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9:02 PM
Regarding this and this: can we close the latter, non-migrated version, rather than the first (with 4 votes at the moment)?

@Kannappan The point of this "jury duty" chatroom is to separate out the "duty" messages from other ones. We should take that discussion to here... =)

The answers to the migrated one won't ever be accepted as the account doesn't exist anymore.

But well I think no harm done, if the question stays open for a while. (Not that I fell particularly good about abandoning question like that, but we have plenty of such questions at MSE.)

@Srivatsan I see.

@KannappanSampath Having brought you here, I should immediately add that now is not a good time for me. With real life things to take care of.

9:04 PM
OK @Srivatsan!

@Srivatsan That is no issue with me!

@Matt Wouldn't it be simpler to ask moderators to merge accounts? I mean only 1 close vote is missing there...
I voted to close already, I am not sure if I will be allowed to vote for reopen there.

@MartinSleziak Sure, is it the same user?
Ideal would be if the one with the close votes disappeared and the one with the answers got attached to the user's account.

@Matt What are the odds that two people would come here with the same question, same wording and same login name.

9:07 PM
Didn't look at the name. Well then...
I'll vote to close. How do I contact a moderator?

1

I'm currently working my way through Harvard's online abstract algebra lectures (if you're interested, you can find them here). The lectures come complete with notes and homework problems. Of course, since I don't actually go to Harvard, I can't hand in the homework assignments to find out if I...

"Are you being rhetorical or do you want to do the math?" :)

@Matt "Flag" option right under the text of the post.

@Matt I believe the correct way is to flag for moderator attention. Then you'll get a window where you can write message explaining what you're asking from moderators.
I don't think it is possible to flag a user, so you probably have to flag question to do this.

@MartinSleziak Man, you are thorough in your instructions. :=)

9:09 PM
In the question whose link I just posted, OP wants us to apparently grade all the problems assigned at Harvard Algebra course in Group Theory! Is it a good idea for the site?

no
people need to learn to check for themselves

@Srivatsan Well I was looking if it is possible to flag a person rather than question. (I wasn't sure.) My guess was that Matt might be looking for that too.
That was the reason why I tried to elaborate that much.

@QED So, someone could add a comment!

I think that question you linked it fine

@MartinSleziak No, you cannot flag a user. Even if you want to merge accounts, the only possibility is to flag a post (question/answer).

9:11 PM
I am willing to as well!

@Srivatsan, nice with the Archaeologist badge.

@Srivatsan, @MartinSleziak Done! Now I'm off, I have something I'm looking forward to reading : )

Bye matt
And thanks!

Bye Matt.
@AsafKaragila Yep. The only one in town. :D

@Srivatsan I know how that's like. I recall being the only Outspoken guy...

9:14 PM
But I got it a while ago...

I only noticed recently.
I now just need to give a bounty on someone else's question, and award it, then I'll have all the feasible bronze badges :-)

Ah, wish you luck.

Well. I am one vote from 200 points and 50 from the limit today. Someone find me a nice question to answer and have five upvotes on my answer ;-)

9:16 PM
You'll need some -- just so that your flags are deemed helpful.

Srivatsan: I guess you have good chance for Taxonomist with martingales and fixed-point-theorems.

t.b.'s flag weight is amazing.

@Srivatsan As long as they are not deemed unhelpful which cuts away brutally 10 points from the current weight I'll be fine.

what's a flag weight?

@MartinSleziak Aha, I must go on a tagging spree now. I will add the [martingale] tag even to set-theory questions.

9:18 PM
It's the way the system knows how valid are your flags.

@Srivatsan lol

I was sure I'll get Taxonomist on [descriptive-set-theory] but I ended up getting it for [examples-counterexamples] which I don't even recall starting :-D

@MartinSleziak But realistically, those are more like long-term investments. I'm sure they will pay off eventually.

When I saw Mariano at the chat today, I thought he wants to talk about algebra retagging. (You mentioned he left a comment at some post you edited, Srivatsan.)

@AsafKaragila Did you get that recently.

9:19 PM
@Srivatsan Not too long ago, a month perhaps.
Huh.. apparently three months ago.

You might have to thank for me for that. Not sure how much I contributed, but I did tag quite a few posts with that tag.
@AsafKaragila Uh oh, not that long ago. My edits were more like a month ago.

@QED Well Piracetam seems to be the nootropic that everyone talks about. Some people say it works incredibly well for them, others say no effect.

@Srivatsan Well, you could add descriptive set theory tags ;-)

I see

@AsafKaragila Sure. :-) Search for Cantor function and tag random posts from the results with that tag.
That may not be that bad an idea, actually.

9:22 PM
I think that my favourite comments correspondence with Gerry was one that I tagged something as desc. set theory and he untagged it saying that it's a technical term and whatnot, then posted another comment that he may have been wrong and he'll let someone familiar with the topic retag. Then I told him I've just finished a course on the topic and I have at least the basic knowledge to say that this is indeed a fitting tag :-P

And quite the long answer too!

I have to go. See you later!

bye

Bye Martin

9:26 PM
@AsafKaragila Ah man. I wanted to vote up your answer this week. I read it and found some bits useful, although I don't have a big picture yet. :/

Which answer? I have almost 400 of them by now :-)

Oh... I think that after this I'll only have +8 and +10 voted questions, no +9...
Oh right, I have another one with +9 on which I am actually working right now :-D How could I have not noticed that??

Hi everyone

Hi

9:28 PM
Has anyone here seen my percolation question?

I saw it

I think I left it around here somewhere

Please don't post a question on the main site and then come here and ask people to look at it, at least not just out of the blue.

I'm not expecting people to rush out to view it =p

I will answer your question if you can help me prove that in the second model of Cohen every infinite set has a surjection onto $\omega$. :-)

9:31 PM
Don't ask a physicist to do maths. He'll use the word prove and make every mathematician in the room cry.

@AsafKaragila Wait, is that true in ZF? [newbie, so be gentle... =)]

What's up with all the physicists coming in here lately?

@AsafKaragila well, I prefer them to ask about induction, rather than topologists...

Does "infinite" mean "can be put in 1-1 correspondence with a proper subset"?

@Srivatsan No, it is completely possible to have a model with an infinite set which has no surjection onto $\omega$.
@Srivatsan No, infinite means cannot be bijected with a finite ordinal.

9:33 PM
@AsafKaragila Oh ok.
And they are not the same thing?

Dedekind infinite means can be bijected with a proper subset, which in ZF is exactly as saying that there exists an injective function from $\omega$ into $x$.
@Srivatsan Not without some choice, as I said you can have this situation. It's one of the basic examples, actually.

@AsafKaragila OK. I suspected that, but off the top of my head, I cannot distinguish proofs using choice vs. proofs not using choice vs. proofs using choice but can be reworded to avoid it.

@Srivatsan Often the proof is not just to be reworded but completely reconstructed.

@AsafKaragila Hm, that would be a fourth possibility. =)

9:39 PM
I "know" (as in, I have seen it written) that it does not use choice. As I said, I cannot even tell them apart.

Heya =)

two books on gereral topology that's ... like ... a representative sample.

@Srivatsan The point is that the theorem's original proof is describing how the function is defined based on the two injective functions.
It's not very complicated but not too simple either. With AC it amounts to a triviality using ordinals.

Any tips to finding the slant asymptote to

$$y = \frac{3y^2 x}{y^3 + x^3}$$ ?

I know what it is, just not how to find it =)

9:41 PM
@AsafKaragila I think I know the overall story. Pair of injective functions $\rightarrow$ bijective can be done without choice. Other situations need some choice.

what is slant asymptote

Eh, an asymptote that is not horizontal or verticall. It looks like \

@N3buchadnezzar Heuristically, you'll get a asymptote along the curve where the denominator will vanish. This means $x^3+y^3=0$ which means $x=-y$.

@Srivatsan Quite.

I meant what's the math definition

9:44 PM
@KannappanSampath So this will give the result of $y = x - 1$ ? =)

http://www.wolframalpha.com/input/?i=plot+y+%3D+3*y^2*x%2F%28y^3%2Bx^3%29
wow that is horrible

Sweet. I am beginning to see the picture of how I'm supposed to prove something complicated. However I don't see immediately the fine details and it seems some combinatorics will be in order.

@AsafKaragila Where do you want to prove something compleicated?

9:46 PM
can you put it in 's?
so I can copy it

@Srivatsan David Roberts' question about Cohen's second model. I want to prove that thing I mentioned not long ago, that every infinite set in that model surjects over $\omega$. I have an idea how to show this, but it's not a trivial idea. It might also be wrong.

oh how did you do that!

@KannappanSampath The asymptote can be expressed as that line =)

Let me write a few notes down, hopefully it will make sense tomorrow :-D

@N3buchadnezzar, do you have a mathematical definition of the thing you're looking for?

9:47 PM
@QED @Srivatsan: I added the stuff we worked out, plus a proof of the roots, here. I think it is pretty much the same as the proof by David Speyer. Should I just remove it, or do you think it is unique enough?

@QED I replaced * by %2A and ^ by %5E.

aha
cool

@QED Perhaps Oblique asymptote is a better word, I will look up a wikipedia article for you. If that suffices =)

9:48 PM
@robjohn I'm not sure if it is unique enough. But I think you explain better than David. =)

@AsafKaragila I just told you the asymptote was $x=-y$ and NOT $y=x-1$

@robjohn I think it's just that seeing such a chain of (in)equalities engages me more than text... =)

@tb also known as URLencoding

No, you told that to someone else!

@KannappanSampath But both my plots and the textbook says otherwise...
In analytic geometry, an asymptote () of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity. Some sources include the requirement that the curve may not cross the line infinitely often, but this is unusual for modern authors. In some contexts, such as algebraic geometry, an asymptote is defined as a line which is tangent to a curve at infinity. The word asymptote is derived from the Greek ἀσύμπτωτος (asímptotos) which means "not falling together," from ἀ priv. + σύν "together" + πτωτ-ός "fallen." The term was introduced by...

9:50 PM
I don't want a wiki link...

@Srivatsan I think that long stretches of text are harder to follow that a fairly easy to read sequence of equations

@N3buchadnezzar You're Norwegian, right? Can you translate me the title "Det Lys Aldri I Livet Orke" please?

@robjohn Hey, we think alike. =)

That sentence makes no sense

I was told that it makes no sense about 5 years ago, it should be broken Norwegian.

9:52 PM
@N3buchadnezzar, this curve is called the Folium of Descartes

You know, the time where Norwegian black metal bands sang in German and German black metal bands sang in Norwegian, and no one was doing that right.

@Srivatsan Equation $(5)$ was the part that took a long time. It took me a long time to get to the point that I realized $\operatorname{Re}\left(\frac{1}{e^{ix}+1}\right)=\frac12$

@AsafKaragila "That" "light" "never" "in" "life" "have the energy to"

@robjohn Isn't that the Riemann hypothesis?

@robjohn One more problem I found with David's answer is that it is very back-and-forth. It is somehow how he discovered the solution, not the solution itself. =)

9:53 PM
@QED Yeah, even my book states it by that name. alas no means to find the asymptote. Or the behaviour of the functions when x becomes arbitarily large.

@AsafKaragila Yeah, right. Gimme my Millennium prize :-p

@robjohn I am not made out of Clay! :-)

@N3buchadnezzar, anyway the first thing you should d obefore trying to solve the problem is get the definitions of all the things in involved formalized and clear

@robjohn Sorry, officially your proof was the second one.

e.g. remember when you got stuck proving that identity about absolute values, the instant you saw the definition (|v| = v . v) you could prove it
it's very often like that

9:55 PM
@Srivatsan Yeah, I had everything but Equation $(5)$ as soon as the question was posted. I had a day of work getting ready for yesterday's 6 hour meeting and then yesterday gone for the meeting. I finally finished $(5)$ this morning

@robjohn Rob: I will go through your answer later. But tell me this: do you have an estimate of the error.

Let $f(x)$ denote my function, I am then looking for a function g(x) on the form
$g(x) = ax + b$
such that
$$\lim_{x\to \pm \infty} [f(x) - g(x)] = 0$$

@robjohn Damn that elusive one.

@Srivatsan I don't, but I will see if I can get one.

9:56 PM
Pardon me not being very good in mathematics, would this definition suffice ?

no

I know, thats part of the problem...

so there's no worry that you can't solve the problem it's not even well posed (given that definition)

@robjohn See: for the given interval $[0,1]$, the error is exponentially small. Why? Because the norm of the polynomial $(x-\frac12)^n$ is constant times $2^{-n}$. But I think the question is more interesting for $[-1,1]$. There plugging in $x^n$ gives a distance of $\frac{\mathrm{const}}{n+1}$. But is the minimum distance much smaller?

I guess you cant call it a function either, more a curve.
Well my curve sort of looks like y = x -1 without the curl at the origo though...

9:59 PM
Hm. Perhaps this new answer will get me over the curve :-D

@N3buchadnezzar, do you see what I mean though

Huzzah. >200 points!

@robjohn I hope I conveyed the motivation nicely. =)

One day closer to Epic.

@AsafKaragila I do

10:00 PM
so you can't really attempt to solve a question that isn't well posed

Huh?

I think QED is addressing N3.

Well here it just states the asymptote... =( en.wikipedia.org/wiki/Folium_of_Descartes

@Asaf Good that you pointed out to the OP from Nepal it is not good to sign his posts with notes such as those!

@N3buchadnezzar, but does the book not define it?
before asking you to find it
assuming you're working from a book

10:03 PM
@Srivatsan Why would the problem be much different on [-1,1] than on [0,1]? For a polynomial $P(x)$ on $[0,1]$, $P\left(\frac{x+1}{2}\right)$ is a function on $[-1,1]$
So we have a factor of $2^n$ on the lead coefficient

@N3buchadnezzar, even if someone claims to have given you "the answer" you have no way of checking it for yourself

Yeah.

@QED Not in terms of generall curves, Just the definition I gave based on functions.

@N3buchadnezzar Thanks. I guess. Regardless to the fact that the title is near meaninglessness the track itself is still pretty awesome.

so just say "the folium is not the graph of any function therefore the definition of 'asymptote' doesn't apply"

10:05 PM
@robjohn Yes, I think that $2^{-n}$ is a distraction. My point is that: let's say that the error for QED's question is $3^{-n}$; that doesn't seem surprising. But that implies that the error for my question is $(3/2)^{-n}$: this seems quite surprising to me. I know this is just psychological or whatever, but it helps bring the relevant question to the fore.
Now, I can simply ask: is the error in $[-1, 1]$ exponentially small or polynomially small? =)
Well, you don't necessarily have to see it the same way. As long as you understood my question, it's fine. =)

@Srivatsan I think the error in the $[0,1]$ problem is $2^{-n}$ times the error in the $[0,1]$ case because of the scaling of the polynomials. Assuming a lead coefficient of $1$.
What is QED's question?

what?

@QED is that better? :-)

@robjohn QED's question is over [0,1]. Mine is over [-1, 1]. So they are just a scaled version of each other.

@Srivatsan yes, I believe so.
Since there is that nice mapping from one to the other.

10:10 PM
no

@QED no to what?

not better
I don't know what to answer

@robjohn OK, @robjohn, I feel that you think the scaling is not a big deal (and I agree). I just wanted to convey that I like to think of [-1,1] more than [0,1]; that's all. I think I blew up the difference a bit; so let's stop this discussion here. =)

@QED I thought you were wondering about "what is QED's problem" which has a negative second meaning. I thought changing to "what is QED's question" would be better.

I know the sense in which you meant it
I don't have any question though
although I don't think he wanted to accept what I was saying

10:13 PM
Me and robjohn are discussing the $L^1$ approximation question, not N3buchadnezzar's question. robjohn has posted an answer to it now.

@QED ah, the first I saw of the question was from you on the chat.

hm
I don't know

@QED I saw the question on the main site after that.
So we were attributing it to you. :-)

attributing what to me?

The polynomial approximation question

10:16 PM
Well I need to hand in these problems, so being picky about notation is not my forte at the moment ^^

I guess I will just ask my lecturer to clearify the question, and how to solve it.

to get adefinition of asmyptote that actually applies to the curve defined in the question
Srivatsan asked about the L_1 distance from x^2 to a line, and I was wondering if that was just one example of a more general question - Approximation theory generalizes it greatly

@robjohn robjohn, that is not N3buchadnezzar's question. I am fairly certain of that. =)

I guess I could define the definition as: Let A describe my curve, and let g, be a function on the form ax + b. Assume that A tends to infinity then the distance between A and g, should tend to zero, as A tends to infinity. If g is a asymptote to A.

10:19 PM
@Srivatsan Yes, I was just checking and it is QED's

I don't think you can say someone owns a question so primitive as this

@Srivatsan But you asked about $x^2$ first, right?

it's a pretty natural thing to ask (hence it being an entire field)

@robjohn Yep, I believe so.

@QED you posted it to the main site. So for the purposes of our discussion, it is your question :-)

10:21 PM
Not the first human mostly (although we still don't have a reference for this problem).

@Srivatsan and that is where we came up with the main attack using the signum function, etc

@robjohn Yes. My interest in this comes a little bit before this particular discussion. One second, let me pull out a link.
@robjohn See this.
This happened 1-2 days prior to my posting the problem in this chat to you. I focused on a special case (thinking that an analytical solution is impossible).
My first instinct was to ask for some lower bound on the distance (an upper bound is easy), even if it wasn't the tightest one. The motivation for that is that the compactness argument naturally gives an ineffective bound, so I was led to wonder what the distance would look like.

@Srivatsan, from Approximation Theory there are some important theorems
oh wait I got this for $L_\infty$ norm
you may not be interested in that

@Srivatsan Unsurprisingly it seems that entirely new techniques are needed for answering such questions. As QED says and as GEdgar (?) pointed out in a comment, approximation theory is all about such questions. This might even be some standard problem somewhere (but nobody has supplied a reference to literature yet).

It seems the Chebyshev polynomials answer both the $L^1$ and $L^\infty$ problems if I was reading correctly.
Is that correct?

10:30 PM
David says so. I do not know the theory of Chebyshev polynomials.

The article on Wikipedia indicated they were solutions for the $L^\infty$ question, and we seem to have shown they are for the $L^1$ question.

@robjohn If that is true, there should be a direct explanation of that, no? Like: if $p$ is the best $L^{\infty}$ approximation, then the best $L^1$ approximation is blah.
After all, $\infty$ and $1$ are like duals or something.

There might be. I haven't thought about it at all.
@Srivatsan See this

@robjohn Hm, thanks.
I have seen this after reading the first answer; anything specific you want me to see?

@Srivatsan Just that is shows these polynomials are the solution of the $L^\infty$ question

10:38 PM
Oh, ok =)

derivaties of chebychev
in the case of L_1 isn't it?

@QED I think the roots are the same as what they claim for the Chebyshev polys
All I computed were the roots of the minimal poly, and they look the same (mapping from $[-1,1]$ to $[0,1]$)

10:59 PM
@QED Ah, looking at the roots listed in the WP article, these are the Chebyshev polys of the second kind, and the ones I linked to Srivatsan are the Chebyshev polys of the first kind.
@Srivatsan: So Chebyshev polys of the first kind solve the $L^\infty$ question and the second kind solve the $L^1$ question.

@robjohn I guess that makes sense because: derivative of cos(nx) will be essentially sin(nx) and vice-versa.

@Srivatsan yeah, I guess so...
It is interesting that that article doesn't mention the $L^1$ question.

@robjohn :=)

11:15 PM
@Srivatsan: Hmm... In David Speyer's answer, he claims that from $\left( 1+a_1 x + a_2 x^2 + \cdots + a_m x^m \right) \sqrt{1-x} = \left( 1+b_1 x + b_2 x^2 + \cdots + b_m x^m \right) + O(x^{n+1})$, he was able to solve for $a_n$ and $b_n$ by solving some linear equations. However, for any sequence $a_n$ we can find a sequence $b_n$ that satisfies that. Am I missing something there?

I don't know what you're talking about... =)
I actually need some time. Could we chat later at night?

@Srivatsan In his answer, he supposedly used that equation to solve for $a_n$ and $b_n$
Sure. I need to do some things, too. later

@rob: Apparently the albatross is now solved. I haven't seen the solution fully.

@robjohn Don't forget to add: "This is a helpful hint, but not quite a proof." =)

11:28 PM
:-)
I just don't think the whole story is in the answer. I don't see how he got the polys from his steps.

Didn't I just say the same thing, in possibly different words?

@Srivatsan Sort of. I wonder where his solution actually came from though.
I would like to see how Mathematica solved those linear equations.

@robjohn What kind of wonderment is that?
As in: why do you wonder that about that solution?

The answer is fine. The solution seems to have a big hole at the point of the equation I copied above.
By the answer, I mean the polynomials he arrived at. By the solution, I mean the way he achieved the polynomials.

Just do yourself a favor and ignore me right now. I am pretty much following that advice myself. =)
[I am doing some three things at one time.]

11:35 PM
I need to be paying some bills right now. I should concentrate on that now, too :-)

@robjohn I eventually figured out what you meant: that's why I deleted my comment.
@robjohn Perhaps you should.

@Srivatsan However, your point was correct, so in case someoe else might read it, I wanted to clarify.

@robjohn I have this annoying habit of being correct. Cannot help it. =)

@Srivatsan It's hard to live with. I don't know how my wife lives with it :-)
not your habit, but mine :-)

Ahem. I was wondering if you are referring to your wife's -- that would be third valid interpretation.

11:50 PM
Interesting: it seems the OP likes Bill's approach here. I guess that whenever we do understand the answer, it looks better than other answers...

am so sick of these polynomial questions

sick? Why so?

especially "prove 9|p(n)"
because induction is a stupid way to deal with them

@QED On the other hand, it is quite a bit non-arbitrary.

more reliable than what?

11:53 PM
I removed that bit on "more reliable".

I just hate it because I feel like everyone is learning something bad

It's not bad. No proof method could be bad, could it? What argument do you have against induction? That it is mechanical?

Very good. The day ends and I have 225 points.
Now to eat something and catch some ZZZ.

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