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August 31

Exterior algebra as Hopf algebra

I became a bit confused about exterior algebras as Hopf algebras. There is an excellent Wiki article about exterior algebra, and there is also one about Hopf algebra. However, they do not answer my question.

A finite dimensional exterior algebra may be equipped with a coproduct and a counit (described in the exterior algebra Wiki entry), with which it becomes a bialgebra. If one chooses a maximal form (volume form), one may introduce a further mapping, the Hodge dual (which depends on the choice of this particular maximal form). It is stated in the exterior algebra Wiki entry that the Hodge dual becomes an antipode map, with which the exterior bialgebra becomes a Hopf algebra. However, it is known that the antipode map for a bialgebra, if exists, then it must be unique. Therefore, it follows, that the choice of this particular volume form, which we use to define the Hodge dual, should be unique, if the resulting map is intended to be an antipode map. (It should be uniquely determined by the product, unit, coproduct and counit maps.) How can we express it then with the product, unit, coproduct and counit?

AndrasLaszlo 09:47, 31 August 2007 (UTC)[reply]

What field of Math?

I sometimes pass time by playing around in Microsoft Excel, creating sets of randomized data and analyzing them. For example, I made a simulation of a coin being flipped many thousands of times, then calculated the longest run of heads or tails, and then tried to find the relationship between total number of coin flips and max length of run. I can experimentally come up with an equation that describes this data, but I wonder if anyone could suggest what areas of math I could look into (given a university calculus education) to try to empirically figure out the mathematical basis behind this relationship (and other like-minded inquiries). Thanks, jeffjon 12:59, 31 August 2007 (UTC)[reply]

probability seemed the obvious answer, and you would also need statistics. Maybe also Probability theory is a good link. The articles may be a bit theoretical - searching the web will no doubt turn up many much simpler introductions.87.102.88.202 13:40, 31 August 2007 (UTC)[reply]

I mostly understand those fields already; let me make my question a little more specific. I already know how to, for example, determine the probability that there will be a run of 3 heads in 10 consecutive coin flips. The next steps are the ones I lack:
1) Given that I can find the probability of a run of 10, 11, 12... heads or tails in a row, how would I translate that to "on average, the max run in X coin flips is Y long"?
2) Given that set of data (i.e. a table of number coins flipped vs. average lengths of maximum runs), how would I determine the mathematical equation that relates X and Y? (Experimentally, it looks like the answer is about

Y=1.45*ln(X)+0.266

, but of course that's an approximation. I'd like to find the exact theoretical relationship.) jeffjon 13:57, 31 August 2007 (UTC)[reply]

I would expect X to be about 2^Y, in which case

Y={\frac {\ln(X)}{\ln(2)}}\approx 1.44\ln(X)

. Gandalf61 14:30, 31 August 2007 (UTC)[reply]

That's a decent rough approximation, but still not exact. jeffjon 17:41, 31 August 2007 (UTC)[reply]

This is related to Markov chains. Let (m, h, t) stand for the random variable giving, in a sequence of n flips, the triple consisting of m = max run length, h = length of final consecutive run of heads, and t = length of final consecutive run of tails. For example, in the sequence TTHHHHHTHTT, (m, h, t) = (5, 0, 2). Constraints: m ≤ n, h×t = 0, m ≥ max(h, t). Given n, there is a probability distribution of (m, h, t) over the thus constrained space of possible outcomes. For reasons of symmetry between heads and tails (assuming a fair coin), P(m, s, 0) = P(m, 0, s), which means we can simplify this to a two-component random variable (m, s), where s = max(h, t). Summing P(m, s) over s then gives the distribution of just the max run lengths. For n = 0, we have P(0, 0) = 1, and for n = 1, P(1, 1) = 1. Given the distribution of (m, s) for n, it is easy to compute the distribution for n+1. If s < m, there is a 50% probability of a transition (m, s) → (m, s+1), and a 50% probability of (m, s) → (m, 1). When s reaches m, this becomes, 50-50, (m, m) → (m+1, m+1) and (m, m) → (m, 1). This is all numerical stuff, where you need a numeric value for n to compute this. Presumably the recurrence relations have some analytical solution, giving P(m, s) as an expression in variables m, s and n, but I don't immediately see a way of getting there. --Lambiam 14:42, 31 August 2007 (UTC)[reply]

Let Y_n be the expected value of the maximum run length for a sequence of length n. Defining a_n := Y_n2ⁿ⁻¹, the sequence a_n is an integer sequence (sequence A102712 in the OEIS). I haven't actually proved it is the same sequence, but equality holds for all 30 entries listed, and the proof may be easy: the entry at the OEIS gives a characterization that looks closely related. It also gives a generating function. --Lambiam 19:41, 31 August 2007 (UTC)[reply]

Extreme value theory is a somewhat related field, since it concerns the distribution of the maximum (or minimum) of a large number of random variables. However in this particular case the number of random variables (run lengths) is small and not even fixed, so you'll probably need to look elsewhere. 84.239.133.38 08:40, 2 September 2007 (UTC)[reply]

Please help me with "Newton Forward Interpolation"

Given Sin 45=0.7071, Sin 50=0.7660, Sin 55=0.8192, Sin 60=0.866, using Newton Forward Interpolation formula find the value of Sin 52? —Preceding unsigned comment added by Raghu.savi (talk • contribs) 18:11, 31 August 2007 (UTC)[reply]

See Newton polynomial. Conscious 18:31, 31 August 2007 (UTC)[reply]