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It may sound too basic to even be a question, but I couldn't find a straight answer in Wolfram Alpha, Wolfram Mathworld or Wikipedia. Several other examples of more complicated functions are given.

In Wolfram Mathworld it is written that

A smooth function is a function that has continuous derivatives up to some desired order over some domain. (...) The number of continuous derivatives necessary for a function to be considered smooth depends on the problem at hand, and may vary from two to infinity.

$f(x) = x$ has derivative 1 of the first order and 0 of second order, so I would say based on this it has at least 2 derivatives. I think it also has an infinite number of derivatives which are also 0.

Another page on Wolfram Mathworld says the following:

A $C^{\infty}$ function is a function that is differentiable for all degrees of differentiation. (...) All polynomials are $C^{\infty}$. (...) $C^{\infty}$ functions are also called "smooth" (...).

Since $f(x) = x$ is a polynomial, I'm concluding that the paragraphs above mean it is also smooth.

user985366
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    I also think they are smooth – Socre Aug 24 '15 at 18:39
  • "Smooth" runs the gamut from continuously differentiable to $C^\infty,$ and even that is probably too narrow a spectrum. There is no one fixed definition. In the literature you'll often see things like "for the purposes of this paper, "smooth" will mean ___," where "____" is a precise definition. – zhw. Aug 24 '15 at 18:55
  • @CarstenS The question is in the title: Is the function f(x)=x smooth? – user985366 Aug 24 '15 at 23:21
  • I made an attempt to answer it based on what I found, and I showed my reasoning to provide something to the question. So yes, the question contains an answer, but I still was not sure if this holds up, and I was still interested in further information and others' input. – user985366 Aug 24 '15 at 23:27
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    @zhw.: Is there any definition of "smooth" that does not apply to $f(x)=x$? – celtschk Aug 25 '15 at 06:32
  • I've never seen one. – zhw. Aug 25 '15 at 16:20
  • @celtschk any such definition must be boring and useless... – MichaelChirico Aug 25 '15 at 18:03

3 Answers3

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A function is smooth is it has derivatives of infinite order. $f(x) = x$ is smooth because it has infinitely many derivatives which are all 0, except for the first one. Polynomials are smooth because eventually their derivatives are 0.

Michael Menke
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    Is it agreed upon that "smooth" = $C^\infty$ diffable? – Fraïssé Aug 24 '15 at 18:44
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    Yes, that's the standard definition – Michael Menke Aug 24 '15 at 18:45
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    @MichaelMenke Not so sure that is standard. – zhw. Aug 24 '15 at 18:57
  • I wouldn't be so confused if it was standard ha. You flip a text book on differential geometry, says "smooth =$ C^\infty$ diffable, you flip a text book on wavelet, says smooth = $C^k$ diffable... – Fraïssé Aug 24 '15 at 19:20
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    @IllegalImmigrant yeah technically smooth is $C^\infty$, but many times for all the calculations we only need our function to be $C^1$ or $C^2$ or something.. so we abuse the notation and we call those function "smooth", which basically comes to mean "it's regular enough to employ all the theorems we need with no headaches" – Ant Aug 24 '15 at 21:56
  • I've heard smooth mean that the first derivative is continuous. – bjb568 Aug 25 '15 at 02:31
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    $x\mapsto x$ satisfies every definition you guys have given. – Akiva Weinberger Aug 25 '15 at 02:36
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    What are "derivatives of infinite order"? – JiK Aug 25 '15 at 10:23
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    @JiK: While he surely meant derivatives of arbitrary order, a straightforfard definition of a derivative of infinite order would be $\lim_{n\to\infty}\frac{\mathrm d^nf}{\mathrm dx^n}$ provided that limit exists. For example, the infiniteth derivative of any polynomial would be $0$, and the infinifteth derivative of $\exp(x)$ would be $\exp(x)$. Indeed, since that infiniteth derivative could well be differentiable itself (as the two examples are), one could identify this infinity with $\omega$ and continue with $\frac{\mathrm d^{\omega+1}f}{\mathrm dx^{\omega+1}}$. and so on. – celtschk Aug 25 '15 at 18:59
  • Typo found: is -> if – user3105485 Aug 25 '15 at 19:09
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Yes, the identity function has derivatives of every finite order, and is therefore smooth. It doesn't matter that most of the derivatives are $0$ everywhere -- being $0$ is a perfectly cromulent way to exist.

hmakholm left over Monica
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You may be having issues with the difference between existence and triviality.

If $f(x)=x$ then

$f(x)=x$ is continuouss

$f'(x)=1$ is continuous

$f''(x)=0$ is continuous

$f'''(x)=0$ is continuous

etc...

So all its derivatives are continuous. On the other hand, take $g(x)=x\times|x|$

$g(x)=x\times|x|$ is continuous

$g'(x)=\frac{|x|}2$ is continuous

$g''(x) = \frac12$ if $x>0$, $g''(x)=-\frac12$ if $x<0$ $g''(0)$ is undefined

Clearly $g''$ is not continuous because of $g''(0)$ not existing, and so $g$ only has two derivatives.

Intuitively, all of $f$'s derivatives have no breaks in their graph ($y=0$ is simply a nice line), while $g''$ graph has a gaping hole in it at $x=0$.

BSteinhurst
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Christopher King
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