# Approximation of a series expansion

In the approximation of a series expansion for the parameter $$\delta$$, I take only the first two terms of

$$\delta=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}z^n}{n^{\frac{3}{2}}}$$

and have

$$\delta\approx z-\frac{z^2}{2\sqrt{2}}$$

This, however, is then approximated in my lecture notes as

$$\rightarrow z \approx \delta+\frac{\delta^2}{2\sqrt{2}}$$

Where I am unsure how to get there, when after solving for $$z$$, I have no term $$\delta^2$$. What further approximation was made here?

Let's try series inversion.

If we have:

$$\delta=z-\frac{1}{2^{3/2}} z^2+\frac{1}{3^{3/2}} z^3-\dots$$

And we assume:

$$z=a_1\delta+a_2 \delta^2+a_3 \delta_3+\dots$$

Then we can substitute:

$$(a_1\delta+a_2 \delta^2+a_3 \delta_3+\dots)-\frac{1}{2^{3/2}} (a_1\delta+a_2 \delta^2+a_3 \delta_3+\dots)^2+ \\ +\frac{1}{3^{3/2}} (a_1\delta+a_2 \delta^2+a_3 \delta_3+\dots)^3-\dots=\delta$$

Since we are only interested in the first and second order terms, let's compare the coefficients in front of $$\delta$$ and $$\delta^2$$:

$$a_1=1 \\ a_2-\frac{a_1^2}{2^{3/2}}=0$$

Which means we have, up to 2nd order:

$$z=\delta+\frac{1}{2^{3/2}} \delta^2+\dots$$