Judging from the length of his spams, AP must have the
smallest micro penis on the planet. What a looser!
Archimedes Plutonium schrieb:
Konig der Wissenshaften
ayatollah potassium schrieb am Montag, 13. Januar 2003 um 16:41:19 UTC+1:
[from sci.math.research]
David Corfield wrote:
I have a few questions about combinatorics and species, and would be grateful for any comments.
The species of permutations on even sets corresponds to the series
(1 - x^2)^ -1 = 1 + x^2 + x^4 + ...
Does it have a square root?
We have (1 - x^2) ^-1/2 . (1 - x^2) ^-1/2 = (1 - x^2) ^ -1
So, equating coefficients of x^2n:
(2n)! = sum over r from 0 to n ( (2n choose 2r) ((2r)!!^2)((2n-2r)!!^2)
Eg. 24 = 9 + 6 + 9, 720 = 225 + 135 + 135 + 225,
but I can't see how the permutations on 4 or 6 elements divides *nicely* this
way.
1/sqrt(1-x^2) = exp( (f(x) + f(-x))/ 2) where f(x) = - log(1-x) = sum (x^n / n)
is the generating function for cycles. Thus,
1/sqrt(1-x^2) is the exp. generating function for permutations with all cycles of even length.
This is the same as a partition of a set of size 2n into two sets of
size n ("even" and "odd") with an ordered pair of bijections ("forward"
and "back") between them. Those bijections, also called matchings,
are what account for the 2r!! terms in your formula.
So I guess your question is, to show that a permutation on a set of size
2n is in bijection with decompositions into two even cardinality subsets
(2n = 2a+2b) and an an all-cycles-even permutation on each. The most
obvious approach doesn't work, as there is no canonical bijection
between permutations with all-odd and all-even cycles on a 2k-element set.
So it may be a question like permutations versus orderings where the
same generating function leads to different species.
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