Now, $\hat{G}$ forms a group as well (easy to check), so in particular, this above sum gives $\sum_{\chi\in\hat{G}} \chi(h) = \sum_{\chi\in\hat{G}} \chi(h)\chi_*(h) = \chi_*(h)\sum_{\chi\in\hat{G}}$, which makes that sum 0
But anyway, so thanks to what I said, we know that $\delta_1(h) = \frac{1}{|\hat{G}|}\sum_{\chi\in\hat{G}} \chi(h)$, but then of course you can express $\delta_g(h) = \delta_1(gh^{-1})$, so in fact $\hat{G}$ spans $\{\delta_g\}$, which then spans $\mathbb{C}\langle G\rangle$
@Semiclassical what information do I need to add to keep it from being a research question? I thought I explained all the prerequisites given in the paper.
(which probably isn't an actual notation used by anyone, but the alternative is $\sum_{\{x:x^n=1\}}1$, which puts the important part in the subscript and thus is bad)
Hi guys, I just have a small question. I am solving a problem involving construction of confidence intervals for the parameter of a uniform distribution using some form of the max and min of 3 iid RVs with that uniform distribution as the pivot
So X1, X2, X3 are uniform (0.5theta, 1.5theta). I want to get a confidence interval for theta using the minimum. First, I used (Xmin-0.5theta)/theta as my pivot. Is that correct?
@LeakyNun So the idea being that the starboard presented a tempting question---what does that mean---and my saying "I'll bite" is to say "I know it's obvious, but I'll ask anyways."
room topic changed to The Nineteenth Byte: General discussion for http://codegolf.stackexchange.com | abandon all work, ye who enter here —aditsu (no tags)
just like how PPCG chat (used to) have a quote in the description
well, we also have "just ask; don't ask to ask" here, but I don't really think that's a quote
If you apply our $a_1 + a_2\chi_2\chi_1^{-1} + \ldots + a_n\chi_n\chi_1^{-1}$ to $h$, we know this will be different in at least one term than if we apply to to $1$
@ALannister Well, each coefficient is still an integer. So that would seem to give an integer polynomial automatically, though not in a canonical way (since adding $p$ to a coefficient doesn't change it as a poly in Z/p)
@Semiclassical but let's say $f = g_{1}g_{2}\cdots g_{k}$ where $f$ and all the $g_{i}$ are thought to be polynomials in $\mathbb{Z}_{p}$ but $h$ is any divisor of $f$ in $\mathbb{Z}$. If my goal were to show that some $g_{i}$is a divisor of $h$ over $\mathbb{Z}_{p}$ wouldn't I have to do some conversions somewhere?
Anyway, I was about to post something on the main MSE page. This is supposed to be an intermediary step on the way to proving Eisenstein's criterion for monic polynomials
Assume that $f = g_{1}g_{2},\cdots, g_{k}$ is a factorization of $f$ into a product of irreducible polynomials over $\mathbb{Z}_{p}$ (i.e., all $g_{i} \in Z_{p}[x]$ and $f$ is viewed as a polynomial over $\mathbb{Z}_{p}$) and let $h$ be any divisor of $f$ over $\mathbb{Z}$.
I need to prove that...
Assume that $f = g_{1}g_{2},\cdots, g_{k}$ is a factorization of $f$ into a product of irreducible polynomials over $\mathbb{Z}_{p}$ (i.e., all $g_{i} \in Z_{p}[x]$ and $f$ is viewed as a polynomial over $\mathbb{Z}_{p}$) and let $h$ be any divisor of $f$ over $\mathbb{Z}$.
I need to prove that...
