The twelfth root of two

Introduction

Prince Zhū Zàiyù (朱載堉; 1536 – 19 May 1611) of the Ming dynasty was a Chinese mathematician, astronomer, geographer, physicist, writer, choreographer, musician and music theorist. He has a place in the universal history of music for being the first to imagine and describe the equal temperament on which the tuning of modern musical instruments is based. To this end he calculated the twelfth root of two and its powers with 24 digits, for which it also occupies a place in the history of Chinese mathematics. He used a 2:5 abacus pair of 81 rods each for this numerical feat.

Of course we are not going to repeat his calculations here to such a degree of precision, we will limit ourselves to a smaller number of digits on a traditional 13-digit abacus,

especially to show the efficiency of Newton's method for cube roots and abbreviated operations, which will allow us to obtain 7-8 digits of the twelfth root of two with such a modest instrument.

Method

Since we can write

We can obtain this twelfth root by chaining two square roots and a cubic root. But first let's remember that

when , if we abuse this expression we obtain

a very imprecise value, but enough to show us that if we start calculating roots directly from two we will spend most of the time dividing by numbers that start with one; which is as uncomfortable as it is unnecessary. To avoid this, we will start by calculating square roots of

so that

What follows is a rather long calculation. It is assumed that the reader is already familiar with obtaining square roots by the method of half remainders (hankukuho) and cubic by Newton's method before undertaking a twelfth root, so that part of the calculations will be omitted and entrusted to the reader for brevity.

1st step, first square root

ABCDEFGHIJKLM

-------------

1250           First root digit is 3

0350           Subtract 3x3=9 from AB

0175           Divide by 2 in place

3175           Enter first root digit in A

...            Continue as usual until the 6th root digit

-------------

3535531380955  

Now the “accelerated phase” begins ...

ABCDEFGHIJKLM

-------------

3535531380955

               Rule 1/3>3+1 (G)

3535533480955

      -15

       -09

        -15

         -15

          -09

3535533320296

           -045

3535533320295  Rule 3/3>9+3 (H)

3535533950295  

       -45

        -27

         -45

          -45

3535533902100

           -27

3535533902097  Rule 2/3>6+2 (J)

3535533906297  

        -1    

         +3    

3535533905597

         -25

          -15

           -25

3535533905329  Rule 3/3>9+3 (K)

3535533905959

          -45

           -27

3535533905911  Rule 1/3>3+1 (L)

3535533905932  

           -15

3535533905931  Rule 1/3>3+1 (M)

3535533905933  Compare to

-------------

2nd step, second square root

ABCDEFGHIJKLM

-------------

3535533905933  First root digit is 5

1035533905933  Subtract 5x5 from AB

0517766952966  Divide by 2 in place

5517766952966  Enter first root digit in A

...            Continue as usual until the 6th root digit

-------------

5946033314921

Accelerated phase:

ABCDEFGHIJKLM

-------------

5946033314921  Rule: 3/5>6+0 on G

5946036314921  cannot subtract 6x9=54 from HI, revise down

     -1

      +5

5946035814921

      -45

       -20

        -30

          -15

           -125

5946035341904  Rule: 3/5>6+0 on H

5946035641904  cannot subtract 6x9=54 from IJ, revise down

      -1

       +5

5946035591904

       -45

        -20

         -30

           -15

5946035544603  Rule: 4/5>8+0 on I

5946035584603  cannot subtract 8x9=72 from JK, revise down

       -1

        +5

5946035579603

        -63

         -28

          -42

5946035572981  Rule 2/5>4+0

5946035574981

         -36

          -16

5946035574605  Revise up

        +1

         -5946

5946035575010  Compare to

-------------

3rd step, cube root

Now we could obtain

But to avoid dividing by numbers that start with one during Newton's method, we are going to multiply the previous amount by 25 but to avoid dividing by numbers that start with one during Newton's method, we are going to multiply the previous amount by 25 (dividing twice by two and multiplying by one hundred) so that.

ABCDEFGHIJKLM

-------------

5946035575    

29730177875    divide by 2 in place

148650889375   divide by 2 in place

memorize or write down 148.650889 elsewhere… The first cube root digit is 5.

ABCDEFGHIJKLM  Divide 148 by 5^2=25 (two digits)

-------------

 148       25  Rule 1/2>5+0 on B

 548       25

 -25

 523       25  Rule 2/2>9+2 on C

 595       25

 59 5      25  revise up D twice

  +2

   -50

 592       25

10+            Add double of first root (5)

1592        3  Divide by 3, Rule 1/3>3+1 (two digits)

3692        3  Revise up B twice

5092        3  Revise up C three times

+3

 -9

5302        3  5.3 is the next root approximation, square it on J-M

5302        9

         +15

         +15

        +25

5302     2809

ABCDEFGHIJKLM

 148651  2809  Enter 148.651 into B-G and divide (4 digits)

               Rule: 1/2>5+0 on B

 548651  2809  

 -40

   -45

 508201  2809  Revise up C twice

 +2

  -5618

 522583  2809  Rule 2/2>9+2 on D

 529783  2809

   -72

     -81

 529 549 2809  Revise up E

   +1

    -2809

 529126812809  Next digit is 9

 52919

106+           Add double of first root (5.3)

158919      3  Divide by 3

…

52973          Next root is 5.297

 ...

ABCDEFGHIJKLM  Divide 148.650889 by 5.297 (9 digits)

-------------

1486508895297  

2427108895297  

2803348895297  

2806170695297  

2806311785297  

280631178      Clear J-M and reuse divisor space

2806321186

28063221266

280632222066

2806322234769

2806322239001

2806322239     Clear remainder

2806322239003209363790824995280347

ABCDEFGHIJKLM  Divide 28.06322239 by 5.297 (8 digits)

-------------

2806322239

2806322239

5157822239

5251882239

5294209239

5297501339

5297924609

5297943421

52979462428

529794643092

5297946454435

529794645      Clear remainder and displace value to the right

ABCDEFGHIJKLM

-------------

  529794645    Add twice the

 +5297         previous root

 +5297

 1589194645

 1589194645 3  Divide by 3

  529731548

ABCDEFGHIJKLM  Multiply by 2 -------------  in place or

  529731548    divide by 5

 1059463096  

 10594631      Round to 8 digits  

Our final result:

compare it to