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May 29

What is 'lakh' and 'crore'? How can I understand Indian articles with strings such as 1,00,00, 1,50,00 and so on ?

sirs i really don't understand this lakh and crore business which has lately become very common on the internet apparently it is some kind of indian custom, indian reckoning please tell me how can you place the comma after two digits only (counting from the front) firstly how can you go beyond lakh and crore how can i reckon a number with eight digits 1,00,00,00 ??? one hundred lakh and how many crore ? if a crore is one tenth of a lakh, hold on, one lakh is just a hundred thousand ? so how much is a crode please?? if this 1,00,00 is a lakh entire sirs please forgive my discursiveness i was trying to read up each and everything on this subject and i could not put my head around it kindly direct me to any pertinent source where i can understand lakh, crore, indian customary numbers 2601:481:80:6E60:6C4B:56AE:F2AB:2844 (talk) 23:55, 29 May 2024 (UTC)

We have articles on Lakh and Crore. Do they answer your question? They seem to be a feature of the Indian counting system, and not common in other countries. --RDBury (talk) 00:27, 30 May 2024 (UTC)

I think I have studied very well and I am understanding it now.surely this will improve my scores in JEE.the invigilators and examiners will be very pleased with my fast reckoning. thank you sirs 2601:481:80:6E60:6846:DCBF:5125:F19D (talk) 05:18, 31 May 2024 (UTC)

If you see commas in unexpected places, the easiest is to ignore them. 12,34,567 is the same number as 1234567.  --Lambiam 06:20, 30 May 2024 (UTC)

Kindly don't be writing just any x y z, how can it be same amount if the crode and lakh is arranged in a different manner, since a crode and lakh have values each of their own, they cannot be mingled around or preponed 2601:481:80:6E60:6846:DCBF:5125:F19D (talk) 05:17, 31 May 2024 (UTC)

They are interconvertible values. One crore is equal to 100 lakh, and also equal to 10 million. So 3.5 crore + 7 lakh = 35000000 + 700000 = 35700000, which you can also write as 3,57,00,000 or as 35,700,000 or as 35 700 000.  --Lambiam 09:27, 31 May 2024 (UTC)

In many parts of Europe we use a decimal comma, not a decimal dot. You ignore those at your peril - if your honorarium is 3,14 Euro per hour, it's quite different from 314 Euro per hour ;-). --Stephan Schulz (talk) 18:17, 5 June 2024 (UTC)

My response was directed at the original poster, 2601:... Before responding, I checked that the IP geolocates to a staunchly decimally dotted area. The JEE, for which they are preparing, also uses decimal points.  --Lambiam 20:56, 5 June 2024 (UTC)

May 30

A proof attempt for the transcendence of ℼ

The proposition "if $a$ is rational then $a$ is algebraic" is comprehensively true,
and is equivalent to "if $a$ is inalgebraic then $a$ is irrational" (contrapositive).

My question is this:
The proofs for the transcendence of $\pi$ are of course by contradiction.
Now, do you think it is possible to prove somehow the proposition "if $\pi$ is algebraic then $\pi$ is rational", reaching a contradiction?
Meaning, by assuming $\pi$ is algebraic and using some of its properties, can we conclude that it must be algebraic of degree 1 (rational) – contradicting its irrationality?

I know the proposition "if $a$ is algebraic then $a$ is rational" is not comprehensively true (${\sqrt {2}}$ is a counterexample),
but I am basically asking if there exist special cases $a\in \mathbb {R}$ such that it does hold for them. יהודה שמחה ולדמן (talk) 18:48, 30 May 2024 (UTC)

There are real-valued expressions $E$ such that the statement "if $E$ is algebraic, $E$ is rational" is provable, but this does not by itself establish transcendence. For example, substitute ${\tfrac {22}{7}}$ for $E.$ Given the irrationality of $\pi ,$ proving the implication for $\pi$ would give yet another proof of the transcendence of $\pi$. I see no plausible approach to proving this implication without proving transcendence on the way, but I also see no a priori reason why such a proof could not exist.  --Lambiam 19:24, 30 May 2024 (UTC)

Also, for a while now I am looking to prove the transcendence of $\pi$ by trying to generalize Bourbaki's/Niven's proof that π is irrational for the $n$th-degree polynomial:

${\begin{aligned}&\theta =0.5\pi \\[5pt]&p(\theta )=a_{0}+a_{1}\theta +a_{2}\theta ^{2}+\cdots +a_{n}\theta ^{n}=0\\[5pt]&f(x)={\frac {{\bigl [}\,p(x)\,p(-x)\,{\bigr ]}^{m}}{m!}}\\&I=\int \limits _{-\theta }^{\theta }f(x)\cos(x)dx=2\int \limits _{0}^{\theta }f(x)\cos(x)dx=2\sum _{k\,=\,0}^{nm}(-1)^{k}f^{(2k)}\!(\theta )\end{aligned}}$

Unfortunately, I failed to show that $I$ is a non-zero integer (aiming for a contradiction).

Am I even on the right track, or is my plan simply doomed to fail and I am wasting my time?

Could the general Leibniz rule help here? יהודה שמחה ולדמן (talk) 12:28, 2 June 2024 (UTC)

I suppose that you mean to define $p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n},$ where the $a_{i}$ are integers, and hope to derive a contradiction from the assumption $p(\theta )=0.$ For that, doesnt'it suffice to show that the value of the integral is non-zero?

I'm afraid I'm not the right person to judge whether this approach offers a glimmer of hope.  --Lambiam 15:40, 2 June 2024 (UTC)

May 31

Meridional Radius of Curvature

Hi y'all.

$M(\varphi )d\varphi ={\frac {(ab)^{2}}{{\big (}(a\cos \varphi )^{2}+(b\sin \varphi )^{2}{\big )}^{\frac {3}{2}}}}d\varphi ={\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}d\beta =??(\beta )d\beta .$

(φ, β = geodetic, reduced latitudes)

If ${\frac {(ab)^{2}}{{\big (}(a\cos \varphi )^{2}+(b\sin \varphi )^{2}{\big )}^{\frac {3}{2}}}}=M(\varphi )$ equals the "meridional radius of curvature", then what does

${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}\,$ equal ("reduced meridional radius of curvature"?) and what is its symbol (_rM(β)? )?

--2601:19C:4A01:7057:4C27:AD22:B7E2:D04A (talk) 15:35, 31 May 2024 (UTC)

I cannot relate the quantity ${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}$ to a radius of curvature. It is the speed of a particle moving along the meridian for ${\dot {\beta }}=1.$

For the radius of curvature of the meridian at reduced latitude $\beta$ I find

${\frac {((a\sin \beta )^{2}+(b\cos \beta )^{2})^{\frac {3}{2}}}{ab}}.$

As far as I know there is no standardized symbol for this. I don't think that the notation $M(\varphi )$ is common either.  --Lambiam 20:18, 31 May 2024 (UTC)

Of course $M(\varphi )$ and ${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}$ are valid (though here written with e and e' instead of a and b), but β is not given its own integral identity, even though ${\frac {(ab)^{2}}{{\big (}(a\cos \varphi _{n})^{2}+(b\sin \varphi _{n})^{2}{\big )}^{\frac {3}{2}}}}\neq {\sqrt {(a\sin \beta _{n})^{2}+(b\cos \beta _{n})^{2}}}$. 2601:19C:4A01:3561:4C27:AD22:B7E2:D04A (talk) 03:23, 1 June 2024 (UTC)

I don't understand what it means that ${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}$ is "valid".

The angles $\beta$ and $\varphi$ are related by

$a\tan \beta =b\tan \varphi .$

Here is a numeric example, randomly generated:

$a$ = 239.2188308713, $b$ = 192.1989786957

$\beta$ = 1.3880315979, $\varphi$ = 1.4233752785

Then

${\frac {(ab)^{2}}{((a\cos \varphi )^{2}+(b\sin \varphi )^{2})^{\frac {3}{2}}}}$ = 292.5274922901

 

${\frac {((a\sin \beta )^{2}+(b\cos \beta )^{2})^{\frac {3}{2}}}{ab}}$  = 292.5274922901

This is not a numerical coincidence. For comparison,

${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}$ = 237.8141595361.

 --Lambiam 09:28, 1 June 2024 (UTC)

Right, ${\frac {(ab)^{2}}{{\big (}(a\cos \varphi )^{2}+(b\sin \varphi )^{2}{\big )}^{\frac {3}{2}}}}={\frac {((a\sin \beta )^{2}+(b\cos \beta )^{2})^{\frac {3}{2}}}{ab}}$ but ${\frac {(ab)^{2}}{{\big (}(a\cos \varphi )^{2}+(b\sin \varphi )^{2}{\big )}^{\frac {3}{2}}}}d\varphi ={\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}d\beta \neq {\frac {((a\sin \beta )^{2}+(b\cos \beta )^{2})^{\frac {3}{2}}}{ab}}d\beta$!

So what is ${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}$, which equals ${\frac {ab}{\sqrt {{\big (}(a\cos \varphi )^{2}+(b\sin \varphi )^{2}{\big )}}}}$?--2601:19C:4A01:650:1123:BA2C:D056:629 (talk) 15:00, 1 June 2024 (UTC)

Writing $R$ for the meridional radius of curvature, a variable that depends on $\beta$ (or, equivalently, on $\varphi$), we have:

${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}=R\,{\frac {d\varphi }{d\beta }}.$

This is the tangential speed of a particle moving along the meridian when $\beta =t+\beta _{0},$ in which case the rhs equals $R\,{\dot {\varphi }}.$  --Lambiam 17:30, 1 June 2024 (UTC)

Okay, so you are saying ${\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}$ is the variable for tangential speed (let's call it "S") and using the chain rule:  M(φ) = S(β(φ))β'(φ) and S(β) = M(φ(β))φ'(β), therefore M(φ)dφ = S(β)dβ.

But:  $M(0)={\frac {b^{2}}{a}}$ and $M(90)={\frac {a^{2}}{b}}$ , while $S(0)=b$ and $S(90)=a$, so S is a radius, not speed (I know, speed here is a calculus thing, not literally "speed", but still) and I should point out S(90-β) = R(β), geocentric radius! --2601:19C:4A01:650:19F3:4CE1:97CE:10D5 (talk) 19:09, 1 June 2024 (UTC)

I also just figured out ${\frac {a}{b}}S(\beta )=N(\varphi )={\frac {a^{2}}{\sqrt {(a\cos \varphi )^{2}+(b\sin \varphi )^{2}}}}$, the prime vertical radius of curvature and conversely, of course, ${\frac {b}{a}}N(\varphi )=S(\beta )$.  --2601:19C:4A01:6E9F:1937:5ABC:70DF:B9B3 (talk) 15:38, 3 June 2024 (UTC)

My $0.02: $\varphi$ plays a favored role in defining the radius of curvature because this latitude defines the normal vector to the meridian and so is directly related to the definition of curvature. The corresponding expression in terms of $\beta$ is useful in carrying out integrals but I don't think it's necessary to invent a name for the integrand. cffk (talk) 19:53, 3 June 2024 (UTC)

Technically, isn't "S" the integrand for geodetic distance (just here focused on the north-south meridian distance, either geodetically or as the parametric version of the plane Pythagorean distance), with respect to σ rather than β? I've seen some articles define the geodetic "s" as the spacetime variable, itself!  --2601:19C:4A01:C40C:A8A9:9580:7416:38DE (talk) 19:29, 4 June 2024 (UTC)

While the terms meridian, latitude and geodesic suggest a problem in spheroidal geometry, everything going on here can mathematically be seen as taking place on a good old planar ellipse. The formula

$\int _{\beta _{1}}^{\beta _{1}}{\sqrt {(a\sin \beta )^{2}+(b\cos \beta )^{2}}}\,d\beta$

gives the elliptic arc length between two points on an ellipse in the standard parametric representation $(x,y)=(a\cos \beta ,b\sin \beta )$ using $\beta$ as the name of the parameter. It is easily seen to be equivalent to the formula given at Ellipse § Arc length.  --Lambiam 04:43, 5 June 2024 (UTC)

June 1

Antiprisms in Higher dimensions.

Antiprism talks about higher dimensions, but only in the context of four dimensional Antiprisms created from a Polyhedron and its Polar dual. Is there any reason not to extend this to, for example, being able to make an n+1 dimensional Antiprism out of the n dimensional cube and the n dimensional orthoplex or the 24-cell with itself? Also would the 24-cell anti-prism defined this way be a uniform 5-polytope or is the fact that all but two of the 4-dimensional facets are octohedral pyramids make it non-uniform?Naraht (talk) 19:51, 1 June 2024 (UTC)

That section also mentions five-dimensional antiprisms:

"However, there exist four-dimensional polyhedra that cannot be combined with their duals to form five-dimensional antiprisms.^[8]"

Apparently, the generalization to higher dimensions is not straightforward.  --Lambiam 04:07, 2 June 2024 (UTC)

However, the fact that it needs to be constructed *probably* means that it doesn't apply to any of the six regular 4-polytopes. Looked at the paper on www.semanticscholar.org . Interesting.Naraht (talk) 16:02, 3 June 2024 (UTC)

June 7

Reduced Dedekind Phi function

Euler totient function $\phi (n)$ (sequence A000010 in the OEIS) has a reduced function: Carmichael lambda function $\lambda (n)$ (sequence A002322 in the OEIS), and $\lambda (xy)=lcm(\lambda (x),\lambda (y))$ instead of $\lambda (x)\lambda (y)$ for coprime x, y, while $\lambda (p^{r})=\phi (p^{r})=(p-1)p^{r-1}$ if p is prime and r>=1, and a composite number n is Carmichael number if and only if $\lambda (n)$ divides $n-1$, and Dedekind psi function $\psi (n)$ (sequence A001615 in the OEIS) should also have a reduced function: $\eta (n)$, and $\eta (xy)=lcm(\eta (x),\eta (y))$ instead of $\eta (x)\eta (y)$ for coprime x, y, while $\eta (p^{r})=\psi (p^{r})=(p+1)p^{r-1}$ if p is prime and r>=1, and a composite number n is Lucas-Carmichael number if and only if $\eta (n)$ divides $n+1$, but I cannot even find the function $\eta (n)$ 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, …). 125.230.9.88 (talk) 02:38, 7 June 2024 (UTC)

$\eta (399)={\text{lcm}}(\eta (3),\eta (7),\eta (19))={\text{lcm}}(4,8,20)=40,$ yet $40\not \mid 398.$ Did you mean to write "if and only if $\eta (n)$ divides $n+1$"? But for any prime $p,$ $\eta (p)=p+1.$ The modified criterion requires $n$ to be composite.  --Lambiam 06:55, 7 June 2024 (UTC)

Sorry, typo, $n+1$ instead of $n-1$, fixed. 2402:7500:92C:A1B7:8906:5B5F:EA02:8E54 (talk) 15:02, 7 June 2024 (UTC)

It is also true for Carmichael numbers, n is a number n is Carmichael number if and only if $\lambda (n)$ divides $n-1$, but for any prime p, $\lambda (p)=p-1$, thus both require that n is composite … I fixed it. 2402:7500:92C:A1B7:8906:5B5F:EA02:8E54 (talk) 15:06, 7 June 2024 (UTC)

Also, you state above that $\lambda (p^{r})=\phi (p^{r}),$ yet, according to both Carmichael function and A002322, $\lambda (2^{3})=2$ while $\phi (2^{3})=4.$ "Your" $\lambda$ function is given as OEIS sequence A011773 (Variant of Carmichael's lambda function).  --Lambiam 13:44, 7 June 2024 (UTC)

Well, I forgot that it is only worked for odd prime p, and not worked for p=2 and r>=3, thus this function is in fact (sequence A011773 in the OEIS). 2402:7500:92C:A1B7:8906:5B5F:EA02:8E54 (talk) 15:04, 7 June 2024 (UTC)

Choosing a point inside a quadrilateral

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)

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)

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)

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)

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.  --Lambiam 07:01, 8 June 2024 (UTC)

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.  --Lambiam 15:43, 8 June 2024 (UTC)

I think I thought – but I was wrong – 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 − 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 − w, satisfies both requirements. For any configuration, there is an interval [w_lo, w_hi] of w-values that guarantee visibility. If we can prove that both endpoints vary continuously with the vertices, a solution is provided by w = (w_lo + w_hi) / 2.  --Lambiam 21:15, 7 June 2024 (UTC)

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)

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)

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.  --Lambiam 15:26, 8 June 2024 (UTC)

June 9

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:

${\sqrt {y}}{\frac {d^{2}y}{dx^{2}}}=e^{x}$

The next was listed as the Langmuir-Boguslavski equation:

${\frac {d}{dx}}\left(x^{n}{\frac {dy}{dx}}\right)={\frac {1}{\sqrt {y}}}$

Finally, there was an equation listed as the Rayleigh equation:

${\frac {d^{2}y}{dx^{2}}}+k{\frac {dy}{dx}}+m\left({\frac {dy}{dx}}\right)^{3}+n^{2}y=0$

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 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)

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.  --Lambiam 06:31, 10 June 2024 (UTC)

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)

@Nerd1a4i, I don't have time to delve into the details, but the first two might have been invented on Wikipedia. Check out the mention at Wikipedia:List of citogenesis incidents. —Kusma (talk) 08:20, 10 June 2024 (UTC)

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)

The Langmuir-Blodgett equation may come from this paper or the "previous papers" cited in footnote 1.  --Lambiam 11:09, 10 June 2024 (UTC)

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)

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 “3/2 power” law – it has a factor $\Delta V_{t}^{3/2}.$ In a second note, doi:10.1063/1.4960396 , 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 Child–Langmuir–Boguslavsky (CLB) equations for cylindrical Langmuir probes", two equations that were solved jointly.

Our article Debye sheath has subsections 2.2 The Bohm sheath criterion and 2.3 The Child–Langmuir law, while Child–Langmuir law redirects to Space charge § In vacuum (Child's law). It remains unclear where Boguslavsky (or Boguslavski) enters the picture and how this relates to the differential equation.  --Lambiam 05:11, 11 June 2024 (UTC)

References

What is the largest number satisfying this condition?

Numbers which contain no repeating number substring, i.e. does not contain “xx” 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, …, 9797, 9898, 012012, 013013, 014014, …, 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)

Off the top of my head, I think there are an infinite number of them. Bubba73 ^{You talkin' to me?} 00:23, 10 June 2024 (UTC)

The decimal representation of a natural number is a word in the regular language A* 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.  --Lambiam 05:56, 10 June 2024 (UTC)

Another question: Are there infinitely many such numbers which are primes? 118.170.47.29 (talk) 07:25, 12 June 2024 (UTC)