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

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 $\map {\Delta f} {\mathbf x} = \map f {\mathbf x + \Delta \mathbf x} - \map f {\mathbf x}$

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 $\map {\Delta f} {\mathbf x}$ such that:

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

Here:


 * 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, \map f {x, y} = \begin{cases}

\dfrac {-3 x y} {x^2 + y^2} & : \tuple {x, y} \ne \tuple {0, 0} \\ 0 & : \tuple {x, y} = \tuple {0, 0} \end{cases}$

which is not continuous at $\tuple {0, 0}$ but $\map {f_x} {0, 0} = \map {f_y} {0, 0} = 0$

--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 $f$ 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:
 * $\ds \lim_{\Delta \mathbf x \to \mathbf 0} \frac {\size {\map f {\mathbf x + \Delta \mathbf x} - \map f {\mathbf x} - \map {\nabla f} {\mathbf x} \cdot \Delta \mathbf x} } {\norm {\Delta \mathbf x} } = 0$
 * --abcxyz (talk) 01:41, 1 October 2012 (UTC)


 * Yes, 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)