@MikeMiller Take a continuous function $f$ with $f(x) \geqslant c > 0$ for all $x\in\mathbb{R}$ that is not differentiable, and take $$F(x) = \int_0^x f(t)\,dt.$$
@MikeMiller Right. And since $F'(x) \geqslant c$, $F^{-1}$ has exactly the same differentiability class as $F$ if we take $f$ bounded (which I forgot to demand above).
Ugh... I'll be upset if my suggested edit gets rejected... Between the time that I pressed submit and the edit was put through, someone else had submitted their own edit... That's not fair, I didn't get any notification that someone else was editing...
@TedShifrin WTF is this " 3 Well, I even checked my junk folder, which is where you belong. Maybe it'll get here in a decade or so. - 2d ago by Ted Shifrin " about - you need to be careful when talking to students like that, you can severely cripple their self esteem.
@TedShifrin I've seen first hand what it can do, and throw-away comments (which was an hilarious burn) coming from a professor to a student can really hurt them.
Hey @TedShifrin!!! I am looking again at the exercise at which I have to find the inflection points of the curve $C=V(x^3-yz^2) \in \mathbb{P}^2(\mathbb{C})$. I found that the point $(0,1,0)$ is a singular point of the curve. Does this mean that $[0,1,0]$ is not a inflection point? Or am I wrong?
@TedShifrin sorry I didn't think it was done in jest. None the less, I'd rather have failed and been told to butt out than not mentioned it had it been what I thought it was (damn that's a convoluted sentence)
I'm all over the place, on one level I'm doing quantum field theory and group-theoretic differential equations, on another level I'm still not happy with basic projective space :(
Let $C=V(F)$ be an algebraic curve. $$F[x,y,z] \in K[x,y,z]$$ The inflection points of $F$ are exactly the non-singular of the curve, that are interesection points with the hessian.
@BalarkaSen I'm in a place called Wales ATM, the internet is so slow (and also most of my books are back in Uni) could you check for me, it literally takes 12 seconds for a google results page to show.
Yes, @evinda, I certainly know how to prove it, but I don't know what you know and what you don't. It takes some knowledge about projective goemetry, conics, maybe cubics.
@AlecTeal you can model a large chunk of basic category theory on the notion of a homotopy http://math.stackexchange.com/a/438697/82615 Think of a category as a way to talk about the set of all paths (arrows) between two points (objects) (a,b) & (c,d) in the plane, and then a functor is just like a continuous map mapping paths to paths, then a 'natural transformation' is just a homotopic deformation of one image into another, as in this picture http://math.stackexchange.com/a/399960/82615 This is all motivated by thinking of the 'action' in theoretical physics as a functor
@AlecTeal If that analogy makes sense, the rest should fall into place, the youtube videos by the Catsters on category theory plus that analogy plus some random pdf's on the internet, plus some browsing of Maclane's book should give you a large overview of category theory :)
@bolbteppa math.ucr.edu/home/baez/qg-fall2004/action.pdf is literally downloading at 3 bytes per second - I could actually ring up a dialup exchange and shout letters down the phone faster.
@evinda: No, because there are lots of different approaches, and I have no idea what you know. You seriously need to work with your fellow students and your professor.
I have a doubt about grad school applications. Some universities have asked me to list the relevant courses which I've done and asked me to list in detail, as much information as possible. How much in detail do they generally want?
@DanielFischer One question: suppose that $A$ is self-adjoint operator in $B(H)$. We can show that $\ker A^n = \ker A$, but, do you know example that $\text{Im} A^{n+1} \subsetneq \text{Im} A^n$? (Im=image, I can say range also...) Maybe shift operator (from my head, but, hm...)
@Jayesh: They're trying to get some idea of the textbook and scope of the courses. Listing which chapters of the text you covered, or the main topics (a few lines should do it) will probably be enough.
@MikeMiller real quick, you know continuous means "the preimage of a set in Y needs to be open in X", is it true that continuous means "U is closed in X if and only if f(U) is closed in Y", or is it a typo?
That's a typo, @Alec. (Such a map is called a closed map.) It should say $f$ is continuous iff for all closed $U \subset Y$, $f^{-1}(U)$ is also closed.
I haven't gotten in any trouble since middle school, when I skipped out on a mandatory after-school review session when I was SICK. I GOT IN-SCHOOL-SUSPENSION FOR IT.
Well, once you get to college, @teadawg, you'll find that we faculty are much more understanding if you let us know (by email) if something unavoidable comes up and you have to miss class. Don't come to us after the fact begging for mercy :)
I can beat that, @Mike. One of my probability students — a math major — was in my office hours getting help on homework, went down to class with me, turned in the homework, then said he didn't feel well and left. So be it. But, after the class was over, he came back and asked me to tell him what he'd missed. When I got angry, he waited until the end of my next class, apologized, and asked me again, asking what he'd done wrong. Sigh
@Cortizol Books have been known to contain errors. I'm not saying this is one, but off the top of my head, I can't imagine how to construct such an example.
@AlecTeal To show that a problem is NP-complete we have to show that it is polynomially transformable to an other NP-complete problem, right?? How do we know which problem we use?? Do you have an idea??
If they were truly mitigating, he wouldn't have been fine, getting help from me the hour before class, @Alec. He's skipped class numerous times, and I've seen him on the way to class. I have no attendance policy, but it's not my job to give my class more than once.
@Jasper, when he walked out of class at the beginning? If he was really sick, why was he right there at the end of the class? And an hour later? Nonsense.
As I said, we're much more understanding if we're alerted to problems ahead of time, or as they happen. Obviously, if someone is hospitalized for surgery or depression, we need to be told that something's going on.
@teadawg1337 On the other hand, while I am very affected by my mental problems, many people seem to think that I am a douchebag for the way my life has turned out. All those 'friends' have long abandoned me.
Well my vague understanding is that Grothendieck basically considered isomorphism classes of categories of curves of genus g with n points removed, asking how the Galois group affects this, and then Drinfeld came along and used the similarity between category-theory descriptions and hopf-algebra descriptions (which he exploited in his quantum groups proof) to re-analyze the problem, that's my vague understanding!
@MikeMiller I urge you to do it, no response teaches them nothing, they could assume "he was too busy to do it", "no" means you disapprove and you are literally refusing to help - but anyway!
@Jasper: One of my best students last year told me, finally, that he was suffering from depression. It took guts for him to tell me, but I was able to be a better teacher and much better adviser once he told me the situation. And I went out of my way to help him when he missed class for a week, let him finish the course late, etc.
Of course, when students asked me to go over a test question in review, I asked if they'd started by reading the posted solution. When they said NO, I said NO. I said I'd discuss any questions they had about the posted solution.
Well, since there's no posted questions, I'm glad to discuss them in office hours tomorrow, or another set of office hours they were invited to arrange with me before the final.
@TedShifrin I wish my university posted solutions, it'd be nice to know if I'm right, or other approaches, I hate "prove that" questions for this reason because it's easy to write crap.
I don't usually post solutions in advanced courses, @Alec, but I will always answer questions in office hours, and I write reasonably detailed comments when I grade the exams.
OK, @bolbteppa, I think this relates to a question I asked my prof a few months ago : You have you dessins of algebraic curves over \bar{Q} and you know that G = Gal(\bar Q/Q) acts on the whole set of dessins transitively.
See my university isn't like that, the lectures recognise their questions and guard the answers... I'm sure I heard this one guy call question 2 "My precious"
Yeah, I have that problem, too. For things like linear independence proofs there are only a few possible problems to give on an exam. A few years ago, for the Multivariable Math class, I started putting exam solutions behind a password :)
I once emailed a Cambridge physics professor for one chapter of notes that was missing on his website. I got no reply. The next day, he removed all the other chapters as well.
@TedShifrin I get it for those type of questions, there's only so much you can do, it's bookwork basically, but sometimes you get a gem of a question with many ways to answer, and some lovely shortcuts, I'd love to see what I might miss.
Funnily enough @bolbteppa I came up with a similar idea for Cayley graphs of groups and asked about them to my professor. Here's the bit I recall from his talk : G acts on the Cayley graph Cay(G). Embed Cay(G) onto the smallest genus surface M. G acts on the vertices and small nbhds of Cay(G) embedded in M.
Err. Now he said something about Nielson embedding theorem and 84(g-1) theorem which I can't recall anymore.
@JasperLoy I've gotten 2 of his, Topological manifolds and smooth manifolds, but you say there'll be a second edition of the Ri[that bloke] manifolds book?
Reception of the Riemannian manifolds book is more mixed than his previous two, which are very well-liked. I don't have any opinion on it since I've never read it.
@bolbteppa Now, here's an interesting idea : You have an algebraic curve f(w, z) over \bar Q. You have the dessin G of this curve. Now note that monodromy of the riemann surface of this algebraic curve acts on the vertices of G ;)
@evinda I don't know. I don't know the definitions, and it looks all just wrong to me because apparently the "DFS" algorithm ignores the fact the it is a directed graph, so the vertex b ought to be unreachable from every other vertex.
@DanielFischer I think that it is actually unreachable from every other vertex. The red edges I added show to the node which is the parent of the node from which the edge begins.
So let me make sure I understand: You have a drawing (dessins) of curves over \bar{Q} which means Q plus roots of polynomials over Q. Consider the non-planar drawings. Are you basically saying the nodes of the dessins are like the fixed points of the Galois group as it acts on this surface, or why are you removing the nodes? I would imagine moduli spaces are used as a way to distinguish the planar (1,3,2) against non-planar (1,2,3) (if that's right?)
@bolbteppa Note that the nodes of the dessins are branch points of the corresponding riemann surface, so they are fixed under the action of monodormy groups
But monodromy groups are precisely galois groups!
so to make the action on the dessins free, you have to chuck out the branched locus
this is what you do in branched galois covers, say. removing the branched locus (the nodes) to make the cover galois
@bolbteppa This is no big deal : you have to make the action of monodromy groups transitive on the sheets of your riemann surface. thus you have to remove little neighborhoods along the branch points.
@DanielFischer I will look at tha algorithm we were talking about yesterday.. :) But could I ask you before that something else? I want to describe an algorithm with time complexity $O(m)$ that, given a set $M$ with $m$ numbers and a positive integer $p \leq m$, returns the $p$ closest numbers to the median element of the set $M$. How could we do this?
I will grant it to @Chris'sis that the fact that $$\int_0^{2\pi}\log|1-e^{it}|dt=0$$ i.e. that $$\int_0^{\pi}\log \sin tdt=-\pi \log 2$$ is relevant to the proof of Jensen's formula.
@BalarkaSen I at least encourage you to read Kline before you actually make that mental decision, being a teenager is too young to close your mind up to such a large area of math, especially since this is the historical source of your material ;)
@evinda To find the median, we need the element in the middle of the array after sorting, or the two elements in the middle if the number of elements is even. Thus basically, at each step you only have one recursive call instead of two, since you don't need to sort the part of the array that comes after or before the median. That pushes the complexity down to $O(m)$, provided you choose the pivot in a way that guarantees that it is not too far away from the median.
@evinda Calculate? We determine it by almost-partially-sorting the array, not by calculation. Unless the number of elements is even, when the median is the mean of the two closest values.
@Sabಠ_ಠ Program a bit. You will occasionally have the question "how do I do this efficiently?", then you go and look if thinking didn't turn up something good.
