@Nick Can't be right. You cannot have $A = 0$, since that would give you a $Cx^2$ in the numerator. $1 = (Ax+B)(x-3) + C(x^2+1) = Ax^2 + (B-3A)x - 3B * Cx^2 + C = (A+C)x^2 + (B-3A)x + (C-3B)$ gives you $A+C = 0$, $B-3A = 0$ and $C-3B = 1$.
@Chris'ssis: If I were seriously asked that question, I'd ping you and have you answer it for me. But I think you can't because the function in the interval is discontinuous.
I'll take that as a no. last I recall he was using the alias grothendieck, or maybe he reverted to user####. if I still used skype I could prolly talk to him again.
If I did want to talk to him, it'd be about math (which I rarely talk about with you, or the many others that have asked why I don't talk to them in chat over the years now...), but I never said I wanted to talk to him again. I suppose it'd be nice to at some point, just don't have any plans to.
@Chris'ssis: So, about integrating $\int^0_{-1} \text{floor}(x)$ .. how do I go about it. Or just link me to something if it's too long and boring to explain.
@blue the galois groups are mimicked by Aut_H(G) for some group G and subgroup H. the galois extensions are mimicked by characteristic subgroups. Claim : if N <= H <= G is a chain of characteristic subgroups, then there is a left-exact sequence 1 --> Aut_H(G) --> Aut_N(G) --> Aut_N(H).
@BalarkaSen try generalizing to more general categories, for fun. (e.g. if the category isn't concrete, how do we define the relative automorphism groups arrow-theoretically? enjoy.)
@BalarkaSen "a lot" is relative; I doubt an actual category theorist would say so. think of extensions L/K as arrows K->L, and relative automorphisms as maps f:L->L that fix the map K->L after postcomposition. then one can try the same exact sequence trick you brought up.
@BalarkaSen from what I understand, nothing: while we often want more from a galois theory than just that small exact sequence, we can generalize its structure to more general categories. search for "galois categories"
@BalarkaSen: The video, I linked it for a reason, $$\int_0 ^{\pi / 6} \sec y \cdot dy = \ln(\sqrt3)(\text{What})^{64} + C $$ Quickly, find $\text{What}$
@Nick when you flag something in chat, it gets echoed across the entire SE network and interrupts a lot of people's business (this includes all high-rep users, not just mods). this math room in particular is known to be the worst offender with frivolous flags.
@MichaelHampton heh, that's what I thought too, but another mod told me it was this room. I guess we're in competition. or maybe I misremembered that chat.
@Nick no, I'm saying the flag was fine, yes we sometimes get superfulous flags, but that one was valid. Mass giberish posts is a problem worth flagging
though a better description would be helpful :)
@BalarkaSen you keep using that word, I don't think it means what you think it means
@Nick yeah, that's why the comment about better explanation. Flagging is the correct thing to do if there is a disruptive behavior that needs attention (either for clean up or for remediation) but if the flag description isn't clear, a whole lot of blue is going to show up to figure out what the heck it actually meant
Sorry, I have to ask again: Why is it only $f(x)$ and not $f(\{x\})$ on the right hand side? $$\int_1^\infty\frac{f(\{x\})}{x^2}dx=\int_0^1f(x)\Big(\sum_{n=1}^{\infty}\frac{1}{(n+x)^2}\Big)dx$$
@blue a positive point about galois theory for groups is that the Fix operator transforms any short exact sequence 1 --> A --> B --> C --> 1 of groups into 1 --> Fix(A) --> Fix(B) --> Fix(C), which apparently is the key object of cohomology theories.
also, the left-exact seq 1 --> Aut_H(G) --> Aut_N(G) --> Aut_N(H) might or might not be related to the theory of derived functors, in the process of extending it to the right.
that's all the interesting stuffs i could think about.
@BalarkaSen: The video, I linked it for a reason, $$\int_0 ^{\pi / 6} \sec y \cdot dy = \ln(\sqrt3)(\text{What})^{64} + C $$ Quickly, find $\text{What}$
@Nick I don't know where you got the constant of integration from, but the 'what' is simply equal to $1$ or $i$. $$ \displaystyle \begin{aligned} \int_{0}^{\frac{\pi}{6}} \sec y \text{ d}y \ \ \overset{u = \sec y + \tan y}= \int_{1}^{\sqrt{3}} \dfrac{1}{u} \text{ d}u = \log ( \sqrt{3} ) \end{aligned} $$
@robjohn as a matter of fact, because I was talking about Barnes G-function, the series $X$ in that proof can also be computed using this function (Barnes G-function).
@BalarkaSen I honestly have no idea. We might as well be going in around in circles, questioning each other as to why we like/dislike things that we just like/dislike!
@Khallil As sets in higher math, they are equal. As sets in elementary math, they may be considered distinct as the two copies of two are taken to be distinct.
@Chris'ssis Abel summation, like integration by part, plays the difference of one sequence off the sum of another. I don't see how this follows from Abel summation, or perhaps I don't see the two functions you are using.
@BalarkaSen Cardinality is one of my issues. If $\{ 1, 2, 2 \} = \{ 1, 2 \}$, then $|\{ 1, 2, 2 \}| = |\{ 1, 2 \}|$ so we might as well be saying that $3 = 2$.
