I have a hard time finding the analytical solution to the following non-linear equation:
$$ (1+x)^{1-p} + p^{\frac{p}{1-p}}x(p-1) - 1 = 0 $$
where $p \in ]0,1[$ and $x > 1$. I would like to have a solution $x$ in terms of $p$ for each fixed $p$ in the specified interval. Of course it's possible that no analytical form of the solution exists. If so, I'd also be happy to hear an argument why that is the case. However, for some values of $p$ the solution is 'nice', for example for $p = \frac{1}{2}$ it's easy to compute that $x = 8$ and for $p = \frac{1}{3}$ I obtained $x = \frac{9}{16} (5 \sqrt{3} + \sqrt{11 + 64 \sqrt{3}}) $.
Any kind of help is greatly appreciated!
This a highly transcendental equation and, for general p, you will need some numerical method.
It effectively happens that for some particular values of $p$, there are explicit solutions (this is the case for $p =\frac 15,\frac 25,\frac 35,\frac 45,\frac 14,\frac 2 4,\frac 34,\frac 13,\frac 23$ but I do not think that we could do anything for, say, $p=\frac 16$.
Let $k=(p-1) p^{\frac{p}{1-p}}$ ($k$ varies between $-1$ and $0$) and $p=\frac m n$ ($m$ and $n$ being integers, $m<n$). The equation write $$(x+1)^{1-\frac{m}{n}}=1-kx\implies (x+1)^{n-m}=(1-kx)^n$$ which makes that we face a polynomial of degree $n$ in $x$. In fact $x=0$ is a solution; so the polynomial of degree $(n-1)$ and we do not know to solve analytically quintic polynomials. Then ... ?
In French, we have an expression which says "This, Sir, is the cause of your daughter's being dumb"
We can have approximations; for example, when $p$ is close to zero, a Taylor expansion gives $$p (x-x \log (p)-x \log (x+1)-\log (x+1))+O\left(p^2\right)$$ If we neglect the term $\log(x+1)$ we have $x\sim \frac ep-1$ which could be a good starting point.
Similarly, when $p$ is close to $1$, we have $$(p-1) \left(\frac{x}{e}-\log (x+1)\right)+O\left((p-1)^2\right)$$ which gives $$x=-e W_{-1}\left(-e^{-1-\frac{1}{e}}\right)-1\approx 4.76$$ which is probably very close to the limit.
Between these two limits, it seems that the function looks like an hyperbola. A quick and dirty nonlinear regression gives as an estimate $$x_0=\frac 1p \left(\frac{373 }{181}p^{2/3}+\frac{1882}{697} \right)$$ which seems to be quite good (as shown below).
We can generate a first iterate of Newton method and get $$x_1=\frac{(x_0+1)^p-p x_0-1}{(p-1) \left(p^{\frac{p}{1-p}} (x_0+1)^p-1\right)}$$
$$\left( \begin{array}{cccc} p & x_0 & x_1 & \text{solution} \\ 0.05 & 59.5967 & 59.4965 &59.4964 \\ 0.10 & 31.4412 & 31.4184 &31.4184 \\ 0.15 & 21.8795 & 21.8791 &21.8791 \\ 0.20 & 17.0246 & 17.0316 &17.0316 \\ 0.25 & 14.0718 & 14.0808 &14.0808 \\ 0.30 & 12.0789 & 12.0876 &12.0876 \\ 0.35 & 10.6389 & 10.6465 &10.6465 \\ 0.40 & 9.54726 & 9.55343 &9.55343 \\ 0.45 & 8.68954 & 8.69423 &8.69423 \\ 0.50 & 7.99670 & 8.00000 &8.00000 \\ 0.55 & 7.42457 & 7.42663 &7.42663 \\ 0.60 & 6.94356 & 6.94455 &6.94455 \\ 0.65 & 6.53306 & 6.53316 &6.53316 \\ 0.70 & 6.17829 & 6.17768 &6.17768 \\ 0.75 & 5.86837 & 5.86720 &5.86720 \\ 0.80 & 5.59508 & 5.59353 &5.59353 \\ 0.85 & 5.35213 & 5.35033 &5.35033 \\ 0.90 & 5.13459 & 5.13268 &5.13268 \\ 0.95 & 4.93857 & 4.93666 &4.93666 \end{array} \right)$$ Starting with this estimate, Newton method will probably converge in a couple of iterations.
Trying for $p=0.23456$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 14.853058 \\ 1 & 14.861762 \\ 2 & 14.861760 \end{array} \right)$$