Mathematics - 2017-03-25 (page 3 of 4)

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6:00 PM

In higher generality you've got some arbitrary normed space, but if you're in infinite dimensions you no longer know that all linear maps are continuous (equivalently, bounded in operator norm)

So you typically add in that as an additional condition on the derivative

right, makes sense!

@Daminark I don't know if it's thanks to you, but suddenly the cloud of confusion is disappearing, haha! I've been postponing to read on differentiability for weeks now, because it looked so intimidating in higher dimensions :P But now it makes perfectly sense why they noted that the difference function $r(h)$ (near 0) is of smaller order than any linear function, because that property was easy to transfer in higher dimensions!

Yup, you got it! With a lot of multivariable calculus, and really analysis in general, a lot of what you see looks like scary formalism but turns out to just be alright once you get used to it

See, hard work does pay off :-)

(eventually)

It can often be a slow burn, though.

6:17 PM

True dat pal, true dat.

I'll say that there's this one thread on r/math which goes "Algebraists of reddit, why do you like algebra?"

The very first answer, which is kinda why I anticipate liking it, is "Lack of epsilons"

About the slowest burn of that sort I can think of is 'completing the square'.

You see it in high school, and instead of remembering it just learn the quadratic formula.

Memorization > Understanding of course

But if you do grad-level physics, it has its revenge the moment you start having to do gaussian integrals a lot.

(Completing the square is the proof of quadratic formula, but nobody cares about that)

6:19 PM

Well, sure.

But you actually just use the completing the square bit when doing gaussian integrals.

Oh huh

Sure, it's used in a lot of other contexts. baby min/max problems

diagonalization

You don't bother writing out the roots, you just directly write "oh hey, this quadratic is just of the form a(x-b)^2+c."

Right.

Main thing with the physics context is that typically you don't just have the one-variable version. Hence the versions here: Gaussian_integral

integrating 1/(x^2+2x+2)

Ya.

6:23 PM

@BalarkaSen It takes a certain level of math maturity to realize these things...which is not usually developed yet in most high school students

The main way that I knew that quadratic formula was true was to just plug the roots in, show they work, and then you know, you can only have 2 roots so...

We derived it by just completing the square last year.

Grade 7?

Yeah.

Ugh I hope the supervisor lets me skip algebra 2 and pre-cal at least...

well ahead of the norm :-)

6:26 PM

How is Sperner ?

What was the hint you gave me again?

Skipping alg 2 isn't a good idea imho @MeowMix there's so much trig to learn there

@skillpatrol I already know trig, however.

Counting the red-green edges

In two different ways

@BalarkaSen Just write "$S_1$ up to $S_4$ are solvable so we can skip this question by Galois theory" on every algebra question in high school ever

3

6:29 PM

@skillpatrol And the calc BC content too

You really need a good command of trig for calc.

Yeah...

Just remember $e^{ix} = \cos(x) + i\sin(x)$ and you're good! :P

@Ted gave me a trig-calc problem and it wasn't oo hard

@Dami that's consequential of the taylor series expansions of $e^x$ at $ix$, right?

Yeah

6:31 PM

Math builds on itself a lot.

@skillpatrol It was "what is the cone of maximal area inscribed in a sphere of radius 1"

Oh, that sounds fun.

Though I tend to just define $\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$ and $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$

Are we allowing a slanted cone? (Not that I know if it makes a difference.

Hmm, @Meow: What about the other one I gave you? :D

6:33 PM

Hi prof @TedShifrin

@Semiclassical Yeah, you can

hi t

Howdy, Demonark, @skull, @Semiclassic, @MikeM

Hi.

Hi @Ted

6:33 PM

Hmm, neat.

Hey @Ted!

Oh, and there's a @Balarka.

But it's trivial that a right cone is of maximal volume

I'm taking a cooking break.

Hi @Ted

6:33 PM

Zach: I'm talking about the paper folding one. It's interesting!

Oh, I'll go rummaging through chat history to find it :P

Salut Ted, comment était le repas hier ?

By the way, Balarka showed me another example of a non-closed subspace

Non, @Astyx, c'est ce soir-ci.

It was the set of continuous function $[0,1] \to \Bbb R$ and his subspace was the set of functions which are polynomials

6:35 PM

Ah très bien :p

Sure, but I wanted you to relate that to $V^{\perp\perp}$ (saying so for the third or fourth time).

And how, if you're given a non-polynomial function on $[0,1]$, you can approximate it arbitrarily close with polynomials

Yes, that's a beautiful theorem, although I'm not fond of the proofs.

Any continuous function.

Well if $V^\perp$ is closed then obviously $(V^\perp)^\perp$ is too

Is that the chebyshev min-max theorem, or am I conflating?

6:36 PM

Did you prove that $V^\perp$ is always closed?

But is $V^{\perp\perp}$ the smallest closed subspace containing $V$?

And how does $(V^\perp)^\perp$ relate to $V$ itself?

@MikeMiller shifted arctangent?

He's done that, I believe, @MikeM. That's what started all this.

But maybe he hasn't actually proved that exercise.

Are there no nice proofs of the Weierstrass theorem ?

I like the proof using approximate identities

6:37 PM

The proof in Rudin is conceptually a nice proof (using approximations of the identity).

:P

Oh lord that proof

oh duh. I checked it myself by differentiating lol

But it's still yucky.

It was great

I don't know what the Weierstrass theorem is.

6:37 PM

But none of us had any clue what Soug was doing the first time around

Stone-W

Oh. So this is just the generality of functions on R?

Stone-W is more general (point-separating algebra mumble mumble).

Tobias Kildetoft

Hi @ted

Alessandro Codenotti

Hi @Ted

6:38 PM

Heya @Tobias.

Hi @Alessandro.

@Ted I was trying to prove that the equilateral triangle is the triangle of maximum perimeter inscribed in a circle. It's ezpz using some trig and calc but I dunno how to do it easily

Good to get in all these hellos. I'm headed back to the kitchen in a few.

(This is a reply to the cone thing)

Nice with Lagrange multipliers (choosing the right independent variables), @Balarka.

Sure, but that's not what I want to use.

6:39 PM

would the variables be the angles here?

Yeah.

So what do you want to use, @Balarka?

Hmm.

Can you do it by Euclidean geometry @TedShifrin?

I can make basic symmetry arguments for area; I don't think I know how to do that for perimeter.

I have a sneaking suspicion I've done this calculation before.

6:40 PM

I guess I don't know how to prove Weierstrass. You can obviously reduce to the case of smooth functions, but I can't actually approximate with the Taylor polynomials except locally.

It's rather deep, @MikeM.

I rather wonder if there's a slick way to do it that exploits the relationship of the law of sines to the radius of the circumcircle.

Sure, but I should still be able to prove it.

@Semiclassical That's what I did.

Ahh.

6:41 PM

BTW, @Balarka: Did you mean minimum perimeter?

I'll confess, I can never remember how to -prove- that relationship.

Yeah, minimum. Sorry.

@Semiclassical It's not hard. You use the perp drawn from the other vertex.

Sounds reasonable.

You get $a/\sin(A) = b/\sin(B)$ first and you use another perp to get $b/\sin(B) = c/\sin(C)$

Well, sure. Proving the law of sines is easy.

6:42 PM

Oh, you mean $= 2R$?

Right.

@TedShifrin hi !

@Ted Yes, I have proved $V \subseteq (V^\perp)^\perp$ if that's what you mean

Err, I am forgetting how to do it too

Anyhow, @Balarka, I honestly have no clue.

OK, I'm headed back to the kitchen.

Hi/bye @Liad.

6:45 PM

@TedShifrin bye :P

@Semiclassical Ah, a'right. Note that it's trivial for right triangles (because the hypotenuse is a diameter).

Then you just fix the base and make a right triangle inside the circle with that exact base.

The angles on the top vertices are same because they are based on the same arc.

So nothing happens.

hmm, neat.

@TedShifrin Fair enough, me neither.

Oh, speaking of neat things, I saw some physics this week that I hadn't before while doing an intro lab.

Or, well, a manifestation that I hadn't expected.

i got this question maybe one of you could help :
$X = [0,1] \ ^ { \Bbb N}$ , is there a metric on $X$ that induce the box topology?
i think no, i defined $A=(0,1] \ ^{\Bbb N}$ , so $\overline A = X$ , and i want to show that there is an element of $\overline A$ that there is no sequence of elements from $A$ that convergences to that element in contradiction to the fact that $X$ is metrizable.

I thought taking $(0,1,0,1\dots$ but i see now that it is incorrect. someone see another element that would work?

6:49 PM

In lab, they were taking cathode-ray tubes (basically just an electron gun shooting a beam of electrons to light up a dot on a screen) and seeing how the beam/dot responds to magnetic fields.

As a very simple example, if you put the north end of a magnet on one side of the CRT, the beam (and therefore the dot) will curve downwards due to the Lorentz force on it.

If you put the north end on the opposite side, the dot will move up. Pretty simple/familiar stuff.

What happens if you put the north poles of two different bar magnets on opposite sides of the CRT?

Two dots ?

Something something Stern Gerlach?

Yeah that

The dot (really a small focused circle of electrons) gets squeezed into an ellipse!

Oh cool

6:52 PM

Moreover, the principal axes of that ellipse are at a 45 degree angle to the magnets.

Oh I guess S-G happens if you use different poles of two magnets on the opposite sides

Nice.

Yeah.

And why is that ?

Spin.

Nope.

I asked a question on Physics SE to find out.

What they suggested (and which I think is right) is that one should look at the combined field of the two north poles.

6:54 PM

Oh, I was answering the wrong question ("Why does S-G happen?")

This is about your phenomenon

No idea about that.

Ah.

Ah right

The answer is here: physics.stackexchange.com/a/321056/55641

And you're saying we don't have symmetry relatively to the plane of symmetry of the magnets ?

If you look at the picture in the middle, the field lines (ignoring direction) in the plane of the magnets are basically just a saddle point.

6:56 PM

Yeah

And the point is that since the electrons are coming down into that picture ('into the page') you have to look at what those field lines do to the electrons.

Huh

Quite interresting

The upshot is that electrons coming in at two of the corners will be pulled in by the Lorentz force, but at the other corners they're pulled out.

So it squeezes the beam into an ellipse in the way observed.

Ah, so you just see what the saddle vector field does to the unit circle if you let the flow run for a little time along it

It shouldn't be hard to see that you turn it into an ellipse

Right.

6:59 PM

But why do we lose symmetry ?

I don't get that

@Astyx Look at y = x. The field acts upwards and downwards along them in the first and the third quadrant respectively.

So the circle becomes an ellipse with y = x as it's major axis.

You mean, why the pattern doesn't stay the same if I reflect along the vertical axis in the that picture? (The field does, of course.)

What about $y=-x$ ?

It acts upwards and downwards on the 2nd and 4th quadrant respectively

@Semiclassical or the horizontal for that matter

7:02 PM

along y = -x, I mean

I think the point is that the Lorentz force (which is the cross product of the magnetic field and the velocity of the electrons) is a pseudo-vector.

It seems to me that what happens in $(x,y)$ is the symmetrical of what happens at $(-x,y)$ compared to $x=0$

And same for $y$

So I don't understand how we lose symmetry

So the point is that it behaves differently w/r/t the reflection symmetry than the velocity vectors or the field lines.

Huh right, that seems convincing

@Astyx Oh, that's a good point. Err

7:04 PM

And if you took the south poles instead, you would rotate the ellipse by 90 degrees then

Right.

Fair enough

Ok, ok, I see.

As evidence of that, suppose you rotate by 180 degrees rather than reflect.

In that case, the pattern does get mapped onto itself.

Doesn't it ?

7:06 PM

I mean, imagine moving the two magnets around the edge until they've each moved to the opposite side.

(I am chickening out of this. Too hard for me.)

Then the diagonal pattern will rotate with it, and a 180 degree rotation of that leaves it unchanged.

Oh I misread you, I agree

So it's the difference between a rotation, which preserves orientation, and a reflection which reverses it.

@Mike So the theorem says that if you have a continuous, complex-valued function $f$ on a closed interval, let's just say $[0,1]$, there is a uniformly converging sequence of polynomials. Now define $g(x) = f(x) - f(0) - x(f(1)-f(0))$, for $x\in [0,1]$, and let it be $0$ elsewhere.

7:07 PM

So some rather nice math/physics altogether.

Ah, you're going to approximate it by subtracting off sequences of explicit interpolations?

Define $Q_n = c_n (1-x^2)^n$, where $c_n$ is defined so that $\int_{-1}^1 Q_n = 1$

Next you'll have a function that's zero at the endpoints, so you'll subtract off $c(1-x^2)$ to kill off the value at 1/2.

Yeah, ok, got it.

Indeed @Semiclassical

I mean, the whole vector-pseudovector thing isn't something I'd show an intro student.

But it is a nice puzzle for someone who knows to take advantage of symmetry.

7:10 PM

Kinda, though note that this process is heavy, like the $P_n$ are defined by convolution

Is it a gram-schmidt thing?

@Astyx Hello, have a nice day :D

Oh hi @Socrates

Long time no see

Hi @Socrates

What's up ?

7:13 PM

Life can be so easy

@Daminark There should be a purely combinatorial approach with the right choice of basis polynomials, I think.

And so hard

To be honest, give me blood and I do Sir...

@Daminark en.wikipedia.org/wiki/…

@Semiclassical By the way I had this question : Are there particle-to-particle forces that are not along the line between those two particles ?

7:18 PM

Oh I like that @Mike

Typically, no. You need to conserve momentum.

However, the Lorentz force is actually an example of something that breaks this.

Suppose I have two electrons at rest. Then there will be an electrostatic repulsion between them, but there's no magnetic field and so no force.

But if I have those same two electrons in motion, even if they're at the same relative displacement from one another, there will be a force.

did somebody smoke at the pc?

Moreover, that force that one of the electrons experiences will be perpendicular to both the magnetic field generated by the other electron -and- to its own velocity vector.

Hence the force is determined not just by the line between the two particles but furthermore their velocities.

That's what I find weird

The probabilistic argument is almost surely much newer than the polynomials.

7:22 PM

Is there a deeper explanation, or is that only something we know from experience ?

(More properly: Momentum is still conserved, but now there's momentum stored in the magnetic fields in addition to the linear momentum of the particles themselves.)

@Astyx no problemo, bro ;)

Yeah, you can give a fuller account in terms of special relativity.

Typically, though, we just say "That's what the Lorentz force does, deal with it" and move on.

Physics discussion on the math chat, math discussions in the physics chat, awesome

Yep.

7:24 PM

Nice example of duality

feel the freedom

And is the force applied to one of the electron via the magnetic field the opposite of the one applied to the other one ?

Haha @bolbteppa

@Daminark Actually I really like that probabilistic argument.

(Do we still have some kind of Newton's third law ?)

damnit, my userprofile takes forever to refresh in the chatterbox

7:27 PM

Hmm.

I'm not sure. The thing is that once you get to cases like this you typically don't try to account for things in terms of 'forces' directly.

Rather, you reformulate Newton's 3rd Law as a statement of momentum conservation.

dafuq is Sobolev regularization

And in that sense, you still have N3L once you account for the momentum stored in the fields.

Yeah, but since the forces do not work, the momentum cannot be modified by these forces

But at level of Newton's 3rd law as a statment about action-reaction forces, I don't think it's unchanged.

What?

Did you type something?

7:29 PM

Oh wait I'm being silly

Guys SORRY for interrupting. I am new here, and two of my questions were left to hold for no reason: math.stackexchange.com/questions/2201633/logarithmic-inequality

I was thinking of it in terms of energy, obviously you change momentum

Ah.

I reposted the problem and it has been termed "duplicate": math.stackexchange.com/questions/2202400/logarithmic-inequality

You do get similar things for energy, though. (You'd better, or electromagnetic radiation wouldn't carry energy.)

7:31 PM

@Sid i forgot the magic word "please"

Please help. It's a normal algebraic question and it is on hold

Reposting the same question is going to get it flagged as a duplicate.

So what should I do

you need to focus

And that's not really linear algebra is it ?

7:31 PM

Go back to your original question, and improve it.

you can make it

user image

They've asked you to add more details etc, in particular what you've tried so far.

Question about @Ted's lecture on directional derivatives:
Shouldn't he write $\begin{align}\lim_{t\to0}\frac{(\Vert \vec a\Vert+\vert t\vert)-\Vert \vec a\Vert}{t} \end{align}$ instead? So the absolute value for $t$.

Apparently, they want me to add my solution. If I knew how to solve it, why would I ask it?

7:32 PM

Alright I think that helped me, thanks a lot @Semi

They're not asking you for your solution, they're asking for you to document your -attempts- at a solution.

@ShaVuklia No, because if t is negative on the LHS of that equation, then the length of the sum is reduced.

I don't know how to start. What way will my solution help

?

@Sid knew like in the past? which day we have today?

@ShaVuklia He takes $(1-t)$ out, which is positive for $t$ small unless I'm mistaken

7:34 PM

Your attempting to solve it will give us a sense of what you know and what would be helpful. In addition, it shows that you've put some amount of work into this problem beyond just typing up the question.

omg, ^ this

A question that says "here's my problem help plz" is going to get closed because it shows no work beyond just the typing.

(the "unless I'm mistaken" is obviously about what he does, not about it being positive)

Oh ok. I solved other peoples questions. I'll just wait I guess. I dont want to confuse the other solvers

Theres nothing I can do to solve it

If you wait, nothing is going to happen. The question will remain closed.

7:36 PM

@Astyx ohh I see it, we take out $(1-\frac{t}{\Vert \vec a\Vert})$ entirely

Do it, do it now

It will not be reopened unless you edit it to an extent that users on the site judge it as having provided sufficient context/details.

@ShaVuklia Otherwise you wouldn't have equality, only an $\le$ sign

yes, thank you

My pleasure

7:39 PM

good manners, i like it

I try my best

Is it just me, or is there not actually a definite question here? math.stackexchange.com/q/2202359/137524

sorry, lost the signal

I think he wants to know about the application of these formulae

But he didn't explicitely ask for them, that's true

Eh. It falls between 'unclear what they're asking' and 'too broad.'

Because "What's an application of this formula" is really broad.

7:46 PM

I'd say both

I don't get the comment also

me neither :P (the comment, I mean)

Yeah, that's wtf.

It's been quite I while since I last answered a question on main, maybe I'll do that now

OK, let's see

8:05 PM

Ugh, this is hard.

combat training, yes

I have to show that the orthogonal complement of the orthogonal complement of a subspace is the closure of that subspace

anyone here know a good book to study PDE

OH I HAVE IT

:D

:D

8:07 PM

The book, or the problem?

the problem!

It's the continuity of our inner product!

Neat.

Wait... inner products have to be continuous, right?

What do you think ?

anything is possible

8:09 PM

Oh, pfft. Obviously, right?

How ?

with love

isn't it weird that we're always living in the past?

@Alucard = Null?

Yup

8:11 PM

yes

=Socrates

the fact that human reaction time is about 250 milliseconds always means that what we're perceiving is always at least one-fourth of a second in the past

That's very insignificant though.

insignificant is a very relative term

So is perception

8:13 PM

it's insignificant to us precisely because it's our very reaction time

^ i like

my bad for taking the chat in a tangential direction

I was just deeply engrossed in thinking how impossible it is actually to dodge a bullet

:D

why even dodge

what is the alternative you're suggesting?

stand your motherfucking ground

8:23 PM

@Meow So why is the inner product continuous ?

Well, it looks like it's because $\langle a\mathbf{v},\mathbf{w}\rangle = a\langle \mathbf{v},\mathbf{w}\rangle$

?

So then, as you go back in forth in the direction of $\mathbf{v}$, it is continuous

That's not really sufficient

What if I show that if you go back and forth in any direction, it's continuous?

8:30 PM

Still not sufficient

smoke some cigarettes, lol

Then how the fuck will I prove continuity?

:]

8|

By calming down first of all :p

:D

8:32 PM

Smoking is bad for your health

@Astyx I was kidding.

oh yes, you get cancer from it

I know @Meow

That being said, is there some inequality you could use ?

Cauchy-Schwarz...?

:D

8:35 PM

Why not, see what it brings

Cauchy-Schwarz is love, Cauchy-Schwarz is life

5

yes

I've been caught unawares by Cauchy Schwarz so often

I mean it's useful but if you choose your line of work carefully enough you can avoid inequalities entirely, which I'd say is a good life choice :P

Jk you can't run from them forever

@Daminark I used Cauchy-Schwarz for like 80 percent of the things in my stat exam. Go away man

:P

8:43 PM

@Balarka How were your exams?

I can solve your problem for you!

Don't do stats!

:P

@Daminark The other option is biology which is shit

Oh yeah how did they go/have they been going?

Exams were pretty good.

And yikes bio is bleh

It's cool but bleh

8:44 PM

Somebody ping me when a wild @Ted appears, please

Agreed. Coolness and crappyness are not mutually exclusive.

lol at @TedShifrin learn to be a vampire or die

8:58 PM

what a boring guy

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