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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:

stats_noob
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  • The Babylonian square root algorithm is just a special case of Newton's method. – PM 2Ring Feb 01 '22 at 06:33
  • 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. – PM 2Ring Feb 01 '22 at 06:38
  • Thank you for your reply! I wonder if the Newton method can be used for decimal exponents? – stats_noob Feb 01 '22 at 06:39

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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.

b00n heT
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  • 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? – stats_noob Feb 01 '22 at 06:40
  • Is there some reference you might have for understanding why the Babylonian method can be generalized this way? – stats_noob Feb 01 '22 at 06:41
  • 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.

     – b00n heT Feb 01 '22 at 06:41
  • I don't have any reference... the generalization becomes quite apparent when computing the limit of the original Babylonian method – b00n heT Feb 01 '22 at 06:43
  • Notice: I've edited the esponent of $x_{n-1}$ as I had a small mistake. – b00n heT Feb 01 '22 at 06:46
  • Thank you so much for your answer! I am trying this with a pen and a paper right now! – stats_noob Feb 01 '22 at 06:48
  • You are welcome. Best of luck. It's easy to implement it in Excel as well if you want :) – b00n heT Feb 01 '22 at 06:49
  • With a good initial $x$, Newton's method converges quadratically, unless the function has duplicate roots. You can get a good initial $x$ for $y=x^3$ by reducing $y$ to a suitable range. Eg, let $m=y/2^{3b}$ with $1/8\le m<1$. Use Newton's to find $x=m^{1/3}$, with initial $x=4m/7+15/32$. Then $(x\cdot2^b)^3=y$. The absolute error in the initial linear approximation is <0.04 – PM 2Ring Feb 01 '22 at 10:39
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    Here's a graph of that linear approximation. The next comment has a small Python demo of this technique. – PM 2Ring Feb 01 '22 at 10:42
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    https://sagecell.sagemath.org/?z=eJxdUsFOg0AQvfMVLzVRqDS0RaM2Yrx5MR78AbMtg25kd8myxCVN_91hoa1xLsDMmzdv3nCBN_pxRl-1UOS-TJli120J1hgXVdYoKOG-IFVjrENlyTcp6pIfUVRSFcAfAzjuk00EDlmhR1FguYEl11mNZRQKF3ghx3TaybYVULgE0xhN2oECQqUgFOMU5pu63k2nS3TNGe0MuL-rnWxqgqmQB2jFvQtCliHHHIsxqTgZ9MaBfYEqOcp5lZqEhWgaa3zItQxeDRsoPGEJqlvuWIWS55Ji2pvsDtcMnGN1m-XrI9c_F0c9xkJCalihPyl-mAwaohzoYoUMfj5fsyqf8Ht-qjdWahdXs708bLD3hxT70h9myQngcV0wy-mbNYttG5fM84gVLe7Ps4bYWhLfR63v1DpjCe1O1KPx06FGn3zKTg4mslHPLIOs2LlwbMeNcV-s76ZN-uFYtRHueKxR9qzfzFJMqWHRvz_JHxwP4u3z5BdGpKvr&lang=python – PM 2Ring Feb 01 '22 at 10:43