# Inverse function of $y=W(e^{ax+b})-W(e^{cx+d})+zx$

I have a simple question for which I am looking for a closed form expression (If there exits one). In other words, given:

$$y=W(e^{ax+b})-W(e^{cx+d})+zx$$

where $W$ is the Lambert $W$ function and $a,b,c,d,z$ are some constants, what is the function $f$, such that $$x=f(y)$$

EDIT : If there exists no closed form solution, I will be happy to see nice arguments supporting this.

$\rightarrow$EDIT2 : As can be seen, we have a solution for the simplified version of this problem. If there exists a solution I have the following ideas to resolve the full version of the problem:

$1$- Is it possible to write $$W(e^{f(x)})=W(e^{ax+b})-W(e^{cx+d})$$

$2$- Having $$y_1=W(e^{ax+b})+z_1x$$ and $$y_2=-W(e^{cx+d})+z_2x$$

where $z=z_1+z_2$, $y=y_1+y_2$ and $f^{-1}(y_1)$ and $f^{-1}(y_2)$ are known functions as already found. Can we say that $f^{-1}(y)=f^{-1}(y_1)+f^{-1}(y_2)$? or can we modify this idea to get somethign useful?

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Why do you think it has a "closed form" expression? – GEdgar Sep 24 '12 at 14:14
@GEdgar To be honest, I dont know if there is a closed form. I am only looking for one, if there exists. Probabily I need to edit the question. Thanks for the comment. – Seyhmus Güngören Sep 24 '12 at 14:36
Could you explain background of this task, if it's not a secret. – KvanTTT Sep 25 '12 at 14:09
@KvanTTT Of course it is not a secret. This problem comes from my research. It is the first step of everything. The inverse funtion defines a decision function. I need this function only in $[0,1]$ though, I need to insert the inverse function afterwards into two probability densities which are parametrized by this decision function as well. Finally I will have the desired densities and the decision rule for my detection problem. However, they are also parametrized by some more parameters for which I will seek a solution. Those equations seem more awful than this problem. – Seyhmus Güngören Sep 25 '12 at 14:21

$y=W(e^{ax+b})-W(e^{cx+d})+zx$

$W(e^{cx+d})+y-zx=W(e^{ax+b})$

$(W(e^{cx+d})+y-zx)e^{W(e^{cx+d})+y-zx}=e^{ax+b}$

$(W(e^{cx+d})+y-zx)e^{W(e^{cx+d})}e^{y-zx}=e^{ax+b}$

$(W(e^{cx+d})+y-zx)\dfrac{e^{cx+d}}{W(e^{cx+d})}=e^{(a+z)x+b-y}$

$W(e^{cx+d})+y-zx=e^{(a-c+z)x+b-d-y}W(e^{cx+d})$

$(e^{(a-c+z)x+b-d-y}-1)W(e^{cx+d})=y-zx$

$W(e^{cx+d})=\dfrac{y-zx}{e^{(a-c+z)x+b-d-y}-1}$

$e^{cx+d}=\left(\dfrac{y-zx}{e^{(a-c+z)x+b-d-y}-1}\right)e^{\frac{y-zx}{e^{(a-c+z)x+b-d-y}-1}}$

Then you can only force to use Lagrange inversion theorem or Lagrange reversion theorem unless the special cases when $a=0$ or $c=0$ or $a-c+z=0$ .

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 If there is no mistake in my problem $z=c-a$. Can we simplify further or even obtain $x$ alone? – Seyhmus Güngören Sep 29 '12 at 17:10

I tried to solve the simplified version of the problem. Here are what I've done so far.

I am given: $$y=W(e^x)+zx\quad\quad(1)$$

From the definition of $W$, for

$$x=re^r$$

We have $$r=W(x)$$

Having $e^{x}=e^{re^r}=te^t\rightarrow t=W(e^x)$ I also have $$y=t+zx$$ Using $x=t+\log(t)$, I have $$y=t+z(t+\log(t))=t(1+z)+z\log(t)$$ taking the exponential of both sides, $$e^y=e^{t(1+z)+z\log(t)}=e^{t(1+z)}t^z$$ Taking the $(1/z)$th power of both sides we have $$e^{y/z}=e^{\frac{t(1+z)}{z}}t.$$ Let $t^{'}=t\frac{1+z}{z}$ then we have $$e^{y/z}=e^{t'}t^{'}\frac{z}{1+z}$$ Accordingly we have $$e^{t{'}}t^{'}=e^{y/z}\frac{1+z}{z}$$ Again using the definition of Lambert $W$ function we get $$t{'}=W\left(e^{y/z}\frac{1+z}{z}\right)$$ Going back to the relation $t^{'}=t\frac{1+z}{z}$, we get $$t=W\left(e^{y/z}\frac{1+z}{z}\right)\frac{z}{1+z}$$ Using the relation $t=W(e^x)=W\left(e^{y/z}\frac{1+z}{z}\right)\frac{z}{1+z}$ in equation $(1)$,$$y=W\left(e^{y/z}\frac{1+z}{z}\right)\frac{z}{1+z}+zx$$ which can be rewritten as $$x=\frac{y}{z}-W\left(e^{y/z}\frac{1+z}{z}\right)\frac{1}{1+z}$$ I checked for some numerical results and it seems that the solution for this result is correct.

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Let's do some special cases.

Is it true that the solution of $$y = \operatorname{W} \bigl(\operatorname{e} ^{x}\bigr) + z x$$ is $$x = \frac{y - \operatorname{exp}{\left(y/z - \operatorname{W} \left({(z + 1) \operatorname{e} ^{y/z}}/{z}\right)\right)}}{z}$$ I state this as a question because Maple doesn't do this by itself.

simplified: $$x = \frac{y}{z} - \frac{\operatorname{W} \left({(z + 1) \operatorname{e} ^{{y}/{z}}}/{z}\right)}{z + 1}$$

SG has written this in an answer. Here is my version, not much different:

Let $\operatorname{W}$ be the Lambert W function, so that (formally) $a=\operatorname{W}(b) \Longleftrightarrow ae^a = b$.

Question: solve $y=\operatorname{W}(e^x)+zx$ for $x$. Answer (at least formally): $$x = \frac{y}{z} - \frac{\operatorname{W} \left({(z + 1) \operatorname{e} ^{{y}/{z}}}/{z}\right)}{z + 1}$$

Explanation. This solution works as long as these steps are reversible:

\begin{gather*} y = \operatorname{W}(e^x)+zx \\ \operatorname{W}(e^x) = y-zx \\ e^x = (y-zx)e^{y-zx} \\ 1 = (y-zx)e^{y-xz-x} \\ e^{y/z} =(y-xz)e^{y-xz-x+y/z} \\ \frac{e^{y/z}}{z} = \left(\frac{y}{z}-x\right)e^{(y/z-x)(z+1)} \\ \frac{(z+1)e^{y/z}}{z} = \left(\frac{y}{z}-x\right)(z+1)e^{(y/z-x)(z+1)} \\ \operatorname{W}\left(\frac{(z+1)e^{y/z}}{z}\right) = \left(\frac{y}{z}-x\right)(z+1) \\ x = \frac{y}{z} - \frac{\operatorname{W}\big((z+1)e^{y/z}/z\big)}{z+1} \end{gather*}

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 $W(1)=0.567$ for this special case what are the parameters $c$ and $d$? – Seyhmus Güngören Sep 24 '12 at 14:42 If you like, $c=d=0$ then use $y+W(1)$ in place of $y$. – GEdgar Sep 24 '12 at 14:50 yes I see, but shouldnt we have another constant term even for $a=1,b=c=d=0$? like $y=W(e^x)+zx+0567$? doest it imply the same simple case? – Seyhmus Güngören Sep 24 '12 at 14:51 ok you are right. It implies the same thing. Sorry. – Seyhmus Güngören Sep 24 '12 at 14:52 I checked in Matlab 2008. It gives no answer. It is strange. I had this question: syms g l d f m h0=solve('g*d-g*(m+l)-l*g*log(g/f)+l*f',g) matlab 2007 solved it newer versions couldnt. – Seyhmus Güngören Sep 24 '12 at 15:02