I will talk now to our Italian guys if it's worth to buy IC tickets in advance
suddenly I saw today that several are sold out. It couldn't happen e.g. in NL for non-high-speed trains
@robjohn: namely for Beijing I only need to reach the hotel - there I'll meet my colleagues, but because of the Italy trip they fly a day earlier than me
I saw Matt enter the chat and then leave. Since the gravatars float in from above and off below, it looked as if he tried to join the chat, but couldn't stick.
@Ilya You can no longer verify this because he's not the owner anymore, but you can see here for example that Kannappan's name is in italics and he's the owner of that room.
If $f$ is analytic and compactly supported -- can I pick a point in the boundary of its support, use that it's analytic to get a neighbourhood on which its Taylor series converges and then since the points is in the boundary the neighbourhood intersects with both the support and outside where $f$ is zero so $f$ is zero on all of that neighbourhood?
The problem is that you speak about non-existent gadgets. Analyticity only makes sense for functions on open sets and if $f$ is analytic on $U$ then the set of its zeroes is closed and discrete in $U$.
It's the identity theorem for analytic functions: if two analytic functions on $U$ coincide on a set with an accumulation point in $U$ then they are the same. If the set of zeroes of $f$ were not discrete in $U$, then $f$ would coincide with the zero function on a set with an accumulation point in $U$, hence it would itself have to be zero.
So I do have to use the identity theorem. I was trying to "see" it without theorems.
I think I'll have to "see" why they coincide if they share an accumulation point.
Thank you.
@robjohn Yes, this chat can be quite a time eater. I don't mind from time to time, after all I enjoy this. But I haven't been enjoying myself as much as I used to. For example, I liked Asaf being around. And then there are a few other factors that prevent me from visiting as often as I used to.
I think it is clear that the set of points where the functions coincide is closed, since analytic functions are continuous, so the problem is to see why it is open.
@MattN I went to a conference, two weekends of travel, no time to access the internet during the week. Lots of drama seems to have happened here in the mean time.
I retagged this question from (statistics) to (algebra-precalculus) and (mathematical-modeling) but am not quite sure it's right. Do we have a tag that is "more elementary" (and perhaps more applied) than algebra-precalculus?
I'm quite happy. I got some homework today, a new piece to work on. And I'm particularly pleased because it's one I like and it's not impossibly difficult : )
@robjohn I removed [statistics] because the OP seemed to have chosen it simply because the calls inputs to the function he wants to construct "statistic"s. It certainly has nothing to do with estimating the parameters of unknown probability distributions.
@robjohn Interestingly, the linked mathematical statistics article claims that the term has a special meaning in Scandinavia, which may be biasing my understanding.
The OP is trying to design a scoring system, and has extremely vague ideas about what he wants from it. He certainly doesn't seem to have any correctness criteria for his scoring system that he could meaningfully use to evaluate it as a statistical model.