Talk:An infinitely differentiable function that is not analytic
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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
- <math>f: x\mapsto e^{-1/x^2} ~ (x\ne0); f(0) = 0<math>
admits continuous derivatives of any order <math>n\in\mathbb{N}<math> in all points of <math>\mathbb{R}<math>, given by
- <math>f^{(n)}(x) = \left\{\begin{matrix} R_n(x)\,f(x) & (x \ne 0) \\ 0 & (x = 0)\end{matrix}\right.,<math>
where <math> R_n <math> is a rational function of the form <math> R_n(x) = P_n(x)/x^{3n} <math>, with <math> P_n <math> a polynomial, such that <math>R_n<math> is well-defined on <math>\mathbb{R}^*=\mathbb{R}\setminus\{0\}<math>.
continuity of f and f(n)
Any function of this form is indeed continuous on <math>\mathbb{R}<math>:
- on <math>\mathbb{R}^*<math>, it is a product of continuous functions,
- in <math>x = 0<math>, <math> R_n <math> is "at worst" equivalent to <math>c/x^{3n}<math>, and for any (also negative) integer <math>m<math>,
- <math>\lim_{x \to 0} x^m e^{-1/x^2} = 0,<math>
- such that <math>\lim_{x \to 0} f^{(n)}(x) = 0 = f^{(n)}(0)<math>, i.e. continuity of <math>f^{(n)}<math> also in <math>x = 0<math>.
proof of the formula for the n-th derivative
For <math>n = 0<math>, we do have <math>f(x) = f^{(0)}(x)<math> as in the above formula, with <math>R_0 = 1 (=P_0)<math>. In order to complete the proof by induction, it remains to show that if <math>f^{(n)}<math> is of the above form for some <math>n<math>, then its derivative is again of the form <math>f^{(n+1)}<math>.
Of course, <math>f^{ (n)} = R_n f<math> implies that <math>f^{ (n+1)} = R_n' f + R_n f' = R_{n+1} f,<math> with (using f' =f×(+2/x3))
- <math>R_{n+1}(x) = R_n'(x) + 2 R_n(x)/x^3 = P_{n+1}(x)/x^{3n+3}<math>
for all <math>x \ne 0<math>, where (for <math>R_n = P_n/x^{3n}<math>) <math>P_{n+1} = x^3 P_n' - 3 n x^2 P_n + 2 P_n.<math>
Thus it remains to consider the difference quotient in <math>x = 0<math>, which has the form
- <math>\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<math>
(using the limit already mentioned previously), i.e. <math>f^{(n + 1)}(0)=0<math>.
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)