How to remove singularity from the solution?

Question

Suppose we have

$f(x)=\frac{1}{sin(x-x0)cos(x-x0)}$,

we know that $f(x)$ is undefined at $x=x0$. Is there any possibility to remove this singularity so that $f(x)$ would be a smooth function for all $x$?

Answer 1

In complex analysis, we know that there are three types of isolated singular points. Brown and Churchill define them as poles of order m, removable singular points, and essential singular points (Section 72).

Furthermore, the point $x_0$ is an isolate singular point "if an infinite number of the coefficients $b_n$ of the principal part are nonzero" (Brown and Churchill, 242). This would be the portion such that $\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+...+\frac{b_n}{(z-z_0)^n}+...$ I'm sure it goes without saying but we can adjust these to the denominator $(x-x_0)^n$.

Let's take a look at $f(x)=\frac{1}{sin(x-x_0)cos(x-x_0)}$. We have three types of isolated singular points for which to test. Logically let's test for the essential singular point first. A reminder: if $x_0$ is this type of point then it has an infinite number of nonzero terms in the principal part (the one which contains negative powers). f(x) is $\frac{1}{sin(x)cos(x)}$ centered about $x=x_0$. Here is a list of commonly used Taylor series approximations. We can rewrite f(x) as $f(x)=\frac{1}{sin(x-x_0)cos(x-x_0)}=[sin(x-x_0)cos(x-x_0)]^{-1}$. Note that once we find the correct approximations for our $x_0$ adjustment and multiply out our resulting Taylor series would become an infinite collection of nonzero negative powers. Since the Taylor series for sine begins at $n=1$, with little-to-no work we find ourselves with a Laurent series.

So to answer your question, we cannot remove the singularity $x_0$ since $x_0$ is an essential singular point.