How to find the Laurent expansion for $1/\cos(z)$ How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$
 A: Note $\frac{1}{\cos z}=-\frac{1}{\sin (z-\frac{\pi}{2})}$.
Let $t:=z-\frac{\pi}{2}$. Then $0<|t|<1$, $\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots$, so
$$
\begin{align}
\frac{1}{\sin t}&=\frac{1}{ t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\dots } \\
                &=\frac{1}{t}\frac{1}{1-\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)}\\
                &=\frac{1}{t}\left[1+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)+\left(\frac{t^2}{3!}-\frac{t^4}{5!}+\frac{t^6}{7!}+\dots\right)^2+\dots\right]\\
                &=t^{-1}+\frac{1}{3!}t+\left[\left(\frac{1}{3!}\right)^2-\frac{1}{5!}\right]t^3+\dots
\end{align}
$$
It seems no closed forms for higher terms.
A: To get a recurrence for the series of the reciprocal, consider
$$
\begin{align}
1
&=\sin(x)\sum_{k=0}^\infty a_kx^{2k-1}\\
&=\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\sum_{k=0}^\infty a_kx^{2k-1}
\end{align}
$$
and apply Cauchy products to get
$$
\sum_{k=0}^n(-1)^k\frac{a_{n-k}}{(2k+1)!}=\left\{\begin{array}{}
1&\text{if }n=0\\
0&\text{otherwise}
\end{array}\right.
$$
So that we get $a_0=1$ and
$$
a_n=\sum_{k=1}^n(-1)^{k-1}\frac{a_{n-k}}{(2k+1)!}
$$
which gives $\{a_k\}$ to be
$$
\left\{1,\frac16,\frac7{360},\frac{31}{15120},\dots\right\}
$$
Then we get
$$
\begin{align}
\frac1{\cos(x)}
&=-\frac1{\sin(x-\pi/2)}\\
&=-\sum_{k=0}^\infty a_k(x-\pi/2)^{2k-1}\\
&=-\frac1{x-\pi/2}-\frac{x-\pi/2}{6}-\frac{7(x-\pi/2)^3}{360}-\frac{31(x-\pi/2)^5}{15120}-\dots
\end{align}
$$
A: Since $$lim_{z \rightarrow \frac{\pi}{2}} (z-\frac{\pi}{2})(\frac{1}{\cos(z)})=lim_{z \rightarrow \frac{\pi}{2}} \frac{1}{-\sin(z)} = -1$$
we know that $\frac{\pi}{2}$ is a pole of order 1 of our function $\frac{1}{cos(z)}$ with residue -1. Therefore we know that our Laurent series looks like
$$\frac{-1}{(z-\frac{\pi}{2})} + \sum_{i=0}^{+\infty}a_0(z-\frac{\pi}{2})^i$$
Since $\frac{1}{\cos(z)}\cos{z}=1$, we can determine the remaining a's out of the laurent series of $\cos(z)$.
$$\left(\frac{-1}{(z-\frac{\pi}{2})} + \sum_{i=0}^{+\infty}a_0(z-\frac{\pi}{2})^i\right)\left(-(z-\frac{\pi}{2})+\frac{1}{6}(z-\frac{\pi}{2})^3-\frac{1}{120}(z-\frac{\pi}{2})^5 + ...\right)=1$$
We find for example that $-a_0(z-\frac{\pi}{2})=0(z-\frac{\pi}{2})$. So $a_0=0$.
You can find the complete series here.
