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I was recently posed the question "solve for $x$ in $x^x-x=1$". The intended answer was $x=0$, assuming that $0^0=1$, but I used brute force and determined another solution, $x\approx1.776775040097$ (which Wolfram Alpha agrees with me on). Is there a closed form or symbolic solution to this - an exact solution? I have tried solving with the super square root (and Lambert W function), but this didn't seem to work out for me. Is there a way to solve it?

e dt
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  • I don't understand what you mean by "What is the reason behind this solution?" $x^x$ increases much more slowly than $x+1$ when $x$ is near zero, so the function $x^x - x - 1$ is negative near zero. However, for large $x$, $x^x$ is much much larger than $x+1$, and so the function is eventually positive. The function is continuous, thus it follows from the intermediate value theorem that the must be a zero in there somewhere... – Xander Henderson Jun 19 '20 at 14:51
  • @Moo: Yes, I used a numerical method to get my numerical answer. – e dt Jun 19 '20 at 14:52
  • With Maple we get $$1.776775040097054697479730744038756748637411034329296139084374015273118658932824770702072786151313524$$ – Dr. Sonnhard Graubner Jun 19 '20 at 14:52
  • @XanderHenderson: When I say "What is the reason behind this solution?" I don't mean "What is the reason that there is a solution?" - you elucidate very well why there must be a solution. In fact, my choice of words there was probably unhelpful; on further reflection, I'm really looking for a closed form solution, or just any symbolic solution. – e dt Jun 19 '20 at 14:55
  • @edt In that case, what makes you think that there is a closed form solution? – Xander Henderson Jun 19 '20 at 14:56
  • @XanderHenderson: I'm not sure there is! See the last sentence of my post. – e dt Jun 19 '20 at 14:58
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    edt - likely to prevent your question to be closed, you should improve the subject line. As I understand your question (after reading the comments) it is more: "is there a closed form for x^x-x-1=0 besides its numerical solution(s) of 0 and approx 1.776" - or the like. – Gottfried Helms Jun 20 '20 at 06:31
  • See also : https://tetrationforum.org/showthread.php?tid=1258 where $x^x = x $ is discussed. – mick Nov 22 '23 at 12:01

2 Answers2

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Consider that you look for the zero's of function $$f(x)=x^x-x-1$$ Its first derivative $f'(x)=x^x (\log (x)+1)-1$ cancels at $x=1$ and the second derivative test $f''(1)=2$ shows that this is a minimum.

Build a Taylor expansion to get $$f(x)=-1+(x-1)^2+\frac{1}{2} (x-1)^3+\frac{1}{3} (x-1)^4+O\left((x-1)^5\right)$$ Using series reversion, then $$x=1+\sqrt{y+1}-\frac{y+1}{4}-\frac{1}{96} (y+1)^{3/2}+O\left((y+1)^2\right)$$ where $y=f(x)$. Making $y=0$, this gives as an approximation $$x=\frac{167}{96}\approx 1.73958 $$ To polish the root, use Newton method starting with this estimate. The iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.739583333 \\ 1 & 1.778584328 \\ 2 & 1.776779132 \\ 3 & 1.776775040 \end{array} \right)$$

Edit

If we make the first expansion $O\left((x-1)^n\right)$ and repeat the inversion series, we generate the sequence $$\left\{2,\frac{7}{4},\frac{167}{96},\frac{175}{96},\frac{160 379}{92160},\frac{3687}{2048},\frac{12144341}{6881280},\frac{110221693}{61931520 },\frac{211659504277}{118908518400}\right\}$$

We can also use $x_0=2$ and use high order iterative methods. For order $4$, that is to say one level after Householder method, we have

$$x=2\,\frac {4575+67460 a+299400 a^2+558920 a^3+463660 a^4+141128 a^5} {6655+86720 a+352260 a^2+615000 a^3+483960 a^4+141128 a^5 }$$ where $a=\log(2)$.

This gives, as another approximation, $x=1.776779506$.

Claude Leibovici
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$$x^x-x=1$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($x^x,x$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.

$$e^{x\ln(x)}-x=1$$

We see, we cannot solve the equation in terms of Lambert W because of the exponent of the exponential.

$$e^{x\ln(x)}=x+1$$ $$1=(x+1)e^{-x\ln(x)}$$ $$(x+1)e^{-x\ln(x)}=1$$ $x\to t-1$: $$te^{(1-t)\ln(t-1)}=1$$

We see, the equation can be solved in terms of Hyper Lambert W.

$$G((1-t)\ln(t-1)e^{-t};t)=1$$ $$t=HW((1-t)\ln(t-1)e^{-t};1)$$ $$x=HW((1-t)\ln(t-1)e^{-t};1)-1$$

So we have a closed form for $x$, and the representations of Hyper Lambert W give some hints for calculating $x$.

see also: exact solution to $x+x^x=3$

Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997

Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119

IV_
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  • See also : https://tetrationforum.org/showthread.php?tid=1258 where $x^x = x $ is discussed. – mick Nov 22 '23 at 12:01