- JAPAN 255
0.00000 00000 33333 35111 11225 39690 66667 28234 77694 79595 875+.
But this value does not fit with the procedure followed by Takebe Kenko; it does
not even yield the correct first approximation. The figure given by Smith and
Mikami appears to represent the value obtained by Takebe Kenko after the first
approximation was subtracted, but with the result multiplied by the square of the
diameter.^13 In appreciating Takebe Kenko's method, the first problem to be solved
is the source of this extremely accurate measurement of the circle. According to
Smith and Mikami (1914, p. 148), Takebe Kenko said that the computation was
given in two other works, both of which are now lost, leaving us to make our own
conjectures. The first clue that strikes us in this connection is the seemingly strange
choice of the square of the arc rather than the arc itself. Why would it be easier to
compute the square of the arc than the arc itself? An answer readily comes to mind:
The arc is approximated by its chord, and the chord is one side of a convenient right
triangle. In fact, the chord is the mean proportional between the diameter of the
circle and the height of the arc, so that in this case it is y/dh = \/l0_5. When we
square it, we get just dh = 10~^5 , which acts as Takebe Kenko's first approximation.
That result suggests that the length of the arc was reached by repeatedly bisecting
the arc, taking the chord as an approximation. This hypothesis gains plausibility,
since it is known that this technique had been used earlier to approximate ð. Since
a^2 = 4(a/2)^2 , it was only necessary to find the square of half the arc, then multiply
by 4. The ratio of the chord to the diameter is even easier to handle, especially since
Takebe Kenko has taken the diameter to be 10. If ÷ is the square of this ratio for a
given chord, the square of ratio for the chord of half of the arc is (l - v^l - x)/2. In
other words, the iterative process χ >-* (l — y/l — z)/2 makes the bisection easy. If
we were dealing with the arc instead of its square, each step in that process would
involve two square roots instead of one. Even as it is, Takebe Kenko must have
been a calculating genius to iterate this process enough times to get 53 decimal
places of accuracy without making any errors. The result of 50 applications yields
a ratio which, multiplied by 100 • 450 , is
0.00001 00000 00333 33335 11111 12253 96833 52381 01131 94822 94294 362+.This number of iterations gives 38 decimal places of accuracy. Even with this plau-
sible method of procedure, it still strains credibility that Takebe Kenko achieved
the claimed precision. However, let us pass on to the rest of his method.
After the first approximation hd is subtracted, the new error is 10~^12 times
0.3333333..., which suggests that the next correction should be 10~^12 /3. But this
is exactly /i^2 /3, in other words h/(3d) times the first term. When it is subtracted
from the previously corrected value, the new error is
10 -19 · 0.17777 77892 06350 01904 76806 15685 4870 +.
The long string of 7's here suggests that this number is 10~^19 times + ^ =
|jj = which is (8/i)/(15d) times the previous correction. By continuing for a few
more terms, Takebe Kenko was able to observe a pattern: The corrections are ob-
tained by multiplying successively by /i/(3d), (8ft)/(15d), (9h)/(Ud), (32/i)/(45d),
(25h)/(33d),.... Some sensitivity to the factorization of integers is necessary to
(^13) Even so, there is one 3 missing at the beginning and, after it is restored, the accuracy is "only"
33 decimal places. That precision, however, would have been all that Takebe Kenko needed to
compute the four corrections he claimed to have computed.