Heya @Alyosha

Hello.

I cannot refuse to not disagree

@Alyosha Wattcha been doing these days?

I mean, any math?

Preparing for exams, missing abstract algebra.

@Alyosha Ah, snap.

Yes. I hope to learn quite a bit over the summer before university so I can sit in on interesting courses, though.

@Alyosha You are doing your final year in school, right?

Yes.

Lucky you.

Why?

I have still got some 4 or 5 years to finish.

The non-math subjects are becoming unbearable.

Ah. Well, school is fine, one can just read interesting stuff in lessons. Exams are the problem.

Isn't the math more unbearable?

@TedShifrin Sure not.

Your actual school classes??!!!

Oh, that.

Yes, it's unbearable.

UH huh.

I got some 16 outta 50 in my prepositional exam this year.

So literally passed but only just.

@Chris'ssis Hi .. are you there ? :D

what is a prepositional exam? prepositions?

@r9m Yeap! Hi :-)

@TedShifrin heh. no, there are exams before the final exams here.

like first and second terminal.

@Balarka: It's as if you were speaking Irdu to me.

@Chris'ssis Wait .. I'll show you sth crazy :-)

@r9m OK

@TedShifrin You have no idea how sick the educational system here is.

It's quite ill here, too, but the brightest survive regardless.

I am not the brightest. Or even bright. The one who gets 16 out of 50 is not at all bright.

You say ill as if it was ever healthy.

@Alyosha I think it's better out there

No, @Balarka, you're plenty bright. You're just stubborn and lazy about stuff you think you don't care about.

Sort of like how I hated history classes in high school (except for senior year) because they were taught by idiots who made us memorize stupid things. Hated, hated, hated.

Perhaps. Regardless, it gets boring poking the education system after a while.

@TedShifrin If you ever had to face a word problem from one of our exams, you won't think that I am being stubborn.

@Chris'ssis $$\int_{-\infty}^{\infty} \frac{1}{(e^x-x)^2+\pi^2}\,dx = \frac{1}{1+x_0}$$ where, $x_0$ is the unique root of to $xe^x = 1$ :D

Just be nihilistic and get on with interesting things.

You essentially have to memorize math there.

You're talking about 16/50 on a maths test, @Balarka?!!

@TedShifrin Yes.

Now that's just absurd.

I got the 16 out of all the algebras. I left all the arithmetic and geometry out.

@r9m lol, that's pretty cute. :-)

That's what I mean. Lazy and stubborn.

shrugs

If you can do all this fancy stuff you do in here, you can do that stuff easily. You just don't care to give it your attention.

I have to give it attention sooner or later. If I can't get a good grade then I am going to fail.

sigh

Right. No excuse not to get 100s on it.

@Chris'ssis lol ?! ;)

@TedShifrin Well, the bright side is I got almost 75-80% on all subjects, except language.

@r9m It looks funny, and it's related to some integrals involving the Lambert W function I saw some time ago in a paper. :D

heya @Kaj

@Kaj Addicted.

Like I said.

LOL

Maybe @Kaj will start avoiding me, just like you, @Balarka.

@TedShifrin I don't avoid you.

@TedShifrin, that's a lot less likely now that I'm done with your classes

And aren't you thrilled, @Kaj :P

@KajHansen Did Ted teach you the unforgivable curses in the classes?

@Balarka: I would, quite sincerely, encourage you to be the best all-round student you can be. In the end it will pay off.

unforgivable curses?

@TedShifrin I am in the last year of my middle-school. So if I don't, I am going to perish.

Yes, @Balarka, and you should excel.

@TedShifrin whenever i talk to you, i get an image of Mad-eye Moody growling.

I dunno what that means.

@TedShifrin, did you see the question Faraad posted on here regarding functions from a sphere to a torus?

@TedShifrin you better not.

Nope, @Kaj.

@BalarkaSen, that is actually somewhat accurate when he's in certain moods.

I'm only in those moods when students who are very capable let me down :( @Kaj

@KajHansen $\Bbb CP^1 \to \Bbb C/\Lambda$?

Seems elliptically fishy.

@KajHansen What a poor drawing.

Not familiar with that notation, unfortunately @BalarkaSen.

@Kaj: That needs to be differential geometry, since he mentions lengths of curves :P

It's not MY drawing :P

I know. I don't know who this studiosus is, but we've crossed paths many times.

@KajHansen That is a good question.

I wonder: If you started with a great circle on a sphere, surely that couldn't still be a simple path once mapped to a torus?

@KajHansen I think not. There are infinitely many great circles in a sphere, contrary to a torus.

What do you mean, "contrary to a torus," @Balarka?

Great circle = geodesic. Just as infinitely many in the torus.

@TedShifrin there ain't infinitely many simple great circles in a torus.

Read what I just wrote.

Great circle means nothing outside of a sphere.

@TedShifrin Map preserves length of curves.

@Balarka: You ain't making sense.

"I was wondering if there was a map from the sphere to the torus which preserves length of closed curves?"

What is true, @Kaj, is that, although there are some closed geodesics on the torus (we can use Clairaut to figure out which, I think), most will not be, and probably most will self-intersect.

What are the chances to win for this guy?

@Kaj: Cute that Faraad changed his SE name once he knew I am on the prowl :D

There are 6 equiprobable cases (p=1/2), but 2 are linked together

@Kaj: $\Bbb CP^1$ is the Riemann sphere (see chapter 8 of my book :) ) ... If $\Lambda\cong \Bbb Z\oplus \Bbb Z$ is a lattice in $\Bbb C$, the quotient group is homeomorphic to a torus. @Balarka is showing off ...

OK, I think I'll be leaving the chat.

@TedShifrin Ha! @cc, I hate that about minesweeper. It would be a beautiful game if the player had perfect information.

so I don't think $(1/2)^4$ is the right answer

@TedShifrin No, I am babbling about things I know superficially.

Yeah, that doesn't become you. :)

grins

But in the meantime, please be a solid top-notch student. You'll be better for it in the end.

@TedShifrin I have to digest and memorise a lot of math to get good grades. That's how the syllabus here goes.

And that's what irritates me.

"memorize" ?

@KajHansen Yes,

They give you very clever and essentially wrong problems and you have to do it precisely the way it;s given in the book.

I do remember having to memorize some stupid rules. E.g. an answer was wrong if a radical was left in the denominator of a fraction.

Mostly, they give you word problems.

Oh, and an answer was wrong if you didn't show every single minute painstaking step.

rolls all six eyes

@KajHansen I am perfectly fine with algebra. If they give the whole paper on algebra I can have full marks.

@KajHansen That too.

oh great, now @Kaj is stealing all my eyes.

LOL

What pains to say that they are essentially teaching us banking instead of math.

Well, @Balarka, admittedly, that's not interesting to a Hardy-Ramanujan like you, but for the average middle/high school student, those are essential skills.

I actually had a reasonably good experience in 8th grade math. We did geometric constructions for a whole day every Friday.

The other grades....not so much.

@TedShifrin Me? Hardy-Ramanujan? I never read that paper, sorry.

You should be able to do it effortlessly with very little thought, @Balarka.

A B . . . C D say A and C have mines, so user must choose B and D consecutively to win, There are only 6 possible events, so $p(\text{win})=\frac 2 6 = \frac 1 3$?

Stop being obstinate, @Balarka.

Yeah, I reckon I have to do it their way.

I should add that we also got MathCounts problem sets in 8th grade. Those required some actual cleverness.

@KajHansen What?

@BalarkaSen, In the USA, MathCounts is one of the more popular competitive math leagues, geared towards middle schoolers. A bit like USAMO I guess?

However if the 2 areas are linked together this probability becomes $\frac 1 2$

@KajHansen WAT

@TedShifrin I reckon I am going to be in serious trouble if I don't get good grades.

Indeed, @Balarka, and it may really ruin your very promising future. So please don't do that.

@BalarkaSen, Yeah. There are regional, state, and national level tournaments. Within these, there are lightening rounds, slow individual rounds, team rounds, and so forth. en.wikipedia.org/wiki/Mathcounts

@Kaj: You need to remember to give Irby a motivational speech before school starts :P

The exam's next month so I have to do a lot of study to do my usual math and preparation for the exam both, @Ted

@KajHansen Math Olympiads are the last thing I will ever sit on.

Well, doing well in school should get priority. Once the exams are passed and past, then go back to your maths fun. @Balarka

@TedShifrin Not worried. not the final exam yet.

You sound like my students who — ultimately — fail, @Balarka.

Just have to obtain 80% or over.

@TedShifrin i am tempted to star that.

@TedShifrin, I need to impress upon him the importance of keeping up with his work too. My 3510 stuff aside, I'm proud that I was able to get perfect scores on all the 3500 Webworks. I reckon that really helped in the end...

@BalarkaSen, I'm not a huge fan of competition math. But at least MathCounts problem sets for homework required actual thought, instead of rote algorithmic "math" like usual middle/high school homework.

Yes, @Kaj, for sure, plus actually keeping the log to study from. I think you're going to turn out to be one of my students who didn't excel in my class but, given that horrid experience, excels for everyone else :P

No, seriously @TedShifrin, I never studied like I am doing this year. In fact, i can;t recall the last time I got 75% in history apart from year.

Well, @Balarka, not that I can inspire you ... but I took pride in being an excellent student all-round, not just in math.

@TedShifrin That is not at all encouraging, unlike my professor who got 9 out of 50 in his algebra exam in his last year.

Speaking of which...I'm taking French next year @TedShifrin.

=P

good, @Kaj, you can turn in your math to me in French. Oh, wait ...

Well, this fall I should say.

@Balarka if you don't do well in your other subjects, you won't get to do math where you want. Practical, real-world things may not be as interesting, but they still matter.

@AndrewG I'm aware of that.

heya @AndrewG ... you got my answer this morning?

just saw it, Ted, thanks :)

i'm still working through these notes

it will take you a while :)

unless you know more than I think

yes. But it's slowly starting to click. Really cool stuff.

I'm glad you're enjoying it :)

I think I'm falling in love with geometry :3

aww beams

@AndrewG grumble

beams more

Seriously, balarka, education is about playing the game to climb a ladder just as much as genuine learning, and as much a that sucks, there's no way around it.

@TedShifrin and @AndrewG filled me with dread of the exam the next month.

If I could go back and master boring stuff to do better on my tests, I would jump at the chance.

well, you need some dread to get motivated, @Balarka. You're too invested in your math ego.

@TedShifrin i think i can do better in all subjects if i really study hard, but my only fear is that i might not get good grades in math.

Well, the math in your school should be a total joke. If you sit down and think, instead of resenting, you should master it in short order.

Well, I guess you are right after all.

I should study hard this month.

It's been known to happen once or twice.

I haven't advised a thousand students for nothing ...

yes, @Balarka.

@r9m I imagine that should work by some complex analysis. The fact that that root appears in answer makes me think of some specific poles ...

heh. ok, i guess a month should be enough for preparation of a 50-total exam.

Make us proud, seriously, @Balarka.

I won't be on here much for several weeks, btw :P

@TedShifrin where are you going?

on vacation to California

Have fun with rain forests there.

no rain forests ... dire, dire drought

@TedShifrin wat?

i heard it was full of rain forsets

or i am reading geography for nothing

no, the state is burning up

@TedShifrin cool. like us.

very not cool :(

the humans have totally ruined the earth ... but I digress

i should bu-byes now. it's midnight and i am thinking of having a bit snacks and reading a storybook.

to relax my mind for preparation.

sleep well, @Balarka.

@TedShifrin i will sleep at 6:00 AM.

that's my usual time.

aren't you back at school?

@TedShifrin extra holidays =P

oh right

i am joking.

the holidays are done for.

don't joke ... I have no sense of humo(u)r.

i have a fever, so can;t go to school.

convenient how you keep having that fever

do your parents ever take your temperature?

@TedShifrin yes, this one is sirius.

serious?

i will be consulting a doctor tomorrow.

damn ... well, you can't do math with a fever. Get healthy.

that's why i am not doing math.

killing time.

i can;t think.

@Chris'ssis ic :D

drink lots of fluids and get sleep

@TedShifrin that's always your opinion.

@r9m Am I on the wrong track?

for pretty much anything.

@Chris'ssis no. you're not.

good night, @Balarka

@Chris'ssis IDK ... but you make perfect sense :-) .. I just saw the problem in AoPS .. haven't tried it yet with pen and paper :|

good bye.

@r9m :D

@DanielFischer HALP-.

@Pedro: I wore him out helping me earlier :P

URGENTLY, @Pedro?

LOL

@DanielFischer Are you busy?

@PedroTamaroff Not really. I just played on your all-caps misspelling.

@DanielFischer That was intentional.

@Pedro likes to misspell.

He thinks it be hip.

It may be nowadays, @Ted.

Yeah, what would we knowz?

@TedShifrin It's actually more of an onomatopoeia.

@PedroTamaroff What do you want help with?

See, I can spell.

bows @Pedro

@DanielFischer A criterion for a sequence to be a Schauder base of a Banach space.

ah, coming from blue's query ...

@TedShifrin What did blue ask?

The condition is as follows, @DanielFischer.

Why for infinite dimensions $F^n$ has dimension $>n$.

Over what?

$F$

:confus:

Dimension as what?

$n\ge\aleph_0$

That's exactly the point :)

Oh, you got me there.

$(x_n)$ is a Schauder basis iff it has dense linear span and there exists $K>0$ such that for any sequence of scalars and all $n<m$

$$\left\lVert \sum_{i=1}^n a_ix_i\right\rVert\leqslant K \left\lVert \sum_{i=1}^m a_ix_i\right\rVert$$

@TedShifrin Explain?

Hey, don't drag me into this. You're talking to @DanielF.

@TedShifrin I am asking about the $F^n$ thingy.

Oh, well, as a vector space, you need an uncountable (Hamel) basis ...

What do you mean by $F^n$?

Suppose you look at $\prod_{i=1}^\infty F$.

Then any (algebraic) basis must be uncountable.

Right, I was guessing you meant the product.

@DanielFischer Do you know the condition? Books says it is due to all mighty Banach.

@PedroTamaroff No, I don't know it. Thinking about the proof.

@DanielFischer I have the proof.

It is in the book.

I am not following a part of it.

Okay, which part?

@DanielFischer Okay.

So, suppose we have those two conditions.

First, by induction on $n$ in $$\sum_{i=1}^n \alpha_ix_i=0$$ one can show the $x_i$ are all linearly independent.

Huh?

Induction on the number of sumands.

If $\alpha_1x_1=0$, evidently $\alpha_1=0$.

Why?

Because $x_1\neq 0$.

Why is that?

We can assume $x_1\neq 0$; else we may just discard it from the putative base.

ROFL

Don't we need some additional hypothesis(es)?

No.

Why should any finite subset be linearly independent from what you've said?

@PedroTamaroff Well, you must assume that none of the $x_n$ is zero. Adding zeros to the family doesn't change the inequality.

@TedShifrin OK. So, suppose now that $\alpha_1x_1+\alpha_2x_2=0$.

Then $\lVert \alpha_1x_1\rVert \leqslant K\rVert \alpha_1x_1+\alpha_2x_2\rVert =0$.

Yeah, I am ok with that as long as we assume all the vectors are nonzero.

Aha.

Well, so now let $S$ be the span of the $x_n$.

@Chris'ssis: I didn't see you come in. Hi there.

I've been busy the last couple of days with mod stuff, and this morning with work.

@robjohn Hi. :-)

work ... what's that? :)

We have maps defined on $S$ by $P_n(\sum_{i=1}^m \alpha_ix_i)=\sum_{i=1}^n \alpha_ix_i$, $n<m$.

@robjohn I see. No pb. :-) I posted a nice series above. Take a look on it when you have some time.

@DanielFischer That is well defined since the $x_i$ are linearly independent.

These are projections.

And they are continuous, with $\lVert P_n\rVert \leqslant K$.

yes

Hence uniformly continuous and from $\overline S=X$; we can extend them to all of $X$.

@Chris'ssis I was looking back. I guess I haven't gotten that far.

ok @Pedro

And this extension is still a proyection, still continuous, and $\lVert P_n\rVert \leqslant K$, keeping the name.

@robjohn It has an interesting closed form.

The book says "It suffices to show $P_nx\to x$", to show this is a Schauder basis.

sure: What's your definition of Schauder basis?

That every $x$ is a sum $\sum_{i\geqslant 1} \alpha_ix_i$ in norm.

so ?

@Chris'ssis just found it.

@TedShifrin I don't see it.

@robjohn Yeap. That's the series.

$\|P_n x - x\| = \|\sum_{k=n+1}^\infty x_k\|$

@TedShifrin WAT.

@PedroTamaroff You need $P_n \circ P_m = P_n$ for $n < m$, and then the word of magic is equicontinuity.

@DanielFischer That holds, yes.

@DanielFischer The $P_n$ are EC because the norms are uniformly bounded, yes?

Yes, and those are the ingredients you need to finish the proof.

Let me think.

@DanielFischer Wait, do you mean to show $P_nx\to x$? I can see that. I cannot see why this implies every $x$ is uniquely a sum in norm.

I get that it should be $$\sum_{k=1}^\infty(-1)^k\frac{\log(k)}{k} -\sum_{k=1}^\infty(-1)^k\frac{\log(k)}{k^2}$$ I've computed the first sum for an answer somewhere... I think I should be able to compute the second sum, too.

@DanielFischer pls raspond

@robjohn It seems it works this way even if there is a convergence issue for this method. :-)

@PedroTamaroff $P_n \circ P_m = P_n$ plus convergence shows that you have well-defined coefficient functionals $\varphi_n$, so that every $x$ has the representation $x = \sum \varphi_n(x)\cdot x_n$. It remains to see the uniqueness of that representation, and if I understand correctly, that's what you don't see?

@Chris'ssis There is no convergence issue.

@DanielFischer No, I didn't realize we had those coefficient functionals.

@robjohn I refer to the convergence of the double series in absolute value.

@Chris'ssis Well, yes, if you take away the $(-1)^{k+n}$ the series does diverge.

@robjohn This point is related to the well-known $$\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{ (-1)^{k+n}}{k+n}$$ where that method doesn't work.

:16467922 Yes. The point is that the coefficient of $x_k$ in $P_n(x)$ is the same for all $n \geqslant k$.

@DanielFischer Wait, but a priori we cannot write $P_n(x)$ as a sum of the $x_k$; but approximate it.

@robjohn anyway, it's good it works! :-)

@PedroTamaroff $P_n$ has finite-dimensional range.

@DanielFischer Oh, DERP.

@Chris'ssis Do you get some better closed form for $\zeta'(2)$, or does your answer have that in it?

Right, $P_n:X\to S\subseteq X$.

Dang it, so silly.

Thanks.

@TedShifrin Maybe you were trying to say something like that above. Dunno.

@Chris'ssis Whatever $A$ is, can you write $\zeta'(2)$ in terms of it?

@Chris'ssis Why do you have a double sum, if at the beginning you have a sum of simple sums (that should give another simple sum)

@robjohn Sure, that is well-known.

@Chris'ssis Okay, then I feel better about my answer.

@Chris'ssis That is also equal to $\Phi$.

@robjohn :D

@PedroTamaroff I doubt anything I wrote before equals that character (alone).

Hey, I'm cramming for a Calculus final and could use a little help please.

@Moshe Calculi hurt, yes.

I got a derivative but now I'm pretty sure I have to simplify it, but I'm at a loss as to how.

@robjohn see $(45)$ here mathworld.wolfram.com/RiemannZetaFunction.html

@Moshe OK.

@robjohn By the way, did you compute that second sum before? (I wonder if you had it posted on MSE)

@PedroTamaroff ^

@Moshe That is correct, only that I'd write $f'(x)$ in the second line, else it looks like $f(x)$ equals the second line too.

@PedroTamaroff Is there a way to simplify?

You can draw a common factor of $(2x-3)^3(x^2+x+1)^4$.

Then work with the rest.

Yes, now you can simplify the rightmost parenthesis.