6:00 PM
@UserX , place it on top of another. And no

@UserX Order by numerator/denominator.

@TheArtist This is getting annoying.

3 mins ago, by Ice Boy
trolling = teasing

@Sawarnik believe me il be a good husband to you...and our horoscope match

@BalarkaSen elaborate on the Cantor and the numerator/denominator part a little more?

6:00 PM
I will report to mods if you continue.
Even if you are troll, or not.
Most probably you are though. Not Anastasiya though.

3 mins ago, by Ice Boy
@Sawarnik put him on ignore

@Sawarnik are you serious? :o learn to be more patient :) il stop
@Sawarnik I stopped :)

@IceBoy Yeah, I ll just do.

@RandomVariable why are you breaking up the numerator? you need to use the fact that the denominator is a factor of the numerator, or you do need to handle the singularity

I thought that he/she would stop after sometime.

6:02 PM
@UserX List all the rationals (with increasing numerator) with denom and numerator that add up to 2, then 3, then 4, etc.

@Sawarnik i said I stopped :)

And cancel out multiplicities.

@Sawarnik save yourself

Ignored.

@Sawarnik really? :'(

6:02 PM
@UserX Yup :D

@UserX 1/1 is teh only rational with numer + denom = 2. 2/1 and 1/2 are the ones with denom + numer = 3.

@Sawarnik but i seriously stopped

@TheArtist you wanna troll me now?
3 mins ago, by Ice Boy
@TheArtist I said YOU ARE A TROLL

@Ice Boy I stopped coz I'm getting ready to go :p

@G.T.R Yes, in a uniform space, the closure of a set is the intersection of all of its uniform neighbourhoods.

6:03 PM
so the list will be 1, 1/2, 2, 1/3, 3, 1/4, 2/3, 3/2, 4, etc. @UserX

@Ice Boy maybe later....gtg now :)

see ya pal

@IceBoy You should ignore him too.

@TheArtist I'll be here if you wanna try it on me

@Ice Boy tell that you won't ignore me ;)
@Ice Boy , sure sure .....bye for now

6:05 PM
later

What the hell am I seeing about Sawarnik and Anastasiya?

@BalarkaSen Just nonsense.

@BalarkaSen They're getting married.

@UserX Oh. I should congratulate.

Thanks for clearing that question out too.

6:07 PM
just some troll's attempt

@UserX NP

@UserX Don't you think it would be illegal?
@Nick Why did you flag once? :O
I have only flagged once too.

@Sawarnik why would it be illegal?

@UserX We are 14.

@UserX underaged wizardry

6:12 PM
@BalarkaSen I am going to leave the 14 club in less than a week now :P
Leave you with Anastasiya :P

Is $\log(e+\pi)$ irrational?

And hey @Balarka How was the Diwali? :)
@UserX Nice question :D

@UserX Open problem, AFAIK

I assume you didn't do anything except
locking yourself in a room.

@Sawarnik math, yes.

6:17 PM
If we prove it's irrationality, can we use the Gelfond–Schneider theorem to prove the trascedence of $e+\pi$?

no, it's open

@BalarkaSen haha lol

I know it's open. But is it equivalent to showing that e+π is a trascedental number?

@UserX no, much more than that
it's stronger in fact

Why?

6:19 PM
Last ping, @Anastasiya-Romanova sorry, whatever that artist troll did [neglecting the small probability that you are artist yourself]. just ignore whatever he said :))

$log(e)$ is algebraic, even though $e$ is transcendental

@BalarkaSen :D

@Sawarnik stop being 14. What are you afraid of? That an online shy asian girl will know that you are into her? So what?

@UserX and you stop too.
@BalarkaSen Why don't you celebrate a little? Its fun :)

@Sawarnik BS

6:23 PM
BS?

@robjohn I was using the fact that $$\int_{-\pi}^{\pi} \frac{\sin nx}{\sin x} \ dx = \int_{-\pi}^{\pi} \text{Im} \frac{e^{inx}}{\sin x} \ dx.$$ But then I moved the $\text{Im}$ outside of the integral and said that $$\int_{-\pi}^{\pi} \text{Im} \frac{e^{inx}}{\sin x} \ dx = \text{Im} \ \text{PV} \int_{-\pi}^{\pi} \frac{e^{inx}}{\sin x} \ dx.$$ But that seemingly doesn't make any sense, even though things somehow work out from there.

Let $\Bbb M$ denote the set of irrational numbers. Is $\Bbb M$ dense in $\Bbb R \setminus M$? I can't formulate this correctly.m.
Nevermind the last question. Fucked up my thoughts.

:P
BS?

@UserX Huh?
Irrationals aren't dense in rational! They don't even overlap!

I know that's not what I had in mind

6:32 PM
I am sorry if anyone got annoyed :) especially @Anastasiya-Romanova and @Sawarnick :) I was just joking :)
Won't repeat it again :)

@TheArtist Do you want to talk about math or not?

@Ice Boy , I do . But have to apologize :/ I didn't think he would tkae it serious :/ can u tell that i Told sorry ?

@RandomVariable If you're extra nit-picky, you need to take the principal value before you move the $\operatorname{Im}$ out of the integral. But meh. Everybody does two steps of that sort at once. It makes total sense.

@TheArtist the proof of that will be shown in your future behaviour

@Ice Boy sure :p
@Ice Boy ok so...oh btw is it allowed to talk mecjanics/applied math here? Coz most people think it's phyiscs

6:49 PM

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...
It is more on topic here^

@DanielFischer It doesn't make sense to me from the definition of the Cauchy principal value since $\frac{e^{inx}}{\sin x}$ not only has a simple pole at $x=0$, but also at $x=-\pi$ and $x=\pi$.

@balarkasen: fyi, i managed to track down the connection i was rambling about yesterday re: Cantor sets, dynamical systems, and the 2-adic norm

@RandomVariable You must take principal values not only at $0$ but also at $\pm \pi$, $$\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert x\rvert < \pi - \varepsilon} \dotsc$$

if it sparks your curiousity (it might not) check out sections 1.5 and 1.6 of Devaney's text on dynamical systems
essentially, they use the 2-adic norm to describe how 'close' two different chaotic orbits of a quadratic map are

@RandomVariable It might be clearer if you integrate say from $-\frac{\pi}{2}$ to $\frac{3\pi}{2}$, then you have two normal simple poles.

6:58 PM
I've got a matrix $A = \begin{pmatrix} 3 & -2 \\ 2 & -2\end{pmatrix}$. I've computed that $A^n = \dfrac{1}{3}\begin{pmatrix} 2^{n+2} + (-1)^{n+1} & 2(-1)^n - 2^{n+1} \\ 2^{n+1} + 2(-1)^{n+1} & 4(-1)^n - 2^n\end{pmatrix}$. But how do I find a polynomial $f$ such that $e^{tA}=f(A)$?

sum it up by the dfn. of the matrix exponential?
though my approach to those kinds of problems is to write $A^2$ in terms of $A$ and $I$ by appealing to the Cayley-Hamilton theorem

@Semiclassical I think i'm partly confused as to what the 't' actually is.
@Semiclassical And f is a polynomial of degree less than 2.

well, it depends on the context. in an ODE context i'd interpret as a $t$, and $e^{t A}$ as an operator that tells you the behavior at later times
in the same sort of way that, if you've got the scalar ODE $y'=-ky$, the factor $e^{-k t}$ converts $y(0)$ to $y(t)=e^{-k t}y(0)$
except more complicated b/c it's a matrix
if you don't have context, though, just treat it like a parameter
i.e. the thing you use to write $e^{t A}$ as a power series
but i really think you'll have an easier time if you think in terms of Cayley-Hamilton

@Semiclassical Ah, later on we use it to solve differential equations so that would make sense.

right. it's really nice for that purpose, since as an operator it converts initial data (at $t=0$, say) into data for $y(t)$

7:11 PM
@BakralaSen ?

@BalarkaSen Smack such fake

Hi people! Please give me a hand!
Say that we have an $n$-dimensional random vector normally distributed: $\mathbf{x} \sim N(\bar{\mathbf{x}}, \Sigma)$. What if we apply PCA on $\Sigma$? What would be the vector that is given by $\mathbf{y}=P\mathbf{x}$, where $P$ is the projection matrix?

@DanielFischer I've never actually seen how to define the Cauchy principal value when the singularities are at the endpoints, although I suspected you could. But integrating over that other interval does make it a lot clearer. I'm not sure if I should edit my answer or not.

Thanks a lot!

@Semiclassical I'm looking in my notes, but i'm not really understanding what they're doing. I've done the following so far:

$P = \begin{pmatrix}2 & 1 \\ 1 & 2\end{pmatrix}, J = \begin{pmatrix} 2 & 0 \\ 0 & -1\end{pmatrix} = P^{-1}AP$

$f(A) = Pf(J)P^{-1} = P\begin{pmatrix}f(2) & 0 \\ 0 & f(-1)\end{pmatrix}P^{-1}$

But I don't understand what i'm meant to do from here, or is this just not what I'm supposed to do?

7:12 PM
I have also aked a question about, but it has been viewed only 3 times ... Here
2

@nullgeppetto upvoted

What's the cardinality of the universal set?

Thanks a lot for the upvotes @Hippalectryon, whoever else!

;)

@RandomVariable If you want to do that by contour integration, I would just convert $$\oint\frac{\sin(nx)}{\sin(x)}\,\mathrm{d}x =\oint\frac{z^n-z^{-n}}{z-z^{-1}}\,\frac{\mathrm{d}z}{iz}$$ that is, $z=e^{ix}$
Of course, then it starts to look like my answer.

7:33 PM
@CareBear For your catalog are you purposely ignoring integrals?

@robjohn That's what's done in another answer to that question and what I was trying to avoid to make the evaluation slightly simpler. I'm just going to change the interval so that their aren't singularities at the end points.

@RandomVariable Then you need to justify the two Principal Value integrals and why their singularities cancel out

@robjohn Does something beyond the fact that they're simple poles need to be said?

@RandomVariable for what?

7:49 PM
@robjohn About why their singularities cancel out.

@RandomVariable It would be a lot better to say something about the residues cancelling

@Alizter No... it's on my to do list.

@CareBear Would you like help?

Sure, I can add you as an editor of the catalog. Do you have an idea for classifying integrals and generating a catalog?

@CareBear Because ultimately there will need to be some sort of sorting convention. Are you familiar with Gradshteyn's tables?
You can find this book online (shamefully) and they have a section clearly defining how the integrals will be sorted.
I believe it is the largest printed table.

7:54 PM
Hm. I used two-tier process; first, SEDE query and then processing with Google script.

@CareBear The problem is it gets particularly complicated with mixed integrals
I'm thinking of an automated venn diagram that groups these integrals
Then one can select which properties it needs
for example integrals that have trig and log
that way we can make the search process more efficient
Let us continue this in another chat

@Alizter What are you trying to do ?

@Alizter Sort ?

@Hippalectryon It is difficult to find integrals on SE. CareBear has a catalogue of sorts.

8:07 PM
Ok

8:38 PM
@Hippalectryon How do I prove that (maybe you recommend me some tools) $$\sqrt{2\sqrt[\Large 3]{3\sqrt[\large 4]{4\cdots \sqrt[\Large n]{n}}}}<2, \space n\ge2$$?

@Chris'ssis screwdriver?

@Hippalectryon lol :-)))

Let me think a bit

OK
Ahhhhhhhhhhhhhhh, it's straightforward ... (no need for pen and paper)

@Chris'ssis Let me thiiiink
I'm trying to draw curves but desmos is stupid -__-

8:49 PM
@Hippalectryon OK

Duh Desmos doesn't like recursive functions. Whatever.
Let's take a sheet of paper.

@Chris'ssis: $$\sum_{n=1}^\infty\frac{\log(n)}{n!}\le\log(2)$$

So, $\displaystyle\prod_{k=2}^\infty k^{\frac{1}{k!}}<2$

@robjohn Yeah
@Hippalectryon aham (I mean "yes")

8:52 PM
@Chris'ssis ?
@Chris'ssis That sounded like 'ahem ahem' xD

@Hippalectryon lol :-)))

Then ye it's obvious xD

What's the next step then? (hmmm ...)

@Chris'ssis In what ? We directly arrive at what @robjohn said

@Hippalectryon well, after that step
@Hippalectryon One needs to prove that $$\sum_{n=1}^\infty\frac{\log(n)}{n!}\le\log(2)$$ holds.

8:56 PM
@Chris'ssis That it is less than $1$ is very easy.

@robjohn that sum is about $0.603783$. Not sure what you mean by very easy.

@Chris'ssis Use $\log(n)\le n-1$ and get a telescoping series

@robjohn it doesn't work.

@Chris'ssis what? It shows the sum is less than $1$
$$\sum_{n=1}^\infty\frac{n-1}{n!}=\sum_{n=1}^\infty\left(\frac1{(n-1)!}-\frac1{n‌​!}\right)=1$$

That is nicely done

9:04 PM
Am I too tired here?
@robjohn I have no idea what you are trying to say ... the right side is $\log(2)\approx 0.693147$

@Chris'ssis I said that getting the sum to be less than $1$ was easy... $\log(2)$ is a bit harder.

@Chris'ssis He says that it's very easy to show that the sum is less than $1$. He doesn't say it's easy to show it's less than $\log 2$.

@Chris'ssis Sleep a bit or you'll become a zombie >:c like me

@robjohn Oh, but I was focusing on the result of the problem, to prove it's less than $\log(2)$.
@DanielFischer Of course, showing that part is very easy.

Good night, any help for this exercise : If $a^n - n$ divides $b^n-n$. I have to prove that $a=b$.
$(a,b)\in \Bbb{N}$

9:07 PM
@MarcGato For all $n$ ? For a specific $n$ ?

@Hippalectryon oops, for all $n$.

I'm pretty sure @robjohn knows the answer :P I'm such a coward

hallo everyone
I need help evaluating this limit
It's an indeterminant form 0/0

@Chris'ssis I can't do anything without pen and paper, even simple stuff.

9:11 PM
I tried to divide by $x^{4}$ but no luck

@RandomVariable hehe, I try to do things without pen and paper for improving, stretching my imagination, I think it's pretty helpful (at least in my case it was so). I force my mind to visualize everything fastly, it's more like an experiment.

@Sabಠ_ಠ What else do you know ?

I tried l'hopital's no luck
I get stuck after differentiating twice
I can't find the trick to it

@Chris'ssis Use $\log n < \frac{n}{e}$ for $n\in\mathbb{N}$ to get $$\sum_{n=1}^\infty \frac{\log n}{n!} < \frac{1}{e}\sum_{n=2}^\infty \frac{1}{(n-1)!} = 1 - 1/e < \log 2.$$
2

Can anyone help?

9:16 PM
@DanielFischer wow, that was awesome ... (very nice)

@Sabಠ_ಠ How else can you write $x^3-3x^2+4$ ?

hmm

@Sabಠ_ಠ Polynomial division by $x-2$? Might be a little tedious, but if your L'Hospital doesn't work for you ...

@Sabಠ_ಠ Try factoring it
Same for the denominator
Then Oooooh magic will appear :D

it's a really long calculation isn't there a simpler way rather than factoring
?

9:18 PM
@Sabಠ_ಠ It's not long
@Sabಠ_ಠ Don't you see obvious roots ?

lemme try factroring and see if I get the answer
@Hippalectryon I see it for the numerator

@Sabಠ_ಠ What ones ?

You see what I mean now by admiring work and people? It's exactly the solution above! Time to learn here some.

x = 2

@Sabಠ_ಠ Is that all ?

9:19 PM
nop
i'm calculating the others

@Sabಠ_ಠ Look at all the 'easy' roots (-2,-1,0,1,2)
Always check those
That will save you some minutes

How do you check these? I never heard about them :S

Check if they are roots

ohh

If they are, it's faster than calculating them by factoring and finding coefficients

9:21 PM
yeah that's how I did it, but I never knew these were called the easy roots

@DanielFischer I remember we studied that $f(x)=\log(x)/x$ in high school first, and we could use it then to compare $e^{\pi}$ and $\pi^{e}$.

@Sabಠ_ಠ That's the name I give them :) it's not official at all

:)

You can call them the magic jellyfish roots if you like it that way :)

So it's the trial and error process?
:P

9:22 PM
@Sabಠ_ಠ Yep, but a fast one

I call it "trial and error"

You wouldn't do that for the root 5sqrt(27)/e

Yea, but I normally go like this, 0, 1, -1, 2, -2

The order doesn't matter

The thing is I don't know the tricks to find limits easily

9:23 PM
@Sabಠ_ಠ Have you found new roots ?

I got a remainder with x-2

@Sabಠ_ಠ There isn't one magic trick that works all the time :)

@Chris'ssis Yes, or generally $x^y$ and $y^x$, that way one can see that the only integer solution with $x\neq y$ is $2^4 = 4^2$ easily, for example.

@Sabಠ_ಠ So, what roots did you find for the numerator ?
Other than 2

9:24 PM
-1

Hence what is its factored form ?

@DanielFischer Indeed.

(x-2)(x+1)(ax+b) right ?

yep

Find a (obvious, =1, it's the leading coefficient) and b

9:24 PM
now long division?

@Hippalectryon $(x-2)(x+1)(ax+b)$

@DanielFischer Oh that mistake :c
Thanks

I get remainder

@Sabಠ_ಠ Don't make the same mistake I did, as @DanielFischer remarked it's (x + 1))
@Sabಠ_ಠ How so ?

9:27 PM
Because I made the same mistake as you LOL

In my mind x = 1
but I should have used x-1
lemme try again
naa
I still get the remainder

Uh
Show us

I expanded (x+1)(x-2)

That makes (x^2-x-2)

9:29 PM
then I used long division and divided the cubic numerator by the expanded thing

:/

We have (x^2-x-2)(x+b)=x^3-3x^2+4
What do you want to divide here ? Just find b
expand (x^2-x-2)(x+b), remember that the coefficients before each power of x must be the same
Done

ohh
but why doesn't long division work?

-2b=4=>b=-2=>x^3-3x^2+4=(x+1)(x-2)^2
@Sabಠ_ಠ It does, you probably made a mistake

9:31 PM
I lack simplicity in my life

@Sabಠ_ಠ Now what can you say about the denominator's roots ?

a =1 b = 1

:O wut b=1 ?

nop
b = 2

9:33 PM
typo :P
:O
(x^2 + 3) (x - 2)^2

+3 ???

yea

OH WAIT
I was thining about the numerator xD

@Sabಠ_ಠ YEEEEE

9:37 PM
:D

Kirby dance
@Sabಠ_ಠ Well done :D

shakes dat ass
Ty :)

No prob :)

I got my finals tomorrow and I'm learning about limits Facepalm

Uh :c

9:38 PM
It's funny how I can do derivatives in 30 seconds though

@Sabಠ_ಠ You'd be scared by @Chris'ssis's limits

ouch
I hope I don't get crazy limits tomorrow

$$\lim\limits_{n \to \infty} \dfrac{1}{\log n} \sum\limits_{k=1}^n \dfrac{1}{k}\tan \frac{\pi k}{2n+1}=\frac{1}{2}$$

hi all

Obvious

9:39 PM
so I finally put a bounty on the question ...

@DanielF I think I'm going to need to get a coffee IV and carry it around with me.

@Sabಠ_ಠ Well I gtg, hope you do well in your tests :)
gone

Ty c ya :P

@DanielFischer well, it'd be cheaper and easier than going to the local coffee shop whenever I got tired.

9:42 PM
@MikeMiller Chew some coffee beans.

That's a fair idea. Maybe I'll bring a few around with me from now on.

hope the bounty does not dissapoint as in the past

@DanielF, @Mike, @Hippa (oh, welcome back)
@Sab ... oh, I thought you'd left

Good day @Ted. How's probability going, still enjoying it?

Limits? @Sab ... That's mostly in first calc ...
yes, @DanielF, although now I'm about to work my second exam to make sure it's easy enough. I'm on pace ... And purposely not saying some stuff in class to see if they read the book when their answers don't agree with the back of the book (I'll give it away tomorrow after they turn in homework :) )
Just discussed Markov and Chebyshev inequalities. After the exam we'll do joint distributions ... the semester is winding down. I've learned a ton. Some of my students have, too :P

9:51 PM
Good.

I'm getting a dozen downvotes ... so it makes me mad, when it's random and comes out of the blue. But I'll get over it :)
How've you been, @DanielF?

@CareBear I saw somewhere at some time that you don't want to use long names anywhere; is that true?
Hi @Ted

the bounty does not seem to improve the amount of views much ...

@TedShifrin All in all not bad. Although I'm a little depressed that decent questions are so hard to find in the last months.

Well, interestingly, diff geo has picked up a bit (although I get downvotes there ... it's like René has deputized friends to downvote me).
@Mike: Are you chatting with Jacob about the diff geo stuff? You should.

9:56 PM
@TedShifrin Yes, he gave me a very good explanation as to why I want my connection to be torsion-free.

Great. I hope you guys will have many fruitful math conversations.

If I recall, the first sentence of his explanation was "Connections with torsion are a bitch to work with."

It was only in the middle of my professional career that I began to understand what torsion really means. Elie Cartan was a smart guy. The books don't really make it clear at all.
LOL ... well, if you want an invariant connection on a Lie group, it typically will have torsion.

@MikeMiller Not excessively long and difficult to use in a sentence, like the infamous ... healthier.

"The Hessian is no longer symmetric" is good enough reason for me, for now, I think.
@CareBear Well... I suppose this is much healthier.

9:58 PM
OK, sure ... @Mike ... although it's not super-geometric.

all of my questions are either trivial or too hard :(

Join the crowd, @mick.

@TedShifrin I guess its the curse of math

@DanielFischer May I recommend my favorites?

Well, in some ways, yes. But you'll run into medium ones, too :)

9:59 PM
@TedShifrin well, I like to hear geometric motivation. we're just not getting any in class, where we just permute long series of equations to get identities I'm not sure why we care about for tensors I'm not sure why we care about!

@DanielFischer my questions ;)

@CareBear You may. I'll see whether I like some of them.