Is there a way to obtain an approximate expression for the square root $\sqrt{\varepsilon}$ of a small number $\varepsilon \ll 1$?
To be more precise, I would like to have an expression which (1) I can easily handle by a mental calculation and (2) does not involve a square root. Of course, I can easily calculate $\sqrt{0.01}$ but I have to admit that I would have to think a bit harder for $\sqrt{0.001}$.
I commonly use Taylor series expansions to calculate approximate results for expressions like $(1+\varepsilon)^\alpha \approx 1 + \alpha \varepsilon$ but this approach obviously fails here since $\sqrt{\varepsilon}$ is not analytic for $\varepsilon = 0$.
$\endgroup$ 72 Answers
$\begingroup$Write $\varepsilon$ as the product of $a$ and $10^{-n}$, where n is an even number. For a simple mental approximation of its square root, take $b$ to be a known square close to $a$ and evaluate:
$$\sqrt{\varepsilon}\approx\left(\sqrt{b}+{{a-b} \over 2 \sqrt{b}}\right)10^{-n/2}$$
Example: $$\sqrt{0.17}=\sqrt{17*10^{-2}}\approx\left(\sqrt{16}+{{17-16} \over 2 \sqrt{16}}\right)10^{-1}={33\over8}10^{-1}=0.4125$$
Which is a fairly accurate approximation of $\sqrt{0.17}=0.412311...$ The error in using this method is visualized below.
Yes, that technically does involve square roots, but if you can mentally calculate the square root of 0.01, I take it the square root of 16 borders on acceptability.
$\endgroup$ 2 $\begingroup$If the goal is not to have a mental process, but rather to approximate $\sqrt{\varepsilon}$ purely in terms of simpler functions, there are several ways to go about it. As you pointed out, there is no Taylor series expansion around $\varepsilon=0$; but if you know that $0<\varepsilon\le 1$, say, you can certainly use the Taylor series expansion around $\varepsilon = 1$ or $\varepsilon = 1/2$, either of which will converge.
An alternative is to use the successive iterates generated by Newton's method applied to $f(x)=x^2-\varepsilon.$ As long as you start with $x_0 \ge \sqrt{\varepsilon}$ (e.g., take $x_0=\max(1,\varepsilon)$), then these iterates will converge monotonically from above. The iterates are defined by $$ x_{n+1} = \frac{1}{2}\left(x_{n} + \frac{\varepsilon}{x_{n}}\right).$$ So your first approximation (assuming $\varepsilon < 1$ for simplicity) is $$ x_1 = \frac{1 + \varepsilon}{2}; $$ your second is $$ x_2 = \frac{1}{2}\left(x_1 + \frac{\varepsilon}{x_1}\right)=\frac{1+\varepsilon}{4}+\frac{\varepsilon}{1+\varepsilon};$$ your third is $$ x_3 = \frac{1}{2}\left(x_2 + \frac{\varepsilon}{x_2}\right)=\frac{1+\varepsilon}{8}+\frac{\varepsilon}{2(1+\varepsilon)}+\frac{2\varepsilon(1+\varepsilon)}{(1+\varepsilon)^2+ 4\varepsilon}; $$ and so on. As seen in the figure below, these iterates converge fairly rapidly, with more reluctant convergence near $\varepsilon=0$.
