Guys, not sure how readable this is, but we're proving the chain rule here. It's mostly symbols tho. My question is; at the end, they divide by $\Vert\vec k\Vert$. How do we know this isn't equal to zero?
A standard trick is to compare the harmonic series to the following series: $$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+\frac{1}{16} + \cdots$$
@SteamyRoot well if is a positive i wanna say it diverges if 0 undefined if -1 i think its 0 less than -1 it seems to converge to some negative value (maybe)
@Ted Okay, so $$ \left\vert\frac{(O(h)}{C\vert h\vert}\right\vert\leq \left\vert\frac{O(h)}{O(h)}\right\vert\to 0. $$ It's not written out neatly, but that's the main idea. I'll write it out now.
You should note that for $a = -1$, that sum becomes $-1 + 1 -1 + 1 - \cdots$, so this series doesn't convergence. For $a = 1$ you're just taking $1 + 1 + 1 + \cdots$ so that clearly diverges.
There is nothing wrong with Akiva Weinberger's post. However, since this post was made in the context of a somewhat more rigorous and formal setting, I'm posting a formal proof that I believe would most likely be considered acceptable in that setting.
Let $z$ and $y$ be extended integers such...
@Faust7 binary operations are covered in a tangential non prerequisite course. Groups were never discussed. Even rings were never defined. I just happened to know of them already (to the professors surprise)
granted, I always thought rings were referring to things like boolean values, set algebra, dog and cat algebra, etc.
i never realized it referred to a classification of number sets
@Ted This is my attempt at the proof; so we have $$ \Vert O(\Vert\vec h\Vert)\Vert\leq C\Vert\vec h\Vert. $$ So it follows that $$ \frac{o(O(\Vert\vec h\Vert))}{C\Vert\vec h\Vert}\leq\frac{o(O(\Vert\vec h\Vert))}{O(\Vert\vec h\Vert)}\to 0. $$ So we see that $o(O(\Vert\vec h\Vert))=o(C\Vert\vec h\Vert)=o(\Vert\vec h\Vert)$.
@BalarkaSen The whole point of the Hopf map is that taking inverse images of stuff is locally multiplication by $S^1$. My guess was [annoying mental pictures] $\implies$ taking inverse images of stuff in the suspension of the Hopf map is also locally multiplication by $S^1$, at least away from the poles where things are probably more annoying.
@Faust7 units and elements with norm 1 are one and the same
@Faust7 it is the entire set that makes the generators though, right? Because technically 1 is an element with norm 1 and it isn't a generator by itself... though if I add $\sqrt{3}$ then 1 and the square root of 3 serves as a generating set. Is one more preferred over the other?
@Sha: I think I would rather pick a function, say $f(h)$, which satisfies $f(h) = O(h)$. If I have another function, $g(h)$ that satisfies $g(h)=o(f(h))$, then I want to show that $g(h) = o(h)$. So for any $\epsilon$, $|g(h)|\le \epsilon|f(h)|\le \epsilon C|h|$ for $h$ small enough.
Then it follows that $|g(h)|\le \epsilon|h|$ for $|h|$ small enough.
@BalarkaSen Also the picture we had for the Hopf map (a torus with two linked circles on it) is, intrinsically, nothing more than two parallel circles on a torus.
that moment when you automate something ordinarily taking you hours a day and you're just sigh time to relax for a second... ok now to find more stuff to do.
(Balarka said he believes me that the preimage of a generic point is a torus. He cited some high-level theorems I don't know. I posted my reason for saying it's the torus a few comments above. @Ted)
@TedShifrin Here is my problem. It is kind of tricky. I might post a question on it actually. You have one triangle in space and several others that share a line segment with it, if you were to form a geodesic which triangle is the one to pick? I assume it has something to do with angles but I am not quite sure.
There is nothing wrong with Akiva Weinberger's post. However, since this post was made in the context of a somewhat more rigorous and formal setting, I'm posting a formal proof that I believe would most likely be considered acceptable in that setting.
Let $z$ and $y$ be extended integers such...
