Background and TopicsThis chapter provides a brief description of methods for finding the roots of perfect powers, then delves into methods for mentally approximating square roots, a general method for mentally extracting square roots to any number of digits that is more powerful than the traditional method, an iterative technique for reciprocal square roots, and extensions of these methods to cube and higher-order roots.

Notes andErrata(If you have any more to contribute, pleaseemailme! This is a compendium of feedback)Asummary table for all chapters is foundprinter-friendlyhere.

Color Code Type MeaningNoteA clarification or elaboration of the text. TypoA simple mistake that does not affect the method presented. ErrorAn error that may affect a method or the reader interpretation of it.

Page Code Explanation GeneralIt is my preference in this book when expressing a decimal value that continues on past the digits that are displayed, that the last digit shown is rounded off, followed by ellipses ( ). For example, a value of log 2 = 0.3010299956 when shown to 5 decimal places will be given as 0.30103 rather than 0.30103 or 0.30102 This helps a great deal when comparing an approximation to an exact value to a certain number of places.

7966577993506607 should read 6657793506607. This does not affect the result. 80,857.14142842857 should read 7.14142842856 This does not affect any result. 82The squared term in the denominator of Equation 7 should not be squared. This does not affect anything else. 83John McIntosh derives Aitken's fourth square root method by a different means. If the midpoint of a and N/a is M, and the deviation of M from a or N/a is D, then N = a x N/a = (M-D)(M+D) = M^2 D^2. Then N^(1/2) = (M^2 D^2)^(1/2) = M x (1 (D/M)^2)^(1/2). If we assume that D/M is small and take the first term of the Taylor series, we get N^(1/2) = M x (1 (D/M)^2/2). The value D/M is the fractional deviation that Aitken refers to, and this formula tells us to reduce M by half the deviation squared, which is what Aitken does in his fourth estimate. 84To avoid any possible confusion between the prime symbol used to represent an assignment, in this case a_n' = N/(a_n)^(p-1), and the prime symbol used to represent the solution of Halley's method, you might think of the last two general equations beginning with f(a_n)' rather than [a_(n+1)]'. The method is used correctly in the text. In the bottom equation, a_n should read a_1 since n=1, and the result should be 7.141442715700. 85In the formula for a_2, 49x50 in the denominator should read 49x51. The result is correct. 89An alternate, more intuitive derivation of the general square root algorithm can be found .hereIn the bottom equation, a_n should read a_0 as it does in the following line. 90In the top equation, -2n-2 should read -2n-4 as it does when that term next appears (at the top of page 91). This does not affect the result. 91-92+ R_(n+1)" at the end of the equations at the bottom of page 91 and the top of page 92 should read - R_(n+1)/a_0 to be consistent everywhere. 93

In the square root example, b4=54 is actually greater than 50 despite my preference for keeping b's less than 50. This does not invalidate the example, as the reasons for limiting the size of b are to ease multiplications and to limit ripple effects to neighboring number groups. In this case the value was very near 50. The same example done with all b's kept less than 50 is found .here 94Another example of finding a square root with the general algorithm from this chapter, and one with more explanation of the steps involved, can be found .hereYet another example of finding a square root with the general algorithm from this chapter can be found .here 97I feel a little guilty about saying that the approximation for the reciprocal square root contains no division (although other authors have no hesitation in doing so), when the 0.5 multiplier is in practical terms a division by 2. 98The result .1400280084023 should read .1400280084025 , which is even more stunningly accurate than the result given. The approximation of 1/(1-.0003) by 1+.0003 uses the first two terms of the Binomial Series expansion for (1+x)^(-1) = 1 x + x^2 - x^3 + This provides an accuracy of .1400280084. The full accuracy of .14002800840025 can also be obtained quite easily by including the third term (-.0003)2 = 9x10^(-8). Multiplying this by 2.8x10^(-5) adds 2.52x10(-13) to achieve the higher accuracy without much effort. 100The result 4.91875 should read 4.91876... In the first equation, a_0 in the denominator inside the parentheses should read a_0^(p-1). The example given is correct. The Chebyshev correction term to subtract for cube roots can also be calculated as (a1-a0)^2 / a0, which is simpler here since (a1-a0) = -.08 is a single digit in this example. See the last section of paper.thisIn the last line of text, a_(n+1) should read a_(n+1)'. This does not affect the derivation. 101In the equation above by the Binomial Theorem, the second term after the division sign should be positive rather than negative. In the text in the middle of the page, 8_p^2 should read 8p^2. The overall derivation is correct. 1044.4979408 should read 4.4979414, which makes the cube root approximation even more accurate. 106To be consistent on rounding the digit before the ellipsis ( ), 4.7328 should read 4.7329 1075.5349164 should read 5.5349164^(1/2). The result of 2.16894364 should read 2.16894354 which now matches the desired 48^(1/5) to the last digit calculated.

Additional Materials Related to Topics in This ChapterSquare Root With Low Number Groups: A paper demonstrating the general square root algorithm is given here for N=51, but unlike the book all the values of b are kept less than or equal to 50, which is recommended in the chapter in order to ease the multiplications and to limit the ripple effects on neighboring number groups.

Another Square Root Example: A paper providing another, more detailed example of the general square root algorithm is given here.

Yet Another Square Root Example: A paper providing yet another detailed example of the general square root algorithm is given here.

An Alternate Derivation of the General Square Root Algorithm: The derivation given in the book of the general square root algorithm is not very intuitive and is rather complicated. Here is an alternate derivation that provides a better visual feel for why the algorithm works as it does. The extension of this algorithm to cube roots is also discussed.

Cited Reference MaterialsA.C. Aitken's fascinating talk given in 1954,

The Art of Mental Calculation; With Demonstrations,which is discussed in the first part of this chapter, can be foundhere. I remember waiting for weeks in the late 1980's to locate the journal and get a copy of it (a University of Minnesota librarian graciously photocopied the article and mailed it to me).Lehmer's 1923 article in which he describes a variation of the general square root algorithm for use with mechanical calculators can be found here. This is not an easy read, and it took me awhile to verify that it is indeed related to the method presented in the book.

In the near future, this section will contain additional relevant sections of some of the references cited at the end of this chapter.

Acknowledgements of Contributors to this PageAgain, thanks goes to John McIntosh for providing a number of the errata and notes you see on this page.