It's evident from the graph but I'm not sure how to show this mathematically. This dispersion relation is supposed to be roughly parabolic
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1What? $$;E(0)=2-4\cos (0)=2-4=-2;,;;E\left(\frac\pi a\right)=2-4\cos\pi=2+4=6$$ The two above are numbers, not parabolas... – DonAntonio Feb 20 '14 at 21:22
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It is parabola shaped at the min and max, but the equation gives numbers the same as you got DonAntonio which is why I was confused, I don't know how to say it's a parabola http://www.wolframalpha.com/input/?i=y%3D+2-+4cos%284k%29+from+-2pi%2F4+to+2pi%2F4 – Kyrios Feb 20 '14 at 22:00
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Almost any minimum or maximum will be approximately parabolic. You can calculate the Taylor series at $(x_0,y_0)$ and get $y \approx y_0+y'(x_0)(x-x_0)+y''(x_0)(x-x_0)^2\dots$ Since you are at a minimum or maximum, you have $y'(x_0)=0$ so that term goes away. The higher order terms are small when you are near the minimum or maximum. In your case of $y=2-4 \cos (4x)$, one minimum is $(0,-2)$ and the Taylor series is $y \approx -2+32x^2-\frac {128}3x^4+\dots$ For $|x| \ll 1$ you can ignore the quartic and higher terms.
Ross Millikan
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