[2022 Fall] Assignment4-1¶

Course: AP3021

In [1]:

import matplotlib.pyplot as plt
import math
import numpy as np
import pandas as pd
import os

import matplotlib.pyplot as plt import math import numpy as np import pandas as pd import os

4-1.1:¶

Locate the first nontrivial root of $sin(x) = x^3$ where x is in radians.

1. Graphical technique (Python)
2. Bisection program (𝑥𝑙 = 0.5, 𝑥𝑢 = 1, 𝑤h𝑒𝑛 𝜀𝑎 < 𝜀𝑠 = 2%) (Python)

In [2]:

# plot y = sin(x) and y = x^3

x = np.linspace(-10, 10, 500)
y1 = np.sin(x)
y2 = x ** 3

plt.plot(x, y1)
plt.plot(x, y2)

plt.xlim(-10, 10)
plt.ylim(-10, 10)

plt.title("Technique of sin(x) = x^3")
plt.grid()
plt.legend(["sin(x)", "x^3"], loc ="lower right")
plt.axhline(y = 0, color='k', linestyle='-')

# plt.savefig("./src/imgs/A4_1_a-1.png", dpi=300)

plt.show()

# plot y = sin(x) and y = x^3 x = np.linspace(-10, 10, 500) y1 = np.sin(x) y2 = x ** 3 plt.plot(x, y1) plt.plot(x, y2) plt.xlim(-10, 10) plt.ylim(-10, 10) plt.title("Technique of sin(x) = x^3") plt.grid() plt.legend(["sin(x)", "x^3"], loc ="lower right") plt.axhline(y = 0, color='k', linestyle='-') # plt.savefig("./src/imgs/A4_1_a-1.png", dpi=300) plt.show()

In [3]:

# plot y = sin(x) - x ^ 3
# set xl = 0.5; xu = 1.0

x = np.linspace(0.0, 1.0, 500)
y = np.sin(x) - x ** 3

plt.plot(x, y)

plt.xlim(0.49, 1.01)
plt.ylim(-0.25, 0.5)

plt.title("Technique of y = sin(x) - x^3")
plt.grid()
plt.legend(["sin(x) - x^3"], loc ="upper right")
plt.axhline(y = 0, color='k', linestyle='-')

# plt.savefig("./src/imgs/A4_1_a-2.png", dpi=300)

plt.show()

# plot y = sin(x) - x ^ 3 # set xl = 0.5; xu = 1.0 x = np.linspace(0.0, 1.0, 500) y = np.sin(x) - x ** 3 plt.plot(x, y) plt.xlim(0.49, 1.01) plt.ylim(-0.25, 0.5) plt.title("Technique of y = sin(x) - x^3") plt.grid() plt.legend(["sin(x) - x^3"], loc ="upper right") plt.axhline(y = 0, color='k', linestyle='-') # plt.savefig("./src/imgs/A4_1_a-2.png", dpi=300) plt.show()

In [4]:

# Bisection Program

def f(x) :
    ans = np.sin(x) - x ** 3
    return ans

def count_ea(new_x_root, old_x_root) :

    if (old_x_root == -1) : # jump out the first data.
        return 9999
    else :
        ea = abs((new_x_root - old_x_root) / new_x_root)
        ea = ea * 100   # turn into percent
    
    return ea

def bisection(x_lowwer, x_upper, es, x_root, iter_max) :
    iter_count = 0
    
    while True :
        last_x_root = x_root
        x_root = (x_lowwer + x_upper) / 2
        iter_count += 1
        ea = count_ea(x_root, last_x_root)
        temp = f(x_lowwer) * f(x_root)

        if (temp < 0) :
            x_upper = x_root
        elif(temp > 0) :
            x_lowwer = x_root
        else :
            ea = 0.0    
            # return x_root

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

# Bisection Program def f(x) : ans = np.sin(x) - x ** 3 return ans def count_ea(new_x_root, old_x_root) : if (old_x_root == -1) : # jump out the first data. return 9999 else : ea = abs((new_x_root - old_x_root) / new_x_root) ea = ea * 100 # turn into percent return ea def bisection(x_lowwer, x_upper, es, x_root, iter_max) : iter_count = 0 while True : last_x_root = x_root x_root = (x_lowwer + x_upper) / 2 iter_count += 1 ea = count_ea(x_root, last_x_root) temp = f(x_lowwer) * f(x_root) if (temp < 0) : x_upper = x_root elif(temp > 0) : x_lowwer = x_root else : ea = 0.0 # return x_root print("count:", iter_count, "root:", x_root, "ea:", ea) if ea < es or iter_count >= iter_max: return x_root

In [5]:

es = 2.0  # 2%
x_lowwer = 0.5
x_upper = 1.0
x_root = -1
iter_max = 500

ans = bisection(x_lowwer, x_upper, es, x_root, iter_max)

print("\nThe approximate ans:", ans)

es = 2.0 # 2% x_lowwer = 0.5 x_upper = 1.0 x_root = -1 iter_max = 500 ans = bisection(x_lowwer, x_upper, es, x_root, iter_max) print("\nThe approximate ans:", ans)

count: 1 root: 0.75 ea: 9999
count: 2 root: 0.875 ea: 14.285714285714285
count: 3 root: 0.9375 ea: 6.666666666666667
count: 4 root: 0.90625 ea: 3.4482758620689653
count: 5 root: 0.921875 ea: 1.694915254237288

The approximate ans: 0.921875

4-1.2¶

Also perform an error check by substituting your final answer into the original equation in 1-2 (calculate true error 可不用打程式)

In [6]:

# Also perform an error check by substituting your final answer into the original equation in (b) 

def count_true_error(true_value, approximation) :
    return true_value - approximation

def count_et(true_value, approximation) :
    true_error = true_value - approximation
    et = (true_error / true_value) * 100

    return et

true_value = 0.9286263  # calculator

et = count_et(true_value, ans)
print("True error:", et)

# Also perform an error check by substituting your final answer into the original equation in (b) def count_true_error(true_value, approximation) : return true_value - approximation def count_et(true_value, approximation) : true_error = true_value - approximation et = (true_error / true_value) * 100 return et true_value = 0.9286263 # calculator et = count_et(true_value, ans) print("True error:", et)

True error: 0.7270201156267075

from the calculator¶

The root of $y = sin(x) - x^3$ is $−0.9286263, 0, 0.9286263$

And from the previous conduction, true value is $0.9286263$ and the approximation is $0.7421875$.

Therefore we get the true percent error is $20.07683822868252\%$

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Last update: 2024-04-27