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December 3rd, 2009, 02:19 PM   #1
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Pi formula

Well, I've done it. I've doubled the speed of the Leibniz series for pi, while finding a formula for the denominators.

8/3 + 8/35 + 8/99 + ...

Let p be the previous denominator.
Let t be the number of terms so far.
n = p + (32*(t-1))

Now, I'm having trouble doubling it AGAIN, and finding formulae for it. Apparently, it still has 32, attached to it.

304/105 + 2,352/19,305 + 6,448/156,009 + ...

So, I need some help, here.
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December 5th, 2009, 03:02 AM   #2
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Your results are correct, but how did you obtain the formula you gave? Why can't you just apply the same method again to obtain a second formula?
In your last series, the nth numerator (starting at n = 0) is given by 16(16(2n + 1) + 3).
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December 10th, 2009, 07:11 AM   #3
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Re: Pi formula

Well, this is how I did it:

8/3 + 8/35 + 8/99 + 8/195 + ...

35 = 3 + 32
99 = 35 + 64
195 = 99 + 96

64 = 32*2
96 = 32*3

35 = 2nd
99 = 3rd
195 = 4th

So:

p + (32*(n-1)) = denominator

And that is how I did it.
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December 11th, 2009, 03:42 AM   #4
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How did you obtain "8/3 + 8/35 + 8/99 + 8/195 + ..." in the first place?
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December 15th, 2009, 05:41 PM   #5
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Re: Pi formula

4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + ... = Pi

4 - 4/3 = 8/3
4/5 - 4/7 = 8/35
4/9 - 4/11 = 8/99
4/13 - 4/15 = 8/195

Therefore:

8/3 + 8/35 + 8/99 + 8/195 + ... = Pi
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December 17th, 2009, 06:41 AM   #6
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If you sum the terms in groups of four instead of two, you get the second series. I gave the expression for the numerator, which is obtainable by working out 4/(8n + 1) - 4/(8n + 3) + 4/(8n + 5) - 4/(8n + 7).
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January 21st, 2010, 01:57 PM   #7
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Re: Pi formula

I found the formula for the numerator! It's this:

Code:
p + (2^(t+9))
Thanks, skipjack, for the other formula!
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January 23rd, 2010, 12:48 AM   #8
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That's incorrect. The correct formula is p + 2048(t - 1), which is easily derived from the result I gave earlier.
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January 23rd, 2010, 10:18 AM   #9
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Re: Pi formula

Whoops! Works for the first 2 differences.
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February 12th, 2010, 10:23 AM   #10
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Re: Pi formula

Okay, this is the finished product.

Starting with n = 0:

16(16(2n+1)^2 + 3)
______________________________________
4096n^4 + 8192n^3 + 5504n^2 + 1408n + 105


Sorry about the double post, it wouldn't let me edit! -_-
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