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Suppose a polynomial $P$ of degree $n$ has $n$ distinct real roots then $P'$ (the derivative of $P$) has $n-1$ distinct real roots.

Proof by Induction:

Base case: For $n=1$, $P_1 (x)=a_0+a_1x, a_1\neq 0$ has $1$ real (distinct) root, $x=\frac{-a_0}{a_1}$. Then $P_1 ' (x)=a_1$ has $1-1=0 $ real distinct root.

Induction step: Assume that the claim holds for some $n\in\mathbb{N}, n>1$. That is, $$P_n (x) = a_0+a_1x+a_2x^2+...+a_nx^n$$ has $n$ distinct real roots implies that $$P_n'(x)=a_1+2a_2x+3a_3x^2+...+na_nx^{n-1}$$ has $n-1$ real distinct roots.

Now, suppose that for $n+1$, $$P_{n+1} (x) = a_0+a_1x+a_2x^2+...+a_nx^n+a_{n+1}x^{n+1}$$ has $n+1$ real distinct roots. Claim is that $P_{n+1}'(x)$ has $n$ real distinct roots.

We know: $$P_{n+1}'(x)=a_1+2a_2x+3a_3x^2+...+na_nx^{n-1} +(n+1)a_{n+1} x^n$$. I know by Induction Hypothesis that $$a_1+2a_2x+3a_3x^2+...+na_nx^{n-1}$$ has $n-1$ real distinct roots.

But I cannot think how to use this fact to argue the case for $P_{n+1}'(x)$.

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    $\begingroup$ Use Rolle's theorem. $\endgroup$ Dec 14, 2013 at 17:09
  • $\begingroup$ Thanks Andreas. Here is my idea: Write $P_{n+1}$ as $$P_{n+1}(x) = (x-r_1)(x-r_2)...(x-r_n)(x-r_{n+1})$$ where $r_i$ are distinct real numbers. Then by Rolle's theorem, I will get $p_i \in (r_i,r_{i+1})$ s.t. $P_{n+1}'(p_i)=0$. In total I will get $n$ such $p_i$, all distinct. Thus, $P_{n+1}'(x)$ has n distinct real roots. Is this fine? $\endgroup$ Dec 14, 2013 at 17:31
  • $\begingroup$ @ugstudent1243 just a question. How do you apply the induction hypothesis? To deduce that $P_{n+1}''$ has $n-1$ distinct real roots you have to know that $P_{n+1}'$ has $n$ distinct real roots. $\endgroup$
    – P..
    Dec 14, 2013 at 17:43
  • $\begingroup$ I guess, I don't need induction-proof after I used Rolle's theorem. $\endgroup$ Dec 14, 2013 at 17:52
  • $\begingroup$ Is my idea correct? I need your confirmation before I can be happy. :D $\endgroup$ Dec 14, 2013 at 18:47

2 Answers 2

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Just draw a picture of $P(x).$ You will see that between any two zeros there is a critical point (UNLESS the zero is degenerate, in which case it is a critical point itself). The way to justify the picture is Rolle's theorem as Andres says.

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Between any two roots of $p(x)$ there'll be a root of $p'(x)$ by Rolle's Theorem , so $p'(x)$ has at least $n-1$ roots and they are distinct , but since $p'(x)$ has degree $n-1$ thus these are only $n-1$ distinct roots of $p'(x)$

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