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All,

I'm working on a problem whereby, given a known quantity, Q, and number of intervals, n, I want to calculate the rate parameter -1 < r < 1 of a finite geometric series. Is there a way to reorder the following equation as a function of Q and n such that r ~ f(Q, n)? :

$Q = \frac{r(1 - r^n)}{1 - r}$

I've read the post at Find parameters of short geometric series, which seems to suggest that there is no exact solution for r, but I just want to confirm this. If there is no precise solution, is there, perhaps, some reasonable approximation that can be expressed analytically?

Cheers,

Aaron

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That equation can be rearranged to give the polynomial

$r^{n+1}-r(Q+1)+Q=0$

Unfortunately, there isn't a general solution for $r$ in terms of radicals (curse you, unsolvable groups). However, it must be noted that there are specific values of $Q$ and $n$ for which it does, especially when $n=4$ since $1$ is always a root.

The best you can usually do is to approximate the roots of this polynomial in $r$. There are many ways of doing this. Since you have a polynomial and a relatively short interval $(-1,1)$ to look at, you can use the bisection method or Newton's method to quickly approximate a root in that interval.

I recommend the former since the latter (depending on your initial guess) might approximate the root at $r=1$ instead of a root in $(-1,1)$. There are more efficient methods for approximating roots of a polynomial, but these two are the easiest to understand.

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You wrote the formula for the sum erroneously, it should be $$ Q = \frac{1-r^{n+1}}{1-r} $$When we have $-1 \lt r \lt 1$, $Q$ is an increasing function of $r$, with limits $Q(-1)=0, Q(1)=n+1$. So to have a well-defined problem, we assume that $0 \lt Q \lt n+1$. Then there is exactly one solution of your equation, which with a little bit of algebra can be found to the root of the polynomial equation $$r^{n+1} -Qr +(Q-1)=0 $$ between $-1$ and $1$. There is exactly one such root, but in general there is no closed form solution for it. But you can find that solution numerically to any approximation that you want! for instance, by interval bisection or by newton's method.

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  • $\begingroup$ thank you for your input. i should have mentioned that i am looking at the special case where my index begins at 1, not zero, in which case the formula I provided is correct: mathworld.wolfram.com/GeometricSeries.html. $\endgroup$ – Aaron Aug 19 '14 at 16:10
  • $\begingroup$ In that case too, my method applies, with a slightly different formula. $\endgroup$ – kjetil b halvorsen Aug 19 '14 at 17:29
  • $\begingroup$ it does, thanks for the great answer. was tough to choose between the two. $\endgroup$ – Aaron Aug 19 '14 at 20:57

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