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For a function with continuous derivatives through order $m+1$ on the interval between $0$ and $x$, the Taylor formula with integral remainder isThis is Taylor's theorem about zero; translation gives the formula about any other point. To prove it, the case $m=0$ is the fundamental theorem of calculus. For $m\ge 1$, integration by parts givesSubstitution into the formula at order $m-1$ proves the formula at order $m$ by induction. The integral estimate, valid also for negative $x$, isFor the smooth function satisfying $f^{(3)}=f$, each fixed derivative is continuous on $[-R,R]$ and hence is bounded there by the extreme value theorem. Thus a bound $M_j$ exists for every $j$. Differentiating the differential equation repeatedly gives $f^{(j+3)}=f^{(j)}$. Only the three functions $f,f^{\prime },f^{\prime \prime }$ are needed: take $M$ to be the maximum of their absolute-value bounds on $[-R,R]$. Then $|f^{(j)}(x)|\le M$ simultaneously for every $j\ge 0$ and $|x|\le R$.

At zero the derivatives consequently repeat the values $1,0,0$. Taylor's theorem therefore givesThe last limit of a sequence follows because the ratio of consecutive majorants is $R/(m+2)$, eventually less than $1/2$. Since $R$ is arbitrary, Taylor expansion from a periodic derivative equation provesThis proves convergence to the function using a remainder bound; it does not assume beforehand that a smooth function equals its Taylor series.