5
$\begingroup$

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

  1. $p(r_i) = 0$, $i = 1,2,...,n,$ and
  2. 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? :))

  • 0
    @Barre: Your proof is wrong. Equations (1) and (2) can be equal for any $n$, if the $r_i$ are selected appropriately. For instance, $(r_1,r_2,r_3) = (0,1,2)$.2012-01-06

2 Answers 2

11

Here's another attempt :

Because $p$ is of degree $n$ and has $n$ distinct roots, we can write

$p(x) = \lambda \prod_{i = 1}^n (x-r_i)$

By differentiating, we have

\frac{p'(x)}{p(x)} = \sum_{i = 1}^n \frac{1}{x-r_i}

And setting $x = \frac{r_1+r_2}{2}$, we get

$0 = \sum_{i = 1}^n \frac{1}{\frac{r_1+r_2}{2}-r_i}$

Notice the first two terms cancel, which leaves

$0 = \sum_{i = 3}^n \frac{1}{\frac{r_1+r_2}{2}-r_i}$

But if you assume $n > 2$, since for $i > 2$, we have $r_i > r_2 > \frac{r_1+r_2}{2}$ we would get

$\sum_{i = 3}^n \frac{1}{\frac{r_1+r_2}{2}-r_i} < 0$

So the only polynomials that can possibly satisfy your conditions are degree $2$ polynomial with positive discriminant (and they all work).

3

I don't think you can conclude what you think you can conclude. You are given $ p(x) \propto \prod_{i=1}^{n} (x-r_i) $ and $ p^{\prime}(x) \propto \prod_{i=1}^{n-1} \left(x - \frac{r_i + r_{i+1}}{2}\right), $ since all the roots of each polynomial are accounted for. By differentiating the first equation you have $ p^{\prime}(x) \propto nx^{n-1} - (n-1)\left(\sum_{i}r_i\right) x^{n-2} + O(x^{n-3}), $ and by inspection of the second equation you have $ p^{\prime}(x) \propto x^{n-1} - \left(\frac{r_1}{2} + r_2 + \dots + r_{n-1} + \frac{r_{n}}{2}\right)x^{n-2} + O(x^{n-3}). $ I think that all you can conclude by comparing the coefficients of the $x^{n-2}$ terms is that $ \frac{n-1}{n}\sum_{i}r_i = \sum_{i}r_i - \frac{1}{2}(r_1 + r_n), $ or that $ r_1 + r_n = \frac{2}{n}\sum_{i}r_i = \frac{2}{n-2}\sum_{1 for $n>2$ (and the equation is automatically satisfied for $n=2$, regardless of $r_1$ and $r_2$).

  • 0
    @Joel: You're right; I edited the solution to correct that.2012-01-07