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.

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)[reply]

July 15

Next Julian period

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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.^[1] --Lambiam 14:04, 18 July 2024 (UTC)[reply]

References

^ Hellmuth Kneser (1950). "Reelle analytische Lösungen der Gleichung $\varphi (\varphi (x))=e^{x}$ und verwandter Funktionalgleichungen". Journal für die reine und angewandte Mathematik. 187: 56–67.

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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 4; for example,

0=.000000..._{4},

{\tfrac {3}{7}}=.123123..._{4},

and

1=.333333..._{4},

This expansion is not unique:

.033333..._{4}=.100000..._{4}={\tfrac {1}{4}}.

Viewing these quaternary expansions as infinite sequences

{(a_{n})}_{n=1}^{\infty }\in 4^{\omega }=\mathbb {N} ^{+}{\to }\{0{,}1{,}2{,}3\},

the function

r:4^{\omega }\to I

for interpreting a quaternary expansion as a real number is unique, 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 for

r^{{-}1}(1)=333333...,

which is the only possible expansion of

1,

avoid sequences ending in an infinite stream of

3

s.

However, using

r,

we can define a bicontinuous pairing function

\pi :4^{\omega }{\times }4^{\omega }\to 4^{\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 quaternary digits and produces a single stream of quaternary digits. It takes two digits at a time, one from each of the two input streams, and outputs two digits on the output stream. Using binary expansions, in which one quaternary digit equals two bits, we can say that for each pair of two bits it produces four bits.

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 11:11, 26 July 2024 (UTC)[reply]

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)[reply]

[1] Hellmuth Kneser (1950). "Reelle analytische Lösungen der Gleichung $\varphi (\varphi (x))=e^{x}$ und verwandter Funktionalgleichungen". Journal für die reine und angewandte Mathematik. 187: 56–67.

[1]