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12:00 AM
$$\sum_{k=1}^{10000}\frac{\mu(k)}{[\frac{10000}{k}]^2}=-26.78027...$$
$$\sum_{k=1}^{10000}\frac{\mu(k)}{[\frac{10000}{k}]^3}=-26.22826...$$
$$\sum_{k=1}^{10000}\frac{\mu(k)}{[\frac{10000}{k}]^4}=-25.72294...$$
$$\sum_{k=1}^{10000}\frac{\mu(k)}{[\frac{10000}{k}]^5}=-25.39717...$$
Why is the sum changing so little when I change the exponent?
Can someone explain this to me?
[x] is the floor function, Im worried there might be a simple reason so I don't wana post it yet
 
user19161
@JonasTeuwen OK, I think you are just being nuts now...
 
@JacobBlack Na.
 
user19161
Then what backstabbing?
 
Alright.
 
12:16 AM
...
 
SKullCharlie
 
:P
the pic does not fit correctly, damn it
 
12:50 AM
Palindrome 47174
 
O.o
 
Screenshot it fast, I am going to upvote!!
 
@OrangeHarvester heh
 
user19161
1:10 AM
0
Q: "Oops! Something bad happened!" while logging in

Jacob BlackIn the past week or so, I have had problems logging in to this site. Each time I try to log in, it says "Oops! Something bad happened!" and I have to try a few times before I am successful. This has never happened to me in the past two years. Is there something that needs fixing?

 
user19161
Guys, my very first question on meta, please upvote it! =)
 
Commented!
 
user19161
Yes, an upvote would be good too. =)
 
user19161
This bug has affected my ability to answer lhf. By the time I log in, the lhf has 9000 answers already. =)
 
Brwose while logged in!!
 
user19161
1:20 AM
I guess the answer might be: this is yet another strange thing that happens to me.
 
user19161
1:30 AM
Hmm, perhaps I can get 20k by the end of March...
 
user19161
@Tim The best complex analysis book I think which is available in English is the two volumes by Eberhard Freitag, translated from German.
 
Hello all!
 
2:17 AM
@amWhy
 
@Ethan Hello!
 
Do you know any analytic number theory?
 
@Ethan Not my specialty, sorry!
 
oh
@Sanchez
@Sanchez are you familear with selbergs proof of the pnt?
 
@Ethan, no
 
user19161
2:19 AM
Hey @sanchez we meet again...
 
Hey @Jacob, or @Jason, or whatever..
 
user19161
@amWhy Are you aiming for 30k?
 
@JacobBlack Yeah...I guess..., that's the next carrot! I see you're well on your way to being "trusted" as a user!!
 
user19161
@amWhy Haha, I might "collapse" in any one of 9000 ways before that!
 
@sanchez hi.
 
2:24 AM
@BenjaLim, hi
 
@Sanchez Hi.
 
Hey
 
user19161
Why are there two copies of hi?
 
@JacobBlack I hear you!!
 
@Jason
 
user19161
2:26 AM
@BenjaLim That no longer pings me! Are you going to scold me again today?
 
@Sanchez Hi
 
lol
@BenjaLim, why do you keep hiing me? lol
 
What's going on guys?
 
user19161
Nothing. I am waiting for you to scold me again.
 
Crap my phone is retarded
 
user19161
2:27 AM
@Sanchez Your LOL will only encourage more pings, so no LOL, LOL.
 
LOLOLOLOL
 
@JacobBlack How nice it is spending the eve of chinese new year computing the integers in a biquadratic field.
@Sanchez I wanted to say I am sorry for not acknowledging you on main. When I said that was "my proof" what I meant to say was "here's a proof I have at hand right now".
 
@BenjaLim, it's okay. I just think it's much better to give people proper credit whenever possible.
 
that is true. I always try to do that.
 
user19161
@Sanchez Speaking about credit, I wanna say this again.
 
user19161
2:31 AM
Sometimes, someone posts a comment as an answer.
 
user19161
It is a simple question with a simple answer.
 
user19161
I think of the same answer and post it as an answer.
 
@Sanchez Maybe you should delete your comment on main now that I have deleted mine.
 
user19161
I hope people are OK with that.
 
@BenjaLim, okay.
 
user19161
2:32 AM
I find it weird that I have to say "As ABC says" even for something as simple as 1+1=2 because he posted the comment or answer 5 seconds before I do.
 
@Sanchez Thanks for deleting the comments.
 
No problem.
 
@Sanchez Hmmm it seems the Yoneda Lemma would be very helpful in future
I have seen people prove stuff like because $\hom(A,B)\cong \hom(C,B)$ therefore $A \cong C$ by Yoneda or something like that
 
user19161
@benja Did you watch Twilight?
 
@JacobBlack What is twilight?
 
2:35 AM
@BenjaLim, it's one of those general nonsense tool, but yea
 
user19161
@BenjaLim A movie with me in it.
 
user19161
@benja Please stop revealing my personal details in this chat.
 
@JacobBlack deleted.
@JacobBlack I wrote that because you said to me you're Jacob Black. I don't believe it.
 
user19161
Anyway, I thought you would like the movie character @benja HAHAHAHA
 
2:39 AM
@BenjaLim, didn't read that in detail, but I've seen it used in other contexts
 
@Sanchez Now I have full motivation to learn it. Otherwise I have to keep checking maps all over the place
 
haha
 
@Sanchez and I like how to check if the map in the link was well - defined one had to do the old analyst's trick of adding and subtracting
 
user19161
So @ben what classes are you taking this semester?
 
@JacobBlack Measure theory, differential geometry and algebraic geometry.
Not too excited about measure theory.
@Sanchez My analysis is a little behind than the algebra
 
user19161
2:42 AM
Hey @anorton
 
@JacobBlack Hey!
 
user19161
@anorton What does your username mean?
 
@Sanchez Hmmm I am little unsure of starting with something like Kempf's book. It seems there aren't a lot of examples in there. Perhaps I should speak to my supervisor to say look at Harris' book for examples?
 
user19161
(removed)
 
user19161
@anorton I know I know.
 
user19161
2:45 AM
You can remove it.
 
Ok. :)
 
user19161
Feel free to keep your secrets.
 
user19161
I have many secrets too. =)
 
@BenjaLim, sure. You can definitely ask him for advice.
 
@Sanchez I always think that looking at concrete examples is good.
 
2:46 AM
@Jacob I figure that if you keep yours, I can keep mine...
 
and you can just skim harris lectures 1 and 2 for some examples
Sure.
 
(I'm 99% sure I know your name, though...)
 
user19161
(removed)
 
(removed)
lol
 
@Sanchez ok.
 
2:48 AM
This is the "removed" chat session...
 
lol
 
user19161
You're all NUTS
 
@Sanchez Did you see some brazilian guys I think solved the willmore conjecture?
 
I heard about that, but that's like a year ago I think.
 
'night, all!
 
2:49 AM
and I don't know anything about those stuff
more like geometric analysis I believe.
 
yea. We have a few brazilian guys here and they speak with the same accent as f. marques
 
haha, nice.
 
@Sanchez yea. I saw the youtube clip I was like waaa
 
user19161
Haha this Will Jagy makes interesting comments on this site.
 
@BenjaLim, god knows :P
 
user19161
2:50 AM
@Sanchez Don't worry, if anyone on this site is nuts, I am.
 
Nice :)
 
user19161
(removed)
 
@Sanchez I was just trying to be friendly :)
 
Sure.
 
Man I hope measure theory is not going to be boring
 
user19161
2:52 AM
Now there are 9000 skulls in this chat.
 
it is not going to be boring
it IS boring
 
@Sanchez Hahahahahahah
Honestly I am not amazed at finding out that the Lebesgue measure of the rationals is zero
 
user19161
That can be done in a calculus course.
 
@JacobBlack really?
 
user19161
Yes, it's just a one line proof.
 
2:54 AM
@JacobBlack really?
 
user19161
Yes.
 
@JacobBlack So establishing countable additivity of the measure is a one line proof?
@Sanchez I respect you that you started with Hartshorne.
 
@BenjaLim, lol
Not really.
 
Sorry what I meant to say was some of the words I saw, I had seen in Atiyah - Macdonald.
@Sanchez Do I need to know a lot about transcendental extensions ?
 
@BenjaLim, no
I suppose you know what transcendental degree, separably generated etc means
You may want to know those, but otherwise no.
 
3:00 AM
I read yesterday that the dimension of a variety is defined to be the transcendence degree of $k(V)/k$.
 
user19161
Why are undergrads so crazy about AG?
 
@JacobBlack You don't understand: it is the basis for many higher things.
 
@BenjaLim, that's one way to define it.
and you removed the other message too quickly, I didn't read it lol
 
user19161
OK, I don't understand. You can scold me again.
 
@Sanchez I don't even remember what I wrote....
damn
 
user19161
3:04 AM
I don't know why Benja keeps scolding me, LOL.
 
@JacobBlack I have studied ANT. I could study class field theory but then many people advised me not to do so.
 
@BenjaLim, Depending on people, I don't find the transcendence degree so intuitive. The definition using chain of irreducible closed subsets is very natural, and I can only interpret the transcendence degree one by Noether normalization
 
@Sanchez how?
 
How what?
 
how do you interpret it using Noether Normalization?
 
3:06 AM
Let's say you have finite type $k$-algebra $A$, and that $k[x_1,\cdots,x_n] \subset A$ is an integral extension
 
ok this means finite dimension $k$ - vector space? Finitely generated $k$ - algebra?
 
no, finitely generated.
 
ok.
 
Integral extension (+ of finite type) behaves like a covering map. The dimension wouldn't change
 
ok......
 
3:07 AM
and one can expect that the dimension of $k[x_1,\cdots,x_n]$ is $n$ (it corresponds to $\mathbb{A}^n$)
 
haven't seen that interpretation before.
yea.
 
If you pass these to function field (suppose that $A$ is a domain)
then it says that $Frac(A)$ has transcendence degree $n$ over $k$
 
right.
 
So here we go
 
user19161
@benja I thought you said you were coming to visit me LOL
 
3:09 AM
@JacobBlack In your dreams I was going to :)
@Sanchez yea?
 
@BenjaLim, one thing that you can probably understand is that finite maps (which is the same as integral + finite type as you probably have read in AM-GM) is quasi-finite. (i.e. finite fibers) For example, If $A \to B$ are maps of finitely generated $k$-algebra, $B$ is finite as $A$-module, then for each prime ideal $p$ in $A$, there are only finitely many primes in $B$ that lie over $p$
 
yea that is right.
 
All I am saying is, Noether normalization gives you $k[x_1,\cdots,x_n] \subset A$. Passing to quotient field says that $n$ is actually the transcendence degree of $Frac(A)$. Final thing is integral extensions of finite types do not change dimension. Put these together make the transcendence degree definition natural (at least to me)
 
In the case that $G$ is the group of automorphisms of some ring $A$
we have $A/A^G$ being an integral extension
$G$ acts transtiviely on primes over a given prime $P \subseteq A^G$.
and so as you say spec has finite fibres.
 
hm
If $G$ is finite, yes probably.
Otherwise I'm not sure.
 
3:17 AM
Ah yes sorry the assumption is that $G$ is finite :)
Sorry of course
@Sanchez Otherwise we cannot do the bullshit of taking things like $\prod_{\sigma \in G} \sigma(x)$
but in any case the inclusion $i : A^G \hookrightarrow A$ may not be a finite map
 
yea
the other way round, and yes.
 
@Sanchez quasi finite
@Sanchez I should go and help my mum out now :)
in preparing dinner :)
 
@BenjaLim, sure. Happy CNY! :)
 
@Sanchez how do you know it's abbreviated CNY?
 
Why not? ;)
 
3:24 AM
Just curious :)
bye!
 
user19161
Haha guys, I am so happy I just integrated something. math.stackexchange.com/questions/298461/…
 
user19161
3:43 AM
This chat is dead.
 
I'm not dead yet.
 
3:56 AM
@BenW. You're not fooling anyone...
3
 
user19161
@robjohn I came back from the grave.
 
user19161
Hey @ora
 
@JacobBlack Hello!
 
user19161
@OrangeHarvester Now someone has to downvote my meta question, sad...
 
@JacobBlack was bound to happen. :P
 
user19161
4:05 AM
@OrangeHarvester Why? Is it a bad question? It is a genuine issue that affects me.
 
user19161
I have to click up to 5 times sometimes. Also, sometimes I need to close and reopen the browser.
 
Do you expect such issues in a particular time period?
 
user19161
No, but it has happened too often the past week.
 
@JacobBlack Well, mostly it is a question of dropped packets. It sometimes happens when the server load increases a lot. Sometimes it is expired cookies. The thing is with so many different configurations possible and race conditions etc. that it is almost impossible to track.
 
user19161
@OrangeHarvester So it's not my fault right?
 
user19161
4:08 AM
I am thinking if I should change some settings...
 
No, I don't think so.
 
user19161
Did you see Will's comment? HAHAHA
 
user19161
I just visited his website.
 
he has a website?
ohh yes
I saw that one too.
 
user19161
4:11 AM
Yes, saw his picture too.
 
cholesterol thing.
 
user19161
robjohn is spying on us.
 
I guess Kap is very famous in US. I have seen his name mentioned by Lehrer too.
 
user19161
He is not talking but he is watching us...
 
user19161
@OrangeHarvester Kaplansky is a famous mathematician.
 
4:13 AM
@JacobBlack Okay. I did not know that.
 
user19161
@OrangeHarvester Well, I ain't too sure, but I see his name many times, that's what I mean. =)
 
user19161
Have you seen Linear Algebra Done Right?
 
user19161
Do you know there is Linear Algebra Done Wrong?
 
4:19 AM
Yes.
 
user19161
HAHAHA
 
user19161
The latter is actually very good too.
 
user19161
Actually the search in amazon is pretty bad.
 
Hmm. I have note seen that. I thought it was some kind of antidote to Axler's book. (Matrices, Determinants and All). I am getting enough of it in Hoffman Kunze.
Amazon search is good, what are you talking about?
 
user19161
I type in the author's first name last name and it doesn't give all his books.
 
4:21 AM
For example?
 
user19161
Peter Petersen.
 
Amazon would not have become good if there search sucked.
Which book does it now show?
 
user19161
I mean in the normal search box, not advanced search.
 
user19161
It doesn't show his latest book, Linear Algebra.
 
user19161
He seems to be writing one on manifolds too, let's hope it is published some day.
 
user19161
4:22 AM
He has Riemannian Geometry already.
 
Hmm. Well, searching for 'linear algebra peter petersen' hangs up the search. That has been something I have observed recently in amazon with some other book too.
 
user19161
Best is use last name plus title
 
May be it does not want to show the results in search because the book is in hot demand or something? Amazon is weird in that way. I am not sure, going to the page of the book, it says only 1 copy left.
 
user19161
No, if it is very popular it WILL show up.
 
user19161
I think it is not well known at the moment.
 
4:26 AM
I have got the stuff! Petersen Linear Algebra.
 
user19161
It used to be notes on his site.
 
Nopes, I got the 2012 edition
 
user19161
It is a course for beginning grad students in UCLA.
 
grad? it says undergrad.
 
user19161
Well, the course is for beginning grad.
 
user19161
4:28 AM
Texts can be classified in any way.
 
user19161
UTM or GTM, not much difference.
 
It makes difference to someone on the outside (like me). :P I would decide which topics to learn based on whether it was UTM or GTM.
 
user19161
Now I can tell you that you should look at the contents page to decide, not the series it belongs to.
 
user19161
Universities vary greatly in their courses.
 
user19161
Some topics can be undergrad here, grad there and others the opposite.
 
4:31 AM
Looking at the content page is fine. But I needed an estimate of how much math graduates are expected to know.
 
user19161
Ah OK, you can ask me about books. =)
 
:P
It was 3 years ago. Now, even I know a lot about basic books. Advanced books I will ask you.
 
user19161
Artin is classified at upper undergrad/beginning grad.
 
user19161
In many US unis, they do mostly basic calculus in the lower undergrad years.
 
user19161
4:35 AM
They don't go so fast as the UK ones.
 
Hmm. Keep telling.
 
user19161
If you do 3 years in say Cambridge as an undergrad, you can pass many qualifying exams for PhD in the US.
 
user19161
Hey @san is back!
 
me? yea
but as you have said, this chat is dead.
 
4:44 AM
long live the chat.
 
lol
 
 
1 hour later…
5:49 AM
hello
how do we write math equations in this website
?
any body there ?
 
latex
2
 
6:06 AM
Let {.} be the fractional part function
$$\sum_{k=1}^\infty\frac{\mu(k)}{k}\{ \frac{k}{4} \}=-\frac{1}{\pi}$$
For all integers $a>1$, and all $s$ with real part greater then $1$

$$\sum_{r=1}^a\gcd(a,r)^s\sum_{k=0}^\infty\frac{\mu(ak+r)}{(ak+r)^s}=0$$
Also$$\frac{12}{d(n)}\sum_{d\mid n}\ln(d)^2-3\ln(n)^2=\sum_{d\mid n}\Lambda(d)^2+2\sum_{d\mid n}\Lambda(d)\ln(d)$$
And$$\sum_{k=1}^n \arctan(2k^2)=\frac{\pi}{2}n-\arctan(\frac{n}{n+1})$$
For most values of n,$$\frac{\pi}{2}=n\arctan(\frac{1}{x})+\arctan(\frac{Re((x-i)^n)}{Im((x+i)^n)})$$
$$\int_{0}^1\frac{\sqrt2}{\sqrt{1-x}}\frac{\sqrt2}{\sqrt{1+\sqrt{x}}}\frac{\sqrt2}{\sqrt{1+\sqrt{\sqrt{x}}}}\frac{\‌​sqrt2}{\sqrt{1+\sqrt{\sqrt{\sqrt{x}}}}}..\ dx=\sqrt{\frac{\pi}{2}}\zeta( 1+\frac{1}{2})$$
Also,$$\sum_{n=0}^\infty \frac{2^n}{2^{2^{n}}+1}=1$$
 
6:43 AM
@EdPegg Can you beat L*****y G*****s with 800k?
 
@Novice How is that answer any more definitive than ... everything else known up to this point?
(Which is to say, how is this news?)
 
@anon
 
Try this one. Have you seen anyone other than Bill D. who does that formatting?
 
@Novice Do you know what "how is this news" means?
 
@anon no.
 
6:54 AM
It means, you're late to the party - this information about the stylistic consistency has been common knowledge for a long time.
What is the point of pinging people with no comment when you know for certain that they are in chat ready to respond to actual questions etc?
 
@anon Have you seen $$\sum_{k=1}^\infty \frac{\mu(k)}{k}\{kx\}=-\frac{\sin(2\pi x)}{\pi}$$
 
It does look cool, yes.
 
It works for all x, lol im having fun inverting fourier series with the mobius function
 

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