# Asymptotic estimate of the form $k=C\log\log n$.

Suppose that we have $$n=k^{\frac{1}{\beta}k^22^{k+12}\log\frac 1{\alpha}}$$ where $\alpha,\beta\in(0,1)$. It is claimed that $$k\sim C\log\log n$$ for some constant $C=C(\alpha,\beta)$. How do one deduce such an estimate?

I am totally lost here, any help is very much appreciated.

I assume logarithms in base $2$. This would not change anything anyway, besides constants.
First, note that $$n=k^{\frac{1}{\beta}k^22^{k+12}\log\frac{1}{\alpha}} = 2^{\frac{1}{\beta}k^22^{k+12}\log\frac{1}{\alpha} \log k} = 2^{\frac{2^{12}\log\frac{1}{\alpha}}{\beta}k^22 ^{k} \log k}$$ so $$\log n = \frac{2^{12}\log\frac{1}{\alpha}}{\beta}\cdot k^22 ^{k} \log k = C'\cdot 2 ^{k + 2\log k+\log\log k}$$ setting $C'\stackrel{\rm def}{=} \frac{2^{12}\log\frac{1}{\alpha}}{\beta}$.
Taking the logarithm again, we get $$\log\log n = k + 2\log k+\log\log k + O(1)$$ and, when $k\to\infty$, the right-hand side satisfies $$k + 2\log k+\log\log k + O(1)\operatorname*{\sim}_{k\to\infty} k$$ which implies $$\log\log n\operatorname*{\sim}_{k\to\infty} k.$$
• Yes you are right, it is actually $\log\frac 1{\alpha}$. – BigbearZzz Feb 23 '17 at 21:25
• So $C(\alpha,\beta)$ is really independent of $\alpha$ and $\beta$? At first I arrive at something very similar but I thought I was wrong since I got $C=1$. – BigbearZzz Feb 23 '17 at 21:29
• Think of it as an approximation of the form $f(x) \sim_{x\to\infty} x$. This is the first term; you can also write it $f(x) = x + o(x)$, or $\frac{f(x)-x}{x} \xrightarrow[x\to\infty]{} 0$. But you good tdo an expansion of that sort to lower order terms: e.g., $f(x) = x+\sqrt{x} + o(\log x)$ (meaning $\frac{f(x)-(x+\sqrt{x})}{\log x}\xrightarrow[x\to\infty]{} 0$ "the function $f$ is well-approximated at $\infty$ by the simpler function $x+\sqrt{x}$, and the remaining term is negligible in front of $\log x$". – Clement C. Feb 23 '17 at 21:39