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.

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)