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July 13

Given the results from powers of tau in the trusted setup ceremony ; the verifying and the proving key, how can I find the point [f] resulting from the trusted setup in Groth16 ?

Moved to the Computing section of the Reference desk —  --Lambiam 13:36, 16 July 2024 (UTC)

July 15

Next Julian period

Moved to the Science section of the Reference desk —  --Lambiam 13:24, 16 July 2024 (UTC).

July 16

In SageMath, how to use GF() on a very large finite field ?

Moved to the Computing section of the Reference desk —  --Lambiam 13:37, 16 July 2024 (UTC)

July 18

Functions whose every derivative is positive growing slower than exponential

Is there any smooth function with the following two properties:

$f^{(n)}(x)>0$, i.e. the nth derivative of f is strictly positive for every x and n.

$\lim _{x\rightarrow \infty }{\frac {f(x)}{b^{x}}}=0$ for every b > 1. The hard case is when b is small.

Functions like $a^{x}$ (for a > 1) are the only ones I can think of with the first property, but none of them has the second property because you can always choose b < a. So I am asking whether there is any function with the first property that grows slower than exponential.

120.21.218.123 (talk) 10:09, 18 July 2024 (UTC)

Wouldn't any power series with positive coefficients that decrease compared to the coefficients of the exponential do? The exponential is $1+\sum _{k=1}^{\infty }{\frac {1}{k!}}x^{k}$, so e.g. $f(x)=1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}$ should do the trick. The next question is whether you can find a closed-form expression for this or a similar power series. --Wrongfilter (talk) 13:02, 18 July 2024 (UTC)

Good thinking. It is of course the case that the first property holds for any power series where all coefficients are positive. Plotting on a graph, I think your specific example doesn't satisfy the second property, but others where the coefficients decrease more rapidly do. 120.21.218.123 (talk) 13:26, 18 July 2024 (UTC)

$1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}>1+\sum _{k=1}^{\infty }{\frac {1}{k{+}1}}{\frac {1}{k!}}x^{k}=$ $1+\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k}=$ $1+{\frac {1}{x}}\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k{+}1}=$ $1+{\frac {1}{x}}\sum _{k=2}^{\infty }{\frac {1}{k}}x^{k}={\frac {\exp x-1}{x}}=$ $\Omega (2^{x}).$  --Lambiam 13:45, 18 July 2024 (UTC)

A half-exponential function will satisfy your requirements. Hellmuth Kneser famously defined an analytic function that is the functional square root of the exponential function.  --Lambiam 14:04, 18 July 2024 (UTC)

References

A variant of Wrongfilter's idea that I think does work:

$f(x)=\sum _{k=0}^{\infty }{\frac {x^{k}}{(k!)^{2}}}$

(taking $0^{0}$ to be $1$).

Numerical evidence suggests that $\log f(x)\sim 2{\sqrt {x}}.$ One might therefore hope that $e^{2{\sqrt {x}}}$ would also work. However, its second derivative is negative for $x<{\tfrac {1}{4}}.$  --Lambiam 22:20, 20 July 2024 (UTC)

And some higher derivatives are negative for even larger values of x. The eighth derivative is negative for $x<6.17$, for instance. 120.21.79.62 (talk) 06:48, 21 July 2024 (UTC)

The fourteenth derivative is negative for $x<20$. That's as high as WolframAlpha will let me go. 120.21.79.62 (talk) 06:54, 21 July 2024 (UTC)

Yes, shifting the graph along the x-axis by using $e^{2{\sqrt {x+c}}}$ won't help.  --Lambiam 11:46, 21 July 2024 (UTC)

July 21

If the sum of the first m factorial numbers is equal to the sum of the first n positive integers

If the sum of the first m factorial numbers is equal to the sum of the first n positive integers, i.e. 1! + 2! + 3! + … + m! = 1 + 2 + 3 + … + n, then (m,n) = (0,0), (1,1), (2,2), (5,17), right? 220.132.216.52 (talk) 20:30, 21 July 2024 (UTC)

The triangular numbers, modulo 19, are reduced to one of 10 possibilities: 0, 1, 2, 3, 6, 7, 9, 10, 15 and 17. The sum 1! + 2! + 3! + ... + m!, for m > 17, modulo 19, is reduced to 8. Therefore no further factorial sums are triangular.  --Lambiam 22:19, 21 July 2024 (UTC)

I looked at mod 7 with about the same result. The left hand side is 5 for m≥6 and the right hand side can never be 5. So you only have to check m from 1 to 5. (Btw, I would count 0! = 1 as a factorial number, so the sums of factorials would be 1, 2, 4, 10, 34, ... . (sequence A003422 in the OEIS)) --RDBury (talk) 22:28, 21 July 2024 (UTC)

Wrote up a quick MATLAB script to find numbers which can be used as modulos to show that the list is finite, it starts: $7,13,14,17,19,21,25,26,28,34,35,38,\ldots$. Obviously if a number appears in the list then all its positive multiples do too. The list of nontrivial numbers starts $7,13,17,19,25,59,67,73,79,83,94,97,\ldots$ GalacticShoe (talk) 03:01, 22 July 2024 (UTC)

Submit to OEIS! —Tamfang (talk) 20:31, 28 July 2024 (UTC)

July 22

Computing the centre of a triangle...

(On math section as it's a geometric/trig problem essentially)

In CSS, color-mix() takes 2 Params.. However I have colors to mix that take 3 or more params.

Whilst with 2 params you can do a simple linear interpolation, based on the weights of the 2 params, I wasn't sure how it could be done for 3.

One approach I had considered was (at least for an RGB blend) is to compute the centrepoint of a triangle in 3D space, where the 3 points of the triangle are the three colors. However that would assume equal weights of each color, I figured

So for a given "triangle" defined by (r1,g1,b1),(r2,g2,g2), (r3,b3,g3) and a mix ratio of w1:w2:w3  compute the centroid(?) of the triangle representing the blended color. ?

Alternatively is there a different math/geometrical technique that is used in actual computer graphics work?
ShakespeareFan00 (talk) 17:18, 22 July 2024 (UTC)

I don't have an actual computer graphics answer, but the interpolation method still works for three points, simply take the weighted sum of points assuming $w_{1}+w_{2}+w_{3}=1$ (if not, then just define new values ${\frac {w_{1}}{w_{1}+w_{2}+w_{3}}},{\frac {w_{2}}{w_{1}+w_{2}+w_{3}}},{\frac {w_{3}}{w_{1}+w_{2}+w_{3}}}$ which do add to $1$.) In other words, you can just take $(w_{1}r_{1}+w_{2}r_{2}+w_{3}r_{3},w_{1}g_{1}+w_{2}g_{2}+w_{3}g_{3},w_{1}b_{1}+w_{2}b_{2}+w_{3}b_{3})$ (or, more concisely, $w_{1}(r_{1},g_{1},b_{1})+w_{2}(r_{2},g_{2},b_{2})+w_{3}(r_{3},g_{3},b_{3})$.) GalacticShoe (talk) 17:45, 22 July 2024 (UTC)

Thanks. I thought I was thinking along the right lines..
In case you are wondering why I asked -s:Page:The_color_printer_(1892).djvu/55 ShakespeareFan00 (talk) 17:52, 22 July 2024 (UTC)

The weighted average makes sense for additive colour mixing, but the colour resulting from pigment mixing is not so easily determined. For example, the colours   and   are complementary. Their sum in RGB colour models is   and their average is  , as grey as it gets. However, mixing red and green paint gives more of a brown colour.^[1] A colour model that is more likely close to that of The Color Printer is the RYB colour model. If the pigments are fully opaque, the subtractive model is adequate, but generally pigments are not fully opaque.  --Lambiam 21:22, 22 July 2024 (UTC)

This is something I actually did! Back in the '80s, I was consulting for a company called ImageSet. One of our specialties was tools to allow people to take the equivalent of screenshots on MSDOS (using a TSR program), and then convert them to matching colors to be printed. We used trilinear interpolation to achieve the desired colors, if I remember right. It was a ridiculous amount of precision, especially considering that every CRT user adjusted the brightness etc. to their own favored wrong colors. --jpgordon^{𝄢𝄆𝄐𝄇} 17:06, 27 July 2024 (UTC)

July 23

Abel's reciprocity relation

Do we have an article that covers Abel's reciprocity relation for differentials of the third kind? Tito Omburo (talk) 18:37, 23 July 2024 (UTC)

Google Books and Google Scholar searches for "Abel's reciprocity relation" do not yield any results,^[2][3] so is this perhaps better known under a different name?  --Lambiam 21:22, 23 July 2024 (UTC)

I've seen it called the "first reciprocity law", but that's a bit non-specific. Tito Omburo (talk) 22:31, 23 July 2024 (UTC)

In this book I find a treatment of "The Reciprocity Theorem" for "Abelian differentials of the second and third kind", and in this one one of what is called "the reciprocity law for differentials of the first and third kinds". I'm only vaguely familiar with the first principles of differential geometry; the limited preview afforded by Google Books does not allow me to understand the notation and see how these two theorems, which superficially look rather different, are related. Both are apparently related to "Abel's theorem" on sums of integrals of certain types of function. We appear to treat only a very simple application of the latter theorem under the name "Abel's identity", so I am afraid the answer to your question is negative.  --Lambiam 09:29, 24 July 2024 (UTC)

Type of curvature?

Where $b'={\frac {a^{2}}{b}}$ (polar radius of curvature {RoC}) and $N(\phi )$ is the normal RoC, what is ${\frac {b'}{N(\phi )}}$ (especially the vertex latitude, ${\frac {b'}{N(\phi _{v})}}$)? Since it involves ${\frac {1}{N(\phi )}}$ it would suggest a type of curvature, rather than RoC—?
Likewise with geocentric latitude, $\psi$, there is ${\frac {b}{R(\psi )}}$ (with $R(\psi )$ being the geocentric radius).
This is based on $N(\phi )\cos(\phi )=R(\psi )\cos(\psi )$ (so, technically, isn't N the geographic prime vertical RoC and R, besides being the geocentric radius, is also the geocentric prime vertical RoC?). -- Kaimbridge (talk) 18:43, 23 July 2024 (UTC)

The ratio ${\frac {b'}{N(\phi )}}={\frac {a^{2}}{bN(\phi )}}$ can be rewritten as ${\sqrt {\frac {1-e^{2}\sin ^{2}\phi }{1-e^{2}}}},$ where $e={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}$ is the eccentricity of the ellipsoid. Eccentricity is a dimensionless quantity, also when the radii $a$ and $b$ are expressed using units of length such as kilometres or miles, and so is this ratio. At the poles ($\phi =\pm {\frac {\pi }{2}}$) it equals $1$ and at the equator ($\phi =0$) it equals ${\frac {b}{a}}.$ If the radii are dimensioned lengths, then so is curvature; its dimension is then the inverse of length. Therefore there is no plausible way to interpret this dimensionless ratio as a form of curvature.  --Lambiam 21:05, 23 July 2024 (UTC)

But I thought the root definition of curvature is that its radius is its inverse, and anything else, here b', is just a modifier.

What about the idea of $N$ and $R$ being different prime vertical RoC (in the case of the foundational, parametric $\beta$ latitude, the prime vertical RoC is just $a$)? -- Kaimbridge (talk) 03:57, 24 July 2024 (UTC)

I do not understand what you are asking. A geometric interpretation of these quantities? Indeed, the radius of curvature is the inverse of the curvature; my point was that if one is dimensioned, so is the other, while the ratio about which the question appeared to be is inherently dimensionless, so it cannot be some type of curvature. I also do not understand what you mean by "just a modifier". It is the prime-vertical radius of curvature at the pole, which you proceed to divide by that at at geodetic latitude $\phi .$  --Lambiam 08:17, 24 July 2024 (UTC)

By "just a modifier", I mean ${\frac {1}{N(\phi )}}$ prime-vertical curvature, so b' is a modifier of that curvature.

Okay, so what about $R(\psi )$ in this situation? Would $R(\psi )$ in ${\frac {b}{R(\psi )}}$ be considered the geocentric prime vertical RoC?

In terms of purpose/context, ${\frac {b'}{N(\phi )}}$ and ${\frac {b}{R(\psi )}}$ are used (as vertex latitudes $\phi _{v}$ and $\psi _{v}$) in the geographic and geocentric (respectively) calculation of geodetic distance (rather than the usual $\beta$ based, parametric calculation, which has a neutral value, ${\frac {a}{a}}$), all using the same, iteratively found, geodetically adjusted longitude difference ($\Delta \Lambda =\Delta \lambda +Amp$). -- Kaimbridge (talk) 06:54, 25 July 2024 (UTC)

A radius of curvature at a given spot is the radius of an osculating circle. It can be viewed as a vector from the centre of that circle to the given spot. If the direction of this RoC, viewed as a vector, is the vertical direction, it is a vertical RoC. (If, moreover, the plane of the osculating circle is perpendicular to the meridian through the given spot, so it intersects the ellipsoid along the east–west direction, it is the local prime-vertical RoC.) Unless the ellipsoid is a sphere, the geocentric radius at any other spot than the poles or equator, viewed as a vector from the centre to that spot, is not in the vertical direction, so it is not the radius of a locally osculating circle and it is not particularly meaningful to interpret it as either vertical or as a RoC, let alone both.  --Lambiam 10:41, 25 July 2024 (UTC)

July 26

I joined the X and Y axes together. What Wikipedia page already mentions the concept?

I had / have this fairly rather long blabbersome brilliant idea. My question is since there is no way that I, with a rather low IQ, could come up with something "new", then my idea certainly must be merely a rehash of some ideas already mentioned on several Wikipedia pages, in fact probably just a sentence on just one page. But which one(s)? Thanks. Jidanni (talk) 04:56, 26 July 2024 (UTC)

Your page is cumbersome to follow, but if I'm correct in interpreting it, you are essentially proposing the use of a pairing function or Hilbert curve. It is not possible to continuously reduce dimension in this manner (more precisely, 1D and 2D space are not homeomorphic). It would help if you would more rigorously formulate the function you are proposing rather than merely using examples.--Jasper Deng (talk) 06:22, 26 July 2024 (UTC)

Wikipedia defines "pairing function" only for natural numbers, but below I use the term for (not necessarily unique) functions wrapping up two values (not necessarily integers) into a single one of the same type.

Letting $I$ stand for the unit interval $[0,1],$ the Hilbert curve can be described as a function $h:I\to I\times I.$ Input one number $z,$ output $h(z)=(x,y)\!:$ a pair of numbers. This function is surjective, which implies that there exists an inverse pairing function $h^{{-}1}:I\times I\to I.$ Being an inverse means that when $h(z)=(x,y),$ we have $h^{{-}1}(x,y)=z.$ Function $h$ is also continuous, but it is not injective. Many output pairs are reached several times; for example, $h({\tfrac {1}{6}})=h({\tfrac {1}{2}})=h({\tfrac {5}{6}})=({\tfrac {1}{2}},{\tfrac {1}{2}}).$ So the inverse is not unique.

Numbers in $I$ can be written in base 2; for example, $0=.000000..._{2},$ ${\tfrac {3}{7}}=.011011..._{2},$ and $1=.111111..._{2},$ This expansion is not unique: $.0111111..._{2}=.100000..._{2}={\tfrac {1}{2}}.$ We can view these binary expansions as infinite sequences $a={(a_{n})}_{n=1}^{\infty }\in 2^{\omega }=\mathbb {N} ^{+}{\to }\{0{,}1\},$ The function $r:2^{\omega }\to I$ given by $r(a)=\textstyle {\sum _{n=1}^{\infty }}a_{n}2^{{-}n}$ interprets a binary expansion as a real number. This function $r$ is continuous and surjective, just like function $h$ before, so it has an inverse. But, as before, function $r$ is not injective, so the inverse is not unique. However, by convention, it has a "canonical" inverse: other than $r^{{-}1}(1)=111111...,$ the only possible expansion of $1$ in the domain of $r,$ avoid sequences ending in an infinite stream of $1$s.

Now, using $r,$ we can define a bicontinuous pairing function $\pi :2^{\omega }{\times }2^{\omega }\to 2^{\omega }$ such that

$h(r(\pi (u,v)))=(r(u),r(v)).$

This means that we can give a "canonical" definition for $h^{{-}1}$ by using $r$ and its canonical inverse:

$h^{{-}1}(x,y)=r(\pi (r^{{-}1}(x),r^{{-}1}(y))).$

The function $\pi$ can be described in the form of a 4-state finite-state automaton that gobbles up two streams of bits and produces a single stream of bits. It takes two bits at a time, one from each of the two input streams, and outputs two bits on the output stream.

I suspect that the "brilliant idea" is akin to this way of pairing $x$ and $y.$ I expect the idea is well known, but perhaps only as folklore, and I doubt that it is described or even hinted at in Wikipedia mainspace.  --Lambiam, edited 10:51, 28 July 2024 (UTC) (originall 11:11, 26 July 2024 (UTC))

July 27

Data estimation with excessive log functions

In health care, I noticed that many estimation algorithms make extensive use of log functions. For example, the ASCVD 10-year risk estimation from "2013 ACC/AHA Guideline on the Assessment of Cardiovascular Risk" sums up a coefficient times the log of age, a coefficient times the log of total cholesterol, a coefficient times the log of HDL, etc... It is a set of coefficients, each multiplied by the log of an attribute. Is this type of function or algorithm the result of a specific type of data modeling? It looks to me like they took a sample data set and correlated the log of each attribute, one at a time, to the outcome and produced a coefficient that represents how correlated the log of that attribute is in the sample set. But, I'm just guessing and I'd prefer to know how this type of function is actually produced. 75.136.148.8 (talk) 10:54, 27 July 2024 (UTC)

I'm not familiar with how this estimator was devised, but model building is an art, especially in cases where the data is noisy and the causal processes are poorly understood. Social scientists routinely use purely linear regression models, because that is what they were taught as students, it is the default model of R, which many use, and everyone else in their field does this. When a variable (independent or dependent) can only assume positive values, it cannot have a normal distribution. This is an indication that pure linear regression may not be the best approach when devising an estimator. So then it is good practice to use a data transformation that makes the observed distribution more normal. I don't know if this is why they did what they did. Another possibility is that they just computed the correlation coefficients and saw they were higher when using a logarithmic scale.  --Lambiam 11:52, 27 July 2024 (UTC)

Are there other triangular numbers with all digits 6?

6, 66, 666 are all triangular numbers, are there other triangular numbers with all digits 6? 218.187.67.217 (talk) 16:42, 27 July 2024 (UTC)

These correspond to solutions of the Diophantine equation $10^{p}=12k^{2}\pm 3k+1\!:$

$10^{1}=12\cdot 1^{2}-3\cdot 1+1$

$10^{2}=12\cdot 3^{2}-3\cdot 3+1$

$10^{3}=12\cdot 9^{2}+3\cdot 9+1$

For each solution, the number $8k^{2}\pm 2k$ is an all-6 triangular number.

I don't expect any further solutions, but neither do I see an argument exhibiting that they cannot exist. The weaker requirement $12k^{2}\pm 3k+1\equiv 0~~({\text{mod}}~10^{p})$ has four solutions for $k$ for each given value of $p,$ corresponding to the final digits $1,3,7,9.$ For example, for $p=10$ they are $8811378991,8647117103,1352882897,188621009.$ The polynomials in $k$ in the rhs of the Diophantine equation are irreducible. It seems that considerations based on modular arithmetic are not going to give further help.  --Lambiam 19:59, 27 July 2024 (UTC)

The discriminant of the quadratic is $9-48(1-10^{p})=48\cdot 10^{p}-39=3(16\cdot 10^{p}-13)$. This needs to be a perfect square for there to be a solution, so we need $16\cdot 10^{p}-13=3k^{2}$ for some integer k. Since $16\cdot 10^{p}$ will get "closer" to being an even perfect square as p approaches infinity, I heuristically wouldn't expect more than a finite amount of solutions to exist.--Jasper Deng (talk) 03:34, 28 July 2024 (UTC)

This gives yet another way of phrasing the problem. Define the recurrent sequence $(a_{p})_{p=0}^{\infty }$ by:

$a_{0}=1,~a_{p{+}1}=10a_{p}+39.$

It goes like this:

$1,49,529,5329,53329,533329,5333329,...$

The first four values are squares. Will the sequence ever hit another square?  --Lambiam 10:05, 28 July 2024 (UTC)

It turns out that because the discriminant is added or subtracted to 3 and then divided by 2a=24 in the quadratic formula, there are even more stringent restrictions: the numerator has to be divisible by 24, so we must have $3k\cong \pm 3\mod 24$ and thus $k\cong \pm 1\mod 8$. That restriction alone would seem to greatly reduce the amount of candidates (only every other odd perfect square satisfies that).--Jasper Deng (talk) 04:49, 29 July 2024 (UTC)

If the sequence $(a_{p})_{p=0}^{\infty }$ ever hits another square $k^{2},$ its square root $k$ will satisfy this requirement. This can be seen as follows. For $p\geq 2,a_{p}\equiv 17~~({\text{mod}}~64)$ since $a_{2}=529\equiv 17~~({\text{mod}}~64)$ and $a_{p{+}1}=$ $10a_{p}+39\equiv$ $170+39=$ $209\equiv 17~~({\text{mod}}~64).$ The only residue classes for $k$ modulo $64$ that have $k^{2}\equiv 17~~({\text{mod}}~64)$ are $9,23,41,55;$ in all four cases, $k\equiv \pm 1~~({\text{mod}}~8).$  --Lambiam 10:13, 29 July 2024 (UTC)

Right. For any modulus m you can use the recursion to easily compute a_p mod m. It's a bit harder, but still possible to then determine if a_p is a quadratic residue mod m. If it isn't then you can eliminate that a_p as a non-square. Do this for a few thousand prime (or prime power) values of m and you have a sieve which only let's though those a_p's that are square and a vanishingly small number of "false positives". (There are going to be some m where all the values of a_p are quadratic residues, but this won't happen if 10 is a primitive root mod m, and this occurs at a relatively constant rate.) This could be implemented in Python (or whatever) fairly easily to eliminate all the non-square a_p's up to some value, say p≤10000. Keep in mind that a₁₀₀₀₀ would have around 10000 digits, but there's no need for multiprecision arithmetic to carry this out. However, all you would be doing is creating a lower bound on the next highest square a_p, you wouldn't actually be proving there are none. (That's assuming the sieve didn't produce an actual square a_p with p≤10000.) It shouldn't be hard to use a probabilistic argument to show that the "expected" number of squares is finite, but this wouldn't be a proof but rather an indication that it's unlikely that there will be additional squares above a given bound. In any case, I couldn't think of anything that would answer the original question better than a somewhat wishy-washy "probably not". --RDBury (talk) 13:10, 29 July 2024 (UTC)

July 30

Axiom of choice, axiom of countable choice, any others?

We have the Axiom of countable choice, which is weaker than Axiom of choice, so is it possible, meaningful or even already done to have an Axiom of aleph 1 choice or Axiom of cardinality of continuum choice, weaker than the Axiom of choice? Also, could the Axiom of dependent choice actually be such an axiom, corresponding to some particular aleph?Rich (talk) 20:20, 30 July 2024 (UTC)