@sunflower facepalm lvl: contravariant

@GustavoBandeira I have 21 what?

@robjohn I'm supposed to continue? I have to leave now. Sorry.

@PeterTamaroff So that for all $x\in[x_{k-1},x_k]$ we have $|f(x)-f(x_k)|\le\epsilon/2$ or $|f(x)-f(x_{k-1})|\le\epsilon/2$

@robjohn Is this because of MVT?

@PeterTamaroff yes, the derivative is less than $M$ and the distance from one of the ends of the intervals is $\frac{\epsilon}{2M}$

Problem of the day (for realzies this time): Suppose $f:\Bbb R^2\to \Bbb R$ is such that for every square $\square ABCD$ in the plane, $f(A)+f(B)+f(C)+f(D)=0$. Is $f$ necessarily identically zero? Justify your answer with either a proof or counterexample. What if we replace the condition of squares with the condition of equilateral triangles $\triangle ABC$?

@robjohn Purrfect.

@anon I will look at it in +2hs

@PeterTamaroff Then you use the convergence of $f_n$ on $\{x_k\}$ to find an $n$ so that for all $k$, $|f_n(x_k)-f(x_k)|\le \epsilon/2$

:sadpanda: is sad.

@sunflower any squares

btw the problem with squares is an old putnam one. the (equilateral) triangle one is my modified version.

what does "there exists nonzero space for triangles" mean?

that was quick

try cubes in R^3 then :-)

@robjohn: Hi! Long time no chat.

@anon: Greetings to you as well.

long time no see

Mr. Spivey

(rhyme time)

I was inactive for most of the past year; now I'm sort of back.

@MikeSpivey hey there!

@robjohn I see you've managed to get yourself elected moderator. Congrats!

@MikeSpivey Yeah, they were losing another mod or two for a while so they brought in mixedmath and me.

@MikeSpivey we were the next two in the election, so they asked us

@robjohn I noticed Zev is taking a break. Anyone else gone?

@MikeSpivey I think that Zev is the only one off now, but some of the others are taking things slower.

@robjohn And I think I read somewhere that you and @anon are the co-owners of the chat room? How did that happen?

Asaf leaving was a good while ago. It transitioned into robjohn being owner. Then when robjohn became mod he made me co-owner.

@anon Thanks. Asaf left? He's still on the site. So he just gave up chat?

@MikeSpivey how long has it been since you were here? When Asaf left, the ownership passed through another to me. When I became a mod, I added anon as an owner.

@MikeSpivey Correct.

@MikeSpivey yes...

Why did Asaf give up chat (assuming that the answer is not something Asaf would rather keep private)?

The last time I remember being on chat was January.

@MikeSpivey Asaf made an off-color remark and was flagged, popular opinion was against him, and he then rethunk his priorities and decided to stay off the timesink here.

@MikeSpivey Asaf is on self-imposed exile

@anon Thanks for the info. I certainly won't argue about chat being a timesink. :)

That being his final motivation is mostly speculation on my part.

The site has changed a lot in the past year. Lots more questions that scroll off the front page fast; it's starting to look like Stack Overflow.

@anon I have no actual information about why he left.

And some longtime users whose answers I enjoyed seeing are inactive: Arturo, J.M., t.b., Byron Schmuland, for example

@sunflower yes. I was going to say "look at cardinalities" at first but then I confused hom(2,A) with hom(A,2) and thus believed |hom(2,A)|=|P(A)| mistakenly so ignored that idea.

@MikeSpivey J.M. is around, but he is a mod on Mathematica

@MikeSpivey Arturo and t.b. are gone for "an unspecified time"

@MikeSpivey I know nothing about Byron

@robjohn: Any particularly interesting questions, answers, or conversations from the past year that I should take a look at? :)

@MikeSpivey being a mod takes some time, so I can see why J.M. is a bit less active.

@MikeSpivey You've been gone for a whole year?

@robjohn I believe it. I certainly don't have the time to run for moderator.

@robjohn From January until last month I only checked the site sporadically and probably averaged around one answer per month during that time.

Hola

@robjohn so are free now?

@JasperLoy No, he's about 30 years older than I am. :) "Spivak" does sound like a Slavic variation on my name, though.

that's what your picture reminds me of

@sunflower can u plz tell me?

maybe we can write it as the system $a-b=\sqrt{ab}$, $a^3+b^3=6$.

i don't get the second step in this answer mathoverflow.net/questions/115691/… where does the rest of the infinite sum go to?

is the infinite summation just missing before the second term?

@sunflower where did u get this>

@sunflower $u^2 - v^2 = uv$ this one

@sunflower ok

$\sqrt[6]{3+x}=u>0\iff \sqrt[3]{3+x}=u^2$, etc etc

@sunflower yes now I understand it

@sunflower I'm trying to get it

@sunflower I'm stucked here<--

@robjohn: Nice catching up again, albeit briefly. I need to go write a final exam in differential equations.

@anon: Nice catching up with you as well.

@sunflower sorry for disconnection I trouble with the modem

@sunflower I got it

@MikeSpivey Ah, so you were not completely gone. I thought I remembered seeing some answers from you in the last year.

@pourjour I am back

@MikeSpivey Nice chatting with you, too. Drop by again soon.

@robjohn thanks but sunflower explain to me what did u say early

@pourjour okay. Sorry I was gone before.

@robjohn no problem thanks anyway

@sunflower I think I need to raise power just to 3th

@sunflower sorry I didn't notice that

@sunflower and thanks for help

@GustavoBandeira Hello to you too!

Sure?

hmm, does showing that it's unbounded imply the limit is infinite?

p(n^2+1), the problem is slightly beyond me but that part isn't clear to me

@sunflower You should go to the Independence Hall. People like that, apparently.

@sunflower Are you good with analysis?

@sunflower That is good,.

@JasperLoy Now he is.

hi everyone!

@Nusha Are you Russian?

@JasperLoy That is offensive on so many levels.

@JasperLoy Then again, there are some levels that are offensive.