Let $f(z)=\frac{1}{1-\cos z}$. I want to calculate the Laurent series of $f$ around $z=0$, up to the coefficient of $z^6$.

It's easy to see that $g(z)=z^2f(z)$ (Defined as its limit at $0$ is holomorphic since $z=0$ is a pole of order $2$. Furthermore, it's easy to see that $g(0)=2$. I can calculate the Maclaurin series of $g(z)$ until $z^8$ and divide by $z^2$ to get the result. Also, since $g(z)$ is an even function, the odd coefficients are all $0$. And since $g(z)$ is of class $C^\infty$ on a small enough disk, I think I can just (In theory) calculate $g^{(n)}(0)=\lim_{z\to 0} g^{(n)}(z)$ by L'Hospital rule.

The problem is that the expression $g^{(n)}(z)$ becomes more and more complicated when $n$ grows, and I must calculate all its even derivatives at $0$.

So I wanted to know if there is a more efficient way than calculating all the even derivatives $g^{(n)}(z)$ until $n=8$, and then the derivatives of the numerator and denominator of each expression until one of them isn't $0$ at $0$. For example, in the denominator there will be a $(1-\cos z)^{2^n}$, which has a zero of order $2^{n+1}$ which must cancel with some zero in the numerator. It's too big a calculation just to use L'Hospital until $n=8$.

There are some related questions about computing a general term of the series; I'm just thinking about a moderately fast way to compute some of the coefficients.


What Szeto described is far superior in this particular case, but generally, if you are interested only in some of the terms of Laurent series, you can do something like this.

As you've said, $z_0 = 0$ is a pole of order $2$, so Laurent series of $f$ is of the form $$\frac{a_{-2}}{z^2} + a_0 + a_2z^2 + a_4z^4 + a_6z^6 + \ldots$$ where I also used that $f$ is an even function.

Taylor series of $1-\cos z$ is $$\frac{z^2}{2!} - \frac{z^4}{4!} + \frac{z^6}{6!} -\ldots$$ so we know that

$$(\,\frac{a_{-2}}{z^2} + a_0 + a_2z^2 + a_4z^4 + a_6z^6 + \ldots\,)(\,\frac{z^2}{2!} - \frac{z^4}{4!} + \frac{z^6}{6!} -\ldots\,) = 1\\ \frac{a_{-2}}{2!}+(\,\frac{a_0}{2!}-\frac{a_{-2}}{4!}\,)z^2 + (\, \frac{a_{2}}{2!} - \frac{a_0}{4!}+\frac{a_{-2}}{6!}\,)z^4 + (\, \frac{a_{4}}{2!} - \frac{a_2}{4!}+\frac{a_{0}}{6!} - \frac{a_{-2}}{8!}\,)z^6 + \ldots = 1$$

giving you recursion

$$\frac{a_{2n}}{2!}-\frac{a_{2n-2}}{4!}+\ldots+(-1)^{n+1}\frac{a_{-2}}{(2n+4)!} = 0,\ n\geq 0$$

with $a_{-2} = 2$.

  • $\begingroup$ I just want to avoid term-by-term multiplication, which, IMO, is tedious. $\endgroup$ – Szeto Jun 3 '18 at 11:29
  • $\begingroup$ @Szeto, well, it is a moderately fast way to compute some of the coefficients, definitely not as tedious as computing derivatives. It's not like you have a lot of choice in general. $\endgroup$ – Ennar Jun 3 '18 at 11:47

By the famous identity $$\cos 2x=1-2\sin^2 x$$, we get $$1-\cos x=2\sin^2 \frac{x}2$$ $$\frac1{1-\cos x}=2\csc^2 \frac{x}2=\frac{d}{dx}\cot \frac{x}2$$

Since the Laurent series of $\cot x$ is well-known, you can easily obtain the Laurent series of $\frac1{1-\cos x}$ by term-by-term differentiation.


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