Solving $1-x = (1-ax)^n$ for $x$ in general for some values of $a$ and $n$.

Question

I have an equation of the form $1-x = (1-ax)^n$ for $0<a<1$ and some value $n \in \mathbb{N}$. When substituting in the values I have for $a$ and $n$ (which are $a=0.002$ and $n=2999$ I cannot seem to solve the equation. So I tried writing it in a more general form to see if there was anything I could do with that, but I cannot think of anything. The only way I can seem to get an answer is by graphing both sides and looking where they intersect, but this is not accurate (it gives $x=1$ as a solution). Is there a trick when solving equations like this? Thank you for all your help.

Answer 1

Consider that you look for the zero of function $$f(x)=1-x - (1-ax)^n$$ Assuming that $n$ is a positive integer $n > 1$, this represents a polynomial of degree $n$ which cannot be solved if $n>4$. So, you need either numerical methods (such as Newton) or approximations.

If you consider the derivatives $$f'(x)=a n (1-a x)^{n-1}-1$$ $$f''(x)=-a^2 (n-1) n (1-a x)^{n-2}$$ the first derivative cancels at $$x_*=\frac{1-\left(\frac{1}{a n}\right)^{\frac{1}{n-1}}}{a}$$ and notice that $f(0)=0$. So, assuming that this is a minimum and that $f(x_*)<0$, ther is a root which is $ > x_*$.

To get a first approximation, you could build a Taylor series around $x=x_*$ to get $$f(x)=f(x_*)+\frac 12 f''(x_*)(x-x_*)^2+O((x-x_*)^3)$$ and ignoring the higher order terms, get an estimate $$x_0=x_*+\sqrt{-2\frac{f(x_*)}{f''(x_*)}}$$

Appliead to the case $a=0.002$, $n=2999$, this would give $x_0 \approx 0.721$ and, using this Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.720871 \\ 1 & 1.009707 \\ 2 & 0.997500 \\ 3 & 0.997494 \end{array} \right)$$ which is the solution for six significant figures.

If you know that the solution is close to $x=1$, you could expand $f(x)$ as a Taylor series at $x=1$ and get $$f(x)=-(1-a)^n+(x-1) \left(a n (1-a)^{n-1}-1\right)+O\left((x-1)^2\right)$$ Using the first term only would give $$x=1+\frac{(1-a)^{n+1}}{a n (1-a)^n+a-1}$$ which, for the worked example, would immediately give $x=0.997494$.