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Let $r\in\mathbb N$ and $f$ be an entire function on $\mathbb C$ such that for every $R\in\mathbb C[z]$, there exist polynomials $P_{i,R}(z)\in\mathbb{C}[z]$ ($0\le i\le r$) not all zero such that, for every $z\in \mathbb C$, one has $$\sum_{i=0}^rP_{i,R}(z)(f+R)^{(i)}(z)=0.$$ Then, $f$ is a polynomial.

Any clue to prove that?

Thanks in advance

joaopa
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  • Nice question, +1. Not sure about the title, though. – Julien Apr 15 '13 at 03:41
  • I guess the assumption is: there exist polynomials not all equal to zero $P_{i,R}$... otherwise, any entire function satisfies that with $P_{i,R}=0$. – Julien Apr 15 '13 at 03:47

1 Answers1

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This is not true. Counterexample: Let $f(z) = e^z$, and let $R$ be a polynomial of degree $n$. Let $r = n+2$, $P_{i,R} = 0$ for $0 \le i \le n$, and $P_{i,n+1} = 1$, $P_{i,n+2} = -1$. Then the sum on the left in your expression turns into $0 + \ldots + 0 + e^z - e^z = 0$.

Lukas Geyer
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