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?
Solve $x^x-x=1$
lambert-wtetration
Related Solutions
$\def\srt{\operatorname{srt}}$ Although there are integral solutions, the super root $\sqrt[n]x_s,\srt_n(x)$ has an expansion using Lagrange reversion:
$$\sqrt[k]x_s=\srt_k(x)= 1+\sum_{n=1}^\infty\frac{\ln^n(x)}{n!}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0$$
Applying general Leibniz rule and $e^z$ Maclaurin expansion each $k-2$ times if $2<k\in\Bbb N$:
$$ \frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^\infty\sum_{n_2=0}^{n-1}\dots\sum_{n_{2k-5}=0}^\infty\sum_{n_{2k-4}=0}^{n-1-\sum_\limits{j=1}^{k-3}n_{2j}}\left.\frac{d^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}{dt^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}e^{(n_{2k-5}+1)t}\right|_0\prod_{j=1}^{k-2}\frac{n_{2j-3}^{n_{2j-1}}}{n_{2j-1}!}\left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_{2m}}\\n_{2j}\end{matrix}\right)\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0$$
where $n_{-1}=-n$. Next, use Kronecker delta and factorial power $m^{(n)}$ in $\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0=\delta_{n_{2j},n_{2j-1}}n_{2j-1}^{(n_{2j})}$. As hinted by Quantile Mechanics $(96)$ to $(97)$, remove the odd indexed sums substituting each $n_{2j}=n_{2j-1}$. We reindex $n_{2j}\to n_j$ and simplify:
$$\begin{align}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^{n-1}\dots \sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j} (n_{k-2}+1)^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-2}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)= \sum_{n_1=0}^{n-1}\dots \sum_{n_{k-1}=0}^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-1}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)\end{align}$$
where $n_0=n$. These links support the last equality. Therefore:
$$\bbox[1px, border: 2px solid red]{\sqrt[k]z_s=\srt_k(z)=1+\sum_{n=1}^\infty \sum_{n_1=0}^{n-1}\dots\sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j}\frac{(-1)^{n_1}\ln^n(z)}{\Gamma \left({n-\sum_\limits{j=1}^{k-2}n_m}\right)(n_{k-2}+1)^{1-n+\sum\limits_{j=1}^{k-2}n_j} n}\prod_{j=1}^{k-2}\frac {n_{j-1}^{n_j}}{n_j!}}$$
where all can be infinite sums or all, except the $n$ sum, can have an upper bound of $n$. Assume improper sums are empty. We find that:
$$\srt_1(z)=z$$ $$\srt_2(z)=1+\ln(z)+\sum_{n=2}^\infty\frac{\ln^n(z)}{n!}(1-n)^{n-1}$$ $$\srt_3(z)=1-\sum_{n=1}^\infty\sum_{k=1}^{n-2}\frac{(-1)^kk^{n-k+1}n^{k-2}\ln^n(z)}{(n-k)!k!}$$
$$\srt_4(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=1}^{n-k-1}\frac{(-1)^k n^{k-1}k^{m-1}\ln^n(z)}{(n-k-m)!k!m!m^{m+k-n-1}}$$
$$\srt_5(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-k-1}\sum_{j=1}^{n-k-m-1}\frac{(-1)^k k^m n^{k-1} m^{j-1}\ln^n(z)}{(n-j-k-m)!m!k!j!j^{j+m+k-n-1}}$$ Shown here
$$\srt_6(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-1-k}\sum_{j=0}^{n-1-k-m}\sum_{p=1}^{n-k-m-j}\frac{(-1)^k n^{k-1}k^m m^j j^{p-1}\ln^n(z)}{(n-1-k-m-j-p)!k!m!j!p!p^{p+j+m+k-n-1}}$$ shown here $$\vdots$$ $$\srt_\infty(z)=\sqrt[z]z$$
Compare both links for verification. The region of convergence is near $|z|=1$. Help is wanted with the region of convergence for the multiple series expansion and other representations of $\srt_k(z)$.
As noted by OP, this can be solved for $C=0$ by squaring both sides and multiplying by $2$, yielding $$ 2x^2 e^{2x^2}=2, $$ or $2x^2=W(2)$, or $x=\sqrt{W(2)/2}\approx 0.652919$. For $C=1$, of course, the solution is $x=1$. More generally, note that the solution is a fixed point of $$ f(x) = \exp\left(-(x-C)^2\right); $$ since this function is a contraction of $\mathbb{R}$ onto $(0,1]$, it has a unique fixed point, and iterating $x\rightarrow f(x)$ (starting at $0$, say) will rapidly converge to the solution for any $C$. In particular, the initial iterate of $e^{-C^2}$ is an excellent approximation for even moderately large $|C|$... it differs from the exact solution by less than $10\%$ for $C=\pm 2$, by about $1\%$ for $C=\pm 2.5$, and by less than $0.1\%$ for $|C|\ge 3$.
Expanding in powers of $x$, we have $$ x e^{(x-C)^2}=xe^{C^2}e^{-2Cx + x^2}=1, $$ or $$ xe^{C^2}=e^{2Cx-x^2}=1+(2Cx-x^2)+\frac{1}{2!}(2Cx-x^2)^2+\frac{1}{3!}(2Cx-x^2)^3+\ldots. $$ This has a self-consistent asymptotic solution of $x \sim e^{-C^2}+\sum_{k=2}^{\infty}b_k(C)e^{-kC^2}$, where the coefficients are polynomials in $C$ and can be found by substitution and equating like terms.
Best Answer
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, $x=1.776779506$.