A special polynomial For $n \geq 2$, let $\phi =|t(t-1)...(t-n)|\; \forall\; t \in [0,n]$.
1/ Show that $\phi$ peaks at a point belonging to $[0,1]$
2/ Evaluate $\dfrac{\phi'}{\phi}$ in accordance with $\displaystyle g(t)=\sum_{k=0}^n \frac{1}{(t-k)}$
Any help would be appreciated.
 A: Well, assuming that I understand you correctly, the first part is the product:
$$\phi_n(t)=\left|\prod_{k=0}^nt-k\right|\implies{\left(\phi(t)=0\iff t-k=0;\\\therefore\phi(t)=0\iff t=k\right)}$$
You can evaluate this product using the falling factorial, provided that you appropriately restrict the domain of $n$ to $n\geq2$:
$$\phi_n(t)=\left|(t)_{n+1}\right|\ :\ n\geq2$$
By definition, this is the same as
$$\phi_n(t)=\left|\frac{\Gamma(t+1)}{\Gamma(t-n)}\right|;\ \ \ \ \ \text{if }t\in\mathbb{Z}, \text{then }\phi_n(t)=\left|\frac{t!}{\Gamma(t-n)}\right|$$
Now, I'm not sure what you mean by "peaks at a point belonging to $[0,1]$," but if you mean that $\phi_n(t)$ has a local maximum at $0\leq t\leq 1$ then you only need to show that $\phi_n(0)=0$, $\phi_n(1)=0$, and $\exists t\in(0,1):\phi_n(t)\neq0$.
From the original product, you know that $\phi_n(k)=0$, where $k\in\left\{x\in\mathbb{Z}\mid0\leq x\leq n\right\}$, so the first two statements are true (you can also show this using the limit of the gamma function in the denominator). For the final statement, consider that $\Gamma(x)$ has no zeros (so the numerator is never zero), and is finite if $x$ is not a negative integer (so the denominator is finite if $x$ is not an integer). Thus if $t\notin\mathbb{Z}$, then $\phi_n(t)\neq0$.
Hope this helps.
As for the second part to your question, I'm not sure how to interpret it. Is $\frac{\phi\prime}{\phi}$ meant to denote $\frac{\partial\phi_n(t)}{\partial t}\frac{1}{\phi_n(t)}$? And if so, what does $g(t)$ have to do with it?
