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with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. This paper extends the phase portrait to three states to represent the nonlinear vehicle dynamics with steering and longitudinal tyre force inputs and consideration of the longitudinal. (1) There is one equilibrium solution of this system – find it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the linearized system. This is a second order systemwhich is autonomous (time does not appear explicitly). from second-order equation to first-order system; what is a phase portrait; direction field of a first-order system; graphing in the xy- tx- and ty-planes; vector notation for a first-order system; semesters > spring 2020 > mth264 > resources > video > linear systems: basics Video | Linear Systems: Basics. Recall the basic setup for an autonomous system of two DEs: dx dt = f(x,y) dy dt = g(x,y) To sketch the phase plane of such a system, at each point (x0,y0)in the xy-plane, we draw a vector starting at (x0,y0) in the direction f(x0,y0)i+g(x0,y0)j. Damped Pendulum. Nonlinear Systems and Stability Autonomous systems and critical points Stability and phase plane analysis of almost linear systems Linearized stability analysis and plotting vector fields using a MSS Numerical solutions and phase portraits of nonlinear systems using a MSS Models and applications: TEXT: Text(s) typically used in this course. The classic Van der Pol nonlinear oscillator is provided as an example. 3 in Third and 3-42 Fourth Quadrant. We illustrate all these cases in the examples below. • Understand the linearized models using the “eigen-techniques” you learned earlier. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. The dynamical equation and the state equation of the system are established. We define | {
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superimposed. The dynamical equation and the state equation of the system are established. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. These states can also be correlated with velocity spectral behaviors. A phase portrait is a geometric representation of the trajectories of a dynamical system in the phase plane. (Hint: Use polar coordinates. The low intensity xed point appears on the phase portraits. 26 Phase Portrait for sand Y1 Magnitudes of 9. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. 50= 1 (c) 8+82+0. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. 1 Phase portraits 72 3. warn(warning_msg, ODEintWarning). Is there a way for plotting phase portraits and vector fields for autonomous system of delay differential equations in. This video deals with. The dynamical equation and state equation of the system are established. Strogatz, Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering, Cambridge: Westview Press, 2000. Phase Portraits of Nonlinear Systems Consider a , possibly nonlinear, autonomous system , (autonomous means that the independent variable , thought of as representing time, does not occur on the right sides of the equations). Damped Pendulum. One- and two- dimensional flows. to sketch the phase portrait. Stable and unstable manifolds of equilibrium points and periodic orbits are important objects in phase portraits. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. I have a set of three differential equations and I want to make a phase portrait of them. For a chaotic system, there will be many distinct loops in a phase portrait, showing that the system | {
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For a chaotic system, there will be many distinct loops in a phase portrait, showing that the system is aperiodic and does not approach a stable. Quiver function is being used for phase portrait plots obtained using ode. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. - A ”limit cycle” is a periodic orbit that trajectories approach. 3 Determining Time from Phase Portraits 29 2. 3 Symmetry in Phase Plane Portraits 22 2. Derive the dynamics of a linear and nonlinear systems. Fig 1: 3-D phase portrait of the Rossler attractor Integration Solver The Runge-Kutta method for a system of ordinary differential equations is explained here. The dynamical variables of the system, in this case the angular position and velocity , are the coordinates defining the system's phase space. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. Sketching an accurate phase portrait for a non-linear system of DEs is time consuming but the series of 3 videos will help with shortening that time with added understanding. Phase Plane Portraits of Almost Linear Systems Interesting and complicated phase portraits often result from simple nonlinear perturbations of linear systems. 2 Bifurcation set and phase portraits of the Hamiltonian system (5). Ask Question Asked 4 years, 7 months ago. Solving 2x2 homogeneous linear systems of differential equations 3. According to Takens, almost all d-dimensional sub-manifolds could be embedded in a (m=2d+1) dimensional space. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. These programs provide animated phase portraits in dimension two and three, i. The tem-poral response of a system need not be the. 2 : Linear analysis of nonlinear pendulum : Mechanical systems model for a pendulum. (a) This plot shows the vector field for a planar dynamical system. , another nonlinear system x_1 = 1 x3 1 x_2 = x1 x22 | {
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the vector field for a planar dynamical system. , another nonlinear system x_1 = 1 x3 1 x_2 = x1 x22 equilibrium points are described by x1 = 1 and x2 = 1 note: the equilibrium points of a nonlinear system can be nite (2 in the previous examples, but any other number is possible, including zero) or in nite, and they can be isolated points in state space Oriolo: Stability Theory for. For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. Phase Plane Analysis 17 2. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360∘. Weak non-linear oscillators and. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. Ott, and A. This suggests that the only. , regularly timed speech with a metronome). By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at 휇 = 0. 6: Phase portraits on the (one-dimensional) centr emanifoldandthebifurcation diagram. First, we cover stability definitions of nonlinear dynamical systems, covering the difference between local and global stability. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. 1 (Saddle) Consider the system x˙ 1 = −x 1 −3x 2 x˙ 2 = 2x 2, x. in weakly-nonlinear systems was investigated. 1 Phase Portraits 18 2. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. It is not restricted to small or smooth nonlinearities and applies equally well to strong and hard nonlinearities. warn(warning_msg, ODEintWarning). Limit Cycles. (1) There is one equilibrium solution of this system – find it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the | {
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system – find it! (2) Linearize the system near this equilibrium, and draw the phase portrait of the linearized system. The aim of this section is to present programs allowing to high- light the slow-fast evolution of the solutions of nonlinear and chaotic dynamical systems such as: Van der Pol, Chua and Lorenz models. x′= x−y, y′= x2 +y2 −1 2. Four possible phase portraits for this system are shown along the right side of the page. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. Many nonlinear dynamic systems have a rotating behavior where an angle defining its state may extend to more than 360∘. doc Author: tien Created Date: 11/15/2002 4:16:10 AM. By varying the initial conditions of the system, it is found. The type of phase portrait of a homogeneous linear autonomous system -- a companion system for example -- depends on the matrix coefficients via the eigenvalues or equivalently via the trace and determinant. Phase portrait. In paper [12], the vibration of a typical two degree of freedom nonlinear system with repeated linearized natural frequencies was investigated systematically. We discovered the system’s rich behavior such as chaos through phase portraits, bifurcation diagrams, Lyapunov exponents, and entropy. The course revises some of the standard phase portrait methods encountered in the Dynamical Systems course in part II and extends these ideas, discussing in some detailed centres, via the use of Lyapunov functions, limit cycles and global phase portraits. 1 Concepts of Phase Plane Analysis 18 2. Keywords: nonlinear dynamics, chaos, electrical circuits. A phase portrait is mapped by homeomorphism, a continuous function with a continuous inverse. Linear approximation of autonomous systems 6. Let us consider the below block diagram of a non linear system, where G 1 (s) and G 2 (s) represent the linear | {
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the below block diagram of a non linear system, where G 1 (s) and G 2 (s) represent the linear element and N represent the non linear element. Existence, uniqueness, and strong topological consequences for two-dimensions. • As much as possible, piece the phase portraits of the linearized systems together to get an approximate phase portrait of the full non-linear system. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. Damped Pendulum. Phase Plane Analysis 17 2. If the stable manifold is of higher dimension, then y 1 =h 1(x,µ ),y 2 =h 2(x,µ )and. There are lots of practical systems which can be approximated by second-order systems, and apply phase plane analysis. (e) Draw the phase plane portrait of the nonlinear system via v and h nullclines. The resonance effects are most pronounced where both the. 4 Phase Portraits and Bifurcations. A numerically generated phase-portrait of the non-linear system Zoomed in near (0,0) Zoomed in near (2,1) The critical point at (2,1) certainly looks like a spiral source, but (0,0) just looks bizarre. The vertical diametric phase distribution of the singly charged OV extracted from this phase portrait @Fig. Saturations constitute a severe restriction for stabilization of system. The long time dynamics are. All chapters conclude with Exercises. Chua}, year={1969} }. We describe the phase portrait for bang-bang extremals. • Understand the linearized models using the “eigen-techniques” you learned earlier. 2 Phase portraits • A phase portrait of an n-dimensional autonomous system x ′ (t) = f (x (t)) is a graphical rep-resentation of the states in x-space. Nonlinear. Plot the maximum amplitude in the frequency. We draw the vector field given at each point (x,y) by the vector. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by | {
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more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. 3 Determining Time from Phase Portraits 29 2. General Calendar. In this research a new graphic. In this section we will give a brief introduction to the phase plane and phase portraits. The phase portrait can indicate the stability of the system. Also, this work showed that the extreme multi-stability phenomenon of the behaviour of infinitely many coexisting attractors depends on the initial conditions of the variables of the system. motion of the system. In physical systems subject to disturbances, the distance of a stable equilibrium point to the boundary of its stable manifold provides an estimate for the robustness of the equilibrium point. In these cases the use of the phase portrait does not properly depict the system’s evolution. 2 Global bifurcation analysis 69 3. When a double eigenvalue has only one linearly independent eigenvalue, the critical point is called an improper or degenerate node. The Poincar´e-Bendixson theorem Any orbit of a 2D continuous dynamical system which stays in a closed and bounded subset of the phase plane forever must either tend to a critical point or to a. , a particular state of the system) over time. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. 7 (2009), 369–403. Draw the phase line of the equations and Answer. , another nonlinear system x_1 = 1 x3 1 x_2 = x1 x22 equilibrium points are described by x1 = 1 and x2 = 1 note: the equilibrium points of a nonlinear system can be nite (2 in the previous examples, but any other number is possible, including zero) or in nite, and they can be isolated points in state space Oriolo: Stability Theory for. 3 Symmetry in Phase Plane Portraits 22 2. For each case, we construct a phase space portrait by | {
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for. 3 Symmetry in Phase Plane Portraits 22 2. For each case, we construct a phase space portrait by plotting the values of the dynamical variables after repeated application of the map (equation (1), followed by (6) and (7)) for a range of initial conditions. Extensibility of solutions 50 §2. Equilibrium points. For optimal bang-bang trajectories with high values of the energy integral, a general upper bound on the number of switchings was obtained. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. doc Author: tien Created Date: 11/15/2002 4:16:10 AM. The phase space portraits of these two systems are shown in figure 6. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. Knowledge of λ1 and λ2, and v1 and v2, en-ables us to sketch the phase portrait near (x∗,y∗). Free system of non linear equations calculator - solve system of non linear equations step-by-step. • Understand the linearized models using the “eigen-techniques” you learned earlier. Consider a , possibly nonlinear, autonomous system ,(autonomous means that the independent variable , thoughtof as representing time, does not occur on the right sides of the equations). So, for a periodic system that obeys the law of energy conservation (e. Let's zoom into these four critical points, and look more closely at the phase portraits near them. If the system is period-n (the same state repeats after n clocks), there will be n points −→ period-n attractor. Students will learn nonlinear differential equations in the context of mathematical modeling. 26 Phase Portrait for sand Y1 Magnitudes of 9. Each set of initial conditions is represented by a different curve, or point. o Equilibrium solution • Exponential solutions o Half-line solutions • Unstable solution • Stable solution • Six important cases for portraits Real Eigenvalues o Saddle point o Nodal sink o Nodal source. An example | {
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cases for portraits Real Eigenvalues o Saddle point o Nodal sink o Nodal source. An example of a non-linear system: the predator / prey model. The phase portrait of a dynamical system can be reconstructed from the observation of a single variable by the method of delays as proposed by [1]. The x-nullclineis a set of points in the phase plane so that. An example of a non-linear system: the predator / prey model. Phase Portraits of Nonlinear Systems. We draw the vector field given at each point (x,y) by the vector. Damped Pendulum. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. 2 Singular Points 20 2. warn(warning_msg, ODEintWarning). : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x’=x+4y, y’=2x−y −5 0 5 −5 0 5 x y Time Plots for ‘thick’ trajectory. Linear and Nonlinear Systems of Differential Equations. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. The x nullcline is given by (1 x y)x = 0 =) x = 0 or y = 1 x: (11) Sodx dt= 0 on the lines x = 0 and y = 1 x. Phase portraits and Hooke diagrams of the proposed driven nonlin-ear system are consistent with empirical observations. Second-Order Systems. Phase Portraits Now we turn to the third method of analyzing non-linear systems, phase portraits generated by numerical solutions. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. two-dimensional phase distributions for OV beams with charges m51 ~a! and m52 ~b!. Local Phase Portrait of Nonlinear Systems Near Equilibria. In fact, if we zoom in around this point, it would look like the case of a node of a linear system (in the sense of Chapter 7). The low intensity xed point appears on the phase portraits. Fig 1: 3-D phase portrait of the Rossler attractor Integration Solver The | {
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on the phase portraits. Fig 1: 3-D phase portrait of the Rossler attractor Integration Solver The Runge-Kutta method for a system of ordinary differential equations is explained here. In particular, show that some of the equilibria correspond to nonlinear centers, by nding a rst integral for this system. Compare the phase portraits of the linear and the nonlinear maps near the origin. Q: Find the phase portrait of this second-order nonlinear system with such differential equation: $$\ddot{x}+0. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. Phase portraits and Hooke diagrams of the proposed driven nonlin-ear system are consistent with empirical observations. What is a Phase Portrait? Above, we have an animated phase portrait, but what is it? A phase portrait, in it’s simplest terms, is when we plot one state of the system against another state of the system. In paper [12], the vibration of a typical two degree of freedom nonlinear system with repeated linearized natural frequencies was investigated systematically. Thus, the equilibrium x = 0 is a saddle, hence unstable, when = 0. 11 (Nonlinear terms can change a star into a spiral) Here's another example. Mindlin, Nonlinear dynamics: A two-way trip from Physics to Math, Taylor and Francis, 1996. John Polking’s pplane: MATLAB, JAVA. The same concept can be used to obtain the phase portrait, which is a graphical description of the dynamics over the entire state space. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. Autonomous Planar Nonlinear Systems. Change parameters in the Linear-2D MAP so as to get the linearization at the origin. Following bifurcation in the system occurs in a range of parameter values g from 0. (c) The interior xed point at (1=3;1=3) is a global attractor. Changes in the dynamics of the orbits in the phase space usually represent variations of the physical | {
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in the dynamics of the orbits in the phase space usually represent variations of the physical parameters that control a non-linear system and consequently are of great importance for any modelling effort. This allowed to obtain exhaustive solution of the control problem comparing to the known results. systems) Suppose (x*,y*)=(0,0) is a linear center for a cont. 28 (1997), 755-778. 26 Phase Portrait for sand Y1 Magnitudes of 9. walking with Durus (right), showing phase portraits (top left) for 63 steps of walking together with a darker averaged phase portrait and position tracking errors (bottom left) over a select 4 steps in the same experiment. phase portrait get from simulink Example 2. Critical (equilibrium) points occur when (˙x,y˙) = (0,0). Key words: vibroimpact motion, unilateral and symmetrical rigid arrester, stereo-mechanical impact theory, phase portrait, two dimensional mapping. , is attracted to infinity. Students will learn nonlinear differential equations in the context of mathematical modeling. 1 Solution curves in the phase plane of the Lotka-Volterra predator-prey model102 6. Solving 2x2 homogeneous linear systems of differential equations 3. Dynamical Systems and Chaos. 13 from the book. In this section we will give a brief introduction to the phase plane and phase portraits. Determine the stability of these limit cycles. Problem 4: Hamiltonian Systems. The phase portrait is a plot of a vector field which qualitatively shows how the solutions to these equations will go from a given starting point. MATLAB offers several plotting routines. By varying the parameters of the equation for the non-linear pendulum and then plotting. Phase Portraits of Nonhyperbolic Systems. Poincaré-Bendixon theorem. φ 1 = phase shift of the fundamental harmonic component of output. the allee due at noon on friday sept 14th, in the box provided (to the. 1) Find all equilibrium points by solving the system 2) Let a standard software (e. One- and two- dimensional flows. | {
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points by solving the system 2) Let a standard software (e. One- and two- dimensional flows. Phase portrait of system of nonlinear ODEs. Save the phase portraits to submit on Gradescope. 2 : Linear analysis of nonlinear pendulum : Mechanical systems model for a pendulum. (4) (Formerly numbered 135A. The fixed points can be classified according to their stability as follows: • If Re(λ1) > 0 and Re(λ2) > 0 ⇒ repeller (unstable node). In this section we will give a brief introduction to the phase plane and phase portraits. Neural Information Processing Systems (NIPS). Smith; Nonlinear Ordinary Differential Equations, 3rd Edition, Oxford University Press, 1999. We illustrate all these cases in the examples below. This page plots a system of differential equations of the form dx/dt = f(x,y), dy/dt = g(x,y). Change parameters in the Linear-2D MAP so as to get the linearization at the origin. Perko, Di erential Equations and Dynamical Systems (Second edi-tion, Springer, 1996). Phase portraits are an invaluable tool in studying dynamical systems. µ< 0 µ< 0 µ< 0 µ x Figure 5. Let us discuss the basic concept of describing the function of non linear control system. Q: Find the phase portrait of this second-order nonlinear system with such differential equation:$$ \ddot{x}+0. These are systems that do not depend explicitly on. Determining time from phase portraits. nonlinear transform of coordinates and uses a full nonlinear system’s model. Planar Almost Linear Systems: Phase portraits, Nonlinear classi- cations of equilibria. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. Fixed points occur at values θ∗ such that 0 = 1 + 2cosθ∗. which can be written in matrix form as X'=AX, where A is the coefficients matrix. 5 Global analysis for hardening nonlinear stiffness (c>0) 72 3. System analysis based on Lyapunov's direct method. Damped Pendulum. (b) This plot includes the solutions (sometimes called | {
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on Lyapunov's direct method. Damped Pendulum. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. The figure shows the manner of convergence of these projected trajectories, which start with different initial conditions. In physical systems subject to disturbances, the distance of a stable equilibrium point to the boundary of its stable manifold provides an estimate for the robustness of the equilibrium point. Autonomous and non-autonomous systems Phase portraits and flows Attracting sets Concepts of stability 2. Lyapunov's direct method. This course introduces the main topics of low-dimensional nonlinear systems, with applications to a wide variety of disciplines, including physics, engineering, mathematics, chemistry, and biology. We also show the formal method of how phase portraits are constructed. The undamped system has analytical solutions, (for both rotary and oscillatory motion), in terms of Jacobian Elliptic functions, that with the phase portrait gives a total qualitative and quantatitive explanation. These variables and their evolution in time produce the phase portrait of the system. Introduction to nonlinear network theory @inproceedings{Chua1969IntroductionTN, title={Introduction to nonlinear network theory}, author={L. The author starts off with an introduction to nonlinear systems, then moves on to phase portraits for 2-D systems, before moving on to advanced concepts of stability theory and feedback linearization. : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x'=x+4y, y'=2x−y −5 0 5 −5 0 5 x y Time Plots for 'thick' trajectory. It may be best to think of the system of equations as the single vector equation x y = f(x,y) g(x,y). The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. 2Switched Nonlinear Systems 1. We first | {
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exponent spectra and phase portraits are investigated. 2Switched Nonlinear Systems 1. We first focused on a nonlinear pendulum (shown in Fig. Phase Portrait for the linearization in Example 6. Phase portrait. Consider the nonlinear system (a) Show that the origin (O, O) is a nonlinear saddle awl plot the phase portrait, including the. 3 Determining Time from Phase Portraits 29 2. Generally, the nonlinear time series is analyzed by its phase space portrait. The motion of the mass is governed by Newton's second law. (iv) Since replacing x by x + 2… gives the same equations the portrait. Lyapunov's direct method. Phase Plane Analysis 17 2. Homework 6. Thompson and H. 1) "For nonlinear systems, there is typically no hope of finding the trajectories analytically. Fixed points, limit cycles, and stability analysis. Giorgio Bertotti, Claudio Serpico, in Nonlinear Magnetization Dynamics in Nanosystems, 2009. 58=0 (b) 8+8+0. Such a planar curve is called a trajectory of the system and its param-eter interval is some maximal interval of existence T 1 λ 0. Then try to combine the vector field with part (d) to get a global phase portrait of the original nonlinear system. I am unable to do for this case. 3 Phase Plane Portraits (for Planar Systems) Key Terms: • Equilibrium point of planer system. John Guckenheimer, in Handbook of Dynamical Systems, 2002. phase portrait (or phase diagram) for asystem depicts its phase space andtrajectories andis ageometricalrepresen- tation ofthe qualitative behavior ofthe system. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. These plots readily display vehicle stability properties and map equilibrium point locations and movement to changing parameters and system inputs. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto ries of a linear system x = ax +by a, b, c, d constants. • A PLL is a control system that generates an output signal whose | {
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x = ax +by a, b, c, d constants. • A PLL is a control system that generates an output signal whose phase is related to the phase of the input and the feedback signal of the local oscillator. Differential equations are used to map all sorts of physical phenomena, from chemical reactions, disease progression, motions of objects, electronic circuits, weather forecast, et cetera. Solve system of nonlinear equations - MATLAB fsolve Nl. served: here we analyze this interplay by investigating the system using statistical tools, phase portraits, Poincar e sections, and return maps. Numerical Construction of Phase Portraits. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. A phase portrait represents the directional behavior of a system of ODEs. 1) As in § 3. Solving 2x2 homogeneous linear systems of differential equations 3. So, for a periodic system that obeys the law of energy conservation (e. Since in most cases it is. 1 Concepts of Phase Plane Analysis 18 2. Phase portrait. Let A= 3 −4 6 −7. { Nonlinear spring-mass system { Soft and hard springs { Energy conservation { Phase plane and scenes. (b) This plot includes the solutions (sometimes called streamlines) from different initial conditions, with the vector field superimposed. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. Its phase is ˇ=2, showing that the polariton resonance is below the laser line. Generally, the nonlinear time series is analyzed by its phase space portrait. Linear stability analysis may fail for a non-hyperbolic fixed point: ( Re(µ 1, 2) = 0, or at least one µ i = 0 ). The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. 1 we draw the phase portrait (or phase diagram), where each point (x,y) corresponds to a specific state of the system. Each set of initial conditions | {
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where each point (x,y) corresponds to a specific state of the system. Each set of initial conditions is represented by a different curve, or point. 2 Phase portraits • A phase portrait of an n-dimensional autonomous system x ′ (t) = f (x (t)) is a graphical rep-resentation of the states in x-space. By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at 휇 = 0. In this research a new graphic. Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). The phase portrait for the reduced dynamics for x is shown in Figure 5. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. Use technology to solve nonlinear programs, including computer programming and graphical analysis. These states can also be correlated with velocity spectral behaviors. (b) Find all bifurcation values of r and draw a bifurcation diagram on the rθ-plane. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. phase portrait (or phase diagram) for asystem depicts its phase space andtrajectories andis ageometricalrepresen- tation ofthe qualitative behavior ofthe system. Chaos of such a system was predicted by applying a machine learning approach based on a neural network. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. The procedure was applied in modeling of self-excitation oscillations for high-speed milling and is based on determination of non-linear self-excitation force and non-linear coefficient. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto ries of a linear system x = ax +by a, b, c, d constants. Consider the nonlinear system dx dt = r − x2, dy dt = x− y. 5 | {
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x = ax +by a, b, c, d constants. Consider the nonlinear system dx dt = r − x2, dy dt = x− y. 5 Summary of stability properties for planar ODE systems. One was the anticipated constancy of the system energy. By varying the initial conditions of the system, it is found. Design of feedback control systems. Overview of nonlinear mechanics of particles and nonlinear oscillations Lagrange’s equations and nonlinear di erential equations Flows on a line and bifurcations Multi-dimensional ows and linear systems Phase portraits, stability, and limit cycles Dissipative systems, reversible systems, Index theory Weakly nonlinear oscillations and two. 2 Phase Plane Analysis. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. 2 $\begingroup$ How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following? \begin{align} \dot{x} &= 2 - 8x^2-2y^2\\ \dot{y} &= 6xy\end{align}. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. HW # 7 Nonlinear Dynamics and Chaos Due: Monday, 95/01/30 1. SKETCH an approximate phase portrait for (6). Nonlinear Systems-lecture notes 4 Dr. 2 Global bifurcation analysis 69 3. As pointed out in @13#,a clear signature for the presence of a phase singularity is a new fringe starting at the location of the singularity. 3 Determining Time from Phase Portraits 29 2. The dynamics of airflow through the respiratory tract during VB and BB are investigated using the nonlinear time series and complexity analyses in terms of the phase portrait, fractal dimension, Hurst exponent, and sample entropy. In addition, if. The real space images in the top row show a portion of an unforced rotating spiral wave pattern [Fig. same'' means that type and stability for the nonlinear problem are the same as for the corresponding linear problem. d) Plot a computer-generated phase portrait to check your answer to (c). Weak non-linear oscillators and. 6: | {
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a computer-generated phase portrait to check your answer to (c). Weak non-linear oscillators and. 6: Phase portraits on the (one-dimensional) centr emanifoldandthebifurcation diagram. Effect of nonlinear terms. Now, if = 0, the system has one equilibrium point, x = 0. Now consider the nonlinear di erential equation = 1 2sin (6) Determine the equilibria of this system and their stability type. 2 Singular Points 20 2. 4 Phase Plane Analysis of Linear Systems 30. A simple example of a map is the Lotka-Volterra system describing two competing populations (e. Its usage is also observed heavily in smart brakes systems of current automotive vehicles. Phase portrait. In this research a new graphic. MATLAB offers several plotting routines. A time series provides information about a large number of pertinent variables, which may be used to explore and characterize the system's dynamics. µ< 0 µ< 0 µ< 0 µ x Figure 5. Day 9 - Two Dimensional Systems - Phase Planes Day 10 - Two Dimensional Systems - Eigenvalues and Eigenvectors Day 11 - Nonlinear Two Dimensional Systems - Jacobian Day 12 - More practice with Two Dimensional Nonlinear Systems Day 13 - Bifurcations in 2-D Systems - Limit Cycles Day 14 - Hopf Bifurcations, Lorenz Equations, Chaos and Fractals. Nonlinear systems - existence and uniqueness theorem, continuous dependence, variational equations. ends up in one of the fixed points at x n = (2n+1)π. freedom and analysis of phase portraits, i. The nonlinear gyroscope model, which is employed in aerospace engineering [24], generally exhibits chaotic behavior. As the initial angle increases, we can see that the shape of the non-linear phase trajectory approaches that of the seperatrix. Albu-Schaffer. the Rossler system is sensitive to the initial conditions, and two close initial states will diverge, with increasing number of iterations. Different initial states result in different trajectories. Here we will consider systems for which direction fields and phase portraits are | {
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trajectories. Here we will consider systems for which direction fields and phase portraits are of particular importance. Based on velocity phase portraits, each of the nonlinear response states can be categorized into one of the three states in the order of increasing chaotic levels: lock-in, transitional, or quasiperiodic. A simple example of a map is the Lotka-Volterra system describing two competing populations (e. Classification of phase portraits. Ott, and A. Reversible Systems (2) THEOREM (Nonlinear centers for rev. The simple pendulum is a great example of a second-order nonlinear system that can be easily visualized by the phase portrait. Autonomous and non-autonomous systems Phase portraits and flows Attracting sets Concepts of stability 2. In sum, we illustrate the revised system’s fit to the kinematics in both noncyclic speech and cyclic tasks (i. phase portrait get from simulink Example 2. The trace-determinant plane and stability. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. Learn more in: Chaotic Attractor in a Novel Time-Delayed System with a Saturation Function. Evolution of a dynamical system corresponds to a trajectory (or an orbit) in the phase space. John Polking’s pplane: MATLAB, JAVA. Conclude: any i. 5 \dot{x}+2 x+x^{2}=0 $$Method 1: Calculate by hands with phase plane analysis. What is a Phase Portrait? Above, we have an animated phase portrait, but what is it? A phase portrait, in it's simplest terms, is when we plot one state of the system against another state of the system. How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following?$$\begin{align} \dot{x} &= 2 - 8x^2-2y^2\\ \dot{y} &= 6xy\end{align}$$I can easily find the equilibria, which are$$\left\{ (0, \pm 1), \left(\pm \frac{1}{2}, 0\right) \right\} The corresponding stable subspace for $\left(\pm | {
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\pm 1), \left(\pm \frac{1}{2}, 0\right) \right\} The corresponding stable subspace for $\left(\pm \frac{1}{2}, 0\right)$ is. Limit Cycles: Recall that analysis of linearized systems. We define the equilibrium solution/point for a homogeneous system of differential equations and how phase portraits can be used to determine the stability of the equilibrium solution. For more information on phase portraits and types of fixed points for linear systems of ODEs, see, for example: S. bifurcation diagrams and phase portraits. Click on the button corresponding to your preferred computer algebra system (CAS). Consider the nonlinear system dx dt = r − x2, dy dt = x− y. Phase portrait generator. Basics : Introduction to the notion of dynamical systems, examples of non-linear systems, Discrete and Continuous time, from one to the other, Poincaré section. Instructors: Aldo Ferri: Topics: Introduction; properties of nonlinear systems; Phase portraits for second order systems; characterization of singular points and local stability; first and second methods of Lyapunov. Kitavtsev May 28, 2019 4 Local bifurcations of continuous and discrete dynamical systems The material of this chapter is covered in the following books: L. ) Lecture, three hours; discussion, one hour. A numerically generated phase-portrait of the non-linear system Zoomed in near (0,0) Zoomed in near (2,1) The critical point at (2,1) certainly looks like a spiral source, but (0,0) just looks bizarre. 4 Global analysis for softening nonlinear stiffness (c<0) 68 3. Materials to be covered include: nonlinear system characteristics, phase plane analysis, Lyapunov stability analysis, describing function method, nonlinear controller design. for x, where F ( x ) is a function that returns a vector value. (reductor and multipliers). Requisites: course 33B. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. For a much more sophisticated phase plane plotter, see the MATLAB plotter | {
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x'=ax+by y'=cx+dy. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. As the initial angle increases, we can see that the shape of the non-linear phase trajectory approaches that of the seperatrix. This allowed to obtain exhaustive solution of the control problem comparing to the known results. Pauses are inserted between setting up the graphs; plotting the linear phase portrait for $$x = 2n\pi$$; adding this behavior to the full phase plane; plotting the linear phase portrait for $$x = (2n+1)\pi$$; adding that to the full phase. 01385 till 0. Click on the button corresponding to your preferred computer algebra system (CAS). 1 Phase Portraits 18 2. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. Consider the homogeneous linear first-order system differential equations x'=ax+by y'=cx+dy. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Paragraphs 4. Around the origin there are periodic orbits corresponding to small oscillations of the pendulum that are called librations. population growth. First, we cover stability definitions of nonlinear dynamical systems, covering the difference between local and global stability. • λ 12==λ 0. Phase plane Analysis: 2nd order nonlinear systems, phase portrait graphical representation, singular points. Just as we did for linear systems, we want to look at the trajectories of the system. It starts in your web browser and you can directly input your equations and parameter values. The book is very readable even though it has a lot of jargon (read heavy mathematics). My professor told us to use a plotter to check our work (the hand-drawn phase portraits) but the one he linked to us won't work on my mac so I am trying to see the plots in Matlab but I don't know how to plot them and would be absolutely grateful for some help (I. And it turns out, with this omega two this was the separatrix case, | {
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grateful for some help (I. And it turns out, with this omega two this was the separatrix case, but that was the intermediate axis case. which can be written in matrix form as X'=AX, where A is the coefficients matrix. The dynamics of airflow through the respiratory tract during VB and BB are investigated using the nonlinear time series and complexity analyses in terms of the phase portrait, fractal dimension, Hurst exponent, and sample entropy. One was the anticipated constancy of the system energy. Sketching Non-linear Systems In session on Phase Portraits, we described how to sketch the trajecto ries of a linear system x = ax +by a, b, c, d constants. Then sufficiently close to (0,0) all trajectories are closed curves. By varying the initial conditions of the system, it is found. Following bifurcation in the system occurs in a range of parameter values g from 0. Pages 486 - 493 cover the five important cases. Here we will consider systems for which direction fields and phase portraits are of particular importance. The following worksheet is designed to analyse the nature of the critical point (when ) and solutions of the linear system X'. 3 Symmetry in Phase Plane Portraits 22 2. The behaviour of the system is investigated through numerical simulations, by using well-known tools of nonlinear theory, such as phase portrait, bifurcation diagram and Lyapunov exponents. Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). Control of bilateral teleoperation systems with linear and nonlinear dynamics. The study of nonlinear dynamics is based on the study of phase portraits, wavelet and Fourier spectra, signals, chaotic phase synchronization, Lyapunov indicators. Keywords Piecewise nonlinear system, rolling mill, non-smooth homoclinic orbit, bifurcation, chaos. Around the origin there are periodic orbits corresponding to small oscillations of the pendulum that are called librations. Planar linear systems - eigenvalues and eigenvectors, | {
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of the pendulum that are called librations. Planar linear systems - eigenvalues and eigenvectors, phase portraits, classification. MATLAB offers several plotting routines. That is, only initial points located on this orbit result in this closed orbit. As before, we use a phase portrait for stability analysis. We will look at three examples, and also reexamine the undamped pendulum that we studied previously using only its vector field. The higher degree of chaoticity in BB relative to VB is unwrapped through the maximal Lyapunov exponent. Phase Plane Analysis 17 2. For optimal bang-bang trajectories with high values of the energy integral, a general upper bound on the number of switchings was obtained. Phase portraits via trace and determinant. In fact, if we zoom in around this point, it would look like the case of a node of a linear system (in the sense of Chapter 7). (561); Notes LS (power series excluded), GS; Handout on phase portraits. Complex eigenvalues, phase portraits, and energy 4. Although the Riccati equation is not generally a Morse–Smale vector field, we are able to show that it possesses suitable generalizations of many of the important properties of Morse–Smale vector fields. If a system is chaotic, there will be an infinite number of points in the phase portrait. Thus, the equilibrium x = 0 is a saddle, hence unstable, when = 0. (reductor and multipliers). Consider the non-linear system dx dt = y dy dt = 2x+(1x2 y )y. (a) Compute the eigenvalues of A. As a result of one more Andronov-Hopf bifurcation more complex quasiperiodic solution is formed in the system—it is torus of dimension three. 2 Constructing Phase Portraits 23 2. 01385 till 0. Animated phase portraits of nonlinear and chaotic dynamical systems allow one to shape globally the state- and time-dependent convergence behaviour ideally suited to the non-linear or time. This approach of linearizing, analyzing the linearizations, and piecing the results together is a standard approach for | {
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analyzing the linearizations, and piecing the results together is a standard approach for non-linear systems. : A = 1 4 2 −1 λ1 = 3 ↔ v1 = [2,1]T λ2 = −3 ↔ v2 = [−1,1]T x'=x+4y, y'=2x−y −5 0 5 −5 0 5 x y Time Plots for 'thick' trajectory. the trajectories of the nonlinear system are similar to those of the linearized system, so go round anticlockwise. One- and two- dimensional flows. 2 Phase Plane Analysis. [2] Consider x′ 1 = 5x1 −x2 1 − x1x2, x′ 2 = −2x2 +x1x2. • Be able to determine the phase plane and phase portraits of a 2 by 2 linear system. By varying the initial conditions of the system, it is found. 2~a!# is. 2 Bifurcation set and phase portraits of the Hamiltonian system (5). Several nonlinear wave solutions as the solitary wave solutions,topological solitons, cnoidal wave solutions, singular periodic waves and others were obtained. φ 1 = phase shift of the fundamental harmonic component of output. 4 Phase Plane Analysis of Linear Systems 30. REFERENCES [1] Berrymann, A. Phase portraits are an invaluable tool in studying…. Putting all this together we see that the phase portrait is as shown below. Assume that r > 0. In a planar system such as this, the nullclines can provide useful information about the phase portrait. The phase portrait is a plot of a vector field which qualitatively shows how the solutions to these equations will go from a given starting point. We show that our model can recover qualitative features of the phase portrait such as attractors, slow points, and bifurcations, while also producing reliable long-term future predictions in a variety of dynamical models and in real neural data. 2 Prey dynamics predicted by the Lotka-Volterra predator-prey model. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. Existence of Periodic Orbits. The Poincar´e-Bendixson theorem Any orbit of a 2D continuous dynamical system which stays in a closed and bounded subset of the phase plane forever must either tend to a | {
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system which stays in a closed and bounded subset of the phase plane forever must either tend to a critical point or to a. Linear stability analysis may fail for a non-hyperbolic fixed point: ( Re(µ 1, 2) = 0, or at least one µ i = 0 ). Damped Pendulum. 3 Symmetry in Phase Plane Portraits 22 2. Solving 2x2 homogeneous linear systems of differential equations 3. Problem: Construct and analyze a phase-plane portrait of a nonlinear system depicted in the following picture (desired value is w = 0), decide which of the equilibrium points are stable and which are not. Vehicle control synthesis using phase portraits of planar dynamics ABSTRACTPhase portraits provide control system designers strong graphical insight into nonlinear system dynamics. Let us discuss the basic concept of describing the function of non linear control system. 6 and the phase portrait for the original system is in Figure 5. System analysis based on Lyapunov's direct method. and sketch the phase portrait on the circle. Then sufficiently close to (0,0) all trajectories are closed curves. Determining time from phase portraits. Normalized phase portraits or cylindrical phase portraits have been extensively used to overcome the original phase portrait’s disadvantages. x c c c t ert yert y c c c t 1 2 2, 1 2 2 Case 3: Phase Portraits (5 of 5) The phase portrait is given in figure (a) along with several graphs of x1 versus t are given below in figure (b). ) Lecture, three hours; discussion, one hour. the behavior of the nonlinear system from various initial conditions. Jordan, Peter Smith, and P. 2 Phase portrait for an example system. phase portrait get from simulink Example 2. -----, Phase portraits of non degenerate quadratic systems with finite multiplicity one, Nonlinear Anal. Students will learn nonlinear differential equations in the context of mathematical modeling. This allowed to obtain exhaustive solution of the control problem comparing to the known results. Unit 2: Nonlinear 2x2 systems. 1 | {
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solution of the control problem comparing to the known results. Unit 2: Nonlinear 2x2 systems. 1 Phase portraits 72 3. In this paper, based on the classic Chua's circuit, a charge-controlled memristor is introduced to design a novel four-dimensional chaotic system. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. A differential equation system has a limit cycle, if for a set of initial conditions, x(t 0) = x0 and y(t 0) = y0, the solution functions, x(t) and y(t), describe an isolated, closed orbit. The low intensity xed point appears on the phase portraits. Consider the nonlinear system (a) Show that the origin (O, O) is a nonlinear saddle awl plot the phase portrait, including the. | {
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Two Circles inside a right angle!
The other day I was playing with Ms Paint drawing circles here and there -I coincidentally drew a circle inside a right angled triangle which I already drew . Strangely A problem struck to my mind and I tried solving it , but I was unable to do so . I put forward the statement of the problem which I managed to frame my self
Problem : The legs of a right angled triangle are of length $a$ and $b$ . Two circles with equal radii are drawn such such that they touch each other and sides of the triangle as shown in the figure . Find the radius of the circle in terms of $a$ and $b$.
Figure - Of course my MS Paint one -
Further Scope- Is there any way to generalize this for other shapes or for any other triangle?
-------EDIT---------------------
Now to make things interesting : Say we have a right angled triangle which is given . Then is there a method by which we can construct those two circles with a straightedge and a compass?
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Perhaps this topic may be useful? mathworld.wolfram.com/Excircles.html – Erik Miehling Mar 22 at 6:25
@ErikMiehling thanks but it isn't helping much! – Shivam Patel Mar 22 at 6:39
@ShivamPatel Can you give an example of a generalization? I don't really know "right pentagons" and "right hexagons". – DanielV Mar 22 at 6:51
Or perhaps a generalization to multidimensional right triangles and hyperspheres would be meaningful? – DanielV Mar 22 at 7:03
Straightedge and compass construction: draw an arbitrary circle with center $C$ that intersects the legs $AC, BC$ at $D, E$, respectively. Locate $F$ on $AC$ such that $CF = 3CA$. Locate $G$ such that $ECFG$ is a rectangle. Draw $CG$. Draw the angle bisector of $A$. The intersection of this angle bisector with $CG$ is the center of one of the two circles. – heropup Mar 23 at 3:44 | {
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Writing $a := |BC|$, $b := |CA|$, $c := |AB| = \sqrt{a^2+b^2}$, and $r = |PE| = |PF|$ (so that $|PD| = 3r$), we have \begin{align} |\triangle ABC| &= |\triangle ABP| + |\triangle BCP| + |\triangle CAP| \\[4pt] \implies \qquad \frac{1}{2} |BC||CA| &= \frac{1}{2} \left(\; |AB| |PF| + |BC||PD| + |CA||PE| \;\right) \\[4pt] \implies \qquad a b &= c r + 3 a r + b r = r ( 3 a + b + c )\\[6pt] \implies \qquad r &= \frac{ab}{3 a + b + c} = \frac{ab}{3 a + b + \sqrt{a^2+b^2}} \end{align}
To address @DanielV's suggestion of generalizing to higher dimensions, consider a right-corner tetrahedron $OABC$, with right corner at $O$ and edge lengths $a := |OA|$, $b := |OB|$, $c := |OC|$. (Note that I'm changing notation slightly from the above.) Let a sphere with center $P$ and radius $r$ be tangent to the faces around vertex $A$, and let a congruent sphere (tangent to the first) be tangent to the faces around vertex $O$. Then $P$ has distance $r$ from faces $\triangle OAB$, $\triangle OCA$, $\triangle ABC$ (the ones touching $A$), and distance $3r$ from face $\triangle OBC$ (the one opposite $A$).
Here's a poor attempt at a diagram:
(In this case, the altitudes from $P$ are color-coded to match their parallel counterparts through $O$. The black altitude is to face $\triangle ABC$.)
Thus,
\begin{align} |OABC| &= |OABP| + |OBCP| + |OCAP| + |ABCP| \\[4pt] \implies \qquad \frac{1}{6}a b c &= \frac{1}{3}\left(\; r\;|\triangle OAB| + r \;|\triangle OCA| + r\;|\triangle ABC| + 3r\;|\triangle OBC| \;\right) \\[4pt] &= \frac{1}{3}r \cdot \frac{1}{2} \left(\; a b + c a + 3 b c + 2\;|\triangle ABC| \;\right) \\[6pt] \implies \qquad r &= \frac{abc}{3bc + ab + ca + 2\;|\triangle ABC|} \qquad (\star) \end{align} | {
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Fun fact: The Pythagorean Theorem for Right-Corner Tetrahedra says that $$|\triangle ABC|^2 = |\triangle OBC|^2 + |\triangle OCA|^2 + |\triangle OAB|^2$$ so that we have $$|\triangle ABC| = \frac{1}{2} \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}$$ and $(\star)$ becomes $$r = \frac{abc}{3bc + ab + ca + \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}}$$
In $4$-dimensional space (where there's an analogous Pythagorean Theorem, as there is in any-dimensional space), we have $$r = \frac{abcd}{3bcd + acd + abd + abc + \sqrt{\;b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2\;}}$$ and so forth.
Incidentally, the matching-notation version of the initial answer is $$r = \frac{ab}{3b + a + \sqrt{\;b^2 + a^2\;}}$$ | {
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-
@qwr: Thanks. :) As for coloring: I just like to highlight the association of a vertex with its opposite edge in a triangle, and other elements get their colors accordingly. I chose not to color-code the altitudes dropped from $P$ to the edges (I had reflexively done so in a "draft" image), in order to tie the congruent segments together visually. – Blue Mar 22 at 8:00
@mathh: I use GeoGebra for my figures these days. – Blue Mar 22 at 12:02
@AJP: "$|BC|$" represents the length of segment $\overline{BC}$. (I often leave out the over-bar to save typing.) By analogy, I use "$|\triangle ABC|$" for the area of $\triangle ABC$ (and then "$|OABC|$" for the volume of tetrahedron $OABC$, etc); this is non-standard, but I like it. :) The symbol ":=" indicates "is defined to be"; writing "$a:=|BC|$" says "I'm going to write '$a$' for '$|BC|$'". Others (even I) might otherwise write "let $a = |BC|$", but using ":=" is ever-so-slightly better, as it distinguishes definition "'$a$' means '$|BC|$'" from relation "$a$ equals $|BC|$". – Blue Mar 22 at 13:10
@ShivamPatel: I added a figure that may (or may not) help with the $3d$ case. Just don't ask me for the $4d$ version. :) – Blue Mar 22 at 14:46
@mathh: In GeoGebra, you can easily mark an angle with an arc via the "Angle" tool; if the sides of the angle happen to be perpendicular, GeoGebra turns the arc into a box. (There's a setting somewhere to turn that behavior off and on.) The little dashes are somewhat less convenient: with the segment selected, you have to open the "Object Properties..." panel and then the "Decoration" tab. (You can likewise "decorate" angles with little dashes and such.) – Blue Mar 22 at 15:28
Assuming the corner of the triangle is the origin of a Cartesian plane, the line of the hypotenuse is $$y = -\frac{a}{b} x + a$$
The center of the second circle is at $\begin{bmatrix} 3r \\ r \end{bmatrix}$. | {
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The center of the second circle is at $\begin{bmatrix} 3r \\ r \end{bmatrix}$.
The radius of the second circle is the directed vector $\begin{bmatrix} ra \\ rb \end{bmatrix}\frac{1}{\sqrt{a^2 + b^2}}$.
Altogether :
$$\frac{rb}{\sqrt{a^2 + b^2}} + r = -\frac{a}{b}\left(3r + \frac{ra}{\sqrt{a^2 + b^2}}\right) + a$$
$$r = \frac{ab}{ 3a + b + \sqrt{a^2 + b^2}}$$
Since Blue showed an elegant volume based approach, I guess I'll try a multidimensional coordinate based approach. Suppose the origin $\begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \end{bmatrix}$ and the points $\begin{bmatrix} \mathcal{l}_0\\ 0 \\ 0 \\ \vdots \end{bmatrix}$, $\begin{bmatrix} 0 \\ \mathcal{l}_1 \\ 0 \\ \vdots \end{bmatrix}$, $\begin{bmatrix} 0 \\ 0 \\ \mathcal{l}_2 \\\vdots \end{bmatrix}$ etc form a multidimensional right triangle. Suppose one hypersphere is tucked in at the origin and one hypersphere is tucked in at the corner along the first axis.
The hypotenuse plane of the triangle is $$\frac{x_0}{\mathcal{l}_0} + \frac{x_1}{\mathcal{l}_1} + \frac{x_2}{\mathcal{l}_2} + \dots = 1 \tag{Plane equation}$$
Or equivalently, using $\circ$ for dot product and $N = \begin{bmatrix} \frac{1}{\mathcal{l}_0} \\ \frac{1}{\mathcal{l}_1} \\ \frac{1}{\mathcal{l}_2} \\ \vdots \end{bmatrix}$, then the plane is: $$N \circ X = 1 \tag{Plane equation with dot product}$$
The center of the second hypershpere is $$c = \begin{bmatrix} 3r \\ r \\ r \\ \vdots\end{bmatrix} = rc_0 \tag{Center of second hypersphere}$$
The vector directed from the center of the second hypersphere to the hypotenuse plane is $$r_2 = r\frac{N}{|N|}\tag{Directed radius to hypotenuse}$$
Altogether: $$N \circ (c + r_2) = 1$$ $$N \circ \left(rc_0 + r\frac N {|N|}\right) = 1$$ \begin{align} r &= \frac{1}{N \circ c_0 + |N|} \\ &= \frac{1}{3l_0^{-1} + l_1^{-1} + l_2^{-1} \dots + \sqrt{l_0^{-2} + l_1^{-2} + l_2^{-2} \dots}} \end{align} | {
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Note this is the same answer as Blue, but with $\mathcal{l}_0\mathcal{l}_1 \mathcal{l}_2 \dots$ factored out of the numerator and denominator.
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+1. It's worth nothing that your multi-dimensional argument effectively re-derives the formula for the distance from point $P(p_0,p_1,\dots,p_n)$ to hyperplane $Q:\;q_0x_0+q_1x_1+\cdots+q_nx_n=q$; namely, $$\text{dist}(P,Q)=\frac{|\;p_0q_0+p_1q_1+\cdots+p_nq_n-q\;|}{\sqrt{\;q_0^2+ q_1^2 +\cdots+q_n^2\;}}$$ Replacing $p_0=3r$, $p_1=\cdots=p_n=r$, $q_i=l^{-1}_{i}$, and $q=1$, and then setting the computed distance equal to $r$, gives the equation you've solved for $r$. (Note: Dropping the formula's absolute value sign makes the distance negative here, since $P$ is on the "origin-side" of $Q$.) – Blue Mar 22 at 22:29
Interesting how many different ways there can be to reach the same result. This is one reason I say that instead of memorizing formulas, students would appreciate mathematics more if they saw the "magic" of how many seemingly unrelated approaches give the same result in the end. – DanielV Mar 22 at 23:38
@DanielV Can you look at the edit please... – Shivam Patel Mar 23 at 3:24
@ShivamPatel Look at youtube.com/watch?v=CMP9a2J4Bqw espcially after 00:58 where it describes what is possible with a compass and straightedge. In short the answer is "yes", by applying what you see in the video to the formulas Blue and I gave you. ...But I doubt there is any short way to do it. Keep in mind that you effectively assume $a=1$ or $b=1$ since the units are arbitrary. – DanielV Mar 23 at 3:37
Alternative solution: | {
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Alternative solution:
Recall that for any triangle $\triangle ABC$, the area of the triangle equals the product of the inradius and its semiperimeter; i.e., $|\triangle ABC| = rs$, where $s = (a+b+c)/2$. Therefore, given legs $a, b$, $c = \sqrt{a^2+b^2}$, and $$r = \frac{2|\triangle ABC|}{a+b+c} = \frac{ab}{a+b+\sqrt{a^2+b^2}}.$$ Draw the tangent line to the two circles at their common point of tangency: this creates a smaller similar triangle with scaling factor $\frac{b-2\rho}{b}$ where $\rho$ is the common radius of the two circles. If $r$ is the inradius of $|\triangle ABC|$ as shown, then $$\frac{\rho}{r} = \frac{b - 2\rho}{b}.$$ Putting all of this together, we find $$\rho = \frac{br}{b+2r} = \frac{a b}{3a+b+c} = \frac{ab}{3a+b+\sqrt{a^2+b^2}}.$$ This method easily generalizes to more than two congruent circles tangent to one side: if $n$ circles are arranged along the leg of length $b$, then it is straightforward to find that $\rho = \frac{ab}{(2n-1)a + b + \sqrt{a^2+b^2}}$.
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What about determining $a,b$ from $r$?
You can't - $a,b$ are not uniquely determined by $r$. Draw the two circles first, then draw the legs a and b as infinite lines. And line tangent to the right most circle will intersect lines A and B creating a right triangle, but for different tangents $a$ and $b$ will be different and r hasn't changed.
You can't even turn this into an interesting question by asking for all the possible ways to do it, because it turns out that these circles don't actually impose any restriction - since any angels can be achieved by the tangent line, any right triangle can. All $r$ does is create a scaling factor. $a$ can take on any value in $(2r,\infty)$ and $b$ the corresponding value in $(4r,\infty)$
suppose we say $a=kb$ for some fixed $k$ then? – Shivam Patel Mar 22 at 6:46
I believe the question was to find the radius, given $a$ and $b$. – N. Owad Mar 22 at 6:48 | {
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# How to find the intersection of two hyperplanes in $n$ dimensions?
I want to find out the intersection of two surfaces that exist in $n$-dimensions. These surfaces are defined by a system of $2$ linear equations.
$$A_1x+B_1y+C_1z+\cdots = D_1$$
$$A_2x+B_2y+C_2z+\cdots = D_2$$
But first, how would these surfaces look like? In $3$-D space, they would be planes, but in dimensions higher than $3$ are these not planes? Can we generalize a way to find the intersection of such linear expressions in some way?
Also can this problem be solved using some tools, preferably Matlab?
• The surfaces are hyperplanes. Solve a linear system of the form $\rm A x = b$, where $\rm A$ is a $2 \times n$ matrix. If $n > 2$, the linear system is underdetermined. If $n=3$, the intersection could be a line. If $n=4$, the intersection could be a $2$-dimensional plane. – Rodrigo de Azevedo Jul 16 '17 at 15:27
• You can use Gaussian elimination to find a parametrization of the affine space that is the intersection, assuming that the intersection is not empty. – Rodrigo de Azevedo Jul 16 '17 at 15:31
• @RodrigodeAzevedo And how do I figure out t1 and t2? – Vidor Vistrom Jul 16 '17 at 15:44
• Suppose we have $$\begin{bmatrix} 1 & 0 & -1 & -1\\ 0 & 1 & -1 & -1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} = \begin{bmatrix} 1\\ 1\end{bmatrix}$$ The solution set is the plane parameterized as follows $$\left\{ \begin{bmatrix} 1\\ 1\\ 0\\ 0\end{bmatrix} + t_1 \begin{bmatrix} 1\\ 1\\ 1\\ 0\end{bmatrix} + t_2 \begin{bmatrix} 1\\ 1\\ 0\\ 1\end{bmatrix} : t_1, t_2 \in \mathbb R \right\}$$ – Rodrigo de Azevedo Jul 16 '17 at 15:48
• @RodrigodeAzevedo: I suggest that you post your solution, perhaps mentioning the reduced row-echelon form of the augmented matrix for the $2\times n$ system, as an Answer. I will certainly upvote even if you mostly transcribe your Comments into the Answer box. – hardmath Jul 16 '17 at 21:49 | {
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Suppose we would like to find the intersection of $2$ hyperplanes in $\mathbb R^n$
$$\begin{array}{rl} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n &= b_1\\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n &= b_2\end{array}$$
In matrix form,
$$\begin{bmatrix} — \mathrm a_1^\top —\\ — \mathrm a_2^\top —\end{bmatrix} \mathrm x = \begin{bmatrix} b_1\\ b_2\end{bmatrix}$$
or, more economically, $\rm A x = b$, where $\rm A$ is a $2 \times n$ matrix. Without loss of generality, we assume that both hyperplanes are in Hessian normal form, i.e., $\| \mathrm a_1 \|_2 = \| \mathrm a_2 \|_2 = 1$.
If vectors $\rm a_1$ and $\rm a_2$ are collinear, then the $2$ given hyperplanes are either parallel or overlapping, neither of which are very interesting. Thus, we assume that vectors $\rm a_1$ and $\rm a_2$ are not collinear, i.e., matrix $\rm A$ has full row rank. Let $\theta := \arccos ( \mathrm a_1^\top \mathrm a_2 )$ be the angle formed by vectors $\rm a_1$ and $\rm a_2$.
The solution set of the linear system $\rm A x = b$ is a $(n-2)$-dimensional affine space. To find this affine space, we must find a particular solution and the null space of $\rm A$. One particular solution would be the least-norm solution, i.e., the one closest to the origin | {
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$$\begin{array}{rl} \mathrm x_{\text{LN}} := \mathrm A^\top \left( \mathrm A \mathrm A^\top \right)^{-1} \mathrm b &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} \| \mathrm a_1 \|_2^2 & \langle \mathrm a_1, \mathrm a_2 \rangle\\ \langle \mathrm a_2, \mathrm a_1 \rangle & \| \mathrm a_2 \|_2^2\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & \cos (\theta)\\ \cos (\theta) & 1\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \frac{1}{\sin^2 (\theta)} \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & -\cos (\theta)\\ -\cos (\theta) & 1\end{bmatrix} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \left(\frac{b_1 - b_2 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_1 \\ | \end{bmatrix} + \left(\frac{b_2 - b_1 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_2 \\ | \end{bmatrix}\end{array}$$
where $\sin (\theta) \neq 0$ is ensured by the non-collinearity of vectors $\rm a_1$ and $\rm a_2$.
To find the $(n-2)$-dimensional null space of $\rm A$, we solve the homogeneous linear system $\rm A x = 0_2$. Introducing an $n \times n$ permutation matrix $\rm P$ with certain desired properties, we reorder the columns of $\rm A$, i.e., $\rm A P P^\top x = 0_2$, and obtain a homogeneous linear system in $\rm y:= P^\top x$
$$\rm A P y = 0_2$$
If $\rm P$ is chosen wisely, then there is a $2 \times 2$ matrix $\rm E$, a product of elimination matrices, that puts $\rm A P$ in reduced row echelon form (RREF), i.e.,
$$\rm E A P = \begin{bmatrix} \mathrm I_2 & \mathrm F\end{bmatrix}$$
where $\rm F$ is a $2 \times (n-2)$ matrix. Hence, the null space of $\rm A P$ is given by
$$\left\{ \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta : \eta \in \mathbb R^{n-2} \right\}$$
and, thus, the null space of $\rm A$ is given by | {
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and, thus, the null space of $\rm A$ is given by
$$\left\{ \mathrm P \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta : \eta \in \mathbb R^{n-2} \right\}$$
Lastly, the intersection of the $2$ given hyperplanes is given by
$$\left\{ \color{blue}{\mathrm x_{\text{LN}} + \mathrm P \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta} : \eta \in \mathbb R^{n-2} \right\}$$
In MATLAB, we can use function rref to find permutation matrix $\rm P$ and matrix $\rm F$. Of course, we can also use function null to find an orthonormal basis for the null space of $\rm A$. | {
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1. ## Need help with a definite integral please
I would appreciate advice on solving the following problem.
Evaluate the definite integral:
integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx
Here is my own attempt.
Step 1: Solve for the generic (indefinite) integral via substitution
Let u = x^2 + 1
So then
du / dx = 2x
2x dx = du
dx = du / 2x
Returning to the original function:
integral 2x (x^2 + 1) dx
= integral 2x (u) dx
= integral 2x (u) (du / 2x)
= integral u du
= integral x^2 + 1
= (x^3 / 3 ) + x + C
Step 2: solve for the definite integral given the stated limits
At upper limit of 2:
((2)^3 / 3) + (2) +C
= (8/3) +2 + C
= (8/3) + (6/3) + C
= 14/3 + C
At lower limit of 1
((1)^3 / 3) + (1) +C
=(1/3) +1 + C
= (1/3) + (3/3) + C
= 4/3 + C
Subtracting the lower limit result from upper limit result:
(14 /3 + C) - (4/3 + C)
=14/3 + C - 4/3 - C
= 10 / 3
But my textbook and calculator both say the answer is (21/2)
What am I doing wrong?
2. Originally Posted by lingyai
I would appreciate advice on solving the following problem.
Evaluate the definite integral:
integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx
Here is my own attempt.
Step 1: Solve for the generic (indefinite) integral via substitution
Let u = x^2 + 1
So then
du / dx = 2x
2x dx = du
dx = du / 2x
Returning to the original function:
integral 2x (x^2 + 1) dx
= integral 2x (u) dx
= integral 2x (u) (du / 2x)
= integral u du
= integral x^2 + 1
= (x^3 / 3 ) + x + C
Step 2: solve for the definite integral given the stated limits
At upper limit of 2:
((2)^3 / 3) + (2) +C
= (8/3) +2 + C
= (8/3) + (6/3) + C
= 14/3 + C
At lower limit of 1
((1)^3 / 3) + (1) +C
=(1/3) +1 + C
= (1/3) + (3/3) + C
= 4/3 + C
Subtracting the lower limit result from upper limit result:
(14 /3 + C) - (4/3 + C)
=14/3 + C - 4/3 - C
= 10 / 3
But my textbook and calculator both say the answer is (21/2)
What am I doing wrong?
Evaluate:
$\int_1^2 2x(x^2+1)dx$ | {
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} |
$\int_1^2 2x(x^2+1)dx$
Ssubstitute $u=x^2+1$ , $dx=\frac{1}{2x}du$:
$\int_{x=1}^2 2x(x^2+1)dx=\int_{u=2}^5 u\ du=\left[u^2/2\right]_{u=2}^5$ $=\frac{25}{2}-\frac{4}{2}=\frac{21}{2}$
RonL
3. Originally Posted by lingyai
I would appreciate advice on solving the following problem.
Evaluate the definite integral:
integral (upper limit 2, lower limit 1) [ 2x (x^2 +1 )] dx
Here is my own attempt.
Step 1: Solve for the generic (indefinite) integral via substitution
Let u = x^2 + 1
So then
du / dx = 2x
2x dx = du
dx = du / 2x
Returning to the original function:
integral 2x (x^2 + 1) dx
= integral 2x (u) dx
= integral 2x (u) (du / 2x)
= integral u du
= integral x^2 + 1
= (x^3 / 3 ) + x + C
Step 2: solve for the definite integral given the stated limits
At upper limit of 2:
((2)^3 / 3) + (2) +C
= (8/3) +2 + C
= (8/3) + (6/3) + C
= 14/3 + C
At lower limit of 1
((1)^3 / 3) + (1) +C
=(1/3) +1 + C
= (1/3) + (3/3) + C
= 4/3 + C
Subtracting the lower limit result from upper limit result:
(14 /3 + C) - (4/3 + C)
=14/3 + C - 4/3 - C
= 10 / 3
But my textbook and calculator both say the answer is (21/2)
What am I doing wrong?
What you did wrong?
That there after Integral u du.
You continued with Integral x^2 +1.
That is wrong.
You were already in the substitution and then you went back to the original, without integrating yet. You should have continued the integration usin the substitution. That is why you chose to use substitution in the first place.
Then in your going back to the original, from INT. u du, the switchback is wrong also.
What is u?
It is x^2 +1
What is du?
It is 2x dx.
INT. u du = INT. x^2 +1
only?
It should have been
INT. u du = INT. (x^2 +1) *(2x dx)
which is the same as the original integrand.
Which would have put you back to square one or the beginning again.
4. Hello, lingyai!
Thank you for showing your work and reasoning . . .
Evaluate: $\:\int^2_1 2x(x^2 + 1)dx$
Did you (or anyone) notice that the integrand is a polynomial? | {
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Evaluate: $\:\int^2_1 2x(x^2 + 1)dx$
Did you (or anyone) notice that the integrand is a polynomial?
$\int^2_12x(x^2+1)dx \;= \;\int^2_1(2x^3 + 2x)\,dx \;= \;\frac{1}{2}x^4 + x^2\bigg|^2_1$
. . $= \;\left(\frac{1}{2}\cdot2^4 + 2^2\right) - \left(\frac{1}{2}\cdot1^4 + 1^2\right) \;= \;(8 + 4) - \frac{3}{2}\;=\;\frac{21}{2}$
5. Originally Posted by Soroban
Hello, lingyai!
Thank you for showing your work and reasoning . . .
Did you (or anyone) notice that the integrand is a polynomial?
$\int^2_12x(x^2+1)dx \;= \;\int^2_1(2x^3 + 2x)\,dx \;= \;\frac{1}{2}x^4 + x^2\bigg|^2_1$
. . $= \;\left(\frac{1}{2}\cdot2^4 + 2^2\right) - \left(\frac{1}{2}\cdot1^4 + 1^2\right) \;= \;(8 + 4) - \frac{3}{2}\;=\;\frac{21}{2}$
I was too busy noticing that:
$
\:\int^2_1 2x(x^2 + 1)dx=\:\int^2_1 \frac{1}{2}\frac{d}{dx}(x^2 + 1)^2dx
$
$
=\frac{1}{2}(x^2 + 1)^2 \bigg|_{x=1}^2=\frac{25-4}{2}=\frac{21}{2}$
but that would not help the OP spotting what they had done wrong
RonL | {
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} |
# Align*: horizontal spacing is inconsistent
I am new to Latex, and I am having some trouble with the align* environment. In the code below, the first two equations are horizontally spaced differently from the second two equations:
\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\dfrac{1}{n}-\dfrac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1, \inf(A)=-1$.\\\\
\textit{Proof}. Suppose for contradiction that $1$ is not an upper bound for $A$, i.e. $\exists n,m \in \mathbb{N} \mid \dfrac{1}{n} - \dfrac{1}{m} > 1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\text{Because }n(m+1)\geq m+1,\text{ we have}\\
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot\\
\end{align*}
$\newline$Suppose for contradiction that $-1$ is not a lower bound for $A$, i.e. $\ex n,m \in \N \mid$\\$\dfrac{1}{n} - \dfrac{1}{m} < -1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\text{Because }m(1+n)\geq 1+n, \text{ we have}\\
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\end{align*}
$\newline\newline$ Claim that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$, or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp$\\$\left|(1 - \dfrac{1}{m}) - 1\right| < \epsilon$. | {
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} |
\begin{align*}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\end{align*}
Hence, for $M:=\left\lceil\dfrac{1}{\epsilon}\right\rceil+1$, $m > M \imp \left| 1 - 1 - \dfrac{1}{m}\right|<\epsilon$. Because $\epsilon$ was arbitrarily chosen, it follows that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$.
$\newline\newline$ Claim that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$, or $\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp$\\$\left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$.
\begin{align*}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \dfrac{1}{\epsilon}\right\rceil+1, n > N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$. Because $\epsilon$ was arbitrarily chosen, it follows that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$.
\end{document}
I'm not exactly sure why this is happening. Perhaps it has something to with the \frac{}{} equations in the second two equations... and if so, how can I fix the spacing so that all the equations are aligned the same? Thank you. | {
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• What do you consider the first and the second equation in your code? It's not very clear... Nov 7 '17 at 4:11
• Welcome to TeX.SE. If your issue is with alignment across separate align*, you need to combine them into a single align* enironment and use \intertext{} for the text that you have between the two align* blocks. Nov 7 '17 at 6:07
• Well, the space between the lines are probably the same, but as you say the lines are higher in the second two align blocks due to the \fracs. But do you really want to change that? You don't need more space between the lines for the first two. (You can of course do it, see e.g. tex.stackexchange.com/questions/2929/…, ) Nov 7 '17 at 8:27
• @Werner The first two equations are the proofs by contradiction. The last two equations are the limit proofs. I want to fix the horizontal alignment since the first two proofs are spaced differently than the second two.
– user147592
Nov 7 '17 at 13:19
• @TorbjørnT. I apologize for not being more clear. I want to fix the horizontal alignment, not the vertical alignment. The first two equations are horizontally aligned differently from the second two, and I want to fix that.
– user147592
Nov 7 '17 at 13:21
You're probably looking for \intertext:
\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\dfrac{1}{n}-\dfrac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1$, $\inf(A)=-1$.
\begin{proof}
Suppose for contradiction that $1$ is not an upper bound for $A$, i.e.\@
$\exists n,m \in \mathbb{N} \mid \dfrac{1}{n} - \dfrac{1}{m} > 1$. | {
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"url": "https://tex.stackexchange.com/questions/400065/align-horizontal-spacing-is-inconsistent"
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\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\intertext{Because $n(m+1)\geq m+1$, we have}
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot\\
\intertext{\indent Suppose for contradiction that $-1$ is not a lower bound for $A$,
i.e.\@ $\ex n,m \in \N \mid \dfrac{1}{n} - \dfrac{1}{m} < -1$.}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\intertext{Because $m(1+n)\geq 1+n$, we have}
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\intertext{\indent Claim that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$,
or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp \left|(1 - \dfrac{1}{m}) - 1\right| < \epsilon$.}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\intertext{Hence, for $M:=\left\lceil\dfrac{1}{\epsilon}\right\rceil+1$,
$m > M \imp \left| 1 - 1 - \dfrac{1}{m}\right|<\epsilon$. Because $\epsilon$
was arbitrarily chosen, it follows that
$\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$.\endgraf\medskip
\indent Claim that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$,
or $\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$.}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \dfrac{1}{\epsilon}\right\rceil+1$,
$n > N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$.
Because $\epsilon$ was arbitrarily chosen, it follows that
$\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$.
\end{proof}
\end{document} | {
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\end{document}
EDIT: Found the error: you have too many & in the last line of the first two align envs
Below is a cleaned up MWE (with the error still present)
• don't use \dfrac in the text, causes excessive line spacing
• don't use \\ or \newline in the text, you never want manual line breaks in the text (this is a very common mistake among new users)
• use \intertext{...} for comments inside align (mathtools provides \shortintertext which has different spacing)
Cleaned MWE
\documentclass{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\frac{1}{n}-\frac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1, \inf(A)=-1$.
\begin{proof}
Suppose for contradiction that $1$ is not an upper bound for $A$,
i.e.
$\exists n,m \in \mathbb{N} \mid \frac{1}{n} - \frac{1}{m} > 1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\intertext{Because $n(m+1)\geq m+1$, we have}
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot
\end{align*} | {
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Suppose for contradiction that $-1$ is not a lower bound for $A$,
i.e. $\ex n,m \in \N \mid \frac{1}{n} - \frac{1}{m} < -1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\intertext{Because $m(1+n)\geq 1+n$, we have}
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\end{align*}
Claim that $\lim_{n=1, m \to \infty} \frac{1}{n} - \frac{1}{m} = 1$, or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp$ $\left|(1 - \frac{1}{m}) - 1\right| < \epsilon$.
\begin{align*}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\end{align*}
Hence, for $M:=\left\lceil\frac{1}{\epsilon}\right\rceil+1$,
$m > M \imp \left| 1 - 1 - \frac{1}{m}\right|<\epsilon$. Because
$\epsilon$ was arbitrarily chosen, it follows that
$\lim_{n=1, m \to \infty} \frac{1}{n} - \frac{1}{m} = 1$.
Claim that
$\lim_{m=1, n \to \infty} \frac{1}{n} - \frac{1}{m} = -1$, or
$\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp$
$\left|(\frac{1}{n}-1) -(-1)\right| < \epsilon$.
\begin{align*}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \frac{1}{\epsilon}\right\rceil+1, n > N \imp \left|(\frac{1}{n}-1) -(-1)\right| < \epsilon$. Because
$\epsilon$ was arbitrarily chosen, it follows that
$\lim\limits_{m=1, n \to \infty} \frac{1}{n} - \frac{1}{m} = -1$.
\end{proof}
\end{document}
Here is a image of the first page (after cleaning), I can see what you mean about the unevenness | {
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Here is a image of the first page (after cleaning), I can see what you mean about the unevenness
• So there isn't a way to fix the unevenness?
– user147592
Nov 7 '17 at 15:14
• @SuhasHoysala See my update, I just left the image and the clean MWE for others to actually see the problem. You have to unnecessary & Nov 7 '17 at 15:27
• I removed the extra & (before the \bot command), but it still was uneven... I'm not sure what I'm doing wrong.
– user147592
Nov 7 '17 at 20:33
• @SuhasHoysala did you also remember to use \intertext as I did in my cleanup? My fix was literally the mwe cleanup I posted and then removing the two excessive & Nov 7 '17 at 20:35
• Yes, I copy-pasted your code verbatim and removed the two excessive &. It still doesn't seem to be fixed.
– user147592
Nov 7 '17 at 21:44 | {
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} |
# How can I write negations for the following statements?
Premise: Let $F(x,y)$ be the statement $x$ can fool $y$, where the domain for discourse for both $x$ and $y$ is all people.
I converted the following corresponding statements:
1) Nobody can fool me
2) Anybody can Fool Fred
3) Everyone can fool someone
to
1) $\neg \exists xF(x,me)$
2) $\forall xF(x,Fred)$
3) $\forall x\exists yF(x,y)$
Now I am trying to negate these and I can't just throw in a not in front of the statements and I don't know how to go about this.
I looked at the negation rules and came up with this but I doubt it's correct:
1) $\neg\exists xF(x,me) \iff \forall x\neg F(x,me)$
2) $\forall xF(x,Fred) \iff \forall x\neg\neg F(x,Fred) \iff \neg[\exists x\neg F(x,Fred)]$
3) $\forall x\exists yF(x,y) \iff \neg\forall x\exists yF(x,y)$
• Well, $\lnot (\lnot Q)$ is $Q$ so the negation of $\lnot\exists xF(x,me)$ would be ... $\exists xF(x,me)$, right? – fleablood Aug 15 '17 at 22:17
• And the negation of $\forall x A$ is $\exists x \lnot A$ so the negation $\forall x F(x, fred)$ is $\exists x \lnot F(x, fred)$. – fleablood Aug 15 '17 at 22:19
• 3 is a little harder. As with 2) is $\lnot(\forall x \exists y F(x,y))$ will be $\exists x (\lnot \exists y F(x,y))$ but we need to find the negation of $\exists y F(x,y)$. The negation of $\exists x Q$ is $\forall x \lnot Q$ so the negation is $\exists x \forall y \lnot F(x,y)$. i.e. There is someone who can not fool anybody. – fleablood Aug 15 '17 at 22:25
• I think you're spot on, @fleablood. Please post your comments in an answer! I'll upvote it. – Namaste Aug 15 '17 at 22:29
• Wait... you aren't actually negating them. You are stating the equivalence of them. You can't "just" put a not in front, but you must start by putting a not in front. – fleablood Aug 15 '17 at 22:30 | {
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} |
Except for what I assume to be a typo on the answer to 3, what you've written is correct (or more accurately, not wrong), although it seems you've missed out the actual negation of the statements which is what you're after.
Essentially, all you need is the rule
$$\neg((\forall x)\quad P(x))\iff((\exists x)\quad \neg P(x))$$
as well as the more basic $\neg\neg P\iff P$.
Let
$$S_1\equiv¬((∃x)\quad F(x,\text{me}))$$
$$S_2\equiv(∀x)\quad F(x,\text{Fred})$$
$$S_3\equiv(∀x)\quad((∃y)\quad F(x,y))$$
Then you're after $\neg S_1,\neg S_2,\neg S_3$.
As has been noted in the comments, the first is simple
$$\neg S_1\equiv\neg\neg((∃x)\quad F(x,\text{me}))$$
$$\neg S_1\equiv(∃x)\quad F(x,\text{me})$$
Your working for the second is helpful and we get
$$\neg S_2\equiv\neg((∀x)\quad F(x,\text{Fred}))$$
$$\equiv\neg¬((∃x)\quad¬F(x,\text{Fred}))$$
$$\equiv(∃x)\quad¬F(x,\text{Fred})$$
Finally, as has been pointed out by @fleablood, the last is slightly harder, but is just applying the above rule twice
$$\neg S_3\equiv\neg((∀x)\quad((∃y)\quad F(x,y)))$$
$$\equiv(∃x)\quad\neg((∃y)\quad F(x,y))$$
$$\equiv(∃x)\quad((∀y)\quad \neg F(x,y))$$
Okay. First to translate:
1) "Nobody can fool me". "Nobody" can be thought of as either "Everyone can not...", or as "There doesn't exist anyone who can..."
So either $\forall x\lnot F(x,me)$ or $\lnot \exists x F(x,me)$ will both be correct and are equivalent. So yes you are are correct.
As for 2) and 3)... I have nothing to add. You did those just fine. (You did 1) just fine as well, but I did have something to add.)
So the negations:
The rules are simple:
A) $\lnot$(That's not true) is, of course, "That is true" so $\lnot(\lnot Q) \iff Q$.
B) $\lnot$(Everbody does something) is, there is at least one person who doesn't do it. So $\lnot \forall x Q \iff \exists x \lnot Q$.
C) $\lnot$(Somebody does something) is. Nobody does something. So $\lnot \exists x Q \iff \forall x \lnot Q$.
So | {
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So
1) $\lnot(\lnot \exists x F(x,me)) \iff \exists x F(x,me)$
or
$\lnot \forall x\lnot F(x,me) \iff \exists x\lnot(\lnot F(x,me)) \iff \exists x F(x, me)$.
i.e. It is not the case that noone can fool me $\iff$ someone can fool me.
2)$\lnot(\forall x F(x, fred)) \iff \exists x \lnot F(x,fred)$.
Not everbody can fool Fred = there is someone who can't fool Fred.
3) $\lnot (\forall x \exists y F(x,y)) \iff$
$\exists x \lnot(\exists y F(x,y)) \iff$
$\exists x \forall y \lnot F(x,y)$
i.e. not(Everbody can fool somebody) = somebody can not fool anybody. | {
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Regular Polyhedra coordinate points generation
Is it possible to get all points on a Polyhedron surface using two surface parameters, say
$\phi,\theta$ spherical co-ordinates?
Just like in ParametricPlot3D, can we start with PolyhedronData["Tetrahedron"] to obtain spatial point positions?. The tip of position vector should cross edges automatically during $\phi,\theta$ sweep.
• You can always make this kind of tricks – Dr. belisarius Dec 29 '14 at 19:22
• @belisarius: fine,but want to append or prepend available Polyhedra. – Narasimham Dec 29 '14 at 19:59
• This is a nice start. – Dr. belisarius Dec 29 '14 at 20:04
• Stick a small sphere around the barycenter. Now parametrize by the spherical angles, using intersection of ray through spherical point (theta,phi) with polyhedron. – Daniel Lichtblau Dec 29 '14 at 23:57
• Clear enough, we need to have direct coordinates as function of ($\theta,\phi, n$). – Narasimham Dec 30 '14 at 7:58
Exploit the fact that the vertices of the dual to a Platonic solid correspond to the centers of the faces of the solid itself. For instance, the dual to a cube is a regular octahedron, and the six vertices of this octahedron are in the directions of the centers of the faces of its dual cube.
Find the normalized directions of the face centers of the cube (for instance) this way:
FaceCenters = Normalize /@ PolyhedronData[PolyhedronData["Cube", "Dual"], "Faces"][[1]];
The {x,y,z} direction in Euclidean coordinates as a function of spherical angles θ and φ are of course:
x[θ_Real, φ_Real] := {Sin[θ] Cos[φ],
Sin[θ] Sin[φ],
Cos[θ]}
Now sweep through all spherical angles, and for each corresponding direction find the nearest face center direction (i.e., the one with the smallest angle to the direction defined by θ and φ). For each such direction, the distance to the surface is 1/Sin[ψ], where ψ is the scalar angle between the candidate direction and its nearest face direction: | {
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} |
SphericalPlot3D[
1/Sin[x[θ, φ].Nearest[FaceCenters, x[θ, φ]][[1]]],
{θ, 0, π}, {φ, 0, 2 π},
PlotPoints -> 100] // Quiet
If you repeat with a Tetrahedron, for instance, you get this:
If you repeat for an Octahedron, you get this:
If you repeat for a Dodecahedron, you get this:
• Wow... I really thought @Narasimhan would have accepted this answer! What more can he possibly want?! – David G. Stork Jan 6 '15 at 0:58
• That was great. My apologies for the inordinate delay earlier – Narasimham Feb 10 '18 at 5:17
• Wow... two months late! Better late than never! – David G. Stork Feb 10 '18 at 6:20 | {
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What is the least number of square roots needed to express $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{100}$?
What is the least number of square roots needed to express $\sqrt{1}+\sqrt{2}+\cdots+\sqrt{100}$ if it must be expressed in the form $a+b\sqrt{c}+d\sqrt{e}+\cdots$ where $a,b,c,d,e,\ldots$ are all integers?
Solution
In order to find the least number of square roots, we must express this number in simplest form. Thus, the numbers in the radicand that are in simplest form that will get counted are numbers with at most one of each prime factor. Then the question simplifies to "How many numbers are there between $1-100$ with at most one of each prime factor?
There are a few ways to proceed from here. I think the best way would be to complementary count. Then we are looking for numbers with at least $2$ factors of each prime, thus multiples of perfect squares. Our perfect squares are $1,4,9,16,25,36,49,64,81$. Thus we have $1+25+11-2+3+2 = 40$ such numbers by the principle of inclusion-exclusion. Thus, the answer is $100-40 = 60$.
Question
How is it that the number being expressed in simplest form will give the least number of radicals? The solution seems to imply that but why is it true?
• You mean expressing it as $a + b \sqrt{c} + d\sqrt{e} + \cdots$ where $a,b,c,\dots \in \mathbb N$ right? Because otherwise the question would be very problematic. – ThePortakal Jan 8 '16 at 3:45
• Yes, of course. Otherwise I wouldn't technically have to use any square roots. – user19405892 Jan 8 '16 at 3:45
• I added the extra condition in the question so that doesn't work as $3+\sqrt{3}+\sqrt{5}$ is not an integer. By least number of square roots, I mean the number of square root symbols. So $\sqrt{3}+\sqrt{3}$ would be two square roots while $2\sqrt{3}$ would be one. – user19405892 Jan 8 '16 at 4:07
• Why are you copying the exact solution here? – GohP.iHan Jan 9 '16 at 11:04
• @GohP.iHan It is actually my solution. – user19405892 Jan 9 '16 at 13:44 | {
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The square root of each square-free number less than $100$ contributes a term to the sum with a surd (a square root that cannot be simplified out), but all the square multiples of the number under the square root ($4$ times, $9$ times, $16$ times, etc.) can be combined with the term for the original square-free number, so they do not contribute any additional terms. For example,
$$\sqrt{11} + \sqrt{44} + \sqrt{99} = 6\sqrt{11}.$$
One way to count the integers whose square roots do not contribute new surds to the sum is as follows:
• $10$ perfect squares in the range $1$ to $100$, inclusive;
• $6$ cases of twice a perfect square in the range $8$ to $98$, inclusive;
• $4$ cases of $3$ times a perfect square in the range $12$ to $75$, inclusive;
• $3$ cases of $5$ times a perfect square in the range $20$ to $80$, inclusive;
• $3$ cases of $6$ times a perfect square in the range $24$ to $96$, inclusive;
• $6$ cases including $2$ cases each of $7$ times a perfect square, $10$ times a perfect square, and $11$ times a perfect square in the range $28$ to $99$, inclusive;
• $8$ cases including $13 \times 4$, $14 \times 4$, $15 \times 4$, $17 \times 4$, $19 \times 4$, $21 \times 4$, $22 \times 4$, and $23 \times 4$.
These add up to $40$ terms of the original sum that either are integers or can be combined with other terms, leaving $60$ unique surds. This is the same as your result, of course.
Here is the sum completely worked out, confirming that $60$ square roots are needed: | {
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\begin{align} \sqrt1 + & \sqrt2 + \cdots + \sqrt{100} \\ =& \quad 1 + \sqrt2+\sqrt3 + 2 + \sqrt5+\sqrt6+\sqrt7 + 2\sqrt2 + 3 + \sqrt{10} \\ & + \sqrt{11} + 2\sqrt3 + \sqrt{13} + \sqrt{14} + \sqrt{15} + 4 + \sqrt{17} + 3\sqrt2 + \sqrt{19} + 2\sqrt{5} \\ & + \sqrt{21} + \sqrt{22} + \sqrt{23} + 2\sqrt6 + 5 + \sqrt{26} + 3\sqrt3 + 2\sqrt7 + \sqrt{29} + \sqrt{30} \\ & + \sqrt{31} + 4\sqrt2 + \sqrt{33} + \sqrt{34} + \sqrt{35} + 6 + \sqrt{37} + \sqrt{38} + \sqrt{39} + 2\sqrt{10} \\ & + \sqrt{41} + \sqrt{42} + \sqrt{43} + 2\sqrt{11} + 3\sqrt5 + \sqrt{46} + \sqrt{47} + 4\sqrt3 + 7 + 5\sqrt2 \\ & + \sqrt{51} + 2\sqrt{13} + \sqrt{53} + 3\sqrt6 + \sqrt{55} + 2\sqrt{14} + \sqrt{57} + \sqrt{58} + \sqrt{59} + 2\sqrt{15} \\ & + \sqrt{61} + \sqrt{62} + 3\sqrt7 + 8 + \sqrt{65} + \sqrt{66} + \sqrt{67} + 2\sqrt{17} + \sqrt{69} + \sqrt{70} \\ & + \sqrt{71} + 6\sqrt2 + \sqrt{73} + \sqrt{74} + 5\sqrt3 + 2\sqrt{19} + \sqrt{77} + \sqrt{78} + \sqrt{79} + 4\sqrt5 \\ & + 9 + \sqrt{82} + \sqrt{83} + 2\sqrt{21} + \sqrt{85} + \sqrt{86} + \sqrt{87} + 2\sqrt{22} + \sqrt{89} + 3\sqrt{10} \\ & + \sqrt{91} + 2\sqrt{23} + \sqrt{93} + \sqrt{94} + \sqrt{95} + 4\sqrt6 + \sqrt{97} + 7\sqrt2 + 3\sqrt{11} + 10 \\ =& \quad 55 + 28\sqrt2 + 15\sqrt3 + 10\sqrt5 + 10\sqrt6 + 6\sqrt7 + 6\sqrt{10} \\ & + 6\sqrt{11} + 3\sqrt{13} + 3\sqrt{14} + 3\sqrt{15} + 3\sqrt{17} + 3\sqrt{19} \\ & + 3\sqrt{21} + 3\sqrt{22} + 3\sqrt{23} + \sqrt{26} + \sqrt{29} + \sqrt{30} \\ & + \sqrt{31} + \sqrt{33} + \sqrt{34} + \sqrt{35} + \sqrt{37} + \sqrt{38} \\ & + \sqrt{39} + \sqrt{41} + \sqrt{42} + \sqrt{43} + \sqrt{46} + \sqrt{47} \\ & + \sqrt{51} + \sqrt{53} + \sqrt{55} + \sqrt{57} + \sqrt{58} + \sqrt{59} \\ & + \sqrt{61} + \sqrt{62} + \sqrt{65} + \sqrt{66} + \sqrt{67} + \sqrt{69} \\ & + \sqrt{70} + \sqrt{71} + \sqrt{73} + \sqrt{74} + \sqrt{77} + \sqrt{78} \\ & + \sqrt{79} + \sqrt{82} + \sqrt{83} + \sqrt{85} + \sqrt{86} + \sqrt{87} \\ & + \sqrt{89} + \sqrt{91} + \sqrt{93} + \sqrt{94} + \sqrt{95} + \sqrt{97} \\ \end{align} | {
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• How do you know that we can't use nested radicals to reduce the number of radicals used? – GohP.iHan Jan 9 '16 at 11:03
• @GohP.iHan The problem statement stipulates that the answer must be in the form $a+b\sqrt{c}+d\sqrt{e}+\cdots$ where $a,b,c,d,e,\ldots$ are all integers. – David K Jan 9 '16 at 13:37
• Ohhh my bad... the author had a different question when I first viewed it. Thank you!! – GohP.iHan Jan 9 '16 at 13:51 | {
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} |
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## AP®︎ Calculus AB (2017 edition)
### Unit 1: Lesson 7
Determining limits using direct substitution
# Limits by direct substitution
AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
Sal explains how you can easily find limits of functions at points where the functions are continuous: simply plug in the x-value into the function! Later we will learn how to find limits even when the function isn't continuous.
## Want to join the conversation?
• I think Sal may have left something out. Correct me if I'm wrong, but on certain occasions, the entire function does not have to be continuous. If it is continuous at the value, a, then it should be fine to use direct substitution even though the function might be undefined elsewhere. This is because it isn't undefined at a so your output will be defined. Also, since your function is continuous at a, the limit there will exist meaning that you won't have to worry about jump discontinuities. Please tell me if this all seems reasonable.
• Because it is a continuous graph, every interval would be continuous provided it is a real number
• how would we know if the problem is to be solved by direct substitution method? or if the function given is continuous? without drawing its graph
• lim h(x) as x ->6 for h(x) =square root of (5x+6) should be 6 or -6 or both?
• y=√x is a function that returns only positive values. So in this context, we are only looking at the principal root, and the limit is positive 6. | {
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Limits are unique: they cannot have multiple values. So "both" would never be an option.
• At , what's the difference between f(x) as x->a, and f(a)?
• One is a limit, the other is an evaluation of the function. If the function is continuous and defined at (in your example), a, then they're equivalent. But you can get some very interesting results if the function is not continuous or not defined.
The limit is basically saying what the function seems to be going to as x gets closer to closer to a, but the function may not be defined at that point.
• Is the limit zero or none since it is continuous?
• The limit is 0 as Sal has demonstrated in the video.
• Can a limit be a fraction?
• yes a limit can be a fraction. For example if I asked you to find the limit as X approaches 2 for the functions (1/X), when you use substitution to find the limit you find that the number it approaches is .5 or 1/2.
• I thought it like this : Lim x-> a f(x) = f(a) if f(x) is continuous at x=a, not vice versa? Can someone tell me why?
• That statement holds in both directions. A function is continuous at a point a if and only if the limit value equals the function value at a. Since the two statements are equivalent, this can be used as a definition of continuity.
• Ok so basically if a function, say f(x), is continuous at x=c, then the lim x-->c = f(x)? Is this why you can find limits of continuous functions by direct substitution?
• Yes, the limit as x->c of f(x) is f(c). This property is equivalent to the epsilon-delta definition of continuity, and it's why we can use direct substitution for most familiar functions.
• By Sal and almost all discussion about relation between continuity and limit: it always says that as long as the function is continuous, there is a limit.
i am wondering if there is a case that the function is continuous but the limit does not exist. | {
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https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Active_Calculus_(Boelkins_et_al)/1%3A_Understanding_the_Derivative/1.7%3A_Limits%2C_Continuity%2C_and_Differentiability
in figure 1.7.2, the right graph, at point x=1. the left hand limit is 2. but the as x approaches 1 from the right, the function keeps oscillating, so the limit does not exist.
so the function is continuous by the graph, but limit does not exist.
am I wrong? or am I give some wrong example?
• Interesting question!
By definition, a function f(x) is continuous at x=x_0 if and only if for every epsilon>0, there exists delta>0 such that whenever x is within delta of x_0, f(x) is always within epsilon of f(x_0).
Though the function might look continuous at x=1, it is really discontinuous at x=1 due to the oscillatory behavior. For example, if we choose epsilon = 1/2, there does not exist delta>0 such that whenever x is within delta of 1, f(x) is always within epsilon of f(1). This is because there are values of x arbitrarily close to 1 (from the right) such that f(x) differs from f(1) by as much as 1 unit.
(1 vote)
• I'm not entirely sure if this statement is always true. But my argument bases on the assumption that the function f(x) = |x| is not continuous at x = 0, which I think is true but correct me if I'm wrong.
So, the limit lim_{x -> 0}f(x) for f(x) = |x| is definitely defined since it approaches 0 from both sides. The function itself is usually defined so that |x| equals 0 at x = 0. Which means that lim_{x -> a}f(x) equals f(a) at a = 0, but as said AFAIK the absolute value function |x| is NOT continuous for x = 0. | {
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And that would disprove the statement that if lim_{x -> a}f(x) equaling f(a) will mean that f is continuous at a (and vice versa).
(1 vote)
• Hey Raphael! I think you're getting continuity and differentiation confused. For a function, f(x), to be continuous, f(c) = lim_{x->c}f(x). This means that f(c) = lim_{x->c-}f(x) = lim_{x->c+}f(x) = lim_{x->c}f(x). As long as this holds true the function will be continuous at the given x-value, c.
Differentiation on the other hand is different. Differentiation requires that a function is continuous at the given x-value along with being absent of cusps, sharp turns (like x=0 for f(x) = |x|), and vertical tangents.
So to answer your question, f(x) = |x| is continuous at x = 0, however, it is not differentiable at x = 0. | {
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# Order of magnitudes comparasions
I have a list of order of magnitudes I want to compare.
My only idea is using calculus methods (limits , integral, etc...) to assert the functions relation.
I need your help with the following.
I need to determine (again by order of magnitude) how to order the following (by asceneding order):
$log(n) / log(log(n))$
$log(log(n))$
$log^3(n)$
$n/log(n)$
$n^{log(n)}$
Is it even possible to use limits on such functions?
I mean if I take $f(n)$ to be $log(n) / log(log(n))$ and $g(n)$ to be $n^{log(n)}$, It seems very hard to find out the limit.
-
There are quite a number of comparisons to be made. It is in most cases relatively straightforward to decide about the relative long-term size.
Let's start with your pair. We have $f(n)=\log(n)/\log(\log (n))$ and $g(n)=n^{\log(n)}$.
For large $n$ (and it doesn't have to be very large), we have $f(n)\lt \log(n)$.
Also, for large $n$, we have $\log(n)\gt 1$, and therefore $g(n)=n^{\log(n)}\gt n$.
So for large $n$, we have $$\frac{f(n)}{g(n)} \lt \frac{\log(n)}{n}.$$ But we know that $\lim_{n\to\infty}\dfrac{\log(n)}{n}=0$. This can be shown in various ways. For instance, we can consider $\dfrac{\log(x)}{x}$ and use L'Hospital's Rule to show this has limit $0$ as $x\to\infty$.
It takes less time to deal with the pair $n/\log(n)$ and $n^{\log(n)}$. If $n$ is even modestly large, we have $n/\log(n)\lt n$. But after a while, $\log(n)\gt 2$, so $n^{\log(n)}\gt n^2$. It follows that in the long run, $n^{\log(n)}$ grows much faster than $n/\log(n)$. | {
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As a last example, let us compare $\log^3(n)$ and $n/\log(n)$. Informally, $\log$ grows glacially slowly. More formally, look at the ratio $\dfrac{\log^3(n)}{n/\log(n)}$. This simplifies to $$\frac{\log^4 (n)}{n}.$$ We can use L'Hospital's Rule on $\log^4(x)/x$. Unfortunately we then need to use it several times. It is easier to examine $\dfrac{\log(x)}{x^{1/4}}$. Then a single application of L'Hospital's Rule does it. Or else we can let $x=e^y$. Then we are looking at $y^4/e^y$, and we can quote the standard result that the exponential function, in the long run, grows faster than any polynomial.
Remark: The second person in your list is the slowest one. Apart from that, they are in order. So you only need to prove four facts to get them lined up. A fair part of the work has been done above.
-
Your answer is very very good and understood. I will work on these tomorrow, make sure I understand anything. Thank you very much! – SyndicatorBBB Nov 3 '12 at 20:49
@Guy: I don't mind your accepting my answer. However, it is useful in general to wait somewhat longer, in order to elicit a variety of approaches. – André Nicolas Nov 3 '12 at 20:54
I commented below. Can you please advise ?Thank you. – SyndicatorBBB Nov 4 '12 at 10:01
For instance: $$\log(n^{\log n})=\log^2(n)$$ And since asymptotically: $$\frac{\log(n)}{\log(\log(n))} < \log^2(n)$$ We know that: $$n^{\log n}$$ Is at least exponentially larger than: $$\frac{\log(n)}{\log(\log(n))}$$
-
How did you calculate this inequalties? – SyndicatorBBB Nov 3 '12 at 20:46
Just observe: a $\log(n)$ cancels out, and $1/\infty$ is obviously smaller than $\infty$. – nbubis Nov 3 '12 at 20:47
Undestood! Thank you very much! – SyndicatorBBB Nov 3 '12 at 20:50
@André Nicolas I really liked the comprasion of the functions with simplier functions.
I was trying to use this technique on some of these functions.
Eventually I found a couple which I was not able to solve using this technique and I don't understand why. | {
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Assume $f(n) = log(log(n))$ and $g(n) = n/log(n)$
Now one might say, $f(n) < log(n)$ and $g(n) > 1/n$
If I look at the ratio/limit of $f(n)/g(n)$, I find out that $f(n)$ is much faster.
I completly understand that I chose unwisely the lower bound of g(n). Does it mean that on such functions I cannot use this technique?
If you "give away" stuff to make things simpler, have to be sensible about it. Replacing $g(n)$ by something that approaches $0$ is unreasonable. For example, could use $f(n)\lt \log(n)$, and keep $g(n)$. How one "weakens" an inequality has to be guided by basic intuition/experience about the sizes of things. The one thing that will be constantly useful is (i) $\log(n)$ is long-run slower than any positive power of $n$ and its twin (ii) For constant $a\gt 1$, $a^n$ is long-run faster than any polynomial in $n$. – André Nicolas Nov 4 '12 at 14:26 | {
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# If all points of a real function with positive values would be local minimum, can one say it is constant function?
During my studies I faced a function $$f:\mathbb{R} \to \mathbb{R}^+$$ with the property: for all $$x \in \mathbb{R}$$ and all $$y$$ in open interval $$(x-\frac{1}{f(x)} ,x+\frac{1}{f(x)})$$ we have $$f(x) \leq f(y)$$. At first I guessed maybe it is necessarily constant function but I could not show this. I tried to generate a counter example for my guess and unfortunately I could not again. would anyone please help me?
• The answer to the question in the title is (of course) no. [Define $f(x) = 2$ for $x \ne 0$ and $f(0)=1$.] But the question in the text is more complicated. – Gerald Edgar Aug 6 '20 at 9:31
• What is $\mathbb{R}^+$? Is $f(x)=0$ allowed? – Wilberd van der Kallen Aug 6 '20 at 9:52
• $\mathbb{R}^+$ is denoted for the open interval $(0,+\infty)$ – M. Reza. K Aug 6 '20 at 10:12
My answer is: Such a function must be constant. Suppose (for puoposes of contradiction) $$f$$ is a nonconstant function $$f : \mathbb R \to (0,+\infty)$$ such that $$\forall x\in\left(a-\frac{1}{f(a)},a+\frac{1}{f(a)}\right),\quad f(x) \ge f(a) .$$
For $$m > 0$$, define $$U_m := \{x : f(x)>m\}$$.
Lemma 1: For all $$m \in \mathbb R$$, the set $$U_m$$ is open.
Proof. Let $$x \in U_m$$. Then $$(x-1/f(x),x+1/f(x)) \subseteq U_m$$. So $$U_m$$ is open.
Define $$\mathcal G = \{(a,b,m) : a < b, f(a)\le m, f(b)\le m, \text{ and }\forall x\in(a,b),f(x)> m\}$$. | {
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Lemma 2: Let $$a,b \in \mathbb R$$ with $$f(a) \ne f(b)$$, and $$m>0$$. Then there exists $$a_1, b_1$$ with $$a < a_1 < b_1 < b$$ and $$m_1 \ge m$$ such that $$(a_1,b_1,m_1) \in \mathcal G$$.
Proof. One of $$f(a), f(b)$$ is smaller, assume WLOG that $$f(a) < f(b)$$. Now $$U_{f(a)}$$ is open, $$a \notin U_{f(a)}$$, and $$b \in U_{f(a)}$$. The maximal open interval $$(c,d) \subseteq U_{f(a)}$$ with $$c is nonempty, in fact $$(c,b) = (c,d) \cap (a,b) \ne \varnothing$$. By maximality, $$f(c) \le f(a)$$. Choose $$e$$ with $$c < e < b$$. For $$x \in (c,e)$$ we have $$f(c) < f(x)$$ so $$c \le x-1/f(x)$$ and thus $$f(x) \ge 1/(x-c)$$. Thus $$f(x)$$ is unbounded on $$(c,e)$$. Let $$m_1$$ be such that $$m_1 \ge m, m_1 > f(c), m_1 > f(e)$$. The open set $$U_{m_1}$$ has $$U_{m_1} \cap (c,e) \ne \varnothing$$ and $$c,e \notin U_{m_1}$$. So let $$(a_1,b_1)$$ be a maximal open subinterval of the open set $$U_{m_1} \cap (c,e)$$.
Lemma 3: Let $$(a_1, b_1, m_1) \in \mathcal G$$, and let $$m > m_1$$. Then there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 \ge m$$ such that $$(a_2, b_2, m_2) \in \mathcal G$$.
Proof: If $$f(a_1) \ne f(b_1)$$, apply Lemma 2 directly. Otherwise, pick any $$c \in (a_1,b_1)$$ and apply Lemma 2 to $$a_1, c, m$$. | {
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Main Proof: We assume $$f$$ is not constant. So there exist $$a_1 < b_1$$ with $$f(a_1) \ne f(b_1)$$. By Lemma 2 there exist $$a_2, b_2$$ with $$a_1 < a_2 < b_2 < b_1$$ and $$m_2 > 2$$ so that $$(a_2,b_2,m_2) \in \mathcal G$$. Then by Lemma 3, there exist $$a_3, b_3$$ with $$a_2 < a_3 < b_3 < b_2$$ and $$m_3 > 3$$ so that $$(a_3,b_3,m_3) \in \mathcal G$$. Continuing recursively, we get sequences $$(a_k), (b_k), (m_k)$$ so that $$\forall k:\quad a_k < a_{k+1} < b_{k+1} < b_k,\quad m_k > k,\quad\text{and } (a_k,b_k,m_k)\in\mathcal G .$$ The sequence $$(a_k)$$ is increasing and bounded above, so it converges. Let $$a = \lim_{k\to\infty} a_k$$. Then $$a_k < a_{k+1} \le a \le b_{k+1} < b_k$$. From $$(a_k,b_k,m_k) \in \mathcal G$$ we conclude $$f(a) > m_k > k$$. This is true for all $$k$$, so $$f(a)$$ is not a real number, a contradiction.
• Your answer was enjoyable and very nice to me. It really engaged me and I am thinking if instead of $\mathbb{R}$ in domain of the $f$, replace a complete metrically convex metric space, does your answer still work? – M. Reza. K Aug 7 '20 at 12:11
• @M.Reza.K for every $x,y$ in the complete metrically convex space there is an isometry $g:[a,b]\to X$ such that $g(a)=x, g(b)=y$. Apply the result to $h=f\circ g$. It follows that $f(x)=h(a)=h(b)=f(y)$. I am interested if we can drop the completeness: the proof seems to work for rational numbers... Perhaps it also works for intrinsic metric spaces? – erz Aug 7 '20 at 12:40
• @erz My proof certainly fails for rational numbers. A decreasing sequence of closed intervals may have empty intersection in $\mathbb Q$. – Gerald Edgar Aug 7 '20 at 13:32
This is not an answer but it greatly narrows down the class of functions $$f$$ with the described property.
1. If $$c>0$$, then the sets $$f^{-1}([c,+\infty))$$ and $$f^{-1}((c,+\infty))$$ are both open.
Indeed, every $$x$$ in any of these sets comes with an open ball of radius $$\frac{1}{f(x)}$$. | {
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Indeed, every $$x$$ in any of these sets comes with an open ball of radius $$\frac{1}{f(x)}$$.
Consequently, the sets $$A_c=f^{-1}((-\infty, c))$$ and $$B_c=f^{-1}((-\infty, c])$$ are both closed. In particular, $$f$$ is lower semi-continuous.
1. Let $$x\in B_c$$. We know that if $$d(y,x)<\frac{1}{c}\le \frac{1}{f(x)}$$, then $$f(x)\le f(y)$$. On the other hand, if $$y\in B_c$$ and $$d(y,x)<\frac{1}{c}$$, then $$d(y,x)<\frac{1}{f(y)}$$, from where $$f(y)\le f(x)$$. Hence, if $$y\in B_c$$ and $$d(y,x)<\frac{1}{c}$$, we get $$f(y)= f(x)$$.
Not only this means that $$f$$ is locally a constant on $$B_c$$, but also that $$f^{-1}(c)$$ is closed.
1. Let us prove the following strengthening of the Baire's theorem: let $$X$$ be a complete metric space, and let $$B_n$$ be a sequence of closed sets such that $$X=\bigcup B_n$$. Then $$X=\overline{\bigcup int B_n}$$.
Indeed, if $$U\subset X\backslash \bigcup int B_n$$ is open and nonempty, it is metrizable with a complete metric, and $$B_n\cap U$$ is closed in $$U$$. Since $$U=\bigcup (B_n\cap U)$$, from the usual Baire's theorem, there is $$m$$ such that $$V=int (B_m\cap U) \ne \varnothing$$. But then $$V=int (B_m\cap U)=int B_m\cap U\subset int B_m \backslash \bigcup int B_n=\varnothing$$. contradiction.
1. So, combining these observation we get the following picture: there is a dense open set $$U$$ such that $$f$$ is locally constant on $$U$$; every level set of $$f$$ is closed, and the distance between elements of $$f^{-1}(c)$$ and $$f^{-1}(d)$$ is at least $$\frac{1}{\max \{c,d\}}$$.
The only way it's not a constant function is when the components of $$U$$ are like the gaps in the Cantor set. So the possible counterexample may be constructed like the Cantor's staircase, but the smaller the component, the larger is the value on it. I was unable to perform this fine tuning though, maybe it's impossible. | {
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# Problems about Countability related to Function Spaces
Suppose we have the following sets, and determine whether they are countably infinite or uncountable .
1. The set of all functions from $\mathbb{N}$ to $\mathbb{N}$.
2. The set of all non-increasing functions from $\mathbb{N}$ to $\mathbb{N}$.
3. The set of all non-decreasing functions from $\mathbb{N}$ to $\mathbb{N}$.
4. The set of all injective functions from $\mathbb{N}$ to $\mathbb{N}$.
5. The set of all surjective functions from from $\mathbb{N}$ to $\mathbb{N}$.
6. The set of all bijection functions from from $\mathbb{N}$ to $\mathbb{N}$.
My thoughts on this:
1. The first set is uncountable by using diagonalization argument.
2. I have read something about it saying this is countable since we can see this as a set of finite sequences of natural numbers. But I have hard time following the argument.
3. The article I read says this is uncountable but I couldn't follow the argument either.
4. For (4),(5) and (6), I am not even sure how to approach problems like these.
Could someone please explain how to approach this kind of problems in general?
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And does it matter when I change all the $\mathbb{N}$ to $\mathbb{R}$?
-
Recall the definition of a function as a set of ordered pairs. That should help. – Potato Dec 3 '11 at 2:27
Maybe you can argue that (2) is countable because any given sequence that is non-increasing eventually stops decreasing at some finite natural number or hits zero and then is forced to be constant at 0. Then come up with a clever way to list all the functions, maybe by listing them first by starting position then by "length before it is constant" or something like that. – tomcuchta Dec 3 '11 at 2:28
@tomcuchta Your idea for (2) is alright, but a small nitpick. The function need not go to $0$ but might stop at $1$ (say). What is true (and important) is that the function is eventually constant. – Srivatsan Dec 3 '11 at 2:39
I removed the (functional-analysis), the (set-theory) and the (proof-theory) tag. Each of these tags is intended for subjects usually taught at an advanced undergraduate or graduate level. – t.b. Dec 3 '11 at 2:42
@t.b.Thanks, I guess I am confused about what function spaces really study. – geraldgreen Dec 3 '11 at 3:12
Another approach to the 6th part.
Let us denote the set of all bijections from $\mathbb N$ to $\mathbb N$ by $\operatorname{Bij}(\mathbb N,\mathbb N)$.
Clearly $$|\operatorname{Bij}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c.$$
Let us try to show the opposite inequality.
For arbitrary function $f\in\operatorname{Bij}(\mathbb N,\mathbb N)$ we define $$\operatorname{Fix}(f)=\{n\in\mathbb N; f(n)=n\}.$$ (That is, $\operatorname{Fix}(f)$ is the set of all fixed points of the map $f$.) Let us try to answer the question, whether for any $A\subseteq\mathbb N$ it is possible to find a bijection ${f_A}:{\mathbb N}\to {\mathbb N}$ such $\operatorname{Fix}(f_A)=A$. Let us consider two cases. | {
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If the complement of the set $A$ is infinite, i.e. $\mathbb N\setminus A=\{b_n; n\in\mathbb N\}$ (and we can assume that the elements $\mathbb N\setminus A$ are written in the increasing order, i.e. $b_1<b_2<\dots<b_n<b_{n+1}<\dots$ ), then we can define a function $f_A$ as follows: $$f_A(x)= \begin{cases} x, & \text{if }x\in A, \\ b_{2k+1}, &\text{ak x=b_{2k} for some k\in\mathbb N},\\ b_{2k}, &\text{ak x=b_{2k+1} for some k\in\mathbb N}. \end{cases}$$ In the other words, we did not move the elements of $A$ and the elements from the complement were paired and we interchanged it pair. Such map is a bijection from $\mathbb N$ to $\mathbb N$ such that $\operatorname{Fix}(f_A)=A$.
Next we consider the case that $\mathbb N\setminus A$ is finite.
If $\mathbb N\setminus A=\emptyset$, then $f=id_{\mathbb N}$ is a function fulfilling $\operatorname{Fix}(f_A)=\mathbb N=A$.
If the set $\mathbb N\setminus A$ is a singleton $\{a\}$, then is is impossible to find a bijection $f$ such that $\operatorname{Fix}(f)=A=\mathbb N\setminus\{a\}$. No element of $\mathbb N\setminus A$ can be mapped on $a$ (since all such elements are fixed). But $a$ cannot be mapped on $a$ since this would mean that $a$ is a fixed point.
But if the set $\mathbb N\setminus A$ has at least two elements, then it is possible to construct such a map. We again assume that the elements $b_k$ are written in the increasing order, i.e. $\mathbb N\setminus A=\{b_0,\dots, b_n\}$ and $b_0<b_1<\dots<b_n$. We put $f_A(x)=x$ for $x\in A$, $f_A(b_k)=b_{k+1}$ for $0\le k<n$ and $f_A(b_n)=b_0$. (I.e. we made a cycle consisting of elements of $\mathbb N\setminus A$.)
This defines a bijection ${f_A}:{\mathbb N}\to{\mathbb N}$ such that $\operatorname{Fix}(f_A)=A$. | {
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This defines a bijection ${f_A}:{\mathbb N}\to{\mathbb N}$ such that $\operatorname{Fix}(f_A)=A$.
The assignment $A\mapsto f_A$ that we described above is a map from the set $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ consisting of all subsets of $\mathbb N$, whose complement is not a singleton, to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This map is injective; since from $f_A=f_B$ we get $A=\operatorname{Fix}(f_A)=\operatorname{Fix}(f_B)=B$.
From the set $\mathcal P({\mathbb N})$ with the cardinality $\mathfrak c$ we have omitted a countable set $\{\mathbb N\setminus\{a\}\in\mathbb N\}$. Therefore the cardinality of this difference $\mathcal P({\mathbb N})\setminus\{\mathbb N\setminus\{a\}\in\mathbb N\}$ is again $\mathfrak c$.
So we found an injection from a set of cardinality $\mathfrak c$ to the set $\operatorname{Bij}(\mathbb N,\mathbb N)$. This yields the opposite inequality $$\mathfrak c\le|{\operatorname{Bij}(\mathbb N,\mathbb N)}|$$ and by Cantor-Bernstein theorem we get $|{\operatorname{Bij}(\mathbb N,\mathbb N)}|=\mathfrak c$.
-
Thanks for doing the details. It pointed out a flaw in my answer. – David Mitra Dec 3 '11 at 14:07
I did not notice that we are using basically the same argument until you pointed it out. I like your answer, but I am out of votes for today. – Martin Sleziak Dec 3 '11 at 14:14
This is not a complete answer. [I am taking $\mathbb N$ to include $0$.] I will give you a proof that there are uncountably many bijections from $\mathbb N$ to itself. This obviously subsumes the questions (4)-(6).
Consider functions $f : \mathbb N \to \mathbb N$ of the following special form: For every $k$, the pair of elements $\{ 2k, 2k+1 \}$ is mapped to itself. However, we have two ways of doing this:
• either: map $2k$ to itself, and similarly for $2k+1$.
• or: map $2k$ and $2k+1$ to each other. | {
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• either: map $2k$ to itself, and similarly for $2k+1$.
• or: map $2k$ and $2k+1$ to each other.
The key point is that for each $k$, you could pick one of the two possibilities independently. Since there are $2$ options for each $k$, you have $2^{\mathbb N}$ options; i.e., there are uncountably many such functions satisfying this restriction. And each such function is a bijection from $\mathbb N$ to itself.
The above needs a little bit of work to make it completely rigorous. In particular, try to formalise what I mean by "$2^\mathbb N$ possibilities".
For (3), the idea is similar. For each $k$, try mapping $k$ to either $2k$ or $2k+1$. Details are left as an exercise.
For (2), does the following hint help? Every non-increasing function $f : \mathbb N \to \mathbb N$ is eventually constant. So the function is uniquely specified if we describe the finite initial "portion" of the function.
Another approach (due to tb) for (2): If you plot the graph of a non-increasing function, it looks like a staircase that goes down as we move from left to right -- a staircase with finitely many steps because you the function can go down only finitely many times. One can also reconstruct the function if we record the upper right corners of each of those steps. Since the latter corresponds to a finite subset of $\mathbb N \times \mathbb N$ (which is countable), it follows that our set of non-increasing functions is countable as well.
-
I'll use this opportunity to mention here a very nice (at least in my opinion) proof that the cardinality of all bijections $\mathbb N\to\mathbb N$ is $\mathfrak c=2^{\aleph_0}$, which is basically the part 6. I making it a community wiki, since the idea is not mine. I believe I've seen this at sci. math but I am not sure about it and I cannot find the link now. (Feel free to add some pointers and links, if you know some.)
HINT: If someone wants to try to develop the idea by himself, the basic idea is Riemann's rearrangement theorem. | {
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Let us denote the set of all bijections from $\mathbb N$ to $\mathbb N$ by $\operatorname{Bij}(\mathbb N,\mathbb N)$.
Clearly $$|\operatorname{Bij}(\mathbb N,\mathbb N)| \le \aleph_0^{\aleph_0} = 2^{\aleph_0} = \mathfrak c.$$
Now we will show the opposite inequality. Let us choose a series $\sum_{n=0}^\infty x_n$, which is convergent but not absolutely convergent (e.g. $\sum_{n=0}^\infty x_n = \sum_{n=0}^\infty (-1)^n\frac1{n+1}$). By Riemann's theorem, for any real number $r$ there exists a permutation $\pi\in \operatorname{Bij}(\mathbb N,\mathbb N)$ such that $\sum_{n=0}^\infty x_{\pi(n)}=r$. If we choose for each $r$ one such permutation, we get an injective map from $\mathbb R$ to $\operatorname{Bij}(\mathbb N,\mathbb N)$. Thus $|\operatorname{Bij}(\mathbb N,\mathbb N)| \ge |\mathbb R| = \mathfrak c = 2^{\aleph_0}$.
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+1, nice approach. I don't think you have to make an answer CW just because it isn't your idea. Writing an answer is contribution enough and one deserves some credit for that. =) – Srivatsan Dec 4 '11 at 0:24
Let \eqalign{ A_1&=\text{the set of all functions from }\Bbb N\text{ to }\Bbb N. \cr A_2&=\text{the set of all non-increasing from }\Bbb N\text{ to }\Bbb N. \cr A_3&=\text{the set of all non-decreasing }\Bbb N\text{ to }\Bbb N. \cr A_4&=\text{the set of all 1-1 functions from}\Bbb N\text{ to }\Bbb N. \cr A_5&=\text{the set of all onto functions from }\Bbb N\text{ to }\Bbb N. \cr A_6&=\text{the set of all bijections from }\Bbb N\text{ to }\Bbb N. \cr }
Consider $A_6$:
$A_6$ may be thought of as the set $\bigl\{\,\pi :\pi \text{ is a permutation of } (1,2,3,\ldots)\,\bigr\}$. | {
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For any permutation $\pi$, we may associated a binary sequence $(x_n)$ by setting $$x_n=\cases{ 0,& \text{if } \pi_n=n\cr 1,& \text{otherwise} }.$$ This induces an onto map from $A_6$ to the set of infinite binary sequences (Edit: not quite, as Martin Sleziak points out in his answer, the binary sequences that have "only one 1" have no element in $A_6$ mapped to them. But, as the set of such sequences is countable, the rest of the argument will go through. Martin also describes nicely how to obtain this map.). Since the latter set is uncountable, so is the former.
So, $A_6$ is uncountable.
From this, it follows that $A_5$, $A_4$, and $A_1$ are uncountable.
Now, consider $A_2$:
If $f$ is a non-increasing function, then $f$ must be eventually constant. Thus, the cardinality $A_2$ would be at most the cardinality of the set $$\bigcup_{j,m} \bigl\{\,(a_i)_{i=1}^\infty : a_i=m, i\ge j\,\bigr \}.$$ But this latter set is a countable union of countable sets and is, thus, countable.
So, $A_2$ is countable.
Now, consider $A_3$:
The cardinality of the set of non-decreasing functions is at least the cardinality of the set $$\{\, A\subset\Bbb N : A \text{ is an infinite subset of } \Bbb N\, \}.$$
Since this latter set uncountable, $A_3$ is uncountable.
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"If f is a non-increasing function, then f must be eventually constant" Why ? it can also be increasing. – GinKin Mar 11 at 12:45
@GinKin I took "non-increasing" to mean $f(b)\le f(a)$" whenever $a\le b$ (that is, (perhaps non-strictly) decreasing). This is standard terminology. What do you mean by "it can also be increasing"? – David Mitra Mar 11 at 13:11
Whoops mixed the terms, I meant that if it doesn't increase then it doesn't have to be eventually constant, it can be decreasing. – GinKin Mar 11 at 13:24
@GinKin But the range is $\Bbb N$. If it were strictly decreasing, it would eventually be constantly $0$. – David Mitra Mar 11 at 13:25 | {
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# The number of solutions of a polynomial system using a Gröbner basis
The following extract from this Wikipedia page explains that we can easily deduce the number of solutions of a polynomial system using Gröbner basis:
Given the Gröbner basis G of an ideal I, it has only a finite number of zeros, if and only if, for each variable x, G contains a polynomial with a leading monomial that is a power of x (without any other variable appearing in the leading term). If it is the case the number of zeros, counted with multiplicity, is equal to the number of monomials that are not multiple of any leading monomial of G. This number is called the degree of the ideal.
Question: Is there a function in Sage computing this number of zeros from a given Gröbner basis?
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Yes, these
monomials that are not a multiple of any leading monomial of G
form a basis of the quotient ring $R/I$ (by the reduction algorithm), also called a normal basis for $I$.
sage: R.<x,y,z> = PolynomialRing(QQ,3)
sage: I = Ideal([x^2+y+z-1,x+y^2+z-1,x+y+z^2-1])
sage: B = I.groebner_basis(); B
[x^2 + y + z - 1, y^2 + x + z - 1, z^2 + x + y - 1]
sage: I.normal_basis()
[x*y*z, y*z, x*z, z, x*y, y, x, 1]
The number of elements in this basis, or the vector space dimension of $R/I$, can also be found as follows:
sage: I.vector_space_dimension()
8
The zero set, or variety, of the ideal is the following:
sage: I.variety(QQbar)
[{z: 0, y: 0, x: 1},
{z: 0, y: 1, x: 0},
{z: 1, y: 0, x: 0},
{z: -2.414213562373095?, y: -2.414213562373095?, x: -2.414213562373095?},
{z: 0.4142135623730951?, y: 0.4142135623730951?, x: 0.4142135623730951?}]
The number of elements is not 8, because we are missing the "multiplicities". Indeed,
False
So perhaps the more interesting quantity is
5
which agrees with the number of zeros without multiplicity.
(There is an easy algorithm to compute the radical of a zero-dimensional ideal.)
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(There is an easy algorithm to compute the radical of a zero-dimensional ideal.)
more
Is it required to compute the Gröbner basis? Does the function normal_basis() recompute the Gröbner basis?
( 2020-01-26 20:05:55 -0500 )edit
The method normal_basis() internally requires a Gröbner basis. If one has not yet been computed, then it computes one. If one has been computed already, then it uses that one; it does not recompute it. Note also that the result / speed depends on the chosen monomial ordering (degrevlex by default). Of course the dimension does not depend on this choice.
( 2020-01-27 03:57:43 -0500 )edit | {
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# Solving a Eigenvalue Problem
$$\textbf{Eigenvalue problem on a unit square \Omega = [0,1]^2 :}$$
Consider the eigenvalue problem with the Dirichlet boundary condition that is, $$-Lu = \lambda u$$ where $$L = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$.
The boundary condition is that $$u=0$$ on $$\partial \Omega$$.
I am computing the $$\textbf{eigenvalues}$$ and $$\textbf{eigenfunctions}$$ of Laplacian numerically in Mathematica. Specially, I am interested about the $$\textbf{multiplicity}$$ of a specific eigenvalue. I already know that Tally or Count can certainly count the occurrances of eigenvalues in the list called vals.
{ℒ, ℬ} = {-Laplacian[u[x, y], {x, y}],
DirichletCondition[u[x, y] == 0, True]};
{vals, funs} =
DEigensystem[{ℒ, ℬ},
u[x, y], {x, 0, 1}, {y, 0, 1}, _];
Tally[vals]
In the eigenvalue problem there are infinite numbers of eigenvalues. So, We can not list all of them. I want to find out $$\textbf{multiplicity}$$ of a specific eigenvalue that counts the total number of occurrances. For example, the multiplicity of the eigenvalue $$5 \pi^2$$ is $$2$$. Similarly, I want to find out the multiplicity of the eigenvalue $$50 \pi^2$$ is $$3$$.
• Edit your question to include your Mathematica code rather than pictures of the code. Feb 15 '21 at 17:51
• Convert your cells to InputForm and look at the Markdown help Feb 15 '21 at 18:24
• Feb 15 '21 at 19:30
With the exact solutions
u[{nx_, ny_}, {x_, y_}] = Sin[nx π x] Sin[ny π y];
and $$\{n_x,n_y\}\in\mathbb{N}^2$$, the eigenvalues are
λ[nx_, ny_] = (nx^2 + ny^2) π^2;
Test:
Assuming[Element[nx | ny, PositiveIntegers],
{u[{nx, ny}, {x, 0}], u[{nx, ny}, {x, 1}],
u[{nx, ny}, {0, y}], u[{nx, ny}, {1, y}]} // FullSimplify]
(* {0, 0, 0, 0} *)
-D[u[{nx, ny}, {x, y}], {x, 2}] - D[u[{nx, ny}, {x, y}], {y, 2}] ==
λ[nx, ny] * u[{nx, ny}, {x, y}] // FullSimplify
(* True *) | {
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The number of pairs $$\{n_x,n_y\}$$ equal to $$\lambda/\pi^2$$ is therefore given by OEIS A063725. You can calculate the multiplicity of a given value of $$i=\lambda/\pi^2$$ directly from the generating function,
g[q_] = (EllipticTheta[3, q] - 1)^2/4;
m[i_Integer?NonNegative] := SeriesCoefficient[g[q], {q, 0, i}]
For example, the multiplicity of $$\lambda=50\pi^2$$ is 3, as you know:
m[50]
(* 3 *)
There is no eigenvalue at $$326\pi^2$$, on the other hand:
m[326]
(* 0 *)
More generally, the $$d$$-dimensional generating function
g[d_, q_] = ((EllipticTheta[3, q] - 1)/2)^d;
allows us to find the multiplicity of the eigenvalue $$\lambda=i\pi^d$$ directly:
m[d_Integer?NonNegative, i_Integer?NonNegative] :=
SeriesCoefficient[g[d, q], {q, 0, i}]
For example, the $$d=17$$-dimensional hypercube has $$1\,416\,786\,753\,216$$ eigenvalues $$\lambda=250\pi^{17}$$:
m[17, 250]
(* 1416786753216 *)
If you want the actual solutions, not just the number of solutions, PowersRepresentations is a good starting point (but it gives solutions containing zeros, which we need to filter out):
s[d_Integer?NonNegative, i_Integer?NonNegative] :=
Select[PowersRepresentations[i, d, 2], Min[#] > 0 &]
For example, the eigenvalue $$\lambda=50\pi^2$$ in $$d=2$$ dimensions can be composed in two ordered ways, as you know:
s[2, 50]
(* {{1, 7}, {5, 5}} *)
Enumerating all permutations of these ordered combinations gives the full multiplicity of 3:
t[d_Integer?NonNegative, i_Integer?NonNegative] :=
Join @@ Permutations /@ s[d, i]
t[2, 50]
(* {{1, 7}, {7, 1}, {5, 5}} *)
In $$d=17$$ dimensions the eigenvalue $$\lambda=250\pi^{17}$$ has 999 ordered solutions,
s[17, 250]
(* {{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 6, 14},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 10, 10},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 15},
...
{3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5},
{3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6}} *) | {
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which then become the aforementioned $$1\,416\,786\,753\,216$$ unordered solutions from t[17, 250].
• We should work with the generating function explicitly because it gives a very efficient recipe, especially in the high-dimensional case. Just look at the given example for the 17-dimensional hypercube: manually summing and counting is going to be very tedious and difficult, whereas the call to m[17, 1000] takes less than half a second. The same is true, to some extent, for your case $d=2$ when you work with large values of $\lambda$. Feb 16 '21 at 8:32
• Sorry, this is not the right place for an introduction to generating functions or Jacobi theta functions. Feb 16 '21 at 9:31 | {
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# Show why the value converges to $$\pi$$
$$a_0=1$$
$$a_{n+1}=a_n+\sin{(a_n)}$$
Explain why the following occurs:
$$a_0=1$$
$$a_1=1+\sin{(1)}\approx 1.841470985$$
$$a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709$$
$$a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333$$
$$a_4\approx 3.141592612$$
$$a_5\approx 3.141592654\approx\pi$$
Note by Jack Han
3 years, 7 months ago
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## Comments
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Let's see here...
The interesting thing I was talking about is the fact that the given series always converges into a value $$a_n$$ such that $$sin(a_n)=0$$ for different values of $$a_0$$. To put it in more exact terms, it always converges to a value of $$a_n$$ such that $$cos(a_n)=-1\Rightarrow a_n=m\pi$$, where $$m$$ is an odd integer. ($$m$$ can also be even, but that is a degenerate case where all terms are same) | {
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I'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.
Consider the function $$f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) }$$
Now see what happens when we take some arbitrary value of $$x$$ (say $$x=1$$)and then do the following repeatedly:
$$x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) }$$ (":=" is the assignment operator )
In the above figure, we can see two points marked. One is red, which represents the first value of $$x$$($$=1$$). The other is brown, and is after one iteration of above step.
We can see that when we do $$x:=x+sin(x)$$, what is actually happening is that $$x$$ is surfing along the slope of the curve $$cos(x)$$. We move the value of $$x$$ down the tangent. Change $$x$$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $$x=\pi$$.
I know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.
In general, series defined as $$a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }$$
Will converge to the nearest value of $$a$$ (nearest to $$a_0$$) such that $$f(a)$$ is minimum, provided the value of $$\alpha$$ is not too large.
- 3 years, 7 months ago
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Good work.
- 3 years, 7 months ago
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Good work! Newton's method for estimating roots.
Pretty much seals the deal. Great solution +1.
- 3 years, 7 months ago
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Yes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1
- 3 years, 7 months ago
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As usual, since the series converges.. this means that when $$n\to\infty$$ ,
$$a_{n+1}=a_{n}$$. But $$a_{n+1}=a_{n}+sin(a_{n})$$
$$\Rightarrow sin(a_{n})=0$$
Now how do we know that $$a_{n}=\pi$$? We know this since $$a_0=1$$ and the series is constantly increasing. Therefore, it converges onto the first value of $$x>1$$ such that $$sin(x)=0$$.
- 3 years, 7 months ago
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Bro , But It is not always true $$\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) }$$ .
- 3 years, 7 months ago
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Why not? Do you have a counter-example?
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But can't the value of zero be discarded? Because here, $$a_{n}=1$$ for all n.
There is no question of convergence at all as all the terms have a fixed value.
- 3 years, 7 months ago
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My Bad , I wrongly co-relate two diffrent situations , Actually this example is given when I got two answwer to an single recursion , which is something Like that $${ a }_{ n+1 }={ \left( \sqrt { 2 } \right) }^{ { a }_{ n } }$$ , I misinterpreate this current situation, So I deleted that irrelevant comment , But Yes this is good recurance , and I'am trying to prove convergence of this recurance. If I got correct than I report you.
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Thanks for joining the party :D
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No, the sequence only increases, only that the rate of increase decreases till it reaches 0 at $$a_n= \pi$$
you can easily see it from the graph if x+sin(x), it is a non decreasing function and reaches inflection point at x= pi
but yes convergence remains unproven, i am trying
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My bad. Sorry I jumped to a wrong conclusion. Thanks for replying. | {
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My bad. Sorry I jumped to a wrong conclusion. Thanks for replying.
But the rate at which the value of $$a_{n}$$ increases isn't the same as the rate at which $$x+sinx$$ is increasing. So how do you directly conclude from the graph?
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it is not the increasing nature that confirms, it is the fact that the graph lies above x=y as well
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I sort of understand what you're saying...But isn't it easier to say this: The increment in the variable in every iteration is $$sina_{n}$$ and as $$a_{n}$$ tends to $$\pi$$, $$\quad sina_{n}$$ becomes smaller and smaller. (I think that is what you were saying initially about rate of change slowing down).
I have deleted my previous comment as it might only muddle things up for someone else.
Also, please check out my response to your post on this problem of mine.
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Indeed it is ,simply that sin(an) is positive, in the range and hence increment is positive or sequence rises :)
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Can someone prove the convergence of this recursion?
@Ronak Agarwal @Mvs Saketh @Azhaghu Roopesh M @Pratik Shastri @Raghav Vaidyanathan @Deepanshu Gupta or anyone else who's interested.
- 3 years, 7 months ago
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This is turning out to be very interesting...
I want to know if we can find a general form for a function $$f(x)$$ such that the series $$a_1,a_2,...$$ defined by:
$$a_{n+1}=a_{n-1}+f(a_{n-1})$$
Converges for a given value of $$a_0$$.
Further, is it true that all of the values of such $$a_n$$ as $$n\to \infty$$ satisfy $$f(a_n)=0$$?
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I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?
- 3 years, 7 months ago
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What are you thinking?
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What are you thinking?
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are you sure you have differentiated it correctly ? there should be atleast one $$x$$ in your $$f'(x)$$ i think
- 3 years, 7 months ago
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If you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded
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Which value converges to pi ??
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# Integration of piecewise function
I was doing a question that involved integration of a piecewise function.
The question, as can be seen above, asks to evaluate the integral from 0 to pi. I worked it out by breaking the function into 2 integrals where the first had lower and upper limits of 0 and pi/2 respectively and adding it to the integral with lower and upper limits of pi/2 and pi respectively.
Now, my question is, how can we use FTC 2 on the first integral if it is not continuous over the domain [0, pi/2] as the pi/2 is not included in the sin (x).
I have tried my best to articulate my question but if there is any confusion, please do let me know.
Is there something I am missing? Can someone kindly clarify.
You can still evaluate the integrals individually and then add them together. There is a jump discontinuity at $$\frac{\pi}{2}$$ due to the interval given in the piecewise, but this point will not impact the sum. That is because the thickness of a single point is intuitively $$0$$.
Consider the point $$c$$ (Note: c is finite):
$$\int_c^{c} f(x) dx= F(c)-F(c)=0$$ by the FTC.
Additionally, note that the function is integrable from that interval.
$$\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 0-(-1) = 1$$
• Hi, yes, I can see why. However, does this discontinuity affect our ability to use FTC 2 as it requires continuity over [a, b]? Aug 14 at 12:30
• It does not because the sum of one point is, intuitively, $0$. In other words, a point has no thickness/width to sum. Aug 14 at 12:34
• Oh ok. Thank you. So I would be correct in saying that in a case like this, there is some flexibility to the theorem? Aug 14 at 12:37
• @Oofy2000 No because $\sin(x)$ is integrable/well defined/continuous from $0 \le x \lt \frac{\pi}{2}$ Aug 14 at 12:41
• @Oofy2000 Added one. Here is a rigorous explanation, if interested: math.stackexchange.com/questions/511197/area-under-a-point Aug 14 at 13:13
First prove this, from the definition of the integral: | {
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First prove this, from the definition of the integral:
Prop. Suppose $$p\in[a,b]$$, and $$f:[a,b]\to\Bbb R$$ satisfies $$f(x)=0$$ for all $$x\in[a,b]\setminus\{p\}$$. Then $$f$$ is integrable on $$[a,b]$$ and $$\int_a^bf=0$$.
Hence, even though $$f(x)-\sin(x)$$ does not quite vanish at every point, it follows that $$\int_0^{\pi/2}(f(x)-\sin(x))\,dx=0,$$hence $$\int_0^{\pi/2}f(x)\,dx=\int_0^{\pi/2}\sin(x)\,dx.$$
(Technicality: Since $$\sin(x)$$ and $$f(x)-\sin(x)$$ are both integrable on $$[0,\pi/2]$$, so is $$f$$. | {
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# Continuous process vs Discrete process!
I encountered two statements in Tibshirani's article Regression Shrinkage and Selection via the Lasso, p.58 (pdf).
1. Subset selection is a "discrete process" which leads to unstable coefficients.
2. ridge regression is a "continuous process" leading to stable coefficients.
My question is, why ridge is called continuous process while subset selection is a discrete process. Also, why ridge leads to stability and the Subset selection to instability.
• 1. Do you have a particular book or something that says this? It might be useful to answerers to have context. 2. Are you asking "what makes it discrete or continuous?" or are you asking "why would subset selection be unstable while ridge regression is stable?" or are you asking "why would discreteness/continuity be described as stable or unstable?" – Glen_b Jul 25 '16 at 1:54
• @ Glen_b, my question is why ridge is called continuous process, and subset selection is discrete process, and why the first one leads to stability and the later to instability. – jeza Jul 25 '16 at 2:03
• You still need to provide some context / a quote. – gung Jul 25 '16 at 17:33
• @ gung, now, it is ok – jeza Jul 25 '16 at 18:42
• Selecting a subset of variables is a binary decision and in that sense discrete. Ridge regression, on the other hand, performs regularization by putting stronger emphasis on certain variables (without merely making the decision to include or remove them) as a result of a modified residual term. A more precise explanation can be found in Wikipedia. – Igor Jul 25 '16 at 21:18 | {
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For subset selection, in its most general form, you pick some metric to optimize and try all possible subsets of variables to include in your linear regression model. Suppose you have $p$ covariates, then there are $2^p$ models, i.e. there are $2^p$ vectors $\beta$ to choose from. Hence, jumping from model to model is a discrete process. You cannot "interpolate" between models.
For ridge regression, you add an $l_2$ penalty to the coefficients, so that the model you choose is $$\min_\beta \| X\beta - y\|^2 + \frac{\lambda}{2}\|\beta\|^2$$ In this case, for any $\lambda \ge 0$ you get a different model (that is solution of $\beta(\lambda)$). In fact, $\beta$ will be a smooth function of $\lambda$: change $\lambda$ a little bit, and your fitted $\beta$ will change a little bit. Hence, there are an infinite number of possible models. Thus, this is a continuous process and you can smoothly go from one model to the other by changing $\lambda$.
This is also the reason why subset selection is unstable, and ridge regression is stable. Suppose you change the metric you use to fit the subset selection process a little bit, and you can end up with a very different model (here with different model I mean what elements of $\beta$ are non-zero). As I noted above, this is not true for ridge regression; a small change in $\lambda$ leads to a small change in $\beta$.
Also worth noting that in general all coefficients in Ridge regression are non-zero for all values of $\lambda$, but they converge monotonically to $0$ as $\lambda \to \infty$. | {
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# Calculate $\int_0^\infty\frac{\sin(x)\log(x)}{x}\mathrm dx$.
Calculate $\displaystyle\int_0^\infty\dfrac{\sin(x)\log(x)}{x}\mathrm dx$.
I tried to expand $\sin(x)$ at zero, or use SI(SinIntegral) function, but it did not work. Besides, I searched the question on math.stackexchange, nothing found.
Mathematica tells me the answer is $-\dfrac{\gamma\pi}{2}$, I have no idea how to get it.
Notice that
\begin{align*} \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} t^{s-1}e^{-xt} \, dt \right) \sin x \, dx \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-tx} \sin x \, dx \right) t^{s-1} \, dt \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{1+t^2} \, dt \\ &= \frac{1}{2\Gamma(s)} \beta\left(\frac{s}{2}, 1-\frac{s}{2}\right) \\ &= \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}. \end{align*}
(This heuristic computation is valid line-by-line on the strip $1 < \Re(s) <2$ in view of Fubini's theorem, and then extends to the larger strip $0 < \Re(s) < 2$ by analytic continuation.) Differentiating both sides, we get
$$\int_{0}^{\infty} \frac{\sin x \log x}{x^s} \, dx \stackrel{(*)}{=} -\frac{d}{ds} \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} = \frac{\pi}{2\Gamma(s)\sin\left(\frac{\pi s}{2}\right)} \left( \psi(s) + \frac{\pi}{2}\cot\left(\frac{\pi s}{2}\right) \right).$$
Now the answer follows by plugging $s = 1$:
$$\int_{0}^{\infty} \frac{\sin x \log x}{x} \, dx = -\frac{\gamma \pi}{2}.$$
Justification of $\text{(*)}$. Let us prove that
$$F(s) := \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx \tag{1}$$ | {
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$$F(s) := \int_{0}^{\infty} \frac{\sin x}{x^s} \, dx \tag{1}$$
is analytic on the open strip $S = \{ s \in \Bbb{C} : 0 < \Re(s) < 2 \}$ and its derivative can be computed by the Leibniz's integral rule. A major issue of $\text{(1)}$ is that the defining integral converges only conditionally for $0 < \Re(s) \leq 1$ and thus raises some technical difficulties. In order to circumvent this, we improve the speed of convergence using integration by parts:
$$F(s) = \underbrace{\left[ \frac{1-\cos x}{x^s} \right]_{0}^{\infty}}_{=0} + s \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx.$$
Notice that the resulting integral is absolutely convergent on $S$.
Now we claim that $g(s) := F(s)/s$ is differentiable and its derivative can be computed by the Leibniz's integral rule. Let $\epsilon$ be such that $\bar{B}(s, \epsilon) \subset S$. Then whenever $0 < |h| < \epsilon$,
\begin{align*} &\frac{g(s+h) - g(s)}{h} - \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{9em} = \int_{0}^{\infty} \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \, dx. \end{align*}
Notice that the integrand is dominated by
$$\left| \frac{1-\cos x}{x^{1+s}} \left( \frac{x^{-h} - 1}{h} + \log x \right) \right| \leq \frac{1-\cos x}{x^{1+\Re(s)}} (\max\{ x^{\epsilon}, x^{-\epsilon} \} + 1) |\log x|$$
This dominating function is integrable. Hence by the dominated convergence theorem, as $h \to 0$, we have
$$\lim_{h \to 0} \frac{g(s+h) - g(s)}{h} = \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$
Plugging this back, we know that $F(s)$ is differentiable with
$$F'(s) = \int_{0}^{\infty} \frac{1-\cos x}{x^{s+1}} \, dx + s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx.$$
Finally, performing integration by part to the latter integral yields the desired conclusion: | {
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Finally, performing integration by part to the latter integral yields the desired conclusion:
\begin{align*} &s \int_{0}^{\infty} \frac{(1-\cos x)(-\log x)}{x^{1+s}} \, dx \\ &\hspace{5em} = \underbrace{\left[ \frac{(1-\cos x)\log x}{x^s} \right]_{0}^{\infty}}_{=0} - \int_{0}^{\infty} \left( \frac{\sin x \log x}{x^s} + \frac{1-\cos x}{x^{1+s}} \right) \, dx \end{align*}
• @JackD'Aurizio, Thank you! – Sangchul Lee Sep 26 '16 at 13:35
• Nothing to thank me for, it is a really well-written answer, credits to you. – Jack D'Aurizio Sep 26 '16 at 13:38
• +1. Not too many people care to write a justification $^{*}$ as you did it ( included me !!! ). – Felix Marin Oct 2 '16 at 8:37
Hereafter the $\ds{\ln}$-branch-cut runs along $\ds{\left(-\infty,0\right]}$ with $\ds{\ln\pars{z}\ \,\mrm{arg}}$ given by $\ds{-\pi < \mrm{arg}\pars{z} < \pi}$.
\begin{align} \mc{J} & \equiv \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x = \Im\bracks{{1 \over t - \ic}\int_{0}^{\pars{t - \ic}\infty} \ln\pars{x \over t - \ic}\expo{-x}\,\dd x} \\[5mm] & = -\,\Im\braces{{t + \ic \over t^{2} + 1}\int_{\infty}^{0}\bracks{\ln\pars{x \over \root{t^{2} + 1}} + \arctan\pars{1 \over t}\ic}\expo{-x}\,\dd x} \\[5mm] & = \Im\braces{{t + \ic \over t^{2} + 1}\bracks{% \int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x - {1 \over 2}\ln\pars{t^{2} + 1}+ \arctan\pars{1 \over t}\ic}} \end{align}
Note that
$\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \left.\partiald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x\, \right\vert_{\ \mu\ =\ 0} = \Gamma\,'\pars{1}\ =\ \overbrace{\Gamma\pars{1}}^{\ds{=\ 1}}\ \overbrace{\Psi\pars{1}}^{\ds{-\gamma}}\ =\ -\gamma\quad}$ where $\ds{\Gamma}$
and $\ds{\Psi}$ are the Gamma and Digamma Functions, respectively. Then, | {
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and $\ds{\Psi}$ are the Gamma and Digamma Functions, respectively. Then,
\begin{align} \mc{J} & \equiv \bbox[8px,#ffe,border:0.1em groove navy]{% \Im\int_{0}^{\infty}\ln\pars{x}\expo{-\pars{t - \ic}x}\,\dd x} = -\,{\gamma \over t^{2} + 1} - {1 \over 2}\,{\ln\pars{t^{2} + 1} \over t^{2} + 1} + {t\arctan\pars{1/t} \over t^{2} + 1} \\[5mm] & = \bbox[8px,#ffe,border:0.1em groove navy]{-\,{\gamma \over t^{2} + 1} + \totald{}{t}\bracks{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}} \label{2}\tag{2} \end{align}
Note that$\ds{{1 \over 2}\,\arctan\pars{1 \over t}\ln\pars{t^{2} + 1}}$ vanishes out when $\ds{t \to \infty}$ and $\ds{t \to 0^{+}}$.
By replacing \eqref{2} in \eqref{1}: \begin{align} &\color{#f00}{\int_{0}^{\infty}{\sin\pars{x}\ln\pars{x} \over x}\,\dd x} = -\gamma\int_{0}^{\infty}{\dd t \over t^{2} + 1} = \color{#f00}{-\,{1 \over 2}\,\gamma\pi} \end{align}
• (+1) That is how I just did this. Then I looked on MSE to see if it's already been discusses and found this page. Well done my friend! – Mark Viola Apr 23 at 3:36 | {
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# All real numbers in $[0,2]$ can be represented as $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$
I would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$.
I'll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted by brackets.
$$2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$$
$$1=\sqrt{2 - \sqrt{2 - \sqrt{2 - \dots}}}=(-)$$
$$0=\sqrt{2 - \sqrt{2 + \sqrt{2 + \dots}}}=-(+)$$
$$\phi=\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \dots}}}}=(+-)$$
$$\frac{1}{\phi}=\sqrt{2 - \sqrt{2 + \sqrt{2 - \sqrt{2 + \dots}}}}=(-+)$$
In general, the expansion can be found by a very easy algorithm:
• take any number in $(0,2)$, square it
• if the result $>2$ write $+$, if the result $<2$ write $-$
• subtract $2$ from the result, square, repeat
If on some step we get $2$ exactly, we just write $(+)$ and the expansion is finished.
Examples:
$$\pi-2=--+-++-+-+++++++-+-+---------+-+--+--+--+++---++++ \dots=1.141592653589793 \dots$$
Basically, $50$ terms of our expansion gave only $15$ correct decimal digits for $\pi$. But considering the expansion can be coded as binary, it's not so bad.
The convergence plot, and two binary plots for this $50$ terms can be seen below:
$$e-1=+-----+++-++-+---++-++++-+---++-+++-++++-++++---++ \dots=1.71828182845905 \dots$$
Do you know any reference about this expansion? Can every real number between $0$ and $2$ be expanded this way?
Is number $2$ special in this case, or can we make a similar expansion using some other number (and other power for the root)?
Edit
Now that I think about it, we can use the general expansion for $x \in [0,a]$:
$$x=\left(a \pm \left(a \pm \left(a \pm \dots \right)^p \right)^p \right)^p$$
$$a=2^{\frac{p}{1-p}}$$
For example:
$$\frac{1}{4}=\left(\frac{1}{4} + \left(\frac{1}{4} + \left(\frac{1}{4} + \dots \right)^2 \right)^2 \right)^2$$ | {
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$$\frac{3}{4}-\frac{\sqrt{2}}{2}=\left(\frac{1}{4} - \left(\frac{1}{4} - \left(\frac{1}{4} - \dots \right)^2 \right)^2 \right)^2$$
etc.
However, this case $p=2,~~~a=\frac{1}{4}$ is not just a random example, it's the only rational expansion of this kind. So I would say it's more important than the titular root expansion.
Edit
An interesting article that connects the nested roots of this kind to Chebyshev polynomials: http://www.sciencedirect.com/science/article/pii/S0022247X12003344
• Wow. Great question, +1 – user223391 Apr 6 '16 at 21:33
• Beautiful idea. One could make a binary-like number system based on it. Let a "bit" code 0 for minus or 1 for plus. I wonder if it could have unique representation contrary to the usual way to write numbers, for example $0.1b = 0.011\dots b = 1/2$. – mathreadler Apr 7 '16 at 8:01
• I wonder if there is some class of numbers which have periodic "number expansions" in that number system. Like the rationals have for ordinary number systems using the division algorithm. – mathreadler Apr 7 '16 at 8:27
• @mathreadler, that would be a subset of algebraic numbers – Yuriy S Apr 7 '16 at 8:30
• the $50$ bits gives $50/\log_{10}(2) \approx 15$ decimals which is just one or two bits short of mantissa of a ordinary double precision floating point number system. – mathreadler Apr 7 '16 at 15:36
Here is a possible explanation. Let $\alpha \in [0, \pi/2]$ and define $\epsilon_1, \epsilon_2, \cdots$ by $\epsilon_i = \operatorname{sgn}( \cos ( 2^i \alpha )) \in \{-1, 1\}$. Here, we take the convention that $\operatorname{sgn}(0) =1$. Then applying the identity $2\cos\theta = \operatorname{sgn}(\cos\theta) \sqrt{2 + 2\cos(2\theta)}$ repeatedly, we have
$$2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ \cdots + \epsilon_n \sqrt{2 + \smash[b]{2\cos(2^{n+1} \alpha)} }}}}.$$
This can be used to show that, with an appropriate definition of infinite nested radical, the following identity | {
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$$2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ 2 + \cdots }}}$$
is true. This shows that any real number between $[0, 2]$ can be written as an infinite nested radical of the desired form. Moreover, if we denote $x = 2\cos\alpha$, then
• $\epsilon_1 = \operatorname{sgn}(2\cos (2\alpha)) = \operatorname{sgn}(x^2 - 2)$,
• $\epsilon_2 = \operatorname{sgn}(2\cos (4\alpha)) = \operatorname{sgn}((x^2 - 2)^2 - 2)$,
and likewise. This explains why signs are determined by OP's algorithm.
• Wow, this is a great explanation, which also shows where $2$ comes from – Yuriy S Apr 6 '16 at 22:47
• @YuriyS, Thank you! And yes, this explains why the range is $[0, 2]$. Unfortunately this seems not work for other cases with $a > 2$ instead of $2$, and I even suspect that the analogous claim may be false. (For an an analogous situation, we can in fact check this. Numbers of the form $\pm \frac{1}{2} \pm \frac{1}{4} \pm \frac{1}{8} \pm \cdots$ represents any real number in $[-1, 1]$, but the range of $\pm \frac{1}{3} \pm \frac{1}{9} \pm \frac{1}{27} \pm \cdots$ is the (shifted) Cantor set.) – Sangchul Lee Apr 6 '16 at 22:59
• Nice answer (+1). This is theme of problems 183 - 185 in the classic Problems and Theorems in Analysis I by G. Polya and G. Szegö – Markus Scheuer Apr 9 '16 at 16:02
• This is incredible. – 6005 Jul 26 '16 at 21:08
Peculiar observation
If we define a binary number $b = b_1b_2\cdots b_n$ with digits mapped to the symbols like this: $$b_k = \begin{cases}0 \text{ if } (-) \text{ at position } k\\1 \text{ if } (+) \text{ at position } k\end{cases}$$ Then if we run the algorithm proposed in the question, looping
x(k) = x(k-1)^2-2;
b(k) = (x(k)>0);
the vector b will get logical values corresponding to bits 1 and 0 of the binary number above and we can calculate it for the linear space of $x\in[0,1]$. If we do this we can then calculate each number as the scalar product $$[1/2,1/4,\cdots,1/2^k]b$$ and if we then plot it, it will look like | {
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Which is kind of a peculiar plot having a bit of a discontinuous and fractal structure. I think the largest discontinuity is around $x = \sqrt{1/2}$ but I have no theoretical explanation why..
edit as pointed out by Sangchul Lee this seems similar to Tent Map
• Good observation. I am not expert of symbolic dynamics, but this is related to the tent map orbits. Basically this is because the string of signs are produced by the doubling transform on the circle via the correspondance I explained in my answer. – Sangchul Lee Apr 8 '16 at 2:58
Your algorithm pretty much shows that there exists an expansion for every number in $[0,2]$.
If we replace $2$ by $a>2$ (and keep the square root), we will fail because we need that squaring a number from the interval $[u,v]$ produces a number that is either in $a+[u,v]$ or in $a-[u,v]$. So we must have $u=0$ and $v\ge a$ and $v^2\le a+v$. The last two imply $a\le v\le 2$.
If we additionally switch to $k$th roots, the condition becomes that $v\ge a$ and $v^k\le a+v$, hence $a\le v\le\sqrt[k-1]2$. | {
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# An example of an algebra but not a sigma algebra?
Can someone give me an example of an algebra but not a sigma algebra, with a $\sigma$-finite measure $\mu$ on it?
• As far as I'm aware, $\sigma$-finiteness is only defined on sigma algebras - how are you defining it here? – B. Mehta Apr 14 '18 at 22:43
• @B.Mehta I understood this to mean a measure on the sigma-algebra generated by the algebra. – Severin Schraven Apr 14 '18 at 22:44
Let $X=[0, \infty)$ and consider $\mathcal{A}=\{ \text{finite union of }[a, b)\}$.
It's clear that $[a, b)\cap [c, d) = [\max\{a, c\}, \min\{b, d\})$ and $[a, b)^C = [0, a)\cup [b, \infty)$.
So $\mathcal{A}$ is an algebra but clearly not a $\sigma$-algebra since it doesn't contain any open interval.
Take the obvious measure $m([a, b)) = b-a$.
• +1 Neat example. – Severin Schraven Apr 14 '18 at 22:58
• Thanks Jacky that’s very understandable! :D – JINGYA HAN Apr 14 '18 at 23:35
Take all subsets of $\mathbb{N}$ which are either finite or have finite complement. It is an algebra, but not a sigma-algebra. For the sigma-finite measure you can pick the zero measure. | {
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• Or for the measure pick counting measure. – GEdgar Apr 14 '18 at 23:01
• Thanks for your example. In this context I don’t think the measure could be replaced by a counting measure since that would not always be finite. But say if we replace the measure by a counting measure, does it mean this measure does not have a unique extension on the whole natural number set, since it would meet the requirement of using Carathéodory Extension Theorem? – JINGYA HAN Apr 14 '18 at 23:34
• @JINGYAHAN We could replace the measure with the counting measure, it would still be sigma-finite as we can write $$\mathbb{N}=\bigcup_{n\in \mathbb{N}} \{n\}$$ and the singeltons have counting measure equal to $1$. – Severin Schraven Apr 15 '18 at 0:26
• @SeverinSchraven oh yes you’re right. I confused the definition. The point is you cannot assign a singleton with an infinity measure. Thanks for your help! – JINGYA HAN Apr 15 '18 at 0:31
• @JINGYAHAN Exactly, we don't want concentration of mass on singeltons. By the way you should accept Jacky's answer (on the left you of the question you can find the respective button), unless you are not yet satisfied with the answers given and want to wait for another one. – Severin Schraven Apr 15 '18 at 0:36 | {
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It is the point which corresponds to the mean position of all the points in a figure. Centroid of an Area • In the case of a homogeneous plate of uniform thickness, the magnitude ∆W is With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Center of Mass of a Body Center of mass is a function of density. So to find the centroid of an entire beam section area, it first needs to be split into appropriate segments. Determine the x - and y -coordinates of the centroid of the shaded area. The centroid of a right triangle is 1/3 from the bottom and the right angle. The Find Centroids tool will create point features that represent the geometric center (centroid) for multipoint, line, and area features.. Workflow diagram Examples. y=2 x, y=0, x=2 Find the coordinates of the centroid of the area bounded by the given curves. An analyst at the Scotland Department of Environment is performing a preliminary review on wind farm applications to determine which ones overlap with or are in view of wild lands. How to calculate a centroid. Beam sections are usually made up of one or more shapes. Centroid by Composite Bodies ! y=x^{3}, x=0, y=-8 The cartesian coordinate of it's centroid is $\left(\frac{2}{3}r(\theta)\cos\theta, \frac{2}{3}r(\theta)\sin\theta\right)$. Solution: First, gather the coordinate points of the vertices. The coordinates of the centroid are simply the average of the coordinates of the vertices.So to find the x coordinate of the orthocenter, add up the three vertex x coordinates and divide by three. For more see Centroid of a triangle. Recall that the centroid of a triangle is the point where the triangle's three medians intersect. It is also the center of gravity of the triangle. Next, sum all of the x coodinates ... how to find centroid of composite area: how to calculate centroid of rectangle: how to find centroid of equilateral triangle: 4' 13 Answers: (X,Y) in The centroid or center of | {
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