let [tex]f: R^n \rightarrow R^n [/tex] be differentiable, with [tex]\sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} = 0 [/tex]. show that [tex]\omega = \sum_{i=1}^n f_i dx_1 \Lambda ... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n [/tex] is exact(adsbygoogle = window.adsbygoogle || []).push({});

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here's what i got so far:

[tex]\omega[/tex] is a n-1 form, since the "^" on the [tex]dx_i[/tex] indicates that this term is omitted. f is defined on [tex]R^n[/tex] which is an open, star-shaped set. because of this, we can use poincare's lemma, which states that on an open starshaped set, every closed form is exact. therefore we only need to show that [tex]\omega [/tex] is closed. that is, dw = 0.

when i write out dw, i get

[tex]dw = \sum_{i=1}^n \sum_{\alpha = 1}^n \frac{ \partial{f_i}}{\partial{x_\alpha}} dx_\alpha \Lambda dx_1 \Lambda... \Lambda \hat{dx_i}\Lambda ... \Lamda dx_n [/tex]

how do i simplify this? how do i make this equal zero? if [tex]\sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} = 0 [/tex], does this mean that the expression i wrote for dw equals zero? i don't really see the connection here, since [tex]\sum_{i=1}^n (-1)^i \frac{ \partial{f_i}}{\partial{x_i}} [/tex] has an extra factor of [tex](-1)^i[/tex]...i'm not sure about the subscripts either. i hope someone can help me with this.

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# Homework Help: Showing a form is exact

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