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07:00
I probably could not S++ every level in that game, even if I put 1000hours in, but I think I could A+ every level in SMB in less than 1000hours
Here's an old version of the best times on everything
I have SS'd every level but one... it burns
Oh seriously!
Hello@Balarka!!!
Well, all the normal levels. It then opens a room with "Megadifficult", "Gigadifficult", etc
I think Giga is the one towards the achievement so that's the 'one' I haven't SS'd
Only one of the crazy ones I've beaten...
Well I dont want to ask this question on main so I am asking it over here
Given two metric spaces X and Y and given that f and g are continuous mappings from X to Y and f(x)=g(x) for every x in A where A is a subset of X. Prove that f(x)=g(x) holds true for every x in $A$closure
@PaulP can i get some help from you?
Is it fine if I just prove f(x)=g(x) for some $x\in A'$ where $A'$ is the set of limit points of $A$
07:06
Hmm, it looks like there might be more about the directional derivatives being linear for differentiable function than meets the eye. I guess we can do something wacky forgetting about what $D_{\vec{v}}\vec{f}$ says about $f$, and concentrate on the $\vec{v}$?
Hello@Balarka Are you free?
@MikeMiller if $f$ is an entire function that is not a polynomial, why does the functoin $g(z)=f(1/z)$ have an essential singularity at $0$?
I am not, @Remember
Okay fine
@Remember hint : draw a picture. don't forget what limit points and continuous functions mean.
07:08
@iwriteonbananas Remember the classification of singularities. Show that the assumption that f is not a polynomial implies |f(z)| does not just blow up as you go to $\infty$.
Since it's also clearly not a finite value at infinity it must not be a removable singularity.
@MikeMiller did you mean $|g(z)|$ does not blow up for $z\to \infty$?
I meant |f|. Equivalently I meant |g(z)| does not blow up as z -> 0.
right, my bad
well, that's what i've been trying to see, but i havent been able to.
07:58
Ahh I still cant get it
@Remember $\{a_n\}$ be a sequence converging to $a$. If $f(a_i) = g(a_i)$ for all $i$, is it ever possible that $f(a) \neq g(a)$, where $f, g$ are continuous functions?
There should be nothing so hard about the exercise once you know why it should be true.
@BalarkaCant you do it by using definition of continuity and idea of limit points?
so why in the world aren't you doing it?
I am trying my best to do it ... I dont like sequences and i dont want it to be done through it..@Balarka
well, probably you don't like sequences because you haven't studied it well enough.
aren't you doing Rudin?
08:09
That might be a reason...
But using open balls and stuff its better at questions in continuity I feel
No simmons@Balarka
ok, if you want to do it by open balls definition, you can do that too.
all you need is the definition of continuous functions, really.
that
$f(B_\delta(x)) \subseteq B_\epsilon (f(x))$?
without appropriate quantifiers, that statement makes no sense.
Okay okay
but yes, you can use the pullback of open sets is open definition to do it.
08:12
Ahh thats one theorem which i proved myself
which theorem?
Balarka, where did you study analysis from?
I didn't study a lot of analysis, @Soham.
The little you have.
did a first few chapters of Rudin once, and that's about it.
08:13
Oh.
If F is continuous then f^-1(G) is open whenever G is open @Balarka
Well with the metric spaces
and all
@Soham When are you going to meet Samik?
Kichhu bole ni specific.
Whose Samik@Soham
He said he'll inform me when he gets back.
@Rememberme A prof
08:17
ah, okay.
How do you guys get to meet prof... !!
I have never met one
@Rememberme I . . . worked it out. ;)
Gugel shows the way.
glares at @Soham
08:18
Wha??
I have a black belt in Google-fu.
we don't want the uni to be filled with 15 y.o. grad students, so shaddup.
"we"
Grad students?!
Why?? and Grad students !!
And correction : I'm 16.
:)
08:19
You're in 11th grade?
Yeah.
:(
Doing a lot of mult calc?
Ahh Soham cmon tell me how do you get to meet these people
Where do you live?
Bangalore
@SohamChowdhury yeah, kind of.
08:21
And I haven't met anyone yet.
@Remember lot of good math guys in IISc. mail one of them, I guess.
You will
@Rememberme Find the websites of the unis close to you and email the people there. Exactly what I did.
Ahh .....
@BalarkaSen CMI is not in Bangalore
08:22
I never plucked up the courage to mail anyone :P
Balarka you are the master.... people call to talk to you :p
@BalarkaSen Spectral sequences guy/gal?
@Soham oh, I mean, I've never mailed anyone to get them to talk to me :)
@DanielFischer if $g$ is an entire function, why does an estimate $|g(z)| \leq | \sqrt{z} |$ force $g$ to be constant?
My schoolteachers did it for me, I think.
08:26
@BalarkaSen You have nice teachers is all I'll say. :)
sure thing.
Anyway. Bunch of slick proofs in the cosets/quotients section.
oh, speaking of it, T. Chow mailed back. finally.
@BalarkaSen Ahh Now I get what you mean
Tim Chow?!
08:27
yeah, author of the article on spectral sequences I was reading.
08:40
why the "hence . . . " bit?
@iwriteonbananas there?
@Balarka?
09:03
@iwriteonbananas If a holomorphic function grows too slowly near an isolated singularity, then the singularity is removable. Here the isolated singularity is $\infty$. You can see that for an entire function by the Cauchy estimates, for $\lvert z\rvert \leqslant R$, we have $$\lvert g'(z)\rvert = \frac{1}{2\pi} \left\lvert\int_{\lvert\zeta\rvert = 2R} \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta \right\rvert \leqslant \frac{1}{2\pi}\cdot 2\pi(2R)\frac{\sqrt{2R}}{R^2} = \frac{2\sqrt{2}}{\sqrt{R}},$$
and letting $R\to\infty$, you see $g'(z) = 0$ for all $z$.
Uh, @DanielFischer, can you help at all?
@SohamChowdhury What with?
why does aH = Hb for some b imply aH = Ha as well?
as cosets?
@SohamChowdhury Since $a \in aH$, from $aH = Hb$ it follows that $a\in Hb$. Also, $a\in Ha$, and the cosets $Hb$ and $Ha$ are either disjoint or identical. Since the are not disjoint, we have $Hb = Ha$.
ah.
thanks.
09:08
You're welcome.
Hey guys
basic question about probability someone asking me -
if A, B are events
is B|A always independent of A?
I think so...
but I'm not certain it makes sense, haven't done prob in a while
09:34
@DanielFischer @SohamChowdhury maths.kisogo.com/index.php?title=Coset for future reference.
09:49
@DanielFischer good point, thanks
@BalarkaSen when you told me to construct a surjective self map on $S^n$ of degree 0, is this what you had in mind? $S^n \to D^n \to D^n/ \partial D^n \cong S^n$ where the 1st map projects everything down (i.e. "forget the last component") and the 2nd map is the quotient map
off topic - what does a weighted sum of 1 have to do with convex hulls? I understand the definition of a convex hull a set of points as the smallest convex set containing the points, but I don't see where a weighted sum of 1 enters the discussion.
10:16
@iwriteonbananas What d'you mean by "projects everything down"?
You mean projecting both of the upper and lower hemispheres onto the disk bounded by the equatorial circle?
But then your map is of degree $2$, not $0$, right?
Ah, cool. I finally understand why $\Bbb Z/n\Bbb Z$ is the One True Notation.
That's exactly how I felt after I learnt quotient groups.
somewhat unsurely raises hand for a high-five
Quotients and subgroups are done (somewhat)! :)
shakes head hopelessly kids these days.
2
@Soham Prove that $Q_8/\Bbb Z_2 \cong V_4$.
you don't really know something until you do the exercises - Balarka
Almost inb4.
10:27
oh, btw, you didn't compute $\mathsf{Aut}(D_4)$.
@BalarkaSen Trying.
after that, you had to do the harder $\mathsf{Aut}(Q_8)$
(fun fact : the latter can be done geometrically)
$Q_8/\Bbb Z_2$ is hand-doable. :P
(I think)
10:30
well, sure.
do you know the first isomorphism theorem yet?
no.
next section.
ok, prove that one yourself.
$f : G \to H$ be a homom. prove that $G/\ker f \cong \text{im} f$
It's not that hard. If you understand quotients, you can do that without much voodoo
of course, you have to show that kernel is a normal subgroup and all
done that.
$\varphi(gng^{-1})$ etc
10:33
good
{1,-1} under multiplication is iso to $\Bbb Z_2$ na?
yeah
do the first isomorphism theorem, then that Q_8/Z_2 \cong V_4 thing follows as a corollary
@BalarkaSen yes, that's what i mean. why would the composition have degree 2? it factors over 0 since $D^n$ has homology 0 for $n\geq 1$
it doesn't factor over anything at the homology level. how do you define degree of maps between non-sphere things?
as for why it's 2, note that fiber over each point has cardinality 2.
@BalarkaSen homology is a functor, so it preserves composition. the composition is a map $S^n\to S^n$ but it factors over $D^n$, which in homology is 0
10:43
oh, nevermind my cardinality of fiber argument. i was lying.
hmm. ok, I agree that your map has degree 0
@iwriteonbananas :P i was confused by your composition of things. actually, the map I had in mind is the same as yours, but I'd have put it in a different way
how would you put it?
namely, take S^n, map it to S^n by sending the upper hemisphere to all of S^n by pinching the equator and then the lower hemisphere to all of S^n similarly by pinching the equator, but compose with the antipodal map before this time.
this has degree 0 because it's nullhomotopic
another equivalent way too put this would be to take a nullhomotopic surjective map $S^1 \to S^1$ (given by the word $aa^{-1}$, say) and suspend $n$ times.
suspension preserves degree (why?)
what does the map given by the word $aa^{-1}$ do?
hmm nvm
exactly what u described above, right?
10:48
yes, precisely
i havent learned about suspensions yet, so i cant answer ur question
it's in Hatcher
well, if A = {1, -1}, $Q_8/\Bbb Z_2 = \{A, iA, jA, kA\}$, and that last set is iso to $V_4$.
works?
why is your way of defining the map the same as the composition i described above?
@SohamChowdhury sure
10:50
ah.
i feel like I "get" why it does. :)
@iwriteonbananas look at what you map does restricted to the upper hemisphere
and then look at what it does to the lower one
it's precisely what I am doing with my map
What are you working on?
true
that's a cool alternative way of seeing it
10:52
@BalarkaSen I want to see a bunch of simple (not the technical way) quotients of groups I know. any idea where I can find a few examples?
yeah, discovery of that got me a compliment from Ted saying my geometric intuition is better than him (although I am still not sure if it was meant as a sarcasm), @iwriteonbananas
@SohamChowdhury Dummit-Foote
loads of examples there
let me dig out my not-so-legal copy.
you're slowly turning into jasper, as already has been said
10:54
are those examples from things I'll know?
oh, well, I'd like to think I'm helping you out in some way.
really.
i was just joking
and it hurts my eyes.
there's that, too. :P
okay. I've opened DF.
well. this is a big book.
in fact, I think I'm going to look over this book to kind of ensure there's nothing incomplete about what I've learnt. here goes.
11:51
it's hard to focus, it's so friggen hot here
i'ts the same tempereature here
are you from kolkata?
yes, I am.
oh, wait.
are you too?
well, that reduces the number of places by one. :P
11:58
indeed indeed....
so you're from the same place as balarka?
aye.
i was actually guessing that you know @Balarka irl before your "hell no".
kolkata is far far away from me
well, that puts a "possibility metric" on the set of places
not a metric actually. but still.
12:01
what are u studying atm?
what are you doing? complex?
no, im procrastinating
because i need to compute this integral
I'm reviewing the group theory I learned (till quotients) from Dummit-Foote.
put the \s in
no, i dont care
i'll just delete it
bananas pls :'(
12:03
it's too ugly
what do you need this for?
analysis on manifolds course
divergence theorem etc.
mm.
@Balarka look at this. $Q_8$ is not the only one.
12:11
hi guys :D
et al
Hello @skillpatrol !!
Could someone of you tell me why it stands that $R_x+F_{W_0, x}=0$ and $-R_y+F_{W_0, y}=0$ at the question physics.stackexchange.com/questions/187539/… ?
@MaryStar Hi pal, how are you?
I'm fine... I am preparing myself for Monday...

How are you? @skillpatrol
@MaryStar Fine thanks :-)
12:39
Hello@Soham
hi.
what's up? still on your break?
No.. no Finished continuity today
with exercises
Think of this @Soham:
Can a mapping be closed or open?
a function can be closed or open, if that's what you mean
Yup mapping~function
user147690
Sad doing past assignments for revision and solving the whole things in 30 min, when you got like 75% the first time around...
12:42
Hello@AlexC
user147690
Hey Sayan
Its better we are studying on our own no pressure of assignments at least @Alexc :p
user147690
@Rememberme Indeed! And I suppose that it is good that I have learned the content then :)
Yes...
So @Soham So what do you think can I say a mapping from two metric spaces X and Y which is continuous open or closed??
@AlexClark I know that feeling :(
@Rememberme Yes, I think I saw that in Bredon. Let me finish revising from DF and then I'll hit that book again.
Can't remember.
12:47
hey everybody i want you to judge me this math.stackexchange.com/a/1308281/202081
Oh okay .... Let me try proving theorems myself
I have been getting some success
Balarka voice That's very good.
is it the right limit ? or is there some tricky way i was not been aware of ?
13:03
wait.
@Balarka, does one get $SO_3$ if one sets $A = -A$ in $SU(2)$?
If so, I think I'm beginning to "get" quotients.
sulphur trioxide?
In mechanics and geometry, the 3D rotation group, often denoted SO(3), is the group of all rotations about the origin of three-dimensional Euclidean space R3 under the operation of composition. By definition, a rotation about the origin is a transformation that preserves the origin, Euclidean distance (so it is an isometry), and orientation (i.e. handedness of space). Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation. Composing two rotations results in another rotation; every rotation has a unique inverse rotation; and the identity...
Topology?
Altop?
Na re.
Just groups.
Oh.... DF has this??
13:07
Aluffi ^_^
gtg cming back in 5min
later pal
@Soham How do you get those small shots
hey@Skill
@Balarka DF had around four examples of quotients. Am I missing something?
Yes ! Yes I have understood first half of your blog @PaulP
13:29
@robjohn I just evaluated another amazing series!!!!!!!!!!!11
$$\sum_{n=1}^{\infty} \frac{H_n}{n^2}\left(\zeta(2)-1-\frac{1}{2^2}-\cdots-\frac{1}{n^2}\right)$$
Great!!!!!!!!!!!!!!!!!!1 Well, I simply love myself (for getting such results)! :-)
You've been doing this $\zeta(2) - \left(\sum_{k=1}^n \frac{1}{k^2}\right)$ thing for a while.
whats ζ(2) ?
@SohamChowdhury That can be written in terms of polygamma, but I like it more the way it looks in this form.
@Agawa001 $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$
ah ok
?
13:35
That's how it should look.
u r editin it too much !!
ok
u hv prblm? i dnt k???????
That's painful :P
i was just bein confused
the actual one is correct
i assume that ζ(1)=e ?
m i wrong ?
yes i think
@Chris'ssistheartist That looks like something we've already done, but I could be mistaken.
@robjohn No, definitely no. That one is very advanced amongst series (especially when only using elementary tools of real analysis).
13:42
@SohamChowdhury I know Q_8 is not the only one
Hello@Balarka
@robjohn We possibly discussed this version $$\sum_{n=1}^{\infty} \frac{H_n}{n}\left(\zeta(2)-1-\frac{1}{2^2}-\cdots-\frac{1}{n^2}\right)$$ that I can presently compute in 3 different ways. I showed you in the past my solution to it. I think r9m added it to his blog.
hi
@SohamChowdhury really? oh well.
come up with your own examples, then, I guess :)
hey.
$\Bbb{R/Z} = S^1$?
I think I saw that in an exercise.
R/Z has two different intepretations
13:45
Hi
one is the standard quotient group thing
@Balarka, does one get SO3 if one sets A=−A in SU(2)?
I pinged you with that.
i guess
@BalarkaSen It's not $\Bbb{R}/2\pi\Bbb{Z}$ ?
13:46
same thing
I think I'm beginning to grok quotients :)
$$\large \text{Thank you Lord for the brilliance!!! :-)}$$
we have a god-fearing lady who is adept in the sciences here, I see. :P
@Chris'ssistheartist have u ever tried/worked on to solve the puzzles of cicada 3301 ? :)
@BalarkaSen ki korli?
13:47
@TheArtist No, never heard of them. :-)
@BalarkaSen For any group generated by a single element, the order of the group is the order of the element ?
@SohamChowdhury what d'you mean?
@Gato yes
what did you do all this while?
@Chris'ssistheartist you should check them out , its pretty cool :)
@SohamChowdhury schoolwork
13:48
i have to start. procrastinating like anything, but there's only a week left.
when does yours reopen?
13th
or 15th, I dunno
15th-e extend korlo toh.
yeah, heard about that
Ours was that from the start too. 1 week :(
are you in eng mid or beng mid?
13:51
eng. you?
maybe
@BalarkaSen forgive me for not knowing, but for how many subjects do you have to appear for exams?
@TheArtist Thanks! :-) I have to finish a book first. :P
@BalarkaSen physical science is nice, I guess.
13:52
@Chris'ssistheartist heheh ok :) oh yes, hows that work going ?
sure is, @Soham
I heard ICSE is going to combine PCB.
we missed it.
anyway.
when do you wake up in the morning?
WBBSE guys after this batch are going to have open-book exams
@TheArtist I'm amazed of myself every day :-)))))))) God really created us very powerful beings. All is fine, I have a lot of marvellous work. :-)
@Chris'ssistheartist God, if he exists, sure has an eye for mathematical beauty.
3
13:54
@SohamChowdhury Well-said!
I think whoever starred that interpreted it differently.
But Occam's razor is awesome. @Balarka, yes, I guessed.
when do you wake up?
sigh. 9 AM. happy?
@SohamChowdhury He exists 100%.
Let's not argue. :)
13:56
@Chris'ssistheartist :) very glad you feel like tat, yes very true :) i wish i felt the same about myself everyday
thats math room theological debates are discussed elsewhere
@BalarkaSen yes :P
@Chris'ssistheartist agreed with you 100% :) I believe that God exists too
@BalarkaSen Wen I look at the canonical surjection between Z onto Z/nZ : $\pi$, as $\pi$ is surjective, then a subgroup of $Z/nZ : K=\pi(\pi^{-1}(K))$. As $\pi^{-1}(K)$ is a subgroup of $Z$ then it looks like $kZ$, therefore $K$ is generated by $\pi(k)$. Now if I denote $\overline{k}=\pi(k)$, I have $\vert K\vert=\vert \overline{k}\vert$, then $\pi(dk)=d\overline{k}$ where $d$ is the smallest integer such that $d\overline{k}=\overline{0}$. How can I finish to prove that $d=n/k$ ?
@BalarkaSen basically mine shifted three hours.
people, cut the starring
13:57
we're having a star-war.
2
couldn't resist.
lol ^^
i'm a filthy hypocrite sometimes.
what does "exist" mean when you are talking about God?
@skill "believe it or I'll punch you"?
@Gato OK, what d'you want to prove, precisely?
13:59
@skillpatrol "exist" about what ?
@BalarkaSen That for any cyclic group there is a unique subgroup of that form with order $n/k$

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