Q:

Tangents of the Means of the Roots of $n$th Order Polynomials

In Stewart's Calculus w. Early Transcendentals 8th Ed. Chapter $3$ Problems Plus Question 26 we establish that the tangent of the mean of any two roots of a third order polynomial passes through the third root. Does this hold for an nth order polynomial? If so how does one go about deriving the product given as $$\prod_{k=0}^n (x-a_k)$$ where $a_k$ is the $k$th root of the $n$th order polynomial. The means of the roots can be taken as $$\mu_{a_0,a_{n-1}}=\frac{a_0+a_1+a_2+\ldots+a_n-1}{a_n}=\frac{\sum_{k=0}^{n-1} a_k}{a_n}$$any help would be appreciated.

A:

This is not true in general. Consider $P(x) := x(x-1)(x-2)(x-3)$ and take the average of $0, 1, 2$ (the first three roots) to check that it fails in this case. 
Edit: This does work with $\deg(P)$ odd and having consecutive roots $\alpha, \alpha+1, \ldots, \alpha + 2n$ for $\alpha \in \mathbb{R}$. Indeed: 
Let $P$ have roots $0, 1, 2, \ldots, 2n$ (WLOG we can start at $0$ -- the case when $P$ has roots $\alpha, \alpha+1, \ldots, \alpha+2n$ reduces to this case). The average of the first $2n$ roots is $\mu := \frac{1}{2n}(0 + 1 + \cdots + 2n-1) = n - \frac{1}{2}$. The tangent line through $\mu$ is 
$$T : y = P'(\mu)x + P(\mu)-P'(\mu)\mu.$$
If it were the case that $T$ hits $(2n, 0)$, then $0 = P'(\mu)2n + P(\mu) - P'(\mu)\mu \implies P'(\mu)(\mu - 2n) = P(\mu)$. Now, 
$$P'(\mu) = \sum_{i=0}^{2n}\prod_{j\neq i}(\mu - j),$$
and so 
$$P'(\mu)(\mu - 2n) = (\mu-2n)^{2}\left( \sum_{i=0}^{2n-1}\prod_{j\neq i}(\mu - j) \right) + P(\mu) = P(\mu),$$
whence
$$\sum_{i=0}^{2n-1}\prod_{j\neq i}(\mu - j) = 0.$$
So if we can show this above equation holds, then our claim holds. And indeed it does: This is really
\begin{align*}
&----\left(n-\frac{3}{2}\right)\left( n-\frac{5}{2} \right) \cdots \left( n - \frac{4n-1}{2}\right) \\
&+\left(n-\frac{1}{2}\right)---\left(n-\frac{5}{2}\right) \cdots \left( n-\frac{4n-1}{2}\right) \\
&+\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)---\cdots \left( n-\frac{4n-1}{2}\right) \\
&\vdots\\
&+\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\left( n-\frac{5}{2} \right)\cdots ----.
\end{align*}
Notice that, for example, the last factor in the first line is $\left(n-\frac{4n-1}{2}\right) = \left(-n+\frac{1}{2}\right)$. So adding the first term to the last yields
\begin{align*}
\left( n-\frac{3}{2}\right) \cdots \left(n - \frac{4n-3}{2}\right)\left(n - \frac{4n-1}{2} + n - \frac{1}{2}\right) = 0.
\end{align*}
Do the same for the 2nd term and the 2nd-to-last term, the same for the 3rd term and the 3rd-to-last term, etc. There are an even number (precisely $(1/2)(4n) = 2n$ many) terms. So this sum is zero, which is what we wanted. 

