I was naively wondering if there is some kind of formula or general result about the zeros of the integral of a polynomial. In particular, I was wondering if, when you fix a positive integer $n$ and define $\mathcal{P}_n:=\{p~\text{is a polynomial s.t. all zeros have order a multiple of $n$}\}$, if you have a formula or general information about the zeros of the integral of the polynomials from $\mathcal{P}_n$. (I don't think this particular case helps much, though)

More particularly, let $I:H(\mathbb{C})\to H(\mathbb{C})$ the integration operator $$ I(f(z)):=\int_{0}^zf(s)ds. $$ If I choose $n\in\Bbb N$ "special" and $m\in\Bbb N$ "special", could I have $I^m{p}\in\mathcal{P}_n$ for (all? which one?) $p\in\mathcal{P}_n$?

  • $\begingroup$ You increase the degree so you might end up with one more root. $\endgroup$ – mvw Sep 11 '18 at 11:06

The zeroes of $f(z)$ are related to the maxima and minima of $\int_0^zf(s)ds$, not the zeroes.

If we set $g(z) = f(z+1)$, then the behaviour of the zeroes of $f$ an $g$ are identical; their graphs are exactly the same except they are sideways translations of one another. However, $I(f)$ and $I(g)$ can have wildly different zero behavious since $I(g)(z) = I(f)(z+1) + c$ for some constant $c$. In other words, the graph hasn't just been shifted sideways, but also vertically.

Since changing $f$ in such a way which didn't really affect how its zeroes behave (degree, the distances between them, etc.) changed the zeroes of $I(f)$ completely, the conclusion must be that you cannot extract much information about the zeroes of $I(f)$ just from information on the zeroes of $f$.

  • $\begingroup$ Convincing enough. Thanks for that. $\endgroup$ – Filburt Sep 11 '18 at 21:57

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