# 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?

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).

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.