Mathematics - 2019-06-06 (page 2 of 2)

« first day (3228 days earlier) ← previous day next day → last day (1783 days later) »

5:00 PM

also jetzt ist zeit für Homologische algebra

MatheinBoulomenos

viel Spaß

Danke :)

Hey, does anyone know a proof of the De Moivre-Laplace theorem that addresses the uniformity? I worked out the details by myself, but I'm not completely convinced by my calculations and the fact that the uniformity usually gets stated, but not addressed in the proofs (at least those I've seen) makes me feel I'm missing something one way or the other.

MatheinBoulomenos

5:17 PM

Hi @Ted

Is there a term for abelian groups of the form $\mathbb{Z}/p_1\mathbb{Z}\oplus\mathbb{Z}/p_2\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}/p_n\mathbb{Z}$ where $p_i$ are all primes?

MatheinBoulomenos

semisimple abelian group

though that also allows for an infinite direct sum

semisimple finite abelian group if you want a finite sum

semisimple, danke schön.

infinite collections might be included, too, at some point but yeah, I'll have to make a distinction at some point.

Subhasis Biswas

5:33 PM

sup

I am going to ask a question that I have no idea about. But I will ask anyway

@AlessandroCodenotti Yeah, the integer-coefficient polynomials are not well-ordered. Is there a name for their order type?

AnalyticHarmony

H

@AlessandroCodenotti On the other hand, the natural-coefficient polynomials are well-ordered, right? Are they order-isomorphic to $\omega^\omega$? And the natural-coefficient polynomial towers to $\varepsilon_0$?

Subhasis Biswas

Is the Dirichlet eta function on $(0,1)$ continuous (i guess being analytic, it is so). Now, the absolute value function is continuous.

Has there been any approach to show that the norm has $0$ only as a lower bound on $(1/2,1)$ but it doesn't attain it anywhere? (intermediate value theorem on $\mathbb{R}$)

@AlessandroCodenotti Moreover, do their (commutative) addition and multiplication have anything to do with the (commutative) Hessenberg addition and multiplication of ordinals?

AnalyticHarmony

5:54 PM

$f:[a,b]\to\mathbb{C}$ continuous on [a,b] and zero outside [a,b]. Then $f$ is uniformly continuous on [a,b] and on \mathbb{R}\setminus [a,b].

Why is $f$ uniformly continuous on all of $\mathbb{R}$?

My attempt was considering a point x of [a,b] and a t outside, then I must choose a $\delta$ such that $|f(x)-f(t)| = |f(x)| < \epsilon$ but I think this is not the way, any help?

There's a mathematical concept I'm thinking of and I'm not sure what the name is, or if it has a name.

If you take a number, and divide it by some number X, and continue dividing the results by X, seeing how many times you can divide until you get to some predefined point

I'm trying to figure out if there's a way to tell what X should be if you want to have a specific number of steps.

Do you want to reach a specific number or get below a specific number? The former is probably not possible in a lot of cases.

Until I get to something that can be rounded to 1.

below 2

@AnalyticHarmony This is wrong without additional hypotheses. Consider the indicator function of any compact interval.

6:12 PM

Well, if $a$ is your starting number, $b$ is your ending number, and $n$ is the number of steps you want, you just want to divide by $\sqrt[n]{\frac{a}{b}}$ assuming these are all positive real numbers.

$\sqrt[n]{\frac{a}{b}}$

I'm not familiar with some of that.

What do the curly braces and its contents mean?

Ah, do you not have latex enabled in chat?

No

How do I do that?

It's the n-th root of a divided by b. Check out the info at the top-right of the screen to get latex enabled in the chat.

I dragged "Rendering On" to my bookmark bar and clicked it, but nothing happened

Oh, there it is

render MathJax

6:17 PM

You've got it working?

Yeah

Excellent

And that formula works fantastic! Thanks!

Is there a name for this?

Uhm, let's call it "Rithaniel's miraculous number finding thing"

(In other words, no. I just came up with the formula just now.)

lol

6:24 PM

It should be noted that diving $a$ by $\sqrt[n]{\frac{a}{b}}$ $n$ times will get you exactly to $b$ and not below $b$. Not sure if that's a relevant distinction in this case or not.

No, that's fine. I put in 1 and it worked for me.

Yeah, also the majority of those numbers will be irrational, which, if this is being used in a computer environment, will lead to rounding errors.

It is an a computer environment, but I'm doing Math.floor() anyway.

Particularly as $n$ gets large or $\frac{a}{b}$ gets small.

In the worst case, just round $\sqrt[n]{\frac{a}{b}}$ up. That should work regardless.

6:27 PM

I have a database of people and I'm producing a heat map of US counties that they live in. Blue for almost none, red for the most populous counties. With this particular distribution, the highest populations are in a small handful of counties. The rest of the country is pretty close to zero. So if I do a linear graph, everything is going to be completely red or completely blue.

If I do it this way, I see a lot more detail.

This is the result, btw i.imgur.com/yhr9mDK.png

Looks pretty nice. The numbers on the left are average people per square kilometer? (Probably could use a legend to indicate)

No, not the average. Just the flat-out number of people in our database.

Ah, fair enough.

It looks like you have pretty steep drop-off around the peak counties.

I figure we're going to still be fiddling with it. I think I need to add another step.

Subhasis Biswas

7:25 PM

@AnalyticHarmony because we only need to worry about the points in $[a,b]$

And continuity on a closed and bounded interval implies uniform continuity

@AnalyticHarmony I don't think this is true for every such case. Take $f:[5,6] \to \mathbb{C}$ such that $f(x)=1$ when $x \in [5,6]$

the extension mapping is not even continuous, let alone uniform continuity

@AlessandroCodenotti , sup

is my counter example correct?

Alessandro Codenotti

Counterexample to what?

Subhasis Biswas

@AnalyticHarmony thIS

Yes, it is. I proposed a similar counter-example above.

Subhasis Biswas

@Thorgott ahaaaaaaaaaaaaaaaaaaaaaaaa

$\chi_{[a,b]}$

awesome

@Thorgott , I just came across a nice problem

7:45 PM

Do you guys ever remember some math fact but find yourself unable to remember exactly where you learned it?

Like, it suddenly occurs to me that I have no idea when or how I was first introduced to the notion of a magma.

Subhasis Biswas

@Rithaniel I am very noob at math. But it does happen.

@Rithaniel I heard the name, I remember. I forgot the concept

What's the problem?

So, you know what a group is? A magma is like a group, but satisfies different axioms.

Subhasis Biswas

@Rithaniel a groupoid

Namely, the binary operation of a magma only satisfies the closure axiom.

Subhasis Biswas

7:48 PM

simple binary operation that obey the closure

Yes, that's another term for it.

Subhasis Biswas

@Rithaniel yeah.

Every group must be a groupoid/magma

Then there are quasigroups which satisfy divisibility and semigroups which satisfy associativity.

Correspondingly loops and monoids when you add an identity to either such type of magma.

Subhasis Biswas

@Thorgott don't solve it. No hints. $f: \mathbb{R} \to \mathbb{R}$ be such that $f(x)=f(x^2), \forall x \in \mathbb{R}$. Then $f$ is constant

@Rithaniel yes. Prologue to group theory

groupoids are generally usually reserved for structures which are like groups except where the binary operation is only partially defined

it shouldn't be exchangeably used with magma

7:51 PM

how is every group a groupoid?

Okay, I've not actually known why there were different terms for the same object.

Subhasis Biswas

@Ultradark closure

That's useful info to have.

a groupoid is devoid of the closure axiom right

Subhasis Biswas

does the loop group has something to do with homotopy classes?

7:52 PM

so I don't see how it can be a group

if a group requires closure

Subhasis Biswas

@Ultradark ahh. That's why we can't interchange with magma

okay

I haven't taken abstract algebra yet though so...

Subhasis Biswas

a groupoid, with additional condition, i.e. closure is called a magma

I haven't taken any higher math courses :/

A loop is not related to homotopy classes, no.

It's a type of algebraic structure satisfying three axioms. Closure, identity, and divisibility.

7:56 PM

yeah that one's a stretch

Subhasis Biswas

In mathematics, a loop group is a group of loops in a topological group G with multiplication defined pointwise. == Definition == In its most general form a loop group is a group of mappings from a manifold M to a topological group G. More specifically, let M = S1, the circle in the complex plane, and let LG denote the space of continuous maps S1 → G, i.e. L G = { γ : S 1 → G | γ ∈ C (...

i was talking about tis

that's different

Subhasis Biswas

i accidentally discovered it

but also has nothing to do with homotopy classes, almost by definition :P

Ah, that's new to me. Now I have something to read.

7:57 PM

its literally the space of all loops (based at some point)

just has a natural group structure if the underlying space is a topological group

Subhasis Biswas

@BalarkaSen no shifting?

wtf does that even mean

@SubhasisBiswas I think that's wrong. Take $f$ to be the indicator function of $0$ (the hypothesis do not allow you to control the function at $0$).

Subhasis Biswas

@Thorgott which one?

The $f(x)=f(x^2)\forall x\Rightarrow f\text{ constant}$ one.

Subhasis Biswas

8:01 PM

@Thorgott can you construct a proper counterexample?

when is the product topology more desirable to equip a topological space with as opposed to the box topology. The product topology is a more course grained topology but I am not quite sure I understand this notion completely

the box topology is never correct

Yes, the indicator function of $0$ is one.

Actually, you can take any such function and change $f(0)$ and $f(1)$ to something arbitrary without violating the property.

Subhasis Biswas

i forgot to mention

$f$ is a continuous function

otherwise that won't work

Ok, then it's true.

I wonder if it holds without the assumption of continuity when restricted to the right sets though.

8:04 PM

@RyanUnger whiteboy released a new song, called "straight white male"

the hook is just

brilliant

Clearly, the values the functions takes on the sets $(-\infty,-1)\cup(1,\infty)$, $\{-1,1\}$, $\{0\}$ and $(-1,0)\cup(0,1)$ can be chosen independently of each other then, but I can't think of a function satisfying the property that is non-constant on these sets.

okay the product topology is apparently more useful in practice

In before there is a counter-example, but only with choice..

Balarka I heard you can figure out any question even if you know nothing about it a priori

8:22 PM

so I guess my question is, what influential work have you done?

Subhasis Biswas

8:41 PM

@Thorgott i believe I have solved it

"BELIEVE"

let me post it

Sure.

Subhasis Biswas

8:53 PM

Suppose, (towards a contradiction) that $f$ is not constant. Then for at least one $b \in \mathbb{R}$, $f(1) \neq f(b)$. First, consider the case where $b \neq 0$. WLOG $b \in \mathbb{R}^+$ [if not, then consider $b^2 \in \mathbb{R}^+$, the rest of the procedure will remain the same since $f(b^2)=f(b)\neq f(1)$]

By the hypothesis, $f(b^{1/2}) =f (b) \neq f(1)$. Iterating the process, $f( b^{ \frac{1}{2^n}})\neq f(1)$. Letting $n \to \infty $, we get either $\lim_{x\to 1^-}f(x)$ or $\lim_{x \to 1^+}f(x) \neq f(1)$ [depending on whether $b<1$ or $b>1$ respectively]. But, by sequential crite…

@Thorgott

:( not good enough?

If $n |k$, is it true that the dihedral group $D_n$ can be embedded into the dihedral group $D_k$ as a normal subgroup?

Subhasis Biswas

@user193319 wow, that's a nice question.

I'm pretty certain it can be embedded as a subgroup (currently working on proving that the proposed homomorphism is surjective).

If $c$ is the rotation of $D_k$, $d$ the reflection, I claim that $D_n \cong \langle c^{n/k},dc^{n/k} \rangle$.

Subhasis Biswas

You are saying that (intuitively), that a larger symmetry contains a smaller symmetry?

I have injectivity...Is it obvious that $\langle c^{n/k},dc^{n/k} \rangle$ contains $2n$ elements.

@SubhasisBiswas Yes, that's correct.

9:04 PM

Looks good to me.

Subhasis Biswas

@Thorgott felt sorta amazing when that idea sparked

@user193319 @Thorgott, what do you think?

Sure, $D_{2k}$ is a subgroup of $D_{2n}$ whenever $k | n$. Just embed the $k$-gon inside the $n$-gon, vertex-by-vertex; the isometries of the $n$-gon are isometries of the embedded $k$-gon as well. There is no reason to expect this is going to be normal (unless it's index $2$ or something). $D_2 \cong \Bbb Z_2$ embeds in $D_6 \cong S_3$, but there is no normal subgroup of order $2$.

That's sort of cheating because $1 | 3$, but I can't be bothered to look for "real" examples.

In the index $2$ case sometimes you can say more, $D_{4n} \cong D_{2n} \times \Bbb Z_2$ if $n$ is odd for example.

Subhasis Biswas

@BalarkaSen is was thinking about this "pinning the objects with each other by vertex"

That's such a complicated way to describe inscribed polygons

Subhasis Biswas

@BalarkaSen dujon ke dujoner kaan dhore daar koriye deoa

does it work for $3$ dimensional symmetries?

9:15 PM

It works everywhere. There is no mystery about it.

Just think for more than 2 seconds and you'll get it

Subhasis Biswas

for example, consider a cube. Will it necessarily contain the symmetries of a square? I was thinking about it. I believe it should. For rotations (planar rotation), it is good. For the symmetries where the figure is flipped around the vertical and horizontal axes, it holds good too

The other cases for the cube are not applicable for the square, since it has lesser degrees of freedom

Yes, $D_4$ embeds in the rotation group of the cube, which is $S_4$.

Subhasis Biswas

that is my intuition. Formulation is what I need to work on.

It's one of the Sylow $2$-subgroups of $S_4$, in fact.

Just take the equatorial square in the cube

Subhasis Biswas

@BalarkaSen got it

9:20 PM

There are three conjugate $D_4$'s in $S_4$, corresponding to the three equatorial squares (xy, yz and xz planes).

Subhasis Biswas

@Thorgott proof good?

@BalarkaSen

@BalarkaSen what does non conjugate mean here?

What are you pinging me for exactly

Subhasis Biswas

@BalarkaSen for the verification.

i still don't know how to reply to my own message

Oh I meant non-normal. They are conjugate.

Subhasis Biswas

@SubhasisBiswas does this work

9:22 PM

$H_1, H_2 \leq G$ are conjugate if there is some $g \in G$ such that $gH_1 g^{-1} = H_2$.

All the Sylow $p$-subgroups are conjugate in a finite group, is what the above demonstrates.

Subhasis Biswas

hmm... "some" $g$

I personally would just prove it directly rather than going by contradiction, but other than that, it's good.

Subhasis Biswas

can you give your line of idea @Thorgott

I sometimes wish I grew up in such a mathematically engrossing environment. Looking back at how I spent my 20 years makes me sad.

@user193319 There doesn't seem to be a small order example I can give you. I checked this, which says there are two conjugacy classes of Klein 4-groups in $D_{16}$. So there you have it, there is no normal subgroup of $D_{16}$ isomorphic to $D_4$.

Ah, very nice. Thanks!

9:26 PM

For $b>0$, $f(b)=f(b^{1/2n})\rightarrow f(1)$ by continuity. For $b<0$, $f(b)=f(b^2)=f(1)$ by the foregoing. Finally, $f(0)=f(1)$ by continuity.

Subhasis Biswas

@Thorgott that's an abridged version

@BalarkaSen I was also trying to find a subgroup of some $D$ that is not normal.

couldn't do it.

I mean, I gave an example. $D_2$ in $D_6$.

Subhasis Biswas

restricted my search with $D8$

But $D_2$ is not really a good example of a dihedral group, was my point, so I was looking for bigger examples.

@SubhasisBiswas Of course that's not going to work, after $D_2$ your only choice is $D_4$ which is index $2$, so normal.

Subhasis Biswas

@BalarkaSen can a line be considered a polygon?

9:29 PM

A bigon, yes.

Subhasis Biswas

aren't we dealing with closed shapes?

@BalarkaSen understood. I keep on missing these things.

It's not a line, it's two vertices and two edges joining them.

Literally what a 2-gon should be

$D_4$ is the symmetry group of that

No, a line isn't a bigon!!

Hi @Ted

But that is one my favorite set-up lines in teaching Gauss-Bonnet: Let bigons be bigon.

hi, a @Balarka.

Subhasis Biswas

9:33 PM

hi @TedShifrin

I was trying to figure out how to make that exact pun.

I used it too many times teaching diff geo. But I did come up with it the very first time :P

@TedShifrin Lol, you do G-B on a bigon?

Probably better to spell it: Let bigons be bygone.

Why not, @Balarka? Take a lune of a sphere, for example.

Yeah it is a manifold with corners, but... :D

9:35 PM

Remember that we do regions with piecewise smooth boundary. You refuse to learn the general theorem.

Why not on a teardrop? (One vertex, one edge joining the vertex to itself)

@TedShifrin Hey I learnt it when I took the differential geometry class this semester!

Yeah, you can also wonder if there's a non-smooth geodesic that crosses itself and look at Gauss-Bonnet for that.

You learned about geodesic curvature and angle excess finally?

Yup.

Yippee.

Oh, someone posted on main a question, allegedly from Jack Lee's Riemannian Geometry book, which I am sure is totally false. The problem was to show that Grassmannians (with the obvious $O(n)$-invariant metric) are isotropic manifolds.

It's very natural. Curvature term on the interior, on the boundary (geodesic curvature), and concentrated curvature on the corners (angle defects), all three terms appear.

9:38 PM

Other than projective spaces and spheres, I'm pretty sure that's wrong.

Unfortunately I ran out of time otherwise I would have given a proof sketch of Gauss Bonnet at the end

@BalarkaSen Indeed. Generalizing Chern's high-dimensional theorem to things with corners, I dunno.

Do you know what isotropic means?

Isom(M) acts transitively?

For every point $p$, there is an isometry fixing $p$ that carries any unit tangent vector to any other unit tangent vector.

No, that's just homogeneous.

Or a little more than that. On tangent vectors maybe?

Got it

9:40 PM

Hullo!

That has to be crap.

$S^2\times S^2$ isn't isotropic, I'm pretty sure.

heya @ÍgjøgnumMeg

Yeah I don't believe it

If it were true, I'd already have known it. :P

Subhasis Biswas

user image

Hey @Ted @Balarka :)

@Subhasis those look like props on the show "Gladiators"

9:41 PM

You can draw just two vertices, and two different arcs joining them, @SubhasisBiswas.

Subhasis Biswas

@SubhasisBiswas I was referring to this

the symmetry group $D_1$. What I had in my mind was the two points being reflected

@TedShifrin that will be more suited.

@BalarkaSen why the corners are of special note? is it because the "smoothness" is violated (like $|x|$) ?

The boundary curve turns abruptly at corners

Subhasis Biswas

@BalarkaSen does it have anything to do with the tangent bundles?

No, it does not have anything to do with a million words you keep throwing up!

Subhasis Biswas

@BalarkaSen I am throwing up not because I am trying to sound smart

I am just interested in the intuition

trying to capture in my mind what is actually going on

9:47 PM

It's better to understand what these terminologies mean before trying to get an intuitive understanding of them

Subhasis Biswas

@BalarkaSen the boundary curve turns abruptly. So in my head I am visualisiing a curve that does not have a tangle at that particular point

like I said $|x|$.

What is a "bundle," by the way? I've seen that word tossed around but not actually seen the definition.

Subhasis Biswas

@Rithaniel a collection of tangents at a point.

In differential geometry, the tangent bundle of a differentiable manifold M {\displaystyle M} is a manifold T M {\displaystyle TM} which assembles all the tangent vectors in M. As a set, it is given by the disjoint union of the tangent spaces of M. That is, T M = ⨆ x ∈ M...

The derivative is discontinuous at that point, sure. But the point is the measure of "turning", which is the concentrated curvature at the corner.

Gauss-Bonnet isn't that hard to understand. You can do the following exercise if you really want to get a hint of what it's about. Take three arcs of great circles which meet at three distinct points, making a "triangle" on a sphere of radius $1$. Say the angles of this triangle are $\alpha, \beta, \gamma$. Suppose the triangle traps an area $A$. Then prove that $(\alpha + \beta + \gamma) - \pi = A$.

This is an exercise in spherical geometry, so beware.

@SubhasisBiswas No.

Subhasis Biswas

@BalarkaSen will try. I will make mistakes, sure. But bear with me. Help this kid learn.

9:53 PM

@Rithaniel Roughly speaking, it's a family of objects parametrized over a topological space $X$.

Subhasis Biswas

@BalarkaSen what about when the "object" is a tangent?

That is very general.

So a vector bundle, eg, is a family of vector spaces $\{V_x\}_{x \in X}$ parametrized over a topological space $X$.

But it's not a random family, there's some regularity.

For example, for every point $p \in X$ you can find a neighborhood $U$ of $p$ small enough such that the family $\{V_x\}_{x \in U}$ looks like the projection $\Bbb R^k \times U \to U$.

Okay, so you wouldn't just have a family of unrelated objects and call it a bundle?

Exactly, no. The last condition is the crucial one.

It's called "local triviality". A trivial family of vector spaces over $X$ would be the projection to the second coordinate $\Bbb R^k \times X \to X$, fiber over each point being the vector space $\Bbb R^k$.

9:56 PM

user image

why dV/dt is negative definite? Shouldn't be negative semi definite?!

A vector bundle is a family of vector spaces over $X$ where for each point you can get a sufficiently small neighborhood where the family becomes trivial.

Ah, okay. Still fairly general, but I believe I understand

any suggestion?

Subhasis Biswas

Apparently, if I try to evaluate this via surface integral, then... it's not going to go very nice (will try though).

What are the prerequisites?

@Rithaniel Yeah, bundles are very general objects. But they're nice. I'd recommend the first chapter of Hatcher's "Vector Bundle and K-Theory" if you want to learn some of it at some point.

The introductory words on chapter 1 are definitely worth going through regardless

He draws some nice pictures and described the tangent bundle of a sphere

@SubhasisBiswas No, no, absolutely no surface integral crap

Subhasis Biswas

10:05 PM

@BalarkaSen was going to end up horribly anyway

a direct application of Gauss Bonnet?

No, this is basic spherical geometry.

Subhasis Biswas

okay. Don't solve this

just tell me the starting point

"Think about it for more than 2 seconds" is the starting point.

Subhasis Biswas

been thinking....

What are you trying to evaluate

Subhasis Biswas

10:08 PM

wait. don't tell anything right now.

Compute the area trapped by a lune on a sphere first.

Subhasis Biswas

@BalarkaSen lune?

i ain't gonna google

Take two great circles. They bound two pairs of bigons, each pair being of equal area.

Compute that area

Subhasis Biswas

ok

You should be able to guess what it is before embarking on complicated computations (a full discussion does require calculus but forget about it for now)

Subhasis Biswas

10:13 PM

$\alpha -\pi$?

What is $\alpha$

Subhasis Biswas

the angle at the intersection of the two great circles

If $\alpha = 0$ then the area is $-\pi$???

Subhasis Biswas

wait, then there is going to be another one. $2\pi - \alpha - \pi$

two pairs

Again, check $\alpha = 0$ :P

Subhasis Biswas

10:18 PM

we are discussing about a sphere of unit radius, right?

Yes

I'm heading to bed. After you come up with the right guess, come up with a proof which does not require integral calculus; just some basic analysis argument.

Night

Subhasis Biswas

morning

this is very analogous to the case of circles... Area enclosed by a circular section is $\theta r^2/2$

Perhaps a good framing question "What portion of the surface of a sphere would be contained in a spherical lune?"

Subhasis Biswas

is the answer $2 \alpha$?

@Rithaniel

Haven't given an attempt at proving it, but that would be my guess, yes.

Next step is to explain your reasoning and try to remember what you did so that you can apply it to other tasks in the future.

Subhasis Biswas

10:39 PM

actually I guessed it with the help of the integral $\int_0^\alpha \int_0^{\pi/2} {\sin (\theta) d \theta d \phi}$. I will try to further explain it analytically

Subhasis Biswas

10:58 PM

I want to carry my analytical argument from here. (not much of a formal "proof"). In two dimension, the figure (circle) is flat and the great circles in this case are its diameters (the $z$ axis being absent, what we see in $2$ dimension, is just a flat line with zero curvature).

Now, as we inflate the figure into a bigger sized (now the $z$ axis appears) ball, every aspect regarding its surface area is multiplied by $4$. So, the section formula turns into $4 \times \alpha /2 =2 \alpha$.

(solid angle)

now, the rest of the problem...

@BalarkaSen didn't melon do a video on that

Yes, he did

@Thorgott how do you know what melon means

@BalarkaSen indeed, brilliantly targeted

Subhasis Biswas

@SubhasisBiswas $2 \alpha$ is the combined area on the front side and the back side. It is analogous to $2 \times sector$ in case of 2D

11:13 PM

I watch him

me too

11:52 PM

@SubhasisBiswas: Note that $\dfrac{\alpha}{2\pi}\cdot 4\pi = 2\alpha$.

« first day (3228 days earlier) ← previous day next day → last day (1783 days later) »

Transcript for

Jun '196

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile