Wikipedia:Reference desk/Archives/Mathematics/2016 July 11

= July 11 =

Dice game
$$x$$ players each roll a $$y$$-sided die (containing numbers from 1 to $$y$$). If they roll a 1 they leave the game. The remaining players roll again, and so on until no players are left. What is the expected number of rolls rounds? (does this problem have a name?) 24.255.17.182 (talk) 01:02, 11 July 2016 (UTC)
 * The problem is easiest when all the players are independent. You did not specify whether you meant the total number of rolls by everyone (whose expectation is x times the expected value of a geometric distribution with $$p = \frac{1}{y}$$, or $$xy$$), or the total number of rounds before everyone is out. The latter case is a bit difficult to compute, I'll revisit it later today probably.--Jasper Deng (talk) 01:24, 11 July 2016 (UTC)
 * Yes, all the players are independent, and I meant the total number of rounds before all the players are out. 24.255.17.182 (talk) 01:26, 11 July 2016 (UTC)
 * Let $$p=\frac1y$$ be the probability to drop out and n be the number of players (that's more natural than x). Denote by $$a_n$$ the result. It is easy to see (by conditioning on the first step) that $$a_n$$ satisfies the recursion
 * $$a_0=0$$
 * $$a_n=\frac{1+\sum_{k=0}^{n-1}a_kp^{n-k}(1-p)^k\binom{n}{k}}{1-(1-p)^n}$$
 * I don't know how to find a closed form, but the first values (for $$y=6$$) are
 * $$a_1=6$$
 * $$a_2=\frac{96}{11}$$
 * $$a_3=\frac{10566}{1001}$$
 * $$a_4=\frac{728256}{61061}$$
 * $$a_5=\frac{3698650986}{283994711}$$
 * I couldn't find a match for this on OEIS. -- Meni Rosenfeld (talk) 14:40, 11 July 2016 (UTC)
 * I found a match for $$p=\tfrac12$$. https://oeis.org/A158466, https://oeis.org/A158467.
 * Apparently this is related to skip lists. It also appears that there is no elementary closed form.
 * There is an expression as a non-recursive sum though:
 * $$a_n = \sum_{k=1}^n\frac{(-1)^{k}\binom{n}{k}}{(1-p)^k-1}$$
 * -- Meni Rosenfeld (talk) 14:49, 11 July 2016 (UTC)

Countable real analysis?
I was wondering if one could get rid of uncountable sets in real analysis by not taking the step to define real numbers. Instead of defining R first (which involves a limiting process but it yields something that does not exist within the countable domain) and then using that, one could try to calculate whatever one wants to calculate and move the analogue of the continuum limit at the end. Count Iblis (talk) 04:03, 11 July 2016 (UTC)
 * There are a many workers who, probably because of some grave moral failing, don't like uncountable sets and have examined such questions. For example, you might look into the work of Errett Bishop.  In general, these systems have not been widely adopted. --Trovatore (talk) 04:11, 11 July 2016 (UTC)
 * Sorry, was this a dig against us morally depraved individuals that aren't stricly "mathematical realists"? I will be sure to add this to my already extensive list of personal failings.  ;-D  Sławomir Biały  (talk) 19:48, 11 July 2016 (UTC)
 * It was a little piece of ironic hyperbole. I often sense this sort of moralism coming in the opposite direction from advocates of restricted ontologies and methodologies, who somehow can't see the beauty of the true thoughts of God.  Turning the tables a little. --Trovatore (talk) 21:20, 11 July 2016 (UTC)
 * Yes, what I've seen so far from reading about such attempts is that it gets quite complicated. My perspective is that when doing math, say you calculate an integral or you prove some theorem, you are only ever going to manipulate a finite number of symbols using a finite number of rules. So, in that sense all of math can be re-interpreted as discrete math. This means that it should be possible to eliminate all concepts referring to uncountable quantities. Count Iblis (talk) 05:58, 11 July 2016 (UTC)


 * The number of concepts that we will ever enumerate is finite, but the number of potential concepts is infinite (e.g. the natural numbers). Similarly, the set of possible limiting processes in your description is necessarily uncountable, even if you choose to avoid referring to it as a set and only ever enumerate a finite number of them.  —Quondum 06:20, 11 July 2016 (UTC)


 * Unless one wants to go down the route of rejecting the law of the excluded middle as in Constructivism (mathematics). Dmcq (talk) 08:32, 11 July 2016 (UTC)
 * Well, that was my first response; Bishop is the earliest constructivist to have come up with a well-developed approach to analysis. Of course it's philosophically misguided, but an interesting system all the same, so I'm told; I've never really gotten around to diving into it.  --Trovatore (talk) 08:59, 11 July 2016 (UTC)
 * Perhaps Skolem's paradox will make them feel happier? Basically just consider uncountable as a word used in the theory rather than anything about real life. Dmcq (talk) 10:49, 11 July 2016 (UTC)
 * I'm not exactly sure what you're suggesting here. The content of Skolem's paradox is really only that first-order logic is not strong enough to capture the notion of countability (or more relevantly, absolute uncountability).  It doesn't imply that it doesn't really exist. --Trovatore (talk) 18:07, 11 July 2016 (UTC)


 * I do see that you get uncountable quantities as soon as you allow countable infinities. So, I guess one needs to resort to the same tricks as we use in theoretical physics where we impose a cut-off, and only at the end take the limit to infinite cut-off. So, this is like ultrafinitism, but then with the twist that we have a one parameter family of ultrafinitistic models with larger and larger cut-offs. Count Iblis (talk) 19:44, 11 July 2016 (UTC)
 * Having a limit to the counting numbers strikes me like the xkcd comic Counting sheep. ;-) Dmcq (talk) 23:43, 11 July 2016 (UTC)
 * See also SMBC. --Trovatore (talk) 23:51, 11 July 2016 (UTC)

Related properties of complex and split-complex numbers
Am I correct in saying that there an equivalent of the Cauchy–Riemann equations that applies to split-complex numbers? My guess is that for a function $$u+jv=f(x+jy)$$ expressible in terms of the algebraic operations, these would be:
 * $$\tfrac{\partial u}{\partial x} = \tfrac{\partial v}{\partial y}$$ and $$\tfrac{\partial u}{\partial y} = \tfrac{\partial v}{\partial x}.$$

Through a change of basis
 * $$u'=\tfrac{1}{2}(u+v), v'=\tfrac{1}{2}(u-v)$$ and $$x'=\tfrac{1}{2}(x+y), y'=\tfrac{1}{2}(x-y),$$

I expect that these equations translate to
 * $$\tfrac{\partial u'}{\partial y'} = 0$$ and $$\tfrac{\partial v'}{\partial x'} = 0 .$$

This reflects the separation of split-complex numbers into the direct sum R ⊕ R.

The split-complex numbers and complex numbers are very closely related; e.g. they are each isomorphic to the Clifford algebra over a one-dimensional vector space, the only difference being the sign of the quadratic form, and (considered as real algebras) complexify to the same algebra. My question is (assuming that I have not confused the above): can we use this insight to reflect this back to somehow split the complex numbers into two noninteracting algebras (as with a direct sum)? —Quondum 04:25, 11 July 2016 (UTC)
 * The first part is correct, yes, and the second part corresponds to using the null basis of $1 + j⁄2$ and $1 − j⁄2$. And this is where I have to stop due to limited knowledge at present, but the conclusion would be surprising since AFAIK there is no way to split C like you can do for 2R, so something must be subtly different for it to work in one case and not in the other. Double sharp (talk) 12:33, 11 July 2016 (UTC)
 * It does seem to be subtle, and I'm trying to see how far one can take the correspondence. C and 2R each embed in 2C, and retain the property that $$\tfrac{\partial f}{\partial\overline{z}} = 0$$ for what I'll call an "expressible function".  In the case of 2R, this strengthens to an algebraic statement, true without the need to appeal to continuity: we can split an expressible function's input and the function itself as follows –
 * f(&alpha;z) = &alpha;f(z), f($\overline{&alpha;}$z) = $\overline{&alpha;}$f(z), or
 * f(z) = &alpha;f(&alpha;z) + $\overline{&alpha;}$f($\overline{&alpha;}$z)
 * where &alpha; is the idempotent element $1 + j⁄2$. The partial differential equations follow directly from these, hence these are stronger statements.
 * The idempotents algebraically decompose any element in 2R over a real basis, but arbitrary R-linear combinations of a basis are not expressible. Conversely, C does not permit algebraic decomposition over a real basis, but arbitrary elements are expressible as an R-linear combination of a basis.  These two observations seem to be flip sides of the same coin: they each prevent an arbitrary R-linear function on the algebra, and must in some sense be related .  This somehow seems to suggests to me that there must be an algebraic statement valid for C that is the equivalent of my equalities using idempotents above, that somehow strengthens the Cauchy–Riemmann equations to a purely algebraic statement.  —Quondum 14:45, 11 July 2016 (UTC)