Power series of tan(x), cot(x), csc(x)

Here’s a little how-to on figuring out the power series of tan(x), cot(x) and csc(x).

Start with the generating function for the Bernoulli numbers:

$$\sum_{k=0}^{\infty} B_k \frac{t^k}{k!} = \frac{t}{e^t-1}.$$

Take only the even powers of t on the LHS:

$$\sum_{k=0}^{\infty} B_k \frac{t^k}{k!} + \sum_{k=0}^{\infty} B_k \frac{(-t)^k}{k!} = \frac{t}{e^t-1} – \frac{t}{e^{-t}-1}$$

\begin{align}
2 \sum_{k=0}^{\infty} B_{2k} \frac{t^{2k}}{(2k)!}
&= \frac{t\left(e^{-t}-e^t\right)}{\left(e^t-1\right)\left(e^{-t}-1\right)} \\
&= \frac{t\left(e^{-\frac{1}{2} t}+e^{\frac{1}{2} t}\right)\left(e^{-\frac{1}{2} t}-e^{\frac{1}{2} t}\right)}{e^{\frac{1}{2} t}\left(e^{\frac{1}{2} t}-e^{-\frac{1}{2} t}\right)e^{-\frac{1}{2} t}\left(e^{-\frac{1}{2} t}-e^{\frac{1}{2} t}\right)} \\
&= \frac{t\left(e^{\frac{1}{2} t}+e^{-\frac{1}{2} t}\right)}{e^{\frac{1}{2} t}-e^{-\frac{1}{2} t}}.
\end{align}

Put \(t=2ix\):

$$2 \sum_{k=0}^{\infty} B_{2k} \frac{(2ix)^{2k}}{(2k)!} = \frac{2ix\left(e^{ix}+e^{-ix}\right)}{e^{ix}-e^{-ix}}$$

$$\sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k 4^k x^{2k-1}}{(2k)!} = \frac{i\left(e^{ix}+e^{-ix}\right)}{e^{ix}-e^{-ix}}.$$

Recall that \(\sin x = \frac{e^{ix}-e^{-ix}}{2i}\) and \(\cos x = \frac{e^{ix}+e^{-ix}}{2}\):

$$\displaystyle \frac{\cos x}{\sin x} = \cot x = \sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k 4^k x^{2k-1}}{(2k)!}.$$

That was pretty easy. How about \(\tan\) and \(\csc\)?

Consider

$$\cot x – 2\cot 2x = \cot x – \frac{\cot^2 x – 1}{\cot x} = \frac{1}{\cot x} = \tan x.$$

This means that

\begin{align}
\tan x
&= \sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k 4^k x^{2k-1}}{(2k)!} – 2\sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k 4^k 2^{2k-1} x^{2k-1}}{(2k)!} \\
&= \sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k 4^k (1-4^k) x^{2k-1}}{(2k)!}.
\end{align}

And finally,

$$\cot x – \cot 2x = \cot x – \frac{\cot^2 x – 1}{2\cot x} = \frac{\csc^2 x}{2\cot x} = \frac{1}{2 \sin x \cos x} = \csc 2x,$$

so

\begin{align}
\csc x
&= \sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k 4^k x^{2k-1}}{2^{2k-1} (2k)!} – \sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k 4^k x^{2k-1}}{(2k)!} \\
&= \sum_{k=0}^{\infty} B_{2k} \frac{(-1)^k (2-4^k) x^{2k-1}}{(2k)!}
\end{align}