# Entire solutions of finite difference equations

Let $(E):\ \sum_{k=0}^n P_k(z)g(z+k)=0$ with the $P_k\in\mathbb C[z]$ a finite differences equation in $\mathbb C$.

Is it true that every any entire solution $g$ of (E) of exponential type $<\pi$ is an exponential polynomial?

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Do you mean $g(z+i)$? –  abx Mar 11 '14 at 18:35
Is my edition correct? –  Alexandre Eremenko Mar 12 '14 at 4:46
It's best to avoid using $i$ as a summation index in contexts such as this where $i$ might be needed as a square root of $-1$ (e.g. one natural thing to try is the Laplace transform, whose inversion involves complex exponentials). Better to sum over $j$, $k$, or $m$. –  Noam D. Elkies Mar 12 '14 at 5:04
Alexandre > Thanks your edit was correct. Following Noam, I changed the summation index $i$ by $k$. –  joaopa Mar 12 '14 at 6:18

The answer is no. Function $g(z)=(e^z-1)/z$ is entire, of exponential type $1$, is not an exponential polynomial, but it satisfies the equation $$(z+2)g(z+2)-(1+e)(z+1)g(z+1)+ezg(z)=0.$$ Ref. J-P Bezivin et F. Gramain, Solutiuons entieres d'un systeme d'equations aux differences, Ann. Inst. Fourier, Grenoble, 43 (1993) 792-814.
Edit. There exists also such equations with entire solutions of smaller growth, of order less than one. This paper has explicit examples with order $1/3$ and $1/5$: K. Ishizaki and N. Yanagihara, Wiman-Valiron method for difference equations, Nagoya Math. J., 175 (2004) 75-102.