Need help solving exponential linear equation used to compare the value of a fixed annuity vs. an increasing annuity

Question

A couple of years ago I came up with a function (shown below) that can be used to determine if/when an increasing annuity will be worth more than an annuity paying a fixed amount. $$f(n)=v^n(\mu +Qn)-\mu$$ where $0<v<1$, $n>0$, and $Q>0$

I am interested in finding the root of this equation, but it seems to be impossible using standard methods. I think it can be transformed so the Lambert W function could be applied, but that isn't exactly a closed form solution that can be computed with a calculator.

The information below hopefully can help to solve it - (or just help to find something else interesting).

1. Since $\lim \limits_{n \to \infty} f(n) = -\mu$, roots > 0 only exist if $\mu>0$.
2. All of the derivatives of the function can be expressed in one formula, where the $p^{th}$ derivative is equal to $$(\ln(v))^{p-1}[v^n\ln(v)(\mu +Qn)+pQv^n].$$ It can then be shown that the root of the $p^{th}$ derivative is equal to $$-\frac{p}{\ln(v)}-\frac{\mu}{Q}.$$

Thanks for your help!

Answer 1

This is still an algebraic transcendental equation for which even if a closed form existed, it would be futile attempting to compute it with a hand-calculator.

I would suggest a Newton-Raphson iteration, which if the function doesn't have a saddle point should converge quite rapidly.

Answer 2

You can get a root using Lambert W function. However, to calculate the numerical value using that expression, you need something that can calculate W function. So this will only be useful if you have that something. BTW, I did find an online W function calculator, which requires Java.

Solving for $n$ such that $0=v^n(\mu +Qn)-\mu$. Substituting $n=x-\frac{\mu}{Q}$ , letting $A=\frac{\mu}{Q}\ln{v}$, and rearranging:

$\mu=v^{x-\frac{\mu}{Q}}(Qx)=e^{x\ln{v}}e^{-A}Qx \Longrightarrow$

$\frac{\mu}{Q}e^A = xe^{x\ln{v}}\Longrightarrow$

$Ae^A = \ln(v) xe^{x\ln{v}}\Longrightarrow$

$W(Ae^A) = x\ln{v}=(n+\frac{Q}{\mu})\ln{v} \Longrightarrow$

$n=\frac{W\left(Ae^A\right)-A}{\ln v} $

When $\frac{-1}{e} < x < 0$ then $W(x)$ yields two values, one below $-1$ and one between $-1$ and zero. Further the expression $y=xe^x$ for $\frac{-1}{e} < x < 0$ maps two different values of $x$ to one value of $y$. One of the two different values of $x$ yielding the same $y$ is less than -1, and the other is between $-1$ and zero. This means that $W(Ae^A)$ maps two different values of $A$, each to two output values. For $-1 < A < 0$ and $-1 < W(Ae^A) < 0$ then $A=W(Ae^A)$. This represents the case that $n=0$, so it can be ignored.

Since I don't know the meaning of the variables, I'm not sure what the other 3 cases represent. But I'm confident that at least one of them is the solution you're seeking.

If you want to post a valid set of variable values in the OP, I can run through the calculations to see which case applies.