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

    Notation notes:

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

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

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

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

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

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

    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 $\displaystyle \omega$ 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 $\displaystyle f$ is continuous)? Then I could just let $\displaystyle G$ 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 $\displaystyle g(x) := f(\sqrt{x})$ so $\displaystyle f(||x||) = g(||x||^2)$, and let G be a primitive for g. Then $\displaystyle {\partial\over\partial x_i} (\frac12 G(||x||^2)) = g(||x||^2)x_i=f(||x||)x_i$...
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