Taylor series of $\csc(x)$

Question

How can I compute the Taylor series of $\csc(x):=\frac{1}{\sin(x)}$ at $0$? When I google this I can see that indeed this series can be computed, the first term being $1/x$ but how can this be if $\sin(x)$ vanishes at $x=0$?

Answer 1

What you found in Wikipedia can't be a Taylor series. In fact there's not such thing as the Taylor series of $\csc(x)$ at $x=0$ since this function is not defined at $x=0$, much less the derivatives of any order.

You probably found the Laurent series(*). Since $$g(x)=x \csc x=\frac x{\sin x}$$ is continuous and differentiable at $x=0$ if we define $g(0)=1$, and it's Taylor series can be found, say $$g(x)=a_0+a_1x+a_2x^2+\cdots$$ at least at some interval containing $x=0$, then $$\csc x= \frac{g(x)}x=\frac{a_0}x+a_1+a_2 x+a_3 x^2+\cdots,$$ which gives the aforementioned Laurent series.

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(*) The Laurent series of a function at $x=a$ is a generalization of the Taylor series for functions $f(x)$ with a pole at $x=a$, that is, functions that go to infinity at $x=a$ but such that $(x-a)^k f(x)$ has a removable discontinuity at $x=a$. If $k$ is the least integer such that this is true, then the Laurent series has the form $$f(x)=\frac{b_{-k}}{(x-a)^k}+\cdots+\frac{b_{-2}}{(x-a)^2}+\frac{b_{-1}}{x-a}+b_0+b_1(x-a)+b_2(x-a)^2+\cdots.$$

Answer 2

An easy way to derive the Laurent series of $\csc x$ is to utilize the identity $$\csc x=\cot(\tfrac x2)-\cot x.$$ Since, $$\cot t=\frac1t-\sum_{n=1}^\infty\frac{2\zeta(2n)}{\pi^{2n}}t^{2n-1}$$ we get $$\csc t=\frac1t-\sum_{n=1}^\infty\frac{2(4^n-2)\zeta(2n)}{4^n\pi^{2n}}t^{2n-1}$$ where $$\zeta(2n)=\frac{(-1)^{n-1}4^n\pi^{2n}}{2(2n)!}B_{2n}.$$