@Ted I did find his "fundamental theorem of calculus" a lot of fun but admittedly I'm pretty bored by Klein geometries. Back to Kobie, I think. Also, side note, glad you enjoyed your TDay.
Should I use that an entire function is trascendental iff it is an essential singularity at infinity?
Something of the sorts?
Here I am given a function that is holomorphic in all the sphere but $-1$ and $2$, I am given the residues at those points and two values of the function, I am asked to determine the function.
oh yeah, scale f(z) by a rational function so it has no zeros or poles, then replace z with 1/z, then multiply by the appropriate power of 1/z to make it holomorphic everywhere, hence bounded hence constant
@MikeMiller My question is, suppose you have $f\in \mathcal O(\Bbb E)$, and suppose there is an annulus $r<|z|<1$ where it is injective. Then it is injective.
intuitively the alternating and symmetric groups on a set should be among the only "natural" ways to make a permutation group on a generic set (without, say, labelling the elements or arbitrarily imposing some choice of ordering). more precisely, there is a functor Sym:Set->Grp, suppose there is another functor F:Set->Grp for which there is a natural transformation F->Sym comprised of injective maps, then must the image FX always be a normal subgroup of Sym(X)?
@MikeMiller Technically, the question asked "Could I have any suggestions or topics to study for handling the situations correctly?", so the comment sort of addresses that. I still don't like its content or tone; flagged as not constructive. The question is pretty unclear too.
If sequence $f_n$ is uniform convergence on $(-\infty,0)$ and it uniform convergence on $[0,+\infty)$ is then $f_n$ uniform convergence on $\mathbb{R}$? I just need is this true or not. I think it's true.
@PedroTamaroff Sorry... we had company and I was gone most of the evening. The way to approach this is to note that any curve, $\gamma$, circling inside the annulus is mapped to a curve, $f(\gamma)$, that does not intersect itself. The Jordan Curve Theorem says $f(\gamma)$ divides $\mathbb{C}$ into the interior, $\Omega$, and exterior, $\mathbb{C}\setminus\Omega$.
@PedroTamaroff for each $a\in\Omega$, the argument principal says that $\frac1{2\pi i}\int_\gamma\frac{f'(z)}{f(z)-a}\mathrm{d}z$ is the winding number of $f(\gamma)$ around $a$, which is $1$ and the number of zeroes of $f(z)=a$ inside $\gamma$.
@bubba, where did you get that about compulsive hand-washing? I met the man and spent a bit of time with him, and never noticed anything of the sort. There was a lot of good about him, even beyond the mathematical brilliance, and a lot to aspire to. — Gerry Myerson3 hours ago
@GerryMyerson -- I got the handwashing thing from here: amphetamines.com/paul-erdos.html. I met him, too, and I must admit I didn't notice the handwashing, either. But, handwashing or not, I still wouldn't want to be him, He struck me as a man without perspective or balance in his life; a mathematician first and a human being second. Not the kind of life I want, and not one I would recommend to others. But, this is all a question of personal choice, of course. — bubba3 hours ago
hi guys, can someone look at this? :) http://math.stackexchange.com/questions/1043131/can-the-probability-of-a-trump-poverty-be-calculated-without-making-case-distinc/1043140#1043140
@eBusiness May I ask your opinion on the tags for that question? Tha fact is the title is quite "complicated", so I'd like to let the question have the best tags
@Nick: I met him several times. He visited UGA regularly to work with Carl Pomerance. I didn't know him well, but most of us would advocate a healthier life.
I need a bit of reassurance my arguments are correct, as this is a case in a continious function space and I am very shaky there when it comes to foundations
Now it's get a bit tricky for me, first I want to show that $A\subseteq dA$, so I take a function $f(x)\in A$. Any open ball centered in $f(x)$ will obviously include $f(x)$, so it has a point in $A$. Let $B(f,\epsilon)$ be any open ball centered in $f$,
The definition we have for nowhere-dense is $Int(ClA)=\emptyset$
Don't need closed, but I need to think about the closure. Since in this case the Closure is $A$, i.e. $A$ is closed, it makes it easier
It's important to point out, as I am prone to falling to such misbelieves by exercises :P
Okay, so we are now observing $B(f,\epsilon)$. I want to say that there exists $g$, so that $g(x)=f(x)$ for almost all $x$, except a certain small segment where it rises above (or below) $M$, and it still would keep the distance less than $\epsilon$
And for the other direction, I assume $f$ is in $dA$, and then I show it's in $A$. If it's not in $A$, then there is some $t$ so that $f(t)>M$ (or $f(t)<-M$, the arguments will be the same in that direction). So I know from anaylsis that there is some segment where $(a,b), b>a$, where $f>M$ for all $x\in(a,b)$, and then I can show the distance with any function in $A$ is positive, denote it by $\epsilon$ and state that the open ball with distance $\epsilon/2$ won't contain any functions in $A$
@N3buchadnezzar There are actually lots of ways to demonstrate it geometrically. A good one is the unit circle, I can't recall the exact method by which you prove the isosceles triangle at $45^{\circ}$ but it must be easy enough to figure out. Also, try superimposing the wavy graphs, any highschooler can do that to see them intersecting at $\frac{\pi}{4} \text{ rad}$
@UserX I'm no expert but doubling the q is doubling the I which means you're doubling the amplitude... idk, possibly the period decreases. (Is the period halved?... idk again)
@JasperLoy There's an evolutionary imperative why we give a crap about our family and friends. And there's an evolutionary imperative why we don't give a crap about anybody else. If we loved all people indiscriminately, we couldn't function.
