Suppose I have a function defined by this recurrence-relation:
$$R(0) = r$$ $$R(n) = R(n-1) * (1+G)d^{n-1}$$
Here r is a base value, G is a base growth rate (G>0) and d is a decay in the growth rate (d<1). A real-world example is r=1000000, G=1, d=0.85.
(For context, this is an equation that appears to govern revenue in year n of a company with revenue r in year 0, which is nominally growing at G per year, because statistically the growth rate of recurring-revenue companies tends to decrease by 85% per year.)
It seems to me that the function can be simplified (or am I wrong?) to:
$$R(n) = r(1+G)^n d^{n(n-1)/2}$$
I can see by graphing this in Excel and using a parabolic-best-fit that in fact this is exactly a parabola.
I can't understand how to compute that transformation with the general equation, and furthermore since that means there exists A, B, and C such that:
$$R(n) = An^2 + Bn + C$$
How can I produce the values of A, B, and C from the parameters r, G, and d?
Please enlighten me!
The function can indeed be simplified like you indicated. In case your simplification is speculative, a brief justification is given in a remark at the end.
The function $R(n)$ is an exponential function, it certainly cannot be expressed as $An^2+Bn+C$ for constants $A$, $B$, and $C$. However, if you define $L(n)$ by $L(n)=\log(R(n))$, then indeed there are $A$, $B$, and $C$ such that $L(n)=An^2+Bn+C$. Te numbers $A$, $B$, and $C$ can be computed readily from your expression for $R(n)$. We get $$L(n)=\log r+\left(\log(1+G)-\frac{1}{2}\log d\right)n +\frac{1}{2}\log d n^2.$$ Since $\log d$ is negative, one might want to let $c=\frac{1}{d}$ and replace $\log d$ by $-\log c$.
Remark: Calculate. We get $R(1)=r(1+G)d^0$, $R(2)=r(1+G)^2d^1$, $R(3)=r(1+G)^3d^{1+2}$, $R(4)=r(1+G)^4d^{1+2+3}$, $R(4)=r(1+G)^5d^{1+2+3+4}$, and so on (the pattern is clear).
So the exponent of $d$ in $R(n)$ is $1+2+3+\cdots +(n-1)$. By the standard formula for the sum of the first $k$ positive integers, the exponent is $\frac{n(n-1)}{2}$. This happens to work also for $n=0$.