Dividing power series (2): More complicated examples

This thread will include more intricate long division problems. We start by deriving the power series for $\tan(x)$ and $\cot(x)$. It is easy to note that $\tan(x)$ and $\cot(x)$ can be obtained by using their very definition. $\tan(x)=\dfrac{\sin(x)}{\cos(x)}$ and $\cot(x)=\dfrac{\cos(x)}{\sin(x)}$. We first expand in series and then divide the two power series using the scheme of long division. This method, while simple, do not reveal the relationship between these two functions' coefficients and the Bernoulli numbers. We will illuminate this point in another thread where we seek to obtain the power series for the generating function of Bernoulli numbers. For now, we contend to just obtain the power series for $\tan(x)$ and $\cot(x)$.

Series representation of $\tan(x)$

Thus $\tan(x)=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}+\dfrac{62x^9}{2835}+\dfrac{1382x^{11}}{155925}...$

Series representation of $\cot(x)$

The function is not defined at $x=0$, since $sin(0)=0$, division by zero is undefined. The function doesn't have a Maclaurin series centered at $0$. Nevertheless, we can still expand it into what is called the Laurent series. The Laurent series is a power series that includes negative degree.

Thus $\cot(x)=\dfrac{1}{x}-\dfrac{x}{3}-\dfrac{x^3}{45}-\dfrac{2x^5}{945}-\dfrac{x^7}{4725}-\dfrac{2x^9}{93555}...$

Series representation of $\sec(x)$

We use the identity $\sec(x)=\dfrac{1}{\cos(x)}$ to derive the Maclaurin series for $\sec(x)$.

Thus $\sec(x)=\color{green}{1}+\dfrac{\color{green}{1}x^2}{2!}+\dfrac{\color{green}{5}x^4}{24}+\dfrac{\color{green}{61}x^6}{720}+\dfrac{277x^8}{8064}+\dfrac{50521x^{10}}{3628800}...$

Long division may give us the series, but it doesn't hint at the fact that $\sec(x)$ is one of the two generating functions of Euler numbers. In fact, since we reduce fractions to their simplest form, this obscure the general formula for the coefficients of $\sec(x)$. The series of $\sec(x)$ is as followed:

$$\sum_{n=0}^{\infty}\dfrac{(-1)^{n}E_{2n}x^{2n}}{(2n)!}$$

The first few Euler numbers are:
$E_0=1$, $E_2=-1$, $E_4=5$, $E_6=-61$, $E_8=1385$, $E_{10}=-50521$, $E_{12}=2702765...$

Thus we can rewrite the series as followed:

$\sec(x)=\color{green}{1}+\dfrac{\color{green}{1}x^2}{2!}+\dfrac{\color{green}{5}x^4}{4!}+\dfrac{\color{green}{61}x^6}{6!}+\dfrac{\color{green}{1385}x^8}{8!}+\dfrac{\color{green}{50521}x^{10}}{10!}...$

Series representation of $\tanh(x)$ and sech(x)

Using the same method above, we can derive the series for $\tanh(x)$ and $\operatorname{sech(x)}$. The difference is that instead of being a positive term series, they become alternating series. The series for $\tanh(x)$ is:

$\tanh(x)=x-\dfrac{x^3}{3}+\dfrac{2x^5}{15}-\dfrac{17x^7}{315}+\dfrac{62x^9}{2835}-\dfrac{1382x^{11}}{155925}...$

The series for $sech$ is:

$\operatorname{sech}(x)=1-\dfrac{x^2}{2!}+\dfrac{5x^4}{24}-\dfrac{61x^6}{720}+\dfrac{277x^8}{8064}-\dfrac{50521x^{10}}{3628800}...$

$$\sum_{n=1}^{\infty}\dfrac{(-1)^nE_{2n}x^{2n}}{(2n)!}$$

Series representation of $\dfrac{x}{\sin(x)}$

At $x=0$, the function has the form $\dfrac{0}{0}$. Using L'Hopital rule to find the limit $$\lim_{x\to 0} \dfrac{x}{\sin(x)}$$, we have $$\lim_{x\to 0} \dfrac{1}{\cos(x)}$$, which is $1$. So the function is defined when $x=0$. By long division, we have:

Thus $\dfrac{x}{\sin(x)}=1+\dfrac{x^2}{3!}+\dfrac{7x^4}{360}+\dfrac{31x^6}{15120}+\dfrac{127x^8}{604800}+\dfrac{73x^{10}}{3421440}...$

Series representation of $\dfrac{1}{\sin(x)}=\csc(x)$

It is now easy to obtain the series representation for $\csc(x)$, we only have to divide it by $x$, the series is:

Thus $\csc(x)=\dfrac{1}{x}+\dfrac{x}{3!}+\dfrac{7x^3}{360}+\dfrac{31x^5}{15120}+\dfrac{127x^7}{604800}+\dfrac{73x^9}{3421440}...$

Again, the function is not defined at $x=0$, so there is no Maclaurin series for $\csc(x)$. What we obtain is the Laurent series.