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Is it possible to find exactly (not numerically) the minimum of the function $$\sqrt{15-12\cos(x)}+\sqrt{4-2\sqrt{3}\sin(x)}+\sqrt{7-4\sqrt{3}\sin(x)}+\sqrt{10-4\sqrt{3}\sin(x)-6\cos(x)} $$ on the interval $[0,2\pi)$ ? Maple 18 and Mathematica 10 fail with it.

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  • $\begingroup$ The derivative of the function cancels at $x=\pi/3$ and the minimum is exactly $6$. The big question is how to prove it. $\endgroup$ Oct 26, 2014 at 10:23
  • $\begingroup$ @Claude Leibovici: Thank you for your interest to the question. $\endgroup$ Oct 26, 2014 at 10:26

1 Answer 1

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http://www.wolframalpha.com/input/?i=sqrt%2815-12*cos%28x%29%29%2Bsqrt%284-2*sqrt%283%29*sin%28x%29%29%2Bsqrt%287-4*sqrt%283%29*sin%28x%29%29%2Bsqrt%2810-4*sqrt%283%29*sin%28x%29-6*cos%28x%29%29
seems that minimum is 6 enter image description here

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  • $\begingroup$ How is this a nonnumerical answer? $\endgroup$ Oct 26, 2014 at 10:21
  • $\begingroup$ @Hagen von Eitzen , I think it can help someone in getting nonnumerical answer $\endgroup$
    – Antony
    Oct 26, 2014 at 10:23
  • $\begingroup$ Numerically, the minimum is attained at approximately $1.5599$ (which is suspiciously close to, but clearly diffrent from $\frac\pi 2$) with a value of about $6.6248$. $\endgroup$ Oct 26, 2014 at 10:24
  • $\begingroup$ Oh, dammit. Forget my comments - my PARI/GP converted to power series too early. $\endgroup$ Oct 26, 2014 at 10:25
  • $\begingroup$ Anwer is 6 ($x=pi/3$) $\endgroup$
    – Antony
    Oct 26, 2014 at 10:27

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