Proof of Taylor's Theorem

Turning the Remainder Formula into a Proof

In “Taylor’s Theorem,” the Lagrange remainder formula was stated and used to estimate approximation errors. The central claim is that the error at \(x\) is determined by the \((n+1)\)st derivative at some point between the center \(a\) and \(x\). This tutorial proves that claim. The key idea is to build an auxiliary function that vanishes at \(a\) to order \(n+1\) and also vanishes at \(x\), then apply Rolle’s Theorem repeatedly.

The proof does not need an explicit formula for the intermediate point. Instead, it establishes that such a point must exist. That distinction matters: the theorem is an existence result, and in most applications the point is not calculated.

A Repeated-Rolle Lemma

Ordinary Rolle’s Theorem starts with equal function values at two endpoints and produces a point where the derivative is zero. For Taylor’s Theorem, the auxiliary function will have not only a zero at each endpoint, but also several zero derivatives at the center. The following lemma describes what repeated applications of Rolle’s Theorem then give.

Proof. We use induction on \(m\). If \(m=0\), then \(\phi(u)=\phi(v)=0\). The function is continuous on the closed segment between \(u\) and \(v\) and differentiable in its interior. Rolle’s Theorem therefore gives \(c\) strictly between the endpoints with \(\phi'(c)=0\).

Now let \(m\geq1\), and assume the lemma holds for \(m-1\). By Rolle’s Theorem applied to \(\phi\) on the segment between \(u\) and \(v\), there is a point \(d\) strictly between them such that \(\phi'(d)=0\). At \(u\), the derivative \(\phi'\) satisfies \(\bigl(\phi'\bigr)^{(j)}(u)=\phi^{(j+1)}(u)=0\) for \(j=0,\ldots,m-1\). Also, \(\phi'(d)=0\). Apply the induction hypothesis to \(\phi'\), with endpoints \(u\) and \(d\). It gives a point \(c\) strictly between \(u\) and \(d\) such that \(\bigl(\phi'\bigr)^{(m)}(c)=\phi^{(m+1)}(c)=0\). Since \(d\) lies strictly between \(u\) and \(v\), this \(c\) is also strictly between \(u\) and \(v\). The induction proves the lemma. \(\square\)

The lemma applies whether \(u<v\) or \(v<u\); Rolle’s Theorem is applied to the segment with those endpoints. Its hypotheses require continuity on that closed segment and differentiability in its interior at each application. These properties follow from the stated differentiability assumptions.

Proof of Taylor’s Theorem

Recall the Taylor polynomial \(T_{n,f,a}\) from “Higher-Order Taylor Polynomials.” By the derivative-matching property established there, \(\bigl(T_{n,f,a}\bigr)^{(j)}(a)=f^{(j)}(a)\) for \(j=0,\ldots,n\). We now prove the Lagrange form stated in “Taylor’s Theorem.”

Proof. Fix \(a\) and \(x\ne a\). Define the real number $$ K=\frac{f(x)-T_{n,f,a}(x)}{(x-a)^{n+1}}, $$ which is well-defined because \(x-a\ne0\). On the interval between \(a\) and \(x\), define $$ H(t)=f(t)-T_{n,f,a}(t)-K(t-a)^{n+1}. $$ The function \(H\) has derivatives through order \(n+1\) there.

At \(t=a\), the derivative-matching property gives \(f^{(j)}(a)-\bigl(T_{n,f,a}\bigr)^{(j)}(a)=0\) for every \(j=0,\ldots,n\). Also, the \(j\)th derivative of \((t-a)^{n+1}\) is zero at \(a\) whenever \(j\leq n\). Consequently, $$ H^{(j)}(a)=0\quad\text{for }j=0,1,\ldots,n. $$ At \(t=x\), the choice of \(K\) gives $$ H(x)=f(x)-T_{n,f,a}(x)-K(x-a)^{n+1}=0. $$ Apply the Repeated Rolle’s Theorem lemma with \(\phi=H\), \(u=a\), \(v=x\), and \(m=n\). It gives a point \(c\) strictly between \(a\) and \(x\) such that \(H^{(n+1)}(c)=0\).

The polynomial \(T_{n,f,a}\) has degree at most \(n\), so its \((n+1)\)st derivative is zero. The \((n+1)\)st derivative of \((t-a)^{n+1}\) is the constant \((n+1)!\). Thus $$ 0=H^{(n+1)}(c)=f^{(n+1)}(c)-K(n+1)!. $$ Rearranging gives \(K=f^{(n+1)}(c)/(n+1)!\). Substituting the definition of \(K\) proves $$ f(x)-T_{n,f,a}(x) =\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}, $$ as required. \(\square\)

The proof also covers \(x<a\): the lemma only requires that \(c\) be strictly between the two endpoints, in either order. The power \((x-a)^{n+1}\) remains signed in the identity. The theorem asserts existence of \(c\); it does not assert that \(c\) is unique.

Seeing the Auxiliary Function at Work

What the Proof Requires—and What It Does Not

The auxiliary function packages the desired identity into two kinds of zeros: the difference between \(f\) and its Taylor polynomial has zero derivatives at \(a\), and the constant \(K\) is chosen to make the auxiliary function vanish at \(x\). Repeated Rolle’s Theorem then turns those zeros into a zero of the \((n+1)\)st derivative. The choice of \(K\) is the step that connects that final derivative equation back to the original remainder.

A useful feature of this proof is that it does not assume \(f^{(n+1)}\) is continuous. The hypotheses require that this derivative exist on the interval. In the repeated-Rolle argument, the functions to which Rolle’s Theorem is applied are continuous on the relevant closed segments and differentiable in their interiors; the needed continuity follows from differentiability of the functions at the relevant points. Continuity of the highest derivative is not an extra requirement.

A common error is to apply Rolle’s Theorem only once and stop after obtaining a zero of \(H'\). That is enough when \(n=0\), but for a higher-order Taylor polynomial the proof needs a zero of \(H^{(n+1)}\). The vanishing derivatives at \(a\) are what allow the Rolle argument to be repeated. Another common error is to assume the intermediate point is the same for different choices of \(x\). The proof makes no such claim: \(c\) may depend on \(f\), \(a\), \(x\), and \(n\).

The resulting identity is the Lagrange form of Taylor’s Theorem already used in “Taylor’s Theorem.” Its proof explains why the next derivative appears and why its evaluation point lies between the center and the target. The next tutorial considers other ways to express the remainder.