Hi

@BalarkaSen got the highest mark in my geometry class.

thanks a lot for your help without talking to you many things I wouldn't have understood.

hey guys wanted to ask whether you know a way to collaborate on exercises, i know there's email but that's very impractical, does anyone know of a tool? (that can also support mathjax?)

Hello chat

hello guys

Hi

hello everybody

anyone know a candidate for my question above?

sharelatex.com

you can learn to tex and work on the same document together

yeah but very buggy

i was thinking about using github and setting mathjax there

don't even know if that's possible

or i heard about gitbooks

don't know whether one can use it

for that purpose

you can use github for a tex document if you want

oh really? do u know of a simple tutorial on that? last time i searched i didn't find something useful

no I dont

I dont work with other people

D:

but my friends tell me that the sharelatex thing works well

overleaf.com is also like sharelatex.com but less buggy but still same problem

@Adeek Good to know.

Congrats, and no problem.

@BalarkaSen do u know anything about my question above?

no

Hi.

Don't spam though. If someone knows something they'll answer.

No worries, just saying that pinging everyone to answer a specific question by a specific user is frowned upon.

I don't know @user153330 Sorry :I

anyway i will try @s.harp suggestions

@Mahmoud no problem

Question I've been fiddling with

@MikeMiller I found here written $\zeta(1)=\infty$ ???

Why are you telling me?

not sure why that's a shock, either. the harmonic series diverges.

@Semiclassical but $\sum_n n$ does too, yet zeta function thinks its -1/12

the harmonic series hella diverges

not all divergent series are the same.

alternatively it's the limit of actually convergent series

unlike sum n

what do you mean the harmonic series is the limit of a convergent series?

a convergent sequence, I think you mean @MikeMiller

@Semiclassical but nobody would have thought that $\sum_n 1/n$ is super hard divergent whereas $\sum_n n$ is only sorta divergent

well, sure.

But also $\frac1N \sum_n^N 1/n$ goes to zero whereas $\frac1N \sum_n^N n$ also diverges

What do you mean by "super hard divergent" ?

eh, that's now got a_n that's not dependent just on n

it's also dependent on N, and that's changing.

@Astyx one method would be to consider the terms I wrote in my last message

oh wait I miswrote

thats not what I should have written

Let $a_N=\sum_n^N 1/n$, then $\frac1N \sum_n^N a_n$ diverges yes or no?

No, I mean that it's the limit $\lim_{\alpha \to 1^+} \frac{1}{n^\alpha}$

ah.

Back, I'm sorry.

you can't approximate the other one similarly

for example if you use that construction with $(-1)^n$ it converges to one half

Cesaro sum, yeah.

I told you because I thought that's non-sense @MikeMiller and I wanted you to help me ?

@s.harp what variable goes to infinity here ?

I think $1/n$ also diverges in this way right, because it the $a_n$ eventually becomes like $\log(n)$ and the sum then the integral which is $N\log(N)$ divide by $N$ you get $\log(N)$ still

@Astyx the $N$

@s.harp Oh right

And how is that more divergent than $n$ ?

well both diverge with this method so :)

Mmmm ... right

:)

divergent series are weird

Semi-convergent are even weirder

must... resist.. temptation... to be mean... to physicists..

@s.harp Nah, that's ok. It's what they are for.

Why would you be mean to them ?

the meanness would be to make fun of a percieved carefree attitude about divergent series, ie most presentations of QFT are:

Physics is the most pure application of Mathematics.

this diverges: Ok big deal just make it zero

Then applied Physics give Chemistry, then applied Chemistry gives Biology ...

that way of thinking of sciences as a "chain" of pureness is not right at all

@Mahmoud It is not nonsense and given a generic question, I would probably rather not think about it.

Also yeah that xkcd is hot garbage.

How can we set a value of something to $\infty$

Infinity isn't a number.

It's a way to make $\Bbb R$ compact

@Mahmoud if it eases your mind you can say that the $\zeta$ function is not defined at $1$, ie that it is a function $U\to\mathbb C$ where $U$ is a subset of $\mathbb C$ that does not contain $1$

You can complete $\Bbb R$ to $\overline{ \Bbb R} = \Bbb R\cup \{\infty\}$

it's not that different from asking what $1/(1-x)$ is at $x=1$.

And still have everything else working fine

@Astyx Well, most things

(except that some operations are undefined, eg $\infty - \infty$)

No it doesn't blow my mind, but I learned too many times to think about $\infty$ as the size of an unending amount, and now we set it to a value ?

Keep in mind, when someone says $\zeta(1)=\infty$, they don't mean that in the same way that they'd mean $\zeta(2)=\pi^/6$

I think infinity can be defined as the lack of a definitive value

@teadawg1337 No, that would be a poor definition

it's just a shorthand for "$\zeta(s)$ diverges and is not well-defined at $s=1$"

Yeah, it's bad

@MikeMiller Tried your question using silly Hatcher chap4-tricks. Didn't work.

In the case of the Zeta function it has a pole at 1, so there actually is a continuous/holomorphic extension if you modify the target space to be $\overline{\mathbb C}$ which is a space that has a structure so that its possible to talk about continuous/holomorphic functions

then the function sends $1$ to the $\infty$ point

@AndrewT If you have any partial progress I'd be interested.

@Semiclassical I think it means more specifially that $\zeta(s)$ diverges to $\infty$ as $s$ goes to 1

well, which infinity? :p

Both

That's the point :p

not just two infinities if you're thinking of zeta(s) in the complex plane (which you should)

Anyway why isn't it defined at $s=1$ ? We can write $\frac 1{1^1}+\frac1{2^1}+\frac1{3^1}+ ...$

Yes but this series is divergent

@Mahmoud That's the harmonic series, which is proven to be divergent.

here's a classic proof of the divergence

You can't actually legitimately write this (until you define it)

note that 1/3 is bigger than 1/4. so that divergent series is at least as big as 1/1+1/2+1/4+1/4+ etc.

Hi.

that was the laziest proof I've ever seen!

Can I ask a question about projective geometry?

similarly, 1/5, 1/6, and 1/7 are all bigger than 1/8.

$$\sum_{k=n}^{2n}{1\over n} \gt {1\over2}$$

so you can take the next four terms and say they're all at least as big as 1/8.

Just ask instead of asking to ask! (There's no guarantee you'll get an answer though)

@Topologicalife yes of course

you can repeat that as much as you want. the thing is, you'll get 1/4 a total of 2 times, 1/8 a total of 4 times, 1/16 a total of 8 times, and so forth

@s.harp I'd argue that the 1300's didn't have the same standards for mathematical rigor :P

so each portion of the harmonic series is at least is big as 1/2.

So it doesn't approach any number, I didn't get it at first because it diverges ''slowly'' if you allow me to use those terms.

@Alessandro whereas if you ask to ask, you can be certain that someone will answer your question with "Don't ask to ask !"

Thanks. If A^n is an affine space of n-dimension, E his associated vector space and P(E) the projective space associated to E, then the set defined by P^n = A^n U P(E) has a projective space structure. My question is: Every projective space is of the form A^n U P(E)? i.e In any projective space we can distinguish between a affine part and other part of the infinity?

it's a very slow growth, yes.

but slow growth can still be unbounded.

Not the slowest growth either

@teadawg1337 I was joking that because of the time spent it looked like that was all he was gonna write :)

I like this, first time I see such divergent series.

as teadawg is alluding to, this proof is an old one: it goes back to Nicolas Oresme, if I remember right, who was a medieval monk

@s.harp I was joking as well

@Topologicalife I don't remember a lot of projective geometry, but the projective space without the hyperplane "at infinity" should be an affine space

Exactly @Semiclassical

@Topologicalife what do you mean A^n U P(E)

@Mahmoud You can prove that $\sum {1\over n \log(n) \log(log(n)) ... \log^{k}(n)}$ diverges for any integer $k$

I mean $\mathbb{P}^n=\mathbb{A}^n \cup{}\mathbb{P}(E)$

here's another weird fact, which is much harder to prove

1 + 1/2 + 1/3 + 1/5 + 1/7 + 1/11 +... (sum of reciprocal primes) is also divergent

"reciprocal" primes ?

reciprocals of primes

Multiplicative inverse

e.g. 1/19

Oh right

Another fun fact, the harmonic series skipping all the terms with a $9$ in the denominator converges

heh, I was just getting there.

the Kempner series is weird

(That is, with 9 as a digit)

I think the way you are working with projective structure is unknown to me, so I cannot help you @Topologicalife

Hi everyone

(So you're omitting everything in the 900s, for example)

Heya, @teadawg @Alessandro DogAteMy :)

it's less strange when you consider how unlikely it is for a really big number to not contain any 9s

Hola!

Hi @Ted! How was your travel?

Hey there @Ted!

@Alessandro Yes, this is a wonderful result. I used to assign it every time I taught calculus with proofs.

May I ask whether somewhat mathematical but mostly formal logic puzzles are accepted here as questions?

$\infty -{1\over9} \in \Bbb R$ then ? :)

It was a very long drive yesterday, but I survived :)

Fun trip ... except for the interference of depressing and scary politics.

^

@Semiclassical from the number of $n$ digits $(1-1/9)^n*10^n$ will not contain a $9$

So when are you moving to Canada?

Hi @TedShifrin

@Alessandro, you are not safe either. I was reading about how dissolution of the EU is imminent. The world is f***ed.

hi @Mahmoud

9/10 chance the first digit has no nine, 9/10 change the second digit has no nine… When you have a lot of digits, it becomes really rare.

@TedShifrin Why would it be imminent ?

I must admit I don't follow politics very closely, but the situation looks grim @Ted

Because right-wing populism is overtaking all countries. I dunno. I just read it. I can't explain it.

This election is a whole mess of 'i dunno'

It's a shame people can't be bothered to go and vote

@s.harp Well, my question is if in any projective space we can distinguish between an affine part and other of the infinity.

@Sejanus: I don't know what you mean by formal logic puzzles. But there's a puzzle tag on MSE.

Going back to maths I don't think I know how to prove that the Kempner series diverges actually @Ted

@Topologicalife: It all looks the same everywhere.

@Alessandro: What is the Kempner series?

But, I think @Alessandro already asked my question

s/asked/answered

The harmonic series skipping the terms with $9$ as a digit @Ted

we might define a frak-tal as: a situation which, at whatever scale you look at it, is really frakked up

Oh, I didn't know it had a name. It converges, silly. :)

@Alessandro en.wikipedia.org/wiki/Kempner_series#Convergence

smacks @Semiclassic for worse-than-Ted humor

hah

Too bad @Balarka isn't here to appreciate your being smacked.

I'm sure he'll see it eventually

Well that was pretty obvious in hindsight

Especially so now that it's starred

Thanks guys :)

@TedShifrin After Brexit, we'll get a Departugal

@TedShifrin @Semiclassical Hey guys

And possibly more puns

Heya @Lozansky ...

either that or a F***-it-al

which I guess is what Trump's election was already

Out-of-luck-sembourg?

Well, DogAteMy, I will appoint you in charge of secessionary humor.

@AkivaWeinberger That one is actually good :)

@Alessandro I'm not sure if this counts as a proof, but most large integers contain all digits from 0 to 9, so the number of terms excluded monotonically increases

Oh my ....

No, that's not a proof of anything, @teadawg.

@TedShifrin What's with the ellipsis...?

…Czechout

I type ellipses a lot in chat.

Meh, it was worth a shot.

I think that makes the result plausible, though.

Probabilistic arguments are often heuristics, but hard to make into rigorous proofs.

Yeah.

@teadawg @Alessandro: I'll give you a hint — You can give a direct upper estimate that establishes convergence.

they already linked me a full proof :( @Ted

Hmm. Now you've got me wondering about a 'probabilistic' series like $\sum_{k=0}^\infty \mu_n/n$

I wish there weren't so many proofs available on line.

One of the reasons i don't like MSE as a complete repository of all answers.

I'll have to find something else to prove for tonight

i.e. if you have $\mu_n$ be a sequence consisting of 0's and 1's with a particular probability distribution

namely, for which probability distributions would such a series be convergent?

Depends on the distribution, obviously, @Semiclassic :P

I posted my question about irreducible polynomials on MSE by the way if someone knows an answer

Oh, I did think about that on the drive, @Alessandro.

yes, I was getting there :P

@Semiclassical It would probably be closely linked to the analogue of the prime counting function (i.e. the function counting the number of $1$s up to some point)

and how quickly that function grows

For finite fields it's obvious. For infinite fields it's far from obvious, because you get an extension of infinite degree already just using quadratic polynomials.

Well, that'd be an obvious test case.

hi @Tobias

Namely, that if you use the asymptotic for the prime counting function, you should find that the series doesn't converge

@TedShifrin Hi

Sooo... anyone know some domain in physics or computer science or whatever that makes use of literally everything in mathematics?

Apart from string theory

Even string theory doesn't use literally everything, I'm pretty sure.

Number theory is a counter example

No, there's some algebraic geometry/algebraic number theory in there, depending on you phrase it.

I'm using the urbandictionary definition of literally ;)

Really ?

Yeah.

I'm impressed :p

@Astyx in string theory you have lots of representations of groups in certain dimensions, and you actually do numbertheory with these dimensions somehow

Good to know

that's exactly my problem @Ted I don't see why couldn't all the algebraic numbers have minimal polynomials with bounded degree

I'm not sure prime numbers have managed to show up in physics yet, though I know people have tried.

Sometimes I think string theory is physicists way of making mathematicians think their research will have practical use at some point

@Alessandro: Well, because you can continue to take $n$th roots and the $(2^n)$th roots are all linearly independent, for example.

@TedShifrin My progress in the field of analysis has pretty much reached an asymptote. I'm thinking it's time to branch out to other fields

I'm pretty sure I've come to this realization several times already

shrug

I wouldn't call this a representative segment of analysis, @teadawg.

Well, not analysis

Special functions

There's plenty of stuff other than tricky series/integrals, despite what Chris'ssis thinks of mathematics.

It's an incredibly niche segment

Like discrete math, it's never appealed to me, in general.

Some things with special functions are hooked into representation theory and harmonic analysis on Lie groups, etc., so it becomes a bit more central.

(Pun intended.)

Uhm. Is there a formal definition of 'differential element' on math?

@Topologicalife: What do you mean?

I mean, in physics, it is usually used the term 'differential element'.

You mean the way physicists, chemists, and engineers think of an "infinitesimal" amount of something?

A description that is correct in many ways is given by "differential forms"

@Semiclassical according to some article by Marcus du Sautoy, there is a paper that links primes to quantum energy states - albeit through the Riemann-Zeta function, so maybe that's cheating

Yes, I was getting there, @s.harp :) They've been essential in almost all of my research, for example.

I think 'differential element' has as synonymous 'very small'

@Topologicalife: Basically, that's all a way of saying "best linear approximation."

What do you do with your differential elements? Do you sum them up to get an integral for example

There is a subject called non-standard analysis, which uses sophisticated results from logic to make "infinitesimal" and "infinitely large numbers" rigorous. But for most of us, the set-up of multilinear algebra and differential forms is the answer.

For example, they take a differential of force (as an very small thing) and integrate it (sum it) to get the functional form of force

I'm not convinced @Ted, why do they have to be linearly independent?

Yes @s.harp :) that is what I just said.

Basically because of the Eisenstein criterion, @Alessandro.

@Ted I've started studying a bit of differential equations. Are there any other fields of mathematics that would somewhat suit the skill sets that I've developed with integrals?

But @TedShifrin, a differential from, 1-k form doesn't need to be small. It is not necessary.

@Topologicalife: Multivariable integration and integration along surfaces (or higher-dimensional submanifolds) can all be made rigorous without any notion of infinitesimal, @Topologicalife.

That is what confuse me.

It's all about best linear approximations in various dimensions, @Topologicalife, and you need small to get a good approximation.

But differential forms per se are just what you need to make sense of integration on curvy things.

Harmonic analysis ... representation theory ... brings in algebra and differential geometry along with the analysis.

I'm gonna leave, goodbye everybody

@TedShifrin If you have time, a small question. A few days ago you said that the zero ring was, "by many good texts, not considered to be a ring". Do you have a reference, or a reason why? I can't seem to find anything.

@SteamyRoot: Mike Artin's algebra book requires rings to have identity and that $0\ne 1$. Copying him, so does my book :P

@TedShifrin So the full ring is also not an ideal?

Most mathematicians want rings to have identity. Only certain algebraists seem to want them without.

Sure, it's an ideal, just not a proper ideal.

Why did you jump to that?

But we should be able to take quotients by all ideals, not just proper ones

What is the difference between an anonymous function and defining a function with respect to symbolic variables?

Well, put your convention in whichever definition you want.

But the thing is - the zero ring is the unique ring where $1 \neq 0$

the opposite

@SteamyRoot: I think the substantive issue is whether we require rings to have $1$. That's where the debate occurs.

@TedShifrin I mean, that is basically the motivation behind the definition of an ideal

Right @Alessandro

@Tobias: I'm not going to argue/fight over this.

ok

Even then? The trivial ring has $1$, no?

I think it's stoooopid.

Hello @Ted!

heya @Fargle. Save me from this nonsense :)

You may send me analysis proofs, by the way.

Good ones, of course.

Oh crap--I still have to clean that up a bit.

@TedShifrin I'll have to look into those sometime. I should probably get familiar with algebra first

I survived my two days of 450-mile drives.

@teadawg: Yup, and start learning differential geometry, too :P

@TedShifrin I trust nothing bad happened?

The drives went fine, although I was annoyed that I had to sit 40 minutes on my way out of Yosemite while they blocked the road for tree maintenance. That added to my 8-hour drive. ... And I got over 50 MPG on the trip yesterday, so I was very happy :)

Efficiency!

At any rate these proofs are from chapter 1--I'll try to pick out the, erm, cool ones.

I still like Spivak's Calculus more than Rudin, but you will learn some more recondite stuff from Rudin, of course. I think he goes overboard with point-set topology in Chapter 2, but you know all that anyhow.

I do have both, if that's your recommendation.

I rather like that he gives the Dedekind construction of $\Bbb R$ in the first chapter.

Rudin took that out after the first or second edition and made it an appendix. Appropriately so.

@MikeM, @Balarka: Have I lost all my marbles, or is the question posed here not totally wrong? See my comment.

@Fargle: There are some Rudin-style questions later in Spivak's book (some put there by me).

I like the Dedekind's cut construction, it makes for an easy proof that $|\mathcal{P}(\mathbb{N})|=|\mathbb{R}|$

I think I can give a much less recondite proof.

@TedShifrin What makes the questions Rudin-style? Silly question, but it begged to be asked.

@TedShifrin You were reffering to Artin's Algebra, right? Because I have it here and it does consider the zero ring as a ring - it is mentioned on p347

Harder than the typical, @Fargle? :)

That makes sense.

I have only the second edition here, @SteamyRoot. So I don't know what that page refers to. His definition of ring on p. 324 explicitly includes $1$ in the definition. (As did the course I took from him in 1971. :P)

@Fargle: It was probably a silly remark. But I added several hundred problems to Spivak, many requiring some serious estimation skills and rather interesting. One, in particular, I did steal from Rudin because I find it so cool.

Several hundred...jeez.

@TedShifrin My definition also includes $1$, that's not the problem

@TedShifrin It is obviously nonsensical. That would imply the Mobius band has no derivations!

Meh. OK. @SteamyRoot. I didn't remember that from Artin's course. And I honestly don't care ...

@MikeM: I was being more geometric, @MikeM, but yeah. I haven't taken the time to read Faraad's non-proof.

Fair enough - as long as there isn't an important reason why it isn't (or shouldn't be) considered a ring I'm satisfied :)

Sure, ALternatively "take a bump function"

@MikeM: My initial response when I read the question was to add a constant to the function and get nonsense.

But I think my comments are more to the point.

I suspect the student totally scrambled up the lecturer's question. What's more troubling is that an answer was posted and supposedly accepted.

OK, I need to go eat lunch. Bye, all.

hello guys

Im just wondering what would be the best way to prove that following sequence is limited ?

(-1)^n * (1-3n) / 2n

bounded ?

Please put it between $ signs

$(-1)^n * (1-3n) / 2n$

like this?

Yes That's better

Do you mean "bounded" ?

im not sure whether we're talking about, Im sorry im not native english speaker

No issue

ill check whether it is the same

What do you know about $(1-3n)\over 2n$ ?

yes, bounded

'over' 2n, whats the meaning?

Oh you don't have mathjax right

(1-3n)/2n

yes :/

its the sequence that is bounded

its lim is -1,5

Okay so it converges

yes

What do you know about convergent series ?

and $(-1)^n$

is bounded either from below and above

but doesnt convergent

Yes so you have M, and M' two reals such that $|(1-3n)/2n| < M$ and $|(-1)^n| < M'$

Right ?

yes

Therefore |(-1)^n (1-3n)/2n| < MM'

And thus (-1)^n(1-3n)/(2n) is bounded

Right ?