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August 13

Another probability question

Hello, this is not homework. It is a conundrum that arose out of something I was trying to work out for myself. If we have only one sample of a random variable, and no other information at all, then the best estimate of the mean of the distribution is the sample itself. It may be a rubbish estimate, but nothing more can be said. However, suppose we also know that the mean is >= 0. With this extra information, can any better estimate of the mean be achieved from a single data point? (If necessary, please define "better" in any sensible way that aids the question.) It seems to me that, in the event that the sample is negative, replacing it with zero would always be a better estimate. However, averaged multiple adjusted samples will no longer converge to the true mean, assuming that positive samples are left alone, which seems undesirable. It's not obvious to me anyway that positive samples have to be left alone. Can anything better be achieved? How about if we also know anything else necessary about the distribution, even down to its exact "shape", except that we do not know the mean, apart from that it is >= 0. What then? Does that extra information help at all? 2A00:23C8:7B0C:9A01:8D5A:A879:6AFC:AB6 (talk) 20:56, 13 August 2024 (UTC)

Adjusting positive samples will do you little good if you don't know how to adjust them – but if nothing else is known about the distribution than μ ≥ 0, there is no information to base an adjustment on. For an extreme case, assume all samples have sample size 1. Unknown to the sampler, the distribution is discrete with possible outcomes {−9, 1}. The sample means are them equal to the single element in each sample. Adjusting them to be nonnegative amounts to left-censoring. If μ is the true mean of the distribution, the average of the adjusted (left-censored) sample means tends to (μ + 9) / 10. For the average of the adjusted sample means to tend to the true mean, the positive sample means should be replaced by 10μ / (μ + 9). But this replacement requires already knowing both the value of μ and the outcome space – in fact, all there is to know about the distribution.  --Lambiam 07:33, 14 August 2024 (UTC)

August 14

Hypothetical US Senate reform

Suppose that the US Senate is reformed to be lightly weighted by population, with each state having either 1, 2, or 3 senators (and keeping the total number of 100 senators constant). What would be a plausible method for determining which states get 1, 2, or 3? 71.126.57.129 (talk) 03:33, 14 August 2024 (UTC)

Representatives are assigned using a greedy algorithm: First, every state is allocated one representative; then whichever state currently has the highest ratio of population to representatives is given another representative; repeat until out of representatives. One could reasonably use the same algorithm for senators, with a maximum number of senators per state allowed.--Antendren (talk) 06:31, 14 August 2024 (UTC)

(ec) You can think of the states as each being one party in a multi-party system, and of the population of each state as being voters for their own party (voting for the party, not for a specific candidate). Several multi-winner voting systems can be used, specifically party-list proportional representation, tweaked to keep the number of seats for each party in a given 1 to n range – in the question as posed n = 3. One possible way is the D'Hondt method, which will assign the seats one by one to the state that is proportionally the least represented and still has fewer than n seats. As you can see at Party-list proportional representation § Apportionment of party seats, there are many other methods.  --Lambiam 06:36, 14 August 2024 (UTC)

One plausible way is to start moving senators from the smallest states to the largest. So pick the smallest state first, take one of their senators and assign it to the largest. Then repeat with the next smallest and largest. Keep going until you run out of states that are small or large enough for it to make sense. This works as if you require the total to be unchanged, with 50 states and 100 senators, the number of states with 1 and 3 must be equal. --2A04:4A43:90FF:FB2D:F067:D97A:B158:E36B (talk) 12:44, 14 August 2024 (UTC)

One approach is to define what is meant by a fair distribution of Senate seats. A reasonable ansatz is to minimize the sum of seats/population subject to the constraints that there are 100 seats total and each state gets between 1 and 3 seats. This is a variant of the knapsack problem. Tito Omburo (talk) 13:15, 14 August 2024 (UTC)

Another theory-based approach is to minimize total dissatisfaction, assuming that the members of a group will feel less happy if the group is underrepresented. Let, for group $i,$ the quantity $p_{i}$ stand for the fraction the group takes up in the total population, and $s_{i}$ for the fraction of the seats allocated to the group (so $\textstyle {\sum _{i}p_{i}=\sum _{i}s_{i}=1}.$) Ideally, $p_{i}=s_{i}=1$ for all groups, but this is rarely posible, given the constraints. For a model of total dissatisfaction, choose some function $f(p,s)$ that represents the dissatisfaction of a group with population weight $p$ and representation weight $s;$ then $\textstyle {\sum _{i}p_{i}f(p_{i},s_{i})}$ stands for the proportionally weighted combined dissatisfaction. Plausibly, $s\geq p,$ meaning that a group is not underrepresented, should imply that $f(p,s)=0;$ furthermore, for fixed $p,$ the dissatisfaction should weakly monotonically decrease with increasing $s;$ conversely, for fixed $s,$ the dissatisfaction should weakly monotonically increase with increasing $p.$ A possible formula is given by $f(p,s)=(p{\dot {-}}s)^{2},$ using truncated subtraction.  --Lambiam 16:55, 14 August 2024 (UTC)

August 15

Can two polygons have the same interior angles (in order) if they are noncongruent and have no parallel sides?

Was thinking about the question of whether the interior angles, in order, determine a polygon up to congruence, when I realized that the answer is obviously no if there are parallel sides which you can contract and lengthen at will without changing any of the angles. Barring these polygons with parallel sides and their lengthenings/contractions, then, are there any polygons that share the same interior angles in order without being congruent? GalacticShoe (talk) 17:30, 15 August 2024 (UTC)

I'm not exactly sure what you're asking, but it seems to me that an example can be obtained by starting with a regular pentagon, and lengthening two adjacent sides, and the third opposite side, and shrinking the remaining two sides, while maintaining all angles. Tito Omburo (talk) 17:37, 15 August 2024 (UTC)

I'm realizing now that this question is embarrassingly simple; if we consider extending polygon sides to lines, with the sides then being line segments between the intersections of adjacent sides, then moving lines always preserves angles. For example, in the pentagon example, the lengthening/shrinking you mentioned is akin to moving two lines further away from the center of the pentagon. GalacticShoe (talk) 17:48, 15 August 2024 (UTC)

Upon further searching online, the answer is very much no yes; see this Math StackExchange post. In particular, there are many polygons where you can simply intersect the polygon with a planar half-space such that 1. the line L demarcating the planar half-space is parallel to one of the sides S, and 2. all sides of the polygon either fall on the interior of the half-space, share a vertex with S and intersect L, or are S. GalacticShoe (talk) 17:39, 15 August 2024 (UTC)

I think you meant to say that the answer is, Yes – they can have the same interior angles and yet be noncongruent (from tetragons on).  --Lambiam 22:25, 15 August 2024 (UTC)

Yup, silly mistake on my part; amended GalacticShoe (talk) 03:51, 16 August 2024 (UTC)

August 17

"Suppose given" mathematically idiomatic English?

"Suppose given a positive $\epsilon$." A little thought shows that this is a grammatical English sentence, but would it be idiomatic in mathematics writing? Also, what is the parallel construction in French? I thought it was "Étant donné", but see this is not so. Tito Omburo (talk) 12:56, 17 August 2024 (UTC)

It sounds Victorian to me. The more usual phrasing would just be given without the suppose. --Trovatore (talk) 18:50, 17 August 2024 (UTC)

Without the imperative verb "suppose", it is no longer a complete sentence. Tito Omburo (talk) 00:19, 18 August 2024 (UTC)

Just "Given a positive $\epsilon$" should be followed by a main clause, as in "Given a positive $\epsilon$, find $M$ such that $x>M$ implies $f(x)<\epsilon .$" In that case, "Étant donné" is just fine, like here: "Étant donné un nombre positif rationnel $\varepsilon ,$ ..."

To me, "Suppose given a positive $\epsilon .$" sounds more ungrammatical than Victorian. Still old-school but sounding more grammatical (to me): "Let a positive $\epsilon$ be given." In French, "Soit donné ..."; see here: "Soit donné un nombre positif (arbitrairement grand) h." Better still in English: "Let $\epsilon$ be a positive number."  --Lambiam 20:25, 17 August 2024 (UTC)

Yes, soit donné is what I was thinking of. Here "suppose" is a verb in imperative mood, and the condition we are supposing is that a positive epsilon [be given]. It's a construction that arises sometimes in mathematics writing, and I think it's grammatical. Tito Omburo (talk) 00:19, 18 August 2024 (UTC)

Do you think the following is grammatical:

"Suppose present a parent."

?  --Lambiam 01:17, 18 August 2024 (UTC)

Why not? Tito Omburo (talk) 01:20, 18 August 2024 (UTC)

Why language is as it is is generally an unanswerable question. (Why is morous not an English word? porosity → porous, so why not morosity → morous?) The object of verbs like imagine and suppose is normally a noun phrase (Imagine a spherical cow.) or a clause (Suppose [that] $a$ is prime.). It can also be a subject + an infinitive (Suppose $a$ to be prime.). CGEL Chapter 4 §5.4 writes that to be can be omitted with a reflexive object (as in Suppose yourself away from home.^[1]), So then we have Suppose + object + predicate. In mathematical writing we find this also with other objects (as in Suppose $a$ prime to $b$, and $a-b$ an odd number.^[2]). Constructions of the form Suppose + predicate + object, with object and predicate swapped, do not fit any of these forms.  --Lambiam 07:52, 18 August 2024 (UTC)

My linguistic intuition is that both these constructions (suppose given ε and suppose present a parent) are — grammatical-ish, in the sense that in some contexts I would accept them (though find them a bit strained and unnatural) and in others I would reject them. --Trovatore (talk) 22:34, 18 August 2024 (UTC)

I suppose it could be OK in a construction like: "Suppose, given a positive $\epsilon$, that all negative $\epsilon$s of lower absolute value divide the original value with a remainder no greater than the square root of the log of $\epsilon$ to base 6". But a grammarian would require the commas around the parenthetic phrase. -- Jack of Oz ^{[pleasantries]} 22:40, 17 August 2024 (UTC)

It doesn't read like a complete sentence to me. It's missing something. For example, "Suppose one is given a positive $\epsilon$." would be a complete sentence, as would "Suppose a positive $\epsilon$ is given." and also "Suppose $\epsilon$ is positive." XOR'easter (talk) 22:29, 21 August 2024 (UTC)

I would go for one of "Given a positive number ⁠$\epsilon$⁠, ..." (part of a sentence); "Let ⁠$\epsilon$⁠ be a positive number...." (complete sentence); or "Suppose ⁠$\epsilon$⁠ is a positive number...." (complete sentence). The original proposal "suppose given" is ungrammatical. –jacobolus (t) 00:56, 22 August 2024 (UTC)

Whether or not it is grammatical, it is certainly not idiomatic 21st century mathematical English. 100.36.106.199 (talk) 01:59, 22 August 2024 (UTC)

Really? Is Robin Hartshorne not a 21st century mathematician? Tito Omburo (talk) 02:02, 22 August 2024 (UTC)

I know you are but what am I? 100.36.106.199 (talk) 02:07, 22 August 2024 (UTC)

Also, Jacob Lurie. Griffiths and Harris. Saunders Mac Lane. Vladimir Arnold. J-P Serre. Nick Katz. All seem to use this construction. Also one can find it in the English translations of Bourbaki. Tito Omburo (talk) 02:14, 22 August 2024 (UTC)

August 18

Recursive matrix

Hi. Has anyone ever considered a matrix of matrices which contains itself as one of the entries? Is there a way to make such a thing make sense?

Duomillia (talk) 00:06, 18 August 2024 (UTC)

Sylvester's construction of Hadamard matrices is recursive. But a matrix which contains itself would normally be called self-referential rather than recursive. Self-referential objects, for example the set A = {A}, are generally considered undefined. In fact there are axioms in place, such as the Axiom of regularity, to rule out such objects, --RDBury (talk) 00:36, 18 August 2024 (UTC)

If the matrix is not homogeneous, it could be some kind of Droste matrix. There are certainly nontrivial rooted trees that have themselves as a subtree – think of the indefinite unfolding of a recursive datatype. It is not clear, though, how such loopy matrices might be useful.  --Lambiam 01:02, 18 August 2024 (UTC)

Is there always a prime obtained as the concatenation of a power of 3*n followed by a 1, if n is a positive integer not == 4 mod 11?

Is there always a prime obtained as the concatenation of a power of 3*n followed by a 1, if n is a positive integer not == 4 mod 11? (If n == 4 mod 11, then all numbers obtained as the concatenation of a power of 3*n followed by a 1 are divisible by 11, thus cannot be prime)

This is the smallest prime obtained as the concatenation of a power of 3*n followed by a 1, for positive integers n<=50, not == 4 mod 11 (could you extend this list to n=100 or n=200?)

1,31
2,61
3,811
5,151
6,181
7,211
8,241
9,271
10,9001
11,331
12,466561
13,23134411
14,421
16,23041
17,1326511
18,541
19,571
20,601
21,631
22,661
23,691
24,194084099617653428060161
25,751
27,811
28,41821194241
29,6585031
30,81001
31,86491
32,751447478108161
33,991
34,1021
35,1051
36,12597121
38,14815441
39,1171
40,1201
41,1231
42,158761
43,1291
44,1321
45,109149939351045840229035237837713687871268015108347375034700496875243120429748722166607421968365088105201721191406251
46,1381
47,198811
49,1471
50,384433593750000000001 220.132.216.52 (talk) 10:24, 18 August 2024 (UTC)

I am not aware of any provable answer to this question. But there's a line of heuristic reasoning that seems to work pretty well for these sorts of questions.

A slightly incorrect, but nevertheless useful, way of thinking about the prime number theorem is that, if you don't know anything in particular about a natural number n, then the "probability" that n is prime is about ${\frac {1}{\log n}}$. I put "probability" in scare quotes because there's no randomness involved; n is either prime or it isn't, and you can figure out which. Still this way of thinking about it seems to have considerable power.

So your question is, for each n, is there a natural number $k>0$ such that $10\cdot 3^{nk}+1$ is prime.

For each such k, the "probability" that this value is prime is about ${\frac {1}{(\log 10+nk\log 3)}}$.

If these outcomes were mutually exclusive for different values of k, we could get the probability that there is some such k by adding them up. Of course they're not mutually exclusive (or I don't see why they should be) but what we're mainly interested in is whether the sum is finite or infinite.

And as it turns out, $\sum _{k>0}{\frac {1}{(\log 10+nk\log 3)}}=\infty$.

So basically we can conclude that, using this slightly unjustified notion of "probability", there is such a k almost surely.

Given that there are only countably many n, and probability is countably additive, and given that you haven't found any counterexample, it seems reasonable to conjecture that for each n there is very likely such a k. --Trovatore (talk) 20:57, 18 August 2024 (UTC)

One problem with this argument is that you're not using the assumption that n is not congruent to 4 mod 11, and in fact there are no primes for such n. Given this, I think it would be prudent to add some additional qualifiers like "barring additional number theoretical coincidences". RDBury (talk) 04:03, 19 August 2024 (UTC)

PS. I'm pretty sure the formula should be $10\cdot (3n)^{k}+1$. I haven't checked but I think your argument still works. --RDBury (talk) 04:14, 19 August 2024 (UTC)

And thus $nk\log 3$ should also be $k\log 3n$. 1.168.139.55 (talk) 05:28, 19 August 2024 (UTC)

Ah, my bad; I misread the question. I thought it was a power of 3ⁿ, then adjoin a final 1 in the decimal representation. --Trovatore (talk) 16:33, 19 August 2024 (UTC)

It occurs to me that I can better motivate the technique of investigating whether the sum is finite or infinite, in the following manner: Let's think of the occurrences for different k not as mutually exclusive, but rather as independent, which makes more sense anyway. Then the probability that there is no such k would be the infinite product $\Pi _{k>0}\left(1-p_{k}\right)$, where we write $p_{k}$ for the "probability" for a particular k, namely approximately ${\frac {1}{\log 10+nk\log 3}}$. This infinite product is zero just in case the sum of the $p_{k}$ is infinite. This fact, or close enough, is presumably treated at our infinite product article. --Trovatore (talk) 22:44, 18 August 2024 (UTC)

See OEIS:A088783. In addition to Trovatore's aforementioned heuristics, primes of the form of a power of $n\geq 2$ followed by a concatenated $1$ (i.e. $10n^{k}+1$) have been found for most small values of $n$, with the exceptions of $n$ congruent to $1\!\!\!{\pmod {11}}$ (which you have already disallowed) or $-1\!\!\!{\pmod {33}}$, both of which have covering congruences, as well as a few sporadic unconfirmed values. The sequence of sporadic values starts $185,338,417,432,537,614,668,743,744,773,\ldots$, with the values divisible by $3$ being $417,432,744,786,\ldots$; in other words, we could almost extend the sequence up to $300$ with the exceptions of $139,144,248,262$, where we don't know if there is such a prime. GalacticShoe (talk) 04:25, 19 August 2024 (UTC)

Well, 537 is also divisible by 3, besides, could you give the list up to n=100? 1.168.139.55 (talk) 05:34, 19 August 2024 (UTC)

My mistake, I was finding divisibility by hand and must've miscounted. In any case, I unfortunately don't have the computing power for it. Based on your list, it would seem you would have a better go of it. GalacticShoe (talk) 05:37, 19 August 2024 (UTC)