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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.

Under the hypotheses of Taylor's theorem with Lagrange remainder, for example $f\in C^{n+1}$ on an interval containing $x$ and $x+h$, there is a point $\xi$ between them such thatThe Lagrange remainder makes this an exact equality, rather than just a truncated Taylor polynomial. For $h=0$ the equality is immediate.

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

The quadratic Taylor polynomial of $f(x)=1/(1+x^2)$ at $x=1$ is $p(x)=(x^2-4x+5)/4$. Multiplying $p-f$ by $4(1+x^2)$ gives $(x-1{)}^4$, proving the exact displayed identity. Thus $p\ge f$ for every real $x$, with equality only at one. In particular, a claim that $f>p$ to the right of one has the wrong sign. The Lagrange remainder also proves this direction there, since $f^{\prime \prime \prime }(x)=24x(1-x^2)/(1+x^2{)}^4<0$ for $x>1$. The exact identity is stronger: it controls the sign globally and shows the fourth-order contact at the expansion point.

Let $n\ge 1$. Suppose $f$ has $n-1$ continuous derivatives on the closed interval joining distinct real points $a,x$, and its $n$th derivative exists in the interior. Then some interior point $\xi$ satisfiesEndpoint derivatives in the polynomial are one-sided restrictions of the continuous derivatives. When $n=1$, the assumptions mean endpoint continuity and interior differentiability, and the result is the mean value theorem.^[ref]

For a proof, subtract the degree-$(n-1)$ Taylor polynomial and a multiple $C(t-a{)}^n$ chosen so that the resulting function also vanishes at $x$. It and its first $n-1$ derivatives vanish at $a$. Repeated applications of Rolle theorem give a point where its $n$th derivative vanishes, yielding $C=f^{(n)}(\xi )/n!$. If $|f^{(n)}|\le M$, the Lagrange remainder is bounded by $M|x-a{|}^n/n!$ without knowing $\xi$.