Traditional Abacus and Bead Arithmetic/Roots/Cube root

Theory Let be the number of which we want to obtain the cube root ; Let's consider its decimal expansion, for example: . Let's separate its digits into groups of three around the decimal point in the following way or, in other words, let's define the sequence of integers and let's build the sequence recursively from and let be the integer part of the cube root of i.e.  is the largest integer whose cube is not greater than . Finally, let us call remainders to the differences For our example we have: 000014564567113278945678977256301245678901277025601243004567890123007701781191995000456789012300000770146949021256⋯ Let's see that, by construction, grows as (three more digits in each step), in fact the sequence , i.e. 0, 400, 456, 456.789, 456.789012, etc. tends to (). By comparison, , as the integer part of the cube root of , grows only as (one digit more each step). As is the largest integer whose square is not greater than we have as above but by definition of , or multiplying by but as  grows only as , the second term tends to zero as .   and so that we have For other numbers, the above factors are: and , where is the number of three-digit groups to the left of the decimal point, negative if it is followed by 000 groups (ex. for , for , etc.). This is the basis for traditional manual cube root methods. Procedure We start with . First digit 112832746451256216734385129729 For . It is trivial to find such that its square does not exceed through the use of the following table of cubes that can be easily retained in memory. In the case of the example it is Rest of digits For , we have as defined above and we try to build in the way: where is a one-digit integer ranging from 0 to 9. To obtain we have to choose the greatest digit from 0 to 9 such that: or if we write . Expanding the binomial we have or The left side of the above expression may be seen as simply the previous remainder with the next three-digits group appended to it. If we evaluate the term on the right for each value of and compare with the left term we have: + 00≤ 113789114911≤ 113789230248≤ 113789346017≤ 113789462224≤ 113789578875≤ 113789695976≤ 1137897113533≤ 113789  ⬅8131552> 1137899150039> 113789 and it is clear that the next figure of our root is a 7 but, how can we proceed in the general case without having to systematically explore every possibility ()? Here KnottKnott, Cargill G. The Abacus, in its Historic and Scientific Aspects Transactions of the Asiatic Society of Japan 14 1886 18-73 https://archive.org/details/in.gov.ignca.26020/page/17/mode/2up distinguishes two different approaches: Preparing the divisor Preparing the dividend Preparing the divisor This correspond with the above expression And it is the strategy usually used with paper and pencil and can also be implemented, of course, on the abacus. In the above expression, if we see the left part as dividend and the parentese as divisor, is the first digit of the division but since we don't know b yet, we approximate it using only the main part of the divisor This gives us a guess as to what the value of might be, but we need: Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed. Obtain the next remainder to prepare the calculation of the next digit of the root. You can see an example in Tone nikki blogSquare root and Cube root using Abacus 2017 Tone? とね日記 https://blog.goo.ne.jp/ktonegaw/e/f62fb31b6a3a0417ec5d33591249451b August 17, 2021, see also Modern approaches below. Preparing the dividend Starting again with we prepare the dividend by dividing (the next three-digits group appended to the previous remainder) by As usual, we don't know and we can't evaluate the parentheses on the right, but we can get a clue about by approximating the parentheses by its main part and use it as a trial divisor. so that After that, we need again: Verify that the value thus obtained is correct, or, where appropriate, correct it up or down as needed. Obtain the next remainder to prepare for the obtention of the next digit of the root by evaluating . Please note that: Divisor 3 is involved in the prepared dividend and this leads to non-finite decimal fractions. The division by not only worsens the above, but also makes the prepared dividend specific to the current step, since the value of evolves with the calculation of the different figures of the result. This did not occur in the calculation of square roots and, as a consequence, the process of obtaining cube roots is much more complicated and requires a complex cycle of preparation-restoration of the dividend that, following Knott, can be represented by the following scheme: a) Divide by . b) Divide by 3. c) Obtain as the first digit of the division of the above by . d) Subtract (Equivalent to subtracting and in ). e) Multiply by 3. f) Multiply by . g) Subtract . In our example (), using traditional division and traditional division arrangement (like Knott does), working the two first digits: + Cube root of 456.7890123 Abacus Comment ABCDEFG  456789 Enter number aligning first group with B -343 -7^3=343  113789 First remainder 7113789 Enter 7 in A as first root digit and append second group 7113789 a) Divide B-F by 7reference_name_A 1 7162554 b) Divide B-F by 3reference_name_B 2 7541835 c) Divide B by A (one digit) 7751835 d) Subtract 7*7=49 from CD 77 2835 e) Multiply CDEF by 3. Add 3✕283 to CDEFG 77  854 f) Multiply CDEF by 7. Add 7✕85 to CDEFG 77  599    -343 g) Subtract 7^3=343 to CDEFG 77  256 New remainder   ... Root obtained so far: 7.7 Notes: reference_name_A 1a) It is unnecessary to extend the division by 7 beyond the current three-digit group. The 4 in G is a division remainder meaning 4/7. reference_name_B 2b) The same can be said of division by 3. It is carried out up to column F and the remainder (1) is temporarily added to column G. The value (5) in said column is a strange hybrid meaning 1/3 and 4/7 . It does not matter, it will be reabsorbed in steps e) and f). Modern approaches Members of the Soroban & Abacus Group modified the technique described by Knott to adapt it to modern soroban useCube Roots 2011 Baggs, Shane Heffelfinger, Totton 算盤 Abacus: Mystery of the Bead http://totton.idirect.com/soroban/Hannu/5_Bead/ August 9, 2021 https://web.archive.org/web/20210801083517/http://totton.idirect.com/soroban/Cube_Root/ August 1, 2021. The result is allegedly faster at the expense of being less compact and requiring an abacus with more rods to store intermediate data. The simplicity of having the result directly substituting the radicand is also lost. You can also find a compilation of modern methods for both square and cube roots in Tone Nikki (とね日記) by a Japanese blogger (Author's name does not appear to be available). Examples of cube roots The following examples are all worked using traditional division and traditional division arrangement. Components of the dividend preparation-restoration cycle are labelled with a), b), etc as detailed above. Cube root of 157464 Abacus Comment ABCDEFG Cube root of 157464  157464 Enter number aligning first group with B -125 Subtract 5^3=125 from BCD   32464 First remainder: 32 5 32464 Enter 5 in A as first root digit and append second group 5 32464 a) Divide C-F by 5 (G will be the division remainder) 5 64924 b) Divide C-F by 3 5216404 c) Divide B by 5 5416404 d) Subtract 4x4=16 from CD 54  404 e) Multiply 40x3 in EFG (adding to remainder in G) 54  124 f) Multiply 12x5 in EFG 54   64 g) Subtract 4^3=64 from FG 54 Remainder 0; Done! Root is 54 Clearly, if the remainder is zero and there are no more (not null) groups to add, the number is a perfect cube and we are done. Root is 54. Cube root of 830584 Abacus Comment ABCDEFG Cube root of 830584  830584 Enter number aligning first group with B -729 Subtract 9^3=729 from BCD  101584 101: first remainder 9101584 Enter 9 in A as first root digit and append second group 9101584 a) Divide C-F by 9 (G will be the division remainder) 9112871 b) Divide C-F by 3 9376232 c) Divide B by 9 (A) 9416232 d) Subtract 4x4=16 from CD 94  232 e) Multiply 23x3 in EFG (adding to remainder in G) 94   71 f) Multiply 07x9 in EFG 94   64 g) Subtract 4^3= 64 from FG 94 Remainder 0; Done! Root is 94 Root is 94. Cube root of 666 Abacus Comment ABCDEFG Cube root of 666  666 Enter 666 in BCD    + (Unit rod) -512 Subtract 8^3=512 from BCD  154 First remainder 8154 Enter 8 in A as the first root digit 8154000 Append 000 as new group 8154000 a) Divide B-F by 8 (A) 8192500 b) Divide B-F by 3 8641662 c) Divide B by 8 (A) 8781662 d) Subtract BxB=49 from CD 8732662 e) Multiply C-F by 3 in C-G 87 9800 f) Multiply C-F by 8 (A) in C-G 87 7840 g) Subtract B^3=343 from EFG 87 7497 Root so far 8.7, Remainder 7.497 Now we continue using Abbreviated operations. We need to divide the remainder (7497) by three times the square of the current root () Abacus Comment ABCDEFGHIJKLM 87 7497 87 7497------ Squaring 87           +49 7^2         +112 2*7*8         +64 8^2 87 7497  7569 multiplying by 3 (adding double)        +14         +10          +12           +18 87 7497 22707 dividing 7497/22707, two digits ... 8733 Root 8.733 (Compare to: ) Cube root of 237176659 (three digits) Abacus Comment ABCDEFGHIJ Cube root of 237176659  237176659 Enter number aligning first group with B -216 Subtract 6^3=216 from BCD   21176659 21: first remainder   21176659 Enter 6 in A as first root digit and append second group 6 21176659 a) Divide B-F by 6 (A) 6 35292659 b) Divide B-F by 3 6117633659 c) Divide B by 6 (A) 6157633659 d) Subtract BxB=1 from CD 6156633659 e) Multiply C-F by 3 in C-G 6116992659 f) Multiply C-F by 8 (A) in C-G 6110196659 g) Subtract B^3=343 from EFG 6110195659 Root so far 61, Remainder 10195 ---------- 6110195659 Append third group 6110195659 a) Divide C-H by 61 (AB) 6116714158 b) Divide C-H 3 6155713678 c) Divide C by 61 (AB) 6190813678 d) Subtract CxC=81 from EF 619   3678 e) Multiply D-H by 3 in D-I 619   1158 f) Multiply D-H by 61 (AB) in D-J 619    729 g) Subtract C^3=729 from HIJ 619    000 Done, no remainder! ---------- Root is 619 Cube root of to eight digits The first triplet 110 is between 64 and 125, so that the cube root of 110 591 is between 40 and 50. First root digit is 4 First digit: + Cube root of 110591: First digit Abacus Comment ABCDEFG Cube root of 110591  110591 Enter number aligning first group with B  -64 Subtract 6^3=216 from BCD   46591 46: first remainder   46591 Enter 4 in A as first root digit and append second group 4 46591 OK 1st digit! Second digit: + Cube root of 110591: Second digit Abacus Comment ABCDEFG 4 46591 a) Divide B-F by 4 (A) 4116473 b) Divide B-F by 3 4388234 c) Divide B by 4 (A) 4868234 d) Subtract BxB=64 from CD 48 4234 e) Multiply C-F by 3 in C-G 48 1273 f) Multiply C-F by 4 (A) in C-G 48  511 g) Cannot subtract 8^3=512 from EFG! Going back (See note at the end) 48  511 -f) Divide C-F by 4 (A) 48 1273 -e) Divide C-F by 3 48 4234 -d) Add 8x8=64 in CD 4868234 -c) Revise down B -1  +4 47T8234 d) Subtract BxB=49 from CD (T=10) 4759234 e) Multiply C-F by 3 in C-G 4717773 f) Multiply C-F by 4 (A) in C-G 47 7111 g) Subtract B^3=343 from EFG 47 6768 OK 2nd digit! Remainder 6768 Third digit: + Cube root of 110591: Third digit Abacus Comment ABCDEFGHIJ 47 6768000 Append 000 to previous remainder 47 6768000 a) Divide C-H by 47 (AB) 4714400000 b) Divide C-H 3 4748000000 c) Divide C by 47 (AB) 4795700000 d) Subtract C^2=81 from EF 4794890000 e) Multiply D-H by 3 in D-I 4792298300 f) Multiply D-H by 47 (AB) in D-J 479 689490 g) Subtract C^3=729 from HIJ 479 688761 OK 3rd digit! Remainder 688761 Fourth digit: + Cube root of 110591: Fourth digit Abacus Comment ABCDEFGHIJKLM 479 688761000 Append 000 to previous remainder 479 688761000 a) Divide D-J by 479 4791437914194 b) Divide D-J by 3 4794793046394 c) Divide D by 479 1d 4799482046394 d) Subtract 9^2=81 from GH 4799473946394 e) Multiply E-J by 3 in E-K 4799142184194 f) Multiply E-J by 479 in E-M 4799 68106330 g) Subtract -D^3=729 from KLM 4799 68105601 Ok 4th digit! Remainder 68105601 Now we finish the calculation using abbreviated operations. We need to divide the remainder (68105601) by three times the square of the current root (4799). The first four digits of the result are appended after the ones already obtained; for instance: + Cube root of 110591: Continue using abbreviated operations Abacus Comment ABCDEFGHIJKLM 4799 68105601 Divide E-M by 4799 479914191623 Divide E-M by 4799 47992957204 Divide E-M by 3 47999857 Compare this to As we can see, we have obtained a result with 7 correct figures. Note: We found above that with root 48 we could not subtract , or we had a negative remainder (-1). This might seem unfortunate since it forced us to undo part of the work and correct the new root figure downwards, but in practice what we find is a fortunate result: the small remainder (-1) tells us that 48 was a excellent approximation (by excess) to the root, opening a new way to solve the problem. In fact, what we have is: or where we can use so that compare to . We could have thus achieved great precision with little effort! From elementary arithmetic to numerical analysis The abacus is currently studied as a traditional art or as a means to develop numerical and cognitive skills in general, it is not expected that in the computer age it will be used as a calculator to solve real world problems. But if that were the case and you had to solve a large number of cube roots (something unusual) you might want to move from traditional methods or basic arithmetic to modern numerical analysis methods and try the Newton-Raphson method. You can find an adaptation of this method to the abacus in jccAbacusNewton's method for abacus; square, cubic and fifth roots 2021 Cabrera, Jesús jccAbacus https://sites.google.com/view/jccabacus/abacus-articles/advanced-abacus-algorithms/newtons-method-for-abacus-square-cubic-and-fifth-roots?authuser=0 August 17, 2021... Appendix: Cubes of two digits numbers + Cubes of two-digits numbers +1 +2 +3 +4 +5 +6 +7 +8 +910133117282197274433754096491358326859209261106481216713824156251757619683219522438930297913276835937393044287546656506535487259319406892174088795078518491125973361038231105921176495013265114060814887715746416637517561618519319511220537960226981238328250047262144274625287496300763314432328509703579113732483890174052244218754389764565334745524930398053144155136857178759270461412563605665850368147270496990753571778688804357830584857375884736912673941192970299 This can help you practice two-digit cube roots. Example: References ../Appendix: Abbreviated operations Square root