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So I have tried solving for $x$ algebraicly using the productlog function but all I was able to do is: $$x\log(x) = W(x\log(x)(x+1))$$

Maybe I could use the square-super root formula $e^{W(\log(x))}$, but I got stuck again at: $$W(\log(x+1))=x$$

I feel like I'm only getting deeper into a rabbit hole rather than getting out here, so is it even possible to represent $x$ (which roughly equals 1.776775) in terms of elementary functions and the Lambert $W$-function?

Gary
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Catman 321
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  • Proving that such equations don't have a elementary solution is often an open question or very very difficult. So I think the best answer you'll get in this site is "Probably not". – jjagmath Mar 28 '22 at 21:00
  • It was partially answered here : https://math.stackexchange.com/questions/3726436/how-to-solve-xx-x-1?rq=1 – SacAndSac Mar 28 '22 at 21:05
  • Don't know whether this helps, but maybe it is inspirating: there has been an engaged article about the solutions/inverse involving $x^x$. This was online in the nineties - as a simple ascii-text. I formatted it into a pdf-file. "Wexzal" (Fantini/Kloepfer 1995(?). https://go.helms-net.de/math/divers/WexZal.pdf – Gottfried Helms Apr 02 '22 at 11:24

1 Answers1

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Lambert function will not be of any use for this problem which is purely numerical.

Consider that you look for the non-trivial zero of function $$f(x)=x^x-x-1$$ The first derivative $$f'(x)=x^x (\log (x)+1)-1$$ cancels at $x=1$ and $f(1)=-1$. So, for an approximation, expand as a series $$f(x)=-1+(x-1)^2+\frac{1}{2} (x-1)^3+\frac{1}{3} (x-1)^4+\frac{1}{12} (x-1)^5+O\left((x-1)^6\right)$$ Using series reversion $$x=1+t-\frac{t^2}{4}-\frac{t^3}{96}+\frac{t^4}{12}+O\left(t^{5}\right)\quad \text{where}\quad t=\sqrt{1+f(x)}$$ Since you want $f(x)=0$, then the approximation $$x \sim \frac{175}{96}=1.82292\cdots$$ which is not too bad. Now, you can polish the root with very few iterations of Newton method $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.822916667 \\ 1 & 1.779309489 \\ 2 & 1.776783063 \\ 3 & 1.776775040 \end{array} \right)$$

If you really want to see something with Lambert, make the expansion around $x=3\Omega$ $(\Omega=W(1))$ since $$f(3\Omega)=-0.231333$$ Using series and then series reversion wuld give $$x\sim 3 \Omega+\frac{-3^{3 \Omega } \Omega ^{3 \Omega }+3 \Omega +1}{3^{3 \Omega } \Omega ^{3 \Omega } (\log (3 \Omega )+1)-1}=1.78456$$

Edit

If you know about Padé approximants, $$f(x)+1=\frac{(x-1)^2+\frac{49}{90} (x-1)^3 } {1+\frac{2 }{45}(x-1)-\frac{16}{45} (x-1)^2+\frac{43}{540} (x-1)^3 }+O((x-1)^7)$$ Let $t=x-1$ and $f(x)=0$ to face a cubic equation $$251 t^3+732 t^2-24 t-540=0$$ which has three real roots. Using the trigonometric method, they are $$t_k=\frac{4}{251} \left(7 \sqrt{314} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{124897}{61544 \sqrt{314}}\right)\right)\right)-61\right)\quad \quad k=0,1,2$$ The one of interest is $t_0$ which, numerically, is $0.776346$ making $x=1.776346$ while the exact solution is $1.776775$.

Claude Leibovici
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