4

I always use the Newton-Raphson Method if I want to find the roots of any equation as follow $$x_{1}=x_{0}-\frac{y_{0}}{y'_{0}}$$

But I don't know how to use this method if the equation takes the following form $$x^{x^{x^{x^{.^{.^{.}}}}}}-a=0$$

where $a=$ positive number

I know some solved cases when $a=2$ so that the $x=\sqrt 2$

Now I want to find the solution when $a=e$

3 Answers3

7

You have $x^a=a$ and hence $x=a^{1/a}$. It converges for all $a\in[e^{-e},e^{1/e}]$.

Spenser
  • 19,469
  • Perhaps it is of interest to do the generalization to complex numbers. Then your mentioned interval becomes a region of convergence called "Shell-Thron-region" (see wikipedia) – Gottfried Helms Jan 24 '15 at 11:42
3

Firstly, we assume that $x^{x^{x^{x^{.^{.^{.}}}}}}$ converges...
lets us define a series $a_n$
$a_0 = x$
$a_{n+1} = x^{a_n}$

also we know that series limit is $a$, $a_n\rightarrow a$,
therefore:
$a=x^a$
$\ln(a)=a\ln(x)$
$\ln(x)=\frac{\ln(a)}{a}=\ln(a^{\frac{1}{a}})$
$x=a^{\frac{1}{a}}$

user90803
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Grigory Ilizirov
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2

$$\begin{align}x^{x^{x^{x^{.^{.^{.}}}}}}-a &=0 \\ x^{x^{x^{x^{.^{.^{.}}}}}} &=a \\ x^a&=a \\ a\log_a x &= 1\\ \log_a x &= \frac{1}{a}\\ x &=a^\frac{1}{a} \\ \end{align}$$

Note that this only works for $a\leq e$, because $a^\frac{1}{a}$ is not a unique-valued function.

Joffan
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