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5:12 PM
@user2597879: I hope that the second answer clarified things. If not, let me know.
 
5:27 PM
@Deathslice Assume there is an iso from your 2nd graph to your 1st. If you look at your 2nd graph and consider the 4 vertices in the center, the only subgraphs of the first graph that are isomorphic to these are the "faces" of the rectangular prism in the first graph. After deciding on how the 4 vertices in the center map to a face, the image of the other 4 vertices is determined. But now these vertices are connected by edges in one graph but not the other, so no such iso could exist.
 
Greetings. guys.
Just curious.. is it correct.
 
@RandomVariable No, I don't see it. I guess $\sum \frac{1}{(z + k\pi)^2}$ gives you something related to the trigamma function. (Was that the second derivative of $\log \Gamma$? I never can remember things I don't use.)
 
Yea, it's correct @Rafee. Did you get that done?
It looks painful. Soo painful.
$e^{i\pi} = \cos \pi + i\sin \pi = -1$
 
Huy
5:43 PM
@Khallil: I'm on Windows 10! Heureka!
 
It only took the best part of half a day! :-P
 
Huy
Yes! At least now I can study peacefully tomorrow and don't have any more distractions. :D
 
@Khallil no.. read it on Mashable.com
 
Why is $\int_0^\infty \frac{x^3}{\exp(x)-1} \, \mathrm{d} x = \frac{\pi^4}{15}$
 
That's pretty revolting in my opinion, @Rafee.
 
5:49 PM
@Khallil Do you know, its uses.. in any application or theory..
In mathematics, Euler's identity (also known as Euler's equation) is the equality where e is Euler's number, the base of natural logarithms, i is the imaginary unit, which satisfies i2 = −1, and π is pi, the ratio of the circumference of a circle to its diameter. Euler's identity is named after the Swiss mathematician Leonhard Euler. It is considered an example of mathematical beauty. == Explanation == Euler's identity is a special case of Euler's formula from complex analysis, which states that for any real number x, where the values of the trigonometric functions sine and cosine are given...
 
You can use it to integrate powers of sines or cosines by expressing them as the sums and differences of exponentials, @Rafee.
 
(Already found it on the home page!)
 
:23092301 impressive
 
okay.. thanks man
 
5:55 PM
(Since the identity is the formula evaluated at $\pi$)
 
oh!, thanks..
I am not that Math-Geeky guys like you..
I was curious, so i posted it here.
:-)
I just wanted to know.. how?
 
Hi @Emrakul! :-)
 
user61230
'ello!
 
@DanielFischer Yes, that's the trigamma function. There is obviously some manipulation going on that I don't understand. It can't possibly be as difficult as it looks. But at least I figured out a simple way to derive it.
 
6:12 PM
@RandomVariable Okay, so you have the partial fraction decomposition $\sum \frac{1}{(z+k)^2}$ if I didn't make any mistake (haven't done it on paper, so caveat lector).
Thus using the periodicity of $\tanh$, we can write the sum of residues of $\frac{\tanh^{2m} z}{z^2}$ as the residue of $\frac{1}{\pi^2} \tanh^{2m} z \psi_1(z/\pi)$ at $\pi i/2$ or so. Now a change of variables $z = w+\pi i/2$ transforms the $\tanh$ into $\coth$.
Looks promising, check the details (signs, constant factors, all that jazz).
 
6:42 PM
@DominicMichaelis It is $\zeta(4)\Gamma(4)$... $\zeta(4)=\frac{\pi^4}{90}$ and $\Gamma(4)=6$
 
@robjohn thanks, didn't see the right way for the geometric series
 
$$\int_0^\infty\frac{x^{s-1}}{e^x-1}\,\mathrm{d}x=\zeta(s)\Gamma(s)$$
 
6:53 PM
@DanielFischer Sorry. I stepped out for a minute to water my garden. Why are you looking at the sum of the residues of $\frac{\tanh 2mz}{z^{2}}$ in the upper half-plane? Is that a typo?
 
@RandomVariable We're integrating over a rectangle $-R,R, R+i\pi N, -R+i\pi N$, don't we? Let $N\to \infty$ and you get the sum of residues in the upper half-plane.
 
@DanielFischer I meant why $\frac{\tanh 2mz}{z^{2}}$ as opposed to $ \frac{\tanh^{2m}(z)}{z^{2}}$.
 
@RandomVariable Oh, because typo. Fixed it.
 
@DanielFischer Do you have the power to edit any of your posts on here no matter how old they are?
 
@RandomVariable There's got to be some compensation for the dirty work mods do ;)
 
Huy
7:55 PM
I missed some MSE chat drama. :(
 
@Huy Whole world's a drama.
I guess the statement above is pretty dramatic in itself :P
 
8:10 PM
I love the starring system. People can anonymously bitch about each other.
 
That comment reeks of fishing for stars ...
So does that one ^
So does that ^
...
:-P
 
Khallilbrot?
2
 
$k_{n+1} = k_n^2 + c$
It looks so cool. Kinda like a Moroccan/Indian design!
 
Huy
8:39 PM
@Khallil: Did you plot that?
 
Good ol' Wikipedia!
 
Huy
@Khallil: I plotted the Mandelbrot set as a project in high school. I failed at programming a proper zoom function. :(
I should try again at some point.
 
Which program did you use, @Huy?
 
Huy
@Khallil: I used Python to write a program.
 
Python's tough!
 
8:50 PM
It is?
 
At least that's what I've heard :-P
 
From whom?
 
Never actually tried it myself!
 
Are these people allowed outside visitors?
 
Friends from other universities.
 
8:51 PM
They're idiots. No one (who isn't an idiot) has said that about Python. Google it.
Python is taught to children in my country (UK) as young as 10
 
Ok ... >_>
 
Huy
Python is probably the easiest programming language to start with.
So for once I have to agree with Alec.
 
I'm rarely wrong.
 
Huy
Good one.
 
Find something I've been wrong about
 
8:56 PM
I can come up with one.
 
[how did you not get that?]
 
huh?
I'm referring to the time you said $\Bbb R^3 - \{(0, 0, 0)\}$ is not simply connected.
 
Did I assert or claim that?
 
I don't recall.
 
It's not something I am confident with so I doubt I claimed it.
 
crl
9:06 PM
Does the series: 1, 11, 21, 1211, 111221, ... have a name?
 
Is there supposed to be a double 2 there?
 
No. That's one thing you're wrong about.
In mathematics, the look-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ... (sequence A005150 in OEIS). To generate a member of the sequence from the previous member, read off the digits of the previous member, counting the number of digits in groups of the same digit. For example: 1 is read off as "one 1" or 11. 11 is read off as "two 1s" or 21. 21 is read off as "one 2, then one 1" or 1211. 1211 is read off as "one 1, then one 2, then two 1s" or 111221. 111221 is read off as "three 1s, then two 2s, then one 1" or 312211...
 
crl
@Khallil Oh nice
 
13 mins ago, by Alec Teal
Find something I've been wrong about
 
@Khallil there's a question mark there.
 
9:08 PM
lol
 
:-P
 
crl
I was trying to see next elements jsbin.com/paxaxo/edit?js,output, it is possible to prove that its only elements are 1, 2, and 3? (that it never has any 4 or higher digits)
 
@Khallil languages are not your thing are they!
 
I'm sorry. I don't understand what you mean.
 
But you do have great Dad-joke skills.
 
crl
9:10 PM
> No digits other than 1, 2, and 3 appear in the sequence, unless the seed number contains such a digit or a run of more than three of the same digit.[3]
 
Sorry took me a while to find picture
I found a picture of your mother @Khallil
 
You made me laugh out loud, @AlecTeal!
I haven't laughed like that in a while!!
 
I've now found the image I wanted, of you:
See, it's to do with whatever that Asian-cartoon thing you linked me
 
tips fedora
 
You have no idea how bad that is
imgur.com/a/WueD3?gallery <--- now I get why it's the symbol of tards.
 
9:20 PM
It's called anime by the way, @AlecTeal!
25 mins ago, by Alec Teal
Find something I've been wrong about
 
I'm sure on some planet that your style is very impressive @Khallil
 
(I don't appreciate you posting pictures of me on here.)
 
Your weak link is: this is Earth.
 
I merely jest, @AlecTeal.
 
But yeah moving on @Khallil is your next come back gonna be "rubber glue back to you" or something?
 
9:23 PM
Don't take it personally!
You've completely seen through me @AlecTeal.
 
As I said, rarely wrong.
 
How ever do you do it?
Ok, this is scary. Are you in my house or something?
 
*douses @Khallil in $\text{CO}_2$
 
Whom should I ask to re open my question? I've already edited it but I seems like nobody has seen it...
 
Link me @francolino
 
9:30 PM
-1
Q: Prove $ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2} = 2 $

francolino Prove by definition that: $$ \lim_{(x,y)\to(1,1)} \frac {x^2+2xy-3y^2}{x^2-y^2} = 2 $$ I've been reading quite a lot about how prove this limit, so I want to show what I've done so far so you can tell me any suggestions (tricks) and even point out any mistake. What I've tried: I want to f...

 
I've flagged it for review @francolino but I can't vote to open
 
Hi guys, there seem to be a number of flags being generated.
 
Thank you a lot Alec. :)
 
Any chance you could avoid doxing each other and p*ssing each other off, that would be just lovely.
4
 
Sure.
 
9:34 PM
If there are any for me, I apologise @Richard. I've been messing with @AlecTeal.
 
@francolino don't take it personally, my most viewed and highest rated question of all time is a repost of something that was closed as off-topic.
8
Q: How to fix: [program name] /usr/lib/x86_64-linux-gnu/libstdc++.so.6: version CXXABI_1.3.8' not found (required by [program name])

Alec TealI recently asked for help No idea as to: `/usr/lib/x86_64-linux-gnu/libstdc++.so.6: version `CXXABI_1.3.8' not found` and it got downvoted and closed (I don't know why) It turns out that "make install" - the make target that installs and implies the target "install-target-libstdc++v3" doesn't ac...

 
@francolino Couldn't you just note the following? $$ \dfrac{x^2+2xy-3y^2}{x^2-y^2} = 1 + \dfrac{2xy - 2y^2}{x^2-y^2} = 1 + \dfrac{2y(x-y)}{x^2-y^2} = 1 + \dfrac{2y}{x+y} $$
 
@Richard I'm sorry also. That one I claimed was Kahl's brother was really creepy.
 
@AlecTeal I don't do it because of the reputations or the views, I just wanted to know if it's right. :P
 
@francolino you've got an answer there, do you have a problem interpreting it?
 
9:37 PM
No worries. Play nice and Mjolnir will stay its fitful slumber
 
@francolino the point was more of "idiots vote to close a legitimate question sometimes"
 
user image
3
 
So everyone, that's what we call "overcompensating"
 
Richard's making a point, because the word "idiots" tends to set off alarm bells.
 
@HDE226868 did you read what started this?
An idiot claimed Python was a tough language!
 
9:39 PM
@AlecTeal Yes. Multiple times.
 
Do you know what Python is?
 
Ridiculous. Not worth my time.
 
WB @BalarkaSen
 
@AlecTeal Yes. I'm practically illiterate when it comes to coding, but I know what Python is and its difficulty level.
 
Uhuh....
 
9:45 PM
yo
 
It doesn't give cause to get into an argument.
 
What argument?
We both know that wasn't really his mother. That was the joke. It's a tomb raider reference
 
@AlecTeal What happened before.
 
@AlecTeal Look, messages were flagged multiple times today, and I (and another mod or two, e.g. @Richard) am/are just trying to defuse the situation. It doesn't matter who started it at this point, just don't call people "idiots". Okay?
 
9:47 PM
What words can I use? "Brat" got removed earlier and that's REALLY mild.
 
I don't see why you feel the need to insult people, @AlecTeal.
 
@AlecTeal How about not using derogatory words?
 
@AlecTeal Don't use that, either. Keep it civil. Keep to the Be Nice Policy.
 
(By the way, I wasn't arguing. Merely accepting your jokes with good faith that they weren't malicious, which I'm sure they weren't!)
 
@Khallil \come closer you lovely aesthetically pleasing person.
 
9:50 PM
@AlecTeal hi pal
 
No need to go overboard in the opposite direction, @AlecTeal!
 
You're a very smart person @Khallil - I think you have a future in cryptography.
 
Okay bye!
Hope you feel better soon.
 
(Sorry if I annoyed anybody!)
(Apologies, @AlecTeal.)
 
9:55 PM
@AlecTeal I understand all of the replies, but I want to you the equality by defnition of limit. That's all.
 
What using $\epsilon-\delta$ (which is what I thought your question was)
 
0
Q: About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $

Chris's sis the artistHere is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known nice real analysis tools for calculating them. At some point I'll add my solution. It's a question for the informative purpose rather than finding solutions, the solution ...

@DanielFischer have you ever met integrals like the one I posted? I'm curious if it's something completely new on market.
 
You want to ask a question like "How do I show [equation] using the definition [continuity here]"
 
@AlecTeal Exactly, I wanted to show using de epsilon-delta.
 
Ask a new question, but include the definition of continuity and say "I want to explicitly show this"
 
9:58 PM
@Chris'ssistheartist I'm the wrong person to ask that, I'm not much into integrals.
 
@AlecTeal I'll remember it, but I think what I'm asking (this time) is clear enough. Or should I edit the post and say that I wanted to show by definition of contuinity?
 
@DanielFischer Well, you have on main a lot of answers with integrals, I don't think you're a wrong person. It matters for me if you met anything like that.
 
New question @francolino - make it clear what metric you use (I think you're using $d((x,y),(x',y'))=\sqrt{(x-x')^2+(y-y')^2}$
 
@AlecTeal Okay, I'll write a new question, but the function isn't continues at (1,1). What should I ask then?
 
It was a typo @francolino put the definition of a limit there.
 
10:02 PM
@Chris'ssistheartist But I do almost exclusively residue theorem when dealing with integrals, hardly ever real analysis.
 
See the second form of "Converging to a" here maths.kisogo.com/index.php?title=Limit it should look like that but not for sequences!
 
@DanielFischer This is less iimportant, the important thing to me is if you met anything like that. In case you recall something like that, let me know. Thanks.
 
@DanielFischer: Do you know a good source for the basic Fredholm theory? (In particular, the analytic Fredholm theorem and its consequences.)
 
@AlecTeal why do you have to act like such a tough guy in this room?
 
@AlecTeal Yes, I know the definition of limit, that's why I wrote "I want to find δ for every ε that verifies:". It's unfair that I can't re open it. u.u
 
10:06 PM
Never use that emote again.
@skillpatrol what?
 
@MikeMiller Nope, sorry. Nothing beyond what Rudin has in his FA.
 
I doubt it makes me nicer.
 
@DanielFischer What tag should I use when I ask for some general information about an integral rather than asking for solutions?
 
@AlecTeal TMI
 
@RandomVariable pls let me know if you met anything similar to the integral in the question here math.stackexchange.com/questions/1378536/…
 
10:07 PM
@skillpatrol FU
 
@AlecTeal OK
 
I had a torsion, very serious!
 
@Chris'ssistheartist Ooof. Apart from integration, real-analysis, and so on - which still apply - possibly .
 
@AlecTeal get some rest pal
 
10:09 PM
I was thinking "Take some codeine"
 
@DanielFischer Great. Thanks.
 
@AlecTeal do it
then sleep
 
@DanielFischer: Fair enough. I suppose the main applications are toward an area you've expressed a lack of interest in before... :)
 
@MikeMiller True.
 
@Chris'ssistheartist I'm not sure if I have.
 
10:16 PM
@DanielFischer do mods have the power to delete rooms?
 
@RandomVariable OK, thanks. If you recall anything similar to it, let me know. I'm just curious to know if this stuff is anything new.
 
@skillpatrol Looks like we do. I haven't tried it yet.
 
@DanielFischer thanks for checking :)
 
@skillpatrol - Yes
 
Is main down? I see questions from 3 hours ago on the front page.
 
10:20 PM
@Richard thanks for the confirmation pal
 
@robjohn Seems fine.
 
@robjohn not for me
 
@robjohn Glad to see you again!!! :-) I posted on main an integral that is very important to me. Have you ever met anything like that? math.stackexchange.com/questions/1378536/…
 
@robjohn It's up for me.
 
@DanielFischer There, now it refreshed again and it shows new questions.
 
10:23 PM
@robjohn Probably a websocket took a break.
 
Perhaps it was a ISP problem.
 
@skillpatrol - I heard that if you press F5, it makes the bad questions fly awaaaaaay....
 
@Richard :D
Hi @KevinDriscoll loooong time no see pal
 
11:17 PM
OK, I have a technical (and not very important) question for you all that's been bugging me for a bit now...
Regarding series, the divergence test (sometimes $n$th term test) says that if the terms of the sequence do not approach 0 (as $n \to \infty$), then the series diverges.
So, does a series whose terms don't go to zero pass the test (I mean, it satisfies the "if" part), or does it fail the test (because it diverges, and nobody really cares about a divergent series)?
 
I think probably most people would say it fails the test.
 
I second that. We typically call the family of tests "convergence tests," which seems to imply failure == divergence.
If I were concerned about ambiguity, though, I'd eschew the words "pass" and "fail" altogether in favor of "this series is divergent by the nth term test."
 
11:37 PM
@David the problem with your suggestion is that I like to highlight "pass" and "fail" in all the tests because it helps them understand.
@David But in the end, I will have to be unambiguous.
 
11:55 PM
The main difficulty that I've seen students have is that they think all these tests are logically equivalent while some of them are "converges if" statements and some are "converges only if" or "converges if and only if". I wish some basic deductive reasoning was taught at the primary or secondary level in the US.
 
@PVAL: Do you know a resource for the basic Fredholm theory, incl. the analytic Fredholm theorem?
 
@PVAL Agreed
 

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