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I was reading about the methods that the Ancient Babylonian Civilization used for approximately calculating square roots:

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  • Could this method be used for approximately calculating any "root" - for instance, could we use this method to approximately calculate the cube of some number "S"?

  • Could this method be used for approximately calculating an exponent with a "decimal argument"? For instance, the square root of "S" can be written as S^0.5 - Could we use this method for approximately calculating S^0.3?

Although there are now more modern ways to approximate these calculations, I am interested in learning about the limitations of these ancient methods that existed far before calculators and computers!

Thanks!

Source:

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  • $\begingroup$ The Babylonian square root algorithm is just a special case of Newton's method. $\endgroup$
    – PM 2Ring
    Feb 1, 2022 at 6:33
  • $\begingroup$ Why do you want an algorithm to approximate the cube of S? It's just S×S×S. If you mean the cube root, then sure, that's easy to do with Newton's method. $\endgroup$
    – PM 2Ring
    Feb 1, 2022 at 6:38
  • $\begingroup$ Thank you for your reply! I wonder if the Newton method can be used for decimal exponents? $\endgroup$
    – stats_noob
    Feb 1, 2022 at 6:39

1 Answer 1

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Yes. The Baylonian method can indeed be generalized easily to $$x_n=\frac{1}{k}\left((k-1)x_{n-1}+\frac{S}{x_{n-1}^{k-1}}\right)$$ to compute any $k-th$ root (although the convergence is very slow). Once computed the $k-th$ root you can then compute the $l-th$ power if you want to calculate $S^{l/k}$

Remark: It's possible to prove that this recursion converges, yet it is quite a mess.

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  • $\begingroup$ Thank you so much for your answer! Can the Newton method can be used for decimal exponents? For instance, in the above formula, can K = 0.127? $\endgroup$
    – stats_noob
    Feb 1, 2022 at 6:40
  • $\begingroup$ Is there some reference you might have for understanding why the Babylonian method can be generalized this way? $\endgroup$
    – stats_noob
    Feb 1, 2022 at 6:41
  • $\begingroup$ Sure $K=0.127=\frac{127}{1000}=\frac{l}{k}$ Now compute the $1000-$th root and its $127$ power. Obviously this does not generalize to irrational exponents. $\endgroup$
    – b00n heT
    Feb 1, 2022 at 6:41
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    $\begingroup$ Here's a graph of that linear approximation. The next comment has a small Python demo of this technique. $\endgroup$
    – PM 2Ring
    Feb 1, 2022 at 10:42
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    $\begingroup$ sagecell.sagemath.org/… $\endgroup$
    – PM 2Ring
    Feb 1, 2022 at 10:43

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