i have no idea what that means. I tried using $x=-\omega^k$, that makes the numerators $0$. I tried using $x=-a^4\omega^k$, because the $a^4$ was in your correct answer, but then the numerators evaluate to $1-a$ instead of $1-1$.
@Benjamin: So I was trying to think of the right characterization of absolutely flat things to use here, and I want to use that absolutely flat local rings are fields
Let's have a quick check to see why this is the case
suppose $R$ is not a field
then take $x \in R$ such that $x$ is not a unit
then look at $(x)$
Oh I forgot to say that $x \neq 0$
then because principal ideals are idempotent, a quick corollary of this is that $(x) = (ax)$ where $a$ is some element in the ring such that $x = ax^2$
@rob: the other two answers did that too. In hindsight, I'm surprised I didn't see that first. But... I didn't. I was really just going for another said and lost Roosevelt analogy
It's funny how $\sec$ and $\csc$ are virtually unheard of here in Denmark while apparently they show up again and again in US education. They are pretty much never discussed in highshool(-equivalent...ish) maths
let's take a step back and reconsider what we know. There is one prime ideal in $(A/R)_m$, which corresponds to a unique prime ideal in $A/R$ contained in $m$
@Gigili You should add that stipulation to your question on the main site. (I don't know whether I'll have any ideas, but I'll give it a few minutes' thought, at least.)
strictly speaking, I only used that if $A$ is absolutely flat, then $A_m$ is a field for any maximal ideal $m$ - perhaps you're thinking to prove that result?
A rule which is not too silly is: have as few enemies as possible. A corollary to that is: don't spend the few spots for enemies on has available on math.SE... it is simply a waste! There are so many people out there who __*REALLY*__ deserve enemyhood!
IOW, right beside the classical «Thou shalt not covet thy neighbour's house,» I'd put «Thou shalt find your enemies among your neighbors, not random people out there on the internet!»
For a good amount of time, it would show my previous grav in some conditions and my new one in others. Conditions were things like: who you were, which browser you used, and what ratings or size settings you had for viewing gravatars. It was rather incoherent.
they are within the integration interval (I think). i have five (very similar) integrals. the first is $\int_0^\infty \frac{1}{x+a}\ dx,\quad(a > 0 \in \mathbb{R})$.
Of course the real smart ones do things like this: \usepackage{xspace} \def\RR{\ensuremath{\mathbb R}\xspace} ... As \RR is locally compact... Take an $x \in \RR$... :)
Or you could fiddle with unicode and unicode-math.sty to just let you type in the glyph ℝ directly in your TeX source :)
ok, back on topic: can someone just tell me how to do that integral. i hate to just ask questions, but i've been working on this problem for 14 hours straight (and 1 week) and need to get it done soon (like very soon - oh, I can't get Maple to do it, either). $displaystyle\int_0^\infty \frac{1}{x+a}\ dx$?
ok. you think this is just a regular Reals integral? My instructions are to use the residue theorem (this is just one of the five which the original integral broke into usingpartial fractions).
@davidwheeler actually, based on what i'm reading now, it looks like i didn't need the partial fractions. the residue theorem just relies on the singularities
@DavidWallace is there a way to use residue theorem without doing partial fractions? i've already done the partial fractions, but i guess i didn't have to.
@DavidWallace is there a way to use residue theorem without doing partial fractions? i've already done the partial fractions, but i guess i didn't have to.
To use the residue theorem, you'd need to do some kind of transformation that turns 0 to infinity into some closed contour. It's not obvious to me how to do that.
