@copper So you survived a scrunched round-trip!

$\ddot{\theta} = \frac{g}{l}\sin\theta$ where $\ddot{\theta} = \frac{d^2\theta}{dt^2}$ is said to have no simple general solution unless $\theta$ is very small. Why is this?

this is for a pendulum btw

it's true that $\sin \theta \approx \theta$ for small $\theta$, but why for large theta is $\int \int \sin \theta dt^2$ not simple?

@Obliv You realize this is meaningless, right?

TIL Michael Artin's father was a famous number theorist.

@TedShifrin sort of, i ended up with some virus (a cold presumably, 2 negative covid tests later) that knocked me out for a few days. was fun, but i am getting two old to be driving & changing location every two days.

not sure what you mean @TedShifrin

Idk, I just assumed that the sin made the antiderivative non elementary

Think about what the ODE $\dot\theta = \sin\theta$ means, and then go to second order.

@copper.hat Changing location? Aren’t you back?

But doesn't it depend on what $\theta(t)$ is? since the ODE is w.r.t. t

@Obliv warm up with $\dot y = y$.

That’s the whole point. You can’t integrate w r t to $t$.

Oh... but if you somehow had $\theta(t) = ...$ then you could right?

Is the difficult part obtaining such a function of $\theta$ w.r.t t?

Have you taken any differential equations course? Or for starters learned separation of variables in calc 2?

I have not, I'm taking diff eq next semester and calc 2 was many years ago. I can brush up on it real quick

I think I understand why it'd be hard. Given some $\theta$, it must be a function of $t$ such that at some $t_1$ and $t_2$, $\theta(t_1) = \theta(t_2)$ and that requires it to have some trig term

and no elementary antiderivatives exist for something like $\sin(\cos\theta)$

and even then, $\theta$ of a pendulum isn't a perfect $\sin$ or $\cos$ function. Because of the way the tangential force to the arc length varies as $\theta$ does, it'd be something much harder.

well at least we know $\theta = \frac{s}{r}$, $\dot s$ is maximal at $\theta = 0$ for some $\theta$ so $\dot \theta$ is maximal at $\theta = 0$

Suppose that $S=\{(0,0)\}\cup \{(x,\sin 1/x): x\in \mathbb R\setminus \{0\}\}$. Let $B_S$ denote a unit ball in S centred at the origin. Is $B_X$ compact?

koro, haven't we done this? or am i blending memories.

we may have computed the closure of S in R^2, or the closure of S intersect {x >= 0} in R^2. that analysis applies here.

It is clear that B_S is bounded. Let $cl_B(A)$ denote closure of A in B. It turns out that $cl_B(A)= cl (A)\cap B$. So the said unit ball is $B(0,1)\cap S$.

Leslie, I think that probably was about topological sine curve being connected but not path connected.

koro i had in mind chat.stackexchange.com/transcript/36?m=62322762#62322762

@leslietownes sorry for the delayed response. I have gone through the linked discussion.

I'm trying to apply that here.

How this inequality come from previous inequality?

My question is why $sup_{k}|x_k|^{q-p}\leq [sum_k |x_k||^p}^{(q-p)/p}$

?

@Unknownx Note that for every $k$, you have $|x_k|^p\le \sum |x_k|^p. \implies (|x_k|^p)^{q-p}\le (\sum |x_k|^p)^{q-p}.$

@TedShifrin i am back, but spent the first few days in bad recovering from what my sister would derisively refer to as a 'man cold'.

not sure how to write the closure of the unit ball :(.

The unit ball is: $U= \{(a,\sin 1/a): |a|\lt 1, a\ne 0 \}\cup \{(0,0)\}$

I want to find its closure in the set S I wrote above.

If this closure is closed in R^2, then by Heine Borel, the closure is compact.

The answer is 'not compact'. So I must show that this closure is not closed in R^2.

So what is a limit point of this closure in R^2 which is not contained in closure U?

@Koro

$|x_k|^p\le \sum |x_k|^p. \implies (|x_k|^p)^{q-p}\le (\sum |x_k|^p)^{q-p}.$

I understood this.

How is $U$ a ball of any sort?

Hi @copper.hat!

Hi @Koro !

Here: B_S is U.

@Koro still not able to deduce the supremum in the left hand side of the inequality.

@Koro Using the usual topology?

@copper.hat yes, induced from R^2.

@Unknownx Now note that for any $k$: $|x_k|^{q-p}\le (\sum |x_k|^p)^{\frac {q-p} p}$.

(then say 'sup to both sides)

how does it come?@Koro

The points $[-1,1]$ are in the closure but not in $S$.

@Unknownx Add a step:

@Unknownx from the last inquality you wrote.

@robjohn Thank you. I got it.

@Koro thank you for your effort as well. :)

@Unknownx that's exactly what I also did :).

@Koro sorry.

$cl (U) \cap S=T$, say. Then how come (1, 0) for example is a limit point of T?

We should have then $(a_n, \sin (1/{a_n}))\to (1,0)$. It is not possible.

@copper.hat

I meant that $\{0\} \times [-1,1]$ is in $T$.

ah, I see.

But (0,1) is not in T. But it is a limit point of T. We take $a_n= \frac 1{(2n+1)\pi/2}$ so that $(a_n, \sin \frac 1{a_n})\to (0,1)$.

Hence T is not closed.

Thanks a lot @copper.hat :).

mm, are you using a goofy definition of 'unit ball'? if you look at the usual metric space ball at (0,0) with radius 1 in S, points of the form (nonzero, 1) will not be in there. this is a very minor and extremely fixable issue, but i'm not sure i understand what went on with the definition of 'unit ball' above.

and don't lose the forest for the trees. because compactness is a topological property, asking whether B_S is compact when regarded as a subset of the metric space S has the same answer as asking whether B_S is a compact subset of R^2, where you can use the heine borel theorem. it might be bounded, but it ain't closed.

so this really was dealt with when you computed the closure, in R^2, of a set similar to B_S. :)

$U=\{(a,b) \in S: a^2+b^2<1\}$

I simplified this to write the U I wrote earlier.

@leslietownes yes :).

Do folks agree that when proving any group table is a Latin square, that we don't explicitly have to prove every group element appears at least once in every row (and column) since showing that any two row elements must be distinct automatically shows every group element must appear at least once in the row by closure? I see some answers on MSE try to show this, but to me it's not necessary.

uh, if you know that a set has some number n of elements, then any list of n pairwise distinct elements of S will have to include every element of S at least once (in fact exactly once). so i agree with that.

i don't know what "some answers on MSE" are saying. i do separately note that however you define 'latin square,' it's a whole lot easier to show that the multiplication table of a group is a latin square, than it is to show that a given square (even if assumed to be a latin square) is the multiplication table of a group. and maybe some of those MSE answers are wrestling with that, or something related.

the multiplication table of a group is a latin square, but is also more interesting than a general latin square, because it also exhibits associativity (a property not very easy to 'see' or check from a table).

@leslietownes Cool, thanks! I am indeed showing that first much easier implication you mentioned. One of those MSE answers I had in mind was this one: math.stackexchange.com/a/2069864/689775

yeah, there is some redundancy there.

i guess maybe there's no redundancy. the asker doesn't specify that it's a finite group. although maybe most people would include finiteness as part of the definition of "latin square," you could imagine not making that assumption.

although, i wonder if the answerer was even thinking about that.

@leslietownes Yeah I was thinking that too! But also thought that perhaps the assumption of the group being finite was implied by the mention of a group table?

yeah, again, seems like a situation where a lot of people would say, when i say 'table,' i'm thinking finite. but others, maybe not.

sounds like a punishment a strict prof might have dished out in the good old days. go stand in the corner and write the cayley table for the integers, and don't come back until it's done.

I don't get it why when det(A-\lambda I)x=0 (26) has solution?

note that there's no x in the second equation, det(A - lambda I) is just the determinant of the matrix A - lambda I. there's a lot of linear algebra going into the explanation of that. different books approach it in different ways.

For example let's suppose B is 3by3 matrix then if det(B)=0 then it implies span of matrix is plane line or point.

also, this is probably being pulled out of a broader context where it makes sense, but, there's an implicit assumption here that lambda can be 'chosen' and isn't just given to you. that's not always met in practice.

so not necessarily null vector.

"span of matrix" could use a little more precision. the span of its columns, for example?

@leslietownes yup that is what i mean sorry for being bit vague

if three vectors v_1, v_2, and v_3 in R^3 span a plane line or point, then one of them can be written as a linear combination of the other two. any way of doing that will give you a vector (a,b,c), not the zero vector, for which a v_1 + b v_2 + c v_3 = 0. which gives you an element (namely (a,b,c)^T) in the nullspace of the matrix with columns v_1, v_2, v_3.

for example if it is v_1 that is a linear combination of v_2 and v_3 you can find p and q with v_1 = p v_2 + q v_3 and then 1 v_1 - p v_2 - q v_3 = 0 so you could take (a,b,c) = (1,-p,-q). and generally if it is v_j that is a linear combination of the other v's, you can find a linear combination of the v_j's, with the coefficient of v_j equal to 1, that is equal to 0.

@leslietownes oh yes that was what i was thinking day before yesterday. Thanks a lot!

should think about dependence

yeah. depending on how you set it up, the tricky part can be seeing that the determinant being 0 actually tells you that. but that's one step of the many moves of logic that relates these concepts.

you have very organized way of thinking :)

most of my thought gets lots in chaos...

@DLeftAdjointtoU Hi :) Thank you for the offer, but I have enough on my plate already, even though it's the Winter break; I have a lengthy assignment to do and a paper to read thoroughly. Good luck and have fun! :)

@user4539917 Emil Artin is more than a number theorist. He is basically one of the founders of modern algebra. For example, the current most common presentation of Galois theory is due to him.

Are there always > $n$ primes in between $n^2/2$ and $n^2$ for large enough $n$?

The fact that $(W,\phi)$ is unique up to canonical isomorphism and the fact that such canonical isomorphism is unique are two separate conclusions, right?

@Peter what do you think of my above question?

@Peter hello

@Shinrin-Yoku Interesting. You can ask a question about it and formulate it as a conjecture , if you have some context (for example a search range or how it is related to the Oppermann conjecture)

Well I did try it upto and it seems to old for n upto 10^10 (ignoring small counter examples)

Oh, $10^{10}$ is a huge search limit considering that you want to find at least $n+1$ primes. I am currently running a quick search until $10^6$ and it already takes a while. Maybe, I programmed it quite inefficient. Seems that the conjecture is true exactly for $n>10$.

To be honest , even the range upto $n=10^4$ is an enormous task. How did you establish this huge search limit ?

I did not try the full search space.

but i still did many

I first thought that this is stronger then Oppermann's conjecture , but in fact it is much weaker. Dusart's bounds or the known upper bound for prime gaps could be already enough for a full proof.

Not a very large complete search limit, but if it helps , for $11\le n\le 3\ 000$ , we can find at least $n+1$ such primes as desired.

@Feynman_00 no, everything that comes after the phrase "unique up to canonical isomorphism" is an explanation/definition of what that phrase means

@Obliv The equation of motion of the simple undamped pendulum involves an incomplete elliptic integral of the first kind. I have some info (& graphs) here: physics.stackexchange.com/a/718837/123208

And this earlier answer just discusses the period of the pendulum: physics.stackexchange.com/a/595082/123208

@Thorgott I'm sorry if I'm drowning in an inch of water here but why does the existence of a canonical isomorphism imply that such isomorphism is unique? Isn't it impossible to have more than one canonical isomorphism?

@Shinrin-Yoku You might enjoy this related discussion RobJohn & I had in August on the number of primes in the intervals $[n^2-n, n^2+n]$ chat.stackexchange.com/transcript/36?m=61882200#61882200

@Feynman_00 that's the definition

or rather, in a more down-to-earth way, it's what is meant by that phrase

if there were multiple such isomorphisms, you wouldn't call any of them canonical

and, for starters, it is best to think of "canonical isomorphism" as just a phrase to mean precisely that which is illustrated there, rather than as some sort of independent concept

Oh, I thought of canonical isomorphisms as isomorphisms independent from the choice of the basis (or any choice to be more general), which was the source of my confusion as I don't think there would be anything wrong with having two different isomorphism not depending on the choice of the basis

@Thorgott I prefer cannonical isomorphisms. They explode real nice.

I mean, intuitively I can see why canonical identification should be unique, on the other hand this looks like a different nuance from the canonical isomorphism intended as independent from the choice of basis which I wrote above, right?

@XanderHenderson heh

(though some use these words interchangeably at times and there's no overarching guidelines for what these terms mean outside standard contexts, so confusion is bound to arise)

if you have enough categorical intuition, you will just start calling things natural or canonical based on gut feeling and you will never be wrong about it, but I understand that's the opposite of helpful and probably makes it look needlessly esoteric

"needlessly esoteric". Category theory in a nutshell. :P

Oh, natural and canonical can be two different things, I see. I guess I'll need to delve more into category theory in the future. Thank you for helping. :) @Thorgott

I made a pi calculator in Python based on $\tan(\pi/12) = 2 - \sqrt 3$, using Carlson's modified AGM atan algorithm. It doesn't converge as fast as the Salamin / Brent / Gauss AGM algorithm, but it's not too shabby. And it can easily be adapted to compute other arctans, even for complex arguments.

sagecell.sagemath.org/…

I also tried using it with binary subdivisions of the angle, but it doesn't help. You still end up using the same number of sqrt calls, and you lose a little bit of precision with the smaller argument.

I saw this somewhere today- $f-\int f=1\implies (1-\int)f=1\implies f=\frac 1{1-\int} 1=1+\int (1)+\int \int 1+\int \int \int 1+...= 1+x+x^2/2+...=e^x.$

Too bad the geometric series doesn't actually converge (the operator $T(f) = \int_0^x f$ on $C([0,x])$ certainly doesn't have norm $<1$).

(Well, it does for $|x|<1$, so that sort of suffices.)

@Feynman_00 This is the first time that I saw the phrase "up to canonical isomorphism". Usually I see " up to a unique isomorphism ".

@Yai0Phah Uh. I've read/heard it a bunch of times

hey @Ted!

I recently found out about the fact that the space of derivations of $C^k$ germs $(0<k<\infty)$ is infinite-dimensional, and a recent answer of yours on a post helped me understand this

thought that was a lucky coincidence!

I think it’s been discussed in here before, too

Ah, I wasn't aware of that

that’s why Spivak calls the Taylor-like expansion for smooth functions the $C^\infty$ trick.

ha yea, that's a good name

@Yai0Phah thank you for pointing that out.