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Thread: Prove that the following is "exact"...

  1. #1
    Junior Member
    Dec 2009

    Prove that the following is "exact"...

    Problem: Let f: \Re \to \Re and \omega = f(||x||) (\sum_{i=1}^n{x_{i}dx_{i}}) \in A^1(\Re^n). Assuming f is continuous, prove that \omega is exact.

    Notation notes:

    A^1(\Re^n) is a vector space denoting the set of 1-forms on \Re^n.

    ||x|| is the length of the vector x, so ||x|| = \sqrt{(x_1)^2+(x_2)^2 + ... + (x_n)^2}.

    So this is how I tried to approach the problem. I need to find a function, say G, such that dG = \omega.

    So \omega = f(\sqrt{(x_1)^2+(x_2)^2 + ... + (x_n)^2})  (x_1dx_1 + x_2dx_2 + ... + x_ndx_n)

    \omega = f(\sqrt{(x_1)^2+(x_2)^2 + ... + (x_n)^2})x_1dx_1 + f(\sqrt{(x_1)^2+(x_2)^2 + ... + (x_n)^2})x_2dx_2 + ... + f(\sqrt{(x_1)^2+(x_2)^2 + ... + (x_n)^2})x_ndx_n

    Then I want to find \frac{\partial G}{\partial x_1}, \frac{\partial G}{\partial x_2}, ..., \frac{\partial G}{\partial x_n}, but I don't know how to take the derivative of f since I don't know what f is - and what I'm taking the derivative with respect to each time is within f.

    You can probably tell I'm not familiar with proofs dealing with sum notation. Any tips?
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  2. #2
    Junior Member
    Dec 2009
    I've been thinking about this, and maybe it would be okay to just integrate all of the last line of \omega and leave it in integral form. Instead of actually trying evaluate the integral, perhaps it's enough to just state that the integral exists (since f is continuous)? Then I could just let G equal that integral and have that complete the proof.

    I don't see it as an ideal solution, but could it be sufficient?
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  3. #3
    Senior Member
    Feb 2010
    Basically, yea. Let g(x) := f(\sqrt{x}) so f(||x||) = g(||x||^2), and let G be a primitive for g. Then {\partial\over\partial x_i} (\frac12 G(||x||^2)) = g(||x||^2)x_i=f(||x||)x_i...
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