I'm basically in a research position right now, but I have a bit of additional job security since I'm technically in the IT department, so if the research doesn't turn out as well, I can shift to more IT projects.
If I can keep up my research output in the long term, I'd like to, since I really like doing this work. However, I'm not a faculty member focusing on research, and I don't call all of the shots when it comes to how our research projects are run.
It's like I'm working as a research assistant, but I can run my own projects periodically, and I'm mainly there to provide data and stats support. I actually am about to submit my first article as first author. Fingers crossed. (It's in something completely unrelated to math/stats.)
My focus is primarily in medical education. The primary aim is to link educational initiatives to educational outcomes, and eventually workforce and patient outcomes.
I honestly love the work, because it uses a lot of my knowledge from my previous education jobs (in K-12 and higher ed institutional research), and it's helped me get an impression of what biostatistics work looks like.
Plus, I really appreciate how the field I'm focusing on is a natural intersection of things that really matter to people.
If $\Bbb R^2$ is embedded on each face of a unit cube, and $\Bbb R^2$ is the orthographic projection of $M,$ bounded by the cube, then is $M=\Bbb R^3?$
@TedShifrin How would you explain the intuition behind choosing the length of the neighborhood $\delta$ so as to show that $$\bigcup_{0 < |h| < 1/n}\left\{x \in \mathbb{R}:\dfrac{f(x+h)-f(x)}{h} > c + \dfrac{1}{k}\right\}$$ is an open set?
(assume $n, k \in \mathbb{N}$)
Oh, and $f$ is some continuous function $\mathbb{R} \to \mathbb{R}$
@TedShifrin So, I am trying to show that $(x - \delta, x + \delta)$ is contained in there so as to show that it's an open set, but I'm having difficulty trying to choose an appropriate $\delta$. $c$ is just a fixed real number.
@TedShifrin So the preimage of an open set is open under a continuous function. If I fix $h$, this would mean that.. I'll have to think about this some more.
@TedShifrin Right, so the preimage... oh. Is it just the fact that $(c + 1/k, \infty)$ is an open set, and because $g$ is continuous, its preimage and hence one of the sets in that union is thus open?
So yet another thing I'm going to have to learn on the fly... $\lim\sup$ and $\lim\inf$ for functions. Why didn't my university use Rudin? I think they're still not using Rudin.
Find some work at the local college doing a tutorial and tutoring math, was surprised they were willing to hire me; considering i dont actually have a degree.
@Clarinet: Most schools that teach Abbott aren't going to try to use Rudin. But the very top, hardest schools in the US still use Rudin (but often offer other sections that use easier texts — e.g., MIT, Berkeley ...).
We have all sorts of Europeans, Indians galore, but you're the first person I recognize ... of course, when people use names like userxxyyyzzz002385 it's hard to know.
I am working on building G^2 smooth spline interpolation algorithms using clothoid segments on the plane and spherical clothoid segments on the sphere and R^3
@TedShifrin I must admit this stuff is confusing for me so I try not to think about it. I think the right pairing is $H^1(M, \mathscr{O}) \times H^0(M, \Omega^1) \to H^1(M, \Omega)$, given by $(\{f_{ij}\}, \omega) \to \{f_{ij} \omega\}$, and then composing with the "trace map" $H^1(M, \Omega) \to \Bbb C$.
This is a nondegenerate pairing, part of the statement of Serre duality, I believe. So $H^1(M, \mathscr{O}) \cong \Omega^1(M)^*$ for Riemann surfaces.
@TedShifrin One question for you: so $$\bigcup_{0 < |h| < \frac{1}{n}}\left\{x \in \mathbb{R}: \dfrac{f(x+h)-f(x)}{h} > c + \dfrac{1}{k} \right\}$$ is open.
Let's consider the upper derivative $\lim\sup_{h \to 0}\dfrac{f(x+h)-f(x)}{h} = f_u^{\prime}(x)$. I'm trying to figure out how to turn that open set into $\{x \in \mathbb{R}: f^{\prime}_u(x) > c\}$. I suspect that what it entails is throwing $\cap_{h=1}^{\infty}$ on that set, and then taking the $\sup$. But I'm not 100% sure with this.
@TedShifrin there is another advantage over polynomial splines, the intrinsic equations are "free" and that makes velocity planning over a clothoid spline simpler (seems paradoxical)
Here's a curvature/plane curves question for you that I made up years ago for my course, @Alexandru. If I tell you the path of the rear wheel of a bicycle, find the (possible) path(s) of the front wheel.
You get a differential equation that almost never can be solved by hand.
$g+1$ is the Lusternik-Schnirelmann category of the surface of genus $g$. For the torus, throw out a meridian and a longitude, and you get one open set $U$. Do it for another essentially distinct pair of meridian and longitude, you get a second open set $V$. $U$ and $V$ cover all of the torus except two points, so you take another chart joining the two.
Well, that problem I gave you is a nice elementary differential geometry problem to set up, but unless there's a clever approach different from mine, it's hard to solve except in obvious cases (line, circle).
@TedShifrin I will need help soon coming up with some asymptotic expansions for some 2d hypergeometric function special forms, that's the question I need to write soon
So it's not hard to track through. If I have a $\bar\partial$-closed $1$-form $\phi$, on my $U_i$ I can write it as $\bar\partial g_i$. I look at the $1$-cocycle $f_{ij} = g_i-g_j$ and it is actually holomorphic. That's the isomorphism.
So we need to do this with your cover and my $\phi=d\bar z$.
Is there any succinct way to say, in a directed graph, "all vertices of the same level at most 2N edges apart" and all vertices of the same level at exactly 2N edges apart" succinctly? I'm trying to think of good function names for siblings/cousins/siblings+cousins when working with a directed graph (tree/dag)
@Lelouch You know from my computation that it's a $\Bbb Z^2$ sitting in $\Bbb C$
In fact $\text{Pic}_0(M) \cong M$, so it's the same lattice that the torus $M$ is a quotient of $\Bbb C$ by
@Lelouch $H^1(M, \Omega)$ is smooth $2$-forms on $M$ modulo $d$ of $(1, 0)$-forms on $M$. So take a smooth $2$-form representative $\omega$, and consider $\iint_M \omega$
If $\Bbb R^2$ is immersed on each face of a unit cube, and $\Bbb R^2$ is the orthographic projection of $M,$ bounded by the cube, then is $M=\Bbb R^3?$ The orthographic projection of $M$ from a face to its opposite is $\Bbb R^2.$
**Theorem**. If $ \{K_\alpha\} $ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\} $ is nonempty, then $\cap K_\alpha$ is nonempty.
*Proof*: Fix a member $K_1$ of $\{K_\alpha\}$ and put $G_\alpha=K_\alpha^c$. **Assume that no point of $K_1$ belongs to every $K_\alpha$.** Then the sets $G_\alpha$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices $\alpha_1,…,\alpha_n$ such that $K_1\subset \cup_{i=1...n} G_{\alpha_i}$. But this means that $K_1\cap K_{\alpha_1}\cap\dots \ca…
wierd what did it contradict
It didn't contradict anything
it said no point of $K_1$ belongs to intersection of $K_\alpha$
B="the intersection of $\{K_{\alpha}\}$ is empty". This is assumed in the first line of the proof ("Assume..."). Then, from this, the proof derives not A, where A="Every finite subcollection of $\{K_{\alpha}\}$ has non-empty intersection". This is the last line ("But this means..").
I'm trying to show that $\Bbb{Z}_m$ is an injective $\Bbb{Z}_m$-module using Baer's criterion. Every ideal in $\Bbb{Z}_m$ is of the form $\overline{d} \Bbb{Z}_m$, where $d$ divides $m$. Let $f : \overline{d} \Bbb{Z}_m \to \Bbb{Z}_m$ be some morphism. I believe I want to find $\varphi : \Bbb{Z}_m \to \Bbb{Z}_m$ such that $\varphi \iota = f$, where $\iota : \overline{d} \Bbb{Z}_m \to \Bbb{Z}_m$ is the inclusion map.
I was able to show that $Im ~f \subseteq \overline{d} \Bbb{Z}_m$.
Initially, I thought $\varphi = id$ might work, but $f$ could move elements in the ideal around.
if $C \subset \mathbb{C}^2$ is an affine plane curve defined as the zero set of a non-constant polynomial in $\mathbb{C}[x,y]$ and $\overline{C}$ is its projectivisation, defined by the homogenization of the polynomial defining $C$ (but not dependent on that polynomial, by the nullstellensatz), is there an accessible way of showing that the projectivisation $\overline{C}$ is in fact the closure of $C$ ?
by closure of $C$ of course we mean as a curve embedded in $\mathbb{P}^2_{\mathbb{C}}$ via the open embedding $\phi_2 : \mathbb{C}^2 \rightarrow \mathbb{P}^2__{\mathbb{C}}$ say, where $\phi_2(x,y) = [x,y,1]$? Without too heavy knowledge of multiple complex variables? The standard proof is apparently via the Weierstrass preparation theorem which is a bit beyond me as i'm pretty rusty with complex analysis beyond the fundamentals
@Thorgott wait a seecccc... You are hypothesizeing A is true but it might be false. You assume B and see B implies not A then conclude B is false? We don't know about A so you can't conclude B is false
@Stupidquestioninc it contradicted finite intersection property, since a finite intersection of that subcollection has nonempty intersection, but you just showed that a finite intersection was empty
@Stupidquestioninc its actually a characterization of compactness of any topological space, $X$, to say that any collection of closed subsets with finite intersection property (such that any finite subcollection has nonempty intersection) has nonempty intersection
@Stupidquestioninc this characterization was famously used to show tychonoffs theorem, for instance, which shows that the arbitrary product of compact spaces is compact
@porridgemathematics I'm surprised that this is hard. It's clear that the closure of $C$ is contained in the projectivization of $C$. The projectivization is obtained from adding points at infinity $\Bbb{CP}^1_\infty \subset \Bbb{CP}^2$ of $C$, so removing any of those from the projectivization of $C$ will make the resulting object non-closed, no?
It seems almost tautological. If you write down the formula for projectivization it should be direct that the added points are limit points of $C$ in $\Bbb{CP}^2$
It's essentially just the definition of compactness dualized by taking complements
"every open cover has a finite subcover" =take complements=> "every collection of closed subsets with empty intersection has a finite subcollection with empty intersection"
and taking complements again reverses the implication
granted, the theorem as stated in Rudin is the contraposition of that latter statement
You can also observe that this must hold by phrasing everything in terms of filter, since families with the fip are exactly those that can be extended to a filter
Good afternoon people, sorry to interrupt. Can someone help me with this question?
I always thought **cylindrical waves** were **3D** waves, but recently I came to know they share the same properties with circular waves (which are **2D**) because they are basically the same thing, only the cylindrical wave is “trivially extended” into the **third dimension**.
Is it correct to say that? What topics of math cover this concept?
just trying to figure out how to make the "2-category of all (in a certain sense smaller) categories" precise with the help of two nested Grothendieck universes
I did some "topology" recently, if classifying the families of topological manifolds that admit a product in the category of topological manifolds counts
@Alessandro I thought so, but it's apparently not. Say you have two universes $ U\in V$. If $\mathcal{C},\mathcal{D}$ are $U$-categories, the functor category $\mathcal{D}^{\mathcal{C}}$ is a $V$-category. But if you do it in terms of classes, then the functor category cannot be constructed unless $\mathcal{C}$ is small. Or so I heard.
But that when you have a universes, and want to work in it, then instead of thinking about all the fuss with classes defined by this and that formula, not being actual elements of the whole universe etc
You can just consider sets to be the elements of the grothendieck universe, and for all practical purposes classes are subsets of that grothendieck uni🅱erse
ah yeah, this is what is meant by saying that a universe gives a "model" of NBG, yes? (tho I don't know what a model is, but I don't wanna know either)
