2
$\begingroup$

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?) $p\in\mathcal{P}_n$?

$\endgroup$
  • $\begingroup$ You increase the degree so you might end up with one more root. $\endgroup$ – mvw Sep 11 '18 at 11:06
1
$\begingroup$

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 $f$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.