Let $\mathcal{M}, \mathcal{T}$ be subsets of a topological space. $\mathcal{M}$ approximates $\mathcal{T}$ iff $\mathcal{T} \subseteq \overline{\mathcal{M}}$.
Let $\sigma$ be ReLU. Let $\mathcal{K} \subseteq \mathbb{R}^m$ be compact. The universal approximation theorem (UAT) says
$$\{ g \circ \si...
I have shown that a function is continuously partially differentiable.
If it holds that $$\frac{\partial^2{f}}{\partial{y}\partial{x}}=\frac{\partial^2{f}}{\partial{x}\partial{y}}$$ forall $(x,y)\in \mathbb{R}^2$ does it follow that the function is twice continuously partially differentiable?
it means it's twice differentiable but those things aren't continuous.
they're still equal everywhere, but they're both discontinuous at 0. i haven't written out the formulas but it seems likely to me that that's what happens.
it's kind of a two-variableization of one of the common examples used to show that derivatives don't have to be continuous.
Ok! And if we can find a point where the equality $\frac{\partial^2{f}}{\partial{x}\partial{y}}(x,y)=\frac{\partial^2{f}}{\partial{y}\partial{x}}(x,y)$ doesn't hold, then does it follow that $f$ is NOT twice partially differentiable?
That link was for my initial question(that the mixedpartial derivatives are equal), or not? If we can find a point where the mixed partial derivatives are not equal, what is then? @TedShifrin
So if the mixed partial derivatives are not equal everywhere, then f is not twice continuously partially differenatiable, right?
CC+transfer is a viable route, most of my friends in late college were transfer students. but it's so much more work. if it isn't financial suicide, far better to take it easy.
The first derivative test for Maxima minima of Differential function States that : at a stationary point across with derivative changes sign from negative to positive moving from left to right is point of minimum and a stationary point across which derivative change is sign from positive to negative moving from right to left is point of Maxima
they make it out as though there are a lot of tiny, smaller things you can plug in and not be just 1 person in a 400 person class. but you will be 1 person in 400 person class.
i was typing this as ted used the same figure. AGAIN.
not for your whole experience but the first year and perhaps even second depending on intended major.
profs you forget about tend not to be good, or really bad.
they're just in the middle
ted mostly it was just real analysis had this reputation of being a 'killer' subject and the required one semester of it was enough for just about everybody.
the second semester wasn't required for anything, it was just more analysis.
so, horrific teachers in the first semester, basically :)
she liked asking gotcha questions where she was clearly expecting someone to give the wrong answer and when they got the right one it visibly annoyed her and she'd switch tacks in some way to make the answer sound nonresponsive. she would have been good at keeping certain people out of universities in the USSR.
rover, never a good sign. but hopefully it was just a sign error or something.
looking back on it, i sorta get it. you're beginning a postdoc and they ask you to teach something that has a reputation for being a 'killer,' and it's not your subject. but that's only an excuse to teach poorly, not to also be captious.
it could be she just had no experience in front of a classroom, or was just phenomenally bad at teaching. i asked about some of the harder sequences and series problems that she didn't assign and was completely blown off.
my wife had a rocky adjustment. got a tenure track job with essentially no teaching experience. the students noticed less than they might have had the subject not been more of a free-for-all than math tends to be
my twisted syntax betraying a desire to phrase everything as indirectly as possible
they do get long. exact expression for an antiderivative of 1/(1 + x^n) is not fun for large n
nothing's wrong, that's just what it is. if you're only working on (-1,1) you can use a series expansion, 1/(1+x^n) has the form 1/(1 - u) with u = -x^n. geometric series formula
2
if i'm wrong, please pipe up, anybody.
i used to give that as an example of how you can write down very simple formulas where the exact, non-series expressions become very hard to come up with. i don't think there's a nice recursion or anything like that. as there sometimes is in the exercises you see in calculus books
i gave that example in office hours, of course. i wouldn't dream of taking that in a classroom setting
if i were writing a textbook i wouldn't try to give the impression that you can integrate rational functions by hand. in theory, yes, and with a good computer, maybe. but anything that involves factoring polynomials is not really suitable for work by hand
except in the very low degree things you see in calculus books
here's one from the vault. the integral of heaviside(sin(x)-(pi/2)*cos(2*x)+1/8) from 0 to 2pi. it's about 3, and this is also clear from the graph of the function without the heaviside( ) around it. but if you're not careful how you input it, a lot of CAS's will mess it up, giving you nonsense like 0 or stuff near 5. sometimes depending on how many digits you tell it to work to.
i never figured out why but i think with some ways of asking a CAS to do a definite integral, it first tries to find a symbolic antiderivative and gets stuck on that.
those parameters aren't magic numbers, they're just the ones i had when i discovered this in something i was trying to do.
even fairly dumb numerical packages get it right or close to right.
i think it has something to do with symbolic integration
@TedShifrin what I was thinking of was one could collapse the part of the one skeleton consisting of the three edges emanating from the apex of the three simplex towards the vertices of the base two simplex, and then the resulting space would be the one skeleton of the two simplex with vertices identified, which would be homotopy equivalent to $S^1 \lor S^1 \lor S^1$ if i'm not mistaken
@epsilon-emperor You need to be familiar with the basic notions of general topology to study algebraic topology, but it's not an extensive background requirement. I haven't read Kosniowski's book, but looking at the table of contents, it introduces all the general topology one would need to know before starting with the algebraic topology, so it's probably fine to read.
i think a more important prereq than point set topology is being comfortable with things like squares/regular n-gons/'most' disk like shapes are topologically all disks
does anyone have any ideas about this? What is $$\mathbb{RP}^{3} = S^3 / \sim$$ where $$\sim$$ identifies antipodal points a covering space of, besides itself of course
hmm, maybe it covers $S^3 / \sim$ where $\sim$ is the equivalence relation generated by identifying $(x_1,x_2,x_3,x_4)$ with all points $(+/- x_1,+/- x_2, +/- x_3, +/- x_4)$
there are certainly other primes that that gcd can be a multiple of (or in fact be). gcd(2^4 - 1, 3^4 - 1) = 5, gcd(2^3 - 1, 3^6 - 1) = 7.
with n and m between 1 and 500, these are the prime gcds you can get: [5,7,11,13,23,31,43,47,59,71,83,103,107,127,151,157,167,179,191,223,227,229,233,239,263,283,311,347,359,367,383,431,439,443,467,479,491,503,601,643,647,683,719,733,743,839,863,887,911,971,983,1021,1093,1399,1471,1609,2281,2441,4057,4513,5113,6553,6563,7753,11113]
i'll stop spamming the chat. i would focus on 2 or 3 and maybe just the question of what that gcd can be. no changing bases of the exponentials.
this seems intuitively obvious but im not sure how to show it: Let $U \subset \mathbb{R}^2$ be an open set not containing $(0,0)$, and suppose $U$ is simply connected. Show that there is at least one ray $\{(r, \theta) : r \geq 0, \theta = \theta_{0} \}$ for some fixed $\theta_{0}$ that does not intersect $U$
yeah ok, that is an easy counterexample.. i actually asked this because I read that on any simply connected open subset of $\mathbb{C}$ we can define a branch of the logarithm that is holomorphic on the open set
but it seems like the spiral provides a counterexample
if the spiral hits the negative axis in two nearby places, the fact that it's a spiral means you don't need to worry about the values of the log being related to one another the way you'd expect if you could go on a straight line from one intersection point to the other
since you know covering theory, the exponential map is a holomorphic covering map $\exp\colon\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$ and picking a branch of the logarithm for a holomorphic function $f\colon U\rightarrow\mathbb{C}\setminus\{0\}$ is equivalent to lifting $f$ through $\exp$, which is possible iff $\pi_1(U)$ is trivial in $\pi_1(\mathbb{C}\setminus\{0\})$ i.e. iff $U$ doesn't contain non-trivial loops about $0$
im guessing we would be integrating $\frac{1}{z}$ along contours from some fixed based points in the spiral/simply connected open set to whatever point we are taking log at?
and the choice of contour doesn't matter as you say because the open set is simply connected
yeah, our perspectives are somewhat dual. I'm making a point about the nature of $\pi_1(U)$ in $\pi_1(\mathbb{C}\setminus\{0\})$ and Ted is making a point about $H_{dR}^1(U)$ in $H_{dr}^1(\mathbb{C}\setminus\{0\})$ (and these are related by the canonical maps $\pi_1(U)\rightarrow H_1(U)\rightarrow H_1(U;\mathbb{R})$ and the de Rham pairing)
yeah, I'm just giving the more "modern" perspective
@porridgemathematics I think most standard complex analysis texts are somewhat outdated in these matters, but a great one that does speak in this perspective is Narasimhan and Nievergelts Complex Analysis in One Variable
@TedShifrin Here are the questions that were asked, all discussions removed. 1. Are $S^2$ with $n$ points removed and $T^2$ with $m$ points removed ever homeomorphic? 2. What's a simple example of a manifold with $H_1 = 0$ but $\pi_1 \neq 0$? 3. Equip $D^2 \setminus \{0, 1/2\}$ with the "Poincare metric", what's the isometry group? 4. Prove that a finite subset of $M_n(\Bbb R)$ which is closed under multiplication has a matrix with integral trace.
Thanks!! I am happy to have done well. They seemed reluctant to bother me too much but also I think mine was the hardest interview of them all, just because they know me.
There are a few arguments. My argument was, first do it for diffeomorphism. For that, look at meridian and longitude and their intersection number, which is 1 in the torus.
@porridgemathematics For any manifold $M$, consider the space of proper rays $[0, \infty) \to M$ upto unbased proper homotopy. This gives a number (number of equivalence classes of such), which is a homeomorphism invariant of $M$. If $M$ is sphere with $n$ punctures, this number is $n$ (rays going into any of the punctures cannot be properly homotoped to a ray which goes into another puncture).
Similarly, for $T^2$ minus $m$ points, this number is $m$
So $n = m$
But now compute fundamental group.
I could have written a textbook about 1 and 2 but they seemed to be aware and moved on very quickly
naively i was thinking, we can draw a loop that avoids removed points of a torus, so that the complement of this slice is still path connected, but it does not seem like this can ever work on the sphere with some number of points removed
My topology/geometry oral qualifying at Berkeley they asked all sorts of things not on the syllabus because they sorta knew me. Kirby asked Thom-Pontryagin, Wu asked by a manifold with connection had to be paracompact. Kirby led me through thecThom-Pontryagin, which at that point I'd never seen. That was fun.
if I remove n points from a compact manifold, then any compact subspace of the resulting space is contained in a compact subspace whose complement has n connected components (by removing some small neighborhoods of the punctures too)
I have shown that a function is continuously partially differentiable.
If it holds that $$\frac{\partial^2{f}}{\partial{y}\partial{x}}=\frac{\partial^2{f}}{\partial{x}\partial{y}}$$ forall $(x,y)\in \mathbb{R}^2$ does it follow that the function is twice continuously partially differentiable?
