I'm looking to solve the following equation in $x$.
$$\frac{Wa}{b} = \left((\frac{a}{b}+x)\Phi(\frac{a}{b}+x)+\phi(\frac{a}{b}+x)\right)-\left(x\Phi(x)+\phi(x)\right)$$
, where $W, a$ and $b$ are constants and known.
As I fail to solve it analytically, I tried a numerical method, the fixed-point iteration. According to wikipedia, this numerical method can be done by the following procedure:
$x_{n+1} = f(x_n)$ for $n=0, 1, 2, \dots$
So I would think that solving the following equation recursively, $x$ would converge as $n$ increases.
$$-\frac{\frac{Wa}{b} - (\frac{a}{b}+x_n)\Phi(\frac{a}{b}+x_n)-\phi(\frac{a}{b}+x_n)+\phi(x_n)}{\Phi(x_n)} = x_{n+1}$$
But it doesn't converge at all. Anyone that is more familiar with numerical analysis and can explain why it doesn't converge? Or what I am doing wrong?
$$\Phi(z) = \int_{-\infty}^z\frac{e^\frac{-x^2}{2}}{\sqrt{2\pi}}\,dx=\frac{1}{2} \left(1+\text{erf}\left(\frac{z}{\sqrt{2}}\right)\right)$$ $$\phi(z) = \frac{e^\frac{-z^2}{2}}{\sqrt{2\pi}}$$ Let $c=\frac a b$ and $k=W \frac a b$ and you look for the zero of $$F(x)=\Bigg[\frac{1}{2} (c+x) \left(1+\text{erf}\left(\frac{c+x}{\sqrt{2}}\right)\right)+\frac{e^{-\frac{1}{2} (c+x)^2}}{\sqrt{2 \pi }}\Bigg]-$$ $$\Bigg[\frac{1}{2} x \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)+\frac{e^{-\frac{x^2}{2} }}{\sqrt{2 \pi }}\Bigg]-k$$ for which the derivative is $$F'(x)=\frac{1}{2} \left(\text{erf}\left(\frac{c+x}{\sqrt{2}}\right)-\text{erf}\left(\frac{x}{\sqrt {2}}\right)\right)$$
Assuming $c>0$ and $F(0) <0$, the first iterate of Newton method is $$x_0= -\frac {F(0)}{F'(0)}$$ and repeat $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}$$
Trying with $c=\pi$ and $k=e$, the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & -0.04927931754 \\ 1 & -0.04835262336 \\ 2 & -0.04835229602 \end{array} \right)$$
Edit
Working with rational numbers and arbitrary pecision, for $c=\frac{3371}{125}$ and $k=\frac{3371}{250}$, Newton iterates are
$$\left( \begin{array}{cc} n & x_n \\ 0 & -26.1701 \\ 1 & -10.2146 \\ 2 & -13.4840 \end{array} \right)$$