0

Say we have the infinite nested radical $$\sqrt[n]{a + \sqrt[n]{a + \sqrt[n]{a + \cdots}}}$$

When $n = 2$, this evaluates to $\frac{1 + \sqrt{4a + 1}} {2}$, which is the positive root of the equation $x^2 - x - a = 0$.

When $n = 3$, this evaluates to $\sqrt[3]{\frac{9 a + \sqrt{81 a² - 12}} {18}} + \sqrt[3]{\frac{9 a - \sqrt{81 a² - 12}} {18}}$, which is the positive real root of the equation $x^3 - x - a = 0$.

When $n = 4$, this evaluates to $\frac{1} {2} \sqrt{\sqrt[3]{\frac{9 + \sqrt{768x³ + 81}} {18}} + \sqrt[3]{\frac{9 - \sqrt{768x³ + 81}} {18}}} + \frac{1} {2} \sqrt{-\sqrt[3]{\frac{9 - \sqrt{768x³ + 81}} {18}} - \sqrt[3]{\frac{9 + \sqrt{768x³ + 81}} {18}} + \frac{2} {\sqrt{\sqrt[3]{\frac{9 + \sqrt{768x³ + 81}} {18}} + \sqrt[3]{\frac{9 - \sqrt{768x³ + 81}} {18}}}}}$, which is the positive real root of the equation $x^4 - x - a = 0$.

In general, nested radicals of degree $n$ with $n > 1$ give the positive real root of the equation $x^n - x - a = 0$.

Is there a formula or process that will output the expression for a given value of $n$? Is an algebraic solution even possible for $n > 4$? I worked out the three examples given here using WolframAlpha and GeoGebra, but what was quick and simple for $n = 2$ was slower for $n = 3$ and quite tedious for $n = 4$, and WolframAlpha doesn't even show an algebraic solution for $n = 5$.

Lawton
  • 1,759
  • 7
  • 18

1 Answers1

1

If $n=5$, $a=1$, the Galois group of $x^5-x-1$ is $S_5$. So a root cannot be expressed in radicals starting with rational numbers by the Galois theorem, so no algebraic solution is possible if $a=1,n=5$.

For arbitrary $a$, $n=5$ and more general solutions see this question.

markvs
  • 19,653