@TedShifrin That's why I said it works for a simple pole, but for higher order poles, the integrals over partial arcs blow up. However, integrals over complete closed circles will cancel all the higher order terms other than the $\frac1z$ term.
@robjohn: Oh yeah, of course. His "wit" addled my brain.
Anyhow, he's put me in my place. I cannot explain and I certainly cannot read his mind, other than to have answered his original question in the first place.
Hello, it is I. I have returned after sleeping, so I tried the problem again and found that $f(x) = 1, g(x) = 1/x$ if $x \neq 0$ and $g(x) = \pi$ if $x = 0$ works. Let $a = 0$, then $\lim_{x \to a = 0} f(x) = 1 = b$, and $\lim_{x \to b = 1} g(x) = 1$, but $\lim_{x \to a^+ = 0^+ } g(f(x)) \neq \lim_{x \to a^- = 0^-} g(f(x))$
@politeproofs You're still missing the main point we were trying to make. And you still insist on using the letter $x$ everywhere, even though I discouraged that.
oh, fun story. I ran into a question for which one of the best sources is from an 1867 "problems and solutions" text...with apparently no other appearances in the literature until a random 1990's physics paper
Suppose that we choose three points independently and uniformly at random on the surface of a unit sphere as the vertices of a triangle, and consider the area of this triangle. Call this random variable $X$.
The area of such a triangle is the sum of its angles minus $\pi$, so by linearity of expe...
I suppose the reason the probability density behaves like I said is that, if you keep making a spherical triangle bigger by moving the vertices away, then eventually the 'inside' of the triangle flips to the outside
and that's precisely when it would exceed the $2\pi$ max
The part I do find interesting/frustrating: There is a known trivariate density function for the angles $A,B,C$ of a spherical triangle
but apparently no one has succeeded in deriving the pdf for the area (which equals $A+B+C-\pi$) directly from the trivariate density function for the angles
@Semiclassic: Someone posted an interesting question — to use the Clairaut relation (describing geodesics on a surface of revolution in angular momentum terms) to prove the spherical law of sines.
Ah, yes. It's all about dot products of cross products (and the law of sines is cross products of cross products, if you don't use the differential geometry).
Anyone knows of a book where Taylor's formula is presented with the little o and not the big o? I found this suggestion, but would be interested to hear if there are more. math.stackexchange.com/a/1169408/642262
But it's interesting how I get attacked for being right by the stupid ones. I didn't think I was rude, other than asking him to explain something carefully. Probably a MAGA person.
I do like Finch & Jones description of one of Exhumatus's steps: "A miraculous substitution $\cos z=\cos \frac{x}{2} \cos\frac{y}{2}$ is due to Crofton & Exhumatus."
actually, under certain circumstances, the partial arcs can be okay. For example, if the $-2$ order term is $0$, you can integrate a third order pole around a semicircle. The $-3$ order term is cancelled.
I have never used the $o$ or $O$ notation when teaching this stuff, although for proving some properties of Taylor polynomials (like the composition of Taylor polynomials is the Taylor polynomial, etc.) it is convenient.
@robjohn: Yes, I imagine there are a bunch of ad-hoc things like that. When the a** posed the question, I assumed he was trying to do contour integration with $z^{1/n}$ or something, but he never was polite enough to explain the origins.
@schn: The point is this. If $f$ is $k$-times differentiable at $a$, then the Taylor polynomial $P$ of degree $k$ of $f$ at $a$ is the unique polynomial with the property that $(f-P)(x)$ is $o(|x-a|^k)$. This is highly important because you can prove something is the T.P. without computing derivatives if you can prove that estimate.
@Ted I asked the question from yesterday about the volume of balls on main and got an answer. I'm currently trying to understand how the first example works and am not quite managing. But there's a second example, which is a very nice geometric picture, you should check it out: math.stackexchange.com/questions/3994809/…
@schn: I am very fond of that. For example, using that, you can easily write down the T.P. of $\sin(x^2)$ of degree $2n$ if you know the T.P. of $\sin(x)$ of degree $n$. So powerful. Yes, this is in Spivak, of course.
lol no @TedShifrin , that was nice haha . I meant like i studied some proofs in number theory like related to chinese remainder theorem . Those proofs are called constructive proofs . Some times to find answer for some problem related to puzzle and maths we require to construct them . How to get good into them
@Thor: Those are excellent answers Kajelad gave you. I should have written down #1 myself. It's easy enough to compute the curvature, but I haven't tried yet.
@TheReal: You mean, then, any proof that is not a proof by contradiction. I call those direct proofs. With regard to the CRT, you mean that there is an algorithm for finding the solution (or the division algorithm gives you an algorithm for finding the GCD). I don't consider these proofs so much. What do you mean, then, by getting good at these? You mean good at doing the algorithms?
I feel like I should know microlocal analysis, because i have the sense that it's connected with WKB stuff and therefore the semiclassical method in QM
@TedShifrin Yeah like how to get good into finding such algorithms ? The person who would have came first with that algorithm , how he approached ? For example consider this problem related codeforces.com/contest/1474/problem/E . It's a problem requiring construction , i want to get good in similar field .
@Ted Like chinese remainder theorem has algorithms which is constructive in nature and similar is problem that i posted in link . I find very unnatural to come up with them
@Thor: So, as @Balarka was thinking with his surface of revolution (which is a way of implementing what Kajelad did as a metric), the curvature is $K=-f''/f$. I don't see necessarily that that will have to blow up.
Take a disk in R^3 and embed it in such a way that adding one point gives you a singularly embedded sphere. What is the minimum value that the integral of K can be for this surface?
I have one more question . I find Topology very comprehensible but abstract algebra like Galois theory very hard to imagine . I mean i understand the proof but it seems i cannot feel it . How to overcome this ?
@TheReal: You keep asking these questions over and over. You have to practice mathematics. You only get better at it with practice. And you must do lots and lots of examples yourself to get intuition. This applies to everything.
@Ted sorry , Actually i asked after long time . I know practice helps in long run , but some people have better way of handling/approaching things . That's why i ask sometimes ...
Lol @ted , I mean i find difficult to feel the proves in Galois theory and other abstract algebra stuff even i understand them , so how should i study them in a way i can appreciate them . Like i feel how mathematician came with such proofs and theory . It feels very unnatural many times . Till Group,ring theory it's fine but after that it's difficult
Anyhow, I am more curious about Kajelad's curvature conjecture.
My comment was actually to @Thor, not to you, TheReal.
Group actions are super important to understand, and that's the right way to look at a lot of Galois theory. Have you taken a serious one-year course in abstract algebra and done lots of exercises? Or are you trying to "self-learn" by reading? That won't do it.
@Ted i have studied I.E Herstein topics in algebra . Till chapter 6 as part of my college course . I even got good grade too . My professor himself says that this part of math is difficult . In fact number of people taking this course are very less.
I find Herstein a bad book. Algebra is too much about tricks and not enough about integrated concepts. I far prefer Mike Artin's book (although in some ways it is more sophisticated).
there's a nice exposition on the historical development of the solvability of polynomials and the related methods at the beginning of Stewart's Galois Theory book
that's one reason I'm curious about the spherical triangle question I referenced: i'd like to know the motivation behind the from-left-field substitution which resolves the question
@Thorgott I've said repeatedly (but you tell me you ignore me when I say stuff you don't understand) that you should blow up the origin in the plane and then you have a chart on that.
This is a manifold again. If you blow up the origin in $\Bbb R^2$, then you have charts $(x,t)$ and $(y,s)$ with $y=tx$ and $x=sy$ ($s,t\in\Bbb R$ with $s=1/t$ on the overlap).
@TedShifrin Well, I have an integral $\int f(x)g(x) dx$ for which f is approximated by its Taylor polynomial. Then there is an integral with an $o(cx)$ term inside of it.
Are you doing a definite integral, @schn? Yes, you can integrate a Taylor polynomial of degree $k$, and you get the T.P. of degree $k+1$ of the antiderivative, and you can estimate the error. Do some concrete examples. This stuff is in Spivak, too.
I added a lot of stuff on this, in fact, to his latter edition.
@Thor: I forgot to ping you with the blow-up definition.
incidentally, is the following shorthand for indefinite integration with an endpoint at all sensible? $\int_a f(x)\,dx$ as shorthand for $\int_a^x f(x')\,dx'$
@TedShifrin Thanks. I'm more wondering if one can treat the $x$ as fixed inside the little o? It is a definite integral, an expectation. The notes are pretty hard to read like yesterday, but could provide more detail if needed.
In Hatcher's proof of excision, he uses barycentric subdivision. The main crux though lies in the idea that using barycentric subdivision you can make sure that you get a small enough simplex which lies in your specified open cover. Though this takes a ton of effort. Are there other ways to do said subdivision which would show this more easily?
Try to do it for something simple like the node, draw the preimage of the curve and the blow up point and you will see how tangent directions get separated at the singularity point
Hmm the thing is in principle clear but slightly ugly to write down. you essentially want to write down the numbers in $A$ and then the identity inbetween, but in a way such that this is bijective. In principle no problem, but A could start with 0,1,2,...,k for some while
And then say $f(2n) = a_n$ and $f(2n+1) = b_n$ and if this $f$ were computable then $f(2n)$ would be computable as well so $A$, as the image of a computable function, would be r.e.
@Astyx to show noncomputability it suffices that we cannot in finite time decide non-membership
@Astyx So in computer science speak, if we have an algorithm that one by one spews out the members of a set, this does not mean that we have an algorithm that in finite time can tell us if any element is a member of the set
because no matter how long we spew out new elements of the r.e. set, we can never definitely conclude that something that hasn't yet shown up is not in the set
Maybee im asking an XY question so let me rephrase. I'm given two MGF, and i've found out that they are MGF of $X=3Y+3$ and $Z=-W+4$, where $Y \sim Poi(a),W \sim Poi(a)$. Now we want $Cov(X,Z)$ which is equal to $-3Cox(Y,W)$. How do i proceed?
Here is the problem : Given a random variable $X$ with $M_X(s)=e^{as+be^{as}-b}$ and another random variable $Z$ with $M_Z(s)=e^{bs +be^{-s}-b}$ find $Cov(X,Z)$
For rings $R \subset S$ and $M$ a left $S$-module, it is obvious that $\mathrm{Ann}_R(m) = \mathrm{Ann}_S(m) \cap R$ for $m\in M$, right? I have this as a problem and it seems deceptively trivial... By taking the intersection you're just restricting to the subset in $R$ annihilating $m$, which is $\mathrm{Ann}_R(m)$...
Well, expand definitions here. $\{r\in R | r^nm=0\} = \{s\in S|s^nm=0\} \cap R$ is what it's saying. It seems trivial to me and maybe I'm missing something obvious.
How can I prove that $y^2-x^3\in \mathbb{C}[x,y]$ is irreducible?
