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ok well that makes sense.
It was the easiest the use though. Im not sure how to translate this http://en.wikipedia.org/wiki/Bellard%27s_formula
To me it just looks like this, it's obviously not right.
n = 0
pi = 1/2^6
while True:
a= -1*n/2^(10*n)*(-2^5/4*(n+1)-1/4*(n+3)+2^8/10*(n+1)-2^6/10*(n+3)-2^2/10*(n+5)-2^2/10*(n+7)+1/10*(n+9))/10*(n+9))
pi+=a
n+=1
print piHey im probably a little old but i wanna learn more math and i've a passing familiarity with python.
I gave it a shot and it's really slow, can anyone else resolve pi faster?
def seq(n):
n = n+2
return [n, n+2]
def findpi():
pi = 0
k = -1
while True:
i = seq(k)
a = (4.0/i[0])-(4.0/i[1])
pi = pi + a
k= k + 4
print pi
findpi()Im pretty sure 3.0 is "the future", aka an experimental version. All the interesing modules I've found are for 2.5+
added more simple
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