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It's now Caturday and the time is 1200 likes O'clock! AMQ is 22.05, Shields are at 87.52 and the gossip ratio fallen slightly, up to 7.02 :) [For the nerds: PI*Gossip Ratio = 22.05 = current AMQ. Looks like we found our third PI!] |
We won't be silenced by those who won't be silenced. Quis custurd pudding et π.
Showing posts with label pi. Show all posts
Showing posts with label pi. Show all posts
Saturday, 5 November 2011
A mislaid π is retrieved!
π, here
Friday, 30 September 2011
Mai oh Mai, C4A found PAI
Facebook stats are not working properly again so the shield power rating is not available. We have a record 157 UK members and we span 17 major cities worldwide. The ratio of Jenny McCarthy Body Count's likes to Cannabis for Autism's likes on 29/09/2011 was a very close approximation of PI (which is = 3.14159265) |
Sunday, 25 September 2011
Look mum, we went from 108 to π in less than 23 iterations ;)
108 = 2^2 x 3^3 or the solution for ((x-1)^(x-1))(x^x) when x = 3
From which, and with no great leap of imagination at all, we can get the expression (3^(1÷3))^−(2^(1÷2)) or [ ((x^(1/x))^-((x-1)^(1/(x-1)) ] which is almost the lower bound of the expansion so with a bit of iteration we find that the lower bound occurs when x = π and n = 2.
(12^(1÷12))^−(11^(1÷11))^−(10^(1÷10))^−(9^(1÷9))^−(8^(1÷8))^−(7^(1÷7))^−(6^(1÷6))^−(5^(1÷5))^−(4^(1÷4))^−(3^(1÷3))^−(2^(1÷2))^−(1^(1÷1)) = 0.841236826
(3^(1÷3))^−(2^(1÷2))^−(1^(1÷1) = 0.771865166
(3^(1÷3))^−(2^(1÷2)) = 0.595775834
(4^(1÷4))^−(3^(1÷3)) = 0.606624308
(5^(1÷5))^−(4^(1÷4)) = 0.63430981
(3.13^(1÷3.13))^−(2.13^(1÷2.13)) = 0.594577953
(3.12^(1÷3.12))^−(2.12^(1÷2.12)) = 0.59462546
(3.11^(1÷3.11))^−(2.11^(1÷2.11)) = 0.594679901
(3.14^(1÷3.14))^−(2.14^(1÷2.14)) = 0.59453723
(π^(1÷π))^−((π−1)^(1÷(π−1))) = 0.594531362
A little bird helped me along the way x
So I guess I'd better figure out what the upper bound is now, though I assume it's =1
I do hope the upper bound is less than 1.
There's something really real about the interval between 0.594531362 and 0.841236826 or slightly higher.
But, alas;
(9999^(1÷9999))^−(9998^(1÷9998)) = 0.99907846
and
(9999^(1÷9999))^−(9999^(1÷9999)) is 1.
If you'd like to know where the original expression goes, the answer is nowhere special with respect to π:
((π−1)^(π−1))(π^π) = 186.271338549
((π−2)^(π−2))((π−1)^(π−1)) = 5.942346492
and know that it tends towards infinity and sub zero (but not less than -1)..
From which, and with no great leap of imagination at all, we can get the expression (3^(1÷3))^−(2^(1÷2)) or [ ((x^(1/x))^-((x-1)^(1/(x-1)) ] which is almost the lower bound of the expansion so with a bit of iteration we find that the lower bound occurs when x = π and n = 2.
(12^(1÷12))^−(11^(1÷11))^−(10^(1÷10))^−(9^(1÷9))^−(8^(1÷8))^−(7^(1÷7))^−(6^(1÷6))^−(5^(1÷5))^−(4^(1÷4))^−(3^(1÷3))^−(2^(1÷2))^−(1^(1÷1)) = 0.841236826
(3^(1÷3))^−(2^(1÷2))^−(1^(1÷1) = 0.771865166
(3^(1÷3))^−(2^(1÷2)) = 0.595775834
(4^(1÷4))^−(3^(1÷3)) = 0.606624308
(5^(1÷5))^−(4^(1÷4)) = 0.63430981
(3.13^(1÷3.13))^−(2.13^(1÷2.13)) = 0.594577953
(3.12^(1÷3.12))^−(2.12^(1÷2.12)) = 0.59462546
(3.11^(1÷3.11))^−(2.11^(1÷2.11)) = 0.594679901
(3.14^(1÷3.14))^−(2.14^(1÷2.14)) = 0.59453723
(π^(1÷π))^−((π−1)^(1÷(π−1))) = 0.594531362
A little bird helped me along the way x
So I guess I'd better figure out what the upper bound is now, though I assume it's =1
I do hope the upper bound is less than 1.
There's something really real about the interval between 0.594531362 and 0.841236826 or slightly higher.
But, alas;
(9999^(1÷9999))^−(9998^(1÷9998)) = 0.99907846
and
(9999^(1÷9999))^−(9999^(1÷9999)) is 1.
If you'd like to know where the original expression goes, the answer is nowhere special with respect to π:
((π−1)^(π−1))(π^π) = 186.271338549
((π−2)^(π−2))((π−1)^(π−1)) = 5.942346492
and know that it tends towards infinity and sub zero (but not less than -1)..
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