One of my old family friends was a carpenter and had one of 'em. I had to figure it out and explain it to him. Plus I had to convince him that you actually got an ellipse by slicing a cone appropriately; he understood it as a slice of a cylinder.
@AkivaWeinberger That's related to the debate I often have with students about how the usual stretched parametrization has nothing to do with polar coordinates.
Take $\Bbb C^3$ with the euclidean norm. and $Y = \{ x_1=2x_2\}$. i have a functional on $Y$ defined by $\phi(x) = x_1-x_2$. i want to extend it to the whole space and preserve its norm , any suggestions how to do that?
@ÍgjøgnumMeg I was just thinking about it because someone told me that math and music theory have a strong correlation. I was playing the drums last night and made some weird offbeats, but then I started wondering what any of that could have to do with math.
Music theory is more descriptive than prescriptive. I mean, if you want to play a billion notes at once and have it sound good, music theory helps tell you what notes to play, 'cause it's very hard to intuit that otherwise. But usually it's just trying to describe what people are already doing.
@Akiva no that's pretty accurate, music theory will tell you what kind of sound certain chords/scales/modes will give you but if you just smash stuff together it will sound robotic
@ÍgjøgnumMeg Yeah, but even so, there's an element of, "Is this really the best way to describe what this bit of the song is doing? Or does it work this way? Or maybe you can hear it either way"
Check out the YouTube channels I mentioned (especially 8-Bit) though
he talks about some chord in a song for about 10 minutes in one of his Q&As because "theoretically" it shouldn't sound good where it sits but for some reason it makes sense contextually
Well when you learn about Fodor's lemma from Jech remember that the exercise is to show that it is actually an iff so it gives a characterization of stationary sets @Leaky
There's a lot of stuff I don't know because I've mostly just been improvising from the beginning. I feel like it's been a bit limiting since I've gotten more interested in composition.
Properly written the converse of Fodor's lemma reads "let $\lambda$ be an uncountable regular cardinal and $S\subseteq \lambda$ such that every regressive function $S\to\lambda$ is constant on a stationary set, then $S$ is stationary"
in Lean a multiplicative group and an additive group are different things...
if you want to talk about a homomorphism from an additive group to a multiplicative group you would need to define it yourself because nobody has found a use for it yet
So I found out that integration can be developed without a single mention of derivatives. Why is my class teaching integrals as if they are actual antiderivatives?
I don't know what that really means. My lessons are being presented as if integrals = antiderivatives. Someone told me that isn't the right way to think about it.
Completely unhelpful to Mary Stars question probably, but how do we count $\Bbb Z^3$ solutions of that diophantine equation if we do have algebraic geometry tools?
We consider $\text{mSpec}(\Bbb{Z}[x,y,z]/(x^2+3y^2-z^2))$ or something?
@MaryStar dividing by $z^2$, this is equivalent to finding rational solutions to $a^2+3b^2=1$, consider the extension $\Bbb{Q}(\sqrt{-3})/\Bbb{Q}$, note that the norm of $a+b\sqrt{-3}$ is $a^2+3b^2=1$, so we're actually solving the norm equation $N_{\Bbb{Q}(\sqrt{-3})/\Bbb{Q}}(a+b\sqrt{-3})=1$. By Hilbert 90, any solution may be written as $a+b\sqrt{-3}=\frac{c+\sqrt{-3}d}{c-\sqrt{-3}d}$ where we can assume that $c,d \in \Bbb Z$ and $\gcd{c,d}=1$. One gets $\frac{c+\sqrt{-3}d}{c-\sqrt{-3}d}=\frac{c^2-3d^2+2cd\sqrt{-3}}{c^2+3d^2}$
if you want actually apply geometry to find a solutions to that problem, note that solutions to $x^2+3y^2=1$ form a conic and there's a group law on the set of rational points on a conic similar to the one on an elliptic curve.
well, i did it dividing limit into $$ \lim_{x \to 2^-} \left ( \frac{2^x - 4}{x-2} + 3x + 6 \right )$$ and noticing that fraction in limit is just $f'(2)$ where $f(x) = 2^x$, so it's $4\log 2$
@MikeMiller it is sufficient to show that the variety is smooth at the generic point (by upper semicontinuity of the rank, this gets smoothness in an open neighborhood of the generic point), this can be checked via looking at the module of differential $\Omega_{k(X)/k}$. So the algebra this boils down to is that $\Omega_{k(X)/k}$ is a $k$ vector space of the dimension equal to the transcendence degree of $k(X)$
this works for $k$ perfect, I think the statement might fail otherwise
the idea is that if you have $x_1, \dots, x_n \in k(X)$ that are algebraically indenpendent and $k(X) / k(x_1,\dots,x_n)$ is separable, then $dx_1,\dots, dx_n$ form a basis of $\Omega_{k(X)/k}$
getting such a separating transcendence basis in characteristic $p$ is not that easy
For a ring requiring that $(R, \cdot)$ is a monoid gives us a non-commutative unital ring; and requiring $(R, \cdot)$ is a commutative monoid gives us a commutative unital ring. Does this sound right? Because I'd think requiring that it's a monoid would only give us a unital ring, commutative or non-commutative.
