6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $b_n$. $\endgroup$ – Winther Oct 3 '16 at 22:31
3
$\begingroup$

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 $$\cot(z)=\sum_{n=0}^\infty (-1)^n\frac{ 2^{2 n}\, B_{2 n} }{(2 n)!}z^{2 n-1}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.