Wikipedia:Reference desk/Archives/Mathematics/2017 December 17

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December 17

What is the expected ratio of long "stick" to short "stick"?

An unit rod is broken into two at a uniformly random position. What is the expected ratio of the long length to the short length?

My calculations is to imagine a random variable x such that x is from 0.5 to 1

Calculate Integral[x/(1-x),0.5,1] but my calculator kept on giving me "MATH ERROR".

Please help. 110.22.20.252 (talk) 02:21, 17 December 2017 (UTC)

You need to use an open interval, else you have a problem at x=1. Bubba73 ^{You talkin' to me?} 02:43, 17 December 2017 (UTC)

Your integral is basically correct, but you cannot simply use the 1 as upper limit, instead you need a limit to 1. Then, the result will approach infinity as the upper boundary of your interval approaches 1 (from below). Gap9551 (talk) 02:56, 17 December 2017 (UTC)

Which would mean that the expected value is infinite. It would still be infinite for the expected value of ratio of the left and right ends. On the other hand if you wanted the expected ratio of the short to long lengths then take the reciprocal of the integrand and you'd get an answer. --RDBury (talk) 03:01, 17 December 2017 (UTC)

Holy molly, you are right. The expected ratio of the short length to the long end is

Integral[(1-x)/x,0.5,1] = 0.193147

So the expected ratio of the long length to the short length is 1/0.193147 = 5.1774 110.22.20.252 (talk) 03:23, 17 December 2017 (UTC)

No, the expected ratio of long to short is infinity, as shown by your earlier integral. The expected value of the reciprocal is not the reciprocal of the expected value.

Also, in your calculation of expected value of short to long, you should double it or add \int_0^{.5} x/(1-x) dx. --108.52.27.203 (talk) 03:29, 17 December 2017 (UTC)

(edit conflict)Use an analytical solution whenever possible: $\int _{1/2}^{1}{\frac {1-x}{x}}dx=\int _{1/2}^{1}{\frac {1}{x}}-1dx=\ln(2)-1/2$.--Jasper Deng (talk) 03:35, 17 December 2017 (UTC)

The average is $2\ln(2)-1$ as was noted below. Ruslik_Zero 14:27, 17 December 2017 (UTC)

The point was to get a closed solution to the integral itself, and leave it to the OP to do the correction of doubling it.--Jasper Deng (talk) 20:22, 17 December 2017 (UTC)

(edit conflict) Erm, that's not how it works; $\operatorname {E} [1/X]\neq 1/\operatorname {E} [X]$ in general. You're also missing a factor of 2 in front of the integral; you've got a uniform RV on $[1/2,1]$ and the PDF of such a thing is $f(t)=2$ over that interval. It would be better to work the integral out on your own rather than relying on the calculator. In this case, it diverges (to infinity), and so your expected value is infinite (undefined). That's a perfectly valid answer. –Deacon Vorbis (carbon • videos) 03:34, 17 December 2017 (UTC)

So the expected ratio of the long length to the short length is Integral[(1-x)/x,0,0.5] + Integral[x/(1-x),0.5,1] which is infinity + infinity.

but the expected ratio of the short length to the long length is Integral[x/(1-x),0,0.5] + Integral[(1-x)/x,0.5,1] which is 0.193147 + 0.193147 or ln(2) - 1/2 + ln(2) - 1/2

I find it hard to wrap my head around the fact that one ratio is finite while the other ratio is infinite. I would have expected that both ratios to be a finite real number. 110.22.20.252 (talk) 11:59, 17 December 2017 (UTC)

The simple explanation is that longer-to-shorter is unbounded, whereas shorter-to-longer is bounded above by 1. The latter "supremum bound" is responsible for the well-behavedness of integration as opposed to differentiation, and when it's not present, things can go awry.--Jasper Deng (talk) 12:13, 17 December 2017 (UTC)

Three questions

Has any non Abelian group structure been defined on an infinite set with cardinal $\aleph _{0}$ or $\aleph _{1}$ (or even more)? Alireza Badali (talk) 15:48, 17 December 2017 (UTC)

Sure. Matrices provide plenty of examples, e.g. 2 by 2 matrices with real entries and non-zero determinant. Or you could fix a set X of arbitrarily large cardinality and take the permutations of X, or the finite support permutations of X.

It's a consequence of the Lowenheim-Skolem theorem of logic that there's a non-Abelian group of every infinite cardinality.--108.52.27.203 (talk) 16:16, 17 December 2017 (UTC)

Thank you so much but what relation does there exist between logic and group theory? Alireza Badali (talk) 18:39, 18 December 2017 (UTC)

Adding the nonabelian requirement doesn't really make things any more difficult: if you have an infinite abelian group $G$, then you can get a nonabelian group of the same cardinality by taking its product with any finite, nonabelian group – say $G\times S_{3}.$ –Deacon Vorbis (carbon • videos) 16:50, 17 December 2017 (UTC)

Clearly, thank you so much. Alireza Badali (talk) 18:39, 18 December 2017 (UTC)

A completely explicit (if not incredibly exciting) example of a group of order ℵ₁ is the free group on ℵ₁ generators. You could replace ℵ₁ by any infinite κ and get the same result. --Trovatore (talk) 03:59, 19 December 2017 (UTC)

Thank you so much, and we can say it is because of group define isn't a set theory concept only or it has some conditions more than set theory concepts, but I mean was in this question could a person make a non Abelian infinitely many group that I took the answer. Alireza Badali (talk) 08:28, 19 December 2017 (UTC)

I'm sorry, I don't really follow what you're saying here. --Trovatore (talk) 08:33, 19 December 2017 (UTC)

Question: Up to isomorphism is $(\mathbb {R} ,+,\cdot )$ the largest field? Alireza Badali (talk) 18:39, 18 December 2017 (UTC)

No. Lowenheim-Skolem implies there's a field of every infinite cardinality. Also, depending on what you mean by largest, the fact that $\mathbb {C}$ is not isomorphic to any subfield of $\mathbb {R}$ might also answer your question.--108.52.27.203 (talk) 21:11, 18 December 2017 (UTC)

Thank you so much, sorry, I'd forgotten $\mathbb {C}$ yes that's right but can we say it is because of Galois theory with Lowenheim-Skolem theorem? Alireza Badali (talk) 22:50, 18 December 2017 (UTC)

Are large cardinals mental abstract imaginary issues and do their existence depend to other things and can not they define independently like numbers? I think large cardinals like $\aleph _{4}$ only can be made by low cardinals step by step with some actions like all functions from $\mathbb {R} \to \mathbb {R}$ or power set of $\mathbb {R}$ or somethings, don't? Alireza Badali (talk) 23:43, 18 December 2017 (UTC)

Alireza, I'm afraid you have a misunderstanding of the meaning of the aleph notation. Unfortunately this is a common error; many of the popularizers make it. It is not known where (say) $\aleph _{4}$ stands in relation to the cardinality of the continuum. So there might be exactly $\aleph _{4}$ real numbers, or more, or fewer. --Trovatore (talk) 00:50, 19 December 2017 (UTC)

Thank you so much but what do you mean by $\aleph _{4}$ real numbers? doesn't #$(\mathbb {R} )=\aleph _{1}$?!! Alireza Badali (talk) 08:43, 19 December 2017 (UTC)

The statement that there are exactly ℵ₁ real numbers is called the continuum hypothesis, or CH. It is not known whether CH is true or false. It is known that it can neither be proved nor disproved from the standard axiomatization of set theory, ZFC. --Trovatore (talk) 08:46, 19 December 2017 (UTC)

Okay everybody might make some sets with some cardinals but my question is only does #$(\mathbb {R} )=\aleph _{1}$? Alireza Badali (talk) 08:58, 19 December 2017 (UTC)

That's what I was trying to tell you. No one knows the answer to that question. You are precisely asking whether CH is true. --Trovatore (talk) 09:16, 19 December 2017 (UTC)

Or maybe what you're really asking is whether ℵ₁ is defined to be the cardinality of the reals? In that case the answer is much clearer. It's just "no". ℵ₁ is not defined to be the cardinality of the reals, though you can be forgiven for thinking otherwise, given the number of popularizers who have made that mistake. --Trovatore (talk) 09:19, 19 December 2017 (UTC)

I wish never asked this question and you answer, now I am exploding from turmoil, I should destroy myself though I can not believe it yet! but you answered my old philosophical question in my mind that is, is number whole of fact? I wish I didn't take your answer. Alireza Badali (talk) 09:29, 19 December 2017 (UTC)

Once again I find myself unable to work out what you mean. I don't see any reason for what I've said to cause you distress, and I do hope you feel better. --Trovatore (talk) 09:34, 19 December 2017 (UTC)

Thank you so much but now I should rest, bye. Alireza Badali (talk) 09:40, 19 December 2017 (UTC)

Anyway, I'm not interested in logic. Alireza Badali (talk) 14:04, 19 December 2017 (UTC)

Should there exits $n\in \mathbb {N}$ such that #$(\mathbb {R} )\leq \aleph _{n}$ anyway, thank you so much and bye. Alireza Badali (talk) 16:17, 19 December 2017 (UTC)

I don't know whether such an n "should" exist (that's a very open-ended discussion, possibly interesting, but hard to nail down). What I can tell you is that you can't prove from ZFC that such an n does exist (and you also can't prove that it doesn't). For example, it's consistent with ZFC that the cardinality of the reals is ℵ_ω+1, which is larger than ℵ_n for every natural number n. --Trovatore (talk) 20:50, 19 December 2017 (UTC)

Dear Trovatore, Thank you so much for your guidance and help, you are great, but logic isn't important for me and I want be far from it what I understand is #(power($\mathbb {N}$))=#($\mathbb {R}$) because of each real number is to form of $a_{1}a_{2}...a_{n}.b_{1}b_{2}...$ but only please say yes or no for this question: is cardinal of points of a line in a flat plane equal to $\aleph _{1}$? Alireza Badali (talk) 07:47, 20 December 2017 (UTC)

The answer to that question, as I've already explained, is "no one knows". --Trovatore (talk) 08:37, 20 December 2017 (UTC)

Thank you so much and bye. Alireza Badali (talk) 13:13, 20 December 2017 (UTC)

Converting sources of uniform random integers

Inspired by the question above... Let's call a "n-generator" (n-G) a source that gives me a random integer between 1 and n (uniform distribution, independent tries). If I want to emulate the behavior of a m-G from a n-G, while minimizing the number of calls to the n-G for each call to the m-G, which procedure minimizes (a) the maximum number of calls (i.e. worse-case scenario), or (b) the average number of calls? (cf. Best, worst and average case) The answer probably depends on the arithmetic properties of both numbers.

My thoughts so far: I think (a) might be infinite in the general case but I am not sure under which conditions; e.g. for n=2, m=3, I feel the probabilities of every path in the procedure will be of the form ${\frac {k}{2^{l}}}$ and no finite sum of such numbers will yield out 1/3. A brutal upper bound for (b) can be obtained if n>m by "draw one, if it is within [1,n] that is your number, else start again". If n<m, one can create a n^k-G (where k is chosen so that n^k>m, i.e. $k=\operatorname {ceil} (\log _{n}(m))$) by drawing k times and using the decomposition in base n, and then use the "draw one and try again if necessary" strategy. Tigraan^{Click here to contact me} 17:04, 17 December 2017 (UTC)

For part (a), if m has any prime factors that n doesn't have, then it there's no upper bound on the worst case, as you noted. But otherwise, you simply need to roll the n-die k times, where k is the smallest power of n such that m divides n^k. Then you can just partition the possible outcomes into m equal-sized groups, and call them 1 through m. –Deacon Vorbis (carbon • videos) 17:42, 17 December 2017 (UTC)

Or, if you like, $k=\max _{p}\left\lceil {\frac {\max\{r\geq 0:p^{r}\mid m\}}{\max\{r\geq 0:p^{r}\mid n\}}}\right\rceil ,$ where the outer max is taken over all primes p, and by convention, 0/0 = 1. –Deacon Vorbis (carbon • videos) 17:49, 17 December 2017 (UTC)

Group theory question

Hello, if $F_{2}$ is the additive group of integers modulo 2 (one and zero), I am considering $F_{2}^{n}$ as a group. What is the name of the subgroup of $F_{2}^{n}$ with an even number of ones? Does this have a name? Robinh (talk) 20:19, 17 December 2017 (UTC)

Quick note: I think $F_{2}$ might be confused with the notation for the free group on 2 generators. I'll use $C_{2}$ here to refer to the two-element group that you're talking about though (you might also see $\mathbb {Z} _{2},\,\mathbb {Z} /2,\,\mathbb {Z} /2\mathbb {Z}$ used as well).

As far as I know, this subgroup of $C_{2}^{n}$ has no special name, but it's not to hard to verify that it's just isomorphic to $C_{2}^{n-1}$ (quick sanity check is that it has to contain half the elements, so the sizes agree at least). I'll leave it to you to write down an explicit isomorphism between the two. –Deacon Vorbis (carbon • videos) 22:39, 17 December 2017 (UTC)

If you think of F_2 as a field, F_2^n is a vector space, and the group you're interested in is a subspace. In the context of coding theory or neighboring fields, this subspace might be called the subspace of even-weight vectors. I doubt there is a standard notation. --JBL (talk) 23:19, 17 December 2017 (UTC)

(OP) thanks guys, I guess the isomorphism is obvious now you mention it. It also works for $C_{r}^{n}$ where the sum of the n numbers is zero modulo r.

Resolved

Robinh (talk) 08:35, 18 December 2017 (UTC)