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November 27

Did they also pay hazard bonuses for working in the heat?

Is it cheaper for UPS to just air condition their warehouses and package vans?

After paying the initial installation fees for the new HVAC systems, how much will it cost for UPS to run air conditioning and maintain their HVAC systems for one year (at least only when the weather is hot?)

And how much did they pay out in heat-related workers comp claims for one year?

How well will UPS come out ahead from simply air conditioning all places and vehicles that need air conditioned? --2600:8803:1D13:7100:BD6D:70D0:30AC:B227 (talk) 01:13, 27 November 2024 (UTC)Reply

This is not a mathematics question. We don’t answer requests for opinions, predictions or debate. Dolphin (t) 04:59, 27 November 2024 (UTC)Reply

The largest prime factor found by trial division

The largest prime factor found by Lenstra elliptic-curve factorization is 16559819925107279963180573885975861071762981898238616724384425798932514688349020287 of 7337+1 (see [1]), and the largest prime factor found by Pollard's p − 1 algorithm is 672038771836751227845696565342450315062141551559473564642434674541 of 960119-1 (see [2]), and the largest prime factor found by Williams's p + 1 algorithm is 725516237739635905037132916171116034279215026146021770250523 of the Lucas number L2366 (see [3]), but what is the largest prime factor found by trial division? (For general numbers, not for special numbers, e.g. 7*220267500+1 divides the number 12220267499+1 found by trial division, but 12220267499+1 is a special number since all of its prime factors are == 1 mod 220267500, thus the trial division only need to test the primes == 1 mod 220267500, but for general numbers such as 3*2100+1, all primes may be factors) 61.229.100.16 (talk) 20:51, 27 November 2024 (UTC)Reply

I don't have an answer, and Mersenne primes have properties that reduce the number of primes that need to be searched, meaning that it doesn't technically need full trial division, but I would nevertheless like to raise two famous examples which I'm fairly sure were done through manual checking:
  1. In 1903, Frank Nelson Cole showed that   is composite by going up to a blackboard and demonstrating by hand that it equals  . It took him "three years of Sundays" to do so, and I'm fairly sure he would have done it manually.
  2. In 1951, Aimé Ferrier showed that   is prime through use of a desk calculator, and I imagine a lot of handiwork.
GalacticShoe (talk) 02:29, 28 November 2024 (UTC)Reply


November 29

In finite fields of large characteristics, what does prevent shrinking the modulus field size down to their larger order in order to solve discrete logarithms ?

In the recent years, several algorithms were proposed to leverage elliptic curves for lowering the degree of a finite field and thus allow to solve discrete logairthm modulo their largest suborder/subgroup instead of the original far larger finite field. https://arxiv.org/pdf/2206.10327 in part conduct a survey about those methods. Espescially since I don’t see why a large chararcteristics would be prone to fall in the trap being listed by the paper.

I do get the whole small characteristics alogrithms complexity makes those papers unsuitable for computing discrete logarithms in finite fields of large charateristics, but what does prevent applying the descent/degree shrinking part to large characteristics ? 2A01:E0A:401:A7C0:68A8:D520:8456:B895 (talk) 11:00, 29 November 2024 (UTC)Reply

Try web search for Lim-Lee small subgroup attack. 2601:644:8581:75B0:0:0:0:C426 (talk) 23:09, 1 December 2024 (UTC)Reply
First the paper apply when no other information is known beside the 2 finite field’s elements and then this is different from https://arxiv.org/pdf/2206.10327. While Pollard rho can remain more efficient, if the subgroup is too large, then it’s still not enough fast. I’m talking about shrinking modulus size directly. 2A01:E0A:401:A7C0:69D2:554C:93AF:D6AC (talk) 15:17, 2 December 2024 (UTC)Reply

The problem with descent methods like the one described in large characteristic is that the complexity is  . This is explained more clearly in [4]. If the characteristic is small, then the problem is O(k) bits, and the complexity id pseudo-polynomial in the number of bits. If the characteristic is large, then the compexity is   which is exponential in the number of bits of q. Tito Omburo (talk) 16:36, 2 December 2024 (UTC)Reply

Are you sure ? I’m not talking about the complexity of the index calculus part like chosing the factor base or computing individual discrete logarithms or the linear algebra phase. Only about the part consisting of shrinking the modulus through lowering it’s degree… 82.66.26.199 (talk) 10:26, 3 December 2024 (UTC)Reply
The paper I cited above finds the intermediate polynomials by a brute force search in small degree, which is thus polynomial in q. The general heuristic is that the search for intermediate polynomials takes about as long as the index calculus phase. But it looks like no one has a really good way to do it. Tito Omburo (talk) 10:45, 3 December 2024 (UTC)Reply
And in my case, elliptic curves are used to lower the degree and thus the modulus instead of brutefore. The person who wrote the paper told me in fact the idea was developed initially for medium characteristics but refused to give details for large characteristics. 82.66.26.199 (talk) 11:06, 3 December 2024 (UTC)Reply
I'm not really convinced that the approach actually works. How are the elliptic curves constructed? The construction given is very implicit. I'm betting that if one actually writes out the details, it involves lifting something from the prime field. Tito Omburo (talk) 16:02, 3 December 2024 (UTC)Reply
Yes. My request to have more details wasn’t well received. https://sympa.inria.fr/sympa/arc/cado-nfs/2024-12/msg00002.html 2A01:E0A:401:A7C0:5985:1F5D:4595:AA09 (talk) 01:48, 4 December 2024 (UTC)Reply

November 30

Linear differential function

Differential 102.213.69.166 (talk) 04:35, 30 November 2024 (UTC)Reply

What are you talking about? hamster717🐉(discuss anything!🐹✈️my contribs🌌🌠) 14:02, 30 November 2024 (UTC)Reply



December 4

How much is this

 

37.162.46.235 (talk) 11:00, 4 December 2024 (UTC)Reply

1 for n even, 0 for n odd. For   see Grandi's series. --Wrongfilter (talk) 11:18, 4 December 2024 (UTC)Reply
I hope this is not homework. Note that (for finite  ) this is a finite geometric series.  --Lambiam 21:51, 4 December 2024 (UTC)Reply


December 6

Is there anything that would prevent peforming Weil Descent on binary curves of large characteristics ?

The ghs attack involve creating an hyperlliptic curve cover for a given binary curve. The reason the attack fails most of the time is the resulting genus grows exponentially relative to the curve’s degree.

We don’t hear about the attack on finite fields of large characteristics since such curves are already secure by being prime. However, I notice a few protocol relies on the discrete logarithm security on curves with 400/500 bits modulus resulting from extension fields of characteristics that are 200/245bits long.

Since the degree is most of the time equal to 3 or 2, is there anything that would prevent creating suitable hyperelliptic cover for such curves in practice ? 2A01:E0A:401:A7C0:28FE:E0C4:2F97:8E08 (talk) 12:09, 6 December 2024 (UTC)Reply

December 7

Mathematical operation navigation templates

RDBury is right, this discussion belongs at Wikipedia talk:WikiProject Mathematics
The following discussion has been closed. Please do not modify it.

If anyone with some mathematical expertise is interested, I'd appreciate some additional input at Talk:Exponentiation#funny table at end. The question is whether our articles on various mathematical operations could use a navigational template (aka "{{Navbox}}"). Our Exponentiation article tried to use {{Mathematical expressions}} for this purpose, but it doesn't really work. I've created {{Mathematical operations}} as a potential alternative, but the categorization and presentation I've created is probably naïve. (The whole effort may or not be worth it at all.) —scs (talk) 00:36, 7 December 2024 (UTC)Reply

Wikipedia talk:WikiProject Mathematics is a better forum for this kind of thing, since it's focused on Wikipedia's mathematical articles. --RDBury (talk) 04:07, 7 December 2024 (UTC)Reply
@RDBury: Excellent point. Thanks. —scs (talk) 13:49, 7 December 2024 (UTC)Reply

December 8

For each positive integer  , which primes   are still primes in the ring  ?

For each positive integer  , which primes   are still primes in the ring  ? When  ,   is the original integer ring, when  ,   is the ring of Gaussian integers, when  ,   is the ring of Eisenstein integers, and the primes in the Gaussian integers are the primes  , and the primes in the Eisenstein integers are the primes  , but how about larger  ? 218.187.66.163 (talk) 04:50, 8 December 2024 (UTC)Reply

A minuscule contribution: for   the natural Gaussian primes   and   are composite:
  •  
  •  
So   is the least remaining candidate.  --Lambiam 09:00, 8 December 2024 (UTC)Reply
It is actually easy to see that   is composite, since   is a perfect square:
 
Hence, writing   by abuse of notation for   we have:
 
More in general, any natural number that can be written in the form   is not prime in   This also rules out                   and    --Lambiam 11:50, 8 December 2024 (UTC)Reply
I'll state things a little more generally, in the cyclotomic field  . (Your n is twice mine.) A prime q factors as  , where each   is a prime ideal of the same degree  , which is the least positive integer such that  . (We have assumed that q does not divide n, because if it did, then it would ramify and not be prime. Also note that we have to use ideals, because the cyclotomic ring is not a UFD.) In particular,   stays prime if and only if   generates the group of units modulo  . When n is a power of two times an odd composite, the group of units is not cyclic, and so the answer is never. When n is a prime or twice a prime, the answer is when q is a primitive root mod n. If n is 4 times a power of two times a prime, the answer is never. Tito Omburo (talk) 11:08, 8 December 2024 (UTC)Reply

Can we say anything special about every pair of functions f,g, satisfying f(g(x))=f(x) for every x?

Especially, is there an accepted term for such a pair?

Here are three simple examples, for two functions f,g, satisfying the above, and defined for every natural number:

Example #1:

f is constant.

Example #2:

f(x)=g(x), and is the smallest even number, not greater than x.

Example #3:

f(x)=1 if x is even, otherwise f(x)=2.
g(x)=x+2.

2A06:C701:746D:AE00:ACFC:490:74C3:660 (talk) 09:31, 8 December 2024 (UTC)Reply

One way to consider such a pair is dynamically. If you consider the dynamical system  , then the condition can be stated as "  is constant on  -orbits". More precisely, let   be the domain of  , which is also the codomain of  . Define an equivalence relation on   by   if   for some positive integers  . Then   is simply a function on the set of equivalence classes   (=space of orbits). In ergodic theory, such a function   is thought of as an "observable" or "function of state", being the mathematical analog of a thermodynamic observable such as temperature. Tito Omburo (talk) 11:52, 8 December 2024 (UTC)Reply
After you've mentioned temprature, could you explain what are f,g, as far as temprature is concerned? Additionally, could you give another useful example from physics for such a pair of functions? 2A06:C701:746D:AE00:ACFC:490:74C3:660 (talk) 19:49, 8 December 2024 (UTC)Reply
This equation is just the definition of function g. For instance if function f has the inverse function f−1 then we have g(x)=x. Ruslik_Zero 20:23, 8 December 2024 (UTC)Reply

IEEE Xplore paper claim to acheive exponentiation inversion suitable for pairing in polynomial time. Is it untrustworthy ?

I just found https://ieeexplore.ieee.org/abstract/document/6530387. Given the multiplicative group factorization in the underlying finite field of a target bn curve, they claim to acheive exponentiation inversion suitable for pairing inversion in seconds on a 32 bits cpu.

On 1 side, the paper is supposed to be peer reviewed by the iee Xplore journal and they give examples on 100 bits. On the other side, in addition to the claim, their algorithm 2 and 3 are very implicit, and as an untrained student, I fail to understand how to implement them, though I fail to understand things like performing a Weil descent.

Is the paper untrustworthy, or would it be possible to get code that can be run ? 2A01:E0A:401:A7C0:152B:F56C:F8A8:D203 (talk) 18:53, 8 December 2024 (UTC)Reply

About the paper, I agree to share the paper privately 2A01:E0A:401:A7C0:152B:F56C:F8A8:D203 (talk) 18:54, 8 December 2024 (UTC)Reply