Decades ago, when the slide rule was the norm, and calculators with the ability to calculate square roots were the cutting edge of technology, I was given an interesting formula. It went like this:

To estimate the $R$th root of $n$ using a simple calculator (where both $R$ and $n$ are positive numbers; fractions are allowed), do the following.

  1. Enter $n$ into the calculator.
  2. Press the $\sqrt{}$ button 12 times (i.e. calculate the 4096th root of $n$).
  3. Subtract 1.
  4. Divide by $R$.
  5. Add 1.
  6. Square the result (usually press the $\times =$ buttons) 12 times (i.e. calculate the 4096th power of the result).

Fascinatingly, this works pretty well.

For example, using this method on my calculator to find $\sqrt[8.5]{586.426}$ gives the answer with an error of just 0.05%.

This means that: $$\sqrt[R]{n} \approx \biggl(\frac{\sqrt[4096]{n} - 1}{R} + 1\biggr)^{4096}$$

Now, to me, the number $2^{12}$ seems arbitrary. So, my question is:

Why does this work?

I've tried muddling through equations to no avail.

EDIT: I've just spotted this question, which seems relevant.

  • $\begingroup$ Love that you included the slide rule link, hahaha! $\endgroup$ – Chase Ryan Taylor Aug 21 '17 at 12:52
  • $\begingroup$ Yes, @ChaseRyanTaylor — I have fond memories of them! Not so much the old log tables. $\endgroup$ – Paddy Landau Aug 21 '17 at 12:57
  • $\begingroup$ Oh my goodness. That doesn't sound too bad if you know the change-of-base formula. Those tables remind me of the trigonometric ones we sometimes used in geometry. $\endgroup$ – Chase Ryan Taylor Aug 21 '17 at 12:59
  • $\begingroup$ Ah, yes, @ChaseRyanTaylor, I had forgotten about those trigonometric tables. So glad that we don't have to use them any more. These are foreign concepts to my children. $\endgroup$ – Paddy Landau Aug 21 '17 at 13:20

You're correct that $2^{12}$ is arbitrary: it's good because it's large and a power of $2$ (so you can mash the square root button to get the $2^{12}$ root). Let the root we have available be the $k$th, so that in your case $k=4096$ (this just makes the calculation clearer), and the formula is then $$ f_k(x) = \left( 1+\frac{x^{1/k}-1}{R} \right)^k $$ We want to show that for large $k$, $f_k(x) \approx x^{1/R}$. Since $k$ is large and $x$ is positive, we can use the exponential's power series to write $$x^{1/k} = e^{(1/k)\log{x}} = 1 + \frac{1}{k}\log{x} + \frac{1}{2k^2}(\log{x})^2 + O(k^{-3}), $$ where $O(g(k))$ means that the term decreases at the same rate as $g(k)$ as $k \to \infty$, so $$ 1 + \frac{x^{1/k}-1}{R} \approx 1 + \frac{1}{Rk}\log{x} + \frac{1}{2Rk^2}(\log{x})^2 + O(k^{-3}) $$ At this point it's probably most convincing to take logs of everything, so $$ \log{f_k(x)} = k\log{\left( 1+\frac{x^{1/k}-1}{R} \right)} = k\log{\left( 1 + \frac{1}{Rk}\log{x} + \frac{1}{2Rk^2}(\log{x})^2 + O(k^{-3}) \right)}, $$ and since $k$ is large, the argument of the logarithm is close to $1$, and we can apply the power series for the logarithm to give $$ \log{f_k(x)} = \frac{1}{R}\log{x} + \frac{1}{2Rk}(\log{x})^2 - \frac{1}{2R^2k}(\log{x})^2 + O(k^{-2}) $$ Hence $$ \log{\left( \frac{f_k(x)}{x^{1/R}} \right)} = \frac{R-1}{2R^2k}(\log{x})^2 + O(k^{-2}). $$ Exponentiating again gives the error in the approximation as approximately $$ \frac{f_k(x)}{x^{1/R}} -1 \approx \frac{R-1}{2R^2k}(\log{x})^2, $$ which will obviously be very small for large $k$ and "ordinary-sized" $R$ and $x$ (or even large $R$, since the error depends on $1/R$).

  • 1
    $\begingroup$ Man, that's clever! Thank you $\endgroup$ – Paddy Landau Aug 21 '17 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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