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For the given natural number $0<n\in\mathbb{N}$, define the function as follows:

$$f\left(r\right)=2\cdot\sum_{i=0}^{n+2}r^{i-n}+1$$

I want to find the minimum for that function in the domain $r\in\left(0,\infty\right)$, I type some parameters in demos and it seems that the minimum is equal to $$r=1.5$$

But I don't know how to prove it formally, and why it's not a function of $n$

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  • $\begingroup$ It is a function of $n$. For example, what if $n=0$? Then the minimum will be at $r=-.5$. $\endgroup$
    – 3rdMoment
    Dec 2, 2022 at 18:47
  • $\begingroup$ I stated that n is natural number so it should be greater then 0 $\endgroup$ Dec 2, 2022 at 18:51
  • $\begingroup$ The naturals include zero, as a definition is common. $\endgroup$
    – Nij
    Dec 2, 2022 at 19:47

1 Answer 1

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Evaluating the geometric sum: $$f(r)=2\cdot\frac{r^{-n}-r^3}{1-r}+1$$ Note that $f(1)=n+3$. This function is extremized when the function $g(r)=\frac{r^{-n}-r^3}{1-r}$ is extremized. Let's work with that instead. Differentiating and equating to zero: $$\frac{\mathrm{d}g}{\mathrm{d}r}=\frac{r^{-1-n}\left(n(r-1)+r+r^{3+n}(2r-3)\right)}{(1-r)^2}=0$$ Simplify, using $r>0$ and $n>0$: $$\implies n(r-1)+r+r^{3+n}(2r-3)=0$$ The minimum may be at the root of this equation. For example, for $n=3$, the minimum is at $r=1.1908$. Examining this interactive Desmos graph confirms this is true. Playing around with it also shows that the minimum is not $r=1.5$, and it really does depend on $n$.

I will note that, in the limit $n\to\infty$, $g(r)\to-\frac{r^3}{1-r}$ so our optimization condition becomes: $$\frac{r^2(2r-3)}{(r-1)^2}=0$$ For $r>0$, this becomes $2r-3=0\Longleftrightarrow r=\frac{3}{2}$. So, in the limit $n\to\infty$, the sum is minimized at $r=\frac{3}{2}$.

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