How to find the Laurent expansion for $1/\cos(z)$

Question

How to find the Laurent series for $1/\cos(z)$ in terms of $(z-\frac{\pi}{2})$ for all $z$ such that $0<|z-\frac{\pi}{2}|<1$

Answer 1

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.

Answer 2

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.

Answer 3

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 $$ \begin{align} \frac1{\cos(x)} &=-\frac1{\sin(x-\pi/2)}\\ &=-\sum_{k=0}^\infty a_k(x-\pi/2)^{2k-1}\\ &=-\frac1{x-\pi/2}-\frac{x-\pi/2}{6}-\frac{7(x-\pi/2)^3}{360}-\frac{31(x-\pi/2)^5}{15120}-\dots \end{align} $$