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How can you find the minimum of $\sqrt{2x} - \left\lceil \frac{\sqrt{1+8x}}{2} \right\rceil$ for positive integer values of $x$?

TMM
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1 Answers1

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Note that as $x \to \infty$, the function approaches 0 from below. You have points of non-differentiability at values with $$\sqrt{1+8x}/2 \in \mathbb{Z} \leftrightarrow x = (4n^2 - 1)/8 \; \exists n,$$ but at other points the function increases; you check this by differentiation. Also the sequence $\sqrt{(4n^2-1)/4} - n$ is increasing for $n \ge 1$, so the values at these non-differentiable points are increasing. To find the minimum, you should therefore need only check the first few positive integer values.

Mr. Chip
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  • That is perfectly making sense. Interestingsly enough, WolframAlpha does not find a global minimum for "Minimize[{(2x)^0.5 - Ceiling[((1+8x)^0.5/2)], x>0},x]". – Axel Kemper Apr 01 '13 at 22:09