# How to obtain an approximate expression for $\sqrt{\varepsilon}$ where $\varepsilon \ll 1$?

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 $\varepsilon = 0$.

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For a very small $\varepsilon$ you have that $\varepsilon$ is very close to $\sqrt\varepsilon$ anyway. If you want something else, take the inverse; take root; take inverse again. – Asaf Karagila Oct 10 '12 at 14:08
If you have $\varepsilon = a \times 10^{-2k}$, where $1 \le a < 10$, then $\sqrt\varepsilon = \sqrt{a} \times 10^{-k}$; otherwise, if $\varepsilon = a \times 10^{-(2k-1)}$, then $\sqrt\varepsilon = \sqrt{10a} \times 10^{-k}$. So $\sqrt{0.001} = \sqrt{10} \times 10^{-2} \approx 0.03$. – mjqxxxx Oct 10 '12 at 14:12
For rough estimate, shift by even number of places until we get something in range $10$ to $1000$. We know approximately the square roots of such numbers. – André Nicolas Oct 10 '12 at 14:32
Thanks for all the comments. All of your answers still use square roots. I am looking for something like $\sqrt{1+\varepsilon}\approx 1+\varepsilon/2$.. if something like this exists. – shark.dp Oct 10 '12 at 14:45
I think most people would tell you that $\sqrt{\epsilon}$ is the simplest function which behaves the way it does near zero; it's the thing you should be approximating other functions with, not the thing that needs approximating itself. Which is why all the answers you're getting involve the square root function again... – Micah Oct 10 '12 at 16:25

## 2 Answers

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.

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Thanks a lot for the answer (and all the comments above). I think I failed in clarifying what my intention was. I rather was looking for an expression like $\sqrt{1+\varepsilon}=1+\varepsilon/2$ and not a mental "procedure" to find the answer. I guess I shouldn't have mentioned the word "mental" at all. I am curious if there is a simple (that is why I mentioned "mental") approximate expression for $\sqrt{\varepsilon}$? Maybe there is no such expression. And thank you for the "mental trick" anyway! – shark.dp Oct 10 '12 at 15:31
@Marcks but the point with $\epsilon$ is that we want to show that something is smaller than it (normally), is there a trick to make the erroric approximation always less then epsilon? – saturatedexpo Jul 8 at 6:18

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 $\varepsilon=0$.

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Thanks a lot for your answer. That goes indeed in the right direction. Sadly, the $x_n$ expressions are more or less "expansions" around $\varepsilon=1$. The convergence is really fast. The Taylor expansions of $\sqrt{\varepsilon}$ and $x_3(\varepsilon)$ are identical up to $(\varepsilon-1)^7$! However, as you pointed out, the convergence for $\varepsilon\rightarrow 0$ is really bad. For $\varepsilon=0.001$ we have for example $x_3(0.001)\approx 0.128$ and $\sqrt{0.001}\approx 0.032$. So it would be really nice to have kind of an "expansion" around $\varepsilon=0$. – shark.dp Oct 10 '12 at 20:43
@shark.dp: If you want the convergence to be more uniform, then you can choose a better, $\varepsilon$-dependent, starting point. For instance, use $x_0=10^{-k}$ for all $\varepsilon \in [10^{-2k-1}, 10^{-2k}]$. – mjqxxxx Oct 10 '12 at 22:48
@shark.dp: But I do get your point. None of these approximations capture the qualitative behavior of the cusp at $0$. – mjqxxxx Oct 10 '12 at 22:49