@AkivaWeinberger $|z - 3 + 2i| \leq 4$, what's the range of |z|?
I'm getting stuck using triangle inequality I wrote$ | |z|−|3-2i| | \leq |z−(3-2i)|$ so $| |z| - \sqrt{13} | \leq |z−(3-2i)| \leq 4 $ or $-4 \leq |z| - \sqrt{13} \leq 4 $ which gives $\sqrt{13}-4 \leq |z| \leq \sqrt{13}+4$ but that lower bound is negative.. ? should I take the lower bound as 0 instead? When I drew a picture, I'm getting the lower limit for |z| to be $ 4 - \sqrt{13}$, what's going wrong here? I'm confused, why does triangle inequality give a different result?
@user76284 Do you know this theorem: if $D$ is a kernel-perfect orientation of $G$ and $f(x)=1+d_D^+(x)$ for all $x\in V(G)$, then $G$ is $f$-choosable
or rather; does an odd cycle $C_n$ have a kernel-perfect orientation? I would think that the cyclic orientation is kernel-perfect, but that can't be true, becuase $\chi(C_n)=3$
anyhow, any ideas why the cyclic orientation of an odd cycle isn't kernel-perfect? I see absolutely no difference with the cyclic orientation of an even cycle for the kernel-perfect-ness
I like that the Cyrillic, Latin, and Greek alphabets all have a capital H-looking glyph but have completely different ideas for what the lowercase version should look like
@AkivaWeinberger I once wrote a massively fuckheaded short story by google translating gibberish from English to Bengali and then back again, and used that for my literature project in school.
in any case, for $C_n$ ($n$ is odd), I was thinking; any isolated points have a kernel, so we don't have to consider those. And we only really need to consider the components of our induced subgraph. Now such a component is a path of lengt $\leq n$. and I would think that each path has a kernel?
@BalarkaSen I remember you said once you're an expert in contour integration. I have 2 methods by contour integration, but not sure if these options are the best. How about comparing what I have to what you have?
Whereas one can only pick $\lfloor n/2 \rfloor$ independent vertices in $C_n$ for $n$ odd, and each vertex has only one successor, and then counting finishes the proof.
@Balarka so I dunno any geometry and thus my use of these words is gonna be sketchy at best, but if this makes any sense, is it true that a Riemannian manifold such that for any two points, there's a unique minimal geodesic between them, that the manifold is simply connected?
@Daminark OK, if $M$ is not simply connected there's a nontrivial homotopy class. I think if $M$ is compact you can always choose a geodesic representative of that nontrivial homotopy class, and do the circle argument there.
So there is a closed geodesic $\gamma$ in $M$ which is not null-homotopic. Mark two points $p$ and $q$ on $\gamma$ such that the two arcs from $p$ to $q$ in $\gamma$ that we get have equal length - you can always do this
This gives two points $p$ and $q$ in $M$ with two distinct - even homotopically distinct - minimal geodesics joining them, contradicting hypothesis on $M$
So, you want to find the minimal k such that $C_n$ is k-choosable? It seems like it's impossible for it to be 1- or 2-choosable--consider the list colorings where every node is the same color, or the same two colors.
@BalarkaSen The idea is that there is always an action minimizer, and the concern is that it's constant; you solve that by being homotopically nontrivial
@ShaVuklia Hrm. My suspicion is that the choice number is always 3, but I don't know how to prove this. It seems like it's far easier to exhibit counterexamples than to demonstrate k-choosability.
I wonder what else can be said about compact Riemannian manifolds with any pair of points having a unique minimal geodesic between them. Not all simply connected manifolds appear as such; eg $S^2$ obviously doesn't
What are the other topological obstructions for such manifolds?
Even for C_3, there are lots of cases to check--where all lists are disjoint (000), where one pair of vertices share one color in the list but no other overlaps (100), and then (extending this notation) 110, 111, 200, 210, 211, 221, 222, 300, 311, 322, 333.
@FarhadRouhbakhsh Or simpler. For sin(pi cos(x))/x^2 you make the change of variable pi/2-x=y. And then you exploit the simple fact that lim_(x->0) sin(x)/x=1.
@BalarkaSen So the question is about how long the geodesics are, mostly
You want enough isometries (preferably for them to act transitively) so that you only need to check one point, but not too many (or it will force new geodesics to exist)
('Cuz compact negatively curved manifolds are covered by R^n (with some bad metric), so are aspherical. Simply connected just says they are topologically R^n)
@XanderHenderson By Cartan-Hadamard theorem, negatively curved 11th grade students students are all aspherical
How can one prove the statement
$$\lim_{x\to 0}\frac{\sin x}x=1$$
without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution.
This is homework. In my math class, we are about to prove that $\sin$ is continuous. We found out, that proving the above statemen...
The issue im having is... $\frac{ds}{d\theta}=\sqrt{r^{2}+(\frac{dr}{d\theta})^{2}}$ ... integrating both sides just gives me $s=f(\theta)$ where in fact im looking to get the summation of ds to give L and then I can form a Riemann sum
honestly Im trying to get my head around the derivation of the arc length of a polar curve
@Waiting Hey little man, i told you that i know limit siny/y is equal to 1. im just asking you after changing variables and simplifying limits where do you find siny/y
