In [this review](http://home.iscte-iul.pt/%7Ejaats/myweb/papers/new_kuras.pdf) of the Kuramoto model, Eq. 14 is obtained by expanding the following integral in powers of $K r$,


$$
r = K r \int\_{-\pi/2}^{\pi/2}\cos^2(\theta) g(K r \sin{\theta}) \mathrm{d}\theta
$$
where $g(\omega)$ is some unknown function (though is later taken to be the p.d.f. of the Lorentz distribution $g(\omega) = (\gamma/\pi)/(\gamma^2 + \omega^2)$, for some parameter $\gamma$). The asymptotic expression they obtain via the expansion is
$$
r \sim \sqrt{\frac{8(K-K\_{c})}{-K\_{c}^{3}g''(0)}}
$$
where $K\_c = 2/(\pi g(0))$. I don't see how you can expand this integral in powers of $Kr$ (I think this means a Taylor expansion in a new variable $z=Kr$ at $z=0$) without knowledge of $g$. Not much information is available, lots of other reviews appear to also omit this step. Is it a straightforward expansion? How is Eq. 14 obtained?



Power series expansion of the order parameter in the Kuramoto model
Presumably (because this is physics) $g$ is analytic in a neighborhood of the origin, so we can Taylor expand.


I think the really important part in deriving this equation for $r$ is the fact that the contribution from the first-order derivative term in $g$ is zero.
Just underneath equation (10), they assume that the frequency distribution is an even function, i.e. $g(\omega) = g(-\omega)$.
This implies that $g'(0) = 0$ and for that matter all the odd-order derivatives of $g$ are zero as well.
You could also observe that $\cos^2\theta\cdot\sin\theta$ is an odd function, so even if $g'(0)$ were non-zero, that term would integrate to zero anyway.


If we Taylor expand to second order in $Kr$ we get:


$$\begin{align}
1 & = K\int\_{-\pi/2}^{+\pi/2}\cos^2\theta\cdot g(Kr\sin\theta)d\theta \\
 & = K\int\_{-\pi/2}^{+\pi/2}\cos^2\theta\left(g(0) + \frac{g''(0)}{2}K^2r^2\sin^2\theta + \mathscr{O}(K^4r^4)\right)d\theta
\end{align}$$


Do you see where to go from here?


