Re: Computing PI -- a simple programming problem Message #20 Posted by Gerson W. Barbosa on 16 Oct 2006, 8:37 p.m., in response to message #19 by Namir
Quote:
Pi = { K1/K2 * [ 1 + sum((-1)^i / (1 + 2i)^11) ]} ^(1/11) for i = 1,2,3,..., n
You're right!
As I mentioned earlier, the table below demonstrates that fewer correct digits are obtained as the number of terms increases. The equation is in the HP-200LX Solver format. As you can see, the series is as slowly convergent as the Gregory-Leibniz series, despite the first three iterations giving twelve correct digits. That's why I said I had a nifty algorithm to better solve the first decimals. The first few decimals I meant :-)
P=(K1/K2*(1+SIGMA(I,1,N,1,(-1)^I/(2*I+1)^E)))^(1/E)
K1 = 14863564800
K2 = 50521
E = 11
N = 1 => P = 3.141592647876891
N = 2 => P = 3.141592653725998
N = 3 => P = 3.141592653581560
N = 4 => P = 3.141592653590661
N = 5 => P = 3.141592653589660
N = 6 => P = 3.141592653589820
N = 7 => P = 3.141592653589787
N = 8 => P = 3.141592653589795
N = 9 => P = 3.141592653589793
Anyway, these series are interesting as they can be expressed in terms of the Dirichlet Beta Function:
1 - 1/3n + 1/5n - 1/7n + ... = Beta(n); n = 1, 3, 5...
Gerson.
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I've just found this site:
http://home.att.net/~numericana/answer/analysis.htm
There we can read
Quote:
...
Every other time, such an integration gives an exact expression for the alternating sum of some new power of the reciprocals of odd integers.
...
which explains it all...
Edited: 16 Oct 2006, 10:28 p.m.
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