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21:00
@robjohn where?
@Chris'ssis I've already solved this one, and what do you mean by natural?
@teadawg1337 Ah, if you solved it then it OK ...
@teadawg1337 Did you ever hear of Catalan-Botez identity?
@teadawg1337 Use that at first. The rest is pretty boring usual stuff.
$$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k}=H_{2n}-H_{n}$$
@teadawg1337 ^^^
@robjohn What i have done is differentiation with respect to $s$ and i have applied the fundamental theorem of calculus of variation . What i am afraid is that i am not getting some nice equations . :(
21:07
Partial fractions and a bit of use of the generating function of the harmonic function complete the job.
@iwriteonbananas ?
@Chris'ssis In this answer
@Theorem When dealing with $L^p$ rather than $L^2$ sometimes the reverse Hölder inequality helps.
@Chris'ssis ??? This arose in my work with polylogs and $\frac18\log^2(2)$... I know nothing about the harmonic function...
Well, identities thereof
It often helps to linearize things similar to $L^2$
@Vrouvrou let $x_0 \in E$ and let $(x_n)_n$ be the sequence described in the assignment. since you know $x_0$ is an adherent value of $(x_n)$, let $x_{n_k}$ be a subsquence such that $d(x_{n_k}, x_0) \to 0$. to prove that $x_0 \in \overline{f(E)}$, it suffices to show that any $B_\epsilon(x_0)$ has nonempty intersection with $f(E)$. suppose toward contradiction that $f(E) \subset B_\epsilon(x_0)^c$. Then $d(x_0, x_{n_k}) \geq \epsilon$ which contradicts convergence
21:13
@Chris'ssis The answer I got is $\frac18\ln^2(2)-\frac14\ln(2)+\frac{\pi^2}{48}$, but I question its validity after seeing your method...
@robjohn In the first step , to find the necessary conditions i just perturbed the minimizer $u^*$ to $u^*+s \phi$ , differentiated with respect to $s$ and equated it to $0$.$s$ is a scalar , $\phi$ is a function that belongs to $H^{1,2}$.
@teadawg1337 I told you above what you should do. Just let me know where you're stuck.
@Theorem yes, but you are trying to get $L^p$ estimates, right?
@Chris'ssis As I've said, I have no experience with using identities of the harmonic function...
@robjohn aah ok , you are talking about my second question , i.e. for which $q$ such minimiser actually exists .
21:16
@teadawg1337 I think you know more than you say you know. I'm very good at noting that. ;)
@Theorem Oh, I didn't realize the first part was a question. It sounded like a statement of things you knew.
@robjohn my first question is if the necessary condition , i.e. the equations i have gotten are correct . :)
I'm on @iwriteonpeoplewhowriteonbanana
bonsoir, les frères Jacques
hi @robjohn, rehi @teadawg, bananas
21:23
@ted hi how is that ?? "les frères Jacques"
Penses-y un peu :P
@TedShifrin G'day
You practicing to go down under, @robjohn? :P
@Chris'ssis $$\frac18\ln^2(2)=\sum_{n=1}^{\infty}\frac{\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}}{‌​16n^2-8n}+\sum_{n=1}^{\infty}\frac{1-4n}{64n^4-64n^3+16n^2}+\sum_{n=1}^{\infty}\f‌​rac{1}{16n^2-8n}$$
ugh, @teadawg, what horrific things are you doing to that poor series?
21:25
heyho mctedster
I'm tired of grading differential geometry homeworks, so I'm having a martini :P
that's a great idea
want one, bananas?
yes, but i'd rather smoke weed
LOL, well, I don't do that, so I can't offer :)
21:27
im just trying to be one of the kewl kids
aren't you too old to be kewl?
@TedShifrin How do you take your martinis?
LOL, that got @David's attention. Very dry, usually on a rock when I'm home, with multiple olives ... and gin, of course.
i dunno :O
@Chris'ssis $$\sum_{n=1}^{\infty}\frac{1-4n}{64n^4-64n^3+16n^2}+\sum_{n=1}^{\infty}\frac{1}{‌​16n^2-8n}=\frac14\ln(2)-\frac{\pi^2}{48}\implies \sum_{n=1}^{\infty}\frac{\sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}}{16n^2-8n}=\frac18\‌​ln^2(2)-\frac14\ln(2)+\frac{\pi^2}{48}$$
21:28
* thinks he's going to ignore teadawg until this passes *
hi @Alizter
leave that poor series alone teadawg
it's never done anybody no harm
@iwriteonbananas I gave up pot decades ago, because everyone drug-tests nowadays, and that stuff (THC) sticks around in your system for ever...
bananas is in a different country, @David
Hey, I'd show the rest of my work but it'd fill up half the screen... I just put the bare minimum required to prove my result
@DavidWheeler dont worry, i have gallons of old urine in the freezer
21:29
@TedShifrin When I was a younger person, I had a neighbor who had brought some hashish back from some arabic country where it was legal
@TedShifrin Are you good?
somehow, bananas, I don't think that will pass, as it were
It still had the brand logo stamped in it
@teadawg1337 Great!
my back is killing me, @Alizter, but I'm ok otherwise, and you?
21:30
@TedShifrin Did you see my probability question? Well it's not mine but you get the idea.
no, @Alizter ... I've been working
it's official: @teadawg has now gone over to the dark side.
morning ted
@DavidWheeler isn't weed very illegal in all those countries?
40 mins ago, by Alizter
@TedShifrin One hundred people line up to board an airplane. Each has a boarding pass with assigned seat. However, the first person to board has lost his boarding pass and takes a random seat. After that, each person takes the assigned seat if it is unoccupied, and one of unoccupied seats at random otherwise. What is the probability that the last person to board gets to sit in his assigned seat?
@iwriteonbananas might be now, this was long ago....
21:31
oh, that's the letter-in-the-envelopes question, @Alizter
I'm sure he's heard the problem before, @Alizter
hmm, more or less
Yes, I get 1/2
good night, @Mike
21:31
@Theorem What is preventing us from simply adding a constant to $u$ to bring down the ratio?
@Ted I was being considerate, I don't think a page and a half of work in my handwriting would translate very well to this chat :P
However my reasoning is far from clear
@DavidWheeler and smuggling it to the USA sounds stoopid
gee wilikers
@Alizter He gets in his assigned seat iff it's the last one available. First peron has $(n-1)/n$ odds of not sitting in that seat; next has $(n-2)/(n-1)$...
@DanielFischer Could you take a look at my question?
-1
Q: Can linear execution time be achieved

evindaThe SELECT algorithm determines the $i$th smallest of an input array of $n>1$ distinct elements by executing the following steps. Divide the $n$ elements of the input array into $\lfloor \frac{n}{5} \rfloor $ groups of $5$ elements each and at most one group made up of the remaining $n \mod 5$ ...

21:32
Yes @Mike
@iwriteonbananas I cannot vouch for my former neighbor's intelligence
multiply
@robjohn I am wondering how thats going to help me .
The famous question, @Alizter, is what's the probability that no one is in the right seat?
Yours is different.
21:33
And that's got the cutest answer.
Yup @Mike ... I even have that one in my algebra book.
This is a chat fest.
what's the probability that ted smacks the passenger that took his seat ?
@Theorem I'm just trying to understand the first part. It looks as if we can simply add a constant to $u$. That doesn't change $\nabla u$ but changes $u$
Combinatorics is all about counting things. Unfortunately, if I can't use my fingers to count, it doesn't go well...
21:33
0, @iwriteonpeoplewhowriteonbanana, unless he's you
@Ted Hello, sorry for ignoring you during that
If it's you, Hippa, 100%.
We're not so different, you and I, @Ted
@TedShifrin Hopefully, you don't know what I look like :D
@teadawg: Did you figure it out by totally elementary means, knowing that $\sum_{k=1}^n 1/k \sim \log(n)$?
21:34
(n-1)/n! as a wild guess
I won't make any obvious comments, Hippa.
Way....yyyy too small, @Alizter :)
I'm not sure I've done your version before, @Alizter.
I need to ponder it.
@robjohn Yes , but i i will still have a factor $\mu(u)$ , which really is giving me trouble .
My not-so-educated guess, @Alizter, is that the answer to yours is $1/n$. I need to think harder to justify it.
I said what it is and why above, @Ted :P
@Ted Using asymptotic series? No
21:39
@TedShifrin 1/n is too small. Thats as if the first pasenger guessed correctly. However there is also when he doesn't but possible the second and third swapped and rest were correct.
But the prob is definitely >1/n
@teadawg: This is totally elementary, just using the sum to approximate the integral.
Integral?? I never used any integration...
@Theorem I guess I am asking if there is anything preventing $\varphi$ from being a constant.
@teadawg1337 You have answers with integrals.
You're using far more arcane things, @teadawg. You should understand that $\sum_1^n 1/k$ looks like $\int_1^n dx/x$.
@Mike: Where is said answer?
21:43
@TedShifrin he gets in his last seat iff it's available. the first guy has probability (n-1)/n of not getting in his seat; the next guy (n-2)/(n-1), ... the second to last guy 1/2
take the (telescoping) product
I actually know nothing about asymptotic expansion... This is all new to me
what am I on about? that's 1/n, not 1/2
Ah, good point. I was misapplying symmetry. I'm in geometry mode now, not probability :P
Ha ha ... My instincts were right, then :)
Indeed. remembers henceforth never to trust Mike
did you trust me before?
@teadawg: What I'm talking about is far more fundamental for you to learn/master/use.
21:44
@robjohn I guess you are not considering $\varphi$ is as a constant .
Yes, $H_n\sim \log n$ is used in NT quite a bit. The bit I have studied anyway.
Its a fun fact
It's used everywhere, @Alizter.
Another way of thinking about it: each assignment of seats is just a permutation of $[n]$. Precisely $(n-1)!$ of those preserve the last one - those are the permutations of $[n-1]$.
@TedShifrin Well I have literally only studied Algebra and NT so thats my everywhere!
@Mike: That's related to the group-theoretic interpretation of the envelope problem I put in my book ... counting fixed-point free permutations.
21:46
mhm
I know the story behind the problems, though not from your book, @Ted
I take it Homotopy doesn't make sense for graphs
You're wise never to look at anything I write, @Mike, just as I'd be wise to ignore you.
what, @Alizter?
You either have rings, joined rings or a vertex
or cycles/rings
I have no idea what you're talking about.
21:47
@Alizter: I was reasoning it out that the problem should be symmetric in the passengers/seats. It should be the same for the first as for the last.
I've gotta stop working on math for the day, there's too much stuff that I didn't know until now and it's overwhelming
@MikeMiller As I understand homomorphic classes or whatever they are called can be morphed into eachother but classes are sepereated by holes and sticky-outy-bits. But in homotopy classes they are just holes.
oy @Alizter
I don't understand what you're saying, but I suspect you don't either.
Probably
21:49
Go back to understanding the airplane problem :P
@TedShifrin I get it now :P
A and R are part of the same homeomorphism class
so are O and D
Then figure out the no-one-gets-the-right-seat version, @Alizter.
@TedShifrin ok, let me incoherently say stuff first
Depends on how you write your R's, but yes.
@Ted Do you think I should shelve my work in analysis until I take a college course to understand the material better? :/
21:51
But then if considering homotopy classes
R and O are in the same class
@teadawg: I think you should know fundamentals before you specialize in arcane things.
As far as I understand misunderstand
To be working on these polylogs and not know what I told you is criminal, IMHO.
@TedShifrin I thought polylog was something WolframAlpha made up
21:52
;)
they're cousins of polliwogs, @Alizter
You're correct, @Alizter, those are homotopy equivalent.
They could be made up as far as I know, since I did learn about polylogs from WolframAlpha...
But you should try to understand the definition - and use that to see why - and not just the intuiting idea that you can collapse lines to points.
And I'm telling you about monologs, @teadawg.
21:53
Mathematics isn't made up, it's discovered. Except polylogs, those are made up.
5
Draw the picture from the integral test and work out why $\lim\limits_{n\to\infty}\sum_{k=1}^n \dfrac 1k - \log(n) \to \gamma$.
are you gone
@iwriteonpeoplewhowriteonbanana
His skeleton is still here, @Ramanewb.
@ted hum... I would prefer him alive, I don't like speaking to a skeleton !
Sometimes it's better, @Ramanewb. Skeleta don't talk back as much as live ones.
21:55
I just feel like such a fool for jumping into something without knowing the underlying fundamentals beforehand...
3
Q: Execution time of function

evindaThe following pseudocodes are given. quicksort(A,p,r) if p<r then q<-partition(A,p,r) quicksort(A,p,q-1) quicksort(A,q+1,r) partition(A,p,r){ x<-A[r] i<-p-1 for j<-p to r-1 if A[j]<=x then i<-i+1 swap(A[i],A[j]) swap(A[i+1],A[r]) return i+1 Wh...

Could anyone help me?
@ted but you can't get cleverer talking to a skeleton
I'm not belittling you, @teadawg. I'm giving you (perhaps unsolicited, but then solicited) advice.
You can't get cleverer talking to Hippa, either, @Ramanewb. :D
@ted talking to him alive, yes. Not to his skeleton !
no, @Ramanewb, talking to him alive just makes everyone exasperated.
21:56
@ted you're not wanna tell me what he's done to you uh ?
Not I :P
@TedShifrin I mean, it explains why I've been having so much trouble... Now I just don't know what the fundamentals are that I should learn before diving back in to where I was :/
@ted "when I get dementia I may forget all the evil things you've done to me, Hippa, but not yet !"
It takes time, @teadawg. Just chill and learn slowly.
You found that on a crypt wall, @Ramanewb?
@Ramanewbie lol....pure gold
21:58
OK, it's time for a nap.
No, it's time for dinner soon, @Mike.
When is your talk?
I have to clean my room tonight before the prospectives see it... :P
it's purely real, he send that @iwriteonbananas
@teadawg1337 dont listen to ted, study all day until the wheels fall off
oh god ... don't let them near all that infestation.
21:59
Friday. Ciprian wants me to make some modifications but I'll be done with those tonight.
@Ramanewbie oh i believe you
Only one is nontrivial to do.
I still remember my first grad school seminar talk, @Mike. Make us proud.
@ted absolutely not, you posted that a few month ago (the message was even starred)

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