# 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

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

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

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

### continuity of f and f(n)

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

${\displaystyle \lim _{x\to 0}x^{m}e^{-1/x^{2}}=0,}$
such that ${\displaystyle \lim _{x\to 0}f^{(n)}(x)=0=f^{(n)}(0)}$, i.e. continuity of ${\displaystyle f^{(n)}}$ also in ${\displaystyle x=0}$.

### proof of the formula for the n-th derivative

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

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

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

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

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

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

## 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.

## correlation with Stone-Weierstrass?

By Stone-Weierstrass, there must be some sequence of polynomials in x^n converging uniformly to the non-analytic smooth function on [-1, 1]. What does this sequence look like? (IT can't approach a Taylor polynomial). I forgot to log in; username nnn9245129.2.56.113 18:04, 16 February 2007 (UTC)

That sequence will look absolutely and completely different than Taylor's polynomial. That two functions are close to each other says nothing about whether their first derivatives are close to each other, same for higher order derivatives. And the Taylor polynomial is about all derivatives. Oleg Alexandrov (talk) 03:22, 17 February 2007 (UTC)

## Requesting a clarification

I arrived here purely by chance and am not a mathematician. The opening sentence says that there are two important types of function. The second sentence states that "One can prove that...". Does this mean that it is possible to prove, or that one of the two function types can be used to prove? I assume it is the former, but the wording is unclear enough that I wish to raise the question. —Preceding unsigned comment added by 87.113.135.124 (talk) 17:38, 30 June 2009 (UTC)

## Doubts about first proof (Which seem to be unjustified)

I have noticed that the proof considers affinely extended real numbers and not real numbers. (I refer to the non-analytic function in the article (the one on top).) exp(-1/x) for x > 0 can lead to strange things when considering just real numbers. (non necessarily correct)

I have been interchanging ${\displaystyle \lim x\downarrow 0}$ and ${\displaystyle 1/\infty }$. There is a subtle difference. This is the difference between arbitrary small and positive (i.e. negligible) and something that is more difficult to imagine.

Why? This is a set-theoretic issue.

Let +0 be 1 / infinity and let +0 be an element of the real set. (Which is not the case......)

It seems that there is a strict lexicographical ordering on the reals. (By quickly looking at it.)

if not a < b and not b < a then a >= b and b >= a meaning that a == b.

I guess that a + (+0) > a. Thus if you are very strict, a + (+0) <> a. In calculus, the == operator is overridden for practical purposes.

In the couter example:

first -1 / x is computed.

Let's take x = +0

We get -1 / x == negative infinity, which is not a real number.

Now suppose that affinely extended real numbers are used. Then exp(negative infinity) = 0, according to mathworld.

Using affinely extended real numbers seems to make things really complicated.

I think that exp(-1 /x) is already non analytic for x = +0.

My understanding of non-analytic smooth functions is that the constants of the power series do not shrink fast enough. As i (term index) gets larger.

As I was told, and what is illustrated on Wikipedia, is that the natrual logarithm seems to diverge in the domain x >= 2. However, this part of the function seems to be smooth. Perhaps the natural logarithm is a better example. — Preceding unsigned comment added by Wouterbu (talkcontribs) 03:45, 30 April 2013 (UTC)

why do you need to consider x=0+ ? the definition only requires to compute exp(-1/x) for x > 0 ; if x > 0 then 1 / x is a real positive number and exp(-1/x) is a real positive number . No need to use affinely extended real numbers anywhere. Mennucc (talk) 09:20, 30 April 2013 (UTC)

Your reasoning is correct. However, as I view it, rounding the infinitesimals is inappropriate here.

Let there be infinitely many reals within the domain [0,+0]. (By the definition of the real set.) It says that x is a positive real number. However, +0 is an element of R^+ (that's what I think).

According to Wikipedia, there is a total ordering on the reals. I find this very difficult, so let me just repeat what I read.

a <= b or b <= a, where a and b are in R Both sides of the disjunction can be true yielding equality.

My guess is that equality means set theoretic equality.

Next we can try this:

Again 0 < x, then according to the axioms there exists an y in between 0 and x. Next there exists an y' in between 0 and y. a y, y' etc. This is a recusion.

What we get is a positive infinitesimal that is > 0. To be precise, by applying axiom 2 an exhaustively you get something like:

x -> x / 2, which is a definition of an infinitesimal. It is still debateble wheter it is allowed to apply the aximom non-finitely many times. Also the axiom does not state that the intermediate value is in the middle.

However, this may just about be the difference between reals and rationals. With rational numbers you are allowed to use finitely many steps, with real numbers, this need not be the case.

Thus, assuming that +0 is a positive real number, (like it is in the IEEE floating point standard), the sign bit is set to 0, indicating positive, then -1/+0 is not a real number (it should give -infinity, which can also be represented by IEEE floating point)

Though using this same IEEE standard, you may get the same solution as described on the page associated to this talk (or NaN). However, I don't think that Taylor's theorem is defined on affinely extended real numbers. — Preceding unsigned comment added by Wouterbu (talkcontribs) 11:03, 30 April 2013 (UTC)

You should read http://en.wikipedia.org/wiki/Archimedean_property where it says "In the axiomatic theory of real numbers, the non-existence of nonzero infinitesimal real numbers is implied by the least upper bound property". Usually in calculus +0 does not exists. +0 exists in "non standard analysis" but that is quite esoteric, never taught in calculus 101 . Moreover IEEE standard is not a good reference to understand how the mathematical real line works. Mennucc (talk) 11:59, 30 April 2013 (UTC)

Actually, +0 is not a positvive number in the IEEE floating point standard. +0 > 0 (0 is either +0 or -0) is false.

Indeed, it is not a good reference anyway. I think that I will read the article about the Archimedian property when I have time.

Every infinitesimal is 0 in span. However, there are different kinds of 0. Not all infinitesimals have the same span. This is because there can be a ratio difference right. dy/dx for example may be unequal to 1. Both dx and dy are 0 spanned infintesimals.

Currently I am under the impression that I do understand. Yesterday I found this:

http://en.wikipedia.org/wiki/Ordinal_number#Ordinals_extend_the_natural_numbers (Again without having read everything.)

Is it a good idea to look at unit infinitesimal as ${\displaystyle {\frac {1}{\omega }}}$? According to another wikipedia page, Leibniz used the definition ${\displaystyle {\frac {1}{\infty }}}$. — Preceding unsigned comment added by 62.131.60.88 (talk) 12:08, 1 May 2013 (UTC)

Having read (part of the article), I am now convinced that my critisicm was wrong. The Archimedean property makes sense. No need to use big infinite like numbers. I have misinterpreted the axiomatization of Tarski. It is only allowed to apply the a rule finitely many times and not infinite. Just as $\infty$ is not a member of the real set, neither is ${\displaystyle {\frac {1}{\infty }}}$. The real set is a field. If ${\displaystyle 1/\infty }$ would be an element of the real set and if ${\displaystyle {\frac {1}{\infty }}\neq 0}$, then ${\displaystyle \infty }$ would also be an element of the real set, thus leading to a contradiction.