Mathematics - 2017-07-29 (page 3 of 3)

gian

23:00

Oh sorry I misunderstood your question @TedShifrin.

user2457324

@Kirill did you have any luck with this? follwing @Ted's hint --> I did the following:
"so I am going to call M_m the last column of M matrix. then, simplifying the last term in the determinant I get: W^{-1} B [M_m \dot M_m] B^T W^{-1}
\dot --> is kind of the element by element product"
not sure if we can use some properties of symmetric matrices to work with the last column?

gian

If $X$ is not open that means that the basis does not cover $X$.

Daminark

I don't think you want to use contradiction here @gian

Ted Shifrin

Demonark: I had trouble saying what it meant for $U$ to be open in every topology containing $\mathcal A$. Can you help?

Bob

I posted the following problem and answer at this URL: math.stackexchange.com/questions/2373595/… I am wondering what the level of difficulity of the problem is. Is it a problem for a 1st year math student? a graduate student?

Daminark

23:02

So, what do you know about the open sets of a topology in relation to a basis @gian?

Lucas Henrique

Woah, this answer, dude.

Ted Shifrin

It would have been perfectly appropriate in the upper-division probability course I taught a few years ago, @Bob.

gian

Well if $U$ is not in the topology generated by $\mathscr{A}$, doesn't that mean there is some other set outside of $\mathscr{A}$ that is involved in an arbitrary union to produce $U$?

Ted Shifrin

First-year, no, too much unknown terminology. Graduate, no.

What does it mean to be in the topology generated by $\mathcal A$, @gian?

Bob

second year? third year? fourth year?

Daminark

23:04

Phrase that in a slightly better way. If you know a set is open in a topology generated by some basis, what can you say about it?

Ted Shifrin

I'd say third, @Bob. I don't know why we're playing this game, but ...

Bob

@ted thanks, I want to know how I am doing.

gian

@TedShifrin It means that it is some arbitrary union of the basis elements.

Daminark

You're at least going in a direction I like

Ted Shifrin

LOL ... but that's very field-specific, Bob, but sure.

OK, @gian. So if NOT that ... what does that mean?

Don't tell me sets outside of $\mathcal A$. Be very specific.

<-- gives the floor back to Demonark and slinks away.

Eric Silva

23:06

i wanna do math but i also wanna cook

what a conundrum

gian

Well if not that then it's not open

Daminark

Well, I'd rather not think about this in terms of a contradictory argument, this is a case where it's clearer to think directly

So, we're looking at the intersection of topologies containing $\mathcal{A}$

Just think about one topology

gian

Right

Daminark

Let's say $\tau$ is a topology on $X$ containing $\mathcal{A}$

What do we immediately know about $\tau$? Just by definition of a topology?

gian

That the elements of $\mathcal{A}$ are open sets.

Daminark

23:09

A bit more than that, what facts do we know about topologies?

Semiclassical

hi chat

gian

Well if you mean the axioms, $\tau$ must contain $X$ and $\emptyset$

Daminark

Keep going

Ted Shifrin

Ah, good, Demonark.

gian

Finite intersections of open sets must be open and unions of open sets must be open

Daminark

23:12

Okay, so we know that elements of $\mathcal{A}$ are open sets in $\tau$, so by what you said, any union of elements of $\mathcal{A}$ should be open in $\tau$, right?

gian

Right

Daminark

But now, what is the set of unions of elements of $\mathcal{A}$?

Ted Shifrin

Demonark: Make sure you address where you're looking at the intersection of all such topologies ... :)

gian

The topology generated by $\mathcal{A}$!

Daminark

Exactly!

@Ted what do you mean by "where"? Is it about the topologies being on $X$?

gian

23:14

So any topology on $X$ that contains $\mathcal{A}$ will have a certain set of open sets. The intersection simply "clears" all other non-common sets.

Ted Shifrin

No, no. You're being too literal. In your line of questioning, how is @gian looking at the intersection of all topologies ... ?

OK, @gian. In fact, an open set in that intersection must be a union of elements of $\mathcal A$.

gian

Yes..

What did I say wrong?

Ted Shifrin

I was more specific :)

Because we can make a topology that does not contain a particular open set that is not a union of things in $\mathcal A$. (Too many "not"s, but ...)

gian

Right but this intersection includes all topologies on $X$, including the one generated by $\mathcal{A}$ for which what you said does not apply.

Ted Shifrin

well, not all topologies on $X$ ...

gian

23:20

Sorry, the ones that contain $\mathcal{A}$

But my point still stands right?

Ted Shifrin

Yes. I still don't like this question, but it's a good lesson in how to say things carefully.

gian

Okay I see. I feel like I made it more difficult than it had to be with the contradiction.

Ted Shifrin

Right.

ABcDexter

Hello everyone.

Akiva Weinberger

Hi @ABcDexter

Daniel Castle

23:22

Hi @ABcDexter

Ted Shifrin

rehi DogAteMy

gian

Thanks for the help. Are you specialized in topology?

Lucas Henrique

Are there two Akivas or two Daniels?

Ted Shifrin

I hope not.

Lucas Henrique

Either way, hi @ABcDexter :)

ABcDexter

23:23

Could anyone tell me how to solve these type of puzzles?

www2.stetson.edu/~efriedma/puzzle/calculator

Ted Shifrin

Were you asking me, @gian? I taught point-set topology a lot of times. But no, I worked mostly in complex geometry and differential geometry.

ABcDexter

Well, i could brute force, but with 4 functions, and 4 operaritons, it would grow exponentially.

Akiva Weinberger

@LucasHenrique As far as I know there's only one of me

Ted Shifrin

One DogAteMy is more than enough :)

ABcDexter

Any help is highly appreciated:. :)

Ted Shifrin

23:25

I yield to the clever puzzle-solvers in here. I'm not one.

gian

Oh okay cool. I still don't know what I want to do lol.

@ABcDexter, which one you need help on?

Ted Shifrin

@gian: You have many years to go before you need to decide.

gian

Mhm.

Akiva Weinberger

@ABcDexter 5->6 seems to be 5+7=12, 12/2=6

6->7 is 6*3=18, 18-11=7

12->13 is 12/2=6, 6+7=13

Ted Shifrin

So, DogAteMy, do you have strategic advice, other than trial and error or just seeing it?

Daniel Castle

23:28

does anyone think it's worth my mathoverflow question on the collatz conjecture on mathoverflow or will it just get booed out with -10 votes

Akiva Weinberger

The ones with only two operations are easy because you don't have a whole lot of options

Ted Shifrin

MO is not patient with questions that don't belong there, Daniel.

Daniel Castle

fair enough

Akiva Weinberger

Hm, "13->14 with 3 buttons"

gian

I don't thin Mathematics Exchange is either :D

Daminark

23:29

@gian actually you need to decide right now on finite group theory

gian

Huh?

Ted Shifrin

@gian: A number of us get fed up with people merely posting their homework questions with nothing else.

Daminark

Oh I'm just kidding

Ted Shifrin

We are not here for people to copy to turn in their homework.

gian

@Daminark I'm only in high school D:

I really like group theory though.

Daniel Castle

23:30

Well I'm not doing it for homework, I was just genuinely interested in the collatz conjecture and was looking into it

Ted Shifrin

I was addressing @gian, not you, Daniel.

Daniel Castle

oh ok

gian

I haven't gotten to rings yet. I'm getting there though.

@TedShifrin

Woops

Ted Shifrin

But MO is really intended for advanced graduate students and researchers ...

gian

@TedShifrin I know.

ABcDexter

23:31

Any possible algorithm?

Akiva Weinberger

Oh I got it, 13->14 in 3 steps is *3, -11, /2

13->39->28->14

user2457324

@TedShifrin Hi -- I added on the result I got from your hint to my Q here: https://math.stackexchange.com/questions/2376054/satisfying-the-following-determinant-inequality
kindly take a look in due course and advise for any simplification. Or if you can suggest someone who may have played around with things like this before, then, that would be much appreciated. thank you again!

Akiva Weinberger

@ABcDexter Well for the 3 steps ones you can look at what things you can get from the start in one move, and what things you can get from the end in one move going backwards

and see if you can bridge anything in the middle

I don't see a general method, though

ABcDexter

other than trying all possible branch till solution in reached. basically a DFS on a directed graph, with given root as the original number, and known depth. od DFS, to reach the leaf (the final desired number)

Akiva Weinberger

One thing we could do is see if there are specific move combinations that are useful

ABcDexter

23:34

@AkivaWeinberger Please go ahead

Akiva Weinberger

Like, +7,-11 and -11,+7 could be useful if we want to move 4 in either direction in two moves

ABcDexter

Okay, and?

Akiva Weinberger

I dunno

ABcDexter

still, your idea is pretty good.

Ted Shifrin

@user2457324: Not the last column, but the group of the last $n$ columns. I don't think your notation with the $m$'s is helpful. Don't you want to break it up into $m$ $m\times n$ matrices?

Akiva Weinberger

23:37

I know we could multiply or divide by 1.5, but that's gonna get annoying with fractions

ABcDexter

maybe i could jump from the required number back to original.

i presume, fractions won't work in case we deal with coprimes

Akiva Weinberger

Hm 19->20 in three steps, 19+7=26, 26/2=13, 13+7=20

ABcDexter

It can not be generalised, have to brute force with calculations.

is this the case?

gian

What's the motivation for defining order topologies?

Akiva Weinberger

Looks like it, I dunno

Next is 26->27 in three steps

ABcDexter

23:40

thanks anyways.

;)

Ted Shifrin

Because we do lots of analysis with $\Bbb R$ and inequalities ... what things generalize? @gian

ABcDexter

*

:)

Ted Shifrin

@gian: In general, I would recommend that you study a book like Spivak's Calculus and learn some rigorous calculus before you do point-set topology. Most of the motivation for topology comes from that.

ABcDexter

bye everyone. take care.

Akiva Weinberger

26/2=13, 13+7=20, 20+7=27

Ted Shifrin

23:41

good luck, AbcD

Akiva Weinberger

@ABcDexter Bye

4->5 with four buttons. Upping the difficulty a bit, probably

ABcDexter

thank you, have a competition coming up in about 3 hours.

Ted Shifrin

Ah.

This kind of puzzle math has just never appealed to me ...

gian

I read through the 3rd edition Spivak

Well not really read; I worked through it.

ABcDexter

for me, it used to, but that was 8-10 years back.

Semiclassical

23:43

Back in middle/high school I was did a bunch of Math Club competition stuff

ABcDexter

anyways, bye.

gian

Is there a team math competition at the high school level?

ABcDexter

May the Force be with You...

Semiclassical

But that's as much a social thing as it is a math thing (not that I was terribly social in high school)

Ted Shifrin

OK, @gian. If you worked a number of the more interesting exercises, then that's good.

Semiclassical

23:43

Depends. Our state had it.

There were a few regional math events at the college level, but not much.

Closest thing I have to 'puzzle math' these days is random problems in the AMM and stuff here

Ted Shifrin

GA has lots of high school competitions. UGA and Tech both sponsor pretty tough ones.

gian

@TedShifrin, I was reading through Little Rubin and I was able to immediately see a lot of the correlations between the analysis and the topology I'm going through now.

Daminark

@TedShifrin :(

Ted Shifrin

OK, @gian, so your work with deltas and epsilons should make you appreciate order ...

Semiclassical

the idea that math competitions somehow correlate with math research is an idea I've never understood.

Ted Shifrin

23:45

It's true, Demonark. I'm not quick and clever and I never have been fascinated by numbers ...

gian

I need to go back and review total orders, brb.

Daminark

Oh are you thinking about the calculator stuff? I dunno if that's my thing, but puzzles in general are nifty

Ted Shifrin

LOL

Semiclassical

I think the closest thing I ever had to 'number fascination' were divisibility rules.

Ted Shifrin

I'm not knocking them, and I know a lot of people enjoy them and are very good at them. AoPS has a big puzzle component to it, I guess. It's just never been my interest/strength.

Semiclassical

23:47

I didn't really understand why they worked for a long time, and that was interesting.

Daminark

Did I ever tell you about polynomials having multiples with only prime exponents?

Akiva Weinberger

Yeah I don't know if I'm gonna continue this

@Daminark Babai, that was a fun one

Ted Shifrin

@Semiclassic: I've told the story in here about how I figured those out when I took Artin's algebra course. The guy down the hall reminded me that he'd figured them all out in 6th grade.

Akiva Weinberger

I mean extremely frustrating

but it felt great when I finally got it

Semiclassical

(I did create a divisibility rule of my own, though. $100^k =2^k$ mod 7. So 123412 mod 7 = 12*2*2+34*2+12 = 5*4+6*2+5 = 55=6)

Akiva Weinberger

23:48

98=7*14, checks out

(98=49+49)

Oh, cool (re: edit, example)

gian

Wow how could I have been so careless to forget a total order

Akiva Weinberger

That's clever

Ted Shifrin

Oh, that's different from the usual 7 divisibility test, Semiclassic.

Figure out a test for 13 next. :) [It's an exercise in my algebra book.]

Semiclassical

I came up with it in high school and was rather proud of it.

Akiva Weinberger

@TedShifrin The '10a+b is a multiple of seven iff a-2b is' test?

Ted Shifrin

23:50

Right.

Did you figure out 13, DogAteMy? If so, what about 17? :P

Akiva Weinberger

That doesn't help you with mods 'cause you multiply the residue by 5 each time,

but it is clever, in that just realizing "you're allowed to multiply the mod by something is time" is clever

Ted Shifrin

as long as it's a unit, yeah.

Semiclassical

I mostly liked that I could structure in blocks:

`12 34 12`
`--------`
` 5 6 5`
` 4 2 1`
`20 12 5`

bah, why isn't that coming out in code form

Akiva Weinberger

Test $\begin{align}a\\\hline b\end{align}$

Semiclassical

oh well. (I mostly like code form because it's in terminal font and therefore spaces things right)

Ted Shifrin

23:52

you probably need a matrix format for that, DogAteMy

Akiva Weinberger

Yay I got it

Semiclassical

$a\over b$

I never knew about \over until recently. not sure why one should use it instead of \frac, though.

Akiva Weinberger

$\begin{matrix}12&34&12\\\hline5&6&6\\4&2&1\\20&12&5\end{matrix}$

Semiclassical

good call.

Akiva Weinberger

There's another command that lets you right-align columns

Semiclassical

23:54

5, 6, 6 is the top line mod 7, to be clear.

Ted Shifrin

it was old TeX, Semiclassic.

Semiclassical

Ah, makes sense.

Daminark

@Akiva So that's probably one of my three favorite linear algebra problems

Akiva Weinberger

$\begin{array}{rrr}12&34&12\\\hline5& 6&6\\4&2&1\\20&12&5\end{array}$

Semiclassical

@TedShifrin As a silly variation on what I was saying, 100 = 104-4=13*8-4 mod 13

Ted Shifrin

23:57

I know how to right-align in usual LaTeX, not in here, DogAteMy.

Semiclassical

so i could do my rule but with powers of -4 instead of +2 for mod 13.

Akiva Weinberger

@Daminark The prime power polynomial problem?

Daminark

Yeah

Akiva Weinberger

The problem isn't really linear algebra

Ted Shifrin

Semiclassic, why is that any better than $10\equiv 3\pmod 7$ and $10\equiv -3\pmod{13}$?

Akiva Weinberger

23:57

(The problem statement, at least)

Semiclassical

Didn't say it was. There's a reason I called it silly :P

Daminark

For the second one, let's say $V$ is an $n$-dimensional vector space, and $T:V\to V$ is a linear operator with $n$ distinct eigenvalues, show that another operator $S$ commutes with it iff it's a polynomial of it

Akiva Weinberger

@Daminark So I can assume $T$ is diagonal?

Wait no

Actually I'm not sure

Semiclassical

Though I guess the reason I like 100 instead of 10 is that it's pretty easy to do two-digit numbers mod 7 mentally.

Akiva Weinberger

The eigenvectors have to be perpendicular to each other, right?

Daminark

23:59

Nope

Akiva Weinberger

Hm true

Well they still form a basis, right?

Semiclassical

at the same time, one ends up with half as many blocks as doing it with 10. so it's about twice as efficient in that regard.

Daminark

You neither need that to be true nor know it, this is over any field

Semiclassical

(still pretty silly, of course)

Daminark

Yeah they do

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$Mathematics$ Mathematics