But if you're looking for an example where the nonempty intersection of two open balls is not an open ball, taking the open disks of radius $1$ centered at $(0, 1)$ and $(0, 0)$ should work in $\mathbb{R}^2$.
look I don't know how you know what ive been reading you crafty little bugger but oh wait I think I cyber begged for someone to buy it for me on here prior to finding the link. never mind carry on good sir
I will admit though that, because I'm aiming for grad school in CS, I think I will have a lot easier time than you all who had to take the subject test in math.
I've heard the math GRE is hard, while the general GRE is basically just an SAT over again...
This might be a stupid question but I just want to make sure. If I want a random integer in the range [min, max] and have (random % (max - min + 1)) + min, which is a usual way to get around modulo bias, is there any property of random I must be aware of other than ensuring that random > (max - min)?
That sounds like it would be difficult if I want a range [big, big + 5], and the number is large enough that I would almost never happen to land on a number in that range no matter how many times I generate random.
My brain must be completely broken right now because that is making no sense to me (doesn't help that I only know C and Bash). So I keep generating random until 0<=random<2**(8*n)?
Even if I wanted an integer between, say, 5 and 10000.
Yeah right now I'm completely failing to think of this abstractly. I think I should just look at how some C library does it or something. Sorry for being so hard to teach!
i got something else i am stuck on, i need to extend the bounded functional $\phi(x_1,x_2,x_3) = x_1-x_2$ while preserving norm. $\phi$ defined on $Y = \{ x_1 = 2x_2\}$, and i found that its norm is $1/\sqrt{5}$.
we are working in $\Bbb C^3$ with the euclidean norm
It's not well-defined: How do you make sense of $\int 1$?
A better starting point is to realize what the integral of a compactly supported smooth function is in the land of distributions. $f \mapsto \int f$ is a continuous functional on $C^\infty_c(U)$, so it is some distribution.
Now recall how a locally $L^1$ function $f$ gives rise to a distribution: as the functional $g \mapsto \int fg$. So the $g \mapsto \int g$ is, as a distribution, the function $1$!
You could set up what you're trying to do by picking a sequence of compactly supported smooth functions $f_n$ so that $f_n \to 1$ uniformly in $C^\infty$, and for a distribution $\delta$ you could try to define $\int \delta = \lim \int f_n$. But as above there's no reason to believe this converges.
@user616128 well the book on algorithmic determination of closed forms for hypergeometric terms and their sums and products @Dair yes that is correct I am aware of the philosophical views of that contributor to the publication, but the algorithms he contributed are in themselves incredibly insightful. So, if I gave you a set of principles each represented by a particular statement ${\{S_1,S_2,S_3,...S_k}\}$
and I tell you that that one of those statements is "Doron is an ultrafinitist", you would immediately reject the whole set as being of no value to contemplate? of course not
@user616128 before I was provide with the necessary link to obtain it free of charge, yes I probably went for a cyber beg when I thought I had to pay for what it is sold at google
@BalarkaSen Interestingly while the dual space to $L^p$ for $1 \leq p \leq 2$ is $L^{p/(p-1)}$, the dual space to $L^p$ for $p > 2$ iirc includes non-measures.
so i got that function sequence $f_n(x) = \sqrt[n]{1 + x^2}$ for $0 \leq x \leq 2$, so is it really so simple? its rather clear that its pointwise convergent $f_n \to 1$ and uniformly convergent cause $|\sqrt[n]{1 + x^2} - 1| \leq \sqrt[n]{1 + 2^2} - 1 = \sqrt[n]{5} - 1 \to 0$
or in case $f_n(x) = \frac{\sin (nx)}{n}$ and $x \in \mathbb{R}$ it's just $f_n \to 0$ and $\frac{\sin (nx)}{n} \leq \frac{1}{n} \to 0$, is it that easy?
@LeakyNun We use the epsilon-delta definition, for all $\epsilon \gt 0$ there exists an $N \in \Bbb N$ such that for all $n \ge N$ and $x \in X$ we have $|f_n(x)-f(x)| \lt \varepsilon$
You're asking whether Zq acts on Zp x Zp. The answer is yes iff there is an element of order dividing q in Gl(2,p). Leaky's calculations are still relevant here.
ok if q>p then if f:Zq -> Aut(Zp x Zp) then ord(f(1)) | q, so f(1)=0 or ord(f(1)) = q. If the latter, then q | p(p-1)^2(p+1), which means q|p or q|p-1 or q|p+1...
There is a normal V_4 subgroup of A_4 given by the disjoint transpositions, its quotient is Z/3 by its order, and there is clearly a section from Z/3 to A_4
Suppose $X$ is a normed space and $x_1,\dots, x_n$ are linearly independent. given $\alpha_1,\dots,\alpha_n$ i need to show there is a bounded functional on $X$ ,call it f s.t. $f(x_i) = \alpha_i$.
so i can define $f$ on $span\{\alpha_i\}$
and by Hhan-Banach , because $f$ is linear and bounded on $span \{\alpha_i\}$ then it can be extended to $X$
@AlessandroCodenotti is it easy that $f $ is bounded? if i define it as $f(x_i) = \alpha_i$ ? it is easy that it's bounded on the $x_i$'s, but i dont see why on $span\{x_i\}$
Okay. I always run into this stupid problem in various contexts (Calculus, Topology, etc.), which should be easy: Given $\delta > 0$, I want to find $0=x_0 < x_1 < ... < x_n = 1$ such that $x_{i+1}-x_i < \delta$. How do I do this? I draw picture after picture and nothing comes to me. I've solved this 'problem' a million times but I can never remember how to do it. If it makes things easier, the $x_i$'s can partition $[0,1]$ into subintervals of equal length.
The Heisy boi is fucking weird because there is a (non-split) central extension 1 -> Z_p -> H -> Z_p x Z_p -> 1, a non-split extension 1 -> Z_p x Z_p -> H -> Z_p -> 1, and a split extension 1 -> Z_p x Z_p -> H -> Z_p -> 1 where those Z_p x Z_p and Z_p sit in completely different and weird fashions in H
@Mike @Leaky Also: If your nonabelian group has order p^2q, and has a subgroup of order p^2, the p-Sylow subgroup is normal. But then n_q is either 1, p or p^2 and is 1 (mod q), so the only possibility is p^2 = 1 (mod q). That happens iff (p, q) = (2, 3).
@MikeMiller So denote an element of H by (a, b, c) where a is the 12-th entry, b is the 13-th entry and c is the 23-th entry. The normal subgroup is given by all the elements of the form (a, b, 0) and the Z_p subgroup is give by (0, 0, c)
@LeakyNun Yes, that's why q does not | p or p - 1, which is key to the calculation
@MatheinBoulomenos I know any extension 1 -> N -> G -> H -> 1 for an H-module N gives me a class in H^2(H; N). Can you build an extension from a class in H^2(H; N) in general? I suppose you can.
@Liad the proof uses the equivalence of norms which is proved by the compactness of the unit ball, so the proof doesn't actually avoid that and so it seems overly complicated
