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.

  • 4
    $\begingroup$ I also think they are smooth $\endgroup$ – Socre Aug 24 '15 at 18:39
  • $\begingroup$ "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. $\endgroup$ – zhw. Aug 24 '15 at 18:55
  • $\begingroup$ @CarstenS The question is in the title: Is the function f(x)=x smooth? $\endgroup$ – user985366 Aug 24 '15 at 23:21
  • $\begingroup$ 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. $\endgroup$ – user985366 Aug 24 '15 at 23:27
  • 5
    $\begingroup$ @zhw.: Is there any definition of "smooth" that does not apply to $f(x)=x$? $\endgroup$ – celtschk Aug 25 '15 at 6:32

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.

  • 1
    $\begingroup$ Is it agreed upon that "smooth" = $C^\infty$ diffable? $\endgroup$ – Carlos - the Mongoose - Danger Aug 24 '15 at 18:44
  • 2
    $\begingroup$ Yes, that's the standard definition $\endgroup$ – Michael Menke Aug 24 '15 at 18:45
  • 2
    $\begingroup$ @MichaelMenke Not so sure that is standard. $\endgroup$ – zhw. Aug 24 '15 at 18:57
  • 6
    $\begingroup$ @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" $\endgroup$ – Ant Aug 24 '15 at 21:56
  • 5
    $\begingroup$ $x\mapsto x$ satisfies every definition you guys have given. $\endgroup$ – Akiva Weinberger Aug 25 '15 at 2:36

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.

  • 1
    $\begingroup$ I had to look that word up. Between this and popularizing schadenfreude, The Simpsons has really influenced the language, hasn't it? $\endgroup$ – Akiva Weinberger Aug 25 '15 at 2:34
  • 7
    $\begingroup$ @columbus8myhw yes, it's certainly embiggened our vocabulary $\endgroup$ – Gregory J. Puleo Aug 25 '15 at 3:30

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


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 $x=0$.

  • $\begingroup$ Funny definition of a "gaping" hole, that is a nothingth of a unit wide :) $\endgroup$ – thepeer Aug 25 '15 at 15:10
  • $\begingroup$ @thepeer Is a unit more or less than the gap of the grand canyon? $\endgroup$ – PyRulez Aug 25 '15 at 17:21
  • $\begingroup$ Does it matter, seeing as we're talking about a nothingth of it? $\endgroup$ – thepeer Sep 1 '15 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.