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

ÍgjøgnumMeg

17:00

also jetzt ist zeit für Homologische algebra

MatheinBoulomenos

viel Spaß

ÍgjøgnumMeg

Danke :)

Thorgott

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

17:17

Hi @Ted

Rithaniel

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

Rithaniel

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

17:33

sup

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

user76284

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

AnalyticHarmony

H

user76284

@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 $\mathbb{R}$)

user76284

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

AnalyticHarmony

17:54

$f:[a,b]\to\mathbb{C}$ 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?

Hypersapien

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.

Thorgott

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.

Hypersapien

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

below 2

Thorgott

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

Rithaniel

18:12

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.

Hypersapien

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

I'm not familiar with some of that.

What do the curly braces and its contents mean?

Rithaniel

Ah, do you not have latex enabled in chat?

Hypersapien

No

How do I do that?

Rithaniel

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.

Hypersapien

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

Oh, there it is

render MathJax

Rithaniel

18:17

You've got it working?

Hypersapien

Yeah

Rithaniel

Excellent

Hypersapien

And that formula works fantastic! Thanks!

Is there a name for this?

Rithaniel

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

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

Hypersapien

lol

Thorgott

18:24

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.

Hypersapien

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

Rithaniel

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.

Hypersapien

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

Rithaniel

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

Thorgott

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

Hypersapien

18:27

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

Rithaniel

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

Hypersapien

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

Rithaniel

Ah, fair enough.

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

Hypersapien

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

Subhasis Biswas

19:25

@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

Thorgott

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

Rithaniel

19:45

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

Thorgott

What's the problem?

Rithaniel

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

Subhasis Biswas

@Rithaniel a groupoid

Rithaniel

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

Subhasis Biswas

19:48

simple binary operation that obey the closure

Rithaniel

Yes, that's another term for it.

Subhasis Biswas

@Rithaniel yeah.

Every group must be a groupoid/magma

Rithaniel

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

Balarka Sen

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

Ultradark

19:51

how is every group a groupoid?

Rithaniel

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

Subhasis Biswas

@Ultradark closure

Rithaniel

That's useful info to have.

Ultradark

a groupoid is devoid of the closure axiom right

Subhasis Biswas

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

Ultradark

19:52

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

Ultradark

okay

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

Subhasis Biswas

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

Ultradark

I haven't taken any higher math courses :/

Rithaniel

A loop is not related to homotopy classes, no.

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

Ultradark

19:56

yeah that one's a stretch

Subhasis Biswas

Loop group

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

Balarka Sen

that's different

Subhasis Biswas

i accidentally discovered it

Balarka Sen

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

Rithaniel

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

Balarka Sen

19:57

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?

Balarka Sen

wtf does that even mean

Thorgott

@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?

Thorgott

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

Subhasis Biswas

20:01

@Thorgott can you construct a proper counterexample?

Ultradark

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

Ryan Unger

the box topology is never correct

Thorgott

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

Thorgott

Ok, then it's true.

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

Balarka Sen

20:04

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

the hook is just

brilliant

Thorgott

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.

Ultradark

okay the product topology is apparently more useful in practice

Thorgott

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

Ultradark

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

Ultradark

20:22

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

Subhasis Biswas

20:41

@Thorgott i believe I have solved it

"BELIEVE"

let me post it

Thorgott

Sure.

Subhasis Biswas

20:53

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?

user193319

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.

user193319

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?

user193319

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

@SubhasisBiswas Yes, that's correct.

Thorgott

21:04

Looks good to me.

Subhasis Biswas

@Thorgott felt sorta amazing when that idea sparked

@user193319 @Thorgott, what do you think?

Balarka Sen

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"

Balarka Sen

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?

Balarka Sen

21:15

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

Balarka Sen

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.

Balarka Sen

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

Balarka Sen

21:20

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?

Balarka Sen

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

Balarka Sen

Oh I meant non-normal. They are conjugate.

Subhasis Biswas

@SubhasisBiswas does this work

Balarka Sen

21:22

$H_1, H_2 \leq G$ 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" $g$

Thorgott

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.

Balarka Sen

@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 $D_4$.

user193319

Ah, very nice. Thanks!

Thorgott

21:26

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.

Balarka Sen

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

Subhasis Biswas

restricted my search with $D8$

Balarka Sen

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?

Balarka Sen

21:29

A bigon, yes.

Subhasis Biswas

aren't we dealing with closed shapes?

@BalarkaSen understood. I keep on missing these things.

Balarka Sen

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

Ted Shifrin

No, a line isn't a bigon!!

Balarka Sen

Hi @Ted

Ted Shifrin

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

hi, a @Balarka.

Subhasis Biswas

21:33

hi @TedShifrin

Rithaniel

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

Ted Shifrin

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

Balarka Sen

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

Ted Shifrin

Probably better to spell it: Let bigons be bygone.

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

Balarka Sen

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

Ted Shifrin

21:35

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

Balarka Sen

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!

Ted Shifrin

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?

Balarka Sen

Yup.

Ted Shifrin

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.

Balarka Sen

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.

Ted Shifrin

21:38

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

Balarka Sen

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

Ted Shifrin

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

Do you know what isotropic means?

Balarka Sen

Isom(M) acts transitively?

Ted Shifrin

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.

Balarka Sen

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

Got it

ÍgjøgnumMeg

21:40

Hullo!

Ted Shifrin

That has to be crap.

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

heya @ÍgjøgnumMeg

Balarka Sen

Yeah I don't believe it

Ted Shifrin

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

Subhasis Biswas

ÍgjøgnumMeg

Hey @Ted @Balarka :)

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

Ted Shifrin

21:41

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 $|x|$) ?

Balarka Sen

The boundary curve turns abruptly at corners

Subhasis Biswas

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

Balarka Sen

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

Balarka Sen

21:47

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 $|x|$.

Rithaniel

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.

Tangent bundle

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...

Balarka Sen

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.

Balarka Sen

21:53

@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?

Rithaniel

That is very general.

Balarka Sen

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 $\Bbb R^k \times U \to U$.

Rithaniel

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

Balarka Sen

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 $\Bbb R^k$.

CroCo

21:56

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

Balarka Sen

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.

Rithaniel

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

CroCo

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?

Balarka Sen

@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

22:05

@BalarkaSen was going to end up horribly anyway

a direct application of Gauss Bonnet?

Balarka Sen

No, this is basic spherical geometry.

Subhasis Biswas

okay. Don't solve this

just tell me the starting point

Balarka Sen

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

Subhasis Biswas

been thinking....

Thorgott

What are you trying to evaluate

Subhasis Biswas

22:08

wait. don't tell anything right now.

Balarka Sen

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

Subhasis Biswas

@BalarkaSen lune?

i ain't gonna google

Balarka Sen

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

Compute that area

Subhasis Biswas

ok

Balarka Sen

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

22:13

$\alpha -\pi$?

Balarka Sen

What is $\alpha$

Subhasis Biswas

the angle at the intersection of the two great circles

Balarka Sen

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

Balarka Sen

Again, check $\alpha = 0$ :P

Subhasis Biswas

22:18

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

Balarka Sen

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 $\theta r^2/2$

Rithaniel

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

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

22:39

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

22:58

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...

Ryan Unger

@BalarkaSen didn't melon do a video on that

Thorgott

Yes, he did

Ryan Unger

@Thorgott how do you know what melon means

skullpatrol

@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

Thorgott

23:13

I watch him

skullpatrol

me too

Ted Shifrin

23:52

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

Transcript for

$Mathematics$ Mathematics