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I've began to reteach myself Algebra and am brushing up on my roots (pun intended). I've been following this website (which makes it simple to refresh your memory), and as I reviewed the root examples I couldn't help but wonder if there was a simple mathematic process for calculating $\sqrt[n]{x}$.

I've reviewed the Wikipedia article covering nth roots and learned of the nth root algorithm:

$$x_{k + 1} = \frac{1}{n} \Biggl( (n - 1)x_k + \frac{A}{x^{n - 1}_k} \Biggr)$$

This algorithm allows us to take an initial guess of $x_0$ and then iterate using the recurrence relation until precision is reached. The other method that stood out was logarithmic calculation in which:

Starting from the equation that defines $r$ as an nth root of $x$, namely $r^n = x$ with $x$ positive and therefore its principal root $r$ also positive, one takes logarithms of both sides (any base of the logarithm will do) to obtain:

$$n \log_b r = \log_b x$$

hence

$$\log_b r = \frac{\log_b x}{n}$$

The root $r$ is then recovered from this by taking the antilog:

$$r = b^{\frac{1}{n} \log_b x}$$


However, in my opninion these seem quite complex and has me wondering; is there a simpler method, that doesn't include guessing, that can solve for $y$ in the following:

$$\sqrt[n]{x} = y$$

I initially attempted with simpler math such as: $\sqrt[n]{x} = \frac{x}{n}$ and $\sqrt[n]{x} = \sqrt{\frac{x}{n}}$ but obviously neither worked except for a few cases such as $\sqrt[3]{27} = \sqrt{\frac{27}{3}}$.

EDIT: I spoke with a friend on the topic and he pointed out the following:

$$\sqrt[n]{x} = x^{\frac{1}{n}}$$

How accurate does this remain across the board? I’ve tried various samples and it seems pretty consistent to me.

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    $\begingroup$ In a way, every possible method involves some guessing and iterative approximation. How else would you arrive at the inifinitely many quite irregular digits $1.414213562373095\ldots$ when looking for $\sqrt 2$? Note that a similar guess-and-approve takes place for example when performing a long division to find $\frac 27=0.285714\ldots$ $\endgroup$ – Hagen von Eitzen Jan 18 at 17:10
  • $\begingroup$ For those old people brought up using common logarithms and with access to a table of logarithms, the second method (with $b=10$) is simple. For younger people with access to a calculator with an $x^y$ button, the calculation can be done directly $\endgroup$ – Henry Jan 18 at 17:34
  • $\begingroup$ @Henry - what if you want 12 digits? My old logarithm table (which I threw away around 1973) had only 4 digits. The CRC table in my bookshelf (which I will throw out today) has only 10 digits for natural logarithms. $\endgroup$ – Hans Engler Jan 18 at 17:51
  • $\begingroup$ How do you define "simple" and "complex"? Otherwise that is opinion based? $\endgroup$ – Somos Jan 18 at 18:08
  • $\begingroup$ @Somos Since my knowledge of mathematics is lacking I would go with the textbook definitions. $\endgroup$ – PerpetualJ Jan 18 at 18:09
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The n-th root algorithm converges quadratically (the number of correct digits is essentially doubled each step) and uses only "simpler" operations such as addition, division, integer powers.

This method is fast, reliable, and accurate and does not require anything except basic arithmetic. You know in advance how many steps are needed to get, say 100 digits. With a decent starting value, you will need about 10 steps.

The log method requires the computation of logarithms and exponentials which is in turn done approximately. So this method has an uncertain amount of computational work associated with it and accuracy control is also uncertain.

"Simplicity" is a subjective criterion and must be balanced against reliability, accuracy, speed. Many (including me) would call the n-th root algorithm simple and elegant.

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    $\begingroup$ (+1) Doubling the precision is a very good thing in addition to the simplicity of the root algorithm. $\endgroup$ – robjohn Jan 18 at 18:38
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Here is a method. It does require an initial guess, though.

The generalized binomial theorm:

$(x+y)^k = x^k + kx^{k-1}y + \frac {k(k-1)}{2} x^{k-2}y^2 + \cdots$

We can see that the power of x drops each term, and the power of y increases one each term. The coefficients are a little bit more complicated. For each we take the previous coefficient, multiply it by the previous power of x and divide it by the current power of y.

How do we use it here.

We make a guess. $a\approx x^\frac 1n. $ It has some error we will call $b$

$x^\frac 1n = (a^n + b)^\frac 1n$

And using the generalized binomial thereom.

$x^\frac 1n = a + \frac {1}{n}\frac {b}{a^{n-1}} - \frac {1(n-1)}{2n^2}\frac {b^2}{a^{2n-1}} + \frac {1(n-1)(2n-2)}{6n^3}\frac {b^3}{a^{3n-1}} + \cdots$

For example

$71^\frac 13 = (4^3 + 7)^\frac 13\\ 4 + \frac {1}{3}\frac {7}{4^2} - \frac {1\cdot 2}{2\cdot 3^2}\frac {7^2}{4^5}+\frac {1\cdot 2 \cdot 5}{3!\cdot 3^3}\frac {7^3}{4^8}- \cdots$

$4 + \frac {7}{48} - \frac {49}{2304}+\frac {1715}{530846}- \cdots$

As you can see that denominator is growing rapidly.

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    $\begingroup$ As you add more terms, each term is approximately $\frac {7}{64}$ of the previous term. $\endgroup$ – Doug M Jan 18 at 19:21

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