# $\sqrt[7]{3+\sqrt[7]{3+\sqrt[7]{3+\dotso}}}$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}}$$\newcommand{\root}[4]{\sqrt[\leftroot{#2}\uproot{#3}#1]{#4}}$$\newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}$$\newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}$$\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}$

Find the value of $$\root{7}{4}{3}{3+\root{7}{2}{1}{3+\root{7}{1}{2}{3+\cdots}}}$$

What I tried? I apply the same steps on these:

$$x=\root{n}{-1}{1}{t+x}$$

$$x^n=t+x$$

$$\underline{x^n-x-t=0}$$

That was where I was stuck.

Any way to solve that kind of equations?

Or how to solve this:

$$x^7-x-3=0$$

Note: I don't want to use numerical approximation.

NOTE TO READERS: The answers are false. It can be solved like explained here: https://arxiv.org/pdf/math/9411224v1.pdf

• I think you mean $x^7 - x - 3 = 0$ Aug 21, 2017 at 9:15
• @MCCCS You need to prove also that your sequence converges. Without this proof you can not get this equation. Aug 21, 2017 at 9:28

It is possible though to use other types of operations, for example, if you add hypergeometric functions, you can solve an equation of general form $x^n - x - a = 0$ with a finite hypergeometric representation. But these are defined in terms of infinite sums, so in another sense are perhaps not much different from your infinite radical you began with in terms of "satisfactoriness". (Arguably the infinite radical is already a very elegant-looking expression of the solution to the polynomial in its own right and the hypergeometric series is rather more sticky.) On the other hand, they are extremely useful operations, and can be manipulated in a crazy number of ways, moreso then I believe the infinite radical can be.
However, we can approximate the solution using $[1,n]$ Padé approximants built around $x=1$ and get the following results $$\left( \begin{array}{ccc} n & x_n & \approx \\ 1 & \frac{13}{11} & 1.18182 \\ 2 & \frac{16}{13} & 1.23077 \\ 3 & \frac{765}{622} & 1.22990 \\ 4 & \frac{3343}{2721} & 1.22859 \\ 5 & \frac{43858}{35695} & 1.22869 \\ 6 & \frac{191739}{156044} & 1.22875 \\ 7 & \frac{419121}{341099} & 1.22874 \\ 8 & \frac{5497012}{4473715} & 1.22874 \\ 9 & \frac{48064227}{39116797} & 1.22874 \\ 10 & \frac{210129473}{171012676} & 1.22874 \end{array} \right)$$