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Math Help - Diffrenciability

  1. #1
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    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?
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  2. #2
    TD!
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    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
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  3. #3
    TD!
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    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)}}<br />
{{\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.
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  4. #4
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    Can you re-define my post into limits?
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  5. #5
    TD!
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    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.
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