Mathematics - 2017-03-01 (page 4 of 5)

Alessandro Codenotti

20:03

Not necessarily

Take $B=\{(a,b):a,b\in\Bbb R\}$, do you agree that $B$ is a basis of the Euclidean topology?

$(1,2)\cup(4,5)$ is a union of elements of $B$ but it's not in $B$

Astyx

Hi to those I haven't seen already today

Akiva Weinberger

Bonsoir (rhymes with Tom Sawyer)

Astyx

Haha

If we have a scalar product in a hyperplane of a vector space, is it always possible to extend it to the space itself ? Do we just need to arbitrarily chose a vector not in the hyperplane and say it's orthogonal to the hyperplane and unitary and voilà ?

Zach Hauk

hi

Astyx

Hi @Zach

Zach Hauk

20:17

i was watching some of @Ted's LA lectures lat night

Ted's favorite equation popped up

Akiva Weinberger

What's his favorite, $Ax\cdot y=x\cdot A^\top y$?

Zach Hauk

yep

oh we also derived law of cosines

Akiva Weinberger

We also have $A^\top Ax\cdot x=\|Ax\|^2$

and thus $\|A^\top Ax\|\|x\|\le\|Ax\|^2$

which might be useful for things

Zach Hauk

fuck there's dried blood all over my foot

Akiva Weinberger

Got to go.

And good luck

Zach Hauk

20:21

adios

Akiva Weinberger

on your dried blood foot situation

Daminark

Hey

Damn :/

Astyx

I didn't know Ted's lectures were that brutal Zach

Alessandro Codenotti

lol

Zach Hauk

Hi @Daminark

There's this one guy who comments on every one of Ted's lectures

I call him "Ted's biggest fan"

Alessandro Codenotti

20:22

Now I kinda want to hear a super aggressive death metal song about linear algebra @Astyx

Astyx

Now you mention it, so do I

Zach Hauk

"My Life is a Null Space (Everything goes to 0)"

Daminark

"My life is a kernel" :P

Zach Hauk

"Fuck You, You Kernel (Stop sending everyone to 0!)"

and more...

include the hit single
"I'm Gonna Cross You With The Zero Vector"

Astyx

"My life has Cauchy sequences that do not converge"

ie it's not complete

Alessandro Codenotti

20:28

Where are Ted's smacks when we most need them?

Astyx

lol

Daminark

smacks in Ted's stead

lol

s.harp

Do you guys sometimes get mails from cranks?

Alessandro Codenotti

I thought that happens only to professors?

Zach Hauk

am i a crank? :P

s.harp

20:40

I've got a mail address on the institute website, I sometimes get emails telling me about this guy gabor fekete and his amazing discoveries and how its a crime he hasnt been acknowledged yet

> Dear Professor,

*Heartiest Greetings and the Best Wishes for the New Year!!!*

*Now all the Quantum Phenomena / Quantum Effects, which remained mystery so
far, can be explained by the New Quantum Theory. Earlier theories could not
explain these mysterious phenomena.*

*New Quantum Theory* developed in 2012 by an Indian scientist ‘*Narendra
Swarup Agarwal*’ explains not only the Wave Particle Duality but also all
the Quantum Phenomena which could not be answered earlier by the Quantum
Theory.

Zach Hauk

gabor fekete???

Alessandro Codenotti

You're a grad student right? Too many people there, I keep forgetting that kind of stuff

s.harp

yes

Alessandro Codenotti

Maybe he got ignored by all the professors already and now he's spamming the grad students

Zach Hauk

i was a crank at one point... when i was 10 or 11

i find it embarrassing that an adult is still not following rigor however

user379685

20:44

How can i prove that limit x->0 $ (1+sin(x))^{(1/x)} $ equals e?

Lozansky

How do you prove that the Schrodinger equation, i.e. $$\frac{d}{dt} \psi(t) = -iH\psi(t)$$ is a Hamiltonian system?

Zach Hauk

use {} to make the power around the whole $(1/x)

s.harp

do you know what $\ln(1+\sin(x))/x$ converges to @user379685

Alessandro Codenotti

Lacking rigour and claiming to have solved the Riemann hypothesis are quite different things @Zach

Zach Hauk

Oh

I never claimed any of that, luckily

user379685

20:45

@s.harp no

Simple

0

Q: Find the probability generating function of a particle that performs a symmetric random walk starting at the origin

A particle performs a symmetric random walk in two dimensions starting at the origin; each step is of unit length and has equal probability $1/4$ of being northwards, southwards, eastwards, or westwards. The particle first reaches the line $x+y=m$ at the point $(X,Y)$ and at the time $T$. Find...

Zach Hauk

I just was fucking with new concepts

s.harp

@user379685 do you know what the power series expansion of $\ln$ around $1$ is?

@Lozansky I'm not sure what you mean with Hamiltonian system, but in the usual sense of the word SE is not a hamiltonian system

Zach Hauk

Umm, @AlessandroCodenotti is it bad if i study two things at once?>

Lozansky

@user379685 $sin(x) \approx x$ for small $x$

Zach Hauk

20:48

I want to continue studying projective geometry while starting Spivak's

(I haven't done any rigorous calculus)

Lozansky

@s.harp Wait hold on

Alessandro Codenotti

It depends on you I guess. I prefer to focus on something when I study stuff on my own, but I also have to take courses in different topics at the same time

Lozansky

$$\dot{q}(t) = \nabla_p H(p(t),q(t))$$
$$\dot{p}(t) = -\nabla_q H(p(t),q(t))$$

@s.harp

Zach Hauk

@AlessandroCodenotti do you think i can handle it?

Alessandro Codenotti

Maybe. But you shouldn't hurry too much, listen to Ted, he knows what he's talking about regarding pedagogy

He's definitely the most experienced teacher around here

Akiva Weinberger

21:01

Does Brouwer's fixed point theorem hold for a torus minus an open disk?

Mike Miller

State the theorem for me?

Every map has a fixed point?

Akiva Weinberger

Yeah

The proof for balls starts by drawing lines from $f(x)$ to $x$ to the boundary

but that doesn't seem to be doable here.

Zach Hauk

isnt brouwers fixed point theorem

where you have like a continuous mapping from a set to itself

and at least one point is fixed under the mapping

like a closed disk

Akiva Weinberger

Yeah, for convex sets in $\Bbb R^n$.

Krijn

(Brouwer is also Dutch, fwiw)

Akiva Weinberger

21:06

It doesn't apply to, say, a circle (rotate) or a sphere (map everything to its opposite point)

but it does apply to a cube or a solid ball in any dimension.

Zach Hauk

circle != disk and sphere != ball, right?

the latter of these contain their interior?

Akiva Weinberger

Yeah.

Zach Hauk

@AkivaWeinberger do you have any more topology questions?

Astyx

How do you call the axis of rotations of order 2 of a tetrahedron ?

As in, what's the geometrical name ?

Akiva Weinberger

@ZachHauk Continuing from the last thing I gave you: Why do the closed sets have to be unbounded?

Recall that the Heine–Borel theorem states that a set is compact iff it’s closed and bounded, so essentially I want you to show that the distance between two disjoint compact sets is nonzero. (Hint: Every continuous function on a compact set attains its minimum.)

Vrouvrou

21:18

hello @Astyx

Zach Hauk

so for example, a finite subset of $\Bbb Z$ is always compact?

Astyx

Hi again @Vrouvrou

Topologicalife

Hi there.

Astyx

Hi Topologicalife

Akiva Weinberger

@ZachHauk Yeah. You should be able to prove that yourself from the definition of compact.

Krijn

21:21

@Pedro I got some pings from a user I don't know about something I did, but can't for the life of me figure out what happened. You seem to have been confused about it as well. Have you found out what the problem was?

Vrouvrou

@Astyx i'm with the same problem, $\min\{n_i\}=0$ when $I$ is infinite , i don't see an other answer

Zach Hauk

its obviously bounded

and its obviously closed too

Topologicalife

I have this function: $$h(x)=\frac{(x-c)^2}{2}+\lambda|x| \mbox{ with } \lambda > 0 \mbox { and } c \in \mathbb{R}$$. I want to prove that there is a minimum in $x = 0$ of value $\dfrac{c^2}{2}$. Note that $h$ is not diferentiable at $x = 0$. How can I do it?

Akiva Weinberger

@ZachHauk From the finite subcover definition, I mean.

Astyx

@Vrouvrou What if $n_i = 3i+2$ ?

Vrouvrou

21:22

i know that it is 2

Astyx

So it's not $0$

Zach Hauk

@AkivaWeinberger start by removing any redundant open sets of the (infinite cover)

Vrouvrou

so what can be the answer for my problem ?@Astyx

Zach Hauk

That is, it's redundant if 1. it covers none of that subset or 2. there exists 2 open sets in the cover already covering that

Topologicalife

I tried to prove $f(c) \leq f(x)$ where $c$ is the minimum, but I'm getting stuck with a part of it.

Astyx

21:24

Well $\min\{n_i\}$ can be anything, for any integer $k$, if $n_i = k+i$ then $\min\{n_i\} = k$

Zach Hauk

Under these conditions, each cover must cover at least one element of that subset not covered by any others

Astyx

So you can't simplify the answer

But that's not an issue

You just need to know there is a min

Akiva Weinberger

@ZachHauk Yup.

Zach Hauk

and so, since there are a finite number of elements, then there must be a finite number of covers

thus it has a finite subcover

Astyx

@Zach Do you know Euclidean spaces ?

Zach Hauk

21:25

what do you mean? what about them?

Vrouvrou

@Astyx the min exists because $N$ i bounded from below

Akiva Weinberger

Yep. Alternatively, take one set in the cover for the first element, one set for the second, one set for the third, etc. That will give you finitely many.

Astyx

@Vrouvrou kinda, that justifies the existence of an inf rather

@Zach Well have you studied them ?

Zach Hauk

that's very vague, sorry

what do you need to know?

Astyx

No, I just want to know what you know about them :p

Zach Hauk

21:28

name an example of a topic in that, please

Akiva Weinberger

@ZachHauk Here's another two examples: Is $\{1,1/2,1/3,\dots\}$ compact? What about $\{1,1/2,1/3,\dots\}\cup\{0\}$? (Prove from definition)

Zach Hauk

by the way

how do you define bounded?

Alessandro Codenotti

I'm a bit sick so I'm not thinking straight and @liad questions gave me a doubt, $A\neq \bigcup_{a\in A} a$ in standard notation, right?

Akiva Weinberger

Hint for the second one: Take an arbitrary open set to cover $\{0\}$. Now you only need to cover the points not in that set.

Zach Hauk

wouldn't you also need to define an ordering relationship?

Alessandro Codenotti

21:29

Just a sanity check

Akiva Weinberger

@ZachHauk It's contained in some sphere

Astyx

@Alessandro right

You need $\{a\}$

Akiva Weinberger

Or, there's an $M$ such that the distance between any point and the origin is less than $M$.

Zach Hauk

but that's introducing a metric

Akiva Weinberger

This is for subsets of $\Bbb R^n$.

Zach Hauk

21:30

oh, okay

Akiva Weinberger

Heine–Borel only applies to subsets of $\Bbb R^n$. In fact, it's false for general metric spaces

Vrouvrou

thank you @Astyx

Astyx

My pleasure

Zach Hauk

@AkivaWeinberger the cover $\{1\}, \{1/2\}, \dots$ has no finite subcover..

Akiva Weinberger

Yup. And those are open in the set, why?

Zach Hauk

21:32

im stupid

Akiva Weinberger

You're close

Krijn

Just wave your hands and make them open

2

Zach Hauk

those aren't open, are they?

Krijn

But wave in the proper way

Alessandro Codenotti

@Astyx just as I wrote him a few minutes ago, thanks, I'm not too sick then :P

Zach Hauk

21:33

@AkivaWeinberger does the cover have to be open in $\Bbb R^n$ or

Alessandro Codenotti

@Krijn "topology: physics edition"

Akiva Weinberger

@ZachHauk ...or in the set? It doesn't matter, they're going to end up the same. But try finding a cover in $\Bbb R^n$ right now

Zach Hauk

$$\cup$$

how do i do the large cup

Krijn

@AlessandroCodenotti No no, here it really works if you wave your hand in the proper way: Hold them together as if you are praying and then open them juuuuuust a little

Akiva Weinberger

@ZachHauk \bigcup $\bigcup$

Alessandro Codenotti

21:35

Fair enough

Krijn

Coincidentally, I TA a class in topology, mostly for students that study both mathematics and physics

Zach Hauk

ok so

i propose

$(1/n - 10^{-n}, 1/n + 10^{-n})$

Akiva Weinberger

Yeah, that'll work

So what about the other set?

$\{0\}\cup\{1,1/2,1/3,\dots\}$

Zach Hauk

notice that

we must also contain 0

which is a limit point

so any open set containing 0 must contain part of this set

let's say we have that the lower bound of this open set is $n$

we have that $n > 0$

thus, take the reciprocal of $n$

and floor it

Akiva Weinberger

hits gas on car

Zach Hauk

21:42

since $\frac{1}{n} > \lfloor \frac{1}{n} \rfloor $

wait

no, i mean

wait, yes

that's what i meant

so ya floor it

and we get a natural

Astyx

You get 0 mainly

Akiva Weinberger

@Astyx What?

Zach Hauk

uhh no

Akiva Weinberger

Oh, $n\ll1$, @Astyx

Astyx

Oh right ...

Krijn

21:43

Just say that there is some $1/n$ for which all smaller $1/k$ are in the open

Astyx

My bad

Zach Hauk

@Krijn good point

therefore

Krijn

Keep it simple

Zach Hauk

there are finitely many $1/k$ greater than $1/n$

Akiva Weinberger

Yeah

Zach Hauk

21:44

and in conjunction with the open set, makes a finite subcover

Akiva Weinberger

Right. 'Cause you never need infinitely many sets (open or not) to cover finitely many points.

By the way, $\{0\}\cup\{1,1/2,1/3,\dots\}$ is known as the one-point compactification of $\{1,1/2,1/3,\dots\}$. 'Cause adding one point makes it compact.

Also, $\{1,1/2,1/3,\dots\}$ is homeomorphic to $\Bbb N$, right? @ZachHauk

The homeomorphism given by taking the reciprocal.

Heya, @Ted.

Ted Shifrin

Heya, DogAteMy.

Astyx

Hi @TedShifrin

Zach Hauk

hi @Ted

Akiva Weinberger

I made Zach show that $\{1,1/2,1/3,\dots\}$ isn't compact but that its union with $\{0\}$ is.

Zach Hauk

21:47

@AkivaWeinberger I don't remember homeomorphic

Ted Shifrin

@Zach: Cuz you haven't learned topology yet :P

Zach Hauk

continuous mapping?

Ted Shifrin

Hi @Astyx

Zach Hauk

@Ted Akiva was giving me a little sneak peek

Ted Shifrin

continuous with continuous inverse.

Akiva Weinberger

21:47

It essentially means "the same" in topology. Technically, it's a continuous bijection (one-to-one map) with a continuous inverse

Or, it's any bijection that doesn't change the open sets.

Zach Hauk

@Ted I'm watching the lectures at a rate of 1 per night. Is that too slow?

Astyx

Did you know the age of miss America was correlated to murders by heat ?

Ted Shifrin

My course has basic topology in $\Bbb R^n$ in it, @Zach.

Akiva Weinberger

I mean, it's faster than the students in that class @ZachHauk

Ted Shifrin

@Zach: That's for you to decide. Also, I sent you an email.

Zach Hauk

21:48

also i just had some frozen berries, which are good

Ted Shifrin

They're probably better eaten thawed.

Akiva Weinberger

My grandma likes frozen grapes.

Zach Hauk

AOL "You Got Mail!" voice plays

Ted Shifrin

I just made a quinoa, veggie, shrimp salad to take to a potluck tonight.

Sorry, DogAteMy, unkosher.

Zach Hauk

@Ted oh is it the psets?

Alessandro Codenotti

21:49

Hi @Ted

Ted Shifrin

Hi @Alessandro

@Astyx Establishing that correlation = causation, clearly.

Akiva Weinberger

/ˈkinˌwɑ/. What a word. Apparently it was borrowed from the Spanish quinua, which took it from the Quechua.

Ted Shifrin

I thought it was American Indian in origin, actually. Silly me.

Zach Hauk

Quixote is an interesting word

Ted Shifrin

Now that is Spanish.

Astyx

21:51

@Ted Obviously this is not a coincidence

Ted Shifrin

Yes, obviously.

Akiva Weinberger

Why we don't spell it kinwa after the modern Quechua spelling, I dunno

Zach Hauk

@Ted Last night I just glossed over the first 4 or 5, and watched most of the Subspaces lectures

and I saw your favorite equation too

Ted Shifrin

@Zach: Have you ever done vector proofs of interesting geometry stuff?

Zach Hauk

I saw the law of cosines proof too

Ted Shifrin

21:52

Favorite formula :P

It's surprisingly useful/important ... even in functional analysis :)

Akiva Weinberger

@ZachHauk In any case, I was gonna say that $\{0\}\cup\{1,1/2,1/3,\dots\}$ (and things homeomorphic to it) is the one-point compactification of $\Bbb N$ (and things homeomorphic to it)

Zach Hauk

oh, so $\Bbb R^2$ isn't compact because it's unbounded

Ted Shifrin

@Zach: It starts slowly. Things get faster and more interesting later ...

Zach Hauk

@Ted You sent me a pdf containing various proofs using linear algebra

and trig

Ted Shifrin

@Zach: But be careful. Because we could put a metric on it in which it is bounded.

Akiva Weinberger

21:53

@ZachHauk Right. Or, you could cover it with lots of open disks of radius 1 or something, which has no finite subcover

Ted Shifrin

That didn't have linear algebra, @Zach. That was trig and beginning calculus stuff with trig.

Krijn

@TedShifrin What was this?

Zach Hauk

oh, nevermind

I read "using linear algebra" instead of "usually linear algebra"

Ted Shifrin

@Krijn: $Ax\cdot y = x\cdot A^\top y$.

@Zach: You should work on your reading.

Zach Hauk

I'll go back to third grade

Ted Shifrin

21:55

Hmm, OK, good idea.

Krijn

@TedShifrin Of all the formulas you could pick

Ted Shifrin

@Krijn: Yeah, but my linear algebra and multivariable math students never forgot it because of that. It's obviously not my universal favorite. Just my favorite at the beginning of the course.

Stokes's Theorem is high up there.

Krijn

Yeah that's a proper fantastic one

So much in so few symbols

Alessandro Codenotti

@TedShifrin is there a topological property that is the right thing to look at rather than boundedness? (Like being completely metrizable rather than completeness)

Zach Hauk

@Ted to be honest when i was skimming the beginning lectures, I wanted to raise my hand because I knew the answer :P

Akiva Weinberger

21:57

@AlessandroCodenotti I'd say compactness

Ted Shifrin

DogAteMy: He means to characterize compactness in a general metric space.

Akiva Weinberger

Oh. Total boundedness, I think

Ted Shifrin

@Alessandro: Look up "totally bounded."

G'night, @MikeM.

Akiva Weinberger

https://en.wikipedia.org/wiki/Totally_bounded_space

It's a metric property, not a topological one, but eh

Ted Shifrin

Sure, @Zach. There will be times (perhaps even in the calculus stuff) that it goes a bit slowly for you.

Akiva Weinberger

22:00

The countable discrete metric space is bounded (everything fits in an open ball of radius 2) but not totally bounded (no finite amount of balls of radius 1/2 can cover it)

In $\Bbb R^n$, the concepts are the same.

Zach Hauk

the taxicab metric is kind of wonky

the unit circle is a diamond

Akiva Weinberger

Got to go

Zach Hauk

adios

Ted Shifrin

Look at the different unit spheres in the $\ell^p$ norm, @Zach.

(Or unit circles in two dimensions.)

Alessandro Codenotti

Hm, $(0,1)$ and $\Bbb R$ are homeomorphic but one is totally bounded while the other isn't

Ted Shifrin

22:02

You have to specify metrics, @Alessandro.

I can pull the usual $\Bbb R$ metric back to $(0,1)$, of course.

Zach Hauk

back to the spectral theorem

Alessandro Codenotti

Sure, but it won't even be bounded then. I guess boundedness just doesn't work well with homeomorphisms after all

Ted Shifrin

Uh huh ...

Definitely not a topological property.

Zach Hauk

i was confused because i thought bounded was a topological property

then i realized we were talking about metric spaces

Astyx

Bye chat

Ted Shifrin

22:07

Bubye

Zach Hauk

one by one they all leave

Lozansky

@TedShifrin Hello

Zach Hauk

@Ted sent back

Krijn

I'm still here tho

Alessandro Codenotti

22:24

I'm leaving too, good night everyone

Krijn

Night

LegionMammal978

> Find the effective yield for a $2000 investment that is worth $4000 after 15 years.

Don't get this. I understand compound interest, etc., but haven't heard of effective yield on an investment.

Daminark

RIP - the chat

Mike Miller

Wht re we doingv

Krijn

Number theory/arithmetic geometry.

Zach Hauk

22:39

RIP ted

Daminark

$\mathbb{shrugs in La^T_eX}$

Zach Hauk

@Ted yeah I'll need a password...

LegionMammal978

@Daminark Huh? I can use $\LaTeX$!

Daminark

Lolol, so "Does ____ in Latex" is a joke I tend to make a lot

Like people know me in part for that joke

And one of my friends bought a used book for a class which actually had "Cries in LaTeX" as an annotation

My reaction: :O

$L^{_A}T_EX!$

Zach Hauk

@Ted If you're worried about security you can send it to my FTP server

because this digital id is not working

Transcript for

$Mathematics$ Mathematics