Taylor expansion question I'm reading this paper, and on pages 10-11 it is asserted that one Taylor-expand the expression
\begin{equation*}
4\left(k_0 + k_1 - 2\sqrt{k_0k_1} \cos\left(\frac{x_1 - x_0}{2}\right)\right) 
\end{equation*}
at order two in $|x_1 - x_0|$, $|\sqrt{k_1} - \sqrt{k_0}| \ll 1$ to obtain
\begin{equation*}
k_0|x_1 - x_0|^2 + 4|\sqrt{k_1} - \sqrt{k_0}|^2 + \mathcal{O}\left(|x_1 - x_0|^2 \cdot |\sqrt{k_1} - \sqrt{k_0}|\right) \, . 
\end{equation*}
I believe "$\mathcal{O}$" is something like little-$o$, though that's not what I'm concerned about here.
My question is how the authors got this Taylor expansion. I understand that one can use the Taylor expansion $\cos x = 1 - x^2/2 + o(x^4)$ to find
\begin{align*}
4k_0 + 4k_1 - 8\sqrt{k_0k_1} \cos\left(\frac{x_1 - x_0}{2}\right)  &= 4k_0 + 4k_1 - 8\sqrt{k_0k_1} + 2\sqrt{k_0k_1}\cdot (x_1 - x_0)^2 \\
&\qquad- \frac{1}{3}\sqrt{k_0k_1} \cdot (x_1 - x_0)^4 + o\left((x_1 - x_0)^2\right) \\
&= 4|\sqrt{k_1} - \sqrt{k_0}|^2 + 2\sqrt{k_0k_1}\cdot (x_1 - x_0)^2 \\
&\qquad- \frac{1}{3}\sqrt{k_0k_1} \cdot (x_1 - x_0)^4 + o\left(\sqrt{k_0k_1} \cdot (x_1 - x_0)^4\right) \, ,
\end{align*}
but I don't understand what the authors do with the $\sqrt{k_0k_1}$ term.
I think it involves the remark that the authors make after: when $|\sqrt{k_1} - \sqrt{k_0}|^2 \ll 1$, one has
\begin{equation*}
k_1 = k_0 + \mathcal{O}\left(|\sqrt{k_1} - \sqrt{k_0}|\right) \, , 
\end{equation*}
but I don't understand why this is true. Perhaps one could plug in this expression in for $k_1$ in $\sqrt{k_0k_1}$, but I'd still like to know where this latter Taylor expansion comes from at least.
Summary of question: where do either of the two Taylor expansions (the ones that aren't mine) come from?
 A: Mixing up a lot of asymptotic notations, here... let's start with Hildebrand's "Short Course on Asymptotics" definitions (there are a lot of variants in the wild, be careful):

*

*$f(x) = O(g(x))$ (for $x_1 \le x \le x_2$) if there is a constant $c > 0$ so that $\lvert f(x) \rvert \le c \rvert g(x)$ whenever $x_1 \le x \le x_2$.
Often $x \to \infty$ is assumed, in which case the condition is $x \ge x_1$ for some $x_1$, or $x \to 0$, in which case the condition is $\lvert x \rvert \le \epsilon$ for some $\epsilon$. Here it is the later.

*$f(x) \ll g(x)$ is Vinogradov's notation for $f(x) = O(g(x))$, or they might write this to say here $\lvert \sqrt{k_1} - \sqrt{k_0} \rvert \ll 1$ is very small.

Under the assumption $k_0$ and $k_1$ are almost equal (second take on $\ll$), you certainly have:
Assume $\sqrt{k_1} = \sqrt{k_0} + d$, where $d$ is small. In that case, by the binomial theorem:
$\begin{align*}
   k_1
     &= k_0 + 2 k_1 d + d^2 \\
   k_1 
     &= k_0 + O(d) \\
     &= k_0 + O(\lvert \sqrt{k_1} - \sqrt{k_0}\rvert)
\end{align*}$
A bit of more careful bounding the expressions should give the above. Hildebrand's course gives a whirlwind tour on manipulating asymptotics (and assproximations via Taylor series).
A: As @vonbrand explained, there are two things mixed together.
For conveniency, let $y=\frac {x_1-x_0}2$ to make
$$B=\frac A4=k_0 + k_1 - 2\sqrt{k_0k_1} \cos\left(y\right) $$ Expand first for small $y$
$$B=\left(k_0+k_1-2 \sqrt{k_0 k_1}\right)+ \sqrt{k_0 k_1}\,y^2-\frac{\sqrt{k_0
   k_1}}{12}  y^4+O\left(y^6\right)$$
$$B=\left(\sqrt{k_0}-\sqrt{k_1}\right)^2+ \sqrt{k_0 k_1}\,y^2-\frac{\sqrt{k_0
   k_1}}{12}  y^4+O\left(y^6\right)$$ Now, expanding
$$\sqrt{k_0 k_1}=k_0+\frac{k_1-k_0}{2}-\frac{(k_1-k_0)^2}{8 k_0}+O\left((k_1-k_0)^3\right)$$ Just combine.
