Well, I don't know enough about America's stance on gay marriage or gay rights, but I will say that I truly hope that it does improve without knowing how bad it was where you lived before.
yeah my sister came into my room and said "eric died" and I said "okay please get out of my room" and she said "no nate, eric died" then I started bawling.
I think that was the most real moment I've ever had with my sister.
@Andreas: I liked it from a young age, and particularly got hooked on proofs and multivariable calculus early in college. I also found out my last year of high school that I could teach better than most of the teachers — and I loved that.
@Dodsy, I'm 18 so I don't know if the 10 years would have been true in my case. You may not have meant it that way, though :D. @TedShifrin Dark secrets revealed... any particular book of Ian Stewart that you would recommend?
I grew up with a father who measured his worth based on how much money he made (though I would never discount his worth or how much he has accomplished)
@Daminark Well, given a simple closed curve $\gamma$ on that plane and hol functions $f, g$ containing that curve, if $|f| < |g|$ along that curve, then $g$ and $f + g$ have the same number of zeroes inside the curve, yea
(Actually Forster proves open mapping theorem, as well as all the relevant theorems beyond the very basics of complex analysis, in Ch 1. Doesn't require Rouch\'e)
He proves a big theorem on the local description of holomorphic maps between Riemann surfaces that does it all
@Daminark You don't actually need open mapping to get inverse function theorem. It's kinda trivial once you know the inverse function theorem for smooth functions :)
Suppose that we have that $\alpha^{\frac{N-1}{p_1}}, \dots, \alpha^{\frac{N-1}{p_k}} \neq 1$ where $p_1, \dots, p_k$ the prime divisors of $N-1$. Then can we deduce that $N$ is a prime or not? And if for some $p_i$ it holds that $\alpha^{\frac{N-1}{p_i}}=1$ do we deduce that $N$ is composite? If so, why?
Notice that the topology on $\Bbb P^1 = \Bbb C \cup \{\infty\}$ is just the usual topology on $\Bbb C$ plus complement of compact subsets of $\Bbb C$ for neighborhoods of $\infty$.
Say you pick a point in $\Bbb{CP}^n$. That corresponds to a copy of $\Bbb C \subset \Bbb C^{n+1}$
Choose a vector in that copy of $\Bbb C$.
That can be written as $(z_0, z_1, \cdots, z_n)$ for $z_i$ complex numbers.
But if I multiply them by $\lambda$ to get $(\lambda z_0, \lambda z_1, \cdots, \lambda z_n)$ for $\lambda \neq 0$ that also corresponds to a vector in that specific copy of $\Bbb C$.
So you set up a "homogenenous coordinates" for $\Bbb{CP}^n$, where every point can be written as $[z_0 : z_1 : \cdots : z_n]$, $z_i$ complex numbers, and in this coordinates $[\lambda z_0 : \cdots : \lambda z_n] = [z_0 : \cdots : z_n]$ for $\lambda \neq 0$.
