Vrftjkry

Laurent for a function is a generalization of the Taylor series. With a Laurent series, however, the powers can be negative. An advantage of the Laurent series over the Taylor series is expanding around singular points for a given function. Looking for a demonstration of this advantage for an unusual function.

Can Laurent's theorem be used on all Taylor theorem expandable functions?

I don't know if it's "unusual", but for example, $f(z) = sin(1/z)$ has an isolated singularity at $z=0$. You can do a Laurent expansion around that point:

$$ sin(1/z) = z^{-1} - frac{z^{-3}}{3!} + frac{z^{-5}}{5!} - ldots = sum_{n=-infty}^infty a_n z^n$$
where $a_n = (-1)^{(n+1)/2}/(-n)!$ for odd $n < 0$, $0$ otherwise.

I'm not quite sure what you're referring to as "Laurent's theorem".
But any Taylor series is a Laurent series where the terms in negative powers happen to be $0$, so wherever Taylor expands the function, so does Laurent.

A Taylor series has a radius of convergence.

Some familiar functions that radius is infinite.

$e^x = sum frac {x^n}{n!}$

But for others is its not

$ln (1-x) = -sum frac {x^n}{n}$

A Laurent series gives you some flexibility in determining where the series will converge and will not converge.

$frac {1}{1-x} = sum_limits{n=0}^infty x^n$ converges when $|x|< 1$

$frac {1}{1-x} = sum_limits{n=1}^infty -x^{-n}$ converges when $|x|> 1$

One place it comes in handy:

If you have contour integral in complex analysis. And your function has a Laurent series representation, and the the contour is inside the convergent part of the Laurent series

$int_gamma f(z) dz = 2pi i (a_{-1})$ where $a_{-1}$ is the coefficient of the $z^{-1}$ term