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I was wondering if there exists a reliable way to extract the $n$th root of a given number. For example, if you have a large perfect square such as 10,404, how would I go about taking its square root by hand?

And what about cube roots? Is there a way to find the cube root of any number by hand?

If there is a way to take the cube and square roots of any number, is there a technique that can be applied to extract the cube or square root of polynomials? What about any root?

I think that, if there is a way to take the $n$th root of any number of polynomial, I would benefit from learning it, since I consider being able to work with roots important to my algebra foundation.

Thanks

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  • $\begingroup$ For the square root, there is a method similar to the usual division by a number. If we know that the number is a perfect square we can restrict the possibilities (ending digits , residue mod 9). For cube roots and higher , I am not aware of an efficient method by hand. $\endgroup$
    – Peter
    Feb 25 at 18:58
  • $\begingroup$ For square roots we can use the algorithm described here under "Digit-by-digit calculation" / "Decimal (base 10)" $\endgroup$ Feb 25 at 19:03
  • $\begingroup$ @StefanOctavian I've seen similar algorithms for higher roots done (somehow you take 3-digits at a time for cube roots.) But they stopped teaching the square-root extraction the year before I would have had to learn it. $\endgroup$
    – B. Goddard
    Feb 25 at 20:38

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If you would like to compute $\sqrt[p]{a}$, the following iterative process will do it to whatever accuracy you please:

First, make a guess and call it $x_0$. Then perform this iteration:

$$x_{n+1} = \frac{(p-1)x_n^p+a}{px^{k-1}}.$$

For instance, if you want $\sqrt[5]{11}$, guess that $x_0=2$. The iteration looks like

$$x_{n+1} = \frac{4x_n^5+11}{5x_n^4}.$$

So we have $$x_1 = \frac{4\cdot2^5+11}{5\cdot 2^4} = \frac{139}{80}=1.7375.$$

$$x_2=\frac{4\cdot1.7375^5+11}{5\cdot 1.7375^4} = 1.631392308.$$

This number to the 5th power is $11.55558806$, so we're getting close.

$$x_3=\frac{4\cdot1.631392308^5+11}{5\cdot 1.631392308^4} = 1.615704970.$$

And $1.615704970^5 = 11.0105\ldots.$

This is an implementation of Newton's Method on the polynomial $x^p-a$.

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