I have been trying to read Asymptotic Analysis by J.D Murray. There is a page in the book, rather, a certain line in this page, that has confused me thoroughly.
What was said on the page before, is that $h$ is a function such that $h$, $h'$, $h''$ are all continuous on an interval $[0,T]$ and a larger interval containing $0$ as an interior point, and $h'(0) = 0, h''(0)<0$ so $h$ has a local maximum at $0$. We may restrict to a small interval $[0,\delta]$ if necessary.
So far it has been established, that we can write $h(t) - h(0) = \frac{1}{2}t^2h''(\theta)$ which is easy to see via noticing that we can write this difference as a Taylor series remainder.
The part on this page that confuses me is when we write $h(t) - h(0) = -s^2$, and now look for a first approximation for $t$ in terms of $s$, so we write $\frac{1}{2}t^2 h''(0) + ... = -s^2$ , then in the next line in the page (2.27) we immediately get $t = (\frac{-2}{h''(0)})^{\frac{1}{2}} + \mathcal{O}(s^2)$. What is the $...$? and how do we go from this to (2.27)?
I put $...$ = $\mathcal{o}(t^2)$ which I think is true, but I still have no idea about this line and the line after it in general... Any insights would be very helpful to me , thanks!