Talk:Smooth function Source: en.wikipedia.org/wiki/Talk:Smooth_function

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Can you quantify or make mathematically precise what you mean by a "large gap"? 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.

Charles Matthews 17:14, 2 Dec 2003 (UTC)

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

Charles Matthews 10:09, 13 Nov 2004 (UTC)

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)[reply]

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 unsigned comment added by 138.210.251.10 (talk) 06:29, 21 February 2014 (UTC)[reply]

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]

I don't think C^ω is actually defined anywhere in this article, which is a little ridiculous considering how often it is mentioned. As far as I can tell, C^k is only explicitly defined in the article for non-negative integers. Eebster the Great (talk) 21:43, 14 March 2009 (UTC)[reply]

This seems to have been resolved since, but I find the passages that say "C^ω therefore analytic" or "analytic therefore C^ω" rather silly, given that they are synonyms. Also I think it should be pointed out more clearly that in spite of appearance, C^ω is not a smoothness condition. Marc van Leeuwen (talk) 13:24, 15 April 2011 (UTC)[reply]

"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. Erasmuse (talk) 03:37, 16 June 2014 (UTC)[reply]

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)[reply]

I was looking at the article to see if ${\displaystyle C^{\infty }}$ 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)[reply]

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 unsigned comment added by 128.100.86.53 (talk) 18:08, 12 February 2010 (UTC)[reply]

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)[reply]

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)[reply]

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 R^m, then f has component functions f₁, ..., f_m. Each of these may or may not have partial derivatives. We say that f is of class C^k if all of the partial derivatives ${\displaystyle {\frac {\partial ^{l}f_{i}}{\partial x_{1}^{l_{1}}\partial x_{2}^{l_{2}}\cdots \partial x_{n}^{l_{n}}}}}$ exist and are continuous, where ${\displaystyle i}$ is an integer between 1 and m, each of ${\displaystyle l_{1},l_{2},\ldots l_{n}}$ is an integer between 0 and k and ${\displaystyle l_{1}+l_{2}+\cdots +l_{n}=l}$.

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