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## Estimating Roots

Published on Sunday, March 29, 2015 in , , ,

3 years ago, I posted a tutorial about estimating square roots of non-perfect squares, including tips and tricks.

Since then, I've wondered if there was a general formula for estimating other roots, such as cube roots, fourth roots, and so on. Reddit user InveighsiveAd informed me that there's a simple general formula very similar to the method I've taught for square roots! Once you pick up the basic idea of this method, you'll be able to astound friends, family, and teachers.

The approach for estimating roots originates from an approach developed by Leonhard Euler, and involves taking derivatives, so I won't delve into the math behind why this works here. I'll focus more on the resulting formulas, which can be used to

The method I taught for estimating square roots basically boiled down to this formula, where a was a perfect square equal to or less than x, and b was equal to x - a:

$\\&space;\sqrt{x}=\sqrt{a+b}\approx&space;\sqrt{a}+\frac{b}{2\sqrt{a}+1}$

With Euler's method, we'll be estimating roots using the same basic approach of breaking up a number into a number which is a perfect power (square, cube, 4th power, etc.) and the difference between that power and the targeted number. The following formula may look scary at first, but it's simpler than it looks:

$\\&space;\sqrt[y]{x}&space;=&space;\sqrt[y]{a&space;\pm&space;b}&space;\approx&space;\sqrt[y]{a}&space;\pm&space;\frac{b}{y&space;(\sqrt[y]{a})^{y-1}}$

y is simply the root we wish to know. For square roots, y would equal 2, for cube roots, y would equal 3, and for 4th roots, y would equal 4. As a matter of fact, I'm not going to concern this article with anything past 4th roots, as this quickly becomes complex. Here are the formulas for square, cube and 4th roots individually:

$\\&space;square&space;\&space;roots:&space;\&space;\sqrt{x}&space;=&space;\sqrt{a&space;\pm&space;b}&space;\approx&space;\sqrt{a}&space;\pm&space;\frac{b}{2\sqrt{a}}&space;\\&space;\\&space;cube&space;\&space;roots:&space;\&space;\sqrt[3]{x}&space;=&space;\sqrt[3]{a&space;\pm&space;b}&space;\approx&space;\sqrt[3]{a}&space;\pm&space;\frac{b}{3(\sqrt[3]{a})^{2}}&space;\\&space;\\&space;4th&space;\&space;roots:&space;\&space;\sqrt[4]{x}&space;=&space;\sqrt[4]{a&space;\pm&space;b}&space;\approx&space;\sqrt[4]{a}&space;\pm&space;\frac{b}{4(\sqrt[4]{a})^{3}}$

These look worse than they really are. Remember that a is always chosen to be a perfect power, so you're working with an easily determined number. If you were going through this process for cube root, and using 729 for a, the cube root of 729 would be 9. So, any where you see the cube root of a, you can mentally replace it with 9, in this example.

Obviously, knowing perfect squares up through 31 will be of help, as in the original method. Knowing the perfect cubes from 1 to 10, as many Grey Matters readers already do, will allow you to estimate cubes of number up to 1,000. Memorizing or being able to quickly calculate perfect 4th powers will allow you to estimate 4th powers up to 10,000!

For those confused by the ± symbol in the equations, it simply means that we're going to choose a to be the closest perfect power, and adjust b accordingly. For example, if we want the cube root of 340, then we'd use 343 (73), and work it out as the cube root of 340 as the cube root of (343 - 3).

Let's estimate the cube root of 340 as a full example. As explained above, we've already broken this up into the cube root of (343 - 3). Your mental process might go something like this:

$\\&space;cube&space;\&space;roots:&space;\&space;\sqrt[3]{340}&space;=&space;\sqrt[3]{343-3}&space;\approx&space;7&space;-&space;\frac{3}{3\times&space;7^{2}}&space;\\&space;\\=7-\frac{3}{3\times49}=7-\frac{3}{147}=7-\frac{1}{49}=6\frac{48}{49}$

How close is 64849 to the cube root of 340? The two numbers are very close, as this Wolfram|Alpha comparison shows!

Colin Beveridge, of Flying Colours Maths has helpfully pointed out that the error in the method will increase as you get approach the geometric mean of two closest consecutive perfect powers. For example, when using this method to find the cube root of 612, which is close to 611 (the approximate geometric mean of 512 and 729), you'll be farther off.

Let's find out exactly how far off we would be. The cube root of 612 could be worked out as (729 - 117), but (512 + 100) is closer, so we'll use the latter. Working this out, we'd get:

$\\&space;cube&space;\&space;roots:&space;\&space;\sqrt[3]{612}&space;=&space;\sqrt[3]{512+100}&space;\approx&space;8&space;+&space;\frac{100}{3\times&space;8^{2}}&space;\\&space;\\=8+\frac{100}{3\times64}=8\frac{100}{192}=8\frac{25}{48}$

Wolfram|Alpha shows that 82548 ≈ 8.52, while the actual cube root of 612 ≈ 8.49. It's off by about 3 hundredths, but that's still a good estimate!

As an added bonus, if you wind up with a fraction whose denominator ends in 1, 3, 5, or 7, you can use the techniques taught in Leapfrog Division or Leapfrog Division II to present your estimate with decimal accuracy! Yes, it's just the same number presented differently, but working out decimal places in your head always comes across as impressive. Personally, I reserve the decimal precision for when I know the root is close to a perfect power.

Try this approach out for yourself. If you have any questions, feel free to ask them in the comments!

### 1 Response to Estimating Roots

9:20 PM

I think that using this method, I can only mentally calculate for up to fourth root since I have memorized the first 100 cubes but it would still be very difficult.

Although it's easy to extract exact 5th roots, I don't think that I could handle the mental strain of raising a number to its fourth power in my head especially if it's very large :)