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18:00
I find pleasure by working questions rather than creating. Is it just me?
@robjohn Galois theory-quinitcs-are solvable-in $\Bbb E$?
I mean are quintics solvable in $\Bbb E$?
Schanuel says no.
What else do we have?
@BalarkaSen in general, I have multiple proofs for the questions I create. (this is just because, maybe, I'm a bit lucky)
@Chris'ssis That's nice.
@BalarkaSen Indeed.
Actually, I don't think you are lucky. You are an expert mathematician, IMO.
@BalarkaSen I give you my honest word I have no background in mathematics.
18:05
I find no good reason to believe that. Even if I'd believe that, it wouldn't change my opinion that you are a mathematician, as well as an expert in the field of closed forms of I&S
@BalarkaSen :-)
@BalarkaSen not yet
@Alizter I have suggested before.
@BalarkaSen I wanna put now an interesting question related to math.stackexchange.com/questions/554003/…
Could we possibly finish this without using dilogarithm at all?
That series is a famous series that also has a name.
18:09
Sos and Cody are on I&S.
@Chris'ssis Unlikely. Well, no, I think.
Sos is a natural genius at these things.
So is Z.
@BalarkaSen My intuition tells me "definitely yes".
brb
@BalarkaSen So what exactly are you doing with sextic equations right now?
@Alizter :-)
@Alizter An extensive research on the solvability of them in the extension that contains much less double variable holomorphs, mainly.
Hello all.
18:14
But that's not all, of course.
After using cuts to create the real numbers, how do you find decimal approximations for things like the square root of two?
You are taking an abstract algebra approach?
@Alizter Not all of it, no.
Uhm, I'm not sure. Oh was that not for me?
I am talking about Galois theory.
18:15
When I get onto Galois theory I will be excited
I am learning abstract algebra at the moment
@Alizter Sure.
@Anthony Use newtons method
Is that the normal method, though?
usually
Does constructing the reals only tell you that there exists a square root of two?
18:17
@Alizter Have you considered quintics before?
@BalarkaSen Really I have never considered past cubics
@Alizter Try quartics. Fun, they are.
I am trying to learn abstract algebra and proving the unsolvability stuff.
@Alizter Where are you now? Ring theory?
@BalarkaSen Yes I am studing rings, domains and fields.
Also field extensions is what I am learning at the moment
18:19
Good. You need a clear concept of field extension to hammer at it.
@Alizter Do you mind me giving an exercise?
What topic?
@BalarkaSen I am not sure what $\mathbb{E}$ is :-)
Field extensions. Have you learned Galois extensions/automorphisms?
@BalarkaSen hehe I barely know what a homomorphism is
@robjohn The-smallest-field-of-$\Bbb C$-generated-by-$\log$-and-$\exp$. Some call it EL, which you must be more familiar with.
@Alizter Basics are rather important.
=)
18:22
Hmm when I say field extensions I mean quadratic extensions on integers because I started the chapter today :P
@Alizter You are pretty advanced for 15.
@Alizter Which book?
Anderson/Feil first course in abstract algebra
I much like D&F. Okay, I gotta go. Byes.
Funny thing today, doing some prep for maths challenge. The only question I got wrong was to do with percentages. :|
@BalarkaSen I am not sure, since I have never seen that before. Is it possible to solve quintics with trigonometric functions as it is with cubics?
18:42
Messing around with mathematica I got very excited. $\int_0^1\frac{\mathrm dx}{\Gamma(x)}\approx W(1)$
but not equal (unless you are a physicist).
@Alizter And $\gamma \sim 1/\sqrt{3}$
@N3buchadnezzar I thought $\sim$ indicated that $x\sim y \iff \lim_{x\to \infty}\frac{x}{y}\longrightarrow1$
Nah thats $\propto$, but $\sim$ has several meaning. Look it up on wiki =)
@Alizter Here en.wikipedia.org/wiki/Tilde#Mathematics. One meaning is "poorly approximates"
Well it is not that bad
Yours is bad however :p
$\log 2 \approx (2/5)^{2/5}$, is also nice alas somewhat unuseful.
@N3buchadnezzar Shhhh ;)
$(1/2)^{1/2} > \log 2 > (2/5)^{2/5}$ :p
18:58
$\displaystyle\int_0^x\frac{\mathrm dt}{\Gamma(t)}=W(1)$ Solve for $y$.
*derp. $\huge x$
@robjohn Of course not, but no proof, alas.
Shcanuel's conjecture guarantees a no, so one must rely on that only.
@BalarkaSen How do you learn so much math at such a young age?
@JasperLoy I hear it all the time, but I neither understand what 'so much' did I learn nor how 'such a young' I am.
Why not ask this to Alizter? He is young too and knows a good deal more than me.
@BalarkaSen I see, when I was 14, I didn't even know calculus.
19:13
@JasperLoy That can be understood =P.
@BalarkaSen Well, I grew up without the internet, mostly.
Yes.
@robjohn One thing can be guaranteed, though.
@BalarkaSen yes?
Quintics aren't solvable in terms of trigonometric functions of $\Bbb Q(\pi)$-rational arguments.
As in such case the quintic must be of cyclic galois group.
The part with arguments $\Bbb Q(\pi)$-algebraically independent is loads tougher.
@robjohn Actually, this is equivalent to my question, as $\sin$ and $\text{arcsin}$ are both EL.
@BalarkaSen That is why I asked.
19:23
Oh, okay.
I think this conjecture is open currently.
@BalarkaSen However, it has been a long time since I have looked at Galois Theory, so I am not a good person to ask about it.
Can any reference be found?
@robjohn This is still transcendental number theory.
@robjohn Probably in grad school right? LOL
@JasperLoy actually, the last year of undergrad.
@robjohn Oh dear! Now we know what they do in grad school, lol.
19:28
I have no idea of the courses, lol
@robjohn
$$\text{Li}_2(x)=\lim_{\epsilon\to 0}\, \frac{1}{\epsilon^2}\, \Bigg\{ {}_2F_1\left(\epsilon, \, \epsilon\, ; \, 1+\epsilon\, ; \, x\right)-1 \Bigg\}$$
@BalarkaSen I try to have as little to do with hypergeometric functions as I can :-)
@robjohn I don't even know what these functions are, lol.
@robjohn I like hypergeometric functions much more than other real analytic function. Even more than Lerch's zeta.
Exercise : Reduce the quartic curve $y^2 = x^4 + 1$ to a cubic form by a $\Bbb C$-rational transformation.
@GabrielR. A topological anti-Riemann, super-Desaregous, ultra spaced, hyperelliptic super-genus example of a A-hypergeometric, semimodular, pseduosemistable Riemann surface.
19:48
that's one hell of an example
@BalarkaSen WTF are you talking?
@BalarkaSen You wrote that paper?
@BalarkaSen You and who?
19:53
it's really groundbreaking though. you deserve fame and medals
@JasperLoy You, of course.
"SOME UNIQUENESS RESULTS FOR FREELY ANTI-HEAVISIDE, CONDITIONALLY
MEROMORPHIC MATRICES"
@BalarkaSen Oh that site is crap, lol.
@JasperLoy "PAPPUS, INTRINSIC SUBALEGEBRAS AND PROBLEMS IN
COMMUTATIVE TOPOLOGY". You wrote that one.
@BalarkaSen That paper tricked me for a while.
From the paper :
"Moreover, the groundbreaking work of J. Loy on isometric isomorphisms was a
major advance"
And the reference : "J. Loy. Introduction to Galois Theory. De Gruyter, 1992."
You wrote a book, Jasper?
19:59
@BalarkaSen I plan to in future.
It is well known in this chat that I plan to write three books.
Really?
On Mathematics?
One on algebra, one on analysis, one on geometry and topology.
@GabrielR. Oohh, I got a catchy one : "SUPER-TRIVIALLY STOCHASTIC CONNECTEDNESS FOR $\tau$-ELLIPTIC, $\tau$-LEGENDRE, LEFT-$p$-ADIC MEASURE SPACES"
20:01
@robjohn Although I have some solutions at hand, I wonder why the way you used yesterday is more powerful than all mines.
@robjohn $\displaystyle \sum_{k=1}^{\infty} \left(\frac1k-\frac1{k+n}\right)$. This makes the difference. Actually, it makes $n$ from denominator disappear.
@JasperLoy I'd love to read the latter. Send that to me please, if you write.
@robjohn ah, wait.
@robjohn done. I was misled by a mistake in a second approach. It seems all works fine. Ready to write it up. :D
@robjohn anyway, that way of writing the harmonic number is pretty useful.
@Chris'ssis Yes, I have used it in conjunction with the standard one to sum several Euler Series.
@robjohn I remember you also used it when you worked on some really nasty series involving harmonic numbers. (it's d**n powerful!!!)
@Chris'ssis Besides it gives an analytic extension of the Harmonic Numbers
20:10
@robjohn yeah
@Chris'ssis I have some identities that allow computation of $H(x)$ for special $x\in\mathbb{Q}$
@Chris'ssis Not too long ago I used $H(1/3)$ in an answer.
@robjohn Really? That is interesting!
@robjohn In a way, these are kind of polygamma identities, right?
@Chris'ssis Yes. $H(x)=\psi(x+1)+\gamma$
this was the answer
Indeed.
@JasperLoy I RCA today
20:17
@robjohn I saw that and upvoted. It's a magical answer!
@N3buchadnezzar Real and Complex Analysis?
if $f = u + i v$ is analytic, does this imply that u and v are harmonic functions?
@robjohn Yeah
@N3buchadnezzar yes! Think Cauchy-Riemann equations
Does that statement count both ways?
@N3buchadnezzar There is a theorem that says that a non-constant harmonic function is the real part of an analytic function, at least locally
20:21
I am looking forward to reading rudin
@robjohn Does that come from your research? The way is d**n clever.
It's standard, I think.
@Chris'ssis It is original work, but I am sure I am not the first to do that.
@robjohn Great! I think is less important who was the first to discover that. It's just nice to discover things on your own.
@robjohn Indeed.
Wait, @robjohn, you do research?
On what?
Euler sums?
20:24
@BalarkaSen just personal work. I am not a professional mathematician any more.
@robjohn Published paper, I presume?
@BalarkaSen no
Oh.
Okay, gotta go.
Latter
@BalarkaSen have fun
20:34
@robjohn What made you retire? If it is not too personal
20:52
Any help on $c)$ i was thinking of using that $u(\cos x, \sin y) = h(\theta)$
Then I think I can look at this as a line integral around the unit circle
For one it has no singularities so it should be zero
secondly I think I can use the gradient theorem here
Use (b)
@N3buchadnezzar didn't retire. I got a job at Apple.
ey
@Zibadawa *scratches head
When is the iPhone 6 coming out?
21:16
@skullpatrol are you going to get one?
21:31
iPhones are for kids, lol.
Meh
I still can not solve it
\begin{align*}
\int_0^{2\pi} h(\theta) \mathrm{d}\theta
=
\int_0^{2\pi} u(\textbf{r}(\theta))\left|\textbf{r}'(\theta)\right|\mathrm{d}\theta
=
\int_C u \mathrm{d}s
\end{align*}
Where $r(x) = (\cos x, \sin x)$.
@robjohn Unlike me, I have retired several times from SE, lol.
21:51
@N3buchadnezzar what are you trying to solve?
59 mins ago, by N3buchadnezzar
user image
22:08
I need more detailed help than what I'm getting in the answer to my question.
It's driving me batty, because I want to learn so badly, and I've been trying and trying to apply the hint the guy who answered me gave me, it's still not working out, and he doesn't seem to respond to follow ups.
Which question, @JessyCat?
Link please.
Okay, let me figure out how to do that
[this thing](link)
Copy and paste it here
22:10
1
Q: Weak Lower Semicontinuity Generalized to any $L_{p}$ space

Jessy CatI am having difficulty with the following proof: Generalize the weak lower semi-continuity of$L^{p}$ norms to all $1\leq p < \infty$; i.e., show that if $u_{n}\to u$ weakly in $L^{p}$, then $||u||_{L^{p}}\leq \inf_{k}||u_{k}||_{L^{p}}$. We were given the following hint: Consider the integral $...

You need to use Hölder (replacing Cauchy-Schwarz) and otherwise Fatou, as before, I think.
@N3buchadnezzar I think that they wrote $h$ wrong... It should be $h(\theta)=\cos(\theta)e^{\cos(\theta)}\cos(\sin(\theta)) - \sin(\theta)e^{\cos(\theta)}\sin(\sin(\theta))$
Hi, when I have $x^2$ how can I say this with words please ?
No , I know you need to use Holder. I'm posting what I've been doing as a comment. Hold on a sec.
@Julien $x$-squared
22:13
@robjohn thank you!
@JessyCat: Using $1/p + 1/q=1$, what is $(p-1)q$? This will show you why $v\in L^q$.
@julien also "the square of x."
"the second power of x"
@skullpatrol So if I would like to say the opposite , Can I say : $x$ is not a square number ?
@N3buchadnezzar are you sure that is right? it isn't the harmonic function above if so.
Yes, you can say that @julian :-)
22:19
@N3buchadnezzar However, using that $h$, it is then simple.
@skullpatrol Thank you!! :)
Thanks for asking :-)
@robjohn Did i do it right?
Turned it into a closed integral with no singularities
@N3buchadnezzar These problems are usually integrating the real part of an analytic function around the unit circle, but the $h$ they give is not the real part of the analytic function they've given above.
@Ted Shifrin, that part I understand now
22:22
@N3buchadnezzar The harmonic function they give above is the real part of $ze^z$
It's the rest of it!
I posted what I've done as a comment now. Reload & it should appear.
@robjohn It is a typo, should have \sin (\sin \theta) at the end
No, you didn't get the $q$-norm of $v$ as you should have!
I realize this. What happened ?
$(|u|^{p-1})^q$? What is $(p-1)q$?
22:28
@Ted Thanks again for talking to Jacob, he shot me an email today.
@Mike, yes, he emailed me to tell me :) And to mumble at me about my plans for this summer.
I hope you told him to speak up.
$\frac{p}{q^{2}}$?
Does someone know when is $tan(\theta)$ is irrational ?
I don't need to tell him that, @Mike. He's not bashful with me :D
22:30
@robjohn nice. You're a very lucky person. :-)
Um, no, @Jessy, unless I'm sloppy.
No, I'm sure it's me. :(
@Ted But you should always tell people who mumble to speak up!
What do you mean, @Julien?
@N3buchadnezzar Then the $h(\theta)=u(e^{i\theta})$
22:31
Yes, @Mike, I knew you meant that.
@Julien You mean when is the result of the function irrational?
Ah, I'm glad my joke was just not funny instead of incomprehensible.
@Ted, I read it wrong. It's p!
Yippee !! So $v\in L^q$!!
Okay, but how does that help with the rest of the problem?
22:33
Well, @Mike, you're so clouded in smoke that it's hard to tell the difference :D
when x is in whole degrees, when is tan(x) irrational?
Well, you were asking why the $v$ they gave you was in $L^q$. So you can apply Hölder as suggested, @Jessy
I need some rest now. Back later.
@Julien tan(45°)=1
@Ted Nah, I threw out what I had
22:33
I tried that, and I didn't get what I was supposed to get on the RHS.
I take advice sometimes, especially when I know it's true anyway.
@Julien I am really unsure, my obvious answer would be when cos(x) is irrational or sin(x) irrational, or when both are irrational and when you divide them you get another irrational.
But I am not sure if that's even slightly helpful
LOL, aw @Mike.
@N3buchadnezzar then $\int h(\theta)\,\mathrm{d}\theta=\mathrm{Re}\left(\int f(e^{i\theta})\frac{\mathrm{d}e^{i\theta}}{ie^{i\theta}}\right)$
Anyhow, I'm sure Jacob will be glad to share thoughts if you have questions.
22:36
@Julien Rather often.
Well, @Jessy, now I'm lost. You're writing $\|u\|_p^p = \int uv$ and applying Hölder. Doesn't the rest work out as before with the sequence?
@Studentmath Hey I read that message you pinged me with earlier.
heya @Karl
@TedShifrin hi, how are you?
Just having trouble typing with my bandaged thumb :) Otherwise, fine, but this week will be deadly.
22:39
@KarlKronenfeld yeah, had to find a specific subgroup of the positive Q's (speaking of irrationality) that is of the ordinal w^2, so I think w^a where a is a natural number is almost certain, question can we get that a to w..
In fact I am working on my exercise, and I would like to know if I can put an infinite number of points on the unit circle which has an irrational distance between all of them
No, because I assume things must have to cancel in order to get just $||u||_{L^{p}}\leq \inf_{k}||u_{k}||_{L^{ p}}$, otherwise, why bother introducing all that v stuff in the first place? So, no, there is algebra I'm having trouble with, I suppose.
@TedShifrin Break it playing tennis?
@Studentmath I looked it up, and it is claimed to be the case.
Why do you care about $\tan\theta$, then, @Julien?
no, @Mike ... gashed it cooking.
22:40
Oh? Any link I can look into? It is thought-provoking I have to admit, it's extremely counter-intutive
On the other hand |N|=|Q| so a lot of things are counter-intutive..
@Ted You might embellish that story. An underhanded opponent putting a blade into the ball has a certain excitement to it.
Everybody knows how you can tell if an elephant's been in your fridge -- footprints in the butter. But how can you tell if the elephant has been trying to follow Szemeredi's original proof of his theorem on original progressions?
Still, @Jessy, use the $v_k\in L^q$ just as we obtained $v$ from $u$?
@Paul: Give it up.
@Ted, I don't get it?
Give what up?
As @Pedro would say, STAHP.
22:42
Stop what?
the jokes?
@Mike: I would expect you to be so so underhanded ... if you were not out of breath :D
@Ted, I thought you were saying you didn't like Szemeredi's proof.
LOL, you read too much into things I say, @Paul ... you and @skull.
22:44
@TedShifrin Because for a finite number of points on the unit circle which has an rational distance between all of them and the proof using the fact that if $\tan\theta$ is rational then this one $\sin \theta$ is that too.
@Ted, You make exactly the opposite mistake. You read too little into the things I say.
@Studentmath I found it on wikipedia. It has no reference, so I am looking elsewhere now
ROFL @Paul.
@Ted Whoa now, I'm not dastardly enough to do that. You'd need to be playing against Dick Dastardly or something.
Being a student, if it is on wikipedia it's true.
22:45
or @Pedro @Mike ... oh, wait, he doesn't need to resort to subterfuge.
Despite what all the teachers say, I have never found something wrong there.
I'm not a particularly light conversationalist. I usually intend a deep meaning behind what I say, even if I fail to make my intentions apparent.
I have @Studentmath.
Did it stay wrong after you did?
I haven't fixed everything wrong I've found, but I've fixed a few things.
@Paul: I didn't mean to stifle your attempts at lightheartedness. My apologies.
22:47
OK, I need to make sur ei don't fail my midterm tomorrow. Adios.
Adios @Mike.
Good luck there.
@Studentmath Its credibility is in question, since I have no idea who made that claim. Anyway Joel David Hampkins reiterates that claim in this answer: math.stackexchange.com/a/37161/67848
@Ted, Davide followed up. He said I should apply the definition of weak convergence right after my first $\leq$, which he says should be an inequality, and that should do the job. Maybe I'm unclear about the definition of weak convergence?
Sorry, meant an equality.
@Ted Well, I meant to refer to my difficulties in following Szemeredi's proof but it might be a bit of a longshot that anyone here is available to help me with that. @Mike, I'm referring to you here??!!
22:48
Lemme try hacking it out a little bit more!
Szemeredi is not @Mike's style of number theory. We have experts on that at UGA, but I'm not one of 'em.
He's right, @Jessy. Remind yourself what the definitions are.
Can put it up as a question I guess, someone can direct us to a good proof.
@Studentmath Well, professor Hamkins gives the proof right in that answer.
@Ted, can I email any of the UGA experts?
Well, I'm sure you can, but I'm not going to tell you to tell them I told you to :P
22:52
Huh, reading it now.
@Ted, I can leave your name out of it if you request that. But an expert would only be useful to me if they are really available. Getting no reply or getting a reply that says "Can't help, I'm busy." won't be so useful to me.
@TedShifrin My claim about $\omega^2$ being the upper limit yesterday was shil bult in its prime form. (1) I found a way to embed things like $\omega^n$ for all $n$, and (2) then it turns out any countable linear order embeds $\mathbb Q$. :)
I'm not supposed to use the definition involving linear functionals, right?
If $\tan\theta$ is irrational then $\sin(2\theta)$ is irrational because we have : $\sin(2\theta)=\frac{2tan(\theta)}{1+tan^2(\theta)}$. It is correct ?
Indeed, @Karl. My gut told me $\omega^k$ should work, so I'm glad.
22:53
@Ted In other words, yes I'd appreciate the names of experts in this area who may be willing to help, and no, I won't mention you in the email.
No, @Julien. That's $\tan(2\theta)$.
@KarlKronenfeld That's really amazing.
Well, no, it's not. Minus error.
Beyond my grasp too. I wonder how far can you reach with N?
@TedShifrin no problem, so it is correct ? :)
22:55
@Paul: Neil Lyall, Akos Magyar (who's currently at UBC but returning to UGA), and Alex Rice (about to take a postdoc at Rochester but currently at Bucknell).
Just the fact that $(\int uv)^{1/p} \leq $ (by Holder) $(\int u)^{1/p} \lim \int v_{k} )^{1/p} $
is it*
I doubt it, @Julien. Try $\theta=\pi/4$.
I would say the upper limit there is certainly w^2, but I am probably wrong (again).. just my intution, still.
Then what? Do I immediately get that this is $\leq \lim \ inf ||u||^{1/p}||v_{k}||^{1/p}$?
22:56
@Ted Stack overflow is also a possibility but there's a $1000 fine for not tex formatting the questions correctly, and I'm not the world's greatest expert in that domain.
No, @Jessy. The $v$'s all need $q$-norms.
@Ted Thanks for the contacts.
Alex is the youngest and hence most likely to respond @Paul.
And you may tell him I sent you to him :P
I must be a psychic because I can sense some Lp -- Lq duality here?
How though??? That's the part I'm having the most trouble with us keeping those q norms and p Norns in there. There must be a trick to it, but nobody will tell me what it is!!
22:58
very psychic, @Paul.
I told you before, @Jessy. Don't say that.
You end up with $q$ norms for $v$ because $q(p-1)=p$.
@TedShifrin I think I said some silly things, in fact I want $\tan(\theta)\in \mathbb{Q}$
To be honest, @Julien, I'm not sure. I'm too busy doing too many things here. I should leave :P

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