Actually, I don't prove Liouville itself, but a stronger result: If $f$ is holomorphic and non-constant, then the image of $f$ is dense in $\mathbb{C}$
I'm pretty sure the the "it looks like z=z^k" bit comes from expanding f around the point as a power series and approximating it by the smallest order terms.
Can someone please explain this: Suppose there are $n$ points in a plane out of which $m$ points are collinear.Number of triangles that can be formed by joining these $n$ points as vertices=$nC3-mC3$.
@PVAL-inactive Right, so there's no class of 2-complexes with every possible group appearing as fundamental group of some 2-complex out of that class such that two of them are homeomorphic iff they have isomorphic $\pi_1$, I think?
@MrAP each triangle is uniquely determined by its endpoints, of which there are $n$. Disregarding the $m$ collinear points, there are $\dbinom n 3$ choices. However, the three endpoints can't be all among the collinear points, so $\dbinom m 3$ choices are rejected.
@Balarka There's no listing of all the simply connected two-complexes, so that for an arbitrary two-complex C there is some computable positive integer N(C) so that if C is simply-connected C is homeomorphic to one of the first N(C) complexes in the list.
@EricSilva The result is actually nontrivial. And it's not good enough for the graph result because the resulting $n+1$-current could be something crazy and leave the cylinder. There might be some better result.
In my functional analysis class, I proved that the completion of a direct sum is the direct sum of it's completions by using the fact that the completion functor is left-adjoint to the forgetful functor.
Sometimes, when I'm bored (e.g. if my functional analysis prof talks about numerics of PDEs), I just pick a random number, check if there is a simple group of that order on OEIS and if not, try to prove that there is no simple group of that order
2
really fun, I would recommend that for killing time
@MatheiBoulomenos You too might find the last of the problems in my exercise set interesting (I don't think Leaky has even really tried starting it yet)
@PVAL-inactive When speaking English with a German accent it is sometimes hard to say a German word with a perfect German accent in the middle of speaking English
Can someone help me with this: Find the number of words which can be formed by taking two alike and two different letters from the word COMBINATION. I am getting $frac{2P2}{2!}*2P2$.
We have that $y-x>0$. By the Archimidean property we have that $n>\frac{1}{y-x}$. There also exists $a,b\in \mathbb{Z}$ such that $a<xn$ and $b>yn$. Why does the last sentence hold? From which property does this hold?
I used this site for more than a year before I saw the chat. The chat is now the main part that I use. I guess that tells us something about something.
