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11:00 AM
The best way to study mathematics is to do exercises, not theory.
 
@beginner Many people do not have critical thinking skills.
 
And I believe it's no problem to hurry keeping a whole bag of solved problems with you :)
 
@WillHunting my dad and mum are both professors but i wont say fields cause they said i can be tracked
 
In this world, it is very painful to be able to think, because many people cannot think, and make decisions that cause others suffering.
 
Oh yeah @beginner?
That's cool
Any of them are number theorist?
 
11:02 AM
@beginner It's good to be a professor, but professors are not necessarily smart people. I know of some rather stupid ones.
 
@WillHunting i think they are smart though
@BalarkaSen no neither
 
K I misread "professors" as "mathematicians"
 
@BalarkaSen oh hehe no
@BalarkaSen i would probably own all my math books if they were
 
Anonymous
@BalarkaSen are your parents professors?
 
no none.
 
11:03 AM
It's very easy to get a PhD these days. Just go online and do some stupid courses, lol.
 
Anonymous
@WillHunting I didn't get you
 
my mum has a masters degree as well and both have bachelors
 
@AshwinGokhale What are you referring to now?
@beginner Most professors I know about have PhDs.
 
they both have phd yes, i just meant on top
 
Meanwhile, here's a related article of Arnold about mathematical education that I take more seriously than the article of Thurston (for reasons which would earn me another smack from Ted) pauli.uni-muenster.de/~munsteg/arnold.html
 
11:04 AM
@BalarkaSen what is residue for modolo arithmatic?
 
Anonymous
@WillHunting how does taking an online course get you a PhD?
 
@beginner I have a Permanent Head Damage.
 
Why can't I tag Random Variable here in chatroom?
 
@AshwinGokhale Well, there are many such programs around these days, not for math though.
 
@beginner I am not going to survey on this. Read up some few chapters of a standard elementary number theory book.
 
11:05 AM
@BalarkaSen can i only ask you to verify things, not to teach?
 
On the other hand, isn't this explained on Dummit-Foote already?
 
Anonymous
@WillHunting I never asked anything about what you were studying?
 
@AshwinGokhale I did a math degree ten years ago.
 
@beginner You can ask me if you get stuck on any problem. I am too busy with my studies to explain elementary concepts which are very well explained in standard books.
 
@BalarkaSen ok i will do that only then thank you
 
11:07 AM
I'll advice you to pick up a book and go through it rather than picking up terms, @beginner
 
@beginner Instead of reading random books proposed by random strangers, you should figure out a course of study first.
 
@WillHunting i will finish dummit and foote before i am '15 or so'
 
@BalarkaSen Advise is the verb and advice is the noun.
 
@Will It's a chat dude. Do you expect everyone to spell correctly? :P
 
@beginner I think you should finish a course in linear algebra first before reading that book.
 
11:09 AM
noun is a thing or a place, verb is a describing word?
 
Anonymous
@beginner why don't you check the schedules of tripos?
 
adjective is a verb noun?
@BalarkaSen did you start on dummit and foote?
 
I started mathematics with elementary number theory, if that's what you mean.
 
@BalarkaSen what was your text book order?
or survey order as you call it hehe
 
@beginner Not all books are surveys. The order of books are hard to recall but I think I studied calculus before number theory.
 
11:10 AM
oh so i am lost then :(. you were doing dummit and foote at my age
but i didnt start them yet
 
Anonymous
Don't compare your start with somebody else's @beginner
 
@beginner How about, say, beginning with elementary number theory? It's a quite self-contained subject.
 
balarka inspires me ashwin
ill be back later
 
Hi @M.N.C.E. Nice to see you here ^^
 
@beginner What you should do is to be less overenthusiastic. It's dangerous. Have you studied set theory?
Ugh covering spaces are hard. @DanielFischer can you help me with something?
 
11:17 AM
@BalarkaSen Maybe.
 
@DanielFischer Munkres wants me to find $p : A \to B$ a covering map, $q : B \to C$ another covering map such that $q \circ p$ is not a covering map. I think I have already said this before but $p : S^1 \times \Bbb N \to S^1 \times \Bbb N$ as $(x, n) \to (x^k, n)$ and the projection $q : S^1 \times \Bbb N \to S^1$ as $k \to \infty$ works. But is there no such example when $A, B, C$ are path connected and locally path connected?
This should induced inclusion of fundamental groups via the maps $\pi_1(p)$ and $\pi_1(q)$
 
@BalarkaSen gimme a problem
 
@UserX Have you studied normal subgroups?
 
@Chris'ssis $$\int_0^{\pi/2}\arctan^3(\sin(x))dx=4\sum_{k,n,m=1}^\infty(\sqrt{2}-1)^{2(k+n+m‌​-1)}\dfrac{\frac{-1}{8k+8m+8n-12}+\frac{1}{8k+8m-8n-4}+\frac{1}{8k-8m+8n-4}-\frac‌​{1}{8k-8m-8n+4}}{(2k-1)(2n-1)(2m-1)}$$
@Chris'ssis (The latex doesn't render properly in chat, just see the source)
 
@DanielFischer I have no idea what to do then. Maybe $q \circ p$ is a local homeomorphism but I can't see this.
 
11:25 AM
@BalarkaSen No connected counterexample springs to mind.
 
@BalarkaSen not really. Just definitions, haven't even seen worked out problems yet.
 
@BalarkaSen A composition of local homeomorphisms is a local homeomorphism.
Proof: ...
 
Might be.
I believe it :P
 
@BalarkaSen Don't believe, prove. Start with the definition.
 
1./ Proof 2./ ???? 3./ Profit
 
11:28 AM
@DanielFischer $f : A \to B$ is a local homeomorphism if for all $x \in A$ there is an nbhd $U$ around $x$ such that $f$ restricted to $U$ is a homeomorphism.
Erm. I am missing something, aren't I?
 
@BalarkaSen Incorrect.
 
Yeah, right. OPEN MAP.
$f(U)$ is open.
 
Yes, the point is that (the restriction of) $f$ is a homeomorphism between open neighbourhoods of $x$ and $f(x)$.
 
So well this is obvious @DanielFischer. $f, g$ be local homeos from $A \to B$ and $B \to C$. Composition of homeos are homeos and $f(g(U))$ is open.
 
@BalarkaSen Not so fast, grasshopper.
You need to restrict the neighbourhoods, generally.
 
11:33 AM
$U_x$ be in $A$. $f(U_x)$ is open and $f|_{U_x}$ is a homeo. Now $f(U_x)$ is an nbhd of $f(x) \in B$
 
Good, and then?
 
So $g|_{f(U_x)}$ is a homeo and $g(f(U_x)) = g \circ f (U_x)$ is open.
 
@BalarkaSen No. You can't conclude that $g\lvert_{f(U_x)}$ is a homeomorphism.
 
Could someone verify my above integral with Mathematica ?
 
@DanielFischer Well $g$ is also a local homeomorphism and $g|_{U_y}$ for all $y \in B$ is homeomorphism by definition.
As $f(U_x)$ is an open nbhd of $f(x) \in B$, it follows that $g|_{f(U_x)}$ is a homeomorphism
 
11:36 AM
@BalarkaSen The point is "for all $y\in B$ there exists a neighbourhood $V_y$ of $y$ such that ..."
 
Ah OK.
 
That neighbourhood can be very very very small.
 
@Venus I don't remember. But I used to keep a notebook (which I still have), and it's in there.
 
In that case assume $U_y \subset f(U_x)$.
 
@BalarkaSen \subset
@BalarkaSen And then?
 
11:42 AM
Ack. We have to restrict the nbhds too.
 
@DanielFischer As small as my XXX.
 
@Venus Hi.
 
@WillHunting smaller, much smaller.
 
@RandomVariable Could you take picture of it & send it to me ^^
Please...
@M.N.C.E. Hi too...
 
@venus Is your boyfriend from Mars?
 
11:46 AM
@Venus I'd rather just post something on the other forum.
 
@WillHunting No, he is an earthling
@RandomVariable Really? Can't wait then. Please let me know when it's done :-)
 
@DanielFischer We have to pick some $V_x$ around $x$ in $A$ such that $f(V_x)$ is inside that $U_{f(x)} \subset f(U_x)$, right?
 
$\prod \limits _1^{p-1}\gamma_v = \prod \limits _1^{p-1}\alpha\gamma_v=\alpha^{p-1}\prod \limits _1^{p-1}\gamma_v\implies a^{p-1}\equiv1\pmod p$ is so clever!!
 
12:26 PM
I decided I will try to sleep and get up early.
But first, Coffee vs study done Anova imgur.com/2PVACRO
Strong statistical evidence that coffee is correlated with study done
 
@BalarkaSen Yes. Typically, one takes the intersection $f(U_x) \cap V_y$, and its preimage in $A$ and image in $C$.
 
Which really means little, since pages done is the real indication of work done: imgur.com/jEexYP6
Anyway goodnight
 
Anonymous
@Committingtoachallenge What were you reading?
 
@AshwinGokhale I was doing Cohn, but I decided I should sleep, since I may be seeing people tomorrow
 
Anonymous
@Committingtoachallenge Good night!I am beginning with Artin now
 
12:32 PM
@AshwinGokhale Best of luck :). Talk tomorrow
 
@Venus I edited my post.
 
@RandomVariable Thanks. I'm stuck in the second line. It never crossed to mind to multiply it by $\dfrac{\cos2x}{\cos2x}$
 
Is @Advaitha 's account suspended?
 
@RandomVariable It's indeed a cute one. Do you have other cute integrals that can be done in a few lines using elementary methods? Maybe you can provide some links
@Integrator Yeah. I had little argument with him. I'm quite surprised he acted like a child & also rude
@Integrator Is he your teacher or lecture?
It looks like someone post a monster
0
Q: Very Messy Integral.

eaxdpiotnyeantialFind: $$\large \left|\int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}}\frac{\ln|x|^x+(\arcsin x)^3(\arctan x)-(1+x^2)\arctan x}{(\ln|x|)^2(x+x^3)}dx\right|$$

 
Anonymous
12:57 PM
@Chris'ssis Have you solved your integral?
 
@Venus No, not really.
 
@Venus No, Not my teacher!
 
1:18 PM
@Venus It's just pretending to be hard.
 
@DanielFischer Why did you delete your answer sensei?
 
@Venus Because I noticed the OP spamming no-effort questions in short succession.
Not a behaviour I condone.
 
He has edited his post
 
@Venus Now I have to think whether to undelete.
 
Haik sensei! Noted
It should be undeleted since he got it from exam
Besides, he seems a nice user by seeing his profile
 
1:25 PM
undeleted
 
I doubt he didn't give any attempt to solve it
HAHAHAHAHAHAHAHA @DanielFischer
@eaxdpiotnyeantial Have a tough day at school buddy? :-)
 
what do you mean?
 
Greetings
 
Good day @Chris'ssis
@eaxdpiotnyeantial See the history chat
 
@Venus Hi
 
1:29 PM
oh it's a week old @Venus
 
@AshwinGokhale Sure. :-)
 
@eaxdpiotnyeantial Ganbatte!!
 
@Venus Is it an english word, sorry english too is my second language
 
@eaxdpiotnyeantial What's your first language?
 
@Venus oh i don't post regularly, i collect them and post them in a go, so that might had made you think that. BTW Hindi.
 
leo
1:32 PM
@Venus what's yours?
 
@Chris'ssis Yup
@Chris'ssis I know. Let me see it first
 
@Venus OK. Check that. I consider that piece a masterpiece that I'll add to my book. :-)
 
@leo You first
@Chris'ssis You're an author? What book?
 
leo
@Venus spanish
 
@leo Hola, como esta senor? :D
 
1:35 PM
@Venus Yeah, I'm going to publish a book about integrals, series and limits, mainly things created by me.
 
@Chris'ssis Sounds great! Let me know if it has been published
 
leo
@Venus ha ha ha, good
 
@leo Malay
 
OK, @DanielF, but that doesn't solve my problem.
 
@leo Where do you live? Madrid? Barcelona?
 
leo
1:38 PM
n should go with ~ so it become ñ
 
@Venus,@Chris'ssis it came in the exam with ^2014, ^2013 and many dirty paper, very tough . :( I scored 120/320
 
@eaxdpiotnyeantial I can relate that
 
Still working on it, and at the same time I do research every day for discovering other great results such that all buyers will totally love my book. Sure, out there will be always criticism ... (for me it's important to be a successful book)
 
@Venus/@Chris'ssis/etc. what are you people currenltly doing, still in highschool like me?
 
@Chris'ssis Don't forget to give a little story for every problem. It'll look good I guess
 
leo
1:40 PM
@Venus Neither of those. Far from them. Look
@Venus nice
 
@BalarkaSen By "solve your problem" you mean finding connected coverings such that $q\circ p$ is not a covering?
 
right
 
@Venus Yeah, I also like the stories for the problems, but sometimes it's hard to find when a problem appeared first.
 
@leo I can't locate in which city do you live by seeing that map
 
leo
@Venus San José, Costa Rica
 
1:42 PM
@BalarkaSen Well, you know that if $q$ is finitely-sheeted, then it's a covering. So take an infinitely sheeted covering.
 
@Venus In his book, Inside Interesting Integrals, Paul J. Nahin talks about the name of some integrals, some of them seem to be very old.
 
@Chris'ssis You're right but at least for the famous problems you should give it
 
@DanielFischer The converse is totally not true.
 
@Venus The book will mainly contain original creations, things that came from my mind. I don't need a book like the rest, but something totally different.
 
@leo Oh my, I forgot many countries in America also use Spanish
 
1:44 PM
@BalarkaSen You mean that if $q$ is infinitely-sheeted, then $q\circ p$ can still be a covering? As if I didn't know.
But, @BalarkaSen if you want to construct a counterexample, that's what you need to start with.
 
@Chris'ssis At least you can give your personal story when those problems came to mind
 
I know, but I was rather thinking on the lines of proving that no counterexample exist, @Daniel
 
leo
@Venus no problem
 
@Venus They come to mind every day, while sleeping, while doing accounting, while tutoring, while going shopping, jogging ...
 
@Chris'ssis Are you below 30? If I may know
 
1:46 PM
@BalarkaSen I think it's a little early to say that.
 
Oh you're thinking about connected.
 
@Venus Around 30.
 
I was assuming path connected and locally path connected stuff, @DanielF
 
@Chris'ssis Still young.
 
@BalarkaSen Yes, and? Still too early to say "doesn't exist". Anyway, trying to construct an example if it fails may give hints as to how one could prove none exists.
 
1:49 PM
Anyway, I use your method to answer integral problem here. I hope you don't mind @Chris'ssis
 
@Venus Please do not use the method I just showed you. That's why I deleted the post.
 
@Chris'ssis I just did. Sorry
I'll delete mine
 
@Venus No need, you can let that one now.
 
Sorry, I can't delete it. It's accepted
 
@Venus No problem. No need to delete it.
 
1:51 PM
Are you sure?
@Chris'ssis Heave a sigh of relief finally
 
hehe, it's OK.
 
Sorry, next time I won't do it again. I promise ^^
 
@Venus Did you see this one?
10
Q: An integral $\int^\infty_0\frac{\tanh{x}}{x(1-2\cosh{2x})^2}{\rm d}x$

M.N.C.E.I would like to enquire about the possible methods of computing the following integral $$\color{blue}{\int^\infty_0\frac{\tanh{x}}{x(1-2\cosh{2x})^2}{\rm d}x=\ ?}$$ A possible way I see of doing this is to apply the substitution $x\mapsto-\ln{x}$, which yields $$-\int^1_0\frac{x^3(1-x^2)}{(1+x^2)...

 
@RandomVariable Does user Yourself have an account on MSE?
@Chris'ssis Yes. I've upvoted too. Why do you ask?
Is there something wrong in David's answer?
Or you have a shorter answer for that problem?
 
@Venus Can we do that in a fast way?
 
1:58 PM
@Chris'ssis I don't know. I can't think of it at the moment. But it seems your question implying so
@DanielFischer If I may know, where do you live sensei?
 
senpai, @Venus
 
@Venus Northern central Europe.
 
@BalarkaSen I respect to him as a guru
 
We all do.
 
@DanielFischer Scandinavia countries?
 
2:02 PM
@BalarkaSen On this site I only respect @robjohn as a guru. Sorry, but this is my personal opinion. It's not OK to speak in the name of all.
 
Sorry if that too personal. You may not answer it
 
@Venus A bit less north.
 
Cute problem, @Balarka
 
@MikeMiller All of my problems are either very interesting or trivially false.
 
@DanielFischer What did you mean by a bit less north? It's Arctic ocean?
 
2:04 PM
I wouldn't go that far, @Balarka.
 
@Venus A bit south of Scandinavia.
 
@DanielFischer Either UK or Germany, but I bet Germany
 
Hmm. Map $\Bbb R \times \Bbb R^{+} \to S^1 \times \Bbb R^{+}$. Then map $S^1 \times \Bbb R^{+}$ to $S^1$ by projection.
Whoops. The composition is a covering space.
Nevermind.
 
@DanielFischer Your family name sounds familiar. I've heard it somewhere
 
@BalarkaSen If you mean to take the discrete topology on $\mathbb{R}^+$, but then it's not connected. If you take the standard topology, $S^1\times \mathbb{R}^+ \to S^1$ is not a covering.
 
2:08 PM
Oh right.
Hrm.
 
@Venus Not exactly a rare name.
 
Why would a sane person take the discrete topology, @DanielFischer?
 
@MikeMiller It can be useful.
 
I am not quite sane, @MikeM
 
@DanielFischer I google it, it shows up this one in Wiki: en.wikipedia.org/wiki/Bobby_Fischer
The Great Bobby Fischer
 
2:10 PM
@Venus The one who catches fish with booby traps?
 
@BalarkaSen Oh.
 
@DanielFischer Is he your ancestor?
 
Arthur Fischer, the mod.
 
@DanielF Fair enough, I suppose.
 
2:13 PM
@Venus Neither one is a relation (within reasonable limits, of course if you go back far enough, we're all related). Nor is the great Ernst Fischer.
 
@Sawarnik I forgot Arthur. The strictest mod in MSE :D
 
LOL. We're bringing up a whole load of Fischers.
 
@DanielFischer Lots of the great Fischers. It strongly indicates you're a Germany. ^^
 
Hehe, he's not a nation.
 
@Venus I'm half of Europe thrown in a mixer.
5
 
2:17 PM
If I have a son I'll give him name Daniel & if a girl, I will name her Daniela :D
@DanielFischer But you do have a nationality right?
 
@Venus No, just a citizenship.
 
@Chris'ssis Did you see my post earlier ?
 
@Hippalectryon Thanks, yeah. I'll check the details as soon as I finish the proof I', working on ... that is ...
10
Q: An integral $\int^\infty_0\frac{\tanh{x}}{x(1-2\cosh{2x})^2}{\rm d}x$

M.N.C.E.I would like to enquire about the possible methods of computing the following integral $$\color{blue}{\int^\infty_0\frac{\tanh{x}}{x(1-2\cosh{2x})^2}{\rm d}x=\ ?}$$ A possible way I see of doing this is to apply the substitution $x\mapsto-\ln{x}$, which yields $$-\int^1_0\frac{x^3(1-x^2)}{(1+x^2)...

@Hippalectryon in one line
 
:O
@Chris'ssis Isn't trigamma $\psi^{(3)}$ ?
Nevermind
 
@Hippalectryon No, that answer is correct.
 
2:28 PM
Indeed. I forgot for a moment that $\psi_0$ was digamma and not gamma
 
@DanielFischer Does that half of Europe contain bits of Western Europe?
If so, then the mixer is bound to blow up at some point.
 
@BalarkaSen I suspect @DanielFischer is a Super Aryan
 
@Venus Arya ?
 
Explanation of the joke : the poles are in Western Europe.
 
2:34 PM
@BalarkaSen Yes, some Dutch and French. Why should that cause the mixer to blow up?
 
@Venus He does, but I don't know if he would want me to share it with anyone. He uses a generic username. Actually, I don't even know what it is off the top of my head.
 
@BalarkaSen Not all Poles. Some are still in Poland.
 
Well the semi-poles are quite dense around the western Europe.
 
2:36 PM
Oh that
 
2:50 PM
whoa whoa whoa
this hawaiian earring stuff is interesting
 
Done
 
@DanielFischer The Hawaiian earring looks homotopy equivalent to wedge of countably infinitely many circles to me.
In other words, $\Bbb C$ minus countably many points on the unit circle.
 
@Chris'ssis Tell me when it's posted :-)
 
@Hippalectryon What do you refer to?
 
2:59 PM
@Chris'ssis When you were saying "Done", wasn't it for the one-line answer ?
 
@Hippalectryon It's a bit more than one line.
 

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