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Let $a_1,a_2,a_3...a_n$ be real numbers such that the polynomial $p(x)=x^n+a_1x^{n-1}+...+ a_{n-1}x+a_n$ has n distinct real roots. Does there exist $\epsilon$ >0 such that for all $b_1,b_2...b_n \in R$ with the property that $|a_j-b_j|<\epsilon$, for all j=1,2,...n the polynomial $q(x) = x^n+b_1x^{n-1}+...+b_{n-1}x+b_n$ has n distinct real roots?

I am momentarily studying integration in $R^n$ but I couldn't make any relation to this one.Maybe it is off topic related to integration. But how do I prove it?

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Let $[-T,T]$ be an interval containing all of the roots of $p(x)$ with $T > 1$.

Now $$|p(x) - q(x)| \le |a_1 - b_1| |x|^{n-1} + |a_2 - b_2| |x|^{n-2} + \cdots + |a_n - b_n| \le |T|^{n-1} \sum_{i=1}^n |a_i - b_i|.$$

If $p$ has $n$ distinct roots, then $p'(x)$ has $n-1$ distinct roots by the mean value theorem. Let $x_i$ be these roots. Now suppose that $m = \min |p(x_i)|$.

If $|p(x) - q(x)| < m/4$ for all $x$, then when $p(x_i)$ is negative, $q(x_i)$ is negative. If $p(x_i)$ is positive, then $q(x_i)$ is positive. This can guide your condition on $\sum_{i=1}^n |a_i -b_i|$. You can then conclude that $q$ has $n$ distinct roots by the intermediate value theorem.

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Yes it exists ONE $\epsilon$ > 0 but I feel the true question is about ALL $\epsilon$ > 0. Let $x_1$, $x_2$ ,....$x_n$ the roots of p and take q having as roots $x_1$ + ${\epsilon_1}$ and all the other the same $x_2$ ,....$x_n$. We have then $b_1$ = $a_1$ + ${\epsilon_1}$; $b_2$ = $a_2$ + ${\epsilon_1}$$S_2$; $b_3$ = $a_3$ + ${\epsilon_1}$$S_3$; ..... ; $b_n$ =$a_n$ + ${\epsilon_1}$$S_{n-1}$ where the $S_i$ are the symmetric basic polynomials of $x_2$ ,....$x_n$.Take the minimum of the right numbers ${\epsilon_1}$$S_i$ as the $\epsilon$ needed and you have your conditions.

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