It's the number following the # in the permalink. I am not sure that's the official terminology but that seems to be a logical name.
@Gigili Your message has the permalink http://chat.stackexchange.com/transcript/message/2546179#2546179. I mean the number 2546179. So I would write :2546179 Hi Gigili.
@Srivatsan If I find something, I will let you know. I myself have been unsettled about the notation, so when I saw the question, I felt I had to figure it out, for myself as well as the OP.
@robjohn Can you do such a decomposition globally? Your answer explains that it can be done at a single point, but it's not clear to me that you can make it work for all points at the same time.
@HenningMakholm That is something I will have to investigate. That was beyond the scope of the question, but it is something that I was wondering myself.
@robjohn The partials themselves only tell us how the function behaves on the axes. We'd need something to guarantee that it is not completely wild off the axes.
@HenningMakholm That is true if we want to expand, but for the decomposition of the partials at a point, I don't think it is necessary. I could be wrong.
@HenningMakholm Yes. I was thinking just about the decomposition into conformal and conjugate conformal once you had the partials, but if we want the conformal etc to really be conformal, then yes, we need continuity.
@HenningMakholm That is exactly what I was thinking about :-)
Does anyone know how to rank answers according to length. I wanna check where mine stands and how much more I should write to get to the top [just in terms of length]... =)
Also, the only thing higher than high school level was MacLane and Birkhoff's Algebra, which I checked out. I strongly suspect the library buyers simply mistook it for an easy book.
@Matt: You can make matrices look good with \begin{pmatrix}
Can someone explain Phira's comment here: meta.math.stackexchange.com/questions/3256? I am not sure I follow the reference (is that a proverb?). Checking the dictionary now..
I think I'll have to look at it tomorrow though. It's late here and I have to be at uni at 9 tomorrow because the buggers concocted a series of "short tests" that you must attend in order to be admitted to the exam. It totally wrecks every other of my Mondays because I have to get up so early : /
Amusing that we used very similar words, @robjohn.
@robjohn I think I understood now. It's not a very natural thing to say, where I come from. I guess it is more common to regard it shameful to fool a person repeatedly, whereas the fooled is usually not chided for getting fooled. =)
I think he's tacitly assuming the "you" instictively takes the ostensible rep count for granted, hence is being fooled once by autorecalc and he learns twice after a manual recalc
@anon Aw, that is a nice interpretation. I am not attaching significance to rep, but the rep got added and taken away for a reason; I want to know that reason, that's all.
@robjohn OTOH the other pov is all too forgiving. There are pros and cons for both.
The question is clear: give an interpretation that justifies (1), (2), (4), but still (3) is incorrect. My stand is that one needs to look no further than the set-of-functions interpretation, provided one interprets equals to mean set containment (either \in or \subseteq -- whatever is appropriate).
@Srivatsan FWIW, I think your answer answers the question as asked. I don't really like that interpretation intuitively (it seems to me to give too short shrift to the usual intuitions about equality and equations), but it appears to be technically consistent.
@robjohn No, my answer does not talk about Theta. I am just saying that for Theta it might not matter whether you take = to mean equality or set containment (inclusion/membership).
@HenningMakholm Ok. You're right. I'm getting ahead of myself. Let me refine the claim: containment is a right way to interpret. It covers all cases and works as you would expect...
It tells us that x^3 + Theta(x^2) = Theta(x^3), and it tell us that the other way isn't true. Check, Check.
Another test is this: we would expect Theta(f) = Theta(g) to imply that Theta(g) = Theta(f). I'm claiming that this also work, thanks to our dear old Theta.
and f+O(h) = O(g) I see as f + any element of the class that doesn't grow any faster than a constant multiple of h doesn't grow any faster than a constant multiple of g
