Given initial and final values, solving for an average growth rate over $n$ periods is trivial, however I'm interested in whether or not it's possible to solve for (or at least approximate) the growth rate given the number of periods and the sum of the values taken in each of the $n$ periods.
For instance, we know that
$$\sum_{i=1}^n (1+g)^{i-1} = \frac{1-(1+g)^n}{1-(1+g)}$$
and therefore if we want this to sum to some number, let's say $b$, then we have the equality
$$\frac{(1+g)^n-1}{g} = b$$
My question is, can you now approximate the $g$ that satisfies this? We of course assume $ g \approx 0$. It seems clear that finding a general analytical solution to this is impossible as it just expands into a higher degree polynomial, the roots of which need to be estimated numerically (as I understand it.) However, if there is a technique for this I'd love to hear it as it has an application to an Excel model I'm building and I'm in a bit of a bind with it.
Best Answer
Consider that you look for the zero of function $$f(g)=(1+g)^n- b\,g+1$$ for which $$f'(g)=n(1+g)^{n-1}-b \qquad \text{and} \qquad f''(g)=n(n-1)(1+g)^{n-2}$$ The first derivative cancels at $$g_*=\left(\frac{b}{n}\right)^{\frac{1}{n-1}}-1$$ So, for a first approximation, develop $f(g)$ as a Taylor series around $x_*$ to get $$f(g)=f(g_*)+\frac 12 f''(g_*) (g-g_*)^2+O((g-g_*)^3)$$
So the first approximation could be $$g_0=g_*+\sqrt{-2\frac{f(g_*)}{f''(g_*)}}$$ Using it for $n=20$ and $b=123456789$, this would give $g_0=1.805$ while the exact solution is $1.599$. Starting from this guess, Newton method will converge very fast as shown below. $$\left( \begin{array}{cc} n & g_n \\ 0 & 1.80522 \\ 1 & 1.69723 \\ 2 & 1.62762 \\ 3 & 1.60177 \\ 4 & 1.59879 \\ 5 & 1.59875 \end{array} \right)$$ We could make a better approximations using $$g_0=g_*+3 \frac{f''(g_*)}{f'''(g_*)}\qquad g_0=g_*+\frac{4 f(g_*) f'''(g_*)}{f(g_*) f''''(g_*)-6 f''(g_*)^2}\qquad \cdots$$ For the worked example, these would respectively give $g_0=1.657$ and $g_0=1.587$.
Specific case where $g \ll 1$
This case corresponds to problems in finance. Consider now the original equation $$b=\frac{(1+g)^n-1}{g}$$ and develop the rhs as a Taylor series built around $g=0$ $$b=n+\frac{n(n-1)}{2} g +\frac{n(n-1)(n-2)}{6} g^2 +\frac{n(n-1)(n-2)(n-3)}{24} g^3+\frac{n(n-1)(n-2)(n-3)(n-4)}{120} g^4 +\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{720} g^5 +O\left(g^6\right)$$
Using series reversion, this will give as an approximation $$g=t-\frac{(n-2)}{3} t^2+\frac{(n-2) (5 n-7)}{36} t^3-\frac{(n-2) \left(17 n^2-44 n+29\right)}{270} t^4+\frac{(n-2) \left(193 n^3-708 n^2+885 n-374\right)}{6480}t^5+O\left(t^6\right)$$ where $t=\frac{2 (b-n)}{n(n-1)}$.
Let us try for $n=20$ and $b=25$. The above formula will give $g=0.022863$ while the exact solution is $0.022854$.
For sure, we could take more terms and have a better accuracy.