User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Characterization of Differentiability
By the way, there seem to be any number of ways to pronounce $\partial$ and $\nabla$, what ways do you hear commonly?


 * Usually, I hear 'del' or 'dau' for $\partial$, and 'nabla' for $\nabla$. I would say the 'informal' definition is not really informal, but rather can be viewed as an approach from a functional analytic perspective (which might be undesirable) and is bound to lead to confusion; therefore, I suggest to go with the second defn. --Lord_Farin 10:24, 1 April 2012 (EDT)

Hey, very interesting! On ProofWiki we have the following definition:


 * $\displaystyle \frac {\mathrm d f}{\mathrm d t} = \sum_{k=1}^n \frac {\partial f} {\partial x_k} \frac {\mathrm d x_k}{\mathrm d t}$

But check out what Larson does as a theorem. (Of course, I am PW-ifying it and expanding it to be more powerful and general than the context he has it in).

Let $f$ as being differentiable WRT $\mathbf x$. Let every $x_i$ represent a differentiable function of $t$.

Then $\exists \Delta f(\mathbf x)$:

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

Let $\Delta t \ne 0$.

Then:

Letting $\begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \to \mathbf 0$ we get $\Delta \mathbf x \to \mathbf 0$ but also $\Delta t \to 0$ because by hypothesis each $x_i$ represents a differentiable function of $t$.

Not 100% sure how Larson knows that $\dfrac {\Delta f(\mathbf x)}{\Delta t} \to \dfrac {\partial f(\mathbf x)}{\partial t}$, though. --GFauxPas 19:19, 1 April 2012 (EDT)

--GFauxPas 19:19, 1 April 2012 (EDT)

What's the best way to incorporate differentiability of functions $\R \to \R^n$ and $\R^n \to \R$? The former can easily be integrated into Definition:Differentiable, but the latter is very different because it's not "differentiable iff the derivative exists". --GFauxPas 20:20, 1 April 2012 (EDT)