Solving $px^n -x + (1-p)=0$

Question

I'm interested in solving the following equation in $[0,1]$ $$px^n - x + (1-p)=0$$

where $p \in [0,1]$ and $n \in \mathbb N -\{1,2 \} $ both constant.

To start with, we can easily see that $x=1$ is a solution and I also know there is another solution in $[0,1]$ for every $n \in \mathbb N -\{1,2 \}$ and for all $p> p_c (n)$ . I've tried using the Horner method with $(x-1)$ we get :

$$(x-1)(px^{n-1} + px^{n-2}+..+ px^2 + px+p-1)=0 $$

So we get $(px^{n-1} + px^{n-2}+..+ px^2 + px+p-1)=0 $ . Then we can do : $$ x^{n-1} + x^{n-2}+..+ x=\frac{1-p}{p}$$

or $$\frac{x^n-1}{x-1}=\frac{1-p}{p}+1 $$

But this doesn't seem to give anything useful. Any ideas on how we can solve this?

Answer 1

If $x\in (0,1)$ then for $n\in\mathbb{N}-\{1,2\}$ large we have $p\cdot x^n\approx 0$. We have $px^n-x+(1-p)=0$ and $p\cdot x^n\approx 0$ implies $$ -x+(1-p)\approx 0 $$ Then $x\approx 1-p$.

Answer 2

Let's write $$ \left\{ \matrix{ f(x) = p\,x^{\,n} - x + \left( {1 - p} \right) \hfill \cr g(x) = {{f(x)} \over {x - 1}} = p{{x^{\,n} - 1} \over {x - 1}} - 1 = \hfill \cr = p\left( {1 + x + \cdots + x^{\,n - 1} } \right) - 1 \hfill \cr} \right. $$

We have that $g(x)$ is continuous, and $$ \left\{ \matrix{ g(0) = p - 1 \le 0\quad \left| {\;p \le 1} \right. \hfill \cr 0 \le g(1) = n\,p - 1\quad \left| {\;1/n \le p} \right. \hfill \cr 0 \le g'(x) = p\left( {1 + 2x + \cdots + \left( {n - 1} \right)x^{\,n - 2} } \right)\quad \left| \matrix{ \;0 \le p \hfill \cr \;2 \le n \hfill \cr} \right. \hfill \cr} \right. $$ so that, for $1/n <p \le 1$ and $2 \le n$:
$g(0)$ is non-positive, $g(1)$ is positive, g(x) is strictly increasing in the interval,
then you will have one (and only one) real root in the interval $[0,1)$, which confirms what you already know.

As for calculating the root, I do not see any other method than to use Newton- Raphson or the secant approximation, starting with $P_0,P_1$ as from the sketch.