Definition talk:Differentiable Mapping/Real-Valued Function/Point

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The new definition is different from the old one. The old definition, from Larson, was:

Let $f: \mathbb X \to \R$ be a real-valued function, where $\mathbb X \subseteq \R^n$.

Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \in \mathbb X$.

Let $\Delta f\left({\mathbf x}\right) = f\left({\mathbf x + \Delta \mathbf x}\right) - f \left({\mathbf x}\right)$

where $\Delta \mathbf x = \begin{bmatrix} \Delta x_1 \\ \Delta x_2 \\ \vdots \\ \Delta x_n \end{bmatrix}$.

We say that $f$ is differentiable at $\mathbf x$ iff there exists some $\Delta f\left({\mathbf x}\right)$ such that:

\(\displaystyle \Delta f \left({\mathbf x}\right)\) \(=\) \(\displaystyle \nabla f\left({\mathbf x}\right) \bullet \Delta \mathbf x + \begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \bullet \Delta \mathbf x\)

where $\begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \to \mathbf 0$ as $\Delta \mathbf x \to \mathbf 0$.


The limit being taken is the limit of a neighborhood
$\nabla$ is the gradient operator
$\bullet$ is the dot product
$\mathbf 0$ is the zero vector with $n$ entries
$\varepsilon_i$ is some real number

And Larson brings this example to show how partial derivatives existing is not a sufficient condition for differentiability:

$f:\R^2 \to \R, f(x,y) = \begin{cases} \dfrac {-3xy}{x^2 + y^2} & : (x,y) \ne (0,0) \\ 0 & : (x,y) = (0,0) \end{cases}$

which is not continuous at $(0,0)$ but $f_x(0,0)= f_y(0,0) = 0$

--GFauxPas (talk) 21:06, 30 September 2012 (UTC)

Yes, the current definition is wrong. I made a note of that on the definition page.
Also, isn't the gradient defined iff $f$ is differentiable? Do you know of a source that asserts that $f$ need not necessarily be differentiable for its gradient to exist? I thought that (in at least one definition) the gradient is defined by the equation you mentioned:
$\displaystyle \lim_{\Delta \mathbf x \to \mathbf 0} \frac {\left\vert{f \left({\mathbf x + \Delta \mathbf x}\right) - f \left({\mathbf x}\right) - \nabla f \left({\mathbf x}\right) \cdot \Delta \mathbf x}\right\vert} {\left\Vert{\Delta \mathbf x}\right\Vert} = 0$
--abcxyz (talk) 01:41, 1 October 2012 (UTC)
Ys, I'm in agreement here. I'll put a note on Linus44's page about it. --prime mover (talk) 06:02, 1 October 2012 (UTC)
Generally I think $C^1$ means continuous, with continuous partial derivatives. Which admittedly isn't what I wrote.
It's not satisfactory to change back to the old definition (above), however. It's not possible to use the gradient operator to define differentiablity when $\nabla f$ is undenfined if $f$ is not differentiable. Moreover this definition is asserting the existence of $\Delta f$, which is already defined and trivially exists.
Adding "continuous at $x$" after "real-valued function" will resolve the problem, if there's no disagreement about convention. --Linus44 (talk) 16:03, 1 October 2012 (UTC)
Scratch that; we do need something stronger (Frechet derivative after all) --Linus44 (talk) 20:46, 1 October 2012 (UTC)