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June 9

zeroes of zeta function

I was watching an explanation of the Riemann Zeta Function and there was a comment that Riemann calculated a few non-trivial zeroes, saw they all had real part 1/2, and because he was a genius was able to make his conjecture. My question is how do you calculate a non-trivial zero of that function? RJFJR (talk) 15:16, 9 June 2014 (UTC)[reply]

Riemann presumably used the Riemann–Siegel formula. Sławomir Biały (talk) 21:46, 9 June 2014 (UTC)[reply]

generators of symmetric group

Hello. I am playing with a subgroup of S₂₄ generated by 12 generators, and I have discovered (via GAP) that the size of the subgroup is 24!, so they generate the whole of S₂₄. If this is the case, I can create any permutation by composing my 12 generators. How can I find the appropriate combination of my generators to effect, say, a single 2-cycle? What techniques are there to do this? Thanks, Robinh (talk) 20:17, 9 June 2014 (UTC)[reply]

I don't know what all the techniques for this there are. But for single 2-cycle case, as a quick exploration, if you haven't already, look at the odd powers of those of the 12 generators {the g(i), 0<i<13} that are odd permutations, and also look at those "g(m)g(n)s" for 0<m<n<13, for which exactly one of g(m) and g(n) is an odd permutation. As you may know, you don't have to check g(n)g(m) if you have already checked g(m)g(n), since they will have same cycle structure.--Rich Peterson199.33.32.40 (talk) 21:12, 10 June 2014 (UTC)[reply]

Thanks for this. Your observation that 2-cycles are odd might well be the key I've missed. But my generators are all 4-cycles or pairs of 4-cycles; I can create 5-cycles quite easily. I've been playing with commutators [that is, x y xinverse yinverse] but with no luck so far. I guess I'm looking for techniques to generate any given permutation from my 12 generators. Best wishes, any ideas welcome, Robinh (talk) 22:40, 10 June 2014 (UTC)[reply]

I guess what I'm really asking is: "are there any systematic ways to investigate combining a fixed set of generators together to create specific permutations, that are more effective than the pretty much random approach I'm taking right now?" Robinh (talk) 22:47, 10 June 2014 (UTC)[reply]

That is a study i know little about. But if you had had instead a set of generators with wellknown or easy to figure out procedures for constructing arbitrary permutations, you would be all set. So if you could get to, from your generators, a good set of 24 single 2 cycles, you could generate anything straightforwardly. But a set of such 2 cycles can be obtained, by conjugating a single 2-cycle by a 24 cycle...if you can find a way to link your 4-cycles together to form a 24-cycle, and then you can also find a 2-cycle, the problem will be reduced to a previously solved one. Hope this helps.Rich (talk) 00:06, 13 June 2014 (UTC)[reply]

(OP) thanks Rich. I've found solitary 3,4,5 and 7-cycles so far. Still no solitary 2-cycles! But I never thought to seek a 24-cycle because I've been trying to get small cycles rather than a maximal length one. Best wishes, Robinh (talk) 00:42, 13 June 2014 (UTC)[reply]