# Laurent series for $\cot (z)$

I'm looking for clarification on how to compute a Laurent series for

$\cot z$

I started by trying to find the $\frac{1}{\sin z}$. I've found multiple references that go from an Taylor expansion for $\sin z$ directly to an expression for $\frac{1}{\sin z}$ but I am unable to follow how they got there. This thread Calculate Laurent series for $1/ \sin(z)$ started to answer my question but I do not understand how to use the given formulas to "iteratively compute" the coefficients, and the example given has several coefficients in place and I'm not sure how they were obtained.

• If we have a series for $f(z) = \sum a_k z^k$ and let $\sum b_k z^k$ be the power-series for $\frac{1}{f(z)}$ then the Cauchy product of these two series must be the power-series for the function $f(z)\cdot \frac{1}{f(z)} = 1$, i.e. $1 + 0z + 0 z^2 + \ldots$. This means that $a_0b_n + a_1b_{n-1} + \ldots + a_n b_0 = 0$ for $n >1$. If you now already have computed $b_0,b_1,b_2,\ldots,b_{n-1}$ then this formula say that $b_n = -\frac{1}{a_0}\cdot[\text{terms you know}]$. This gives us an iterative prodedure to compute all $b_n$. – Winther Oct 3 '16 at 22:31

If you are looking for a trucated series, you could start from $$\tan(z)=z+\frac{z^3}{3}+\frac{2 z^5}{15}+\frac{17 z^7}{315}+\frac{62 z^9}{2835}+O\left(z^{11}\right)$$ which makes $$\cot(z)=\frac 1{z+\frac{z^3}{3}+\frac{2 z^5}{15}+\frac{17 z^7}{315}+\frac{62 z^9}{2835}+O\left(z^{11}\right)}=\frac 1z \frac 1{1+\frac{z^2}{3}+\frac{2 z^4}{15}+\frac{17 z^6}{315}+\frac{62 z^8}{2835}+O\left(z^{10}\right)}$$ and perform long division to get $$\cot(z)=\frac{1}{z}-\frac{z}{3}-\frac{z^3}{45}-\frac{2 z^5}{945}-\frac{z^7}{4725}+O\left(z^9\right)$$ If you want the infinite series consider that $$\cot(z)=\frac 1 {\tan(z)}=f(z)=\sum_{i=0}^\infty a_iz^{i-1}$$ what you can rewrite as $$1=\tan(z)\sum_{i=0}^\infty a_iz^{i-1}$$ that is to say $$1=\left(\sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} z^{2n-1}\right)\times \sum_{i=0}^\infty a_iz^{i-1}$$ For simplicity, let us define $$b_n=\frac{B_{2n} (-4)^n (1-4^n)}{(2n)!}$$ in order to solve $$1=\sum^{\infty}_{n=1}b_nz^{2n-1} \times \sum_{i=0}^\infty a_iz^{i-1}$$ Developing, we get $$1=a_0 b_1+a_1 b_1 z+ (a_2 b_1+a_0 b_2)z^2+ (a_3 b_1+a_1 b_2)z^3+ (a_4 b_1+a_2 b_2+a_0 b_3)z^4+ (a_5 b_1+a_3 b_2+a_1 b_3)z^5+ (a_6 b_1+a_4 b_2+a_2 b_3+a_0 b_4)z^6+(a_7 b_1+a_5 b_2+a_3 b_3+a_1 b_4)z^7 +\cdots$$ Now,we need to solve, for the $$a_i$$'s the equations $$a_0 b_1=1$$ $$a_1 b_1=0$$ $$a_2 b_1+a_0 b_2=0$$ $$a_3 b_1+a_1 b_2=0$$ $$a_4 b_1+a_2 b_2+a_0 b_3=0$$ $$a_5 b_1+a_3 b_2+a_1 b_3=0$$ $$a_6 b_1+a_4 b_2+a_2 b_3+a_0 b_4=0$$ $$a_7 b_1+a_5 b_2+a_3 b_3+a_1 b_4=0$$ This does not make much problem (using successive eliminations for example).
This leads to the infinite series $$\cot(z)=\sum_{n=0}^\infty (-1)^n\frac{ 2^{2 n}\, B_{2 n} }{(2 n)!}z^{2 n-1}$$