If a polynomial of degree $n$ has $n$ real roots, then its derivative has $n-1$ real roots. ($n \geq 2$)
I wanted to prove "If a polynomial of degree $n$'s roots are all real, then its derivative only has real roots. ($n \geq 2$)" with an inductive method, but I'm not sure if this reasoning is acceptable.
Let degree-$n$ polynomial be $P_n(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$. Note that coefficients $a_i \in \mathbb{R}$ are arbitrary because $P_n(x)$ only has real roots.
Without loss of generality, let's suppose we are counting all repeated real roots individually, so that $P_n(x)$ could have $n$ real roots and $P'_n(x)$ could have $n-1$ roots.
Let $B_n=\{b_1,...,b_{n-1}\}$ be a set of roots of its derivative.
First, when $n=2$, $P_2(x)=a_2 x^2+a_1 x +a_0$.
Then, $P'_n(x)=2a_2x+a_1$, so $B_2=\{\frac{-a_1}{2a_2}\}$. And since $\frac{-a_1}{2a_2} \in \mathbb{R}$, the statement is true for this case.
Suppose that when $n=k$, we have $k-1$ real roots of $P'_k(x)$.
Note that $P'_k(x)=ka_kx^{k-1}+(k-1)a_{k-1}x^{k-2}+...+a_1$
and $B_k=\{b_1,b_2,...,b_{k-1}\}$.
Consider when $n=k+1$.
$P'_{k+1}(x)=(k+1)a_{k+1}x^k+ka_kx^{k-1}+(k-1)a_{k-1}x^{k-2}+...+a_1$.
Note that $P'_{k+1}(x)$ has same roots as roots of the equation "$-(k+1)a_{k+1}x^k=ka_kx^{k-1}+(k-1)a_{k-1}x^{k-2}+...+a_1=P'_k(x)$".
Since $"-(k+1)a_{k+1}"$ is a real coefficient and , $-(k+1)a_{k+1}x^k$ and $ka_kx^{k-1}+(k-1)a_{k-1}x^{k-2}+...+a_1$ are drawable on cartesian plane. This means, the roots are all real numbers.
As shown in the induction reasoning, we conclude that the statement is true. [End].
But, even though I put the word "arbitrary" at first, I think the relations between each coefficient might be different when $n$ changes.
For example, we might have [$a_3x^3+a_2x^2+a_1x+a_0$]'s [$a_2,a_1,a_0$] that they could never be used in [$a_2x^2+a_1x+a_0$] because polynomials must have real roots only.
So, I think I need a stronger condition that could make the reasoning much clearer.
Can I get some suggestions to improve this induction method? Or, if there is any new idea to prove the statement, it would be really appreciated as well. Thank you.
$\endgroup$ 21 Answer
$\begingroup$Let $\{b_1 < b_2 < \dots < b_m\}$ be the distinct real roots of $P(x)$ and $k_i$ be the multiplicity of $b_i$. Then by the hypothesis we have $k_1 + k_2 + \dots + k_m = n$. By Rolle's theorem we know that the derivative of $P(x)$ has at least $m-1$ distinct real roots (at least one on every interval of the form $(b_i, b_{i+1})$). Now a root of $P(x)$ with multiplicity $k$ is a root of $P'(x)$ with multiplicity $k-1$ (it's easy to see that if you write $P(x)$ as $(x-\rho)^k \cdot s(x)$ for some polynomial $s(x)$ and take the derivative). In total, $P'(x)$ has at least$$m-1 + (k_1 - 1) + \cdots + (k_m - 1) = m-1+n-m = n-1$$real roots and since it has degree $n-1$ it has exactly $n-1$ real roots.
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