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Consider a real polynomial $f$ of degree $d$ which has $d$ real roots not necessarily distinct. In general, can we accomplish the following?

  1. For every $\epsilon>0$, can we perturb each coefficient of $f$ by less than $\epsilon$ and guarantee real, distinct roots?

I just need one such perturbation to work. I know that a priori, not every perturbation will be nice (read: the number of real roots does not vary continuously like it does in the separable case). For example if $f=x^{2}$, this has $0$ as a root of multiplicity $2$ but if every perturbation of the constant term in the positive direction, there won't be be any real roots at all. But any $x^2+bx+c$ with $b^2>4c$ will be a perturbation that yields two real roots so we've solved 1) for $f=x^2$.

I think this will not be difficult and can be done. Just consider $f=c\prod(x-r_i)^{k_i}$ and reduce to the case of a single $(x-r_i)^{k_i}$. Recenter root to be $0$ so that we have $x^d$ and do this directly. This last part needs argument but I think can be done.

  1. How many coefficients do I need to perturb to achieve objective 1?

For example, in most cases, it is not possible to do just by perturbing the constant coefficients. The geometric intuition is that a degree $d$ real polynomial with distinct real roots will have $d-1$ local extrema none of which occur at roots. However for those with only $k<d$ distinct real roots have fewer extrema or have the extrema at roots. How do we perturb these to gain these additional extrema or to move existing extrema at roots to take advantage of them to get distinct roots). Ideally, this should translate into something about the relative sizes and signs of the coefficients.

Thank you for any comments, solutions or references to the literature that you can provide.

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  • $\begingroup$ I wonder if it's possible to prove that a random perturbation of each coefficient in the range $(-\epsilon, +\epsilon)$ has all roots distinct with probability 1? It is a stronger result which proves your question, and it seems intuitively plausible. $\endgroup$ – fractal1729 Jun 19 '17 at 22:29
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    $\begingroup$ @fractal1729 A random perturbation of the coefficients will indeed break the condition of rood equality, but in general has a non-zero probability of causing a double root to become two imaginary roots, so you will not get $d$ distinct real roots with probability $1. $\endgroup$ – Mark Fischler Jun 19 '17 at 22:37
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    $\begingroup$ @fractal1729 The example of $f=x^3$ shows that you cannot do it with one coefficient, right? This has a repeated real root at zero but any constant coefficient perturbation will have only one real root and two complex roots. Proof by picture. $\endgroup$ – Suana Jun 19 '17 at 22:48
  • $\begingroup$ @Suana $f = x^3$ is $1x^3 + 0x^2 + 0x + 0$, so aren't there still four coefficients? Mark's idea makes sense, I missed the condition that the roots have to be real. Thanks guys! $\endgroup$ – fractal1729 Jun 19 '17 at 23:47
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You can always disambiguate $n$ double roots by perturbing $n$ coefficients.

In general, a perturbation that changes a polynomial of degree $d$ having $d-k$ roots (some of which have multiplicity of $2$ or more) to one that has $d$ distinct real roots requires perturbing $k$ coefficients, and these can always be chosen to be the last $k$ (the constant term, then the linear term, and so forth).

However, there are cases where fewer coefficients will suffice. In particular, whenever $d < 3k-1$ you can "remedy" a polynomial with a deficiency of $k$ roots, by perturbing fewer than $k$ coefficients.

The proof is constructive. For the case of $k$ roots of multiplicity $2$, choose for each double root a sufficiently small quantity having the same sign as the second derivative at that root. Then form a polynomial of degree $d$ passing through the chosen values at the respective roots. Add that polynomial to the original and you get a perturbed polynomial with $d$ real roots. If the perturbation is larger than the allowed change, then divide the added polynomial by some sufficiently large positive number.

For example, consider $$ P(x) = x^8-8x^7-8x^6+160x^5-86x^4-872x^3+768x^2+720x-675=0 $$ which has double roots at $x=-3, x=1, x=5$ and thus a deficiency of $3$ real roots.

Now add $$ f(x) = -\frac1{80}(x^2-2x-7) $$

$$ \bar{P}(x) = x^8-8x^7-8x^6+160x^5-86x^4-872x^3+\frac{61439}{80}x^2+\frac{28801}{40}x-\frac{53993}{80}=0 $$ which has real roots at $$ \{ -3.00285, -2.99715, -0.99998,+0.99012,+1.00988,+2.99998,+4.99715,+5.00285\} $$

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protected by J. M. is a poor mathematician Apr 3 at 9:19

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