Landau notation on $f(n) = \frac{1}{4n \tan(\frac{\pi}{n})}$

Question

Does someone would have time to show me how to use the Landau notation "Big O"? A useful example could be on $f(n) = \frac{1}{4n \tan(\frac{\pi}{n})}$.

Answer 1

$f(x)=O(x)$ as e.g. $x\to \infty$ means that there exists a constant $M>0$ such that $|f(x)|\leq Mx $ for sufficiently large $x$.

Lets analyze your example. First note that the powerseries of cotangent is $\cot x= \frac{1}{\tan x}= \frac{1}{x}-\frac{1}{3}x -\dots$ where the rest of the terms involve powers $x^m, m\geq 3$. Therefore, for $x$ close to zero these terms are very small comparing to $x$, so you can write $\cot x= \frac{1}{x}- \frac{1}{3}x + O(x^3)$. This means that there exists a constant $M>0$ such that $$|\cot x-(\frac{1}{x}- \frac{1}{3}x)| \leq M|x|^3$$ for sufficiently small $x$.

In your case $f(n)=\frac{1}{4n\tan(\pi/n)}=\frac{1}{4n}\cot(\pi/n)= \frac{1}{4\pi}- \frac{\pi}{12n^2}+O(1/n^4)$ where we have an extra power in the $O$ coming from the term $1/4n$.

Answer 2

One has, by the Taylor series expansion, as $x \to 0$, $$ \frac{1}{1+x}=1+O(x),\qquad \tan x=x+O(x^3),\tag1 $$ then $$ \frac{x}{\tan x}=\frac{x}{x+O(x^3)}=\frac1{1+O(x^2)}=1+O(x^2)\tag2 $$ Hence, as $n \to \infty$, we have $\dfrac{\pi}n \to 0$, and $$ \begin{align} f(n) := \frac{1}{4n \tan(\frac{\pi}{n})} =\frac1{4\pi}\cdot \frac{\frac{\pi}{n}}{\tan(\frac{\pi}{n})} =\frac1{4\pi}\cdot \left( 1+O\left(\frac1{n^2}\right) \right) =\frac1{4\pi}+O\left(\frac1{n^2}\right) \tag3 \end{align} $$ as wanted.