@mirgee There are no magical ways to become good at something, but you need to struggle a lot, to read books, to learn from the others, to perform your own research on things, to create problems that arises from your own questions.

I also struggled and suffered to understand things that sometimes were much over my head. I've never given up.

For instance, from robjohn I've learned a lot of very nice things, perhaps it is the person from that I've learned the most beautiful things so far.

@Hawk if the local extrema of $f$ is attained at a point $s$ outside $[-1,1]$, that means $f$ is pretty much monotone in the interval $[-1,1]$, so the extrema is attained at the end points, ie $|f(x)| \le max \{f(1),f(-1)\} \le 1 < \frac32$ .. on the other hand if $f'(s)=0$ for some point $s \in [-1,1]$, so $2as+b = 0$, that is $|\frac{b}{2a}| = |s| \le 1$, now $max |f(x)| = |\frac{b^2}{4a} - c| \le |\frac{b}{2a}|.|\frac{b}{2}| + |c| \le |\frac{b}{2}| + |c|$,

... now its easy to show $|a|,|b|,|c| \le 1$, so the conclusion follows

@Chris'ssis I think you are being too humble. I have been observing you on this site for some time, and with each of your answers, each of your posts on this chat my jaw dropped :) To me, your skills are godlike.

@Chris'ssis Also, I am glad to hear it's normal to struggle :)

@Chris'ssis Will do my best to be like you :D

@mirgee Yeah, it's really normal to struggle. There is no other way. Learning is a continuous process that lasts a lifetime. :-)

@robjohn sorry, I just saw your question ...

@robjohn You here?

@r9mYou are becoming too mathy these days.

@robjohn that hypergeometric series is related to a question you solved in the past $$\sum_{k=0}^{\infty} \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}\frac{k! n! m!}{(k+n + m + 2)!}$$

@robjohn you posted here that mind-blowing answer mathematica.stackexchange.com/questions/39228/…

@Sawarnik why did Hawk put himself in the middle of Bay of Bengal ? .. :P

@Sawaria define mathy :P

@r9m Haha :D And he put Robjohn into a remote island in Canada, before he fixed himself :D:D

@Rommo Mathy means you doing too much maths and losing the fun element!

@Sawarnik y u think so ? :P

@r9m Especially when SuperSis comes around, and she goes and you keep quiet.

@robjohn here is my way there. I'm not sure if it's something related to my way.

@r9m And infact he didn't know he had put himself somewhere before I pointed out he's drowning in the sea!

@Sawarnik :P .. LOL .. you did it kya ?

@robjohn one more time, sorry, but I saw your question with much delay.

@r9m No no! But then he said he is in a navy submarine, and put him in ISI when he joins, so he currently may be in the sea itself.

@Sawarnik put him in ISI HQ at Pak .. when he joins ISI kol :P

ROFL

@r9m Do you think I know where is the HQ in Pak!

@Sawarnik :P .. okay then put him in some desert in pak :P .. he will eventually find his way to ISI HQ Pak .. and mark it himself :P

hi

I was looking at math.stackexchange.com/questions/765803/… where I think the current answer is wrong

can anyone help?

If $f:\mathbb R\rightarrow \mathbb R$ is a function satisfying $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb R$, then $f(x)=x\cdot f(1)$ How is that derived?

@Sush with just that much info $f(x) = xf(1)$ holds for $x \in mathbb{Q}$

@Sush Does it really hold? The only continuous solutions are $ax+b$ which I think doesn't satisfy as well.

@r9m, No!

How did you get it?

@Sush en.wikipedia.org/wiki/… This may help.

My book says that $f(x)=x\cdot f(1)$ holds for $x \in \mathbb{R}$

@r9m

@Sush for that to be true $f$ has to be continuous at atleast one point in $\mathbb{R}$

@Sush Take $f(x)=x+1$. Then $2x\neq x+1$.

Sorry.

hmmm!

I made a mistake.

r9m is right! See the link.

@Sawarnik, i can't see that link, as slow connection!

My memory is so weak :(

@Sush Oh no.

@Hawk, Hi, bye.

@Sush haha, hi,bye...

@Sush Bye!

Sush, sush, sushi.

@Sawarnik don't start this...you will again get banned...

@Hawk Oh -_- :(

@Hawk Do you know why is earth not a sphere?

Because God created it that way

@user127001 No. -_-

@Sawarnik that sounds like a physics question....

@Sawarnik but the earth is not a sphere because it is a geoid

@Hawk It is.

@Hawk Why so?

Because science

@user127001 Yes, science .. ?

Yea, science!

@Sawarnik not sure, but maybe due to the rotation...

maybe some lateral stretching force comes into effect...

but, that is not at all my concern...i am not interested in physics...

@Hawk Yay!

@r9m... I saw your solution, but these rigorous derivations gives me fever...

earthscience.stackexchange.com/questions/108/… .

@Hawk Which one?

see above, he posted a solution to me...

Oh, I answered the physics question correctly!!!

@Hawk Hope you recover from it soon.

@Hawk is there any point in the proof that you didn't understand or like to clarify ?

Q: What does it mean for a function to not have a total derivative?

@r9m I will tell you later...doing a paper...

Crap.

I confused $\Gamma(N)$ with $\Gamma_0(N)$

sorry I got disconnected

I don't think the answer to math.stackexchange.com/questions/765803/… is correct. Can anyone help?

@Pedro I wonder if there is a generalization to the problem I gave.

For what conditions on the group $G$ with all elements satisfying $(xy)^n = y^n x^n$ is $G$ abelian?

@r9m I have an inequality for you. Want it?

@Sawarnik let me see it :)

Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$$ @r9m

@Sawarnik wow .. makes me wanna cry :P

@Sawarnik I wonder if you are done with set theory.

@BalarkaSen I don't think so.

@Sawarnik Then I wonder what you are really doing these days.

@BalarkaSen Wasting time.

I suspected.

@BalarkaSen I think I am in one of my stints of inactivity. Although I have been trying introductory real analysis.

@r9m Happy crying :)

@BalarkaSen Orly, where?

Everywhere.

@BalarkaSen Like, in the simplest case?

Closed and open intervals.

We had this conversation before.

Also, mapping.

Functions, i.e.

@BalarkaSen I know that.

@BalarkaSen And maybe this too.

@Sawarnik OK. Then why do you say you don't know set theory?

@BalarkaSen I know only the first 2 chapters of NCERT 11th, cardinals and uncountable and blah blah, i don't.

Oh, those. I don't count them on fundamental set theory, which is just what we need.

There is a teensy difference between ignorance and apathy, man.

Hmm, apathy .. new word for me.

@BalarkaSen Ok, I studied just enough [for example I skipped the whole real number chap] for starting limits.

@Sawarnik It seems so.

But what really is in the real numbers chapter?

@BalarkaSen Second chap here, matstkipbjm.files.wordpress.com/2011/03/…

And much of the 1st chap as well.

@r9m Ok.

I was out.

@Sawarnik I will try it later pal :P

@Sawarnik Does completeness mean topological completeness?

:|

is it considered rude to @someone who you think might be able to answer a question?

or is this a meta question?

@user2179021 Generally no.

@user2179021 Are you talking about doing so in the chatroom, or on main? If you have a friendly relationship with them already (e.g. have chatted a lot here), it might be ok to ping them if you sincerely think they'd enjoy the question. Otherwise, I wouldn't recommend it. The couple of times I've seen it done (typically to high-profile users), the response hasn't been stellar...

@Sawarnik On main

@anorton Ah ok.. Then maybe I made a mistake. It's for this question math.stackexchange.com/questions/765803/…

which has one answer which I think is incorrect

@user2179021 I agree with @anorton

@user2179021 Oh, then I second Anorton's views.

thank you for the answer

I have noticed that sometimes if there is a long mathematical looking incorrect answer questions often then get ignored

Well, there's certainly a difference between you pinging someone for help with your own question, and you pinging someone for help with someone else's question. (The case you mention seems to be the latter...) So, I wouldn't worry about it...

@anorton Oh ok thanks. My other problem is that I am not sure how to ping someone :)

@user2179021 Pinging = @

@anorton I mean you need a single word name to put after @

Oh! Just ignore the spaces.

but how do you find out what that is?

so no hyphen?

No hyphen. For example, I could do @PedroTamaroff

math.stackexchange.com/users/62773/will-nelson would be willnelson then

(sorry, Pedro, but you walked in at an opportune time. ;) )

not will-nelson?

Yes, that would be willnelson

and do capitals matter?

No

ok thank you

this has been very helpful!

My pleasure. (signing off now...)

I would love to know the answer to that question as well!

not enough probability experts about

if only Did chatted here :)

I notice that impossibly hard limit questions get answered very quickly but moderately hard probability questions are left unanswered sadly

Is it bad that I only have 5 votes left today, and I still have several hours of activity left? :)

Intense

@meer2kat ?

@Sawarnik yar?

@meer2kat 'yar' ka matlab?

OK, anyone here with expertise in computational Galois theory?

A question I wanna propose for a contest $$\sum_{n=1}^{\infty} \arctan\left(\frac{2n}{n^4+n^2+1}\right)$$

@Chris'ssis Is this related to one of those telescoping voodoos?

@BalarkaSen Yeah.

Is rainfall an important factor in the long term carbon cycle?

Fun q: Which deadly cyclone is visible in the famous Blue Marble photo en.wikipedia.org/wiki/File:The_Earth_seen_from_Apollo_17.jpg ?

can someone tell me what the symbol ~ (ind over it) means? i.imgur.com/jnW1fyu.png

I don't know the context here, @EwokNightmares

@BalarkaSen Starting out from simple things I was able to get identities like the ones of Ramanujan.

@Chris'ssis Ramanujan's elementary analytical identities are none of my interests. =)

@BalarkaSen statistics

@Chris'ssis The true impact of the works come essentially from algebraic number theory.

$$\displaystyle \prod_{n \geq 1}\left( 1+e^{-(2n-1)\pi \sqrt{19}}\right)=\frac{1}{3} \left(27-3 \sqrt{57}\right)^{1/3}+\frac{\left(9+\sqrt{57}\right)^{1/3}}{3^{2/3}}$$

The left side is the so-called Weber modform $\mathfrak f(\sqrt{-19})$

I see.

@Chris'ssis These are all derived from Ramanujan class invariants, which generates the class function field of $\Bbb Q(\sqrt{-n})$.

@BalarkaSen Nice.

blahblahblah

flagged (joking)

@MickLH Ok. But I won't delete.

@Sawarnik say what?

@meer2kat If I can understand Spanish, then you may well understand Hindi :)

$$\sum_{n=1}^{\infty} \arctan\left(\frac{4 n}{n^4+3 n^2+8}\right)$$

@Chris'ssis please leave it up long enough for me to get out of bed !

no chatjax on my phone

@MickLH OK :-)

thanks :)

I'm able to generate infinitely many series of this type that are simply so sweet.

@Sawarnik what does "ka" mean?

@meer2kat of.

@Sawarnik yar of matlab?

@Sawarnik that doesn't make any sense

@meer2kat what does "matlab" mean? The pronunciation is quite different than what you are thinking.

@Sawarnik i assumed you meant the program, matlab

@meer2kat 'matlab' means 'meaning'. no, not the program.

@Sawarnik is it an actual word in hindi? not the math program?

hmmmm

and i assume yar has some meaning too?

@Chris'ssis that's a beauty

@meer2kat No you said yar.

@meer2kat thank you!

@Sawarnik so 'yar' of meaning. still not making sense of it.

@meer2kat if you translate it correctly, it would be 'meaning of yar?' or 'what does yar mean?'.

@Sawarnik oh didn't notice the question mark. yar is like yes. but...pirate-y.

If we have an integral domain $R$ and we want to find a function $f$ such that it's a Euclidean domain, could we not use that $f(r)=\text{order}(r)$, where the order of $r \in R$ is its order for either the $+$ or the $*$ operation?

The only drawback I see is that it only applies to rings such that all elements have finite order.

Or this one $$\sum_{n=1}^{\infty} \arctan\left(\frac{4 n}{n^4-5 n^2+8}\right)=\frac{3\pi }{4}$$

@Chris'ssis .....you gave the answer :(

@meer2kat Were you gonna solve it?

lol it's different from the original series by a coefficient

@Sawarnik nah LOL

@meer2kat there are $2$ arctan series above where I didn't give the answer.

@Chris'ssis eh effort :P nah i'm @ work. can't do math right now

It's all about that $\frac{x-y}{1+xy}$-trick.

Oh wut

@BalarkaSen What is that?

@Saw yeah, deleting that was a good idea. I've been doing calculus for six years. i know how to solve these things kid

If I have pochhamer's symbol all over the place, did I go the wrong direction?

@Sawarnik It's a well-know arctan trick.

Try taking the tan of $x-y$

@meer2kat hmm .. let SuperSis give you some nightmare questions.

@meer2kat You've got nothing to prove, you don't even have to play into that game lol

@MickLH I'm just letting him know

@BalarkaSen Oh, I don't remember a single trigonometric identity. Except, $\sin^2x+\cos^2x=1$ of course.

@Sawarnik Go to the hell and rot, geometer.

@BalarkaSen Aw cut him a break he's like 14

@BalarkaSen 1. I am not a geometer. 2. There is no place called hell. 3. How can I rot myself?

@MickLH :D

Dear lord. Here we go guys...

@Sawarnik 1. No comments. 2. Of course, SF exists. 3. I dunno, bake yourself in a frying pan or something?

@Chris'ssis Have you ever got into elliptic whatnots? I find them pretty beautiful.

@BalarkaSen Only a bit. I'm going to be there soon ...

@Chris'ssis I see. Good luck!

It's a wonderful branch of mathematics.

@BalarkaSen I didn't see many people interested in it. Many are really afraid of it.

@BalarkaSen From what little I've seen, I agree.

@Chris'ssis I can give you a lot of peoples interested and working in it.

I guess I did go in the wrong direction lol, these Pochhamer symbols are getting ridiculous

It's a branch in Number theory.

And with lots of open problems.

@Chris'ssis The amateurs are afraid.

@BalarkaSen I figured out an elliptic thing the other day

Not the professionals who work over it.

What is necessary to understand it? So far to me it just looks like an extension of field theory.

@Alyosha Well, a bit complex algebraic geometry, I think.

At least that's all of the prerequisites I have.

Otherwise, there are geometric approaches.

Like, for example, hyperbolic geometry.

@BalarkaSen I see.

I'm sure it's been done a million times, but you can come up with a closed form for the Arithmetic - Geometric mean in terms of an elliptic integral

@Chris'ssis Modular forms have many impacts on all over mathematics, most of them discovered in the 20th century.

@MickLH I hate elliptic integrals, to be frank.

It's the dullest thing I have ever seen in mathematics.

I just thought it was ironic that it came up, I literally just started learning about them maybe a week or two ago

Exploring simple algorithms with complex results

I even ended up with a new square root engine out of it for my software :D

Is it correct that transformations in hyperbolic geometry are all Mobius transformations?

blahblahblah

Apparently there's a bijection between the Lorentz (i.e. relativistic) transformations and the Mobius, but I don't know.

I am back.

@Alyosha Transformations?

Do you mean those preserving geodesics?

I don't know, I read something to that effect somewhere and wanted to check in what sense it was true.

@Alyosha Well, the connection with Moebius transformation is more easily seen if you look at the geodesics in hyperbolic plane.

I am not really a geometer, so I can onyl give you an algebraic description of the model.

I've not studied geometry much at all, is there an equation that hyperbolic geodesics satisfy?

Of course.

Which?

The geodesics are precisely the translates of the x-coordinate (from the fundamental domain, but let's leave it for now) and the other geodesics are the half-arcs, precisely the ones you get by spherical inversion.

So you get $x = 0$ and $(x - a)^2 + y^2 = c^{-2}$ as geodesics.

The isometries here are precisely the moebius transformations, if that's what you mean.

I.e., takes geodesics to geodesics.

$c$ being the curvature?

all of the semicircles in the upper half plane with diameter on the x-axis are geodesics

@Alyosha $a, c$ are constants.

well, parameters

@seaturtles Yes.

you can vary a,c to get different geodesics

@seaturtles Yup.

Is the geodesic definition as 'the line with shortest distance' be used (with the Euler Lagrange equation and the hyperbolic metric) to prove the same thing?

Yes.

That's how they are derived.

How's the metric defined?

The usual hyperbolic metric $ds^2 = (dx^2 + dy^2)/y^2$

Is the elliptic-geometry metric similar?

I don't know. Never read about them.

Okay, thanks.

I am merely a newbie in hyperbolic geometry.

Well, at least now you know that you're not the newest newbie.

=)

@seaturtles You geometer?

nah

Shakehands! We are both communists!

Anyone here who is not a communist?

encrypted-tbn0.gstatic.com/…

Not the best Tintin, but I like that bit.

@Alyosha Glad to here you are a communist too.

=P

@BalarkaSen Me.

@seaturtles So may I assume you are a pure algebraist?

ish

Now you are confusing me.

Confusometer.

@seaturtles Algebraic geometry then.

That's my final guess.

$$\sum_{n=1}^{\infty} \frac{e^{-(2n-1)} \sin(2n-1)}{2n-1}=\frac{1}{2}\arctan\left(\frac{\sin(1)}{\sinh(1)}\right)$$

Anyone here?

@N3buchadnezzar No.

@N3buchadnezzar unfortunately.

@N3buchadnezzar Yes.

@N3buchadnezzar Limbo.

Mmm, I find something a bit strange

@N3buchadnezzar Life? Existence? 42?

I got a futile attempts badge! Yahoo. Now I can retire from stackexchange as the purpose of my life has been achieved.

Guys, guys, there's the PSA looking at us.

@N3buchadnezzar Sawarnik's childishness?

@BalarkaSen teehee oh well

@BalarkaSen There's the NSA looking at us.

@bal @saw it's the circcclllleee of liiiiiife it's the wheeeel of fortuuuunnne it's a leap of faith

Nonsense. Are you gonna post or what, @N3buchadnezzar?

@meer2kat ? and ? and ?

@Sawarnik it's a song. get in the game

@Chris'ssis $-\Im\left( \int \sum_{n \ge 1}e^{-(2n-1)x}dx\right)_{x=(1+i)}$? I fear I've overlooked convergence issues, though it seems superficially kosher.

Oh noes I have to go.

@DanielFischer People should ask more functional analysis questions.

Especially basic ones (that I can answer) : )

@meer2kat Since you have done 6 years of calculus, and you are of 19 years old age, so means you have started out at 13?

Where $s = it + \sigma $

@Sawarnik we've already discussed this.

@MattN. Aiming for the silver badge?

@meer2kat orly?

Hi @mee

@user127001 hello reasonable being!

@Sawarnik yar

@DanielFischer I haven't decided yet.

Now this question is in some way "obvious", because $e^x = \lim_{n\to \infty} (1+x/n)^n$

Hi @use

It would be cool though.

Hi @Sawarnik

If it is so obvious, why does one need a heavy proof relying on pointwise convergence?

Apparently Jordan's curve theorem is very hard to prove.

@N3buchadnezzar Just because the sequence of integrands converges pointwise doesn't mean the sequence of integrals converges.

@meer2kat no.

@Sawarnik yerp

@Alyosha It is. It's doable without advanced machinery for "sufficiently nice" curves.

@meer2kat i don't think so. gimme a proof.

@DanielFischer Right, so we need to use uniform convergence

@Alyosha something like that should work.

@Sawarnik 42.

@DanielFischer Broken link aside, is this comment helpful in any way?

Right? =)

@N3buchadnezzar Problems ahead. a) you can't integrate from $0$ to $\infty$, as $\left(1-\frac{t}{n}\right)^n$ grows as $t^n$ in absolute value. b) uniform convergence isn't sufficient, you need dominated convergence or so.

@meer2kat That is not a proof.

@Sawarnik That is not a proof of a lack of proof

@Sawarnik Nor is anything you say or do. Move on.

@MickLH thanks man

np :)

@N3buchadnezzar dollar dollar bill, yall

@MattN. Not really, but it was well-intentioned, exposing a class of nowhere dense sets with positive measure.

@N3buchadnezzar The domain of integration has (in the limit) infinite measure.

Oh well.

blahblahblah

Yeah I get it =)

What is this?

damnit so familiar....... where

@Sawarnik gosh....ugh

Etiquette Guidelines | $\LaTeX$ support for chat

@DanielFischer lol

there ya go

The emperor has spoken :-)

@robjohn you have to see this one $$\sum_{n=1}^{\infty} \frac{1}{n^2} \cos\left(\frac{\pi}{n+\sqrt{n^2+1}}\right)=\frac{\pi^2}{12}$$

It's AWESOME!

It's really hard for me to help people who are struggling with 6-1...

@Mike 6-1?

6-1=5

@skullpatrol You are correct.

@Mike 6 + (-1)

Try to be patient :-)

Wow. For the first time ever, I just got (temporarily) kicked out of the review queue because I did 20 close-vote reviews in a day. :)

Wow :)

Heyyyy is that the guy who hates random strings of text in the star queue?

@skullpatrol Patience is my best quality here. The actual problem is when I can't help.