Mathematics

Topologicalife

12:01 AM

@AkivaWeinberger it has two real roots since $f(0) = -1$ and because we showed that $f'(x) > 0$ for all $x$.

Right?

Hi @Semiclassical

Ted Shifrin

12:15 AM

@Topologicalife: That looks like a Spivak question. Think using Rolle's Theorem to show at most two roots.

Hi @Semiclassic, DogAteMy

Topologicalife

Yeah @TedShifrin

But he already solved it.

I was asking just a detail about his/her proof.

Ted Shifrin

Oh, ok.

Topologicalife

I think because we showed that all the roots are in $[-1,1]$, and $f(0) = -1$ and $f$ is increasing for all $|x| > (0,1)$, then $f$ must have two roots.

It makes sense on my mind :D

Err.

I messed up that msg.

I meant: I think because we showed that all the roots are in $[−1,1]$, and $f(0)=−1$ and $f$ is increasing for all $x\in(0,1)$, then $f$ must have two roots.

Maks

@TedShifrin Could you explain a theorem to me ?

More than the theorem, a consequence of it

Ted Shifrin

Like what?

Maks

12:26 AM

There's a theorem of subspaces

Ted Shifrin

I dunno what you're referring to.

Maks

that say is V and W are subspaces with dimention $n < \infty$
Such that $ V = \langle b_i \rangle$ and $ W = \langle c_i \rangle$
Then there's only one linear transformation T:V -> W such that $T(b_1) = c_1$ for all $i$

Ted Shifrin

Those are bases for the respective subspaces. OK, sure.

Zach Hauk

hi!

Maks

With that theorem

If they gave you any linear transformation

Ted Shifrin

12:29 AM

hi @Zach

Zach Hauk

just had some really good meatballs

Maks

And the vectors they give you on the domain are a basis of all the space on the domain, then,by theorem, you know its a LT

For example

Ted Shifrin

Well, no, you define it on other vectors to make it be a LT

Maks

If I'm asked if there is a LT such that $T(1,0) = (1,2,3)$ and $T(0,1) = (0,4,2)$ by theorem I can say yes, because $(1,0)$ and $(0,1)$ are a base of $\mathbb{R}^2$

Zach Hauk

@Ted Akiva told me how he took the AP calc BC test right before high school

Ted Shifrin

12:31 AM

Yeah, actually, you need the $v_i$ to be a basis for $V$, but the $w_j$ can be anything.

Maks

@TedShifrin Yay, exactly

Now, why ? I dont see the relation

Zach Hauk

do you think I should ask the administration about doing something like that?

Ted Shifrin

@Zach: Only take it if you're going to ace it. But what does it gain you at your high school?

Zach Hauk

I don't know yet

to ace it i would only need to do some review

Ted Shifrin

I would recommend learning Spivak's book and knowing way more than the BC ... well. But that's just me.

I taught a high school kid the Spivak course and then he took the BC for the formalities ...

Zach Hauk

12:33 AM

i mean

Spivak before anything else?

like, before i continue with algebra and stuff

Mike Miller

Do whatever you want man

Ted Shifrin

Depends on your life plan. I still don't see the point of being a graduate student before you go to college. But it's not my life.

Maks

@TedShifrin BC ?

I have the hoffman

Oh, I thought you were talking to me, sorry

Zach Hauk

@Ted I don't know where to go even

I don't have anyone to really talk to about math besides the people in this chat, so aside from what I have right now, I don't know what to study or what to do

Ted Shifrin

@Maks: A linear transformation is uniquely determined by where it sends basis vectors in the domain.

@Zach: We had discussions about this months ago. I am more in favor of building solid foundations, but my opinion is not universal.

Zach Hauk

12:38 AM

I mean, specifically?

sorry, as in, should I be studying Spivak's after this?

Ted Shifrin

I told you to learn Spivak and a rigorous multivariable calculus course before all this algebra.

Zach Hauk

You did? sorry, i don't recall.

Ted Shifrin

Yeah, you weren't much interested in my thoughts.

Zach Hauk

Are your lectures rigorous enough?

Ted Shifrin

As you know, math is all about the exercises you do.

5

Zach Hauk

12:44 AM

@Ted am I allowed to use a computer for the "diagonalizing quadratic forms" question?

Ted Shifrin

I told you to skip that question once you know what's going on.

Zach Hauk

how do you mean?

Semiclassical

rehi

Ted Shifrin

It's just straightforward computation. So do one or two if you want.

Semiclassical

I think it's not a bad exercise, once you've done a few by hand, to figure out how you'd get Mathematica to do the computation.

Zach Hauk

12:47 AM

oh, I misread

thought you said "Skip that question until you know what's going on"

For #6 i just looked at cross sections and so far I have 1 sheeted hyperboloid, 2 sheeted hyperboloid, paraboloid, and ellipsoid

are there any others I'm missing?

Ted Shifrin

Are you getting that from the math that you've learned here?

Yes, there are degenerate cases for sure.

Topologicalife

Is there a proof that norms on R^n are equivalent, which is "allowed" to use some topological notions like compatness, but "not allowed" to use other topological notions like continuous functions?

Ted Shifrin

Nope, continuity is essential.

Topologicalife

Why?

Zach Hauk

I looked at all the cases for $ax_1^2 + bx_2^2 + cx_3^2 = 0$

Ted Shifrin

12:52 AM

You needn't have $0$, @Zach.

Zach Hauk

(which you can obtain by diagonalizing your quadratic)

Mike Miller

@Topologicalife How do you talk about equivalence without continuous maps?

Zach Hauk

oh, nevermind :P

so when $a,b,c > 0$, we have an ellipsoid obviously

Ted Shifrin

Not with a $0$ on the right.

Semiclassical

Not with anything nonpositive, for that matter.

Zach Hauk

12:54 AM

oh

Topologicalife

@MikeMiller well, for example, saying they generate the same topology.

Zach Hauk

with a 0 that form isn't an ellipsoid

Mike Miller

@Topologicalife Which is nothing more than saying that the identity map in both directions is continuous.

Semiclassical

With zero there's hardly any solutions in the first place. (So long as we're over R^3, I mean.)

Zach Hauk

ok, tell you what, multiply both sides so that the constant is positive, that will make things much easier.

Topologicalife

12:55 AM

Or if there are constants $a$, $b$ such that $a \|x\|_1 \leq \|x\|_2 \leq b\|x\|_1$

Semiclassical

You can probably make things even simpler: multiply both sides so that the constant is 1.

unless it's 0.

^

...hrm.

Which is a case to look at.

Semiclassical

12:55 AM

Yeah, fair point.

Zach Hauk

so, we know that an ellipsoid is a possible one, though

Ted Shifrin

Everything you listed was ok. I'm not convinced you've listed everything.

Topologicalife

@MikeMiller I'm talking about continuity if I say that tere are constants $a$, $b$ such that $a \|x\|_1 \leq \|x\|_2 \leq b\|x\|_1$?

Mike Miller

Yah.

Zach Hauk

now, for one negative $a,b,c$, we have something like $ax_1^2 + bx_2^2 - cx_3^2 = 1$ (i'll check the zero case in a bit)

so notice that when $x_3 = 0$, we have an ellipse

Ted Shifrin

12:58 AM

BTW, you're not going to get paraboloid from this question.

NO. That is totally wrong.

Zach Hauk

b-b-but

Topologicalife

@MikeMiller why? I don't see notions of continuity there.

Zach Hauk

wait, how is that wrong? I'm just taking the xy-plane slice?

Semiclassical

Probably better to say: The $x_3=0$ cross section is an ellipse.

Ted Shifrin

It's still a surface, @Zach.

Zach Hauk

1:00 AM

?

but, it's in the plane

Mike Miller

@Topologicalife Those are equivalent to saying that the identity maps $(X, \| \cdot \|_1) \to (X, \| \cdot \|_2)$ and vice versa are continuous.

Zach Hauk

the $x_1$-$x_2$ plane

Ted Shifrin

Zach, no, it is not.

This is a fundamental error.

Topologicalife

@MikeMiller right.

Thank you.

Zach Hauk

then how do i take a cross section?

Ted Shifrin

1:01 AM

I'm asking you what the surface is ... not what its various cross-sections are.

Zach Hauk

but if i look at the cross sections i can analyze the surface... right?

Semiclassical

I don't think he was claiming that...

Zach Hauk

he = who? @Semiclassical

Ted Shifrin

This is very false.

You said "when I have $c=0$ I have an ellipse."

Semiclassical

You.

Zach Hauk

1:03 AM

i edited it

to mean $x_3 = 0$

Ted Shifrin

It just displayed my comments in reverse order.

So, again, when $c=0$, what is the surface?

And also you did see my comment that you can't get paraboloids.

Zach Hauk

oh, it's some weird degenerate infinite cylinder thing

@TedShifrin yeah i saw

and I realize why :P

Ted Shifrin

It's a right elliptic cylinder, yes.

OK, I'm outta here for now.

Zach Hauk

alright

im gonna analyze these surfaces

sorry for being a hassle lol

@Akiva want to analyze some surfaces with me

Akiva Weinberger

Don't really have the time at the moment

Zach Hauk

1:14 AM

heh, i was kidding

Adeek

hi everyone

I was wondering let us say I have the following definition for projective modules

P is said to be projective if whenever we have a surjective morphism $g : B \rightarrow C$ and $f : P \rightarrow C$ then f factors through B.

why are free moudles projective using this definition ?

hi @TedShifrin

Adeek

1:33 AM

nvm I figured it out

Zach Hauk

hi Adeek

Adeek

hi @ZachHauk

Zach Hauk

1:46 AM

im just classifying some surfaces

Secret

Political use of mathematics: Using metric geometry to give a notion of district compactness in helping to combat gerrymandering

Pretty hard to become a juror nowadays. They need to know the law, statistics, and now geometric group theory

3

Daminark

2:03 AM

Hey everyone!

Zach Hauk

hi @Daminark

Simple

use generating function to show that for a symmetric random walk $P(S_1S_2,\dots,S_{2n}\neq 0)=P(S_{2n}=0)$ for all $n\geq 1$

@Semiclassical do you mine give me a hint to do this problem

Daminark

How's it going?

Semiclassical

Not sure what that's supposed to mean. S_n is the position after the nth step?

Zach Hauk

i jus thave a headache

Daminark

2:14 AM

:(

Hope you feel better

Simple

@Semiclassical yes

Zach Hauk

i'm classifying surfaces which are the solutions of setting a quadratic form to a constant

ZION

Hi i'm new here also can someone give me tips to improve my mathematical writing ability:math.stackexchange.com/questions/2129016/…

Semiclassical

Mmkay.

And it's step size of +/-1 I presume

Simple

It is symmetric random walk, I think so

Semiclassical

2:16 AM

It doesn't terribly matter what the step size is, come to think of it, since you only care about zero vs. nonzero.

Simple

Since this is symmetric, the probability of returning to 0 sometime is equal to 1?

Semiclassical

I think so.

Well, suppose we denote $P(S_1,S_2,\cdots,S_{2n}\neq 0)=p_n.$

Then $p_{n+1}=p_n\cdot P(S_{2n}\neq 0)=p_n(1-P(S_{2n}=0))$.

Simple

yes

Semiclassical

Which, if that's supposed to be $p_n$ as well, means we'd need $p_{n+1}=p_n(1-p_n)$.

I'm not convinced that helps much, though.

I mean, $p_n^2$ is kinda a pain as far as GFs go.

Simple

I am still not sure how to apply generating function

Semiclassical

2:23 AM

Nor I, if I'm honest. I was hoping this would lead to one deriving the GF for the symmetric random walk, and then work backwards to deduce the above from the GF.

Simple

$P(S_1,S_2,S_3,\cdots,S_{2n}\neq 0)=\sum_{k>n}f(2k)$?

Semiclassical

Yeah.

Simple

?

Semiclassical

Convinced myself what I wrote was not quite right.