How was this approximation of transcendent equation solution found? I have an equation for $\xi$:
$$\xi\gamma=\cos\xi,$$
where $\gamma\gg1$. I've tried solving it assuming that $\xi\approx0$ and approximating $\cos$ by Taylor's second order formula:
$$\xi\gamma\approx1-\frac{\xi^2}2,\tag1$$
then I get
$$\xi\approx2\left(\sqrt{\gamma^2+1}-\gamma\right).\tag2$$
The book I read gives the approximate solution of it as
$$\xi\approx\frac1\gamma\left(1-\frac1{2\gamma^2}\right).\tag3$$
When I tried to understand how they got this, my first thought was like "they made another simplification and assumed $\xi\approx\frac1\gamma$ in $(1)$". I thus thought that this approximation would be worse. But when plotting these solutions as functions of $\gamma$, I found that $(3)$ in fact converges much faster to numerical solution than $(2)$!
So I now wonder: how did they get $(3)$?

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{}$
\begin{align}
\xi = {\cos\pars{\xi} \over \gamma}
\end{align}

It leads to the iteration:
  $$
\xi_{n + 1} ={\cos\pars{\xi_{n}} \over \gamma}\,,\qquad \xi_{0} = 0
$$

$$
\xi_{1} = {1 \over \gamma}\,,\qquad\xi_{2}={1 \over \gamma}\,\cos\pars{1 \over \gamma}\approx\color{#66f}{\Large{1 \over \gamma}\pars{1 - {1 \over 2\gamma^{2}}}}
$$
since $\ds{\cos\pars{x} \approx 1 - \half\,x^{2}}$ when $\ds{x \approx 0}$.
