Lab 2–CPS106
This lab gives you more practice on dealing with iterations and approximations in python.
(资料图)
1) Write a program that converts a given positive integer to its binary representation.
Hint: write the program using a for loop. To get the number of bits in a given integer, use the function bit_length() in python. For example, for a = 983265982365, _length() will be 40.
Test your program with following integers:
a = 87324834345,
a = 65234242,
a = 576762341243968.
Compare your results with the built-in function bin(int a) in python.
2) Let f(x) = x**5 + 52*x**3 - 29 be a function. This function has only one root in the real numbers. Write a bisection search algorithm to find that root up to two digits of accuracy, ., for epsilon = as in the class.
3) Solve question 2 using the Newton-Raphson method. Compare the number of iterations with the bisection search.
代码
2-2 bisectionSearch
# -*- coding: utf-8 -*-
"""
Created on Sat Jan 7 10:50:46 2023
@author: pc
"""
def f(x):
return x**5 + 52*x**3 - 29
#for i in range( 10 ):
# print(f(i))
i = 0
f1 =
while(i < 3 ):
i = i + 1
f1 = f(f1)
print( f1 )
d =
xn =
for i in range(1,10):
xn = xn * (2-xn*d)
print("iteration %d: 1/d=%" % (i, xn))
#bisection search
left =
right =
epsilon =
if f(left) > 0:
print("error!")
else:
counter = 0
while left < right:
mid = float((left + right) /2)
counter += 1
if f(mid) < :
left = mid + epsilon
if f(mid) > :
right = mid - epsilon
print("iteration %d: l=% r=%" % (counter, left, right))
2-2-2 bisectionSearch
# -*- coding: utf-8 -*-
"""
Created on Sun Jan 8 09:02:36 2023
@author: pc
"""
def ComputerFunction(x):
return x**5 + 52*x**3 - 29
a = 0
c = 2
b = a + (c - a)/2
epslon = b - a
count = 0
while epslon > :
if ComputerFunction(a) * ComputerFunction(b) < 0:
a = a
c = b
b = a + (c - a)/2
else:
a = b
c = c
b = a + (c - a)/2
epslon = b - a
count+=1
print(a)
print("The number of iterations is:",count)
2-3 Newton-Raphson
# -*- coding: utf-8 -*-
"""
Created on Sat Jan 7 23:58:10 2023
@author: pc
"""
def f(x):
return x**5 + 52*x**3 - 29 #'''定义 f(x) = (x-3)^3'''
def fd(x):
return 5*x**4 + 52*3*x**2 #'''定义 f'(x) = 3*((x-3)^2）'''
def newtonMethod(n,assum):
time = n
x = assum
Next = 0
A = f(x)
B = fd(x)
print('A = ' + str(A) + ', B = ' + str(B) + ', time = ' + str(time))
if f(x) == :
return time,x
else:
Next = x - A/B
print('Next x = '+ str(Next))
if abs(A - f(Next)) < : #1e-6: #
print('Meet f(x) = 0,x = ' + str(Next)) #'''设置迭代跳出条件，同时输出满足f(x) = 0的x值'''
else:
return newtonMethod(n+1,Next)
newtonMethod(0, 1) #'''设置从0开始计数，x0 = '''
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