[2022 Fall] Assignment4-5¶

Course: AP3021

4-5-1¶

Use the Newton-Raphson method to find the root of $f(x) = e^{-0.5x} (4 - x) - 2$

4-5-2¶

Employ initial guesses of (a) 2, (b) 6, and (c) 8. Explain your results. (Python+解釋) Hint: Think about the problems of this method.

In [1]:

import math

import math

In [2]:

def f(x) :
    e = math.e
    ans = ((e ** (-0.5 * x)) * (4 - x)) - 2

    return ans

def f_prime(x) :
    e = math.e
    ans = (-0.5 * (e ** (-0.5 * x)) * (4 - x)) - (e ** (-0.5 * x))

    return ans

def newton_raphson(x0, es, iter_max) :
    iter_count = 0
    iter_count_list = []
    x_root = x0
    print("x0 =", x0)
    print()

    while True :
        last_x_root = x_root

        try :
            x_root = last_x_root - (f(x_root) / f_prime(x_root))
        except :
            print("total use", iter_count, "times.")
            return "Divergence"

        iter_count += 1
        iter_count_list.append(iter_count)
        if x_root != 0 :
            ea = abs((x_root - last_x_root) / x_root) * 100        

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

        if (ea < es or iter_count >= iter_max) :
            print("total use", iter_count, "times.")
            return x_root

def f(x) : e = math.e ans = ((e ** (-0.5 * x)) * (4 - x)) - 2 return ans def f_prime(x) : e = math.e ans = (-0.5 * (e ** (-0.5 * x)) * (4 - x)) - (e ** (-0.5 * x)) return ans def newton_raphson(x0, es, iter_max) : iter_count = 0 iter_count_list = [] x_root = x0 print("x0 =", x0) print() while True : last_x_root = x_root try : x_root = last_x_root - (f(x_root) / f_prime(x_root)) except : print("total use", iter_count, "times.") return "Divergence" iter_count += 1 iter_count_list.append(iter_count) if x_root != 0 : ea = abs((x_root - last_x_root) / x_root) * 100 print("iter time:", iter_count, ",ea =", ea, "root:", x_root) if (ea < es or iter_count >= iter_max) : print("total use", iter_count, "times.") return x_root

a. $x_0 = 2$¶

In [3]:

x0 = 2
es = 0.01
iter_max = 500

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

x0 = 2 es = 0.01 iter_max = 500 print("\nThe approximate ans:", newton_raphson(x0, es, iter_max))

x0 = 2

iter time: 1 ,ea = 609.9293556607687 root: 0.2817181715409549
iter time: 2 ,ea = 63.73755414477369 root: 0.7768868450453745
iter time: 3 ,ea = 11.888408438696143 root: 0.8817078789285671
iter time: 4 ,ea = 0.4510949099401444 root: 0.8857032411666447
iter time: 5 ,ea = 0.000627839236345159 root: 0.8857088019940234
total use 5 times.

The approximate ans: 0.8857088019940234

b. $x_0 = 6$¶

In [4]:

x0 = 6
es = 0.01
iter_max = 500

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

x0 = 6 es = 0.01 iter_max = 500 print("\nThe approximate ans:", newton_raphson(x0, es, iter_max))

x0 = 6

total use 0 times.

The approximate ans: Divergence

c. $x_0 = 8$¶

In [5]:

x0 = 8
es = 0.01
iter_max = 500

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

x0 = 8 es = 0.01 iter_max = 500 print("\nThe approximate ans:", newton_raphson(x0, es, iter_max))

x0 = 8

iter time: 1 ,ea = 93.39913842615294 root: 121.19630006628846
iter time: 2 ,ea = 100.0 root: 7.212131452880089e+24
total use 2 times.

The approximate ans: Divergence

---

Last update: 2024-04-27