A Way of Approximation of a Cube
Root
By Mohammad H. Poursaeed
Abstract. Differential calculus can be used to approximate cube roots. In this note, however, we present a
method of approximating cube roots that is more accurate than the method of differentials and does not require
advanced mathematical information.
Introduction
The approximation of a cube root is sometimes introduced as
an application of differential calculus [1]. In this note, we adopt
an intuitive interpretation of the problem and employ a method
of approximating a cube root whose precision is greater than the
classical differential method.
Approximation of a Cube Root
Similar to the process of approximating a square root, which
is done according to the area of a square, the approximation of a
cube root, can be worked out as follows:
Suppose that we are going to determine the cube root of a
number a. Without loss of generality, we can assume that a is a
positive number. Thus, in fact, we seek the measure of the side of
a cube having volume a. Now, if the side of the cube has measure
b, we have:
√
a = b3 ⇒ b = 3 a.
[19]
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Alabama Journal of Mathematics
If a is not a perfect cube, then we can consider two cubes with
sides of given measure such that one of them is inscribed within
the cube having volume a, while the other circumscribed about
the cube whose volume is a. As a result, the length of the side
of the cube having volume a is bounded by the lengths of the
sides of the inscribed and circumscribed cubes. For example, to
determine the cube root of 28, we can consider two cubes with
sides of length 3 and 4 as the inscribed and circumscribed cubes
respectively. Consequently, the cube root of 28 is a number between
3 and 4. Similarly, we can be more discerning in our choice of lower
and upper bounds. For example:
3
3
(3.01) = 27.27 < 28 < 29.791 = (3.1) .
Thus, we establish that the cube root of 28 is between 3.01 and
3.1.
Now if we let k be the length of the side of the cube (inscribed
or circumscribed) with volume nearest to a, and we let x be the
difference between the length of the side of this cube and length of
the side of the cube having volume a, then we have:
x = k − b ⇒ b = k − x,
and
a = b3 = (k − x)3 = k3 − 3k2 x + 3kx2 − x3 .
(eq. 1)
x
k
b
Cube of Volume a and circumscribed cube.
As you see, the computation of the cube root of a is equivalent
to determining the solution of f (x) = −x3 +3kx2 −3k2 x+k3 −a =
0. However, if the inscribed and circumscribed cubes are chosen
such that the lengths of their sides are very close to the length of
the side of the cube having volume a, then the value of |x| is very
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21
small. Therefore, x3 will be considerably smaller in magnitude still.
Thus, for small x, f (x) ≈ 3kx2 − 3k2 x + k3 − a, and our problem
can be reduced to that of finding the roots of the corresponding
quadratic function g (x) = 3kx2 − 3k2 x + k3 − a.
This presents a small problem — while f (x) has only one root
(of multiplicity 3), g(x) has two roots. Which of these roots is the
approximation of b? Since x (the root of f (x)) is assumed to be so
close to zero, the root of g(x) that we seek is logically the smaller
of the two roots. So if x0 is root of g (x) of lesser magnitude, and
x∗ is the only root of f (x) , then we will have:
g (x0 ) = 0, f (x∗ ) = 0,
⇒
√
3
a = k − x∗ ≈ k − x0
k
g (x) = 3kx − 3k x + k − a = 0 ⇒ x1 , x2 = ±
2
2
2
3
k
⇒ x0 = −
2
r
4a − k3
.
12k
r
4a − k3
12k
(eq. 2)
In order to establish the validity of eq. 2, it is necessary to
show that 4a − k3 > 0.
Case 1: When the volume of the inscribed cube is the closest to the value of a, we have:
k3 < a ⇒ k3 < a < 4a ⇒ 4a − k3 > 0.
Case 2: When the volume of the circumscribed cube is the
closest to the value of a, we have a < k3 . If we suppose
that 4a − k3 < 0, then either:
k3
k3
k3
k3
<
or
<a<
.
8
4
8
4
In either case it can be shown that:
a<
¯ 3
¯
¯k
¯
¯ ¯
¯ − a¯ < ¯k 3 − a¯ = k3 − a.
¯8
¯
In this case, the volume of the inscribed cube with sides
of length k2 is closer to a than the volume of the circumscribed cube and should be used instead of the circumscribed cube.
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Alabama Journal of Mathematics
Therefore, we can assume that 4a − k3 ≥ 0. Thus,
√
k3
k
k3
>
⇒ 3a> .
4
8
2
Putting equations 2 and 3 together, we have:
4a − k3 ≥ 0 ⇒ a ≥
√
k
3
a≈ +
2
r
(eq. 3)
4a − k3
.
12k
Example: Approximating the value of
√
3
28, we have:
33 = 27 < 28 < 64 = 43 ⇒ k = 3
√
3
3
⇒ 28 ≈ +
2
s
(4) (28) − 27 ∼
= 3.036591.
(12) (3)
√
One will note that our approximation differs from 3 28 by less
than 2 × 10−6 . This is√significantly better than the result obatined
when approximating 3 28 using differentials.
We now show that in the general case, if k is the length of
the side of the inscribed cube, the suggested method yields a more
precise approximation that the result obtained when using differentials.
√
3
Note that as the graph of the function h (x) =
x is con√
3
cave (i.e., concave down), the approximate value of x calculated
through the use of differentials is always more than the actual value.
That is:
√
√
a − k3
3
a<k+
≈ 3 a.
2
3k
On other hand, according to eq. 2, it turns out that x0 is
negative and therefore:
3
3
f (0) = k3 − a < 0, f (x0 ) = g (x0 ) − (x0 ) = − (x0 ) > 0.
Therefore, by the Intermediate Value Theorem, the only root of
f (x) is in the interval (x0 , 0) . Thus:
x0 < x∗ < 0 ⇒
√
√
3
a = k − x∗ < k − x0 ≈ 3 a.
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23
√
The point is that the approximate value of 3 a using our method
is also greater than the actual value. Thus, we can easily demonstrate that the following inequality is correct which, in turn, indicates that, compared with the method of differentials, our method
is the more accurate of the two.
¯
¯
√
√
3
3
¯
¯
3
3
−
a
a
¯k + a−k
¯ = k + a−k
2
3k
3k2 −
>
k
2
+
q
4a−k3
12k
¯
¯
¯ k q 4a−k3 √ ¯
√
3
3
¯
− a = ¯2 +
a¯¯ .
12k −
References
[1] L. Leithold, The Calculus with Analytic Geometry, fifth
edition, Harper & Row Publishers, Inc., 1986.
Department of Mathematics
Lorestan University
Khorramabad, Iran
Poursaeed M. H.
P.O.Box 68135-465
Fax: +98 661 22782
poursaeed_mh@yahoo.com
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Alabama Journal of Mathematics
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