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?

  • 1
    $\begingroup$ Would help to know what $\Phi$ and $\phi$ are. Also defining $f(x) = x \Phi(x) +\phi(x)$ and $c = a/b$ turns the problem into $W = [f(x+c)-f(x)]/c$. Did this come from some sort of mean value problem? $\endgroup$ Aug 1, 2022 at 20:12
  • $\begingroup$ $\phi$ and $\Phi$ are commonly used as the standard normal probability density function and standard normal cumulative distribution function, respectively. These symbols are used in introductory textbooks in probability and statistics. See Devore's Probability and Statistics for Engineering and the Sciences, Eighth Edition, pages 153-154, or Wackerly, et. al's Mathematical Statistics with Applications, Seventh edition, pages 178-179. The formulas are commonly available on the internet: en.wikipedia.org/wiki/Normal_distribution. $\endgroup$ Aug 3, 2022 at 18:45

1 Answer 1


$$\Phi(z) = \int_{-\infty}^z\frac{e^\frac{-x^2}{2}}{\sqrt{2\pi}}\,dx=\frac{1}{2} \left(1+\text{erf}\left(\frac{z}{\sqrt{2}}\right)\right)$$ $$\phi(z) = \frac{e^\frac{-z^2}{2}}{\sqrt{2\pi}}$$ 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 $$\left( \begin{array}{cc} n & x_n \\ 0 & -0.04927931754 \\ 1 & -0.04835262336 \\ 2 & -0.04835229602 \end{array} \right)$$


Working with rational numbers and arbitrary pecision, for $c=\frac{3371}{125}$ and $k=\frac{3371}{250}$, Newton iterates are

$$\left( \begin{array}{cc} n & x_n \\ 0 & -26.1701 \\ 1 & -10.2146 \\ 2 & -13.4840 \end{array} \right)$$

  • $\begingroup$ So the derivative is equal to $\Phi(x+c)- \Phi(x)$, right? But it seems that the iterations can be quite unstable for certain starting values of $x$. Makes it not always converge...any comments on that? $\endgroup$ Aug 2, 2022 at 9:13
  • $\begingroup$ @Steven01123581321. Could you give me a couple $(c,k)$ of your choice ? $\endgroup$ Aug 2, 2022 at 9:15
  • $\begingroup$ $(c, k) = [(26.968, 13.484), (10.787, 10.716)]$ $\endgroup$ Aug 2, 2022 at 9:23
  • $\begingroup$ @Steven01123581321. I did not have any trouble to find $-13.484$ for the first $\endgroup$ Aug 2, 2022 at 10:01
  • $\begingroup$ For the first one, I have for the first iteration $26.170...$ and then the derivative is $0.0$, so my next iteration fails...Can you share your results at the first iterations? $\endgroup$ Aug 2, 2022 at 12:24

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