= Taylor expansion from a periodic derivative equation
{c}
{title2=$f^{(r)}=f$}
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 $f,f',\ldots,f^{(r-1)}$. 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 $\sum_{k\ge0}x^{3k}/(3k)!$.
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