Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Questions:

  1. I know you can define a directional derivative on some subset of $\mathbb R^n$, but what can be said about an arbitrary set of points, $S$? What are the most general criteria $S$ must satisfy in order to define a directional derivative at each point in $S$? I suspect $S$ must define a differentiable manifold, but I quickly get lost in the terminology when I read the Wikipedia article on the subject.
  2. What about a generalized notion of a directional derivative? Say a set of points, $S$, must satisfy a set of criteria, $C$, in order to define a directional derivative, $D$, at each point in $S$. Is there any generalized notion of a directional derivative, $D'$, that applies to a set, $S'$, that satisfies a set of criteria, $C'\subset C$, such that $D'$ reduces to $D$ when the remaining criteria, $C-C'$, are applied to $S'$? In other words, is there a generalized directional derivative that the usual directional derivative is a special case of? If so, what criteria must apply to a set of points for there to be a well defined generalized directional derivative?

My understanding is that in many branches of math, there are many equivalent ways of defining things. If there is any chance these questions can be answered with relations between points and neighboring points, I would prefer that.


My math background:

My background in math (aside from "engineering math") is one course that used "baby Spivak" as the text book, and that was two years ago. Some of the intuition stuck, but the most of the rigor and definitions I learned have since faded.

share|improve this question
    
Somewhat related: en.wikipedia.org/wiki/… –  Willie Wong Jan 18 '13 at 9:15
add comment

2 Answers

up vote 1 down vote accepted

An extended comment:

It all really depends on what you want out of the directional derivative. Namely: what are the properties of the directional derivative that you hope to preserve in this generalisation process?

At its rawest, a directional derivative assigns an object $D_vf(p)$ to the triple $(f,p,v)$ where $f$ is a function on some set $S$, $p$ is some element in the set $S$, and $v$ is some concept which we will call a "direction". So all we need to be able to "define" something that "looks like" a directional derivative are the following:

  • The domain $S$ which is a set
  • The codomain $K$ which is a set
  • A family of sets $T_p$ where the parameter $p\in S$, representing the possible directions at each point $p$. We let $T$ denote the disjoint union of all the $T_p$, so that each element of $T$ can be associated with a point $p\in S$ and a direction $v\in T_p$.

Then given $S,K,T_p$, we can say that "a directional derivative" is a mapping $$ K^S\times T \to K $$ sending a function $f: S\to K$ (or, $f\in K^S$) and a vector $(p,v)\in T$ to the "value of the directional derivative" $D_vf(p)$.

The above tells us absolutely nothing about the operation, besides the type of objects it acts on and the type of objects it returns.


Maybe you want the directional derivative to tell us "whether a function increases or decreases or stays constant" in a given direction.

To do so, we need first a notion of "size". That means that the value of the function (its codomain), should admit some sort of order, which we denote by $\leq$.

Secondly, we need a way to associate the directions $T_p$ actually to some notion of "close by points". One way of doing this is to associate to make directions "curves in $S$". On a curve we need to have notions of "relative closeness" to the point $p$. So we model after our intuition of what a curve is and take $(O,\prec)$ a totally ordered set to measure the degree of closeness. And we say identify each directional vector $X\in T$ with a function $X: (O,\prec) \to S$.

Then we can say that $f$ is increasing in the direction of $X$ if there exists some element $o\in O$ such that for all $o' < o$, $f\circ X(o') \geq f(p)$, where $p$ is the base point of the vector $X$.

Note that no notion of differentiable manifold is required here.


What you are asking here is a question similar to those in reverse mathematics. And in these kinds of investigations, you cannot just give the name of the object you want defined, because if you want "precisely the directional derivative on $\mathbb{R}^n$", it is trivially true that you will need precisely all the setup of functions on $\mathbb{R}^n$ and differentiability and what not, since otherwise the object would be different. So to ask a "how to generalize" question, it is of utmost important you state what are the properties you want to keep from the concept. Otherwise the question cannot admit a well formulated precise answer.

share|improve this answer
    
I actually tried to (informally) generalize a directional derivative by using the notion of a direction here: pathintegral.org/blog/?p=582 I did not mention it because I did not want to look like I was just advertising myself or something. I was hoping that with some formal definitions, I would be able to prove it was equivalent to some generalized notion of a derivative known to be equivalent to the standard one in $\mathbb R^n$. Thank you for the thorough answer. I learned something about generalizations too--namely that the $C'$ in my question need not be unique! +1 –  Feynman Jan 18 '13 at 16:03
    
I took a quick glance at your blog. I think the intuition behind what you wrote almost necessarily leads you to either a Riemannian (if you insist on the metric) or smooth (if you don't) manifold. Some additional conditions may be needed, but your post used continuity and "infinitesimal linearity", which would tie into what Ittay Weiss wrote in his answer. –  Willie Wong Jan 21 '13 at 8:59
add comment

The derivative of a function $f$ at a point $p$ is the best linear approximation of the function at that point. Thus, in order to have a notion of derivative in the most general context the very least that you need is a notino of linearity. So, at the point where you want to take the derivative it needs to make sense to speak of a linear mappings. Indeed the context of differentiable manifolds provides just that since at every point in a differentiable manifold there is the tangent space which is a linear space so you can speak of linear transformations.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.