Evaluating Cube roots of fractions

Question

For example : $\sqrt[3]{\frac {21}{37}}$ If I evaluate it in a scientific calculator, the answer would be $0.8280 \dots$. But is there another way to solve it instead of using the calculator? And why the answer just had to be in decimals? Can't the answer be in fractions?

Answer 1

Well, most calculators output decimals instead of symbolic expressions, so it's not surprising that that's what you got. Symbolically, there's not much that you can do... you can write it as $\frac{\sqrt[3]{21}}{\sqrt[3]{37}}$ if you wish, but I'm not convinced that's a meaningfully better way to write it. This article details one method for computing cube roots by hand.

Answer 2

First, you won't be able to write the answer as a fraction because if $(a/b)^3 = 21/37$, then we have $37a^3 = 21b^3$. Therefore the numbers $37a^3$ and $21b^3$ must have the same prime factorization. But the number of times $7$ appears in the prime factorization of $37a^3$ is a multiple of $3$, whereas the number of times it appears in that of $21b^3$ is one more than a multiple of three.

Now to calculate a few decimal places, we use the fact that $21/37 \approx (5/6)^3$. Specifically, $$\sqrt[3]{\frac{21}{37}}= \frac{5}{6}\sqrt[3]{1-\frac{89}{4625}}.$$

For $n \geq 1$ and $0 < x < 1$, Taylor's formula gives us $$(1-x)^{1/3} = 1 - \frac{1}{3}x + \binom{1/3}{2}x^2 - \dots +(-1)^{n-1}\binom{1/3}{n-1}x^{n-1} + \varepsilon_n(x),$$ where the (necessarily negative) error term $\varepsilon_n(x)$ satisfies the inequalities $$(-1)^n \binom{1/3}{n}x^n (1-x)^{1/3 - n} \leq \varepsilon_n(x) \leq (-1)^n \binom{1/3}{n}x^n.$$ (See Theorem 7.7 in Calculus, Vol. 1, by Apostol.) We can weaken this inequality slightly to the more convenient form $$(-1)^n \binom{1/3}{n}\left(\frac{x}{1-x}\right)^n \leq \varepsilon_n(x) \leq (-1)^n \binom{1/3}{n}x^n.$$

Now say we want to know the value of our cube root to within $10^{-9}$. We look for a value of $n$ such that the left-hand side in the above inequality is less than $10^{-9}$ for $x = 89/4625$. We have $x/(1-x) < 0.02$, so $(-1)^5\binom{1/3}{5}[x/(1-x)]^5 < (22/729)(2 \times 10^{-2})^5$, which is less than $(6/5) \times 10^{-9}$, showing that $n = 5$ will provide enough precision.

Returning to Taylor's formula, we now have $$\frac{5}{6}\left[1 - \frac{1}{3}x - \frac{1}{9}x^2 - \frac{5}{81}x^3 - \frac{10}{243}x^4 - \frac{22}{729}\left(\frac{x}{1-x}\right)^5\right] \leq \sqrt[3]{\frac{21}{37}}\leq \frac{5}{6}\left[1 - \frac{1}{3}x - \frac{1}{9}x^2 - \frac{5}{81}x^3 - \frac{10}{243}x^4 - \frac{22}{729}x^5\right], $$ where $x = 89/4625$, yielding $$ 0.827953329397 \leq \sqrt[3]{\frac{21}{37}} \leq 0.827953329405 $$ or an accuracy better than $10^{-11}$. Compare this with the actual value $\sqrt[3]{21/37}= 0.827953329403...$

Granted, this method is not easy to carry out without a calculator, but it shows how such a calculation can be done.

Answer 3

In the same spirit as user49640's answer, knowing that the solution is close to $\frac 56$, we could build the simplest Pade approximant of function $x^3-\frac{21}{37}$ around $x=\frac 56$.

This would give $$x^3-\frac{21}{37}\approx \frac{\frac{89}{7992}+\frac{6893 }{3330}\left(x-\frac{5}{6}\right)}{1-\frac{6}{5} \left(x-\frac{5}{6}\right)}$$ from which the solution is $$x=\frac{68485}{82716}\approx 0.827953$$

Another way would be using Newton method for which the iterates would be $$\left( \begin{array}{ccc} n & x_n & \text{approx} \\ 0 & \frac{5}{6} & 0.833333 \\ 1 & \frac{6893}{8325} & 0.827988 \\ 2 & \frac{982489486039}{1186648388775} & 0.827953 \end{array} \right)$$

Answer 4

If the numbers in the cube root are themselves perfect cubes, eg: $\sqrt[3]{\frac{343}{27}}=\frac{7}{3}$, as $7^3=343$ and $3^3=27$ or you can rewrite them as perfect cubes, then you'll be able to get exact answers as fractions or integers out of them. However, more often than not most numbers aren't well behaved like this, and the best you can do is an approximation.

For calculators it's a lot easier to compute decimals instead of fractions so that's what they do, but if you wanted to do a fractional approximation, you could use a power series of $\sqrt[3]{x}$ centered at $c$: $$\sqrt[3]{x}=\sqrt[3]{c}\sum_{n=0}^\infty{\left[\left(\frac{c-x}{c}\right)^n\prod_{m=1}^n{\left(\frac{3m-4}{3m}\right)}\right]}$$

Where you'd select $c$ as the closest perfect cube larger than $x$, where $x$ is the number you want to approximate. To actually compute it by hand instead of letting the sum range from $0$ to $\infty$, just cap it off and calculate the first couple terms. The more terms you calculate the better the approximation obviously. For example to find $\sqrt[3]{24}$, you'd let $x=24$, and $c=27$, since $27$ is the next largest perfect cube bigger than 24: $$\sqrt[3]{24}=\sqrt[3]{27}\sum_{n=0}^\infty{\left[\left(\frac{27-24}{27}\right)^n\prod_{m=1}^n{\left(\frac{3m-4}{3m}\right)}\right]}=3\sum_{n=0}^\infty{\left[\left(\frac{1}{9}\right)^n\prod_{m=1}^n{\left(\frac{3m-4}{3m}\right)}\right]}$$ And then just start calculating terms in the sequence, starting by plugging in $n=0$, then $n=1$, $n=2$, and so on. In this case you would get: $$\sqrt[3]{24}\approx3\left(1-\frac{1}{27}+\frac{1}{243}+\frac{5}{6561}+\cdots\right)\approx\frac{6350}{2187}$$ It's important to note that this is just an approximation, and that the root isn't really equal to the fraction. But since every other answer here, and really even what the calculator spits out are just approximations as well, I guess it doesn't matter. The main difference is that this approximation will give you fractions, as opposed to the standard decimal methods.