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

June 9

Calculating diminishing returns formula for game stat caps

lta –fifteen thousand two hundred twenty four (talk) 07:46, 9 June 2026 (UTC) The following discussion has been closed. Please do not modify it.

Hi everyone, I'm trying to design a custom RPG mod and need some help figuring out the exact math for a stat curve.

Right now, I have a "Haste" stat that increases a character's attack speed. If a player has 0 Haste, their attack cooldown is 1.0 second. I want to implement a system of diminishing returns so that no matter how much Haste a player stacks, the attack cooldown can never drop below 0.2 seconds (effectively a hard asymptote at 0.2).

I know I can use a hyperbola for this, but I'm struggling to parameterize it. Ideally, I want 100 points of Haste to bring the cooldown down to 0.6 seconds, while 500 points should bring it down to around 0.3 seconds, scaling smoothly in between.

What is the standard algebraic formula to express this kind of curve where $y$ approaches a minimum limit as $x$ approaches infinity, given these specific data points?

Thanks! ~2026-33106-46 (talk) 08:09, 4 June 2026 (UTC)

I'm betting this is for an RPG; in my experience devs use all sorts of complex expressions for curves like this, including piecewise defined functions that mere mortals will struggle to understand. So I don't think there is a "standard" formula, but if it were me then I'd use a Linear fractional transformation, in other words a function of the form (ax+b)/(cx+d). Their graphs, aside from some degenerate cases, are hyperbolas with a vertical and horizontal asymptote. It turns out these are uniquely defined by three values, and you've specified three values L(0) = 1, L(100) = .6 and L(∞) = .2. I get L(x) = (x+500)/(5x+500); this has L(500) = 1/3, which I hope is close enough to your desired value of .3.

To determine values of a, b, c, d from the three given points. First, note that by dividing the top and bottom by a constant, we can assume any value for a, so take a=1. That gives (x+b)/(cx+d). From L(0) = 1 we get b/d = 1, d=b, and we have (x+b)/(cx+b). From L(∞) = .2 we get 1/c = .2, c=5, and L(x) = (x+b)/(5x+b). From L(100) = .6 you get 100+b = .6(500+b), 100+b = 300+.6b, .4b=200, b=500. You can use a similar calculation to work out an expression with L(0), L(∞), and L(some other value) all given. It should be noted that fitting a curve to a set of values is more of an art than a science, and there are many possible tools and approaches you could use, but as I said, this is the approach I would use. --RDBury (talk) 09:24, 4 June 2026 (UTC)

June 8

Decimal Place Value

lta –fifteen thousand two hundred twenty four (talk) 07:46, 9 June 2026 (UTC) The following discussion has been closed. Please do not modify it.

What digit is in the hundredths place in the number (9.823)? ~2026-33778-68 (talk) 23:57, 8 June 2026 (UTC)

Assuming the dot in "9.823" stands for the decimal separator and the number can also be written as ⁠+9823/1000⁠, it is equal to

9 + ⁠8/10⁠ + ⁠2/100⁠ + ⁠3/1000⁠,

or

9 units + 8 tenths + 2 hundredths + 3 thousandths.

So the hundredths place is occupied by a digit "2".  ​‑‑Lambiam 04:42, 9 June 2026 (UTC)

June 10

Prime numbers

Why are prime numbers distributed randomly? ~2026-34141-00 (talk) 11:27, 10 June 2026 (UTC)

They aren't in the usual way one would mean, naively distributed randomly with distribution matching the prime number theorem. Cramér's conjecture, in its strong form, turned out to be false, because the behavior of primes in small intervals is not random in this sense (see, e.g., James Maynard (mathematician)). I once attended a lecture by Peter Sarnak, where he commented that the random thing is not the prime numbers, but the Möbius function (suitably understood), in the context of Chowla's conjectures (see, e.g., this ). But it's not clear to what extent number theory as refined our understanding of the Möbius in the last decade or so, it might be that it too is no longer regarded as "random" in the relevant sense. Sławomir Biały (talk) 11:40, 10 June 2026 (UTC)

Randomness is an illusion. Certainly in mathematics. And arguably even in physical reality, see superdeterminism. When things appear random to us, it is because we are finite beings. So our ability to make distinctions, acquire information, and do computations is very limited. But to God, everything is perfectly orderly and predictable. JRSpriggs (talk) 14:01, 10 June 2026 (UTC)

This is hard to answer because in one sense it's obviously false (prime numbers are defined deterministically) but it does seem to work in a lot of cases with only a few tweaks to the probabilities for "local" (mod-p for small primes p) factors; many of those are still only conjectures but the numerical evidence is usually strong.

See for a better exposition than I can give. Sesquilinear (talk) 23:47, 10 June 2026 (UTC)

The sieve of Eratosthenes can be used to generate the prime numbers. The lucky numbers are generated by a sieve that is a twist on the sieve of Eratosthenes. The lucky numbers have no known interesting number-theoretical properties, which puts them in the non-academic playpen of recreational mathematics, but they have attracted some interest because their distribution gives the same appearance of randomness as the primes. It is, in fact, a plausible conjecture that the appearance of randomness of the distribution of the prime numbers can be explained as a general property of sets of numbers generated by some class of sieves, unrelated to any interesting number theory. The notion of "appearance of randomness" is, however, hard to formalize. Combined with the lack of applicability of number theory, it does not look like a promising topic for study.  ​‑‑Lambiam 07:48, 11 June 2026 (UTC)

"...it does not look like a promising topic for study." could be very famous last words.Rich (talk) 08:35, 12 June 2026 (UTC)

Note that I did not write, "it is not a promising topic for study". I offered my assessment about its appearance because I think it explains the reluctance of mathematicians to take deep dives following up on this study from 70 years ago.  ​‑‑Lambiam 13:40, 12 June 2026 (UTC)

June 13

A question on subgroups of the orthogonal group

When can you realize a group $G<O(n)$ (strict subgroup inclusion) as the symmetry group of an actual set of points in ${\mathbf {R}}^{n}$?

Clearly, if $G$ acts transitively on $S^{n-1}$, this is impossible. Because if it does, then if your putative point set contains some point, then it actually contains all points at the same distance from the origin, which means your putative point set is a union of $n$-spheres and it actually has symmetry group $O(n)$. By exactly the same logic, if you can find some group $H$ such that $G<H\leq O(n)$ such that every $G$-orbit is also a $H$-orbit (i.e. the sets of orbits are identical), then $G$ cannot be realized as a symmetry group of a set of points in ${\mathbf {R}}^{n}$, because any set of points that would have $G$ as its symmetries would have more symmetries (the ones in $H\setminus G$). But can anything else be said without having to explicitly exhibit said bigger group or prove its nonexistence? And does anything change if you relax things from ${\mathbf {R}}^{n}$ to ${\mathbf {R}}^{m}$ where possibly $m>n$? Double sharp (talk) 12:50, 13 June 2026 (UTC)

I think without more structure of the subgroup or set, it would be very difficult to give a better criterion than the obvious one you already state. Subgroups of O(n), with no extra structure (e.g., that they are closed, for example) can be quite wild. Sławomir Biały (talk) 13:44, 13 June 2026 (UTC)

A useful example might be G a two dimensional rotation group. If G is the cyclic group generated by a 90° rotation, then the orbit of a single point (not the origin) would be a square, and the symmetry group of the square is the dihedral group of order 8. To get G as the symmetry group of a set of points you'd have to have at least 2 orbits or 8 points. It can be done, say {(1, 0), (-1, 0), (0, 1), (0, -1), (1, 2), (2, -1), (-1, -2), (-2, 1)}, but it requires more points than there are elements in G. --RDBury (talk) 14:03, 14 June 2026 (UTC)

Huh. I feel a bit stupid now, because this establishes a flaw in my original argument: just because every $G$-orbit is also an $H$-orbit doesn't actually mean that a union of $G$-orbits will necessarily have all of $H$ as symmetries. Once you put in the first point in this $C_{4}$ example, you've established where the reflection axes could be if you wanted to extend it to $D_{8}$, so you can add another square without its reflection to get a point set with essentially "chiral square symmetry". I think it still works when $H=O(n)$ itself, since then you really do have no choice but to make a bunch of concentric spheres, but it seems the situation's even more non-obvious and hard to say anything useful for. (Unless I suppose we restrict it and, say, impose closedness of the group.) Double sharp (talk) 05:45, 15 June 2026 (UTC)

Finite subgroups of O(n) are closed. ~2026-28259-76 (talk) 10:29, 15 June 2026 (UTC)

Question didn't specify finiteness. Sławomir Biały (talk) 10:31, 15 June 2026 (UTC)

Yes that’s obvious? Maybe my point wasn’t clear. I thought Double Sharp was saying (in effect) “maybe if we impose closedness things will be easier” and I was just pointing out that this specific example (as well as the entire finite case) is already closed. ~2026-28259-76 (talk) 10:45, 15 June 2026 (UTC)

Agree, but it would be nice if there were some clarity on this point. Sławomir Biały (talk) 10:52, 15 June 2026 (UTC)

Yes, I meant "if we impose closedness in general things may be easier". Of course, the finite example RDBury and I were talking about is closed. Double sharp (talk) 13:14, 15 June 2026 (UTC)

• Every finite group in $O(n)$ is the symmetry group of some set of points in $R^{n}$. For a finite subgroup $G$, choose a basis $v_{i}$ so that the norms of the $v_{i}$ are all different, and $\langle v_{i},(a-I)v_{j}\rangle \neq 0$ for all $v_{i},v_{j}$ and all $a\in G$, $a\neq I$. Then $G$ is exactly the symmetry group of the set $X$ = union of orbits $Gv_{i}$. Indeed, if $hX=X$, then for each $i$, $hv_{i}=g_{i}v_{i}$ for some $g_{i}\in G$. On the other hand $\langle v_{i},v_{j}\rangle =\langle hv_{i},hv_{j}\rangle =\langle g_{i}v_{i},g_{j}v_{j}=\langle v_{i},g_{i}^{*}g_{j}v_{j}\rangle$. So $\langle v_{i},(I-g_{i}^{*}g_{j})v_{j}\rangle =0$ which violates the choice of $v$'s unless $g_{i}=g_{j}$.
• In general, given a closed subgroup $G$, there is a subset whose symmetry group is exactly $G$ if and only if $G$ is equal to the intersection of the stabilizer of all of its orbits (which we here term "orbit closed"). That is, G is "orbit closed" provided: $h\in G\iff h(Gx)=Gx,\forall h\in O(n),x\in \mathbb {R} ^{n}$. For one direction, take a countable dense subset $u_{i}$ in $S^{n-1}$, and consider the set $X=\{0\}\cup \bigcup _{i}2^{-i}Gu_{i}$. If $hX=X$, then $hu_{i}\in Gu_{i}$ and by compactness of $G$ and density of the $u_{i}$, $hx\in Gx$ for all $x$. So, assuming $G$ is orbit-closed, $h\in G$. (For the other direction, if $G$ is the symmetry group of some set $X$, then $G$ is orbit closed, for if $h$ preserves every orbit of $G$, then it must preserve $X$ and therefore $h\in G$.)
• There is maybe a more computational criterion. Still, for a closed subgroup $G$ consider the ring of $G$-invariant polynomial functions on the coordinates $\mathbb {C} [x_{1},\dots ,x_{n}]^{G}$. This is a finitely-generated polynomial ring in a set of basic invariants by the Hilbert-Nagata theorem, $\mathbb {C} [x_{1},\dots ,x_{n}]^{G}=\mathbb {C} [f_{1},\dots ,f_{r}]$, where each $f_{i}$ is a polynomial such that $g^{*}f_{i}=f_{i}$. An element $h\in O(n)$ preserves every $G$-orbit if and only if if preserves every element of this ring, i.e., $h^{*}f_{i}=f_{i}$ for every generator $f_{i}$.
• To make this practical, I would suggest first looking at the invariants of the connected component of the identity $G_{0}$, which can be given as the polynomials that are annihilated by the Lie algebra of $G$. This can be found by Grobner basis methods. Determining the algebra $(\mathbb {C} [x_{1},\dots ,x_{n}])^{G_{0}})^{G/G_{0}}$ is beyond my ken, I'm afraid. But it should at least be possible to determine any additional elements of the Lie algebra ${\mathfrak {so}}(n)$ which annihilate $\mathbb {C} [x_{1},\dots ,x_{n}])^{G_{0}}$, and thus rule out any larger connected orbit-closure. (Importantly, the finite group argument does not generalize to the case where $G_{0}$ is not the identity, as the example of $G=G_{0}=SO(n)$ shows.) Sławomir Biały (talk) 13:22, 16 June 2026 (UTC)

  Nice! ~2026-28744-62 (talk) 20:54, 16 June 2026 (UTC)

  Very nice indeed. Thank you!! Double sharp (talk) 09:17, 17 June 2026 (UTC)

June 17

Plot with change in scale of axis

Note axes

I'd been corresponding with PJRay (talk · contribs) about File:Timeline_of_Superconductivity_from_1900_to_2015.svg which has the axis scales switching from 10 K to 50 K, and 40 years to 5 years per tick, though there are double dashes implying a broken axis. Is it considered acceptable, or is there a better way to indicate this?

An alternative using an inset

An alternative might be to draw the whole plot at a fixed scale, and add an inset magnifying the crammed area.

Thanks, cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 22:33, 17 June 2026 (UTC)

The science section is a more appropriate venue for what is considered good practice in presenting scientific plots.

One thing I can say is that the axis breaks are not visually prominent and that I wouldn't have noticed them without explicit warning; also, I think that they normally represent a gap and that a discontinuous change of scale is highly unusual.

It would make sense to use a logarithmic scale for the temperature scale.  ​‑‑Lambiam 04:42, 18 June 2026 (UTC)

I agree with a logarithmic scale for temperature and a (wider) linear scale for time in a single graph. JRSpriggs (talk) 13:24, 18 June 2026 (UTC)

FWIW, the classic text in the field is Edward R. Tufte's The Visual Display of Quantitative Information. (You can check this out on archive.org btw.) My point here is that this is not a field of study you can master in an hour, and I certainly don't claim to be an expert. That said, the broken axes in the original version do seem sketchy to me and the consensus seems to be that they are rarely a good idea. I'd also question whether you need all the information in the chart; perhaps leaving out all the points except "record holders" and leaving out the labels would remove the clutter without seriously affecting what the chart is trying to say. The details might be better presented in a table. Lambiam makes a good point here; scientists, who deal with this kind of issue all the time, might be better able to answer this question more easily than mathematicians. It's really a question of effective communication rather than mathematics. --RDBury (talk) 08:36, 19 June 2026 (UTC)

PS. On using broken axes in charts, this addtwodigital post seems to cover the main criticisms and discusses different alternatives. Again, it's a question of communication. What point is the chart trying to make? What is the target audience for the chart? You have to answer this questions before asking that is the "best" format. --RDBury (talk) 08:59, 19 June 2026 (UTC)

Thanks, everyone. I'll move the question to the Science Desk. cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 15:06, 19 June 2026 (UTC)

June 20

Busy Beaver vs Tree

Which is larger between bb(tree(n)) and tree(bb(n)) ?

I would have thought bb wins as it grows faster than any calculable function, but then I realised that would only apply to bb(tree(n)) vs tree(tree(n)).

Any thoughts everyone?

Duomillia (talk) 02:36, 20 June 2026 (UTC)

The article you link to doesn't show any function called tree() in all lower case. It defines TREE(), and I think I've seen Tree() attested as well.

But in any case, whatever you mean by tree(), why exactly do you think it's computable? That seems extremely unlikely to me. --Trovatore (talk) 03:02, 20 June 2026 (UTC)

According to our Busy beaver article, the notation BB is ambiguous. Using the score function Σ, we have:

1. Σ(TREE(1)) = TREE(Σ(1)) = 1.
2. Σ(TREE(2)) = 6, whereas TREE(Σ(2)) = TREE(4), which is unimaginably much larger than the already unimaginably large TREE(3).

 ​‑‑Lambiam 05:26, 20 June 2026 (UTC)

Are there any notions that drop associativity from ordinary category?

Inspired by the discussion in Rewriting the PDIFF article, I looked into whether there are categories that do not assume associativity. For example, there is a notion called a semicategory, which does not assume identities. So I continued searching for references, I then looked for a notion that if there was a notion that does not assume associativity, and it seems there is a notion called neocategory. While semicategory is already a niche notion, neocategory feels like a provisional name and seems to an even more niche notion. Is there a standard notion of a ordinary category without associativity?--SilverMatsu (talk) 03:19, 20 June 2026 (UTC)

Presumably you still want a composition? Then maybe a (directed) graph with a binary operator on morphisms (such that the binary operator is not necessarily defined for all pairs of morphisms). After all, a category is a (directed) graph with a binary operator (composition) that satisfies some conditions where the vertices of a graph form a class. —- Taku (talk) 04:10, 20 June 2026 (UTC)

If there is such a notion, it is not a "standard" notion. The difference between these neocats and arbitrary partial magmas seems to be that the former have some form of left and right identities, which can be seen as being in a one-to-one correspondence with the nodes of a directed graph.  ​‑‑Lambiam 05:03, 20 June 2026 (UTC)

Thank you for the example. Yes, I should have said that the composition does not satisfy the associativity. Indeed, a small category is a generalization of a monoid, and roughly speaking, a category is a directed graph equipped with a monoid-like composition. Thank you for explaining the difference between partial magma and neocats. There might also be a notion that weakens the identity axioms of neocats.--SilverMatsu (talk) 05:52, 20 June 2026 (UTC)

Well I'm just learning about category theory but I can see problems withtrying to turn the octonions into an ordinary category. If you do a search on 'octonion category' though there are some results I think I'll leave till I know a bit more! NadVolum (talk) 20:45, 20 June 2026 (UTC)