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September 22

How can we briefly characterize a given set of vectors, as "linearly dependent - every proper sub set being linearly independent", while we only refer to the vectors rather than to their set?

For example: S is a set of the following vectors:

A=(1,1,0),

B=(1,0,0),

C=(0,1,0).

Note: A=B+C, and B=A-C, and C=A-B, so the set S is linearly dependent.

Using A,B,C only, i.e without using S, what's the shortest description, claiming that the set S is linearly dependent but every proper sub set of S is linearly independent? HOTmag (talk) 13:30, 22 September 2024 (UTC)[reply]

@HOTmag: I'd just say 'S has k linearly independent elements' (in the give example k=2). --CiaPan (talk) 14:46, 22 September 2024 (UTC)[reply]

Is your response a suggestion of rephrasing my question?

If it's intended to be an answer, then please note: My condition requires to be "using A,B,C only, i.e without using S". Additionally, where does your description claim, that S is linearly dependent? HOTmag (talk) 14:56, 22 September 2024 (UTC)[reply]

One way to characterise the set is "a set of vectors, any one of which can be written in terms of the others in a unique way". The set is just A, B and C, i.e. any property of them is a property of the set of them. --2A04:4A43:900F:F4C3:49F4:4EFB:C442:608F (talk) 15:15, 22 September 2024 (UTC)[reply]

Since my condition requires to be "using A,B,C only, i.e without using S", so I guess you mean the following: "Each vector, can be written as a unique linear combination of the other vectors". Thanks. HOTmag (talk) 15:52, 22 September 2024 (UTC)[reply]

Assuming the vectors are

v_{1},v_{2},\dots ,v_{k}\in R^{n}

, form the matrix

V

whose columns are

v_{1},\dots ,v_{k}

. The stated condition is then:

the matrix of $k\times k$ minors of $V$ is zero, and
each of the columns of the matrix of $(k-1)\times (k-1)$ minors of V has a non-zero entry.

- Tito Omburo (talk) 17:36, 22 September 2024 (UTC)[reply]

Your description, both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:41, 23 September 2024 (UTC)[reply]

You can say, "each of the sets {A,B}, {A,C} and {B,C} is linearly independent". --Lambiam 21:05, 22 September 2024 (UTC)[reply]

You also have to add that the set {A,B,C} is linearly dependent, but then the description - both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:40, 23 September 2024 (UTC)[reply]

There is a unique linear combination generating 0, and in this linear combination all coefficients are nonzero.2404:2000:2000:8:FDE8:8311:95E3:654D (talk) 00:00, 23 September 2024 (UTC)[reply]

Yes, and by symbolic notation the description even becomes shorter. Thanks. HOTmag (talk) 08:42, 23 September 2024 (UTC)[reply]

September 23

Converting map scale into height and width

How can I calculate the height and width in meters of a future geographic map of the world based on its desired scale in cm and km? 212.180.235.46 (talk) 15:35, 23 September 2024 (UTC)[reply]

For a map of the world, you will have to specify which map projection it will use, and then to which part(s) of the projection the scale is to apply (and for some projections, how much of the globe to include). Mapping a spherical surface onto a plane surface cannot be done with a constant scale factor. Eg: you might specify a Mercator projection, between latitudes +/-85deg, with a scale of 1:10,000,000 at the equator, which would give you a map about 4m wide along the equator, and (judging by the illustration for Mercator) slightly more than that high. Other projections will vary considerably. -- Verbarson ^talk_edits 18:07, 23 September 2024 (UTC)[reply]

My own map with van der Grinten projection and equatorial scale of 1:20 000 000 for example is 1.18 m high and 1.94 m wide. Out of curiosity, I started to think about dimensions of imaginative humongous world maps. With that as benchmark, if my calculations are correct, a 1:160 000 scale world map, for example, would be 147 m long and 242 m wide. 212.180.235.46 (talk) 20:35, 23 September 2024 (UTC)[reply]

The scale of a map is the ratio between the distance between two points on the map and that between the corresponding points on the ground. Any map projection necessarily distorts the distances, so the scale you get if you take New York and San Francisco as the two points is generally substantially different from what you get if you take Berlin and Moscow. For maps of the whole Earth it is non-trivial to decide on a "nominal scale". In spite of the name, the van der Grinten projection is not a geometric projection in which points on the map are found by following rays projected along a straight line out of a shrunk globe. The equatorial and central-meridional scales are well defined, though. The length ℓ_eq of the equator is approximately 40,000 km. If the equatorial scale is chosen to be 1 :

S

_eq, the width of the map is ℓ_eq /

S

_eq. So for

S

_eq = 160,000 this comes out as about 250 m. The length ℓ_mer of a meridian, measured from one pole to the other, is about 40,000 km. Letting

S

_cm stand for the central-meridional scale factor, the height of the map will be ℓ_mer /

S

_cm. If these scale factors are chosen to be equal, the map will have a height of half its width. For a circular map as shown in the article,

S

_cm = 2

S

_eq. --Lambiam 10:14, 24 September 2024 (UTC)[reply]

September 26

Given the optimal ate pairing e(A,B)=y is to possible to determine I and J such as e(I,J)=2y or even e(I,J)=3y?

Simple question : let’s say I have a pairing friendly curve having a very large trace, and that I have a pairing with points A∈ $G_{1}$ and B∈ $G_{2}$ such as the optimal ate pairing e(A,B)=y, then is it possible to fully determine 2 point I and J such as e(I,J) equals a target multiple of the finite’s field element y ?

Or does it requires full pairing or full generalized Miller’s inversion and thus would be impossible in practice on a curve like bn254 ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 15:51, 26 September 2024 (UTC)[reply]

September 27

Uses of Pascal's tetrahedron

The following is a use of Pascal's triangle:

To find how many ways there are to make a total of $n$ circles all black or red, the formula is just $2^{n}$ . For example, there are $2^{6}=64$ ways to make a group of 6 circles, all black or red, classified by whether each circle is black or red. An example is black-red-black-red-black-red.

But how about finding the number of ways to make a group of 6 circles, all black or red, classified by how many are black or red. To find out how many ways there are to make a total of $n$ circles all black or red classified by how many are black and how many are red, you use the n+1th row of Pascal's triangle. For $n=6$ , this means we use the seventh row, which is $1,6,15,20,15,6,1$ . This means that there is one way to color 6 circles where all of them are black, 6 where 5 are black and one is red, 15 where 4 are black and 2 are red, 20 where 3 are black and 3 are red, 15 where 2 are black and 4 are red, 6 where one is black and 5 are red, and one where all 6 are red.

How about a similar application for Pascal's tetrahedron?? Here is the seventh layer of the tetrahedron:

1 6 6 15 30 15 20 60 60 20 15 60 90 60 15 6 30 60 60 30 6 1 6 15 20 15 6 1

Just as the seventh row of Pascal's triangle can be used for the classification of ways to make 6 circles all of which are black or red classified by how many are black and how many are red, it is likewise true that the seventh layer of Pascal's tetrahedron can be used for... Georgia guy (talk) 14:06, 27 September 2024 (UTC)[reply]

It counts how many ways there are to make 6 circles all of which are black, red, or green, classified by how many are black, how many are red, and how many are green. For example, if you want there to be 2 of each colour, you get

{6 \choose 2,2,2}={6 \choose 2}{4 \choose 2}{2 \choose 2}=90

ways, which is exactly the middle entry of this layer. Double sharp (talk) 14:55, 27 September 2024 (UTC)[reply]

I would call 1, 6, 15, ... the sixth row, and the top row, with a single 1, the zeroth row. That's what Wikipedia's does in the articles I've seen at least. Anyway, this should be in the article Pascal's pyramid, and can be further extended to Pascal's simplex. The article on the Multinomial theorem is relevant here as well. --RDBury (talk) 17:25, 27 September 2024 (UTC)[reply]

RDBury, is the top row "1" not really a row?? Georgia guy (talk) 21:02, 27 September 2024 (UTC)[reply]

Well, you can call it whatever you want and index it as you like; I'm not going to get into the philosophy of rows. But the formulas are simpler if you start with the top "1" as row 0. --RDBury (talk) 00:55, 28 September 2024 (UTC)[reply]

RDBury, does that statement parallel the statement that trigonometry is simpler if you use radians as opposed to degrees?? Georgia guy (talk) 01:04, 28 September 2024 (UTC)[reply]

The OP was calling what most people call row 6 "the seventh row" and I thought that was worth pointing out for future reference. Getting caught up in what counts as a "row" and whether statements are parallel is a matter for philosophy and linguistics. --RDBury (talk) 02:20, 28 September 2024 (UTC)[reply]

September 28

Bitcoin price rigging

I have been day trading bitcoin for almost a year now. I read articles from bitcoin news to try and guess when the dips and dives will happen and so far it has helped. I just read and article about bitcoin possibly losing freedom to governments and large institutions being able to rig the price. This has happened recently with the German government of Saxony selling off a hoard of seized coin, The Mt. Gox dispersal, and the US dept of Justice mysteriously moving 30/230k of its seized hoard two days after Trumps bitcoin speech in Nashville raised the price. https://mpost.io/u-s-unloads-2-billion-in-bitcoin-from-silk-road-seizure/

Is it possible for a whale (large bitcoin hodler) to make smaller transactions in a short time to move the price for a larger transaction at a later time? On the upswing this is called 'pump and dump' and 'poop and scoop' on the down swing. Both are illegal in most market trades.

This post may fit better in Humanities where finance is listed or IT where crypto may belong. 2604:3D08:5E7A:6A00:D94:3638:168B:18A0 (talk) 08:38, 28 September 2024 (UTC)[reply]

I'm pretty sure that this isn't a math question. But the article on Pump and dump does explicitly mention cryptocurrencies. --RDBury (talk) 13:22, 28 September 2024 (UTC)[reply]

The reason I posted in math is because I am wondering if a small buy or sell in a quick time frame will actually move the price enough for traders like me who watch the minute candle scale. I use MetaTrader 4 where the price moves in microseconds every time a buy/sell happens. Is there a formula for volume needed to move the price in a small time frame or article about time/volume/price/ ratio calculations?

Pump and dump, Bear raid, Short and distort, and Uptick rule, all help to explain how it is possible for whales to control the price. Fear_of_missing_out#Investing is the main cause of up-spikes since I have been investing and most are followed by dives. https://www.weforum.org/agenda/2024/08/explainer-carry-trades-and-how-they-impact-global-markets/ The carry trade crash the 1st week of August caused another huge dive. Foreign_exchange_market#Carry_trade 2604:3D08:5E7A:6A00:D94:3638:168B:18A0 (talk) 17:48, 28 September 2024 (UTC)[reply]

The time needed to move the needle with high-frequency trading is a combination of the latency of the connection between the computers of the flash traders and those of the exchanges, which depends on the current state of the networking technology and the physical distance between the traders and the exchange, as well as any regulations enforcing a time lag. It then will take time before small-time traders can see the needle having moved. A mathematical model will require too many parameters to be of practical use. --Lambiam 08:51, 29 September 2024 (UTC)[reply]

September 29

Is this a possible Groth16/ZkSnark verifier‑side simplification ?

Hello,

The verification algorithm is already simple but I was thinking about some costly environments like blockchains having low block limits. This might be naïve thinking but I was wondering at possibility : normally the prover gives 3 elliptic curves points to the verifier A ; B ; C When public inputs are used C/the inputs vector is split.

But as a simplification part, why not completely ditch the C parts of the proof when public inputs are used ? That way, the verifier would have to compute 1 pairing in less for verifying the proof. I’m meaning e(C,verifying_key_part). It seems to me the requirement to pair with public inputs would still ensure the security of the system… Is it because skipping that pairing would allow to forge public inputs ? As far I understand, a malicious prover would still have to satisfy all constraints of the quadratic arithmetic program and thus would have to use public inputs satisfying constraints. Or is it because it would be impossible to rework the protocol to have the prover being able to produce proofs that verify ?

Or even maybe both of the assumptions above ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 11:51, 29 September 2024 (UTC)[reply]

September 30

How to find max(x^y) for each r where x + y = r and x, y are positive integers

Drawing fractal canopy diagrams had me wondering for a given SVG file size, I could add more branches or more depth. Below are two examples:

To make the tree as dense as possible, I wish to maximise the number of leaf nodes. In other words, how can I find max(x^y) for each r where x + y = r and x, y are positive integers?

I realise I can use calculus but I'm unsure what equation I should differentiate.

Thanks, cmɢʟee⎆τaʟκ 00:50, 30 September 2024 (UTC)[reply]

Unfortunately, this is actually quite complicated. We can rewrite the problem as

\max(x^{r-x})

, and consider all real

x

instead of just integers. Notice that

\ln(x^{r-x})=(r-x)\ln(x)

. Because

\ln

is monotonic over

x

,

x^{r-x}

is maximized when

(r-x)\ln(x)

is. Its derivative is

-\ln(x)+(r-x)/x

, which is

0

when

r=x+x\ln(x)

. This is a nice closed-form expression, but it's for

r

in terms of

x

. Inverting it is complicated, but Wolfram Alpha gives us

x=r/W(er)

where

W

is the Lambert W function. While this is sort of a closed-form expression, it's still unfortunately annoying to work with since

W

is implicitly defined as

W(z)=w\Leftrightarrow z=we^{w}

. In any case,

x=r/W(er)

is irrational for all positive rational

r\neq 1

. The proof of this is annoying so I've put it below, but ultimately what it implies is that when

r>1

, as far as I can tell, the best you can do is either round

x

for convenience, or take floor/ceiling of

x

and compare values to get the max over integers.

Proof that

r/W(er)

is irrational for positive rational

r\neq 1

Suppose $r\in \mathbb {Q} ^{+}\backslash \{1\}$ , and let $w:=W(er)$ .
$W(er)=0\Rightarrow r=0$ and $W(er)=1\Rightarrow r=1$ , so $w\neq 0,1$ as well.
By definition, $er=we^{w}$ . Since $w\neq 0$ , we can rearrange to get $r/w=e^{w-1}$ .
Lindemann's 1882 theorem implies that if $k$ is nonzero rational, then $e^{k}$ is not only irrational, but transcendental as well.
If $w$ is rational, then since $w\neq 1$ , $e^{w-1}$ is transcendental, while $r/w$ is rational, which is contradictory.
$w$ must be irrational and is nonzero, so $r/w$ is irrational.
We conclude that $r\in \mathbb {Q} ^{+}\backslash \{1\}$ implies $r/W(er)$ is irrational.

GalacticShoe (talk) 02:41, 30 September 2024 (UTC)[reply]

Here is a numerical recipe (Newton's method) for solving

r=x+x\ln x

for real-valued

x

:

Set $x=9$
Iterate the replacement $x=x^{*}$ until convergence, where
$x^{*}={\frac {x+r}{2+\ln x}}.$

In practice (

r<90

), two iterations will bring you close enough; then test

\lfloor x\rfloor

and

\lceil x\rceil

to get the optimal integer value. --Lambiam 10:17, 30 September 2024 (UTC)[reply]

Thanks so much, @GalacticShoe and @Lambiam. I thought analytically I was heading into a dead end. Guess solving it iteratively is still the best for small values. cmɢʟee⎆τaʟκ 11:45, 30 September 2024 (UTC)[reply]

P.S. It seems there's an interesting trend:

x=1 for r=2	(1 term):	1¹.
x=2 for r=3 to 4	(2 terms):	2¹ and 2².
x=3 for r=5 to 7	(3 terms):	3², 3³ and 3⁴.
x=4 for r=8 to 11	(4 terms):	4⁴, 4⁵, 4⁶ and 4⁷.

x changes when r is the x–1th triangular number + 2. Serendipity? cmɢʟee⎆τaʟκ 17:06, 30 September 2024 (UTC)[reply]

Unfortunately, serendipity. The number of

r

for each

x

is the sequence OEIS:A108414. Although it starts with

1,2,3,4

, it quickly levels out. The inverse triangular number function you're looking for is

\lceil ({\sqrt {8r-7}}-1)/2\rceil \approx {\sqrt {2r}}

, while

x=r/W(er)

grows faster than

r/\ln(r)

, which in turn grows faster than

{\sqrt {2r}}

, hence the number of terms slowing down in growth. GalacticShoe (talk) 03:46, 1 October 2024 (UTC)[reply]

Note that, unlike for T_n, these first differences in the r-values for which x changes do not only always increase but may even decrease. --Lambiam 03:58, 1 October 2024 (UTC)[reply]

After some testing, it seems that rounding suffices at least for r < 100, possibly more. To simplify it, applying Newton's method twice, we can get this approximation which maximizes

x^{y}

for these values of

x+y=r

:

x=round\left({\frac {9+5.197225r}{2.373852+4.197225\ln(9+r)}}\right)

GalacticShoe (talk) 04:14, 1 October 2024 (UTC)[reply]

Thanks so much, @Lambiam and @GalacticShoe. I much appreciate the time and effort you've put into it. Cheers, cmɢʟee⎆τaʟκ 20:17, 2 October 2024 (UTC)[reply]

October 3

Prime Pythagorean triples

Are there infinitely many Pythagorean triples where both of a and c are prime?? (Using the rule that a is always the short leg and b is the long leg, I conjecture [please disprove me if possible] that there are no Pythagorean triples where b is prime.)

Properties:

c-b is always 1.
Except in the case of 3,4,5, and 5,12,13 b is always a multiple of 60.

Georgia guy (talk) 23:38, 3 October 2024 (UTC)[reply]

The formula for Pythagorean triples, assuming the numbers are relatively prime, is p=x²-y², q=2xy, r=x²+y², where x>y>0, x and y are relatively prime, and x and y are not both odd. Your a and b are p and q, possibly in a different order, and your c is r. Suppose b is prime. Then b cannot be q since q is even, so b is p and p > q. Further, x-y is a factor of p, and since p is prime, x-y = 1. So we get p = 2y+1, q=2y(y+1). Then p>q implies 2y²-1 < 0, which is impossible for y>0. As a further result of this proof, the only Pythagorean triples with a prime side are of the from 2y+1, 2y(y+1), 2y²+2y+1, which includes the 3, 4, 5 and 5, 12, 13 examples. It also generates at least a few more: 11, 60, 61; 19, 180, 181; 29, 420, 421. I think you're right about b being a multiple of 60; I haven't written out a proof, but I don't expect it would be that difficult. I suspect there are infinitely many triples where a and c are prime. This amounts to saying there are infinitely many primes p so that (p²+1)/2 is also prime. Statements like this are usually difficult to prove though. For example it's unknown if there are infinitely many primes p such that p+2 is prime. Unless there is a congruence or set of congruences, or a polynomial factorization which can prove it easily then the likelihood is that it's extremely difficult; there's seldom a middle ground. --RDBury (talk) 04:23, 4 October 2024 (UTC)[reply]

For al such triples with 5 < p < 100000, q ≡ 0 (mod 60). --Lambiam 07:55, 4 October 2024 (UTC)[reply]

It is easily seen that (q mod 60) = 0 iff (y mod 15) ∈ {0, 5, 9, 14}. In each of the 11 other cases, by elementary modular arithmetic, at least one of p and r is divisible by at least one of 3 and 5. This includes the two triples (3, 4, 5) and (5, 12, 13). --Lambiam 08:24, 4 October 2024 (UTC)[reply]

A somewhat more interesting issue is the divisibility properties of p, q and r when they aren't restricted to primes. That q is divisible by 4 is trivial. Either p or q is divisible by 3 (but not both) is seen by considering the 9 pairs of remainders of x and y mod 3. The x%3=y%3=0 is eliminated because x and y are assumed relatively prime. With a similar case breakdown, either p, q or r is divisible by 5. So if neither p or r are divisible by 3 or 5, then q is divisible by 60. This idea breaks down for divisibility by 7, but one can say that r is never divisible by 7, nor by any other prime s with s%4=3. --RDBury (talk) 13:31, 4 October 2024 (UTC)[reply]

For reference, the OEIS sequence of hypotenuses for such triples is OEIS:A067756. GalacticShoe (talk) 13:57, 4 October 2024 (UTC)[reply]

October 5