Talk:Smooth function

message: Can you quantify or make mathematically precise what you mean by a "large gap"? Phys 15:48, 2 Dec 2003 (UTC)

message: Any objections to moving this to differentiable function? That is currently a redirect to derivative. --MarSch 30 June 2005 16:07 (UTC)
:Infinitely differentiable is not the same as (one time) differentiable.--Patrick June 30, 2005 20:58 (UTC)
::smooth functions are a very special case of differentiable functions, but there is no such article yet. I think it would be easy to expand this once it is moved there.--MarSch 1 July 2005 10:09 (UTC)
Differentiabillity class could also be merged there. In the intro here C^1 and all that is explained so I think it would be a natural move. --MarSch 1 July 2005 10:12 (UTC)
:Yes, I think merging is good. However, differentiable function sounds like there is one type of differentiability, so perhaps differentiability or differentiability of functions or the existing name differentiability class is best.--Patrick 1 July 2005 14:38 (UTC)
:There should be a separate article "Differentiability Classes" and all the stuff explained in the intro about C^1 etc should go there, since to define a smooth function you would have to define differentiability classes first; the latter are more fundamental. --Stephen 9 March 2008 16:52 British Time  —Preceding unsigned comment added by 217.44.113.21 (talk) 16:53, 9 March 2008 (UTC)
thread_title:  moving this to [[differentiable function]] 

message: I don't think Cω is actually defined anywhere in this article, which is a little ridiculous considering how often it is mentioned. As far as I can tell, Ck is only explicitly defined in the article for non-negative integers. Eebster the Great (talk) 21:43, 14 March 2009 (UTC)
replies: text: This seems to have been resolved since, but I find the passages that say "Cω therefore analytic" or "analytic therefore Cω" rather silly, given that they are synonyms. Also I think it should be pointed out more clearly that in spite of appearance, Cω is not a smoothness condition. Marc van Leeuwen (talk) 13:24, 15 April 2011 (UTC)

text: The definition of Taylor series needs derivatives of every order. Therefore one may talk of Taylor series only for smooth functions. In other words, a function is smooth if and only if it has a Taylor series. D.Lazard (talk) 10:43, 16 June 2014 (UTC)
thread_title:  C<sup>ω</sup> 

message: I was looking at the article to see if C^\infty functions form a vector space.  And, this article told me they form a Frechet space, which means nothing to me.  Of course, I can click the link and find out that it is a topological vector space.  Wouldn't it make sense to specifically say the space is a vector space?  StatisticsMan (talk) 04:32, 3 February 2010 (UTC)
thread_title:  Vector space 

message: Let g be some function then if a function f is of class Cr and g composed f is of class Cr then g is also of class Cr. Is this true and if this is true should not a theorem be in the article stating this and for which a proof should be provided.  —Preceding unsigned comment added by 128.100.86.53 (talk) 18:08, 12 February 2010 (UTC)
replies: text: Such things are obviously false (regardless of the order of composition); take f constant for instance. Marc van Leeuwen (talk) 13:21, 15 April 2011 (UTC)
thread_title:  Question/Theorem 

message: This may be an easy question to answer (or maybe not), but why all the interest in the class of C^k functions?  Why not just k-times-differentiable functions?  I understand that there are functions, such as x^2*sin(1/x), which are differentiable but not C^1, but... so what?  What parts of the theory become cleaner by excluding such examples?  (I guess theorems like Clairaut's theorem would be one example.)  Anyhow, I think the article should definitely say why it became the standard thing to insist that a function is not just differentiable k times, but continuously so. Kier07 (talk) 02:00, 5 March 2010 (UTC)
thread_title:  Why C^k? 

message: Shouldn't be the paragraph be like this:
Let n and m be some positive integers.  If f is a function from an open subset of Rn with values in Rm, then f has component functions f1, ..., fm.  Each of these may or may not have partial derivatives.  We say that f is of class Ck if all of the partial derivatives \frac{\partial^l f_i}{\partial x_{1}^{l_1}\partial x_{2}^{l_2}\cdots\partial x_{n}^{l_n}} exist and are continuous, where i is an integer between 1 and m, each of  l_1, l_2, \ldots l_n  is an integer between 0 and k and  l_1+l_2+\cdots +l_n=l .
''
?''
--Wohingenau (talk) 13:15, 3 July 2012 (UTC)
thread_title: Multivariate differentiability classes -- wrong indices ??