Mathematics - 2017-04-29 (page 2 of 5)

Semiclassical

4:05 PM

@Daminark that sort of reminds me of this video: youtube.com/watch?v=137Ei0C3Vdg

though the players in that don't exactly 'win'

Daminark

Kek

Oh so I think we've reached the high point in analysis

$(1+\frac{1}{10})^2 \to 0$ as $2\to \infty$

This appeared on the board as we did Vitali coverings

Ted Shifrin

Greetings, Demonark.

Balarka Sen

@Daminark Except the limit should be e^(1/5)

But yep, very deep stuff

Ted Shifrin

4:20 PM

Oh, happy un-un-sleep, @Balarka.

Mike Miller

conference is selling a bunch of JDG stuff

gonna spend all my savings

Ted Shifrin

Oh oh, @MikeM. Should I start saving you food scraps?

Mike Miller

No, it's ok. I can dumpster dive.

Anonymous

@BalarkaSen Is there any general method to find shortest distance of a point from a general 2 degree curve? (When parameterization is difficult)

LegionMammal978

Question: Given a number $n$ and its prime factorization, can one find a divisor $d|n$ such that $|d-\frac nd|$ is minimized? In my case, there are too many divisors to simply enumerate

Balarka Sen

4:22 PM

@Ted that sounds like a new element in the periodic table

@blue Yeah, minimization

Ted Shifrin

@blue: All I can suggest is Lagrange multipliers.

Balarka Sen

lagrange optimization

Anonymous

Haven't heard of Lagrange multipliers :P

Anonymous

Gotta look it up

Ted Shifrin

Very powerful.

Anonymous

4:23 PM

Lagrange multiplier

In mathematical optimization, the method of Lagrange multipliers (named after Joseph Louis Lagrange) is a strategy for finding the local maxima and minima of a function subject to equality constraints. For the case of only one constraint and only two choice variables (as exemplified in Figure 1), consider the optimization problem maximize f(x, y) subject to g(x, y) = c. We assume that both f and g have continuous first partial derivatives. We introduce a new variable (λ) called a Lagrange multiplier and study the Lagrange function (or Lagrangian or Lagrangian expression) defined by ...

Ted Shifrin

You basically want to find where the vector from a point on your curve to the fixed point is normal to the curve. That's it.

Anonymous

Interesting.

Anonymous

@TedShifrin Right!

Daminark

Hey @Ted! Went to SHS and it turns out there was just some bruising from that fall, nothing serious happened

Ted Shifrin

OK, Demonark, great. I'm glad you got checked. I'm sure you're relieved and can stop worrying now.

Balarka Sen

4:25 PM

@blue The idea is just that say $P$ is your point and $y = f(x)$ is your curve. Blow up a circle centered at $P$; when it touches the curve, the normal to the curve is a multiple of the normal to the circle.

Anonymous

Ow! It's on Khan Academy! (youtube.com/watch?v=BSKtQcLQLWU) Watching it right away! :)

Daminark

And @Balarka what? 9/10^n should go to 0, no?

Anonymous

@BalarkaSen Ah, yes. Makes sense

Ted Shifrin

It's also on my lectures, but Khan is probably a more suitable level for you.

Balarka Sen

Lagrange multipliers is infinitely useful in real life

in exams or whatever

Ted Shifrin

4:26 PM

@Balarka: More directly, it's locally just the Pythagorean Theorem.

"exams or whatever" = "real life"??!! :D

Semiclassical

hi @chat

Balarka Sen

That's all that's left of the real life of JEEers

@blue can tell you about that

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Hi @Alessandro, @Semiclassic

What's JEE, @Balarka?

Semiclassical

lagrange multipliers are fun

though I tend to default to them when I shouldn't.

Ted Shifrin

4:27 PM

Yeah, but the problem Zach made up led to a horrible mess even with Lagrange multipliers.

Anonymous

@BalarkaSen Well :P Sorta true :D

Ted Shifrin

(triangle with vertices on concentric circles)

Daminark

You do use it for the spectral theorem so that's helpful

Semiclassical

I remember that.

Balarka Sen

@Daminark (1+1/5*1/x)^x -> e^(1/5) as x -> infinifty

Ted Shifrin

4:28 PM

I am not minimizing the importance of LM, Demonark. I can prove Spectral Theorem without that, of course.

Balarka Sen

The jokemeister didn't understand the easy joke

Semiclassical

I'm tempted to use Mathematica to test that problem numerically.

Daminark

I mean that's all I know that uses it. And how else do you prove it?

Ted Shifrin

@Semiclassic: Zach's question? I got a numerical solution from LM + Mathematica.

Semiclassical

Right.

Ted Shifrin

4:29 PM

Demonark, quite easily if you do the hermitian case (which specializes to the real case).

Semiclassical

I basically just mean "I wonder what a contour plot would look like."

Ted Shifrin

You prove using the hermitian inner product that eigenvalues must be real. Then it's just induction, as with the proof you learned.

Daminark

Oh yeah I guess in finite dimensions you can use FTA to get that eigenvalues exist

Ted Shifrin

Contour plot will be difficult, @Semiclassic, as you have to watch two points moving on concentric circles.

Balarka Sen

@TedShifrin It's a national exam in India qualifying in which gets you ticket to IIT's, aka the largest institutes in engineering in India, and medical institutes and etc. Every year millions of STEM students take that exam.

Semiclassical

4:30 PM

Well, just parametrize it by the angles.

Balarka Sen

In short, JEE is death.

Ted Shifrin

No, no, not FTA. You need to know they're real, Demonark. That's critical.

Semiclassical

What radii of circles, though?

Ted Shifrin

@Semiclassic: He had 1,2,3.

Semiclassical

mmkay.

Ted Shifrin

4:31 PM

@Balarka: That sounds like the exam ordeal that @Astyx is going through in France ...

Semiclassical

I'll take the first vertex to be at x=1, then.

Ted Shifrin

Right, @Semiclassic. It reduces immediately.

Daminark

I mean knowing that any eigenvalues would need to be real is one thing, but establishing that you have eigenvalues at all would need FTA or Lagrange multipliers, no?

Balarka Sen

France has stuff like that? I didn't know

Anonymous

@BalarkaSen Umm, medical students don't take JEE. They take the NEET. :P

Ted Shifrin

4:32 PM

@Demonark: I grant the FTA. That's really not the point.

Balarka Sen

@blue I see. It's death and that's the main point anyway

Ted Shifrin

The point with LM is you're exhibiting a real eigenvector (with the multiplier being the eigenvalue).

Anonymous

@BalarkaSen Lol, you are over-dramatizing it :P The scary part is the 1.5 million applicants for the exam. The exam itself is not so scary :D

Daminark

That's fair

Balarka Sen

it's very scary from my perspective anyway

SoumyoB

4:34 PM

wait so @Astyx is a medical student? o.0

Semiclassical

Daminark

Though if you're doing things in Hilbert spaces, LM do become vital

:P

Ted Shifrin

No, no, @SoumyoB. He wants to go to good math universities.

SoumyoB

ohh

Ted Shifrin

Are you using angles as parameters, @Semiclassic?

Semiclassical

4:35 PM

Right. Horizontal is angle with axi-s of point on radius-2 circle

Vertical is same but for radius 3.

Daminark

But yeah aside from spectral theorem and economics, what are Lagrange multipliers good for?

Semiclassical

And I'm plotting the area.

@Daminark It's useful in statistical mechanics for similar reasons as econ.

Balarka Sen

It's useful anywhere you want to minimize/maximize

Spectral theorem is not really the point of Lagrange, but of course an essential application

Ted Shifrin

Demonark: Physics (mechanics) is full of them. But truly every max/min problem you want to do is constrained, and it's a lot better to do LM than to eliminate variables and get horrendous messes. You'd know this if you'd done any concrete homework problems.

SoumyoB

well I find this a pretty strange phenomenon, competition seems to be written in the very roots of human civilization; millenia ago in the prehistoric ages we were competing for food and survival against predators, now we're competing over who can get the top seats in exams

Semiclassical

4:37 PM

Actually, I just realized that I plotted the signed area.

Hence why the lower-right part of it is negative.

Balarka Sen

@SoumyoB It's still competing over food and survival, in one sense.

Semiclassical

(I used the determinant formula for the area of a triangle.)

Balarka Sen

Nothing has changed much, primally :P

Ted Shifrin

@Semiclassic: So are you seeing the MAX in that picture?

Daminark

I mean I haven't seen too many contexts firsthand in which optimization is important, aside from our professor saying "Economists, pay close attention!"

Semiclassical

4:38 PM

Somewhere around (2.2,4.5)?

SoumyoB

nah I think this is more of a competition for status (mostly). At least people here do it mostly for status "HEEYYY LOOOK!! My son got into the IIT's!!"

Daminark

But that's neat

SoumyoB

most of them could have managed to get basic human needs met either way, I think

Semiclassical

More precisely, Mathematica gives area of 1.22621 when $(\theta,\phi)= (2.38821,4.20459)$.

I'll make a plot of the scenario that corresponds to.

Ted Shifrin

I didn't convert my solution into angles, but that sounds reasonable.

Anonymous

4:42 PM

@SoumyoB I hardly ever find an IIT student boasting that they are from IIT (very rare). Once they get in they realize that it's pretty crap after all. A 6 hour exam can't decide your life. It's the parents and relatives who boast mostly about their kids. (Well that's true for nearly all countries I suppose :P)

Ted Shifrin

In the US it's the GRE that drives people nuts (for grad school in math).

SoumyoB

@blue that's exactly what I meant when I said people mostly do it for status

@TedShifrin oh it does drive people nuts here too

Daminark

@Ted we did actually have some optimization problems on our part that week, but they were very long, it was due the day off the midterm (which we were told wouldn't include them), and we had already burned 40 hours on that pset, so most of us skipped the LM problems

Ted Shifrin

Uh huh. Uh huh.

Semiclassical

kinda neat.

Ted Shifrin

4:43 PM

I had been hoping for an elegant solution, but there wasn't one, @Semiclassic.

Astyx

The medical students' exams are better centralised, they only have 800 tick boxes to check or so, and they're done, it lasts 2 days I think. As to wether this is better, that's another question ..

Ted Shifrin

Probably explains why no one has this problem in a textbook :D

Astyx

And also hi Ted, SoumyoB

Ted Shifrin

No, it doesn't look isosceles. There's a right angle going on there, as I mentioned when I worked on it, but not an angle of the triangle.

Semiclassical

The fact that it looks to have the same x-coordinate for the two vertices is interesting.

SoumyoB

4:44 PM

it's so funny that there are coaching centers for the admission exams for getting into IIT's, and these coaching centers have their own admission exams, and now I'm hearing that there are coaching centers opening up to train students for the admission exams of the coaching centers that will train them for the main admission exam

Semiclassical

so that one of the sides is perpendicular to the x-axis.

Ted Shifrin

I don't think that's true, either, @Semiclassic.

Balarka Sen

@Soumyo #Inception

Ted Shifrin

Hi @Astyx

You saw that I took your name in vain earlier.

Semiclassical

It seems to be so looking at the numbers.

But, going off of laptop for now.

SoumyoB

4:45 PM

a time will come when right after the baby is born he/she will have to decide whether he/she wants to do engineering or medical

Ted Shifrin

@Semiclassic: It didn't look that way when I solved it in cartesian coordinates, but I didn't keep the file.

Astyx

I almost became a medical student actually. But I loved maths too much :p

Daminark

@SoumyoB more like, when the baby will decide whether his/her offspring will do engineering or med school

Balarka Sen

@Astyx welcome to the dark side

Sha Vuklia

Let $f\colon\mathbb R^n\to\mathbb R$ be continuous and homogeneous of degree $\alpha>0$. Show that $f$ is differentiable at $\vec 0$ with derivative $\vec 0$ if $\alpha>1$.
I've shown that $f(\vec 0)=0$. I can show that all partial derivatives are $0$. Let $\vec u\in\mathbb R^n$, then
$$
\lim_{t\to 0}\frac{f(t\vec u)}{t}=\lim_{t\to 0}t^{\alpha-1}f(\vec u)=0.
$$
However, here I chose $\vec u$ fixed, while I actually need $\vec h$ to be arbitrary. How can I do that?

Astyx

4:47 PM

Lol, I regret nothing !

SoumyoB

@Astyx ibb.co/mFNCvk

Daminark

We may not have cookies, but we do have lemmanade @Astyx!

Sha Vuklia

@Astyx I studied medicine for half a year, but I dropped out of uni, because it sucked so much:p

TheGreatDuck

@Secret i saw you tagged me with some result regarding my question. You didn't solve it, did you? I'm not sure what to make of what you wrote?

Semiclassical

Hmm

Daminark

4:48 PM

@Sha That sort of thing is exactly why I don't like how Europe makes people decide what they want to major in so early

Ted Shifrin

@Sha. You're not understanding what your computation is, are you? $\vec h = t\vec u$.

Astyx

And grapes, abelian ones ! @Daminark leaves

Ted Shifrin

In fact, you want to allow $\vec u$ to vary over the unit sphere.

Sha Vuklia

@Ted ah, and then by continuity, $f$ assumes a maximum?

Ted Shifrin

Demonark: Worse — I could not have majored in math(s) and French simultaneously in Europe, as I understand it.

Right, @Sha.

SoumyoB

4:51 PM

I'd love to learn French some day

Ted Shifrin

I'm going to have to recall a lot of French and German very soon.

Ran Wang

I somehow ended up in marketing first....... and I really hate it...... then I did a graduate study on mathematics.....

(Not to say marketing is bad....)

Eric

hi chat

Ran Wang

hi @Eric

Daminark

Also, if you do A-Levels in Britain, you only end up taking 3 classes in your last two years of high school, so those kinda need to be decided based on your intended major

SoumyoB

4:52 PM

I've heard most Asian languages have a 'staccato' characteristic or comprise very rough sounds but French is one of those rare languages that have a 'legato' characteristic or are very soothing in sound

Ran Wang

Hmmmm that is an interesting way to put it. I know Chinese and Japanese are like that

Ted Shifrin

Hi @Eric

Daminark

Hey @Eric!

Ran Wang

For example, that is pronounced "thate" in chinese and "thato" in Japanese

Because we cannot do "t"

Eric

@SoumyoB it's a liaison, it's also present in the dialect of portuguese I speak, it sounds pretty, but people not where I'm from never know what I'm saying

SoumyoB

4:54 PM

I think Indians and most Asians too have thick accents precisely because of that

in comparison, everyone loves a French accent

Ran Wang

Right~

Balarka Sen

i like russian accent

Ted Shifrin

@Eric: When I've seen Brazilian movies, it's always seemed to me that elements of Portuguese sound Russian. So weird.

Ran Wang

For the chinese though, i think the problem is simply we don't have access to enough real english

Most people learn English from other Chinese....... therefore......

Daminark

Vodka for you comrade! @Balarka

Eric

4:55 PM

@Ted I never noticed that, but my girlfriend said the EXACT same thing when she watched Brazilian films with me, so it's definitely there

Ran Wang

I actually like French accent....

Italian ones can be nice too

SoumyoB

@Eric I wish my language comprised of liaisons too!

Eric

@SoumyoB as with anything it has pros and cons, I love it because I'm proud of my regional culture, but spelling is a nightmare

Daminark

I am rather fond of the Greek accent, as heard from precisely 2 people

My dad's boss in NY, and Sougi

Eric

my landlord is greek and when I first met him he asked me and my roommates if any of us knew greek, of course two of us knew classical greek, so it was a funny moment where we tried to have a conversation and it was just not working out

Balarka Sen

4:59 PM

i'd love to get an american "on-high" accent

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$Mathematics$ Mathematics