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3:00 AM
but the fact that it so explicitly measures the euler char just from some random vector field is pretty crazy
 
exactly
 
draw a couple vector fields on the sphere, too, it's a fun little exercise
(I'm not so good at drawing vector fields on surfaces of higher genus)
 
@KarlKronenfeld i'm not totally sure yet, but i can tell you that I've been studying all of this out of the first chapter in a book called "Applied Finite Group Actions." So, I'm betting I'm going to get to it before long.
 
haha ok
 
@KarlKronenfeld You can read the table of contents here
 
3:06 AM
It is certainly an endofunctor as you described and not a functor $(Set,\cong)\to (Set,\to)$? (If you get my drift)
 
In particular i'm looking forward to reading the applications towards combinatorial designs.
 
@AlexanderGruber species can be applied to designs?
hrm
 
@blue looks that way.
Maybe I should write about some of this species stuff in the blag.
 
@KarlKronenfeld haha what the hell
so apparently even if $X$ has the fixed point property $I \times X$ doesn't have to
 
3:18 AM
a counterexample is given in Bredon: $\mathbb{HP}^2 \vee S(\mathbb{CP}^2) \vee S(\mathbb{CP}^2)$, where the latter is the reduced suspension of the complex projective plane and the former is the quaternionic projective plane
how incredibly pleasant, i suppose.
 
oh my gosh... combinatorial species are supported by Haskell
 
4:23 AM
the Preservation of Favoured Races in the Struggle for Life supported by Hitler
 
 
2 hours later…
5:55 AM
@KarlKronenfeld want to try a problem?
 
@skullpatrol ...
 
6:53 AM
This statement from wikipedia:
"In mathematics, a base (or basis) B for a topological space X with topology T is a collection of open sets in T such that every open set in T can be written as a union of elements of B."

It it saying that every open set in T can be written as the union of ALL the elements of the base? Or is it saying that every open set in T can be written as the union of SOME of the elements of the base?
 
no, it would be nonsense if every open set could be written as a union of all elements of the base (what would that mean? the union of all elements of the base is a single set: the whole space)
it's the latter
 
Yes :)). I agree. I thought so too. I just had to be sure.
 
here's an alternative definition of base; you might want to check that the two def'ns are equivalent
 
@MikeMiller Which proof? I mean, what're you referring to?
 
$\mathcal B \subset \mathcal P(X)$ is a base for a topology on $X$ if $\bigcup \mathcal B = X$ (that is, the basis covers $X$) and if $B_1, B_2 \in \mathcal B$, then for any point $x \in B_1 \cap B_2$, there is a basis element $x \in B_x \subset B_1 \cap B_2$
looking at your wikipedia article that's actually the first thing they list under properties
 
7:02 AM
Oh I just had a look at that theorem.
 
prove it yourself!
 
So that's an alternative definition
 
@BalarkaSen I'm not aware that Arnold wrote more than one topological proof of abel-ruffini. but to clarify, the article I glanced at outlined it as considering the solvability of the monodromy groups of appropriate riemann surfaces.
anyway, bedtime
 
Goodnight. Thanks Mike.
 
@eXtremiity It's the definition I was taught
 
7:03 AM
@MikeMiller Oh, you're referring to topological galois theory.
Yeah, they prove that the monodromy of a quintic Riemann surface is not solvable.
And that only stuffs in radical extensions of $\Bbb C(z)$ has solvable monodromy.
 
@eXtremiity you like topology?
 
How can I use that $\mathbf R^2\setminus S^1$ has two connected components to prove that $\mathbf R^2\setminus \partial K$ has two connected components when $K\subseteq \mathbf R^2$ is compact convex with non-empty interior (special case of the Jordan curve theorem). I know that if $K$ satisfies these requirements we have $K\cong \overline{B(0,1)}$ and $\partial K\cong S^1$ and assume I can apply this in some way but I can't quite see how.
 
@ComTruise. I love analysis. I really enjoy it. If I had a magnificent memory, I would love it even more.
How about you?
 
@MichaelCorleone Does that mean the interior of $K$ is homeomorphic to an open unit ball?
 
@ComTruise Yes I think so
 
7:15 AM
Sometimes topology is really boring.
Mostly because it throws away pieces of informations about angles.
 
Well then $\mathbb R ^2 \setminus \partial K \simeq \mathbb R ^2 \setminus S^1$ right???
 
I'm not sure how
 
the interior portions are homeomorphic and the exterior portions are homeomorphic
just construct the homeomorphism piecewise
 
The interior portions are homeomorphic I agree, but I don't see how the exterior portions are homeomorphic
In any case, thanks for the help. I'll give it a try
 
that would be more than necessary... you have $\mathbb R ^2 \setminus \partial K=int(K)\cup \mathbb R ^2 \setminus K$
you know int(K) is connected
so just show $\mathbb R ^2 \setminus K$ is connected
using convexity and compactness of $K$ I think it is easy to show it is path connected, hence connected
 
7:26 AM
Thanks for the suggestion
 
just a suggestion, there are probably multiple ways of going about this
 
8:21 AM
@r9m are you there ?
 
8:33 AM
Hey @G.T.R
 
GTR. Your polynomial trick is really cool.
Do you think you could provide me with the method :p ?
I'd like to impress my classmates.
 
r9m
@G.T.R am now :)
 
@r9m. Watch the latest Naruto?
 
r9m
@eXtremiity I'm downloading it now :)
very poor internet speed :( .. Its killin me
 
It's a great episode!
I'm sure you will enjoy it.
 
r9m
8:42 AM
gr8 .. but no spoilers plz :P
 
:D :D :D ! ofc not :). enjoy!
 
@eXtremiity well you can try this one as well math.stackexchange.com/questions/811503/another-polynomial-game
@BalarkaSen good day
 
@G.T.R. That's different to where I give you f(x) and f(y) and you tell me the polynomial, right?
 
@eXtremiity yes, very different (much harder)
 
Interesting.
 
8:49 AM
@eXtremiity for the other game, the crucial point is that coefficients are positive
the strategy is to ask for $P(1)$ first
 
Ahh :)
 
this gives you an upper bound on the greatest coefficient of the polynomial
and say $P(1)=x$
 
Upper bound?
Hmmm
 
yes, $P(1)$ is the sum of the coeffients. Since they are positive, $P(1)$ is greater than the biggest coeffient of $P$
 
Ah, ok yes - the smallest upper bound. The supremum.
well perhaps not exactly, but I understand what you mean. Then what do you do?
 
8:55 AM
then ask for any $P(y)$ where $y>P(1)$ ($y$ integer)
 
r9m
@G.T.R nah ... easy enough :-) .. I've got a similar variant but with entries of a square matrix, where the objective of the first players is to make the determinant $0$, and that of the second is to stop the first palyer :) ..
.. this is pretty much the same as the polynomial one :)
 
The trick is that $P(y)=\sum_0^n a_ky^k$, and this is looks a lot like writing numbers in base $y$
 
Cool!
 
@r9m nice one... is it on main ?
 
r9m
@G.T.R idk .. it is from a local competition :)
 
8:58 AM
@eXtremiity next thing you do is you convert $P(y)$, which is given in base $10$ to base $y$ and that gives you the coefficients
@r9m game theory is non-existant in French education, pretty sad
 
r9m
@G.T.R oh .. I took a game-theory course by force :P .. against my instructor's advice :P
 
Hmmm, I see. And what are the conditions on the polynomial? The coefficients have to be positive?
 
@eXtremiity yes, it's the key point
@eXtremiity you want to guess my polynomial ?
 
I have not exactly understood your procedure. Well the last step that is.
Can I give you polynomial images and you run me through it?
The powers can be any positive integer? Or do you require consecutive powers?
Actually, don't worry :). I should be studying. Games can wait :p
 
@eXtremiity fine, I have to go as well
 
r9m
9:05 AM
@eXtremiity exams ahead ?
 
Thanks though :).
Yeh, my last exam is on Monday.
Real Analysis >_>
though.
 
r9m
oh ,,. Best of luck !! :D
 
Thanks ^_^.
 
r9m
@G.T.R are you on fb btw ? :P
 
@Hippalectryon Heh. Do you think it would please him? : )
 
9:13 AM
Interesting polynomial question: math.stackexchange.com/questions/840496/… Would love to get some quick help there ;)
 
9:28 AM
@r9m I am, what's your profile url?
@r9m done
 
r9m
9:43 AM
@G.T.R did you just send me fb request ?
 
@r9m no, not me. I sent you an email
@r9m sure
@r9m done
 
recognizing semidirect products are a hell bit confusing.
completely confuzzled
 
r9m
10:16 AM
@robjohn can you remove the chat line above with my mail address ? :) .. It was too late to delete (I got disconnected there)
 
Greetings
hmmm, I've just created a new pack of cute questions.
 
@r9m Starred :P
@r9m A quick question for you, prove that no continuous real to real function can take each value in $\mathbb{R}$ twice. In case you haven't heard of it already :)
 
r9m
@Sawarnik that's simple :) .. do you know how to prove it ? :-)
 
10:40 AM
@r9m Yesterday I computed (you should give it a try) $$\int_0^1 (\log(1-x) \log(1+x))^2 \ dx$$ I love this integral (I found a d**n clever way)
 
r9m
@Chris'ssis okay :D :)
@Chris'ssis elementary way ? :-)
 
@Sawarnik By twice do you mean 'twice' or 'exactly twice' ?
 
Hello @TedShifrin
 
r9m
@Hippalectryon he possibly means exactly twice :)
 
@Hippalectryon He means all continuous real to real function is injective.
 
10:49 AM
@r9m Well, this is a discussable point. ;)
 
@Chris'ssis That's weird
 
Maybe
 
@Chris'ssis The $\frac{1}{2}\left(\log^2(2)-\zeta(2)\right)$ in the sum does not depend of $n$, why don't we take it out ?
 
$n$ is the index, @Hippa
 
@Hippalectryon are you kidding? :-)
 
10:51 AM
Oh wait it's not a sum it's a series
-_-
Nvm
 
@Hippalectryon How are you? (btw)
 
@Chris'ssis fine :)
 
Got 50 minutes before my next class
 
@Hippalectryon and .. who are you? :-)
 
10:53 AM
@Studentmath Hey ;)
 
@Chris'ssis No idea :P
 
@TedShifrin Heyy Prof. Ted :)
 
@Ilan heya, checking it out
 
@Studentmath You go work on your exams! lol
@Studentmath I'll check the answer in some minutes, going back to studying ;)
 
@r9m Sorry to take so long. I was away from chat and didn't hear the ping.
 
r9m
10:54 AM
@robjohn Thanks a lot :D
 
I can't believe I am making basic group theory mistakes.
 
r9m
@Chris'ssis nice opportunity here .. ;) .. who are you super sis ? (I'm aware super heroes don't reveal their true identity .. but still :P )
 
@Ilan did you mean 'also there's a $u$ so that $T^{n-1}(u)=0$? not $\neq$?
 
@r9m you made me laugh out loud :-)))))))))))
 
r9m
LOL
 
10:57 AM
@IlanAizelmanWS Seems good to me
 
I am forgetting my math.
sigh I think I forgot most of what I knew of group actions.
 
@Sawarnik it's the extreme/intermediate value theorem applied multiple times
 
I know what you're thinking @Hippa
.
 
@BalarkaSen :D
 
I think my Graph Theory Prof. thinks I am retarded
 
r9m
11:02 AM
Bite him :P and prove his point :P
 
@Hippalectryon Heh.
That was a classic one, really.
 
@Student discrete math are the toughest IMO
 
@BalarkaSen Best meme i've ever donc :3
 
@r9m Better, do a tongue twist hex. Langlock
 
@G.T.R I love them the most - I get 95+ in every work (besides one where I got 90) and I pretty much know everything right. But he is cynical (funny cynical) when he corrects my mistakes.
 
11:05 AM
@r9m that integral I showed you has amazing connections with some important series that may help you evaluate some integral-titans. The point is to bring it to the useful form since otherwise is good for nothing (for this you need to put some efforts).
 
r9m
@Chris'ssis I have to undergo crush the minions plan .. I'll approach the titans eventually :P
 
@r9m :-))) Well, each integral of this type has a lot of things to say ...
@robjohn have you seen the integral above? It's really crazy awesome!
 
Anyone who is interested : Find a reducible (in $\Bbb Z[x]$) polynomial $f(x)$ with all coefficients in $\{0, \cdots, 9\}$.
Looks easy, but is extremely hard.
 
r9m
 
@Studentmath Got it sorted out now, thx! :)
@Hippalectryon Thx 2.
 
11:17 AM
@IlanAizelmanWS :)
 
In 2011, the OPERA experiment mistakenly observed neutrinos appearing to travel faster than light. Even before the mistake was discovered, the result was considered anomalous because speeds higher than that of light in a vacuum are generally thought to violate special relativity, a cornerstone of the modern understanding of physics for over a century. OPERA scientists announced the results of the experiment in with the stated intent of promoting further inquiry and debate. Later the team reported two flaws in their equipment set-up that had caused errors far outside their original confi...
;)
 
@Chris'ssis Nooooo
 
r9m
@Chris'ssis ya .. ;)
 
@Chris'ssis @r9m We know it's an error :3
Too bad
xD
 
Actually, let me do a correction : *coefficients in $\{0, ..., 10\}$. Bonus : Why doesn't the previous one work?
 
r9m
11:23 AM
@Hippa that meme is just a silly joke with 'Pics or it didn't happen' :P
 
@r9m ye i know
 
I needed help with finding the derivitive of $f(x)=\int^{x^2}_{0}(1+t^2)^{-3}dt$
I tried dividing the limits by x (and other places), but couldnt get an answer, help?
 
r9m
In calculus, Leibniz's rule for differentiation under the integral sign, named after Gottfried Leibniz, tells us that if we have an integral of the form : \int_{y_0}^{y_1} f(x, y) \,\mathrm{d}y then for x in (x0, x1) the derivative of this integral is thus expressible : {\mathrm{d}\over \mathrm{d}x} \left ( \int_{y_0}^{y_1} f(x, y) \,\mathrm{d}y \right )= \int_{y_0}^{y_1} f_x(x,y)\,\mathrm{d}y provided that f and its partial derivative fx are both continuous over a region in the form [x0, x1] × [y0, y1]. Thus under certain conditions, one may interchange the integral and partial di...
 
@r9m I have to use the first fundamental theorem of calculus
 
@VibhavPant Do you know the derivative of $g(y) = \int_0^y (1+t^2)^{-3}\,dt$?
 
11:27 AM
(this is from Apostol)
@DanielFischer yes
$g'(y)=(1+y^2)^{-3}+C$ IIUC
 
@VibhavPant So does writing $f(x) = g(x^2)$ help?
 
oh
/me tries
$(1+x^4)^{-3}$?
oh wait
chain rule
Thanks @DanielFischer
 
Right, chain rule.
 
Also
Isnt $\int^{x^2}_0(1+t^2)^{-3}dt=x\int^x_0(1+(tx)^2)dt$?
(dividing limits and friends by x)
 
That works too.
 
11:33 AM
looks like Im wrong somewhere
So, if $f(x)=x\int^x_0(1+(tx)^2)dt$, how would I evaluate $f'(x)$
 
Well, you are left with an integral in there after differentiating. If you evaluate that integral, it should come out correctly.
 
$$x^{11} + 8x^6 + 9x^5 + 10x^4 + 8x^3 + 7x^2 + 3x + 3 \\ = (x^9 + 3x^8 + 6x^7 + 9x^6 + 9x^5 + 8x^4 + 6x^3 + 4x^2 + 2x + 1)(x^2 - 3x + 3)
$$
 
@Chris'ssis this integral?
 
This is the gigantic polynomial, coefficients positive and smaller than 10 and prime at $x = 2$
 
@BalarkaSen Did you prove the unicity ?
 
11:37 AM
@DanielFischer So the first fundamental theorem only works if the function under the integral is a function of only t $t$? (assuming $\int^x_0 g(t)dt)$
 
@Hippalectryon Unicity?
I don't even know what that is.
 
@BalarkaSen you said the so it's unique ight ?
 
@Hippalectryon Oh, no, presumably it's not unique. In fact, I think there is a polynomial of degree 10 whereas mine is 11.
 
@robjohn Yeah.
 
@VibhavPant If the integrand also depends on $x$, you need the Leibniz rule, involving an integral of the derivative of the integrand with respect to $x$. It works, but in cases like this, it's more complicated.
 
11:41 AM
Ah
Thanks!
 
In fact, @Hippa, I believe $\{0, ..., 9\}$ problem is still open.
 
12:00 PM
Question about the Jordan's curve theorem - what about the points sitting on the curve itself? They are both inside and outside, neither..?
 
@Studentmath Neither. They are in the boundary of both regions.
 
@Daniel Thanks!
 
@r9m Yes, its very simple :)
@Hippalectryon Exactly twice.
 
Hoy @Sawarnik
 
Aoy @Balarka
 
12:09 PM
Got any interesting problem, @Sawarnik?
 
@BalarkaSen Isn't it very old?
 
yep.
 
If this is what topology is like, god save me...
This is the most rigirious proof I've ever read
And there are some creepy proofs in Graph Theory
 
point set topo is boring, yes.
 
@Studentmath Which?
 
12:15 PM
Jordan Curve Theorem
Well, restricted version of it
but still
 
Ah.
 
@BalarkaSen Disagree. It can be fascinating.
 
It's not trivial, that proof.
 
It's very intutive
 
@DanielFischer =(
 
12:16 PM
but the proof is terryfying
terrifying
terr..
Scary.
 
Hi Everyone
 
@Sawarnik Do you have any number theory?
 
No.
 
Anything else?
Functional eqn stuffs, @Sawarnik?
 
@Chris'ssis I have an idea... it uses an identity that I have used before...
 
12:26 PM
@robjohn What identity?
 
@BalarkaSen let $f:[-1,1]\to \mathbb R$ such that $f(x^2-1)=2xf(x)$.prove that $f=0$
 
@Chris'ssis $\log(x)=\lim\limits_{n\to\infty}n(x^{1/n}-1)$
 
Dan
Anyone here good at stochastic processes?
 
r9m
@Chris'ssis Have you seen this one ? :D $ \displaystyle \lim\limits_{n \to \infty} \sum_{k = 1}^n\frac{(-1)^{k-1}\cdot\sqrt{k}}{(n-k)!\cdot(n+k)!}$
 
@robjohn Does it really work?
 
12:33 PM
I didn't see that. @Sawarnik
@G.T.R Hmm.
 
@BalarkaSen I gave this link, open on your own risk: math.stackexchange.com/questions/184043/…
 
@r9m Is it that series by Knot? I saw it somewhere on this site I think.
 
No, no, I don;t want to see that.
 
r9m
@Chris'ssis Knot ?! :o what is that ?
 
Dan
If I know the autocorrelation of some process $v(t)$, can I say something about the distribution of $\int_0^t v(\tau)d\tau$?
 
12:35 PM
So IVT?
 
@Sawarnik come on, don't spoil
 
Is it continous? @BalarkaSen And then also, what does IVT do here?
 
r9m
@Sawarnik How fast are you at tracking problem sources ? :P
 
Sorry.
My bad.
 
Yes @BalarkaSen $f$ is continuous
 
12:36 PM
^You didn't mention that :O
@r9m :D Many of these questions I have already seen before, its a matter of finding the links :D
 
@r9m Me neither.
 
@r9m Are you sure what you wrote is correct?
 
r9m
@Chris'ssis ya .. :D well I am less interested with the limit .. but with the n-th term of the sequence :-)
 
@r9m Ok, prove that there is a continuous real to real function can take each value in R thrice :) Have you seen it as well?
 
@r9m Well, that's another story. :-)
 
r9m
12:43 PM
@Sawarnik there are lots of piece wise linear functions with that property :)
 
@r9m Right, genius.
 
1:01 PM
...silence...
 
Dan
@r9m How can that be? (I am probably being really dense)
 
r9m
@Dan Its simple ... wait lemme draw it in MS-paint :-)
 
Dan
Wait I think I got it. Something like an increasing saw-tooth
 
r9m
@Dan ya .. right :-)
 
1:42 PM
@r9m If considering some integers powers in numerator (instead of that radical), we get some nice formulae. Perhaps, it might work then to pass from the integers to reals ... (I did something like that in the past)
@r9m hmmm to better understand me ...
@r9m $$\sum_{k=0}^{n} \frac{(-1)^k}{(n-k)! k! (2k+1)!}$$
 
r9m
@Chris'ssis okay :)
 
1:57 PM
@r9m let me know if you recognize the nice form of it.
 
r9m
@Chris'ssis thumbs up (y) :D .. I'll look into it :-)
 
@r9m You did? ... Ah, OK :-)
@r9m These questions, both yours and mine, can be done by playing with some partial fractions. (I hope I'm not wrong in your case) $$\sum_{k=0}^{n} \frac{(-1)^k x}{(n-k)! k! (2k+x)!}=\frac{2^n}{(x+2)(x+4)\cdots (x+2n)}$$
For the case above, set $x=1$ and you're done.
 
r9m
@Chris'ssis :-) right-o-right .. but what about the $\sqrt{k}$ on the top ? :o
 
@r9m Perhaps you initially consider $k^m$, where $m \in \mathbb{N}$ ... get a formula in the right that also contain $m$ (of course) ... and then note you can extended that $m$ from $\mathbb{N}$ to $\mathbb{Q}$. Set $m=1/2$ and you're done.
(it's an idea only)
 
r9m
2:14 PM
@Chris'ssis ya ... I was trying on the same line .. :o but Idk how to proceed
integer powers on top seem to be easier to handle .. how the heck am I supposed to get fractions there .. curses the problem :(
 
@r9m Does the solution here help you? math.stackexchange.com/questions/828630/…
:-)
bbl
 
r9m
@Chris'ssis yes .. residues help a lot !! :P .. but I'm onto single variable calculus (Calculus I) proof for this one (logarithmic derivative and chain rule) .. if I succeed I'll post :D
 
3:04 PM
@r9m Would you like some functional eq stuff?
 
sigh I always feel my proofs are too messy, not rigirious enough, too rigirious..
It's so annoying.
 
c c
How do you prove an improper integral exist (for a function defined on $[0,\infty)->C$? Cauchy criterion?
 
3:27 PM
(r(e(m(o)v)e)d)
 
@r9m another out-of-the-blue integral inequality... Let $f\in C^1([0,\pi], \mathbb R) $ such that $\int_0^\pi f=0$. Prove that $\forall x\in [0,1], |f(x) |\leq\sqrt{\frac{\pi}3 \int_0^\pi {f'}^2}$
 
3:57 PM
@G.T.R Huh,
@G.T.R My instinct would be to use FTC.
Also, use Cauchy Schwarz.
And introduce some sines and cosines.
You want to prove $$f(x)^2\leqslant \frac{\pi}3\int_0^{\pi}f'^2$$
 
@Pedro I did that but I can't make any use of the hypothesis $\int_0^\pi f=0$
 
r9m
4:13 PM
@G.T.R ah .. well we can begin by looking at Lagrange Identity for $\displaystyle \int_0^x t^2\,dt \cdot \int_0^x f'(t)^2\,dt - \left( xf(x) - \int_0^x f(t)\,dt \right)^2$
by CS inequality it is always non-negative ..
idk .. differentiate this wrt $x$ and check the increasing/dicreasing nature of this expression .. might work .. my idea was at $x=\pi$, we get $f(\pi)^2\leqslant \frac{\pi}3\int_0^{\pi}f'^2$, so maybe we can proceed along a similar line
.. now excuse me .. If I have a football freak rumbling inside my tummy .. BBL :P
 
I wonder how I can get to the thickness of the peterson graph
 

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