# coding: utf-8
# # Fixed Point Iteration
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import numpy as np
import matplotlib.pyplot as pt
# **Task:** Find a root of the function below by fixed point iteration.
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x = np.linspace(0, 4.5, 200)
def f(x):
return x**2 - x - 2
pt.plot(x, f(x))
pt.grid()
# Actual roots: $2$ and $-1$. Here: focusing on $x=2$.
# We can choose a wide variety of functions that have a fixed point at the root $x=2$:
#
# (These are chosen knowing the root. But here we are only out to study the *behavior* of fixed point iteration, not the finding of fixed point functions--so that is OK.)
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def fp1(x): return x**2-2
def fp2(x): return np.sqrt(x+2)
def fp3(x): return 1+2/x
def fp4(x): return (x**2+2)/(2*x-1)
fixed_point_functions = [fp1, fp2, fp3, fp4]
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for fp in fixed_point_functions:
pt.plot(x, fp(x), label=fp.__name__)
pt.ylim([0, 3])
pt.legend(loc="best")
# Common feature?
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for fp in fixed_point_functions:
print(fp(2))
# All functions have 2 as a fixed point.
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z = 2.1; fp = fp1
#z = 1; fp = fp2
#z = 1; fp = fp3
#z = 1; fp = fp4
n_iterations = 4
pt.figure(figsize=(8,8))
pt.plot(x, fp(x), label=fp.__name__)
pt.plot(x, x, "--", label="$y=x$")
pt.gca().set_aspect("equal")
pt.ylim([-0.5, 4])
pt.legend(loc="best")
for i in range(n_iterations):
z_new = fp(z)
pt.arrow(z, z, 0, z_new-z)
pt.arrow(z, z_new, z_new-z, 0)
z = z_new
print(z)
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