Finding coefficients of laurent series for $\frac 1{1-\cos z}$ about zero

Question

I know I'm wrong, but I fail to see why I'm wrong. My goal is to try and find the terms for the Laurent series of $f(z)=\frac{1}{1-\cos(z)}$ but I'm surely off.

$$\begin{align} f(z)&= \frac{1}{1-\cos(z)} \\ &= \sum_{n=0}^\infty \cos(z)^n \\ &= \sum_{n=0}^\infty \left( \sum_{j=0}^\infty \frac{(-1)^jz^{2j}}{(2j)!}\right)^n \end{align}$$

Under nothing that I wrote, can $z$ have a negative exponent. However, it is obvious that it should, both by looking at the function and checking my intuition over at Wolfram Alpha.

Where is my reasoning wrong?

Answer 1

As this is a second order pole, you only need consider $\sum_{n=-2}^{\infty} a_{n}z^{n}$. Now, check the limits of $[z^3\cdot f(z)]''/2!$ and $[z^2\cdot f(z)]'/1!$ as $z \rightarrow 0$ and we get $1/2$ and $0$ respectively. Then, evaluate $\lim\; 1/(1-\cos(z)) - 2/z^2 \rightarrow \frac{(1/4!)}{(1/2!)} = 1/6$ as $z \rightarrow 0$. This is as many terms as I went. But there is no noticeable pattern as far as I can see at a first glance.

Answer 2

The comments have explained why your proposal doesn't work, and leads to nonsense like a constant term of $\sum_{n=0}^\infty (1)^n = \infty$.

When $z\approx 0$, we have $\cos z = 1 - \frac{z^2}2 + \frac{z^4}{4!} - \dots$, and so $$ \frac1{1-\cos z} = \frac1{\frac{z^2}2 - \frac{z^4}{4!} + \dots} = 2z^{-2} \left(1 - \sum_{j\geq 1} \frac{2(-z^2)^j}{(2j)!} \right)^{-1} = 2z^{-2} \sum_{n\geq 0} \left( \sum_{j\geq 1} \frac{2(-z^2)^j}{(2j)!} \right)^n$$ I don't see a particularly snazzy way to say what this generating function counts.

Answer 3

My answer is \begin{multline*} \frac{1}{1-\cos x}=\frac{1}{2\sin^2\frac{x}{2}} =\frac12\biggl(\frac{2}{x}\biggr)^2\Biggl(\frac{\frac{x}2}{\sin\frac{x}2}\Biggr)^2 =\frac{2}{x^2} \Biggl\{1+\\ \sum_{q=1}^{\infty}(-1)^q\Biggl[\sum_{k=1}^{2q}\frac{(2)_k}{k!} \sum_{j=1}^k(-1)^j\binom{k}{j} \sum_{m=0}^{2q}(-1)^{m}\binom{2q}{m} \biggl(\frac{j}{2}\biggr)^{m} \frac{S(2q+j-m,j)} {\binom{2q+j-m}{j}}\Biggr]\frac{x^{2q}}{(2q)!}\Biggr\}, \end{multline*} where $|x|<2\pi$ and $S(2q+j-m,j)$ denotes the Stirling numbers of the second kind. The last step comes from employing Theorem 4.1 in the paper

1. Feng Qi and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Applicable Analysis and Discrete Mathematics 18 (2024), no. 1, in press; available online at https://doi.org/10.48550/arXiv.2204.05612.

Answer 4

From \begin{equation} \begin{aligned}\label{sin-square-recip-ser} \biggl(\frac{x}{\sin x}\biggr)^2&=1+2B_2x^2-4\sum_{k=2}^{\infty}(-1)^k\Biggl[\bigl(2^{2k-1}-1\bigr)B_{2k}\\ &\quad-\sum_{j=1}^{k-1}\binom{2k}{2j}\bigl(2^{2j-1}-1\bigr)\bigl(2^{2k-2j-1}-1\bigr)B_{2j}B_{2k-2j}\Biggr]\frac{x^{2k}}{(2k)!}\\ &=1+\frac{x^2}{3}+\frac{x^4}{15}+\frac{2 x^6}{189}+\frac{x^8}{675}+\frac{2 x^{10}}{10395}+\frac{1382 x^{12}}{58046625}+\dotsm \end{aligned} \end{equation} for $x\in(-\pi,\pi)$, we acquire \begin{align*} \frac1{1-\cos x}&=\frac12\biggl(\frac2x\biggr)^2\Biggl(\frac{\frac{x}2}{\sin\frac{x}2}\Biggr)^2\\ &=\frac12\biggl(\frac2x\biggr)^2\Biggl\{1+\frac{B_2}{2}x^2-4\sum_{k=2}^{\infty}(-1)^k\Biggl[\bigl(2^{2k-1}-1\bigr)B_{2k}\\ &\quad-\sum_{j=1}^{k-1}\binom{2k}{2j}\bigl(2^{2j-1}-1\bigr)\bigl(2^{2k-2j-1}-1\bigr)B_{2j}B_{2k-2j}\Biggr]\frac{x^{2k}}{2^{2k}(2k)!}\Biggr\} \end{align*} for $x\in(-2\pi,2\pi)$.

References

1. Xue-Yan Chen, Lan Wu, Dongkyu Lim, and Feng Qi, Two identities and closed-form formulas for the Bernoulli numbers in terms of central factorial numbers of the second kind, Demonstratio Mathematica 55 (2022), no. 1, 822--830; available online at https://doi.org/10.1515/dema-2022-0166.

Answer 5

By definition of cosine $$\frac{1}{1-\cos z}=\frac{-2e^{iz}}{(e^{iz}-1)^2}\tag1$$ On the other hand $\frac{t}{e^t-1}=\sum_{n=0}^\infty\frac{B_n}{n!}t^n$ gives $$\frac{-2e^t}{(e^t-1)^2}=\sum_{n=0}^\infty\frac{2(n-1)B_n}{n!}t^{n-2}\tag2$$ From $(1),(2)$ $$\frac{1}{1-\cos z}=\sum_{n=0}^\infty\frac{(-1)^{\frac n2+1}2(n-1)B_{n}}{n!}z^{n-2}$$ Due to $(1-1)B_1=0$ and the fact that other odd indexed Bernoulli numbers vanish, letting $n=2k$ we have $$\frac{1}{1-\cos z}=\sum_{k=0}^\infty\frac{(-1)^{k+1}2(2k-1)B_{2k}}{(2k)!}z^{2k-2}$$