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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! $\endgroup$ Aug 21, 2017 at 12:52
  • $\begingroup$ Yes, @ChaseRyanTaylor — I have fond memories of them! Not so much the old log tables. $\endgroup$ Aug 21, 2017 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. $\endgroup$ Aug 21, 2017 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. $\endgroup$ Aug 21, 2017 at 13:20

2 Answers 2

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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 $\endgroup$ Aug 21, 2017 at 12:57
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As you point out, $2^{12}$ just seems to be arbitrary and is used to denote a number that is sufficiently large. So we can reach the required result by just taking limit to infinity instead of the factor $2^{12}$. i.e. We need to prove that,$$\sqrt[n]{x} = \lim_{k \to \infty}\left(1+ \frac{x^{1/k} - 1}{n}\right)^k$$ which is not a hard nut to crack.


Here's the computation of the limit. Consider,$$\begin{align}y &=\lim_{k \to \infty}\left(1+ \frac{x^{1/k} - 1}{n}\right)^k\\& = \color{blue}{\lim_{k\to \infty} \left(1 + \frac{x^{1/k} - 1}{n}\right)}^{{\color{blue}{\left(\frac{n}{x^{1/k} -1}\right)}}\cdot \left(\frac{x^{1/k} - 1}{n}\right)\cdot k}\tag{1}\\& = \lim_{k\to \infty}\exp{\left(\frac{x^{1/k} - 1}{n}\right)\cdot k}\tag{2}\\& = \exp{\left(\color{blue}{ \lim_{k\to \infty}\frac{x^{1/k} - 1}{1/k}}\right)\cdot \frac1n} \\&= \exp\left(\log(x)\right) \cdot \frac1n\tag{3}\\& = \exp \left(\log(x^{1/n}) \right) \\& =\boxed{\sqrt[n]{x}}\end{align}$$


Explanation:

$(1)$ Here I made some adjustments in the exponent term to make it in the form of a standard limit.

$(2)$ Notice that, it is of the form $\lim_{x\to\infty} \left(1 + \frac1x\right)^{x} = e.$

$(3)$ Here we can simply use $\lim_{x\to 0} \frac{x^a - 1}{a} = \log (a)$.

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  • $\begingroup$ You first claim has a $k$ in it, which I guess should be an $n$. $\endgroup$
    – Sahaj
    Jan 21 at 12:07
  • $\begingroup$ Yes. A typo. Fixed it. Thanks! :) $\endgroup$
    – Utkarsh
    Jan 21 at 12:10
  • $\begingroup$ @Sahaj BTW our rep points just differ by a 1 lol $\endgroup$
    – Utkarsh
    Jan 21 at 12:13
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    $\begingroup$ Excellent, thank you! $\endgroup$ Jan 21 at 15:14

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