Let us say we are given an abelian variety $X$. Denote the torsion points by $T$. Assume we have some variety $V$ of some dimension $d$ in $X$. We know that $T$ is dense in $X$. What can be said about the density of $V \cap T$ in $V$?
I've been drinking homemade macchiatos all morning.
(Because cost came up above: I pay about $90 for a 5 lb bag of beans (including shipping) and use about half an ounce of ground coffee per shot, which comes out to around 60¢ per macchiato, I guess. Also, for those that care, I get my 5 lb bag, weight it out into 8oz portions, which I bag and freeze; I put a new bag of beans into the hopper of my grinder about once every week to 10 days.)
The median of a data set is the $a$ which minimizes $\sum |y_n - a|$, and the mean is the $a$ which minimizes $\sum (y_n - a)^2$. What minimizes $\sum \left((y_n - a)^2 + |y_n - a|\right)$?
I have reason to suspect it's also the mean (or at least very close), but I get tripped up considering the derivative of this last expression.
There's not going to be any analytical expression for this. Remember that the derivative of $\sum |y_n-a|$ is a sum of $\pm 1$, so you will get the mean only when those add up to $0$. Thus, you need an even number of data points, and there have to be equal numbers to the left and right of the mean.
The deviation from the mean will have to cancel those $\pm 1$, in general, to get a critical point.
@copper What is lasso regression?!
@shin Sadly, the lecturer in the MIT ODE course recently died. He was a good friend of mine.
That's what I was worried about. Interestingly, for the data sets I've tried, the mean gets you pretty darn close, and the problem itself wants the integer that minimizes that quantity for a data set of integers.
So the minimizer is definitely NOT at a critical point, @Fargle. The mean kills the derivative of the first term, but we get a net +4 derivative from the second term, I believe.
Minimizing among integers is totally NOT amenable to anything calculus.
It's just totally discrete math and/or quadratic programming.
I agree, but my suspicion---possibly ill-founded---was that solving the minimization problem generally would mean I'd only have to check very close to the analytic solution.
$n$ points in the plane. We want to cover them all with disjoint unit disks
For example, if they're all very close to each other, we can use a single disk; if they're all very spread out, we can give each point its own disk. But with an intermediate spacing it's not clear.
(a) Show that there is an $n$ and an arrangement for which this is impossible. (b) Show that for $n=10$ this is always possible.
SAJW. they'd found a youtube video on the n = 10 case. i don't think they'd explicitly raised the possibility of a configuration where you can't do it. there was static about whether the coins could cover more than one point.
someone linked a math.se thread on it. people goof up in various ways in introducing the probabilistic model that a lot of people use to handle the problem.
the world's greatest link on this problem has yet to be written.
i didn't mean to actually criticize it as old news. everything old is new. i'd never heard of the problem until then.
@AkivaWeinberger Depends on how you define the square root function. Though I do think that $\sqrt{x}\sqrt{y}$ is as good as it is going to get, assuming that you mean $\sqrt{\cdot} : [0,\infty) \to \mathbb{R}$.
What course/subject covers exterior algebras, cohomology, chains, etc? I'm missing large swaths of foundational knowledge to breach the topics I'm looking into at the moment in the way I'm looking into them.
I would imagine that it is commonly taught at the advanced undergraduate or graduate level. E.g. I would expect those topics to come up in a second year of undergraduate algebra (if such a thing exists).
I would also expect it to be in a somewhat specialized "topics" class, maybe.
@leslietownes Here's one for you, then. A point is chosen from inside an equilateral triangle, and line segments join it to the vertices, so that the angles at the point are $x$, $y$, and $z$.
The line segments joining the point to the vertices are rearranged to form a new triangle. (So each side of this new triangle has the same length as one of the line segments from the point to one of the equilateral triangle's vertices.)
Suppose given $PSL_2(\mathbb{F}_7)$ and its conjugacy classes with the number of each elements in each conjugacy class and only one 3 dimensional character
using this, how do I find the irreducible characters of dimension 6,7 and 8?
@AkivaWeinberger That isn't at all surprising, as every real number is an equivalence class of Cauchy sequences of rational numbers. So every real number (computable or not) is the limit of a sequence of rationals. And there is no reason why we couldn't choose that sequence to be increasing (so that the limit is the supremum).
@Axoren Pick your favorite uncomputable number, $\alpha$. Choose any rational $a_1 < \alpha$. For each $n$, choose a rational number in the interval $(\frac{1}{2}(a_{n-1}+\alpha), \alpha)$. Then $(a_n)_n$ is increasing, bounded above by $\alpha$, and $\sup a_n = \alpha$.
This appeals to the density of $\mathbb{Q}$ in $\mathbb{R}$.
Consider $PSL_2(\mathbb{F}_7)$. Suppose all we know are its conjugacy classes with the number of each elements in each conjugacy class and only one 3 dimensional character.
Using this, how do I find the irreducible characters of dimension 6,7 and 8?
It is easy to construct another 3 dimensional i...
because the sequence can be out of order, so if a number is in the image we'll eventually know, but if a number is not in the image we might never find out
I would rather have one question with several answers than many questions with several answers. Whether or not the original answer to the original question is garbage, all of the answers should be in one place.
i usually do a quick dupe search if i'm interested in maybe solving a problem and it feels like 'someone has to have asked this before.' usually at least one answer is OK, or across a group of answers someone can piece together how to fit a slightly different, but not materially different, set of hypotheses.
In this particular case, I am torn, because (a) the original question is not very good and (b) the original answer is (I am told---I didn't bother to read it) garbage.
I struggle a little bit here when I don't know who is replying to what. Do I just need to be here more often and get the hang of it, @leslie, @Ted, @Xander? (or any one of the above.)
@TedShifrin That is your prerogative, but your opinion runs counter to the consensus opinion of the majority of the users who engage with meta in order to give an opinion, and I am supposed to enforce that consensus opinion with my diamond.
@TedShifrin Yeah, the original question is, I think, garbage. Which makes things easy in this case.
Well, I think a lot of people have the misconception confusing equal derivatives with equal values of the function. So that is an interesting mistake to explain.
My interest in socratic teaching/learning has been pushed out with the desire to have a compendium of perfect answers. I just don't think that's education. It's an encyclopedia.
Let the place die along with democracy in this country.
Please understand I was not complaining when I asked about who is responding to what. I assume it is someone consecutive in flow, but this chat often has parallel conversations going on.
@TedShifrin I agree. But the original goal of the SE network is to build something which more similar to an encyclopedia than a tutoring or education service.
@TedShifrin In all fairness, Socrates would not do others work or thinking for them. I think chat rooms, particularly this one, is structured better for socratic teaching.
I certainly have noticed that in my main field of interest — diff geo — the level of questions has sunk abysmally to the point where people are asking stuff below the level of introductory textbooks ... repeatedly and repeatedly.
In any event, there are other places on the internet which are more purpose built for Socratic exploration (reddit seems to have active mathematics communities, and there is always Quora). Math SE is shooting for a different niche.
i feel a little bad for the guy who is going to have a reply ping on his 6-year-old wrong answer. but it does make the encyclopedia better to have a comment explicitly flagging the error.
@amWhy: In the old days, I used to be quite successful on main with certain OPs engaging them in a sort of conversation. Eventually, either they or I would write up an answer. But then it started happening that people swooped in, ignoring the conversation, and just posted an answer to get rep. I realize I'm in a minority on this, so again I should just disappear.
@leslie I have no compunctions about calling out the garbage (but I did so politely).
@TedShifrin "But then it started happening that people swooped in, ignoring the conversation, and just posted an answer to get rep." Yes. That is a huge problem, and I wish it would stop. This was meant to mitigate that problem (to some extent).
@Xander There are some truly interesting questions that have been asked (that would be closed in an instant today) on which I had to work numerous hours to get a solution. But they were certainly lacking context. In fact, I even asked the OP "where did this come from?" and got no response. But I found it interesting enough to solve anyhow.
@TedShifrin I understand. I was here in the "old days" too. It was a different place then. But it was a format not adaptable to the increasing droves of people. Which was not necessarily, given the interruption to your method, a bod thing.
Isn't $1/(2^n)$ computable for every finite $n$, so it satisfies Akiva's definition of a computable sequence, yet the sum of all of them sums to $1$, so it's bounded, it is also always increasing, but the supremum is $lim_{n\rightarrow\infty} 1/(2^n)$, which is uncomputable?
I see, that's what I'm missing, so if it never repeats, it may as well include a subset of all $\mathbb N$ which if it included all of it would converge for sure.
@TedShifrin "Context" is a compromise between the folk who want to answer homework questions, and the folk who want high quality questions on the site. It doesn't make anyone happy. If we could all agree that homework questions are terrible, and should not be allowed on the site, then I am sure that we could relax the context requirements.
One of my favorite homework questions posts is one I gave an answer that was so ridiculous they couldn't pass it in but it was right. math.stackexchange.com/a/1817756/187120
Normally these things are checked by decrementing some number of bins, and making changes somewhere else. If you have to decrement indefinitely, then you wont halt.
The point isn't that the operation can't be done by a computer, the point is that for this specific input, the output couldn't be described by any (input-less) computer
I thought something like that might be the case, my memory on these things is a little hazy since I learnt about it all mid year and haven't used it since.
If I have a sequence $[0,1,\dots,\infty)$, is the least upper bound of that sequence computable? Does it even have one?
It was pretty fun finding out that Fibonacci arose from me deriving the product recurrence relation
But then I imagine a kid trying to pass that in on their homework and it looks like $a_n = (-1)^{F_{n}} = (-1)^{\frac{(1 + \sqrt{5})^{n} - (1 - \sqrt{5})^{n}}{2^{n}\sqrt{5}}}$
The instructor is gonna give them such a side eye.
But you can't say it's wrong
@AkivaWeinberger So we're not saying that the supremum is uncomputable, only that in general, the program computing the supremum does not exist.
@TedShifrin Honestly, I could be convinced to remove the entire mechanic of "accepting" an answer. An upvote means that one person found your work useful. The green check means the same thing, but the "one person" is the original asker. I think that there is a good faith argument to be had about whether or not that one person should matter more than anyone else.
Mind you, I'm not really putting forward an argument (one way or another), but I think that there are good arguments against the mechanic of "accepting an answer" (and good arguments in favor).
Well, we hope the OP will expend the effort to follow up and see if he/she understands and learns from the answer. Random people (as that example I linked to shows) will upvote without the remotest sense of critical thinking.
@TedShifrin Okay. Just letting you know. I added my downvote as well, so now down to four. Ted, I am not posting to upset you. Perhaps I should leave. And I agree with your last comment I see before leaving.
@Axoren Hm? The supremum of this sequence is uncomputable. The sequence $a_n=\sum_{i=0}^n 2^{-c_i}$ is a computable sequence of rationals iff $c_n$ is a computable sequence, but $\sup a_n=\sum_{i=0}^\infty2^{-c_i}$ is a computable real iff $\{c_n:n\in\Bbb N\}$ is a computable set, and there exist sets that are one but not the other
For an example
Consider the set $\{|x^3+y^3+z^3|:x,y,z\in\Bbb Z\}$.
It was an open problem until recently whether $33$ was in that set
As it turns out, (8,866,128,975,287,528)³ + (–8,778,405,442,862,239)³ + (–2,736,111,468,807,040)³ = 33, and there are no smaller solutions
That set is computably enumerable, because I can write a program that enumerates all triples $(x,y,z)$ and outputs the sums of their cubes. It will eventually list all elements of that set, but not in order
However, that set might(!) not be computable, because it's possible that there does not exist an algorithm that does the inverse problem: given a number, is it in that set?
What I was saying is that we're specifically saying that taking $c_n \to \sup a_n$ is the uncomputable part. This doesn't say that $\sup a_n = \alpha$, then $\alpha$ cannot be the output of a turing machine.
@TedShifrin Well, then I'll say you upset me, after letting you a few of us were working on it, your response; sometimes with your flippant broad strokes, and comment I mentioned was not appropriate. Perhaps I'm too sensitive, and it's all my fault for saying this. You have an enormous role here, and many of us love your dedication to what you do here. But you are too quick to degrade others. like you've done here today. Talk is cheap. Mutual respect goes a long way.
For this $a_n$, $\alpha$ cannot be the output of any Turing machine. For other $a_n$, it might be
This isn't a complete proof because I haven't given you an example of a set that is computable but not computably enumerable
That sum-of-cubes thing is conjecturally an example, but presumably you want a proven example
There does exist one: assign a number to each Turing machine (called Gödel codes); then consider the set of Gödel codes of Turing machines that halt
Another one: assign a number to each sentence in the language $(0,1,+,\cdot,<,\forall,\exists,\land,\lor,\lnot)$ (called Gödel codes also); then the set of Gödel codes of statements that can be proven from the Peano Axioms is also known not to be computable, but is computably enumerable because you can enumerate all proofs
The conjecture that the set of provable statements was computable was called the "Entscheidungsproblem" ("decision problem"); it was proven false in 1933 (I think) by Alan Turing
and also independently by Alonzo Church
and also I think they both showed that the halting problem is unsolvable
This is the part that I'm not getting. $\alpha$ is a number first and the supremum of $a_n$ secondarily. You're saying that $\alpha$ cannot be the output of any Turing machine because we can't solve $\sup a_n$.
