Taylor expansion from a periodic derivative equation ($f^{(r)}=f$)

For a smooth real function satisfying the displayed equation with positive integer $r$, differentiating makes $f^{(j+r)}=f^{(j)}$. On any compact interval, all derivatives are therefore bounded by the maximum of the bounds of $f,f^{\prime },\dots ,f^{(r-1)}$. The Taylor formula with integral remainder consequently has error at most $M{R}^{n+1}/(n+1)!$, which tends to zero uniformly for $|x|\le R$. Hence the function equals its globally convergent Taylor series. For $r=3$ and initial derivatives $(1,0,0)$, it is ${\sum }_{k\ge 0}{x}^{3k}/(3k)!$.