@Chris'ssis You still working on the $A(1,1)$ series?

@robjohn Yes.

@Chris'ssis I haven't looked any further for another way

@robjohn I hope to get something.

@Chris'ssis I hope so, too. I would like to see it.

@robjohn I shouldn't work anymore today. I do too many mistakes. I drop everything for some hours.

@Chris'ssis I need to get some sleep, too. I don't know if I will for a while.

Oh, @robjohn is out.

@Chris'ssis here?

@N3buchadnezzar Yes

@Chris'ssis Tips? Seems simple, but can not quite see the solution in my head.

Well the answer is obvious, but I do not know a rigorous way to prove it.

@N3buchadnezzar Didn't I evaluated it in one of my answers?

You did? I found it in my notes from three years ago, without a solution :p

@N3buchadnezzar you need to use the fact that $\displaystyle I_n=\frac{I_{n-1}+I_{n+1}}{2}$

@Chris'ssis This forces I to be integer?

@N3buchadnezzar No. All integrals are of the form $I_n=n \pi$

@N3buchadnezzar just prove the relation above (easy job) and then you have a recurrence relation. Then, you're done.

You still have to solve the recurrence relation

@N3buchadnezzar Geez. I mean you didn't get my point.

:p

I know I_n = n \pi , you can do that in your head from looking at the equation

$$\displaystyle \int_0^\pi \frac{1 - \cos nx}{1 - \cos x} \,\mathrm{d}x=\frac{\displaystyle \int_0^\pi \frac{1 - \cos (n-1)x}{1 - \cos x} \,\mathrm{d}x+\int_0^\pi \frac{1 - \cos (n+1)x}{1 - \cos x} \,\mathrm{d}x}{2}$$

$$=\int_0^{\pi} \frac{1-\cos nx \cos x}{1-\cos x} \ dx$$

$$=I_n+\int_0^{\pi} \cos nx \ dx=I_n$$

$$I_1=\int_0^{\pi} \ dx =\pi$$

$$I_2-I_1=\int_0^{\pi} \frac{\cos(x)-\cos(2x)}{1-\cos(x)} \ dx =\int_0^{\pi}(2\cos(x)+1) \ dx =\pi$$

@N3buchadnezzar Hence, you're done.

$$I_n=n\pi$$

q.e.d.

Seems to hold =)

@N3buchadnezzar Yeap ;)

I think you are missing a minus sign in the third line

I mean $I_n - \int \cdots = I_n$

It does not matter since the integral is zero but.. =)

@N3buchadnezzar ? $$\int_0^{\pi} \frac{1-\cos nx +\cos nx -\cos nx \cos x}{1-\cos x} \ dx=\int_0^{\pi} \frac{(1-\cos nx) +\cos nx(1 -\cos x)}{1-\cos x} \ dx$$

@N3buchadnezzar there is no missing sign.

hi

someone on the main site mentioned a bound of $2n^2$ for the $n$th prime. Anyone got a proof or an idea?

or any kind of simple upper bound on the $n$th prime? Specifically, I could use one whose $n$th roots tend to $1$

hi, sorry for this, but I have a very simple and stupid question. Vectors are represented with a small arrow over them (e.g. $\vec{v}$. Is there any notation for tuples? something like $\tup{t}$.

Why are guys always doing integrals?

@robjohn I didn't take any break .... lol :-)

@Chris'ssis okay...

@robjohn I gave up the work on that series. I'm working now on some products. It's over my head an elementary solution to it (for the moment I mean).

@robjohn Yeah, that's true. I had very very good days and solved a lot of things, but then I got a bit depressed with it. So, I'll try it again after some time.

@Chris'ssis there are definitely good days, and there are definitely days that are not so good.

@Chris'ssis Usually, you'll come across something that will spark a new approach. Then you will be back at it :-)

@robjohn Exactly. :-)

@robjohn Here is a funny limit. Did you see it? math.stackexchange.com/questions/580945/…

is there a natural way to project a real-matrix $M$ onto the set of matrices with eigenvalues between $0$ and $1$?

if all your eigenvalues are positive, multiply your matrix by a sufficiently small scalar

but that won't project it uniformly, nevermind

actually i want to accompany this by a projection onto trace $k$ matrices

i think this will not be a pseudo projection

er

i think this WILL BE a pseudo projection

i want to do a convex relaxation of the set of projection matrices

which have eigenvalues either 0 or 1

@robjohn Hi , @Chris'ssis Hi

Could anyone tell me how I can indent text for equations without \tag? That is, without the brackets and also without causing the equations to automatically centre?

I got a question on this (which is I guess not worth a post): en.wikipedia.org/wiki/Convolution#Convolution_of_measures here it says the the covolution is again a finite measure . Someone knows why ?

Check that the definition is satisfied.

@Chris'ssis You know, on that limit, I am not so inclined to use anything but Stirling since it seems to be tailored for Stirling.

@Complexanalysis Hey there

@robjohn Agree 100%.

@Complexanalysis Hello :-)

@robjohn @Chris'ssis wats up ?

@Complexanalysis A bit tired. I go to bed for a few minutes, I cannot think anymore.

Back a bit later.

@Chris'ssis Ok

@anon

Morning

HAI

I am wondering what can we say about a group, whose order is divisible by a prime $p$ and whose subgroups have all indices divisible by $p$.

indices?

@robjohn That day, they came to deliver my books but I was not in, and they left a note for me to go to the post office to collect them, so that's how it works at least in my case.

Yes, $[G:H]$s, @Mike.

We can say $G$ has nontrivial center, for example.

From the class equation.

Guess that is the important thing.

Oh, gotcha. Was thinking indices like $G_p$ lol

What's that?

$G_p$?

@pedro You seem to be studying many topics at the same time, lol.

The subscript of something is called the index

So I thought you just meant the subscript was a prime number. Hence much confuse

@JasperLoy How many did you count?

@PedroTamaroff I didn't count, just relying on my intuition.

Intuition...?

Or my subconscious mind.

Meh.

@PedroTamaroff If the group is of the order $p^n$ then it has a non trivial centre . But for a group with arbitrary order ..

@Complexanalysis Yes, sure.

@PedroTamaroff Your statement doesn't directly follow from class equation .

It does.

$|G|=|Z|+\sum |G:C(x_i)|$, $p\mid |G|$ and $p\mid |G:C(x_i)|$ for each $i$, $x_i$ noncentral representative.

@PedroTamaroff So we have something like $n = |Z(G)| +\sum m_i p$

Whence $p\mid |Z|$, so $Z\neq 1$.

Yep.

@PedroTamaroff ah ok , we already know that $Z(G)$ has at least one element namely $e$

@PedroTamaroff Hi Pedro. I was talking to you a couple of days ago regarding the Thomae's function problem link: math.stackexchange.com/questions/669359/…

OK?

@Complexanalysis Yes, but that doesn't prevent $Z$ from being trivial!

@PedroTamaroff I have not been able to successfully establish exact points where function would be continuous and discontinuous....can you help me with this. I am just transitioning from 1 variable analysis to two variable one and am unable to come up with the proof...should I start of with assuming an arbitrary y and verify continuity for different x

@PedroTamaroff Sorry, I really tried after your help but just couldn't do it. This is the first time i'm dealing with multi variate analysis

@PedroTamaroff why?since $p| n$ and $p|\sum m_i p$ it ought to divide $|Z(G)|$

@Complexanalysis Yes, sure, so $Z$ has more $pk$ elements, so $>p$ elements, but saying $1\in Z$ doesn't mean $Z$ is not trivial.

@PedroTamaroff surely

@JasperLoy So everything was good.

@Pedro "Show that a normal subgroup of $H$ is normal in $G$ when H is a direct summand of $G$."

hi

I need help getting started on this. If im given a power series $\sum_{i=1}^{\infty}a_kx^k$ with radius of convergence $R>0$ then what is the radius of convergence of $\sum _{i=1}^{\infty} a_k*x^{2k}$?

@PedroTamaroff the group can't be finite (consider the index of a p-sylow). I think the indices of its p-sylows (even infinite groups necessarily have them) would also need to be infinite. It's an interesting question.

@anon what question as you referring to ?

click the gray arrow at the beginning of my message

(gray/grey w/e)

@anon got it .

one must decide if one wants to consider infinite cardinals divisible by p, or if one wants to impose a more structural generalization, like $\forall H<G~\exists x\in G~x^p\in H$

hmm, any set partition into p-power-size cells admits a "p-adic" cardinality. wonder if that could be applied to G-set theory for p-groups G.

Got you thinking? The group should be finite else Lang is speaking bull!

@anon

Good afternoon @Pedro

@PedroTamaroff lang generally is

@PedroTamaroff The order would be a power of $p$.

@DanielFischer Good evening

@Complexanalysis Bon soir, Buena sera, good evening, n'Abend, make your pick.

@DanielFischer I would have preferred Konbanwa :D

thats japanese

@Complexanalysis Sorry, Japanese has gone to get some cigarettes.

@DanielFischer haha

I doubt it will ever come back, now that it's gone for so long.

The only Japanese I remember is "Kiyoshi Oka".

I know just few word @DanielFischer

Ay~

How can I should that if C is countable, then R\C is countable?

JESUS this site is busy

Hi

How can I find all integer $x\neq 3$ such that $x-3|x^3-3$

@AlexanderGruber algebra is hard

Hi @TedShifrin

hi @Mike ... We're about to get snow and an inch of ice here. May be stuck in a cold house without power for days ... Of course, school was canceled, too.

Without power?

But how will you charge your iPad? :)

I won't do nothing!

And can't drive, either ...

Oh boy.

Hope you've got a fireplace...

I do, but not much wood.

Hey, go stock up before the ice lands.

shrug I can wear sweaters and wrap up in blankets.

And California only has to worry about a major drought -- we had one of those in the late 70s when I was in grad school.

We don't worry about those. We just live with them :D

Well, I remember worrying some in my days there.

You weren't from a desert, which is the kicker here.

I prefer dessert to desert ...

Don't we all?

Some people love it on the desert ... and the camels sure do.

Eh, those people aren't normal.

you don't think humps are normal?

If I say yes, will you call me bigoted against hunchbacks?

Unrelatedly: I got an unexpected email. Mind if I ask (elsewhere) for advice?

No comment on the first. No problem on the latter.

@Ted We're hunkered down here. Groceried up. Hopefully we won't lose power here in the big city, otherwise we're snuggling for warmth.

Yeah, this is where my remote location (with 2 acres of trees) is not so pleasant, @Kevin.

Oh well, I have plenty of gin.

I approve @TedShifrin

@TedShifrin Not enough alcohol, that doesn't burn well.

I'm not burning it, silly @Daniel.

I'm getting drunk :P

@TedShifrin Well, you are when you work out the next day.

@Mike: You don't know me well enough. "Work out" is not in my vocabulary ... other than walks and tennis.

@TedShifrin Then it's okay. Although I'm more of a Whisky man.

I have scotch for you, then, @Daniel.

Well, the gin's not too bad for you.

At least you're not getting drunk on beer.

@TedShifrin It'll be very old until I get to Georgia.

I have gout, @Mike, so I rarely touch beer, and only very expensive beer :P

Ouch. My dad has that, too. It's pretty rough for him.

Runs in the family... gulp

I developed it after surgery #1, @Mike.

It hasn't affected him much lately, which is good.

Well, I take a pill every day. It works.

But I remember times when I was in high school and he would just be unable to move for basically a day, whimpering. :(

Well, I could see that dealing with you day in, day out might make one whimper :D

Yeah, that's what he told me too.

See, we're not as dumb as we look :P

Hi Ted Hi Mike

Hi Paul

Ted, how did you settle on your branch of mathematics?

Well, his colleagues had already taken all the good ones.

;)

@Mike, how are your applications going? "Well, his colleagues.." Does that refer to me?

hi @Paul

I think choosing a specialisation is often extremely difficult because it's normal to find oneself equally good at all the subfields one has encountered.

I think I realized when I took a differential topology class from Guillemin how much I liked geometric stuff. And then I liked differential geometry when I took it my first year in grad school and I just asked Chern if he'd be my adviser (he was on my algebra qual committee, ironically).

@PaulEpstein I was just joking that his branch of math is "bad". Not at all seriously. And my applications are all in, it's just the waiting game now.

@Ted I agree that differential topology sounds extremely interesting. My problem was that I was impeded by the foundations. There was too much stuff I had to just accept, without being given time to check out the proofs of the underlying results.

@Ted I might give it another go.

well, @Paul, I love multivariable analysis :P

First, Spivak's calculus of manifolds, then perhaps something else, then diff top.

As I've said before, I'm only 47, lots of time left.

Indeed.

@Ted, lots of time left for you too -- plenty of people doing good work in their eighties.

Intersection theory and degree theory lead to all sorts of beautiful results ... used in many parts of mathematics. And it ties in beautifully with complex analysis and several complex variables, too. I was always excited by the interweaving of different fields in math ... Still am.

No, @Paul. NO more work for me.

@Ted, you've given up research for teaching?

I haven't done research (other than talking with other people) in over 10 years.

@Ted, there are some mathematicians who publish very little but are renowned because their conversations inspire many. Some names come to mind but it's risky to give them here where they can be googled.

LOL ... I know a few of those, @Paul.

I suppose it's something of a credit to the community that people can get recognition that way -- others are honest enough to say they got good ideas from someone else even if these occurred in an informal conversation which no one recorded.

Normally, I'm very critical of the academic maths community, but perhaps unfairly so.

And my perceptions must be out of date because I left academic maths around 1994.

@Mike If you get accepted by your first choice but rejected by your second choice, will you be disappointed by the rejection even though you preferred the first choice so didn't miss out on anything?

Geez, @Paul, don't make him more analytically nervous.

@Ted, he's got nothing to be nervous about.

That doesn't stop people.

@PaulEpstein That's a secret.

@Mike, not a particularly well-kept one.

No, I mean the answer to your question.

gets another martini

How can I prove $\sum_{i=1}^n \frac{1}{i^3}<\frac{3}{2}$ for all n via induction?

@user4140 Well, first you would prove a base case and then you would prove that if it's true for $n-1$, then it's true for $n$.

I think @user4140 knows that much, @Mike. It seems non-obvious to me ... nor do I have any idea why one would choose to do it that way.

oh, of course, how silly of me

thanks, on to the next question.

But we can use the integral test to give a bound on the partial sums, and prove that bound by induction, perhaps?

Glad to be of help.

lol :p

I'm kidding, I allready tries doing that.

Anyway, you wouldn't prove it directly. I would suggest proving it's bounded above by some formula that depends on $n$, and then show that this nicer formula is bounded above by $3/2$.

I have to do it by induction

Thanks for rephrasing what I suggested, @Mike.

@TedShifrin I try my best.,

that doesn't sound like induction.

You would prove it's bounded above by this formula by induction.

In fact, by the integral test, all the partial sums are less than $1/2$, so this is too crude for that.

Wait. I'm full of it. I do get $3/2$.

Yeah, that's the right tool here. So here's an inductive proof.

$1 < \frac 32$

So maybe this isn't so silly. The $n$th partial sum is bounded by $\frac32-\frac12(n-1)^{-2}$. Can one prove such a thing by induction?

I was going to write something silly but it seems like too much work now.

Being silly is often too much work.

I've decided I won't be funny ever again.

Not to say I was before.

Yeah. What I said is a valid approach. Prove the statement $a_n < \frac32-\frac1{2(n-1)^2}$ by induction.

Done giving it away.

@Mike, I want that sworn testimony notarized.

@TedShifrin Not a chance.

GAH! Question move so fast here. I feel like if I don't get an answer before I'm off the fornt page (like 15 minutes) I'll never get one

If it makes you feel better, @Kevin, I've pretty much quit answering 'em.

@Ted Sadly not

Well, @Kevin, I do what I can.

@KevinDriscoll Do like me and Jasper, answer the easy questions you see.

I answered a silly 10th grade geometry question today.

I asked a question today I knew the answer to because I thought it would get me easy points. :P

Stop that, @Mike.

It wasn't a bad question @Ted!

@user4140: Have you worked on what I suggested?

@TedShifrin If $G = H \times K$ for normal subgroups $H$ and $K$, and $N$ is a normal subgroup of $H$, do you have any idea how I'd prove that $H$ is a normal subgroup of $G$?

So, @Mike, I spent 10 hours grading papers Friday and Saturday, only 2 came to pick them up on Monday, and of course class is canceled all week.

It's easy to reduce this to $kNk^{-1} = N$ for $k \in K$, but not obvious what to do with that. I've yet to use that $H \cap K = \{e\}$.

@user4140 It's pi ^ 2 / 6 by the way

@TedShifrin Fun!

@PaulEpstein No it's not. His question is $\zeta(3)$

@TedShifrin Hmm, I don't know why I asked. I can get free points for that question.

You mean $N\times \{e\}$ is a normal subgroup of $G$?

No, $N$ is a subgroup of $G$. (This is an internal direct product.)

Reread your question.

But I mean, sure, same thing.

Oh ... OK

You typed $H$, not $N$.

Oh... oops.

Of course $H$ is a normal subgroup by def'n.

Yeah, so $(h,k)(n,e)(h,k)^{-1} = (hnh^{-1},e)$. Done.

Yeah I'm not seeing it.

Ugh.

Goddamnit.

withdraws compliments on @Mike

I see it now. :P

@TedShifrin Rightmost term should be $e$ in the end

Thanks. Fixed.

But anyway, I was translating to the language $G=HK$, and trying to show that $khNh^{-1}k^{-1} = N$... and so $kNk^{-1} = N$

Living in the direct product makes it easier :P

Yeah, but equivalent argument.

Remember, to have the internal direct product, $H$ and $K$ still have to commute.

That's trivial and completely obvious and something I hadn't realized.

Well, see, I know something.

:D

Well, that question was too easy. I need to think up another one.

I thought you were done with grading.

This is for myself.

@Mike I don't want to answer the easy questions. That just seem to incentivize more and more people to post questions like "How do I do Integral 31335?"

The students in the algebra class I grade for are busy trying to understand what is and isn't a group.

I know @Kevin, but sometimes you can tell there's a different attitude in the questioner.

@KevinDriscoll This site is going that way and it's not coming back. I gave up.

I post one-line hints and then leave and hope for some free points.

Oh yeah I'm not trying to say all easy quesitons are equivalent

Well, @Mike, I'm going to take a lot longer to get to 20K than it took to get to (almost) 15K.

But there are still occasional interesting, more substantial/advanced questions that pique my interest.

@TedShifrin You should start posting really bizarre integrals.

I just feel like I can't get answers because the people interested in approximation stuff and applied ideas here are swamped by all the other questions

Mine will just never get seen

That's not true, @Kevin. But I know nothing about approximation theory.

How do I solve $$\int_0^{\pi\sqrt{3}}\frac{\sin^{-1}(x)}{\ln(\sin(x)}dx$$

It may not be true

@mike dollar signs

But I do know that my questions disappear into the later pages VERY quickly

I don't know how anyone would find them

You'll get like 50 upvotes.

Label with analysis?

Just keep making up integrands and posting! Maybe change the bounds!

That question is ridiculous, @Mike. As @Pedro would say, STAHP.

It's like free points.

Maybe I should propose some more specific tags related to approximate methods

Numerical analysis for sure, @Kevin.

There isn't even an "Asymptotic expansions" tags

Special functions do show up in harmonic analysis and applications, @Mike. Don't be such a snoot.

I have to say that an integral involving implicitly defined functions really got me hooked 6 months ago or so, @Mike. I finally got it.

@TedShifrin Noooo

Let me find it.

I didn't see what @Mike said, but I deal with trying to compute odd integrals of Gamma functions and Bessel functions regularly

Though I don't post them here

@KevinDriscoll I was being needlessly hostile towards questions where people are looking for an exact form for an integral.

Ah okay

I actually never considered posting such questions here

but perhaps I should, given the interest

not so easy to find even questions I've answered ... grr

are they not on your answered tab?

Yes, @Kevin, but that goes on 14 pages or so. @Mike: Here it is.

@Ted AH RIGHT! Over 15000

Not quite, @Kevin.

Rep I mean

I know.

OH woppos, ya almost

I would upvote your effort but..... I try nto to upvote things I havent read and understood

Stick to your principles. But that integral one I just linked is a truly interesting problem.

@TedShifrin I'm about to send another thing.

OK, I need to eat dinner :P

And I need to go to art history.

Indeed.

My sister would scold you if you didn't.

@TedShifrin I got 32 points from my question. Hooray!

Hey ted

I liked this impicit integral ^^