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Precalculus Mathematics Homework Help
Geometric Proof of Vector Triple Product: Find Coefficients b and c
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[QUOTE="Harrisonized, post: 3632687, member: 346109"] Okay. So it turns out that I took a wrong step here: [A×(B×C)].[proj A]=0 Instead of using proj A, I can just use A. [A×(B×C)].A=0 [bB+cC].A=0 b(B.A)+c(C.A)=0 Therefore, the constants b and c are m(C.A) and -m(B.A), respectively, and it turns out that m=1 (by any vector we choose to test what k is). ----- Since [A×(B×C)]=(C.A)B-(B.A)C, I obtain the following relation: ∇×∇×C = ∇(∇.C)-(∇.∇)C = ∇(∇.C)-∇.(∇C) (This isn't really precalc anymore.) Hence, the curl of the curl of a function exists in ℝ[sup]n[/sup], since the gradient is defined as ∇ = (d/dx[sub]1[/sub], d/dx[sub]2[/sub], ... , d/dx[sub]n[/sub]) and the divergence is defined as the dot product of ∇ and a vector of the same size. As you can see, one question leads right into another. This still isn't a coursework question (or even remotely related to anything I'm doing in class), but that doesn't mean it's inferior to coursework questions. I hope you people of PF can actually try (gasp!) -responding- this time instead of completely ignoring my threads. :( [h2]Homework Statement [/h2] Why is the curl of the curl of a function C in ℝ[sup]n[/sup] defined under the pre-existing definition of the gradient, while the curl of C is only defined when the function is in ℝ[sup]3[/sup]? In symbols, let the curl of the curl of C be written as: (∇×)[sup]2[/sup]C How come (∇×)[sup]2n[/sup]C (where n∈[B]Z[/B]) is defined in ℝ[sup]n[/sup], but not (∇×)[sup]2n+1[/sup]C? Does there exist a process ((∇×)[sup]2[/sup])[sup]1/2[/sup] such that after two iterations on C, ∇(∇.C)-∇.(∇C) is obtained? That's the equivalent of asking if there exist operations (∇×) and (∇×)[sup]-1[/sup]. If so, how would I go about finding out what they are, and are they generalizable to ℝ[sup]n[/sup] such that the curl is defined for ℝ[sup]n[/sup]? [h2]Homework Equations[/h2] I want to avoid all tensor use. [h2]The Attempt at a Solution[/h2] Let C = (x[sub]1[/sub], x[sub]2[/sub], x[sub]3[/sub]) Let ∇×C in ℝ[sup]3[/sup] be (v[sub]1[/sub], v[sub]2[/sub], v[sub]3[/sub]), where v[sub]1[/sub] = (d/dx[sub]2[/sub])x[sub]3[/sub]-(d/dx[sub]3[/sub])x[sub]2[/sub] v[sub]2[/sub] = (d/dx[sub]3[/sub])x[sub]1[/sub]-(d/dx[sub]1[/sub])x[sub]3[/sub] v[sub]3[/sub] = (d/dx[sub]1[/sub])x[sub]2[/sub]-(d/dx[sub]2[/sub])x[sub]1[/sub] Then it makes sense to generalize the curl by: ∇×C = (v[sub]1[/sub], v[sub]2[/sub], ... , v[sub]n[/sub]) v[sub]1[/sub] = (d/dx[sub]2[/sub])x[sub]3[/sub]-(d/dx[sub]3[/sub])x[sub]2[/sub] v[sub]2[/sub] = (d/dx[sub]3[/sub])x[sub]4[/sub]-(d/dx[sub]4[/sub])x[sub]3[/sub] . . . v[sub]n-1[/sub] = (d/dx[sub]n[/sub])x[sub]1[/sub]-(d/dx[sub]1[/sub])x[sub]n[/sub] v[sub]n[/sub] = (d/dx[sub]1[/sub])x[sub]2[/sub]-(d/dx[sub]2[/sub])x[sub]1[/sub] Is there a way to write this using ∇ and (∇.) only? Since this is a vector function that is arrived at from a vector function, both operations must be used at least once. The divergence produces a scalar, and the gradient produces a vector. Therefore, C = ∇F for some function F. But it's known that ∇×(∇F) = 0. So there's obviously a contradiction. -Is super confused- [/QUOTE]
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Geometric Proof of Vector Triple Product: Find Coefficients b and c
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