Tim*_*ras 4 python approximation python-3.x continued-fractions
我是Python的新手,被要求创建一个程序,该程序将输入作为非负整数n,然后使用连续分数的前n + 1个项来计算e的近似值:
我试图破译这个问题,但无法完全理解它所问的一切。我不是在寻找确切的答案,而是希望有一个例子可以对我有所帮助。
这是确切的问题,
下面是我之前对连续分数所做的代码。
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)
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没有更多信息,使用e的简单连续分数展开可能是一个好主意,如Wikipedia所示:
e = [2; 1,2,1,1,4,1,1,1,1,1,1,1,1,10,1,1,12,1,1,...
可以使用简单的列表推导轻松创建此序列。
为了评估简单的连续分数扩展,我们需要以相反的顺序处理列表。
以下代码将在Python 2或Python 3上运行。
#!/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()
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输出
Terms = 12
Continued fraction: [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8]
Fraction: 23225 / 8544
Float 2.718281835205993
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向程序传递参数20,以使用Python浮点数获得最佳近似值:2.718281828459045
e 值可以表示为以下连分数的极限:
e = 2 + 1 / (1 + 1 / (2 + 2 / (3 + 3 / (4 + 4 / (...)))))
初始形态2 + 1 /落在主要形态之外,但之后它会继续如图所示。您的工作是对此进行深入评估n,此时您停止并返回到该点的值。
确保以浮点数进行计算。