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CTK Exchange
alexb
Charter Member
2508 posts |
May-31-10, 06:15 PM (EST) |
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2. "RE: Vector Sum"
In response to message #0
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Well, it is absolutely obvious if n is even. For n odd, with one vertex on the x-axis, it is as obvious that the y-component of the sum is zero. (Both are obvious in view of the available symmetry.) But now rotate the whole thing by 360/n, moving each vertex to the next. The sum won't change but the new y-component will again be zero. So you have that the projection of the sum on two independent vectors is zero, hence the sum itself is zero.
In any event, think of these vectors as complex numbers:s = 1 + a + a^2 + ... + a^(n-1) a being the primitive root of unity. Then s(a - 1) = a^n - 1 = 0. But a ≠ 1, so s = 0. |
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