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To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.

1)Take the number $X$, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e. $\sqrt{\sqrt{\sqrt{\sqrt{....X}}}}$

2)next, subtract $1$, divide by $3$ (for cube root, and any number $n$ for $n$th root), add $1$.

3) Then square the resultant number (say $c$) 13times (or 10 times if you had taken out root 10 times) i.e. $c^{2^{2^{....2}}}=c^{2^{13}}$. This yields the answer.

I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.

For finding the log, the method is similar:-

1)Take 13 times square root of the number, subtract 1, and multiply by $3558$. This yield s the answer.

Why do these methods work? What is the underlying principle behind this?

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  • $\begingroup$ For step 3, squaring $c$ for 13 times is $c^{2^{13}}$. $\endgroup$
    – peterwhy
    Oct 29, 2013 at 8:40
  • $\begingroup$ @peterwhy Added in the question $\endgroup$
    – stochastic13
    Oct 29, 2013 at 8:44

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Let's use these classical formulae :

$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$ $$\ln\,x=\lim_{n\to\infty}n\left(x^{1/n}-1\right)$$

to get (replacing the limit by a large enough value of $n$ : $N=2^{13}$) : \begin{align} \sqrt[3]{x}=e^{\left(\ln x/3\right)}&\approx \left(1+\frac {\ln x/3}N\right)^N\\ &\approx \left(1+\frac {N\left(x^{1/N}-1\right)}{3\,N}\right)^N\\ &\approx \left(1+\frac {\left(x^{1/N}-1\right)}3\right)^N\\ \end{align}

Concerning the decimal logarithm we have : $$\log_{10}\,x=\frac{\ln\,x}{\ln\,10}\approx \frac N{\ln\,10}\left(x^{1/N}-1\right)$$

For $N=2^{13}$ we may (as indicated by peterwhy) approximate the fraction with $$ \frac N{\ln\,10}=\frac {2^{13}}{\ln\,10}\approx 0.4343\times 8192\approx 3558$$

Hoping this clarified things,

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    $\begingroup$ So by similar reasoning, the $3558$ in the second question comes from $2^{13}/\ln10=3557.7\ldots$. $\endgroup$
    – peterwhy
    Oct 29, 2013 at 9:41
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    $\begingroup$ @RaymondManzoni If I understood correctly, for the first method, any number of times the square root operation can be performed, greater the number greater the accuracy. For the second method (taking log), we can use a different number, but we then also have to change $3558$ to another appropriate number.? $\endgroup$
    – stochastic13
    Oct 29, 2013 at 11:34
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    $\begingroup$ Theoretically you can use any number of square root operations for the first method, but if you are using a pocket calculator to actually compute the square roots repeatedly, you will end at 1 after some 10 steps and the cube root will be zero. But even if you stop earlier the results get worse. For the real world compuation there is an 'optimal' number of steps. $\endgroup$
    – gammatester
    Oct 29, 2013 at 11:51
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    $\begingroup$ @SatwikPasani: The relative error for the cube root should be nearly $\dfrac{\left(\ln(x)/3\right)^2}N$ that is : $$\left(1+\frac {\left(x^{1/N}-1\right)}3\right)^N\approx \sqrt[3]{x}\cdot\left(1+\frac{\left(\ln(x)/3\right)^2}N\right)$$ so that a larger $N$ (especially for values of $x$ near to $1$) should give you better approximations. But as indicated by gammatester this will work only if you didn't 'consume' the available precision with the successive square roots (i.e. with a value of 1.000000000107 and a precision of $12$ digits you can't get much better than a three digits precision). $\endgroup$
    – Raymond Manzoni
    Oct 29, 2013 at 14:13
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    $\begingroup$ Concerning the second method adding one more iteration ($m=14$) will have to be compensated by doubling the multiplicative constant to nearly $7115$ and so on for larger $m$. In general after m iterations you'll have to use the value $\dfrac {2^m}{\ln\,10}$ instead of $3558$. $\endgroup$
    – Raymond Manzoni
    Oct 29, 2013 at 14:53

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