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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.

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  • $\begingroup$ 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
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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 $1/R$).

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

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