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I was solving a problem where I needed to find the value of $n\in [0,60] \cap \mathbb N$ for which $$f(n)=n \prod_{k=0}^{n-1}\frac{60-k}{60}$$ is maximized.

Somewhat irresponsibly, I computed the value of the expression for $n=1,2,...,9$, saw that it decreases after $n=8$ and assumed that $n=8$ gives the maximum value.

Not only does this assume that the value of $f$ "increases then decreases," allowing only one candidate for a maximum, but it also doesn't shed any light on how I would solve the question or graph the function in the general case.

Therefore, I have 3 questions:

(1) How would I find the maximum (or maxima, if unsure of how many extrema there would be) analytically?

(2) Is it trivial to show that $f$ "increases then decreases"?

(3) How might I sketch $f$ (by hand) ?

Can the answers for the above questions be generalized for the general case function $g(n)=n \prod_{k=0}^{n-1}\frac{\alpha-k}{\alpha}$ for $n\in [0,\alpha] \cap \mathbb N$?

David G. Stork
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actinidia
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1 Answers1

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Partial answer, points (1) and (2). $$g(n+1)=(n+1)\prod\limits_{k=0}^{n}\frac{\alpha-k}{\alpha}= n\prod\limits_{k=0}^{n}\frac{\alpha-k}{\alpha} + \prod\limits_{k=0}^{n}\frac{\alpha-k}{\alpha}=\\ g(n)\frac{\alpha-n}{\alpha}+\frac{g(n)}{n}\frac{\alpha-n}{\alpha}=\\ g(n)\frac{\alpha-n}{\alpha}\frac{n+1}{n}$$ To have a growth, we ask for $$\frac{\alpha-n}{\alpha}\frac{n+1}{n}\geq1 \tag{1}$$ because that will make $\color{red}{g(n+1)\geq g(n)}$. This happens for: $$(\alpha-n)(n+1)\geq \alpha n \Leftrightarrow \\ \alpha n+ \alpha - n^2 -n \geq \alpha n \Leftrightarrow\\ n^2 +n - \alpha \leq 0$$ or $$n \leq \frac{-1+\sqrt{1+4\alpha}}{2} \tag{2}$$ As a result, the growth finishes at around $\left \lfloor \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor$ or $\left \lfloor \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor +1$. For $n >\left \lfloor \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor +1$ we have $$\frac{\alpha-n}{\alpha}\frac{n+1}{n}<1 \tag{3}$$ which means $\color{red}{g(n+1)<g(n)}$ and the sequence will start decreasing, thus the maximum is reached at around $\left \lfloor \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor$ or $\left \lfloor \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor +1$

For $\alpha=60$ we have, from $(2)$, that $n\approx 7$ or $8$, which can be seen here. For $\alpha=160 \Rightarrow n\approx 12$ or $13$

rtybase
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  • Why "$\left \lfloor \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor$ or $\left \lfloor \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor +1$" instead of $\left \lceil \frac{-1+\sqrt{1+4\alpha}}{2} \right \rceil $ ? – actinidia Mar 07 '18 at 14:56
  • Because it's not 100% accurate, for $\alpha=60$, $(2)$ gives $n\approx7.26$, flooring is $7$, but maximum is achieved at $8$. For $\alpha=160$, $(2)$ gives $n\approx 12.159$ , flooring is $12$ and maximum is achieved at $12$. – rtybase Mar 07 '18 at 15:05
  • But in both of those cases, doesn't ceiling do the job? – actinidia Mar 07 '18 at 15:06
  • What I meant to write is, I am not sure if ceiling always takes over flooring. – rtybase Mar 07 '18 at 16:54