问题
I am new to Python and was asked to create a program that would take an input as a non-negative integer n and then compute an approximation for the value of e using the first n + 1 terms of the continued fraction:
I have attempted to decipher the question but can't exactly understand everything it is asking. I am not looking for an exact answer but hopefully an example to help me on my way.
This is the exact question
Below is a code I have done with continued fractions before.
import math
# Get x from user
x = float(input("Enter x = "))
# Calculate initial variables and print
a0 = x//1
r0 = x-a0
print("a0 =", a0, "\tr0 =", r0)
# Calculate ai and ri for i = 1,2,3 and print results
a1 = 1/r0//1
r1 = 1/r0 - a1
print("a1 =", a1, "\tr1 =", r1)
a2 = 1/r1//1
r2 = 1/r1 - a2
print("a2 =", a2, "\tr2 =", r2)
a3 = 1/r2//1
r3 = 1/r2 - a3
print("a3 =", a3, "\tr3 =", r3)
回答1:
Without further information, it's probably a Good Idea™ to use the simple continued fraction expansion of e, as shown in Wikipedia:
e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, ...]
This sequence can easily be created using a simple list comprehension.
To evaluate a simple continued fraction expansion we can process the list in reversed order.
The following code will work on Python 2 or Python 3.
#!/usr/bin/env python
''' Calculate e using its simple continued fraction expansion
See http://stackoverflow.com/q/36077810/4014959
Also see
https://en.wikipedia.org/wiki/Continued_fraction#Regular_patterns_in_continued_fractions
Written by PM 2Ring 2016.03.18
'''
from __future__ import print_function, division
import sys
def contfrac_to_frac(seq):
''' Convert the simple continued fraction in `seq`
into a fraction, num / den
'''
num, den = 1, 0
for u in reversed(seq):
num, den = den + num*u, num
return num, den
def e_cont_frac(n):
''' Build `n` terms of the simple continued fraction expansion of e
`n` must be a positive integer
'''
seq = [2 * (i+1) // 3 if i%3 == 2 else 1 for i in range(n)]
seq[0] += 1
return seq
def main():
# Get the the number of terms, less one
n = int(sys.argv[1]) if len(sys.argv) > 1 else 11
if n < 0:
print('Argument must be >= 0')
exit()
n += 1
seq = e_cont_frac(n)
num, den = contfrac_to_frac(seq)
print('Terms =', n)
print('Continued fraction:', seq)
print('Fraction: {0} / {1}'.format(num, den))
print('Float {0:0.15f}'.format(num / den))
if __name__ == '__main__':
main()
output
Terms = 12
Continued fraction: [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]
Fraction: 23225 / 8544
Float 2.718281835205993
Pass the program an argument of 20 to get the best approximation possible using Python floats: 2.718281828459045
As Rory Daulton (& Wikipedia) mention, we don't need to reverse the continued fraction list. We can process it in the forward direction, but we need 2 more variables because we need to track 2 generations of numerators and denominators. Here's a version of contfrac_to_frac which does that.
def contfrac_to_frac(seq):
''' Convert the simple continued fraction in `seq`
into a fraction, num / den
'''
n, d, num, den = 0, 1, 1, 0
for u in seq:
n, d, num, den = num, den, num*u + n, den*u + d
return num, den
回答2:
The value e can be expressed as the limit of the following continued fraction:
e = 2 + 1 / (1 + 1 / (2 + 2 / (3 + 3 / (4 + 4 / (...)))))
The initial 2 + 1 / falls outside of the main pattern, but after that it just continues as shown. Your job is to evaluate this up to n deep, at which point you stop and return the value up to that point.
Make sure you carry out the calculation in floating point.
来源：https://stackoverflow.com/questions/36077810/continued-fractions-python