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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.

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    This equation often comes up in [the context of annuities](http://www.financeformulas.net/Annuity_Payment_Formula.html). For general exponents $p$ it has no "closed form" solution in the sense of the quadratic or cubic formulas. However there are special cases (like $p=0.5$) where such formulas apply. In the general case you want to get a good initial guess for the $x$ that gives $y$ and use an iterative root-finder (Newton method or false position, for example). – 2017-02-03
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    While I did not find a really good treatment of this on Math.SE, [this previous Question](http://math.stackexchange.com/q/1794181/3111) is certainly relevant. – 2017-02-03
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    @hardmath Is there some way to get an approximate solution given that there is no closed form solution. I am just wondering. I don't want to iterative root-finder because it will complicate things further for me. – 2017-02-03
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    For certain cases (e.g. $p = 1,0.5,0,-1$) the equation is equivalent to a low-degree polynomial (first- or second-degree) and the usual root formula can be applied. An approximate solution is therefore available by interpolating between the values of $p$ for which those solutions are known. It might be helpful to give more context, so that Readers can appreciate your reluctance to use an iterative root-finder. – 2017-02-03
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    @hardmath I am trying to compare convergence of several different progressions. For such reasons, I am interested in getting only approximate convergence time (or the order of convergence). Therefore approximate estimates would suffice for me. I think interpolation should give me partial solutions. Thanks so much for sharing the relevant resources. – 2017-02-03
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    Do you know anything a priori about the signs and magnitudes of $A,B$? Given some sample coefficients I would put together an example calculation that serves to illustrate various ideas floating around this Question. – 2017-02-03

2 Answers 2

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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

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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 $x^*$.

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    Note that the necessity of a numerical approach was discussed with the OP in the Comments below the Question, and reluctance to pursue this to any accuracy was expressed by OP owing to the nature of the application. Nevertheless your discussion of bracketing an initial guess for the root is a welcome addition. – 2017-02-03
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    @hardmath: such late additions by the OP should appear as question updates, not as comments. – 2017-02-03
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    They are not late additions, they occurred within the first hour of the OP (in response to my requests) and well before you posted your Answer. However you are correct that ideally all the clarifications should be incorporated into the body of the Question; I will urge @learner to do so. – 2017-02-03
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    @hardmath + @Y: I am thankful for both the answers. They helped me solve my problem. I can accept only one of the answers although both of them were equally helpful. Sorry for the creating the confusion. I will updates questions from now on. – 2017-02-04