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Given the function:

\begin{equation} Λ_{c,d,r}(x)=rx^{c-1}\left[ 1-\frac{1-(1-c)r}{rd}\ln x \right]^{d} \end{equation} I would like to prove that its inverse function exists and is defined as: \begin{equation} \mathcal{E}_{c,d,r}(x)=e^{-\frac{d}{1-c}\left[ W_k\left( B(1-x/r)^{1/d} \right) -W_k(B)\right]} \end{equation} where $d,r \in \mathbb{R}$, $c \in \mathbb{R}^{+}-\{0\}$, $B$ is a constant defined as: \begin{equation} B=\frac{(1-c)r}{1-(1-c)r}e^{\frac{(1-c)r}{1-(1-c)r}} \end{equation} Moreover, $W_k$ is the $k$-th branch of the Lambert-W function which is a solution of the equation: \begin{equation} x=W(x)e^{W(x)} \end{equation} (we are only interested in the branches k=0 and k=-1 which represent real solutions.) My problem is that I am not able to actually do the work needed in order to acquire the Lambert-W form while attempting to solve the $\mathcal{E}=Λ^{-1}$. I have tried both ways (from $Λ$ to $\mathcal{E}$ and the inverse) since I know the expression of the function which I am actually looking for but no luck yet.

Defining $α \equiv -(1-c)/d$ in order to get away with less typing, we now have: \begin{align} & \mathcal{E}_{c,d,r}(x)=e^{-\frac{1}{α}\left[ W_k\left( B(1-x/r)^{1/d} \right) -W_k(B)\right]} \Leftrightarrow \mathcal{E}^{a}_{c,d,r}(x)e^{W_k(B)}=e^{W_K\left(B(1-x/r)^{1/d}\right)}\Leftrightarrow \\ &\log \mathcal{E}^{a}_{c,d,r}(x)+W_k(B)={W_K\left(B(1-x/r)^{1/d}\right)} \Leftrightarrow \\ & W_k^{-1}\left( \log \mathcal{E}^{a}_{c,d,r}(x)+W_k(B) \right)=B(1-x/r)^{1/d} \Leftrightarrow \\ & \frac{W_k^{-1}\left( \log \mathcal{E}^{a}_{c,d,r}(x)+W_k(B) \right)}{B}=\left( 1-\frac{x}{r}\right)^{\frac{1}{d}} \Leftrightarrow \left[ \frac{W_k^{-1}\left( \log \mathcal{E}^{a}_{c,d,r}(x)+W_k(B) \right)}{B} \right]^d=1-\frac{x}{r} \end{align} Therefore: \begin{equation} x=r\left[1- \left[ \frac{W_k^{-1}\left( \log \mathcal{E}^{a}_{c,d,r}(x)+W_k(B) \right)}{B} \right]^d \right] \end{equation} Now, by making use of the Lambert-W property: \begin{equation} x=W(x)e^{W(x)} \Leftrightarrow W^{-1}(x)=W^{-1}\left[ W(x)e^{W(x)} \right] \end{equation} My guess is that I am close to the end now, by starting with $Ε$ and building my way up to $Λ$. But I am still not able to find out somehow what the $W^{-1}_k(x)$ would be. And I can see it is of vital importance in order to finish this one. Any ideas or suggestions are most welcome.

Thanks!

UPDATE

Ok so after forcing myself to become familiar with the Lambert W, I continued where I left it from and ended up with the following:

\begin{equation} Λ_{c,d,r}(x)=rx^{c-1}\left[ x^{1-c}-\left[ \frac{1-(1-c)r}{rd}\ln x \right]^d \right] \end{equation} But I am not able to see now how to get to the final form of it. I would like the whole bracket to be to the power of $d$ and also the term $x^{1-c}$ should not be there, according to the $Λ$ which I would like to have.

If anyone could help, I would really appreciate it. Thank you again.

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    $\begingroup$ If $W^{-1}_k(u)$ is the $k$th branch of the inverse of the Lambert W function, then it is the same as $ue^u$, the inverse of the Lambert W function by definition. $\endgroup$ – Simply Beautiful Art May 4 '16 at 12:11
  • $\begingroup$ I would start with $\mathcal{E}$ because it looks easier to manipulate into the other. $\endgroup$ – Simply Beautiful Art May 4 '16 at 12:14

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