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? :))

  • $\begingroup$ @WNY: I don't think the $r_i$ are given. $\endgroup$
    – joriki
    Jan 6 '12 at 16:20
  • $\begingroup$ inspecting 1 and 2 gives $(2/n)\sum r_i=r_1+r_n$ which holds more often than you say. i think the problem must be more involved (ie you cant just compare the first two coefficients) $\endgroup$
    – yoyo
    Jan 6 '12 at 16:32
  • $\begingroup$ @Barre: Your proof is wrong. Equations (1) and (2) can be equal for any $n$, if the $r_i$ are selected appropriately. For instance, $(r_1,r_2,r_3) = (0,1,2)$. $\endgroup$
    – TonyK
    Jan 6 '12 at 16:36

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).


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 $$ r_1 + r_n = \frac{2}{n}\sum_{i}r_i = \frac{2}{n-2}\sum_{1<i<n}r_i $$ for $n>2$ (and the equation is automatically satisfied for $n=2$, regardless of $r_1$ and $r_2$).

  • $\begingroup$ Thank you for pointing this out to me. Indeed all I've proven is $r_1 + r_n = \frac{2}{n}(r_1 + .... + r_n).It seems I'll need to start over or at least look at another coefficient. $\endgroup$ Jan 6 '12 at 17:10
  • $\begingroup$ Are you sure about the condition $r_1 + r_2 = 0$ for $n = 2$ ? In that case I think you find the empty equation $\frac{1}{2} (r_1+r_2) = \frac{1}{2} (r_1+r_2)$, which is because all degree $2$ polynomials work. $\endgroup$
    – Joel Cohen
    Jan 6 '12 at 18:39
  • $\begingroup$ @Joel: You're right; I edited the solution to correct that. $\endgroup$
    – mjqxxxx
    Jan 7 '12 at 1:29

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