> ... being skilled in math could indicate having a more open-minded approach to new ideas, but might also be associated with lower levels of politeness. psypost.org/2023/08/…
The Question:
In what ways, if any at all, might it be possible to move away from and towards a point simultaneously?
Motivation:
I read in "Dialetheism and Its Applications", edited by Young et al., in the section written by Priest, that
[m]oving towards something and moving away from it (in ...
also you could just fix a point on the real line and take a flow in the + direction, then to the left of the point there's an influx and to the right of the point there's an outflux corresponding to 'moving towards' and 'moving away'
@Shaun Honestly, this seems like a kind of meaningless semantic game. Until you define "moving towards" and "moving away from", the question is kind of meaningless.
To me, these terms have fairly natural definitions: let $x(t)$ denote the position of a particle at time $t$. For simplicity, assume that $x$ is differentiable. Then the particle is moving towards a point $y$ if $\frac{\mathrm{d}}{\mathrm{d}t} d(x(t),y) < 0$, and moving away from a point $y$ if $\frac{\mathrm{d}}{\mathrm{d}t} d(x(t),y) > 0$, where $d$ is whatever metric you are using to measure distance.
Or, perhaps, the particle is moving towards $y$ on the interval $[a,b]$ if $d(x(a),y) > d(x(b),y)$ (if you don't want to assume differentiability, or even continuity, I suppose).
Per either of these definitions, simultaneously moving towards and away from a point is impossible.
@XanderHenderson @Shaun What about motion on a circle? As $\theta$ increases from $0$ you move away from $(1,0)$ but are also moving toward it (the long way).
i thought about it sometimes when teaching. you see it on MSE, too. "how do i solve X" "here's [a straightforward approach]" "no not that way, like, what is a clever way of solving it if you didn't know that way" "so you were so hung up on looking for a clever way of solving it, or ascertaining some kind of divine intent about the ways you were not supposed to solve it, that you didn't bother solving it" "yes"
I may be completely off here, but would the generators $a=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, $b=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ be a simple example?
there's that picture of how SL(2,Z) acts on the upper half plane, with all the nested semicircles. isn't it reasonably clear from a deep enough dive into that?
it obviously isn't that clear to me since i don't remember it. i think we did this in a complex analysis class. and i think one time, on some homework, i showed that something contained a copy of F_2 by making something act on the plane or something like it with a picture.
ted: this is the guy who periodically breaks into leslie's house
This is inspired by
Prove that $x$ is an integer if $x^4-x$ and $x^3-x$ are integers.
For which positive integers $m$ and $n$
do $x^m-x$ and $x^n-x$
being integers
imply that
$x$ is an integer.
$x$ is assumed to be real.
I don't really have an idea
of what to do
except for the idea that
this prob...
> Let $$R=\sup \left\{\left|x\right|\ge 0:\sum a_nx^n \text{ converges}\right\}.$$ If $R=0$, then the series converges only for $x=0$. If $R>0$, then the series converges absolutely for every $x\in \mathbb R$ with $|x|<R$, since it converges for some $x_0\in\mathbb R$ with $|x|<|x_0|<R$.
Why are we guaranteed that there exists an $x_0$ such that $|x|<|x_0|<R$?
In general, if $M = \sup(A)$ and $x < M$, then there must be some $a \in A$ such that $x < a \le M$. If $A$ does not have a maximum, then the inequality is strict, i.e. $x < a < M$.
@XanderHenderson do we know if $\left\{\left|x\right|\ge 0:\sum a_nx^n \text{ converges}\right\}$ has a maximum or not? If we don't, wouldn't it be more correct to write $|x|<|x_0|\leq R$, since this includes $|x|<|x_0|<R$?
The point of this theorem (lemma? observation?) is that if a power series converges for some $x_0$, then it will converge for any other $x$ with $|x| < |x_0|$. The goal is to find the "biggest" value $R$ such that the series converges for any $x$ with $|x| < R$. Once you have that "biggest" value, then you know some nice things:
(1) the series will converge for all $x \in (-R, R)$, and
(2) the series will diverge for all $x \in (-\infty, -R) \cup (R, \infty)$.
At $-R$ and $R$, the series may or may not converge. You need to do some extra work if the behaviour at the endpoints is actually important to whatever problem you are working through.
@冥王Hades I think Xander was trying to say that you being rude most likely comes from you being immature. You being proud of this as a means of boosting your own ego might be a little misguided. Besides reading a statistic as something which applies to you as a person is similar to reading tarot cards...
@sunny I've been pondering on this some more, and I definitely think stating $|x|<|x_0|\leq R$ is more clearer/correct than $|x|<|x_0|<R$ . From the definition of the supremum $R=\sup (A)$ and the fact that $|x|<R$, we only know that there is some $|x_0|\in A$ such that $|x_0|>|x|$. Since $R$ is the supremum, we have $|x_0|\leq R$, i.e. not $|x_0|< R$.
sorry to drag this out, but my point was this: if $A=\{1,2,3\}$, then $2.8<3$, but $2.8<3<3$ is incorrect. It should be $2.8<3\leq 3$. Of course, the non-strict inequality also holds when $A=(0,3)$,i.e. when the set has no maximum.
I was reading about the term, "Significant" digits in the book, "Numerical Mathematical Analysis " by J Scarborough.
The definition of the significant digits were given as:
A significant figure is any one of the digits $1, 2,3,... 9;$ and $0$ is a significant figure except when it is used to fix ...
@robjohn Ok, but are my understanding and inferences mentioned in OP correct?
If it is so, I would be very glad to accept your answer, if you think about posting your comment as an answer with the suitable added inputs. Thank you!
(I am talking about the inference and understanding written in bold in my post)
@robjohn ChatGPT says this, "Your understanding of significant digits is mostly correct. Let's break down your questions:
1. **Regarding $7800$ and $7878$:** Your reasoning is valid. In the number $7800$, the zeros are considered placeholders, and the significant figures are $7$ and $8$. Zeros used in this way are not considered significant.
2. **Regarding $46300$:** You've grasped the concept well. The ambiguity arises because without additional context, it's not clear whether the trailing zeros are significant or merely placeholders. They could represent actual precision or be used to h…
This is inspired by
Prove that $x$ is an integer if $x^4-x$ and $x^3-x$ are integers.
For which positive integers $m$ and $n$
do $x^m-x$ and $x^n-x$
being integers
imply that
$x$ is an integer.
$x$ is assumed to be real.
I don't really have an idea
of what to do
except for the idea that
this prob...
@Jakobian For a real power series, yes. If you expand to complex power series, then only on some disc about zero.
And note that, in this case, "degenerate interval" means $\{0\}$ (which can be thought of as the closed interval $[0,0]$, I suppose).
@sunny Again, the goal is to show that a power series converges on an interval. At the end of the day, it will actually be possible to show that if $|x| < R$, then there exists some $x_0$ with $|x| < |x_0| < R$. So, yes, it is more accurate to start off with the weak inequality, but, at the end of the day, you can get the strict one.
But, again, it doesn't matter.
This is analysis. No two things are ever really equal. Everything is always up to some epsilon. It is best not to get too precious about strict vs weak inequalities. It almost never matters.
@sunny A priori, you know that $R$ is the supremum of some set---call that set $A$.
By definition of "supremum", if $x \in A$, then $R \ge x$.
Moreover, if you pick some $y < R$, either $y \in A$, or there exists some $x \in A$ such that $y < x$.
Moreover, if $A \ni y < R$, then (by definition of the supremum) there is some $x \in A$ with $y < x \le R$.
So, long story short, if $R = \sup(A)$ and $|x| < R$, then there is some $x_0$ with $|x_0| \le R$ such that $|x| < |x_0| \le R$.
For the strict inequality, properties of the supremum are not really relevant. Rather, you can prove that if $|x| < R$, then there exists some $x_0$ with $|x| < |x_0| < R$. But this follows from the fact that the set on which the power series converges is an interval. So the strict inequality is a property of intervals, not suprema.
@leslietownes Can I buy a copy for $-5$ lesliecoin?
@geocalc33 Anything with "Riemann" or "zeta" in it is likely to get closed fast.
@geocalc33 My PhD advisor once thought that he had a geometric approach to RH, via fractals. I don't think it panned out, but some interesting mathematics came out of the program.
@XanderHenderson ok, so $|x| < |x_0| < R$ follows from the fact that the set on which the power series converges is an interval. Note to myself; I must show the set is an interval :)
Am I right in thinking this self-normalizing result about free products actually holds more generally for amalgamated free products? groupprops.subwiki.org/wiki/…
I'm reading through the proof and it seems like the proof easily generalizes to the amalgamated setting.
@geo “The Riemann Zeta-Function and the One-Dimensional Weyl-Berry Conjecture for Fractal Drums”, Proceedings of the London Mathematical Society (3) 66, No.1 (1993), pp. 41-69, (with C.Pomerance). and “The Riemann Hypothesis and Inverse Spectral Problem for Fractal Strings”, Journal of the London Mathematical Society (2) 52, No. 1 (1995), pp. 15-35, (with H.Maier). might be of interest.
I've not looked at either paper in years, so I can't remember exactly which is which. There is also one on the Weyl-Berry conjecture which might be relevant.
(I don't have a citation for that one handy, but Google should help. I think that was was also co-authored with Helmut.)
> I'm having trouble with the full interpretation of (1) geometrically. It's unclear to me what the Mellin transform of the sum over that spectrum of the family of metrics means geometrically. Is there any nice geometrical interpretation of this? Does a geometric interpretation extend to the critical strip?
I believe that Michel would say that it measures the "geometric oscillation" of the set.
Very, very roughly speaking, this is related to the roughness, periodicity, and scaling of the boundary.
@冥王Hades hard drive failure? Then sue the hard drive manufacturer. Sounds like @leslietownes volunteers to write your legal complaint. Which jurisdiction should you file, Tokyo or USA?
@GratefulDisciple yeah, the header somehow ruined the platter completely. No idea how that happened. I was going to switch to SSDs completely but I was putting it off because I’m lazy
and now I payed the heavy price of losing all the data
@leslietownes it’s Seagate
I purchased the hard drive in the US, so probably US? I don’t know anything about legal matters unfortunately
I’m gonna try to see how much of that was stored in the cloud, most of these game services have their own cloud storage solutions for this reason, I doubt I’ll get all of it back though
@冥王Hades How did you know that the head ruined the platter? It's unlikely, more that the heads are damaged and the platters could probably be transferred to another drive of the same model. Get a quote from a data recovery company. At any rate, if the platters DO get damaged, the damage should be local, and there are still lots of sectors that can be saved. Yes, it's expensive, but you seem to be able to afford it.
@冥王Hades Just kidding about the lawsuit. They only provide replacement when the drive is in warranty.
@GratefulDisciple I just performed an open surgery on the drive, the platter is ruined. Now it’s just my hypothesis that the header did it, because I can’t see any other cause. I’m gonna take it to a large Seagate lab to see how much of the data can be recovered regardless
Several years ago, I was driving my daughter somewhere, and Teen Spirit was on the radio. I was trying to convince her that it was a great song. At the end of the song, the DJ came in with station identification, which ended with "...YOUR home all the best OLDIES."
The Question:
In what ways, if any at all, might it be possible to move away from and towards a point simultaneously?
Motivation:
I read in "Dialetheism and Its Applications", edited by Young et al., in the section written by Priest, that
[m]oving towards something and moving away from it (in ...
