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Taylor's theorem expands about $a$ in powers of $h=x-a$. With the usual differentiability assumptions, its finite form with Lagrange remainder iswhere $\xi$ lies between $a$ and $x$. If the remainder tends to zero as $n\to \infty$, this yields the Taylor series $f(x)={\sum }_{k\ge 0}{f}^{(k)}(a)(x-a{)}^k/k!$. Smoothness alone does not guarantee that infinite-series equality.

One standard finite form of Taylor's theorem is the following. Suppose $f$ has continuous derivatives through order $m+1$ on an interval containing $a$ and $x$. Then, if $x\ne a$, there is some $\xi$ strictly between them such thatThe final term is the Lagrange remainder. At $x=a$ the identity holds directly with zero remainder. The finite expansion requires no assumption that an infinite Taylor series converges to $f$.

By Taylor's theorem with Lagrange remainder, for an increment $t$,where ${\xi }_t$ lies between $x$ and $x+t$. The given third derivative bound implies $|R_x(t)|\le |t{|}^3/6$.

Apply the preceding Taylor's theorem at $t=h$ and $t=-h$, with $h>0$. The even-order terms cancel in the central finite difference, givingBoth Lagrange remainders have magnitude at most $h^3/6$, so the triangle inequality gives the sharp central-secant derivative error