递归序列$ x_n = \ sqrt {2} $，$ x_ {n + 1} = \ sqrt {2x_n} $

Question

我如何在python中创建一个脚本，它来自递归序列的值：

$x_1 = \sqrt{2}$, $x_{n+1} = \sqrt{2x_n}$ http://www.sciweavers.org/upload/Tex2Img_1392861864/render.png

Answer 1

X = [sqrt(2)]
for i in range(1,10):
    X.append(sqrt(2*X[i-1]))

Answer 2

这是一个缓慢的解决方案。假设n>> = 1

import math
def recursive(n):
    if n = 1:
        math.sqrt(2)
    return math.sqrt(2*recursive(n-1))