# How prove that polynomial has only real root.

Let this polynomial $f(x)=\displaystyle\sum_{i=1}^{n}a_{i}x^i,\;\;a_{i}\in \mathbb{R}$ have only real roots. Prove:

The polynomial $g(x)=\displaystyle\sum_{i}^{n}C_{n}^{i}a_{i}x^i$ has only real roots where $C_{n}^{i}=\dfrac{n!}{i!(n-i)!}$

My idea: let $x_{1},x_{2},\cdots,x_{n}$ was $f(x)=0$ real root,and we have

$$x_{1}+x_{2}+\cdots+x_{n}=-\dfrac{a_{n-1}}{a_{n}}$$ $$x_{1}x_{2}+x_{1}x_{3}+\cdots+x_{1}x_{n}+x_{2}x_{3}+\cdots+x_{n-1}x_{n}=\dfrac{a_{n-2}}{a_{n}}$$ $$\vdots$$ $$x_{1}x_{2}\cdots x_{n}=(-1)^n\dfrac{a_{1}}{a_{n}}$$

and I find a book has this theory:

if Polynomial $f(x)=\displaystyle\sum_{i=0}^{n}a_{i}x^n(a_{0},a_{n}\neq 0),a_{i}\in R$ only have real roots, then we have $$\Delta_{1}=(n-1)a^2_{n-1}-2na_{n-2}a_{n}\ge 0$$ and $$\Delta_{2}=(n-1)a^2_{1}-2na_{2}a_{0}\ge 0$$

• This is trivial for $n=0,1$ and it is easily true for $n=2$ by discriminant considerations. So it's worth trying. Good question, +1. – Julien Mar 23 '13 at 12:52
• Thank you ,I think this is good question, – math110 Mar 23 '13 at 13:03

This is problem 719 in the exercises book by Faddeev and Sominski. Here is their proof.

Denote by $D$ the set of all real univariate polynomials all of whose roots are real.

Lemma 1 If $f\in D$ and $\lambda \in {\mathbb R}$, then $f'+\lambda f \in D$ also.

Proof of lemma 1. The case $\lambda=0$ follows from Rolle’s theorem, so assume $\lambda \neq 0$. Denote by $x_1<x_2< \ldots <x_r$ the distinct roots of $f$, and denote by $m_i$ the multiplicity of $x_i$ in $f$. Then $\sum_{k=1}^r m_k=n$ where $n$ is the degree of $f$. Consider the rational fraction $g=\frac{f'}{f}$. Its poles are the $x_k$, and $g$ is surjective $(x_k,x_{k+1}) \to {\mathbb R}$ for each $k\lt n$. Also, on the border, $g$ is surjective $(-\infty,x_1) \to (-\infty,0)$ and surjective $(x_n,+\infty) \to (0,+\infty)$. We deduce that there is a $y_k\in (x_k,x_{k+1})$ with $g(y_k)=-\lambda$ and that there is an yet another $y$ satisfying $g(y)=-\lambda$ (so $y$ will be $\lt x_1$ if $\lambda$ is negative and $\gt x_n$ if $\lambda$ is positive). Let us now count all the roots we have found for $f'+\lambda f$ : we have $y,y_1,y_2, \ldots y_{r-1}$, plus the $x_k$ with multiplicities $m_k-1$. This makes up a total of $n$ real roots and concludes the proof.

Lemma 2 If $f,g\in D$ with $g=\sum_{k=0}^n {\gamma}_k x^k$, then $h=\sum_{k=0}^n {\gamma}_k f^{(k)} \in D$ also.

Proof of lemma 2. Let $\lambda_1, \ldots ,\lambda_n$ denote the roots of $g(-x)$, so that $g(x)=\prod_{k=1}^n (x+\lambda_k)$. Iterating lemma 1 above, we see successively that $F_1=f’+\lambda_1f, F_2=F_1'+\lambda_2 F_1, \ldots$, etc, up to $F_{n}=F_{n-1}'+\lambda_n F_{n-1}=h$, are all in $D$.

Lemma 3 Let $f\in D$, $f=\sum_{k=0}^n a_k x^k$, and $m$ be a positive integer. Then $\sum_{k=0}^n \frac{m!}{(m-n+k)!} a_k x^k \in D$ also.

Proof of lemma 3. Use lemma 2 with $g(x)=x^m$, and multiply by $x^{n-m}$ if needed.

Main theorem Let $f\in D$, $f=\sum_{k=0}^n a_k x^k$. Then $\sum_{k=0}^n \binom{n}{k} a_k x^k \in D$ also.

Proof of main theorem. Using lemma 3 with $m=n$, we see that $h_1=\sum_{k=0}^n \frac{n!}{k!} a_k x^k \in D$. So $x^nh_1(\frac{1}{x}) \in D$ also, which means that $h_2=\sum_{k=0}^n \frac{n!}{(n-k)!} a_k x^k \in D$. Using lemma 3 again with $h_2$ in place of $f$ and $m=n$, we see that $h_3=\sum_{k=0}^n \frac{n!}{k!} \frac{n!}{(n-k)!} a_{n-k} x^k \in D$. Then, divide by $n!$.

• Thank you very much! my frend. Have you this book Download? thank you – math110 Mar 25 '13 at 15:35
• @math110 No, this book is not downloadable on the web. – Ewan Delanoy Mar 25 '13 at 16:32
• @EwanDelanoy You are saying that the binomial expansion is in D, so that by the main theorem the polynomial with the squared coefficients is also? Does this tell us that the roots are negative? It is awesome what people have proved. – Betty Mock Dec 1 '13 at 0:21
• But in the last paragraph in h_2 , a_k should be a_{n-k} I think – Idele Dec 4 '15 at 1:12
• @hctb Corrected, thanks – Ewan Delanoy Dec 4 '15 at 7:32