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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 π. Show all posts
Showing posts with label π. Show all posts
Saturday, 5 November 2011
A mislaid π is retrieved!
π, here
Friday, 30 September 2011
It's a good look, but we can read right between your lines, Orac.
A statement of fact cannot be insolent. Unless that fact, should it become public, is a threat to certain activities and interests.
Orac has a particular Modus Operandus, and a very easy task. It is his self appointed role to de-construct fallacy in medicine. Something he does expertly, wittily, and with great precision and thoroughness.
He and his commentariat have infinite patience and capacity with their adversaries in the anti-pashmina movement. No argument or admission goes un-addressed.
Anything Th1Th2 says is given a certain treatment.
Not so, Jacob and Slartibartfastibast who, with their respective offerings of cannabis therapy for autism and the neanderthal admixture hypothesis for autism...
Why no Orac treatment? Could it be that these two hypotheses are a threat in some way to Orac and friends?
Orac has a particular Modus Operandus, and a very easy task. It is his self appointed role to de-construct fallacy in medicine. Something he does expertly, wittily, and with great precision and thoroughness.
He and his commentariat have infinite patience and capacity with their adversaries in the anti-pashmina movement. No argument or admission goes un-addressed.
Anything Th1Th2 says is given a certain treatment.
Not so, Jacob and Slartibartfastibast who, with their respective offerings of cannabis therapy for autism and the neanderthal admixture hypothesis for autism...
Why no Orac treatment? Could it be that these two hypotheses are a threat in some way to Orac and friends?
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)..
Friday, 10 June 2011
4 and 20 black birds escape from the all the πs
That pie is open. That's what she said. Cannabis saves lives, cannabis saves babies, teenage girls and grandmothers love hearing this the most!
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