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12:04 AM
@robjohn this question is basically a duplicate of my recent MO question, even ignoring that he quoted my final paragraph verbatim. is that grounds for closure, even though MO is a different site?
 
@MikeMiller it is grounds for suspension for plagiarism...
 
No body answered my question in Computer Since. Can I copy it to here?
2
Q: How to reduce the cost of search based on previous BFS?

Ilya_GazmanI got an unweighted, undirected graph, with $N$ vertices, where each vertex has degree $K$. In my case its a grid with dynamic obstacles. My goal is to output a map, based on given location on the grid, with the shortest path to all the cells on the grid, some thing like this: 4 3 2 3 3 2 1 2 ...

 
@robjohn actually, everything that user does seems bizarre. See for instance the comments on his only answer.
 
@Ilya_Gazman You should see if the mods there will migrate the question here. If not, and you still want to ask the question here, ask it here with links between the question here and the original on CS.
@MikeMiller at least he gave proper attribution there.
 
@robjohn Does this question fits here?
 
12:09 AM
@Ilya_Gazman It seems better suited to CS. It does run the risk of being closed as off-topic.
 
tnx
 
@robjohn I left a comment
the first half of his question is a the first paragraph of a wikipedia article
 
@MikeMiller Oh, that is bad, too.
 
@robjohn I think he just doesn't have much of a grasp on the english language
 
@MikeMiller that seems clear from his reply to your comment
 
12:17 AM
certainly I don't think any of his posts have been malicious, probably rather it's a misunderstanding of how stackexchange works
anyway I'm stepping out of the fray, House isn't going to watch itself
 
@MikeMiller have fun
 
 
4 hours later…
leo
4:02 AM
I remember some post about the irrationality of radical roots of rationals, does someone have some pointers?
 
4:19 AM
I have what's maybe a dumb question: In math.stackexchange.com/questions/268635/… Why is the integral shown of (1/2-x) equal to the integral of x?
 
4:32 AM
@user939259 they each calculate the area of a triangle, only one is the mirror image of the other
 
ah thanks
 
of course, you could always verify the old-fashion way (actually compute both and see they are equal), but that wouldn't explain why the equality is written there
 
Yeah I verified that they were equal, but the way it was written made it seem like there was some obvious identity I should have known
 
Greetings
@robjohn Interesting.
 
@Chris'ssis I am writing up something with a bit more detail, but it is not much longer.
 
4:36 AM
@robjohn OK
 
5:26 AM
@AlexanderGruber love the new hat
 
@MikeMiller thanks. I'm a wizard.
 
The uncertainty in probability can kill me..
 
@AlexanderGruber you should edit your ufl picture like that.
 
@MikeMiller I grew a weird old man moustache for my UF ID picture
 
5:47 AM
hahahaha
i wish i could do cool pranks like that.
 
@r9m I also found huge improvements to my way.
 
6:10 AM
@robjohn can you correctly plot it with Mma? It shows me nothing for the negative values of $x$. This one -> (E^x x^2 PolyLog[2, E^(-x)])/(-1 + E^x)^2
 
6:30 AM
If I have a 3 vectors in R3 and I want to check if they are an orthonormal base I first need to see if they a base, then orthogonal base and then check if they can be orthonormal base?
@Studentmath Oi! I'm worried about not succeeding in the test :(
 
@Ilan it's normal, but you'll see, if you study hard enough all will be fine. I am feeling pukish just imagining my first test is in like two weeks.
And correct, I think..
If they are orthogonal, you can turn them to orthonormal with a simple process
 
6:45 AM
badung adung
 
7:20 AM
@skullpatrol I was called again and suggested to rethink the salary range ... (they said what I ask do not fit what they can offer)
(unbelievable)
 
@chris's sis it has then become a matter of price then, right?
 
@skullpatrol To tell the truth, I failed the last few interviews because of that ... (yeah, a matter of price)
 
Oh, I see.
If you like the work, give it a try.
With the added experience you can ask for more $ latter :-)
 
I couldn't do something I don't like. Yeah, I'm a good specialist in what I do.
@skullpatrol Well, it's not a matter of experience since on the last job I was considered the best in EU for a certain position. So, I go there with a lot of experience.
When I went there I knew nothing, and in 2 years not only that I learned all I had to learn, but I also developed a lot some special techniques in manufacturing and had great result. I cannot sell myself for nothing. :-)
(I learned alone all since no one there knew how to approach that area)
Anyway.
Here is another cute question. Evaluate
$$\int_{-\infty}^{\infty} \frac{x^2 e^{x} \operatorname{Li}_2(e^{-x})}{(e^x-1)^2} \ dx=\zeta(4) i^2\space (i^2=-1)$$
I have the meal now and then I write up the proof for this one that is awesome.
hehe, I have 2 approaching ways, but one is particularly nice. I love this stuff.
brb
 
7:50 AM
Wait ... it's $$\int_{-\infty}^{\infty} \frac{x^2 e^{x} \operatorname{Li}_2(e^{-x})}{(e^x-1)^2} \ dx=2\zeta(4) i^2\space (i^2=-1)$$
 
8:00 AM
@AlexanderGruber Pfffft
This is unbelievable. You're a mathematician, @Alexander!
 
Where does the definition $$\sum_{y:x \le y \le z}\mu(x,y)=\delta(x,z)$$ intuitively come from (where $x,y,z\in\text{a Poset}$)? I don't see how it's related to the usual $\mu(n)$.
 
@Alyosha What's $\mu(x, y)$?
 
The generalised mobius function.
 
@Alyosha Can you give a short explanation of what it is? My net is so slow today that I can't even open up PDFs.
Could it be that it is defined that way?
 
http://i.gyazo.com/19c770a401dc0f87978ecb0c089a7a8d.png

Can you load this?
 
8:09 AM
Ah, so it's defined that way.
Well, it makes sense, as $$\sum_{n|x} \mu(n) = \delta(x)$$
 
Ah, OK.
Do you have an idea for the general proof of the theorem?
 
Let me see.
I am not comfortable much with ordered stuffs, but isn't that just a consequence of sum manipulation?
 
I think it must be, I'll have a go later today (after a physics exam).
Also, Balarka, do you know whether Lang is a good place to learn Galois for the first time?
 
8:56 AM
\o
 
9:16 AM
There are some probability questions you can right away feel what you have to do, and some you can just stare at them and feel clueless
 
r9m
@Chris'ssis :D (y)
 
@r9m See my last integral above. I'm preparing the proof right now.
 
r9m
@Chris'ssis ya looks nice :-) .. I'll try it too :)
 
Glad you like it.
@r9m Think of a brilliant way (of course).
 
r9m
9:33 AM
@Chris'ssis lemme se if I can manage it some way ... brilliant way .. I have to think about that .. :|
 
@Chris'ssis PolyLog[2,x] is not real for x>1.
 
@Chris'ssis I have written everything up in MathJax, but it is too long and uses linebreak arguments to even up the line spacing for chat. I can post it as an answer, or I guess generate it as a PDF or PNG
 
@robjohn That's sure. I was deceived by its form.
@robjohn Well, I don't want to post that series on main. It's better to generate a PDF I think. :-)
@robjohn Now I also have 2 solutions to it.
 
@Chris'ssis what is wrong with putting it on main?
 
9:40 AM
@robjohn Well, I want to publish it first. After that, I can post it.
@robjohn And for that double series I have now $4$ solutions. (my second solution to that question avoids the use of that double series)
 
@robjohn, how does $$ \sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} = \sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k} $$ hold?
I can't understand how k goes from 0 to m+n in LHS?
 
hey there. Sorry for chiming in with off topic stuff. I need a small answer to a pronunciation question. How would you say $\mathring x$? I mean that in the way as $x'$ is usually read as "x prime". For context: it is the base point of a piecewise linearisation or the midpoint in an integration scheme.
 
If it a coincidence that the index of the vector field produced by a dipole and the index of a point charge is equal to $n$, where both fields behave $\sim \frac{k}{r^{n+1}}$ (i.e. $n=2$ for dipoles, $n=1$ for a point charge)?
 
@Sush Just think of the LHS as the sum of $\binom{n}{k}\binom{m}{r-k}x^r$ over all $r$ and $k$. The binomial coefficients are non-zero when $0\le k\le n$ and $0\le r-k\le m$.
@Sush And think of the RHS as the sum of $\binom{n}{k}\binom{m}{r}x^{r+k}$ over all $r$ and $k$. The binomial coefficients are non-zero when $0\le k\le n$ and $0\le r\le m$.
then it is just a change of indices, as long as all the non-zero terms are included
 
That is, if we have a vector field $V$ due to a collection of charges and the sum of its indices is $n$ (not counting the index at infinity is that's nonzero), must the field fall off asymptotically as $r^{-1-n}$?
 
I think I have it right finally, but the solution seems so 'easy to reach' it just feels wrong.. hate it.
 
@Alyosha actually the dipole is acting as the difference of a positive and negative monopole separated by a small distance. In effect, this gives the resultant field to be the derivative of the monopole field, and the derivative of $k/r^n$ is $-nk/r^{n+1}$
 
10:20 AM
@robjohn Nice approach (but a bit tough :D)
 
@Chris'ssis I just applied the same methods that I've used in other Euler Sum problems.
None of the first three formulas is new. I've used them before. The partial fractions was the critical step.
 
@robjohn Indeed. Those partial fractions are the key.
 
If I have 5 lads in a competition, and the hosts asks N questions one after the other, E[N]=15, Var(n)=2. Every time the hosts asks a question, each kids answer it, each answer, question, and everything else is independent of the other. The probability that a lad answers correctly is 0.4. We mark by X the number of questions where at least one lad answered them correctly, and want to find E[X], Var(X).
 
@robjohn thanks for the proof. I think this a pretty hard question. I doubt it might be given on a contest (like Putnam).
 
So.. if I place indicator $Y_i$ which is 1 if at least one of the lads answered correctly, 0 otherwise, Than E[X]=E[Y]*E[N], and E[Y] would simply be $P(Y=1)=1-(0.6)^5$, no..?
 
10:27 AM
@Chris'ssis I agree that it is difficult. The expansion into partial fractions itself is cumbersome for a contest.
 
@robjohn But things will never be the same from now on since we've gained a lot of experience with this one. :-)
 
Gah, it just feels wrong, I don't know how to even check myself.
 
@robjohn That's why a question like this one is very precious - one may learn a lot of things working on it. :-)
 
@Chris'ssis I find it amazing that what you got as $2\zeta(4)$ I got as $\frac23\zeta(2)^2+\frac13\zeta(4)$
 
@robjohn hehe :-)
 
10:32 AM
both are $\pi^4/45$
 
@robjohn Indeed.
@r9m are you done? :D
@robjohn have you seen that integral above?
 
@Chris'ssis this integral?
 
@robjohn Yes.
 
@robjohn how are you with E(expectation) in probability?
I would guess this question is extremely simple, I just fear my approach is extremely off, it feels too simple
 
11:37 AM
@robjohn did you manage to look over my whole solution to that series question?
 
@Alyosha I never tried Lang, but I am a bit partial to modern textbooks.
What are your backgrounds on abstract algebra? @Alyosha
 
12:41 PM
@Chris'ssis I've gotten through a few pages. Without motivation, it is a hard read. There are derivations of formulas that I know will be used later, but they are a page or two in length of pretty serious integration. There are places that I can see simplifications using partial fractions, too.
 
OK ..
 
is the opposite of a serious integration a funny derivative :-)
 
@skullpatrol Maybe .. :-)
 
I've just finished the proof to $$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n^2} \left(1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\right)$$
Now I wanna try my luck with $$\sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n^3} \left(1+\frac{1}{2^3}+\cdots+\frac{1}{n^3}\right)$$
 
1:05 PM
@robjohn maybe this one is not hard to read
 
@Chris'ssis: I have updated the PDF (at the same URL) using $\zeta(2)^2=\frac52\zeta(4)$ to get the same result as you did.
 
@robjohn Well, this is less important since the work is correct. :-)
 
:16024966 I believe I computed that same sum around the beginning of the year...
 
@robjohn yeah, I think so ... (I did many things in the meantime)
brb
 
:16024966
$$
\begin{align}
\sum_{k,n=1}^\infty\frac1{n^2k(n+k)^2}
&=\frac12\sum_{k,n=1}^\infty\left(\frac1{n^2k(n+k)^2}+\frac1{nk^2(n+k)^2}\right)\\
&=\frac12\sum_{k,n=1}^\infty\frac1{n^2k^2(n+k)}\\
&=\frac12\sum_{k,n=1}^\infty\left(\frac1{n^3k^2}-\frac1{n^3k(n+k)}\right)\\
&=\frac12\zeta(2)\zeta(3)-\frac12\sum_{k,n=1}^\infty\frac1{n^4}\left(\frac1k-\frac1{n+k}\right)\\
&=\frac12\zeta(2)\zeta(3)-\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}
\end{align}
$$
It was that one, then I used this answer to finish it off.
 
1:18 PM
@robjohn This is the natural way to tackle it. In that proof you saw, in the last 4 pages there is a solution I did in the past by integration. Of course, I can come up with better things (what I already did).
It's important to know how to nicely tackle these series.
 
 
1 hour later…
2:28 PM
"Look, kids : Zombies!" "Ewww, Zombies" throws a tomato "Take that, filthy Zombies"
Now, now, none of us would want to be treated like Zoo animals, would we?
 
2:57 PM
@robjohn let me know if you find a way to evaluate that integral (it's very nice)
 
3:11 PM
Guyz im in exm nd relly ned hlp, wht teh area of circle?
 
-1/12
2
 
@Studentmath haha
 
@Studentmath Wait I thought 1+2+3+4+... = -1/12
 
@MikeCon94 indeed. that's also the area of a circle, a sphere and a torus.
 
Always here to help
 
3:26 PM
Ah great thanks guyz
Could you explain how it is the area of a circle?
 
@MikeCon94 sure. circumscribe a square of area 1 inside a circle and then add off other squares in the sides. keep sprouting these till you get 1 + 2 + 3 + 4 + ...
which is -1/12
 
Just don't forget to adjust units to fit.
Bal, Probability is stabbing me to death
 
this method is known as "balarka-studentmath ascent". named after an indian and an isreali quantum physicists @MikeCon94
 
Haha
 
@AJHenderson Not by any chance another flag?
 
3:31 PM
nah, investigating something, but not that originated in here
 
Phew.
 
This one is interesting. Throwing a dice 60 times, X is the number of even result's outcomes, Y the number of odd result's outcome. Find $E[X^2]$ (easy with Momentum generating over bionmal RV), and $Var(3X-2Y)$ <-- No idea.
 
@BalarkaSen I know group theory but no ring theory.
Did you count Lang as modern?
 
3:58 PM
I never know if what I do is right or wrong in probability. I lack any confidence, and for a good reason. No idea what to do.
 
@MikeCon94 I don't know how things are there, but at UCLA, you are not supposed to access the internet during a test. I am guessing that your instructor would not want you getting help on your exam.
 
Just answered a lhf.
 
4:16 PM
Gah, I am doing something wrong and I can't figure out what.
 
@Studentmath That means it is time to go to bed, lol.
 
It's 19:17 here :/
Plus I always do something wrong and can't figure out what, ain't a symptom for me.
I think I might be dumb after all
I can't figure out how to do that $Var$. Trying to split it up and use some addition in Variance rules gets me to horrid things, as X,Y are dependent on each other.
 
@robjohn you've been trolled I think
 
@G.T.R why do you say that?
 
Yeah, I doubt the lad has been serious.
 
4:32 PM
@robjohn that's a lot to be typed during an exam chat.stackexchange.com/users/119371/mikecon94?tab=recent
 
1007,10017,100117,1001117,... what is general term of this sequence?
 
@cort 1 is inserted each time?
 
@G.T.R Yes
 
Wow, 2 black squares and 1 blue square.
 
Let party start ^_^
 
4:36 PM
@Cortizol $n\geq 1, U_n = 10^{n+2}+7+\sum\limits_{k=1}^{n-1}10^{k}$ ?
 
@Hippalectryon That is right, but how to express $U_{n+1}$ in terms of $U_n$?
 
@Cortizol $U_{n+1} = 10 U_n - 53$
 
I hope i did it right -_- so messy
 
@DanielFischer Thank you
 
Oh no it's wrong
:C
 
4:46 PM
Linear transformation$ T : R^9 /rightarrow R_[x] $ is not isomorphism when it is the zero transformation. right?
 
@DanielFischer I would like to ask you a silly question. How long will you take to read Rudin's PMA if you had nothing else to do and it is the first time you are seeing the material? Would one month suffice, or two, or more?
 
Anyone?
 
What's $R^0$ for you?
Oh wait it's nine.
 
hello
 
I know what you will say next, lol.
 
5:00 PM
@robjohn can you help me with continuity please or someone else
 
@IlanAizelmanWS An isomorphism is also a bijection, so...
 
@Vrou ask
 
@DanielFischer how many points do you get on a per-day basis from old (i.e. >3 days) answers that people upvote?
 
@MikeMiller I get almost 0.
 
You also delete your account twice daily.
 
5:02 PM
I have some problem with the sound in Mint 17, so I am going to use Ubuntu 14.04 instead
@MikeMiller I have not deleted it for a long, long time, lol
 
@Studentmath this is my question
 
I want to read Godement's Analysis 1,2,3,4 but only the first two have been translated into English, sigh.
 
French is not hard to read
 
Still choosing between F and G to learn.
I don't think my brain can handle both
 
@Studentmath i dont understand where is used the continuity
 
5:06 PM
If I get your question, the given answer has it well
 
but where is the continuity of $u\mapsto u^{±}$
 
The continuity, if I have it right - unsure, is used at the same step as well, to be able to reach that conclusion
 
i dont understand
@Studentmath ?
 
@G.T.R T'as pas passé CCP toi non ?
 
@Studentmath are you here ?
 
5:24 PM
@Vrou I'm unsure how to explain it. Perhaps someone else could do it better.
 
ok
@MikeMiller can you help me ?
 
@Alyosha Well, you need a good deal ring theory. Skim through Dummit-Foote first.
@Alyosha I don't think Lang is modern. I much prefer Artin, though it has a bit geometric perspective. So go through D&F first and then do Artin. You also might as well look up some introductory book like Stewart's Galois theory.
 
Seems I can't escape from $E[XY]$, so question if anyone can help me: Given a fair dice is rolled 60 times, X being the number of times it landed on an even side, Y being the number of times it landed on an odd side. What would be E[XY]? Obviously both are the completing binomial RV with n=60, p=1/2, yet..
No idea how to approach it.
 
@Hippalectryon non
 
@MikeMiller I just saw an example of a special case of Eichler-Shimura. Wonderous. I always imagined what implication the modularity problem has.
 
5:36 PM
I get it should be 0, but I am unsure how to explain it.;
 
5:46 PM
@JasperLoy I don't know. And it's too late to test, I already saw the material.
@MikeMiller Hard to say. Few. Most days I get zero or one upvotes for old answers, sometimes half a dozen or so.
 
5:59 PM
@DanielFischer Me too, lol.
 
Anyone mind claryfing it up?
 
Someone ping me to test sound.
 
@JasperLoy Pong, clap.
 
@JasperLoy no
 
Oh no, the ping still sounds like a clap after I installed Ubuntu, lol.
Now we know why Windows is still in business, lol.
 
@BalarkaSen your
 
Yeah. I know.
Typos are annoying.
 
Particularly to @Jasper.
 
@Jasper @Jasper @Jasper. Did you get an applause?
 
@BalarkaSen Only one sound.
 
6:13 PM
@G.T.R Aparemment c'est possible de réussir l'X/ENS et de rater CCP :c
 
@BalarkaSen How much ring theory do I need to be comfortable with (of course there will be some that I can just learn as it is needed in Galois)?
 
@BalarkaSen Hello.
 
@Alyosha Ideals and whatnots.
 
I know of them a bit from NT.
 
Fine then.
 
6:15 PM
@Alyosha Why don't you just learn rings first, doesn't take too long.
 
@JasperLoy I don't think so.
 
They don't seem interesting unless applied to something like Galois or NT.
 
@BalarkaSen Well, I am very patient.
 
@BalarkaSen I asked you if pure mathematics had applications in the classical real-life. You accepted that pure math fields didn't. I'm not sure why you are turning this into a debate.
 
To learn interesting things, it is necessary to learn boring things first.
 
6:16 PM
@Alyosha Well, yeah, if you go to commutative algebra then you'll see that they are interesting in there own.
@JasperLoy Very true.
@ParthKohli Did I?
 
My motivation for studying anything is its closeness to real-life.
 
I mean you can just get an algebra textbook and read it from start to end, how long will that take?
 
@BalarkaSen You did.
 
@ParthKohli Didn't I give you a bunch of counterexamples?
 
@ParthKohli Alas, many of us do not study enough applied math to answer you, but math sure has many applications.
 
6:18 PM
You said something along the lines of "if you're talking about 'applications' in their classical sense, no, pure math doesn't have them."
 
@ParthKohli I don't recall.
 
@BalarkaSen I do.
 
@JasperLoy Loads.
@ParthKohli I disagree.
 
@JasperLoy You have to accept that applied mathematics has 100x applications as pure mathematics does.
 
@ParthKohli The terms pure and applied are artificial. Applied is just pure being applied.
 
6:19 PM
@ParthKohli Pure math has loads of application to applied math.
I have given you counterexamples. Stop arguing.
 
You have given counterexamples. But you are treating exceptions as rules.
 
@ParthKohli "treating exceptions as rules"?
 
@BalarkaSen Pure mathematics is fun, it's interesting, and it means everything to you. But it does not appeal to me because I don't feel connected to it.
You do know what I'm talking about, right?
 
That's fine. But it is not necessary that your feeling has anything to do with applications in real-world! =D
You do realize that, right?
 
I'm not denying the existence of applications in the real world, no - you're misunderstanding me.
Join me at the Root.
 
c c
6:25 PM
@Studentmath isn't $Var(3X-2Y)=Var(X)$ since Y = N-X
 
6:50 PM
Hey guys, I'm having some trouble with Graphics calculations, can anyone help?
The plane V passes through the origin. The vector (2,1,2) stands perpendicular to V. How can I formulate a Vector equation for the plane V?
 
@DanielFischer is it true that for differentiable functions, uniform continuity implies lipschitz?
 
@G.T.R No, for differentiable functions, Lipschitz is equivalent to a bounded derivative. Consider $x^2\sin \frac{1}{x^2}$ on $[-1,1]$. It's uniformly continuous, but not Lipschitz.
 
@G.T.R Uniform continuity of what?
 
Hola @Pedro.
 

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