Talk:Non-analytic smooth function

How it is ill-behaved : added formula for n-th derivative with proof

I just wonder if formalizing the proof is really a good thing, or rather a liability. I sort of liked the older approach, which was driving the point home while keeping the discussion informal. Is the new proof a bit less approachable for non-mathematicians? Oleg Alexandrov 17:50, 18 Apr 2005 (UTC)

Sorry for the delay in reply. Maybe you're right (I was somehow afraid when I saw how it grew big...) But in fact I wanted to have the formula written down for further reference (I didn't remember that it was on x^3n, e.g.). Could we avoid complete annihilation by moving it to somewhere else (subpage /proof ?). — MFH: Talk 16:30, 21 Apr 2005 (UTC)

Ugh - three problems with that. No need to divide the page; the use of subpages A/X has been deprecated for several years. What was the third? Doesn't really matter ... might have been that the copy-pasted page makes little sense. Charles Matthews 17:03, 21 Apr 2005 (UTC)

Actually we have a smooth function page as it is, and I think this page was only made to link to from list of mathematical examples. That's OK - the result is very important. But no need to chop everything so fine. Why not put proofs on smooth function, and leave this less formal - Oleg's point is good. Charles Matthews 17:06, 21 Apr 2005 (UTC)

I came across the subpage and immediately began the process of merging it back into the main article, since subpages are long since obsoleted in the article namespace (Wikipedia:Subpages). However, now that I see this discussion here, it might not be appropriate for me to just paste it right in since there's dispute on the subject. So I'm putting it here into talk:, below. Bryan 05:56, 5 Jun 2005 (UTC)

f is smooth on R

Let us prove in the sequel that the function

$f:x\mapsto e^{-1/x^{2}}~(x\neq 0);f(0)=0$

admits continuous derivatives of any order $n\in \mathbb {N}$ in all points of $\mathbb {R}$, given by

$f^{(n)}(x)=\left\{{\begin{matrix}R_{n}(x)\,f(x)&(x\neq 0)\\0&(x=0)\end{matrix}}\right.,$

where $R_{n}$ is a rational function of the form $R_{n}(x)=P_{n}(x)/x^{3n}$, with $P_{n}$ a polynomial, such that $R_{n}$ is well-defined on $\mathbb {R} ^{*}=\mathbb {R} \setminus \{0\}$.

continuity of f and f⁽ⁿ⁾

Any function of this form is indeed continuous on $\mathbb {R}$:

• on $\mathbb {R} ^{*}$, it is a product of continuous functions,
• in $x=0$, $R_{n}$ is "at worst" equivalent to $c/x^{3n}$, and for any (also negative) integer $m$,

$\lim _{x\to 0}x^{m}e^{-1/x^{2}}=0,$

such that $\lim _{x\to 0}f^{(n)}(x)=0=f^{(n)}(0)$, i.e. continuity of $f^{(n)}$ also in $x=0$.

proof of the formula for the n-th derivative

For $n=0$, we do have $f(x)=f^{(0)}(x)$ as in the above formula, with $R_{0}=1(=P_{0})$. In order to complete the proof by induction, it remains to show that if $f^{(n)}$ is of the above form for some $n$, then its derivative is again of the form $f^{(n+1)}$.

Of course, $f^{(n)}=R_{n}f$ implies that $f^{(n+1)}=R_{n}'f+R_{n}f'=R_{n+1}f,$ with (using f' =f×(+2/x³))

$R_{n+1}(x)=R_{n}'(x)+2R_{n}(x)/x^{3}=P_{n+1}(x)/x^{3n+3}$

for all $x\neq 0$, where (for $R_{n}=P_{n}/x^{3n}$) $P_{n+1}=x^{3}P_{n}'-3nx^{2}P_{n}+2P_{n}.$

Thus it remains to consider the difference quotient in $x=0$, which has the form

$\lim _{x\to 0}{\frac {f^{(n)}(x)-f^{(n)}(0)}{x}}=\lim _{x\to 0}{\frac {P_{n}(x)}{x^{3n+1}}}\,e^{-1/x^{2}}=0$

(using the limit already mentioned previously), i.e. $f^{(n+1)}(0)=0$.

Page title

Isn't the page title a little long? Enochlau 17:21, 21 Apr 2005 (UTC)

Yes, but not infinitely. Charles Matthews 17:36, 21 Apr 2005 (UTC)

I also find it long. Of course not inifinitely, since this is an imaginary concept and does not exist in reality. It also is of no practical use, e.g. everything is always smooth up to the measuring precision.

But to have proofs on a separate page where they could conveniently be edited would be much more useful imho... why not leave it there, even if it is not considered as part of the encycopedia? are the Talk pages considered as part of the encyclopedia? — MFH: Talk 4 July 2005 13:49 (UTC)

It is not a good idea to have subpages to a page. Yesterday I cleaned up all the subpages from the list of mathematical topics. You could put your proof from the subpage as a separate article a rigurous proof that an infinitely differentiable function that is not analytic (which becomes even longer :) Otherwise you can just paste it here on the talk page, but then you cannot put a link to it from the main article. Cheers, Oleg Alexandrov 4 July 2005 15:26 (UTC)

Maybe I don't understand what your point is. Are you trying to say that the schetched proof from this article should be on a separate page? Oleg Alexandrov 4 July 2005 15:28 (UTC)

Sheaf of smooth functions flasque?

In point 3.2 it is stated that the sheaf of smooth functions is flasque. This is not true (example: you cannot extend 1/x from the half line to the whole line). But my defintion of flasque comes from wikipedia, so i dont know if it is the right definition.

Maybe what the author meant was that with holomorphic functions the restriction operation is injective (i.e. extension of holomorphic functions are unique) in contrast to smooth ones.