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)
 * A slightly broader perspective has led me to think of differentiability as a local property. As such, we need open neighbourhoods to work with. Therefore it may be (in a formal sense) void to talk about differentiability without referring to the topology. Paradoxically, I think it might be the case that a function is diffable on $[0..1]$ despite not being so on $[0..1+\epsilon)$ for any $\epsilon >0$. But I may be stretching it. --Lord_Farin 16:25, 7 December 2011 (CST)
 * From the point of view of the original source work, from which Primitives which Differ by a Constant originally came, "differentiable on $[a..b]$" wasn't used. That latter came with a later edit. I have put that proof back to where it originally was. As it stands we have a definition for differentiability on an open interval which is logically consistent, and a definition for diff'ability on a closed one, which can only be applied in one direction. All my experience in real analysis has given me this "common sense" approach to work with.
 * Clearly there is a wider context in which diffability on a closed interval makes sense - but on the real number line (which is where we are limited to until we properly develop the mathematics behind these more advanced, more abstract concepts) we can only have the definitions as they apply on that line.
 * Larson may well have a specific reason for needing to define the derivative on the endpoint of an interval - but on the other hand he may well just be loose. Without seeing it and studying where he's going with it I can't tell.
 * LF: feel free to add an analysis of the paradoxical result above.
 * As for $h(x) = \sqrt{1-x^2}$, it's diffable at $0$ but not at $1$ because it's not defined for $x > 1$.--prime mover 16:51, 7 December 2011 (CST)

I am going to sleep first, but in the mean time think of it in a similar manner as continuity on a closed interval, cf. the function $g$ above for an example with continuity for what might seem paradoxical. --Lord_Farin 16:55, 7 December 2011 (CST)

Problem...
1. I agree with considering vector-valued functions as ordered tuples.

2. I was going to add differentiability $\R^n \to \R^m$ here so I can define the Jacobian, and then do differentials much more simply (in the standard way) in this case. There's a slight problem that should be resolved first -- the operator $\nabla$ is used to define differentiability, but $\nabla f$ isn't defined in the first place until we know $f$ is differentiable. The definition, in effect, should be asserting the existence of $\nabla f$, not of $\Delta f(\mathbf x)$. If there's no objection I'll reorder this to avoid circularity, and change it in accordance with {improve} thingy.