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June 29

Number of ways to split N people into k groups of at least m

(Inspired by the random number game above.)

The number of ways to partition N elements among exactly k sets is the Stirling number of the second kind. What if I impose that each set must be at least of size m? (For bonus points: what if it must be at least of size m, and at most of size M?)

I thought of the following recursion: denoting that number by $P_{m}(N,k)$ , look over all possible sizes j of the group in which the first element is, then we must group j-1 other elements among the rest with it, and we have $P_{m}(N-j,k-1)$ ways to arrange the rest. You need to fix the first element in the group you split off, otherwise different sequences of splits might lead to the same final partition and you would have multiple counting. Hence $P(N,k,m)=\sum _{j=m}^{N-m}C(j-1,N-1)P_{m}(N-j,k-1)$ . The initialisation is fairly trivial: $P_{m}(N,1)$ is 0 if N<m, 1 otherwise. Assuming the reasoning is correct, that formula looks like the kind of stuff where one could get a non-recursive formulation... Tigraan^{Click here for my talk page ("private" contact)} 12:48, 29 June 2021 (UTC)[reply]

See associated Stirling numbers for your first question.--2406:E003:855:9A01:B15B:27D4:E79:599E (talk) 14:54, 29 June 2021 (UTC)[reply]

July 2

E. T. Jaynes and probability interpretation

I keep hearing about E. T. Jaynes having proposed some novel interpretation of probability and written wondrous books and other publications about it. His biography doesn't say much, and the linked pages mention a radical form of Bayesianism without saying what is radical. Is there a TLDR about this? Is it something I should study, if I'm interested in probability and statistics as topics in math? At the moment I have a little bit of understanding of classical probability (I mean à la Kolmogorov) and almost none about statistics. To the extent that I comprehend what Bayesianism is, it doesn't seem like a mathematical topic, but maybe I'm wrong. Thanks. 2601:648:8200:970:0:0:0:23D0 (talk) 08:35, 2 July 2021 (UTC)[reply]

Jaynes is mentioned near the end of the article on Bayesian probability. That article summarizes of Bayesianism if you want details. I think it is a mathematical topic, though perhaps the philosophy of mathematics would be a better classification. I'd say you should definitely know what Bayesianism is, as well as other Probability interpretations, if you're going to study probability and statistics. You can do computations without knowing the underlying philosophy, but without the philosophy the results are just numbers with no interpretation. --RDBury (talk) 09:25, 3 July 2021 (UTC)[reply]

This review of Jaynes's book Probability Theory: The Logic of Science gives a capsule description of his "Principle of Maximum Entropy". In a nutshell, where Baysesianists of the subjective persuasion freely allow the practitioner to select a prior depending on one's whims or superstitions, the maximum entropy principle assigns a prior in an objective way. I don't know how this compares to other objective approaches to Bayesianism (such as those based on Cox's theorem). Ming Li and Vitányi have established a link between the maximum entropy principle and Kolmogorov complexity.^[1] --Lambiam 09:50, 3 July 2021 (UTC)[reply]

Cox's theorem is the foundation of Jayne's book - as is noted in Cox's theorem#Interpretation and further discussion. It's not clear the Jayne is claiming originality for this novel interpretation of probability, but he does offer a lucid exposition of it. catslash (talk) 15:04, 4 July 2021 (UTC)[reply]

However, while Cox's postulates provide no guidance regarding the selection of the prior, this is the essence of Jaynes's contribution. --Lambiam 17:27, 4 July 2021 (UTC)[reply]

July 3

Semi-cyclic numbers

Look at the first few pieces of information in the Details section of Cyclic number. It reveals that 076923 is not a cyclic number despite having half of its first few multiples meeting the criterion rather than all that a cyclic number would require. In this case, the answer is yes, no, yes, yes, no, no, no, no, yes, yes, no, yes. (Moreover, all the no's are permutations of the digits 153846.) Are there any other semi-cyclic numbers in base 10?? Georgia guy (talk) 00:44, 3 July 2021 (UTC)[reply]

Probably an infinite number. The next is I think 32258064516129. Why? It's the next repeating decimal in this table where the period (n - 1) / 2, where n is the prime numerator of the fraction generating the decimal. n is 31, so the fraction is 1/31. And it looks like it works similarly. The first few products are 32258064516129, 64516129032258, 96774193548387, 129032258064516 – yes, yes, no, yes. After that the next is 1/43, and I imagine they continue after that.2A00:23C8:4588:B01:E12C:4F21:2832:E66C (talk) 05:16, 3 July 2021 (UTC)[reply]

See OEIS sequence A097443; for a prime p on this list, the number is n=(10^(p-1)/2-1)/p. Note that p=3 (n=33), is a special case because of repeated digits. The multiples you get correspond to the quadratic residues mod p. OEIS also has similar sequences for a third, a fourth, ... a thirteenth; see the Crossrefs section. All these sequences seem infinite, but I have no idea what the status is on proving it. --RDBury (talk) 08:46, 3 July 2021 (UTC)[reply]

July 5

In search of a finit Calculus?

I have a question: how do followers of finitism actually define calculus?
There should be a problem with the fact that there is such a small symbol there, right?--82.82.76.143 (talk) 17:02, 5 July 2021 (UTC)[reply]

There is no precise definition of finitism in the form of a set of rules of reasoning that are accepted by all mathematicians who consider themselves finitists. The essential step in setting up calculus is the notion of limit. But you also need a notion of a real number. One possible approach is that of constructivism. We can define a real number as a machine that, when presented with a positive rational number

ε

, will produce two rational numbers

a ε

and

b ε

such that

0 \leq b ε - a ε \leq 2ε

. Moreover, the machine is such that its guaranteed that, if

ε < δ

, then

a δ \leq a ε \leq b ε \leq b δ

. Equivalently, using interval notation,

[a ε, b ε] \subseteq [a δ, b δ]

. The intuition is that the real number is "caught" in intervals that can be shrunk as much as you desire. This is not the whole story; we also need to define a notion of equivalence between such interval machines. For example, the constant machine for which

[a ε, b ε] = [0, 0]

for all

ε > 0

is equivalent to one with

[a ε, b ε] = [-ε, ε]

. Anyway, limits can now be defined as (classes of equivalent) interval-producing machines. This is not essentially different from the concept of a Cauchy sequence, but (IMO) more intuitive. Obviously, no machine can be constructed for Chaitin's constant, but a constructivist will probably not agree that such a number exists. --Lambiam 22:17, 5 July 2021 (UTC)[reply]

(edit conflict) N J Wildberger offers a YouTube course on what he calls "algebraic calculus". He's an avid finitist, so I assume this version of calculus follows finitist principles. But keep in mind that a lot of effort went into defining limits, continuity and derivatives without resorting to infinitesimals or "actual" infinities, a project that was completed in the late 19th century. This is turn was driven by criticisms of many people, notably George Berkeley in The Analyst, who found the notion of infinitesimals rather far fetched. Berkely's work, in turn, was a tu quoque attack on criticisms of his earlier work as a Christian apologist. --RDBury (talk) 22:24, 5 July 2021 (UTC)[reply]

July 6

How do I say the Nth root of a number?

I asked this question in the linguistic desk. But the math desk is probably a much better place for this question. How do I say:

${\sqrt[{n}]{a}}$

when n equals to -3, -2, -1, 1.5, or 2.5? - Toytoy (talk) 05:19, 6 July 2021 (UTC)[reply]