TITLE: Approximate $\sqrt{e}$ by hand
QUESTION [19 upvotes]: I have seen this question many times as an example of provoking creativity. I wonder how many ways there are to approximate $\sqrt{e}$ by hand as accurately as possible. 
The obvious way I can think of is to use Taylor expansion. 
Thanks

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\root{\expo{}_{n}} = x_{n}}$. With Newton-Rapson: $\ds{x_{n + 1} = \half\pars{x_{n} + {e \over x_{n}}}\,,\quad x_{0} = 2}$

$\ds{\large\tt\mbox{With}\quad \color{#66f}{\Large n = 3}}$:
  \begin{align}&{\LARGE\sqrt{\expo{}}}\approx
\frac{1}{2} \left\{\frac{1}{2} \left[\frac{1}{2} \left(2+\frac{e}{2}\right)+\frac{2 e}{2+\frac{e}{2}}\right]+\frac{2 e}{\frac{1}{2} \left(2+\frac{e}{2}\right)+\frac{2 e}{2+\frac{e}{2}}}\right\} = x_{3}
\approx 1.648721295
\end{align}

$$
\mbox{Relative Error} = \verts{{x_{3} \over \root{\expo{}}} - 1}\times 100\ \%  =1.48\times 10^{-6}\ \%
$$