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.
Euler totient function (sequence A000010 in the
OEIS) has a reduced function:
Carmichael lambda function (sequence A002322 in the
OEIS), and instead of for coprime x, y, while if p is prime and r>=1, and a composite number n is
Carmichael number if and only if divides , and
Dedekind psi function (sequence A001615 in the
OEIS) should also have a reduced function: , and instead of for coprime x, y, while if p is prime and r>=1, and a composite number n is
Lucas-Carmichael number if and only if divides , but I cannot even find the function in OEIS (it should start with (start from n=1) 1, 3, 4, 6, 6, 12, 8, 12, 12, 6, 12, 12, 14, 24, 12, 24, 18, 12, 20, 6, 8, 12, 24, 12, âŠ).
125.230.9.88 (
talk) 02:38, 7 June 2024 (UTC)reply
yet Did you mean to write "if and only if divides "? But for any prime The modified criterion requires to be composite. Â --
Lambiam 06:55, 7 June 2024 (UTC)reply
Is it possible to choose a point strictly inside a simple quadrilateral such that all four sides (and hence the whole interior) are visible from the point? The chosen point is required to be a smooth function of the the vertices, reasonably middling and free from any nastiness (such as depending on the labeling of the vertices). Thanks,
catslash (
talk) 16:58, 7 June 2024 (UTC)reply
All four sides would obviously be visible from any random point inside a quadrilateral that isn't concave. I suspect your question is more complex that that but you haven't expressed it optimally. --
Jack of Oz[pleasantries] 19:25, 7 June 2024 (UTC)reply
If you take each of the four edges and consider the half-planes on the interior sides of each edge, then the intersection of all half-planes is the precise set of points which can "see" all sides. Although I haven't formalized a proof, I'm confident that this set is always nonempty (and also a convex quadrilateral) if the original quadrilateral is non-trivial.
As for a smooth function, I'm sure that there is some definition of the center of a convex quadrilateral that, applied to this set of points, would suffice.
GalacticShoe (
talk) 20:02, 7 June 2024 (UTC)reply
On reflection, since convex quadrilaterals can only take one specific form, with one concave vertex, two "wingtips", and a "head", it's very easy to describe this shape. It's just the quadrilateral formed from the concave vertex, the head, and the two intersections of opposite sides. As for the center, you can use the intersection of diagonals of this internal quadrilateral.
GalacticShoe (
talk) 22:34, 7 June 2024 (UTC)reply
When just one angle of the original quadrilateral is very close to 180°, then so is one angle of the convex inner quadrilateral, at the same vertex, and the intersection of its diagonals is very close to that vertex. In the degenerate case that the angle is equal to 180°, as when a concave quadrilateral is continuously transformed into a convex quadrilateral, the point of intersection coincides with the vertex. Using the
vertex centroid or
area centroid of the convex inner quadrilateral as the chosen point avoids this. Â --
Lambiam 07:01, 8 June 2024 (UTC)reply
In general, for any area A enclosed by a
Jordan curve, the set P(A) of panoptic points is convex. It is the same as A iff A is convex. If the boundary is a polygon, P(A) is empty, a single point, a line segment, or the area enclosed by a polygon. Â --
Lambiam 15:43, 8 June 2024 (UTC)reply
I think I thought â but I was wrong â that the midpoint of the shortest diagonal satisfies the visibility criterion, but it is not a continuous function of the locations of the vertices. I conjecture that there is a continuous function w from these locations, ordered cyclically, to the unit interval, where w(v2, v3, v4, v1) = 1 â w(v1, v2, v3, v4), invariant under similarity transformations, such that the weighted mean of the diagonal midpoints (v1 + v3) / 2 and (v2 + v4) / 2, with weights w and 1 â w, satisfies both requirements. For any configuration, there is an interval [wlo, whi] of w-values that guarantee visibility. If we can prove that both endpoints vary continuously with the vertices, a solution is provided by w = (wlo + whi) / 2. Â --
Lambiam 21:15, 7 June 2024 (UTC)reply
The midpoint of the shortest diagonal does not work. Consider (0, 0), (1, 4), (0, 3), (-1, 4) taken in that order. The diagonal (1, 4), (-1, 4) is shorter than (0, 0), (0, 3), but its midpoint lies outside the quadrilateral. Of course if both diagonals lie inside the quadrilateral then it's convex and any interior point will do. --
RDBury (
talk) 02:06, 8 June 2024 (UTC)reply
The vertex centroid of the (convex) panoptic quadrilateral seems to do the job. Many thanks. If there is a practical way of choosing a point on the
Newton line of the original quadrilateral, that would also be very interesting.
catslash (
talk) 12:57, 8 June 2024 (UTC)reply
At least one of the midpoints of the two diagonals is contained in the panoptic quad, so the panoptic part of the Newton line is nonempty. It should be easy to determine the endpoints of the (possibly degenerate) panoptic segment of this line and take its midpoint. If the original quadrilateral is convex, this is a trivial exercise. Otherwise, you have to determine where the extended sides adjacent to the concave vertex cut the segment connecting the midpoints short. As with many problems in
computational geometry, the hardest part is not the computations themselves but to get the case distinctions correct and complete.
I think though the result will in extreme cases be markedly less "middling" than the vertex centroid. Â --
Lambiam 15:26, 8 June 2024 (UTC)reply
June 9 Information
Source for Langmuir-Blodgett, Langmuir-Boguslavski, and weird Rayleigh equations?
While cleaning up
List of nonlinear ordinary differential equations and citing all the ones listed, there were three that puzzled me to no end. The first was listed as the Langmuir-Blodgett equation:
The next was listed as the Langmuir-Boguslavski equation:
Finally, there was an equation listed as the Rayleigh equation:
I just want to know if anyone recognizes these or has sources for them. The first two I could only find mention of in a footnote of an old edition of a differential equations handbook, which itself cited no sources for these and they do not appear in the more recent edition of the handbook as far as I can tell, and the last one looks neither like the regular
Rayleigh equation (which is notably linear) or the variant of the
Van der Pol equation which is sometimes called the Rayleigh equation (and both of these drown out any search results for this equation). The two equations named after Langmuir I also checked in plasma physics textbooks for, as I vaguely recall that Langmuir worked on plasma, but I could not find mention in the two introductory books I checked. The closest I could get were sources like this one[1] but I can't seem to tell if the given equation is equivalent, and the source they cite is O. V. Kozlov, An Electrical Probe in a Plasma, which I cannot find online. (There's also
Langmuir-Blodgett film but no differential equation is mentioned in that article.) These have been plaguing me, and the editor who added them hasn't edited in six years so no dice there. Any help would be appreciated!
Nerd1a4i (they/them) (
talk) 19:57, 9 June 2024 (UTC)reply
Differential equations with the names Langmuir-Blodgett and Langmuir-Boguslavski are given
here, without further explanation of reference. The origin of the former is possibly an equation presented in a joint publication by Langmuir and Blodgett many years before the technique was developed to make Langmuir-Blodgett films. Â --
Lambiam 06:31, 10 June 2024 (UTC)reply
Yes, that was the handbook that was the one other source I saw - the other edition of the same handbook I was referencing didn't mention these.
Nerd1a4i (they/them) (
talk) 01:46, 11 June 2024 (UTC)reply
That was me adding it to the list of citogenesis incidents as that was what I believed to be true at the time. I then realized I should ask here.
Nerd1a4i (they/them) (
talk) 01:45, 11 June 2024 (UTC)reply
The Langmuir-Blodgett equation may come from
this paper or the "previous papers" cited in footnote 1. Â --
Lambiam 11:09, 10 June 2024 (UTC)reply
Thanks, I'll try to go through that paper and see if it's got the right equation. Much appreciated for finding a fresh starting point! I don't suppose you have any leads for the other two?
Nerd1a4i (they/them) (
talk) 01:47, 11 June 2024 (UTC)reply
Here is an unresolved lead. In
doi:
10.1063/1.4948923 the authors refer to "the Boguslavsky-Langmuir equation for a cylindrical probe under floating potential", which is not a differential equation but is called a â3/2 powerâ law â it has a factor In a second note,
doi:
10.1063/1.4960396Â , the same authors call this a law that "for cylindrical probe under floating potential corresponds to the Child- Boguslavsky-Langmuir (CBL) equation". The abstract mentions "the Child-Boguslavsky-Langmuir (CBL) probe sheath model", and a later note by partially the same authors,
doi:
10.1063/1.5022236, mentions "the Bohm and ChildâLangmuirâBoguslavsky (CLB) equations for cylindrical Langmuir probes", two equations that were solved jointly.
What is the largest number satisfying this condition?
Numbers which contain no repeating number
substring, i.e. does not contain âxxâ for any nonempty string x (of the digits 0~9), i.e. does not contain 00, 11, 22, 33, 44, 55, 66, 77, 88, 99, 0101, 0202, 0303, 0404, 0505, 0606, 0707, 0808, 0909, 1010, 1212, 1313, 1414, 1515, âŠ, 9797, 9898, 012012, 013013, 014014, âŠ, 102102, 103103, 104104, ⊠as substring. Are there infinitely many such numbers? If no, what is the largest such number?
2402:7500:92C:2EC4:C50:24C1:2841:C6B5 (
talk) 23:25, 9 June 2024 (UTC)reply
Off the top of my head, I think there are an infinite number of them.
Bubba73You talkin' to me? 00:23, 10 June 2024 (UTC)reply
The decimal representation of a natural number is a word in the
regular languageA* over the alphabet A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The numbers satisfying the condition that their decimal representation avoids the pattern XX correspond to the
square-free words of that language. As you can read in the article, there are even
infinitely long square-free words. Â --
Lambiam 05:56, 10 June 2024 (UTC)reply
Another question: Are there infinitely many such numbers which are primes?
118.170.47.29 (
talk) 07:25, 12 June 2024 (UTC)reply
Since there is no discernible logical relation between being non-repeating and being prime, the answer is almost certainly yes, although it may be difficult or impossible to prove this. The number of non-repeating numbers up to is where . One can expect a fraction of to be prime. Â --
Lambiam 08:42, 12 June 2024 (UTC)reply
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.
Euler totient function (sequence A000010 in the
OEIS) has a reduced function:
Carmichael lambda function (sequence A002322 in the
OEIS), and instead of for coprime x, y, while if p is prime and r>=1, and a composite number n is
Carmichael number if and only if divides , and
Dedekind psi function (sequence A001615 in the
OEIS) should also have a reduced function: , and instead of for coprime x, y, while if p is prime and r>=1, and a composite number n is
Lucas-Carmichael number if and only if divides , but I cannot even find the function in OEIS (it should start with (start from n=1) 1, 3, 4, 6, 6, 12, 8, 12, 12, 6, 12, 12, 14, 24, 12, 24, 18, 12, 20, 6, 8, 12, 24, 12, âŠ).
125.230.9.88 (
talk) 02:38, 7 June 2024 (UTC)reply
yet Did you mean to write "if and only if divides "? But for any prime The modified criterion requires to be composite. Â --
Lambiam 06:55, 7 June 2024 (UTC)reply
Is it possible to choose a point strictly inside a simple quadrilateral such that all four sides (and hence the whole interior) are visible from the point? The chosen point is required to be a smooth function of the the vertices, reasonably middling and free from any nastiness (such as depending on the labeling of the vertices). Thanks,
catslash (
talk) 16:58, 7 June 2024 (UTC)reply
All four sides would obviously be visible from any random point inside a quadrilateral that isn't concave. I suspect your question is more complex that that but you haven't expressed it optimally. --
Jack of Oz[pleasantries] 19:25, 7 June 2024 (UTC)reply
If you take each of the four edges and consider the half-planes on the interior sides of each edge, then the intersection of all half-planes is the precise set of points which can "see" all sides. Although I haven't formalized a proof, I'm confident that this set is always nonempty (and also a convex quadrilateral) if the original quadrilateral is non-trivial.
As for a smooth function, I'm sure that there is some definition of the center of a convex quadrilateral that, applied to this set of points, would suffice.
GalacticShoe (
talk) 20:02, 7 June 2024 (UTC)reply
On reflection, since convex quadrilaterals can only take one specific form, with one concave vertex, two "wingtips", and a "head", it's very easy to describe this shape. It's just the quadrilateral formed from the concave vertex, the head, and the two intersections of opposite sides. As for the center, you can use the intersection of diagonals of this internal quadrilateral.
GalacticShoe (
talk) 22:34, 7 June 2024 (UTC)reply
When just one angle of the original quadrilateral is very close to 180°, then so is one angle of the convex inner quadrilateral, at the same vertex, and the intersection of its diagonals is very close to that vertex. In the degenerate case that the angle is equal to 180°, as when a concave quadrilateral is continuously transformed into a convex quadrilateral, the point of intersection coincides with the vertex. Using the
vertex centroid or
area centroid of the convex inner quadrilateral as the chosen point avoids this. Â --
Lambiam 07:01, 8 June 2024 (UTC)reply
In general, for any area A enclosed by a
Jordan curve, the set P(A) of panoptic points is convex. It is the same as A iff A is convex. If the boundary is a polygon, P(A) is empty, a single point, a line segment, or the area enclosed by a polygon. Â --
Lambiam 15:43, 8 June 2024 (UTC)reply
I think I thought â but I was wrong â that the midpoint of the shortest diagonal satisfies the visibility criterion, but it is not a continuous function of the locations of the vertices. I conjecture that there is a continuous function w from these locations, ordered cyclically, to the unit interval, where w(v2, v3, v4, v1) = 1 â w(v1, v2, v3, v4), invariant under similarity transformations, such that the weighted mean of the diagonal midpoints (v1 + v3) / 2 and (v2 + v4) / 2, with weights w and 1 â w, satisfies both requirements. For any configuration, there is an interval [wlo, whi] of w-values that guarantee visibility. If we can prove that both endpoints vary continuously with the vertices, a solution is provided by w = (wlo + whi) / 2. Â --
Lambiam 21:15, 7 June 2024 (UTC)reply
The midpoint of the shortest diagonal does not work. Consider (0, 0), (1, 4), (0, 3), (-1, 4) taken in that order. The diagonal (1, 4), (-1, 4) is shorter than (0, 0), (0, 3), but its midpoint lies outside the quadrilateral. Of course if both diagonals lie inside the quadrilateral then it's convex and any interior point will do. --
RDBury (
talk) 02:06, 8 June 2024 (UTC)reply
The vertex centroid of the (convex) panoptic quadrilateral seems to do the job. Many thanks. If there is a practical way of choosing a point on the
Newton line of the original quadrilateral, that would also be very interesting.
catslash (
talk) 12:57, 8 June 2024 (UTC)reply
At least one of the midpoints of the two diagonals is contained in the panoptic quad, so the panoptic part of the Newton line is nonempty. It should be easy to determine the endpoints of the (possibly degenerate) panoptic segment of this line and take its midpoint. If the original quadrilateral is convex, this is a trivial exercise. Otherwise, you have to determine where the extended sides adjacent to the concave vertex cut the segment connecting the midpoints short. As with many problems in
computational geometry, the hardest part is not the computations themselves but to get the case distinctions correct and complete.
I think though the result will in extreme cases be markedly less "middling" than the vertex centroid. Â --
Lambiam 15:26, 8 June 2024 (UTC)reply
June 9 Information
Source for Langmuir-Blodgett, Langmuir-Boguslavski, and weird Rayleigh equations?
While cleaning up
List of nonlinear ordinary differential equations and citing all the ones listed, there were three that puzzled me to no end. The first was listed as the Langmuir-Blodgett equation:
The next was listed as the Langmuir-Boguslavski equation:
Finally, there was an equation listed as the Rayleigh equation:
I just want to know if anyone recognizes these or has sources for them. The first two I could only find mention of in a footnote of an old edition of a differential equations handbook, which itself cited no sources for these and they do not appear in the more recent edition of the handbook as far as I can tell, and the last one looks neither like the regular
Rayleigh equation (which is notably linear) or the variant of the
Van der Pol equation which is sometimes called the Rayleigh equation (and both of these drown out any search results for this equation). The two equations named after Langmuir I also checked in plasma physics textbooks for, as I vaguely recall that Langmuir worked on plasma, but I could not find mention in the two introductory books I checked. The closest I could get were sources like this one[1] but I can't seem to tell if the given equation is equivalent, and the source they cite is O. V. Kozlov, An Electrical Probe in a Plasma, which I cannot find online. (There's also
Langmuir-Blodgett film but no differential equation is mentioned in that article.) These have been plaguing me, and the editor who added them hasn't edited in six years so no dice there. Any help would be appreciated!
Nerd1a4i (they/them) (
talk) 19:57, 9 June 2024 (UTC)reply
Differential equations with the names Langmuir-Blodgett and Langmuir-Boguslavski are given
here, without further explanation of reference. The origin of the former is possibly an equation presented in a joint publication by Langmuir and Blodgett many years before the technique was developed to make Langmuir-Blodgett films. Â --
Lambiam 06:31, 10 June 2024 (UTC)reply
Yes, that was the handbook that was the one other source I saw - the other edition of the same handbook I was referencing didn't mention these.
Nerd1a4i (they/them) (
talk) 01:46, 11 June 2024 (UTC)reply
That was me adding it to the list of citogenesis incidents as that was what I believed to be true at the time. I then realized I should ask here.
Nerd1a4i (they/them) (
talk) 01:45, 11 June 2024 (UTC)reply
The Langmuir-Blodgett equation may come from
this paper or the "previous papers" cited in footnote 1. Â --
Lambiam 11:09, 10 June 2024 (UTC)reply
Thanks, I'll try to go through that paper and see if it's got the right equation. Much appreciated for finding a fresh starting point! I don't suppose you have any leads for the other two?
Nerd1a4i (they/them) (
talk) 01:47, 11 June 2024 (UTC)reply
Here is an unresolved lead. In
doi:
10.1063/1.4948923 the authors refer to "the Boguslavsky-Langmuir equation for a cylindrical probe under floating potential", which is not a differential equation but is called a â3/2 powerâ law â it has a factor In a second note,
doi:
10.1063/1.4960396Â , the same authors call this a law that "for cylindrical probe under floating potential corresponds to the Child- Boguslavsky-Langmuir (CBL) equation". The abstract mentions "the Child-Boguslavsky-Langmuir (CBL) probe sheath model", and a later note by partially the same authors,
doi:
10.1063/1.5022236, mentions "the Bohm and ChildâLangmuirâBoguslavsky (CLB) equations for cylindrical Langmuir probes", two equations that were solved jointly.
What is the largest number satisfying this condition?
Numbers which contain no repeating number
substring, i.e. does not contain âxxâ for any nonempty string x (of the digits 0~9), i.e. does not contain 00, 11, 22, 33, 44, 55, 66, 77, 88, 99, 0101, 0202, 0303, 0404, 0505, 0606, 0707, 0808, 0909, 1010, 1212, 1313, 1414, 1515, âŠ, 9797, 9898, 012012, 013013, 014014, âŠ, 102102, 103103, 104104, ⊠as substring. Are there infinitely many such numbers? If no, what is the largest such number?
2402:7500:92C:2EC4:C50:24C1:2841:C6B5 (
talk) 23:25, 9 June 2024 (UTC)reply
Off the top of my head, I think there are an infinite number of them.
Bubba73You talkin' to me? 00:23, 10 June 2024 (UTC)reply
The decimal representation of a natural number is a word in the
regular languageA* over the alphabet A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The numbers satisfying the condition that their decimal representation avoids the pattern XX correspond to the
square-free words of that language. As you can read in the article, there are even
infinitely long square-free words. Â --
Lambiam 05:56, 10 June 2024 (UTC)reply
Another question: Are there infinitely many such numbers which are primes?
118.170.47.29 (
talk) 07:25, 12 June 2024 (UTC)reply
Since there is no discernible logical relation between being non-repeating and being prime, the answer is almost certainly yes, although it may be difficult or impossible to prove this. The number of non-repeating numbers up to is where . One can expect a fraction of to be prime. Â --
Lambiam 08:42, 12 June 2024 (UTC)reply