# [Tutor] How to calculate pi with another formula?

Isr Gish isrgish at fastem.com
Fri Oct 29 22:54:01 CEST 2004

```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
>>
>> _______________________________________________
>> Tutor maillist  -  Tutor at python.org
>> http://mail.python.org/mailman/listinfo/tutor
>>
>>
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>

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