# Sequence related to solutions of the equation $x^x=c$

A couple years ago I remember repeatedly pressing $\sqrt{1+ans}$ into my calculator to be astonished that my calculator gives me an answer approaching the golden ratio. I was astonished, and dug deeper into this problem. I realized that the limit of the sequence given by the recurrence:

$$f(x_n)=x_{n+1}$$

Is, if the limit exists, a solution to:

$$f(x)=x$$

And applying this I realized I could solve almost every algebra problem. So here was one that I came across:

$$x^x=c$$

Easily such equation can be rearranged as:

$$x=\sqrt[x]{c}$$

Where now the problem becomes finding the limit of the sequence:

$$x_{n+1}=\sqrt[x_n]{c}$$

This method worked for some $c$ and $x_1$ like $c=2$ and $x_1=1.5$. But the problem is for some $c$ the sequence seems to alternate according to my observations, like $c=100$ (I tried $x_1=3.5$ and others but it doesn't seem to matter as long as it is positive). I ask anyone if they can help me figure out for what $c$ will this sequence converge for every given $x_1>0$.

• It is very interesting that you are eager to learn and observe such details in high school. The field of Numerical Analysis is fascinating specially with a computer or calculator at hand. – NoChance Jan 4 '16 at 6:09
• Thank You, I really appreciate the comment. I drives me forward away from doubts that I may have as math student. – Ahmed S. Attaalla Jan 4 '16 at 6:25
• Fixed-Point iterations method may help you with this. The video in the link gets interesting after few minutes of its start about 6th minute or so. It uses Sin(x), but the concept is what is important. You will find plenty of other references about this. The topic is usually discussed under Numerical Methods. youtube.com/watch?v=CD7kadPr_JQ – NoChance Jan 4 '16 at 7:05

What you discovered is called a fixed point iteration.

A fixed point $x$ of a function $f\colon X\rightarrow X$ is a point satisfying the equation $$x=f(x).$$

A fixed point iteration is defined by the sequence \begin{align*}x_{1}&=\text{given}\\x_{n+1}&=f(x_{n})\text{ for }n>0.\end{align*} We say this iteration converges when $x_{n}\rightarrow x$ for some $x$. The Banach fixed point thereom (a.k.a. contraction mapping principle) gives sufficient conditions for when a fixed point exists, is unique, and can be computed by a fixed point iteration (i.e. as the limit of the $x_{n}$). You should read the article on Wikipedia to familiarize yourself with the ideas.

For simplicity, let's now assume $X=\mathbb{R}$ instead of an arbitrary complete metric space, as is usually treated in the statement of the Banach fixed point theorem. One of the consequences of the theorem is as follows: let $Y\subset X$ such that $f$ is continuously differentiable on $Y$, $\sup_{Y}|f^{\prime}|\leq K$ for some $K<1$, and $f(Y)\subset Y$, then $f$ has a unique fixed point on $Y$ which can be computed by a fixed point iteration initialized in $Y$ (i.e. $x_{1}\in Y$).

In your first example, you have $f(x)=\sqrt{1+x}$. Let $Y=[0,\infty)$. Noting that $|f^{\prime}(x)|<1$ on $Y$ (check) and $f(Y)\subset Y$ trivially, it follows that for any $x_{1}\in Y$, $x_{n}\rightarrow x$ where $x$ is the unique fixed point of $f$ on $Y$. In fact, this fixed point must satisfy $x=f(x)=\sqrt{1+x}$, or equivalently, since $x\geq0$, $x^{2}=1+x$. This quadratic equation has two roots: $x=1/2\pm\sqrt{5}/2$. The positive root is, as you pointed out, the golden ratio $1.61803\ldots$

Your other problem, involves $f(x)=\sqrt[x]{c}$. It is not so clear under which conditions this satisfies the Banach fixed point theorem. In fact, as you noticed, you can find examples in which the iteration is nonconvergent.

Do not despair though, there are better ways to compute solutions to your equation. One that comes to mind is Newton's method, which you should take a look at.

In fact, a solution to your problem is given by Lambert's W function as $$x = \frac{\ln c}{W(\ln c)}.$$ Values of the Lambert W are, in fact, often computed by Newton's method, or higher order Newton's methods.

Newton's method for $g(x)=x^{x}-c$ is given by \begin{align*}x_1&=\text{given}\\x_{n+1}&=f(x_{n})\text{ for }n>0\end{align*} where $$f(x)=x+\frac{cx^{-x}-1}{1+\ln x}.$$ Since $g^{\prime\prime}(x)\geq0$ for $x>0$ and $f(x)>1/e$ for $x>1/e$ (check), it follows that Newton's method converges whenever $x_{1}>1/e$ and $x=f(x)$ has a solution with $x>1/e$ (see this). This is true at least when $c\geq1$, so that we can conclude:

Newton's method for $x^x - c = 0$ converges for $c \geq 1$ and $x_1 > 1/e$.

Here's some MATLAB code:

c = 100.; % Value of c
x = 1.; % Initial guess
while 1
x_new = x + (c*x^(-x) - 1)/(1 + log(x));
if abs (x_new - x) < 1e-12; break; end
x = x_new;
end
disp (x) % Solution

• +1 It's not so hard to show that the fixed point of $f_c(x)=\sqrt[c]{x}$ is attractive for $e^{-1/e}<c<e^e$. – Mark McClure Jan 4 '16 at 5:35
• But the function we want is $f_c(x)=\sqrt[x]{c}$, @MarkMcClure – Thomas Andrews Jan 4 '16 at 5:45
• @ThomasAndrews Typo :) – Mark McClure Jan 4 '16 at 5:45
• Before I accept would I solve $|f'(x_1)|<1$ where $f(x)=c^(1/x)$ to ensure convergence? @MarkMcClure and Par – Ahmed S. Attaalla Jan 4 '16 at 6:37
• @AhmedS.Attaalla Close - You need $|f_c'(x_0)|<1$, where $x_0$ is the fixed point of $f_c$. That at least guarantees that there is a neighborhood about $x_0$ such that every initial seed $x_1$ in that neighborhood leads to a sequence that converges to $x_0$. – Mark McClure Jan 4 '16 at 6:56