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)

Several Variables
I need tof ind a better name for functions of the form $f: \R^n \to \R$. Larson calls them "functions of several variables" but in his book there's less need for precision than PW because the context is clear.

Larson's Def'n:
 * $f$ be a "real function of several variables?", i.e., the domain is a subset of $\R^n$ and the codomain is a subset of $\R$.


 * $f$ is said to be differentiable at some $\mathbf x_0$ iff $\exists \Delta f (\mathbf x)$:

such that $\varepsilon_1, \varepsilon_2, \cdots, \varepsilon_n \to 0$ as $\Delta x_1, \Delta x_2, \cdots, \Delta x_n \to 0$ all at the same time. Figure out the best way to present this clause as to make sure the reader understands they all have to approach at the same time.

Perhaps I should stipulate first that the partials need all exist, or else there's nothing to talk about...

Explanation, going either before the def'n, after the def'n, or maybe even on a seperate page.

To understand the motivation for this, consider a differentiable real function $y = f(x)$.

That is, iff the real function is differentiable, $\varepsilon \to 0$ as $\Delta x \to 0$.

--GFauxPas 08:51, 1 April 2012 (EDT)


 * I would say that, rather than writing '$\Delta x_1, \Delta x_2, \cdots, \Delta x_n \to 0$ all at the same time', it would be a good idea to write $\Delta \mathbf x \to 0$ and using a definition or reference to the limit of a vector-valued function. This allows to more closely resemble the one-dimensional case. Maybe even use $\underline \varepsilon$ as a vector, and write $\underline \varepsilon \cdot \mathbf x$ instead of the sum $\sum_{i=1}^n \varepsilon_i \Delta x_i$. Definition of the gradient of $f$ (vector of partial derivatives) can help to further reduce the summations. --Lord_Farin 09:32, 1 April 2012 (EDT)


 * $f: \mathbf x \mapsto f(\mathbf x)$ is differentiable iff $\exists \Delta f(\mathbf x)$ such that:

?

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

Also, for the definition of gradient, would this definition be too informal?

? Rather than

would that be better?--GFauxPas 09:59, 1 April 2012 (EDT)


 * 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)