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3:02 PM
Function $f: \Bbb{R}^2 \to \Bbb{R}^2$ that is continuous almost everywhere (except at $(1, 0)$)? Hints?
 
@DonLarynx Just make up a piecewise function
 
$f(x) = 1 if x \neq 1, 0 if x = 1$
but that's R
 
@Gigili Hey, congratulations.
 
I have a question for people here and it is not particularly mathematical in nature.I will really appreciate it if someone could help me here. In general, is it very normal for undergraduate students to be made a party to department politics?
 
It's not even normal for graduate students to be let in on such things.
 
3:13 PM
@MikeMiller, to take a concrete example, Prof. X does not like Prof.Y and X perceives me to be close to Y,so in a private conversation X erupts at me and accuses me of being always shielded by Y?
@MikeMiller Note that X does not even mention what I am being shielded against(and I am not sure what I am being shielded against myself!)
 
ugh
that's pretty terrible
 
No, that's absolutely unacceptable behavior on the part of Prof. X.
 
@MikeMiller, so this isn't normal?I was getting used to the notion that profs are humans and are no angels and these things happen.
 
You should not be a part of this.
Yes, professors are human. They make mistakes. But this should not be the sort of mistake they make.
 
it's not necessarily shocking at the level of human nature. but it is entirely inappropriate
 
3:17 PM
Hi guys, I was wondering if someone could point me to a book for learning and understanding (weak and strong) induction thoroughly; including the write-up. :D
 
@MikeMiller, Also take this example: I cannot (and do not want to) speak to X. But when I need to, I drop by X's office(I don't email X for reasons I will mention) and ask if he is free to talk to me, he replies with a no and a rude monologue saying why he is not free . The conversation seems to take us to asking me to go to asking me who among the faculty are researchers and have a publication in the last two years.(Ours is not a very researchy university)
And the same conversation seems to happen again and again.
You have never encountered professors like that?
(sorry, if i am ranting endlessly)
@Semiclassical, I am so happy to see X being bashed!(wonderful way to relieve frustration). :D
 
No. He very well could be busy; a rant in response is unnecessary. In any case, I am not equipped to give you advice on how to deal with this, since I am neither party to this situation, nor at your university. Speak to someone there.
2
 
@user96343 Did you try brownies? People like brownies.
Or neapolitan ice-cream. It has layers.
 
Not just any layers, but three of them!
 
@PedroTamaroff They work on humans. This one insists on correcting my phrase "group of people" to "set of people".
 
3:24 PM
@user96343 Well, tell him humans are enriched with a an operation: screwing each other.
 
that's a relation
 
@PedroTamaroff LOL.
 
@Gigili I was going to add this in a comment, but it was too long by 49 characters:
$$
\begin{align}
&\begin{bmatrix}
d_{11}&d_{12}&d_{13}\\
d_{21}&d_{22}&d_{23}\\
d_{31}&d_{32}&d_{33}
\end{bmatrix}
\begin{bmatrix}
x_{11}&x_{12}&x_{13}\\
x_{21}&x_{22}&x_{23}\\
x_{31}&x_{32}&x_{33}
\end{bmatrix}\\\\
&=
\begin{bmatrix}
d_{11}&0&0\\
d_{21}&0&0\\
d_{31}&0&0
\end{bmatrix}
\begin{bmatrix}
x_{11}&x_{12}&x_{13}\\
0&0&0\\
0&0&0
\end{bmatrix}\\\\
&+
\begin{bmatrix}
0&d_{12}&0\\
0&d_{22}&0\\
0&d_{32}&0
\end{bmatrix}
\begin{bmatrix}
0&0&0\\
x_{21}&x_{22}&x_{23}\\
0&0&0
\end{bmatrix}\\\\
&+
 
Well, with enough brownies and neapolitan ice-cream you can always undo the screwing, @MikeMiller.
So it's a group, under appropriate hypotheses.
Like having enough brownies and ice-cream.
 
But the relation is a subset of $\text{People} \times \{People}$, not a map from that to $\{People\}$
 
3:26 PM
@PedroTamaroff who's the identity?
 
@MikeMiller, I thought so. I haven't done so for fear of igniting a war in the department.
 
@robjohn JFK?
 
@MikeMiller OK, you win Mike.
 
@PedroTamaroff, I may turn out to be the gg'.
 
@user96343 You should tell him it's none of his business, you don't want to be a pawn.
 
3:38 PM
@DonLarynx I wish I could. Isn't that slightly tricky?
 
it's worth pointing out that there is an Academia Stack Exchange. they might have more concrete suggestions for how to handle the situation, either on chat or their main site
 
@PedroTamaroff the group $(\frac{\Bbb{Z}}{m\Bbb{Z}})^{\times}$ is $\{\overline(k) : (k,m)=1\}$ ?
 
@Semiclassical, thank you. I appreciate that. I already it, though. I am not sure I want to create a profile there again.
@MikeMiller, thanks for giving an opinion and letting me know that this is not normal.
 
@user96343 We have some drama here as well, like I know one professor's research student is not allowed to attend conferences where another professor they don't like is attending at the same time lol
 
@ᴇʏᴇs lol.
 
3:46 PM
It kind of sucks for the student because the other professor has a lot more funding and goes to a lot of good conferences that the student can benefit from
 
Sigh, bad things happen. All the student can do is come online and anonymously(?) rant about it and feel good when others seem to be even mildly supporting her. :D
 
@MikeMiller You mean what is the inclusion?
 
Yes. Similarly, what's the inclusion $M_1 \hookrightarrow M_2$? Or more basic, $S^2 \hookrightarrow T^2$?
 
well, obviously there are no such inclusions as S^2 \hookarrow T^2.
hm. actually, i think my inclusion is screwed up.
 
Quite.
 
3:56 PM
i take it back :p.
 
Grazie.
In any case, I never was able to prove that any map would have to kill something in first homology. I had a geometric argument but it relied on a (false) fact. Which was a shame.
 
@Mike D'you know of a good way to determine whether or not any manifold $M$ retracts onto it's boundary $\partial M$?
 
It doesn't.
 
ok, and i dunno of a proof. homology doesn't seem to help.
 
You need Lefschetz duality. I think you can probably do it with Stokes' theorem too, but I don't have the argument in mind.
Of course, if you're doing so, you're restricting to smooth manifolds. But that's a restriction I'm willing to make. :P
 
4:06 PM
everything cool seem to be related to cohomology.
 
That's just because everything cool that you don't know yet requires harder math. The same will be true when you know cohomology and the basic duality theorems.
 
@BalarkaSen sorry for yesterday
I was typing the question and just found the proof @BalarkaSen sorry
 
i've always felt that homology was less cooler than fundamental group and covering space stuff
probably partially because it's an easy-to-compute invariant, unlike $\pi_1$.
 
That it's easy to compute is why we can do anything at all.
Anyway, modulo $\pi_1$, homology captures all of homotopy theory anyway.
 
really? the only relevant bridge between homology and homotopy i know of is Hurewicz, and that's seemingly not at all enough to compute homotopy in general.
 
4:12 PM
it seems like that's a pretty big modulo
 
A map $X \to Y$ between simply connected spaces that induces an isomorphism on homology is a weak homotopy equivalence.
 
that sounds new and interesting.
 
You'll learn it later.
 
"hopefully"
 
Well, I didn't include "unless you get into a tragic car accident", or "unless you suddenly quit math", as these things are always implicit.
 
4:15 PM
a term which i kind've wish i knew the meaning of (only kind've, since it's not clear if it's at all relevant to the stuff i like to do) is "group cohomology"
i've seen that occasionally in connection to certain physics things, but never in a terribly clear way
 
Hatcher refers to Brown's book on group (co)homology, you might take a look if you're interested. i haven't read it, so be warned.
 
thanks
 
speaking of homotopy, i remembered that $[A, \Omega B] = [\Sigma A, B]$ thingy. i have one or two ideas in mind, but haven't had the time to think about it.
 
It's almost definitional, minus some technicalities (which you deal with by invoking the definition).
 
i think what is relevant here is to look at the lines $\{x\} \times [0, 1]$ on $\Sigma A$. these are all loops.
 
4:20 PM
Careful. Reduced suspension.
That's the technicality.
 
yeah, except the line $\{x_0\} \times [0, 1]$ you're pinching in $SA$ to get $\Sigma A$.
 
(on the off chance anyone is curious, the physics i have in mind is discussed on an nLab page)
 
Try drawing the arrow in the other direction first if you're having trouble. Take a map into the loop space and make it a map out of the suspension.
 
on a similar note, i haven't a clue what K-theory is :/
 
Every m X n matrix over the field F is row-equivalent a row-reduced matrix. @BalarkaSen any ideas on this
 
4:31 PM
@BalarkaSen The usual proof for the disk generalizes to manifolds with boundary if you know a little bit about differential forms and Stokes's Theorem. (It's an exercise in my book, btw.)
@Remember: Gaussian elimination works over any field.
 
@Rememberme what's a row-reduced matrix again?
row-reduced echelon matrix, you mean?
 
yes
 
Yup @BalarkaSen
 
@Remember take the first row, subtract this row times appropriate constant from all the other rows and finally multiply the first row by reciprocal of the a_11 term so that you get the first term of the first row as 1 and all the others 0 below.
now take the second row, do the same as above.
can you prove that you end up with a row reduced echelon matrix after doing all this?
(some nonzeroness assumptions are needed to do what i have suggested, but i have skipped them here. you should explicitly prove what happens for those cases)
 
I think so @BalarkaSen
 
4:37 PM
prove, don't think.
 
Okay.....let me give it a try...
 
@Balarka: You do like to give orders.
 
@Mike the map $[\Sigma X, Y] \to [X, \Omega Y]$ seems to be the one obtained from restricting a map $\Sigma X \to Y$ to the lines $x \times [0, 1]/0 \sim 1$, but i haven't verified if it's a surjection. i'll check later.
i gotta go
@TedShifrin who doesn't? :P
 
Just write a map in both directions and show that they're inverse.
 
@Mike, I guess you've decided to ignore me for a few months. But, greetings.
 
4:43 PM
morning, ted (or whatever time of day it is for you now)
 
hi, @Semiclassical. It's afternoon. :)
 
gotcha
it's just about noon over here now, so i should've imagined as much
 
yup ... although I'll mess you up when I switch coasts.
 
hah
is that your plan in the near future?
 
Yup.
 
4:46 PM
gotcha. good luck with all the requisite packing etc
 
Afternoon, everyone.
Does anyone here know about active constraints and convex optimization?
Without visualization, how can one determine the active constraint of a minimization problem?
 
5:05 PM
Hey everyone
 
5:20 PM
@BalarkaSen by an appropriate constant you mean any constant c or something else
 
Hey guys
 
How do you know that there are infinitely many numbers?
 
@TedShifrin I've done very little from that Polytechnique exam (very disappointed). Today was the 6 hour long math exam for the Ens Paris, even worse
 
6 hour exam !!
@LeGrandDODOM?
hi guys
 
5:35 PM
@KarimMansour yup. here is an example from the 2013 session ens.fr/IMG/file/concours/2013/MPI/13_mp_math_d_u.pdf
 
Hello!! At a Sturm-Liouvillen problem how do we know that the two eigenvectors that we have found are linear independent??
 
I guess that day you need to sleep really well and pack some food and have really ready brain @LeGrandDODOM
 
@KarimMansour yeah, good guess ;)
 
but that is insane
 
Do you have an idea about the question I asked above?? @robjohn
 
5:43 PM
@Rememberme by appropriate, i mean "some constant", not "any constant"
 
6:36 PM
let $G \approx <a,b | b^{-1}a^3b = a^5 >$. let $H$ be a finite group and $\varphi: G \to H$ a homomorphism. then $g=a^{-1}b^{-1}a^{-1}bab^{-1}ab \in \ker \varphi$
how do i use the fact that $H$ is finite?
does this imply that $\varphi(a)^n = 1$ for some $n$?
 
dear god that's hideous
 
Heyo
 
I'm wondering what to do with this old question: math.stackexchange.com/q/470782/20792
 
6:47 PM
it seems fine; what about it?
 
None of the answers are that satisfying...I don't feel like I've gotten to the bottom of anything
 
@iwriteonbananas yikesees. where did that come from?
 
@BalarkaSen tutor gave it to me in a tutorial today
 
you have to prove that that horrendous element is inside the kernel for any finite group $H$? that's just false, so i guess the question must be something else
 
@BalarkaSen: Why should that be false?
ah, @Pureferret. If there's something you can expand on from your particular question - maybe narrow it down to a single, specific question, that you felt went particularly unanswered, you should post a new one. I don't think you'll get anything further out of editing the old one, or anything.
 
6:51 PM
@BalarkaSen i'd be very suprised if the question is flawed
 
(In particular, 'what's the longest known Collatz sequence' is probably just going to get a number someone computed, and it seems you find that unsatisfying; just be careful not to ask something that's probably open!)
 
consider $H \cong G$ and $\varphi$ to be the identity map
 
why do you think $G$ is finite?
 
Damn, Mike beat me to it. Hello, all.
 
ah, fair.
 
6:52 PM
@MikeMiller Yeah just thinking that I guess I presumed the real question was obvious
 
i didn't notice that little detail
 
considering there's only one relation, its abelianization is infinite, so it seems unlikely the group itself is finite.
 
@AlexWertheim hi!
 
hello
 
Hello, @Balarka. :) How goes it?
 
6:53 PM
pretty good, i guess
 
Still marching through Hatcher?
 
@MikeMiller can u give me a hint?
 
gonna study some multivariable calculus, taking a break from algebraic topology. the long-term goal is to study diff topo.
@AlexWertheim nah, have only done homology so far
 
Very strange to see someone learn algebraic topology before multivariable calculus. But obviously it is working well for you. :)
 
indeed, but it can't be helped now.
 
6:55 PM
@MikeMiller Thanks formulating question now
 
you have to understand that i've never went through any serious mathematical course-y stuff, @Alex. i am doing the best i can by studying most of the things by myself.
 
@iwriteonbananas I don't know how to do it and don't want to figure it out, so no.
 
of course, the people in here has been a great help :)
 
my feelings are hurt
:P
 
@Balarka: sure. That remark wasn't meant to be pejorative at all, so I hope that it didn't come off that way. In case it's not clear, I find your accomplishments very impressive. :)
 
6:59 PM
Hello.
 
@Owatch Hello to you too
 
Lol
 
no, no, i knew you didn't mean anything. i guess i am just used to taking everything as a sneer, never receiving much of an encouragement before; neither from people in here nor from the people around me.
 
Where's Tee Dog?
 
@MikeMiller done
0
Q: What causes long sequences of consecutive 'collatz' paths to share the same length?

PureferretI asked Longest known sequence of identical consecutive Collatz sequence lengths? some time ago, but I don't feel like it really got to the bottom of things. See, in the answers lopsy find a sequence in the range of $596310 ... 596349$ and makes a heuristic argument: There's nothing special ...

 
7:05 PM
@iwriteonbananas it looks insane.
what kind of tutor would give his students that problem?
 
@BalarkaSen if $H$ is finite we have $\varphi(a)^n = 1$ and $\varphi(b)^k = 1$ for some $n,k$ right?
 
yes, of course.
 
oh it was true
 
yes, it was true, @JC574
 
$ag$ is conjugate to $a$
 
7:09 PM
ok i've proved that $n$ cannot be a multiple of $3$. does that help?
 
maybe.
 
please i need the idea how to begin: math.stackexchange.com/questions/1243845/…
do not do -1 if you know the unswer tel me about it please i know the proof only when the dimention N=1
 
i have no idea what to do
 
Does anyone know when a question is put off cold?
off hold*
-2
Q: The real part of the sum $(i-1)+(i-1)^2+(i-1)^3....+(i-1)^{2013}$?

John DoeI'm not sure how to go around this one. Factorizing doesn't seem to work and there isn't a clear pattern to work by that I see. EDIT: So apparently I need to add context and stuff. I removed the checkmark from the answer because the answer for this question is one of these options (according to ...

 
@John: users have to vote to reopen the question. I've just cast the first nomination. You'll need four others to vote as I have for it to be reopened. (I don't recommend soliciting those votes here, though, unless it's really been too long.)
 
7:22 PM
@AlexWertheim Alright, thanks. Just curious. I think I've added enough substance that it is not a question without context.
Annoyed that such a dumb problem has kept me busy for hours
 
@AlexWertheim So, what have you been studying lately?
anything fun?
 
@Balarka: depends on your definition of fun. :) But I certainly think so.
Been learning topology with a manifold-oriented (hehe) approach from Lee's "Introduction to Topological Manifolds". Still working on commutative algebra from AM. A bit of representation theory from Serre's "Linear Representations of Finite Groups". That mostly covers it as of late.
 
ah, i don't know any of those except a bit of commutative algebra :)
they look fun enough to me!
what does Lee's topological manifold cover?
 
AM is great, a lot of fun. They've all been a bit slow going, partially because I'm still at my job, partially out of sluggishness, and I'm also trying to be pretty thorough.
 
AM has great exercises, but it's a bit dry. That's partially why I gave it up.
I have to come back to it in near future.
 
7:28 PM
Hmm, about four chapters on some elementary point-set stuff. Another two on some prepatory-looking stuff for algebraic topology. And then the remaining five on basic algebraic topology (below the level of Hatcher, I'd say).
 
oh, i thought you were doing manifold calculus.
that must be the book on smooth manifolds.
 
It is a bit dry, but the material is very exciting. I love books like AM.
 
@AlexWertheim Sounds nice! You'll have loads of fun with them.
 
No, actually, this is in some ways the precursor to that book.
(If you're referencing Lee's book "Introduction to Smooth Manifolds", that is.)
 
@AlexWertheim Eh, it just reminds me of Rudin often.
I'd love to have a book with loads of examples and loads of rant.
 
7:31 PM
I can see the comparison, though I find Rudin's proofs less enlightening on the whole, honestly.
 
@Alex Which chapter are you doing, right now? In A-M?
 
Way too much "rabbit-out-of-a-hat" style for my taste. I can almost always see where AM are going, even if their methods are slick.
 
yes, A-M is systematic.
 
I'm not far. I covered all of chapter 1, and did (almost) all the exercises. Now I'm working my way through chapter 2.
 
the flow of exercises lead you to something useful.
@AlexWertheim okay. how did you draw the spectrums? :)
 
7:33 PM
There were some topology gaps I had which made some of them difficult to understand/appreciate, so I'll have to come back to them.
With gusto? ;)
 
@AlexWertheim ah, but Zariski topology is the most important thing in chapter 1!
it's a very interesting language for algebraic varieties.
 
Sure, @Balarka. I did most of them, just not all of them. I'll get to them later, don't you worry. :)
 
@AlexWertheim hah, no, i mean, for example, how did you sketch Spec Z[x]?
(<--- has a nice illuminating method upon the sleeve)
 
@Balarka: I know what you mean, just having a little fun. Truth be told, I wasn't really sure what to draw. My drawing didn't look anything really like Mumford's treasure map, that's for sure. But I just spent a little time computing the prime ideals, and then drew a standard cartesian system. Put points at all prime numbers along the "x"-axis, and put points above primes $p$ at various irreducible polynomials (mod $p$) along the "y"-axis, if that makes sense.
 
I like "with gusto". My standard response is "very carefully"
 
7:41 PM
I'd love to hear an illuminating method, lol.
@Mike: that's really, really weird. One of my best friends from college says the same.
(Ok, maybe it's not that weird, but I find it very coincidental.)
 
@AlexWertheim aww, mumford's treasure map was the trick upon my sleeve.
 
I'll be honest, I've never entirely understood that map.
 
what do you not understand, in particular?
 
@JohnDoe wasn't that answered?
 
it's a sketch of Spec Z[x] as an affine scheme, not just as a topological space
 
7:44 PM
I get most of the points, which actually look kind of similar to what I drew. I don't know what the squiggles over the primes represent.
 
@robjohn Well, it was, but the answer had a mistake and using his method I couldn't get one of the right answers in my textbook. Probably my mistake but I've been on it for hours so I've reopened the question
 
@JohnDoe There were 4 reopen votes, so I could vote to reopen it.
 
@AlexWertheim it represents how much information every ideal contains about the space. more precisely, it represents how large the localization of Z[x] at those ideals are.
 
I'd be grateful if you wanted to do that :)
 
you'll understand those when you get to chapter 2 (or was it chapter 3? i forgot).
 
7:46 PM
I'm also not sure what the loop represents (starting at (2,x+1)).
 
@JohnDoe I already voted. It is reopened.
 
Thanks :)!
 
which loop?
 
@Balarka: ah, neat. I look forward to being able to make sense of that - I don't think I'm quite there yet (at least I hope not, haha).
The hairpin like curve.
 
oh, that.
 
7:54 PM
@AlexWertheim Maybe he's me, and you never noticed?
 
it depicts the intersection of the prime ideal (x^2 + 1) with the ideals (2), (3), etc.
bah, internet connection.
 
@MikeMiller: damn, I am really unobservant. Either that, or that's quite a trick you pulled. =P
 
bad internet connection is responsible for half the nerd-suicides going around all over the world.
 
Hmm, interesting. That makes more sense, thanks!
 
@Alex But, there is more to the curve than you think there is. It's actually an arithmetic version of branched covering maps.
I don't really know about that stuff, but you see, when you pull up a prime ideal of Z to a prime ideal of Z[i], it splits. so these are kinda stuff like covering maps.
the real precise-version of this is complicated, i guess. goes back to grothendieck.
 
7:59 PM
Hmmm. I know some of what you're talking about, but I don't know what a covering map is, let alone a branched covering map, or an arithmetic version of a branched covering map. I'll have to keep reading.
 
yeah, you'll get to covering maps when you do algebraic topology.
branched coverings are silly names for maps which are covering maps away from a finite number of points.
 
I can't find my coffee mug :(
 
use a ffee mug instead.
 
Or drink straight from the tap. #hardcore
 
8:03 PM
havent showered since thursday, time to shower
 
i sometimes don't shower for a month. waste of time, really.
 
@AlexWertheim: I considered, but this is a carafe, and I can't control the output that well.
 
hi guys
 
@MikeMiller: have you considered an IV drip? Faster into the bloodstream that way, too.
 
@BalarkaSen ?? are you joking
 
8:06 PM
haha, @Alex
 
I suppose that's harder to track down than a mug though, isn't it. :)
 
@KarimMansour nope. i do that every winter
 
@AlexWertheim: I get good coffee, precisely because I enjoy drinking good coffee. So there are some negatives to that approach.
 
xD
@MikeMiller what is the heaviest caffe you drank ?
 
well, time to switch to opium, @Mike
 
8:07 PM
lol
 
What do you mean by heaviest?
 
strongest
that is after you drank it it kept you awake for a long period of time
 
I mean, I guess tautologically an espresso by caffeine per volume. I prefer dark roasts, so I get less caffeine per cup.
 
I prefer tea over coffee. Much more refreshing, and less strong.
 
actually caffe is good for solving problems but not for studying
like for studying it will make you very hyper
atleast for me that is what happens.
 
8:10 PM
i guess Mike is ignoring me for making that lame co-objects joke. :p
i just couldn't resist.
 
Do people really put other people on ignore in chat here?
 
I haven't seen anything worth responding to, is all.
 
sometimes, @Alex, yeah.
 
I don't
 
@AlexWertheim: If someone aggravates, annoys, or otherwise upsets me, I do, so as to avoid being further aggravated, annoyed, or upset. It's worked fine for me thusfar.
In particular, you don't get pinged anymore, if the person in question pings people problems whenever they enter.
 
8:13 PM
@MikeMiller: that makes sense. Sometimes I get a bit rattled by some of the stuff posted here, and make the mistake of responding instead of ignoring things. I should know better.
Ah, that's another good reason.
 
Ok, lunchtime's over. I should get back to work. See ya.
 
See ya!
 
please what is the relation between u and $\nabla u$ how i can write u(x) with $\nabla u(x)$ ?
 
hey guys
somebody familiar with polard rho factoring algorithm?
 
8:31 PM
I drink coffee like water, I don't even feel a buzz.
But it's one very tasty drink
 
@LeGrandDODO: C'est ridicule, alors. If we gave any sort of entrance exam in the US, almost no one would/could major in math. Hmm, what a nice thought :P
hi @AlexW :)
 
Hello @ted.
 
hi Jasper
 
hi @TedShifrin
 
hi @Karim
 
8:41 PM
I think its a good idea though to have such an exam
 
hi @Ted
 
it really tests you out and gives you enough time too to think
 
well, I like long exams, but 30+ questions for 6 hours is crazy. Even mature mathematicians can't function like that.
hi @Balarka
 
i am feeling sick. i guess i won't be able to go school tomorrow.
what a happy thought.
 
8:43 PM
@ted At first I thought maybe I would not come to this room anymore, but maybe I should still come. =)
 
I thought you told me you were out of school now
 
out of school?
 
I'm probably not coming much at all any more, Jasper ...
You told me last week you were done with school, @Balarka. I thought it was long vacation.
 
@TedShifrin That is bad news for me.
 
@TedShifrin did I? well, the school was off for a week, but it wasn't that long
 
8:44 PM
oh, a week, gotcha
 
i think the time for me to beg for your book is closing in. :p
 
Begging is illegal here.
@KarimMansour You mistook Sayan for me that day. I would never ask Balarka a math question.
 
oh I see
 
Hi @TedShifrin
@JasperLoy Are you back now
 
hi mr eyeglasses
 
8:53 PM
no fun alg top question for me to answer today. blah.
 
@ᴇʏᴇs I don't know, we'll see.
 
are you guys familiar with polard rho algorithm?
 
i vaguely recall that.
 
@ᴇʏᴇs Hey you did not answer me in the other room!
 
it is a way to factor numbers @BalarkaSen
but there is something I don't understand
 
8:56 PM
yes, i know.
 
I don't understand how choosing a polynomial will help us
if we don't use random at random
 
@TedShifrin I sent Barry Simon an email regarding his 5 books today, lol.
 
don't be too surprised if you don't get a response
 
lol
ofcoz he wouldn't reply
 
@TedShifrin You know him?
 
8:58 PM
nope
 
I think
 
@TedShifrin Do you ever get humorous e-mails in regards to your books
 
@TedShifrin I won't be surprised, because most people do not respond to my emails.
 
humorous? not really ... mostly people wanting answers ...
 
Conway is pretty good in responding to e-mails concerning his works
 
8:59 PM
Most of my emails are rubbish to people, just like what I say in chat.
 

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