Closed form for solution of $\ln(x)\ln(x+1)=1 $

Question

Is it possible to find a close form for the solution of this equation (maybe with the use of Lambert W function)?

$$ \ln(x)\ln(x+1)=1 $$

Answer 1

A separate solution using Lambert function.

If we expand $$f(x)=\log(x)\log(x+1)-1$$ as series around $x=0$ and just use the first term, the solution is $$x=\frac{1}{W(1)}=\frac{1}{\Omega}=\omega$$ Expanding again around $x=\omega$ $$f(x)=-(1-\log (\omega ) \log (1+\omega ))+\sum_{n=1}^\infty (-1)^{n+1}\,a_n \,(x-\omega)^n$$

Defining $$a_n=\frac{\log (\omega +1)}{n \,\omega ^n}+\frac{\log (\omega )}{n\,(\omega +1)^n}-\frac {b_n}{n!\,\omega ^{n-1} \,(\omega +1)^{n-1}}$$

the first coefficients are $$\left( \begin{array}{cc} n & b_n \\ 1 & 0 \\ 2 & 2 \\ 3 & 6 \omega +3 \\ 4 & 22 \omega ^2+22 \omega +8 \\ 5 & 100 \omega ^3+150 \omega ^2+110 \omega +30 \\ 6 & 548 \omega ^4+1096 \omega ^3+1214 \omega ^2+666 \omega +144 \\ 7 & 3528 \omega ^5+8820 \omega ^4+13104 \omega ^3+10836 \omega ^2+4704 \omega +840 \\ \end{array} \right)$$

Truncating to any order, we can use power series reversion and have the exact coefficients. Trying, the numerical values are

$$\left( \begin{array}{cc} k & x_{(k)} \\ 1 & 2.3050703\\ 2 & 2.3033257 \\ 3 & 2.3083386 \\ 4 & 2.3068490 \\ 5 & 2.3073554 \\ 6 & 2.3071730 \\ 7 & 2.3072415 \\ 8 & 2.3072150 \\ 9 & 2.3072255 \\ 10 & 2.3072212 \\ 11 & 2.3072230\\ 12 & 2.3072223 \\ 13 & 2.3072226 \\ 14 & 2.3072224 \\ 15 & 2.3072225 \\ \end{array} \right)$$ This is extremely slow for quite few decimal places.

Answer 2

The number $$x=2.3072224797671340050176675978910511364349575707671\cdots$$ surprizingly close to $\log(10)$ is known in the $ISC$ to be precisely the zero of function $$f(x)=\log(x)\,\log(x+1)-1$$

What is interesting is, if $x_0=\log(10)$, by Darboux theorem, Newton method should converge to the solution wtithout any overshoot of the solution since $$f(x_0)\times f''(x_0) =0.00013878 ~~>~0$$

The convergence is very fast $$\left( \begin{array}{cc} n & x_n \\ 0 & \color{red}{2.30}25850929940456840179914546843642076011014886288 \\ 1 & \color{red}{2.30722}19378768308564732222822813490367349349200691 \\ 2 & \color{red}{2.3072224797671}265755214115256783516726665701256023 \\ 3 & \color{red}{2.3072224797671340050176675978}896545894505276172324 \\ 4 & \color{red}{2.3072224797671340050176675978910511364349575707671}\\ \end{array} \right)$$

We can have totally explicit formulae using one single iteration of Newton-like method of order $n$

$$\left( \begin{array}{cc} n & x_{(n)} & \text{method} \\ 2 & \color{red}{2.30722}193787683085647322 & \text{Newton}\\ 3 & \color{red}{2.30722247}865584766172127 & \text{Halley}\\ 4 & \color{red}{2.30722247976}941421518848 & \text{Householder}\\ 5 & \color{red}{2.3072224797671}2983685267 & \text{no name}\\ 6 & \color{red}{2.3072224797671340}1249354 &\text{no name}\\ 7 & \color{red}{2.3072224797671340050}0429 &\text{no name}\\ 8 & \color{red}{2.3072224797671340050176}9 & \text{no name}\\ 9 & \color{red}{2.30722247976713400501767} &\text{no name}\\ \end{array} \right)$$

Edit

If you are not afraid by long coefficients,build the $[2,2]$ Padé approximant (have look here) $$P_{(2,2)}=\frac {a_0+a_1 t+a_2 t^2 }{1+b_1 t+b_2 t^2 }\qquad \text{where} \qquad t=x-\log(10)$$ and, at the price of a quadratic equation $$x=\color{red}{2.307222479767134}51$$