Heron's Cube Root Method

 

 

As is my nature, I love to roam through old math journals, and recently I found a 1920 article on a beautiful hand calculation for cube roots by Heron from his long lost Metrica.  Until nearly the end of the Nineteenth Century, Math Historians were writing that the ancient Greeks could not calculate square roots (they seldom even mentioned the thought of a cube root) because, "We can't find the evidence so they didn't know it." The discovery of Heron's Metrica in 1897 provided the evidence with several examples of square roots, and one single example of a Cube root.

This is a pretty accurate cube root for 2000 years ago, accurate to the third decimal place, and an error in the root of a little less than 0.0013  that's the total error. If you read it carefully, a question that has plagued all who have studied it.  What is the origin of the 100 added to 180 to get 280?  Some early researchers thought it was the original number, 100, and dismissed Heron's method as not very accurate for many numbers.  But this article provides an explanation that is, in the words of the author, closer than seven digit logarithms for large numbers (he didn't define).  

After doing a few by hand, I "cheated" and built a spreadsheet to compute them and give the error.  But if you try a few by hand, you will agree that it is way ahead of that thing they call the cube root algorithm in YouTube videos.  So I'll let You in on the secret, (but don't tell the other kids!).

Heron's method begins with the surrounding roots of the number.  For the 100 example he used, the lower root bound is 4, and the upper is 5, that is, the number 100 is bound by 64 and 125.   Now we need to determine what Heron called the excess, and the deficit.  The excess is 5^3 - 100 or 25.  The deficit is 100 - 4^3 = 36.  

Then we form the numerator of the fractional part of the solution, multiply the deficit by the upper root bound, 5, producing 180.  

Now for the denominator, we take the 180 from the numerator, and add the product of the excess times the lower bound, producing the unexplained 100.  To get our final root estimate, we add the lower root, 4 to this fraction.  My calculator gives 4.642857 but for the actual cube root of 100 it gives  4.6415888..   That's a total error of a smidge more than 00126.  If you cube the approximation you get 100.08199.   

I ran a spreadsheet for numbers from 10 to 330 and notice a few quirks I believe.  The error seems greater when the deficit is more than the excess.  Which actually makes 100 look worse than average.  If you try 90 instead, where the excess and deficit are 35 and 26, almost the opposite of 100 you get much greater accuracy.   For 90, the true value is 4.481404747, and the estimate is 4.481481481, and the total error is 0.0000076734.  The cube of the estimate is 90.0046  (What we country folk call Pretty Dang Good!)

It also seems that as the numbers get bigger, the errors get slightly smaller, and I can believe the 7 digit logarithm quote.  

So if you are driving down the road and factoring the license plate number in front of you is trivial (Oh, Come On, you know you all do that!)  try taking the cube root in you head.  Fun on the Highway .... (I missed what turn off??????)

ADDEDUM, With a Hat Tip to Keith Raskin;

So if you only read really old journals, you may miss something really important.  Case in point, the 2017 Bulletin of Parnas Mathematical Society, which is in Brazil (Brasil)   It

The Bulletin has an extension of Heron's root taking method for any odd root.   I've only just begun playing with it, but I'll add the method here, and my results for 5th roots.  

The system is much the same, but with a power used in the multiplication of the roots and the excess and deficiency, and the first new task is to find out the power k we want to use in those products.  The key is to let the root you seek, n, be equal to 2k+1; so for the fifth root, since 2k+1 = 5 gives us k=2, we want to square the lower and upper bounds of the root when we multiply.  

Example   Find the fifth root of 100 (which your calculator will tell you is 2.51188...  ) .   

The bounding roots are 2 and 3.  The deficit is 100 - 2⁵ = 68, and the excess is 3⁵- 100 = 143.  Now we proceed as before, except the numerator will be 3² times the deficit, 3² x 68 = 612.

 Now for the denominator we find the product we add to the numerator, which is the square of the lower root bound, 2, times the excess, 143, 2² x 143 = 572.  So the fractional part of our estimate will be 612 / (612 + 572) ... which is 153/296, or as a decimal, 0.51689.  Adding our lower root bound we get 2.051689. An error of 0.00500.