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I am trying to show the following:

$$ O(e^{n(\cos n^{-2/5}-1)}) = O(e^{-Cn^{1/5}})$$

The problem I'm having is I'm trying to get a hang of asymptotic notation, and I can't quite figure out how to rigorously (in kinda painful detail) to see how to connect the dots.

I can see that I can expand the inside and write: $\cos n^{-2/5}-1 = -\frac{(n^{-2/5})^2}{2}+O(n^{-8/5})$...and then if I multiply this by $n$ I would get $-\frac{n^{1/5}}{2}+O(n^{-3/5})$.

I don't see how to go from $O(e^{-\frac{n^{1/5}}{2}+O(n^{-3/5})})$ to what I want rigorously, step by step.

Thanks!

MathStudent
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1 Answers1

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Note that $$ O\left(e^{-\frac{n^{1/5}}{2}+O(n^{-3/5})}\right) = O\left(e^{-\frac{n^{1/5}}{2}}\cdot e^{O(n^{-3/5})}\right) =\\ O\left(e^{-\frac{n^{1/5}}{2}}\cdot e^{O(1)}\right) = O\left(e^{-\frac{n^{1/5}}{2}}\cdot O(1)\right) = O\left(e^{-\frac{n^{1/5}}{2}}\right) $$ So, in particular, we would take $C = \frac 12$

Ben Grossmann
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