There's something called the Natural density for natural numbers which measures how dense a subset of the natural numbers is. Is there anything like that for a subset of the algebraic numbers?
Say we start at point $a$ on the real number line, move right and left haphazardly for a while, and end at point $b$. The net distance, a state function, is the distance between $a$ and $b$. The total distance, a path function, is the number on our pedometer. For the net distance, the exact differential $dx$ is a tiny move along the $x$-axis, the entity that shows up from the very beginning of Calculus.
For the total distance, however, I don't understand what the inexact differential $đx$ actually represents. Does it have the same geometric interpretation as $dx$?
In cohomology you often have things like $A\overset d\to B\overset d\to C$ where $A$ and $B$ are vector spaces (or abelian groups or modules) and $d\circ d=0$
Recall that $\omega + 1$ has $0,1,2,3,4,5,...,\omega$
starting from any finite number and add any finite number a finite number of times, the result is always finite, meaning that the distance between $\omega$ and any finite starting point $\alpha$ does not reduce
(will need to figure out how to define distance between ordinals, because of the fact that they do not commute under addition)
Let $A$ be a quaternion algebra over a field $K$ of characteristic zero. By the famous Skolem-Noether theorem, every $K$-algebra automorphism $\varphi \colon A \to A$ is of the form
$$
\varphi(x) = gxg^{-1},
$$
for some invertible element $g \in A^\times$.
Question: Does it follow that also eve...
@Secret You could think of them as surreals, where addition and multiplication are commutative and subtraction and nonzero division are defined
(There's also the "natural sum" $\alpha\oplus\beta$ and the "natural product" $\alpha\otimes\beta$, which is the same as the restriction of surreal addition and multiplication to the ordinals)
yeah, natural sum can capture that, though the surreal $\omega -1 $is not the same as $\omega \setminus \{1\}$, it is not even of the same order type if I recall
(The distance between two points is defined to be the infimum of the arclengths of all paths between the two points, and the diameter is defined to be the maximum distance between two points)
(In essence it's because Riemannian manifolds are defined intrinsically without reference to an ambient space)
This can be fixed easily by having $c+\omega = \omega$
Meanwhile $\omega +c$ can stay. This is consistent with $\omega + \alpha$ remains unchanged for every $\alpha < \omega$
Thus with this new system, it is now possible to have a new pathway, besides taking one step at a time from a finite number towards $\omega$ (which means one can never reach $\omega$ in any finite number of steps), or to take $\omega$ steps at once and thus going from $n$ to $\omega$ to $\omega 2$
It is now possible to take e.g. $5$ steps to reach from $n$ to $\omega$, by choosing a suitable element $c$ that is between finite numbers
However, I am still mapping the partial order (I suspect there are actually incomparable elements, but I so far yet to found any) of this system. The problem of allowing these strange "finite" elements $c$, is that the ordering becomes more and more strange the higher up the hierarchy I go
(and as you might have noticed, because $a < b \implies ca < cb$ fails in general, associativity of addition is actually broken. But then, this is not a big deal, because in the usual ordinal arithmetic, $\omega+ \omega$ is really $(1+1+\cdots)+(1+1+\cdots)$ and hence there is no associativity here, thus what we done here is basically defining something that behaves like "$\frac{\omega}{5}$"
Now, this system is still impredictive (cannot be defined in terms of encoding of natural numbers), because one still cannot reach $\omega$ from any combination of term involving $c,2c,3c,4c$. To even hop from one $c$ to another level of $c$, $c$ steps need to be taken
I don't know if there is even a predicative way to define infinity, or anything that can reach infinity in finite steps
I've read that the Bott Periodicity Theorem is one of the most celebrated theorems in all of (algebraic?) topology. So correct me if I'm wrong but Bott's Periodicity Theorem shows that the homotopy groups of the classical groups are all periodic, and it gives rise to a homomorphism $J : \pi_i(O(n)) \to \pi_{n+i}(S^n)$ and this tells us some things about the homotopy groups of spheres, but which part of this make it into such a celebrated result?
Because for example framed cobordism tells us more about the homotopy groups of spheres (I think) because we get an explicit isomorphism
Plus even though we get the $J$ homomorphism from the Bott Periodicity Theorem, does it help us actually calculate any homotopy groups of spheres? Spectral sequences seem a more pwerful tool to calculate the homotopy groups of spheres in some cases
I do not know enough algebraic topology to know what I don't know about these specific topics, but if someone could point out what it is that I'm missing that would be very helpful
if f_n are non-increasing real-valued functions and we know that their sum converges to a finite real number everywhere (denote the sum by f), then is $\lim_{x\uparrow a}f(x)=\lim_{n\to\infty}\lim_{x\uparrow a}f_n(x)$?
that uparrow means left-side limit
and a is an arbitrary real number
sorry there was a mistake, I'll try again, the question is: is $\lim_{x\uparrow a}f(x)=\sum_{n=0}^\infty\lim_{x\uparrow a}f_n(x)$?
which is the same as $\lim_{x\uparrow a}\sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty\lim_{x\uparrow a}f_n(x)$, so can we exchange the order of the limit and the sum?
Is it possible to find natural number solutions for something like $3a^2+2b^2+3c^2+2d^2+4f^2+2g^2+3h^2+2k^2+3l^2 = 8m$, where all the variables are different from each other?
I just found a way to generate pythagorean quadruples.
A pythagorean quadruple is an equation in the form $a^2+b^2+c^2=d^2$. You can say that $d=m^2+n^2+o^2$, $c=m^2-n^2+o^2$, $b=m^2+n^2-o^2$, and $a=m^2-n^2-o^2$, where m, n, and o are different natural numbers, in order to generate values for a, b, c, and d that work.
Let's say I have a group $G$ acting on a set $X$ in two different ways, but both produce the same orbits for all $x \in X$ are the actions equivalent in some way then?
@Perturb: My suggestion is that this arises only by relabeling elements of your set $X$ in such a way as to preserve the orbits. So you choose any bijection of $X$ that maps orbits to orbits and one group action is the composition of the other with this bijection.
friggin woke up at 6:15 telling myself I'd just get through it on saturday... just hit the snooze like 12 times and woke up two hours later still tired.
@TedShifrin Checks window to see if Ted is stalking me
@TedShifrin Sorry for the long wait, so consider $C_3$ as $\langle (123) \rangle$ which is a subgroup of $S_4$ acting on $X = \{1, 2, 3, 4\}$. This produces orbits $\Omega_1 = \{1, 2, 3\}$ and $\Omega_2 = \{4\}$ and if we consider $C_3$ as $\langle (132) \rangle$ we get the same orbits. Perhaps this is completely wrong in any case since they both are the natural action of $S_4$ on $X$
yeah, there were quite a few fliers last time I was there.
but at least i have energy this morning. Just going to get more coffee and work on my programming projects and I'm kind of interested in investigating non-standard analysis.
a fun question for the chat: n people wear a different hat each; I collect the hats, shuffle them, and then distribute the hats back so that each person gets a hat; what is the expected number of people getting their original hats?
I wonder if there exists an example of a sequence of non-decreasing functions $f_n:\mathbb{R}\to\mathbb{R}$ so that $f(x)=\sum_{n=0}^\infty f_n(x)$ is defined and finite for all $x$ and a point $a\in\mathbb{R}$ for which $\lim{x\uparrow a}f(x)\neq\sum_{n\to\infty}\lim_{x\uparrow a}f_n(x)$.
You can think of each hat permutation as an element of $S_n$, and then count the number of elements with 1 fixed point, then number of 2 fixed points, ect.
First things I note is that $f$ is non-decreasing too so that $\lim{x\uparrow a}f(x)$ exists (and $\leq f(x)$) but could it be the case that the limit $\sum_{n\to\infty}\lim_{x\uparrow a}f_n(x)$ does not even exist at all?
Well, a matrix is said to be PSD if it has no negative eigenvalues. (there's other equivalent definitions but that's the simplest one)
You can show, without too much trouble, that if M is PSD then the matrix elements satisfy the following inequality: $$|xy-zw|\leq \sqrt{1-x^2}\sqrt{1-y^2}+\sqrt{1-z^2}\sqrt{1-w^2}$$
So that's something I know how to do.
What I don't know how to prove is the following: Suppose $x,y,z,w$ satisfy that inequality. Show that there exists $u,v$ such that $M$ is PSD.
@TedShifrin Hi. I just got offered a job as an IT-consultant. I will have a phone meeting with their HR department on Wednesday and arrange stuff like when I start and wages and all that stuff.
@TedShifrin I remember after taking an intro linear algebra class I didn't appreciate (well, I did but not that much) it, but then literally every other math class was like: Lol, so here we're going to use linear algebra in some way you probably would not have expected.
@Spencer1O1: You're not ready for most of them. You need to learn single-variable calculus before you do the (harder) multivariable stuff. But most of the linear algebra you can have fun with.
@Semiclassical Also, it feels artificial af, which is part of the issue I had with it originally... but you slowly realize it's actually super friggin fundamental to so many things.
@ÉricoMeloSilva it is, but linear algebra is particularly stark insofar as it can show up in high school, lower-division college, upper-division college, graduate school...
@Spencer1O1 If you're gonna brute force it, you're gonna have a bad time. That problem has been extensively studied. Norvig has ran an exhaustive search for the answer utilizing optimized distributed programs...
@PaulPlummer That'd be cool for sure! There was a famous problem in group theory that turned out to be independent of ZFC but I can't remember its name
Let $A$ be a quaternion algebra over a field $K$ of characteristic zero. By the famous Skolem-Noether theorem, every $K$-algebra automorphism $\varphi \colon A \to A$ is of the form
$$
\varphi(x) = gxg^{-1},
$$
for some invertible element $g \in A^\times$.
Question: Does it follow that also eve...
