# Diffrenciability

• January 20th 2006, 11:34 AM
ThePerfectHacker
Diffrenciability
My Calculus book defined a function $z=f(x,y)$ to be diffrenciable at point $(x_0,y_0)$. If and only if, the increment $f(x_0+\Delta x,y_0+\Delta y)$ can be expressed as:
$f(x_0+\Delta x,y_0+\Delta y)$= $f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y$ $+\epsilon_1 \Delta x+\epsilon_2 \Delta y$
as $\Delta x\rightarrow 0, \Delta y\rightarrow 0$ then $\epsilon_1\rightarrow 0,\epsilon_2\rightarrow 0$.
My problem is that I understand it but I was hoping for a more rigorous definition. Can someone give one in terms of limits?
• January 21st 2006, 04:44 AM
TD!
What do you mean, with a limit? Like how the derivative is defined for just 1 variable?
This definition actually uses limits, since the limit of e1 (c.q. e2) for Δx (c.q. Δy) going to 0 has to be 0. How does this definition "lack rigor"?

You may generalize this definition to functions $f:\mathbb{R}^n \to \mathbb{R}$.

Then the function f which is defined on a neighborhood of $\vec a \in \mathbb{R}^n$
is differentiable in a if there exists a vector $\vec m \in \mathbb{R}^n$ with

$f\left( {\vec a + \vec h} \right) = f\left( {\vec a} \right) + \vec m \cdot \vec h + \varepsilon \left( {\vec h} \right)\left\| {\vec h} \right\|$

and $\mathop {\lim }\limits_{\vec h \to \vec 0} \varepsilon \left( {\vec h} \right) = 0$

Note that this is a multi-dimensional limit and m is exactly the gradient of f in a.

This definition can then also be generalized to functions $\vec F:\mathbb{R}^n \to \mathbb{R}^m$
• January 21st 2006, 09:46 AM
TD!
I'll do that last thing I said, the generalization for $\vec F:\mathbb{R}^n \to \mathbb{R}^m$

I'll give the notation as above as well as in "one limit", they're completely equivalent.

A function $\vec F:\mathbb{R}^n \to \mathbb{R}^m$ defined on a neighborhood of a is said to be differentiable in a if there exists a lineair transformation $\mathcal{L} :\mathbb{R}^n \to \mathbb{R}^m$ with

$\vec F\left( {\vec a + \vec h} \right) = \vec F\left( {\vec a} \right) + \mathcal{L}\left( {\vec h} \right) + \vec E\left( {\vec h} \right)\left\| {\vec h} \right\|$

and $\mathop {\lim }\limits_{\vec h \to \vec 0} \vec E\left( {\vec h} \right) = 0$

or, equivalently, if

$\mathop {\lim }\limits_{\vec h \to \vec 0} \frac{{\vec F\left( {\vec a + \vec h} \right) - \vec F\left( {\vec a} \right) - \mathcal{L}\left( {\vec h} \right)}}
{{\left\| {\vec h} \right\|}}=0$

If that limit exists, we have $\mathcal{L} = D\vec F\left( {\vec a} \right) = \vec F'_{\vec a}$

Perhaps you were looking for something like that? The same thing can be done for the definition of the derivative in my first post.
• January 21st 2006, 02:23 PM
ThePerfectHacker
Can you re-define my post into limits?
• January 21st 2006, 02:31 PM
TD!
Quote:

Originally Posted by ThePerfectHacker
Can you re-define my post into limits?

Your case falls under my last definition, take n = 2 and m = 1.