Welcome to the mathematics section
of the Wikipedia reference desk.

Select a section:

WP:RD/MA

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.

May 3

Can Carmichael number be Lucas-Carmichael number?

Also, varying the signs, there are four different sequences for similar numbers:

squarefree composite numbers k such that p | k => p-1 | k-1
squarefree composite numbers k such that p | k => p-1 | k+1
squarefree composite numbers k such that p | k => p+1 | k-1
squarefree composite numbers k such that p | k => p+1 | k+1

the 1st sequence is Carmichael numbers, and the 4th sequence is Lucas-Carmichael numbers, but what are the 2nd sequence and the 3rd sequence? Are there any number in at least two of these four sequences? If so, are there any number in at least three of these four sequences? 61.224.150.139 (talk) 05:07, 3 May 2024 (UTC)[reply]

According to Lucas–Carmichael number, it is unknown whether there are any Lucas–Carmichael numbers that are also Carmichael numbers. GalacticShoe (talk) 05:52, 3 May 2024 (UTC)[reply]

2. The sequence is OEIS:A208728, and it starts

15,35,255,455,1295,2703,4355,6479,9215,\ldots

3. The sequence is OEIS:A225711, and it starts

385,2737,6061,6721,17641,24769,25201,\ldots

GalacticShoe (talk) 06:21, 3 May 2024 (UTC)[reply]

Do all numbers in any of these four sequences except 15 and 35 have at least three prime factors? 61.224.150.139 (talk) 06:41, 3 May 2024 (UTC)[reply]

Yes. You can show that:

If $p-1,q-1|pq-1$ , then $p-1|q-1$ and $q-1|p-1$ , implying $p-1=q-1\Rightarrow p=q$ , which is disallowed.
If $p-1,q-1|pq+1$ , then $p-1|q+1$ and $q-1|p+1$ , implying either $p=q-2$ or $p=q+2$ (since they can't be equal.) The rest of this proof is left to the reader since I don't feel like writing it down, but based on the fact that $n|(n+2)^{2}\Leftrightarrow n|4$ , it can be shown that $pq=15,35$ only.
If $p+1,q+1|pq-1$ , then $p+1|q+1$ and $q+1|p+1$ , implying $p+1=q+1\Rightarrow p=q$ , which is disallowed.
If $p+1,q+1|pq+1$ , then $p+1|q-1$ and $q+1|p-1$ , implying $p\leq q-2$ and $p\geq q+2$ , which is not possible.

GalacticShoe (talk) 07:52, 3 May 2024 (UTC)[reply]

Are all of these four sequences infinite? If so, do all of these four sequences contain infinitely many terms with exactly 3 prime factors? Also, do all of these four sequences contain infinitely many terms which are divisible by a given odd prime number? 2402:7500:943:D56F:909B:9877:85C8:AFAA (talk) 02:02, 5 May 2024 (UTC)[reply]

@GalacticShoe: 49.217.60.214 (talk) 05:02, 6 May 2024 (UTC)[reply]

It is known that there are infinitely many Carmichael numbers, and it was recently (Wright, 2016) proven that there are infinitely many Lucas–Carmichael numbers. Unfortunately, I am unsure of the other two sequences, although Wright's paper might have more information for someone more mathematically literate than I. GalacticShoe (talk) 17:14, 10 May 2024 (UTC)[reply]

If Dickson conjecture is true, then do all of these four sequences contain infinitely many terms with exactly 3 prime factors? 2402:7500:900:DEEB:B513:C07E:8EF3:8275 (talk) 04:13, 11 May 2024 (UTC)[reply]

In order for a sequence to be in both 1. and 2., this would require that all prime factors

p|k

satisfy both

p-1|k-1,p-1|k+1\Rightarrow p-1|2\Rightarrow p=2,3

. The only squarefree composite number that is only composed of

2,3

is

6

which can easily be seen to not be in either sequence. Similarly, 3. and 4. would require all prime factors

p|k

to satisfy both

p+1|k-1,p+1|k+1\Rightarrow p+1|2

which does not hold for any primes

p

. Since 1. and 2. cannot coexist, nor can 3. and 4., this means that no number occupies three or more of the four sequences. GalacticShoe (talk) 06:25, 3 May 2024 (UTC)[reply]

Carmichael numbers are the numbers n such that

\lambda (n)

divides n-1, where

\lambda (n)

is the Carmichael lambda function (also called reduced totient function, since it is the reduced form of the Euler totient function) (sequence A002322 in the OEIS), and Lucas-Carmichael numbers should be the numbers n such that

\eta (n)

divides n+1, and this

\eta (n)

should be a reduced form of the Dedekind psi function, use the same reduce rule as the Carmichael lambda function to the Euler totient function (i.e. use the least common multiple in place of the multiplication for

\lambda (mn)

and

\eta (mn)

with coprime m, n), but I cannot even find this function in OEIS (it should start with (start from n=1) 1, 3, 4, 6, 6, 12, 8, 12, 12, 6, 12, 12, 14, 24, 12, 24, 18, 12, 20, 6, 8, 12, 24, 12, …) 2402:7500:943:D56F:909B:9877:85C8:AFAA (talk) 02:13, 5 May 2024 (UTC)[reply]

@GalacticShoe: 2402:7500:900:DEEB:B513:C07E:8EF3:8275 (talk) 04:09, 11 May 2024 (UTC)[reply]

Factorial & primorial on wikipedia

I know the factorial notation n!. Recently I cam across 5# which I was unfamiliar with. Not knowing its name, (primorial), it proved hard to track down. I searched in wikipedia and Google for "n#" which seemed like the best bet. Both converted it to "n" and reported stuff about the 14th letter of the alphabet. I then thought maybe it is related to factorial, so I looked at wikipedia factorial (n! redirects to factorial on wikipedia, so that works if/when you don't know the term "factorial".)

So is there a way of making n# findable on wikipedia? If so how? -- SGBailey (talk) 21:21, 3 May 2024 (UTC)[reply]

Search for "#" and find Number sign#Mathematics. —Kusma (talk) 22:25, 3 May 2024 (UTC)[reply]

@SGBailey: "#" is WP:FORBIDDEN in page names. I'm actually impressed that a search on "#" gives Number sign as the only result. I examined redirects and saw the similar Unicode characters ﹟ and ＃. Thinking that it may help search, I have redirected N﹟ and N＃ to Primorial. The "go" feature of the search box ignores "#" in "n#" and goes directly to N, but if you force a real search on n# then the third or fourth (it varies) result for me is now "Primorial (redirect from N﹟)". PrimeHunter (talk) 19:57, 6 May 2024 (UTC)[reply]

Excellent - Thank you. -- SGBailey (talk) 18:00, 10 May 2024 (UTC)[reply]

May 5

Origin of notion that there are ב sub 2 many "curves"

(Sorry for awkward heading -- I couldn't get it to put the ב before the 2 because of some strange artifact of RTL rendering.)

I've seen in several places the claim that, as there are $\aleph _{0}$ natural numbers and ${\mathfrak {c}}=2^{\aleph _{0}}$ (sometimes improperly given as $\aleph _{1}$ ) real numbers, there are some greater number of "curves" (sometimes given as f or, again improperly, $\aleph _{2}$ ). Most recently I was reminded of it at our article on George Gamow's (generally excellent) book One Two Three... Infinity.

The usual complaint about these popularizations, a very valid one, is that they uncritically give these cardinalities as aleph numbers in a way that works only if the generalized continuum hypothesis holds. But there's another, quite serious, problem: The claim that there are more "curves" than real numbers is correct only if you have an extremely liberal notion of what constitutes a "curve".

One reasonable notion is that a "curve" is the image of the real line or the unit interval under a continuous function from the reals to Rⁿ (or similar space), but there are only $2^{\aleph _{0}}$ such functions, and therefore the same number of curves.

My best guess is that someone was taking "curve" to mean the graph of an arbitrary function. But these are not typically curves according to any obvious natural-language meaning; they're just scattered points in the plane.

So, question, what's my question? Does anyone know where this idea originated? Was it Gamow, some other popularizer, multiple sources? And what if anything should we do to clean up the text in our One Two Three... Infinity article? I'm thinking an explanatory footnote but ideally I'd want a source directly speaking to the misconception. --Trovatore (talk) 20:58, 5 May 2024 (UTC)[reply]

This Math Stack exchange entry is relevant, but it doesn't seem to cover what you're asking. One problem is that the statement is true by Wikipedia standards; you could cite the book. You would need a reliable source, such as a published article somewhere, to say it was wrong/vague/misleading in order to state that in our article. At the moment the article points out that you'd need the GCH to say what's in the book, but I guess that's supposed to be "common knowledge" (at least among mathies). — Preceding unsigned comment added by RDBury (talk • contribs) 23:49, 5 May 2024 (UTC)[reply]

This article aims to classify various subsets of the function space 𝐹(ℝ,ℝ) from a constructive-mathematics perspective. The Introduction states: "mathematicians have made numerous attempts to focus on special subsets of this vast vector space (e.g., all real-valued continuous functions [5])", where the cited text is:

Pugh, C.C. Real Mathematical Analysis, 1st ed.; Undergraduate texts in mathematics; Springer Science Business Media: New York, NY, USA, 2002; pp. 223–225.

The latter is available as a pdf here. The article itself denotes this subset as 𝐶(ℝ,ℝ) and concludes in Proposition 4 that 𝐶𝑎𝑟𝑑(𝐶(ℝ,ℝ)) = 𝑐. But this is of course outside the paradise that Cantor created for you. --Lambiam 07:09, 6 May 2024 (UTC)[reply]

The reply given by a fellow Wikipedian to another Math Stack exchange question appears to imply that this also holds within the paradise. --Lambiam 07:23, 6 May 2024 (UTC)[reply]

Cardinality of the continuum § Sets with cardinality of the continuum also lists, without citation, "the set of all continuous functions from

\mathbb {R}

to

\mathbb {R}

". --Lambiam 07:30, 6 May 2024 (UTC)[reply]

Discussed at Stack Exchange. Basically it's because a continuous function from

\mathbb {R}

to

\mathbb {R}

is uniquely determined by its values at rational points. AndrewWTaylor (talk) 16:11, 6 May 2024 (UTC)[reply]

I feel a bit of sympathy for him making those mistakes but he should have had a mathematician read through that chapter. NadVolum (talk) 16:40, 6 May 2024 (UTC)[reply]

Well, it's kind of the publisher's job to do fact checking. The statement was still in the 2012 Dover edition, so there have been multiple chances to fact check since the original 1947 publication. --RDBury (talk) 19:12, 6 May 2024 (UTC)[reply]

In general I'm skeptical of active attempts to use Wikipedia to correct readers' mathematical misconceptions — too much like righting great wrongs, and can easily become a POV magnet (like the old "What mathematics is not" section that once appeared in our mathematics article).

This one irks me, though, and tempts me to go back on that reasoning. I guess it's slightly personal, because I had internalized this bit about the cardinality of the set of curves, and (embarrassingly) didn't get it corrected till grad school. I had figured out for myself that there were only continuum-many analytic functions, because they're determined by the coefficients of the power series, but I conjectured that there were 2^𝔠 many C^∞ functions, and someone had to set me straight on that.

I think it's not just Gamow (whose book, I want to re-emphasize, is a big net positive). I've been trying to remember where else I might have seen it. I thumbed through Lilian Lieber's Infinity (which is a book that heavily influenced me) and didn't find it there. --Trovatore (talk) 20:00, 6 May 2024 (UTC)[reply]

Just a guess, but the article mentions "What is Mathematics?" by Richard Courant and R. Robbins as a source, and of all the sources it seems the most mathematical. It's a "Text to borrow" on Internet Archive so if you create an account you can view it for free. --RDBury (talk) 10:42, 7 May 2024 (UTC)[reply]

There is a parallel in the treatment of cardinal numbers between Gamow and Courant & Robbins up to the point where the latter write (p. 85): "A similar argument shows that the cardinal number of the points in a cube is no greater than the cardinal number of the segment." After that, they muse briefly on the fact that this is counterintuitive since the correspondence does not preserve dimension, but that this is possible because it is not continuous. That ends their treatment of cardinal numbers. Earlier they note (p. 84): "As a matter of fact, Cantor actually showed how to construct a whole sequence of infinite sets with greater and greater cardinal numbers." They even sketch the proof, but do not pursue the question of mathematical objects of higher cardinality than the continuum that are of interest by themselves. --Lambiam 13:21, 7 May 2024 (UTC)[reply]

Digression: In fact it's challenging to come up with an object of larger cardinality that might naturally be considered by non-set-theorists. One possibility is βN, the Stone–Čech compactification of the natural numbers. I believe this is mentioned in an exercise in Folland's Real Analysis. --Trovatore (talk) 18:53, 7 May 2024 (UTC) [reply]

There's a relevant MathOverflow question about finding cardinalities beyond that of the continuum outside set theory. βN is given as an answer, but maybe the most elementary one offered is the set of all field automorphisms of C. But the answers do kind of make me agree with Gamow's surely intended point that it's difficult to find natural objects of size beyond 2^c, though not with his actual assertion. :) Double sharp (talk) 15:21, 8 May 2024 (UTC)[reply]

May 6

Find

Given x=3+2√2, find √x - 1/√x 171.79.74.205 (talk) 17:01, 6 May 2024 (UTC)[reply]

the contradictory part is that in the end, you get (√x-1/√x)^2 = 4, which will give you ±2; but √x which is √3+2√2 can be written as √(2-√1)^2 which is 2-√1 hence √x - 1/√x = -2 171.79.74.205 (talk) 17:10, 6 May 2024 (UTC)[reply]

"... can be written ... ". No, it can't. Not sure if this is an honest question or just trolling. --RDBury (talk) 18:43, 6 May 2024 (UTC)[reply]

I'm not sure what they that is about but can I suggest that 1/(a+√b) = (a-√b)/{(a+√b)(a-√b)} might help? NadVolum (talk) 20:33, 6 May 2024 (UTC)[reply]

Clarify, please. Do you mean:

√x - (1/√x)
(√x - 1)/√x
√(x - 1)/√x
something else...?

I'd suggest using LaTeX/MathJax code within <math>...</math> tags to format the expressions like

{\frac {{\sqrt {x}}-1}{\sqrt {x}}}

etc. Please see WP:MATH for more info. --CiaPan (talk) 10:12, 7 May 2024 (UTC)[reply]

Sorry I misread the question and answered the wrong thing. But the square root of 3+2√2 is plus or minus 1+√2 and the original answer of ±2 is correct. NadVolum (talk) 11:07, 7 May 2024 (UTC)[reply]

√x is usually taken to mean the positive square root when x is positive. At least that's the notation used in Square root. That would make the answer 2. --RDBury (talk) 16:58, 7 May 2024 (UTC)[reply]

Only if you say the square root, and nobody has said that. It also doesn't matter whether the original √2 is positive or negative. NadVolum (talk) 17:07, 7 May 2024 (UTC)[reply]

The issue here is not a lack of the definite article, but the meaning of the symbol

{\sqrt {^{~}}}.

Conventionally, when

a

is a real number,

{\sqrt {a^{2}}}

denotes the same as

|a|,

so

{\sqrt {2}}

is definitely positive. --Lambiam 18:15, 7 May 2024 (UTC)[reply]

The answer is 2 if the meaning of "√x−1/√x" is (√x) − (1/√x). But note that the question uses "√3+2√2" with the meaning √3 + 2√2. --Lambiam 18:25, 7 May 2024 (UTC)[reply]

Is this a homework problem? GalacticShoe (talk) 16:05, 7 May 2024 (UTC)[reply]

Possibly but they tried to check their solution. NadVolum (talk) 16:58, 7 May 2024 (UTC)[reply]

May 8

What is the term for...

What is a proper term for an ordered set of values {a, b, c}, where each value can be independently selected from an allowed set for that value. Each allowed set may be different (but need not be distinct). For example:

a from {a₁, a₂, a₃}
b from {b₁, b₂}
c from {c₁, c₂, c₃, c₄}

The set {a, b, c} could then have any of 24 possible values.

A more concrete example would be all upper/lower/mixed case variations of the word "dog" - "dog", "doG", "dOg", "dOG", "Dog", ....

I saw this called a permutation, which I'm pretty sure is not correct. I was thinking combination, but this doesn't quite seem to match the definition. Is there some standard term that is a good fit for this? Tom N talk/contrib 20:21, 8 May 2024 (UTC)[reply]

See Cartesian product. --Trovatore (talk) 20:33, 8 May 2024 (UTC)[reply]

Come to think of it, though, the direct answer to your question would not be "Cartesian product" but rather "element of the Cartesian product". I don't know that there's a snappy one-word term for this. --Trovatore (talk) 20:42, 8 May 2024 (UTC)[reply]

Collatz Conjecture

From time to time I noodle around the Collatz conjecture, in an attempt to bring my "special insight" into an apparently simple question that has stumped the best and the brightest. Most of the work I've seen seems to be based on trying to find a counter-example, which would render the conjecture null and void. They've tested every number up to some ridiculous number of trillions, so far without any luck, but they take the view, quite rightly, that a giga-zillion examples do not prove the general proposition, and the elusive counter-example could be just round the corner, so they keep searching, and trying new ways to attack the problem.

I've never seen any work that started at the opposite end: the number 1. That is, take 1, and ask "What could produce 1?": Answer: only 2. Repeat. This results in the series 1, 2, 4, 8, 16. Then it starts getting interesting, because 16 could be derived from either 5 or 32. (Being odd, 5 produces 16, because 3*5+1 = 16; and being even, 32 produces 16, because 32/2=16.) Then 5 and 32 can be investigated separately, and so on. The tree quickly sprouts new branches and it just gets more bushy the further we go.

The question in my mind is: Can it be shown that every integer must belong to this tree? If so, would that not prove the Conjecture? Or, if it could be shown that not all numbers are captured, even if we could not identify any specific examples, would that not disprove it?

Yours simplistically, Jack of Oz ^{[pleasantries]} 22:51, 8 May 2024 (UTC)[reply]

Nice way of re-thinking the problem! I agree that if you could prove that the tree hits every positive integer, or that it doesn't, then you've solved the problem. I seriously doubt that it hasn't been tried, but it's a good example of things to try. --Trovatore (talk) 23:07, 8 May 2024 (UTC)[reply]

The specific approach is mentioned in Collatz conjecture § In reverse, which also has a diagram of the first 21 layers of the tree. A more parsimonious representation is achieved by considering that it suffices to show that all odd positive integers are reached. The onset of the odd tree is shown in the very first image in the article; a more extensive one, not shown in the article, is found here --Lambiam 05:41, 9 May 2024 (UTC)[reply]

Well, there you go. Thanks, Lambiam. (At least I should be given credit as the independent co-discoverer of this idea.) -- Jack of Oz ^{[pleasantries]} 07:46, 9 May 2024 (UTC)[reply]

Here's another exploration of the Collatz tree. Double sharp (talk) 11:27, 11 May 2024 (UTC)[reply]

Problem: Give positive integer n, how many natural density of positive integers reach n in their Collatz (3x+1) sequence? Of course, for n = 1, 2, 4, 8, 16, the natural density is 100%, but for n = 32 and 5, what will be the answers? For n divisible by 3, the answer is 0%, since only numbers of the form n*2^k reach n, and I think that the answer for 13 and 80 (which are the two numbers before 40) should be equal. 2402:7500:942:8E8F:A4D8:9B73:8E52:1E7B (talk) 07:47, 10 May 2024 (UTC)[reply]

You can eliminate this by calculate (for example) all positive integers <= 2^16 = 65536, how many positive integers reach 32 (or 5, or 13, or 80)? 2402:7500:900:DEEB:B513:C07E:8EF3:8275 (talk) 04:15, 11 May 2024 (UTC)[reply]

May 10

About abundance and abundancy

Let s(n) = (sequence A001065 in the OEIS)(n) = sigma(n)-n = sum of divisors of n that are less than n

Give integer k, should there be infinitely many positive integers n such that s(n)-n = k?
Give positive rational number k, should there be infinitely many positive integers n such that s(n)/n = k?

2402:7500:942:8E8F:A4D8:9B73:8E52:1E7B (talk) 07:41, 10 May 2024 (UTC)[reply]

The answer to 1. is no. If a number has is composite, then it is completely determined by its set of proper divisors (in particular, it is the product of the smallest prime factor and the largest proper divisor.) By definition

s(n)-n=m

if and only if there is a partition of

m

into unique numbers such that the elements of the partition are precisely the proper divisors of

n

. There are a finite amount of possible partitions of

m

, and thus a finite number of partitions which produce the proper divisors of some number

n

, and as long as the partitions in question are not just the set

\{1\}

(i.e. the partition produced by primes), all such partitions/sets of proper divisors completely determine some unique

n

. Thus for

k\neq 1

there are a finite number of

n

satisfying

s(n)-n=m

. GalacticShoe (talk) 17:28, 10 May 2024 (UTC)[reply]

The smallest values of

n

such that

s(n)-n=m

are given in OEIS: A070015, while the largest values of

n

such that

s(n)-n=m

are given in OEIS: A135244. GalacticShoe (talk) 17:32, 10 May 2024 (UTC)[reply]

Well, I meant s(n)-n = sigma(n)-2*n, not sigma(n) - n (which is s(n) itself), s(n) is (sequence A001065 in the OEIS), while sigma(n) is (sequence A000203 in the OEIS), they are different functions. 2402:7500:900:DEEB:B513:C07E:8EF3:8275 (talk) 04:09, 11 May 2024 (UTC)[reply]

See OEIS:A033880. GalacticShoe (talk) 18:45, 11 May 2024 (UTC)[reply]

Well, so should there be infinitely many such positive integers n? 49.217.136.82 (talk) 07:42, 14 May 2024 (UTC)[reply]

Unfortunately I have no idea, you're gonna have to check the sources in that OEIS listing. GalacticShoe (talk) 08:38, 14 May 2024 (UTC)[reply]

May 11

Dirac delta function

The Dirac delta is a notorious real-valued "function" that is infinite at x=0 and zero everywhere else. In real analysis it is treated as a generalized function (Schwartz distribution). Disclosure, I don't know what those really are, but their construction involves bump functions, which are continuously differentiable at all orders but are zero outside of a region.

In the complex plane of course, any continuously differentiable function is analytic so it must be either constant or unbounded, amirite? So there are no complex bump functions with those properties.

So, is there a complex version of the Dirac delta, and how is it mathematically "handled"? Thanks. 2602:243:2008:8BB0:F494:276C:D59A:C992 (talk) 00:03, 11 May 2024 (UTC)[reply]

The second illustration in the Dirac delta article shows it as the limit of sequence of zero-centered normal distributions, which do not have compact support; this works as well for most applications. So bump functions are not essential. Nevertheless, I don't think this will help in attempting to define a complex version. --Lambiam 06:41, 11 May 2024 (UTC)[reply]

Actually a more fundamental question: are Fourier series and Fourier transforms important in complex analysis? This is where the delta function comes up in the real case, more or less. 2602:243:2008:8BB0:F494:276C:D59A:C992 (talk) 08:02, 11 May 2024 (UTC)[reply]

The theory of Fourier series was developed well before Dirac came up with his delta function. It only plays a role in the theory of the Fourier transform for a purely periodic signal, not perturbed by any noise, something not found in actual practical applications. Even then, the delta function simplifies the presentation, but can be avoided using a mixed representation. I don't see how any of this can be generalized to deal with functions on the complex domain. --Lambiam 16:26, 11 May 2024 (UTC)[reply]

May 13

Prediction of analogous spatial distributions

Hi guys,

I wish to do a specific GIS/spatial analysis task and I suspect I'm using the wrong search terms in my attempts to figure out existing approaches and tools. Here's what I'm trying to do:

I have a raster data set describing the spatial distribution of a metric in the summer of year X (X1) as well as in the spring (X2). I further have data for the distribution of the same metric in summer only of year Y (Y1). I now want to extrapolate summer raster data of Y1 to Y2. To this end, I want to fit a model to the relationship X1 <-> X2 and then use this model for the prediction Y1 -> Y2. A number of spatial and temporal covariates are available.

I'm guessing that what I'm looking at here would be spatial regression followed by prediction. For the "predict distribution from spatial model" bit, it seems that some flavour of kriging would be suitable, but what I first need is essentially a model describing the transition between two distributions of the same metric (not the spatial relationship of one metric to another in the same space) - seems like a different problem? - To be implemented in QGIS and/or R, if any tool-specific recommendations come to mind.

Cheers! --Elmidae (talk · contribs) 12:08, 13 May 2024 (UTC)[reply]

When you make a scatter plot between the X1 and X2 data, do the points seem to lie on a curve with not too much noise? And are the extreme Y1 values not far outside the range of the X1 values? If so, you can simply try curve fitting with a low-degree polynomial and use the curve to read off plausible estimates for the Y2 values. The spatial aspect is then actually irrelevant. It may be relevant for smoothing the observed values before doing anything else. See if this helps with getting a clearer curve. If the metric is necessarily positive, it may further be helpful not to use X1 and X2 directly but to plot instead log(X2) against log(X1). Kriging only plays a role if the summer raster of the years is not the same. --Lambiam 13:09, 13 May 2024 (UTC)[reply]

Calculus question

Hello there, I'm not sure if this is the right place to ask a calculus question seeming the activity here is quite slim.

A cylindrical tank with a radius of 5 meters is being filled with water at a rate of 3 cubic meters per minute. The tank initially contains 10 cubic meters of water.

Write an expression for the volume of water in the tank as a function of time t in minutes.
Determine the rate at which the water level is rising in the tank when the depth of the water is 2 meters.
At what rate is the water level rising when the tank is half full?

Im lost on the second on here, someone help? GoodHue291 (talk) 22:55, 13 May 2024 (UTC)[reply]

Is the the axis of the cylinder horizontal? If so, then more information is needed, such as the length of the tank. catslash (talk) 23:20, 13 May 2024 (UTC)[reply]

Yeah, as presented this question is problematic, since for a horizontal axis you need the length, which isn't given, but for a vertical axis the rate, height, pretty much everything is linear. GalacticShoe (talk) 01:28, 14 May 2024 (UTC)[reply]

Please do your own homework.

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. - Arjayay (talk) 09:01, 14 May 2024 (UTC)[reply]

I put this question here so people can guide me to solving it, they're not going to do it for me. GoodHue291 (talk) 20:32, 14 May 2024 (UTC)[reply]

"The rate at which X is changing" means "the derivative of X with respect to time" where "X" in this case is "the water level". So first you're going to need to find an expression for the water level as a function of time L = f(t) and then work out the derivative of that function to get dL/dt = f'(t). You're also going to need to know the time t at which to evaluate f' which, assuming you know how to convert volume to water level, you can work out using the expression you found in part one. 2A01:E0A:D60:3500:61F0:5F9A:48A:C8D6 (talk) 09:32, 14 May 2024 (UTC)[reply]

May 14

Worst case performance of a randomized primality test?

I am trying to get a idea about how efficient something like this might be. Let's say we have a number N composed of B binary bits, and we now want to generate a "certificate of primality" using the following method: First we choose a "confidence" parameter C which indicates the degree of statistical certainty desired. Then, (1) Select a random number R < N. (2) Take the GCD of N with R; if it is not 1 then N is composite, otherwise we consider R to be a "witness" of N's (possible) primality. (3) Repeat until the level of confidence reaches C. Only question is, how to determine the number of iterations needed until C is satisfied? Earl of Arundel (talk) 16:39, 14 May 2024 (UTC)[reply]

That is usually what is done and its a good way to do it - but it can fall foul of problems like Pseudoprimes. so a better test is done tham the straightforward Fermat's little theorem test. See also AKS primality test for why a full test isn't normally done. The probabalistic tests linked from there give an estimate of their effectiveness - which is very good indeed. NadVolum (talk) 17:29, 14 May 2024 (UTC)[reply]

Yes of course, I am currently using the Miller-Rabin primality test which does exhibit a very good "average performance". However my question was more along the lines of "how much less efficient would it be to do randomized GCD tests instead"? Because surely even *that* would produce a bona fide "certificate of confidence" (albeit perhaps much slower than otherwise). Now I do know that the probability of any two random variables being comprime is

6/\pi ^{2}

, I just can't seem to figure out how to "interpolate" that into a reliable (if inefficient) probabalistic test. Earl of Arundel (talk) 18:16, 14 May 2024 (UTC)[reply]

If the number submitted to the famous EoA primality test may have been selected by an adversary, it can be the product of two almost equal primes. The probability that a single random test detects its non-primality is then about

2/{\sqrt {N}}.

This means that you need about

{\frac {1}{2}}{\sqrt {N}}\log {\frac {1}{p}}

independent random GCD tests before you can assert with confidence

1-p

that the number is prime. For example, for

N=9875191871

you need some 228,818 random tests to achieve a 99% confidence level. Straightforward primality testing by trying successive odd numbers as divisors while skipping multiples of

3

requires

{\frac {1}{3}}{\sqrt {N}}

trial divisions, for the example 33,124. --Lambiam 19:29, 14 May 2024 (UTC)[reply]

Wow, so not even close to being nearly as efficient as the humble trial division algorithm. (I don't know why that surprises me, it is more or less a "scatter-shot" approach to primality testing.) Of course Miller-Rabin blows both of those out of the water. Not only is the required number of iterations relatively low, the best-case performance WRT detecting composites is typically excellent. (I think it even outperforms AKS on average.) Very elegant formulation of the lower-bound there, by the way. I wouldn't have thought it could be reckoned so precisely. Kudos! Earl of Arundel (talk) 20:26, 14 May 2024 (UTC)[reply]

One reason some primality tests run more efficiently is that they only tell you what you want to know, is the number a prime? If you want to know more than that, a factor if it's composite, that will take longer; you're factoring the number instead of just showing that it's composite. A GCD test would produce a factor so it's bound to be less efficient. (Of course, someone could come up with a factorization algorithm that's as fast as the fastest primality test, but that hasn't happened so far, and that's why non-factoring primality tests still have a niche.) --RDBury (talk) 03:58, 15 May 2024 (UTC)[reply]

May 15

How many such binary operations exist in a set with n elements?

There are four possible properties for a binary operation:

So, in a set with n elements, how many such binary operations (which are closed) exist?

Satisfy property 1
Satisfy property 2
Satisfy property 3
Satisfy property 4
Satisfy properties 1 and 2
Satisfy properties 1 and 3
Satisfy properties 1 and 4
Satisfy properties 2 and 3
Satisfy properties 2 and 4
Satisfy properties 3 and 4
Satisfy properties 1, 2, and 3
Satisfy properties 1, 2, and 4
Satisfy properties 1, 3, and 4
Satisfy properties 2, 3, and 4
Satisfy all four properties

2402:7500:92D:FD81:F115:AC09:9228:B1A8 (talk) 10:58, 15 May 2024 (UTC)[reply]

Case 1, just idempotence, is easy. A binary operation on a set of $n$ elements (a finite magma) can be completely described by the $n^{2}$ entries of the $n\times n$ operation table. Idempotence fixes the $n$ entries on the diagonal. For each of the remaining $n(n-1)$ entries there are $n$ choices, so there are $n^{n(n-1)}$ distinct tables.
Case 2, just commutativity, is also easy. The $n$ entries on the diagonal can be chosen freely, as can the ${\tfrac {1}{2}}n(n-1)$ entries of the triangle below the diagonal; the upper triangle is thereby fixed. So the number equals $n^{{\frac {1}{2}}n(n+1)}.$
For case 4, just cancellation, there is a one-to-one correspondence with the Latin squares of order $n$ . See the section Number of $n\times n$ Latin squares.

There is no simple formula for this case, and I suppose also noy for most, if not all, other cases. Some have been tabulated; for case 3, the number of finite semigroups, see OEIS: A023814. --Lambiam 13:51, 15 May 2024 (UTC)[reply]