Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\varepsilon = 0$.

share|cite|improve this question
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
up vote 3 down vote accepted

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.

Error plot

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.

share|cite|improve this answer
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 $\varepsilon=0$.

enter image description here

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.