Is it possible to solve for $r$ in this equation?

$$ v = \frac{1 - r^n}{1 - r} $$

I think it's not... I spent some time trying to wrestle with various substitutions and couldn't get there. I can solve for $r$ practically by writing a program to interpolate but I'm wondering if I'm missing something.

If it's not solvable, is there something about the form of the equation I should be able to recognize to know it's not solvable?

(Trivia: This is actually for a choose-your-own-adventure or "branching" novel; wondering whether I can solve for "choices per chapter" r given a number of chapters v and an average thread length n.)

  • $\begingroup$ You are tacitly asking whether $\sum_{m=1}^{n-1}r^m=v$ has a solution. It has $n-1$ solutions. If you are asking whether one can find those solutions in closed form, the answer is yes for $n-1\le 4$. I doubt that $r^n-vr+v-1=0$ has a closed form solution. $\endgroup$
    – Mark Viola
    Oct 31, 2022 at 19:49
  • $\begingroup$ This is equivalent to solving a polynomial equation, as we have $$1-r^n=(1-r)(1+r+r^2+\cdots+r^{n-1})\\ \implies v=\frac{1-r^n}{1-r}=1+r+r^2+\cdots+r^{n-1}$$ $\endgroup$ Oct 31, 2022 at 19:51
  • $\begingroup$ @AndrewChin That expression is a tautology. It is true for all $r$. $\endgroup$
    – Mark Viola
    Oct 31, 2022 at 19:52
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/4456022/… $\endgroup$
    – boojum
    Nov 1, 2022 at 0:15
  • $\begingroup$ Yes, math.stackexchange.com/questions/902720/… answers my question as well. And teaches me this is a geometric sum. Thank you. Approximation is the way to go. $\endgroup$
    – tunesmith
    Nov 2, 2022 at 17:53


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