Mathematics - 2016-12-22 (page 3 of 6)

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5:00 AM

oh, heading to.

G'night

Yes, like the noodles. My train is at union station. I'i in little tokyo.

Pissed off layman

cya

@Pissedofflayman that username is inappropriate for this site.

Pissed off layman

Why?

it contains foul language

Pissed off layman

5:06 AM

It's in the dictionary, no?

it is still inappropriate language

I see nothing wrong in it

Don't see why it should offend anybody in the universe

it's not appropriate for children

Change to @MaoTseTung

el-o-el, @Sophie

Pissed off layman

5:08 AM

I didn't mean to offend you @TheGreatDuck

you didn't offend me

i am notifying you and the moderators that the username violates site guidelines

Pissed off layman

You said it was foul.

yes, foul language

which is not allowed in the site guidelines

i never said the foul language bothered me.

10

Q: Do we classify the word "piss" as swearing on SO?

Well, I've always thought and always been brought up on the fact that the word "piss" is in fact swearing. I took it out of this question, then asked the OP not to swear. He came back and insisted it wasn't swearing, after googling it seems this differs in different parts of the world. What is ...

Pissed off layman

Then why notify anyone @TheGreatDuck

5:10 AM

quora.com/Is-pissed-off-a-bad-word-in-America

@Pissedofflayman because it violates the rules...

If you read a math book and it's really bad you can call it worse than Bourbaki

@TheGreatDuck Technically speaking he's not even a member of math.SE

@BalarkaSen I said the site guidelines. Not MSE's guidelines. The whole site's guidelines.

So you're reporting him on behalf of and as a representative of all of stackexchange?

5:16 AM

Don't twist this around on me. I am not claiming to be a representative of stack exchange.

I think @TheGreatDuck is now a pissed off layman

I reported him to the moderators for having an inappropriate username

and told him I did so

Pissed off layman

Thanks for your opinion @TheGreatDuck

Y'all can drop this and move on.

3

Pissed off layman

I have.

5:18 AM

@hichris123 thank you. I see no reason why everyone needed to jump on me for notifying a guy I reported him. :p

@Pissedofflayman I'd advise you change your username. I'd also advise you stop spamming chatrooms with the same video. Take a look at chat.stackexchange.com/faq while you're at it too.

I'm tempted to star that as helpful but I don't know if that would be appropriate. :)

Well, try not to stir up too much trouble. :)

Hmm I came of a proof that isn't very elegant.

hey @KajHansen here ?

I am

5:21 AM

@hichris123 I'm not sure how I'm stirring up trouble, but okay. I'll leave.

Proof of what?

@KajHansen I am studying from allufi (For algebraic flavoured category theory) decided to solve some of its problems.

@KajHansen the problems is to define epimorphism and show it is the same as surjection.

I know very little about cat theory, lol

good

@KajHansen this thing is for sets. So, it shouldn't be that weird.

@KajHansen $f : A \rightarrow B$ is epimorphism iff for all sets $Z$ and for all $\alpha_1,\alpha_2 : B \rightarrow Z$ such that $\alpha_1 \circ f = \alpha_2 \circ f$ implies $\alpha_1 = \alpha_2$

One side is easy if f is surjective then it has right inverse so it is an epimorphism.

5:25 AM

How is "right inverse" defined?

Is it ok for the "right inverse" to not be a function in the strict sense?

fine so far

@KajHansen should be a function

$g : B \rightarrow A$ is a right inverse for $f : A \rightarrow B$ iff $f \circ g = id_B$

I.e. we could have more than one thing mapping to the same element if $f$ is a surjection

right inverses aren't unique.

a right inverse just picks one element of each fiber

5:26 AM

oh ok, so there could be a bunch

yeah exactly as @arctictern says

got it

For the other side we do it as follows

like the embedding of R in R^2 is a right inverse for the projection R^2->R

(I prefer the terms postinverse and preinverse myself)

Hi @arctictern

5:26 AM

This sounds very similar to some "foundational" stuff on my first point-set set

hi

suppose f:A->B is not a surjection and see what happens

Suppose $f : A \rightarrow B$ is a epimorphism. Suppose f isn't surjective. That is there exists $b \in B$ such that there exists no $a \in A$ that gets mapped to it. Pick arbitrarily fixed function $\alpha_1 : B \rightarrow A$ construct $\alpha_2 B \rightarrow A$ as follows $\alpha_2$ is same as $\alpha_1$ however $\alpha_1(b) \neq \alpha_2(b)$. Then $\alpha_1 \circ f = \alpha_2 \circ f$ but $\alpha_1 \neq \alpha_2$

that is what I did @arctictern

I want a direct approach though

I am not fully content with this prove by contradiction.

The above contradicts the fact that $f$ is an epimorphism so f must be surjective.

I know I'm not really supposed to be in chat right now for causing trouble earlier but I have to ask: what are you guys talking about?

@artic It seems obvious that a direct proof would exist would be trivial however I can't think of one

That's good @Adeek

5:30 AM

I want a direct proof though @KajHansen

what was the trouble?

notice that those compositions only have domain $A$

So the fact that they send $b$ to different places is immaterial

yeah

it is not obvious that a direct proof should exist. you have the morally correct proof.

Yeah I see

Pissed off layman

5:32 AM

you didn't stir up any trouble @TheGreatDuck you stated your opinion and I thanked you for it.

@Pissedofflayman that moderator told me to leave...

Exercise: find a counter-example to the following statement:

> If $f(x,y)$ is a polynomial with real coefficients, then the image of $f$ is a closed subset of $\mathbb{R}$.

That's a bit of a stretch

But it's over now

I thought maybe I can construct somehow $\alpha_1$ and $\alpha_2$ from $B \rightarrow f(A)$ somehow and prove that $f(A) = B$ somehow using fibers or something like that

Pissed off layman

Whatever, that's their job @TheGreatDuck

5:33 AM

@arctictern would you like to check this question ?

I would like to hear maybe your input about this stuff if you have anything to add

How can we prove that $f:\Bbb Q \to \Bbb Q$, $f(x)=x^3-5x$ is injective?

@arctictern math.stackexchange.com/questions/2067997/…

@Pissedofflayman I better leave just to be sure. I don't want to get in worse trouble. I've been getting in trouble around here a lot lately. They said one more mess up and I'd be banned...

@KajHansen allufi is pretty good I am reading it for my winter break reading and problem solving

That's great to hear!

Pissed off layman

5:35 AM

Cya @TheGreatDuck :-)

cya

I probably won't be back for a day or two

I should let things cool off

I know most of you are pretty upset with me

Pissed off layman

Nah

I'm not convinced the closed subset thing is true @DHMO

:/

There's always going to be boundedness problems

It seems

I don't have a ctrexample though

Pissed off layman

You are The Greatest Duck @TheGreatDuck :D

5:38 AM

...

@Pissedofflayman I suppose you're skullpatrol?

No, I'm just a troublemaker that everyone hates.

goodbye

Pissed off layman

Who? @BalarkaSen

I would almost bet money that the image of every polynomial $f(x,y)$ is closed

@Pissedofflayman A user from this chat. Nevermind if you're not.

5:42 AM

@kaj it isnt true

Meh.

Pissed off layman

(removed)

The range is of the form [a, \infty)

How many Knights can you put on a chessboard such that none of them threaten each other?

@kaj could be (-\infty,a] or (-\infty,\infty) as well

5:44 AM

@Sophie I can see a way to get 20.

Every polynomial I can think of is unbounded either above or below

I have that configuration too

4 in each corner and 4 in the center

Yup.

@Kaj Polynomials are unbounded. But who knows if you can get it to miss a point or two?

is there non-trivial clopen subsets of R?

5:45 AM

No

it's connected

Yeah, that's what I thought @BalarkaSen. But very basic stuff that "seems" ok and works in \mathbb{R} fails in \mathbb{R}^2

So I wanted to be careful

It's a polynomial in x,y though

so I think a lot of the usual weirdness is forbidden.

Fair enough

I wish I could remember the one property I'm thinking of. I've even seen an algebraic proof of the fact, but can not wrap my head around wtf is going on.

@DHMO suppose his set contains an element, since it is open it has to contain a neighbourhood of this point, then it has to contain the limit points of this neighbourhood, etc... Then it has to contain the whole of R

Visually

5:47 AM

But also, so long as there's any $x$ dependence, you can evaluate it along the real x-axis

@KajHansen i want to watch some videos about bounding stuff by below or above.

and along there $f(x,y)$ is just a polynomial in $x$.

Yeah, you can always fix a variable and get a normal poly in the other

@Null get some inequalities book

Actually, one trivial counterexample: $f(x,y)=$const is a degree zero polynomial in x,y

5:48 AM

lol, gonna have to be more specific than that @Null. Finding bounds on things is a very widespread problem throughout math

@Semiclassical And the image is closed

@Sophie thanks

@Null Cauchy-Schwarz Master Class.

@Semiclassical, still closed

is a single point a closed set?

5:49 AM

Yes

I forget my definitions.

Welp.

Closed \implies complement open

Real line with one point removed is open

fair enough.

@Balarka Of course he's skullie.

Yeah.

5:50 AM

@Semiclassical surely enough that one is googable. And tomorrow the rest of the internets :)

Meh.

I have a configuration with 32 Knights. Put one in each white square

I realized but I won't argue if he denies. Way too tedious a conversation.

Maybe @MikeM has insight? We're trying to construct a polynomial $f(x,y)$ that has an image that isnt closed

I'm with Kaj right now. I don't see any way for a polynomial function to have a non-closed image.

5:51 AM

@Semiclassical there is one

Do you know it?

of course

:/

but i dont think i should spoil it

I'm dubious.

5:52 AM

I can understand

This is going to sound weird

Are there polynomials that have infinitely many minima? SURELY not.

Without loss of generality, I can assume it depends on $x$. (If it only depends on $y$, then I can swap the names of the variables; if it depends on neither, it's a constant map and therefore has closed image.)

@KajHansen Nope

Take derivatives

that'd imply that they have infinitely many zeroes of the first derivative.

@KajHansen what about NO minima?

5:54 AM

Any linear or constant function.

Yeah, that was going to be my next question

32 is optimal, if my proof is correct

Something that's asymptotic somehow

constant ones have minima - it's just the constant itself. $f(x) = x$ has none

let's continue restricting ourselves in R

5:55 AM

True. No local minima, though.

But is the point simply that any linear $f(x,y)$ will cover the entire real line rather than a closed subset?

Yes

That's no good for a counterexample

Not that it needs to be linear. $f(x,y)=x^2-y^2$ will do the same.

We have to have something bounded above or below

@Semiclassical $x^2 + x^3$. You can just make one.

Ok, let's start by figuring out sufficient conditions for f to be bounded above or below

5:57 AM

I meant that a constant function has no local minima.

Every term has even degree should work

of course it does - it's the constant itself :)

People are still restricting themselves to R

Hrm. Then what the heck am I trying to say?

What do you mean @DHMO ?

5:58 AM

guys, R + R is not R^2

You think we don't know that?

The image is in R tho

f(x,y) does not mean f(x)+g(y)

we know

@BalarkaSen the fact that we are still not having xy implies that

5:59 AM

No, it implies that we weren't bothering to state examples like that.

I have been thinking of polys where x and y variables are mixed together

sure

Ok, listen to this reasoning guys:

(If this turns out to hinge around something where I can reasonably invoke this XKCD, I'm going to be annoyed)

If we want the image to not be (-\infty, \infty), every term in this poly needs to have even degree and have the same sign, right?

6:02 AM

If f us unbounded above and below hen the image is closed. Suppose f>=M for all x,y then we're trying to find a polynomial such that there's no f(x,y)=M but f gets arbitrarily close but that isn't possible

Now that I thought about it, I'm not sure same sign is necessary

There are concave-up parabolas in R with negative terms

are we doing anything fun

Define fun

We're trying to find a proof of something that we've been told is false but not provided a counterexample of.

I gotta agree with @Sophie

6:03 AM

I remain dubious.

cool I'll stay away

I'm going to be annoyed if this is coming down to everyone not agreeing on definitions

lol

@kaj Quite. Hence the XKCD link above.

Nah, DHMO probably has a valid thing in mind. It's probably in MO somewhere

this kind of questions are all in MO

The argument used over 1 variable can be generalised to more but I don't think you should listen to me

I think if f is any continuous function and f\to\infty whenever |x| is large then the image of f is closed

6:05 AM

I have had that exact thought and had my intuition shat upon in the past

@kaj the counterexample is perfectly valid

The line of thinking I had earlier was to consider the set of images R->R parametrized by the particular value of y.

e.g. f(x,0), f(x,1), etc.

the image is (0,infty)

The image of f(x,y) should then be the union of all those images.

I agree

6:08 AM

So all one would need to establish is that every such image is bounded.

@kaj @semi @soph when should i announce the counterexample?

Can an uncountable union of closed sets be not closed?

a countable union can fail to be closed

Problem with what I've just said is that the image could degenerate to a point.

I'm fine with you announcing it. If it turns out to be a valid counterexample I'm going to be shocked; if it turns out to hinge on a technicality, I'm going to be irritated.

@Semiclassical not necessary; consider [2^(-n-1), 2^(-n)]

the union is (0,1]

6:10 AM

Those images are indexed by Z, not R.

@Sophie (0, 1) in R is uncountable union of all the points in it

I'll announce the counterexample after ten minutes

@Sophie I agree with you that as long as you go to infinity near infinity the image is closed. But it's not clear that this is true for multivariable polynomials.

@DHMO I think this is a little less inspiring when it's something you got off MO or whatever as opposed to having independent personal interest.

^

@MikeMiller but it is inspiring to others

6:13 AM

In the same way that knowing a mosquito is around is inspiration to move somewhere else.

@Semiclassical feel free to index by R.

@MikeMiller well of course i cannot come up with such examples

I can give lots of polynomial maps R^2 --> R^2 with bad image. But I can't get one down to R.

Yeah.

@bal i would be interested to know your examples

I think the only way is if the behaviour at infinity is weird

6:17 AM

Is the domain of the polynomial the whole R^2 or not?

the whole R^2

Well standard things are like (x, y) maps to (x, xy). That's "pinches" the whole y-axis.

I'm dubious that behavior at infinity will help. If we were talking about mappings to RP^1, sure.

Failure of imagination, perhaps.

rawr

@bal any more?

6:19 AM

Nah. Also please ping me by full name.

sorry on mobile

so should I say it now?

I'd vote yes.

I hope it is a damn good counterexample

Yes

^

6:21 AM

$f(x,y)=(1-xy)^2+x^2$ has image $(0..\infty)$

...hnggggg

that is based on the fact that xy=1 has no solution when x=0

yeah.

but has a solution otherwise

cute though. and now I remember where I have seen this

6:22 AM

on math overflow of course

dang, I was working on (1-xy)^2, but didn't see how to fix it

the big list of false beliefs

Oh it's the behaviour at infinity indeed

I have seen it on one of Thurston's questions.

Not disappointed

6:23 AM

^

that's a very nice example/idea!

My proof assumed $f\to\infty$ when |(x,y)| is large but that isn't true for multivariate polynomials

In quite satisfied with the counterexample

Frustrated, but I can't call that unfair.

@Sophie I told you that :)

can any f(x,y) have image [a,b]?

6:26 AM

No

polynomials are unbounded

No because it grows unbounded

well, the silly constant polynomials...

No unless constant

why?

Fix a $y$. Make $x$ large.

6:27 AM

ok y=0 what next

aka "restrict to a line on which it's not constant"

then you just need to know it for single variable polynomials

which is elementary

thanks

how is ordering on Q defined?

Tediously.

2

Though if you just mean the condition for rational x,y to have x<y

how is Q defined

p/q < p'/q' \iff pq' > qp'

6:32 AM

@MikeMiller ZxZ+/~

Assuming any "negative" signs are included in the numerator with the p's, so that the multiplication there doesn't result in sign change

huh?

I see normal p divided by normal q is less than those negatives and such and such

@KajHansen That's taken care of by p in Z, q in Z+

The proof of FLT for $n=3$ is underwhelming. The one for $n=4$ is much prettier

One could probably make a fairly interesting talk on FLT that just focuses on how hard it is to prove various cases, e.g. what's the smallest $n$ for which there's no elementary proof.

6:38 AM

That would be cool

Now try the one for $n=17$ @Sophie

Better yet, $n = 23$

math.stackexchange.com/questions/2068174/…. Wut

can I define a "power series" $\displaystyle\prod_{q\in\Bbb Q}(x-q)$?

Hm, I think I wanted to say 37

@DHMO No.

6:41 AM

Oh right it was 37, not 23

But I think irregular primes have positive density so...

@mike how is it different from sinx=(x-pi)(x-2pi)...

the integers aren't dense in the reals. the rationals are.

i see

It's better to start by understanding what a power series in. Then it will be clear what a power series isn't.

(And that product expansion is wrong.)

6:45 AM

$$\frac{\sin x}{x} = \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{4x^2}{\pi^2}\right)\cdots$$

(1-z^2/pi^2)(1-z^2/4pi^2)...

haha

lol

@TheGreatDuck My response to that "is RH ill-posed" question: youtube.com/watch?v=FwiYNYlqJL0

is that a rickroll

No.

No$^{\omega_0}$

6:50 AM

Jun 28 '15 at 15:22, by Ted Shifrin

NOOOOOOOOOO

@KajHansen I'm somewhere over 50% confident he's trolling.

lol

...

It's a variation on Poe's law. Can't tell the difference between sincere idiocy and trolling.

7:05 AM

The post might be sincere crackpottery. OP kinda appears to have some mission, as if it to bait contention to later throw out their non-mainstream point.

Hi @Brody

Then again, might not be. I've noticed some woo-woos on internet science forums have used this format. Why one would think it effective I don't know.

'Ello @Balarka

wth, literally everyone has a hat on

lol

7:37 AM

yo what's up?

Pissed off layman

chillin' you?

yeah chillin' too

cat

Pissed off layman

mouse

cheese

Pissed off layman

7:51 AM

mold

@Semiclassical i am currently not supposed to be in chat. Please do not ping me.

ham

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