The Analytic Atavar

Idiosyncratic Musings of a Retrograde Technophile

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Location: Chandler, Arizona, United States

Monday, February 12, 2007

Rooting for Old Squares

It is not commonly known that there exists an algorithm for manually calculating square roots which is somewhat similar to long division. This method was taught in public high schools until at least the 1930's, when my father learned it, and later passed it on to me. The existence of such a method was never mentioned in my high school in the 1960's and it is certainly unknown and ignored today with the availability of modern calculators. It is best illustrated by example, and I arbitrarily decided to find the meaningless square root of the current year number, 2007, as shown and explained in what follows:

(ii) (iii) (iv) (v) (vi) (vii) (viii)
 
4   4.  7   9   9   5   5   ...
20^ 07.  00^ 00^ 00^ 00^ 00^ ...    (i)
- 16      (ii)
4   07  
84 -3   36      (iii)
71   00  
887 - 62   09      (iv)
8   91   00  
8949 -8   05   41    (v)
85   59   00  
89589 - 80   63   01    (vi)
4   95   99   00  
895985 -4   47   99   25      (vii)
47   99   75   00  
8959905 - 44   79   95   25      (viii)
3   19   79   75  

i)  Split the number into 2 digit groups in both directions from the decimal point (indicated by the carat ^).
ii)  For the leftmost group (20), find the largest integer whose square is less than or equal to the group (4); write it above the group; square it (16), write below, and subtract to find the remainder. Bring down the next group of digits.
iii)  Double the digit of the root (2x4=8) and write it next to the remainder. Find the largest integer ? such that 8? x ? ≤ 407; in this case 4, which is written above. Multiply 84 x 4 = 336, write below the remainder, subtract, and bring down the next group of digits.
iv)  Double the digits of the root (2x44=88) and write it next to the remainder. Find the largest integer ? such that 88? x ? ≤ 7100; in this case 7, which is written above. Multiply 887 x 7 = 6209, write below the remainder, subtract, and bring down the next group of digits.
v)  Double the digits of the root (2x447=894) and write it next to the remainder. Find the largest integer ? such that 894? x ? ≤ 89100; in this case 9, which is written above. Multiply 8949 x 9 = 80541, write below the remainder, subtract, and bring down the next group of digits.
vi)  Double the digits of the root (2x4479=8958) and write it next to the remainder. Find the largest integer ? such that 8958? x ? ≤ 855900; in this case 9, which is written above. Multiply 89589 x 9 = 806301, write below the remainder, subtract, and bring down the next group of digits.
vii)  Double the digits of the root (2x44799=89598) and write it next to the remainder. Find the largest integer ? such that 89598? x ? ≤ 4959900; in this case 5, which is written above. Multiply 895985 x 5 = 4479925, write below the remainder, subtract, and bring down the next group of digits.
viii)  Double the digits of the root (2x447995=895990) and write it next to the remainder. Find the largest integer ? such that 895990? x ? ≤ 4959900; in this case 5, which is written above. Multiply 8959905 x 5 = 44799525, write below the remainder, subtract, and continue as required.
The process terminates when the remainder is zero and all remaining groups are also zero, otherwise it can be continued indefinitely.

Why does this curious process work? The doubling might raise the suspicion of some relation to Newton's method, but this is incorrect. An analysis can begin by considering the successive steps as generating a sequence of approximations to the square root, i.e.: {x1, x2, x3, x4, ...} = {40, 44, 44.7, 44.79, ...} and examining their error:
2007 - 402 = 407
2007 - 442 = 71
2007 - 44.72 = 8.91
2007 - 44.792 = 0.8559
...
Thus, on comparison with the example calculation above, the method correctly produces, in the form of the remainder, the error after each step. More generally, consider some number, n, and a series of approximations to its square root, {x1, x2, x3, x4, ...}. For some approximation, xi, the error is (n-xi2), the next digit of the approximation would be (xi+1-xi}, and the doubled multiplicative term is then (2xi+(xi+1-xi)), so the remainder for the (i+1) approximation is calculated by the method as:
(n-xi2)-(2xi+(xi+1-xi))·(xi+1-xi) = (n-xi2) - (xi+xi+1)·(xi+1-xi) = (n-xi2) - (xi+12-xi2) = (n-xi+12)
This shows that the method rests on an algebraic identity which correctly produces the remainder assuming the correct digit of the next approximation is found. The origin or author of this method is unknown to me, but it seems likely to have followed the introduction of arabic numerals.

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