Make sure you remove raise NotImplementedError() and fill in any place that says # YOUR CODE HERE .

Fixed Point Iteration

Fixed point:

A number is called a fixed point to function g(x) if g()=. Using fixed points are a nice strategy to find roots of an equation. In this method if we are trying to find a root of f(x)=0, we try to write the function in the form, x=g(x). That is,

f(x)=xg(x)=0

So, if is a fixed point of g(x) it would also be a root of f(x)=0, because,

f()=g()==0

We can find a suitable g(x) in any number of ways. Not all of them would converge; whereas, some would converge very fast. For example, consider Eq.6.1.

f(x)xg(x)g(x)=x5+2.5x42x36x2+x+2=x5+2.5x42x36x2+x+2=x52.5x4+2x3+6x22(6.2)

again,

f(x)=x5+2.5x42x36x2+x+2=0

6x2x2xg(x)=x5+2.5x42x3+x+2=16(x5+2.5x42x3+x+2)=16(x5+2.5x42x3+x+2)=16(x5+2.5x42x3+x+2)(6.3)

Similarly,

2.5x4x4xg(x)=x5+2x3+6x2x2=12.5(x5+2x3+6x2x2)=12.5(x5+2x3+6x2x2)4=12.5(x5+2x3+6x2x2)4(6.4)

B. Complete the code below

For this example we will use a couple of g(x) function to find out which one converges faster.

import numpy as np

import pandas as pd

import matplotlib.pyplot as plt

from numpy.polynomial import Polynomial

f = Polynomial([2.0, 1.0, -6.0, -2.0, 2.5, 1.0])

g1 = Polynomial([-2.0, 0.0, 6.0, 2.0, -2.5, -1.0])

def g2(x):

p = Polynomial([2.0, 1.0, 0.0, -2.0, 2.5, 1.0])

return np.sqrt(p(x)/6)

def g3(x):

p = Polynomial([-2.0, -1.0, 6.0, 2.0, 0.0, -1.0])

return np.power(p(x)/2.5, 1.0/4.0)

a1 = 0.8

g1_a = []

a2 = 0.8

g2_a = []

a3 = 0.8

g3_a = []

# YOUR CODE HERE

raise NotImplementedError()

xs = np.linspace(-2.5, 1.6, 100)

ys = f(xs)

dictionary = {

'x': xs,

'y': ys

}

# Test case:

plt.axhline(y=0, color='k')

plt.plot(xs, f(xs), label='f(x)')

plt.plot(xs, g1(xs), label='g1(x)')

plt.plot(xs, g2(xs), label='g2(x)')

plt.plot(xs, g3(xs), label='g3(x)')

plt.legend()

if len(g1_a) > 0:

root = np.array([g1_a[len(g1_a)-1], g2_a[len(g2_a)-1], g3_a[len(g3_a)-1]])

plt.plot(root, f(root), 'ro')

print(pd.DataFrame({'g1(x)':g1_a, 'g2(x)':g2_a, 'g3(x))':g3_a,}))

CO Fixed_Point_lteration.ipynb File Edit View Insert Runtime Tools Help Changes will not be saved * Share + Code + Text A Copy to Drive Connect Editing Q Make sure you remove raise Not ImplementedError() and fill in any place that says # YOUR CODE HERE, as well as your NAME, ID, and SECTION below: [] NAME = ID = "" SECTION = " Fixed Point Iteration Fixed point: A number & is called a fixed point to function g(x) if g(E) = . Using fixed points are a nice strategy to find roots of an equation. In this method if we are trying to find a root of f(x) = 0, we try to write the function in the form, x = g(2). That is, f(x) = 2 g(x) = 0 So, if & is a fixed point of g(2) it would also be a root of f(T) = 0, because, f() = &- 9(E) = 6 -&=0 We can find a suitable g() in any number of ways. Not all of them would converge; whereas, some would converge very fast. For example, consider Eq. 6.1. f(x) = x + 2.524 22 6.2 +x+2 2-g(x) = 25 +2.524 223 6x2 +x+2 g(x) = -25 2.5x4 + 22? + 6x 2 (6.2) again, f(x) = 25 +2.524 223 6x2 +x+2=0 622 =25 +2.524 22 +2+2 1 B a(zB + 2.524 22 +2+2) CO Fixed_Point_Iteration.ipynb File Edit View Insert Runtime Tools Help Changes will not be saved * Share + Code + Text A Copy to Drive Connect Editing 1 Q 6x2 = 25 +2.524 223 +2+2 1 22 (25 +2.5x4 223 +2+2) 6 /1 (zB +2.524 222 +x+2) 6 = B g(2) = (25 +2.524 22 +2+2) (6.3) 6 Similarly, 2.52* = -25 +22c? + 6x2 2 2 1 -25 +223 + 6x2 2-2) 2 1 24 2.5 1 (-25 + 2x3 + 6x2 2-2) 2= V 2.5 g(x) = 1 2.5 -2.25 +223 + 6x2 1 2) (6.4) B. Complete the code below For this example we will use a couple of g(2) function to find out which one converges faster. B. Complete the code below For this example we will use a couple of g(2) function to find out which one converges faster. [] import numpy as np import pandas as pd import matplotlib.pyplot as plt from numpy - polynomial import Polynomial f = Polynomial([2.0, 1.0, -6.0, -2.0, 2.5, 1.0]) g1 = Polynomial([-2.0, 0.0, 6.0, 2.0, -2.5, -1.0]) def g2(x): p = Polynomial([2.0, 1.0, 0.0, -2.0, 2.5, 1.0]) return np.sqrt(P(x)/6) def g3(x): p = Polynomial([-2.0, -1.0, 6.0, 2.0, 0.0, -1.0]) return np. power(P(x)/2.5, 1.0/4.0) a1 = 0.8 g1_a = [] a2 = 0.8 g2_a = [] a3 = 0.8 g3_a = [] # YOUR CODE HERE raise Not ImplementedError() [] xs = np.linspace(-2.5, 1.6, 100) ys = f(xs) dictionary - { 'x': xs, 'y': ys } [] # Test case: plt.axhline (y=0, color='k') plt.plot(xs, f(xs), label='f(x)') plt.plot(xs, g1(xs), label='g1(x)') plt.plot(xs, g2(xs), label='g2(x)') plt.plot(xs, g3(xs), label='g3(x)') plt.legend() if len(g1_a) > 0: root = np.array([g1_a[len(g1_a)-1], g2_a[len(g2_a)-1], g3_a[len(g3_a)-1]]) plt.plot(root, f(root), 'ro') print(pd.DataFrame({' g1(x)':g1_a, 'g2(x)':g2_a, 'g3(x))':g3_a,})) []