8:00 PM
Oh no, I meant the combinatorical thingy @Daniel @Mike
@Nikolaj the formula in the lniked question has a name
Let me find it, second..

Sorry, @Nikolaj, I don't know that stuf

@MikeMiller No problem :)

@MikeMiller Ah, you were only concerned with that implication. That is true, if $a_n$ converges, the arithmetic mean also converges, to the same limit.

Duh, of course. I proved that at some point.

@seaturtles Did you find any counterexample?

8:04 PM
It is a version of Bayes @Nikolaj. Pretty sure it follows right from it.

@Studentmath That's what I thought too, but for some reason I can't find one that looks like it

Well, you also have the multipication law
for exclusive and exhaustive events
You add that up with Bayes and you get what you are looking for

@AlexanderGruber That guy looks handsome in the extra fittings.
@MikeMiller What's so funny about that?

@Nikolaj simply recall that adding all these B's up is the total sample, and you get the multipcation law along with Bayes

8:08 PM
@BalarkaSen insufferable pedantry is always fun

@Studentmath So I have to take the multiplication law for conditional probabilities and then that ends up with Bayes?

@MikeMiller I find it quite painful.

For excluisive probabilities. And yes
If you still struggle with it once I am done with this exercise, I will take another look and try to give you a better hint, but I am sure you'll manage by then

Hi

My combinatorics got it wrong sigh

8:12 PM
Haha @AlexanderGruber I see your new pic.

darn
I was going for mortarboard today but nobody's posting interesting questions
I'll watch more House

I am thinking of deleting my account again.
Not many lhf these days.

@BalarkaSen nope

@JasperLoy Don't. It's not worth the trouble.

8:14 PM
@seaturtles I thought not. I have come up with a sketch of proof with covering space theory.
Interested?

Please allow me to moan. I am very sad with the way my life has unfolded, very very sad.

@BalarkaSen sure, probably unfamiliar territory though
@MikeMiller good show I thought. too bad cuddy wasn't in the final season.

I think $\binom{10}{1,9}*\binom{15}{2,13}$ would give too little options though.

I don't know if or when I will recover from my mental illness.

Certainly too little.. sigh

8:16 PM
@Studentmath Thanks. I'm having some trouble with understanding it, I think I'll just try and do the metric one first.

@seaturtles $\Bbb C(z)$ is the function field of meromorphic functions over $\Bbb P^1$ so $\text{Gal}(L/\Bbb C(z))$ is something like a monodromy of the ramified covering $X \to \Bbb P^1$ which is galois away from the ramification points, $X$ being a Riemannsurface. Then we have another such covering $X_0 \to \Bbb P^1$ and since they have the same monodromy $G \cong \text{Gal}(L/\Bbb C(z)) \cong \text{Gal}(L'/\Bbb C(z))$, we have $\Bbb P^1 \cong X/G$ and $\Bbb P^1 \cong X_0/G$.
Hence $X/G \cong X_0/G$. But Professor Ted reckons this might or might not imply $X \cong X_0$.
And Alexander reckons that this is a classical galois theory problem which is in Issacs.

@Nikolaj not sure what you mean with metric one

@Studentmath Different assignment. Nevermind it, I can figure that one out. Just the probability one giving me trouble.

I will take a look into proving it soonish if you still have trouble

@Studentmath Someone answered my question and the name was law of total probability. I'll try and do a proof of it once I'm done with the metric one. If you're still here could you potentially proofread it?

8:30 PM
@robjohn these days I'm going to improve that proof and reduce it to $5$ parts. Well, the improvement mainly refers to that double series in the fifth part. To naturally get those dilogarithms (without using some software) one need to cleverly manipulate things there.

@DanielFischer do you know a lower bound (other than AM-GM) for $$\sum_{k=1}^{n}|a_k|^2$$ where $a_k$ are complex numbers ? I'd like something that could force $\prod a_k$ to appear

@seaturtles it's got its moments. this episode is about a blogger. i hate bloggers.

@robjohn I'm glad I managed to do that. Of course, during the time I might come up with other improvements, but for the moment the biggest one is referring to the double series (in the fifth part).

@G.T.R No immediate idea. If you work with $\prod a_k$, you should probably demand that $a_k \neq 0$ for all $k$ to get something nontrivial.

@Nikolaj sure

8:34 PM
@G.T.R $$\frac1n\sum_{k=1}^n|a_k|^2 \ge\left(\frac1n\sum_{k=1}^n|a_k|\right)^2 \ge\left(\prod_{k=1}^n|a_k|\right)^{2/n}$$

@DanielFischer Here math.stackexchange.com/questions/829625/… I think it's the way to go there
@robjohn not sharp enough unfortunately

Anyone can help me with some basic combinatorics? Can't believe I get stuck with it again..

@Chris'ssis The one that I'm working on?

@robjohn sure

@Studentmath Someone decided to post the entire proof so, nevermind the proofreading. How good are you at metrics?

8:41 PM
Rusty

@Studentmath Not done with my proof yet.

Hey @Ethan

@JasperLoy Hi.

@ParthKohli Hello. I feel very sad today.

@JasperLoy Oh, what occurred?

8:44 PM
@ParthKohli Nothing new, just the usual problems.
@Ethan Why did you remove those?

@JasperLoy Oh, what kind of problems?

no one responded

@ParthKohli Nvm, too long to mention.

@JasperLoy Hmm.
@JasperLoy Have you watched Requiem for a Dream?

@ParthKohli Nope.

8:49 PM
@G.T.R As soon as $n \geqslant 6$, you have $x_k \geqslant \lvert x_k - e^{2\pi i/n}\rvert$ for all $k$ with $x_k \geqslant 1$. So you'd only need to look at $2 < n < 6$ closer.

@JasperLoy OK. Since you haven't, don't watch it now.

...that's what is hard to decide. It affects you so much that you can't stop thinking about it for months. But you still can't decide whether you love it or hate it.
Such is the storyline.

@ParthKohli I think I must watch it then

@JasperLoy You may feel a little sick watching it. Stay careful.

8:53 PM
hmm
i want to nap but i'm not quite tired enough to

Think I have it right now.

If I have 25 marbles, 10 blue 15 red, spread around a circle, and I want to find the probability that a blue marble has red marbles on either side of him along the circle. The total number of options obviously are $\binom{25}{15,10}$. The number of options in this case is $\binom{3}{2,1}*\binom{22}{13,9}$. At least it leads me to a nice probability..

9:10 PM
@DanielFischer Bien vu

@robjohn I just found a fantastic improvement to that question.

@Chris'ssis to the question or the answer?

@robjohn I'm referring at the solution to that question. Well, I only need to put things on paper carefully and see that everything is nice.

@Chris'ssis If my way works out, it will be pretty simple.

9:30 PM
@robjohn I found a way that seems to avoid the use of the double series in the $5$th part.

Evening all
Does someone care to help me solve a simple trig equation? :)
$$\cot{\theta} = 3\sin{2\theta}$$

@Ell have you tried double angle formula for $\sin 2\theta$

I have, I was just about to say
I have solved it for 4 solutions, but my method involves dividing by $$\cos{\theta}$$ in $$\cot{\theta}$$ but the problem is, won't I lose solutions if I do this?

@Ell okay .. so you assume $\cos \theta$ is non zero when you divide by it .. that gives a solution ... the other solution is when $\cos \theta = 0$

@DanielFischer it's quite lucky that when $x_k\leq 1$, $\lvert x_k - e^{2\pi i/n}\rvert ^{2/n} \leq 1$.

9:37 PM
@r9m ah okay, thank you! :)

@G.T.R Not for $n = 3$ or $n = 4$, and not necessarily for $n = 5$.

@DanielFischer $n\geq 6$ was supposed :P But it's still striking, and quite sharp

sigh combinatorics...

@DanielFischer the schwarzian is bizarre

@MikeMiller In how far?

9:44 PM
It is the strangest damned expression I've ever seen and somehow it determines precisely Moebius transforms
and not for some obvious reason

@MikeMiller Not rational functions, Möbius transformations.

@DanielFischer Was editing just as you said that.
But nonetheless.

@r9m thanks for that link, but honestly, I think that limit can be computed elementarily, with the high school knowledge.
@r9m Now I've been working on some amazing stuff and cannot focus on other stuff.

@Chris'ssis okay ... cool :D !!

@r9m I (only) look at that and know the answer. How I did it? $$\sum_{n=1}^{\infty} \frac{1}{n^2} \left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)$$

9:49 PM
@robjohn it there a way to edit an answer and the thread not to be bumped to the front page ?

@Chris'ssis irk ...I don't get it .. look at which one ?

@G.T.R i don't think so, except if it's quickly after it has been edited already

@r9m I mean you only look at that series, and without doing anything else (referring to pen & paper), you are able to say the answer. How is that possible?

@Chris'ssis okay you mean .. so you have found a way of becoming Sir. Lancelot with defeating that series(dragon ?) ? :D cool !! how did you do that ? :)

@r9m I'll tell you that but not now. Just think of it for a while. It's a very precious series. :-)

9:58 PM
LAST LAYER RUBIKS SKIP ^^ 1/15552 chance
I think I have gotten 4 or 5 of those ever

@Chris'ssis okay ... I'm in deep mud ... one certain problem is tearing me apart while dancing on my nerves ... gimmie a few more minutes so that I can calm my 'ead and think :P

@r9m I hope it's real analysis

@G.T.R um .. Analysis all right ... but complex :} .. its a part of a proof that I can't get ... the author is acting all breezy cool by omitting important details ... leaving me in a pool of pain :(

@r9m Well, that's real analysis ;)

Gah, I am sure I am getting this wrong..

10:08 PM
@DanielFischer indeed .. my 'pain' is anything but imaginary :P

@r9m What is the proof about?

10:30 PM
@DanielFischer sorry for late reply (I had a blue screen and some 'memory dump error' that I have never seen in my life b4) .. sub harmonic functions :) .. all theorems are indicated trivial or left as an exercise :P lol

@r9m Ah. The lazy way. Delegate the work to the reader.

@Chris'ssis $\zeta (2)^2 - \dfrac{3}{4}\zeta (4)$ ? :D

subharmonic functions are a little horrifying
I was slightly satisfied with the conceptual statement that subharmonic functions are 2-dimensional analogues of convex functions

@r9m It might be ... :-) How you did it?

but I never got a lot out of them

10:38 PM
@MikeMiller You are a little horrifying.

@R9m @DanielFischer speaking of harmonic, consider a man at the center of a circular trampoline. Let it be as if a punctual mass were at the center of the trampoline. Why is the function $\mathbb R^2\to \mathbb R$ that describes the surface of the trampoline harmonic?

if I'm only a little horrifying I've truly failed

@MikeMiller I have a problem for you.

@PedroTamaroff I have too many of my own problems
for instance, I still need a haircut

Suppose $M$ is a module that is generated by $n$ elements and contains a linearly independent subset of size $n+1$. Then $M$ contains linearly independent subset of size $m$ for any $m>n$.

10:41 PM
I believe it

You're not supposed to believe it.

But I do.

You're supposed to prove it.

@G.T.R It isn't, it has a local minimum at the centre. You mean it's harmonic in the punctured disk?

10:42 PM
Sure, but why can't I believe something before I prove it?

@DanielFischer yeah, it's physics, so feel free to make every hypothesis you want

@Chris'ssis the series is $\displaystyle -\sum\limits_{n=1}^{\infty} \dfrac{1}{n^2}\int_0^1 \sum\limits_{i=1}^{n} x^{i-1}\ln x \,dx$ :) .. && I have sos's blog open in another tab :P

@PedroTamaroff Is the proof constructive?

@MikeMiller I haven't tried a constructive proof.

Ah, in that case I'll go get my haircut.

10:46 PM
Oh, OK.

@r9m No need for integrals ... Well, you broke the rules and put things on (virtual) paper ... :-)))). Just use your imagination a bit and you're done :D

@G.T.R Physics? You mean forces and such?

@Chris'ssis come on ... gimmie a break :| .. maybe cannons ... but I killed the bug all right ?! :P

@DanielFischer the guy who did the problem on the blackboard had one argument (not cogent to me though) : every point f(z) is the mean of all the f(z') for $z'$ in a small disk around $z$. Hence harmonicity...

@r9m OK :-)
I'm out for some sleep ...

10:49 PM
@G.T.R If that is so, then it's harmonic. But why should that be so?

@Pedro I'm going to try to write up some notes eventually about that thing I FB'd you about. but nobody's ever gonna read them :D

@DanielFischer I don't know. The guy couldn't explain that either. But when it's true, why should the function be harmonic? (where does Laplacien pops up?)

@G.T.R A continuous function with the mean value property is harmonic. That follows from the maximum principle and the solution to the Dirichlet problem.

@MikeMiller Maybe I will.

@DanielFischer I'll stop being hartnäckig for tonight and give up on physics. Good night

10:57 PM
@G.T.R Bonne nuit.

@DanielFischer Wrong language

@N3buchadnezzar No language is the wrong language.

Well good night in german is "Ich hoffe, dass ein Feuer wird ihnen umbringen"

Close, but not quite.

11:04 PM
OP is not accepting my answer .. :P what else does he want ? decryption of uchiha stone tablet ? :P lol

@r9m You must be new to this.

@r9m Just give him your soul

@MikeMiller I feel unsatisfied if the answer is not accepted or if there is no indication (eg a comment ...) that the OP has read my post :|
@N3buchadnezzar LOL .. is there more that I could add here ?

@G.T.R nope
Using partial fractions, we get
\begin{align} &\small\frac1{(j+n)^2(k+n)^3}\\ &\small=\color{#00A000}{\frac1{(k-j)^3(j+n)^2}}\color{#C00000}{-\frac3{(j-k)^4(j+n)}}+\frac1{(j-k)^2(k+n)^3}\color{#00A000}{-\frac2{(j-k)^3(k+n)^2}}\color{#C00000}{+\frac3{(j-k)^4(k+n)}} \end{align}
The red terms cancel each other and the green terms partially cancel each other. Therefore,

\begin{align}
\sum_{j,k,n=1}^\infty\frac1{(j+n)^2}\frac1{(k+n)^3}
&=\sum_{\substack{j,k,n=1\\j\ne k}}^\infty\color{#C00000}{\frac1{(j-k)^2(k+n)^3}}-\color{#00A000}{\frac1{(j-k)^3(k+n)^2}}\\

11:19 PM

@N3buchadnezzar I choke with laughter !!!! LOL

@robjohn Yeah, it applies to other sites if you have 666 points too.

@N3buchadnezzar ah! I don't think I ever ended up at that exact rep.

@robjohn Too bad =/ Had any other interesting numbers you can remember?

@N3buchadnezzar I was 111,111 a bit ago

11:22 PM
Thats nice
123456 next then
Or perhaps another palindrome

Does that change the rep?

@N3buchadnezzar it dropped me 15 points temporarily

@robjohn totally worth it !! lol

@r9m did you see my less than a page proof of Chris'ssis 8 page answer?

11:25 PM
One of the perks of living in Norway, is an abundance of candle lights, churches and goats.

@robjohn terrific .. I'm reading it :D

@r9m I will add a bit of explanation when I write it up off chat

I do not see quite how the red terms cancel, but I have not had the chance to look at it closer off hand.

@N3buchadnezzar when you add them over all $j,k,n$

Yeah I see it now

11:34 PM
@robjohn I must say. Mother of God.Period ..

@r9m It's just standard series manipulations. I need to write it up with more explanation...
it is pretty dense as it sits

@robjohn okay :)

@r9m I spent almost all of my MSE time today on that, and it isn't even a main question :-)
but I think it is worth it.

11:49 PM
@robjohn thats a spectacular proof nonetheless :-) (my bro mistook your avatar for spiderman .. he's pretty young)

@r9m Is your screen adjusted so that orange looks red? I can see the eyes evoking that impression, though

@robjohn its closer to orange ... its possibly the eyes of the avatar :)

@r9m yeah, I get that