Lab 2–CPS106

This lab gives you more practice on dealing with iterations and approximations in python.

1)  Write a program that converts a given positive integer to its binary representation.

(相关资料图)

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 = '''