12:00 PM
@Sawarnik sigh kepp doing olympiad problems like this an you'll get to a natural boundary through which you can't go anymore.
@robjohn it's not continuous, and @Hippa claims it's nowhere continuous.

@BalarkaSen It is nowhere continuous.

@Sawarnik I haven't proved it yet.

Besides sos440 gave a one line proof that no such function exist at all!

sos404 is a genius.
witt beyond measure.
I know him, he is in I&S

what is "I&S"?

12:04 PM

Integral and series forum.

@BalarkaSen And on FB too, for your disappointment :P

@Sawarnik thanks

@Sawarnik Well, seemingly there are very less nerds in the world.

@BalarkaSen, @Sawarnik: Which problem were you discussing on?

12:08 PM
Why isn't option D correct here?

@Sawarnik So what are you reading right now? You're studying something, right?

So, you guys are studying for JEE? All the best.

@user63477 Not at all.
22 mins ago, by Sawarnik
$f:\mathbb{R}\rightarrow\mathbb{R}$
$\forall (x,y)\in\mathbb{R}^2,\ f(x+y^2)-f(x)\geq y$

@BalarkaSen. Then?

@user63477 Having fun.
I'm just doing pass time mathematics.

12:11 PM
@BalarkaSen: Great!
What do you guys do?

I am a high schooler, I believe @Sawarnik is one too.

This internet ... ufff.
@user63477 No :) But what about you?

I'm done with my 12th. Had taken off from stack exchange. So, kinda lost touch.

@user63477 What will you do now?

Will take BS degree in some IIT.

12:16 PM
26 mins ago, by Sawarnik
$$\int_1^{\frac{\pi}2} x \cos(\frac1{x}) dx>\frac12$$
@user63477 So you qualified?

Results are tomorrow.

Oow!
Good luck.

Yeah! Thank you. How about you?

Where are you expecting to go?
@user63477 I haven't decided anything yet.

Only few of the IIT's has got BS Physics degree. Like 4 IIT's. Kharagpur, Kanpur, Chennai and Roorkee.

12:19 PM
Oh.
Are you from Delhi btw?

No. I'm from Mangalore.

Oh ok.
@user63477 Do you think you ll qualify?

Yeah. I wouldn't be saying all that if I wasn't confident .

Nice :)

@Sawarnik: Any nice problems you got in here?

12:24 PM
@user63477 Which type of question you want?

@BalarkaSen I have shown such an $f$ cannot exist.

@Sawarnik: Math, obviously.
Anything will do.

Ok, the functional equations going on here?

Integrals? Do you have em?

12:27 PM
Yeah! You got the problem?

Means?

Did you crack it?

@user63477 Not me, but I think its cracked.

@robjohn, why not option d true here :

$$\int_1^{\frac{\pi}{2}}x\cos\frac1xdx=0.520013..$$

12:34 PM
accio @robjohn

@Darksonn: Got a nice proof?

@user63477 Ok, a little adaption of this will give you it, math.stackexchange.com/questions/835852/… :)

@Darksonn You don't have computers.
Just pen and paper.

@Sawarnik Up for some elementary number theory?

12:38 PM
@BalarkaSen: Awesome. Share here.

@Sawarnik No idea then

@BalarkaSen You can give the question, but don't think that I could solve it!

@Sawarnik: I didn't like the answers over there. Apparently. I will try the problem.

:)
Coming in 5 mins..

@Sawarnik @user63477 You are familiar with the fact that $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$?

12:42 PM
Yes. Coming in 5mins...

What?
Proof via induction, or some other method?

@Studentmath Try induction.
@Sawarnik @user63477 Now prove a generalization : $\tau(k)$ be the number of divisors of $k$. prove that $$\sum_{k | n} \tau(k)^3 = \left ( \sum_{k|n} \tau(k) \right )^2$$
@MartinSleziak You are disqualified.
=p

Disqualified from what? From MSE?

@MartinSleziak No, from looking at the problem I posed above.

Sorry, I've already seen it. And I know a solution.

12:45 PM
I knew that. That's why I disqualified you.
You always know a solution.
=p

I mean I know the solution, which you are alluding to (using the above sum). Probably there are many other ways to prove this.

@MartinSleziak I know only one.
@MartinSleziak STAHP. Don't spoil.

@BalarkaSen That's certainly not true. In this case, I was lucky enough to see this before.

Well, I have a follow up that you can try : Can you find a multiset with sum of cube of elements equal to square of their sum, not belonging to the divisor class above? @MartinSleziak

That one I do not know.

12:50 PM
@MartinSleziak Well, I believe no one before me and some two other person posed this question anywhere.
So definitely, it's somewhat new.

4 mins ago, by Martin Sleziak
But, of course, I will respect your request not to post spoilers.

@MartinSleziak The above problem? Finding a multiset not in the divisor class?

No, the original question (the easier one).

@Sush what answer do they give?
@BalarkaSen yes?

I think that multiste question could be nice question for the main site

12:54 PM
Meh internet connection
@robjohn Can I look at your approach?
@MartinSleziak Well, a little coding reveals a lot but there is more to it. More to these things.

@BalarkaSen to what?

@robjohn there exists no such function
@Martin Shall I reveal?

I have no problem with that. Wouldn't you prefer to post it on the main rather than in chat? (So that more people can profit from it.)

@BalarkaSen I showed you this
56 mins ago, by robjohn
@BalarkaSen by that condition $f$ is monotone increasing, therefore differentiable almost everywhere. However, by the condition given it is nowhere differentiable.

@MartinSleziak Well, I have no question for now. Only elementary stuffs, I wonder whether they are of general interest.
Nevertheless, I'll just give one such multiset : $\{1, 2, 2, 3, 5\}$
One of infinitely many.
@robjohn Aha.
how does monotone increasing implies differentiable almost everywhere?

1:04 PM
This book has a chapter named Differentiability of monotone functions. They call it Lebesgue's theorem.
IIRC there was some problem with the presentation of the proof in this book. But I'm not sure anymore.
The authors of the book refer to Riesz-Nagy, where the proof did not have the same issues. (I hope I remember this correctly.)

@BalarkaSen: It isn't differentiable everywhere.

But simply by searching for something like Lebesgue theorem differentiable or Lebesgue theorem differentiation you should find plenty of sources with that proof.
Maybe you can throw the word "proof" into the search query.
This result also have its own Wikipedia article Lebesgue differentiation theorem. (Which is not surprising.)
Nevertheless, if you expected some kind of one-line proof, I do not think it is that easy.
Doh'! That Wikipedia article is about another theorem of Lebesgue.... :-(
I should really check twice before I post something hrere.

1:21 PM
Greetings
@robjohn I created a lot of new series! :-) With the ones newly created (this morning) I might publish a book.
:D
I'll send one to you in private. (just to check some details)
Gezzz, it looks so marvellous ...
I hope to have (like) $100$ new series till the end of the day.

1:53 PM
Hello. Can someone shorty write why the union of lower limit topology and the upper limit topology isnt a topology?

The lower limit topology is the one generated by intervals of the form $[a,b)$?

yes, $(a,\infty)|a\in R$

If the union is a topology, then singletons are open.
So it would have to be discrete.
BTW to render curly brackets, you have to use backslash: \{...\}.

i see thanks :)

It seems as a sufficient argument. Singletons are open neither in the lower limit nor in the upper limit topology.
Hence the union is not a topology.

2:44 PM
well

@MartinSleziak can we say its bacause ie $(-\infty,5)\cup(6,\infty)$ and $(-3,\infty)\cup(-\infty,7)$ are in in $\tau_1 \cup \tau_2$ but intersection of these two subset $(-3,5)\cup (6,7)$ isnt in $\tau_1 \cup \tau_2$

it is because space...ace...ace...ace...ace...ace...ace
I need some nice examples of linearly ordered topological spaces
$\mathbb R$, $\omega$, $\omega_1$, the long line $L=\omega_1 \times [0,1)$, ...
I need more

3:11 PM
Hi Everyone

hallo

@JohnDoe Hey.

I initially ended up in 'upper room' by mistake, a stack exchange chat room where the discussions are regarding the difference between evangelicals and protestants...
@AntonioVargas How's it going?

Pretty good. Just woke up.
You?

3:18 PM
why am I having trouble thinking of more linearly ordered topological spaces

Alright been awake for some time...

I need some interesting ones

@AntonioVargas Can I ask an analysis question...Do you know what kind of mapping there exists between R^n and its dual, I know that every member of the dual maps to a row vector in R^n but I'm not sure what the properties of this mapping is, do you have any idea?
It seems like it should be bijective at least.

accio @robjohn. I just googled and read that it is called Lebesgue differentiation theorem. Yuck real analysis!

30.68, 26.65, 24.86
YEs!
One handed

3:24 PM
@robjohn are you around?

@BalarkaSen how's that yuck ? .. that is a very nice theorem .. imho :P

@r9m real analysis is analysis of really bad functions. complex analysis, on the there hand, is my preference!

@JohnDoe Is the transpose such a mapping?

@BalarkaSen ic .. btw if real analysis is analysis of bad functions .. how far is Complex functions from that ? :P
@Chris'ssis Hi .. super sis :-)

3:27 PM
@r9m Hello :-)

complex analysis isn't really analysis of complex functions @r9m.
also, take a killing curse from me (avada kedavra!) for supporting real analysis

@BalarkaSen There are some interesting books on analytic functions and entire functions though.

@AntonioVargas not sure...do you know what I am referring to?

@AntonioVargas Analytic functions are beautiful.

@Chris'ssis how are you doing ? :)

3:28 PM
@JohnDoe Only a little :)

@r9m Very well, I'm very productive today, thanks. :-) How about you?

Analysis makes me sick

@ComTruise Analysis is good, though.
I mean, fine by me.

@BalarkaSen I just started reading a book I found recently .. van rooij and tschikof .. very nice book :P

I think very useful, but it is not fun to me

3:29 PM
@Chris'ssis unproductive as usual :(

@r9m Never heard of it. Real?

@BalarkaSen ya :P .. nice book .. I'm liking it as lot :P

Yuck.

@r9m :-)))

@r9m i guess you won't like my philosophy about mathematics at all. i am not much open minded as you people are.

3:31 PM
@AntonioVargas I'm just saying that any function from the dual space (R^n)' can be mapped to some row vector [a_1,...,a_n] in R^{n}. So I was just interested as to what properties this mapping between R^n and and its dual R^n' has...
@ComTruise What do you prefer?

I like fun fun topology

@r9m number theory!
complex analysis has much more application in NT than real analysis, though i won't say none.

very abstract topology

discrete fourier transform of primes is always a good thing to think your head round.
@ComTruise i like algebraic topology, though mostly for the algebra.

3:33 PM
algebra I like too

@BalarkaSen Aww, drastic change.

@Sawarnik I won't deny it. Rudin is so cool.

@JohnDoe Ohh I see what you mean now. Yes the mapping should be bijective and linear, i.e. if $\lambda$ is the mapping then $\lambda(a_1 \mathrm{e}^1 + a_2 \mathrm{e}^2) = a_1 \lambda(\mathrm{e}^1) + a_2 \lambda(\mathrm{e}^2)$
But I don't really know much more than that. It's not my field.

Rudin is the only good thing about Analysis

That's topologist talk!
I always liked hard analysis.

3:36 PM
@lyme What I meant was that $\{1\}=(0,1]\cap[1,2)$ is an intersection of two sets from $\tau_1\cup \tau_2$, but it does not belong to $\tau_1\cup \tau_2$ itself.

@BalarkaSen Me too.

Okay thanks @AntonioVargas

@com Irreducible.
In general there are loads of galois groups. direct product of cyclic groups, i.e.!

what?

@ComTruise the ping is a link. click it.

3:39 PM
@BalarkaSen Do you have a pdf?

@Sawarnik sure.

it's in blah

Oh let me check.

ok thanks :)

3:41 PM
@ComTruise we are abusing the ping system ever since anon taught us how to do that (or was it someone else?)

Can Galois theory apply to topology?

@ComTruise Yes!
Try Alekseev's book.
Khovanskii for the general idea.
Topological galois theory.

@BalarkaSen The first result is Intro to DNA Analysis by Norah Rudin anyways :D

Rudin "principles of mathematical analysis"

I know.

3:44 PM
@Com Oh, sorry, not Khovanskii
Zoladek "toplogical proof of abel ruffini theorem"
i confused.

someone had too much time on their hands

@Com Do you know a bit about Riemann surfaces? i.e., what they are?

yes a little

then Alekseev should be no problem for you.
it's actually arnold's original notes.

3:49 PM

he taught highschoolers in france about abel-ruffini, starting all the way to galois theory, topology, monodromies, etc. @ComTruise
that's precisely the notes of those classes.
@Sawarnik who the hell is author?
Wow, look at reference [6]. Euclid wrote a book on commutative number theory!
not that i know of any noncommutative number theory...

@r9m My research led me to some discoveries done by Lewin. So, here is a nice result $$\int_0^{\pi/2} \int_0^{\pi/2} \operatorname{Li}_2(-y^2 \tan^2(x)) \ dx \ dy=\pi^2\log\left(1+\frac{\pi}{2}\right)+2\pi \log(2+\pi)+\pi^2 \operatorname{Li}_2\left(-\frac{\pi}{2}\right)-2\pi\log(2)-\pi^2$$

@BalarkaSen Did you understood it?

@Sawarnik no, i am too thick to get it.

@BalarkaSen Did it make sense?

3:57 PM
my simple mind cannot grasp the great ideas, though that's not an argument against making sense.
@Sawarnik it isn't something to make sense out of, right?

great editor, great man.
i gotta go.

bye :)

why are there so few interesting linearly ordered spaces?

@Chris'ssis wow !! :D

4:08 PM
there are plenty of linearly ordered spaces, @ComTruise
pick a linearly ordered set and give it the order topology :)

yes but what are they

I have no idea what that means

$\mathbb R$, $\omega$, $\omega_1$

those are special cases of what I just said

the long line $\omega_1 \times [0,1)$
but I need more

4:23 PM
@r9m That's pretty easy, but knowing how to cleverly use that in some solutions to titans, it's simply critical.

4:36 PM

@Chris'ssis I am now.

5:03 PM
:16146516 okay

@Hippalectryon Ted has said he's at odds with his publishers, yes.
@robjohn Hello!
Could you help me with something?

@PedroTamaroff okay

13 hours ago, by Pedro Tamaroff
@seaturtles Suppose we're given $n$ integers $a_1,\ldots,a_n$. Then there exists $1\leqslant k\leqslant \ell\leqslant n$ such that $a_k+a_{k+1}+\ldots+a_\ell$ is divisible by $n$.

@PedroTamaroff so some consecutive subsequence

@robjohn I should use the "generalized" pigeonhole principle.

5:15 PM
@PedroTamaroff It does look like a pigeonhole problem

Suppose we have finite sets $A,B$, and let $R$ be a relation on $A\times B$. There are is $x\in A$ such that $\#{\rm cl}(a)=\#\{b\in B:a\sim b\}\geqslant \# R/\#A$.

hi

I should thus set up a relation properly.

I was wondering if it is on topic to ask about the greatest living mathematicians in a particular field?

Vietoris lived to be 110 years old.

5:17 PM
@eleanora Just make sure its not very opinion based :)

@Sawarnik How would you do that?

@PedroTamaroff Consider the sums $a_1$, $a_1+a_2$, $a_1+a_2+a_3$, $\dots$, $a_1+a_2+a_3+\dots+a_n$ mod $n$

@robjohn Ah, OK.

@eleanora I don't know...it depends on what exactly you want to ask and the phrasing of the question.

We have $n$ sums and $n-1$ classes.
So two fall in the same class.

5:19 PM
@PedroTamaroff either two are the same and their difference is $0$ mod $n$ or each is different and that means one of them must be $0$ mod $n$

(Assuming none is divisble by $n$, that is)

@PedroTamaroff I guess most people don't know about Vieta ?

@PedroTamaroff It's a good answer, but I would have replaced the question with a statement about the $n^\text{th}$ roots of unity.

Is $(-\infty, a) = \bigcup_{n \in \mathbb{N}}[a-n, a)$?

@sonicboom What do you think?

5:30 PM
sorry, edited to put in the closed bracket

@robjohn Come again?

@sonicboom To me it would be $]-\infty,a[$ (though i'm not a pro)

@PedroTamaroff I've noticed that people don't like things like asking "what are the $a_k$?"

@robjohn But the point is the OP should realize the $z_i$ are the $n$-th roots of unity.

@sonicboom pick an element of either side. Can you show it is an element of the other?

5:32 PM
@Hippalectryon What is ]a,b[? Is it a kind of interval :O

@Sawarnik I'm used to write $[a,b[$ for what you'd write $[a,b)$

@PedroTamaroff Yes... Rather than ask a question either state they are or tell them to show that they are "show that the $a_k$ are $n^\text{th}$ roots of unity"

@Sawarnik $a$ included, $b$ excluded

@Hippalectryon Bourbaki style, but mostly not used I think.

@Hippalectryon Ooh, why do you use this type of notation?

5:34 PM
@PedroTamaroff @Sawarnik Used in France :D

@PedroTamaroff I've just noticed it works better.

We don't use $()$ for intervals in France

And its so confusing your [a,b] means (a,b) ? .. uuf

@robjohn What does?

@Sawarnik It's the opposite
@Sawarnik $(a,b]$ means a excluded and b included right ?

5:35 PM
@Hippalectryon: Your non-standard notation is confusing things...so I'll put it again - Is $(-\infty, a) = \bigcup_{n \in \mathbb{N}}[a-n, a)$?

@PedroTamaroff Uh ?

Yes it is.
Prove it by proving the double inclusion.
Or make a drawing and be done with it. =)

Ooooh wait

@Hippalectryon Yes.

I thought $\infty$ was included -_- so nvm n@PedroTamaroff @sonicboom
@Sawarnik Therefore $[a,b[$ is $[a,b)$
@Sawarnik And $[a,b]$ remains $[a,b]$

5:37 PM
Oh ok!
But its still odd.

-_-
To me it's your's that's odd lol

Ok one last question: Is $[a, a) = \emptyset$? It certainly looks like it to me!

write down what $[a,a)$ literally means and you'll see it is

It means all x in X
...wait
pressed enter by accident.
It means all x in X where x >= a and x < a...obviously the empty set!

5:39 PM
@PedroTamaroff Suggesting something rather than asking a question, even if the suggestion is itself asking them to do something.

there you go

@robjohn Ah, OK.

God I'm rusty, been on the beer for two months solid...just come off it two days ago..sigh..

I wonder how beer tastes like.

It tastes great but I'm sick of it now, back to maths!

5:44 PM
@robjohn Damn, I suck at this.
Suppose we choose $101$ numbers in $[1,200]$. Then there is $x\neq y$ such that $x\mid y$.

@Hippalectryon Guess who has the highest average question score here?

@Sawarnik what do you mean average question score?

@Sawarnik me :D (not)

@robjohn If you have asked 2 question one with score of 6 and other with 4, then the average score is 5.
@Hippalectryon Guess, guess.

5:55 PM
@Hippalectryon My rep? ... its Chris'sis who has as it as 8 (!), Pedro comes fifth or so with 5 :)

@Sawarnik Lol

@Hippalectryon Beyond me :D

@Sawarnik eww