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:
| \(\ds \map {\Delta f} {\mathbf x}\) | \(=\) | \(\ds \map {\nabla f} {\mathbf x} \bullet \Delta \mathbf x + \begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \bullet \Delta \mathbf x\) |
where $\begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \to \mathbf 0$ as $\Delta \mathbf x \to \mathbf 0$.
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 if and only if $f$ 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.