Laurent expand \(\tan(z)\) at \(0\) using this method to compute \(1/\cos(z)\) and taking the product \(\sin(z) \cdot {1\over \cos(z)}\).
Solution
Example. To find the coefficients of \(z^m\) for \(m \leq 5\) in the power series expansion of \(\tan z=\sin z / \cos z\) about \(z=0\), we calculate as follows, again using the notation \(\mathcal{O}\left(z^m\right)\) for terms involving powers \(z^k\) for \(k \geq m\) : \[ \begin{aligned} \frac{1}{\cos z} & =\frac{1}{1-\left(z^2 / 2 !\right)+\left(z^4 / 4 !\right)+\mathcal{O}\left(z^6\right)} \\ & =1+\left(\frac{z^2}{2 !}-\frac{z^4}{4 !}+\mathcal{O}\left(z^6\right)\right)+\left(\frac{z^2}{2 !}-\frac{z^4}{4 !}+\mathcal{O}\left(z^6\right)\right)^2+\mathcal{O}\left(z^6\right) \\ & =1+\frac{1}{2} z^2+\frac{5}{24} z^4+\mathcal{O}\left(z^6\right) \end{aligned} \] so that \[ \begin{aligned} \frac{\sin z}{\cos z} & =\left(z-\frac{z^3}{3 !}+\frac{z^5}{5 !}+\mathcal{O}\left(z^7\right)\right)\left(1+\frac{1}{2} z^2+\frac{5}{24} z^4+\mathcal{O}\left(z^6\right)\right) \\ & =z+\frac{1}{3} z^3+\frac{2}{15} z^5+\mathcal{O}\left(z^7\right) . \end{aligned} \] The end result can be checked by differentiating \(\tan z\) five times. Note that \(\tan z\) is an odd function, so that only odd terms appear in the power series.