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For the equation $-x^{-2}-(1-y)^x\ln(1-y)=0$ for $0<y<1$ and $x > 1$, I'm trying to solve for $y$ in terms of $x$, but not sure if I'm doing it right.

Let $1-y=a$, we get 　$-x^{-2}-a^x\ln(a)=0$ 　=> 　$x^{-2}=-a^x\ln(a)$

Let $a=e^{xt}$, we get 　$-x^{-2}=e^{({xt})^x}\ln(e^{xt})=(xt)e^{{x^2}t}$　

=> 　$-x^{-1}=(x^2t)e^{{x^2}t}$ ......(eq1)

$W()$ is a Lambert $W $function. Apply $W()$ to both side of eq1:

=> $W(-x^{-1})=W((x^2t)e^{{x^2}t})$　=>　$W(-x^{-1})＝x^2t$

=> $t＝x^{-2}W(-x^{-1})$

=> $a=e^{xt}=e^{-x^{-1}W(-x^{-1})}$

=> $y=1-a=1-e^{-x^{-1}W(-x^{-1})}$

So......,am I doing it right?

Welan
  • 99

1 Answers1

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$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$

You have a typo at the last couple of lines. They should be

\begin{align} t&=x^{-2} \W(-x^{-1}) ,\\ a&=\exp(xt)=\exp(x^{-1}\W(-x^{-1})) ,\\ y&=1-a=1-\exp(x^{-1}\W(-x^{-1})) . \end{align}

And it is essential to complete the solution, with the analysis of the argument of $\W$ function. In this case it is $-x^{-1}$ and since you are expecting that $x>1$, the argument would be negative, hence to have real solutions, it must be in a range $[-1/\e,0]$, so $x$ must be at least $\e$. For such $x\ge\e$ there are two real solutions,

\begin{align} y_0&=1-\exp(x^{-1}\Wp(-x^{-1})) ,\\ y_1&=1-\exp(x^{-1}\Wm(-x^{-1})) , \end{align}
where $\Wp$ is the principal branch and $\Wm$ is the other real branch of the Lambert $\W$ function.

Both $y_0$ and $y_1$ are inside the range $(0,1)$ as desired, or more accurately,

\begin{align} y_0&\in(0,1-\exp(-1/\e)] ,\\ y_1&\in[1-\exp(-1/\e),1-\exp(-4/\e^2)]\approx[0.3078,0.418] . \end{align}

$\endgroup$

g.kov
  • 13,581