Mathematics - 2017-02-19 (page 2 of 7)

Balarka Sen

03:00

matrix of second partials

Zach Hauk

It's a matrix whose elements are second partial derivatives

Akiva Weinberger

Ah. Would it be symmetric almost always, then?

Zach Hauk

always

by

that rule

whatever it's called. it says partial x of partial y = partial y of partial x

too lazy to LaTeX right now :/

Akiva Weinberger

I think there are some exceptions, let me see...

Balarka Sen

@ZachHauk There's a beautiful geometric meaning to that discriminant fact I am not sure if you know.

Zach Hauk

03:01

@Balarka what would that be?

Balarka Sen

@Akiva Whenever stuff's C^2

Akiva Weinberger

Continuous second partial derivatives

Balarka Sen

Without continuity of mixed partials it does fail

(@Zach: Exercise)

Zach Hauk

ah yes

Akiva Weinberger

Example

Zach Hauk

03:02

this is a blast to the past of multivariable lol

why does the Hessian / Second Derivative Test even work?

Balarka Sen

I am about to tell you that.

@ZachHauk Locally, how many "geometrically unsimilar" quadratics can there be?

heather

just a quick rewind: you were doing multivariable calc in 6th grade on partial derivatives and matrices and thought it was easy?

Balarka Sen

What does picture of $ax^2 + bxy + cy^2 = 0$ look like in general?

Zach Hauk

a conic?

hmm

ellipse / point

Balarka Sen

Ok. There are three distinct local pictures of a conic. Tell me those.

Zach Hauk

03:05

hyperbola, parabola, ellipse is what you're looking for?

Balarka Sen

Parabola and ellipse are locally not very different, are they?

Akiva Weinberger

Speaking of, I was reading a book where the first half was about conics in the complex projective plane

Balarka Sen

Around a point both look like upside down cups

Akiva Weinberger

where there's only one conic

Zach Hauk

yeah

Balarka Sen

03:05

Also you mean paraboloid

Akiva Weinberger

@BalarkaSen Or right-side up cups...

Zach Hauk

uhh, does the degenerate case where a conic a line count?

Balarka Sen

That's the second picture @Akiva speaks of.

Upside down paraboloid, right side up paraboloids

Akiva Weinberger

Why paraboloid? There's no $z$ here

Balarka Sen

For a hyperboloid what does a local picture look like?

Akiva Weinberger

03:06

You said $ax^2+bxy+cy^2=0$

Did you mean $=z$?

Balarka Sen

Oh, I am sorry. Yes, graph of that in R^3, not level set.

I apologize.

So we're talking of conic surfaces, @Zach, just to clarify

Zach Hauk

alright.

Balarka Sen

Want me to tell about the local picture of hyperboloids around some interesting points? :)

I also mean one-sheeted hyperboloids

Zach Hauk

oh,

the ones that kind of look like lamps

Balarka Sen

Otherwise the two sheeted ones have no interesting local pictures - we discussed the two already

@Zach Right

Do you know a name for that picture?

Zach Hauk

03:09

uhh no

Balarka Sen

Something that's related to horses...

Zach Hauk

oh

saddle

Balarka Sen

Righto.

Zach Hauk

OH

concave, convex, saddle

Balarka Sen

So three pictures. (1) Upside down paraboloid (2) right side up paraboloid (2) Saddle.

OK.

What's the condition on $b^2 - 4ac$ for the graph of $z = ax^2 + bxy + cy^2$ to give three of those pictures?

Zach Hauk

03:12

$b^2-4ac < 0$ is number 2 i think?

Akiva Weinberger

@BalarkaSen You wrote (1), (2), and (2)

Balarka Sen

Hm, no, not quite.

@Akiva. Grr. Make that (1), (2) and (3)

So, @Zach, which (2) did you mean? :P

Zach Hauk

3

Akiva Weinberger

...Why

Zach Hauk

that's basically saying $b^2 < 4ac$

Akiva Weinberger

03:14

Do we set $y=1$ or something?

Balarka Sen

Sorry, that's not right.

I take it back.

Zach Hauk

:9 so i'm wrong

Balarka Sen

$b^2 - 4ac < 0$ is not the saddle.

Akiva Weinberger

So the other one?

Zach Hauk

$b^2-4ac > 0$?

Brody

03:15

@heather Those feelings aren't unique to you, lol

Balarka Sen

Right. Easy to check: $z = xy$.

Can you explain why?

Zach Hauk

ax^2 + bz + cy^2

?

Akiva Weinberger

Looks like a smiley parabola on the line $y=x$, looks like a frowny parabola on $y=-x$

Balarka Sen

No, I mean, $a = 0, b = 1, c = 0$.

Akiva Weinberger

so it's a saddle 'cause it's got both

Balarka Sen

03:16

$z = 0 x^2 + 1 xy + 0 y^2$.

@Akiva has it right.

pilko

Here is some gook looking hyperboloid : goo.gl/images/znygFs but I cannot find any saddle on it.

Zach Hauk

locally

Balarka Sen

@pilko Near any point on the the "equatorial circle"

Akiva Weinberger

Oh, yeah, @pilko. We want "hyperbolic paraboloid", not "hyperboloid"

Brody

A pringle chip

Akiva Weinberger

03:18

@BalarkaSen $z=xy$ is a hyperbolic paraboloid, not a hyperboloid

Balarka Sen

@Akiva I am aware. But locally it still looks like $z = xy$

Akiva Weinberger

Locally.

Yeah.

Balarka Sen

Ofc. We are talking about the local picture, not the global. :)

Zach Hauk

ok so our equation was $z = xy$. when y = x, $z = x^2$, and when $y = -x$, we have $z = -x^2$

Balarka Sen

But thanks for reminding me the terminology. Good audience.

Zach Hauk

03:19

that's why?

Akiva Weinberger

Yeah, pilko's picture has negative curvature (aka "saddle") throughout

Balarka Sen

Aha @Zach.

The point is one direction goes down and one direction goes up.

Anyway, OK. So $b^2 - 4ac > 0$ implies saddle. What about the other two?

(also, note that det of hessian of $ax^2 + bxy + cy^2$ is $4ac - b^2$ :P I bungled a sign)

Daminark

Oh messing up signs is annoying

Balarka Sen

Totally man

Daminark

By the way hey everyone!

Zach Hauk

03:22

$b^2-4ac < 0$ implies local min or max

Akiva Weinberger

Hi

Balarka Sen

@Zach Which when?

heather

@Brody as someone in 8th grade who's struggling to understand parts of single variable calculus and set theory, I'm in awe =)

Zach Hauk

uhh

Akiva Weinberger

@BalarkaSen Did we show that in general? Or just for this case

Zach Hauk

03:23

@heather then again, you're doing that much more rigorously, no?

last time i heard your text was spivaks...

Akiva Weinberger

@heather Calculus varies wildly by difficulty. Parts are harder than others.

Daminark

When we defined manifolds as being locally the graph of a smooth function, as well as the 0 set of a smooth function, we had to prove the equivalence of associated definitions of the tangent bundle

Balarka Sen

@AkivaWeinberger You can come up with a similar argument as yours, with an up direction and a down direction.

But whatever man! Let's get to the good stuff first and handwave technicalities.

Daminark

And I messed up a sign, so I was so confused

heather

@ZachHauk eh, I've looked at it a bit, but I'm also using YouTube videos and such =P

@AkivaWeinberger I'm not in the hard parts =)

Daminark

03:24

Dear lord when I was in 8th grade I did algebra 1

Balarka Sen

@Zach Hint: Complete square.

Zach Hauk

should i be using the $y = x$ thing again?

Balarka Sen

Oh, @Akiva, completing square gives a proof of that in general. So let's do it.

Paul Plummer

That is why I don't put signs on anything. Also why I don't mess around with indexing things.

Daminark

I mean I dropped a negative sign and that gave me $2\delta_j h_k$ instead of $0$

Yeah indices occasionally short-circuit my mind as well

Zach Hauk

03:26

@heather you calling me smart takes away from @Balarka and @Akiva... who are much smarter than me, and about the same age

Akiva Weinberger

@PaulPlummer "OK, so these terms cancel maybe..."

Brody

@heather it's definitely peculiar for the age/level. regardless, there are many things in both the univariate and multivariate flavors that can be approached lightly with basic computation in mind

heather

@BalarkaSen, you're in 8th grade?

Zach Hauk

well no

heather

@AkivaWeinberger, you said you were in 11th?

Zach Hauk

03:27

i think he's about 15

or 16, not sure

heather

geesh....

Daminark

Wait a second, @Balarka I thought you were like, a grad student or something

Akiva Weinberger

@heather Yeah

I think Balarka is 16 or 17

heather

i think i need to leave before I melt from awe =P

Akiva Weinberger

Like, roughly my age but a bit younger

Daminark

03:28

At minimum

Balarka Sen

@Zach Nah. You just have to complete the square.

Akiva Weinberger

So either 11th or 10th grade? Or whatever the Indian equivalent is

Balarka Sen

11th.

Paul Plummer

No I just leave it as an exercise to the reader @AkivaWeinberger

heather

@BalarkaSen ...

Akiva Weinberger

03:28

OK, so we're in the same grade

heather

you guys are amazing

Akiva Weinberger

Thank you (also let me say I'm bad at receiving compliments)

heather

@ZachHauk all of you are crazy smart.

Balarka Sen

I only pretend to be.

Zach Hauk

umm this isn't completing the square but notice that either both $a$ and $c$ must be negative or positive, since their product (times 4) is greater than $b^2$, which must be positive

Brody

03:29

^what Heather said

Akiva Weinberger

@BalarkaSen That's a great answer

Zach Hauk

perhaps it depends on the negativity of positivity of $a$, or $c$?

Balarka Sen

@Zach On $a$, yes.

Zach Hauk

why not on $c$?

Balarka Sen

$ax^2 + bxy + cy^2 = a(x + \frac{b}{2a} y)^2 + \frac{4ac - b^2}{4a} y^2$ or something if I did algebra right.

Zach Hauk

03:32

ok sorry i gotta turn on mathjax

Harry Evans

Good evening again!

Balarka Sen

If $4ac - b^2 > 0$ of course that's a paraboloid. Upside-down if $a<0$, downside-up if $a > 0$

Also it's very clear from that that if $4ac - b^2 < 0$ it's a saddle because you get your two up/down directions.

Point being $z = x^2 - y^2$ is a saddle

Zach Hauk

oh, sorry. to be honest, @Balarka, I didn't know you meant completing the square of the actual polynomial

w.r.t. either variable

DHMO

@BalarkaSen What is so special about $\Bbb C$ that makes the fundamental theorem of algebra work?

Balarka Sen

Yeah, I was being vague.

Will Njundong

03:34

How many are taking French or business (or both)?
Mathematically, how do I tackle this logic?

Zach Hauk

i guess i'm having an off day. or an off week. or an off year.

Balarka Sen

@DHMO It's algebraically closed.

DHMO

@BalarkaSen I thought that's what I'm asking?

Balarka Sen

Did I give a circular answer

DHMO

yes

Balarka Sen

03:35

lol.

Akiva Weinberger

So, TL;DR is, after you make an appropriate change of coordinates, it's either $z=x^2+y^2$ or $z=x^2-y^2$ (depending on the sign of $b^2-4ac$)?

Will Njundong

I'm thinking I need to add French takers, business takers and people who do both, together. I sthat correct?

Zach Hauk

no, depending on $a$ I think.

@BalarkaSen what about $b^2 = 4ac$?

Balarka Sen

Could also be $z = -x^2 - y^2$ @Akiva. This is called "diagonalizing the quadratic form".

Akiva Weinberger

Oh, so $\pm z=x^2+y^2$ and $\pm z=x^2-y^2$

The point is that the former is a cup and the latter is a saddle

cup/cap/thingy

Daminark

03:37

@DHMO I think it's mostly that $\mathbb{R}$ comes really close to doing it

Zach Hauk

oh @BalarkaSen then it looks like there's a whole line of critical points

DHMO

@Daminark heh?

Akiva Weinberger

@BalarkaSen Sounds linear algebra-y. Is this $z=x^\top A x$ for a symmetric matrix $A$, which you're orthogonally diagonalizing?

Daminark

Odd degree polynomials already have a real root by intermediate value theorem

Akiva Weinberger

Never mind, it should be fine.

Daminark

03:38

So I guess it does make sense in some intuitive/philosophical way that a 2 degree extension is all that's necessary

Do you know the proof of FTA?

Akiva Weinberger

$\begin{bmatrix}x&y\end{bmatrix}\begin{pmatrix}a& b/2\\b/2&c\end{pmatrix}\begin{bmatrix}x\\y\end{bmatrix}$?

${}=ax^2+bxy+cy^2$?

DHMO

Can $[\Bbb R:X]=2$ be solved?

Balarka Sen

@AkivaWeinberger Yeah, but the latter two are the same so whatever

Akiva Weinberger

Maybe with the axiom of choice @DHMO

Zach Hauk

i just went to get a piece of bread from a ziploc bag, but little did i know how humid the bag was

Balarka Sen

03:40

Sorry, got disconnected.

@AkivaWeinberger Yeah.

Daminark

@DHMO Does that mean finding a field such that $\mathbb{R}$ is a degree 2 extension?

Balarka Sen

You can find this on Ted's book, called LDL decomposition.

Daminark

My knowledge of algebra is like, baby group theory + some linear algebra here and there

Akiva Weinberger

You should learn Sylow's theorems,

Daminark

That, I know

DHMO

03:41

@Daminark yes

Akiva Weinberger

and not really understand them and then forget what they are. At least, that's what I did.

DHMO

@AkivaWeinberger I'm serious

Akiva Weinberger

@DHMO So am I...

Do you know what a Hamel basis is?

Balarka Sen

@ZachHauk OK, so determinant of Hessian > 0 with a > 0 means the point on the qd form is minimum, determinant of Hessian > 0 with a < 0 means the point on the qd form is maximum, and determinant of Hessian < 0 means the point is a saddle.

Daminark

Lol my REU paper last summer was a proof of the Sylow theorems, along with some applications

DHMO

03:42

@AkivaWeinberger yes

Balarka Sen

if it's = 0 it's inconclusive. You can't say much.

DHMO

@BalarkaSen any idea?

Akiva Weinberger

@DHMO So get a Hamel basis for $\Bbb R$ that includes $\sqrt2$. Let $X$ be the field generated by everything in the Hamel basis except for $\sqrt2$. Does that work?

Balarka Sen

This applies in general for any C^2 function; this is called the second derivative test.

Akiva Weinberger

Would this get $[\Bbb R:X]=2$?

Balarka Sen

03:43

The point is a C^2 function can be locally approximated by not only a tangent space but a tangent quadratic.

If you go down this rabbit hole long enough you'd run into Morse theory

Zach Hauk

Well at least I understand the parabola case

DHMO

@AkivaWeinberger I am doubtful about this

Balarka Sen

I think Akiva's construction has a merit to it.

Zach Hauk

paraboloid*

Balarka Sen

But I am not thinking hard about it

Zach Hauk

03:49

I suppose I should be off to work on my application. thanks for the explanation @Balarka. have a nice night, math.SE (or day)

Balarka Sen

No problem. Have fun.

Zach Hauk

i'll try not to stress out too much ;]

Brody

What is $z=ax^2+bxy+cy^2$ about?

Balarka Sen

If you ever want to know more about this Ted can send you stuff; he has a whole chapter on this in his book

Akiva Weinberger

When is the origin a saddle point

@Brody

Will Njundong

03:50

Hello, is anybody here proficient with sets and logic?

Akiva Weinberger

Maybe?

DHMO

@WillNjundong just ask your question

Brody

What object/graph does the equation plot, I mean? A paraboloid?

Akiva Weinberger

@Brody Yeah. Sometimes an elliptic paraboloid, sometimes a hyperbolic paraboloid:

This has illustrations

Brody

@AkivaWeinberger Thanks. I don't think I've seen this form before.

Akiva Weinberger

03:52

You can do a real linear change of coordinates to get it to the forms in the Wiki

($z\mapsto-z$ is linear...)

Brody

I'm poor with coordinate systems and graphs and such

Will Njundong

There is a group of 191 students, of which 10 are taking French, business and music; 36 are taking French and business; 20 are taking French and music; 18 are taking business and music; 65 are taking French; 76 are taking business; 63 are taking music.
How many take French or business (or both)?
How I have it in mind: https://goo.gl/photos/LCPuei8j8krqX6wr8

Daminark

@Brody When I was in physics last year, coordinates messed me up a good deal

Also not being able to visualize things in motion whatsoever

Brody

I'm poor with math generally, lel. And this is a particular example

Akiva Weinberger

@WillNjundong Remember that the 36 taking French and business include the 10 taking French, business, and music.

So there's only 26 who take French and business and not music.

Daminark

03:56

Lol, with analysis I'm able to fake it, and what I've done in algebra I'm not terrible

Brody

@Daminark What sort of physics and coordinates?

Balarka Sen

@DHMO About your FTA question from before. I don't have a good answer, but the point is C is "a lot more connected" than any other field you'll see in day to day life.

DHMO

@BalarkaSen "day to day life"

Daminark

Just like, in general, last year I had a course which did mechanics first quarter, E&M second quarter, and waves third quarter

Brody

Call me petty, but for some reason I like $ax^2+2bxy + cy^2$ more

not that it effectively makes any difference

Daminark

03:58

My biggest problem was not being able to visualize moving things

Balarka Sen

@DHMO :P

Daminark

But also I never really learned much about cylindrical or spherical coordinates

Or any kind of parametrization

Balarka Sen

But really, there is a proof of FTA using the topology of C, which is a lot better than say R

Because you can disconnect R by removing a point. You can't do that with C

Daminark

In my analysis class we did 2 proofs of FTA

Well, 2 and a half

Balarka Sen

But what happens is C - point has nontrivial fundamental group. That's the key to the proof of the fundamental theorem of algebra

Akiva Weinberger

03:59

("a" proof)

Transcript for

$Mathematics$ Mathematics