_given_k  = [ posterior( n, observed_k, p ) for n in range(0,nmax) ]
cp_n_given_k = numpy.cumsum(p_n_given_k)
for n in xrange(0,nmax):
print n, p_n_given_k[n], cp_n_given_k[n]

-


text:
# [OS X TeX] Dental problem

Alain Schremmer Schremmer.Alain at verizon.net
Fri Feb 18 16:54:38 CET 2005

I have tried to insert @{} in various places but with no luck.

Below is an example (without the body text even the first two tables do
not line up) to show why I would like it to work. Given that this is not
for print but for uploading under GPL it does not really matter but I
think I am beginning to catch the "typesetting" bug.

Regards
--schremmer

\documentclass[11pt]{book}
\usepackage{paralist}
\setdefaultleftmargin{5mm}{6mm}{7mm}{8mm}{}{}

\begin{document}
This is body text text text text text text text text text text text text
text text text text

\noindent
\setlength\tabcolsep{5.5pt}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Cleveland
& Franklins
& Hamiltons
& Washingtons
& Dimes
& Cents
& Mills
\\
\hline
\end{tabular}
\begin{asparaenum}
\def\indent{\hglue 5.3mm}

\item%1
This is body text within asparaenum

\noindent
\setlength\tabcolsep{5.5pt}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Cleveland
& Franklins
& Hamiltons
& Washingtons
& Dimes
& Cents
& Mills
\\
\hline
\end{tabular}
\end{asparaenum}%1, 2, 3.
\begin{compactitem}
\item
This is body text within compactitem

\noindent
\setlength\tabcolsep{5.5pt}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Cleveland
& Franklins
& Hamiltons
& Washingtons
& Dimes
& Cents
& Mills
\\
\hline
\end{tabular}
\end{compactitem}
\end{document}

Piet van Oostrum wrote:

>>PD> I think I've never seen an un-indented tabular environment. I usually set
>>PD> \hsize*{-5mm} or such -- except someone knows the exact name ...
>>
>>
>
>It is half of the intercolumn space (\tabcolsep), which can be suppressed
>by putting @{} in front of the column spec.
>
>
--------------------- Info ---------------------
Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
& FAQ: http://latex.yauh.de/faq/
TeX FAQ: http://www.tex.ac.uk/faq
List Post: <mailto:MacOSX-TeX at email.esm.psu.edu>


text:
Documentation

# nlgreyest

Estimate nonlinear grey-box model parameters

## Description

example

sys= nlgreyest(data,init_sys) estimates the parameters of a nonlinear grey-box model, init_sys, using time-domain data, data.

example

sys= nlgreyest(data,init_sys,options) specifies additional model estimation options.

## Examples

collapse all

z = iddata(y,u,0.2,'Name','Two tanks');

The data contains 3000 input-output data samples of a two tank system. The input is the voltage applied to a pump, and the output is the liquid level of the lower tank.

Specify file describing the model structure for a two-tank system. The file specifies the state derivatives and model outputs as a function of time, states, inputs, and model parameters.

FileName = 'twotanks_c';

Specify model orders [ny nu nx].

Order = [1 1 2];

Specify initial parameters (Np = 6).

Parameters = {0.5;0.0035;0.019; ...
9.81;0.25;0.016};

Specify initial initial states.

InitialStates = [0;0.1];

Specify as continuous system.

Ts = 0;

Create idnlgrey model object.

nlgr = idnlgrey(FileName,Order,Parameters,InitialStates,Ts, ...
'Name','Two tanks');

Set some parameters as constant.

nlgr.Parameters(1).Fixed = true;
nlgr.Parameters(4).Fixed = true;
nlgr.Parameters(5).Fixed = true;

Estimate the model parameters.

nlgr = nlgreyest(z,nlgr);

Create estimation option set for nlgreyest to view estimation progress, and to set the maximum iteration steps to 50.

opt = nlgreyestOptions;
opt.Display = 'on';
opt.SearchOptions.MaxIterations = 50;

z = iddata(y,u,0.1,'Name','DC-motor');

The data is from a linear DC motor with one input (voltage), and two outputs (angular position and angular velocity). The structure of the model is specified by dcmotor_m.m file.

Create a nonlinear grey-box model.

file_name = 'dcmotor_m';
Order = [2 1 2];
Parameters = [1;0.28];
InitialStates = [0;0];

init_sys = idnlgrey(file_name,Order,Parameters,InitialStates,0, ...
'Name','DC-motor');

Estimate the model parameters using the estimation options.

sys = nlgreyest(z,init_sys,opt);

## Input Arguments

collapse all

Time-domain estimation data, specified as an iddata object. data has the same input and output dimensions as init_sys.

If you specify the InterSample property of data as 'bl'(band-limited) and the model is continuous-time, the software treats data as first-order-hold (foh) interpolated for estimation.

Constructed nonlinear grey-box model that configures the initial parameterization of sys, specified as an idnlgrey object. init_sys has the same input and output dimensions as data. Create init_sys using idnlgrey.

Estimation options for nonlinear grey-box model identification, specified as an nlgreyestOptions option set.

## Output Arguments

collapse all

Nonlinear grey-box model with the same structure as init_sys, returned as an idnlgrey object. The parameters of sys are estimated such that the response of sys matches the output signal in the estimation data.

Information about the estimation results and options used is stored in the Report property of the model. Report has the following fields:

Report FieldDescription
Status

Summary of the model status, which indicates whether the model was created by construction or obtained by estimation.

Method

Name of the simulation solver and the search method used during estimation.

Fit

Quantitative assessment of the estimation, returned as a structure. See Loss Function and Model Quality Metrics for more information on these quality metrics. The structure has the following fields:

FieldDescription
FitPercent

Normalized root mean squared error (NRMSE) measure of how well the response of the model fits the estimation data, expressed as a percentage.

LossFcn

Value of the loss function when the estimation completes.

MSE

Mean squared error (MSE) measure of how well the response of the model fits the estimation data.

FPE

Final prediction error for the model.

AIC

Raw Akaike Information Criteria (AIC) measure of model quality.

AICc

Small sample-size corrected AIC.

nAIC

Normalized AIC.

BIC

Bayesian Information Criteria (BIC).

Parameters

Estimated values of the model parameters. Structure with the following fields:

FieldDescription
InitialValuesStructure with values of parameters and initial states before estimation.
ParVectorValue of parameters after estimation.
Free

Logical vector specifying the fixed or free status of parameters during estimation

FreeParCovarianceCovariance of the free parameters.
X0Value of initial states after estimation.
X0CovarianceCovariance of the initial states.

OptionsUsed

Option set used for estimation. If no custom options were configured, this is a set of default options. See nlgreyestOptions for more information.

RandState

State of the random number stream at the start of estimation. Empty, [], if randomization was not used during estimation. For more information, see rng in the MATLAB® documentation.

DataUsed

Attributes of the data used for estimation — Structure with the following fields:

FieldDescription
Name

Name of the data set.

Type

Data type — For idnlgrey models, this is set to 'Time domain data'.

Length

Number of data samples.

Ts

Sample time. This is equivalent to data.Ts.

InterSample

Input intersample behavior. One of the following values:

• 'zoh' — Zero-order hold maintains a piecewise-constant input signal between samples.

• 'foh' — First-order hold maintains a piecewise-linear input signal between samples.

• 'bl' — Band-limited behavior specifies that the continuous-time input signal has zero power above the Nyquist frequency.

The value of Intersample has no effect on estimation results for discrete-time models.

InputOffset

Empty, [], for nonlinear estimation methods.

OutputOffset

Empty, [], for nonlinear estimation methods.

Termination

Termination conditions for the iterative search used for prediction error minimization. Structure with the following fields:

FieldDescription
WhyStop

Reason for terminating the numerical search.

Iterations

Number of search iterations performed by the estimation algorithm.

FirstOrderOptimality

$\infty$-norm of the gradient search vector when the search algorithm terminates.

FcnCount

Number of times the objective function was called.

UpdateNorm

Norm of the gradient search vector in the last iteration. Omitted when the search method is 'lsqnonlin' or 'fmincon'.

LastImprovement

Criterion improvement in the last iteration, expressed as a percentage. Omitted when the search method is 'lsqnonlin' or 'fmincon'.

Algorithm

Algorithm used by 'lsqnonlin' or 'fmincon' search method. Omitted when other search methods are used.

For estimation methods that do not require numerical search optimization, the Termination field is omitted.


text:
## Engineering Mechanics: Statics & Dynamics (14th Edition)

$T=2\pi\sqrt{\frac{mK_{\circ}^2}{C}}$
We can determine the natural period of oscillation as follows: $\omega_n=\sqrt{\frac{K_{eq}}{m_{eq}}}$ $\implies \omega_n=\sqrt{\frac{C}{mK_{\circ}^2}}$ Now, $T=\frac{2\pi}{\omega_n}$ Thus: $\implies T=2\pi\sqrt{\frac{mK_{\circ}^2}{C}}$


text:
## 10.3 Variable selection property of the lasso

You may ask why lasso ($$L_1$$ penalty) has the feature selection property but not the ridge ($$L_2$$ penalty). To answer that question, let’s look at the alternative representations of the optimization problem for lasso and ridge. For lasso regression, it is identical to optimize the following two functions:

$$$\Sigma_{i=1}^{n}(y_{i}-\beta_{0}-\Sigma_{j=1}^{p}\beta_{j}x_{ij})^{2}+\lambda\Sigma_{j=1}^{p}|\beta_{j}|=RSS+\lambda\Sigma_{j=1}^{p}|\beta_{j}| \tag{10.3}$$$

$$$\underset{\beta}{min}\left\{ \Sigma_{i=1}^{n}\left(y_{i}-\beta_{0}-\Sigma_{j=1}^{p}\beta_{j}x_{ij}\right)^{2}\right\} ,\ \Sigma_{j=1}^{p}|\beta_{j}|\leq s \tag{10.4}$$$

For any value of tuning parameter $$\lambda$$, there exists a $$s$$ such that the coefficient estimates optimize equation (10.3) also optimize equation (10.4). Similarly, for ridge regression, the two representations are identical:

$$$\Sigma_{i=1}^{n}(y_{i}-\beta_{0}-\Sigma_{j=1}^{p}\beta_{j}x_{ij})^{2}+\lambda\Sigma_{j=1}^{p}\beta_{j}^{2}=RSS+\lambda\Sigma_{j=1}^{p}\beta_{j}^{2} \tag{10.5}$$$

$$$\underset{\beta}{min}\left\{ \Sigma_{i=1}^{n}\left(y_{i}-\beta_{0}-\Sigma_{j=1}^{p}\beta_{j}x_{ij}\right)^{2}\right\} ,\ \Sigma_{j=1}^{p}\beta_{j}^{2}\leq s \tag{10.6}$$$

When $$p＝2$$, lasso estimates ($$\hat{\beta}_{lasso}$$ in figure 10.3) have the smallest RSS among all points that satisfy $$|\beta*{1}|+|\beta_{2}|\leq s$$ (i.e. within the diamond in figure 10.3). Ridge estimates have the smallest RSS among all points that satisfy $$\beta_{1}^{2}+\beta_{2}^{2}\leq s$$ (i.e. within the circle in figure 10.3). As s increases, the diamond and circle regions expand and get less restrictive. If s is large enough, the restrictive region will cover the least squares estimate ($$\hat{\beta}_{LSE}$$). Then, equations (10.4) and (10.6) will simply yield the least squares estimate. In contrast, if s is small, the grey region in figure 10.3 will be small and hence restrict the magnitude of $$\beta$$. With the alternative formulations, we can answer the question of feature selection property.

Figure 10.3 illustrates the situations for lasso (left) and ridge (right). The least square estimate is marked as $$\hat{\beta}$$. The restrictive regions are in grey, the diamond region is for the lasso, the circle region is for the ridge. The least square estimates lie outside the grey region, so they are different from the lasso and ridge estimates. The ellipses centered around $$\hat{\beta}$$ represent contours of RSS. All the points on a given ellipse share an RSS. As the ellipses expand, the RSS increases. The lasso and ridge estimates are the first points at which an ellipse contracts the grey region. Since ridge regression has a circular restrictive region that doesn’t have a sharp point, the intersecting point can’t drop on the axis. But it is possible for lasso since it has corners at each of the axes. When the intersecting point is on an axis, one of the parameter estimates is 0. If p > 2, the restrictive regions become sphere or hypersphere. In that case, when the intersecting point drops on an axis, multiple coefficient estimates can equal 0 simultaneously.


text:
## Saturday, November 12, 2016

What dice you need; whose dice to buy.

Knowing most readers would sooner give advice than receive it, we state our opinions softly.

## Which Dice You Need

The original edition of the game called for two sets of polyhedral dice and 4 to 20 six-sided dice. Polyhedral dice aren't hard to get these days, so no reason to limit yourself to two sets. Also, modern gamers will want percentile dice.

As a general principle, our group likes to resolve a character's fate with a single cast of the dice. If you find yourself rolling a die repeatedly and summing the numbers in your head, you don't have enough dice! Or so we think.

Another group custom is we expect the host to provide dice for everybody. We stopped carrying dice bags around almost thirty years ago. Possibly related: Jay sometimes calls other players "dice bags" when he doesn't like their style of play.

With that said, here is what I set out for the players:

For the record—and for the search engine—the image shows eighteen d6 with pips, two d100, three d4, three d6 with digits, three d8, three d12, and three d20.

I also keep a set of polyhedral dice behind the screen for sealing the players' fate in secret.

## Whose Dice to Buy

### Chessex

My fave manufacturer right now is Chessex, founded in 1987. They sell dice in sets of seven like the one above.

Chessex sells dice with swirls and speckles. The speckled dice were introduced in 1995. They also sell transparent dice. None of these are to my taste. Sometimes I have to poke around to find the solid-colored, opaque dice with high-contrast paint that I like. I want dice that are easy to read.

Chessex dice are made from a hard thermoplastic. They are cast from molds, painted, and then polished in a rock tumbler. The polishing removes paint from everywhere except the indented numerals. The polishing also rounds the edges.

By the way, note the distinct way Chessex numbers the d4. Not many manufacturers use this style.

There are some other traits that can be used to recognize Chessex dice. One is that opposite sides of a Chessex die sum to a consistent number. In the case of a d20, opposite sides always sum to 21.

If you take a Chessex d20 and look at the 1-face, it shares edges with the 7-face, 19-face, and 13-face. I'm not sure if Chessex dice have always had this arrangement, but molds are expensive, so manufacturers don't change them often.

Finally, I'll mention some traits which Chessex dice have and which in my mind all good dice should have.

6s and the 9s should be distinguished. Chessex dice use underlines. However, the underline on the Chessex d20 is so thin it doesn't take paint well.

The d20 should be numbered from 1 to 20, and not 0 to 9 twice.

In a set of percentile dice, the tens die should be distinguished from the ones die with an extra zero digit on each face.

### Koplow

I have several sets of Koplow dice. Koplow used to sell sets where each die was a different color. I like these sets because they make it easier to find the die I want.

The dice in the above picture were sold as two sets: a set of 6 polyhedral dice, and a set of 4 "value dice".

Koplow dice are manufactured the same way as Chessex dice. The edges are more rounded than Chessex dice.

Koplow Games was founded in 1974. They make more than dice and I'm not sure when they started making polyhedral dice—I bought a few sets in 2005.

Traits for recognizing Koplow dice: opposite faces on Koplow dice don't always sum to the same value. In my set, the d6 and the d10 do, but the rest of the dice don't. Oddly, among the "value dice", the ones die does, but the the tens, hundreds, and thousands dice don't.

On a Koplow d20, the 1-face shares edges with the 6-face, 13-face, and 10-face.

6s and 9s are underlined, except on the d20, where a lower right dot is used.

I bought the "value dice" because I wanted the percentile dice. The ones, tens, and hundreds dice together make a set of per mille dice. Is such a set useful? The Treasury of Archaic Names contains a table with 1000 entries, but most players will find both the hundreds and thousands die to be superfluous.

## Collecting Dice

I'll mention a few more manufacturers. Most of these are no longer in business, but good to know about if you are interested in old dice.

### Creative Publications

Creative Publications started selling polyhedral dice in 1972. They were the only supplier of polyhedral dice—at least in the United States—when the original edition of the game came out in 1974. TSR resold them at prices ranging from \$1.49 to \$3.00. The dice were included with the Holmes Basic set. In fact, most dice were probably sourced by TSR directly from manufacturers in Taiwan Hong Kong.

The Creative Publications dice were made from a soft plastic that didn't wear well.  The dice would lose their color and the d20 would become nearly spherical. Players took to calling them "low impact" dice when harder dice became available.

The Creative Publications d20 was numbered 0 to 9 twice. Opposite faces have the same number. The 1-face shares edges with the 5-face, 9-face, and 6-face. Creative Publications dice are easy to identify because the same color was always used for each shape.

Creative Publications was located in Palo Alto, CA. They became a supplier of dice to gamers by accident, as they were primarily a publisher of educational titles. One of these is "Polyhedra Dice Games: For Grades K to 6" from 1978. The book contains 40 games, each of which requires one of the polyhedral dice for play. Each game requires a skill such as counting, adding, or distinguishing odd from even. Otherwise they are simple games of chance and not interesting to adults. The cover shows a sea of dice from the collection of Dale Seymour; a few of the Creative Publications dice are mixed in.
TSR also sold sets of two d20s numbered 0 to 9 twice. They were used for percentile rolls—the pink die was the tens digit. I'm not sure if Creative Publications ever sold sets with a pink die in them.

### Gamescience

Gamescience has been run by Lou Zocchi since 1973. They are still in the dice-making business. Most of their dice are sold unpainted; an ultra fine Sharpie works well for inking in the numbers.

Gamescience dice are not polished, so they may have burrs left over from the molding process. The molding industry calls these burrs flash. Some people sand them down with fine sandpaper.

Gamescience made a name for itself by introducing dice made out of harder materials. In fact, if I'm not mistaken, they popularized the terms "high impact dice" and "low impact dice". They also introduced dice in a number of new shapes, the most important being the d10, which they starting manufacturing in 1980. Update: In a 2015 interview, Lou Zocchi credited TSR with introducing the first d10 circa 1980. These were the dice in the Moldvay basic box set. Zocchi also says these were the first sets with a d20 numbered 1 to 20. Gamescience was the first company to introduce a d10 numbered 00, 10, ..., 90 for the tens digit in a percentile role. The die was advertised in the April 1990 issue of Dragon magazine:

Gamescience was an early distributor of TSR products. Zocchi and Gygax were acquainted; both wrote for Guidon Games. As a reseller of TSR games, Zocchi started ordering dice from Creative Publications. Zocchi was dissatisfied with the price and the availability, so he started making his own dice—initially just a d20—in 1974.

The original Gamescience d20 was numbered 0 to 9 twice. Early gamers would ink half of the numbers with a different color so a number from 1 to 20 could be generated. In 1980 Gamescience invested in a mold where half of the faces had + next to the numbers. The dice with + signs aren't common; d20 dice numbered 1 to 20 became the standard soon after.

Some traits useful for identifying Gamescience dice: the d4 are truncated tetrahedra. The d20 has a small capital G on the 1-face in addition to the numeral 1. The d20 1-face shares edges with the 11-face, the 19-face, and the 18-face.

Opposite faces on the Gamescience d20 sum to 21, and opposite faces on the d6 sum to 7. The other dice don't have faces that sum to a consistent number.

### The Armory

The Armory was a gaming shop in Baltimore, better known as a dealer in miniatures. I've read that the Armory got started in the dice business by inking or painting Gamescience dice and re-selling them. This was back in the 1970s.
Eventually they had their own molds made. The earliest Armory dice used the letter A instead of the numeral 1 on the 1-face of the d4, d8, and d20. As far as I know this was not done on the d6 and the d12.
The Armory introduced a d30 die around 1982. The older d30 is numbered 0 to 9 thrice, each digit appearing with a plus sign, minus sign, or no modifier. The newer d30 is numbered 1 to 30. The dice were supported with a book called "The Armory's 30 Sided Dice Gaming Tables". It could be used with any of the fantasy RPGs of the time.

In later years, they seem to have outsourced manufacturing to other companies. In 1998 they merged with Chessex. Dice aren't being sold under the Armory brand anymore.

### TSR

TSR resold Creative Publications dice for many years. Update: the dice in the above picture come from two different molds. The blue dice came with the first printing of the Moldvay basic set. The green and orange dice came with the first printing of the Mentzer basic set. For a guide on how to distinguish them, see this post and this post.

By 1981 TSR was having their own dice manufactured. Pictured on the left below are dice from the Moldvay basic set; on the right are dice from the Cook expert set.

The dice did not come not painted or inked in. Crayons were provided to color in the numbers.

The dice seem too small to me. The dice in my expert set are a mix of colors, but I think that is atypical. I've seen all-brown and all-orange sets.

The dice in the first printing of the Mentzer basic set were sold separately in a blister pack as "Dragon Dice". They are unrelated to the collectible dice game, also called "Dragon Dice", which TSR introduced in 1995.


text:
Search by keywords or author
• Vol. 2, Issue 6, 066003 (2020)
Haihui Zhao1、†, Xinhou Chen2, Chen Ouyang3、4, Hangtian Wang1、5, Deyin Kong2, Peidi Yang2, Baolong Zhang3、4, Chun Wang3、4, Gaoshuai Wei3、4, Tianxiao Nie1、5、*, Weisheng Zhao1、5、*, Jungang Miao2, Yutong Li3、4, Li Wang3、*, and Xiaojun Wu2、6、*
Author Affiliations
• 1Beihang University, Fert Beijing Institute, Advanced Innovation Center for Big Data and Brain Computing, School of Microelectronics, Beijing, China
• 2Beihang University, School of Electronic and Information Engineering, Beijing, China
• 3Chinese Academy of Sciences, Institute of Physics, Beijing National Laboratory for Condensed Matter Physics, Beijing, China
• 4University of Chinese Academy of Sciences, School of Physical Sciences, Beijing, China
• 5Beihang University, Beihang-Goertek Joint Microelectronics Institute, Qingdao Research Institute, Qingdao, China
• 6Wuhan National Laboratory for Optoelectronics, Huazhong University of Science and Technology, China
• show less
Fig. 1. Schematic diagram of the polarization tunable terahertz emission from $Bi2Te3$. (a) Femtosecond laser pulses, horizontal linear polarization (HLP), vertical linear polarization (VLP), left-handed circular polarization (LCP), and right-handed circular polarization (RCP), are incident onto the topological insulator $Bi2Te3$ and produce polarization tunable terahertz waves. (b) Macroscopic helicity-dependent photocurrent and only unidirectional spin current can be generated. (c) Microscopic electronic transition under circularly polarized laser pulse illumination (Video 1, MP4, 15.5 MB [URL: https://doi.org/10.1117/1.AP.2.6.066003.1]).
Fig. 2. Linearly polarized terahertz emission and its shift current mechanism. (a) Schematic diagram of the experimental setup for linearly polarized terahertz emission. Linearly polarized laser pump passes through a half-wave plate of angle $α$ and is incident onto the sample emitting linearly polarized terahertz wave. The incident angle of the pump laser, $θ$; the azimuthal angle of the topological insulator, $φ$. (b) Far-field polarization trajectories [$Sx(t)$, $Sy(t)$] of the radiated terahertz waves obtained experimentally from 10-nm-thick $Bi2Te3$ when the pump laser polarization was nearly normal incidence while the sample azimuthal angle was varied. (c) The terahertz peak amplitude of $Sy(t)$ as a function of the azimuthal angle exhibits a 120-deg period. (d) Pump fluence dependence of terahertz peak amplitude from $Bi2Te3$ with different thickness of 5, 8, and 10 nm. (e) Schematic of the shift current generated from the electron transfer along the Bi–Te bonds. (f) Parameters in the $y$ axis extracted through the symmetry analysis of surface state using Eq. (1). (g) $Ip$ denotes the laser intensity profile. Fitting the shape of the shift current (blue line) to the photocurrent in the material (green line).
Fig. 3. Generation of elliptically and circularly polarized terahertz beams. (a) Experimental layout for circularly-polarized terahertz wave generation. Linearly polarized laser pump passes through a quarter-wave plate of angle, $α$, producing elliptically or circularly polarized pump laser beams. The emitted terahertz wave polarization can be elliptical or circular. The incident angle of the pump laser, $θ$; the azimuthal angle of the topological insulator, $φ$. (b) Far-field detected elliptically polarized terahertz polarization trajectories when the pump laser polarization was fixed while rotating the sample azimuthal angles. (c) Production of circularly polarized terahertz waves when fixing the azimuthal angle while rotating the quarter-wave plate for the pump laser pulses. (d) Experimentally observed terahertz component $Sx(t)$ polarity reversal depends on the pump laser helicity. (e) Helicity-independent component $Syz(t)$ from $Bi2Te3$. $δ+$ and $δ−$ mean left-handed and right-handed elliptically polarized terahertz, respectively.
Fig. 4. Macroscopic analysis of PGE. (a) The terahertz peak amplitude of $Sx(t)$ and $Syz(t)$, as a function of the quarter-wave plate angle, respectively. (b) and (c) The time-domain signals for the parameters $C(t)$, $L1(t)$, $L2(t)$, and $D(t)$ in the $x$ axis and $yz$ plane extracted using Eq. (3). (d) Spin-momentum-locked states selectively excited by spin-polarized pump laser form unidirectional spin currents. (e) and (f) The corresponding Fourier transformed spectra of the time-domain signals in (b) and (c).
Fig. 5. Arbitrary manipulation of various terahertz polarization states. Radiated terahertz chirality was consistent with that of the pump laser chirality. (a) LCP pump laser pulses can produce left-handed elliptically polarized terahertz waves and (b) right-handed elliptically polarized pump light can generate right-handed elliptically polarized terahertz beams. (c) Linearly polarized terahertz waves can be generated in $Bi2Te3$ pumped by linearly polarized pump laser beams.
Copy Citation Text
Haihui Zhao, Xinhou Chen, Chen Ouyang, Hangtian Wang, Deyin Kong, Peidi Yang, Baolong Zhang, Chun Wang, Gaoshuai Wei, Tianxiao Nie, Weisheng Zhao, Jungang Miao, Yutong Li, Li Wang, Xiaojun Wu. Generation and manipulation of chiral terahertz waves in the three-dimensional topological insulator Bi2Te3[J]. Advanced Photonics, 2020, 2(6): 066003


text:
# Validating that only certain values are unset

I am checking data output in some testing code, and only some fields can be None, but I think this can be better:

for key in data[0].keys():
self.assertIn(key, data_model)
if data[0][key] is None:
self.assertTrue('rate' in key or 'percentage' in key or 'ratio' in key)
continue
self.assertIsInstance(data[0][key], data_model[key])

Mostly I don't like the ... or ... in the assertTrue

A few things:

• you could store data[0] in some variable with a relevant name.

• you could iterate over both keys and values with iteritems() /items()

• you do not need continue here. A simple else would do the trick and is more explicit.

• The line you don't quite like could be replaced by something like :

self.assertTrue(any(w in key for w in ('rate', 'percentage', 'ratio')))

The resulting code would be :

first_data = data[0]
for key, val in first_data.items():
self.assertIn(key, data_model)
if val is None:
self.assertTrue(any(w in key for w in ('rate', 'percentage', 'ratio')))
else:
self.assertIsInstance(val, data_model[key])

• Thanks - any was exactly what I wanted, just hadn't thought of it. Also good suggestion about items(), and I agree about the else – user3610360 Jun 30 '16 at 14:35