How to solve equation $ x=W(a+bx^{n})+1 $?

Question

How i can resolve the equation $x=W(a+b x^n)+1$, where $W$ is the Lambert $W$ function?

thanks

Answer 1

I don't think it will be easy to solve. Since $x = W(y)$ if and only if $y \ge - \dfrac 1e$ and $y = xe^{x}$, you will have to solve $$a+bx^n = (x-1)e^{x-1},\quad a + bx^n \ge - \dfrac 1e.$$ Are you looking for an exact solution?

Answer 2

??HINT??

Lagrange inversion of: $$\lambda(a+bx^n)e^{-x}+c=x$$ gives: $$x=c+\sum_{k=1}^{\infty}\lambda^k\frac{D_z^{k-1}\left[(a+bz^n)^ke^{-kz}\right]_{z=c}}{k!}$$ Using binomial expansion: $$x=c+\sum_{k=1}^{\infty}\sum_{j=0}^{k}\lambda^k\frac{a^{k-j}b^j \left.D_z^{k-1}z^{nj} e^{-kz}\right|_{z=c}}{j!(k-j)!}$$

The derivatives $D_z^{k-1}z^{nj} e^{-kz}$ can be represented as hypergeometric polynomials.

Remark/ Edit in response to comment of Randa Jaouadi If we place $b=0$ the equation to solve becomes : $$\lambda a e^{-x}+c=x$$ and the series solution reduces to (warning: $b=0$ so only terms with $j=0$ survive: when $b=0$ $b^j=\delta_{j0}$ ...):

$$x=c+\sum_{k=1}^{\infty}\lambda^k \frac{ a^{k} D_c^{k-1} e^{-kc} }{k!}$$ $$x=c+\sum_{k=1}^{\infty}\lambda^k \frac{ a^{k} {(-k)}^{k-1} e^{-kc} }{k!}$$ $$x=c+\sum_{k=1}^{\infty} -\frac{ (-\lambda ak e^{-c})^{k} }{k k!}$$ $$x=c+W\left( \lambda a e^{-c} \right)$$

We have employed the well-known series development of the principal real branch of Lambert $W$ : $$W_0 (x) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}\ x^n = x - x^2 + \frac{3}{2}x^3 - \frac{8}{3}x^4 + \frac{125}{24}x^5 - \cdots $$ See: http://en.wikipedia.org/wiki/Lambert_W_function