imported>Tom.Reding: -redundant class param

2025-02-04T13:30:31Z

-redundant class param

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== Untitled ==

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

I believe there are a number of [[mathematical analysis]] ways. If Taylor's theorem is breaking down as an infinite series expansion, something in the remainder term is blowing up. Smooth says the Fourier transfrom drops off at infinity faster than any polynomial - one can ask for more than that. I'm pretty sure there are classes of functions between smooth and analytic that have been studied, as whole scales (are they called quasi-analytic?). Obviously it's very striking how different the zero sets are, since any closed set can be the zero set of a smooth function.

[[User:Charles Matthews|Charles Matthews]] 17:14, 2 Dec 2003 (UTC)

Seems protecting the semi-open intervals from interference from Wiki-syntax busibodies also damages the format.

[[User:Charles Matthews|Charles Matthews]] 10:09, 13 Nov 2004 (UTC)

== moving this to [[differentiable function]] ==

Any objections to moving this to [[differentiable function]]? That is currently a redirect to [[derivative]]. --[[User:MarSch|MarSch]] 30 June 2005 16:07 (UTC)

:Infinitely differentiable is not the same as (one time) differentiable.--[[User:Patrick|Patrick]] June 30, 2005 20:58 (UTC)

::smooth functions are a very special case of [[differentiable function]]s, but there is no such article yet. I think it would be easy to expand this once it is moved there.--[[User:MarSch|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. --[[User:MarSch|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.--[[User:Patrick|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 [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/217.44.113.21|217.44.113.21]] ([[User talk:217.44.113.21|talk]]) 16:53, 9 March 2008 (UTC) 
Currently, "Differentiability Classes" redirects here to Smooth Function, but it should be the other way around. Smooth functions are a special case of differentiability class. It doesn't make sense to redirect the generalization to the most extreme example. — Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/138.210.251.10|138.210.251.10]] ([[User talk:138.210.251.10|talk]]) 06:29, 21 February 2014 (UTC) 

== Mistake about the definition of function ==

It is unfortunately a classical mistake. "f(x)" is no function, the function that would correspond is called "f". Or x|->f(x). I can't give further information because my English isn't good enough to describe mathematical objects. Bête spatio-temporelle [my name]

== Cω ==

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. [[User:Eebster the Great|Eebster the Great]] ([[User talk:Eebster the Great|talk]]) 21:43, 14 March 2009 (UTC)

: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. [[User:Marc van Leeuwen|Marc van Leeuwen]] ([[User talk:Marc van Leeuwen|talk]]) 13:24, 15 April 2011 (UTC)

"The function f is said to be of class Cω, or analytic, if f is smooth and if it equals its Taylor series expansion around any point in its domain." Is it redundant to say f is smooth? Are there functions which equal their Taylor expansions but aren't smooth? If so, perhaps one should be provided. [[User:Erasmuse|Erasmuse]] ([[User talk:Erasmuse|talk]]) 03:37, 16 June 2014 (UTC)
: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. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 10:43, 16 June 2014 (UTC)

== Vector space ==

I was looking at the article to see if <math>C^\infty</math> 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? [[User:StatisticsMan|StatisticsMan]] ([[User talk:StatisticsMan|talk]]) 04:32, 3 February 2010 (UTC)

== Question/Theorem ==

Let g be some function then if a function f is of class '''C''''''r''' and g composed f is of class '''C''''''r''' then g is also of class '''C''''''r'''. 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 [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/128.100.86.53|128.100.86.53]] ([[User talk:128.100.86.53|talk]]) 18:08, 12 February 2010 (UTC) 

:Such things are obviously false (regardless of the order of composition); take ''f'' constant for instance. [[User:Marc van Leeuwen|Marc van Leeuwen]] ([[User talk:Marc van Leeuwen|talk]]) 13:21, 15 April 2011 (UTC)

== Why C^k? ==

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. [[User:Kier07|Kier07]] ([[User talk:Kier07|talk]]) 02:00, 5 March 2010 (UTC)

===Multivariate differentiability classes -- wrong indices ??===
''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 '''R'''''n'' with values in '''R'''''m'', then ''f'' has component functions ''f''1, ..., ''f''''m''. Each of these may or may not have [[partial derivative]]s. We say that ''f'' is of '''class ''Ck''''' if all of the partial derivatives <math>\frac{\partial^l f_i}{\partial x_{1}^{l_1}\partial x_{2}^{l_2}\cdots\partial x_{n}^{l_n}}</math> exist and are continuous, where <math>i</math> is an integer between 1 and ''m'', each of <math> l_1, l_2, \ldots l_n </math> is an integer between 0 and ''k'' and <math>l_1+l_2+\cdots +l_n=l </math>.

''
?''
--[[User:Wohingenau|Wohingenau]] ([[User talk:Wohingenau|talk]]) 13:15, 3 July 2012 (UTC)

Talk:Smooth function - Revision history

imported>Tom.Reding: -redundant class param