probably a silly question, but.. say a power series has interval of convergence $(-a,a)$ (i.e. it converges in $(-a,a)$ and nowhere outside this set). Does this necessarily mean the power series does not convergence uniformly on $(-a,a)$?
[[regarding my question above (I can't seem to reply to my own message)- never mind, I found the answer. Indeed it can't converge uniformly in $(-a,a)$.]]
@CalvinKhor right but I meant to ask about when the convergence is strictly in $(-a,a)$ i.e. there is no convergence in the endpoints. Abel says if there is convergence in the endpoints, say in $-a$, then there will be uniform convergence in $[-a, b]$ for any $b<a$
@CalvinKhor assume by contradiction there is uniform convergence in say $(0,a)$. so by Cauchy's criterion for uniform convergence there's some $N$ such that for all $n>N$ and $m\in\mathbb{N}$ the absolute value of $\sum_{k=n}^m a_k x^k$ is less than any given $\epsilon$. Let $x\to a^-$ (this is a finite sum) so that the absolute value of $\sum_{k=n}^m a_k a^k$ is $\leq\epsilon$ so that $\sum_{k=1}^{\infty} a_k a^k$ converges contradicting the assumption that there is no convergence in the endpoint
2
I probably shouldn't use this bad notation $a_k a^k$ :)
I actually found this argument (in a slightly different form) in my lecture notes that discuss Abel's theorem. I think it can be changed a little to work without Cauchy's criterion, I have to think about that
I'm gonna go think about how I can get rid of Cauchy's criterion for uniform convergence in this proof. My students didn't learn the basic Cauchy's criterion for convergence of sequences and series in the preceding course and so I'm trying to avoid anything that resembles it everywhere I can. I probably won't show them this result about uniform convergence anyway, but just in case someone asks about it.
it is indeed an odd omission. I mean they haven't even discussed Cauchy sequences of real numbers. They might have mentioned it, but they glossed over the concept and haven't for instance seen a proof that over the reals a sequence is convergent if and only if it is Cauchy.
these are engineering students, I don't think they will learn about banach spaces
So, I would like to know whether 3 dimensional T tetrominoes can tile 6x6xn regions. I know for a fact that 6x6x6 and 6x6x4 are both possible, because I've constructed an example for both, and I'm quite confident that I have an argument that proves impossibility for odd n. This just leaves 6x6x2, and so far I have not been able to construct a tiling, nor have I been able to construct a coloring scheme and measure to prove it impossible
@WojciechKulma transposition just moves things around, but it does so in a consistent manner, and matrix subtraction is element wise, so as long as A and B are the same size, that should be true regardless?
Let $f:I\to \mathbb{R}$ and $I\subseteq \mathbb{R}$ If $\lim_{x\to a}f(x)=L$ Then is it possible to prove that: There exists a function $g:\mathbb{R^+}\to\mathbb{R^+}$ such that, $\forall \epsilon\in\mathbb{R^+} \forall x \in I ( 0 < |x-a| < g(\epsilon)\Rightarrow |f(x)-L|<\epsilon)$
@Prithubiswasleftmse yes, the construction is as follows. Pick any $\epsilon>0$. By definition of the limit, there exists $\delta>0$ such that the condition you wrote is true with $g(\epsilon)$ replaced with $\delta$. Define $g(\epsilon):=\delta$; done
Here is one way that may work: For every e>0, there exists a $d_e<1$ such that ''you know what". Define $D_e$={all possible $d_e$}. Then choose $\sup D_e/2$.
Does this work?
since sup is unique by definition, you get a unique d.
I don't know enough foundational maths to say if what Koro did is Kosher without choice. I know weird things happen with sets: is it possible to define the set of all possible d_e without choice?
For a continuous function $f : [0,1] \to \mathbb{R}$, I want to show that $$\lim_{n \to \infty} \int^{1/\sqrt{n}}_0 f(x)ne^{-nx} dx = f(0)$$
I'd considered that $$ne^{-\sqrt{n}}\int^{1/\sqrt{n}}_0 f(x) dx \leq \int^{1/\sqrt{n}}_0 f(x)ne^{-nx} dx \leq n\int^{1/\sqrt{n}}_0 f(x)dx$$ but I am not sur...
My point is this, (1) If you accept the standard axioms, there is absolutely no need to use completeness or anything extra: a function is something that eats an input and gives you a single output. (2) If you are worried about what axioms are needed, I am not the right person to talk to.
(1a) a function having a single output is not the same as there being a unique number delta that can ever be assigned to epsilon. This would be the uniqueness of the function, and in this particular case the function cannot be unique anyway
(1b) a function having a single output for each input should be made more precise in terms of ordered pairs but I guess you knew that
Hello everyone. Can I ask something about probability theory? So in particular I have a question to an exercise which I started to solve but I can't finish it. Therefore I postet a question on the page and thought maybe there is someone who can help me.
You need to show $\sum_n P(|X_n|≥ a √n) = \infty$. LHS is $\sum_n P(|X_n|^2 ≥ a^2 n)$. I think there should be some identity relating $E(X)$ with $\int P(X>t) dt$ maybe
@CalvinKhor I really just care about understanding this: when you have $f(x)$ equals some inverse integral transform and you do a substitution on $x$ then the resulting integral doesn't converge
my example is $e^{-z}$ being equal to the cahen mellin integral (the inverse mellin transform of the gamma function). substituting $z=n^s$ and summing over $n$ you get a function on the LHS and a convergent integral on the RHS. The RHS allows you to get a formula for the LHS. I'm assuming that if I do a substitution and the RHS doesn't converge then it's probably not useful to get a formula
so then you could ask whether the kernel I mentioned above converges as a distribution
@CalvinKhor I thought too, but complex functions have additional properties that probably contribute to a different structure....
@CalvinKhor I could be wrong.... But since the domain is an open one, and the function is $r$ times differentiable there, so it's holomorphic.... it seems to me that a property to be taken into account...
I believe you will find something for the general case in Leoni’s a first course in sobolev spaces
All the sets for all r and holder coefficients coincide. I don’t know what keeping track of the holder constant is good for. Where is this from? I would continue reading as if C=R^2
I was wondering about this... If a function F satisfies \lVert F\rVert_{r,\alpha}<\infty, then will it also satisfy \lVert F\rVert_{r+1,\alpha}? In this case, it would be necessary to guarantee that D_\alpha F^{(r+1)} is finite too...
@Mrcrg maybe there is something subtle with the boundary. The Cauchy integral formula argument I was thinking of requires the function to be analytic on a set larger than the domain of interest
@CalvinKhor Well, I was trying to prove it and I couldn't, now I've been trying to think of an example for some time, I still haven't been able to. I'll try for a while longer, if not, I'll post the question...
@CalvinKhor I quickly looked at the book you suggested, I didn't find what I'm looking for, later I'll look more carefully...
@CalvinKhor Thanks for the references, I'll take a look. Partington's notes do not appear to have reference to the space
I found the book "Holomorphic Operator Functions of One Variable and Applications", where is defines this space (so at least I know it exists somewhere else... hehe), but seems to be concerned with other things, but still interesting...
@CalvinKhor Thank you for your help! A professor of mine said that this is an important space, but I couldn't find any reference (he didn't provide it to me either)... It seems to me that it might be a better known space in physics... who knows...
Note that I had to raise the limit a couple of points, but this edit would maintain the title height and give the improvement that the edit was trying for
Can someone please take a quick look at my recent question concerning a volume of a certain pyramid using a surface integral: math.stackexchange.com/q/4450561/144766
The idea should be relatively simple, I'm just looking for a confirmation that I'm not missing something major here
@mechanodroid I certainly do not understand why you are using a surface integral rather than just usual limits of integration on a triple integral. I also don't understand why you've set it up this way rather than putting the cone point at the origin (although perhaps spherical coordinates doesn't seem so natural with your problem here).
@micsthepick So, I left a script running overnight to check the 6x6x4 case, but I had to kill it abruptly this morning, because it had consumed too much memory and didn't respond to a keyboard interrupt. I suspect that the 6x6x2 case is not too much difficulty, however I cannot verify that there is just a bug in my code that causes it to return 0
@TedShifrin I agree that it is grotesque, but I know I try many things on the way to finding a nice path, so I want to explain why it won't work before they get in a bad habit.
@copper I had one that arrived during the height of Covid. I delayed, but then realized that my back/neck couldn't take sitting on a hard chair 8 hours a day for however many days — even one hour would be difficult, so I asked for a permanent health dismissal and got it.
I was rejected numerous times by lawyers in Athens, GA, but the first time I got called after I moved to the adjacent county, I got nabbed and was the fore-person of the jury. But never got called again until I moved here.
@copper.hat once, my wife being an attorney actually tainted me. The other times, I expressed my belief that police will testify to support each other, regardless of facts. I have seen them be untruthful in court more than enough.
at least some of the lying is well intentioned. they're not usually trying to railroad people as much as they believe their coworkers and don't want to see someone get away because of a lack of evidence. some of my friends would disagree with my assessment.
i used to keep a diary with lots of details. years later i would tell a story and (sometimes) later check the facts as i had recorded them. my recollection was frequently wrong in many aspects. so, testimony without some corroboration always seems suspect to me. (to be fair, we are talking about 3 decades ago.)
a good investigation can often find some other evidence. it says something when a prosecutor is concerned about the bare minimum amount of evidence necessary to avoid losing on appeal.
because that's a tiny amount of evidence.
there's some famous case, i don't think it's in california. someone saying "i did not do that" and the jury not believing that was sufficient evidence. there was basically nothing else in the record.
which is why you don't take the stand in your criminal case.
@TedShifrin Thanks, I wasn't sure how to set up the limits for the triple integral. Namely, I don't know how to find the limits for the $(x,y)$ variables, I would have to project $S$ to the $xy$-plane and this projection is certainly not a square in general. Could you help me?
@robjohn Thanks a lot for your observation, I've replied to you in a comment about a possible fix.
I find amusing both of you found my approach grotesque when that was basically the only thing that came to mind since I don't know how to set up the triple integral (at least it seems it would be more complicated than this).
@mechanodroid It's not a standard triple integral set-up from your approach because you're using a cone (with its vertex on the z-axis) rather than a cylindrical shape. My approach would be, as I said, to put the vertex at the origin and turn the picture upside-down. It's most natural to work in spherical coordinates and give your surface as $\rho = f(\phi,\theta)$.
The only trickiness is that that surface is limited by the cone over the square (in the plane $z=H$ now). So the four "faces" come from $x=a, z=H, -a\le y\le a$ and so on. So we have to translate those four faces into the spherical coordinate picture, but it should be doable.
