Section 6.4 Taylor’s Theorem

Taylor’s theorem explains how a sufficiently differentiable function is approximated by its Taylor polynomial. Its proof is a repeated application of Rolle’s theorem from Theorem 5.3.2, so it is a genuine outgrowth of the Mean Value Theorem (Theorem 5.3.3).

Subsection 6.4.1

Let \(f\) be a function with derivatives up to order \(n\) near \(x_0\text{.}\) The \(n\)-th Taylor polynomial of \(f\) at \(x_0\) is

\begin{equation*} P_n(x;x_0)=\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k. \end{equation*}

We say that \(x_0\) is a zero of multiplicity at least \(m\) of \(g\) if

\begin{equation*} g(x_0)=g'(x_0)=\cdots=g^{(m-1)}(x_0)=0. \end{equation*}

Lemma 6.4.1.

Let \(g\) have \(n+1\) derivatives on an interval \(I\text{.}\) Suppose \(x_0 \in I\) is a zero of multiplicity at least \(n+1\) of \(g\text{,}\) and suppose \(x \in I\) is another zero of \(g\text{.}\) Then there exists a point \(c\) strictly between \(x_0\) and \(x\) such that

\begin{equation*} g^{(n+1)}(c)=0. \end{equation*}

Proof.

Since \(g(x_0)=g(x)=0\text{,}\) Rolle’s theorem (Theorem 5.3.2) gives a point \(c_1\) between \(x_0\) and \(x\) such that \(g'(c_1)=0\text{.}\) Because \(x_0\) is still a zero of multiplicity at least \(n\) for \(g'\text{,}\) Rolle’s theorem applied to \(g'\) on the interval with endpoints \(x_0\) and \(c_1\) gives a point \(c_2\) between them such that \(g''(c_2)=0\text{.}\)

Repeating this argument yields points \(c_1,c_2,\dots,c_{n+1}\text{,}\) each lying strictly between \(x_0\) and \(x\text{,}\) with

\begin{equation*} g'(c_1)=0,\quad g''(c_2)=0,\quad \dots,\quad g^{(n+1)}(c_{n+1})=0. \end{equation*}

Taking \(c=c_{n+1}\) completes the proof.

Theorem 6.4.2. Taylor’s Theorem.

Suppose \(f\) has derivatives up to order \(n+1\) on an open interval containing \(x_0\) and \(x\text{.}\) Then there exists a point \(c\) between \(x_0\) and \(x\) such that

\begin{equation*} f(x)=P_n(x;x_0)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}. \end{equation*}

The final term is called the Lagrange remainder.

Proof.

The case \(x=x_0\) is immediate, so assume \(x \ne x_0\text{.}\) Choose a constant \(K\) so that the function

\begin{equation*} g(t)=f(t)-P_n(t;x_0)-K(t-x_0)^{n+1} \end{equation*}

satisfies \(g(x)=0\text{.}\) Since \(P_n\) matches the first \(n\) derivatives of \(f\) at \(x_0\text{,}\) the point \(x_0\) is a zero of multiplicity at least \(n+1\) of \(g\text{.}\)

By Lemma 6.4.1, there exists a point \(c\) between \(x_0\) and \(x\) such that \(g^{(n+1)}(c)=0\text{.}\) But the polynomial \(P_n\) has zero \((n+1)\)-st derivative, so

\begin{equation*} g^{(n+1)}(t)=f^{(n+1)}(t)-K(n+1)!. \end{equation*}

Evaluating at \(c\) gives

\begin{equation*} 0=f^{(n+1)}(c)-K(n+1)!, \end{equation*}

hence

\begin{equation*} K=\frac{f^{(n+1)}(c)}{(n+1)!}. \end{equation*}

Substituting this value of \(K\) into the definition of \(g\) and using \(g(x)=0\) yields the formula.

When \(n=0\text{,}\) Taylor’s theorem reduces to the Mean Value Theorem. So it is best viewed as a higher-order version of the same idea.

There is another useful expression for the remainder term. It is especially well suited to error estimates and to proving power-series expansions of familiar functions.

Proposition 6.4.3. Integration by Parts.

Assume the basic properties of the definite integral. If \(u,v \colon [a,b] \to \R\) are continuously differentiable, then

\begin{equation*} \int_a^b u'(t)v(t)\,dt = u(b)v(b)-u(a)v(a)-\int_a^b u(t)v'(t)\,dt. \end{equation*}

Proof.

By the product rule (Proposition 5.1.6),

\begin{equation*} (uv)'(t)=u'(t)v(t)+u(t)v'(t). \end{equation*}

Integrating both sides from \(a\) to \(b\) and applying Corollary 5.4.11 to the function \(uv\text{,}\) we get

\begin{equation*} u(b)v(b)-u(a)v(a) = \int_a^b u'(t)v(t)\,dt + \int_a^b u(t)v'(t)\,dt. \end{equation*}

Rearranging gives the result.

This is a convenient special case tailored to the present argument. A more general integration-by-parts formula, together with a change-of-variables theorem, appears later in the integration chapter; see Proposition 7.4.1 and Theorem 7.4.2.

Theorem 6.4.4. Taylor’s Theorem with Integral Remainder.

Assume the basic properties of the definite integral. Suppose \(f \in C^{n+1}(I)\text{,}\) where \(I\) is a closed interval with endpoints \(x_0\) and \(x\text{.}\) Then

\begin{equation*} f(x)=P_n(x;x_0)+\frac{1}{n!}\int_{x_0}^x f^{(n+1)}(t)(x-t)^n\,dt. \end{equation*}

The final term is called the integral form of the remainder.

Proof.

If \(x=x_0\text{,}\) the formula is immediate. If \(x \lt x_0\text{,}\) then the same argument applied on the interval with endpoints \(x\) and \(x_0\) gives the result after reversing the limits of integration. So it is enough to treat the case \(x_0 \lt x\text{.}\)

For each \(k=0,1,\dots,n\text{,}\) define

\begin{equation*} R_k(x;x_0):=\frac{1}{k!}\int_{x_0}^x f^{(k+1)}(t)(x-t)^k\,dt. \end{equation*}

When \(k=0\text{,}\) the Newton-Leibniz formula (Corollary 5.4.11) gives

\begin{equation*} R_0(x;x_0)=\int_{x_0}^x f'(t)\,dt = f(x)-f(x_0). \end{equation*}

Now let \(1 \le k \le n\text{.}\) Apply Proposition 6.4.3 with

\begin{equation*} u(t)=f^{(k)}(t) \qquad \text{and} \qquad v(t)=\frac{(x-t)^k}{k!}. \end{equation*}

Then \(u'(t)=f^{(k+1)}(t)\) and

\begin{equation*} v'(t)=-\frac{(x-t)^{k-1}}{(k-1)!}. \end{equation*}

Therefore

\begin{equation*} R_k(x;x_0) = \left[\frac{f^{(k)}(t)(x-t)^k}{k!}\right]_{t=x_0}^{t=x} + \frac{1}{(k-1)!}\int_{x_0}^x f^{(k)}(t)(x-t)^{k-1}\,dt. \end{equation*}

Since \((x-x)^k=0\text{,}\) this becomes

\begin{equation*} R_k(x;x_0) = -\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k + R_{k-1}(x;x_0). \end{equation*}

Starting from \(R_0(x;x_0)=f(x)-f(x_0)\) and applying the recurrence successively for \(k=1,2,\dots,n\text{,}\) we obtain

\begin{equation*} R_n(x;x_0) = f(x)-\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k = f(x)-P_n(x;x_0). \end{equation*}

Rearranging gives the desired formula.

Corollary 6.4.5.

Suppose \(f\) has derivatives of all orders on an interval containing \(x_0\) and \(x\text{.}\) If there exists a constant \(M>0\) such that

\begin{equation*} |f^{(n)}(t)| \le M \end{equation*}

for every \(n \ge 1\) and every \(t\) between \(x_0\) and \(x\text{,}\) then the Taylor series of \(f\) at \(x_0\) converges to \(f(x)\text{.}\)

Proof.

By Taylor’s theorem,

\begin{equation*} f(x)-P_n(x;x_0) = \frac{f^{(n+1)}(c_n)}{(n+1)!}(x-x_0)^{n+1} \end{equation*}

for some point \(c_n\) between \(x_0\) and \(x\text{.}\) Hence

\begin{equation*} |f(x)-P_n(x;x_0)| \le M\frac{|x-x_0|^{n+1}}{(n+1)!}. \end{equation*}

By Proposition 2.2.6, \(|x-x_0|^{n+1}/(n+1)! \to 0\text{.}\) Therefore \(P_n(x;x_0)\to f(x)\text{.}\)

Example 6.4.6.

Assume the standard calculus facts that \((e^x)'=e^x\) and that \(e^x\) is continuous. Fix \(x_0 \in \R\text{,}\) and choose a closed interval containing both \(0\) and \(x_0\text{.}\) By the Extreme Value Theorem (Theorem 4.4.4), \(e^x\) is bounded there by some constant \(M\text{.}\) Since every derivative of \(e^x\) is again \(e^x\text{,}\) the previous corollary gives

\begin{equation*} e^{x_0}=\sum_{n=0}^\infty \frac{x_0^n}{n!}. \end{equation*}

Because \(x_0\) was arbitrary, the exponential function is analytic on all of \(\R\text{.}\)

Theorem 6.4.7. Binomial Series.

For each \(\alpha \in \R\text{,}\)

\begin{equation*} (1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n}x^n \qquad \text{for } |x|<1, \end{equation*}

where

\begin{equation*} \binom{\alpha}{n} = \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}. \end{equation*}

Proof.

We omit the proof. The ratio test shows that the series has radius of convergence \(1\text{,}\) and term-by-term differentiation leads to a differential equation satisfied by both sides.

Example 6.4.8.

Taking \(\alpha=1/2\) gives

\begin{equation*} \sqrt{1+x} = 1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac{5}{128}x^4+\cdots \qquad \text{for } |x|<1. \end{equation*}

So \(\sqrt{1+x}\) is analytic at \(0\text{.}\)

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