Learning in Stereo: Math + Python

Kirby Urner urner at alumni.princeton.edu
Sun Apr 2 20:33:37 CEST 2000


>  >>> pi(100)
>  3.04936163598
>
>  >>> pi(10000)
>  3.14149716395
>

Of course you can push this slowly converging series even
further by rewriting f(n) to multiply with long integers:

 >>> def f(n):
       return 1.0/long(x)*x

 >>> pi(100000)
 3.14158310433
 >>> pi(500000)
 3.14159074373 

Never use this as an efficient way to get PI!!

Here's a continued fraction method:

def pi(depth):
   # use continued fraction of user-supplied depth to refine pi
   n = 1.0
   return 4.0 * 1.0/(1 + 1.0/pifract(n,depth))

def pifract(n,depth):
   if n<depth:
      return 2 + ((2*n+1)**2)/pifract(n+1,depth)
   else: return 2.0

And here's another fast converger using every other 
Fibonacci number:

def pifib(n):
   # return approximation for pi using
   # arctan(1) = 4 * SIGMA [ arctan(1.0/Fib(2i+1))]
   sum = 0
   for i in range(1,n+1):
      sum = sum + math.atan(1.0/fibo(2*i+1))
   return 4*sum

meaning you need and algorithm for the Fibonacci's, like:

fibcache = {}          # thanks to Tim Peters
def fibo(n):
    # return nth fibonacci number (recursive, cached)   
    if n == 0: return 0L
    elif n == 1: return 1L
    elif fibcache.has_key(n): return fibcache[n]        
    else: 
      result = fibo(n-1) + fibo(n-2)
      fibcache[n] = result
      return result

or 
      
root5   = 5.0**0.5
gold    = (1+root5)/2.0 # same as phi, named to not conflict with phi()
tau     = 1/gold

def fibo(n): # thanks to David Zachmann
    # return nth fibonacci number as a floating point
    return (gold**n - (-tau)**n)/root5
    

Then of course there's always:

  >>> import math
  >>> math.pi
  3.14159265359

  :-D

Kirby




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