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I am trying to find an estimate in terms of $\alpha\in(0,1)$ of how large $n$ should be so that the below inequality is satisfied for all large $n$. $$ n^{\alpha}\ge \ln(n^2)$$ Is there a good way to go about this?

GA-Student
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  • It depends what you mean here. Any $\alpha > 0$ will work for sufficiently large $n$. Are you trying to estimate what $n$ is needed to solve this? If so, the problem is nonlinear and likely requires a root-finding algorithm. – Gregory Mar 20 '21 at 02:48
  • As an aside, you can show that there is an $\alpha_c$ such that $\alpha \ge \alpha_c$ this inequality is satisfied for all $n > 0$. Perhaps you can build some asymptotic expression off of this. – Gregory Mar 20 '21 at 02:51
  • Yes, I would like to have some analytical lower bound on $n$ (which clearly is dependent on $\alpha$). The $\alpha$ here is fixed. – GA-Student Mar 20 '21 at 02:51
  • What makes you think an analytical solution exists? I doubt (aside from maybe special functions) if this can be given a nice closed form solution. – Gregory Mar 20 '21 at 02:52
  • I am not sure unfortunately. I was hoping there might be which is why I posted the question. May I ask if you meant we cannot get an analytical lower bound on $n$? – GA-Student Mar 20 '21 at 02:54
  • Nonlinear functions in general require root-finding methods. You are essentially interested in finding when $n^\alpha = 2 \ln n$ (the last possible crossing in fact). Take a step back, how would you solve $e^x = x$? Is there an analytical solution for this? – Gregory Mar 20 '21 at 02:57
  • No, there isn't. :\ – GA-Student Mar 20 '21 at 03:03

2 Answers2

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For the equality$$n^{\alpha}= \log(n^2)$$ let $x=n^2$ and solve $$x^{\frac \alpha 2} =\log(x)\implies x=\Bigg[-\frac{2 }{\alpha }W\left(-\frac{\alpha }{2}\right)\Bigg]^\frac 2 \alpha\implies\color{blue}{n_*=\Bigg[-\frac{2 }{\alpha }W\left(-\frac{\alpha }{2}\right)\Bigg]^\frac 1 \alpha}$$ where $W(.)$ is Lambert function.

$$n^{\alpha}\ge \log(n^2) \qquad \text{if} \qquad n\le n_*$$

Claude Leibovici
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One approach is graphically. $\ln x$ is negative for $x < 1$, and climbs slowly after that. A power function $x^\alpha$ passes through the origin for $\alpha > 0$ and rises fairly rapidly. For example, if $\alpha = \frac{1}{2}$, then I don't believe the curves ever intersect. You might use desmos.com to let $\alpha$ vary.

Another approach is calculating specific examples. If $n = 1, \ln n = 0$ and $n^\alpha = 1$ for any \alpha. If $n = 2$, then $2^\alpha \geq \ln 2$ can be solved for \alpha. I expect any intersections will happen at low values of $n$.

RobertTheTutor
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  • Actually, the inequality has to be satisfied for all large $n$ and $\alpha$ is an already given fixed positive real. – GA-Student Mar 20 '21 at 02:48