Find the smallest natural number $n$ such that there exists a polynomial $P(x)$ satisfying :
$P(x)=a_{2n}x^{2n}+a_{2n-1}x^{2n-1}+...+a_0$ has a real root and $2014 \le a_i \le 2015 $ ($ $with $ 0\le i \le 2n )$
Here's all i did :
Let $A$ = $max \frac {|a_k|}{|a_{2n}|} , k = 0,1,...(2n-1)$
$P(x) = a_{2n}x^{2n}(1+ \frac{a_{2n-1}}{a_{2n}x} +...+\frac{a_0}{a_{2n}x^{2n}}) $
Let $ P(x)=0.$ If $|x| \le 1 \Rightarrow |x| \le 1+A $
If $|x| > 1 \Rightarrow 1 \le A ( \frac{1}{|x|} + \frac{1}{|x^2|} +...+ \frac{1}{|x^n|} ) \le \frac {A}{|x|.(1-\frac{1}{|x|}) } \Rightarrow 1\le \frac {A}{|x|-1} \Rightarrow |x| \le 1+A $
So if $x_0$ is a real solution of $P(x)$ , then $|x_0| \le 1 +max \frac {|a_k|}{|a_{2n}|} , k = 0,1,...(2n-1) $
$\Rightarrow |x_0| \le 1+ \frac{2015}{2014} $
$2014 \le a_i \le 2015 \Rightarrow x_0 < 0 \Rightarrow x_0 \ge -1 -\frac{2015}{2014}$
This is all i did , i think we need one more evaluation from $x_0$ to find the minimum value of $n$ . I have no more ideas. Looking forward to getting help from everyone. Thanks very much .
The question Polynomial with bounded coefficients and real root did not solve my problem . Please don't close my question !