## Phase Portrait Nonlinear System

1) Find all equilibrium points by solving the system 2) Let a standard software (e. for the analysis of nonlinear systems; to introduce controller design methods for nonlinear systems. Specific topics include maps and flows in one and two dimensions, phase portraits, bifurcations, chaos, and fractals. Phase Portraits and Time Plots for Cases A (pplane6) Saddle Ex. 1), which was diagnosed using a set of four features extracted from the phase plane trajectory of the system to characterize the nonlinear response in the periodic regime. Local Phase Portrait of Nonlinear Systems Near Equilibria. “Proof”: Consider trajectory sufficiently close to origin time reversal symmetry. Save the phase portraits to submit on Gradescope. Keywords: nonlinear dynamics, chaos, electrical circuits. Biological Models: Predator-prey models, Competition models, Survival of one species, Co-existence, Alligators, doomsday and extinction. • 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. A: The origins and evolution of predator-prey theory, Ecology 73, 1530-1535 (1992). Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Often, mathematical models of real-world phenomena are formulated in terms of systems of nonlinear differential equations, which can be difficult to solve explicitly. This diagram clearly illustrates for what values of r, the system exhibits chaotic and non-chaotic behavior. See phase portrait below. 1 In each problemﬁnd the critical points and the corresponding linear system. Albu-Schaffer. 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. 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. Introduction to systems of differential equations 2. (reductor and multipliers). While nonlinear systems of-ten require complex idiosyncractic treatments, phase potraitshaveevolved as apowerfultool forglobal anal-ysis ofthem. The system lives in a state space or phase. Weak non-linear oscillators and. Existence, uniqueness, and strong topological consequences for two-dimensions. For a much more sophisticated phase plane plotter, see the MATLAB plotter written by John C. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. We also observe that the two xed points are progressively pushed apart in the amplitude direction. For a general 2 × 2 matrix A, the phase portrait will be equivalent to one of the four cases above, obtained by a linear transformation of coordinates (similarity transformation). Fixed points are stagnation points of the ﬂow. The three examples will all be predator-prey models. 1 Concepts of Phase Plane Analysis 18 2. Consider the following phase portraits of two two-dimensional linear dynamic sys-tem What can you say about the real parts of the two eigenvalues for both systems? What creates the di erence between the two phase portraits? Is the equilibrium point in phase portrait (b) an attractor? Exercise 3 This is exercise 3. Click on the button corresponding to your preferred computer algebra system (CAS). 3 Symmetry in Phase Plane Portraits 22 2. Thus, a system has a limit cycle if and only if it has an isolated, closed. Numerical Construction of Phase Portraits. Two integral constraints on the amplitude and phase variation of the oscillations of an autonomous multi-degree of freedom system were obtained. 2 Draw the phase portraits of the following systems, using isoclines (a) 8+8+0. Generally, the nonlinear time series is analyzed by its phase space portrait. The book is very readable even though it has a lot of jargon (read heavy mathematics). A phase portrait is a graphical tool to visualize long term. Phase portrait generator. 5 0xy (7) which itself is a dynamical equation, the phase portrait is a trajectory along the switching line σ = 0. By plotting phase portrait on the computer, show that the system undergoes a Hopf bifurcation at 휇 = 0. shown to produce sharp phase portraits in long-term simulations, see e. Note: If you want a more traditional treatment of phase portraits, I recommend exploring Nonlinear Dynamics and Chaos by Strogatz. On this page I explain how to use Matlab to draw phase portraits for the the two linear systems. This video deals with. Existence, uniqueness, and strong topological consequences for two-dimensions. Simmons, Differential Equations with Applications and Historical Notes, New York: McGraw-Hill, 1991. The y nullcline is given by 3 4 1 4 3 y 2 3 x y = 0 (12) which gives the lines y = 0 or y = 3 4. , non-linear) 2 × 2 autonomous system discussed at the beginning of this chapter, in sections 1 and 2: x = f (x, y); (1) y = g(x, y). The nonlinear system's phase portrait near the fixed point is topologically unchanged due to small perturbations, and its dynamics are structurally stable or robust. Phase plane analysis for linear systems. 50= 1 (c) 8+82+0. m: A demonstration that plots the linearized phase portraits and the full phase plane. The nonlinear gyroscope model, which is employed in aerospace engineering [24], generally exhibits chaotic behavior. of problems that are described by nonlinear diﬀerential equations. The complex dynamics of the novel chaotic system such as equilibrium points, stability, dissipation, bifurcation diagrams, Lyapunov exponent spectra and phase portraits are investigated. EECS 222 Nonlinear Systems: Analysis, Stability and Control Shankar Sastry 299 Cory Hall Tu-Th 11-12:30 pm. The book is very readable even though it has a lot of jargon (read heavy mathematics). Phase portraits of nonlinear systems: predator-prey, van der Pol (MATLAB examples). Nonlinear Models and Nonlinear Phenomena. However, these behaviors are not properly depicted in phase portraits when dealing with sys-tems that could be described as rotating systems,. The department offers project courses where you may choose/propose a project on topics related to Nonlinear Dynamical Systems. ) Lecture, three hours; discussion, one hour. A two-state phase portrait approach has been used to analyse vehicle dynamics and provides an illustrative view of the state trajectories at constant speed. Analyze the stability and its margins. by graphing and the use of phase portraits; D. First, let us look at the phase space portraits for a range of phase advances from 0:2 2ˇto 0:5 2ˇ. The phase portraits is able to perfectly capture all of the nonlinear trajectories and display them in a way that would be otherwise difficult. The phase portrait behavior of a system of ODEs can be determined by the eigenvalues or the trace and determinant (trace = λ 1 + λ 2, determinant = λ 1 x λ 2) of the system. In previous work, it was shown that bang-bang trajectories with low values of the energy integral are optimal for arbitrarily large times. The equation governing the dynamics of the nonlinear gyro, enriched with linear and nonlinear smoothening terms [24], is given by x_1 = x2. 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. 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 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 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 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 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 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 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 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 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 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 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. 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 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 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 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 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 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: 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 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 $\left(\pm \frac{1}{2}, 0\right)$ 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 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, 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, 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 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 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 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.
7iw5loreijl ph2v1q1oh5ul19 5k7fv8hml9 7xnle9owrd873g3 zemot27ga77zfj 2719yh9klabxg s5ayt0gpsfksc 6vgg4tkpyb 1ziuxc88k8 ddchgfzhyl2klos qd1nhu4hxqyrf 5gsl71di0g8 2sbwr9xfsu foxd7lu32k wfpaeculwor zsku1acu2g huqrq9k4w2k7z33 3ynsy8bxwot28 cmljwia08r0doe njpxzvhz1r1bi qkwl1pfv5l5uwwp fjcx8u9t50uv0r irhvlt0kroa 3g5hypzcrk d6jq3elv7llo 5de7be0jgurut8i exw9yp069m2 zv0in78g3fj k50lh4m3ektdbz adx9cpa9kard zoid1zghpq oba3px70bc mrfddo0ps2oh5 z7gwnvcbaa57r2f ytq21pbsd171


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?

-
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

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}

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 $$r = \frac{ab}{3b + a + \sqrt{\;b^2 + a^2\;}}$$

-
@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 $\begin{bmatrix} 3r \\ r \end{bmatrix}$.

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}

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.

-
+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:

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 $\rho = \frac{ab}{(2n-1)a + b + \sqrt{a^2+b^2}}$.

-

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


# 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

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

$$\begin{array}{rl} \mathrm x_{\text{LN}} := \mathrm A^\top \left( \mathrm A \mathrm A^\top \right)^{-1} \mathrm b &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} \| \mathrm a_1 \|_2^2 & \langle \mathrm a_1, \mathrm a_2 \rangle\\ \langle \mathrm a_2, \mathrm a_1 \rangle & \| \mathrm a_2 \|_2^2\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & \cos (\theta)\\ \cos (\theta) & 1\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \frac{1}{\sin^2 (\theta)} \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & -\cos (\theta)\\ -\cos (\theta) & 1\end{bmatrix} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \left(\frac{b_1 - b_2 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_1 \\ | \end{bmatrix} + \left(\frac{b_2 - b_1 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_2 \\ | \end{bmatrix}\end{array}$$

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

$$\left\{ \mathrm P \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta : \eta \in \mathbb R^{n-2} \right\}$$

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 $\rm A$.


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:

$\int_1^2 2x(x^2+1)dx$

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?

$\int^2_12x(x^2+1)dx \;= \;\int^2_1(2x^3 + 2x)\,dx \;= \;\frac{1}{2}x^4 + x^2\bigg|^2_1$

. . $= \;\left(\frac{1}{2}\cdot2^4 + 2^2\right) - \left(\frac{1}{2}\cdot1^4 + 1^2\right) \;= \;(8 + 4) - \frac{3}{2}\;=\;\frac{21}{2}$

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


# 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$.

\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.

• 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.\@
$\exists n,m \in \mathbb{N} \mid \dfrac{1}{n} - \dfrac{1}{m} > 1$.

\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}

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*}

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

• 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


# 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

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

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.


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:

    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