[Tutor] How to calculate pi with another formula?

[Isr Gish]

How can I get the Decimal module without downloading the whole install for Python 2.4

All the best,
Isr


-----Original Message-----
   >From: "Gregor Lingl"<glingl at aon.at>
   >Sent: 10/29/04 2:27:11 PM
   >To: "Dick Moores"<rdm at rcblue.com>
   >Cc: "tutor at python.org"<tutor at python.org>
   >Subject: Re: [Tutor] How to calculate pi with another formula?
     >Hi Dick!
   >
   >Accidentally I just was tinkering around with the new
   >decimal module of Python2.4. (By the way: it also works
   >with Python 2.3 - just copy it into /Python23/Lib)
   >
   >The attached program uses a very elementary (and inefficient)
   >formula to calculate pi, namely as the area of a 6*2**n-sided
   >polygon (starting with n=0), inscribed into a circle of radius 1.
   >(Going back to Archimedes, if I'm right ...)
   >
   >Nevertheless it calculates pi with a precision of (nearly)
   >100 digits, and the precision can be  arbitrarily enlarged.
   >In the output of this program only the last digit is not correct.
   >
   >import decimal
   >
   >decimal.getcontext().prec = 100
   >
   >def calcpi():
   >    s = decimal.Decimal(1)
   >    h = decimal.Decimal(3).sqrt()/2
   >    n = 6
   >    for i in range(170):
   >        A = n*h*s/2  # A ... area of polygon
   >        print i,":",A
   >        s2 = ((1-h)**2+s**2/4)
   >        s = s2.sqrt()
   >        h = (1-s2/4).sqrt()
   >        n = 2*n
   >       
   >calcpi()
   >
   >Just for fun ...
   >
   >Gregor
   >
   >
   >Dick Moores schrieb:
   >
   >> Is it possible to calculate almost-pi/2 using the (24) formula on 
   >> <http://mathworld.wolfram.com/PiFormulas.html> without using (23)?
   >>
   >> If it's possible, how about a hint? Recursion?
   >>
   >> Thanks, tutors.
   >>
   >> Dick Moores
   >> rdm at rcblue.com
   >>
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   >>
   >>
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