[2022 Fall] Assignment4-4¶

Course: AP3021

4-4-a¶

Fixed-point iteration¶

to determine a root of $f(x) = -0.9x^2 + 1.7x + 2.5$ using $x_0 = 5.0$. Perform the computation until 𝜀𝑎 is less than 𝜀𝑠 = 0.01%. Also perform an error check of your final answer(plot the 𝜀𝑎 as the iteration growing)

In [1]:

import math
import matplotlib.pyplot as plt

import math import matplotlib.pyplot as plt

In [2]:

def g(x) :
    ans = math.sqrt((1.7 * x + 2.5) / 0.9)

    return ans

def g(x) : ans = math.sqrt((1.7 * x + 2.5) / 0.9) return ans

In [3]:

def fix_point(x0, es, iter_max, ea_list, iter_count_list) :
    x_root = x0
    iter_count = 0

    while True :
        last_x_root = x_root
        x_root = g(last_x_root)
        iter_count += 1
        iter_count_list.append(iter_count)

        if (x_root != 0) :
            ea = abs((x_root - last_x_root) / x_root) * 100
            ea_list.append(ea)

        print("iter time:", iter_count, ",ea =", ea)
        if (ea < es or iter_count >= iter_max) :
            return x_root

def fix_point(x0, es, iter_max, ea_list, iter_count_list) : x_root = x0 iter_count = 0 while True : last_x_root = x_root x_root = g(last_x_root) iter_count += 1 iter_count_list.append(iter_count) if (x_root != 0) : ea = abs((x_root - last_x_root) / x_root) * 100 ea_list.append(ea) print("iter time:", iter_count, ",ea =", ea) if (ea < es or iter_count >= iter_max) : return x_root

In [4]:

x0 = 5
es = 0.01
iter_max = 500
ea_list = []
iter_count_list = []

print("\nThe approximate ans:", fix_point(x0, es, iter_max, ea_list, iter_count_list))
# print(ea_list)
# print(iter_count)

x0 = 5 es = 0.01 iter_max = 500 ea_list = [] iter_count_list = [] print("\nThe approximate ans:", fix_point(x0, es, iter_max, ea_list, iter_count_list)) # print(ea_list) # print(iter_count)

iter time: 1 ,ea = 43.019388386838855
iter time: 2 ,ea = 14.140958348733184
iter time: 3 ,ea = 4.6679854963306715
iter time: 4 ,ea = 1.541483060629453
iter time: 5 ,ea = 0.5090321399985364
iter time: 6 ,ea = 0.16809099521169962
iter time: 7 ,ea = 0.05550608802781297
iter time: 8 ,ea = 0.01832887042630157
iter time: 9 ,ea = 0.006052438895087782

The approximate ans: 2.8601897496224673

In [5]:

# plot

x = iter_count_list
y = ea_list

plt.plot(x, y)

plt.xlim(0, 10)
plt.ylim(0, 50)

plt.title("Growing of ea by fixed point iteration")
plt.grid()
plt.legend(["ea"], loc ="upper right")

plt.savefig("./src/imgs/A4_4_a.png", dpi=300)

plt.show()

# plot x = iter_count_list y = ea_list plt.plot(x, y) plt.xlim(0, 10) plt.ylim(0, 50) plt.title("Growing of ea by fixed point iteration") plt.grid() plt.legend(["ea"], loc ="upper right") plt.savefig("./src/imgs/A4_4_a.png", dpi=300) plt.show()

4-4-b¶

the Newton-Raphson method¶

to determine a root of $f(x) = -0.9x^2 + 1.7x + 2.5$ using $x_0 = 5.0$. Perform the computation until 𝜀𝑎 is less than 𝜀𝑠 = 0.01%. Also perform an error check of your final answer(plot the 𝜀𝑎 as the iteration growing)

In [6]:

def f(x) :
    ans = -0.9 * x ** 2 + 1.7 * x + 2.5

    return ans

def f(x) : ans = -0.9 * x ** 2 + 1.7 * x + 2.5 return ans

In [7]:

def f_prime(x) :
    ans = -1.8 * x + 1.7

    return ans

def f_prime(x) : ans = -1.8 * x + 1.7 return ans

In [8]:

def newton_raphson(x0, es, iter_max, ea_list, iter_count_list) :
    iter_count = 0

    while True :
        next_x = x0 - (f(x0) / f_prime(x0))
        x_root = next_x

        iter_count += 1
        iter_count_list.append(iter_count)

        ea = abs((x_root - x0) / x_root) * 100
        x0 = x_root
        ea_list.append(ea)
        

        print("iter time:", iter_count, ",ea =", ea)

        if (ea < es or iter_count >= iter_max) :
            return x_root

def newton_raphson(x0, es, iter_max, ea_list, iter_count_list) : iter_count = 0 while True : next_x = x0 - (f(x0) / f_prime(x0)) x_root = next_x iter_count += 1 iter_count_list.append(iter_count) ea = abs((x_root - x0) / x_root) * 100 x0 = x_root ea_list.append(ea) print("iter time:", iter_count, ",ea =", ea) if (ea < es or iter_count >= iter_max) : return x_root

In [9]:

x0 = 5
es = 0.01
iter_max = 500
ea_list = []
iter_count_list = []

print("\nThe approximate ans:", newton_raphson(x0, es, iter_max, ea_list, iter_count_list))

x0 = 5 es = 0.01 iter_max = 500 ea_list = [] iter_count_list = [] print("\nThe approximate ans:", newton_raphson(x0, es, iter_max, ea_list, iter_count_list))

iter time: 1 ,ea = 46.0
iter time: 2 ,ea = 17.108052750727655
iter time: 3 ,ea = 2.209254593287688
iter time: 4 ,ea = 0.03644225367328207
iter time: 5 ,ea = 9.913886563630006e-06

The approximate ans: 2.8601044055074283

In [10]:

# plot

x = iter_count_list
y = ea_list

plt.plot(x, y)

plt.xlim(0, 5)
plt.ylim(0, 50)

plt.title("Growing of ea by Newton Raphson")
plt.grid()
plt.legend(["ea"], loc ="upper right")

plt.savefig("./src/imgs/A4_4_b.png", dpi=300)

plt.show()

# plot x = iter_count_list y = ea_list plt.plot(x, y) plt.xlim(0, 5) plt.ylim(0, 50) plt.title("Growing of ea by Newton Raphson") plt.grid() plt.legend(["ea"], loc ="upper right") plt.savefig("./src/imgs/A4_4_b.png", dpi=300) plt.show()

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Last update: 2024-04-27