Suppose $0<m_1<...<m_n$ and $m_i \equiv i (\mod 2)$. Prove that the following polynomial has at most $n$ real roots. ($\forall 1\le i \le n: a_i \in \mathbb R$). \[a_0+a_1x^{m_1}+a_2x^{m_2}+...+a_nx^{m_n}.\]
Problem
Source: Iran 3rd round 2012-Algebra exam-P1
Tags: algebra, polynomial, induction, function, modular arithmetic, algebra proposed
21.09.2012 21:14
Just to be precise: Here and below every real root is counted with its multiplicity. Also $a_i \neq 0,\, 0 \leq i \leq n$, because otherwise the assertion is not true. Induction on $n$. Let the assertion is true for all indices less than $n$. We will prove it for $n$. We use the following well known proposition: If $f(x)$ is a differentiable function in some interval $(a,b)$ and has $k$ roots in it then $f'(x)$ has at least $k-1$ roots in $(a,b)$. Now let on the contrary assume that there exist $0<m_1<\ldots <m_n ,\, m_i \equiv i \pmod 2$ and $a_i \in \mathbb{R}$, such that $ P(x) = a_{0}+a_{1}x^{m_{1}}+a_{2}x^{m_{2}}+...+a_{n}x^{m_{n}} $ has at least $n+1$ real roots. Then $P'(x)$ has at least $n$ roots. (1) $P'(x) = a_1m_1x^{m_1-1}+a_2m_2x^{m_2-1}+\ldots+a_n m_n x^{m_n-1}= x^{m_1-1}Q(x)$ where $m_1-1$ is even, $ Q(x) = b_0 + b_1x^{r_1}+b_2x^{r_2}+\ldots+b_{n-1}x^{r_{n-1}} $, $ b_i \neq 0 $ and $r_i \equiv i \pmod 2,\, 1\leq i \leq n-1$. Now we want to apply the induction hypothesis on $Q(x)$ but there is a slight technical problem with the root $0$ of $P'(x)$. So in order to proceed we need the following assertion. Let $x_1 <0< x_2$, $P(x_1)=P(x_2)=0$, then there exists $\xi,\, x_1< \xi < x_2, \, \xi \neq 0$, such that $Q(\xi)=0$. Proof: WLOG assume P(x) take possitive values on $(x_1,x_2)$, then take $\xi,\, P(\xi) = \max_{x\in (x_1,x_2)} P(x)$. Obviously$ P'(\xi)=0$, It remains to see that $\xi \neq 0$. Notice that $P'(x)$ don't change its sign in some neighbourhood of $0$ (because $m_1-1$ is even). If assume that $\xi=0$ then it is not possible $P(0)$ to be the max value of $P$ in $(x_1,x_2)$ $\square$ It is time for counting. Let $x_1 \leq x_2 \leq\ldots \leq x_k <0 <x_{k+1}\leq \ldots \leq x_{n+1}$ are real roots of $P$ taken with their multiplicities. Then $Q(x)$ has $k-1$ roots in $(x_1,x_k]$, $n+1-k-1$ roots in $[x_{k+1},x_{n+1})$ and one root in $(x_k, x_{k+1})$. It means $Q(x)$ has at least $n$ real roots which contradicts with the induction hypothesis.
24.09.2012 17:13
This Problem Copied from AMM. Nice solutions of it is available by using Dsecart,s rule of sign!