@MathyPerson You should aim at finding a relation of $d_n$ in terms of $d_k,k<n$.

@Behaviour I liked how he evaded answering quid, because he couldn't just say that quid had no clue of mathematics.

@PedroTamaroff ok. thanks.

@Behaviour Thanks.

Or is it that $D_n$ has $\sigma^n$ instead.. nope, they share $\alpha$

@DanielFischer Given that you haven't publicly declined to nominate, may I assume that you have not yet decided not to nominate yourself?

@PedroTamaroff But is the way I did it wrong?

@MikeMiller Pondering.

@DanielFischer There is a theorem that a powerseries $\sum a_kx^k$ is a rational function iff the $a_k$ satisfy a linear relation $a_{n+k}=b_{k-1}a_{n+k-1}+\cdots b_0 a_n$

@MaryStar Hmm... they don't exclude that $E$ could be finite, do they? Do they want a continuous measure?

No offense to Pedro, but I think your name would be the best on the list of nominees thusfar, if you decide to add it.

@PedroTamaroff Looks credible.

@MaryStar Okay... they said Lebesgue measurable, so I assume $m$ or $\mu$ is Lebesgue measure.

@MikeMiller I don't know where to take offense from that

@DanielFischer The proof is tortuous.

@PedroTamaroff One direction is fairly straightforward, I think.

@DanielFischer Sure.

@PedroTamaroff I just called Daniel a better moderator candidate than you. To some, those could be fighting words.

@MikeMiller Oh. I don't deny that.

You know how it goes... take the list of candidates. Order by reputation. The top three will be elected.

@robjohn So, what am I supposed to do??

@Behaviour Why would one like to be a moderator for free? Just asking.

@Behaviour Looking at the nominations so far, that wouldn't be the worst method.

@MaryStar If they don't tell you something about $E$, then it may be hard to define a function that is in one space but not another...

@MaryStar In most generality, we'd need to have a decreasing sequence of sets whose measures are non-zero but whose measures tend to zero...

decreasing meaning each set is contained in the last.

@Behaviour I guess some would do it for the need of power over the others, for some kind of fame, but my needs couldn't ever be such needs. If I go out now and ask someone to help me with something, after the job that someone will expect some payment from me. It's interesting how people behave in general.

@Behaviour Note that amWhy and Andre Nicolas (highrep users) were quite downvoted in the nominations.

On the other hand, amWhy doesn't have a good reputation.

@PedroTamaroff That's meta voting, it reflects the opinions of meta population. Election results are determined by the main site population, most of which is barely aware of meta or of any issues on the site at all.

Most of the nominees so far are examples of this.

@Behaviour Agreed.

Asking questions and answering questions is enough to pay back to the community in my opinion. Well, it's fair to help the others since you are also helped more or less. Yeah, there is a lot of stuff one can learn from. If there is room to improve that amount stuff, it's OK.

@MaryStar Then build a function defined on the "annuli" that is the difference of one set from the last

it is not hard to construct a function then

Yu!

@robjohn Ok... I will think about it...

the admiral's arrived

@Alexander I really enjoyed that paper you linked few weeks ago :) I didn't understand everything completely, but it was real fun

Now I go to sleep with a question in mind

@Studentmath Oh yeah? :) The graphs and groups one?

Am I the first one that evaluated $$ \int_0^1 \left(\frac{\operatorname{li}(x)}{x}\right)^4 \ dx$$?

@Chris'ssis Yes.

Indeed, generalisation of the SLLN

@PedroTamaroff If you plan to become a moderator, I wanna know to change my e-mail address. I don't trust you.

I'm out.

Ouch.

@Chris'ssis I would never do anything with that.

I'm a man of honor.

You two will end up marrying. I am back to my Algebra

Hrm, I am trying to show $D_{2n}$ is isomorphic to $D_n \times Z_2$. So, they are both normal subgroups of $D_{2n}$, and I know their orders fit - so according to something I proved before, all that's left to do is to show their intersection includes only $1$, or that $D_{2n}=D_nZ_2$. But I don't think either is true.

@Studentmath Were you assigned that task?

Yeah

Exercises in the book

@Studentmath It has some pretty cool ideas in it, makes me want to make some sort of computer program to test things out.

I really loved the application on cryptography @Alexander

A nail in braid-based crypto's coffin :)

@Pedro though it doesn't necesserly mean it's not true, if neither holds. Could be I am looking at the wrong characterisations.

But then again, I think $D_n \cap Z_2$ must be $1$, I just don't see why.

@Studentmath Can you find a normal subgroup of $D_{2n}$ of order $2$?

$Z_2$, or well $<\alpha>$, the cyclic group generated by reflection

@Studentmath Proof?

Well, if I look at $\sigma^{-i} \alpha \sigma^{i}$ it's the same as $\alpha \sigma^{2i}$. Oh wait I am stupid.

Hi @Studentmath @Pedro @Alex ... Someone wanna help me grade homeworks?

@TedShifrin OK.

@Studentmath =O

But shouldn't there be one? The Sylow-2 subgroup... not necesserily normal.

@TedShifrin No.

@Ted I am afraid of commitment

@Studentmath A Sylow subgroup doesn't necessarily have order $2$.

I specifically omitted you, @Mike. Shaddup!

@Ted But then who will grade my students' homeworks?

@Pedro well $n$ is odd, so it must have order 2 - no?

Finals, then, @Alex ;)

@Studentmath You never said $n$ was odd!!!

It's not true otherwise!

If $n$ is odd, then any group of order $2$ has trivial intersection with one of order $n$.

This is so simple!!!!

But wait, why is it normal though?

I lost my confidence in its normality by now

I didn't say it was. I asked you.

Leaves y'all to dihedrate by yourselves ...

@TedShifrin What course is this?

Probability?

screeches

No, mult ... lagrange multipliers plus linear algebra proofs with projections and inner products

@Ted I was just about to suggest that if I could be of use, e-mail me, but I really doubt I could be of use in such a case :P

@TedShifrin But I don't know Lagrange Multipliers!

You can help with prob finals next week, @Studentmath :)

I've told you a dozen times to learn em, @Pedro.

Will be glad, besides few questions I encounter over here I rarely practice it enough

You've had the discussion about Lagrance Multipliers at least 20 times

Ugh... I'm stuck on the integral that Venus posted earlier... $$\int_0^{\infty}\frac{x}{\sqrt{e^x-1}}\mathbb{d}x =4\int_0^{\frac{\pi}{2}}\log(\sec\theta)\mathbb{d}\theta$$

Yeesh, @Pedro, everyone with a math degree from my undergrad knows those. Lots of people without a math degree too.

I'll send you my final so you can write the answer key for me, @Studentmath :D

@Teadawg: You have some theory to learn for my exams!

@Pedro isn't any subgroup which is single of its order is Normal?

@MikeMiller Yes. I'm a failure.

@Studentmath Not necessarily.

Ah but the sylow-2 subgroup isn't necessarily only $<\alpha>$

Yes @Studentmath

@Studentmath Necessarily

Agreed @Pedro

@teadawg1337 That's equal to $A$, the Ary-Lukewicz-Shonda constant.

$$4\int_0^{\frac{\pi}{2}}\log(\sec\theta)\mathbb{d}\theta=2\int_0^{\frac{\pi}{2}‌​}\operatorname{Li}_1 (\sec^2\theta)\mathbb{d}\theta$$

Ok, I have too much work. Bubye.

Bye @Ted

G'night @Ted

I like not having impending deadlines... I can get some work done now.

$$=\int_0^1\frac{\operatorname{Li}_1(w)\sqrt{\operatorname{Li}_0(w)}}{w}\mathbb{‌​d}w$$

@Pedro Never heard of it....

Actually, I don't think $<\alpha>$ is Normal.

Maybe I should just show a specific isomorphism and be done with it.

Haha, got it! Didn't look up the Ary-Lukewicz-Shonda constant or w/e

@teadawg1337 there is a nice expansion of $\log(\cos(\theta))$ that might give you that integral: $$\int_0^{\pi/2}\log(\sec(\theta))\,\mathrm{d}\theta=\frac\pi2\log(2)$$ as you already know :-)

Oh! $<\sigma^n>$ is normal!

$$\int_0^1\frac{\operatorname{Li}_1(w) \sqrt{\operatorname{Li}_0(w)}}{w}\mathbb{d}w=\pi\log(4)$$

And of order 2

what does it mean when a paper says "Communicated by (someone's name)" below the title?

was the author incapacitated or something?

@Pedro is that the A-L-S constant?

do Stephen Hawking's papers have this I wonder?

@TomCruise here you go buddy

oh ok

so the author is alive and well :)

possibly anyway

This is more common in other fields, where the communicating author is the one to which communication about the article should go; often this is an older faculty member who's not moving any time soon (so that the contact information would stay accurate). This is somewhat of a holdover from when it was harder to find someone's contact information than it is now.

@robjohn Did you give the question a shot?

@robjohn If not, no problem, just get something new, it seems I get nowhere.

When there is no idea in your head, it's good to ask other people :)

I am off for the night, good night all!

@Studentmath Good night!

@FreeMind I could not find a way without using contour integration. Are you sure one exists?

@robjohn There is no way as far as I know, I want to approximate it not finding exact solution!

@robjohn You're free to put in Fourier or Taylor series.

I hope I dont get Lou Gehrig's disease

@TomCruise What was the disease?

@TomCruise In addition, tell me what happened that night at Downtown :)

LOL

haha

nothing interesting actually

that is the disease Stephen Hawking has

apparently you can also get it later in life which I didn't know

@TomCruise Well, now, tell me about the night at downtown :D

@TomCruise So nobody gave shit about your chiseled body LoL!

haha, I wish there were something to report

saw a local band...

somehow they resisted.

@TomCruise Bars and clubs suck!

@TomCruise Sometimes they're pretty boring!

yeah I rarely go anymore

maybe once a month

Ugh... This question can be easily explained using polylogarithms, specifically $\displaystyle \operatorname{Li}_{-1}(z)$, but I feel that this is beyond the poster's expertise...

I don't see any way in which one can gain more understanding by thinking about that in terms of polylogarithms instead of Taylor series.

It makes sense to me using polylogarithms, but you're right about Taylor series being simpler

I should get off for the day, I'll be staying up all night if I do anything else involving calculus

@MathyPerson Hint Suppose you have a number $n>6$, and take a composition $k_1+\cdots+k_j=n$ with largest part at most $6$ (this is what you're counting!). Then $k_j=k$ is a number in $1,\ldots,6$, so you obtain a composition of $n-k$ with largest part at most $6$. You need to show this establishes a bijection when $k$ ranges through $k=1,\ldots,6$.

Alternatively, consider the series $1+\alpha+\alpha^2+\cdots$ with $\alpha=x+x^2+\cdots+x^6$.

The coefficient of $x^n$ is the number of ways we can write $x$ as a sum of $1,2,3,4,5,6$. Why?

This means the polynomial is $1-x-x^2-x^3-x^4-x^5-x^6$.

@KajHansen You missed a nice combinatorial question.

When @Pedro

?

@KajHansen Count the number of compositions of $n$ with largest part $6$.

You do this more generally, though.

Well, not that.

Ah, so what you are talking about above?

Just get the generating function.

You can get a nice recurrence relation though.

Hi everybody! I'd be grateful if you could help me with this: math.stackexchange.com/questions/1058066/…

@Larara I don't even.

Just when I am feeling confident I come across this lemma that takes almost a entire page to state. I call this an induction from hell.

:(

where do I even begin

When making a math poster, I don't understand what material to put in the 'Methods' field rather than the 'Discussion' field. To me, they are the same

@TomCruise Sounds ugly.

@PedroTamaroff yeah there are often these nasty lemmas needed to get to bigger and better things

but this one is trying my patience

Take it easy.

Looks nasty, but try to understand the big picture, don't lose sleep on the nifty details. They are important, of course, but everything in its own time.

luckily I know the professor who came up with it, so maybe he can explain it to me quickly

it leads to an interesting result though...

spaces in which the cofinite subsets are precisely the nontrivial connected subsets.

in other words, even though the spaces are big and connected, every connected subset is essentially a singleton or the whole space.

very beautiful

@PedroTamaroff what about the papers you review? Do you have to check every last detail?

Can someone please explain series to me?

you know what a sequence is?

n+1 = {1,2,3,4...}?

I guess not.

think of it as a list of real numbers

$\{a_n\}$ where for each $n$ $a_n$ is a real number

anyway, you start with a sequence

from that you get a sequence of partial sums

I'll give you a specific problem and perhaps you'd be able to help me solve it:

ok

theta (infinity)(k=1) (1+3^k/2^k)

$\Sigma _{k=1} ^\infty (1+3^k)/{2^k}$

Yup

Could you help me step by step on how to solve that?

do you use chatjax?

Yup, I have it, but I don't know the commands, hence I wrote it out like that.

ok

well, $(1+3^k)/2^k \geq 3^k/2^k=(3/2)^k$

now what do you know about $\Sigma _{k=1} ^\infty (3/2)^k$ ?

where did the 3^k/2^k come from?

I am just comparing your terms to (3/2)^k

do you believe that comparison?

I meant in your initial equation:

do you believe that?

I don't know, that's why I'm here asking.

I don't understand how it works.

can you tell me anything about $\Sigma _{k=1} ^\infty (3/2)^k$?

have you studied geometric series?

Yes, but I didn't understand anything and the teacher was as unhelpful as one could be.

ok, well a series of the form $\Sigma_{k=?} ^\infty r^k$, where $r$ is a real number, is called a geometric series

if $|r|<1$ then this series converges. Otherwise it diverges.

so what can you say about $\Sigma (3/2)^k$?

it diverges

ok

and it diverges to $\infty$ in fact

Now use my comparison to tell me about your series

what's confusing me is that it's (1+3^k)/(2^k)

the k is connected to it and it's all in a paranthesis.

to see that $\Sigma (3/2)^k$ diverges to $\infty$, just note that for each $k$ we have $(3/2)^k$>1... so you're adding up infinitely many things >1...

The 1 is what's confusing me.

Since $\Sigma (3/2)^k$ diverges to $\infty$ and $(3/2)^k \leq (1+3^k)/2^k$, we know $\Sigma (1+3^k)/2^k$ diverges also.

So you can basically ignore the 1?

if you add 1 on top the number increases

yes pretty much

just look at a simpler series

you should learn something called the Comparison Test

the ratio test, the comparison test, the integral test.

and nothing on the internet can helpfully explain it.

Let's look at this one for example:

[\sum_{k=1}^{\infty}3*\frac{ 4^k }{ 7^k }]

you want to see if it converges or diverges?

What about $\Sigma 4^k/7^k$?

that diverges cause it's greater than 0

no

hint: it is a geometric series.

oh wait no, it's less than 1, so it converges

ok

but if k = 1, then it would diverge, cause you would add 1 to that equation (4/7)

correct?

a convergent series multiplied by a number (3 in this case) also converges

it doesnt matter where k starts

when does it matter?

if you want to find the SUM of the series it matters

but if you just want to see converge/diverge it doesnt matter

Ok, one more, if you don't mind:

This one's a little tougher

ok

sum_{k=1}^{infty}5*(\frac{ 2 }{ 3 })^{k}-\frac{2^{k-1}}{7^k}

Why don't these equations show up?!

put dollar signs around it

$\sum_{k=1}^{\infty}5*(\frac{ 2 }{ 3 })^{k}-\frac{2^{k-1}}{7^k}$

thanks

$\sum$

use "\" in front of sum and infty

there we go! thanks again ;)

try to break it into simpler series

@TomCruise Where did you get I review papers?

just a guess

if you are a phd student

or professor

2/3 so that converges regardless of the 5*, right?

same with 2/7

@TomCruise I'm neither.

@DemCodeLines yes

but wouldn't 5 multiple with 2, making it 10/3 which would be greater than 1 and as a result, diverge?

$5\cdot (2^k/7^k)\neq (5\cdot 2)^k/7^k$

ok, so it's always only the variables with the k that matter, nothing else?

constant multiples don't have an effect on convergence of series

ok what if the second one diverged?

or what if instead of 2^k/7^k, it was 10^k?

it would diverge

why?

i mean constant multiples outside of everything else have no effect

like the 5

if they're inside (5*2^k), then that would just become (10^k), which would then diverge in that equation, right?

yes

because 10/7>1

ok, but you said it would diverge if one part of that equation converged and another part diverged. Why would the entire equation diverge if there is a part that converges?

here you can use the comparison test to see that

if you don't mind, could you show me how? As I said, internet doesn't help

you want to show $\sum_{k=1}^{\infty}5*(\frac{ 2 }{ 3 })^{k}-\frac{10^{k-1}}{7^k}$ diverges?

You said it would diverge instead of converge, so I wanted to know how.

I guess not.

What do pi and xi stand for in Σpixim, used in the knappsack problem solution?

I need one more anti-anti-integral mod candidate.

@robjohn @Integrator ^^^^^ shouldn't someone teach him ???

@robjohn , I was just trying to help him to get an answer because he was getting downvotes :o and I was very polite :o I don't understand this people -.-

^^^ the comment section....

@TheArtist I've already voted to close the question as off-topic, Just wait, I'm reading comment section!

hi

math.stackexchange.com/questions/1059379/… has zero votes :( Is it unclear/uninteresting?

@TheArtist Relax Man!. I've edited the question, Do we need a moderator for everything? :)

@user2179021 I'm not good at probability, so can't tell anything!!

@Integrator shame.. so who here is ?

@Integrator but he's been selfish :o I'm not saying moderators should do something...just someone needs to teach him about this site, he has so many misconceptions about how the site works. And I tried to explain to him , he doesn't understand me, so someone needs to explain him.

@Integrator it's no longer about someone else editing it for him , it's that he can do it but he doesn't, and talks about how it's simple to him so he doesn't care about others...

why probability-theory for my question?

@user2179021 ^^^ I edited your question to help you out, because you said it hasn't got enough attention....now it's on the top question list :)

@user2179021 I hope you get answers :)

@TheArtist thanks!

maybe smoeone could upvote it too :)

@user2179021 there you go!

thanks :)

I feel rich now

Hi @mrf what's up?

not much, just checking a thing

@mrf checking what?

@TheArtist What's up?

Hello @robjohn !! Do you mean that we should use the following theorem: "If $A$ is measurable then $\forall \epsilon>0 \exists F \subset A$ closed with $m(A \setminus F)<\epsilon$." ??

@Integrator Nm bro ....wbu

hi

@Integrator wow he has commented that he is going to keep on posting images in the future

@TheArtist You've right to vote, and right to flag! Use it wisely!

@TheArtist Please excuse my English!

Greetings

hi @Chris'ssis

@user2179021 hi

I won't ask you my usual question as I now know you don't do prob or combinatorics :)

@user2179021 You might like to know I developed these days a cominatorics problem by Ramanujan (don't aks for details). :-)

@Chris'ssis oh! Well then maybe I will tell my problems :)

@user2179021 Now I'm pretty involved in some integrals involving the logarithmic integrals.

Actually I study one that seems pretty interesting!

Is this question of interest? math.stackexchange.com/questions/1052876/…

it's basically an integral in disguise :)

@user2179021 Which integral?

@Chris'ssis the log of the product is a sum and in the limit the sum looks like an integral

@user2179021 If I have a fruitful idea I'll let you know.

thanks

@Chris'ssis Presumably related to partition function?

That's the most famous combinatorial work of Ramanujan I know of.

What can we say about $$\int_0^1 \operatorname{li}(x) \operatorname{li}(1-x) \ dx$$?

@BalarkaSen I'll let you know sometime later. I also plan to publish an article based upon that.

@Chris'ssis are you working in a uni?

Do let me know, @Chris'ssis. I'm interested in anything except integrals :P

@user2179021 No, as I already told you, I have no background in mathematics, I couldn't work in an uni.

@BalarkaSen Are you an integral haters? It's bad you miss so much beauty there.

@Chris'ssis why not?

@user2179021 I have no math degree.

@Chris'ssis oh I see. I am not sure that is a strict requirement for a uni job

@Chris'ssis I mean publications trump all

@Chris'ssis I am an analysis hater, in general. I know that there's a lot of stuff in there but each to his own ;)

@user2179021 Well, with publications there is no concern about having a math background, but for getting a uni job one needs to have some math background I think. :-)

@Chris'ssis I don't agree. If you have good publications already they will take you seriously whether or not you have a math degree

@DanielFischer @robjohn How would you tackle this one? $$\int_0^1 \operatorname{li}(x) \operatorname{li}(1-x) \ dx$$

@Chris'ssis if you have no publications and no math degree then you have no chance of course

@user2179021 I already have some results that look like those by Ramanujan, seriously speaking (I don't care too much what the others think of them, I consider them totally amazing)

@Chris'ssis did you publish?

@user2179021 Yeap. I'm waiting for some articles to be out (published).

@Chris'ssis You are using $\mathrm{li}(x)=\int_0^x\frac{\mathrm{d}x}{\log(x)}$?

@robjohn yeap.

@user2179021 Sincere people really appreciate my work, but that doesn't mean that I only want to be appreciated, since the criticism is a heathy part of anyone's evolution. Actually, the positive criticism is mandatory! The one that allows you to become amazing on your way!

@Chris'ssis Here's one. $$\lim_{a\to\infty}a^4\int_0^\infty\frac{\cos(ax)}{1+x^3}\mathrm{d}x$$

@robjohn It's a different way of writing my integral?

@Chris'ssis No, just something to think about while I do your integral :-)