WebNov 7, 2024 · Fixed point iteration with While or Do Loop. I need to write a while or do loop to perform the iteration x n + 1 = C o s ( x n) with initial value x 0 = 1 and stops when the absolute value of the difference between two consecutive iterations is x n + 1 − x n < ϵ , where ϵ = 10 − 16. Finally print the final value x n + 1, displaying 16 ... WebApr 8, 2024 · Mathematica can easily add the vertical line. The range of this function is 1 to 3. Then the command calls for Mathematica to create a straight vertical gridline at x=2. None is part of the command that tells Mathematica to just make it a straight dark, non dashed line.. If you're actually using Plot (or ListPlot, etc.), the easiest solution is to use …
plotting - Henon Map Fixed Points Plot versus Iterations and Plot …
WebJun 4, 2016 · plots = Plot [q [x], {x, 0, 1}, Epilog -> {Directive [ {Thick, Red, Dashed}], line1, line2, Green, PointSize [0.02], Point [ {1/3, q [1/3]}], Black, Dashing [0], Text [Framed … WebJun 30, 2016 · and one can see the period two cycle (red and green are the points that repeat themselves) for a certain value of $μ$. For a 2D system, in our case the Henon map, period-$2$ cycle means that the system: $$ 1)x_1=y_2+1-αx_2^2,\quad y_1=β x_2 \\ 2)x_2=y_1+1-αx_1^2, \quad y_2=β x_1 $$ has a unique solution and that this solution … clock tower chiropractic
plotting - How to plot one point - Mathematica Stack Exchange
WebSuppose we have the following simplified system of two ordinary differential equations: x ˙ ( t) = x ( t) 2 + 2 y ( t) y ˙ ( t) = 3 x ( t) The system has a hyperbolic fixed point the origin. Hence there exits a stable and an … WebApr 10, 2024 · In this command sequence, the independent variable is x and the range is 0 to 2 π. For Plot, after entering the function that you wish to graph, you separate the equation and add {independent variable, lower bound, upper bound}. In this example, we are just plotting a function using Mathematica default capabilities. WebMay 5, 2024 · A fixed point is when x n no longer changes, so x n+1 =r x n e -xn becomes x = r x e -x and if x is nonzero that leaves 1 = r e -x. This is solved to give x = log (r) (or x = 0 if it ever hits zero during its evaluation). So x = log … boda real en inglaterra