5. Iterative Computation: Exercise

5. Iterative Computation: Exercise#

Name:

\[ % Latex macros \newcommand{\mat}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\w}{\boldsymbol{w}} \newcommand{\x}{\boldsymbol{x}} \newcommand{\y}{\boldsymbol{y}} \]
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

1. Newton’s method in n dimension#

Newton’s method can be generalized for \(n\) dimensional vector \(x \in \Re^n\) and \(n\) dimensional function \(f(x)={\bf0} \in \Re^n\) as $\( x_{k+1} = x_k - J(x_k)^{-1}f(x_k) \)\( where \)J(x)\( is the *Jacobian matrix* \)\( J(x) = \mat{\p{f_1}{x_1} & \cdots & \p{f_1}{x_n}\\ \vdots & & \vdots\\ \p{f_n}{x_1} & \cdots & \p{f_n}{x_n}} \)$

  1. Define a function that computes $\( f(x) = \left(\begin{array}{c} a_0 + a_1 x_1^2 + a_2 x_2^2\\ b_0 + b_1 x_1 + b_2 x_2\end{array}\right) \)$ and its Jacobian.

def f(x, a, b, deriv=True):
    """y[0] = a[0] + a[1]*x[0]**2 + a[2]*x[1]**2\\
    y[1] = b[0] + b[1]*x[0] + b[2]*x[1]
    also return the Jacobian if derive==True"""
    y0 = 
    y1 = 
    if deriv:
        J = 
        
        return np.array([y0, y1]), np.array(J)
    else:
        return np.array([y0, y1])

  Cell In[2], line 5
    y0 =
         ^
SyntaxError: invalid syntax

a = [-1, 1, 1]
b = [-1, 1, 2]

f([1,1],a,b)

  1. Consider the case of \(a = [-1, 1, 1]\) and \(b = [-1, 1, 2]\) and visualize parabollic and linear surfaces.

%matplotlib notebook

x = np.linspace(-2, 2, 25)
y = np.linspace(-2, 2, 25)
X, Y = np.meshgrid(x, y)
XY = np.array([X,Y])  # (2,25,25) array
Z = 
ax = plt.figure(figsize=(8,8)).add_subplot(projection='3d')
ax.plot_surface(X, Y, Z[0])

  1. Implement Newton’s method for vectors.

def newton(f, x0, *args, target=1e-6, maxstep=20):
    """Newton's method. 
        f: should also return Jacobian matrix
        x0: initial guess
        *args: parameter for f(x,*args)
        target: accuracy target"""
    n = len(x0)  # dimension
    x = np.zeros((maxstep+1, n))
    y = np.zeros((maxstep, n))
    x[0] = x0
    for i in range(maxstep):
        y[i], J = f(x[i], *args)
        if   < target:
            break  # converged!
        x[i+1] = 
    else:
        print('did not coverge in', maxstep, 'steps.')
    return x[:i+1], y[:i+1]

  1. Test how it works from different initial guesses.

newton(f, [0,1], a, b)

newton(f, [1,1], a, b)

2. Bifurcation and Chaos#

A value of \(x_k\) that stays unchanged after applying a map \(f\) to it (i.e. \(x_k = f(x_k) = x_{k+1}\)) is called a “fixed point” of \(f\).

Let us consider the logistic map $\( x_{k+1} = a x_k(1 - x_k) \)$

  1. Plot \(x_{k+1}=ax_k(1-x_k)\) along with \(x_{k+1}=x_k\) for \(a=0.5, 2, 3.3\).

What are the fixed points of these maps?

  1. A fixed point is said to be “stable” when nearby values of \(x_k\) also converge to the fixed point after applying \(f\) many times; it’s said to be “unstable” when nearby values of \(x_k\) diverge from it.

Draw “cobweb plots” on top of each of the previous plots to visualize trajectories. Try several different initial values of \(x_k\).

Are the fixed points you found stable or unstable?

How is the stability related to the slope (derivative) of \(f(x_k)=ax_k(1-x_k)\) at the fixed point?

3: optional) A bifurcation diagram is a plot of trajectories versus a parameter.
draw the bifurcation diagram for parameter \(a\) \((1 \le a \le 4)\), like below:
bifurcation

Hint:

  • Use the logistic() and iterate() functions from the previous lecture.

  • For each value of \(a\), show the trajectory (i.e., the values that \(x_k\) took over some iterations) of the map after an initial transient.

  • Since \(x_k\) is 1D, you can plot the trajectory on the y axis. For example, take 200 points in \(1 \le a \le 4\), run 1000 step iterations for each \(a\), and plot \(x\) after skipping first 100 steps.

3. Recursive call and fractal#

Draw the Sherpinski gasket as below.

shelpinski

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