@Daminark anyway, I'm pleasantly surprised to see that Sylow subgroups are used all the time in my group cohomology course. We're using it while proving stuff that gives us really difficult results in ANT (class field theory)
@anakhronizein again it was probably pretty orthogonal to the math stuff, the pythagoreans probably werent very popular considering they were a bunch of weirdos
I never really grokked the Sylow theorems when I was supposed to study finite group theory; I've got a somewhat better appreciation now that I've wised up about them.
@Daminark basically it allows one to reduce certain statements about general finite groups to statements about $p$ groups which are way easier to handle
there were a couple hundred iirc, but given that a lot of their values dont align with a lot of ancient greek values i think it's probably @anakhronizein
iirc they were pretty ok with women engaging in intellectual life which was pretty progressive for greeks
I was doing an REU, and the apprentice program started off with a 5 week class on the basics of discrete and LA. I was under the impression that our paper had to relate to it, and a person I knew was very into algebra at the time so he reeled me in, so I was originally gonna do something about group actions on graphs
To do so, I read on group actions, but then I decided to instead go the direction of Sylow. So I've def got a soft spot for it
@MoziburUllah Not that I recall. Every interlocutor in Plato's dialogues that I can remember is a man. Though Plato does say that women can ahve a role in running the State in The Republic
@TedShifrin I obviously know how to shift the interval of integration (i.e. by changing the variable), but I am not sure what you mean by rescaling the interval of integration, if not to let $2\pi = N$, once the interval is shifted from $[-\pi, \pi]$ to $[0, 2\pi]$.
@KevinDriscoll A function $f$ is periodic, with period $N$, if $f(x + N) = f(x)$. Now, you're asking me if it is possible to change the "wavelength" of the periodic function?
During these days I've slept a really reduced number of hours, so I am so slow...
@nbro Ok (you switched notation above to $f$ with period $N$) but its immaterial. Suppose $f$ has period $2 \pi$, how can you make a $g$ out of that $f$ that has period $N$?
@KevinDriscoll You're asking me something like rescaling the argument of $f$, like if my function was $\sin$, then $\sin(2x)$ would be a rescaled version of sine. Is that what you mean?
$\Bbb Z/4\Bbb Z$ acts transitively on $\Bbb Z/2\Bbb Z$ by addition. Then, $G_1$ is $\{0,2\}$. I think you mean it acts on the set of left cosets of $G_x$ @Daminark
@LeakyNun: I think thats true of a lot of people apart from Grothendieck who apparently just thought abstractly naturally - but that might be just hero worship.
@orbit-stabilizer: Personally speaking, anything I understand is concrete and what I want to try to understand is abstract until it becomes concrete for me.
@nbro No, letting $N = 2\pi$ is choosing a particular period, namely $2 \pi$. Rescaling $x \mapsto \frac{2 \pi}{N} x$ is just a rescaling, and it tells you what to do with the Fourier integral, namely make the same change of variables
@MoziburUllah I think the thinking behind that saying was that any sufficiently abstract statement will be general enough so that you can find a simple example that will help you understand it.
To say that, you need first to make the change of variable in $c_n$, i.e. you first need to change the interval of integration of $c_n$ and then you say that you create the new function $g$ out of $f$, right?
@orbit-stabilizer: I prefer continental philosophy as opposed to analytical because its like taking time away from being analytical with maths/physics - or so I thought.
@orbit-stabilizer: I've read one essay by Nagel - whats it like to be a bat - so long ago that I only recall the title; I haven't read anything by Singer. Whats his first name?
@KevinDriscoll So, you would start by saying something of the form: "If g has a period of $N$, then $f(x) = g(\frac{N}{2\pi} x)$ has period of $2\pi$"?
Other than Zizek I cant really name a contemporary popular philosopher who Id describe as continental. But my background is very solidly in the analytic tradition
I won't try to reproduce his argument because I don't know it well enough. But hes a Utilitarian so questions about rights to life and such arent going to be central to him
On his faculty page, Mr. Singer argues: “Newborn human babies have no sense of their own existence over time. So killing a newborn baby is never equivalent to killing a person, that is, a being who wants to go on living. That doesn’t mean that it is not almost always a terrible thing to do. It is, but that is because most infants are loved and cherished by their parents, and to kill an infant is usually to do a great wrong to its parents. “Sometimes, perhaps because the baby has a serious disability, parents think it better that their newborn infant should die. Many doctors will accept thei…
It does sound rather utilitarian...I can see why the Christianity would have great problems about accomodating that view. I'm not sure that I accept new borns don't have any sense of their existence over time; its more likely that its a more primitive sense than a wholly developed one that a fully developed person has.
It might also be integral to the learning process for a new born that this is the case too.
I love reading about things that I don't necessarily agree with - most of the time, it turns out that it's usually more nuanced than I originally thought.
I'm pro-choice too; I think one of the first questions I asked on Phil.SE was to think of abortion as a kind of blood sacrifice in order to make a comparison with the Incas - I think I was bit annoyed with Dawkins at the time at suggesting the Incas were an obviously primitive people.
For integral computations it's often quite important. Oftentimes you'll need to change variables when you're trying to integrate, and recognizing when you can do a trigonometric one is highly useful
@KevinDriscoll But I still don't understand why you would start from $g$ and not from $f$, given that we want to prove something about $g$, given facts about $f$.
@nbro You want to prove that there exists an expansion for any $g$. So you have to start with a hypothetical $g$, and then construct an expansion for it.
@nbro If you start with $f$, then what you are saying is that you can get a series for any $g$ which can be generated form some f, but then you would have to separately prove that any $g$ can be generated from soem $f$
Once you gonvert $g$ into a new function periodic on $[-\pi. \pi]$, you can apply the theorem you already hvae
@nbro Exactly. Which is why you start with $g$ and then use it to construct a function periodic on $[-\pi, \pi]$ which you can then apply your known expansion to
@KevinDriscoll So, now we have a function $f(x) = g\left(\frac{N}{2\pi} x \right)$ with period of $2\pi$ and our theorem applies to function with that period, the problem is that the domain starts from $0$, whereas in our theorem the domain of $f$ starts from $-\pi$.
We can now have a change of variable in $c_n$ of $f$
Yea, I just want everything to be on the given domain for $g$, because in your construction $h(-\pi) = f(-2 \pi) = g(-N)$, but just to be consistent we want to keep things for $g$ on $[0,N]$
@nbro Yea, exactly
So now we have $h(x) = f(x+\pi) =g(\frac{2 \pi}{N} (x+\pi))$ and we can apply our known theorem to $h(x)$
@KevinDriscoll I have a question not regarding this specific exercise. Why do they use two different variables in $f(x)$ and $c_n$, i.e. why in one case they use $x$ and in the other they use $t$?
@nbro t is just a 'dummy' variable here. They can't use x because then you'd have say, $f(x) = \sum_{n=-\infty}^{\infty} \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-i n x} \ dx$
and then when you plug in, say, $x = 0$, you end up plugging in $x=0$ inside the integral as well
Sorry I dropped an exponential in my expression for the fourier series of $f$, but you get the idea
