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Math Help - Show that f is differentiable

  1. #1
    Member thaopanda's Avatar
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    Show that f is differentiable

    Let f : R^3 \rightarrow R^2 be given by f ( x_1, x_2, x_3) = ( x_3, x_1). Show that f is differentiable at X_0 = (0,0,0) and find df( X_0).

    I figured to use the definition so:

    f(X) = f( X_0) + df( X_0)(X- X_0) + o(|x- X_0|) as X \rightarrow X_0

    f( X_0) goes to 0 and X- X_0 is just X
    So I'm left with:
    f(X) = df( X_0)(X) + o(|X|) as X \rightarrow X_0

    But I'm not sure how to find df( X_0)
    Doing so would that f is differentiable at X_0 and find it at the same time, yes?
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by thaopanda View Post
    Let f : R^3 \rightarrow R^2 be given by f ( x_1, x_2, x_3) = ( x_3, x_1). Show that f is differentiable at X_0 = (0,0,0) and find df( X_0).

    I figured to use the definition so:

    f(X) = f( X_0) + df( X_0)(X- X_0) + o(|x- X_0|) as X \rightarrow X_0

    f( X_0) goes to 0 and X- X_0 is just X
    So I'm left with:
    f(X) = df( X_0)(X) + o(|X|) as X \rightarrow X_0

    But I'm not sure how to find df( X_0)
    Doing so would that f is differentiable at X_0 and find it at the same time, yes?
    The correct way to think of differentiation is that it is a linearisation process. In elementary calculus, you have a function f(x), whose graph is some sort of curve, and when you differentiate it you approximate the curve by a straight line, namely the tangent to the curve.

    For a function f:\mathbb{R}^3\to\mathbb{R}^2, when you differentiate it you are approximating f by a linear map from \mathbb{R}^3 to \mathbb{R}^2. But linear maps between vector spaces are described by matrices. For a linear map from \mathbb{R}^3 to \mathbb{R}^2, you write the elements of \mathbb{R}^3 and \mathbb{R}^2 as column vectors, and the linear map is given by a 2\times3 matrix. The formula for the derivative matrix Df(X_0) is that it satisfies the equation f(X) = f(X_0) + Df(X_0)(X-X_0) + o(|X-X_0|).

    For this example, f takes the vector X = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\in\mathb  b{R}^3 to the vector f(X) = \begin{bmatrix}x_3\\x_1\end{bmatrix}\in\mathbb{R}^  2. You want to find Df(0), which will be a 2\times3 matrix such that \begin{bmatrix}x_3\\x_1\end{bmatrix} = Df(0)\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} + o(|X|). As it happens, this function f is already linear, so the o(|X|) term disappears, and \begin{bmatrix}x_3\\x_1\end{bmatrix} = \begin{bmatrix}0&0&1\\1&0&0\end{bmatrix}\begin{bma  trix}x_1\\x_2\\x_3\end{bmatrix}. So Df(0) = \begin{bmatrix}0&0&1\\1&0&0\end{bmatrix}.

    As you already said, this process of finding Df(X_0) automatically includes a proof that f is differentiable at X_0.
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  3. #3
    Member thaopanda's Avatar
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    Thank you very much! That made it very clear for me finally
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