Fixed-Point Iteration
http://iailab.kaist.ac.kr/
Industrial AI Lab at KAIST
Source
- By Prof. David J. Malan from Harvard University
- By Prof. Erik Demaine from MIT online lecture
Table of Contents
1. Fixed-Point Iteration¶
1.1. Numerical approach¶
For the given equation:
$$f(x) = 0 \implies x = g(x)$$
Remark: always achievable
$$ \begin{align*} x & = f(x) + x = g(x)\\ x &= -f(x) + x = g(x) \end{align*} $$
Goal: numerically find the solution of $x = g(x)$
Main idea:
Make a guess of the solution, $x_k$
If $g(x_k)$ is 'nice', then hopefully, $g(x_k)$ will be closer to the answer. If so, we can iterate
Iteration algorithm
- choose an initial point $x_0$
- Do the iteration $x_{k+1} = g(x_k)$ until meeting stopping criteria
Convergence check (or analysis)
$\quad$ Let $r$ be the exact solution, $r = g(r)$
$$ \begin{array}{l} x_{k+1} &= g(x_k)\\ \text{error} :&\quad e_k &= x_k - r\\ &\quad e_{k+1} &= x_{k+1} - r = g(x_k) - g(r)\\ &&= g'(\eta)(x_k - r) = g'(\eta)e_k , \quad \eta \in (x_k, r)\\ \\ &\implies \lvert e_{k+1}\rvert &\leq \lvert g'(\eta)\rvert \lvert e_k \rvert\\ \end{array} $$
$$\text{If } \lvert g'(\eta) \rvert <1, \text{error decreases (iteration converges)}$$
1) Example of $x = \cos(x)$
# Computational Thinking on how to calculate cos(x) = x
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
x = np.linspace(-2*np.pi, 2*np.pi, 100)
y = np.cos(x)
plt.figure(figsize = (8,6))
plt.plot(x, y, linewidth = 2)
plt.plot(x, x, linewidth = 2)
# plt.xlim(-2*np.pi, 2*np.pi)
plt.axvline(x=0, color = 'k', linestyle = '--')
plt.axhline(y=0, color = 'k', linestyle = '--')
plt.legend(['cos(x)','x'])
plt.axis('equal')
plt.ylim([-1,1])
plt.show()
# naive approach
x = 0.3
print (np.cos(x))
print (np.cos(np.cos(x)))
print (np.cos(np.cos(np.cos(x))))
print (np.cos(np.cos(np.cos(np.cos(x)))))
print (np.cos(np.cos(np.cos(np.cos(np.cos(x))))))
print (np.cos(np.cos(np.cos(np.cos(np.cos(np.cos(x)))))))
print (np.cos(np.cos(np.cos(np.cos(np.cos(np.cos(np.cos(x))))))))
# better way
x = 0.3
for i in range(24):
tmp = np.cos(x)
x = np.cos(tmp)
print (x)
# better way
x = np.zeros((24, 1))
x[0] = 0.3
for i in range(23):
x[i+1] = np.cos(x[i])
print (x)
# better way
x = 10
for i in range(24):
x = np.cos(x)
print (x)
2) example of $\sqrt{2} \;(x^2 = 2)$
# Use an idea of a fixed point
x = 2
for i in range(10):
x = 2/x
print (x)
x = np.linspace(-3, 3, 100)
y = 2/x
plt.figure(figsize = (8,6))
plt.plot(x, y, linewidth = 2)
plt.plot(x, x, linewidth = 2)
plt.axvline(x=0, color = 'k', linestyle = '--')
plt.axhline(y=0, color = 'k', linestyle = '--')
plt.legend(['2/x','x'])
plt.axis('equal')
plt.ylim([-1,1])
plt.show()
# How to overcome
# Use an idea of a fixed point + kind of *|damping|*
x = 3
for i in range(10):
x = (x + 2/x)/2
print (x)
x = np.linspace(-4, 4, 100)
y = (x + 2/x)/2
plt.figure(figsize = (8,6))
plt.plot(x, y, linewidth = 2)
plt.plot(x, x, linewidth = 2)
plt.axvline(x=0, color = 'k', linestyle = '--')
plt.axhline(y=0, color = 'k', linestyle = '--')
plt.axis('equal')
plt.ylim([-1,1])
plt.show()
Think about why it gives different results.
1.2. System of Linear Equations¶
$$ \begin{align*} 4x_1 - x_2 + x_3 &= 7 \\ 4x_1 - 8x_2 + x_3 &= -21 \\ -2x_1 + x_2 + 5x_3 &= 15 \end{align*} $$
import numpy as np
import matplotlib.pyplot as plt
# matrix inverse
A = np.array([[4, -1, 1], [4, -8, 1], [-2, 1, 5]])
b = np.array([[7, -21, 15]]).T
x = np.linalg.inv(A).dot(b)
print (x)
This solution only possible for small size problems. There are many iterative methods for large problems.
$$ \begin{align*} 4x_1 - x_2 + x_3 &= 7 & x_1 &= \frac{1}{4}x_2 - \frac{1}{4}x_3 + \frac{7}{4}\\ 4x_1 - 8x_2 + x_3 &= -21 & \implies \quad x_2 &= \frac{1}{2}x_1 + \frac{1}{8}x_3 + \frac{21}{8}\\ -2x_1 + x_2 + 5x_3 &= 15 & x_3 &= \frac{2}{5}x_1 - \frac{1}{5}x_2 + \frac{15}{5} \end{align*} $$
- In a matrix form
$$\begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} = \begin{bmatrix} 0 &\frac{1}{4} & - \frac{1}{4}\\ \frac{1}{2} & 0 & \frac{1}{8}\\ \frac{2}{5} & -\frac{1}{5} & 0\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} + \begin{bmatrix} \frac{7}{4}\\ \frac{21}{8}\\ 3\\ \end{bmatrix} $$
- iteration
$$ \begin{bmatrix} x_1^{(k+1)}\\ x_2^{(k+1)}\\ x_3^{(k+1)}\\ \end{bmatrix} = \begin{bmatrix} 0 &\frac{1}{4} & - \frac{1}{4}\\ \frac{1}{2} & 0 & \frac{1}{8}\\ \frac{2}{5} & -\frac{1}{5} & 0\\ \end{bmatrix} \begin{bmatrix} x_1^{(k)}\\ x_2^{(k)}\\ x_3^{(k)}\\ \end{bmatrix} + \begin{bmatrix} \frac{7}{4}\\ \frac{21}{8}\\ 3\\ \end{bmatrix} $$
# Iterative way
A = np.array(([[0, 1/4, -1/4 ],
[4/8, 0, 1/8],
[2/5, -1/5, 0]]))
b = np.array([[7/4, 21/8, 15/5]]).T
# initial point
x = np.array([[1, 1, 2]]).T
A = np.asmatrix(A)
b = np.asmatrix(b)
x = np.asmatrix(x)
for i in range(20):
x = A*x + b
print (x)
Remark) try this one
$$ \begin{align*} x_1 &= -3x_1 + x_2 - x_3 + 7\\ x_2 &= 4x_1 - 7x_2 + x_3 + 21\\ x_3 &= 2x_1 - x_2 + -4x_3 + 15 \end{align*} $$
$$\begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} \leftarrow \begin{bmatrix} -3 & 1 & -1 \\ 4 & -7 & 1 \\ 2 & -1 & -4 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ \end{bmatrix} + \begin{bmatrix} 7\\ 21\\ 15\\ \end{bmatrix} $$
Convergence check
$$x \leftarrow Ax + b$$
$$\begin{align*} x_{k+1} &= Ax_k +b\\ &= A(Ax_{k-1} + b) + b = A^2X_{k-1} + Ab + b\\ & \vdots\\ &= A^{k+1}x_0 + A^kb + \cdots + Ab + b \end{align*}$$
# think about why this one does not work
A = np.array(([[3, 1, -1 ],
[4, 7, 1],
[2, -1, -4]]))
b = np.array([[7, 21, 15]]).T
# initial point
x = np.array([[1, 2, 2]]).T
for i in range(10):
x = A.dot(x) + b
print (x)
# stability, check eigenvalue of A
A = np.array(([[3, 1, -1 ],
[4, 7, 1],
[2, -1, -4]]))
np.linalg.eig(A)
# stability, check eigenvalue of A
A = np.array(([[0, 1/4, -1/4 ],
[4/8, 0, 1/8],
[2/5, -1/5, 0]]))
np.linalg.eig(A)
%%javascript
$.getScript('https://kmahelona.github.io/ipython_notebook_goodies/ipython_notebook_toc.js')