0

As titled, I want to prove (or disprove) $\log_{\cos(\frac{\pi}{n})} (\frac{1}{e})= \Theta(n^2)$. It seems to be true by plotting this function.

However, I do not know how to make use of the $\cos(\frac{\pi}{n})$ to show the function is indeed $O(n^2)$ and/or $\Omega(n^2)$.

William Zheng
  • 428

2 Answers2

2

By a change of base, $$\log_{\cos(\pi/n)}(1/e) = \frac{\ln(1/e)}{\ln\cos(\pi/n)} = - \frac{1}{\ln \cos (\pi/n)}.$$ Our goal is to show that there exist constants $0<a<b$ such that the following holds for all large $n$: $$an^2 \le - \frac{1}{\ln \cos(\pi/n)} \le bn^2.$$

The third-order Taylor polynomial of $f(x) = -\ln \cos x$ is $p_3(x) = \frac{x^2}{2}$, so for large $n$ we have $$\left|-\ln \cos(\pi/n) - \frac{\pi^2}{2n^2}\right| \le \frac{1}{n^3},$$ which yields $$\frac{\pi^2}{4n^2}\le \frac{\pi^2}{2n^2} - \frac{1}{n^3} \le -\ln \cos (\pi/n) \le \frac{\pi^2}{2n^2} + \frac{1}{n^3} \le \frac{\pi^2}{n^2}$$ for all sufficiently large $n$.

angryavian
  • 89,882
1

You just need to apply logarithm properties. Indeed $$ \log_{\cos(\pi/n)}\bigl(e^{-1}\bigr)=\frac{\ln\bigl(e^{-1}\bigr)}{\ln \cos(\pi/n)}=\frac{-1}{\ln \cos(\pi/n)}. $$ Now we can work out the asymptotic behavior as $n\to\infty$. We have $$ \cos\Bigl(\frac{\pi}{n}\Bigr)=1-\frac{\pi^2}{2n^2}+\frac{\pi^4}{24n^4} + O\Bigl(\frac{1}{n^6}\Bigr) $$ that yields \begin{align} \log_{\cos(\pi/n)}\bigl(e^{-1}\bigr) &= \frac{-1}{\ln\Bigl(1-\frac{\pi^2}{2n^2}+\frac{\pi^4}{24n^4} + O\bigl(\frac{1}{n^6}\bigr)\Bigr)} \\ &= \frac{-1}{-\frac{\pi^2}{2n^2}+\frac{\pi^4}{24n^4} - \frac{1}{2} \bigl(-\frac{\pi^2}{2n^2} + O\bigl(\frac{1}{n^4}\bigr)\bigr)^2+ O\bigl(\frac{1}{n^6}\bigr)} \\ &= \frac{-1}{-\frac{\pi^2}{2n^2}-\frac{\pi^4}{12n^4} + O\bigl(\frac{1}{n^6}\bigr)} \\ &= \frac{2n^2}{\pi^2}\frac{1}{1+\frac{\pi^2}{6n^2}+O\bigl(\frac{1}{n^4}\bigr)} \\ &=\frac{2n^2}{\pi^2}\Bigl[1-\frac{\pi^2}{6n^2}+O\Bigl(\frac{1}{n^4}\Bigr)\Bigr] \\ &=\frac{2n^2}{\pi^2}-\frac{1}{3}+O\Bigl(\frac{1}{n^2}\Bigr). \end{align}

dvdgrgrtt
  • 296