Let $f(x)$ be a polynomial in $\mathbb{R}[x]$ such that $f$ depends holomorphically on some real parameters. For a choice of those parameters, $f$ is written in the form $$x^3\left(a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\right),$$ where $a_i$ are real numbers. $f$ has a zero of order $3$ at $x=0$ and suppose that it has no other zeros on $(0,1]$.
I would like to prove (or at least give a justification) that the order of the zero is preserved under a small perturbation of the real parameters, and also that no other zeros appear on the interval $(0,1]$.
I've checked this with Mathematica software, but a more rigorous statement would be great. I'm not sure if Rouche's Theorem can help here.
Any advice is very welcome.
You might think of a small perturbation of $g(x):=a_nx^n+\ldots+a_0$?
That works, as you can see here:
We have $f(x)=x^3g(x)$, where $g$ has no zeros at $x=0$. You get $\epsilon:=\inf_{x\in[0,1]}|g(x)|>0$, because $f$ has no zeros on $(0,1]$ and therefore $g$ hasn't too plus continuity of $g$ and compactness of $[0,1]$.
Now you can perturb $g$ a little bit to $\tilde g$ such that $\|g-\tilde g\|_{[0,1]}<\epsilon$.
If you consider $\tilde f(x):=x^3\tilde g(x)$, then you get $$ |\tilde f(x)|=x^3|\tilde g(x)|=x^3|g(x)-(g(x)-\tilde g(x))|\geq x^3\left||g(x)|-|g(x)-\tilde g(x)|\right|>x^3|\epsilon-\epsilon|=0. $$ Hence, $\tilde f$ has no zeros on $(0,1]$. Now we have to check that $\tilde g$ has no zero at $x=0$. But that can be seen by $$ |\tilde g(0)|=|g(0)-(g(0)-\tilde g(0))|\geq ||g(0)|-|g(0)-\tilde g(0)||>|\epsilon-\epsilon|=0. $$