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Math Help - Implicit Function Theorem for f(x1,...,xm, y1,...,yn) = 0

  1. #1
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    Implicit Function Theorem for f(x1,...,xm, y1,...,yn) = 0

    This is not a problem to be solved, but I could not understand something important from the text. Here it is.

    Let m,n be positive integers, a \in E^m, b \in E^n, and let f_1, f_2, ...,f_n be continuous real-valued functions on an open subset of E^{m+n} that contains the point (a,b) with f_1(a,b) = ... = f_n(a,b) = 0. Suppose that for each i,j=1,...,n

    \frac{\partial f_i(x_1,...,x_m,y_1,...,y_n)}{\partial y_j} exists and is continuous on the given open subset and that the n \times n determinant

    det (\frac{\partial f_i}{\partial y_j}(a,b)) is not zero.

    Then there exists open subsets U \subset E^m, V \subset E^n with a \in U and b \in V such that there is a unique function \varphi :U \rightarrow V such that f_i(x, \varphi (x)) = 0 for each i = 1,...,n

    ....

    There is some line written such that "the continuous function on U \subset E^m whose value at x is det (\frac{\partial f_i}{\partial y_j}(x, \varphi (x))) is not zero at a

    Question

    1. Why does such a continuous function on U whose value at x equals det (\frac{\partial f_i}{\partial y_j}(x, \varphi (x))) ?

    I have no idea why a continuous function has value equals to det of something ?

    2. It says "whose value at x and "not zero at a". What is this? Why it mentioned two points , x and a ?

    If one could explain without using linear algebra, I would be thankful.

    Thank you very much.
    Last edited by armeros; March 9th 2011 at 05:58 AM.
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  2. #2
    Senior Member Tinyboss's Avatar
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    It's just telling you to consider the function that maps x\mapsto\mathrm{det}(\frac{\partial f_i}{\partial y_j}(x,\varphi(x))), which happens to be continuous. That's important because if the determinant is nonzero at a, then it is nonzero in a neighborhood of a, by continuity.
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