How to Estimate Deep Roots in Mental Math

One of the most difficult challenges in mental calculation is for deep roots—meaning cube roots, 4th roots, 5th roots (etc.), where the answer is not an integer. For example, the 7th root of 8500 is 3.64204712…

For square roots, there are simpler methods, such as:

For cube roots, there is also a difficult method to calculate them to as many digits of accuracy as you like.

For deep roots, there is a method involving logarithms, which gets the answer to about 4–5 digits of accuracy, but it extremely difficult to perform. This does however also work for crazy roots and powers, such as \(\sqrt[7.3]{100}\) and \(7.7^\pi\), in case you ever need to solve those!

This article will show the easiest method for solving deep roots, although it is less accurate than the methods described above.

Estimation Method for Deep Roots

As an example, let’s calculate the following 5th root: \(\sqrt[5]{200} = 2.88539981…\). In this case, the relevant power is n = 5, for the 5th root.

You might find it useful to first read the article on estimating square roots, as the method is similar, but more complex.

Choose a nearby guess, \(G\). For example, \(\sqrt[5]{200} \approx 3\), as we know or calculate that \(3^5 = 243\).
Calculate the difference between your guess, raised to the power, and the number you want to deep root. In this example, \(3^5 – 200 = 43\).
Estimate the gradient—using the calculus formula \(n \times G^{n – 1}\). In this case, the gradient is approximately \(5 \times 3^4 = 405\).
Divide the difference by the gradient to obtain the correction. In this case, \(\frac{43}{405} \approx 0.106\)
Add or subtract the correction to the original guess to get the estimate of \(3 – 0.106 = 2.894\).

This was only accurate to 2 significant figures—nearly 3!

You can see that this method is similar to, but less reliable than, the method for square roots.

Improvements to Accuracy

There are two main ways to improve the accuracy. The first is to use a closer guess. In the example above, we could have used 2.9 rather than 3. However, this makes the calculation of the difference \(2.9^5 – 200 = 5.11149\) very difficult!

The other way is to use an alternative estimate of the gradient. The best value is actually \(ng^{n-1}\), where \(g\) is an unknown value approximately halfway between the original guess and the true value of the nth root.

Of course, this is not very helpful unless you know the true value! To succeed, you need to be creative with estimations. Here is an example of how I might solve the previous question more accurately.

My guess is 3, and my difference is 43, as already calculated.
My gradient is \(5 \times 3^4 \approx 400\)
My estimate is therefore \(3 – \frac{43}{400} \approx 2.9\)
Now I have this rough estimate, I’ll recalculate the gradient and proceed more accurately, taking 2.95 as the midpoint between my guess and the true value.
My updated gradient is about \(5 \times 2.95^4 \approx 5 \times 3^2 \times 2.9^2 \approx 5 \times 9 \times 8.4 = 42 \times 9 = 378\)
My updated correction is \(\frac{43}{378} \approx 0.1138\)
My final estimate is \(\sqrt[5]{200} \approx 3 – 0.1138 = 2.8862\)

This is more that 10× as accurate as the previous estimate. Note that the most difficult part is still the division by 378, although you could approximate this in various ways instead.

Finally, I repeat that this method only gives accurate estimates when you choose an initial guess that’s close to the true answer. Otherwise, I recommend the logarithm method, despite its complexity!

Mathematics of this Method

The related article on estimating square roots includes an explanation of why it works, using calculus. The same reasoning explains the deep algorithm, except that the derivative of \(y = x^n\) is, in general, \(n x^{n-1}\). The square roots only considers the case when \(n = 2\) and the gradient is \(2x\).