Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

February 14

Could eversion be a functor?

After reading Sphere eversion, which I didn't understand(not the article's fault), I wonder if there is a class of objects including the 3 particular spheres, and maybe tori that can be everted and form a category, in which eversion would be a functor from that category to itself.Rich (talk) 16:16, 14 February 2025 (UTC)

Presumably, in a categorical treatment, the objects would be self-intersection-free immersions of bounded and connected 2-dimensional smooth orientable manifolds in three-dimensional space. The morphisms would be, presumably, regular homotopies. Some of these morphisms preserve orientation; the others are "disorienting" – they flip the orientation around. Some morphisms are endomorphisms: their domain and codomain are the same object. Eversions are then the disorienting endomorphisms, a subclass of the morphisms. This subclass is not particularly interesting, categorically speaking, since the composition of two eversions is not an eversion. We can form a subcategory whose objects are the evertible objects, but we cannot restrict the morphisms to just the eversions, because the identity morphisms would thereby be excluded.

An endofunctor of this category should map morphisms to morphisms; eversions, being themselves morphisms, live at a lower categorical level. I see no obvious way of defining another notion of morphism between evertible objects in which an endofunctor might correspond to the notion of eversion.  ‑‑Lambiam 06:05, 15 February 2025 (UTC)

February 17

Infinitely many pairs of consecutive primes with given last digits

For which pairs $(d_{1},d_{2})\in \{1,3,7,9\}\times \{1,3,7,9\}$ is it known that there are infinitely many primes ending with the digit d₁ for which the next prime ends with the digit d₂? Note that there are 16 possible pairs to consider.

If one arranges the 16 sets of pairs of consecutive primes into a $4\times 4$-grid, then it is easy to see that at least one set in each row and one in each column must be infinite (by Dirichlet's theorem on arithmetic progressions). Perhaps, all 16 sets might turn out to be infinite.

More generally, the problem could be generalized to bases other than 10. In this case, for base b, there are φ(b)² possible combinations for the mod-b remainders (or equivalently, the last digits in base b) of pairs of consecutive primes greater than the largest prime dividing b, where φ(b) is Euler's totient function. Another possible generalization would be to consider triplets or n-tuples of consecutive primes. In this case, there are φ(b)ⁿ possible combinations for the mod-b remainders (or equivalently, the last digits in base b) of n-tuples of consecutive primes greater than the largest prime dividing b. The corresponding conjecture would then be that all of the resulting φ(b)ⁿ sets of n-tuples of consecutive primes are infinite. GTrang (talk) 05:27, 17 February 2025 (UTC)

For what it's worth, solving b=5 would also solve b=10 since all primes after 2 are odd. It seems reasonable to expect that 16 entries in the grid to be infinite; primes seem to behave randomly in every way they might be expected to behave randomly, but proving the corresponding conjectures usually proves elusive. Dirichlet's theorem is an example of this, since even though the behavior in question is very simple, the proof is very complex. The case b=2 is trivial, but other than that I think a more realistic question is if there has been any research or partial results on the question. --RDBury (talk) 08:44, 17 February 2025 (UTC)

PS. For b=4 it's not hard to see that Dirichlet's theorem implies the (1, 3) and (3, 1) entries must be infinite. Otherwise there would be a point after which the last digits would be all 1 or all 3. It's not hard to find data on-line, for example here for b=10 and here for b=4. I imported the b=4 data into a spreadsheet and found the entries in the grid for the first 10000 primes. The results were (1, 1):2053, (1, 3):2930, (3, 1):2931, (3, 3):2084. As you can see the (1, 3) and (3, 1) entries are significantly larger than the (1, 1) and (3, 3), and from what I can tell this doesn't seem to a transient effect; when you restrict the range to 9000 to 10000 the imbalance doesn't go away. I have no explanation for this. A similar experiment with the entries in the b=10 grid also appear to be significantly imbalanced. --RDBury (talk) 09:30, 17 February 2025 (UTC)

A (1, 3) pair can be twins (e.g. 11 – 13 or 4241 – 4243) and a (3, 1) pair can be just 8 apart (e.g. 683 – 691 or 8753 – 8581), whereas (1, 1) and (3, 3) require a prime gap that is at least 10. In the beginning of the sequence of prime gaps (OEIS A001223) larger gaps are relatively rare. Around 10000 the average gap is about log(10000) ≈ 9.2, but the imbalance may eventually disappear when the average gap is substantially larger than 10.  ‑‑Lambiam 11:02, 17 February 2025 (UTC)

Right. I was thinking b = 4, so to me a (1, 3) pair would be 17, 19. If the "probably" of a number being prime is p, then the "probability" of a (1, 3) or (3, 1) pair is p + (1-p)²p + (1-p)⁴p ... = p/(1-(1-p)²). = p/(2p-p²) = 1/(2-p). Similarly, the "probability" of a (1, 1) or (3, 3) pair is (1-p)/(2-p). These do converge to 1/2 as p→0, but seeing as p is roughly inversely proportional to the number of digits (by the prime number theorem), the convergence will be very slow and the difference will still be noticeable even going up to the 10000th prime (6 digits). It would be interesting to compare different ranges of primes to see if the amount of imbalance actually matches what the prime number theorem predicts, but I'm satisfied for now. --RDBury (talk) 15:23, 17 February 2025 (UTC)

For which quadruples $(d_{1},d_{2},d_{3},d_{4})\in \{1,2,3,4,5,6,7,8,9\}\times \{1,3,7,9\}\times \{1,2,3,4,5,6,7,8,9\}\times \{1,3,7,9\}$ there are infinitely many primes beginning with the digit d₁ and ending with the digit d₂ for which the next prime begins with the digit d₃ and ends with the digit d₄? 49.217.57.194 (talk) 18:52, 19 February 2025 (UTC)

I think we have no theory that helps to settle positive statements of this nature. I believe, nevertheless, that there are infinitely many such pairs when $d_{3}{-}d_{1}\in \{0,1,-8\}$ and otherwise probably none at all. This is based on a heuristic using the average density given by the prime number theorem, but some cases can be definitely excluded by Bertrand's postulate (which in spite of the name is a theorem). Others are excluded by Cramér's conjecture, which however has not been proved.  ‑‑Lambiam 20:40, 19 February 2025 (UTC)

Leibniz rule for antiderivatives

The general Leibniz rule is a formula for computing higher derivatives of products:

$(fg)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}g^{(k)}.$

This is analogous to the binomial theorem, and in fact, as pointed out in the article, the binomial theorem can be proven from the special case f=exp(ax), g=exp(bx). But the binomial theorem can be generalized to negative n using the binomial series. So it seems reasonable (at least to me) to generalize the Leibniz rule to find a formula for (fg)⁽ⁿ⁾ when n is negative, where f^(-k) is interpreted to mean a kth antiderivative of f. An antiderivative is only defined up to a constant, but this can resolved by stipulating, say, f^(-k)(0) = 0. For example, we have:

$(fg)^{(-1)}=f^{(-1)}g-f^{(-2)}g^{(1)}+f^{(-3)}g^{(2)}-f^{(-4)}g^{(3)}\dots$

when the series converges; this is iterated integration by parts. (Note that if g is polynomial then the sum is finite and convergence is not an issue.) Similarly:

$(fg)^{(-2)}=f^{(-2)}g-2f^{(-3)}g^{(1)}+3f^{(-4)}g^{(2)}-4f^{(-5)}g^{(3)}\dots$

and more generally:

$(fg)^{(-k)}=f^{(-k)}g-{k \choose 1}f^{(-k-1)}g^{(1)}+{k+1 \choose 2}f^{(-k-2)}g^{(2)}-{k+2 \choose 3}f^{(-k-3)}g^{(3)}\dots \ .$

This seems relatively straightforward to prove, given sufficient hand-waving on convergence and arbitrary constants, and useful since I happened to need a formula for (xⁱ exp(x))^(-k), so it seems like this is the kind of thing that would be well known and documented. But my searches have not turned up anything; I get a lot of results for the Leibniz integral rule instead, and that's very different. So do these formulas look familiar to anyone? --RDBury (talk) 10:54, 17 February 2025 (UTC)

Here integration by parts is called "the nearest possible approach to a general theorem for finding the antiderivative of the product of two functions". So your Leibniz-rule generalization would apparently be a surprise to the author.  ‑‑Lambiam 20:55, 19 February 2025 (UTC)

Thanks. The first formula is implied by the information given in Integration by parts#Repeated integration by parts. That section doesn't have citations, though that can be forgiven if the information is "common knowledge". But I'm not sure that the first formula would count as a more "general theorem". --RDBury (talk) 14:01, 20 February 2025 (UTC)

Doesnt the formula for $(fg)^{(-(n{+}1))}$ follow by applying the formula for $(fg)^{({-}1)}$ to the terms in the formula for $(fg)^{(-n)}$?  ‑‑Lambiam 19:35, 20 February 2025 (UTC)

Yes. Like I said, the formula is straightforward to prove. I was just saying that given that, and that it's a time saver in certain circumstances, it's odd that it doesn't seem to be well known. In fact I kind of expected to be in Wikipedia. RDBury (talk) 21:22, 20 February 2025 (UTC)

February 22

Field extension as quotient

Let $F$ be a field, $R=F[x_{1},\dots ,x_{n}]$ (note: finitely many variables), and $M$ be a maximal ideal of $R$. So $R/M$ is a field extension of $F$. Is it necessarily an algebraic extension? Antendren (talk) 23:23, 22 February 2025 (UTC)

Yes. Let $x$ be a generator of a transcendental extension of $F$ in $R$. Then $x$ is a polynomial in the $x_{i}$ and the ideal generated by $M$ and $x$ is proper, i.e., $M$ is not maximal, contradiction. Tito Omburo (talk) 00:20, 23 February 2025 (UTC)

Why is it proper? And how are you using that there are only finitely many variables, since it's not true otherwise?--Antendren (talk) 00:31, 23 February 2025 (UTC)

The ideal is proper because it does not contain F (x is transcendental over F). Are you certain it's not true for infinitely many variables? Tito Omburo (talk) 00:45, 23 February 2025 (UTC)

I don't see why x being transcendental makes the ideal proper. Could you give the details?

Yes, it's not true for infinitely many variables. Let $F=\mathbb {Q}$, let $K$ be an extension by a single transcendental element, and let $\{\alpha _{i}:i\in \mathbb {N} \}$ list the elements of K. Define a homomorphism from $F[x_{1},x_{2},\dots ]$ to $K$ by $x_{i}\mapsto \alpha _{i}$, and let M be the kernel.

Note that in this case, some $x_{i}$ corresponds to a transcendental element, and some $x_{j}$ corresponds to its inverse, so $M$ contains $x_{i}x_{j}-1$, meaning that $(M,x_{i})$ isn't proper.Antendren (talk) 00:59, 23 February 2025 (UTC)

To ask this question another way: Suppose $K$ is a transcendental extension of $F$. As a vector space, $K$ is not finitely generated over $F$. As a field, it might be (${\mathbb {Q} }(\pi )$ over $\mathbb {Q}$). What about as a ring?--Antendren (talk) 11:09, 24 February 2025 (UTC)

February 25

Volumes of intersecting cubes

Cube A is 1 unit on each side, with a body diagonal connecting points p & q. A cube B is then constructed with edge pq. As cube A spins along edge pq, does the volume of the intersecting cubes remain constant (at 1/4 unit cubed) or does it vary? And if it does vary, what are the maximum and minimum.Naraht (talk) 02:47, 25 February 2025 (UTC)