Solving a non-polynomial equation

Question

After solving an arithmetic geometric progression, I get the following form,

$$y = A + B x + C x^p $$ where $p \in [-1, 1]$ and $A, B, C, p$ are constants.

I now want to solve for $x$ (or the iteration) at which a desired $y$ value is reached. How should I solve for $x$ in terms of $y$. I searched for solution of nonhomogeneous equations and had no success.

Answer 1

If $p = 0$, then we have a first-degree polynomial to solve:

$$ x = \frac{y-A-C}{B} $$

and again if $p = 1$, then we also solve a first-degree polynomial:

$$ x = \frac{y-A}{B+C} $$

For the intermediate case $p = 0.5$ the problem is equivalent to solving a quadratic (we assume for simplicity that $B,C\gt 0$ and $y\gt A$):

$$ x^{1/2} = \frac{-C + \sqrt{C^2 + 4B(y-A)}}{2B} $$

More such special cases can be treated as polynomial equations, as in particular $p=-1$ can also be solved by applying a quadratic formula.

Under what conditions can we show that the solutions $x = f_p(y)$ are a smooth family with respect to the exponent parameter $p$? Clearly we should focus on solutions where $x\gt 0$ so that $x^p$ is well-defined for all $p\in [-1,1]$.

In addition we should hope to exclude cases with multiple solutions $x$ for the same $y$. It may be that with more detailed information about $A,B,C$ and $y$, such an exclusion may be proven. For example:

Proposition

If $B\gt 0$ and $y\gt A$, then for any $0 \lt p \lt 1$, there exists exactly one positive real solution $x \gt 0$.

Proof:

Since $p \in (0,1)$, we can apply the generalized Descartes rule of signs to the non-polynomial equation:

$$ Bx + Cx^p + A - y = 0 $$

in which there is exactly one change of signs, regardless of the sign of $C$.

QED

Answer 2

For general $p$ the equation has no closed-form solution and a numerical approach will be required.

If you rewrite it in the form

$$x^p=ax+b$$ this shows that you want to find the intersection between a power law and a straight line. Depending on the coefficients, there will be up to two zeroes.

Taking the derivative,

$$px^{p-1}=a$$ and if $a>0$, we have a tangent to the curve at the point

$$x^*=\sqrt[p-1]{\frac ap},\\y^*={x^*}^p.$$

Then if $ax^*+b>y^*$ there are two roots, on either side of $x^*$.