How to solve $M=\frac{t}{\ln(1+t)} - \frac{t^3}{(1+t) (\ln(1+t))^2 }$

Question

Does anyone have any ideas on how to solve $M=\frac{t}{\ln(1+t)} - \frac{t^3}{(1+t) (\ln(1+t))^2 }$ where $M$ is a constant and $t = \frac{a}{x}$, $a$ is a constant and $x$ is the variable. Here $ 0 < x \leq X_{max}$. I wonder if there is a closed-form solution for the formula above.

Answer 1

As mr_e_man commented, it would be surprising that a closed form exist (even using special functions).

If $M$ is a large negative number, using Taylor, $$\frac{t}{\ln(1+t)} - \frac{t^3}{(1+t) (\ln(1+t))^2 }\approx -\frac{t^2}{\log^2(t)}$$ and then an approximate solution is given by $$t=-\sqrt{-M}\,\, W_{-1}\left(\frac{\sqrt{-M}}{M}\right)$$ where $W(.)$ is Lambert function. The Wikipedia page provides series for a shortcut evaluation of $W(x)$ when $x$ is small or large.

This is not too bad as shown below $$\left( \begin{array}{ccc} M & \text{estimate} & \text{solution} \\ -100 & 35.7715 & 39.3988 \\ -200 & 57.2358 & 60.9711 \\ -300 & 74.7152 & 78.5249 \\ -400 & 89.9951 & 93.8614 \\ -500 & 103.811 & 107.723 \\ -600 & 116.556 & 120.506 \\ -700 & 128.470 & 132.453 \\ -800 & 139.713 & 143.725 \\ -900 & 150.399 & 154.437 \\ -1000 & 160.612 & 164.674 \end{array} \right)$$ making Newton method converging in very few iterations.

If $M$ is a positive number, using Taylor again, $$\frac{t}{\ln(1+t)} - \frac{t^3}{(1+t) (\ln(1+t))^2 }=1-\frac{t}{2}-\frac{t^2}{12}+O\left(t^3\right)$$ which could be a way to get an estimate for $0 \leq M \leq 1.5$.

In any manner, a numerical method (such as Newton) would be required to polish the root.