tea.mathoverflow.net - Discussion Feed (Taylor's theorem and the symmetric group)

Andrew Stacey comments on "Taylor's theorem and the symmetric group" (17028)

2011-11-11T13:32:22-08:00

since as far as I know some SE2.0 sites support latex and others don't.

Absolutely correct. They tried supporting it on TeX-SX and we complained vociferously so they turned it off.

The nForum runs on the same underlying software as this place and has LaTeX support (better than MO as well). So the two statements really are completely independent.

quid comments on "Taylor's theorem and the symmetric group" (17020)

2011-11-11T11:08:26-08:00

So the second statement is a necessary, but not sufficient, condition for the first statement to hold.

This seems like an overstatement to me, since as far as I know some SE2.0 sites support latex and others don't. Yet, what was stated, seems like a useful heuristic to me.

(Sorry, I could not help, but continue on this. )

Qiaochu Yuan comments on "Taylor's theorem and the symmetric group" (17018)

2011-11-11T10:57:13-08:00

Well, I assume Steven Gubkin assumed that meta was based on the same software and therefore would automatically have LaTeX support. So the second statement is a necessary, but not sufficient, condition for the first statement to hold.

Andrew Stacey comments on "Taylor's theorem and the symmetric group" (17009)

2011-11-10T23:44:37-08:00

LaTeX isn't supported on meta; this meta is completely independent of SE software.

Those two statements are completely independent of each other. LaTeX (or rather, some subset thereof) could very easily be supported on this forum. However, there are reasons not to.

Qiaochu Yuan comments on "Taylor's theorem and the symmetric group" (17001)

2011-11-10T16:20:50-08:00

Yes. I have an answer and I'm sure Todd Trimble and others will as well; I think this is a fine question.

LaTeX isn't supported on meta; this meta is completely independent of SE software.

DavidRoberts comments on "Taylor's theorem and the symmetric group" (17000)

2011-11-10T15:19:44-08:00

This possibly links to combinatorial species, which use the groupoid of finite sets and bijections, and which is equivalent to the coproduct of S_n for all n.

simoncfr comments on "Taylor's theorem and the symmetric group" (16999)

2011-11-10T13:53:59-08:00

I think this sounds interesting.

quid comments on "Taylor's theorem and the symmetric group" (16998)

2011-11-10T13:52:06-08:00

Regarding your edit: latex is simply not supported on meta.

Steven Gubkin comments on "Taylor's theorem and the symmetric group" (16997)

2011-11-10T12:59:16-08:00

Here is a question I am thinking about asking. My worry is that the question is really just mathematical free association, but if there is a really crisp answer to the question I would love to hear it. I feel like there would be a lot to gain from a good answer, but I have a feeling that everyone will just think it is nonsense.

Anytime I see an $n!$ in some formula, my instinct is to look for the symmetric group on $n$ letters coming in somewhere. I have never done this seriously with the $n!$ in Taylor's theorem.

Question: Is there some way to see the $n!$ in Taylor's theorem coming naturally from a symmetry group?

Possible lead:

Here is a definition of $f^{(n)}(a)$ which does not depend on finding earlier derivatives: Let $g: \mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $g(x_1,x_2,x_3, ..., x_n)$ is the lead coefficient of the unique $n^{th}$ degree polynomial passing through $(a, f(a)), (x_1, f(x_1)),(x_2, f(x_2)),...,(x_n, f(x_n))$. Then $f^{(n)}(a)$ is $1/n!$ times the limit of $g(x_1,x_2,x_3, ..., x_n)$ as $(x_1,x_2,x_3, ..., x_n)$ approaches $(a,a,a,...,a)$. Could the $1/n!$ be related to the symmetry of $g$ under exchange of coordinates?

EDIT: Is there a reason my latex isn't working?