# Gravitational field of a disc. Quadratic aproximation

I calculated the gravitational field above a disc positioned in the $x$,$y$ plane with its centre at $x=y=0$. The answer is: $$2G\frac{M}{R^2}\left(1-\frac{z}{\sqrt{R^2 + z^2}}\right)$$ with $M$=Mass of the disc, and $R$=its radius. My expectation is that if I take the limit $z>>R$, then the gravitational field of the disc should follow the $\frac{1}{z^2}$ dependency, because the disc will look like a point more or less. I tried the Taylor series, but it's pretty clear it won't work, because the first and second derivatives go to zero at infinity. I inserted $\left(1-\frac{z}{\sqrt{R^2 + z^2}}\right)$ in Wolfram alpha and it gives me the expected approximation: $\frac{R^2}{2z^2}$ and says Laurent series. Can someone explain me how to calculate this approximation and how to know which method (Taylor or Laurent series) when to apply?

In terms of a Taylor series, $R/z$ is the small parameter you expand in
$$\frac{z}{\sqrt{R^2+z^2}}=\left(1+\left(\frac{R}{z}\right)^2\right)^{-1/2}=1-\frac{1}{2}\left(\frac{R}{z}\right)^2+\dots.$$
If you think about expanding in terms of $z$ around $z=0$, it doesn't make sense in terms of ordinary Taylor series, since there is a singularity at $z=0$. But a Laurent series is like a Taylor series but also includes singular terms like $1/z^2$.