In the previous two videos, we have mentioned and explained why the three methods work.
We know that it works for square roots, but what about cube roots?
Let’s first focus on long division. In the last video, we have explained why this method
works using specific numbers. In general, let’s denote the square root we want to
calculate as b, and then we use a to approximate the first digit, which would then give us
the remainder r, where r = b - (10a)^2. Since we want the next digit to be x, that means
(10a+x)^2 would be close to b, and so we can rewrite r to be x(20a+x), which is why we
double the quotient, write the blank and so on. So what about cube roots? Again, let’s
denote the cube root we want to calculate as b, and then we use a to approximate the
first digit, which would then give us the remainder r, where r = b - (10a)^3. Since
we want the next digit to be x, that means (10a+x)^3 would be close to b, and so we can
rewrite r to be x times some mess over here, which is our new algorithm for cube roots!
Let’s use some concrete example, 18. Since 18 is between 2^3 and 3^3, we write down 2
in the first digit. Write down the quotient cubed, and subtract like you do in division.
Now we need to use the algorithm we derived here. Substitute a = 2, and we want to find
out x so that this expression is just under 10000. It seems quite difficult here, but
here is a computational trick: Ignore the x^2 term for now because it is not going to
largely affect the entire product, and divide both sides by 10, we want to find x such that
6x times twenty x just under 1000. Then simply test out values of x from a larger value until
it is just under 1000. In this case, it is 6. Now you substitute it back to the original
algorithm, and do the subtraction process. But now you can imagine that implementing
the same algorithm again is very very troublesome, so while long division is technically able
to generalize to cube roots, it will no longer be a practical method.
For continued fraction, it will be even worse for cube roots. Remember how we painfully
obtain the continued fraction expression for sqrt 3? That is already quite good because
it is periodic, and we can stop the process after a certain amount of steps. But continued
fraction for cube roots is never periodic. We can prove this pretty easily. Let’s say
a number n can be written as a periodic continued fraction with period p. Let’s say the repeating
part is x. Then by writing x in full, and replace this repeating part by x, we have
this sort of monstrous equation. However, don’t worry, because we can unveil the entire
continued fraction on the right from the bottom. Eventually, we will end up with the equation
x = (A+Bx) / (C+Dx), where A, B, C, and D are some constants. Rearranging, we have a
quadratic equation in x. That means for a number to have periodic continued fraction
expression, it has to be a root of a quadratic equation. This is definitely not the case
for cube roots. With the non-periodicity, and much higher difficulty to rationalize
fractions with cube roots, using continued fractions to approximate cube roots is basically
impractical. So is some generalization of Babylonian method
somewhat practical? Yes! We have in fact two different generalizations, based on two different
explanations offered in the last video. Let’s focus on the graphical, or Newton’s method
first. In the previous video, we know we can derive this iteration formula from definition
of derivative. In this case the graph would be y = x^3 - a, so we can substitute the values
of f(x_0) and f’(x_0) respectively. With some manipulation, we get this formula. That
means we can use this iteration to calculate cube roots. However, there is a non-calculus
method as well. We first start off with this inequality, where we specify the distance
between a_0 and cube root of a to be smaller than some error bound epsilon, and cube the
inequality. However, there is a new unwanted term with cube root of a^2. Don’t worry
because we can first multiply both sides by cube root of a divided by a_0. Now notice
that these two boxed expressions can add up to cancel out each other, so we add them up
to get this inequality. This is now close to what we want. Multiply both sides by a_0
divided by (a+2a_0^3), and we have this ultimate inequality. Now this boxed expression here
would be a much better approximate than a_0, so we write this as a_1. So this means that
from the non-calculus method, we have this iteration formula. These two separate methods
yield the same iteration formula in square roots, but not in cube roots. So, since they
are different, can we tell which one is more efficient in calculating cube roots? To make
the comparison fair, we use both methods to approximate cube root of 2, using the initial
guess as 1. Now only with the first few iterations, it is very clear that the non-calculus method
is the winner, as it can produce 20 correct digits within only three iterations.
So now, among the three methods to calculate square roots, only Babylonian method can have
a generalized version for cube roots that is practical for use. See if you can generalize
Babylonian method to even higher order of roots! This is the end of a trilogy about
surds. The next video is going to be a standalone one. Subscribe if you want more of these math
videos. See you next time!