While reading a paper in physics i came across asymptotic behavior of $\cos(\sqrt{4n+1}x)-\cos(\sqrt{4n+\alpha}x)$ and it was written this is equal to $O(n^{-1/4})$ for any real $\alpha$. I tried to prove this by considering the taylor expansion, but i couldn't get any result.
Any help is appreciated
paper:http://www.sciencedirect.com/science/article/pii/S0375960111002970
Answer
$$\begin{align} \cos{(\sqrt{2 n+1} x)} - \cos{(\sqrt{2 n+\alpha} x)} = -2 \sin{\left ( \frac{\sqrt{2 n+\alpha}-\sqrt{2 n+1}}{2} x \right )} \sin{\left ( \frac{\sqrt{2 n+\alpha}+\sqrt{2 n+1}}{2} x \right )}\\ \sim -\frac{\alpha - 1}{2} x \frac{\sin{(2 \sqrt{n} x)}}{2 \sqrt{n}} & (n \rightarrow \infty) \end{align}$$
No idea where you got the $O(n^{-1/4})$ thing.
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