Wikipedia:Reference desk/Archives/Mathematics/2011 June 21

= June 21 =

Limit of an increasing sequence
Is ω1 the smallest limit ordinal which cannot be expressed as a limit of an increasing sequence of ordinals? (sequence = indexed by natural numbers) --COVIZAPIBETEFOKY (talk) 00:58, 21 June 2011 (UTC)
 * Yes. Any countable ordinal $$\alpha$$ can easily be expressed as such a limit: fix $$a_n$$ a numbering of the ordinals less than $$\alpha$$.  Define $$b_0 = 0$$, $$b_{n+1} = \max(a_n, b_n+1)$$.  Then it's easy to check that $$\alpha$$ is the limit of the $$b_n$$.  You might also be interested in reading about regular cardinals and cofinality.--Antendren (talk) 04:11, 21 June 2011 (UTC)
 * Nice Antendren! I had a lot of trouble doing a related question in topology but your construction makes it all so simple. Money is tight (talk) 08:08, 21 June 2011 (UTC)

Thanks for the nice explanation and interesting links. I ran across cofinality before, but it didn't make sense to me then; now it does. --COVIZAPIBETEFOKY (talk) 12:47, 21 June 2011 (UTC)

Algebra Question
I've mentioned this a long time ago, but I need some clarification. Let ƒ : (Rk,0) → (R,0) be a function germ. Let Ok denote the local ring of all function germs (Rk,0) → (R,0). Let Jƒ denote the Jacobian ideal generated by the partial derivatives of ƒ, i.e. Jƒ = (∂ƒ/∂x1,&hellip;,∂ƒ/∂xk). The local algebra of ƒ is defined to be the quotient Ok&thinsp;/&thinsp;Jƒ. The quotient of a ring by an idea gives a vector space, in this case a real vector space. It's finite dimensional if, and only if, ƒ has an algebraically isolated critical point at 0 in Rk.

I don't understand the use of the word algebra. Clearly it's an R-vector space, so let's forget addition.
 * What's the binary operation on Ok&thinsp;/&thinsp;Jƒ, is it just multiplication modulo Jƒ?
 * Usually we talk of an algebra over a field, or over a commutative ring. What's the field and/or commutative ring here?
 * The local algebra Ok&thinsp;/&thinsp;Jƒ is a ring itself, so why call it an algebra?
 * If it is an algebra, then should I use a Gothic letter, or is that just for Lie algebras?

Sorry for all the questions, but I've got rings, ideals, vector spaces and algebras attached to the same object, and I'm getting confused. — Fly by Night  ( talk )  18:38, 21 June 2011 (UTC)


 * Multiplication in Ok/Jƒ is just the usual multiplication in Ok modulo Jƒ as you said. Ok/Jƒ is an R-algebra (as is Ok) and the R-action is just boring old scalar multiplication.  Any algebra is a ring so calling Ok/Jƒ an R-algebra implies more structure: basically that the R-action on Ok/Jƒ plays nicely with the ring operations. Rckrone (talk) 01:33, 22 June 2011 (UTC)
 * Oh, okay, I see. What about the Gothic letters? Is that just for Lie algebras? — Fly by Night  ( talk )  02:34, 22 June 2011 (UTC)
 * I haven't noticed a general convention like that in the algebra I've seen, but I can't speak for all areas. I'm not too familiar with germs, so I don't know what the notation should look like there.  I think the gothic letters are used in particular for Lie algebras to distinguish them from the associated Lie group (which is a different object). Rckrone (talk) 03:04, 22 June 2011 (UTC)
 * That sounds about right. The use normal face letters with a subscript letter for the function germ. For some reason the use a curly \mathscr{O} for the local ring. It's a real cross-over of areas. I don't know anything beyond undergraduate abstract algebra, but I need to use it to analyse functions with applications to geometrical problems. It's a real mixed bag. Thanks again. — Fly by Night  ( talk )  20:38, 22 June 2011 (UTC)

Twenty heads in a row problem
A few days ago this problem was discussed, see here It turns out that the approach tried by many involving estimating the average number of rows and use the Poisson distribution does work and you do get an accurate answer that way. This works as follows.

Consider rows of a length of exactly r heads. The probability of a given coin throw being at exactly the start of such a row is 2^[-(r+2)], note that the coin throw right before and right after the row have to be tails. The probability that a coin throw is on any arbitrary position of such a row is thus r 2^[-(r+2)]. If we throw the coin N times, then the expected number of coin throws that are on such a row is N r 2^[-(r+2)]. Since there are always r coin throws that are on the same row, there are N 2^[-(r+2)] different rows. Here we have neglected all the issues regarding different rows interfering with each other. The probability of a row appearing someone is quite small, so we can ignore that to first approximation.

Then summing N 2^[-(r+2)] from r = 20 to infinity yields N 2^(-21) for the expected number of rows of heads of length of at least 20. Then the Poisson distribution implies that Exp[-N 2^(-21)] is the probability that there are no such rows. Count Iblis (talk) 19:07, 21 June 2011 (UTC)


 * So the probability of getting twenty heads in a row if you flip a coin one million times is 1&minus;e&minus;10 62&minus;21 = 0.379256396. Thanks a lot Count Iblis, you cracked the nut! Bo Jacoby (talk) 22:12, 21 June 2011 (UTC).


 * A slightly more precise version of Count Iblis' calculation is this. A row of 20 heads, beginning at the very first flip, requires 20 heads and then 1 tail, and the average number of such rows is 2&minus;21. When beginning at one of the 999979 flips from flip number 2 through flip number 999980, it requires 1 tail and then 20 heads and then 1 tail, and the average is 2&minus;22. When beginning at flip number 999981 it requires 1 tail and 20 heads, and the average is 2&minus;21. The average number of rows of exactly 20 heads, if you flip a coin one million times, is 2&minus;21 + 999979&middot;2&minus;22 + 2&minus;21 = 999983&middot;2&minus;22. Similarly the average number of rows of exactly 21 heads, if you flip a coin one million times, is 999982&middot;2&minus;23, and the average number of rows of at least 20 heads, if you flip a coin one million times, is &Sigma;undefined999982 (999983&minus;i)&middot;2&minus;22&minus;i ≃ 999982&middot;2&minus;21, and the probability of getting twenty heads in a row if you flip a coin one million times is 1&minus;e&minus;999982&middot;2 &minus;21 = 0.379251068. Bo Jacoby (talk) 21:07, 22 June 2011 (UTC).


 * That's interesting! The challenge is in such problems is more to systematically approximate the solution than to find an exact solution... Count Iblis (talk) 14:43, 24 June 2011 (UTC)