Let $P \in \mathbb R[x]$ be a degree-$n$ polynomial with real coefficients such that $P(a) \neq 0$, where $a$ is real. If $P'(a) = P ''(a) = 0$ then prove that $P$ cannot have all roots real.

Can someone suggest a possible solution using Rolle's Theorem. All I could gather was that $P'(x) = 0$ has a repeated root by Rolle's Theorem. But I am stuck after this

  • If $P(x)$ can have $n$ real roots, then by Rolle's theorem you would get $P^{(n)}$ has some real root, which is impossible, cause $P^{(n)}$ is a nonzero constant. – xbh Aug 16 at 7:25
  • @Arthur Sorry about the ambiguity. If $P$ has $n$ real roots and $P'(a)= P''(a)=0$ where $P(a)\neq 0$, then $P'$ would have $n$ roots [count multiplicity], then $P^{(n)}$ would have root. Am I correct now? – xbh Aug 16 at 7:31
  • @xbh If you can prove that $P'$ has $n$ real roots, then you don't need to go to $P^{(n)}$ to prove contradiction; $P'$ has degree $n-1$ and therefore cannot have $n$ roots. I think the main point of this exercise, however, is proving that $P'$ has at least $n$ roots in this case. You've just pointed it out as though it's a triviality and then spent a few lines proving the part which is (relatively) trivial. – Arthur Aug 16 at 7:39
  • @Arthur Thanks. I do not know what the OP has learned, so I wrote a few more lines. If s/he knew about the fact, then my lines are truly not necessary. – xbh Aug 16 at 7:44
up vote 25 down vote accepted

Assume that $P$ has degree $n$ and let $x_1,x_2,\dots,x _n$ be all its roots (repetitions are allowed). Then $P(x)=c\prod_{k=1}^n (x-x_k)$, and if $x$ is not a root of $P$ we have that $$\frac{P'(x)}{P(x)}=\sum_{k=1}^n \frac{1}{x-x_k}.$$ After taking the derivative we obtain $$\frac{P''(x)P(x)-(P'(x))^2}{(P(x))^2}=-\sum_{k=1}^n \frac{1}{(x-x_k)^2}.$$ Finally by letting $x=a$ (which is not a root) we get a contradiction: $$0=\frac{P''(a)P(a)-(P'(a))^2}{(P(a))^2}=-\sum_{k=1}^n \frac{1}{(a-x_k)^2}<0$$ where the right-hand side is negative because $a, x_1,x_2,\dots,x _n$ are all real.

Sketch of proof: Assume all roots of $P$ are real, and let $x_1\leq x_2\leq \ldots\leq x_n$ be the $n$ roots (with repetition if $P$ has repeated roots). What does Rolle's theorem say about the roots of $P'$? How many roots does $P'$ have (counted with multiplicity)? Can $P'$ have a repeated root which is not one of the $x_i$?

I guess $P$ is not constant, otherwise the statement is false: all the roots of the constant $a$ polynomial are real, as everything holds for the elements of the emptyset.

Translate the polynomial by $a$, i.e., $Q(x):= P(x-a)$. Then the conditions can be rephrased to $Q$ equivalently as follows: $Q(0)\neq 0$, $Q'(0)=Q''(0)=0$. In other words, $Q(x)= a_nx^n+ \cdots +a_3x^3 + a_2x^2+a_1x+a_0$, where $a_0\neq 0$ and $a_1=a_2=0$.

Write the Viete formulas for the roots $x_i$:

$\prod x_i= (-1)^na_0\neq 0, \prod x_i \cdot \sum 1/x_i=0$, and $\prod x_i \cdot \sum\limits_{i\neq j} 1/(x_ix_j)=0$.

Put $y_i=1/x_i$ (possible, as $0$ is not a root, as the constant term is nonzero), then after simplification, you obtain $\sum y_i= \sum\limits_{i\neq j} y_iy_j=0$, but then $\sum y_i^2= 0$. So if these are real numbers, then all the $y_i$ are zero, a contradiction.

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.