Taylor series for $f(x)=\cot(x)$

Question

Trying to expand $f(x)=\cot(x)$ to Taylor series (Maclaurin, actually). But I keep "adding up" infinities when using the formula. (Because of $\cot(0)=\infty$) Could you perhaps give me a hint on how to proceed?

Answer 1

The function $\cot x$ is not continuous at zero, and therefore has no power series around zero.

If you know complex analysis, you should look for the Laurent series of $\cot z$ at $z=0$ instead.

Answer 2

Since $f$ is not defined at $0$, its Maclaurin series is undefined.

On the other hand, the pole of $f$ at $0$ is simple, and it's not hard to compute that the residue of $f$ there is $1$, so one can compute a Maclaurin series for $\cot x - \frac{1}{x}$ there, namely, $$\cot x - \frac{1}{x} \sim -\frac{1}{3} x - \frac{1}{45} x^3 - \cdots .$$ Of course, one can rearrange this and write $$\cot x \sim \frac{1}{x}-\frac{1}{3} x - \frac{1}{45} x^3 - \cdots$$ (as for Taylor series, $\sim$ must be suitably interpreted.) This is an example of a Laurent series, or roughly, an analog of a Taylor series allowing negative powers of $x - a$.

Answer 3

The correct answer is that $x = 0$ is not in the domain of $\cot(x)$. Continuity is a property of elements of the domain of the function.