Bakhshali FormulaDate: 12/15/2002 at 04:35:00 From: Murtaza Khaliq Subject: I have a formula for calculating sq-roots by hand I have been going through the site and have noticed that people have been asking for a formula for calculating square roots. I came across this "Bakhshali formula" only a few weeks ago myself. I hope it is useful. Q = (A^2 + b) = A + b/2A - (b/2A)^2/[2(A + b/2A)] In the case of a non-square number, subtract the nearest square number, divide the remainder by twice this nearest square; half the square of this is divided by the sum of the approximate root and the fraction. This is subtracted and will give the corrected root. Example: Taking Q = 41, then A = 6, b = 5 and we obtain 6.403138528 as the approximation to 41 = 6.403124237. Hence we see that the Bakhshali formula gives the result correct to four decimal places. More info can be found in the article at the MacTutor History of Mathematics archive at St. Andrews University (select The Bakhshali manuscript): History Topics http://www.gap-system.org/~history/Indexes/Indians.html Thanks. Date: 12/16/2002 at 15:11:56 From: Doctor Peterson Subject: Re: I have a formula for calculating sq-roots by hand Hi, Murtaza. In copying the formula, you lost the square root signs; I'll rewrite it as sqrt(Q) = sqrt(A^2 + b) = A + b/(2A) - (b/(2A))^2/[2(A + b/(2A))] It's interesting to compare an ancient method like this to those of modern times and other historical periods. I see a strong resemblance between this and the divide-and-average method covered in our FAQ, which is a special case of Newton's method, and was known in ancient Egypt: Square/cube roots without a calculator http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html Newton's Method and Square Roots http://mathforum.org/library/drmath/view/52644.html Let's try applying the divide-and-average method to Q = A^2 + b. Our first guess will be A, so we divide Q by A and average that with A: B = (A + Q/A)/2 = (A + (A^2 + b)/A)/2 = (A + A + b/A)/2 = A + b/(2A) Now let's repeat the process a second time, dividing Q by B and averaging: C = (B + Q/B)/2 = (A + b/(2A) + (A^2 + b)/(A + b/(2A)))/2 Note that (A + b/(2A))^2 = A^2 + b + (b/(2A))^2 so that A^2 + b = (A + b/(2A))^2 - (b/(2A))^2 and (A^2 + b)/(A + b/(2A)) = A + b/(2A) - (b/(2A))^2/(A + b/(2A)) This gives C = A + b/(2A) - (b/(2A))^2/[2(A + b/(2A))] which is exactly Bakhshali's formula! So what he did amounts to repeating the traditional algorithm for two steps. No wonder the page you referred to says that it is faster than Newton's algorithm; it should take half as many steps to get the same accuracy (assuming you apply it iteratively). I'm not sure whether it is any easier than just doing that explicitly; but it is certainly interesting to see! Thanks for calling our attention to it. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/