I'm interest in finding the bound (upper and lower) of the positive root ($x>0$) of the following expression: $$ F(x) := A(p-1)x^{p+1} + A(p+1)x^p - (p-1)x - \frac{p-1}{p+1},$$ where $A \ge 1$ and $p \ge 2$ are given real numbers. Note that $p$ may not be an integer.
I looked at the following:
$$F(0) = -\frac{p-1}{p+1} < 0 \quad \text{and} \quad \lim_{x\to\infty} F(x) = \infty$$ to conclude that there is a positive root at $x = x_0$. The root is also unique since $F(x)$ is convex for all positive $x$.
Is there anyway to find the estimate of $x_0$ (asymptotic form is fine, upper bound is preferred) in terms of $p$ and $A$?
Thank you.
We know that
$$F(1)=2Ap-(p-1)-\frac{p-1}{p+1}=\frac{(2A-1)(p+1)p+2}{p+1}>0$$
which tells us the root is in the interval $(0,1)$.
The first derivative is given by
$$F'(x)=A(p^2-1)x^p+A(p^2+p)x^{p-1}-(p-1)$$
while the second derivative is given by
$$F''(x)=A(p^3-p)(x+1)x^{p-2}\ge0$$
which proves this is convex on $[0,\infty)$.
This proves secant lines will overestimate between the two intersections while tangent lines will always underestimate. Using the secant line intersecting at $x=0$ and $x=1$ and the tangent line at $x=1$ gives us
$$S(x)=[2Ap-(p-1)]x-\frac{p-1}{p+1}$$
$$T(x)=[A(2p^2+p-1)-(p-1)](x-1)+2Ap-(p-1)-\frac{p-1}{p+1}$$
$$S(x)\le F(x)\le T(x)\tag{$\forall x\in[0,1]$}$$
$S$ and $T$ have roots at
$$x_S=\frac{p-1}{(p+1)[2Ap-(p-1)]}$$
$$x_T=1-\frac{(2A-1)(p+1)p+2}{(p+1)[A(2p^2+p-1)-(p-1)]}$$
respectively. This bounds the actual root in the interval $(x_S,x_T)$, and repeating this process can give increasingly tighter bounds, though after this first iteration they are messy due to the exponents.
One can further extend the above by noting that the third derivative is positive. By Taylor's theorem, the quadratic
$$T_2(x)=T(x)+\frac12F''(1)(x-1)^2$$
satisfies $T_2(x)\ge F(x)$ for $x\le1$, allowing us to improve the lower bound without involving complicated exponents.
If one is willing to drop bounds and go for approximations in general, then a pseudo-secant-Taylor quadratic can be formed from
$$S_2(x)=T(x)+\alpha(x-1)^2$$
where $\alpha$ is such that $F(0)=S_2(0)$. This avoids using the second derivative and may prove to be a useful approximation of the root.