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

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

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

Is there any smooth function with the following two properties:

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

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

Functions like ${\displaystyle a^{x}}$  (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 ${\displaystyle 1+\sum _{k=1}^{\infty }{\frac {1}{k!}}x^{k}}$ , so e.g. ${\displaystyle f(x)=1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}}$  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

${\displaystyle 1+\sum _{k=1}^{\infty }{\frac {1}{k}}{\frac {1}{k!}}x^{k}>1+\sum _{k=1}^{\infty }{\frac {1}{k{+}1}}{\frac {1}{k!}}x^{k}=}$  ${\displaystyle 1+\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k}=}$  ${\displaystyle 1+{\frac {1}{x}}\sum _{k=1}^{\infty }{\frac {1}{(k{+}1)!}}x^{k{+}1}=}$  ${\displaystyle 1+{\frac {1}{x}}\sum _{k=2}^{\infty }{\frac {1}{k}}x^{k}={\frac {\exp x-1}{x}}=}$  ${\displaystyle \Omega (2^{x}).}$   --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

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

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

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

Numerical evidence suggests that ${\displaystyle \log f(x)\sim 2{\sqrt {x}}.}$  One might therefore hope that ${\displaystyle e^{2{\sqrt {x}}}}$  would also work. However, its second derivative is negative for ${\displaystyle x<{\tfrac {1}{4}}.}$   --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 ${\displaystyle x<6.17}$ , for instance. 120.21.79.62 (talk) 06:48, 21 July 2024 (UTC)Reply

The fourteenth derivative is negative for ${\displaystyle x<20}$ . 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 ${\displaystyle e^{2{\sqrt {x+c}}}}$  won't help.  --Lambiam 11:46, 21 July 2024 (UTC)Reply

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: ${\displaystyle 7,13,14,17,19,21,25,26,28,34,35,38,\ldots }$ . Obviously if a number appears in the list then all its positive multiples do too. The list of nontrivial numbers starts ${\displaystyle 7,13,17,19,25,59,67,73,79,83,94,97,\ldots }$  GalacticShoe (talk) 03:01, 22 July 2024 (UTC)Reply

(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 ${\displaystyle w_{1}+w_{2}+w_{3}=1}$  (if not, then just define new values ${\displaystyle {\frac {w_{1}}{w_{1}+w_{2}+w_{3}}},{\frac {w_{2}}{w_{1}+w_{2}+w_{3}}},{\frac {w_{3}}{w_{1}+w_{2}+w_{3}}}}$  which do add to ${\displaystyle 1}$ .) In other words, you can just take ${\displaystyle (w_{1}r_{1}+w_{2}r_{2}+w_{3}r_{3},w_{1}g_{1}+w_{2}g_{2}+w_{3}g_{3},w_{1}b_{1}+w_{2}b_{2}+w_{3}b_{3})}$  (or, more concisely, ${\displaystyle w_{1}(r_{1},g_{1},b_{1})+w_{2}(r_{2},g_{2},b_{2})+w_{3}(r_{3},g_{3},b_{3})}$ .) 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

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

Where ${\displaystyle b'={\frac {a^{2}}{b}}}$  (polar radius of curvature {RoC}) and ${\displaystyle N(\phi )}$  is the normal RoC, what is ${\displaystyle {\frac {b'}{N(\phi )}}}$  (especially the vertex latitude, ${\displaystyle {\frac {b'}{N(\phi _{v})}}}$ )? Since it involves ${\displaystyle {\frac {1}{N(\phi )}}}$  it would suggest a type of curvature, rather than RoC—?
Likewise with geocentric latitude, ${\displaystyle \psi }$ , there is ${\displaystyle {\frac {b}{R(\psi )}}}$  (with ${\displaystyle R(\psi )}$  being the geocentric radius).
This is based on ${\displaystyle N(\phi )\cos(\phi )=R(\psi )\cos(\psi )}$  (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 ${\displaystyle {\frac {b'}{N(\phi )}}={\frac {a^{2}}{bN(\phi )}}}$  can be rewritten as ${\displaystyle {\sqrt {\frac {1-e^{2}\sin ^{2}\phi }{1-e^{2}}}},}$  where ${\displaystyle e={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}}$  is the eccentricity of the ellipsoid. Eccentricity is a dimensionless quantity, also when the radii ${\displaystyle a}$  and ${\displaystyle b}$  are expressed using units of length such as kilometres or miles, and so is this ratio. At the poles (${\displaystyle \phi =\pm {\frac {\pi }{2}}}$ ) it equals ${\displaystyle 1}$  and at the equator (${\displaystyle \phi =0}$ ) it equals ${\displaystyle {\frac {b}{a}}.}$  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 ${\displaystyle N}$  and ${\displaystyle R}$  being different prime vertical RoC (in the case of the foundational, parametric ${\displaystyle \beta }$  latitude, the prime vertical RoC is just ${\displaystyle a}$ )? -- 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 ${\displaystyle \phi .}$   --Lambiam 08:17, 24 July 2024 (UTC)Reply

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

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

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