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
This works great, and if I change the precision to, say, 2000, and the range to 2000, I get pi accurate to the 1,205th digit (this took 66 minutes, with psyco employed), when I compare with the pi pages on the web.
Now to my new question. I have an artist friend who knows an artist who needs pi expressed in base 12. I don't know how many digits he needs, but I think he'll take what he can get. Is there a way to use math.log(x, base) with the decimal module to accomplish this? Or is there another way? Or is there no way?
Thanks,
Dick Moores [EMAIL PROTECTED]
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