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May 23

Differential Equation

Does the differential equation x² * d²x/dt² = k have a name? (All I've figured out about this so far is that I don't remember enough about differential equations. I'm not getting anything on solving it except errors.) Thank you. RJFJR (talk) 02:45, 23 May 2024 (UTC)

As to a solution, you could guess that one might be some power of $t$ and accordingly substitute $x=At^{p}$ (where $A$ and $p$ are constants), then solve for $p$ and then solve for $A$. catslash (talk) 09:02, 23 May 2024 (UTC)

Then, since nothing in the equation depends on the absolute value of $t$, you could apply an arbitrary time-shift to get a slightly more general solution $x=A(t-t_{0})^{p}$. catslash (talk) 09:20, 23 May 2024 (UTC)

You're right. Thank you. p=2/3, I was not expecting that. I appreciate it. RJFJR (talk) 14:14, 23 May 2024 (UTC)

This is an autonomous second order equation. If you write ${\ddot {x}}-{\frac {k}{x^{2}}}=0$ and multiply with ${\dot {x}}$, you find ${\frac {d}{dt}}({\frac {1}{2}}{\dot {x}}^{2}+{\frac {k}{x}})=0$ so ${\frac {1}{2}}{\dot {x}}^{2}+{\frac {k}{x}}=C$ for some constant $C$. Solve for ${\dot {x}}$ and you get a separable first order equation. —Kusma (talk) 14:35, 23 May 2024 (UTC)

Thank you. I need to dig out the old textbook and start reading. RJFJR (talk) 19:25, 23 May 2024 (UTC)

catslash's solution is a one-parameter family (indexed by $t_{0}$) of very nice solutions, but in general you should be able to solve the initial value problem for any initial values of $x$ and ${\dot {x}}$, so you'll get a two parameter family. It is easy to show that the solution exists; you can get an implicit formula from Mathematica or other symbolic computation software. —Kusma (talk) 12:17, 24 May 2024 (UTC)

Autonomous system (mathematics)#Special case: x″ = f(x) gives $t$ as a two-parameter function of $x$, but this function looks uninvertable except for the choice of the parameter $C_{1}$ which makes it correspond to my guessed solution. catslash (talk) 22:43, 25 May 2024 (UTC)

May 27

Szekeres Conjecture

Is there something like Szekeres Conjecture, which is different from Erdős–Szekeres theorem? ExclusiveEditor _{Notify Me!} 19:00, 27 May 2024 (UTC)

The sum will never reach 2

I saw a reference to Zeno's paradoxes#Dichotomy paradox in a comic strip. The article does not mention the sum of 1, one half, one quarter, and so on. Where is that sum?— Vchimpanzee • talk • contributions • 22:34, 27 May 2024 (UTC)

The sum is 1 if you sum an infinite number of terms. Bubba73 ^{You talkin' to me?} 23:42, 27 May 2024 (UTC)

Would you believe 2? --142.112.143.8 (talk) 03:59, 28 May 2024 (UTC)

I was hoping to link the sum I was asking about from the Zeno's paradox article.— Vchimpanzee • talk • contributions • 14:37, 28 May 2024 (UTC)

Well, Google actually gave me 1/2 + 1/4 + 1/8 + 1/16 + ⋯.— Vchimpanzee • talk • contributions • 14:37, 28 May 2024 (UTC)

Our article does not mention a sum, but an infinite regression of tasks. Each task has a subtask that must be completed before the whole task can be completed. This is (in Zeno's analysis) as impossible as the task of enumerating all unit fractions in order of magnitude, so that 1/1000 has to come before 1/999 – you can't even start.  --Lambiam 15:44, 28 May 2024 (UTC)

Do you think the Zeno article should mention the other?— Vchimpanzee • talk • contributions • 18:58, 28 May 2024 (UTC)

To be more precise, the section Dichotomy paradox does not mention this sum. Elsewhere, in the introductory paragraph of the section Paradoxes, it is stated that Zeno's paradoxes are often presented as an issue with the sum of an infinite series, although none of the original ancient sources has Zeno discussing the sum of an infinite series. In my opinion it can be given a place in the analysis of the Achilles and the Tortoise paradox, but not so for Zeno's dichotomy paradox.  --Lambiam 20:08, 28 May 2024 (UTC)

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)