# Is f(x) = x smooth?

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.

• 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
• @zhw.: Is there any definition of "smooth" that does not apply to $f(x)=x$? – 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.

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

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

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 $x=0$.

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