[Python]我的递归有什么问题？

Question

I want to performe

  [√(1/2+1/2*√1/2)]  ---------(P1)
  [√(1/2+1/2*√(P1)]  ----------(P2)
  [√(1/2+1/2*√(P2)]  
  etc.

to find out the P(nth)term

I have this for now

from math import *

n=eval(input('fjeowijo'))
i=sqrt(1/2+1/2*(sqrt(1/2)))

def P(n):
  for i in range(n):        
      g=sqrt(1/2+1/2*i)
      i=sqrt(1/2+1/2*i)
      print(g)

P(n)

When I enter 1 for n, the result is 0.7071067811865476, which is only equal to the part " sqrt(1/2) ". Why?

Answer 1

def P(n):
  i = 0
  g = sqrt(0.5+.5*sqrt(0.5)
  while(i < n):
    g = sqrt(0.5+0.5*g)
    print(g)

Could it be what you're looking for ?

Answer 2

In case you want to make it recursive do

def P(n):
    if n <= 0:
        return 0.5
    else:
        return sqrt(0.5*(1+sqrt(P(n-1))))

which works out as

>>> P(1)
0.9238795325112867
>>> P(2)
0.9902490907008235
>>> P(3)
0.9987774031336784
>>> P(4)
0.9998471169658009

As @JoshRomRock points out, python generally limits the maximum depth of the recursion (the default max depth is platform-dependent). In case of CPython, the default limit is 1000.

In case you want to alter such limit do:

import sys
sys.setrecursionlimit(5000)

Note: in this case, the precision offered by the floating point representation may well be the biggest hurdle to the calculation. Please see the official docs for further info on what to expect using floats.

Second note: in case the P function is going to be used several times and/or in multiple points of the code of a program, as a library function, it would make sense to implement it using Memoization . Please see here for a few examples in python.