Definition talk:Differentiable Mapping

I'm tempted to move some of this into Definition:Derivative so as to match the rest of the exposition of this subject. As I planned it, "Differentiable" was a minor (but important) side-issue, i.e. a condition under which the derivative was valid. Same applies to "analytic". The real meat of the issue is that thing called the Derivative.

I'll get round to it later, I've got a busy week this week, I might not get much done. --Matt Westwood 06:39, 27 January 2009 (UTC)

... okay, I think it still needs work: the definition that I learned was that the domain of $\C$ on which the function is defined needs to be a Definition:Region (in the topological sense of being open). I may also introduce the concept of a neighbourhood. As I say, it's all in the plan - I'll be coming back to this entire area and tightening it up in time. --Matt Westwood 07:46, 27 January 2009 (UTC)

Closed Interval
Where did your definition of differentiability on a closed interval come from? I don't believe it necessarily holds at the end points.

Define $f: \R \to \R$ as:


 * $\forall x \in \R: f(x) = \begin{cases}

x & : x \le 1 \\ 1 & : 1 < x < 2 \\ x - 1 & : x \ge 2\end{cases}$

Unless I'm not mistaken, at the endpoints of $[1..2]$, $f(x)$ has a different derivative depending on whether you come at it from the left or right. As such, that means $f$ does not have a derivative at those points.

Intuitively, you can't find the derivative of a curve where there's a sharp corner in it (like here). So you can't say that $f$ is diff'able at those points. Or am I missing something crucial? --prime mover 14:30, 7 December 2011 (CST)


 * I got it from Larson. Even though $f$ doesn't have a derivative at the end points, it has one sided derivatives. It's the same problem with continuity on a closed interval, no?:


 * Define $g: \R \to \R$ as:


 * $\forall x \in \R: g(x) = \begin{cases}

-\pi & : x \le 1 \\ 1 & : 1 < x < 2 \\ \text{a billion} & : x \ge 2\end{cases}$


 * $g$ is continuous on $[1..2]$, according to the definition here, even though it's not continuous at the endpoints. --GFauxPas 14:43, 7 December 2011 (CST)


 * Continuous yes, but not differentiable. I might be prepared to grant you semi-differentiable, which will need a separate page. Sorry, but I'm prepared to stick my neck out and say I believe Larson is over-simplifying to the point of being incorrect. His definition leads to contradictions.
 * Anyone else care to join in on this one? --prime mover 14:51, 7 December 2011 (CST)
 * It's not my intent to defend nor to attack Larson's views, I'm just trying to fill in what "differentiable on $[a..b]$" means (such as here). I don't have any emotional attachment to the issue. Is the real function: $h(x) = \sqrt{1-x^2}$, $0 \le x \le 1$ differentiable on $[0..1]$? --GFauxPas 15:58, 7 December 2011 (CST)