What is the Taylor series of $\frac{1}{\sin(z)}$ about $z_0 = 1$?

Question

This was a exam question so I know it cannot take too long to write out the proof. Only I cannot see an answer.

I would imagine you write $\sin(z) = \sin(1+(z-1)) = \sin(1)\cos(z-1) + \sin(z-1)\cos(1)$ and then use the everywhere-defined Taylor series for $\sin$ and $\cos$ to write $\frac{1}{\sin(z)}$ as the reciprocal of a power series. Then you manipulate it into the form $\displaystyle \frac{1}{1-P(z)}$ where $P$ is a power series and then invert using the geometric series formula. Only my $P$ looks horrible and thus the condition for convergence $|P(z)|<1$ is impossible to compute.

Another series which I cannot do but which I imagine could be done by similar methods is $\displaystyle \frac{1}{2\cos(z) -1}$ about $z_0 = 0$.

Any tips?

Answer 1

Using the Taylor series expansion for $\csc(x)=\frac{1}{\sin(x)}$, we may write write $$\frac{1}{\sin(z)}=\csc(z)=\sum_{k=0}^{\infty}\csc^{(k)}(1)\frac{(z-1)^{k}}{k!}.$$ This series will have radius of convergence $1$ since the poles of $\csc(x)$ are precisely at the the zeros of $\sin(x)$, all of which occur at integer multiples of $\pi$.

Now, this might not be satisfying at all, but this is likely the best that you can do. Ideally we would want to specify the coefficients $\csc^{(k)}(1)$ exactly (by giving a power series expansion we are doing precisely that) however they are extremely messy. The only way I can think of writing them without derivatives involves the Bernoulli numbers and powers of $\sin$ and $\cos$ evaluated at $1$.

Answer 2

The function $z\mapsto 1/\sin(z)$ is meromorphic and has simple poles at points of $\pi\Bbb{Z}$. Thus it has a power series expansion $\sum_{n=0}^\infty a_n(z-1)^n$ around $z_0=1$, with radius of convergence $R=d(1,\pi\Bbb{Z})=1$. Now to determine the coefficients we may can use the identity $1=\sin(z)\left(\sum_{n=0}^\infty a_n(z-1)^n\right)$ in the neighborhood of $z_0=1$. Or, by setting $z=1+t$: $$ \left(\sin(1)\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}t^{2k} +\cos(1)\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}t^{2k+1}\right) \left(\sum_{n=0}^\infty a_n t^n\right)=1 $$ So, the coefficients $\{a_n\}$ may be obtained inductively by the formula $$\eqalign{ a_0&=\frac{1}{\sin(1)},\cr a_n&=-\sum_{k=1}^n\frac{(-1)^{\lfloor k/2\rfloor}}{k!}\delta_ka_{n-k},\ } $$ where $\delta_k=\left\{ \matrix{\cot(1)&\hbox{if $k$ is odd,}\phantom{z}\cr 1&\hbox{if $k$ is even.}} \right.$

Answer 3

By the Faa di Bruno formula and some properties of the Bell polynomials of the second kind, we acquire \begin{align*} \biggl(\frac1{\sin z}\biggr)^{(n)} &=\sum_{k=0}^n\frac{(-1)^kk!}{\sin^{k+1}z} B_{n,k}\biggl(\cos z,-\sin z,-\cos z,\sin z,\dotsc, \sin\biggl[z+\frac{(n-k+1)\pi}{2}\biggr]\biggr)\\ &=\frac{1}{\sin z}\sum_{k=0}^n \sum_{\ell=0}^k\frac{\binom{k}{\ell}}{(2\sin z)^{\ell}} \sum_{q=0}^\ell(-1)^q\binom{\ell}{q}(2q-\ell)^n \cos\biggl[(2q-\ell)z+\frac{(n-\ell)\pi}2\biggr]\\ &\to\sum_{k=0}^n \sum_{\ell=0}^k\frac{\binom{k}{\ell}}{2^\ell(\sin1)^{\ell+1}} \sum_{q=0}^\ell(-1)^q\binom{\ell}{q}(2q-\ell)^n \cos\biggl[2q-\ell+\frac{(n-\ell)\pi}2\biggr], \quad z\to1 \end{align*} and \begin{gather*} \biggl(\frac{1}{2\cos z-1}\biggr)^{(n)} =\sum_{k=0}^n\frac{(-1)^kk!}{(2\cos z-1)^{k+1}} 2^k B_{n,k}\biggl(-\sin z,-\cos z,\sin z,\cos z,\dotsc, \cos\biggl[z+\frac{(n-k+1)\pi}{2}\biggr]\biggr)\\ =\sum_{k=0}^n\frac{2^k\cos^kz}{(2\cos z-1)^{k+1}} \sum_{\ell=0}^k\binom{k}{\ell}\frac{(-1)^\ell}{(2\cos z)^\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^n \cos\biggl[(2q-\ell)z+\frac{n\pi}2\biggr]\\ \to\cos\frac{n\pi}2 \sum_{k=0}^n 2^k \sum_{\ell=0}^k\frac{(-1)^\ell}{2^\ell} \binom{k}{\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^n, \quad z\to0\\ =\begin{cases}\displaystyle (-1)^m\sum_{k=0}^{2m} 2^k \sum_{\ell=0}^k\frac{(-1)^\ell}{2^\ell} \binom{k}{\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{2m} & n=2m\\ 0, & n=2m+1 \end{cases} \end{gather*} for $m,n\ge0$. Therefore, we have \begin{multline*} \frac1{\sin z}=\frac1{\sin1}+\sum_{n=1}^\infty\Biggl\{\sum_{k=0}^n \sum_{\ell=0}^k\frac{\binom{k}{\ell}}{2^\ell(\sin1)^{\ell+1}}\\ \times\sum_{q=0}^\ell(-1)^q\binom{\ell}{q}(2q-\ell)^n \cos\biggl[2q-\ell+\frac{(n-\ell)\pi}2\biggr]\Biggr\}\frac{(z-1)^n}{n!}, \quad |z-1|<1 \end{multline*} and \begin{equation*} \frac{1}{2\cos z-1} =\sum_{m=0}^\infty(-1)^m\Biggl[\sum_{k=0}^{2m} 2^k \sum_{\ell=0}^k\frac{(-1)^\ell}{2^\ell} \binom{k}{\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{2m}\Biggr]\frac{z^{2m}}{(2m)!}, \quad |z|<\frac{\pi}6. \end{equation*} All formulas, notations, concepts, and knowledge can be found in the following two papers.

References

1. F. Qi, Derivatives of tangent function and tangent numbers, Appl. Math. Comput. 268 (2015), 844--858; available online at https://doi.org/10.1016/j.amc.2015.06.123.
2. Feng Qi, Da-Wei Niu, Dongkyu Lim, and Yong-Hong Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, Journal of Mathematical Analysis and Applications 491 (2020), no. 2, Paper No. 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.