Let $S$, $D$ and $r$ be non zero positive real numbers, where $0 < r < 1$, $D \geq S$, and let $k$ be a positive integer, where $k \geq 2$.

Find a general solution for $r$ given that: $$\frac{r^{k-1}(1-r)}{(1-r^{k})} \geq \frac{S}{D}.$$

I don't want to give anybody the idea that I haven't tried to solve this before putting it out there. I've tried shuffling around exponents, logarithmic identities and Mathematica :D but with no luck. It may be that I'm just missing something simple.

I'm not expecting a full solution but maybe some insight or a direction to take, but full and partial answers are all welcome.

  • $\begingroup$ I also wonder if it even has a general solution? Would be possible that it doesn't? If I or somebody else hasn't provided a solution in a couple of days, maybe it's worth trying to prove that it doesn't have one. $\endgroup$ – Buca Hajdini Mar 4 '20 at 18:50

Let $b = \frac{S}{D}$ Then we have $$\frac{r^{k-1}-1+1-r^k}{1 - r^k} \geq b $$ $$\frac{r^{k-1}-1}{1 - r^k} \geq b - 1 $$ If $c=b-1$ then it may be rewritten as $$c r^k + r^{k-1} - (c+1) \geq 0$$ which has no explicit solution for $k>4$, as it would be a polynomial of degree greater than $4$.

  • $\begingroup$ I see how you got the first two inequalities, but I can't seem get the third, albeit I only looked at it for a short while and it's pretty late where I live. And another question: what can we say about the domain of $r$ which would satisfy the inequality given in my question? $\endgroup$ – Buca Hajdini Mar 4 '20 at 20:39
  • $\begingroup$ So when I woke up and glanced at the third inequality again and realized how to get there quite easily, but my final question still stands: what does is say about the domain of $r$ which would satisfy the inequality? $\endgroup$ – Buca Hajdini Mar 5 '20 at 4:06
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    $\begingroup$ When $S$ tends to $0$, that is $c$ tends to $-1$, then the whole domain $(0, 1)$ seems to fulfill the inequality for any $k$.We can easily verify that $r = 1$ always satisfies the inequality. However when we have fixed $c$ and $k$ tends to infinity, the solutions in the domain seem to disappear, e.g. for $c = 0.99$, there seem to be no solutions in the domain for $k \geq 100$. Also for a fixed $k$, as $c$ tends to $0$, the solutions disappear from the domain, e.g. if $k = 2$, there are no solutions in the domain for $c < -0.5$. $\endgroup$ – NikoWielopolski Mar 5 '20 at 6:35

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