7
$\begingroup$

The lambert W function is defined to be the inverse of $f(x)=xe^x$, and the equation $x^x=k$ can be solved fairly easily using the function: $$x^x=k$$ $$\ln(x^x)=\ln(k)$$ $$x\ln(x)=\ln(k)$$ $$e^{\ln(x)}\ln(x)=\ln(k)$$ $$\ln(x)=W(\ln(k))$$ $$x=e^{W(\ln(k))}$$

My question is, can $ x^{x^x}=k$ be solved using the lambert W function, and if so, can power towers of any height be solved using some iterative process?

$\endgroup$
1
  • $\begingroup$ You've made an unediable (edit too short) typo. $x\ln(x) = \ln(k)$. $\endgroup$
    – quapka
    Commented Nov 23, 2014 at 20:48

1 Answer 1

Reset to default
2
$\begingroup$ 
can $x^{x^x}=y$ be solved using the lambert W function, and if so,

Using Maurer's notation, the equation ${^n}x=y$ is solvable using Lambert's $W$ function only for $n=2$.

If we loosely define "transcendence degree of an equation" to be the height of the highest exponential tower with $x$ on top, contained in the equation minus one, then $W$ solves ONLY equations with (transcendence) degree at most 1 (that is, with exponentials of total height 2), but NOT all such equations.

The equation $x^{x^x}=y$ is written as ${^3}x=y$, consequently it has (transcendence) degree 2 (tower with 3 x's minus 1)

If memory serves right, the most general equation solvable by $W$ is something like:

$$a\cdot x^m\cdot d^{b\cdot x^n+c}=y$$

which as you can see can be reduced to having degree 1 at some point in the algebraic solution process, by taking $n$-th roots on both sides of the equation.

Elaborating on the second point, note that adding one plain $x$ to the equation above and it fails solvability. That is, the equation

$$a\cdot x^m\cdot d^{b\cdot x^n+c}+x=y$$

for example, is now NOT solvable by $W$, although it is of the same transcendence degree as the one above, so the definition of "transcendence degree" of an equation, needs to be made a liitle more rigorous, by specifying some additional configuration constraints, which vary from case to case.

can power towers of any height be solved using some iterative process?

Sure they can. Omitting details about existence, the following equivalence holds:

$${^n}x=y\Leftrightarrow x=^{\frac{1}{n}}y$$

and this is the $n$-th order super root of $y$, which, after taking care of domains and ranges, can be approximated using backwards numerical iteration.

Such a method, for example, was presented in Maple by Robert Israel, above, long time ago on sci.math.

$\endgroup$
7
  • $\begingroup$ You just pretty much stated that it cannot solve equations with a transcendence degree of higher than one, but is there any reason/proof of why? $\endgroup$
    – ASKASK
    Commented Nov 23, 2014 at 21:48
  • $\begingroup$ @ASKASK Sure: It's because the Lambert is the inverse of $x\cdot\exp(x)=y$ and not the inverse of some $x\cdot \exp^{(n)}(x)=y$, with $n>1$. Isn't it obvious that the limitation comes from the definition of $W$? $\endgroup$
    – user127032
    Commented Nov 23, 2014 at 21:55
  • $\begingroup$ Well actually you can solve $x^{x^{x+1}} = a$: write this as $y^y = a$ where $y = x^x$, and use Lambert twice. $\endgroup$
    – Robert Israel
    Commented Nov 23, 2014 at 23:32
  • 1
    $\begingroup$ @ioannisgalidakis I still don't see what your point is... just because the definition only has n=1, why does that mean that it can't solve equations of higher n? Robert Israel provided a good counterexample. $\endgroup$
    – ASKASK
    Commented Nov 23, 2014 at 23:36
  • $\begingroup$ @ASKASK Yes, Robert gave an excellent counterexample. Maybe he is hinting towards a stronger definition than what I called "transcendence degree" by showing a certain pattern in the grouping of the power tower, which allows $W$ to be applied iteratively all the way down to a closed form solution. Which, if true, would be wonderful. $\endgroup$
    – user127032
    Commented Nov 24, 2014 at 0:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .