I would like to obtain a closed-form solution for the equation $x^a = 1 - \exp(-x)$, in which $x$ is the (real strictly positive) unknown and $a$ is a real positive parameter.
So far, I have tried to play around with Lambert W function and with the generalization proposed by Mező and Baricz, and cited in recent posts of this forum (On the generalization of the Lambert W function with applications in theoretical physics" István Mező, Árpád Baricz http://arxiv.org/abs/1408.3999) but with limited success.
I have no reason to believe that such a closed-form solution exists but, before giving up and going to numeric solvers, I wanted to have your opinion about the existence of a closed-form solution. Thank you.
I do not think that closed form solution of the positive root of equation $$f(a,x)=x^a -(1 - e^{-x})=0$$ could be obtained.
It seems to me that, beside the trivial $x=0$, the existence of a positive solution requires $a >1$. In such a case $f(a,0)=0$, $f(a,1)=\frac 1e$. The derivative of the function cancels for $$x_*=(a-1) W\left(\frac 1{(a-1) a^{\frac{1}{a-1}}}\right)<1$$ where appears Lambert function. and this point corresponds to a minimum of the function ($f(a,x_*)<0$).
The positive root is smaller than $1$. Expanding the function as a first order Taylor series around $x=1$, an approximate solution is $$x=\frac{e a-2}{e a-1}$$ Using the simplest Pade approximant would give an estimate of the solution $$x=\frac{\left(e-2 e^2\right) a^2+5 e a-3}{\left(e-2 e^2\right) a^2+3 e a-1}$$ Increasing the degree of denominator, we should get $$x=\frac{\left(e-6 e^2+6 e^3\right) a^3+\left(6 e-18 e^2\right) a^2+17 e a-4}{\left(e-6 e^2+6 e^3\right) a^3+\left(3 e-12 e^2\right) a^2+8 e a-1} $$ From any of these estimates, Newton method will converge in very few iterations.
For example, for $a=\pi$, the above estimates are respectively $0.867369$, $0.840271$ and $0.835240$ while the solution is $0.834187$.
Edit
Rewriting the equation $$a\log(x)=\log(1-e^{-x})$$ and remembering that the solution of $$a\log(x)=b+c x$$ is given by $$x=-\frac{a }{c}W\left(-\frac{c e^{\frac{b}{a}}}{a}\right)$$ we can develop $\log(1-e^{-x})$ as a Taylor expansion at $x=x_0$ to get $$\log(1-e^{-x})=\log \left(1-e^{-{x_0}}\right)+\frac{x-x_0}{e^{{x_0}}-1}$$ that is to say $$b=\log \left(1-e^{-{x_0}}\right)-\frac{{x_0}}{e^{{x_0}}-1}$$ $$c=\frac{1}{e^{{x_0}}-1}$$ and so obtain a very good approximation of the solution using Lambert function.
For example, using $a=\pi$, $x_0=1$, the estimate is $0.838721$ but using $x=\frac{e \pi-2}{e \pi-1}$ the estimate is $0.834420$