whatever you do, don't append that to other stories on paywalled news websites (after removing referral info such as the "?smid=..." from ted's link, which can sometimes screw it up)
Leslie: You are savvy in the ways of the web. Perhaps on my computer, instead of on my iPad, I could be cleverer. The gift link definitely did not post.
@XanderHenderson Nope Sir. I think am a bit late to the party. I request you to validate whether my procedure was legit or not! I will be greatly helped by this. I need some clarifications from someone in the field, and presently I feel you're the best person for it.
If you're interested, I can repost the pics of the texed matrrial.
Question. I'm working on proving the Schroder Bernstein THeorem in steps. The first step has me showing that $A_1 \supset C_1 \supset A_2 \supset C_2 \supset \dots$ where $C \subset A$ and there is an injection $f: A \to C$. We also defined $A_1 = A$, $C_1 = C$, and for $n >1$ recursively defined $A_n = f(A_{n-1})$ and $C_n = f(C_{n-1})$.
So with ALL of that in place my simple question came from me doing the proof. I'm doing it via induction, but the only thing that is bothering me is why is $f(A_1) = A_2 \subset C$ explicitly and not $f(A_1) = A_2 \subseteq C$?
@ThomasFinley People are getting irritated because you're asking them to do the work for you. There is a difference between a focused question and going over a few pages of notes to interpret them.
that's what the site MSE itself is for, not the chat
for my own clarification: i wanted to write out the projections onto the hyperplanes $E:x_2=x_3$ and $F:x_3=2x_4$ in $\mathbb{R}^4$, along with the projection onto their intersection ($x_2=x_3=2x_4$)
what's confusing me is that projecting onto E and then onto F doesn't seem to be the same as projecting onto their intersection
but i think the issue is that the hyperplanes E,F aren't orthogonal ( (0,1,1,0).(0,0,1,2)=1 not zero)
the basic instance of this (Dirac's three-polarizer paradox) amounts to the point that projecting a 2D point onto the x-axis, followed by projecting it onto the y=x line, does not put you at the origin
semi: the strong limit of (PQ)^n will project onto the intersection of the ranges of P and Q
so you had the right idea, you just didn't iterate it infinitely many times
theres maybe also some background vibe in there of how things like strong/weak limits appear in mathematical models of quantum mechanics. even in finitely generated situations, 'the algebra generated by [bleh]' isn't large enough to have a complete lattice of projections, you need limits
well, i think the issue is that the pictures i keep drawing are of a particular 3D cross-section of my 4D convex set
and then i was projecting within that 3D cross-section onto a plane
which is simple enough if i'm dealing with the 3D coordinates, but i think i wasn't being careful enough with how that'd work in the original 4D setting
If $R$ is a commutative ring with unity and $a,b \in R$ then if there exists elements $p,q$ such that $ap+bq=1$ then there also exists elements $m,n$ such that $a^2m+b^2n=1$.
There is still something odd tho. the projection onto $x_2=x_3$ is given by $$p_1(x)=\left(x_1,\frac12(x_2+x_3),\frac12 (x_2+x_3),x_4\right)$$ while the projection onto $x_3=2x_4$ is given by $$p_2(x)=\left(x_1,x_2,\frac25 (2x_3+x_4),\frac15(2x_3+x_4)\right)$$
@XanderHenderson This actually reminds me of this: scottaaronson.blog/?p=304 It's a list of 10 signs that a claimed mathematical breakthrough is wrong. And the first sign is that the authors don't use Tex.
so their composition is $$p_2(p_1(x))=\left(x_1,\frac12 (x_2+x_3),\frac25 (x_2+x_3+x_4),\frac15 (x_2+x_3+x4)\right)$$ this is indeed not idempotent and thus not a projection. nothing strange there
but compare this with the projection onto the plane $x_2=x_3=x_4$: $$p_3(x)=\left(x_1,\frac13(x_2+x_3+x_4),\frac13(x_2+x_3+x_4),\frac13(x_2+x_3+x_4)\right)$$
if i restrict to the first,fourth coordinates and rescale the latter, then $p_2(p_1(x))$ and $p_3(x)$ are equivalent
at the level of matrices, the point seems to be that $P_1P_2 e_1=P_3 e_1$ and $5P_1 P_2 e_4=3P_3 e_4=e_2+e_3+e_4$
(that order of P1,P2 is intended)
i can't decide if that's interesting or not
i guess the following is somewhat interesting. If I consider $P_1 P_2-t P_3$, then generically it does have one zero eigenvalue. but when $t=1,3/5$ then you get a second zero eigenvalue
the set itself is defined like this. draw some spherical quadrilateral and take the four exterior sides. these give four angles; take their cosines to get some point in [-1,1]^4
@TedShifrin For some reason I did not get notified about your NYT article message. It is behind a paywall and I have a fair idea what it might be about.
i know the symmetry of my set has something to do with this, b/c if it's invariant under permutations then in particular it's invariant under permuting the last three coordinates
not entirely, but it is convex. the way i know to show it is that it's the projection of a larger set (the one you get if you take all six arcs not just four)
and this larger set is itself can be viewed as a section of the cone of 4-by-4 PSD matrices
but that cone is convex, therefore the section is convex, therefore the projection is also convex
having to juggle between projection and intersection is part of what i get mixed up on at times
@TedShifrin Yep, that was the event I was thinking of. Handled poorly all around. The noose part was a complete non sequitur (I know how it came about, it was not a noose, but everyone was ready to jump at anything.)
The problem is that kids have widely varying maturities and, can do really hurtful & stupid things.
okay, i think i did find some nice way to sum up why my combinations is weird: $P_1 P_2-tP_3$ generically has four linearly-independent eigenvectors, but loses one when $t=1,3/5$
@Jakobian They're not quite to my taste, but Come A Little Closer is an interesting pop-rock hybrid. Coil are certainly creative, but those songs are a bit too dark for me. I appreciated Holocene, though.
@XanderHenderson It's a classic skit (& a great song), but unfortunately that clip is blocked in Australia.
FWIW, Richie Castellano is a current member of Blue Öyster Cult. He's also involved with various other projects. I discovered him through Band Geeks, who cover a broad range of rock styles, but they're especially good at prog. Here's their cover of Siberian Khatru by Yes.
@CowperKettle Probably. ;) I guess I could set it up on my phone...
@Jakobian Hey, they're teenage girls. Why not sing about kittens & going for coffee? :) I think its a cute song, and I think their song-writing skills are improving. They managed to get out of Ukraine safely, but the drummer's house got half-demolished.
> The surname Chebyshev has been transliterated in several different ways, like Tchebichef, Tchebychev, Tchebycheff, Tschebyschev, Tschebyschef, Tschebyscheff, Čebyčev, Čebyšev, Chebysheff, Chebychov, Chebyshov (according to native Russian speakers, this one provides the closest pronunciation in English to the correct pronunciation in old Russian), and Chebychev, a mixture between English and French transliterations considered erroneous.
ajay: by googling it
i dont' mean to be rude, but it's difficult for me to read that image on this screen, and this is probably a well-known induction about a famous family of polynomials
I am trying to prove a something regarding Chebyshev polynomials. Given the polynomials $T_n(x), n = 0, 1, \ldots$ which are recursively defined by
$$\begin{cases} T_0(x) = 1\\ T_1(x) = x \\T_n(x) = 2x T_{n−1}(x) − T_{n−2}(x), & \text{for } n \geq 2\end{cases}$$
I want to show that
For every $n...
@Ajay en.wikipedia.org/wiki/… has most of the details: "From this trigonometric form, the recurrence definition can be recovered by computing directly that the bases cases hold, [...] and that the product-to-sum identity holds"
Inter-Universal Teichmüller Theory lies at least six levels of abstraction away from anything that can be described in layperson’s terms. It doesn’t have “practical examples” in any acceptable sense of the word “practical”.
On the extended real number line, does $\pm\infty=\pm\infty$ and does $\pm\infty\leq\pm\infty$? The reason for the question is the answer by @robjohn here. Without assuming $a_n$ or $b_n$ being bounded, we may have $\pm\infty\geq\pm\infty$ in $(2)$ and $(3)$. Moreover, in $(3)$, we are distributing the limit operation, but this is only allowed if the sequences converge (infinity does not mean they converge).
@sunny I mean, yes, the ordering on $[-\infty, \infty]$ is defined in such a way that $-\infty$ is the smallest element and $\infty$ is the largest. You are basically adding largest and smallest element to $\mathbb{R}$
hey guys, quick question : how can I find the point of intersection between two vectors? My origin is at $(0, 0, 0)$, and my object of interest is at $(x, y, \theta)$. I want to find a point $(a, b)$ such that I can move from $(0, 0, 0) -> (a, b, 0) -> (a, b, \theta) -> (x, y, \theta)$
@Astyx by the point of intersection, I mean if I were to draw a line from the origin, and another line from the object (x, y, \theta), at which point would these lines coincide
@sunny You don't need $a_n, b_n$ to be bounded in $(2)$. Moreover, the assumption of $\limsup_n a_n, \limsup_n b_n$ being non-zero implies that the $\sup$s considered are positive (or infinite), so there's no multiplication by zero
@TedShifrin Analyzing the URL you shared, none of the URL parameters (smid, referringSource) have unique information, so I doubt that's the correct link. I notice that in a typical NTTimes article (viewed on the website, desktop view) there are two buttons: "Share full article" with a gift icon, and "Share options" with a right arrow icon. The URL from "Share full article" icon is completely different; it has a "unlocked_article_code" URL parameter with more than 300 character long value.
@TedShifrin But make sure you only click the "Share full article" icon when you really want to share, clicking it automatically decreases your quota of 10 shares per 30 day rolling period.
@Jakobian I see. Does $\lim_{n\to\infty} (a_nb_n)=\lim_{n\to\infty} a_n\lim_{n\to\infty} b_n$ hold even if one of the limits (or both) is infinity, provided the other isn't $0$?
@Astyx, thanks for the help. I am still a bit confused on how to solve the equations inside a function though - I guess I'd start with a brute-force method, assuming some constant value for t_1 and t_2, and solve for v_2.
@ThomasFinley yeah but no one owes you an explanation or help. and the website is for help yet it doesn't mean any individual person or group of person owes you any help. they help when they feel like it and don't otherwise, and don't need to offer explanations for doing either because the explanation is that they feel like it or don't
Or people just don't know either, so it's a good practice to try and answer your own questions yourself
I remember I sometimes used to post questions and forget about them without trying much, hoping for someone to give me a hint But you should just motivate yourself and attempt solution, again and again
sometimes it takes multiple tries
I know this is a bit off-topic from what everyone was trying to say
@Jakobian for instance, there is a precise definition of $\lim_{n\to\infty} x_n=\infty$, but what is the definition of a sequence equaling infinity for all $n$? The sequences in $(2)$ may be of such kind, i.e. if $x_n$ is unbounded then $y_n=\sup_{n\geq k}x_n=\infty$ for all $n$.
Correction; it should be $y_k=\sup_{n\geq k}x_n=\infty$ for all $k$.
@sunny so as long as you just restrict your concept of infinity, specifically in the extended real numbers, to just an indicator of what are the valid ways of manipulating it, you'll be able to use it to its full extent and develop intuition, without needing to ask deeper questions
I'm not sure what you mean by definition of "$x_n = \infty$ for all $n$"
this is not a definition, but a statement
$\pm\infty$ shouldn't be thought of in an algebraic way, but as the extended reals having an order
so again not sure what you're talking about, possibly some wrong kind of intuition
we have some algebraic rules for $\pm\infty$ to help us with all the identities, but that's another topic imo
anyway the precise definition of $\lim_n x_n$ is as follows
for all real $M$, there is $N$ such that for all $n\geq N$, $x_n\geq M$
this makes sense, why? Because the extended reals have an order, and $\infty$ is the largest element in that order
We can say whetever $\infty \geq 2$ or $\infty \leq 2$
what can't be directly translated is meaning of $\lim_n x_n = a$ for some $a\in\mathbb{R}$ But we can say that e.g. $|\pm \infty-a| = \infty$, by adding rules that $a\pm\infty = \pm\infty$ and $|\pm \infty| = \infty$, which intuitively make sense
thus this again can be translated into language of limits when $x_n$ are real numbers
The definition of $\overline{\mathbb{R}} = \mathbb{R}\cup \{-\infty, \infty\}$ is that of a totally ordered set, where $\infty, -\infty\notin \mathbb{R}$, $\mathbb{R}$ has its usual order, $x\leq \infty $ for all $x\in\overline{\mathbb{R}}$ and $-\infty\leq x$ for all $x\in\overline{\mathbb{R}}$
In other words, we add maximum and minimum to $\mathbb{R}$, just as we could with any other totally ordered set
using this order, we can define what limits are, thus it make sense what $\limsup, \liminf$ etc. is
what might not make sense is algebraic operations, and we add those according to what identities involving limits we want to reflect by those operations
for example, if $x_n\to \infty$ and $y_n\to\infty$, we have $x_ny_n\to\infty$
we want to be able to state an identity $\lim_n x_n \lim_n y_n = \lim_n x_ny_n$ in this case
thus it makes sense to define $\infty\cdot\infty = \infty$
we choose what those operations on infinities mean based on the identities of limits for real numbers
that's why for example $0\cdot\infty$ is left undefined
If $x_n\to 0$ and $y_n\to\infty$, then you can't say anything about the limit of $x_ny_n$, or if it even exists
we defined $\pm\infty$ because we want to convey the idea of what it means for a sequence to get arbitrary large, in either direction
the issues you point out, tell us that maybe it's useful to consider $\overline{\mathbb{R}}$ and its limits on its own
general idea, if you see something that should be defined but isn't, then define it so that it makes sense
if that's impossible, then leave it as a case that someone didn't think about
people might be sloppy with this, because they don't start from ground up, they are already perfectly aware of the extended real line, its limits, its order, and everything in between
I've been watching Alex Kontorovich's lectures on Analytic number theory, and he often references a "prime at infinity" or a similar sounding concept. For example, in lecture 15 at 45:21 he says:
So there are local obstructions that are not just the prime at infinity.
For an example that doesn'...
in herstein's book is there any mention of subrings concept?
I searched it up, but unfortunately haven't come accross it till now.
Hello @TedShifrin, I hope this message finds you well. I wanted to touch base regarding our recent interactions on the Math Stack Exchange. I appreciate the insights and advice that you and others have shared, and I'm thankful for the collaborative atmosphere that this community provides.
I wanted to clarify that my intention in my previous messages was solely to seek validation for my approach and to receive some guidance from someone experienced in the field. I value the diverse perspectives within this community, and your input has been instrumental in my learning process.
Let $f:U\to \mathbb R$ be a smooth function & let $\alpha: I\to U$ be an integral curve of $\nabla f$. Show that if $\beta: \tilde I\to U$ is such that $\beta(s_0)=\alpha(t_0), \|\beta'(s_0)\|=\alpha'(t_0)\|$ then $d/dt (f\circ \alpha) (t_0)\ge d/dt (f\circ \beta) (t_0)$.
@copper.hat :-)
U is open set in R^{n+1}. Partial answer: $\alpha$ is an integral curve of $\nabla f$ so $\nabla f(\alpha(t))= \alpha'(t)$. So $(f\circ \alpha)'(t)= \nabla f(\alpha(t)). \alpha'(t)=|\nabla f(t)|^2= |\alpha'(t)|^2$. $d/dt (f\circ \alpha)(t_0)= |\alpha'(t_0)|^2 =|\beta'(s_0)|^2$. Not sure how to go from here?
Recently, I used a claim for solving a problem. The claim was, "If G is a group, and H is subgroup then, there exists a unique element a in H, such that all the multiples of a actually form all of H"
I am trying to prove a something regarding Chebyshev polynomials. Given the polynomials $T_n(x), n = 0, 1, \ldots$ which are recursively defined by
$$\begin{cases} T_0(x) = 1\\ T_1(x) = x \\T_n(x) = 2x T_{n−1}(x) − T_{n−2}(x), & \text{for } n \geq 2\end{cases}$$
I want to show that
For every $n...
I've been watching Alex Kontorovich's lectures on Analytic number theory, and he often references a "prime at infinity" or a similar sounding concept. For example, in lecture 15 at 45:21 he says:
So there are local obstructions that are not just the prime at infinity.
For an example that doesn'...
