I've been lately having fun with some Putnam problems (http://www.math.harvard.edu/putnam/) and I would like to see how todays problem can be solved and for somebody more experienced to check my attempted solution.
Find all real polynomials $p(x)$ of degree $n \geq 2$ for which there exists real numbers $ r_1 < r_2 < ... < r_n$ such that
- $p(r_i) = 0$, $i = 1,2,...,n,$ and
- p'(\frac{r_i + r_{i+1}}{2}) = 0, $i = 1,2,...,n-1$,
where p'(x) denotes the derivative of $p(x)$
(messy) attempt involving checking coefficients;
Any polynomial of degree $n$ with roots in $r_1, r_2, ...,r_n$ will have the following form:
$p(x) = c(x-r_1)(x-r_2)...(x-r_n)$
Expanding this we must get that the coefficient of $x^{n-1}$ is:
$-c(r_1 + r_2 + ... + r_n)$.
This would imply that the coefficient of $x^{n-2}$ of the (monic equivalent) derivative p'(x) is:
$\frac{(1-n)}{n}(r_1 + r_2 + ... + r_n)$ (1)
So the derivative function must have same coefficient in front of $x^{n-2}$, and from description we get that it is supposed to have following form:
p'(x) = (x-\frac{r_1 + r_2}{2}) (x-\frac{r_2 + r_3}{2})... (x-\frac{r_n-1 + r_n}{2})
Expanding this will yield $x^{n-2}$ term with following coefficient:
$-(\frac{r_1}{2} + r_2 + r_3 + ... + r_{n-1} + \frac{r_n}{2})$ (2)
By inspecting equations (1) and (2) we see that they are equal in case of $n = 2$ and never else. So the answer would be the set of polynomials of degree 2?
(btw, how do I make the numeration of equation look nice? :))