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3:05 PM
Is it true that the elementary row operation of switching the $i$-th and $j$-th rows is implemented by the matrix $P_{ij} = I + E_{ij} + E_{ji} - E_{ii} - E_{jj}$? If so, what's the easiest way to argue that $P_{ij}$ is invertible?
 
Hello!!

We have that $\mathbb{Z}_{14}^{\star}=\{1,3,5,9,11,13\}$. From these elements we get the cyclic subgroups $\langle 1\rangle, \langle 13\rangle, \langle 3\rangle=\langle 5\rangle, \langle 9\rangle=\langle 11\rangle$. How can we show that $\mathbb{Z}_{14}^{\star}$ has no other subgroups?
 
3:28 PM
@MaryStar $(\Bbb Z/14\Bbb Z)^\times = (\Bbb Z/2\Bbb Z \times \Bbb Z/7\Bbb Z)^\times = (\Bbb Z/2\Bbb Z)^\times \times (\Bbb Z/7\Bbb Z)^\times = C_6$
in particular it is generated by $3$
 
You mean that $\mathbb{Z}_{14}^{\star}$ is a cyclic group that is generated by 3, right? @LeakyNun
 
sure
 
So we have that $ \langle 3\rangle=\langle 5\rangle=\mathbb{Z}_{14}^{\star}$. But how do we know that there are only the cyclic subgroups that I wrote above and no more? @LeakyNun
 
subgroups of cyclic groups are cyclic
 
Ah and since these are the only one that are created by the elements of $\mathbb{Z}_{14}^{\star}$ there are no other, right? @LeakyNun
 
3:40 PM
right
 
Great! Thank you!! @LeakyNun
 
what's the probability Logan Paul loses against KSI?
 
I have also an other question...

Let $Z = (\mathbb{Z},+)$ and $G = (M,\star)$ an arbitrary group. To show that for all $a\in G$ the map $\phi_a:Z\rightarrow G$ defined by $\phi_a(k)=a^k$ is an homomorphism from $Z$ to $G$, we have to show that $\phi_a(m+n)=\phi_a(m)\star \phi_a(n)$ for $m,n\in Z$. Why do we have to show also that $\phi_a$ is total ? Is it because we are given a map and we have to show first if it well defined?

Also I want to determine the range of $\phi_a$. It is the set $\{\phi_a(k)\mid k\in \mathbb{Z}\}=\{a^k\mid k\in \mathbb{Z}\}$, right? At the solution it is equal t
 
what do you mean by total
also the solution is wrong about k >= 0
 
4:02 PM
Here is the defintion: https://en.wikipedia.org/wiki/Partial_function#Total_function

At the given solution the range is then equal to the cyclic group generated by a.

@LeakyNun
 
> Total function is a synonym for function. The use of the adjective "total" is to suggest that it is a special case of a partial function (specifically, a total function with domain X is a special case of a partial function over X). The adjective will typically be used for clarity in contexts where partial functions are common, for example in computability theory.
so no, you shouldn't use the word total here
and the solution is correct
the group generated by $a$ is $\{a^k \mid k \in \Bbb Z\}$
also who told you to show that the function is total?
 
At the given solution it is shown that the function is total: $a^k$ is for all k ∈ Z an element of G = (M,∗), since G = (M,∗) is a group and so closed.

Why do we have to show that? @LeakyNun
 
because the solution is stupid that's why
 
In that way they want to show that the given map is well defined, or not?
 
yes
 
4:12 PM
Ok. Thank you!
 
vzn
@MaryStar hi werent you working on a CS Msc awhile back?
 
@vzn Yes
 
vzn
@MaryStar are you in the middle of that?
 
@vzn I have finished, 2 years ago
 
vzn
@LeakyNun interested in those two esp connections to computability theory which seem to be existent but murky
@MaryStar did you do a Msc thesis? are you working on some math degree now?
 
4:19 PM
@vzn I thought there is no computable model of ZFC
 
vzn
@LeakyNun ah might have cited a recent paper on that in here by Yedidia-Aaronson. are inaccessible cardinals "outside of ZFC" in some sense? trying to remember some of this
 
I don't know in what sense
 
vzn
@LeakyNun lol so when are you going to start the "indoctrination"? :P
 
right after the message you quoted
 
vzn
so is it over already? missed it? :| :P
@user21820 there is some line of research crosscutting CS + math in applying SAT solvers to arbitrary problems that leads to insight not directly seen via other approaches.
 
4:25 PM
@vzn Yes, I did My Msc Thesis it is about Decidability in differential algebra (My Msc was in Maths of CS). Now I am teaching students and I want to deepen my knowledge also in some other topics of mathematics.
 
vzn
@MaryStar nice itd be great if you could post a link sometime... am interested in (un)decidability etc myself, have blogged on it etc eg vzn1.wordpress.com/2016/01/22/…
 
@vzn Ok
 
@vzn At least from what I know, the state-of-the-art SAT-solvers are all merely heuristic, and hence you don't get much qualitative insight into problems that you apply them to, since the point of using them is merely to solve the problem faster than a naive brute-force approach.
 
vzn
@LeakyNun inaccessible cardinals reminds me of paris-harrington thm & wonder if there is any connection en.wikipedia.org/wiki/Paris%E2%80%93Harrington_theorem
 
@vzn I'm not the guy to ask this to :)
 
vzn
4:29 PM
@user21820 "heuristics" can be a tricky term. they sometimes outperform brute force searches because they can analyze "hidden structure" in the search space. there are nonw a few (relatively isolated/ scattered) cases of remarkable insights/ proofs related to them.
 
Heuristics that are tailored to a problem will in general be better than the best SAT-solvers. The advantage of using a SAT-solver is simply that you just have to find an efficient way of translating the problem to a SAT problem.
 
If X is a non-trivial topological vector space over F = set of reals , then the topology of X can never be discrete. Help me prove this please.
 
@vzn Citation needed. =)
 
vzn
@user21820 sometimes SAT solver heuristics outperform problem heuristics. heuristics are human inventions, SAT heuristics relate to machine learning in some sense.
 
@vzn That just implies that less effort went into the tailored heuristics than into the heuristics that underlie the SAT-solver, and hence does not invalidate my point.
 
vzn
4:32 PM
@user21820 there are at least 2 notable cases from last few yrs let me dig them up.
 
Sure.
 
vzn
@user21820 your point has some validity but the area is full of (deep/ hidden/ unexpected) nuances.
 
Yes my point is not something amenable to proof.
It is more of a personal heuristic observation. =P
 
vzn
@user21820 its an area of active research. imagine extremely sophisticated patterns being detected in the SAT formulation that a human couldnt detect, etc, there are graph-based analyses of the resulting SAT structures. have been looking/ dabbling into this for years...
 
@vzn That still suggests that a heuristic approach is better, just not being done 'efficiently' in the past. For example, humans can use graph visualization tools to help them understand the underlying graphs, while automatic analyzers will usually be unable to. If humans fail to use the tools available to an automated system, then the comparison would not be a 'fair' one.
 
4:37 PM
@Shobhit what does non-trivial mean?
 
Note that I'm not saying that it is always better to tailor heuristics. After all, SAT-solvers have been around a long time and a lot of effort has gone into them.
 
vzn
@user21820 ok the 2 relatively recent examples were the erdos discrepancy problem (since resolved by Tao) + pythagorean triples search. was recently reorganizing my links so have them around vzn1.wordpress.com/2016/09/16/… vzn1.wordpress.com/2014/02/14/…
 
So the benefit in tailoring heuristics must be weighed against the upfront costs.
@vzn Thanks! I'll take a look.
 
vzn
@user21820 not exactly disagreeing, but another way to see it is that the two approaches are complementary. machine learning is playing an increasing role across all sciences, and think it is making inroads into math, but more gradually there than hoped by me.
 
Yeap. I don't think we actually disagree.
 
4:39 PM
maybe $X = {0}$ @LeakyNun
 
Another example is gradient descent. It is one of the most common first-line approaches to attempting to solve 'somewhat' continuous optimization problems. Iterative improvement is similar. These very generic techniques will work to a certain extent, depending on the 'topography' of the potential function on the search space. But if the topography is more discontinuous, then heuristics become increasingly superior, for the same investment in human intelligence.
 
vzn
@user21820 theres the trees vs forest. gradient descent is far more than what anyone realizes. on a higher level its involved in self driving cars, conquest of Go by deepmind, etc... so it sometimes comes down to how the problem is formulated by humans.
 
@vzn Go is played well only because a ton of processing power was chucked at it. That merely strengthens my point.
When a computer that has the same rate of energy consumption as a human beats a professional Go player, I will sit up and take notice.
 
vzn
@user21820 there is some point to be made but its subtle. the point is that a relatively primitive technique formulated centuries ago (on single variate functions!) can be applied in entirely new ways, not all has been squeezed out of it yet, surprises, even very big ones, likely remain.
 
@vzn This is true, but a lot of these convergence results (that prove that certain optimization algorithms on certain classes of problems will always converge on the optimal solution as time goes to infinity) are simply too slow in practice, even though they have one benefit over trying to tailor heuristics: you don't have to think to use them.
As I said, I will take notice only when computers become more intelligent than humans at the same power consumption.
 
vzn
4:47 PM
@user21820 "you dont have to think to use them." bug or feature? :P
 
@vzn I was trying to give one advantage of such techniques, so... feature.
 
vzn
@user21820 there are some reasonable estimates that supercomputers (GPU clusters, cloud computing etc) are getting close to human hardware computational capability.
 
=P
@vzn What do you mean by that? I said power consumption. Not computational capacity.
 
vzn
@user21820 lol human brain consumes ~20W less than a typical hairdryer :P
 
That is exactly what I'm driving at. Our limitations are because we are just TOO slow.
 
vzn
4:50 PM
@user21820 "limitations"? thought you just asserted humans can outdo gradient descent :P
 
@Ted I'm a bit stuck. I don't get the $x_n$ thing. Does that depend on $N$? Does the interval grow smaller as $n$ grows larger since $x_n=\frac{1}{\sqrt{n}\to0$ for $n\to\infty$?
 
vzn
@user21820 so even if AGI is achieved, you will not be impressed or "take notice" if it takes more than 20W? :P
 
@vzn Correct in a sense. More precisely, if you use a gradient descend algorithm with some mechanisms to get out of local minima, such as using simulated annealing, then we have theoretical proof that it converges to the global optimum, and you will observe its performance gradually improving over time.
In contrast if you put a human against the same problem, that human will have to spend a certain amount of time for each improvement to his/her heuristic. So performance will increase only in jumps, and depending on significant time and effort, but eventually will outdo the generic optimization algorithm.
@vzn The comparison will be by total energy usage. So you can give the computer more power but less time.
 
vzn
@user21820 to some degree maybe an apples vs oranges comparison... personally think an (eventual) AGI algorithm will probably incl some gradient-descent like logic, how about that?
 
@vzn Perhaps.
 
4:59 PM
PhD stuff: molecules keep rolling to the wrong saddle points
ugh
I will need a heavy dose of inaccessibles to refresh my mind, perhaps even a little bit reading the QED logic book
 
vzn
@user21820 another one of my fave refs, undeservably obscure gem, Decomposing Satisfiability Problems by Herwig citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.108.1116
 
5:17 PM
@Oskar: In the case of your original question, convergence is uniform on all of $\Bbb R$ because $M_n=1/\sqrt n\to 0$. Therefore, given any $\epsilon$, choose $N$ so that $1/\sqrt N<\epsilon$ (i.e., $N>(1/\epsilon)^2$). Then for any $n\ge N$, $|f_n(x)|\le M_n\le M_N < \epsilon$.
 
hi @TedShifrin
need your help :P
 
Try this one, @Oskar. What about $f_n(x) = nx/(x^2+n^2)$? Converges pointwise to what $f$? Is it uniform on all of $\Bbb R$? Where will it be uniform?
hi @Shobhit
 
hi @Ted
 
hi Leaky
 
0
Q: About tensor product isomorphisam

Ninja hatoriContinuing this post Why is this isomorphism $M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$ an isomorphism of $M$ - algebras? Suppose that $L/K$ is a finite separable extension of fields and let $M$ denote the Galois closure of $L$. Let $\textrm{Hom}_K(L,M)$ denote the set of all $...

 
5:31 PM
If X is a non-trivial linear topological space over F = set of reals then the topology of X can never be discrete. @TedShifrin
we have just started LTS and i know its definition
 
Usual topology on $\Bbb R$, I assume?
 
yes
 
So any $1$-dimensional subspace has the (induced) topology that $\Bbb R$ does.
 
@Secret the QED book by Tao?
 
5:38 PM
gamifying math is the newest trend in math-ed
 
Could anyone help me understand what these stars are and who are the omega? I suspect there is pullback, but I'm not sure. math.stackexchange.com/questions/2894245/…
 
@Takashi: Yes the star is pullback. But you need to go way back in your text and see first of all what $G$ is, note that $\omega^i$ are left- (or right-) invariant $1$-forms on $G$ adapted to the geometry of $\mathbf x$. Think of adapted frames geometrically.
 
@TedShifrin any clues?
 
Say $\mathbf x$ is an immersion of your $U$ into $N$. Then you want the first $m$ elements of your frame to be tangent to the image and the remaining elements of the frame to be normal to it. The $\omega^i$ are the coframe dual to this frame.
@Shobhit. I gave you one immediately.
 
oh, i didn't knew that was for me. Let me think.
yes, ofcourse it does.
 
5:54 PM
Well, if you show that, then you're done. :)
@Takashi: $G$ is the Lie group of all frames (say $O(n)$ if $N=\Bbb R^n$), and the point is that $e$ is mapping to an appropriately adapted frame as I explained.
 
25 mins ago, by Ted Shifrin
So any $1$-dimensional subspace has the (induced) topology that $\Bbb R$ does.
 
What's your point, Leaky?
 
@TedShifrin je ne comprends pas. peut-etre nous avons definitions differents de topological vector space
 
You're saying that if you have a TVS $V$, the mapping $\Bbb R\to V$ given by $f(t) = tv$ is not continuous? ($v\in V$ fixed)
 
la fonction est continue, mais V peut encore etre discrete?
ah
 
6:01 PM
Leaky: This question is Shobhit's, so let's stick to English, please.
And the answer to your last question is ... ?
 
it can't be discrete beacuse the function wouldn't be continuous
 
Indeed. So what in the definition of $V$ should tell you it should be continuous?
 
that scalar multiplication $\Bbb R \times V \to V$ is continuous
 
Aha. :)
 
I was confused with topological groups
because they can be discrete
 
6:07 PM
Of course.
@Shobhit: Have you been paying attention?
 
yes
still thinking
 
Well, we just gave it away, didn't we? :)
 
you might have, but i haven't understood it yet.
 
So tell me what the definition of TVS says.
 
vector addition and scalar multiplication function are continous
 
6:16 PM
Hi everyone
 
hi
 
@TedShifrin
 
Hi one @Alessandro
 
and only?
 
ok, @Shobhit. So I'm suggesting you fix $v\in V$ and look at the subspace it generates. Think of the mapping from $\Bbb R$ to $V$ given by $f(t)=tv$. You just told me it is continuous.
 
6:19 PM
Is there another?
 
Now that Alessandro is venerable, he must be priceless.
 
he certainly is valuable
 
Ah, but he's now almost antique, so venerable as well.
 
the subspace is just $tv$ as you wrote, what should i think here? definition?
@TedShifrin
 
I'm certainly more antique
 
6:22 PM
We won't start playing the comparison game, @MikeM.
@Shobhit: Yes, the image of $f$ is the $1$-dimensional subspace spanned by $v$. I don't know what you're asking.
 
yes it is the one dimensional space, so? @TedShifrin
 
@Ted I'm looking into it now. I have another question that I'm a little bit stuck with. I'm supposed to show that $$\sum_{n=1}^\infty\frac{1}{n^s}$$ is convergent, for $\epsilon>0$ the variable $s$ belongs to $1+\epsilon\le s<\infty$. I know that series are uniformly convergent if their partial sums are, but that's about as far as I come.
 
Note that if multiplication $\Bbb R\times <v>\to<v>$ is continuous so it's its restriction to $\{1\}\times<v>$
 
@Oskar: Look at the Weierstrass M-test.
@Alessandro: I'm fixing $v$ and varying $t$. Why are you doing that?
But we want to think about the subspace (or induced) topology on that one-dimensional subspace, @Shobhit. If the topology on $V$ were discrete, then .... ?
 
@TedShifrin because I wasn't thinking
 
6:27 PM
You're trying to confuzle me, @Alessandro. Wait 'til you're my age. You'll see how it goes.
 
I did. I considered $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6},$$ but $\epsilon>0$ may be smaller than one, and $$\sum_{n=1}^\infty\frac{1}{n}=\infty.$$
 
@Oskar: Don't you know something about when $\sum 1/n^p$ converges? For which $p$?
(In calculus in the US we learn the Integral Test, which helps for this. I don't know about your curriculum.)
 
It was a long time ago. I'll have a look at it.
@Ted I got $x_n=n$ and $M_n=1/2$. Here, $M_n$ doesn't depend on $n$. The function converges pointwise to zero. It isn't uniform on all of $\Bbb{R}$, but I'm unsure as of where it will be.
 
OK, @Oskar (probably $|x_n| = n$). So you can't be uniformly convergent on any unbounded interval. What if you take a finite interval, say $[-100,100]$?
 
Why not on an unbounded interval?
 
6:42 PM
Because then you will hit points where the function is $1/2$, and so, for small $\epsilon$ you're messed up. Remember my suggestion to draw an $\epsilon$-fence around your limit function. The graphs of $f_n$ must stay inside that fence.
 
The problem for me is that I don't really know how to plot the function. It's nontrivial how e.g. $f_n(x)=e^{-nx^2}$ looks, and WolframAlpha doesn't really help.
 
You should be able to sketch these graphs using common sense (for fixed $n$). You can do WolframAlpha, even, if you specify some different $n$.
 
Okay.
 
You've already found the critical points, so using basic differential calculus you should sketch the graphs.
(Plus you know asymptotes, etc.)
 
You mean like showing whether or not it's a maximum or minimum?
Do I get it right that if we have a bounded interval, and $n$ is bigger than the interval, then we won't hit $M_n$ within the interval?
 
6:47 PM
Right.
 
How are we sure that just because we don't hit the maximum we have a uniform convergence?
 
If there are no critical points in your interval, then the max/min occur at the endpoints. So check the values at the endpoints.
 
Yeah
 
I just answered your question before you asked it.
 
Sketching?
I feel like I'm close to getting how it all works.
Sorry for being slow.
 
6:49 PM
Well, sketching helps to understand. I teach everything (even algebra) with pictures. :P
But you can see it analytically. What is $f_n(100)$?
 
That's great! I love Tristan Needham's book on Complex Analysis called Visual Complex Analysis.
$f_n(100)=\frac{100n}{100^2+n^2}$.
 
OK, and what's an easy upper bound for that?
 
buongiorno @AlessandroCodenotti
 
$\frac{100}{100^2+1}$
 
Buonasera at this time
 
6:52 PM
No, no, in terms of $n$. We're trying to understand large $n$. :P
 
Oh! Of course.
 
@AlessandroCodenotti Alessandro is getting picky in his old age. :P
 
@AlessandroCodenotti come va gli p-adic numeri
@Ted non so che tu parli italiano :o
 
For large $n$ it of course goes to zero.
 
Leaky wants to win a prize for speaking every possible language in here.
 
6:53 PM
If f(x)f(-x)=1 can we conclude anything about the integral \int _{-5} ^{+5} \frac{dx}{1+f(x)}?
 
@Ted :(
 
So then for large enough $n$ you're within the fence and you're done, @Oskar.
 
Leaky veut gagner un prix pour parler tous les langues possibles ici
 
Okay. So around conclusion is that it's uniformly convergent for bounded subsets of $\Bbb{R}$?
 
@tatan Cute question. Yes, we can.
@Oskar, yup.
@tatan: Is there something obvious you can do to rewrite the integral on $[-5,0]$?
 
6:55 PM
@TedShifrin Show me some guiding light please
 
Okay. So back to when we had $x_n=\frac{1}{\sqrt{n}}$ and $M_n=\frac{1}{2\sqrt{n}}$. Then, as $n\to\infty$ we have $x_n\to0$ and $M_n\to0$. Thus, we have all of $\Bbb{R}$ as uniform convergence.
 
We can if f(x) is even right?
 
Right, @Oskar.
 
If you want me to be picky the conjugations and articles have some issues
 
@tatan: Yes, of course, but what integration technique can you use to turn an integral over $[-5,0]$ into an integral over $[0,5]$?
 
6:57 PM
Anyway I haven't been doing much but celebrating today
 
@AlessandroCodenotti please be picky
 
@Alessandro: If you aren't going to celebrate, you'll have to cry about getting old :P
 
oh right happy birthday
 
So, it's all about what $M_n$ is, and, in relationship, what $x_n$ is. If $M_n$ becomes small as $n$ becomes big we're okay. If it doesn't, but $x_n$ depends on $n$ and grows with it we have it for bounded subsets, but what if it goes to zero? What happens if $x_n$ is a constant and $M_n$ is as well?
 
buon compleanno, as google translate says
 
6:58 PM
@TedShifrin I don't know ;-(
 
@Oskar: Analogously, if the $x_n\to 0$, say, and the $M_n$ stay fixed, say, then you need a domain that will stay away from $0$. Again, you look at the endpoint of such an interval. Try things like $|x|\ge a>0$.
@tatan: How about a $u$ substitution?
 
no u
 
@TedShifrin Maybe.. but I am not still getting what we are trying to do
 
hi @Daminark
 
Hey everyone, and happy birthday Alessandro!
 
7:00 PM
hi Demonark
@tatan: You want to turn it into a sum of integrals over $[0,5]$.
 
@TedShifrin Then?
 
@LeakyNun come va con i numeri p-adici?*
 
Does it make sense to consider 0 and $\aleph_0$ the first two inaccessible cardinals, if we remove the uncountability condition?
 
Well, do it, @tatan.
 
@AlessandroCodenotti so p-adic is italian now
 
7:02 PM
@LeakyNun Non sapevo che parli Italiano
 
Hey, do you use a trick to write quickly latex here, or do you plainly write them?
 
thanks :)
@AbdullahUYU we just write them
 
@LeakyNun OK.
 
@user76284 well we have the condition that they be limit ordinals
so no 0
 
In general we avoid personal pronouns when possible (che parli rather than che tu parli)
 
7:04 PM
@user76284 in fact the only reason we need uncountable is to have infinity
 
@TedShifrin If we let g(x)=f(x)(-x)=1 then g(-x)=g(x)... So, g(x) is even. Now, so the integral is =2\int _{0}^{5}\frac{dx}{1+f(x)} right?
 
because V[aleph[0]] models ZFC - infinity
@AlessandroCodenotti may I seek advice from you
as a student of set theory
I talked to my professor about algebra vs set theory
 
@Ted Ah! Now I understand completely. Such a wonderful feeling.
 
and they said algebra has a brighter future
 
@user23571113 I have not gotten married.
 
7:06 PM
so I wonder if you have anything to say about that
 
@tatan: I want you to substitute to change the integral limits ... I don't know what you're doing with what you wrote.
Congrats, @Oskar :)
@Jasper !!! Howdy, stranger.
 
Hello @TedShifrin and @LeakyNun.
 
long time no see
 
@tatan: If you set $g(x) = -xf(x)$, then $g(-x) = x/f(x)$. I don't follow at all.
Very simply: What transformation on $x$ turns $[-5,0]$ to $[0,5]$?
 
u=-x
 
7:08 PM
OK, so now rewrite the integral on $[-5,0]$ in terms of $u$.
 
Thank you, @Ted. Your help is very important for me. I don't think that I'd have gotten this otherwise.
 
@LeakyNun I don't really have the knowledge or the perspective to comment on that
 
@Oskar: You see? We teachers can be useful after all :P
 
@AlessandroCodenotti ma sceglievi set theory
 
Haha! Yes! Indeed. I do think that this understanding would've come from a lecture though. I profoundly believe in the asking questions back and forth between the student and the professor as we just did.
 
7:10 PM
so maybe you have the experience to comment on that?
 
wouldn't have*
 
Hai scelto*
 
Well, I've taught this particular material a lot of times and I think my students got (most of) it from my lecture. Of course, you only truly get it when you do your own exercises.
 
@TedShifrin we get -\int _0 ^5 \frac{du}{1+f(-u)} right?
 
Yeah! I've started trying to have a go at a couple of exercises before I go to the lectures. That way I can relate to what the professor is saying more. I can try to solve the problem in the break, and if it doesn't work ask afterwards if there's time.
 
7:13 PM
No minus sign, @tatan. Why?
 
I don't think I can really say I made a decision yet, I'm also going to take class field theory and algebraic stuff in the next semester
 
@Oskar: It helps some students to read ahead for sure.
 
I just find set theory particularly interesting
 
Yeah!
 
@AlessandroCodenotti yes, I also find set theory interesting, and so did my professor
 
7:14 PM
Maybe that's what can make passive learning active.
 
@TedShifrin dx=-du right? I don't see why it shouldn't be there
 
@AlessandroCodenotti they said they walked around for three months asking staff members about which stream to specialize in
 
What is actually the intuition behind the shuffle algebra?
 
before deciding to specialize in number theory
because the major problems in set theory have already been solved
 
The chat's such a party right now. I love it.
 
7:15 PM
I mean what is the relation to the hopf algebra structure
 
and my professor "would strongly advise against" me specializing in set theory
 
@TedShifrin ;-( I didn't get it yet
 
It's not a smart choice career wise most likely
 
@tatan: The original limits were $-5$ and $0$, so you get $\int_5^0$.
 
I'm just learning what I like now, I'll think about those things more seriously when I'll need to
 
7:20 PM
Yes
Originally it was from [-5,+5]
 
@AlessandroCodenotti I see
well I'm sure Asaf Karagila thought about it before making that decision
but unfortunately they have a habit of never going to chat
@AlessandroCodenotti also, when I need to
 
@Ted so I emailed my REU mentor and he got me a lot of the info you mentioned (basically sent me a huge list of places being like "Yeah this place has this guy, this place has this guy but he may not be an ideal advisor, etc")
 
@LeakyNun True, thanks
 
We're doing just the part from $-5$ to $0$, @Tatan. So it switches to $+$ the integral from $0$ to $5$.
 
@Daminark ideal advisor
 
7:22 PM
Verb conjugation in Italian is hard, but I shouldn't underestimate the English one either :P
 
oh, that's very useful, Demonark. Great!
 
@Daminark Do you enjoy being in the REU? Hi.
 
I'm only just realizing how much nuance there is in this kinda decision. But yeah he also said to consider Canada actually
 
@Daminark Are you looking for a place to continue your studies after you graduate? Or why?
 
@TedShifrin Okay it does... now?
 
7:24 PM
@JasperLoy hey, how've you been? And I'm not quite in the REU at the moment, finished it a couple weeks ago and finished the final draft of my paper a week ago
 
@Daminark interesting. my uk professor told me to consider us, and your us professor told you to consider canada
 
Now write down your sum of two integrals and use what you know to simplify. @tatan
 
Well, he only really mentioned Toronto and McGill, saying that around the UCLA/Michigan level they should be serious considerations (also he's a grad student here who went to undergrad in Canada)
@Alessandro this is for grad school
 
my parents studied there, lol
 
Demonark: And personality issues come into play, too. Are you going to be happy at a cut-throat place where faculty are hard to talk to? If so, that's useful. If not, high-pressure places like Princeton aren't ideal.
 
7:26 PM
@TedShifrin How do I operate with u and x simultaneously?
 
Lee
 
Demonark: You should check to see if Granville is going to be at U. Montreal. He's a superstar analytic number theorist.
@tatan: Once you've rewritten the integral, change the $u$ back to $x$. You can use any letter you want for the integral. :P
 
Lee's Introduction to Riemannian Manifolds will finally be published next month.
 
I continue to be less enthusiastic than you, Jasper. :P
 
@Daminark I see
 
7:27 PM
also we have fundamental theorem of algebra and of arithmetic, when will we get one for analysis
 
Note that the second edition has been renamed from Riemannian Manifolds: An Introduction to Curvature.
 
Demonark: I think he went back to Britain, but someone told me he might be returning to Montreal.
 
@TedShifrin OK, but it seems the length has almost doubled, from 200+ to 400+ pages.
 
I'm curious to see how the content/approach has changed, Jasper.
Lee writes very verbose books.
I've never taught out of them explicitly, though.
 
random question for the chat: what will you be doing after 10 years?
3
 
7:29 PM
math
 
I am very happy I got 40 points on English SE for saying that a tree good for climbing is called climable.
 
Hmm, I'm not sure when it comes to pressure. What's the advantage of such a place over a more relaxed environment? I'm often very worried when I'm about to approach people but then I just send the email anyway
 
I've never heard of such a word (seriously). And no "b"?
 
@TedShifrin Oh that was a typo.
Climbable.
 
Demonark: Personally, I think you should think about it and talk to students at the place about the atmosphere and pressures ...
Oh, ok, @Jasper. Much happier. :P
Some people are prone to depression during graduate work, and it becomes a serious consideration how the grad students have camaraderie, faculty care/don't care, etc.
@Leaky: I'll probably be deceased :P
 
7:31 PM
:(
don't say that
 
@tatan: So did you get it?
 
I'll probably be 4 years out of grad school at that point? So likely either broke postdoc or rich finance person or software engineer :P
 
I don't care what I do as long as I live somewhere warmer and still do math.
 
I will say one thing about Princeton in particular that I'm not too fond of is that they seem to have fewer classes to do earlier on and want you out of there in 4 years (they'll let you stay for 5 but the pressure for 4 is still eh)
 
You can have nightmares about being molested by uniform convergence proofs, @Oskar :P
 
7:34 PM
@LeakyNun broke postdoc, crazy cat lady or both
 
It used to be 3 years, no ifs, ands, or buts. And, yeah, they expect you to know all first-year graduate material on your own.
 
Haha! Yeah, @Ted.
Where did you do your undergrad, Ted?
 
Somehow I'd rather spend the first year or so taking some fun classes and just exploring a bit instead of taking a geodesic path to a thesis, you know?
 
take a year off and travel
 
I agree with you, Demonark. I'm just alerting you to things to think/ask about.
@Oskar: MIT.
 
7:37 PM
every now and then I think about how I'll come up with anything original to write a thesis with
 
What kind of math do you want to do in grad school @Dami?
 
I feel like moving to America. Sweden is so mundane.
 
Most of us currently in America would much rather not be here at the moment.
 
Because of Trump?
 
nods
and because of 30% of the citizens whom he has unleashed ...
 
7:40 PM
Is it that bad? I don't follow the news.
 
Also how does grad school compare to the european bachelor+master+phd system?
 
Isn't it grad school=master+PhD?
 
in the US people tend to go straight from bachelor to grad school (doctorate), doing a masters along the way automatically. Some go to do just masters. Some try for doctorate, don't succeed, get masters.
 
I heard that he has decreased funding to some universities.
 
7:41 PM
I'm not sure just yet. Most likely something in the general area of algebra/NT, though I'd definitely like to learn other stuff too, especially topology. Probably shouldn't foresake analysis completely either. And... Yeah
 
also how did a Hong Kong news make it to the world
 
Congress has made things interesting. State legislatures don't like to pay for education, either.
OK, I need to go eat lunch. Back later.
 
@TedShifrin So in grad school you both take courses and do research?
 
Few things to report on?
 
@TedShifrin May god give you a long life.
 
7:45 PM
I agreed. You seem to be a great guy.
I agree*. Nothing has changed. Haha! I hate it when I make typos.
 
@AlessandroCodenotti since he's out, yeah you do. PhD is 5 years, functionally the union of master's and PhD there
Though somehow the American system apparently gives a bit more research time
 
Ah, I see
 
@TedShifrin Yep
 
At least the guy I talked to said that some European postdocs are envious
 
Typical uk phds are 3 years i think (some 4)
 
7:54 PM
In Sweden they are five years long.
 

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