Section 6.3 Analytic Functions

A function \(f\) is said to be analytic at\(x_0\) if there exists a power series

\begin{equation*} \sum_{n=0}^\infty a_n(x-x_0)^n \end{equation*}

that converges to \(f(x)\) on some open interval containing \(x_0\text{.}\) We say that \(f\) is analytic on an open set \(U \subseteq \R\) if it is analytic at every point of \(U\text{.}\)

Thus analytic functions are exactly those functions that locally agree with power series. The next result shows that such a representation, if it exists, is forced to be the Taylor series.

Theorem 6.3.1. Analytic Functions Are Given by Their Taylor Series.

Suppose \(f\) is analytic at \(x_0\text{.}\) Then \(f\) is infinitely differentiable on some neighborhood of \(x_0\text{,}\) and there exists \(r>0\) such that

\begin{equation*} f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \qquad \text{for } |x-x_0|<r. \end{equation*}

In particular, the power series representation of \(f\) about \(x_0\) is unique.

Proof.

Since \(f\) is analytic at \(x_0\text{,}\) there is a power series

\begin{equation*} f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n \end{equation*}

on some interval \(|x-x_0|<r\text{.}\) By Theorem 6.2.2, we may differentiate term by term on this interval. Repeating this process \(k\) times gives

\begin{equation*} f^{(k)}(x) = \sum_{n=k}^\infty n(n-1)\cdots(n-k+1)a_n(x-x_0)^{n-k}. \end{equation*}

Setting \(x=x_0\) leaves only the term with \(n=k\text{,}\) so

\begin{equation*} f^{(k)}(x_0)=k!a_k. \end{equation*}

Hence

\begin{equation*} a_k=\frac{f^{(k)}(x_0)}{k!}, \end{equation*}

and substituting this into the original series gives the desired Taylor expansion. Since each coefficient is uniquely determined by \(f\text{,}\) the representation is unique.

The theorem shows that analytic functions are automatically \(C^\infty\text{.}\) The converse is false: a function may be infinitely differentiable and still fail to agree with its Taylor series.

Example 6.3.2.

Define

\begin{equation*} f(x)=\begin{cases} e^{-1/x} & \text{if } x>0,\\ 0 & \text{if } x\le 0. \end{cases} \end{equation*}

Then \(f\) is infinitely differentiable on \(\R\text{,}\) and every derivative of \(f\) at \(0\) is equal to \(0\text{.}\) Therefore the Taylor series of \(f\) at \(0\) is the zero series. However, \(f(x)>0\) for every \(x>0\text{,}\) so \(f\) does not agree with that Taylor series on any neighborhood of \(0\text{.}\) Thus \(f\) is not analytic at \(0\text{.}\)

The previous example shows that a smooth function may fail to agree with its Taylor series even when that Taylor series converges everywhere. There are also smooth functions whose Taylor series at a point does not converge anywhere except at that point.

Example 6.3.3.

Consider

\begin{equation*} f(x)=\int_0^\infty e^{-t}\cos(t^2x)\,dt. \end{equation*}

One can show, by differentiating under the integral sign, that \(f\) is \(C^\infty\) and that

\begin{equation*} f^{(n)}(0)=\begin{cases} 0 & \text{if } n \text{ is odd},\\ \pm (2n)! & \text{if } n \text{ is even}. \end{cases} \end{equation*}

Consequently the Taylor coefficient \(f^{(n)}(0)/n!\) has absolute value \((2n)!/n!\) for infinitely many \(n\text{.}\) Since

\begin{equation*} \frac{(2n)!}{n!}=(n+1)(n+2)\cdots(2n)\ge (n+1)^n, \end{equation*}

the \(n\)-th roots of these coefficients tend to \(\infty\text{.}\) The root test therefore shows that the Taylor series of \(f\) at \(0\) has radius of convergence \(0\text{.}\) So this Taylor series diverges for every \(x \ne 0\text{.}\)

Theorem 6.3.4. Borel-Peano Theorem.

For every formal power series

\begin{equation*} \sum_{n=0}^\infty a_nx^n \end{equation*}

there exists a function \(f \in C^\infty(\R)\) whose Taylor series at \(0\) is exactly

\begin{equation*} \sum_{n=0}^\infty a_nx^n. \end{equation*}

In other words, every formal power series occurs as the Taylor series of some smooth function, whether or not that series converges. For a readable account, see [8] and the historical note [7].

Further examples and discussion of non-analytic smooth functions can be found in [6].

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