递归序列$x\u n=\sqrt{2}$，$x{n+1}=\sqrt{2x}$

Question

如何在python中创建一个脚本来延迟递归序列中的值：

$x_1 = \sqrt{2}$, $x_{n+1} = \sqrt{2x_n}$ 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]))